Naked Science Forum

On the Lighter Side => New Theories => Topic started by: Dave Lev on 24/11/2018 08:30:27

Title: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/11/2018 08:30:27
Further our discussion about new matter creation at the excretion disc of the Milky way:
Please see: https://www.thenakedscientists.com/forum/index.php?topic=75261.40

I wonder why any new particle/Atom/Molecular is drifting outwards from the excretion disc of the milky way.
Let's look at the following formula for gravity force:
F = G * M * m / R^2
M = The mass of the SMBH
m = The mass of the particle/Atom/Molecular.

Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.

So how can we justify the drifting outwards mechanism of any new particle/Atom/Molecular?

Could it be that something is missing in the following formula:
F = G * M * m / R^2
Could it be that over time there is a change in the gravity force? (Even if it is only a very minor change).
Do you know that the American continent is drifting away from the European continent by only 1.5 cm per year.
In the past they were fully connected.
So, just based on that small change per year, we have got the vast ocean between those two continents after long time.
We know that:
1. The moon is drifting away from the Earth (by about 1.5 cm per year)
2. The Earth is also drifting away from the Sun.
3. All the planets in the solar system are drifting away from the Sun.
4. All/Most of the moons are drifting away from their host planet.

But objects can also drift inwards.
There are plenty of orbital objects (satellites...) around the Earth that drift inwards and eventually fall down.
Hence, could it be that gravity force must be changed over time?
Actually, if we think about it, it is clear that nothing can stay the same forever. Although there is no friction in space, it seems to me that even gravity can't stay the same forever.
There must be some small change in gravity force over time (even very small change over very long time)
So, could it be that we have to use the following formula:
F = G * M * m / R^2 +/- F(t)
F(t) = represents the change in the gravity force over time.
If the object is in orbital cycle which is too close to the host, the value of F(t) must be positive.
However, if the radius is long enough, the value must be negative.

So, with regards to our solar system:
All the planets are located far enough from the Sun.
Therefore, the value of F(t) for each one of them must be negative. Hence, over time (long enough), the gravity force on each planet is decreasing. In order to compensate that decreasing in gravity force, the planet must drift outwards.

In the same token, could it be that each particle in the plasma is located long enough from the SMBH.
Therefore, its F(t) is negative.
If so, the total gravity force on each particle in the plasma is decreasing over time.
That decreasing gravity force drifts away the particle from the center.

Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Ophiolite on 24/11/2018 14:08:54
The drifting of the continents is a very poor analogy for orbital mechanics.
The movement of the moon away from the Earth is a consequence of an exchange of angular momentum.
Any outward drift of planets can be acocunted for by the reduction of mass of the sun, although resonances with other planets could play a part.
Artificial satellites crashing to Earth are a consequence of air resistance.

Summary: your speculations are unfounded and unnecessary.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/11/2018 13:15:05
Thanks

With regards to Earth Moon orbital radius:

In the following article there is good explanation about Tidal friction:

http://curious.astro.cornell.edu/physics/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate

"The Moon's orbit (its circular path around the Earth) is indeed getting larger, at a rate of about 3.8 centimeters per year.
The reason for the increase is that the Moon raises tides on the Earth.
This effect stretches the Earth a bit, making it a little bit oblong. We call the parts that stick out "tidal bulges."
Tidal friction, caused by the movement of the tidal bulge around the Earth, takes energy out of the Earth and puts it into the Moon's orbit, making the Moon's orbit bigger (but, a bit paradoxically, the Moon actually moves slower!)."
Let's focus on the following: "..takes energy out of the Earth and puts it into the Moon's orbit, making the Moon's orbit bigger..."
So, due the orbital cycle some energy is used to pull the "Tidal buges".
Therefore,  the Moon's orbit is bigger.
However, this bigger orbital radius of the moons means by definition less gravity force between the Earth to moon.
Therefore:

At t=0
The current radius is: 384,400 km (The distance between earth -moon, assuming a perfect cycle)
The gravity force is:
F = G * M * m / R^2
therefore we can say that at t = 0:
F(0) = G * M * m / R^2

One year from now at t = 365 days:
It is clear that the distance should be
384,400 km + 3.8 cm.
That by definition proves that the gravity force between the Earth Moon is decreasing.

So, why can't we write the following formula:

F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)

F(t) = the gravity force at time t.
M(t) = the host mass at time t.
m(t) = The orbital object mass at time t.
R(t) = the radius at time t.
Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference).
F(0) = the gravity force at time 0.

Hence:
At t = 0
Δ(0) = 0

If t = 365 (for example -one year)
F(365) = G * M(365) * m(365) / R(365)^2 = F(0) - Δ(365).

R(365) = R(0) + 3.8 cm = 384,400 km + 3.8 cm.

I don't see any contradiction between this formula to your explanations.
Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/11/2018 21:04:15
I don't see any contradiction between this formula to your explanations.
Do you agree with that?
Sure, I agree (F is changing, not G), but you contradict your OP where you say:
Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.
Here you assert that force is constant (instead of decreasing as in the moon example), and that R is constant, instead of increasing as in that example.

Yes, angular momentum is being transferred from Earth to the orbit of the Moon, and that shoves it to a higher radius, which slows down its linear speed.  Total mechanical energy is constantly being lost.

Your OP seemed to suggest otherwise.  Sure, it was about some atoms orbiting the galaxy, but small particles are hardly expected to follow a fixed-radius circle about the central gravitational source.
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 26/11/2018 22:14:42
Further our discussion about new matter creation at the excretion disc of the Milky way:
Please see: https://www.thenakedscientists.com/forum/index.php?topic=75261.40

I wonder why any new particle/Atom/Molecular is drifting outwards from the excretion disc of the milky way.
Let's look at the following formula for gravity force:
F = G * M * m / R^2
M = The mass of the SMBH
m = The mass of the particle/Atom/Molecular.

Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.

So how can we justify the drifting outwards mechanism of any new particle/Atom/Molecular?

Could it be that something is missing in the following formula:
F = G * M * m / R^2
Could it be that over time there is a change in the gravity force? (Even if it is only a very minor change).
Do you know that the American continent is drifting away from the European continent by only 1.5 cm per year.
In the past they were fully connected.
So, just based on that small change per year, we have got the vast ocean between those two continents after long time.
We know that:
1. The moon is drifting away from the Earth (by about 1.5 cm per year)
2. The Earth is also drifting away from the Sun.
3. All the planets in the solar system are drifting away from the Sun.
4. All/Most of the moons are drifting away from their host planet.

But objects can also drift inwards.
There are plenty of orbital objects (satellites...) around the Earth that drift inwards and eventually fall down.
Hence, could it be that gravity force must be changed over time?
Actually, if we think about it, it is clear that nothing can stay the same forever. Although there is no friction in space, it seems to me that even gravity can't stay the same forever.
There must be some small change in gravity force over time (even very small change over very long time)
So, could it be that we have to use the following formula:
F = G * M * m / R^2 +/- F(t)
F(t) = represents the change in the gravity force over time.
If the object is in orbital cycle which is too close to the host, the value of F(t) must be positive.
However, if the radius is long enough, the value must be negative.

So, with regards to our solar system:
All the planets are located far enough from the Sun.
Therefore, the value of F(t) for each one of them must be negative. Hence, over time (long enough), the gravity force on each planet is decreasing. In order to compensate that decreasing in gravity force, the planet must drift outwards.

In the same token, could it be that each particle in the plasma is located long enough from the SMBH.
Therefore, its F(t) is negative.
If so, the total gravity force on each particle in the plasma is decreasing over time.
That decreasing gravity force drifts away the particle from the center.
Do you agree with that?
1. I dont understand -- where did the new particle come from?
2. How do u know that it is drifting away?
3. Alby said that gravity aint a force -- its a bending of spacetime.
4. When u say that gravity cant stay the same for ever -- do u mean that big G changes with time? -- if so then the most logical way of correcting the formula is to apply an additional term or two directly to big G -- adding a stand-alone term onto the end of the formula is a highly unlikely solution.
5. U mention a mechanism -- Einsteinians dont worry about a mechanism for gravity -- they say that gravity is a delusion created by mass somehow bending the fabric of spacetime etc.
6. The replies in this thread are ignoring a critical problem, the speed of gravity.
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy. Einsteinians give two different silly reasons re why gravity is instantaneous at distance, or appears to be instantaneous (but i wont go into that here).
7. A spiral is an ever increasing or decreasing circular motion -- hencely the name spiral galaxy sort of answers the question. LOL.
8. The Newtonian formula is a worry -- it might not be accurate re spiral galaxies, ie for matter spread out in a flattish plane. -- or it might be accurate if there is dark matter in the galaxy.
9. And then there is the question of the force of dark energy opposing the action of the bending of spacetime by pushing things (like your new particle) away from each other or outwards from the center of the bigbang universe or something.
10. The modern Einsteinian dark age of science will soon end, & real science will be enjoy a rebirth.
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 27/11/2018 01:11:55
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy.
Actually, it can be shown that if gravity travels at c or some other finite speed, then orbiting objects tend to spiral out from each other, violating conservation of energy.

This article shows the basic concepts of the argument: http://www.flight-light-and-spin.com/proof/instant-gravity.htm
Yes u must get an outwards spiral not inwards.
That link is interesting.
I have a problem because i believe that gravity has a finite speed of say well over 20 billion c. The thing is -- any finite speed must result in an outwards spiral -- & a violation of conservation of energy. But i dont believe that gravity has an infinite speed. There must be a mechanism that balances the gain in speed by a loss of speed -- i will have a think.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/11/2018 14:25:51
Sure, I agree (F is changing, not G)
It isn't drifting.  It is being pushed away.
Thanks Halc

So, the corrected formula is as follow:
F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)

F(t) = the gravity force at time t.
M(t) = the host mass at time t.
m(t) = The orbital object mass at time t.
R(t) = the radius at time t.
Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference).
F(0) = the gravity force at time 0.

With regards to the Tidal friction:
Δ(t) is the estimated lost of gravity force due to Tidal friction.
In the Earth/Moon orbital cycle, the "tidal bulges" is very clear (few cm.).
If the Moon mass will be decreased, while the Earth mass will be increased, we might not notice this - "tidal bulges".
However, do you agree that the tidal effect is still valid?
Hence, do you agree that even if we will decrease the Moon mass into a single Atom while we increase the earth mass into a SMBH, the tidal friction will still be there and the Atom will be pushed away?

 
Artificial satellites crashing to Earth are a consequence of air resistance.
So, can we assume that at any orbital system without air resistance, the orbital object is losing energy over time?
Hence, do you agree that in those kind of orbital systems, as the objects are losing energy over time, they are also losing gravity force over time and therefore they are pushed away from their host?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/11/2018 15:54:36
So, the corrected formula is as follow:
F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)

F(t) = the gravity force at time t.
You're being inconsistent with your notation.  In the OP you defined F(t) as "the change in the gravity force over time".

You might as just well say that the correct formula for force F at a given time is F = G * M * m / R2 as per your OP.  Over time, any of those things might change, especially R if the path of m is not a circle about mass M, so F would then not be constant over time.

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Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference).
...
With regards to the Tidal friction:
Δ(t) is the estimated lost of gravity force due to Tidal friction.
Force isn't lost due to friction.  Force is reduced due to the object being less nearby (increase of R).  Friction reduces kinetic energy.  It doesn't reduce force or momentum.

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In the Earth/Moon orbital cycle, the "tidal bulges" is very clear (few cm.).
More like a few meters.

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If the Moon mass will be decreased, while the Earth mass will be increased, we might not notice this - "tidal bulges".
However, do you agree that the tidal effect is still valid?
You're asking if smaller objects still produce tides?  Yes they do.  The ISS for instance produces an immeasurable tide on Earth, and one that puts an accelerating force on the rotation of Earth, as opposed to a drag like the moon has.  That's because the ISS is inside the geosync radius.

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Hence, do you agree that even if we will decrease the Moon mass into a single Atom while we increase the earth mass into a SMBH, the tidal friction will still be there and the Atom will be pushed away?
For the atom to be pushed away, it needs to orbit slower than the rotation of the larger object, and the large object needs to be fluid enough to exhibit tides. A solid cold Earth would not push the moon away since there is no medium for tides.  You need an atmosphere or ocean or flexible mantle for the effect.  So not sure if tides can be raised on a black hole.

Anyway, short story is that things near black holes spiral in due to the bending of space.  It takes acceleration to stay out of them.  Nothing (not even light) can orbit closer than 1.5x the radius of the event horizon.  Your basic tidal force analysis is a Newtonian one, not taking relativistic effects into account.  It works for classical masses and orbital speeds.

 
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So, can we assume that at any orbital system without air resistance, the orbital object is losing energy over time?
From what?  The moon is gaining energy.  Low orbit (below geosync) objects might lose energy from the tidal drag, but high orbit objects gain mechanical energy from tidal trust.  A steel moon in low orbit would eventually fall to Earth.  It needs to be strong enough to not break up from being inside the Roche limit.

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Hence, do you agree that in those kind of orbital systems, as the objects are losing energy over time, they are also losing gravity force over time and therefore they are pushed away from their host?
If they're losing energy, they're falling closer to Earth, which increases the gravitational force on them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/11/2018 06:39:40
You might as just well say that the correct formula for force F at a given time is F = G * M * m / R2 as per your OP.  Over time, any of those things might change, especially R if the path of m is not a circle about mass M, so F would then not be constant over time.

I fully agree that if there is a change in M or m there is a change in F which impacts the R.
However, assuming that there is no change in M or m. why there is a change in R? If you want to push it away, you must set external force.
Friction reduces kinetic energy.  It doesn't reduce force or momentum.
I don't understand how the Kinetic energy changes R while you claim that it doesn't reduce force.
for example:
Based on the following formula (assuming that there is no change in the mass M or m):
F1 = G * M * m / R1 ^2
F2 = G * M * m / R2 ^2
Hence
F1 / F1 = 1/(R1/R2)^2
If R2 is bigger than R1 than:
F1 is bigger than F2.
Therefore, by definition there is a change in the gravity force.
So how the kinetic energy changes the gravity force while we say that the kinetic energy doesn't reduce the gravity force?
Don't you agree that there is a contradiction in this statement?

Can you please explain why cold Earth would not push the moon away since there is no medium for tide?
I don't understand how the matter in mass effects the tidal friction and eventually the kinetic energy.
Can you prove it by real formula or is it just hypothetical idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/11/2018 06:02:32
A potato on a table has constant gravitational force acting on it, because the table holds the R constant, putting the potato in a circular path.  Take away the table, and the R starts decreasing.  The path is no longer a circle, and F goes up.  Neither M nor m has changed.
It needs no external force.  There already is one, as computed by F.

Thanks Halc.
I do appreciate all your efforts.

However, I'm not fully sure with regards to that example.

The main difference is the forwarded motion of a body in space (let's call it forwarded force)!!!

So, before continue our discussion, let's see what causes an orbit to happen (in a perfect conditions)?
http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/1-what-causes-an-orbit.html
"Orbits are the result of a perfect balance between the forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space, such as a large planet or star. An object with a lot of mass goes forward and wants to keep going forward; however, the gravity of another body in space pulls it in. There is a continuous tug-of-war between the one object wanting to go forward and away and the other wanting to pull it in."

Forwarded force - the forwarded motion of a body in space.
Gravity force - the amount of force that pull it in.

Therefore - as long as there is a balance between those two vectors, the object will keep its current radius.
Technically, if there is no air resistance, no tidal, no any external force, no change in mass, the radius will be kept forever.
However, if the forwarded force is greater than gravity force - the object continues moving through space.
If gravity is greater than forwarded force, objects collide.

Hence, it seems to me that the only way to change the orbital radius is by changing the gravity force or the Forwarded force (In perfect conditions).

Do you agree with that?
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 29/11/2018 11:49:25
I repeat what i said earlier.........
6. The replies in this thread are ignoring a critical problem, the speed of gravity.
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy. Einsteinians give two different silly reasons re why gravity is instantaneous at distance, or appears to be instantaneous (but i wont go into that here).
Title: Re: How gravity works in spiral galaxy?
Post by: guest39538 on 29/11/2018 11:55:43
Electric Universe ,  that  simple ,  N is  attracted to N , I am right and science lies .
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/11/2018 16:37:55
No.  There is no forward force.  Any object might have a trajectory that puts it on something other than a perfect circular orbit.
So for an object already in a perfect circular orbit (you don't say this, but it needs to be qualified), the only way to change the orbital radius is by changing the gravity force or by applying a secondary force to one of the objects.
Thanks for the clarification. fully clear.
However, what do you mean by: " applying a secondary force to one of the objects"?

The former can be changed by altering the mass of the primary mass.  Altering the mass of the orbiting thing makes no difference.  I can take half the moon away (halving the gravitational force acting upon it) and it will have zero effect on its orbit.  But take away half of Earth, and the moon will need to orbit further out.
The issue with the mass is also clear. Although, in this example, there is no change in mass.

But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbit
I can't understand this idea.
1. Why tides put cumulative thrust on the moon?
2. If tides put a cumulative thrust on the moon that actually adds to its momentum, why this extra momentum can't be considered as a force? (Don't you agree that extra momentum should increase the speed and therefore it should increase the force)
Please look at the following diagram:
http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
Do you agree that:
1. The Centrifugal force is a direct outcome of the moon speed?
2. In order to keep the moon in the same radius its Centrifugal force must be equal to its gravitational force?
3. In order to increase R (while keeping it in a perfect orbital cycle) we must decrease at the same moment and at the same amplitude the Centrifugal force and the gravitational force?

Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/11/2018 21:00:36
However, what do you mean by: " applying a secondary force to one of the objects"?
Something other than the gravitational force given by that formula.  For the potato, maybe it is the table putting a upward force on it preventing it from falling.  An asteroid hits the moon, altering its orbit.  Tidal force is tangential to the regular gravitational pull of the Earth.  All these forces are something else, and hence can alter the orbit it would have otherwise taken.

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Quote from: Halc
But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbit
I can't understand this idea.
1. Why tides put cumulative thrust on the moon?
It is always pushing in the direction it is already moving, adding to the mechanical energy of the moon.  Tidal forces on the ISS push against its motion and reduce the mechanical energy.  The orbit drops due to tidal resistance, but it drops even more due to friction.

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2. If tides put a cumulative thrust on the moon that actually adds to its momentum, why this extra momentum can't be considered as a force?
Momentum is not force.  Linear momentum is mV, and cumulative thrust adds to mechanical (potential + kinetic) energy, not to momentum, since the velocity of the moon actually slows as it moves to a higher orbit.  Total angular (not to be confused with linear) momentum (mass*radius*tangential-velocity) of the Earth/moon system is preserved, but angular momentum of Earth's spin is transferred to angular momentum of the Earth/moon system as a whole.  That momentum is part of both of them, not just the moon.

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(Don't you agree that extra momentum should increase the speed and therefore it should increase the force)
Linear moment goes down as the moon slows down.  Momentum and force are not necessarily related, as this example shows.
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Please look at the following diagram:
http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
That diagram is really bad science.  Centrifugal force does not exist in the inertial frame implied by that diagram.  If it did, it would counter gravity force and the path of Earth would not curve.  It exists in a rotating reference frame, but in such a frame, Earth would be stationary and have no rotational speed.  The arrow to the left would have zero magnitude. 

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Do you agree that:
1. The Centrifugal force is a direct outcome of the moon speed?
OK, first, there is no moon in that picture, but it works if you put Earth where the sun is and the moon going around.  Then:  No... Centrifugal force is a pseudo-force and does not exist, else the moon would not accelerate towards Earth.

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2. In order to keep the moon in the same radius its Centrifugal force must be equal to its gravitational force?
In a rotating reference frame, yes.  The two forces cancel and the moon remains stationary.  In an inertial frame, no.  Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.

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3. In order to increase R (while keeping it in a perfect orbital cycle) we must decrease at the same moment and at the same amplitude the Centrifugal force and the gravitational force?
You can't keep it in a circular orbit if you increase R.  Maybe a helix, but circles come back to the same point.  That said, I know what you mean.
Increasing R requires a tangential thrust adding mechanical energy to the orbit.  The gravitational force will drop as R increases, and not until then.  You want to add energy, so you need to push in the direction of movement.  Altering the force tangential to the motion does not add energy, it just curves the path elsewhere.

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Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?
Tides push with the rotation (to the left in your picture).  Gravity pushes tangential to the motion (down in your picture).  The former adds energy and thus increases R.  The latter adds no energy, so it merely bends the path from following the straight line it would otherwise follow in the absence of gravitational force.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/11/2018 18:56:40
Thanks Halc

1. Centrifugal force
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Please look at the following diagram:
http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
That diagram is really bad science.  Centrifugal force does not exist in the inertial frame implied by that diagram.  If it did, it would counter gravity force and the path of Earth would not curve.
I don't understand why you disqualified the centrifugal force.
http://scienceline.ucsb.edu/getkey.php?key=4569
It is stated:
"All massive objects in our universe are attracted to each other through a force known as gravity. If gravity were the only force acting between the sun and earth, the two bodies would indeed collapse on one another. Therefore, there must be other forces acting on this system. Remember that the earth, and all the other planets in our solar system, revolve around the sun. Because the earth is in a rotational orbit, there is another force acting on the planet. This force is known as centrifugal force. The direction of the centrifugal force on the earth is opposite the direction of the gravitational force (see the diagram below), so it prevents the sun and earth form collapsing into each other."

What is wrong with this explanation?
If it helps, let's ignore the Sun/Earth/Moon system. Just think about any orbital system with host in the center and an object in a perfect orbital cycle.


In an inertial frame, no.  Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.
How could it be that only the force of gravity exists?
In the following article:
http://scienceline.ucsb.edu/getkey.php?key=4569
It is stated in answer 1:
"the earth has two main forces: inertia, to keep moving straight, and the gravity, to pull it to the sun. It moves in an average of the two directions, as in this image, and moves in a circle around the sun."
It is stated specifically that Inertia is force. Hence, the Earth is in an average of two directions.  (or two forces).

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Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?
Tides push with the rotation (to the left in your picture).  Gravity pushes tangential to the motion (down in your picture).  The former adds energy and thus increases R.  The latter adds no energy, so it merely bends the path from following the straight line it would otherwise follow in the absence of gravitational force.

Can we prove this explanation with mathematics?

For example:
Assuming that the current radius is R1
Than
F1 = G M m / R1 ^2
If the former adds energy and thus increases R
Than, at the radius = R2
F2 = G M m / R2 ^ 2
As R1 bigger than R2 then F2 is lower than F1.
Hence, there is a decrease in the gravity force
ΔF = F1 - F2.
So, how energy can increase R, therefore it changes the gravity force, but it doesn't consider as force.
This is a real enigma for me.

There is another key issue (in an ideal Earth/Sun system).
Let's assume that the tidal sets Energy.
This energy Push the Earth further away from the Sun.
However, due to inertia, the Earth keeps its orbital velocity.
Hence, in this new location, the updated gravity force will be too low to hold the Earth in the orbital path.
Therefore, the Earth will be disconnected from the Sun.

So, do you agree that in order to keep the Earth in the orbital cycle around the Sun, two actions should be taken at the same moment and with full synchronization?
1. Increases radius - R
2. Decreases inertia velocity - V 

Hence, how Tidal (or any sort of energy) can push the earth farther away from the Sun while it also decreases the inertia velocity at the same moment and in full synchronization?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/12/2018 00:18:01
Thanks Halc

1. Centrifugal force
Quote from: Halc
Centrifugal force does not exist in the inertial frame implied by that diagram.  If it did, it would counter gravity force and the path of Earth would not curve.
I don't understand why you disqualified the centrifugal force.
Did you read my reply there?  If there was a second force countering gravity, the net force on Earth would be zero and it would not accelerate at all, but travel in a straight line per Newton's 1st law.  You've not replied to that.

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http://scienceline.ucsb.edu/getkey.php?key=4569
That looks like a question posed to a room of students, and a list of some of their answers.  None of them sound very clear, but the 2nd and 5th one are blatantly wrong, and the 1st one is also wrong.

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It is stated:
"All massive objects in our universe are attracted to each other through a force known as gravity. If gravity were the only force acting between the sun and earth, the two bodies would indeed collapse on one another. Therefore, there must be other forces acting on this system.
Wrong.  Nothing else (EM, or nuclear forces) is acting between Sun and Earth.  The other forces are needed to form both of them in the first place (to form matter at all for instance), but not to keep them in orbit.

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What is wrong with this explanation?
What is wrong is what I already explained in the part of mine you quoted.  Gravity is the only force at work in that diagram, which can be seen by the fact that all objects are accelerating in it.  Gravity force causes the sun to accelerate down, and that force has an equal and opposite reaction acting on the sun, not on Earth.  The arrow belongs there, but it is not centrifugal force, and it should be attached to the sun, which is where the upward gravitational force is being applied.


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Quote from: Halc
In an inertial frame, no.  Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.
How could it be that only the force of gravity exists?
Only gravity is a significant force in orbital motion.  There are other forces like EM, but they're immeasurably small.

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In the following article:  [same link]
You need to find some better articles.  I hope the university isn't pushing these answers as science.  Instead it seems to be a page showing why education is needed, illustrating typical answers from students before taking some class.
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It is stated in answer 1:
"the earth has two main forces: inertia, to keep moving straight, and the gravity, to pull it to the sun. It moves in an average of the two directions, as in this image, and moves in a circle around the sun."
It is stated specifically that Inertia is force.
Yea, it does.  That's part of why answer 1 is wrong.  Inertial is mass, not force.  I cannot accelerate a rock with inertia.  F=ma remember?  If inertial was a force, it would cause mass to accelerate all by itself.

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Can we prove this explanation with mathematics?

For example:
Assuming that the current radius is R1
Than
F1 = G M m / R1 ^2
If the former adds energy and thus increases R
Than, at the radius = R2
F2 = G M m / R2 ^ 2
As R1 bigger than R2 then F2 is lower than F1.
Hence, there is a decrease in the gravity force
ΔF = F1 - F2.
All correct.  The moon exerts less gravitational force when it is further away.

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So, how energy can increase R, therefore it changes the gravity force, but it doesn't consider as force.
R is not a function of energy.  But the energy is being applied as forward thrust here which pushes the moon into a higher orbit, just like a rocket expends fuel to get a satellite to a higher orbit where it moves slower.  Low orbit objects orbit at around 9 km/sec, but it takes more energy to get them uphill to say geosync where they move at only 3 km/sec.  Same thing is happening with the moon, being pushed uphill just like the rocket does to the satellite.
Earth has a similar trivial thrust pushing it away from the sun.

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There is another key issue (in an ideal Earth/Sun system).
Let's assume that the tidal sets Energy.
This energy Push the Earth further away from the Sun.
However, due to inertia, the Earth keeps its orbital velocity.
Hence, it this new location, the updated gravity force will be too low to hold the Earth in the orbital path.
Therefore, the Earth will be disconnected from the Sun.
????
It isn't an inertia thing.  The tide (from the spin of the sun) pushes Earth, which makes it move a bit to fast to stay in the circular path it was in before, so the circle widens a bit.  Earth moves a tiny bit further away, and slows down because it is now moving up hill out of the gravitational well.  This new slower speed exactly balances the smaller gravity at this new radius.

You seem to be neglecting the fact that things slow down when moving away from a gravitational source.  Kinetic energy is being lost to acquired gravitational potential energy.

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So, in order to keep the Earth in the orbital cycle around the Sun, two actions should be taken at the same moment and with full synchronization:
1. Increases R
2. Decreases V
So, how Tidal can control on those two vectors with full synchronization?
Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.  Throw a ball in the air and watch V go down as the ball rises.  A thrown ball is in orbit, but an orbit that typically intersects the ground, so it appears to be a parabola to an observer.  It is in fact an elliptical path.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/12/2018 06:33:49
Earth moves a tiny bit further away, and slows down because it is now moving up hill out of the gravitational well.
V goes down as R goes up.  Throw a ball in the air and watch V go down as the ball rises.
You seem to be neglecting the fact that things slow down when moving away from a gravitational source.  Kinetic energy is being lost to acquired gravitational potential energy.

Thanks
Based on Newton first law:
https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
First law:   In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"
Let's start by asking the following:
What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?
If I understand it correctly, the Earth will "continue to move at a constant velocity".
Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?
Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?
Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.

As we increase R we actually decrease the gravity force.
Gravity force is the power which holds the Earth in the orbital cycle around the Sun.
It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.

It is clear that as Tidal increases R it actually decreases the gravity force.
However, based on your explanation, Tidal doesn't set any force. Therefore, by definition it doesn't set any negative force which slows down the Earth velocity as it increases R.
Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.
Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.
That increasing radius, decreases the gravity force more and more.
Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?
At some point it must be totally disconnected from the sun gravity and  "continues to move at a constant velocity" as explained by Newton first law.

Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.

Can you please explain this issue?

Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.

Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/12/2018 14:36:19
Based on Newton first law:   In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"
Let's start by asking the following:
What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?
If I understand it correctly, the Earth will "continue to move at a constant velocity".
Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?
There is nothing moving away from a gravitational source in your example.  You took gravity away.
You can do this by considering two small rocks at the same positions and velocities of the earth/sun system.  They’re too small to have significant gravity pull between them, so they each continue to move pretty much in a straight line.

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Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?
Newton’s law says that nothing does.  It is called velocity BTW.  ‘Inertial velocity’ isn’t something different, or if it is, you need to tell me what you mean by that.
So yes, I agree with Newton’s law.  Earth will not change its velocity unless acted upon by a force.


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Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.
It actually pushes the Earth tangential to the orbit, which is perpendicular to a vector away from the sun.  If it pushes or pulls, it’s a force.  That’s what force is.

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As we increase R we actually decrease the gravity force.
Gravity force is the power which holds the Earth in the orbital cycle around the Sun.
Force is not power.  Gravity is the force which holds the Earth in the orbital cycle around the Sun, yes.

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It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.

It is clear that as Tidal increases R it actually decreases the gravity force.

However, based on your explanation, Tidal doesn't set any force. Therefore, by definition it doesn't set any negative force which slows down the Earth velocity as it increases R.[/quote]Tital force is a force.  I didn’t say it ‘doesn’t set any force’.  It pushes with the motion, so it actually directly acts to increase speed of Earth, but that velocity slows as the Earth pulls further out of the gravity well.  The net effect is a slower orbital speed.

It’s like a coin in one of those funnel machines, slowly spiraling faster and faster into the center.  You give the coin some thrust, and it moves further away from the center and ends up going slower than before you gave it that push.  In the absence of interference, friction is a force against its motion, and yet the coin is speeding up as it goes deeper into the gravity well.


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Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.
Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.
Less gravity means the Earth has already moved further away from the sun.  You seem to be confusing cause and effect here.  That gravitational force doesn’t go down until the Earth has already moved further out.

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That increasing radius, decreases the gravity force more and more.
Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?
No, away from gravity is up hill.  Surely you know this.  Downhill gets you closer to the source of the gravity (the sun in this case).

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At some point it must be totally disconnected from the sun gravity and  "continues to move at a constant velocity" as explained by Newton first law.
Given enough thrust, that is true, but the tidal thrust decreases with distance, so I don’t think it is possible for one object to spin away one of its satellites in isolation.  But things are not in isolation.  Given enough time and high spin, the Earth could theoretically push the moon beyond Earth’s hill radius and the moon would depart Earth and go into its own orbit about the sun.  In isolation (just sun and Earth say), there is no hill radius.

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Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.
It doesn’t.  It is a positive force tangential to the orbit, pushing forward, not out.  If it was a negative force, things would slow down and drop to a lower orbit.
You seem to continue to envision the tides being an outward thrust instead of a forward one.  That would not have any cumulative effect since it would not affect speed and hence no energy transfer.

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Can you please explain this issue?
I have, and you don’t seem to read what I say.
Tidal forces push forward, not out.  It doesn’t directly increase radius.  The increased speed is too much for the current orbit, so it moves further out, which slows the system since that is moving against the primary gravitational force.

Consider at a comet that passes the sun at the same radius as Mercury, but much much faster.  That’s what tide forces do is make something faster.  So the comet moves to a higher orbit, and in doing so, it ends up nearly stopped as it gets so far away from the sun.  Moving away from the sun slows it down.

Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.

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Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?
Tides create a bulge on the sun that resist its spin, putting negative torque on the sun.  Torque is FR (force * radius).  That force is balanced by positive torque on the sun/Earth system, so same math but much greater R and much less F.  This conserves angular momentum.

So that F is the thrust, pushing in the direction of orbit, not outward.  That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.  That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.  Potential energy E = force*distance, so compute the weight (not mass) of Earth (using the gravitational formula GMmR2  to get the force, and multiply that by the say millimeter it moves away.  From that you get a chunk of energy E.  Kinetic energy is mv2, so the reduction in v from the orbital change is Δv = E√(1/r) where r is the change in orbital radius ΔR and E is the kinetic energy lost to potential energy.  That negative Δv is greater than the positive Δv you get from enough tidal thrust to increase Earth orbit by a millimeter, so net effect is a slower Earth.

I chose one millimeter ΔR as my example, but it takes perhaps 1000 years for the solar tides to do that.  A somewhat larger effect pushing us away is that the mass of the sun is decreasing faster than new material falls in from deep space, so gravity is slowly decreasing.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/12/2018 19:33:13
So that F is the thrust, pushing in the direction of orbit, not outward.  That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.  That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.


Thanks again for your great effort!
I do appreciate.

However, would you kindly explain why the Earth is losing kinetic energy due to Tidal?

It was stated that the tidal trust speeds up the orbital velocity of the Earth.
That speeds up the planet (Lets assume by Δv), which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.
That takes the Earth further away from the sun, so it gains potential energy.
So far so good!

However, with regards to the kinetic energy.
The kinetic energy of the Earth just before speeding up the planet is: mv^2.
After increasing the speed by  Δv, the  updated kinetic energy should be: m(v+Δv )^2

Therefore, by definition due to the increase in speed, we have found that the Earth is gaining higher Kinetic energy.

So, would you kindly explain why you assume that the Earth "loses kinetic energy"?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/12/2018 16:36:40
And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

That's an interesting idea.
The formula for Kinetic emery is: Ek = mv^2
The formula for potential energy is: Ep = GMmR^2
I could not understand why an increasing in potential energy (Ep) should decrease Kinetic energy (Ek).
Can you please explain the rational of this idea for orbital system?
Do we have any formula which shows the energy transformation?
Title: Re: How gravity works in spiral galaxy?
Post by: Janus on 02/12/2018 21:09:28
And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

That's an interesting idea.
The formula for Kinetic emery is: Ek = mv^2
okay
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The formula for potential energy is: Ep = GMmR^2
No.  That is the formula for gravitational force.  Potential energy would be that force integrated with respect to R, or
Ep = -GMm/R

Thus the total energy of an object in orbit is Et = mv^2/2-GMm/R

The total energy of an object in orbit is conserved assuming no additional energy is added.  Thus with an elliptical orbit, as the object climbs from periapis to apapis,  r increases, requiring a decrease in v in order to conserve energy.

If you add KE, the object will begin to climb, exchanging KE for gained PE.   Not only is the KE added lost, but some of the initial KE must be given up too.  If the added KE is just a one time shot, the object goes into an elliptical orbit with a apapis further out than the present orbit.  If it is a continuous addition, like in tidal acceleration, the object will slowly climb outward.  The semimajor axis of the orbit will increase, and total orbital energy can also be expressed as Et = -GMm/2a, where a is the semi-major axis.   But average orbital velocity is V= sqrt(GM/a), so as a increases, the total energy goes up, but v goes down, meaning that KE makes up a smaller part of the total energy.


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/12/2018 05:32:29
Thanks Janus


The formula for potential energy is: Ep = GMmR^2
No.  That is the formula for gravitational force.  Potential energy would be that force integrated with respect to R, or
Ep = -GMm/R

Thus the total energy of an object in orbit is Et = mv^2/2-GMm/R

The total energy of an object in orbit is conserved assuming no additional energy is added. 

Thanks Janus

Now it is fully clear to me.

Et = Kinetic energy (Ek) + potential energy (Ep) = mv^2/2+ (-GMm/R)
However, as "The total energy of an object in orbit is conserved assuming no additional energy is added."

Et = constant for a given orbital cycle.
Therefore, if we increase the potential energy (Ep) we by definition decrease Ek

Let's verify this idea by starting at point 1
Hence:

Et1 = Kinetic energy (Ek1) + potential energy (Ep1)

Et1 = m(v1)^2/2+ (-GMm/(R1))

In our case, the tidal set a  thrust that increased the kinetic Energy by ΔEk and therefore we have got an increased   velocity by (Δv).

Now we can claim that the new Total energy (Et2) due to additional energy which had been added is:

Et2 = Et1 + ΔEk
Et2 = Ek1 + ΔEk + Ep1

This is the new starting point (just before increasing the radius)."
At that moment we can claim that:

Et2 = Ek1 + ΔEk + Ep1 = constant
Et2 = Ek2 + Ep1 = constant
Et2 =  m(v1+Δv)^2/2+ (-GMm/(R1)) = constant

However, due to velocity increase, the radius must be increased let's call it (ΔR)
Hence,
Due to the radius increase:

Ep2 = (-GMm/(R2)) = (-GMm/(R1 + ΔR))
That increase in the Ep must now decrease back the Ek
So, let's call it Ek3

Et2 = Ek3 + Ep2 = constant
Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))

Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))

Hence

m(v3)^2/2+ (-GMm/(ΔR )) = m(v1+Δv)^2/2

m(v3)^2/2- m(v1+Δv)^2/2 = GMm/(ΔR)

m(v3)^2/2- m(v1+Δv)^2/2 = GMm/(ΔR)

m(v3- v1- Δv)^2/2 = GMm/(ΔR)

If so, we need to proof that this new v3 is fully correlated to the expected new orbital velocity due to the gravity force at R2 = R1 + ΔR.

Do you agree with that?

Conclusion:
We have actually increased the Ek by ΔEk (due to tidal) and therefore got an increased velocity = Δv.
This increased Δv set an increased radius = ΔR.
This increased ΔR increased the Ep by ΔEp.
Now we expect that this ΔEp will help us not just to decrease the Ek by ΔEk (in order to get back to v1), but also more than that in order to achieve our goal that v3 must be lower than v1.
Even in a perfect system, we can't request to get back all the energy that had been invested.
So, it is very challenging to expect that the ΔR which was a product of Δv will help us to cancel completely the Δv.
Now, we expect that this ΔR will help us to gain move than just Δv.
Is it realistic?

Even if we assume that it is realistic, how do we know that we get a perfect new velocity?
Too low is also problematic.
We must get a very specific velocity to meet the new gravity force (at R2 = R1 +ΔR).

So, don't you think that it is too challenging goal?
In any case, only if we set full calculation we can verify if it works or not.

Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 03/12/2018 05:58:06
Conclusion:
We have actually increased the Ek by ΔEk (due to tidal) and therefore got an increased velocity by Δv.
This increase Δv set an increased radius ΔR.
This change in the radius increased the Ep.
Now we want that this new Ep will help us not just to eliminate that ΔEk which we have added (due to tidal), but also more that that in order to achieve our goal that V3 must be lower than V1.
Actually, lower velocity by itself is not good enough.
Too low is also problematic.
We must get a very specific velocity to meet the new gravity force.
Is it a realistic goal?
Only if we set full calculation we can verify if it works or not.
I havent followed all of the math, but is the new gravity force due to change in R or in G or in M or in m?
If it is due to change in G then i fear that this might introduce some problems. It might depend on how one defines mass. Its complicated.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 03/12/2018 23:11:07
Et = Kinetic energy (Ek) + potential energy (Ep) = mv^2/2+ (-GMm/R)
However, as "The total energy of an object in orbit is conserved assuming no additional energy is added."

Et = constant for a given orbital cycle.
Therefore, if we increase the potential energy (Ep) we by definition decrease Ek

Let's verify this idea by starting at point 1
Hence:

Et1 = Kinetic energy (Ek1) + potential energy (Ep1)

Et1 = m(v1)^2/2+ (-GMm/(R1))

In our case, the tidal set a  thrust that increased the kinetic Energy by ΔEk and therefore we have got an increased   velocity by (Δv).
If you're including tidal thrust, Et is no longer constant.  That thrust is adding total energy to the orbit, taking energy (and more) away from the sun.  Momentum is conserved in that transfer, but plenty of energy is lost to heat.  But total energy of the orbit Et is increasing.

That thrust is continuous, not one-time, and therefore it causes a gradual rise in orbit, and hence a decrease in orbital speed, not an increase.  Ek goes down, Ep goes up.  If it were a one shot thrust, Ek would indeed momentarily go up, sending the Earth into an elliptical orbit where R varies, but average speed still is lower and average potential energy is still higher.

You can get the same effect with two thrusts, 180 degrees apart (half a year), one to push the far end of the orbit up, and a second one to re-circularize the orbit once there.  But a continuous thrust like tidal thrust actually puts the Earth into a continuous spiral outward, ever slowing.


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Now we can claim that the new Total energy (Et2) due to additional energy which had been added is:

Et2 = Et1 + ΔEk
Et2 = Ek1 + ΔEk + Ep1

This is the new starting point (just before increasing the radius)."
OK, This sort of works, but it represents a one-shot thrust all at once, not a continuous thrust.  Sort of like Earth getting hit by an asteroid.  Yes, the figures are correct.  R has not changed yet.  The Earth is suddenly just moving a smidge faster.  Tidal thrust doesn't work this way so you know.

From reading below, you define Ek2 as the new kinetic energy just after the momentary thrust, but not the kinetic energy of the new orbit (which varies) or the average kinetic energy of that orbit.  But Ep2 is defined confusingly as the (average??) potential energy of the new orbit, not the potential energy associated with Ek2 just after the thrust.  That is quite confusing since 2 is not matched up with 2.
You treat R2 like a constant, so I am assuming that R2 is the gravitational average radius of the new elliptical orbit.  By 'gravitational average radius', I mean the radius at which the Ek and Ep are the same as it would be if the orbit were circular.  For tides, that works.  For momentary jolts of thrust, the orbit will not be circular like that, but the calculations still work if R2 is defined this way.

Rockets tend to put satellites into orbit with such discreet thrusts separated by coasting, rather than continuous thrust which is harder to engineer.

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At that moment we can claim that:

Et2 = Ek1 + ΔEk + Ep1 = constant
Et2 = Ek2 + Ep1 = constant
Et2 =  m(v1+Δv)^2/2+ (-GMm/(R1)) = constant
What do you mean by = constant?  The total stays the same, but both kinetic and potential energy are changing as the R changes, so those terms are not going to be constant.

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However, due to velocity increase, the radius must be increased let's call it (ΔR)

Hence,
Due to the radius increase:

Ep2 = (-GMm/(R2)) = (-GMm/(R1 + ΔR))
That increase in the Ep must now decrease back the Ek
So, let's call it Ek3

Et2 = Ek3 + Ep2 = constant
Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))

Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))
I followed all that.  The last line seems to be a copy of the line above it.

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Hence

m(v3)^2/2+ (-GMm/(ΔR )) = m(v1+Δv)^2/2
The algebra failed here.  (-GMm/ΔR) is a massive figure, enough negative energy to form a black hole.  It represents the energy you'd get by lifting Earth from well within the Schwarzschild radius of the sun.

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If so, we need to proof that this new v3 is fully correlated to the expected new orbital velocity due to the gravity force at R2 = R1 + ΔR.

Do you agree with that?
The scenario you've described puts the thing into an elliptical orbit, but yes, if you compute the figures for the circular average R2, I agree with that.
I've discarded the mathematics beyond the algebra error above since it is all wrong after that.

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Conclusion:
We have actually increased the Ek by ΔEk (due to tidal)
Due to a momentary increase of v actually.  Tides never do that since thrust is continuous, and Ek never goes up, not even momentarily.  The whole path just forms a spiral pattern.

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and therefore got an increased velocity = Δv.
This increased Δv set an increased radius = ΔR.
This increased ΔR increased the Ep by ΔEp.
Now we expect that this ΔEp will help us not just to decrease the Ek by ΔEk (in order to get back to v1), but also more than that in order to achieve our goal that v3 must be lower than v1.
Even in a perfect system, we can't request to get back all the energy that had been invested.
Pretty much you can.  There is no friction except in the transfer of energy from sun to Earth, where much energy is lost to the friction slowing the spin of the sun, and more lost to slowing of spin of Earth.  Both apply thrust similarly.  The moon/Earth system only has significant  friction on the Earth side since the moon is tidal locked.
But in terms of orbits, the friction is negligible.  You can assume there is no loss.

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So, it is very challenging to expect that the ΔR which was a product of Δv will help us to cancel completely the Δv.
Now, we expect that this ΔR will help us to gain move than just Δv.
Is it realistic?
Yea, sure.

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Even if we assume that it is realistic, how do we know that we get a perfect new velocity?
Too low is also problematic.
It isn't perfect now, so don't expect it to be perfect.  Today, the Earth is moving faster than it should and its orbit (while still moving inward) is accelerating outward.  It should bottom out at max speed in 30 days and start moving out again.  The orbit is hardly circular, but for the sake of our computations, you can assume one.

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We must get a very specific velocity to meet the new gravity force (at R2 = R1 +ΔR).
If you want a circular orbit, yes, but you started out circular and gave it a single jolt thrust.  It will indeed be going that perfect velocity at that perfect R2, but it won't be moving tangential at that time, so R is not going to stay at R2.  Such is the non-circular orbit you've put it into.

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So, don't you think that it is too challenging goal?
In any case, only if we set full calculation we can verify if it works or not.
You're not actually computing ballistics here, so your equations should need to bother with the elliptical orbit part.  Yes, R2 can be computed, and the v3 out there should be perfect, and not too difficult to compute.  You don't have a figure for how much energy you intend to add to your system via tidal thrust.  Not sure if you need it.  You just have an arbitrary v2 that is bigger than v1.  That is enough to define the new orbit it seems.  You don't even need to know the mass of Earth since the same orbit change will happen to a small pebble given that same Δv.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/12/2018 19:20:24
Thanks Halc

Do appreciate all your valid remarks.
I will verify the mathematics.

However, it seems to me that there is a fundamental problem with the concept of Tidal Friction.

let's start by understanding the meaning of Friction by Google: "The resistance that one surface or object encounters when moving over another".

In the following article it is stated:
http://www.physics.mcgill.ca/~crawford/PSG/PSG11/204_97_L11.9_tidfric.html

"Tides stretch the oceans, and to a small extent, the solid mass of a planet or satellite. In one complete rotation, the planet material keeps deforming and relaxing. This takes energy away from the rotation, transforming it into heat.
In effect, this is a frictional loss, like a giant brake on the planet. Over the centuries, the moon's rotation on its own axis has slowed until it presents essentially the same face to the earth."

So, the tidal friction is like a giant brake or some sort of resistance.
As an example - If we drive a car and press the brakes, we transform some of the kinetic energy into heat and slow down the velocity of the car.
Is there any possibility to increase the velocity of car by pressing the brakes?
In the same token, tidal friction is considered as a giant brake, which transforms some of the energy into heat.
I assume that the source of the energy is coming from the kinetic energy of the object. Therefore, if we transform some of this energy into heat, than there must be less kinetic energy. Less kinetic energy means less velocity.
So, how could it be that a brake/resistance system (like a tidal friction) can increase the orbital velocity instead of decreasing it?
If someone will tell us that by pressing the brakes in the car, the brakes in turn push the car forward and therefore it increases the velocity, would we accept this answer?
So, is there any chance that we are missing the real impact of the Tidal Friction?
I really can't understand how any sort of resistance or brakes can increase the velocity.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/12/2018 23:17:00
However, it seems to me that there is a fundamental problem with the concept of Tidal Friction.

let's start by understanding the meaning of Friction by Google: "The resistance that one surface or object encounters when moving over another".
Yes.  The quote you give from the article describes tidal friction quite well.

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Is there any possibility to increase the velocity of car by pressing the brakes?
Sure.  Have the road moving faster than the car.  Pressing the brakes speeds up the car closer to the speed of the road.  That's what's going on with tidal friction.  The spin of Earth is the road, so the friction is the brakes, speeding up the moon.  The Earth spins once per day, far faster than the 30 days it takes for the moon to go around.  Until those two match, the moon will continue to move away.  Once they match, the moon will start to get closer again as both speed in lock step.

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Therefore, if we transform some of this energy into heat, than there must be less kinetic energy. Less kinetic energy means less velocity.
It is angular velocity in this case, but yes.  The Earth is always slowing its spin due to tidal friction.  That's why they keep having to add leap-seconds now and then.  The day used to be about 10 hours long.  It is 24 hours now.  If the system is left alone long enough, it will max out at 1440 hours, and then start to shorten again as the moon actually accelerates the spin of the Earth.
I don't think it will be left alone anywhere near that long.  The sun will swallow us before it all happens.

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So, how could it be that a brake/resistance system (like a tidal friction) can increase the orbital velocity instead of decreasing it?
The road is moving faster than the car, so the car is speeding up as it hits the brakes.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/12/2018 05:42:00
Thanks Halc
It is angular velocity in this case, but yes.  The Earth is always slowing its spin due to tidal friction.  That's why they keep having to add leap-seconds now and then.  The day used to be about 10 hours long. 
That is fully clear to me.
However, that phenomena is due to the impact of the gravity force on the object itself.
Based on Newton's law of universal gravitation:
https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
So, every particle at the earth attracts every other particle in the Universe due to gravity force.
That is the ultimate answer for the tidal friction on any orbital system.
However, Newton didn't specify any information about the matter in the object or how it behaves.
It can be a gas/water/metal Star/Planet/Moon or even made of rice. In the formula we only look at the mass.
It can spin in ultra velocity or stay locked
Therefore, there is no proof for the following statement by Newton gravity formula:
The spin of Earth is the road, so the friction is the brakes, speeding up the moon.
If we think that the spin of the Earth can speed up the moon, than by definition we must change the Newton's law of universal gravitation so it will also include that impact.
How can we claim that Newton's law of universal gravitation is fully correct if there is no reflection in the formula for the type of the matter or the spin of the objects/particles?
Based on the current Newton's law of universal gravitation, "every particle attracts every other particle in the universe with a force..."
So, any particle on Earth might go/down or spin faster/slower/locked due to the local impact of gravity force (tidal activity).
However, based on Newton's law of universal gravitation that local activity in one object (Earth) can't have in turn any impact on a particle in another object (Moon).
If we believe that the spin of a particle in one object can affect the velocity of a particle at a far end object than:
Why don't we change the Newton's law of universal gravitation formula in order to represent this breakthrough understanding?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/12/2018 15:52:04
Thanks
Based on Wiki the highest point of the tidal bulge is 54 Cm:
https://en.wikipedia.org/wiki/Tide
"The theoretical amplitude of oceanic tides caused by the Moon is about 54 centimetres (21 in) at the highest point"

The near bulge is ahead of the moon, and thus pulls the moon forward.  The far bulge rotates to behind the moon and pulls it backwards.  The near one is closer, so that forward pull is greater than the backwards pull.  Net effect is gravitational thrust on the moon.
The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.
We can try to calculate the impact of each bulge.
In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.
However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?
Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?
If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.

In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
In our case,
Tidal forces cause a bulge on the near and far sides of Earth.
If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.
Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.
Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?
I would assume that less than Pico cm.
So, I wonder how this Pico cm can set the requested extra thrust on the moon which is needed to push the moon away from Earth?

However, based on your following answer, you don't like the idea of using the Earth as a point-mass.
Exactly.  And the thrust on the moon can be explained if you apply the law that way.  Earth is not a point-mass.  Point-masses cannot be susceptible to tidal friction.
Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.
Is it enough?
However, as the bulges are moving with the moon orbital cycle, than the moon gets this tinny extra gravity force constantly (So, there is no temporary/transient thrust).
Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?
The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.
This effective Earth mass (M+ΔM) should increase the velocity of the Moon.
So, instead of orbiting at velocity v, it will orbit at v +Δv.
However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.
Hence, I don't see any transient thrust which temporarily increases the velocity. 
If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?
Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/12/2018 21:07:50
Thanks
Based on Wiki the highest point of the tidal bulge is 54 Cm
Sounds like an ideal tide: If Earth was entirely covered with deep water and barely rotating, the friction would be minimized and the tides would have that amplitude.  Sounds about right.

I live 150 km inland of the Atlantic and we get tides higher than that even here.  Water tends to pile up when hit by continents and forced to go a different way.  The Atlantic has a good size for resonance and the tides are naturally higher than that theoretical 54 cm.  The Pacific doesn't resonate as well, and the tides there are lower in most places.

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The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.
We can try to calculate the impact of each bulge.
In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.
However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?
Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?
Neglected?  By what?  No real clue what you mean by that choice of words.
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If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.
The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.
I would have run the computations in terms of torque, but you method works as well.

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In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
That link treats all objects as point masses.  That isn't useful in describing tidal effects.  It shows a wrench, but makes no effort to show the torque put on a thrown wrench due to tidal forces.

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If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.
Yes, I agree with that.  Tidal forces are not about center of mass.  They're about deviations from it.

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Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.
Doesn't follow.
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Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?
None.  You need to consider what each bulge does as a force acting on the moon CoM.
I put two identical masses on either side of you, one nearby to the front, and the other further away to the rear.  You accelerate towards the front one because the force from that one is greater, given the smaller separation.  That's what is pulling the moon forward.

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Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.
They don't change the total gravity.  But they change the direction of it.  It is no longer straight towards the center of Earth because the Earth is neither a point mass nor a sphere.

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Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?
You want to dump more mass on Earth?  Sure, that would increase the gravitational force and pull the moon closer in.  This is assuming that new mass comes from outside the orbit of the moon.  A trillion low orbit satellites falling out of the sky will have zero effect since they were already effectively part of mass of Earth.

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The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.
This effective Earth mass (M+ΔM) should increase the velocity of the Moon.
Because it is pulled down to a new lower orbit, yes.  Even if it stayed at its current orbit, it would need to move faster, but dumping new mass on Earth doesn't give it the new energy it would need to accelerate like that, so it just fall to that lower orbit.

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So, instead of orbiting at velocity v, it will orbit at v +Δv.
However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.
What?  It isn't constant gravity force.  You just added mass to M, which changes the force.

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Hence, I don't see any transient thrust which temporarily increases the velocity.
The additional velocity in this case is gained by falling to a lower orbit, just like you personally gain velocity when falling off a ladder.  What do you mean by transient?  The higher velocity from adding mass will be permanent, or it will at least last until the mass is taken away again, or until the tides slow it down over time.

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If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?
Transient?  Are we still talking about the case where mass was added to Earth?  You kind of lost me with this question.


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Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?
I don't think we're talking about adding mass any more.  Not sure when that was abandoned.

There is thrust on the moon, which pushes it uphill, slowing all the way.  There is never a transient increase of velocity like you'd get with a instantaneous boost of momentum from say an asteroid hit.  The whole thing can be visualized with a static force diagram showing all the forces acting on the moon at any one moment, the sum of which is a vector slowing the moon.

This is best done in the simplified case of a non-elliptical orbit.  In the actual elliptical case, the speed of the moon goes up and down in cycles as the separation varies.

If there was a prize given to the known object with the most circular gravitational orbit, I wonder which object would get the prize?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/12/2018 07:58:18
Neglected?  By what?  No real clue what you mean by that choice of words.
The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.
I would have run the computations in terms of torque, but you method works as well.
O.K
Let's set a calculation.
The main tidal impact is on oceans.
I assume that it also doesn't work from pole to pole.
The pick is 54 cm.
Somehow it seems to me that if we take the two bulges and try to spread their total mass over the whole planet, it won't be higher than few centimeters.
However, just for the calculation let's assume 10 Cm or 1 Decimeter.
The radius of the Earth is 6,371 km =
r = 6,371,000 m or 63,710,000 decimeters.
The volume:
V (Earth) = V(r = 63,710,000) = 4/3 π r^3 = 4/3 π 63,710,000^3 = 4/3 π 2.586 10^23
V(Bulges) = V(r = 63,710,001) - V(r = 63,710,000) = 4/3 π (2.58596615 - 2.58596603) 10^23 = 4/3 π 0.00000013 10^23
Hence, the ratio is:
V(Bulges) / V (Earth) =0.00000013 / 2.586 = 5.02 10 ^(-8)
V (bulge) = 5.02 10 ^(-8) /2 * V (Earth) = 2.51 10 ^(-8) * V (Earth)
M = Earth mass
Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) M
Please also be aware that we mainly discuss on water in the bulges. However, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.
In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.
Now, we might think to add those two points of mass at the sea level of the planet, one at the front side and the other at far end side from the moon.
This is not fully correct. If we move closer to the pole, the effective radius is shorter and therefore the impact of the gravity bulges is lower.
However, for this calculation let's assume the worst case and set those two bulges points of mass at the at the maximal distance between them (2 x R).
The Gravity force of the Earth is:
F = G M m / R^2
m = Moon Mass
The Earth to the moon distance, R= 384,400 km
F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) =  G M m 6.99 *  10^(-12)
Bulges distance to the moon:
The bulges don't point directly to the Moon.
Therefore, the effective distance is less than full Earth Radius.
However, I use again the full radius just as a worst case.
Hence:
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) =  G Mb m * 10^(-9) * 4.49 * 10^(-4)
= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)
The ratio in the gravity force between the Bulge and the Earth is:
F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 *  10^(-12) = 0.643 * 10^(-9).
Hence
F(Bulge Total) =  0.643 * 10^(-9) * F(Earth) = G M m 0.643 * 10^(-9) /R^2

Conclusions:
I have calculated the gravity force impact of the Bulges.
It seems to me as a very minor gravity force (comparing to the Earth gravity force).
It might be even weaker than one big chain of mountains.

So, why do you call it "thrust"?
Do you still consider that based on this minor gravity force those bulges can push the moon away from us?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/12/2018 12:51:08
Neglected?  By what?  No real clue what you mean by that choice of words.
O.K
Let's set a calculation.
See, you never tell me what 'neglected' meant, so I am left unable to parse the prior post.

Quote
Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) M
However, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.
Agree, so closer to 1e-8
Quote
In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.
Arguably so, since said mountains cover much less area, but are much taller than 54 cm and are made of rock.  They don't have much a gravitational difference since they're made of lighter rock and float upwards on the mantle.  The Earth mass is just denser on average between the mountain ranges.

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F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) =  G M m 6.99 *  10^(-12)
I got more like 6.77e-12, not 6.99, which was the inverse of the 142 distance to the nearest bulge.
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Bulges distance to the moon:
The bulges don't point directly to the Moon.
Therefore, the effective distance is less than full Earth Radius.
However, I will use the full radius just as a worst case.
I would have used a nice round 6000 km, but fine.  Not like we know the actual numbers here.  I like how you're going about it.
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Hence:
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
I think you mean R(rear) here.
Quote
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) =  G Mb m * 10^(-9) * 4.49 * 10^(-4)
or about 4.5e-13GMbm.  Seems a bit early to be subtracting the two force values? That makes little sense.  F total is just the sum of F(front) + F(rear), and this is only the F component perpendicular to the orbit.  We need to compute the tangential component to get the tidal thrust, and for tangential, F total is indeed the difference between the two tangential components, not the sum.  So OK, you're computing a difference here, but you've not yet done the vector trigonometry to compute the forward and resisting thrust.  OK, so you have a difference of full force here, so it is indeed valid to do the trig on just that one force.  I don't see the vector arithmetic anywhere in your post.

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= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)
OK, I think you substituted the value computed for Mb here.
Quote
The ratio in the gravity force between the Bulge and the Earth is:
F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 *  10^(-12) = 0.643 * 10^(-9).
Hence
No, you subtracted the bulge forces.  The ratio of bulge force to the rest-of-earth would be computed from the sum of the bulge forces, not the difference.

No matter.  That ratio is irrelevant to the tidal thrust.  The main Earth mass plays no role.

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F(Bulge Total) =  0.643 * 10^(-9) * F(Earth)

Conclusions:
I have calculated the gravity force impact of the Bulges.
You need to compute the tangential impact.  Go to just before you subtracted F(rear) from F(front) and compute the tangential force of each bulge separately, and then subtract those.  To do this, you need to assign how far the bulges are pushed off-center by the friction of the spin of Earth.  Let's make it 1000 km.  Each bulge is 1000 km from the line connecting the centers of mass of the two bodies.  That puts a tangential component to the force, and that tangential component is the thrust.  The main mass of Earth is centered on that COM line, so it plays no role at all.

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It seems to me as a very minor gravity force (comparing to the Earth gravity force).
It might be even weaker than one big chain of mountains.

So, why do you call it "thrust"?
Because it pulls in the direction of motion (energy increase), not tangential to it (acceleration without energy increase).  Yes, it is a very minor force compared to the main component, but it is always forward, so the effect is cumulative forever.  The main force is always balanced in all directions, so the cumulative effect is zero after each month.

Yes, the mountains may mass more than the bulges, but they have no cumulative effect since they are ahead as often as behind the COM line, so the net effect is zero.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/12/2018 13:48:22
I will try to illustrate some of the vector work, using the guess that friction pushes the bulges 1000 km off center.

The Earth to the moon distance, R= 384,400 km
...
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
Those forces are respectively GmMb * 7.00e-12 and  GmMb * 6.55e-12

So we've defined two triangles will 1000km on the short side and 378029 or 390771 on the other.
F(front) was computed at GmMb * 7e-12, so the forward thrust from that bulge is
F(forward) = GmMb * 1.8517 e-14   
F(backward) = GmMb * 1.6762 e-14
The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.

Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/12/2018 21:09:49
Because it pulls in the direction of motion (energy increase), not tangential to it (acceleration without energy increase).  Yes, it is a very minor force compared to the main component, but it is always forward, so the effect is cumulative forever.  The main force is always balanced in all directions, so the cumulative effect is zero after each month.
Thanks for all your excellent remarks.

However, I still wonder why you insist to call it "Thrust".
The meaning of Thrust by Google is : "the propulsive force of a jet or rocket engine." "Push (something or someone) suddenly or violently in the specified direction."
I couldn't find any sort of "engine" in the Bulges activity. It is just increases the gravity force.

Never the less, I think that I understand the source for your statement:
Please see Figure 7.24 in the following article:
https://lifeng.lamost.org/courses/astrotoday/CHAISSON/AT307/HTML/AT30706.HTM
"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.) Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."
So, if I understand it correctly,  as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"
Is it correct?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/12/2018 22:58:30
However, I still wonder why you insist to call it "Thrust".
Because it is forward, in the direction of motion.  I turn the wheel of my car and it accelerates it to the left, but that isn't thrust, and so the car turns without gaining speed.  But the engine supplies force in the forward direction, adding mechanical energy to the car as it does so, resulting perhaps in more speed, or perhaps just helping the car up a hill.  The car might be slowing, but the engine is still providing thrust/energy.  So we call that force thrust, and not the force that turns the car left.

Did you see my post 40?  It shows how to compute the thrust.

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The meaning of Thrust by Google is : "the propulsive force of a jet or rocket engine." "Push (something or someone) suddenly or violently in the specified direction."
Well, it might be kilotons of thrust, but it still isn't very violent.

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I couldn't find any sort of "engine" in the Bulges activity. It is just increases the gravity force.
No, it doesn't.  Overall force is the same, but tides change the direction of that force.

Quote
Never the less, I think that I understand the source for your statement:
Please see Figure 7.24 in the following article:
Oh lovely.  They show the bulges dragged off to either side like I've been describing.  They'd be straight at the moon if there was no friction.

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"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.)
So much for my 1000 km estimate I bet.
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Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."
So, if I understand it correctly,  as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"
Is it correct?
Yes.  I've been saying that for how many posts now?  Energy increase (thrust) to the moon, and energy decrease to Earth's rotation rate, as the comment states.  The decrease is greater than the increase.  Momentum is preserved here, but not energy (2.5 terawatts), almost all of which is lost to friction heat.  Only about 1/20th of that energy (around 120 gigawatts) is transferred to the moon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/12/2018 06:25:48
Thanks
So,  we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".
In order to understand the real impact of that offset is, let's verify the following conditions:

1. There is no offset (0 degree) and the bulges are pointing directly to the moon
Do you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?
2. The offset is maximal (90 degree). The bulge are located at/almost the poles. therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?
3. The offset is 60 degree.
Cos (60) = 0.5
Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)
In this case, the effective distance to the moon is as follow:
R(front) =  384,400 km - 3,000 km = 381,400 Km
R(rear) =  384,400 km + 3,000 km = 387,400 Km.
So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.
I wonder if it is related to the idea that there are two bulges.
Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?
In the same token, if we will set only the rear bulge, can we assume that the thrust will be of zero?
So, only if we have them both, we get the impact of thrust?
I have read your following answer in thread 40 and couldn't understand the source of the thrust:
I will try to illustrate some of the vector work, using the guess that friction pushes the bulges 1000 km off center.
So we've defined two triangles will 1000km on the short side and 378029 or 390771 on the other.
F(front) was computed at GmMb * 7e-12, so the forward thrust from that bulge is
F(forward) = GmMb * 1.8517 e-14   
F(backward) = GmMb * 1.6762 e-14
The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.

Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.
In order to understand the idea.
Let's eliminate the Earth.
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/12/2018 14:25:37
So,  we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".
In order to understand the real impact of that offset is, let's verify the following conditions:

1. There is no offset (0 degree) and the bulges are pointing directly to the moon
Do you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?
No thrust, right.  The difference between their force is irrelevant then.
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2. The offset is maximal (90 degree). The bulge are located at/almost the poles.
No, 90° eastward and westward, not at the poles.  The bulges move around the equator, not from pole to pole.  Not exactly.  The orbital plane is about level with the solar system plane, not with the tilt of Earth's axis, so each bulge actually moves a ways north to south and back twice a day, regardless of offset angle.
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therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?
Still zero since they're equal and opposite.

I think I used about 10 degrees in my example.  Possibly still too much.
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3. The offset is 60 degree.
Cos (60) = 0.5
Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)
In this case, the effective distance to the moon is as follow:
R(front) =  384,400 km - 3,000 km = 381,400 Km
R(rear) =  384,400 km + 3,000 km = 387,400 Km.
So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.
I wonder if it is related to the idea that there are two bulges.
Did you read my description of the vector arithmetic in post 40?  Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.

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Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?
I cannot comment on that since it would change the center of gravity of Earth and hence not really be a bulge.
Thrust from each bulge is very much not zero.  They are in opposition, so the net thrust is much less.

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I have read your following answer in thread 40 and couldn't understand the source of the thrust:
Draw a vector diagram and work out the components (downward and forward) of the total force.

Do you know about vector arithmetic?  Surely there are some sites that inform well.  This is a really simple case.

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In order to understand the idea.
Let's eliminate the Earth.
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

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I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
It is only gravity, yes.  Work out the forces in 2 dimensions, not as a scalar.  Force is a vector, not a scalar.

If I am on a skateboard going down a slope, I speed up.  That means gravity force is providing me with thrust, increasing my speed.  Is it so unimaginable that it might do this?  If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will).  If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/12/2018 06:19:12
Thanks
I'm quite confused with the answers about the offset/thrust.
Please let me know if the following is correct:
The Thrust is a direct outcome of the offset. (While the offset was a direct outcome of the tidal friction).
Based on the offset phase you have calculated the thrust in thread 40:
The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.
Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.


Please be aware that I have specifically asked to set them at any offset as you wish.
So, does it mean that in a general case of offset there is no thrust?
If there is no thrust due to the offset, how can we use this idea for tidal friction?

It is only gravity, yes.  Work out the forces in 2 dimensions, not as a scalar.  Force is a vector, not a scalar.
If I am on a skateboard going down a slope, I speed up.  That means gravity force is providing me with thrust, increasing my speed.  Is it so unimaginable that it might do this?  If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will).  If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.
I like the idea of skateboard.
This shows that we can convert gravity force into thrust.
Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.
However, we can't limit the skateboard idea just for tidal friction explanation.
If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.
This is a breakthrough concept in gravity.
Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.
Our scientists believe that it is feasible to get thrust due for tidal friction offset as follow:

"1. There is no offset (0 degree) and the bulges are pointing directly to the moon
No thrust, right.  The difference between their force is irrelevant then.
 2. For offset 60 degree
Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.
 3. The offset is maximal (No, 90° - eastward and westward, not at the poles)
Still zero since they're equal and opposite."

Therefore, we should set a formula that represents this understanding.
So, can we set a formula (or graph) for thrust per Offset tidal friction phase?

However, If thrust works for tidal offset idea, it should also work for the example which I have offered.
If it doesn't work for this example, than it should not work also for tidal offset.
How can we come with idea for a special case and close it only for that case?
Do you agree that if we can't open the idea for any offset in orbital system, than we might have a problem with this idea?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/12/2018 14:44:58
However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

Please be aware that I have specifically asked to set them at any offset as you wish.
So, does it mean that in a general case of offset there is no thrust?
If there is no thrust due to the offset, how can we use this idea for tidal friction?
There is thrust in that case, but the forces will move your objects around, so the thrust might last minutes at best, after which the pieces will rearrange and produce negative thrust.  It all cancels out.  You've removed the large middle piece which is the source of the inertia and friction.

Anyway, if it is not considered a dynamic system (no time involved), that momentary arrangement of matter does indeed put a forward component of force on the moon.  I think perhaps that is what you were trying to ask.

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I like the idea of skateboard.
This shows that we can convert gravity force into thrust.
Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.
However, we can't limit the skateboard idea just for tidal friction explanation.
If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.
This is a breakthrough concept in gravity.
Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.
I beg to differ.  It all follows from Newtonian mechanics.  Kepler's laws are specifically about orbits and derive from Newton's equations, and he was quite aware of tidal effects.

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Our scientists believe that it is feasible to get thrust due for tidal friction offset as follow:

"1. There is no offset (0 degree) and the bulges are pointing directly to the moon
No thrust, right.  The difference between their force is irrelevant then.
 2. For offset 60 degree
Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.
 3. The offset is maximal (No, 90° - eastward and westward, not at the poles)
Still zero since they're equal and opposite."

Therefore, we should set a formula that represents this understanding.
So, can we set a formula (or graph) for thrust per Offset tidal friction phase?
We already did that.  The linear offset is computed from the sin of the angular offset, which is closer to 10 degrees than to 60.  The distance of each bulge to the moon is the mean distance R ± cos(angular offset).
That linear offset produces a forward and retrograde force component on the moon.   Forward thrust is total force(GMmr², where M is the bulge mass) * linear offset / distance of bulge to moon.

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However, If thrust works for tidal offset idea, it should also work for the example which I have offered.
If it doesn't work for this example, than it should not work also for tidal offset.
It does work.  It's just that if there is time involved, all the pieces move around and don't maintain that steady thrust.  That was my complaint.  You said the two bulges were in orbit about each other.  Orbit is a dynamic behavior, and has nothing to do with forces of just 3 masses in these specific places.  Two orbiting objects with the mass of our tidal bulges will not put any significant force on the moon.  In fact, the moon will simply exit the system due to its linear velocity.

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How can we come with idea for a special case and close it only for that case?
Do you agree that if we can't open the idea for any offset in orbital system, than we might have a problem with this idea?
Remove time from your example.  Then it is not an orbital system, but merely a bunch of masses at specific locations, with forces acting between each of them.  You can close it for that case.  With time eliminated, we need not consider anything's velocity or acceleration.  Only forces need be computed.
Ever take a statics class in college?  That's what the course is about:  Forces in the absence of time, and finding balances between them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/12/2018 16:06:09
Thanks Halc

The information about tidal friction is fully clear.
 
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/12/2018 16:00:34
I assume that by now we all understand how the tidal friction set the thrust which is needed to push away the Moon from the Earth.
However, the water bulges in oceans are very unique for the Earth/Moon system.
In one hand, this is the only planet in the solar system with so much water and on the other hand, the moon is relatively big enough to form the bulges.
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
They don't have water. They don't have relatively big moon around them. They don't form those water bulges.
So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/12/2018 18:08:14
However, the water bulges in oceans are very unique for the Earth/Moon system.
In one hand, this is the only planet in the solar system with so much water and on the other hand, the moon is relatively big enough to form the bulges.
The water makes a nice difference, yes.  The size of the moon is irrelevant.  If it was just an automobile up there, it might generate a trillionth of the gravity, a trillionth of the tidal effect on the ocean, and hence a trillionth of the force, which would result in the exact same acceleration.  The automobile would move to higher orbits at the exact same pace as does the moon, but the Earth would have a trillionth of the friction, so the spin would not degrade at a measurable rate like we get from our big honkin moon.

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We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.

The rings of Saturn is a nice example of a moon that did just that.  Too close, and tidal forces pulled it within the Roche limit where it breaks up.
Another example might be Venus, whose moon, if it had ever had one, would probably have gone around the normal way, causing it to drop its orbit that is retrograde to the spin of Venus itself.

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They don't have water. They don't have relatively big moon around them. They don't form those water bulges.
So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?
It works with the planet itself.  Even the relatively solid moon has strain in the crust from tidal forces from Earth, and those forces are really small since the moon is tide-locked, but it still rocks back and forth with its elliptical orbit.

Anyway, the sun is hardly solid and each of the planets form a tide on it.  The big planets are all very liquid and susceptible to significant tides.  The amazing story is Pluto and Charon, both rocks pretty much free of stuff that would exhibit tidal strain, and yet the two of them have managed to become mutually tidal locked already.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/12/2018 06:46:34
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.
Thanks
So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.
Now there is new idea.
All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".

1. Would you kindly explain this idea?
2. Can we prove it by mathematical calculations/formula?
3. Why our moon isn't pushed outwards due to this idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/12/2018 12:24:33
So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.
I think I said it was relevant to all of them.  They're all getting pushed out, except for the two exceptions I mentioned, both of which resulted in no moon.

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Now there is new idea.
All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".
Not a new idea.  Been around for at least 300 years.  The ones that don't do that don't survive, so the only ones left in our solar system are the ones that go slow, and in the same direction.

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1. Would you kindly explain this idea?
If they orbit faster than the primary, the tidal bulges will lag behind the orbit, and the resulting forces will put a braking action on the object.  Similarly if the orbit is retrograde (faster or slower), the force is braking, not thrust.  Energy is lost, so the moon falls down into the planet.
The ISS is slowly dropping its orbit due to this effect since it goes around every 90 minutes, 16x faster than the rotation of Earth.  Left alone long enough, and even absent friction from being too close to the atmosphere, it will eventually fall.

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2. Can we prove it by mathematical calculations/formula?
Yea, the same calculation used to compute the thrust force on the moon.  That calculation yields a negative number for a low or retrograde orbit.

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3. Why our moon isn't pushed outwards due to this idea?
Thrust is forward and positive, so the orbit of the moon increases.  So it is 'pushed outward' at a pace of a few cm per year.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/12/2018 13:02:09
Quote from: Halc
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.
Both our statements are wrong here, so I need to correct this.
I looked at Jupiter's moons as an example, and most of them are in fact not being pushed out, but have diminishing orbits.
Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.
The next 13 moons including all the popular ones have growing orbits.  That's just 13 of the 63 known moons.
The outer 48 moons all have, for some reason, retrograde orbits.  I suspect the reason is that they're all captured objects, and a free object will get a speed boost when passing a gravitational source in the direction of its orbit (not in the direction of its spin).  All the Voyager spacecraft have achieved the majority of their energy from such gravitational boosts.  The effect is to be flung away all the harder, not to be captured.  But if the object passes in front of the primary and outbound, it slows, possibly enough to be captured.  I think this is how each of the moons was captured.  Each needs to come in at greater than escape velocity and lose enough of it to drop below escape velocity.
I find that odd that Jupiter has so many retrograde moons, but none for any of the other planets.

All the moons have been there long enough to become tidal locked.  All those retrograde moons are so far out that their tidal effects result in negligible braking, so their orbits will probably take many billions to trillions of years before they fall into Jupiter.

Mars has 2 moons, and Phobos is dropping (seemingly faster than tidal effects can explain), and is expected to crash in a mere 40 million years.
Saturn has destroyed its last low-orbit moon, forming the current rings.  The lowest one is outside Saturn's geosync orbit and has positive tidal thrust.  All of Saturn's 60 moons have growing orbits.
Uranus has 27 moons, 11 of which are in low orbit and 'falling'.
Neptune has 14 moons, 5 of which are in low orbit and 'falling'.

None of these have retrograde moons.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/12/2018 17:01:33
Thanks for the explanation about the moons.

However, I have no clue why if the Moon orbits in one direction/periods it will be pushed outwards, while if it orbits in the other direction/periods, it should be pulled inwards.
In tidal friction, the offset of the bulges sets the Thrust?
In this case, there is no offset.
So, how can we get a negative or positive thrust without any offset?

Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.

Can you please explain why the orbital direction/periods can set a thrust?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/12/2018 05:51:20
Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.
Can you please explain how the orbital direction sets the offset?
Why orbital periods that are shorter than the primary can set a negative offset?
Do we have any drawing for that idea (as we have for the water bulges on Earth)?
So, how do we get any sort of offsets due to orbital direction or orbital periods?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/12/2018 12:39:58
Quote from: Halc
Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.
Can you please explain how the orbital direction sets the offset?
It is friction.  If my car is on a road, the brakes act as friction.  If the road is moving forward faster than the car, hitting the brakes will accelerate (positive thrust) to a higher speed.  If the road is slower than the car, or moving at any speed in the opposite direction, the brakes will slow the car down (negative thrust).

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Why orbital periods that are shorter than the primary can set a negative offset?
Because the friction of the slower primary drags the bulge to the rear, a negative offset.

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Do we have any drawing for that idea (as we have for the water bulges on Earth)?
They all look the same.  Bulges.  Offsets.  A negative offset puts the bulges on the opposite side.  A zero offset (which you only get with geosync orbits) are directly towards and away from the orbiting thing, and result in zero friction and zero thrust.

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So, how do we get any sort of offsets due to orbital direction or orbital periods?
Gravity tries to pull the offsets straight at the orbiting thing, but the spin of the primary in this case pushes the offset to one side or the other, depending which way the ground under the orbiting object appears to be moving.

If I look at Earth from a forward pointing ship in low orbit (400 km), things tend to appear in front of the ship and disappear behind it, just like the view from an airplane.   So friction with the ground below is going reduce my velocity, so steps are taken to avoid this friction as much as possible.  The airplane has optimal aerodynamics to minimize this drag.  The ISS is high enough to minimize atmospheric drag, but there's little it can do about tidal drag.

Same ship at geosync (36000 km):  The Earth below is now stationary.  I see the exact same spot below whenever I look.  Friction with that would produce zero thrust.

Same ship further out, (100000 km).  Now the Earth features appear from behind me and rotate away to the front.  Earth is spinning faster than my orbit.  Friction with that would push me forward (positive thrust).  The moon's orbit is in this last category (beyond 36000 km).
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/12/2018 18:38:25
Because the friction of the slower primary drags the bulge to the rear, a negative offset.
There are nine planets in the solar system.
We all know that there are bulges in our planet.
However, do we see any sort of bulge at any other planet in the solar system?
If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?
Please advice if you agree with the following:
Tidal friction is based on Bulges.
If there is no bulges there is no offset. If there is no offset there is no thrust. If there is no thrust there is no energy to push or to pull the moon.
So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/12/2018 19:30:31
There are nine planets in the solar system.
We all know that there are bulges in our planet.
However, do we see any sort of bulge at any other planet in the solar system?
If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?
I look at a picture of Earth and see no bulge.  They're not exactly pronounced.  Point is, the tidal forces produce stress on the bodies orbiting each other, and those stresses produce strain of one sort or another, and changing strain is movement that produces heat from friction.  These forces are strong enough to have tide-lock (nearly??) every moon of every planet.  Water isn't necessary.  The tides are quite significant in amplitude on planets/stars that are not rocks.  Venus has a thick atmosphere to drag around.  Pluto and Mercury are the only planets with nothing but solids to work with, and both those have become tide locked with the most significant gravity source nearby.

Please advice if you agree with the following:
Tidal friction is based on Bulges.
If there is no bulges there is no offset. If there is no offset there is no thrust. If there is no thrust there is no energy to push or to pull the moon.[/quote]
None of these assert that the deformation needs to be measurably confirmed from a distance, so yea sure.
Not all deformations manifest as something that can be classified as a bulge.  I can put a foam cube out there spinning, and tidal forces will deform it, but the corners will always stick out the furthest, and hence a given deformation will not necessarily be a bulge.  Even Earth has mountains that 'bulge' out far further than does the water.

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So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?
No.  The forces involved do not need confirmation in order to exist.  Their effects are measurably confirmed.  Most of the moons nearby their primaries have measurable changes to their orbits, Phobos probably being the top of the list, despite being very unlikely to ever produce a measurable deformation of Mars' atmosphere or crust until it physically hits them.

Are you just posting here as a denier of such forces?  You'd have to explain how one object can put stress on another object without producing strain.  That would be quite a piece of new physics to propose.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/12/2018 20:56:17
Thanks Halc.

Water isn't necessary.  The tides are quite significant in amplitude on planets/stars that are not rocks.  Venus has a thick atmosphere to drag around.  Pluto and Mercury are the only planets with nothing but solids to work with, and both those have become tide locked with the most significant gravity source nearby.
Why water isn't necessary?
Please remember that the tidal on Earth is based on water.
https://en.wikipedia.org/wiki/Tide
"The theoretical amplitude of oceanic tides caused by the Moon is about 54 centimetres (21 in) at the highest point"
If I remember correctly, at land the tidal is just few cm (2-3 Cm?).
The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.
Did we try to measure the Tidal impact on other planets?
If I understand it correctly, the offset is also due to water.
Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.
With regards to gas planets -
Let's look at Jupiter as it is the biggest planet in the solar system:
https://solarsystem.nasa.gov/moons/jupiter-moons/overview/?page=0&per_page=40&order=name+asc&search=&placeholder=Enter+moon+name&condition_1=9%3Aparent_id&condition_2=moon%3Abody_type%3Ailike
"Jupiter has 53 named moons and another 26 awaiting official names. Combined, scientists now think Jupiter has 79 moons."
It Radius is: 69,911 km
It's biggest moon is Ganymede:
https://www.google.com/search?q=Biggest+Jupiter+MOON&oq=Biggest+Jupiter+MOON&aqs=chrome..69i57j0l5.6920j0j8&sourceid=chrome&ie=UTF-8
"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.
Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.
I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter
Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.
There is one more issue.
For some moons we believe that there is a negative thrust.
This negative thrust should pull those moons inwards.
This is the theory.
However, did we measure if they are pulled inwards?
What is the chance that they will not be so cooperative with our theory?
Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?




Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/12/2018 00:58:08
Why water isn't necessary?
Please remember that the tidal on Earth is based on water.
Water has the greatest drag on Earth's spin since it has a lot of that, but atmosphere and crust also contribute.
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If I remember correctly, at land the tidal is just few cm (2-3 Cm?).
Sounds like strain to me.  It takes energy to do that, and that energy is lost to heat.

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The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.
Really?  The ratio is about 1.2%, far less than the 11% ratio of Charon to Pluto.  OK, Pluto isn't exactly a planet.

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Did we try to measure the Tidal impact on other planets?
Apparently you don't read my posts.

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If I understand it correctly, the offset is also due to water.
Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.

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Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.
I think those particular figures have quite been verified.  Earth tide offset isn't a fixed figure.  It varies all over the place due to geographic features and ressonance.  It's the friction that counts, and that friction is greatest in shallow areas like around England.


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"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.
Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.
Big time, yes.  All the easier for Jupiter to push it along.  Easy to push small things.

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I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter
Probably, yes.  The tide raised by the ISS is waaaay less than that millimeter, but the ISS is super light, so it accelerates (negative) just as much as a large thing would at that altitude.  That acceleration rate has little to do with the mass of the orbiting thing.'

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Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.
They orbit at different periods, so there is zero chance of this.  Only a moon's own tides affect that moon, not the tides of other bodies, which have random offsets and thus cancel completely in the long run.

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There is one more issue.
For some moons we believe that there is a negative thrust.
Like Phobos and most of Jupiters moons, yes.

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However, did we measure if they are pulled inwards?
Jupiters outer moons are so far out that the tides might not have measurable impact.  The two inner ones very much do have measurable orbit degradation.  Phobos has massive degradation, and has only some tens of millions of years left in its life.

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What is the chance that they will not be so cooperative with our theory?
Pretty much nill.  It isn't exactly an untested theory.

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Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?
Making up your facts I see.  We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.  Most of Jupiter's outer moons have only been discovered in the last 15 years, which is not much time to measure the trivial orbital changes put on them at their very distant and very eccentric orbits.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 14:34:25
For some moons we believe that there is a negative thrust. This negative thrust should pull those moons inwards. This is the theory. However, did we measure if they are pulled inwards?
Like Phobos and most of Jupiter's moons, yes
Jupiters outer moons are so far out that the tides might not have measurable impact.  The two inner ones very much do have measurable orbit degradation.  Phobos has massive degradation, and has only some tens of millions of years left in its life.
Making up your facts I see.  We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.


Why do we ignore the distance between the moon/rings to the planet?
https://en.wikipedia.org/wiki/Phobos_(moon)
"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, closer to its primary body than any other known planetary moon. "

So, Phobos orbits 6000 Km from the surface of Mars while the radius of mars is about 3400 km. so the ratio is 1 to 1.76.
As an example:
The Earth Radius is 6370 Km. If Phobos would orbit around the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.
That orbital radius is lower than the altitude range of navigation satellites:
https://en.wikipedia.org/wiki/Medium_Earth_orbit
"The most common use for satellites in this region is for navigation, communication, and geodetic/space environment science.[1] The most common altitude is approximately 20,200 kilometres (12,552 mi)),"

If I understand it correctly, we expect that a navigation satellite should be pulled inwards.
If so, why is it so big surprise that Phobos is also pulled inwards?

With regards to Jupiter moons/Ring:
https://en.wikipedia.org/wiki/Rings_of_Jupiter
"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]
So, Jupiter radius is 71398 Km.
Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km.
So, don't you see a similarity between satellite around the Earth, Phobos around Mars and Main ring around Jupiter?
Don't you agree that with tidal or without it, all of them must eventually fall down?
Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???
Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?

With Regards to Europa:
Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.
https://www.space.com/15498-europa-sdcmp.html
Distance from Jupiter: Europa is Jupiter's sixth satellite. Its orbital distance from Jupiter is 414,000 miles (670,900 km). It takes Europa three and a half Earth-days to orbit Jupiter. Europa is tidally locked, so the same side faces Jupiter at all times.
The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "
So, It is located far enough from Jupiter. Ratio of about 1:10.
It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).
So, I don't know what kind of information we can extract from Europa:
Does it have tidal bulges? Do those bulges set the positive/negative offset? Do we know if it is pushed outwards or pulled inwards?
I assume that the answer is - No, we don't know as it is too far away to measure.
So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 14/12/2018 15:38:59
Why do we ignore the distance between the moon/rings to the planet?
I don't ignore that.  Jupiter's two innermost moons are falling because the distance from them to the planet is less than the geosync radius of Jupiter.

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As an example:
The Earth Radius is 6370 Km. If Phobos would orbit the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.
A commercial airplane is normally fly at about 10,000 Km.
If it is expected that airplane should come/fall down, why is it so big surprise that Phobos is also coming/falling down?
Airplanes fly at 10 km, not at 10000.  The ISS is only about 400 km up.
Airplanes fall because gravity accelerates them downward, and their orbits are much much smaller than the radius of Earth, so they'll hit the ground without effort to keep them aloft.  The ISS needs no wings because its orbital path does not include the ground in its way.

Anyway, phobos falls for the same reason that an airplane slows if it runs out of fuel: Friction with some moving slower than itself.

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With regards to Jupiter moons/Ring:
https://en.wikipedia.org/wiki/Rings_of_Jupiter
"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]
So, Jupiter radius is 71398 Km.
Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km. (Same distance at commercial airplane on Earth).
No airplanes there.  Rings form when moons pass below the Roche limit and are torn apart by tidal forces.  That ring on Jupiter has Metis in it as well, and Adrastea is probably a chunk torn off Metis, and the ring is all the shrapnel from that destruction.

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So, don't you see a similarity between airplane orbit around the Earth, to Phobos around Mars and Main ring around Jupiter?
Phobos is already starting to break up, so it will form a ring around Mars.  Plenty of similarity there.  Airplanes do no fly in outer space.  A suborbital ballistic airliner might (they don't have any right now), but not anywhere near that high up.

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Don't you agree that with tidal or without it, all of them must fall down?
Without tides, neither Phobos nor Adrastea nor an airplane orbiting at 10000 km will ever fall down.  With tides, all of them will eventually.

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Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???
Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?
You seem to want to simply dismiss any explanation as 'not proof'.  It is easy to demonstrate violations of conservation of angular momentum and make an infinite energy engine if tides produce no thrust on moons.  They already have power generators that harness tidal energy.  Where do you think it comes from then?

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With Regards to Europa:
Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.
Europa is tidally locked, so the same side faces Jupiter at all times.
The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "
So, It is located far enough from Jupiter. Ratio of about 1:10.
It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).
Those tides below the ice have managed to halt Europa's spin, so I'd hardly call that minimal.  The forces putting thrust on its orbit are the tides raised by Europa on Jupiter's atmosphere, not those raised on Europa.  Yes, a moon's own spin contributes to this, but most moons have lost that spin already.

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Does it have tidal bulges?
A permanent deformation is not tidal, so no.  The deformation on Earth that makes sea level at the equator a larger radius than at the poles is not considered a bulge because it is permanent.  Tides are strain, some kind of back and forth motion that requires energy to maintain.  The bulge on Europa is permanent, which is why I said 'no matter'.

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Do those bulges set the positive/negative offset?
The offset of Europa's bulges averages zero.  That's because it is tide locked.
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Do we know if it is pushed outwards or pulled inwards?
Outward.  Geosync of Jupiter is around 170,000 km, and Europa is beyond that, and moving with the spin.  So thrust is positive.

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I assume that the answer is - No, we don't know as it is too far away to measure.
The answer is yes because it would violate conservation laws for it to be otherwise.
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So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?
The laws of physics work everywhere, not just where humans confirm them in court.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 15:56:02
Airplanes fly at 10 km, not at 10000
By the time that I fixed my message, I have got your answer.
So yes, you are absolutely correct.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 16:28:53
The laws of physics work everywhere, not just where humans confirm them in court

Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
Do you agree that this law contradicts the whole idea of tidal friction, as there is no positive or negative thrust if we set the whole mass at the center?
We are using this Newton's Shell Theorem in order to explain many aspects of the Universe.
So, if the laws of physics work everywhere, why this law can't be used here?
Why I can't use it in order to prove that there is no positive or negative thrust due to offset?




Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 21:18:04
Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
Right, but the theorem works only for a very regular object, a sphere in particular.  All kinds of funny things can be done if the objects are irregular.  I can take two objects and put the centers of gravity very close to each other and actually get them to repel each other.  I just can't do it with spheres.

How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.
http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html
"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).
Since the mass M includes the mass of the suppermassive black hole at the galactic center, M is guaranteed to be much greater than M*, the mass of a single star."
So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.
Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.
However, when we come look at the Earth, suddenly it is not a regular object.
Is it real?
How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.
Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun), while for a real regular object (as the Earth) we claim that it is not regular enough.
Sorry, that isn't the way that we have to work in science.
If the Earth isn't regular enough for the Newton's Shell Theorem, than by definition the matter in the orbital radius of the Sun shouldn't be considered as a regular.
How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun, while we reject this idea just because it contradicts our theory about tidal friction?
Therefore, If the none regular matter in the orbital radius of the Sun should be considered as a regular, than by definition the Earth with or without the Bulges should be considered as a real regular.
Don't you see that severe contradiction???
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 01:44:35
How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.
http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html
"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).
That quote is not Newton’s shell theorem, which concerns spherical objects.  That law is a similar one showing how orbital speeds are directly related to the mass of the material orbited, and the orbital radius.  The moon follows that law.  The law says nothing about additional forces resulting from the non-uniformity of the local gravitational field. Stars are flung out of their normal orbits all the time due to close encounters with passing objects.  The law you quote doesn’t prevent that.

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So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.
The calculation of the speed of the orbit if it is a reasonably circular orbit, yes.  It is not a calculation of the local forces making changes to that orbit.

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Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.
However, when we come look at the Earth, suddenly it is not a regular object.
The galaxy isn’t regular either.  Again, consider stars being diverted out of their orbits, which happens a lot, especially in the crowded places near the halo.

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Is it real?
How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.
Earth is a very small object and far less symmetric than the galaxy, just like a rock in my yard is hardly a sphere.  Our planet has this giant moon, making it even less symmetric.  Apples to Oranges to try to compare that to the far more uniform galaxy.  Even then, it takes only one passing rouge star to disrupt our entire solar system.  We’re just luck we’re so far out that those kinds of events are really rare.

BTW, the galaxy is not a uniform sphere.  Not even close.  Newton’s law just doesn’t apply, but the 2nd lay you quote above still does.

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Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun)
There is no spherical object inside the orbital radius of the sun.  The sphere is a mathematical one.  All the crap inside this radius, and not the stuff outside.  That stuff is anything but a sphere.  It resembles more of a middle-heavy pancake.
The Earth on the other hand is actually a sort of sphere, but with significant deviations from being symmetric.  The galaxy below us has no such significant deviations until one pass by at least.

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How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,
We didn’t.  We used that 2nd rule about only the matter within the orbit contributing to the velocity.  You can use that rule on the Earth/moon system as well.  It works just great.  It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS.  That’s what the rule says.

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while we reject this idea just because it contradicts our theory about tidal friction?
It doesn’t go into tidal forces at all.  Tidal forces are not a violation of the rule.  The orbital speed of the moon never deviates from what the rule predicts.  It is at all times based on the (essentially fixed) mass of the Earth and its satellites, which is the list of all the stuff inside the Moon’s orbit.
The rule has no requirement that the distribution of that matter be regularly arranged.  Newton’s law does require that uniformity.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 05:26:43
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How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,
We didn’t.  We used that 2nd rule about only the matter within the orbit contributing to the velocity.  You can use that rule on the Earth/moon system as well.  It works just great.  It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS.  That’s what the rule says.
What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
If so, this law isn't relevant for our discussion.
It doesn't say anything about the mass in the sphere of orbital cycle of an object:

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
Please advice.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 07:51:02
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What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.
The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.
I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.

The "second rule" is just a direct outcome from the "first rule" which is the Newton's Shell Theorem.
With all the respect to ohio-state (and I have a respect...), they can't just invent new rules for gravity.
So, the second statement is not a second rule, it is just a logical outcome from Newton's Shell Theorem.
They also explain how to extract M (M = mass inside star's orbit (in solar masses) ) from kepler law.
"Each star in the disk is on a very nearly circular orbit, anchored by all the mass enclosed within its orbit, whether it's luminous or not. Thus, the amount of mass within a star's orbit can be determined from Kepler's Third Law:"

Hence, by Kepler law we can calculate the total mass which is requested to meet the orbital velocity of the sun around the galaxy, while Newton’s First Theorem tell us that: "M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit)" and therefore, we also can understand that:
"Since the mass M includes the mass of the suppermassive black hole at the galactic center, M is guaranteed to be much greater than M*, the mass of a single star."
In any case, if that was not clear enough, please see the following statement:
http://users.math.cas.cz/~krizek/cosmol/pdf/B102.pdf
Newton’s First Theorem - If the density distribution of a ball of mass M is spherically symmetric, then the size of the force between the ball and a point mass m, that lies outside the interior of the ball, is given by the left-hand side of (1), where r is the distance between the point and the center of the ball."
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 11:25:06
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"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
Please advice.
I don't see the relevancy of that law to this portion of the discussion.  It says that the center of gravity of something like Earth follows a smooth curve even if the various parts (like your mailbox) don't.
This law is very relevant
It actually confirms that the shape of the object is none relevant for its central point of mass.
As long as all the masses in the object are fully connected the center of mass of this object is none relevant with its shape. (Spanner, dog, cat or even elephant).
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"A set of masses connected by springs will follow a path such that its center of mass moves along the same path that a point mass of the same total mass would follow under the influence of the same net force."
So, even if that spanner has an offset, it won't set any extra thrust. In our calculation we just need to focus on its center of mass.
In the same token, as the bulges are fully connected to the Earth, they are part of the whole Earth body. Therefore, in our calculation we need only to follow the center of the mass of the earth.
In other words - We won't get any extra negative or positive thrust due to mountains, oceans or any sort of bulges with or without offset while the whole masses are connected together.
That law by itself proves that the idea of thrust due to bulges is totally incorrect.
Conclusion -
Based on the following laws (each one by itself):
1. Newton's Shell Theorem
2. Newton's second law
I have proved that the shape of the object can't issue any sort of extra thrust. Newton didn't specify that idea in his laws. Therefore, the hypothetical idea of extra thrust due to the offset (or special shape of the object - with or without bulges) is incorrect.
Our scientists must find better idea why all the planets and all the far enough moons are drifting outwards.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 13:10:03
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What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.
The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.
I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.
The "second rule" is just a direct outcome from the "first rule" which is the Newton's Shell Theorem.
With all the respect to ohio-state (and I have a respect...), they can't just invent new rules for gravity.
So, the second statement is not a second rule, it is just a logical outcome from Newton's Shell Theorem.
I'm not disputing the rule.  It is a known thing, yes.  I just don't know the name of it, or how it is derived from Newton's sphere law.

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Hence, by Kepler law we can calculate the total mass which is requested to meet the orbital velocity of the sun around the galaxy, while Newton’s First Theorem tell us that: "M is the total mass in a sphere of radius a, centered on the galactic center.
You're applying the 2nd rule there, not the first theorem.  The galaxy is not a sphere nor is it spherically symmetric, so the first theorem does not immediately apply.  The nameless law does however, so we're good.

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Newton’s First Theorem - If the density distribution of a ball of mass M is spherically symmetric, then the size of the force between the ball and a point mass m, that lies outside the interior of the ball, is given by the left-hand side of (1), where r is the distance between the point and the center of the ball."
Slightly generalized version, and thus a better one, yes.  It doesn't require a sphere, but merely something spherically symmetric.  But the galaxy isn't, so this law doesn't directly apply.   The galaxy is roughly modeled as a disk, and the force vector of a point out of the plane of a disk does not point to the center of mass of the disk like Newton's theorem says it must if it were spherically symmetrical.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 14:07:14
This law is very relevant
It actually confirms that the shape of the object is none relevant for its central point of mass.
I didn't say otherwise.  The shape of Earth for instance, with its bulges, has no effect on Earth's center of mass.  I agree with that.

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As long as all the masses in the object are fully connected the center of mass of this object is none relevant with its shape. (Spanner, dog, cat or even elephant).
All the others, yes, and even a spraying garden hose, but not the cat, which has properties outside physics.

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http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"A set of masses connected by springs will follow a path such that its center of mass moves along the same path that a point mass of the same total mass would follow under the influence of the same net force."
OK.  That one is pretty obvious too, since it derives directly from the computation of center of mass.  The springs are not even necessary.  It can be a collection of 13 random stars from nowhere in particular, and that collection of 13 unrelated stars will have a center of mass that will follow this rule.  What it doesn't mean is that an object at that center of mass will follow the same path.
What it especially doesn't mean is that any force exerted by that collection of spring-attached objects is going to be the same as if all the mass was concentrated at their center of gravity.  This is very easy to demonstrate.

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So, even if that spanner has an offset, it won't set any extra thrust.
The rules don't say that at all.  They talk about net force and those net forces acting on the center of gravity of each object in question.  So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.

The Earth/Sun system could be considered one such object which is not a sphere, and thus it puts massive thrust on another object (the moon), all because of its offset.  According to your assertions, that should not happen, and the moon should wander off around the sun on its own in a nice circular orbit around the center of mass of the combined Earth/Sun object.

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In our calculation we just need to focus on its center of mass.
Wrong!!  Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum.  None of the laws above describe angular effects on the tumbling spanner and such.

I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground.  The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin.  That rules is inadequate for the situation being described.

The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on.  You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical.

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Conclusion -
Based on the following laws (each one by itself):
1. Newton's Shell Theorem
2. Newton's second law
You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

This is the sort of proofs I'm seeing from you.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 15:27:26
Thanks Halc.
I really appreciate all your efforts.
However, please try to use some solid evidences (mathematics, Newton/kepler.. laws) in order to prove your statements.

The rules don't say that at all.  They talk about net force and those net forces acting on the center of gravity of each object in question.  So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.


Can you please prove that a spanner can put a net force on an object just by pointing to some offset?

You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

That example is not clear to me.
If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.
Therefore, the Sun has no impact at all about my location.
If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.

With regards to the following message:
"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"
This is a severe mistake.
If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.
So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.
Newton didn't specify even one word about the impact of spinning velocity.
This is a new idea which had not been confirmed. It seems to me as a new idea which we have invented just in order to offer a nice answer for: Why the Moon and the earth are drifting outwards?
If you believe that there is a possibility to slow down the earth orbital velocity "because Earth's spin velocity cancels some of its orbital velocity at noon", than please prove it.

Wrong!!  Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum.  None of the laws above describe angular effects on the tumbling spanner and such.

I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground.  The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin.  That rules is inadequate for the situation being described.

The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on.  You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical
We do not discuss about rapidly spinning pool of noodle.
We discuss about Sun/earth/Moon orbital cycles.

If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.
I still can't understand how any sort of offset in bulges can set any sort of thrust.
Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.
Please try to prove this idea by mathematics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 18:23:08
So far we have just discussed about the impact of the orbital cycles of Planets and Moons.
However, what about the stars? What about our star - the Sun?
Why all our scientists are positively sure that during all his life time the Sun had to keep the same orbital radius?
How could it be that all the moons and Planets are drifting outwards, (or inwards based on unproved tidal idea) while the sun is fixed at the same radius?
How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
What makes our star so unique that it had to stay so far away from the monster at the galaxy center?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 22:05:16
Can you please prove that a spanner can put a net force on an object just by pointing to some offset?
I didn't say it could do it by pointing to an offset.
The spanner is tumbling, and that tumblingis slowing measurably.  It couldn't do that if it was treated as a point mass as you are attempting to do.  You can't put torque on a point.

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Quote from: Halc
You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.
That example is not clear to me.
If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.
You're not integrated with Earth since you are completely detached from it. Seatbelts are cheating.  I'm just considering the gravity of the sun on your orbit.  According to that 2nd law, or at least according to the way you are using it, nothing outside that R sphere has a net effect on my motion, therefore, since Earth is entirely outside that sphere, its gravity doesn't affect me.

I know it's wrong, but that's how you're interpreting the rule: by oversimplifying and not considering deviations from the uniformity.  The Earth is a huge deviation.  Deviations matter.  The tidal bulges are deviations, and the change the direction of the force vector acting on the moon just as Earth changes the force vector acting on me at high noon.  But you cherry pick which deviations matter and which do not, depending on what purpose you desire.  That's very fallacious reasoning.

I haven't figured out your purpose in all this.  Clearly not here to learn.  You seem bent on finding contradiction in simple orbital mechanics, but not sure why.

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Therefore, the Sun has no impact at all about my location.
If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.
So rule #2 suddenly doesn't matter when you find it inconvenient?

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With regards to the following message:
"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"
This is a severe mistake.
If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.
My velocity relative to the sun varies daily, fastest at midnight and slowest at noon.  Somewhere in between is proper orbital speed at this radius.  At midnight the speed is too high, and at noon it is too low, as per Kepler's 3rd law.  The speeds are easy to compute:  About 30 km/s ± about 0.5 km/s speed when the sun is directly below and above me respectively.

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So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.
Newton didn't specify even one word about the impact of spinning velocity.
This is a new idea which had not been confirmed.
What, that your speed relative to the sun is lower at noon?  That's pretty obvious I'd think.

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If you believe that there is a possibility to slow down the earth orbital velocity
My speed, not Earth's speed.  I'm the one that suppose to fall closer to the sun because I'm moving slower than Kepler says I should.
Of course Earth speed is slowest every full moon, for the same reason.  The difference isn't as much as ±0.5 m/sec.

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We do not discuss about rapidly spinning pool of noodle.
OK, you don't know what a pool noodle is.  It's like the spanner, but more susceptible to angular forces.

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If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.
That was demonstrated many posts ago.  I should the vector math quite a ways back.  The force vector on the moon is not perpendicular to its motion, therefore there is work being performed (energy transfer).

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I still can't understand how any sort of offset in bulges can set any sort of thrust.
Understanding it would cause your argument to fall apart, therefore nothing I can say will make you understand it.  I'm fine with that.

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Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.
Please try to prove this idea by mathematics.
OK, there is a spanner suspended exactly through its center of gravity on a frictionless axis.  It should be able to spin freely, and be balanced.
I give it a permanent stationary offset of 45°.  The spanner is 40 cm long, 20 on each side of the axis.  The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it.  The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force. It will start to rotate to vertical because the bottom is attracted more than the top, being closer to Earth.  That is a torque force being put on an object despite its center of gravity never moving or the force on that center of gravity ever changing.
You can get rid of the axis suspending the spanner if you just put it in orbit and let its speed hold it up there.

The numbers get more significant if the spanner is a bit longer, like say over 12000 km long, even if the 12000 km spanner has an offset far less than 45°
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2018 06:49:42
Thanks
Great example
I give it a permanent stationary offset of 45°.  The spanner is 40 cm long, 20 on each side of the axis.  The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it.  The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force.

However, I really don't understand why there is a difference in forces.
You discuss about 20 cm.
Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km. 
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:
F = M a
In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"Newton's Second Law for a System of Particles:
The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.
The center of mass of a system of particles can be determined from their masses and locations."
So, Our dear Newton set a complicated calculation in order to get the outcome of:
M a = F
Newton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.
So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/12/2018 13:24:06
However, I really don't understand why there is a difference in forces.
I modeled the spanner as two masses separated by 30 cm (I did not put all the mass at the ends of the 40 cm spanner), and applied Newton's F=GMm/r² to get the force, and then F=ma to get the acceleration of spanner that sets it in motion.
In addition, if the Earth under it was entirely stationary and even perfectly spherical at first, it too begins to rotate by application of Newton's conservation of angular momentum law.  The moon does not do this to the Earth because unlike the spanner, it has no net offset and thus exerts no net torque directly on Earth.  All Earth's torque is due to friction, and our spanner example does not involve friction.  We can put some air around it so it stops rotating after a while.

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Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km. 
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:
F = M a
Yes, the motion of the center of mass will follow that law, just like it says.  I never said otherwise.

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In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"
We need to come up with different names for all these laws, because you're calling them all Newton's second law of something, which is confusing.  Newton's second law typically refers to the second law of motion, which is F=ma.  So lets just call the one above the 'extended object law' and this one below the 'law for a system of particles'.

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http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"Newton's Second Law for a System of Particles:
The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.
The center of mass of a system of particles can be determined from their masses and locations."
That's not the law you quoted before.  It's just a statement of the fact that the center of mass of any set of particles (connected or not) can be computed.  This lacks any description of how that computation is done (which isn't hard), and it doesn't mention net forces like the last time you mentioned it.  Maybe this is a different law.
Anyway, this wording just says there is a COM and that is it computable.  Hardly a revolutionary statement, and not something that I would call a 'law'.
The 'law' does not say that the set of particles will exert gravitational force as if they were all at the center of mass, or that the center of mass will follow the same path as the object with all its mass concentrated there.  This is very easy to demonstrate with just a two-particle object.
And yet I find you here misrepresenting these laws and attempting to do just that.

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So, Our dear Newton set a complicated calculation in order to get the outcome of:
M a = F
Newton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.
He said no such thing. That does not follow from any of the laws you've quoted.

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So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).
Particles are never connected.  Two things cannot touch.  None of Newtons laws about sets of objects require connectivity for this reason.  The center-of-mass of the Earth-moon system follows an ellipical path around the sun, which the center of mass of just Earth does not because the moon is yanking back and forth each month.  So treating the two as one unit gives a far smoother curve, even though the two are not physically connected.

This is not always the case.   Earth and Venus could be connected with a thin spidery thread and the center-of-mass of the single object would very much obey that law you quote (you quoted a much better version of it in a prior post), but that COM does not follow the same path as would an object that was actually all at that spot, and does not generate a gravitational field that is identical to a single mass at the COM.

Point is, the connection is not necessary.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2018 15:59:18
Particles are never connected.  Two things cannot touch.  None of Newtons laws about sets of objects require connectivity for this reason.
This is a sever mistake.
All particles on earth are fully connected due to gravity.
In the following law it is stated clearly:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other."
It is also stated:
"Since nothing we have done addressed whether the particles are connected or not, this result generally applies to a system of discrete particles or to an extended object consisting of connected mass elements."
So please - why don't you agree with those simple and clear evidences???

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Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km. 
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:
F = M a
Yes, the motion of the center of mass will follow that law, just like it says.  I never said otherwise.
So, you agree that the center of mass of any kind of object at any shape (even if it is 20,000 Km spanner) must follow that law.
If so, you agree that there is only one center point of mass for any kind of object.
Therefore, there in nothing to disagree.
With only one center point of mass there is no way to get any extra thrust.
I hope that you agree with that.

Conclusions:
I didn't invent those laws.
You are more than welcome to call them at any name. However, those laws prove that there is one center of mass to any shape of object and at any offset.
Therefore, the assumption that there is an extra thrust due to offset contradicts those fundamental laws for gravity.
You didn't offer any alternative mathematic calculation in order to reject those laws and support the hypothetical idea of extra thrust due to bulges offset.
Hence, our scientists must look for better explanation why all the planets and moons are drifting outwards.
I'm specifically using the words "drifting" as I have proved that there is no extra thrust that can push them outwards.
I have no more questions about tidal friction.

With your permission, I would like to focus now on the Sun orbital cycle.



Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2018 19:14:23
I hope that you agree with the following:
1. The spanner has a center point of mass.
2. If we won't set an external force of the spanner there is no extra thrust.
3. In our case the spanner represents the Earth.
If you agree with all the above let's move to the following:

You claim that:
If I put force on the spanner, or net force on the system of completely disconnected objects, that will move the center of gravity, which is thrust.

Now
1. Let's use the idea that all the particles on Earth are connected.
In this case, we will assume that the tidal is so strong that it coverts totally the shape of the earth from ball shape into spanner. (Instead of just two bulges)
So, what?
The Earth in a spanner shape has exactly the same center point of mass as the Earth in ball shape (as all the particles are connected).
So, that external force (tidal) didn't change at all the location of the center point of mass (although it changed the earth shape).
Therefore, there is no way to set extra thrust on the moon.
Do you agree with that?

2. Let's assume that the particles on earth aren't connected.
So, the tidal is so strong that it split the earth into two totally separated objects.
Each object is represented by center point of mass. We can set those two center points of mass exactly where the bulges are located.
In this case I agree that the gravity force between each point to the moon is not equal. and the sum of their gravity force is different from just one central point of mass.
However, even in this case, there is no extra thrust.
(Assuming that we can hold those two points of center of mass at a constant distance and offset with regards to the moon, than the moon will orbit around those points without getting any sort of extra thrust.

3. We will split the Earth into infinite separated points of mass while we hold them at the same spot and the same offset from the moon.
Even in this case there is no thrust on the moon. However, the moon will orbit around infinite number of center of mass points instead of only one or two.

Maybe it is my limited understanding, but somehow I really don't see any way to transfer any thrust from the Earth to the moon due to tidal.
With your permission, let's move on.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/12/2018 04:20:28
Thanks halc

Let's agree that we don't agree on the idea that there is a constant thrust on the moon due to tidal bulges offset on the Earth.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/12/2018 15:09:07
With regards to the Sun

Why are we so sure that during all of his life time the Sun had to keep the same orbital radius?
How could it be that all the moons and Planets are drifting outwards, (or inwards based on the tidal idea) while the sun is fixed at the same radius?
How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
What makes our star so unique that it had to stay so far away from the monster at the galaxy center?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/12/2018 21:18:16
With regards to the Sun

Why are we so sure that during all of his life time the Sun had to keep the same orbital radius?
Well not at the exact same radius since it is under the effect of all the local stars and such.  But it is moving around the galaxy in a pretty good imitation of a circle (having done around 20 laps so far), and if it had been seriously deflected by some large passing object, it would very unlikely be deflected onto this fairly perfect orbit.  So it seems to have been free of that sort of thing the whole time.  That's a large reason why we're here: We've had a nice stable environment that wouldn't have existed if that large mass had gone through and disrupted the solar system.

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How could it be that all the moons and Planets are drifting outwards, (or inwards based on the tidal idea) while the sun is fixed at the same radius?
We have no proof that the radius has been fixed all this time.  The clean orbit tells us that we've never had a large close encounter.  The stars that have had one typically become halo objects.
As for tides, you can't raise a tide on a black hole, and I don't think the tides raised by Sgr-A on the sun have any significant impact on its orbital radius.  Tidal forces follow a inverse-cube law, and that means Sgr-A probably has less tidal effect on the sun than I do.

Tides cannot pull our sun in.  The solar system has positive angular momentum (inclination +63°), so if any of that is affected by Sgr-A induced tides, it will push us further out.  Even if we had negative angular momentum, there isn't enough of it to drop us anywhere close to Sgr-A.

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How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
4 million stars in 15 billion years is one every 4000 years.  Hardly a hungry monster.
Spacetime is bent near the SMBH, and things cannot orbit at all near them.  You need to move at lightspeed to orbit at a radius of 1.5 the radius of the event horizon.  Anything inside that is dragged in (or would need thrust to stay out), and stuff outside that but close still spirals in.  These are relativistic effects, not Newtonian effects.

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If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
A lot of that mass was already there before it formed into stars, but I suppose it all was outside the event horizon at some distant point, just like our solar system was all spread out before gravity pulled all that matter close enough together to make a star.

Things at the center of the galaxy are relatively congested.  Two stars pass close to each other and one goes faster and the other loses kinetic energy.  The slow star falls closer to the SMBH and the fast one shoots out and becomes a halo object.  Few things have a clean orbit that close in.

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So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
They're not.  There is no history we can consult.  We seem to be on the road (a clean orbit) still, so that's evidence that we've not left it, because it is hard to get back if you fall off the road, or at least it takes a lot more than 20 orbits. That's pretty strong evidence, but not proof.

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What makes our star so unique that it had to stay so far away from the monster at the galaxy center?
Most of the stars out this far are like that.  We're in the suburbs, and we're quite ordinary, not unique at all.  The big ones burn up faster, and the little ones burn much slower.  Scientists are not claiming that all the stars out here are heading for Sgr-A except us.
Look near Cassiopeia constellation for our doom.  That's the close encounter that will likely fling us out of our nice circular orbit in less time that it took life to evolve a human.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/12/2018 05:09:31
There is no history we can consult.  We seem to be on the road (a clean orbit) still, so that's evidence that we've not left it, because it is hard to get back if you fall off the road, or at least it takes a lot more than 20 orbits. That's pretty strong evidence, but not proof.
The solar system has positive angular momentum (inclination +63°),

The Sun has currently positive angular momentum (inclination +63°). In other words, it is moving upwards from the Galactic Disc lane while it orbits around the center of the galaxy.
It is expected that once it gets to the pick it should get down and cross the galactic lane downwards.
Hence, the Sun is Bobbing up and down while it orbits the center of the galaxy.
This is not unique for the Sun. Actually all the stars in the galaxy bobbing up/down, in/out or in between.
If I understand it correctly, the Sun should move up and down at least four times before it set one complete orbital cycle.
So, how can we call it "clean orbit"?
Do you think that Newton or keler would accept that bobbing orbital activity as a "clean orbit"?
Can you please find one planet or moon that is bobbing up and down while it orbits around its main host (four times per cycle)?
How can we ignore that incredible positive angular momentum???
Actually you have already offered the answer for that bobbing activity"
Well not at the exact same radius since it is under the effect of all the local stars
So, you even claim that "it is under the effect of all the local stars".
Hence, why don't we accept the idea that what we see is due to the gravity impact of all the local stars?
We all know that gravity works locally.
As we stay on Earth, we are under the gravity force of the earth (Not the Sun, Not the moon and not under the gravity force of the whole galaxy ).
The moon also works under the gravity of the Earth (Although the Sun gravity on the Moon is stronger by at least twice)
The Earth (or actually - the Earth/moon center of mass) works under the gravity of the Sun (While it ignore the gravity of the whole galaxy, and so on.)
So, why the Sun doesn't orbit under the effect of all the local stars???
In other words, what is the real host of the Sun?
If the Sun goes up and down, could it be that it actually orbits around some sort of a center of mass which is the equivalent center of all the local stars? So, could it be that it doesn't directly orbit around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
It is the same phenomenon as the Moon orbits around the Sun (12 times per cycle).
If we shut down the light at the Earth, we might see that the moon orbits around the Sun while it's bobbing inwards and outwards. (As the Moon orbits almost horizontally to the earth/sun disc). If the Moon was orbiting vertically around the earth, we would see that it is bobbing up and down as it orbits around the Sun. Almost identical to the Sun bobbing activity.
So, why do we reject the idea that the bobbing activity shows clearly that the Sun doesn't orbit directly around the center of the galaxy, but around some local center of mass which had been set by the "effect of all the local stars"?

If you don't agree with this idea:
Would you kindly show the formula of gravity which can support that strange bobbing activity or "clean orbit" of the Sun?
I assume that you might claim that it is bobbing due to the gravity of the galactic disc lane. If so, please prove this hypothetical idea by real mathematics based only on Newton and kepler.

In my opinion, this bobbing activity is the smoking gun which we are looking for.
I really can't understand how can we ignore so important activity.
Once we agree that the Sun orbits around a local center of mass (Which is "under the effect of all the local stars" - as the Moon/sun orbit), we will get a clear answer for the Spiral galaxy enigma.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/12/2018 15:10:48
Quote from: Halc
The solar system has positive angular momentum (inclination +63°),
The Sun has currently positive angular momentum (inclination +63°). In other words, it is moving upwards from the Galactic Disc lane while it orbits around the center of the galaxy.
Inclination is the tilt angle and has nothing to do with where we are going.  0° means the axis of the solar system is the same as the axis of the galaxy.  It isn't, it is off by 63°, which means it is more on its side than not.  Any inclination over 90° would give that solar system a negative angular momentum relative to the galactic axis.

What I said has nothing to do with us moving towards the center.  As I've said, our path is on average somewhat circular: the eccentricity is low enough to put us in the disk somewhere at a fairly narrow range of radius, but is otherwise pretty meaningless.  We take a wobbly path around the galaxy and have a speed somewhere between 225 and 250 km/sec most of the time.  The forces that change that are not tidal.

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It is expected that once it gets to the pick it should get down and cross the galactic lane downwards.
I know of nobody (except you perhaps) that expects this sort of thing to ever happen.

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Hence, the Sun is Bobbing up and down while it orbits the center of the galaxy.
This is not unique for the Sun. Actually all the stars in the galaxy bobbing up/down, in/out or in between.
So, how can you call it "clean orbit"?
You're right, its not very clean.  But we're in the plane and moving generally at the velocity needed to be at this current radius, so that's clean enough.

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Do you think that Newton or keler would accept that orbit as a "clean orbit"?
The orbit is described as being non-Keplarian, so no.

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If I understand it correctly, it should move up and down at least four times before it set one complete cycle.
Can you please find one planet or moon that is bobbing up and down while it orbits around its host (four times per cycle)?
Earth bobs up and down around 13 times each lap.  4 times?  No, I cannot think of one with that cycle.  I'd not heard of it for the galactic orbit.  Is there something nearby that we go around every 50 million years or so?  Is that a known thing?

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How can you ignore that incredible positive angular momentum???
Same way I ignore the moon's incredible change in positive solar angular momentum every 14 days.  It all averages out in the end.

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Actually you have already offered the answer for that bobbing activity"
Well not at the exact same radius since it is under the effect of all the local stars
So, you even claim that "it is under the effect of all the local stars".
Yes, that's right.  We don't exactly orbit Sgr-A, which is tiny compared to the mass of the galaxy inside our orbital radius.  Not at all like the Earth, dominated by that one huge mass in the middle with nothing but small objects giving minor but regular deflections from that perfect elliptical path.


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The Earth works under the gravity of the Sun (While it ignore the gravity of the center of the galaxy, and so on.)
Earth accelerates due to the galaxy exactly as much as does the sun.  Gravity is never ignored.  We are inside the hill radius of the sun, so the galaxy in general isn't going to separate the two of us.  A single passing object might.

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So, why the Sun doesn't orbit under the effect of all the local stars???
The sun is within the hill radius of no 'local' dominant mass.  Perhaps there is a cluster that holds together with mutual attraction, but then the orbit of the cluster still goes around the galaxy.  Earth has also made 20 trips around the galaxy, despite the continuous change in velocity.

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In other words, what is the real host of the Sun?
Don't know.  Don't know if there is one before the galaxy as a whole.  Something must hold the galactic arms together, but that's not really an orbit.


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If the Sun goes up and down, could it be that it actually orbits around some sort of a center of mass which is the equivalent center due to all the stars/SMBH in the galaxy?
That would be orbiting the galaxy, and it wouldn't go up and down due to that, except for normal orbital eccentricity just like Earth and every other planet.  Mercury radius changes an awful lot each orbit, but that is no indication that it is orbiting some mass other than the sun.

If there is some local mass around which all the nearby stars cling, then what really counts is the orbital speed and radius of the center of mass of that local group, and not the individual velocity of any particular object that is part of that group.

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So, could it be that it is not directly orbits around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
Exactly.

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It is the same phenomenon as the Moon orbits around the Sun (12 times per cycle).
If we shut down the light at the Earth, we might see that the moon orbits around the Sun while it's bobbing inwards and outwards. (As the Moon orbits almost horizontally to the earth/sun disc).
Right, but that bobbing in and out doesn't mean the moon has any chance of falling into the sun.  OK, it will, but only because the sun will come out here at grab it, not because the moon will fall in.

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So, why do we reject the idea that the bobbing activity shows clearly that the Sun doesn't orbit directly around the center of the galaxy, but around some local center of mass which is under direct "effect of all the local stars"?
I didn't assert that one way or the other.  The local masses are pretty well known, and I don't know if our neighbors are always our neighbors, or if they're just nearby right now.  A good text would say.  I'm pretty sure this sort of thing is known. 
Not sure if our motion relative to this local mass would be considered 'orbital' since there is no central mass to stabilize it.  It is sort of a 3-body problem with similar masses: the movements of the masses is chaotic, not at all following the rules of orbital mechanics.  Any one of the bodies might get flung away from the group, bringing the remaining members of the group closer together.  New masses might come from outside and lose enough speed to become new members of the group.

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If you don't agree:
Would you kindly show the formula of gravity which can support that strange bobbing activity or "clean orbit" of the Sun?
A=GM/r²
That describes the acceleration of any body given any number of masses spaced around it.

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Prove your hypothetical idea for that bobbing activity by real mathematics based only on Newton and kepler.
Integrate the formula above and you get the bobbing activity.  Kepler doesn't much come into play here.  It's all that simple Newton formula.

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In my opinion, this bobbing activity is the smoking gun which we are looking for.
Actually, I'm not really sure what you're looking for here.
Some single central mass (closer than SGr-A) around which we orbit?  No, we're not within any object's hill radius, so our local motion is not really best described as 'orbital'.

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Once we agree that the Sun orbits around a local center of mass (Which is "under the effect of all the local stars" - as the Moon/sun orbit), we have got the answer for the Spiral galaxy enigma.
Enigma?  The 'bobbing' is quite expected.  There are masses nearby, but none close enough for an orbit.  A local center of gravity may well come into play, loosely holding a group of sister stars together, but our motion around that is chaotic, not orbital.
I call them sister stars because most of the the local ones are probably born of the same parent supernova, which explains their somewhat similar mutual velocity as a group.  We're a 2nd generation solar system.  The 1st generation ones lack planets like our inner ones.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/12/2018 19:49:46
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So, could it be that it is not directly orbits around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
Exactly.
Wow

If we agree that the sun orbits around a local center of mass while this local center of mass orbits around the galaxy, than we have to agree that all the stars in the spiral galaxy has a similar orbital motion.
Based on that, now we can get better understanding about the Spiral galaxy.
For example -
When we look at the nearby stars, we might see that each one is moving at a different direction and velocity.
We might think that very soon they all will get out from the spiral arm.
This is incorrect. all the nearby stars will stay with us. Each one of them orbits around it's unique center of mass, while the center of mass is moving with the arm. and that center of mass is a direct product of all the other nearby stars...
Let's look at S2.
We think that it orbits around the SMBH. That is big mistake.
If we look carefully, we also might find that it's orbital cycle isn't clean.
In other words, it also orbit around some center of mass with this center of mass orbits around the SMBH.
This is a very important issue.
Let's take the example of the Moon/sun orbital cycle.
If we ignore the earth, than just by monitoring the orbital cycle of the moon and its mass estimated, we might think that the Sun mass is very low.
So, we can't really extract the real mass of the sun directly from the moon orbit.
We must first find the mass of the earth, and then extract the Sun mass.
In the same token, if we want to extract the real mass of the SMBH we must first find the estimated mass of the S2 center of mass point and based on that data we can extract the real value of the SMBH.
We might find that the mass of the SMBH is significantly higher than our faulty calculation.
Do you agree with that?
 
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/12/2018 17:05:16
Orbital speed is a function of 'reduced mass', which for any relationship except a binary star, is approximated by the mass of the primary.
V = √(Gμ/R) where G is the gravitationl constant, μ is the reduced mass of the pair of objects, and R is the radius.

Reduced mass μ for two objects is: μ = (m1-1 + m2-1)-1 which is pretty much the mass of the sun for Earth's orbit, and SGr-A (plus a couple hundred dark objects) for S2. 

That means that the moon would orbit at about the same speed as Earth if it was by itself, and S2 would orbit at the same speed even if it was the only visible part of some larger object that it orbited.  None of these objects have enough mass to affect their respective reduced masses.


Yes.
You are absolutely correct.
Sorry for my mistake.
However, with regards to the SMBH:
Why do we ignore the accretion disc?
Based on our measurements, the velocity of the plasma in that disc is 0.3 c (speed of light)
If we know the radius of the accretion disc (at the verified velocity), we can easily extract the real mass of the SMBH.
So, why do we insist on S2 which doesn't have a clean orbit instead of the accretion disc which has a perfect cycle orbit?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/12/2018 06:15:07
There is a lot of research going into dark matter, and these sorts of things help them estimate the MACHO component of dark matter (things that don't show light like Mars).
O.K.
As you have mentioned that issue - Dark Matter

Why Newton didn't add that brilliant idea of dark matter in his formulas?
Why he didn't say that any orbital system is based on dark matter?
Our science communities try to prove their hypothetical ideas by using Newton laws. As they have failed to explain the orbital velocity of our Sun, they came with this idea.
Now they even try to set more effort in order to show that this none realistic idea is realistic.
Let me give you an example -
Let's assume that we want to swim in a pool without water.
So, I will tell you that there is water, but the water is dark water.
You can't see it, you can't feel it, you can't smell it or drink it, but it is there because I said that it there. Would you believe me???

So, this is my personal opinion:
Dark matter is a solid proof for the failure of our scientists to show how spiral galaxy really works.
As they have failed to understand the real impact of none "clean orbit", as they have failed to understand the real impact of the Earth/moon "Drifting outwards", as they have failed to understand the real impact of the Ultra high magnetic power around the accretion disc, they couldn't explain how the gravity really works in Spiral galaxy.
I can do it.
I can explain how the whole universe works without any need for dark matter, for dark energy and for any sort of dark magic.
Based on Newton law I can easily explain:
How spiral galaxy really works?
Why there are spiral arms in spiral galaxy?
Why the stars in spiral arms have almost constant orbital velocity at any radius?
Why we see that all the far end galaxies are moving away from us at almost the speed of light, while our observable universe seems to be full with matter.
If we will come back 100,000 billion years (or even 10^Billion years) from now, we will see a similar universe with same density and almost the same numbers of galaxies per observable universe. We might not find our solar system, but there will be many similar.
Again - no need for dark matter or dark energy - just Newton law and simple common sense.
However, in order to do it I need the following basic understanding:
1. "Drifting" outwards
It's very difficult to verify a "drifting" outwards of few cm per cycle while the orbital radius is 1 Arc.  However, all the objects (assuming that they are far enough) are drifting outwards from their host. That is correct for all the objects including: Moons, planets, stars... There is no need for tidal support to explain this Phenomenon.
Therefore, any orbital cycle is by definition spiral shape cycle. So, even if it drifts 1 micro meter each cycle, it is still has spiral orbital shape.
2. Clean orbital cycle.
All the orbital cycles in the Universe must be clean. There is no room in our universe for none clean cycle (unless there is an interruption.)
So, the Sun must set a very clean orbital cycle around its center of mass. That must be correct for any star in the universe even if it is very massive star or object (unless there is no interruption).
3. The magnetic power around the accretion disc, push any matter upwards (or downwards) at a speed of 0.8 c (speed of light) that prevents from any gas cloud, star or even atom to drift inwards in order to be eaten by the SMBH monster.

That's all I need.
Based on those three elements, I can easily explain how our universe works without any sort of dark magic.
If you agree I will show how it works.


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 09:02:04
Dark Matter
Newton didn't need to.  His formulas do not apply only to matter that emits light.  Jupiter will orbit the future white dwarf that the sun will become at the same radius as the black dwarf it will be even later on.  The latter is dark matter.

Based on Wiki:
https://en.wikipedia.org/wiki/Dark_matter
"Dark matter is a hypothetical form of matter that is thought to account for approximately 85% of the matter in the universe, and about a quarter of its total energy density."
"Many experiments to directly detect and study dark matter particles are being actively undertaken, but none has yet succeeded."
This is my opinion about the dark matter:
The galaxy rotation problem - We don't know how to explain that problem based on our current hypothesis/understanding how spiral galaxy works.
Hence, instead of reconsider our hypothetical ideas about spiral galaxies and some other issues, we have found a brilliant idea of dark matter.
However, unfortunately we didn't find any evidence for that unrealistic idea: "Many experiments to directly detect and study dark matter particles are being actively undertaken, but none has yet succeeded."
Therefore, we have two options:
1. Continue to hold our none realistic ideas about spiral galaxies and continue to look for that unrealistic dark matter.
2. Open our mind to different ideas which can perfectly explain the galaxy rotation problem and all the other unsolved issues without any need to dark matter.

However, if I understand it correctly - Our science community reject the idea that they might have a problem with understanding how spiral galaxy really works. Based on their point of view, they only need to find the evidence for that dark matter.
So, how long do we have to wait until they will claim - "Sorry, there is a chance that if we don't find it - it is not there?
One year, 10 Year? Or one billion years?
Why you don't even give a chance to look for a different concept about our universe?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 16:02:05
So is there a model supplied by such open-minded thinking?  Trust me, such a thing would be warmly greeted if it worked.
Trust me.
There is a model that works by 100%.
However, I need your cooperation and good willing.
For just one moment try to forget all the wrong understanding/hypothetical ideas that we have about the galaxy (Including: Age of the Star/galaxy/Universe, dark matter/dark energy and so on).

There are three elements which are needed in this model.

1. The impact of gravity force due to local mass - The impact of gravity force due to local mass is stronger than the impact of gravity force of a very far end object. Therefore, the moon had selected to orbit around the Earth instead around the Sun, while the gravity force of the Sun/Moon is stronger by at least twice than the Earth/Moon gravity.
So, even if we shut down the light on Earth , the moon will continue to orbit around the Earth, while their mutual point of mass orbits around the Sun.
The outcome is as follow:
The moon orbits around a virtual host point (as we can't see the earth) while this virtual host point orbits around the Sun. So, the orbital cycle of the moon sets a clean orbital cycle around a virtual host point (Earth - which we can't see).
In the same token:
Every star in the galaxy must orbit around some host Point. It might be something that we see or something that we can't see. However, any star (at any size) must set a clean orbital cycle (in ellipse shape or a perfect cycle). If we can't see that host point, let's call it virtual host point. So, while the star orbits around that virtual host point, the host point orbits around the center of the galaxy. Therefore, we might think that the orbital cycle of a star around the galaxy is not clean.
With regards to our Sun
Please look at the following motion of the solar system in the galaxy:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
http://www.biocab.org/coplanarity_solar_system_and_galaxy.html


The broken gray line shows exactly the virtual host point of the Sun while it orbits around the galaxy.
So, the Sun sets a clean orbital cycle around its virtual host point. This host point is a direct outcome of the local mass gravity (Due to the nearby mass in the Orion spiral arm).
So, the total orbital motion of the Sun is 217 Km/sec however, it also moves locally upwards/downwards to the galactic disc lane at 5-7 km/sec while it moves inwards/outwards at 20 km/sec.
Each star in the spiral arm has a similar orbital motion. Therefore, when we look at the nearby stars we see the following:
http://www.basicknowledge101.com/photos/2015/perception-of-time-rotation-of-galaxy.png
We might think that they move randomly relative to each other, but in reality, each one orbits around its virtual host point, while all their host points stay together in the arm and set a nice orbital motion of about 217 Km/sec around the galaxy.

Summery -
The orbital cycle around a virtual host point is a key element in my explanation.
Please try to accept this idea as is.
If you totally can't agree with that, you are more than welcome to prove it by mathematics.
However, please don't tell a story why this idea isn't logical based on our current understanding about the Galaxy
Agree?
If you have no objection - we will set the next element.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/12/2018 17:06:29
So is there a model supplied by such open-minded thinking?  Trust me, such a thing would be warmly greeted if it worked.
Trust me.
There is a model that works by 100%.
However, I need your cooperation and good willing.
You don't need that at all if you have a good model that works, and not just a set of assertions that all things behave in a manner that they don't.  I see why you're denying the tidal forces and such.  It apparently conflicts with your assertions.

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For just one moment try to forget all the wrong understanding/hypothetical ideas that we have about the galaxy (Including: Age of the Star/galaxy/Universe, dark matter/dark energy and so on).
We must forget all of Newton's laws as well.  Shall we go back to 'impetus'?
You've supplied no laws to replace those, so all I see are requirements for this dream model of yours, but no actual model that meets those requirements.

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In the same token:
Every star in the galaxy must orbit around some host Point. It might be something that we see or something that we can't see. However, any star (at any size) must set a clean orbital cycle (in ellipse shape or a perfect cycle). If we can't see that host point, let's call it virtual host point.
This for instance violates Newton's laws, since F=GMm/r² doesn't work anymore.  Force actually goes down as distance from that host point decreases.  Sometimes it goes up.  Depends where you are.  That's what Newton's laws say anyway, but you seem to assert that these virtual host points have force of their own, not the objects themselves.  What makes some objects contribute to the force of a host point and other not?

 So, while the star orbits around that virtual host point, the host point orbits around the center of the galaxy. Therefore, we might think that the orbital cycle of a star around the galaxy is not clean.

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Summery -
The orbital cycle around a virtual host point is a key element in my explanation.
Please try to accept this idea as is.
I can accept that, sure.

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If you totally can't agree with that, you are more than welcome to prove it by mathematics.
On the contrary, you need to prove, with mathematics, that it works.  So far I've seen no model, just a list of requirements.

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However, please don't tell a story why this idea isn't logical based on your our current understanding about the Galaxy
Agree?
No problem.  It seems so far not to defy its own rules, so not illogical.  Current understanding of the galaxy has been discarded, as requested.  But don't go quoting that understanding then.  That would be mathematically unsound.
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If you have no objection - we will set the next element.
Have at it!
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 18:00:07
This for instance violates Newton's laws, since F=GMm/r² doesn't work anymore.  Force actually goes down as distance from that host point decreases.  Sometimes it goes up.  Depends where you are.  That's what Newton's laws say anyway, but you seem to assert that these virtual host points have force of their own, not the objects themselves.


The whole idea is based on Newton law!!!
I can easily prove it.
However, in order to prove it we must understand the following:

With regards to our Sun
Please look at the following motion of the solar system in the galaxy:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
http://www.biocab.org/coplanarity_solar_system_and_galaxy.html
The broken gray line shows exactly the virtual host point of the Sun while it orbits around the galaxy.
So, the Sun sets a clean orbital cycle around its virtual host point. This host point is a direct outcome of the local mass gravity (Due to the nearby mass in the Orion spiral arm).
So, the total orbital motion of the Sun is 217 Km/sec however, it also moves locally upwards/downwards to the galactic disc lane at 5-7 km/sec while it moves inwards/outwards at 20 km/sec.
Each star in the spiral arm has a similar orbital motion. Therefore, when we look at the nearby stars we see the following:
http://www.basicknowledge101.com/photos/2015/perception-of-time-rotation-of-galaxy.png
We might think that they move randomly relative to each other, but in reality, each one orbits around its virtual host point, while all their host points stay together in the arm and set a nice orbital motion of about 217 Km/sec around the galaxy


What makes some objects contribute to the force of a host point and other not?
Well, each star gravity force is affected by all the stars in the galaxy (or even in the Universe). However, as gravity works locally, the main gravity force impact is due to all the nearby stars. Therefore, the stars gather together is spiral arms. Each virtual host point sets a gravity force with all the other virtual host points of the nearby stars. They actually hold each other while they set a very high orbital motion. Therefore, a spiral arm is THE key request for their orbital velocity.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 20:39:57
Second Element - Drifting outwards

All the stars/planets/Moons/ are drifting outwards from their host points.
In the same token each virtual host point drifts outwards from the center of the galaxy.
This element is key request for our understanding, although It is quite difficult to prove/disapprove it as we only have verified the Earth/moon and Sun/moon orbital system.
The main idea is that any orbital cycle set a spiral cycle (even if it is perfect cycle of ellipse).
This is the only issue which I think that Newton have missed in his formulas.
Newton didn't consider the impact of long time (Billion years).
He actually verified the forces at a limited time frame (years).
So, the Sun must drift outwards from the Center of the galaxy.
In the same token all the stars are drifting outwards from the center.
Even the new atoms in the accretion disc are drifting outwards.

There is no need for tidal to set a drifting outwards movement.
This is a normal mechanism of gravity.
That gives us the following important understanding: Nothing drifts inwards to the center of the galaxy.


Third Element - New Atom/Molecular creation at the center of the galaxy.
There is very powerful magnetic field around the accretion disc which prevents from any matter to drift inwards to the SMBH.
That proves that the accretion disc is actually excretion disc.
I have discussed this issue deeply in  the following thread:
What is needed to create new atoms in the Universe?
https://www.thenakedscientists.com/forum/index.php?topic=75261.40

So, new matter is created at the accretion disc and drift outwards.
As they get to the magnetic field around the accretion, they are boosted upwards at ultra high speed.
Then the new hot molecular clouds fall down to the disc lane and set the famous gas clouds that we can see from our location.
However, those hot new atoms/molecular in this gas cloud don't orbit directly any more around the SMBH.
From now on they orbit around a virtual center of mass in the gas cloud. While this virtual center of mass of the gas cloud orbits around the SMBH.
Therefore, if we look carefully at those gas clouds, we might see that the atoms/molecular orbit around the center of the gas cloud.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/12/2018 21:22:55
The whole idea is based on Newton law!!!
Nonsense.  You threw that out when you generalized a two-body problem property (nice clean orbits) to n-body problem, which has never been solved.  If some object wanders close to this host-point, it will not orbit it (at high speed since it is so close) as you seem to assert, but rather not accelerate much at all, as predicted by Newton.  Sum of the forces is a small value, so only a tiny acceleration towards that center of mass.  Only in the two body case is there guaranteed to be a mass there when you approach it, and the body will orbit at high speed as a low orbit thing does.

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I can easily prove it.
However, in order to prove it we must understand the following:

With regards to our Sun
Please look at the following motion of the solar system in the galaxy:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
Don't understand what that one is trying to depict.  It is a picture minus any description.

It shows the solar system trajectory as the straight line in the middle, with a star on it.  Then there is this nice clean blue helix that is the 'apparent motion' of the solar system.  Apparent to what if the middle line is the actual solar system?
It has some orthogonal arrows labeled with speeds, like they're velocity components of something, but not specified.

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The broken gray line shows exactly the virtual host point of the Sun while it orbits around the galaxy.
It is labeled 'solar system', not 'virtual host'.  I think it is perhaps the linear mass of the galactic arm that we go around, not a point at all.  If it didn't attract like that, the arms would soon dissolve into a more uniform cloud.
The difference is easy to tell:  If it was an orbit, the inclination would not change.  If it is motion around the arm, it will always be on an axis parallel to the arm.  I think it is the latter, and that makes it not an orbit.  There is no virtual host point, which would give us a random permanent inclination.

Anyway, the 217 km/s in that picture is too much for the apparent mass of the galaxy.  If you're not discarding Newton, then the solar system is moving way too fast at this radius.  You've made no attempt to resolve that contradiction without positing more mass than what we can see, which you seem to find offensive.  The funny host-point thing doesn't change the 217 figure one bit.

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So, the Sun sets a clean orbital cycle around its virtual host point. This host point is a direct outcome of the local mass gravity (Due to the nearby mass in the Orion spiral arm).
An arm isn't a host point, nor does it act like one.  But yes, the dynamics seem to work like that, yes.

Each star in the spiral arm has a similar orbital motion. Therefore, when we look at the nearby stars we see the following:
http://www.basicknowledge101.com/photos/2015/perception-of-time-rotation-of-galaxy.png[/quote]
Wonderful.  That shows us having the clean orbit and all the other stars moving randomly in non-clean directions, none of them having any net effect on us.

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We might think that they move randomly relative to each other, but in reality, each one orbits around its virtual host point, while all their host points stay together in the arm and set a nice orbital motion of about 217 Km/sec around the galaxy
If they all moved around the line (not a point) this way, the motion would be less random than depicted, just like the planet motions are hardly random within our solar system.


What makes some objects contribute to the force of a host point and other not?
Well, each star gravity force is affected by all the stars in the galaxy (or even in the Universe). However, as gravity works locally, the main gravity force impact is due to all the nearby stars. Therefore, the stars gather together is spiral arms. Each virtual host point sets a gravity force with all the other virtual host points of the nearby stars.[/quote]How does a nearby star's host point differ from its own position?  Each has a different host point?  Of what would that be the center of mass?
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They actually hold each other while they set a very high orbital motion. Therefore, a spiral arm is THE key request for their orbital velocity.
The spiral arm goes around a 217 km/s at this radius, which is in conflict with Newton's prediction for a galaxy with our apparent visible mass.

Going offline for a bit now and then over the holidays.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/12/2018 04:45:48
If they all moved around the line (not a point) this way, the motion would be less random than depicted, just like the planet motions are hardly random within our solar system.
Yes, that is perfectly correct.

An arm isn't a host point, nor does it act like one.  But yes, the dynamics seem to work like that, yes.
Thanks


The spiral arm goes around a 217 km/s at this radius, which is in conflict with Newton's prediction for a galaxy with our apparent visible mass.
No, there is no conflict.
We need to understand one key issue: Gravity works locally!!!

Please remember that the moon orbits around the earth, while the gravity force of the Sun/Moon is stronger more than double than the Earth/Moon gravity force. This by itself is enigma. I will explain the root for this orbital motion later on

So, the SMBH has no real impact on the Sun orbital velocity. This velocity is a direct outcome of local gravity. Hence, each section of the galaxy is affected by the local mass/gravity force.
I will explain/prove it later on.

How does a nearby star's host point differ from its own position?  Each has a different host point?  Of what would that be the center of mass?
Yes, each star has its own unique virtual host point. This issue will be clear after getting better understanding about the structure of spiral galaxy and how Newton gravity works at each section of the galaxy..
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/12/2018 17:26:03
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We need to understand one key issue: Gravity works locally!!!
Please remember that the moon orbits around the earth, while the gravity force of the Sun/Moon is stronger more than double than the Earth/Moon gravity force. This by itself is enigma.
Is it?  The moon accelerates more due to the sun than it does due to Earth.  At no time does it accelerate away from the sun, even when between the two.  When it is there, the moon accelerates away from Earth, just like Newton says it should.
Do you agree that the gravity force of the Sun/Moon is much stronger than Earth/Moon?
If so, How can we explain the idea that the moon had selected to orbit around the earth instead around the Sun while the Sun/Moon gravity is much stronger?
What Newton would say about it?


The sun 'orbits' the galaxy, not the SMBH.  The latter does have an impact, but nowhere near enough to give the solar system its speed.  Apparently not even the galaxy has enough.
Speed is approximated by GM/r where M is mass of the spherical thing orbited.  Plug in the mass of the SMBH into that, and you get a figure far lower.  But the galaxy is not a spherical mass, so orbital speed actually increases with radius at some ranges.
Yes, sure.
If we ignore how spiral galaxy really work, than yes, you are fully correct.
Hence, as our scientists ignore completely the real impact of spiral arms, and as they also ignore the great impact of local mass gravity, they have found that even the galaxy can't support the orbital velocity of the Sun. But this is a severe mistake.
How can we ignore the spiral shape of our galaxy?
How can we assume that stars are getting in and out and cross the spiral arms while they orbit around the center of the galaxy?
We are missing the whole idea of spiral shape and local mass gravity.
Our Sun is not there by itself. It is a severe mistake to verify the orbital velocity of the Sun based only of its mass.
Where is the impact of the local mass? Where is the impact of all the nearby stars???
Our sun orbits around a virtual host center. This virtual host center is there due to all the nearby masses. So it is some sort of long chain of gravity forces which holds each other while they all orbit around the galaxy.
We can see it quite clearly in the following diagram of the milky way:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Off course, it is only schematics. We don't see the bridges and branches between the arms.
But it shows the gravity force connection in each arm.
In order to get better understanding, let's look at the last point of mass at the Persues arm. This last point of mass has only the one in front it. So, It doesn't see the whole galaxy. The galaxy can't hold this point of mass in the orbital velocity by itself. (So you are fully correct in the following statement - "Apparently not even the galaxy has enough.")
Therefore, only the gravity force between this last Point of mass to the one in front, still holds it in the arm and keeps its orbital velocity.
In each point of mass there might be few thousands of stars.
Hence, each point of mass in this chain holds itself by gravity force with the one in front, while it also holds the one in the back.
Hence - Gravity Works Locally!!!
That is the Key element of spiral galaxy.
Once you understand that, you understand how spiral galaxy works.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/12/2018 22:33:28
Do you agree that the gravity force of the Sun/Moon is much stronger than Earth/Moon?
If so, How can we explain the idea that the moon had selected to orbit around the earth instead around the Sun while the Sun/Moon gravity is much stronger?
It orbits both of them actually.  Whether something orbits object X or not isn't just a function of the force that X puts on it.
This is part of the 3-body problem, which has no known general solution.

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Quote from: Halc
[Orbital] speed is approximated by GM/r where M is mass of the spherical thing orbited.  Plug in the mass of the SMBH into that, and you get a figure far lower.  But the galaxy is not a spherical mass, so orbital speed actually increases with radius at some ranges.
Yes, sure.
If we ignore how spiral galaxy really work, than yes, you are fully correct.
You described it the same way, with this host point orbiting the galaxy at the same speed, unexplained by the visible mass.

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Hence, as our scientists ignore completely the real impact of spiral arms, and as they also ignore the great impact of local mass gravity, they have found that even the galaxy can't support the orbital velocity of the Sun. But this is a severe mistake.
How can we ignore the spiral shape of our galaxy?
How can we assume that stars are getting in and out and cross the spiral arms while they orbit around the center of the galaxy?
We are missing the whole idea of spiral shape and local mass gravity.
Our Sun is not there by itself. It is a severe mistake to verify the orbital velocity of the Sun based only of its mass.
Where is the impact of the local mass? Where is the impact of all the nearby stars???
Nobody ignores any of that, nor do they claim that the sun follows a clean orbit about the galaxy without that wiggle around the arms.

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Our sun orbits around a virtual host center. This virtual host center is there due to all the nearby masses. So it is some sort of long chain of gravity forces which holds each other while they all orbit around the galaxy.
... at an unexplained speed still.  Your hypothesis made no attempt to explain that.

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We can see it quite clearly in the following diagram of the milky way:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
That's a nice map.  Cool.  Hadn't seen one like that before.

In order to get better understanding, let's look at the last point of mass at the Persues arm. This last point of mass has only the one in front it. So, It doesn't see the whole galaxy. The galaxy can't hold this point of mass in the orbital velocity by itself. (So you are fully correct in the following statement - "Apparently not even the galaxy has enough.")
Therefore, only the gravity force between this last Point of mass to the one in front, still holds it in the arm and keeps its orbital velocity. [/quote]If that happened, the arms would contract into chunks instead of stretch into these pinwheel arms. If the mass in front of it played a role as you describe, it would be accelerating, not going fast.  Clockwise speed (as viewed in that diagram) is due to force towards the galaxy, not towards the stuff in front of it.  The 'one in front' contributes not at all to that.  Forces to the front would push that 'last point' outward, not forward, just as forward tidal forces make the moon orbit higher, not faster.
Anybody with familiarity of orbital mechanics knows this.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/12/2018 17:26:56
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Our sun orbits around a virtual host center. This virtual host center is there due to all the nearby masses. So it is some sort of long chain of gravity forces which holds each other while they all orbit around the galaxy.
... at an unexplained speed still.  Your hypothesis made no attempt to explain that.

The explanation is quite simple and based on a normal "drifting outwards" activity in gravity.

Please look again at the following diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Please try to focus at the Green spiral arm.
If we look directly from the Sun to the SMBH, we can see that it cross Green arm at 6KPC (let's call it star B) while if we look at the other side we can see that the green Arm starts from the edge of the Bar  (let's assume that the distance from the center is 3KPC and call it star A)
So, we have two points (or stars) on the same diameter line and on the same arm.
One point is at a distance of 3KPC and the other is at 6KPC.
We know that all the stars have almost the same velocity - V.
Hence, after time T, both will cross the same distance.
S = T * V
Both of them are drifting outwards while they orbit around the galaxy.
So, let's assume that S1 Represents the distance in the green spiral arm from Star A to Star B.
Therefore, both of them will have to cross the same distance = S1.
However, while Star A set 180 degree from the Bottom point (R=3KPC) to the Top (R=6KPC) , Star B will set only about 90 degree from Top point (R=6KPC) to the left point (R=9KPC).
That is a very simple explanation why all stars in the disc orbit at the same velocity while each one stay on the same arm and why we get the unique spiral shape of the arm.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: andreasva on 25/12/2018 17:32:13
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Could it be that over time there is a change in the gravity force?

This is precisely what is happening over time, in my view. 

It's quite a lengthy theory, so I'll try to be brief.

The universe is infinite, not finite.  The Big Bang never happened.  We are not expanding, or accelerating.  Dark Matter is more than likely a myth.

To understand it, you have to start from 0, and work your way in.  Physics chose the middle, and is trying to work its way out simultaneously in opposite directions.  It's too complicated and confusing.  We got a lot wrong.

You also need to consider basic fundamental equality's in math, and how that applies to the universe. 

First, the whole of the universe can only be described with 3 values. 

0
ι1ι


Looking at 0, it is exactly what it implies, nothing.  Should the universe ever become nothing, that's all it could ever be, because 0=0.  0 is a finite value.  It does not equal anything else but 0.  It cannot spontaneously become something else.  0≠1. Clearly we are not 0, because we exist.  I would also conclude, that the potential to be 0 is infinite, because 0, or absolute 0, is theoretically impossible.  Our existence is the empirical evidence that proves conclusively that 0 has never been, and never will be the state of the universe.  Should the universe ever reach a state of 0, it would reach a perfect equilibrium, and would remain that way forever and always. 

Our universe is, > 0

The next value for the universe would be 1, but in the context of the universe its value would be absolute.  The whole of the universe would be a continuous emptiness, void of all substance.  Like 0, 1 is a finite value.  1=1.  And like 0, if the universe ever achieved this state, it would remain that way forever and always.  Also like 0, it cannot spontaneously become something else.  1=1.  Once again, our existence is the empirical evidence that proved conclusively that ι1ι has never been a state of the universe. 

Our universe is, < ι1ι

These are basic fundamental equality's in math.  They don't change.  If they somehow did change, e=mc^2 would be meaningless.  Math wouldn't work.

Between 0 and 1, lies an infinite number of variables.  Or more to the point, an infinitely variable condition, which makes us an infinitely variable analog state.  We are analog.  The entire universe can be described in wave theory, which is analog.  We have always been in this state, because ∞ = ∞, and cannot be anything else.

Einstein defined half of the puzzle, but he was working from 1 to C.  Quantum mechanics is working on the other half,  from C to 0.  They are inherently different, but similar.  I'm not sure the variables directly translate though, so there is some incompatibility.  The universe is also infinite, not finite, which means the constants are also infinitely variable.  There are no physical constants, just virtual constants.  We're bound to C, so it all seems pretty stable.  We're really looking at the universe from 1 to C, in a finite manner, as Einstein defined.  However, the reality is slightly different, 0<C<1, in an infinite manner.   

The big bang is pseudoscience.  The big bang takes the sum total of the entire universe and compacts it into a singularity, removing physical properties of the universe in the process.  Gravity, gone, C gone.  space-time, gone.  They've created an imaginary state of energy, a singularity, with a value of 1, and then wrapped it in nothing, or 0.  The big bang essentially claims 0=1.  0 and 1 cannot coexist.  It's wrong.  Not too surprising given the original source.  It was developed with a preconceived notion of a beginning, where there is none.

The universe is founded on two basic directions, in and out.  It is a 1-dimensional trip, in a 3-dimensional universe.  We move 3-dimensionally towards 0, but it is a destination beyond reach.  Finite values are not physically possible in our universe.  We are more or less a digital representation of an underlying analog state.  No, we aren't holograms, although I can see where it may be applicable to some extent.

What's really happening in the universe is that we're moving inward, probably at a constant rate of -C, for a lack of better terminology.  Really, there's no explanation in physics for what I'm describing.  So excuse my explanations.

As we move inwards towards a smaller state, gravity weakens between distant galaxies, and they fall back.  The gap widens over time.  It looks like expansion and acceleration, but it's mass receding inward and away. 

In the early part of our existence, our region of space was dominated by matter, and was very close together.  Gravity was spread out, or diversified.  Localized effects were less pronounced.  As we (galaxies) drift apart, the effect of gravity from adjacent galaxies diminishes, while local gravity increases relative to the local mass within it.  Not sure increases is the right choice of wording.  More like, it becomes more locally concentrated or focused.  That effect is in a constant rise the further we recede from other galaxies.

In a spiral galaxy, our motion is not precisely circular.  We're literally spiraling inward towards the center, but we're also receding inwards at the same time, giving the illusion on constancy. 

I suspect our most recent upcoming effort to detect Dark Matter will fail. 

So, given your original statement above, absolutely.  Gravity changes over time.  It gradually becomes more locally focused as the gap between galaxies widens.   

Sorry for the quick explanation.  It's pretty rough, but I hope you can get the gist of it. 

And as always, I could be wrong.  I doubt it though. 
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/12/2018 04:07:56
I am unaware of any such normal outward drift.  I was hoping you would describe where the energy would come from to do that.
There is no need for extra energy in order to set the drifting outwards activity.
This is a normal outcome of "Gravity Friction".
We know that there is a friction at almost any activity.
Somehow, we assume that there is no friction in gravity.
We had long discussion about this issue.
You call it "Tidal friction" and I call it "Gravity friction".
You think that there are moons that drifts inwards and I claim that all the moons drifts outwards due to Gravity friction.



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Both of them are drifting outwards while they orbit around the galaxy.
So you assert, but nothing you've said supports that.  It would require energy, and you need to explain the source of that energy.
As I have stated, all the stars drifts outwards due to gravity friction. So, they actually losing energy instead of gaining energy.
Let's set a simple calculation:
F = G M m / r^2
It is clear that as we increase the radius we decrease the gravity force and vice versa.
I can promise you that all the moons which you think that are drifting inwards - all of them are drifting outwards!!!
In any case, as we can't prove this issue, there is no need to argue about it.
We have agreed that we do not agree on this issue.


I think you have A and B moving outward, but no force has been identified that would justify that assertion.  'A' ends up at top at 3KPC and in the same time, B ends up on the right point at 6KPC, assuming they move at the same speed.  That would have the effect of stretching the green line.  Things move clockwise in that picture remember.  You seem to be under the impression that things go the other way.
Yes, you are absolutely correct.
I have tried to explain the basic idea of velocity adjustment while we freeze the spiral stracture.
In reality it is much more complex.
There are several elements to consider.
1. The location of the star. Is it in the Arm, in the ring, in the bulge, in the bar or outside the disc.
Currently, we only focus on the spiral arm. Starting from the ring to the last edge of the arm – (in the galactic disc). Therefore S2 isn't relevant to our current discussion. I will discuss about it later on.
2. The orbital velocity of the virtual host point of the star (at the arm) - V1
So, yes, all the stars orbit in clockwise direction (in the diagram). However, in order to keep a similar velocity at any distance (we will discuss later on about the differences in the velocities), they actually move outwards and backwards in the spiral arm. This is very critical point and I'm not sure that my explanation is fully clear.
So, the whole idea is that as a star is located outwards from the ring, we would expect it to have a faster orbital velocity (if it was a rigid disc), However, as a star drifts outwards it also set a backwards movement. This movement keeps the star in the arm while it orbits at a similar velocity as a star that is located closer to the center.
Please think about it and let me know if the point is clear to you.
3. The velocity of the star around its host point - V2
A star which is located close to the center has a shorter radius to its virtual host point. This radius increases over time due to gravity friction. Therefore, the orbital velocity of star around its virtual host point is faster when a star is located (in the arm) inwards to the galactic center and of course it is slower when a star is located further away from the center.
4. Measured orbital velocity.
The Orbital velocity of the virtual host point (V1)+ its orbital velocity around its host point (V2) sets the total orbital velocity of a star. so, if  V1 and V2 are in the same direction, the total orbital speed that we might see is V1 + v2. However, When they move in the opposite directions we see V1 - V2.
Therefore, when we measure the orbital velocity of stars we might find some variations in the orbital velocities even at the same radius although their real Virtual host point velocity is identical.
In the same token the Virtual host point of all our nearby stars might have exactly the same V1 orbital velocity. However, due to V2 we see them moving in different directions

Sort of....   We're pretty slow actually (~217).  Things further out have been measured at 260.
Yes, that is very normal.
At the edge of the arm (remember the last green point of mass at the green arm), the orbital velocity gets to its maximal pick. The Gravity force can't hold it for long time. Therefore, after some time this point of mass has to be disconnected from the arm and drifts outwards from the galactic disc. (I will discuss about that activity later on)


 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/12/2018 17:35:13
There is no need for extra energy in order to set the drifting outwards activity.
Then you violate Newton's conservation laws.  You are asserting magic.  An object further away from a gravity well has more energy than one deeper in.  There is no normal drift to a higher energy state.  A force is required to do it, and you've not identified that force.
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This is a normal outcome of "Gravity Friction".
We know that there is a friction at almost any activity.
There is no such thing as gravitational friction.  There is physical friction, with two objects rubbing, changing kinetic energy into heat.  There is also EM friction like how they stop railroad trains, which doesn't involve physical contact, which would wear out the parts in perhaps one day.  The result is the same: heat, and loss of kinetic energy.  You've described an increase of energy that comes from nowhere by magic.  Friction creates heat.  Tidal friction is between water and the ocean floor, and it results in heating of the water.  There is no friction between the moon and Earth since the two are separated by space.

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Somehow, we assume that there is no friction in gravity.
High gravitational acceleration of massive objects radiates energy away in the form of gravitational waves.  That is a form of gravitational friction.  The Earth going around the sun loses energy at a rate of I think ~200 watts in this manner.  The result of this tiny friction causes an immeasurable loss of orbital radius, not an increase.  So yes, there is gravitational friction of a sort, but it has the effect of reducing radius, not increasing it.

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We had long discussion about this issue.
You call it "Tidal friction" and I call it "Gravity friction".
Tidal friction slows the spin of Earth.  It is between water and the ocean floor, and all that friction energy is lost to heat.  The energy/momentum transferred to the moon is not from friction.  The energy comes from the angular energy of Earth.  You've identified no such source of energy that accounts for stars gaining potential energy by moving out of the gravitational well.

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You think that there are moons that drifts inwards and I claim that all the moons drifts outwards due to Gravity friction.
Yes, you claim that, despite the violation of Newton's laws.  Physics doesn't work by making self-contradictory claims, especially ones that make assertions contrary to empirical observation.

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As I have stated, all the stars drifts outwards due to gravity friction. So, they actually losing energy instead of gaining energy.
Let's set a simple calculation:
F = G M m / r^2
It is clear that as we increase the radius we decrease the gravity force and vice versa.
You seem to be equating force with energy.  Do you know the difference?

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I can promise you that all the moons which you think that are drifting inwards - all of them are drifting outwards!!!
In any case, as we can't prove this issue, there is no need to argue about it.
The people that measure the opposite effect will appreciate your promise.  Apparently their very real measurements to the contrary don't count as proof that your assertion is quite wrong.

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Yes, you are absolutely correct.
I have tried to explain the basic idea of velocity adjustment while we freeze the spiral stracture.
In reality it is much more complex.
There are several elements to consider.
1. The location of the star. Is it in the Arm, in the ring, in the bulge, in the bar or outside the disc.
Currently, we only focus on the spiral arm. Starting from the ring to the last edge of the arm – (in the galactic disc). Therefore S2 isn't relevant to our current discussion. I will discuss about it later on.
2. The orbital velocity of the virtual host point of the star (at the arm) - V1
So, yes, all the stars orbit in clockwise direction (in the diagram). However, in order to keep a similar velocity at any distance (we will discuss later on about the differences in the velocities), they actually move outwards and backwards in the spiral arm. This is very critical point and I'm not sure that my explanation is fully clear.
That they do, as does Earth in its orbit, but the net effect is an average radius that doesn't change absent any force applying a net forward force which would account for the increase of energy.  That force would need a reaction force decreasing the energy of something else, per Newton's 3rd law.  If you don't identify that equal and opposite reaction, then your drift assertion violates Newtonian law.  This host point adds zero net force since it is sometimes in one direction, and sometimes in the opposite direction.  That adds up to a wiggle with no net change in speed or radius.

You continuously assert force by magic, without reaction.  Newton would not agree with that.

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So, the whole idea is that as a star is located outwards from the ring, we would expect it to have a faster orbital velocity (if it was a rigid disc), However, as a star drifts outwards it also set a backwards movement. This movement keeps the star in the arm while it orbits at a similar velocity as a star that is located closer to the center.
Please think about it and let me know if the point is clear to you.
The disk of the galaxy is anything but rigid.  All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.  This is not the case of a rigid object, which would be able to spin any any number of different angular rates and still keep its components at some fixed radius.
So I don't really care how stars would move if the galaxy was a rigid disk, since it isn't one.  No, I don't really get what you're trying to convey above.  If a star stays near its arm, it is because the nearby mass of the arm attracts it.

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3. The velocity of the star around its host point - V2
A star which is located close to the center has a shorter radius to its virtual host point. This radius increases over time due to gravity friction. Therefore, the orbital velocity of star around its virtual host point is faster when a star is located (in the arm) inwards to the galactic center and of course it is slower when a star is located further away from the center.
Assuming there is such a host point, you discard Newtonian mechanics with the drifting assertion, and then assert that both gravitational potential energy (from moving away from the gravity source) and kinetic energy (from asserted greater orbit velocity out there) appear as from nowhere, totally violating energy conservation laws.  Oh yes, and even more energy from heat from 'gravitational friction'.  My, but you do pull a lot of energy from nowhere.

Let's see, you also assert that orbital velocity about the host point is a function of the distance of that star or host point from the galactic center, which makes no sense.  The galaxy affects speed around the galaxy, not speed around the host point.  The speed around the host point has no obvious formula since you've not supplied one.  Kepler's laws are of no use here since you're not describing a Kepler orbit.

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4. Measured orbital velocity.
The Orbital velocity of the virtual host point (V1)+ its orbital velocity around its host point (V2) sets the total orbital velocity of a star.
We have a second host point now?  That's fine, but you just now introduced this.  Before you said that the one host point orbited the galaxy as your picture depicted.

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so, if  V1 and V2 are in the same direction, the total orbital speed that we might see is V1 + v2. However, When they move in the opposite directions we see V1 - V2.
V usually refers to velocity, not speed, in which case it is always V1 + V2.  This simplifies the math considerably.

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Therefore, when we measure the orbital velocity of stars we might find some variations in the orbital velocities even at the same radius although their real Virtual host point velocity is identical.
Right.

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Quote from: Halc
Sort of....   We're pretty slow actually (~217).  Things further out have been measured at 260.
Yes, that is very normal.
At the edge of the arm (remember the last green point of mass at the green arm), the orbital velocity gets to its maximal pick.
The first one near the core also moves much faster, even than the 260 figure.  I don't have numbers since I don't know specific objects.  The 217 figure for our speed is just the length of a circle of 10KPS radius divided by the ~200 million years it takes for us to go around one lap.  From this speed the acceleration can be computed needed to keep our host point on that circular path, and from that acceleration the mass required to account for that acceleration can be computed.  Your host point idea doesn't change that acceleration at all, so you (in denial of additional mass) have failed to explain that acceleration.  Nothing you propose explains the force necessary for the observed acceleration.  All I see is assertions that are inconsistent with laws that have been known for many centuries.

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The Gravity force can't hold it for long time. Therefore, after some time this point of mass has to be disconnected from the arm and drifts outwards from the galactic disc. (I will discuss about that activity later on)
If the arm was moving too fast, it would not be at the radius from the galaxy it is now.  We would already be further out.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/12/2018 21:25:52
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
If that is the case, than how could it be that we get the spiral shape from that random orbital activity?
How many spiral shape you can get due to this random orbital activity? Is it 0.000...0001 preset or 0.0..1 present?
Do you know that: "Spiral galaxies make up roughly 72 percent of the galaxies that scientists have observed, according to a 2010 Hubble Space Telescope survey"
https://www.space.com/22382-spiral-galaxy.html
So, do you really believe that if "All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them", they can set a spiral shape in 72 percent of the galaxies that scientists have observed???
If we see that 72% of the galaxies are spiral, don't you think that it couldn't be due to random phenomenon?
So how can we explain that incredible number of spiral galaxies?
How can we ignore the impact of the whole spiral galaxy shape?


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/12/2018 05:56:28
Dear Halc

In one hand you claim:
Mass attracts other mass, so spinning material apparently tends to gather into arm. 
However, this is exactly what I say. Mass attracts other mass.
So, mass gravity works locally. Therefore, the mass in the Spiral arms attracts all the stars in the Arm.
Therefore, do you agree that all the stars in the Arm are directly affected by the gravity force of the Arm?

But you have stated:
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
How could it be that in one hand you claim that - Mass attracts other mass in spiral arms, while on the other hand you claim - All objects are disconnected?
Do you agree that there is a contradiction?

Can you please explain how the following Milky way spiral galaxy had been formed?
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Please explain the following:
1. Bulge:
Why it has a ball shape? Why there is no disc shape in the Bulge
2. Bar:
What is the source for that bar shape?
3. Ring
What is the resone for the Ring shape? Why it has a disc Shape instead of ball shape as we see in the Bulge?
What is the gravity impact of the Ring? Why the ring is so narrow? Why the ring doesn't continue all the way to the far end of the galaxy?
4. Spiral arm
How could it be that after the ring we see spiral arm?
Please explain why the spiral arms had been set in a flat disc and why it has this unique shape?
5. End of the Disc shape.
Why at some point, the spiral arm shape is ended? Why it doesn't last longer?
How could it be that at some point the Dark matter can't hold the stars in the disc?
Why after the far end edge of the spiral arms we see stars far above and below the galactic disc?

Please look at the following orbital velocity:
http://lempel.pagesperso-orange.fr/courbe_rot_voie-lactee.jpg
http://lempel.pagesperso-orange.fr/matiere_noire_pas_si_noire_uk.htm
How could it be that our scientists don't try to match between the spiral structure to the measured velocity???
Can you please explain why we see that measured velocity at each segment of the galaxy?

Perhaps our solar system formed arms before each arm stabilized into rings and then planets.  I don't know.  I'm not an expert.  The spiral arms seem to be a fairly stable structure since many (but not all) semi-large galaxies of similar age seem to have them.

If our scientists don't know the answers for some questions, why do you take it for granted that no one can find the real answer for Spiral galaxy enigma?
Instead of focusing on the negative aspects, why don't you help me to develop the positive ideas about the galaxy?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/12/2018 17:54:14
Mass attracts other mass.
So, mass gravity works locally. Therefore, the mass in the Spiral arms attracts all the stars in the Arm.
Therefore, do you agree that all the stars in the Arm are directly affected by the gravity force of the Arm?
Sure, but not just the mass of the arm.  All stars are directly affected by all mass.  The nearby portions of our own arm certainly have more effect than the distant parts, and the nearby parts of other arms have more effect than distant parts of our own arm.  Acceleration of the sun or a soap bubble in space can be given by a = ∑GM/r² for every object everywhere, which means the sum of acceleration due to each and every object everywhere, not just the objects in our own galactic arm, which is a pretty pathetic arm actually, not one of the main ones.

I agree to most of what you say, but gravity does not work locally.  I cannot think of any object however non-local that has mass but does not exert any gravitational influence on me.

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But you have stated:
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
How could it be that in one hand you claim that - Mass attracts other mass in spiral arms, while on the other hand you claim - All objects are disconnected?
Disconnected as in not attached to each other via steel cables or other means by which their trajectories might be different than what gravitational forces predict.  Two rocks attached with a string trace a different path than the same two disconnected rocks that are influenced only via their mutual gravity.

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Can you please explain how the following Milky way spiral galaxy had been formed?
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
The site would explain it better than me reading the site and deciding how much of it is known and how much is conjecture.  The link is a picture of the current configuration, and gives zero information as to the formation of that configuration.

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Please explain the following:
1. Bulge:
Why it has a ball shape? Why there is no disc shape in the Bulge
I think the bulge is the halo, which is mostly made of objects that have been displaced from their normal paths by close encounters with significant gravitational sources, thus putting them on random trajectories that take them out of the plane of the disk.
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2. Bar:
What is the source for that bar shape?
No clue.  I didn't read that far.  I think you need to go further than the wiki summary to get a good explanation of the current thinking behind that.
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3. Ring
What is the resone for the Ring shape? Why it has a disc Shape instead of ball shape as we see in the Bulge?
Again, I don't claim to have better answers than what you see on the site.  It is not a ball because it is not displaced, but normal matter that has been pulled into disk shape as does any cloud of material that rotates and condenses via mutual gravitational attraction.  A galaxy in many ways is formed similar to a solar system, but in massive slow motion, and thus far greater local gravitational effects.  Unlike a typical solar system, there is no dominant mass that cleans up the orbit of everything else.  A SMBH that is on the order of 1% of the total mass hardly seems up to the task.

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What is the gravity impact of the Ring? Why the ring is so narrow? Why the ring doesn't continue all the way to the far end of the galaxy?
The impact to us is negligible as is all mass further out and not nearby.  It has impact to any material more distant than the ring since it is part of the mass around which that material orbits.
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4. Spiral arm
How could it be that after the ring we see spiral arm?
The picture doesn't show the ring.  The site mentions it, and puts it around 50 KPCS, and the furthest arm in the picture seems to be about a third that distant, so I don't see an arm beyond the ring.
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Please explain why the spiral arms had been set in a flat disc and why it has this unique shape?
Any rotating collection of loose material under mutual attraction is going to flatten into a disk, just like any gas cloud that collects into a solar system.  Interaction friction slows much of motion outside of the plane of rotation, but does not decrease the angular momentum of the collection.

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5. End of the Disc shape.
Why at some point, the spiral arm shape is ended? Why it doesn't last longer?
The arms must be stretched as the inside points go around more often than the outside ones, so it has to get homogeneous after a while, but then again, arms can merge with larger neighbors as ours is doing, so maybe that explains how the arms persist even after all the laps we've taken.  I don't claim to know.

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How could it be that at some point the Dark matter can't hold the stars in the disc?
Who claims that?

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Why after the far end edge of the spiral arms we see stars far above and below the galactic disc?
Who knows?  New gas probably gets pulled in by the galaxy, so surely there is material out there to form stars.  A lot of the answer would be told by the trajectory of these outlier stars.  Stuff further out has had far less forces that tend to pull the material into a disk, just like our solar system's outer material like the Oort cloud is not arranged as a disk.

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Please look at the following orbital velocity:
http://lempel.pagesperso-orange.fr/courbe_rot_voie-lactee.jpg
http://lempel.pagesperso-orange.fr/matiere_noire_pas_si_noire_uk.htm
How could it be that our scientists don't try to match between the spiral structure to the measured velocity???
Don't know what you mean here.  Nothing on that chart is labeled 'spiral structure'.

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Can you please explain why we see that measured velocity at each segment of the galaxy?
Don't know what it depicts.  It shows the disk at this radius moving at about 160 km/s, quite a bit slower than the 217 figure quoted in earlier places you've linked.  It shows apparently stars moving slower than free gas, possibly because a lot of the gas is higher speed due to being ejected in explosions?  Just a guess there.

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If our scientists don't know the answers for some questions, why do you take it for granted that no one can find the real answer for Spiral galaxy enigma?
Instead of focusing on the negative aspects, why don't you help me to develop the positive ideas about the galaxy?
You've not presented any alternative explanations.   You just assert things that are in contradiction with the observed accelerations.  The scientists at least have theories that explain the observed accelerations, however distasteful you find those theories.  There has been no mathematics to back the sort of things you've been asserting, so it is not even wrong.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/12/2018 15:29:11
Quote from: David
Mass attracts other mass.
So, mass gravity works locally. Therefore, the mass in the Spiral arms attracts all the stars in the Arm.
Therefore, do you agree that all the stars in the Arm are directly affected by the gravity force of the Arm?
Sure, but not just the mass of the arm.  All stars are directly affected by all mass.  The nearby portions of our own arm certainly have more effect than the distant parts, and the nearby parts of other arms have more effect than distant parts of our own arm.  Acceleration of the sun or a soap bubble in space can be given by a = ∑GM/r² for every object everywhere, which means the sum of acceleration due to each and every object everywhere, not just the objects in our own galactic arm, which is a pretty pathetic arm actually, not one of the main ones.

I agree to most of what you say, but gravity does not work locally.  I cannot think of any object however non-local that has mass but does not exert any gravitational influence on me.
Please look at the following article:
https://www.space.com/19915-milky-way-galaxy.html
"Spiral arms are like traffic jams in that the gas and stars crowd together and move more slowly in the arms. As material passes through the dense spiral arms, it is compressed and this triggers more star formation," said Camargo.

So, our scientists belive that stars don't stay at spiral Arms but they just move slower at the arms.
Therefore, all the stars actually cross the arms while they orbit around the galaxy.

Based on this idea, the Sun has to cross all the arms several times - as the orbital velocity of the Sun is faster than the orbital velocity of the Arm.
However, in the arm, the orbital velocity is lower than outside the arm.
So, based on this unrealistic idea, once we are outside the arm, we should speed up.

It is stated:
"The solar system travels at an average speed of 515,000 mph (828,000 km/h). Even at this rapid speed, the solar system would take about 230 million years to travel all the way around the Milky Way."
So, what is the expected velocity outside the arm?
How could it be that we get a spiral shape due to this explanation?
Do you think that Newton would accept the idea of several orbital velocities changing per one cycle?
How can we agree with this totally unrealistic idea about spiral arms and changing orbital velocities?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/12/2018 18:02:04
Please look at the following article:
https://www.space.com/19915-milky-way-galaxy.html
"Spiral arms are like traffic jams in that the gas and stars crowd together and move more slowly in the arms. As material passes through the dense spiral arms, it is compressed and this triggers more star formation," said Camargo.

So, our scientists belive that stars don't stay at spiral Arms but they just move slower at the arms.
Therefore, all the stars actually cross the arms while they orbit around the galaxy.
The article didn't say that.  It explains that the stuff moving through the arms is slowed by friction, and thus tends to more or less stay in the arms.  That seems to be why there are arms and not a homogeneous fog without structure.
Nothing in that article suggests that all stars do a certain thing.
The article says that stars and gas are slowed (by friction) to a lower speed relative to the arm, not relative to the galaxy.  The whole galaxy moves at about 217 km/sec at this radius, and things moving at a different speed tend to match the local speed due to collisions with the thick material.

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Based on this idea, the Sun has to cross all the arms several times - as the orbital velocity of the Sun is faster than the orbital velocity of the Arm.
If the sun orbital velocity around the galaxy was faster than the arm, the sun would move permanently away from the galaxy and the arm, moving perhaps closer to the next arm out.  But both are going at about 217 km/sec around the galaxy.  Yes, the sun is not centered on its arm, and thus moves in a sort of orbit around that mass, but that doesn't make the sun faster around the galaxy than the arm.

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However, in the arm, the orbital velocity is lower than outside the arm.
So, based on this unrealistic idea, once we are outside the arm, we should speed up.
You seem to be making up nonsense facts.  If you move away from mass, your speed goes down, just like the speed of a ball slows if you throw it upward.  The idea is indeed unrealistic.  The article suggests no such thing.

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It is stated:
"The solar system travels at an average speed of 515,000 mph (828,000 km/h). Even at this rapid speed, the solar system would take about 230 million years to travel all the way around the Milky Way."
So, what is the expected velocity outside the arm?
That is orbital speed around the galaxy, and you showed a graph of that vs radius.  The galactic orbital speed outside the arm is almost constant nearby, so it will be the same outside of the arm.
The first quote you give is speed relative to the arm material (for material passing through it say from the side).  That relative speed would be slowed due to the friction.

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Do you think that Newton would accept the idea of several orbital velocities changing per one cycle?
If you orbit 6 things, you have 6 orbital speeds, any of which can change due to changes in distance or friction forces or something.  The moon orbits Earth, the sun, the arm, the galaxy, and probably other things, all with different speeds and radii.  Most of those speeds are fairly constant over the period of one cycle, but not completely.

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How can we agree with this totally unrealistic idea about spiral arms and changing orbital velocities?
I didn't see any suggestion of changing orbital velocities of galactic arms.  The speeds are quite stable.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/12/2018 19:07:54
Quote from: David
Please look at the following article:
https://www.space.com/19915-milky-way-galaxy.html
"Spiral arms are like traffic jams in that the gas and stars crowd together and move more slowly in the arms. As material passes through the dense spiral arms, it is compressed and this triggers more star formation," said Camargo.

So, our scientists belive that stars don't stay at spiral Arms but they just move slower at the arms.
Therefore, all the stars actually cross the arms while they orbit around the galaxy.
The article didn't say that.  It explains that the stuff moving through the arms is slowed by friction, and thus tends to more or less stay in the arms.  That seems to be why there are arms and not a homogeneous fog without structure.
Nothing in that article suggests that all stars do a certain thing.
The article says that stars and gas are slowed (by friction) to a lower speed relative to the arm, not relative to the galaxy.  The whole galaxy moves at about 217 km/sec at this radius, and things moving at a different speed tend to match the local speed due to collisions with the thick material.
Well, it is stated: "As material passes through the dense spiral arms"
How could they stay at the arm if they passes through?
What do you understand from the idea of "passes through"?
In any case, if stars stay at the Arms (all the time from day one), than why is it?
Does it mean that they hold each other by gravity force?
If so, that exactly the whole idea which I have presented.
If they don't hold each other by gravity force, why they stay at the Arm?


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/12/2018 05:19:09
Quote from: Dave
In any case, if stars stay at the Arms (all the time from day one), than why is it?
Does it mean that they hold each other by gravity force?
If so, that exactly the whole idea which I have presented.
I never disagreed with that.  Yes, gravity (and friction) seems to be what holds them together, just like those two effects are what hold a solar system together.
So, you agree that the stars hold each other in the arm due to gravity (or friction if you like)
Therefore, do you agree that in this case each star must obey to the local gravity force in the Arm?
It is similar to the Earth/sun gravity force.
The Earth doesn't care about the gravity forces on the Sun. It is just holds/orbits the Sun by gravity force and follows it where ever it goes.
So, as we do not set the calculation why the Earth orbits around the galaxy, we also shouldn't worry why the sun orbits around the galaxy.
The Sun holds itself in the spiral arm.
Where ever the spiral arms goes - the Sun goes.
So, we have to find why/how the spiral arm orbits the galaxy and not why the Sun orbits the galaxy.
Only if the Sun is disconnected from the arm, than we must look for dark matter.
As long as is connected in the arm due to gravity, we must look for answers about the arm
We might find that Dark matter is needed for the Arm, but there is no need to find an explanation for the Sun - as long as it is connected to the arm by gravity.
In the same token - as long as the Earth is connected to the Sun by gravity force, we do not need to find an explanation why the Earth orbits around the galaxy.

As I have stated – Gravity works locally:
Another example -
An asteroid orbits around the Moon. Hence, the moon holds this asteroid by gravity.
The moon orbits around the Earth. Therefore, the Earth holds the Moon which holds the Asteroid by gravity.
So, we can continue at higher hierarchy. But at the end we might find that the asteroid orbits the Galaxy.
However, as long as the asteroid is connected by gravity to the moon, we do not try to find a direct explanation why it orbits the galaxy at supper high orbital velocity (with reference to the galaxy)
In the same token, as long as the Sun is connected to the arm by gravity force, we do not need to find a direct explanation why it orbits the galaxy.
As I have already explained - The sun orbits around a local virtual host point. Hence, we can say that the Sun holds itself to this virtual host point by gravity force (Local gravity force).
However, this virtual Sun' host point holds itself in the spiral arm by gravity force.
Therefore, at this phase, we must find the answer for – "Why the Arm orbits around the galaxy (and not why the sun orbits around the galaxy)"?
Do you agree with that?


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/12/2018 14:05:07
You keep asking this.  Each star's velocity and acceleration is determined by the sum of the force vectors acting upon it.  This is all things, not just local ones.

Yes and No.
Yes - based on theoretically point of view.
No - based on real evidences.
Please look at our moon.
The Gravity force of the Sun/Moon is stronger by more than a twice with regards to the Earth/Moon.
If the Moon had to chose its orbital path based on the sum of the force vectors acting upon it, it will had to orbit around the Sun instead around the Earth, however - this isn't the case.
So, even if the Sun/Moon gravity is stronger than Earth/moon gravity, the moon prefers to orbit around a local host (Earth).
Therefore, as I have stated, Gravity works locally.
In the same token - we can say about the Earth.
Our scientists claim that  more than 100 Billion Sun mass is needed to be at the orbital center of the Sun (in the galaxy) in order to hold the Sun in its orbital path.
If we try to calculate the Galaxy/Earth gravity force (based on that mass), we should find that it is much higher than Sun/Earth gravity force.
So, Based on a pure sum of the force vectors acting upon the earth, it had to ignore the Sun and start running around the galaxy. But again - this isn't the case.
Hence, as the Earth had selected a local mass (our Sun) instead of the galaxy, and as the Moon had selected the Earth instead of the Sun; we have to claim the following:
Each object's velocity and acceleration is mainly determined by the local mass and not by the sum of the force vectors acting upon it.
The sum of the the force vectors could set a tidal. No more than that (Assuming that there is no collision)!!!
 
The arm is not an object like the sun is to Earth.  The arm is a smear of matter, all moving at different orbits about the galaxy.  Some of it goes around in much less time than parts further out.  It isn't one thing that the sun follows.
We have already discussed this issue.
Each star orbits around a local virtual host point. Therefore, although we see that all the stars are moving in different directions, their host points are very stable with regards to each other.
Hence, The Arm Is an ARM. It holds all the Virtual host points at the arm due to local gravity force.
This is the key element of spiral arm.
If the arm can't hold the stars, than by definition the stars must cross the arm and get out of it.
You can't just say that the stars stay at the arm but the arm doesn't hold them by gravity.
This is none realistic statement.
In the same token - as long as the Earth is connected to the Sun by gravity force, we do not need to find an explanation why the Earth orbits around the galaxy.
Sorry - this is incorrect. We can't take it for granted. We need to find an explanation why the Earth orbit around the Sun although its gravity force is lower than the Gravity force of the galaxy.
I will give you a tip - Gravity Works locally.
This is the ultimate answer for:
Why the moon holds/orbits around the Earth instead of the Sun?
Why the Earth Holds/orbits around the Sun instead of the galaxy?
Why the Sun orbits around a virtual host point that holds itself in the arm due to Local gravity force?
If you still don't want to accept the great impact of "local gravity force" than try to explain why the moon orbits around the earth instead of the Sun.
With minor edits, sure.  We must ask why the local portion of the arm orbits around the galaxy at the speed it does. Not why it goes around, but why so fast? It accelerates at about 2e-10 m/sec² and the mass of the matter we see in the galaxy is not enough to explain that acceleration.

We know why the arm orbits:  Things just do when there is a mass to orbit, such as the material that makes up the inner portion (stuff closer than us) of the galaxy.  But orbital speed is very much a function of the mass of that material, and there apparently isn't enough of it.

The answer is very simple - Gravity works locally.
If you agree with that - you have got the answer!!!
There is no need for dark matter in the galaxy to hold the Sun in the orbital path around the center.
All we need is - Gravity works locally in the arm (and everywhere).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 31/12/2018 17:21:05
Quote from: Halc
Each star's velocity and acceleration is determined by the sum of the force vectors acting upon it.  This is all things, not just local ones.
Yes and No.
Yes - based on theoretically point of view.
No - based on real evidences.
The theory was based on that real evidence.  If you have evidence to the contrary, that would counter 4 centuries of physics.

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Please look at our moon.
The Gravity force of the Sun/Moon is stronger by more than a twice with regards to the Earth/Moon.
If the Moon had to chose its orbital path based on the sum of the force vectors acting upon it, it will had to orbit around the Sun instead around the Earth, however - this isn't the case.
Your understanding of orbital mechanics needs a lot of work.  The strongest force acting on the moon is from the sun, and that means that at all times, the moon accelerates towards the sun.  That's the implication of what I've said above.  I did not say the moon cannot orbit Earth.  The moon has no choice or preference about this.  It moves exactly as per forces as described by Newton.

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Our scientists claim that  more than 100 Billion Sun mass is needed to be at the orbital center of the Sun (in the galaxy) in order to hold the Sun in its orbital path.
If we try to calculate the Galaxy/Earth gravity force (based on that mass), we should find that it is much higher than Sun/Earth gravity force.
How do you figure that?  No scientist claims this.  It is in fact 7 orders of magnitude lower.

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So, Based on a pure sum of the force vectors acting upon the earth, it had to ignore the Sun and start running around the galaxy. But again - this isn't the case.
Wrong several ways.
First, the Earth does orbit the galaxy, having gone around nearly 20 times.
Secondly, there is no law that says a greater force will strip Earth away from the sun since the sun would also be subject to that force. This is why the moon orbits Earth despite Earth exerting less force upon it than the sun.
Thirdly, the sun contributes more to Earth's force vector than all other forces combined.

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Each star orbits around a local virtual host point.
So you claim, but you've provided no mathematics for this claim.  How much force is applied by this virtual host point?  How might that be computed?  It needs to work for more than two bodies.  The motion of a two-body system can be reduced to a one-body case and thus has a solution.
Your claim is not similarly backed and is thus no more than a hand-waving assertion.
This is typical of armchair physicists.  All magical claims but no mathematics or predictions.

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Therefore, although we see that all the stars are moving in different directions, their host points are very stable with regards to each other.
Hence, The Arm Is an ARM. It holds all the Virtual host points at the arm due to local gravity force.
This is the key element of spiral arm.
If it was the ARM, it would be the host point, and all the orbiting stuff would orbit that point.  It doesn't.  An arm is a smear with all different angular velocities about the galaxy.  This has been measured.

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If the arm can't hold the stars, than by definition the stars must cross the arm and get out of it.
By definition?  What definition would that be?
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You can't just say that the stars stay at the arm but the arm doesn't hold them by gravity.
This is none realistic statement.
But I didn't say that.

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Sorry - this is incorrect. We can't take it for granted. We need to find an explanation why the Earth orbit around the Sun although its gravity force is lower than the Gravity force of the galaxy.
Even if the galaxy did exert this greater force, Earth would still probably orbit the sun.

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I will give you a tip - Gravity Works locally.
F = GMm/r².  That is not local.  That says there is force at any distance.

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There is no need for dark matter in the galaxy to hold the Sun in the orbital path around the center.
All we need is - Gravity works locally in the arm (and everywhere).
How do you explain the acceleration of the arm?  That's the thing you keep evading, post after post.
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 31/12/2018 20:59:49
The answer is very simple - Gravity works locally.  If you agree with that - you have got the answer!!!   There is no need for dark matter in the galaxy to hold the Sun in the orbital path around the center.  All we need is - Gravity works locally in the arm (and everywhere).
Gravity is the due to the acceleration of aether (ie aetheons) into mass where the aether is annihilated.
But aether does not possess any quantum mass or quantum anything -- but it is good at transmitting force to & from tween quantum mass -- the transmission being by way of reverberations of pulses that travel at over 20 billion c (hencely 1 gravity-second is equal to more than 634 light-years)(taking less than 236 sec to cross the 150,000 LY diameter of the Milky Way)(praps less than 1 sec). 

Hencely the shape of the disposition of cosmic mass in the Milky Way & near the particular stars of interest is important re the action & effect & magnitude of gravity & of big G especially re orbits (moreso than it might appear by the naive use of Newton's equation) -- & the disposition of nearby galaxies has an effect.
In other words gravity is due to tension in the aether, but this is a 3D thing, & hencely the attraction tween two stars needs other stars in every direction otherwise their mutual (2D) attraction will be limited.

When Cavendish did his big G experiment he had the mass of the Earth beneath & the mass of the building next to & above.  This ruined his results.  And nowadays silly Einsteinians try to make use of big G in a galactic environment, & need to invent Dark Matter.  No, Cavendish's big G works best (sometimes) on Earth, in a lab.

The centrifuging of aether near a spinning disc or in&near a spinning spiral galaxy gives a quasi-gravity that allso helps towards bringing about a 1/R relationship rather than the naive  1/RR Newtonian.  And quasi-gravity is a major reason why galaxies (& solar systems)(& Saturn's rings) like to have a disc shape.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/01/2019 07:25:59
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Please look at our moon.
The Gravity force of the Sun/Moon is stronger by more than a twice with regards to the Earth/Moon.
If the Moon had to chose its orbital path based on the sum of the force vectors acting upon it, it will had to orbit around the Sun instead around the Earth, however - this isn't the case.
Your understanding of orbital mechanics needs a lot of work.  The strongest force acting on the moon is from the sun, and that means that at all times, the moon accelerates towards the sun.  That's the implication of what I've said above.  I did not say the moon cannot orbit Earth.  The moon has no choice or preference about this.  It moves exactly as per forces as described by Newton.
O.K
We have to ask
"The sun attracts the moon with a force twice as large as the attraction of the earth on the moon. Why does the moon not revolve around the sun?"

Please look at the following article:
https://www.quora.com/The-sun-attracts-the-moon-with-a-force-twice-as-large-as-the-attraction-of-the-earth-on-the-moon-Why-does-the-moon-not-revolve-around-the-sun
"Here the moon and earth form a system, which is like a Binary system . If two astrologers rotate around their center of gravity together, then it is called binary system"
The answer is: "Binary system"
So, this binary system is a "local gravity force" that gives the moon/earth system the possibility to orbit around the Sun.
Hence, the noon is not their by itself, as the earth is not their by itself.
Now they both can orbit around the Sun.
But it is clear that without setting a binary system with the Earth - Just based on pure gravity force - the moon will prefer to orbit around the Sun.
It will never ever set a binary system with the earth if it was orbiting around the Sun.
Few words about binary-and-multi-star-systems:
https://stardate.org/astro-guide/binary-and-multi-star-systems
"A binary is a pair of stars that orbit each other. A multi-star system consists of three or more stars. The stars in a binary or a multi-star system all formed from a single cloud of gas and dust, so they are true “siblings.”"
The Earth & the Moon are real "siblings". This is a key element. I will explain later on why the current concept of how the Moon had been created is absolutely unrealistic:
https://en.wikipedia.org/wiki/Giant-impact_hypothesis
"The giant-impact hypothesis, sometimes called the Big Splash, or the Theia Impact suggests that the Moon formed out of the debris left over from a collision between Earth and an astronomical body the size of Mars, approximately 4.5 billion years ago"
I will also explain how the earth and the moon had been formed "from a single cloud of gas and dust, so they are true siblings". and why they set this binary system due to gravity force.
But in order to explain it, we must know how our galaxy really works.
The spiral arm is a key element in our discussion.
Therefore, understanding the gravity force between the objects in the arm is vital.
So, the Earth/moon is a binary system which holds them both while they orbit around the Sun.
In the same token, the arm is a multi-star-system.
Each star in the galaxy has its own virtual host point. Together, those host points set that "multi-star-system" which holds all the neaby stars in the arm.
So, the sun is not there by itself, (as the moon is not there by itself)
There is a local system that is bounded by gravity force.
Therefore, I call that Binary/multi-star-system as "local gravity force" which is the base for what we see.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/01/2019 14:46:43
We have to ask
"The sun attracts the moon with a force twice as large as the attraction of the earth on the moon. Why does the moon not revolve around the sun?"
As I've said, it does go around the sun, once per year, just like Earth.  So you want to know why the two (Earth and moon) don't separate due to the greater gravitational force of the sun.

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Please look at the following article:
https://www.quora.com/The-sun-attracts-the-moon-with-a-force-twice-as-large-as-the-attraction-of-the-earth-on-the-moon-Why-does-the-moon-not-revolve-around-the-sun
Ouch.  This is why I rarely go to quora for answers.  I counted well over a dozen serious errors in that article.  This gem for example:
"Earth's mass and size are big so we do not understand the rotation of the earth. But this is not the case even between Jupiter and Sun. Here the sun does not turn around like Jupiter, and the Sun also does not turn around Jupiter!"
Apparently Earth is too large to understand if it orbits the sun or not, but Jupiter is small enough that it doesn't???  What????  It doesn't help that the guy apparently lacks English skills, but really, what was he trying to say with that statement?

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"Here the moon and earth form a system, which is like a Binary system . If two astrologers rotate around their center of gravity together, then it is called binary system"
Do they have to be astrologers?  What if two physicists rotate around their common center of gravity?
If the two astrologers were anywhere near a significant mass like the moon, they would not orbit each other.  Binary systems are not always stable, and the one described there is incredibly unstable.  We need to ask why the Earth-moon system is more stable than the pair of astrologers.

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The answer is: "Binary system"
So, this binary system is a "local gravity force" that gives the moon/earth system the possibility to orbit around the Sun.
Sure.  They're the only two significant masses in that system.  The sun's gravity does not act to separate them.  The non-uniformity of the sun's gravity does, and that non-uniformity is far less between the Earth and moon that is the acceleration of the moon due to Earth.  That's why the two stay together.

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Hence, the noon is not their by itself, as the earth is not their by itself.
That article denies that the Earth is not by itself.  The orbit of Earth is depicted as a perfect circle, apparently unaffected by the pull of the moon.  Yet another mistake.  It also shows the moon curving away from the sun at times, another mistake if the sun's force is greater than the Earth's.  We can forgive these things since the diagrams are clearly not to scale.

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But it is clear that without setting a binary system with the Earth - Just based on pure gravity force - the moon will prefer to orbit around the Sun.
It will never ever set a binary system with the earth if it was orbiting around the Sun.
Not true.  It is orbiting the sun, and yet it forms a binary pair with Earth.  The sun's gravitational field is too uniform at this radius to separate the two, just like the pull of the galaxy is far too uniform to pull the arms apart.  The pull from the sun goes up as the radius drops.  Venus has more acceleration than does Earth for this reason.  Not so with the galaxy according to your graphs.  The net acceleration (proportional to net force) actually goes down as the radius decreases.  This seems to be why galaxies form arms but new solar systems don't seem to.  Maybe they do at first.  Not like we can see that.

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Few words about binary-and-multi-star-systems:
https://stardate.org/astro-guide/binary-and-multi-star-systems
"A binary is a pair of stars that orbit each other. A multi-star system consists of three or more stars. The stars in a binary or a multi-star system all formed from a single cloud of gas and dust, so they are true “siblings.”"
We are true siblings with many (all?) of the stars closest by, but we're not really a multi-star system.  Maybe we are.  Not sure.  The rule is not always true.  Some multi-star systems are made up of stars that didn't come from the same cloud of dust.  Maybe one system captured a passing star from somewhere else, a sort of adopted star, not a true sibling.  A single star probably cannot capture another, but any multi-star system is quite capable of it.

The stability of systems with more than two significant masses is low.  There are no 'host points', so the paths are unpredictable, and such systems are capable of ejecting stars from the family in a way that a binary system would not.

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The Earth & the Moon are real "siblings". This is a key element. I will explain later on why the current concept of how the Moon had been created is absolutely unrealistic:
Well, other moons on other planets are not real siblings in that they formed from the same material.  So their history of being siblings doesn't explain the orbit or lack of it.  Yes, we went up there and found the moon to be the same stuff as us, thus forming the idea that it is a chunk of Earth knocked off or spun off.

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"The giant-impact hypothesis, sometimes called the Big Splash, or the Theia Impact suggests that the Moon formed out of the debris left over from a collision between Earth and an astronomical body the size of Mars, approximately 4.5 billion years ago"
I will also explain how the earth and the moon had been formed "from a single cloud of gas and dust, so they are true siblings". and why they set this binary system due to gravity force.
Being formed by a whack is quite different than forming separately from a common cloud.  The latter hypothesis is unlikely.  Still, Pluto has a moon even more disproportionately large, and that one doesn't appear to be the same material.  Unclear on the history of those two.

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But in order to explain it, we must know how our galaxy really works.
The spiral arm is a key element in our discussion.
Therefore, understanding the gravity force between the objects in the arm is vital.
So, the Earth/moon is a binary system which holds them both while they orbit around the Sun.
In the same token, the arm is a multi-star-system.
Each star in the galaxy has its own virtual host point.
Together, those host points set that "multi-star-system" which holds all the neaby stars in the arm.
You've not explained this host point mechanics, and thus it is just fantasy.  It can be trivially demonstrated with a simulation of 3 or more stars, none of which will exhibit motion about a host point. You can deny the Newtonian forces upon which that simulation would run, but you've not described new laws to replaces the ones you're denying.

Sure, arms hold themselves together via gravity.  That is pretty simple since force (towards galactic center) goes up at the outside edge of the arm, and force (towards galactic center) goes down on the inside edge.  This is very different from Earth/moon where the sun's pull is greater, not less, at the inside edge.  So arm stability can be explained without this host-point nonsense.

None of how the arms hold together is relevant to the question of the acceleration of the arm itself, which is totally unexplained in your posts.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/01/2019 06:11:37
Sure, arms hold themselves together via gravity.  That is pretty simple since force (towards galactic center) goes up at the outside edge of the arm, and force (towards galactic center) goes down on the inside edge.
Dear Halc
If you agree with that, than this is all is needed to explain the spiral arm structure.
As I have stated, each star is connected in the arm by Newton gravity force.
All the stars have a similar orbital speed. (More or less)
Therefore, for any time frame, they all cross the same distance.
In order to achieve it, as they drift outwards, they also drift backwards.
That activity sets by definition the spiral shape of the arms.

That is the whole idea about spiral arms.
Now, let's try to find what our scientists have to say about: "How Our Milky Way Galaxy Got Its Spiral Arms"
https://www.space.com/24642-spiral-galaxies-milky-way-shape-explained.html
Dated - February 12, 2014
"The researchers found that the universe was a very chaotic place in its infancy. The first galaxies were disks with massive, bright, star-forming clumps and little regular structure. To develop the nice spiral forms seen today, galaxies first had to settle down, or "cool," from the previous chaotic phase. This evolution took several billion years."

First contradiction - "Massive disc galaxies" - If our universe was "very chaotic place in its infancy" how could it be that we have got immidiatly disk galaxies with massive, bright, star-forming clumps and little regular structure."
Why those massive disc galaxies had been developed in the chaotic Universe? How long it should take to set the first massive disc galaxy in that unclear process? How could it be that all the billions spiral galaxies had been formed from this chaotic Universe?

"Gradually, the galaxies that were to become spirals lost most of their big clumps, and a central, bright bulge would appear; the smaller clumps throughout the galaxy would begin to form indistinct, "woolly" spiral arms.
These arms would only become very distinct arms once the universe was about 3.6 billion years old. At that age, as the galaxies had a chance to settle down, the turbulence decreased, and new stars would form in a much quieter disk. "We can see the transition from the early chaotic state to the modern, relaxed state," said Bruce Elmegreen.
These first spiral galaxies were either two-armed structures or had thick, irregular spirals with some remaining clumps. More finely structured, multi-armed galaxies like the Milky Way galaxy and its neighbor Andromeda appeared much later, when the universe was 8 billion years old."

Second contradiction - "Age" - We see spiral galaxies at the most far end of the Universe.
The estimated age of many mature spiral galaxies is more than 13.2 Billion years.
So, how could we see today very mature spiral galaxies at estimated age of only 0.6 billion years, if based on this article we need several billions (3.6 Min) and also with the assumption that we have got "Massive disc galaxies" almost immediately from the chaotic Universe?

Third contradiction – "Random process" - They don't show exactly how do we get this spiral arms from  massive disc galaxy. They just say that "Gradually, the galaxies that were to become spirals lost most of their big clumps,.."
This is a random activity by definition. They don't say why they lost the big clumps and how they really got their spiral arms based on Newton gravity.

Conclusions:
I have set a simple explanation why 400 Billion spiral galaxies have got their spiral shape from day one based on Newton gravity force.
Our scientists (even at 2014) try to find an explanation which is not correlated to the universe age time frame. This random activity can't be the base for all the billions spiral galaxies that we see in our universe.
I have set my explanations about the spiral arms structures which is much more relaibale from the last "story" from our scientists.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/01/2019 13:04:57
If you agree with that, than this is all is needed to explain the spiral arm structure.
It is far more complicated than either of us can explain.  You have a bowl of cake batter with a bar of dark batter across the middle.  You stir once or twice and you get a spiral structure something like the galaxy.  You stir 40 times and that structure fades to either super-thing spirals or just one color.  How are the arms still there after 40 laps with all the parts moving at different angular speeds around the galaxy?  Our own arm seems to be one of the ones that is being thinned to the point of nonexistence, possibly explaining why we've not been disturbed by the material in a thicker arm which would have disrupted life formation.

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As I have stated, each star is connected in the arm by Newton gravity force.
All the stars have a similar orbital speed. (More or less)
Too much speed, but yes, more or less the same.  Not the same angular speed, which would be needed to keep the arms intact indefinitely, but just similar linear speed.

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Therefore, for any time frame, they all cross the same distance.
Exactly, but the stars close to the center of the galaxy have far less distance to go, so they go around far more often than the stars at the far ends of the arms that might yet to have gone around even 10 times since they have so much further to travel to achieve one lap.

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In order to achieve it, as they drift outwards, they also drift backwards.
That activity sets by definition the spiral shape of the arms.
That drifting outwards and back defines the clumping of the arms.  The spinning at different angular velocities of each end defines the spiral shape, just like the cake batter.

The speed at which everything moves is unexplained by your assertions.  You never address this issue.

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Now, let's try to find what our scientists have to say about: "How Our Milky Way Galaxy Got Its Spiral Arms"
Your last link was hardly to a scientist.  This one here is at least a magazine article which presumably has some standards beyond what quora obviously has.
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https://www.space.com/24642-spiral-galaxies-milky-way-shape-explained.html
Dated - February 12, 2014
I like this one, reporting on doing the research the correct way:  Look into the past as see how the shapes evolved.  It's sort of like watching an animation of our own galaxy over billions of years, except all the frames have been dropped on the floor and need sorting.
The article is more about how the research was done than what the results of that research were.

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"The researchers found that the universe was a very chaotic place in its infancy. The first galaxies were disks with massive, bright, star-forming clumps and little regular structure. To develop the nice spiral forms seen today, galaxies first had to settle down, or "cool," from the previous chaotic phase. This evolution took several billion years."

First contradiction - "Massive disc galaxies" - If our universe was "very chaotic place in its infancy" how could it be that we have got immidiatly disk galaxies with massive, bright, star-forming clumps and little regular structure."
Why those massive disc galaxies had been developed in the chaotic Universe? How long it should take to set the first massive disc galaxy in that unclear process? How could it be that all the billions spiral galaxies formed from this chaotic Universe?
Hard to parse your comments here.  You're suggesting a contradiction, but I don't see one identified.  Higher density matter tends to group due to its mutual attraction.  That forms galaxies.  If the matter has any angular momentum, then the clump will naturally form a disk shape, just like what happens with new solar systems forming out of a gas cloud.  The disk shape is the only way to preserve angular momentum of a compressing structure.

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Second contradiction - "Age" - We see spiral galaxies at the most far end of the Universe.
The estimated age of many mature spiral galaxies is more than 13.2 Billion years.
So, how could we see today very mature spiral galaxies at estimated age of only 0.6 billion years, if based on this article we need several billions (3.6 Min) and also with the assumption that we have got "Massive disc galaxies" almost immediately from the chaotic Universe?
Nothing said that very mature spiral galaxies were of age 0.6 billion years.  That would be a young galaxy still lacking a mature spiral structure.  From where are you getting this claim of a spiral galaxy that is only 0.6 BY old?

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Third contradiction – "Random process" - They don't show exactly how do we get this spiral arms from  massive disc galaxy. They just say that "Gradually, the galaxies that were to become spirals lost most of their big clumps,.."
This is a random activity by definition. They don't say why they lost the big clumps and how they really got their spiral arms based on Newton gravity.
Missing details is not a contradiction, just not a full description.  Such details might not be clear.  They only looked closely at about 41 galaxies to get an idea of the various stages of the process.  Nobody has a video of the evolution of any one galaxy as it goes through the process.

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Conclusions:
I have set a simple explanation why 400 Billion spiral galaxies have got their spiral shape from day one based on Newton gravity force.
You have no such thing.  You just assert it, and the data from the Hubble image clearly shows that assertion to be wrong since no young galaxy has arms, and thus they're not there from 'day one'.

The article you link suggests no explanations whatsoever.  It just reports data: This is what apparently happens.  Given that, find a theory that explains it, but no such theory seems to be proposed in that article.  You have asserted having an explanation, but none is provided.  You just assert the current observed picture, and fail to explain simple observed speeds of the arms.  So your explanation falls completely short of the current views on galaxy dynamics, which are admittedly only best models at the time.  Those models are always changing, especially since none of them really provides a full explanation.

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You can accept it or reject it.
And so I have.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/01/2019 13:31:19

The speed at which everything moves is unexplained by your assertions.  You never address this issue.

Yes, I can explain it.
However, I need to see the real data.

Please look at the "Observed rotation curve" in the following article?
http://ircamera.as.arizona.edu/astr_250/Lectures/Lecture_22.htm
Observed rotation curve:
We see that at about 0.2 KPC the orbital velocity is maximal - 260 Km/sec
Than it goes down to about 190 Km/sec at 3KPC.
From this point it goes up to 210 Km/sec at 4KPC
and to 230 Km/sec at 7 KPC.
Than it starts to go down again to 200 Km/sec at 10KPC
Goes up to 235 Km/sec at 13 KPC and stay at that velocity at any further distance.
Is this correct?

If so, how could it be that we don't see the real orbital velocity near the SMBH?
We know that the orbital velocity of the plasma is 0.3 c (speed of light)
At the end of this article it is stated:
"Consider a star whose proper motion has been measured to be equivalent to 1000 Km/sec and which lies only 0.01 pc from SgrA*.
So, why we don't see the real velocity at the center?\
Why it shows that the orbital velocity at the center is almost Zero?

Why they also give this information without any connection to the galaxy shape?
I would expect to see measured velocities per radius in the following segments:
1. Accretion disc (What is the Min radius and maximal radius. What is the orbital velocity at each radius?
2. Bulge - Radius range and velocities
3. Bar - Radius range and velocities
4. Ring - Radius range and velocities
5. Disc/spiral arms - Radius range and velocities
6. From the end of the disc - Radius and velocities
Would you kindly direct me to the real measurements of orbital velocities Vs radius in the Milky Way? (From 0 - 20K Pc)
I will give full explanation once I have the real data.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/01/2019 17:43:44


Thanks
Your link above seems to be fairly in line with similar graphs I've seen.
So as you agree on this graph, let's set the correlation with the structure.
In order to do so, please look at the following diagram of the Milky way structure:
http://www1.ynao.ac.cn/~jinhuahe/know_base/astro_objects/galaxies/Milky-Way-Galaxy-files/logarithmic-spiral-pattern-before2001.JPG
We can see clearly that the radius of the ring is 3Kpc.
Now, please remember that the minimal orbital velocity was exactly at that radius.
Than it goes down to about 190 Km/sec at 3KPC.
So, the orbital velocity at the ring is 190 Kpc.
That shows that at the end of the bar, the orbital velocity is minimal. That evidence contradicts your following message:
The ends of the bar are moving faster than anything else, just like the ends of the thrown spanner are moving faster than the rest of it.
So, the end of the bar has the minimal orbital velocity.
This by itself must set a big red light. How could it be???
In any case, the spiral arms starts exactly at that ring (3KPC).
We don't see the end of the arms but we can assume that it ends at about 15 KPC (at about 45K light year).
Let's focus on the Bar and the bulge:
As we move inwards from the ring the orbital velocity is increasing.
It almost goes to 260 Km/s at 0.5KPC.
I get that at about 0.5 KPC.
However, as we get closer to the center from that radius, the orbital velocity gets down.
At about 0.1 KPC the orbital velocity is almost Zero!!!
This is a very important segment in the galaxy.
There is a meaning for that activity.
Our scientists focus only on the orbital velocity of the stars in the arms, while they ignore completely other important segments..
That shows that their knowledge in spiral galaxy is very poor.
I can explain each segment.
However, do you agree by now that there is high correlation between the galaxy structure and the orbital velocity?
Do you agree that your following message is incorrect?
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Why they also give this information without any connection to the galaxy shape?
It seems there is little connection.  Objects at radius X move at the same speed regardless of being in an arm, near one side or the other, or between them.  The arm might affect regular wobble in and out, but not the tangential speed around the galaxy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/01/2019 05:29:31
No, it means that not a lot of pop-science web sites focus on that part.  I haven't tried much to actually see what the consensus is about how things move in that inner 3 kps, and how the bar keeps its shape.  Such a thing is fairly common, hardly a weird anomaly of just our own galaxy.
Please look at the following image from NASA (which is identical to the one that you have offered)
https://solarsystem.nasa.gov/resources/285/the-milky-way-galaxy/
It shows clearly that at 3KPC there are two sides of the ring
The one which is closer to the Sun is called: Near 3KPC Arm
The one which is farther from the Sun is called: Far 3KPC Arm
Together they set the ring.
The end of the Bar is directly located just between those arms on each side.
How can you claim that:
The arms seem not to be continuous tubes of stellar material, but rather chunks arranged sort of like fallen dominoes...
Well I don't see it from that data.
The ring I think is way outside, like maybe 40 kps or something, a halo of gas not particularly dense enough to form luminous objects.
The data is clear. It is time for you to agree with the evidences.
There is a clear ring at 3KPC!!!
This ring has a great impact on the activity of the galaxy.
I will explain it later on.
But first - do you agree that :
1. There is a ring at 3KPC?
2. The orbital velocity at that ring is minimal - 190 Km Sec?
3. The Bar ends exactly at that ring?
4.The arms starts to form exactly from that Ring?
Why do you disqualify even the clear evidences which me & you represent?
Why are you so negative?
Would you kindly look again at the evidences and try to understand that spiral galaxy is not just spiral arms?
Would you kindly try to be more cooperative and help us to highlight the real activity at spiral galaxy?



Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/01/2019 18:01:00
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The one which is closer to the Sun is called: Near 3KPC Arm
The one which is farther from the Sun is called: Far 3KPC Arm
Together they set the ring.
No, they set the inner portions of a spiral.
If you want to call it a ring, fine, but the two arms do not connect.  Each spirals out into the two major arms.  The image clearly shows this.  I notice both arms get noticeably thicker at the point halfway around when the pass relatively close by the opposite end of the bar from which they appear anchored.

No - None of them spirals out. Our scientists call them 3KPC as they stay at 3KPC.
Yes - They are connected at their edges to other spiral arms, but they together set a ring or almost a ring.
I hope that the following image can give you better overview
https://crossfithartford.com/dummies_diagram_of_plant_the_milky_way.php
We can see the two 3KPC arms.
One is called near 3KPC arm and the other Far 3KPC arm.
Why do we call them both 3KPC arm?
Don't you think that the name of the 3KPC represents their radius???
It also seems that both of them are connected/almost connected.
In the following image it is quite clear:
https://solarsystem.nasa.gov/resources/285/the-milky-way-galaxy/
We see that in one side they are connected at a point which is written as galactic bar, while on the other side it is called long bar. At those points we also see the connections to the main spiral arms.
In the image it looks like elliptic ring shape.
However, don't you agree that the mane - 3KPC arm proves that this arm is located all the way at a radius of 3KPC?
They both stay there by themselves.  None of them is part of another spiral arm.
Therefore, if one arm covers the upper half/almost half of the cycle, while the other arm covers the other half/almost half of the cycle, don't you agree that in the total we get a ring/almost ring
You might call it as upper 3KPC arm + lower 3KPC arm.
However, even if you don't see a perfect ring, it is still a ring (or almost a ring...)
We know that the orbital velocity goes down to its minimal value exactly at 3KPC.
Therefore, this ring is a key element in the activity of the galaxy.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/01/2019 19:55:46
Please look at the following diagram about: NGC 1398: AN UTTERLY PERFECT SPIRAL GALAXY

https://www.syfy.com/syfywire/ngc-1398-an-utterly-perfect-spiral-galaxy

We clearly see a ring at the center (at the edge of the bar which isn't fully clear).
It is also stated:
"NGC 1398 is a barred spiral, with that rectangular-shaped feature running across the core and ending at the inner "ring." This is a common feature in spirals — the Milky Way has a big one — and they can affect how the stars and gas move around the galactic center (I describe this in the earlier post on NGC 1398)."
I have no idea what is the radius of that ring or its width, but it shows that there is "inner ring" in barred spiral galaxy as at the Milky Way.
It is quite clear that this "inner ring" has a significant effect on "how the stars and gas move around the galactic center" as I will explain later on.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/01/2019 05:35:30
As I said, I can consider that a ring if you wish, but I see a clear spiral there
Great!  Other people call it a ring. Let's go with that.

Thanks!!!
So, the ring has almost a perfect cycle shape around the Bar. (Although if we look carefully inside the ring, we see clearly that it has a spiral shape.)
Once we agree on that ring, we can now start looking at the structure/velocity of barred spiral galaxy


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/01/2019 07:22:50
Let's start with the following question:

What kind of gravity force that ring could generate?
Let's assume that there is no SMBH at the center.
Let's also assume that we set Billion of stars at a perfect ring of 3KPC and hold them ALL at their current position for very long time.(Please don't tell me what will happen to the ring after one minute. This is a theoretical question about stars ring)

What kind of gravity force we might find inside the ring?
Do you agree that all the stars together will set an equivalent gravity force on any object which is located inside that ring?
Therefore:
1. If an object is located directly at the center - Do you agree that the impact of the gravity force from the stars in the ring at a distance of 3KPC in one side should cancel the impact of the stars in the other side of the ring? Therefore, the net gravity force on that object is - Zero?
2. If an object is located at 1.5KPC from the Center (Let's set it close to 12 (in clock) - The impact gravity force of the stars in 3 and 9 should cancel each other? However, the distance to 12 is now 1.5 KPC while the distance to 6 is 4.5KPC.
Therefore, do you agree that the net gravity force in the direction of 12 is quite strong?
3. If an object is located very close to 12 (let's say at 2.9 KPC). - The impact of the gravity force in 12 is almost maximal (as the distance is just 0.1KPC). However, the impact of 6 is minimal (as the distance is 5.9KPC). Remember that 3 cancel the impact of 9.
Therefore, do you agree that the impact of the ring gravity force is higher as the object is closer to the ring?

 
 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/01/2019 13:57:12
Let's start with the following question:

What kind of gravity force that ring could generate?
Let's assume that there is no SMBH at the center.
Let's also assume that we set Billion of stars at a perfect ring of 3KPC and hold them ALL at their current position for very long time.(Please don't tell me what will happen to the ring after one minute. This is a theoretical question about stars ring)
Without reading further, a hoop like that is in equilibrium and can orbit itself, with or without a central object.  The mass within (not part of) the ring will contribute to higher speed of the ring, but a ring by itself will have some minimum speed.  It isn't totally stable, but stable enough, just like our asteroid belt, which has a central mass.  I can think of no example of a ring on its own.

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What kind of gravity force we might find inside the ring?
A hollow sphere would have a uniform field.  A body inside one would be weightless, but a ring will attract to itself.  I would fall outward down to the nearest part of the ring if I found myself stationary within it somewhere.
A stationary 'me' would have no orbit or radius, so this isn't even in violation of Newton's law that says mass outside my radius has no overall effect on my orbital speed.

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Do you agree that all the stars together will set an equivalent gravity force on any object which is located inside that ring?
I don't know what you mean by 'set an equivalent gravity force' here. Equivalent to what?

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Therefore:
1. If an object is located directly at the center - Do you agree that the impact of the gravity force from the stars in the ring at a distance of 3KPC in one side should cancel the impact of the stars in the other side of the ring? Therefore, the net gravity force on that object is - Zero?
At the center, yes.  Net of zero by symmetry.

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2. If an object is located at 1.5KPC from the Center (Let's set it close to 12 (in clock) - The impact gravity force of the stars in 3 and 9 should cancel each other?
There will be no net pull to either side, yes, again by symmetry.  The stars at 3 and 9 on the ring are both 'south' of the object that is 1.5kpc north of the center, so there is a net pull towards the center from those two stars.

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However, the distance to 12 is now 1.5 KPC while the distance to 6 is 4.5KPC.
Therefore, do you agree that the net gravity force in the direction of 12 is quite strong?
Stronger than any of the others, yes.  You need to integrate the vectors on the entire ring to get the net force on the object.  Taking these single samples will not work.

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3. If an object is located very close to 12 (let's say at 2.9 KPC). - The impact of the gravity force in 12 is almost maximal (as the distance is just 0.1KPC). However, the impact of 6 is minimal (as the distance is 5.9KPC). Remember that 3 cancel the impact of 9.
Yes on the 12max/6min, but no on the 3/9 cancelling.  Almost the entire ring pulls the 2.9kpc object downward.  This is why there is zero net force in a sphere.  To compute the net force of the ring, you need to integrate at least the vertical force vector over the entire ring.
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Therefore, do you agree that the impact of the ring gravity force is higher as the object is closer to the ring?
Not because of your faulty argument above, but yes, it is.  Your argument above would have there being a net force in a sphere, and we know that's false.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/01/2019 20:11:54
Without reading further, a hoop like that is in equilibrium and can orbit itself, with or without a central object.  The mass within (not part of) the ring will contribute to higher speed of the ring, but a ring by itself will have some minimum speed.  It isn't totally stable, but stable enough, just like our asteroid belt, which has a central mass.  I can think of no example of a ring on its own.
Thanks
The activity of the ring is clear. It is also clear that it has a great impact on any object which is located inwards (or outwards).
Let's start by looking inwards:
Let's set back  the SMBH at the center of the galaxy.
We have already found that the impact of the Ring at the center is virtually Zero.
Therefore, any object which is located very close to the center is directly affected by the SMBH. It might be a star, gas cloud or even Atom.
However, as we move further away from the center, the gravity force of the ring is more relevant.
Please remember that the SMBH mass is estimated to be in the range of few millions Sun mass. However, the total estimated mass of the ring is in the range of several billions of sun mass.
Based on that knowledge, let's look again on the orbital velocity at the range of Zero to 3KPC (about 10K LY).
We know that the orbital velocity of the accretion disc is 0.3 speed of light.
The orbital velocity of S2 which is located at a distance of 2-10 Light days from the SMBH is 5000 Km/s
If we get to a distance of 0.1 KPC (326 LY) from the center the orbital velocity gets down to 150 Km/s.
However as we get further away from this point, the orbital velocity stars to increase.
At 0.5 KPC the orbital velocity gets to its pick of almost 250 Km/sec.
So far so good.
We can assume that this increase in the orbital velocity is directly affected by the ring.
However, as we go further away from the center and get closer to the ring the orbital velocity goes down again to its minimal value at the ring - 190Km/s. This is unexpected phenomena.
Why the orbital velocity goes down again as we get closer to the ring?
In order to answer this question - Let's assume that we have only one object which orbits inside the ring.
So, we have the ring, we have the SMBH but only one star which orbits there.
We can easily calculate the estimated orbital velocity of this object at any radius from the accretion ring up to 3KPC.
However, as there are millions (or billions of stars) in the bulge + in the Bar, they also set a gravity force on each other.
Therefore, this could be the answer why we see that the velocity is going down as we get closer to the ring.
There is another factor.
The ring is connected to spiral arms.
So, it is used as a frame which holds the spiral arms and set their orbital velocity.
Therefore, it is clear that the ring is loaded by those spiral arms. This load could also have some impact on the orbital velocity of the ring and the orbital velocity of stars inside the ring.
Do you agree with that?
Please also be aware to a very critical issue.
There is no spiral shape inside the ring!!!
The spiral shape is very clear in the ring itself and in the spiral arms.
Therefore, it is clear that the ring has a significant impact of the spiral arms.
I will discuss about it later on.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/01/2019 08:16:54
You treat the ring and arms as solid objects, talking about loads being put on them and such.  They're not.  They're collections of detached objects which don't alter paths if loads are put on other parts.  You can't alter the speed of the ring by putting a force on part of the ring.  That works only if it is an object.
I disagree with you,
The ring is an object by itself as the Earth is an object by itself.
The Earth is made by many different of Atoms and molecular that are all bonded by gravity force.
In the same token, the Ring is made by many Stars and dust that are all bonded by gravity force.
In one hand you agree that spiral structure start exactly at the ring:
https://phys.org/news/2018-01-image-hubble-barred-booming-spiral.html
agree that the spiral seems to end at 3 kps, not spiraling all the way into the center.  There are some galaxies that do that, but they're in the minority.
But on the other hand you don't see the great impact of the ring.
Why is it?
Let's look carefully at the following image:
https://en.wikipedia.org/wiki/Spiral_galaxy#/media/File:PIA19341-MilkyWayGalaxy-SpiralArmsData-WISE-20150603.jpg
https://phys.org/news/2018-01-image-hubble-barred-booming-spiral.html
https://www.syfy.com/syfywire/ngc-1398-an-utterly-perfect-spiral-galaxy
We see clearly that the arms are connected exactly at the opposite sides of the ring.
In some galaxies it is quite difficult to see the ring, but we see clearly that spiral arms are connected exactly at the opposite sides of the ring and directly to the bar.
This connection between the Arm to the Bar is very important.
It shows that the spiral arm is an extension of the bar.
So, we can think about it as a long arm. Outside the ring we have the spiral shape. Inwards the ring we have the bar.

If we look carefully we in all the spiral galaxies in the Universe - we can also see that all of them have a very symmetrical view.
Spiral arms starts exactly from the opposite sides of the ring - The arm at one side is very similar to the arm in the other side. So in all the galaxies we mainly see two main arms. We might see other arms in the galaxy, but all of them are connected to those two main arms (or near by the conection point)  in a symmetrical view.
We can also ask: Is it feasible to have a spiral galaxy without a ring?
The answer is clearly - Yes, but a spiral galaxy without a ring is quite different from a one with the ring
Please look at the following image:
http://hubblesite.org/image/1636/news_release/2005-01
Please see the ratio between the bar radius to the arm maximal radius:
Do you agree that the ratio is about 1:3?
In the Milky Way the bar gets to 3 KPC while the spiral arms gets to 45 KPC.
So the ratio in the Milky way (with a ring) is 1:15.
Hence, as the ring is more developed, the ratio is higher.
Do you agree that without a ring (or with minimal ring) there is a maximal load (or maximal spiral arm size/radius) that spiral galaxy can take?

Therefore, we have to ask the following:
1. How could it be that the bar is connected to the ring while at the same spot of the connection we clearly see the starting point of the spiral arms? Our scientists only try to explain the orbital velocity in the arm, but they ignore completely the Bar. How could it be that they don't see that the bar is directly connected to the arm?
2. Why there is no spiral shape in the Bar?
3. How could it be that at the ring, the orbital velocity is at its minimal value (190Km/s)?
If we move inwards to the bar (let's assume in the Milky Way - 500 Light years inwards from the ring), the orbital velocity is higher (at about 200 Km/s), while if we move outwards into the spiral arms at the same distance, the orbital velocity is higher again (at about 200 Km/s).
4. Why spiral galaxies have a symmetrical view?
5. What is the real impact of the ring?





Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/01/2019 14:41:23
Quote from: Halc
You treat the ring and arms as solid objects, talking about loads being put on them and such.  They're not.  They're collections of detached objects which don't alter paths if loads are put on other parts.  You can't alter the speed of the ring by putting a force on part of the ring.  That works only if it is an object.
I disagree with you,
The ring is an object by itself as the Earth is an object by itself.
The Earth is made by many different of Atoms and molecular that are all bonded by gravity force.
In the same token, the Ring is made by many Stars and dust that are all bonded by gravity force.
If that were true, I would immediately drop straight through and fall to the core of the Earth because that is where gravity is trying to take me.  The fact that it doesn't happen means that the Earth is bonded by forces other than just gravity, forces that do not exist between the various objects in the ring.

Why should I continue this thread when you post something that obviously wrong?

Quote
We see clearly that the arms are connected exactly at the opposite sides of the ring.
In some galaxies it is quite difficult to see the ring, but we see clearly that spiral arms are connected exactly at the opposite sides of the ring and directly to the bar.
This connection between the Arm to the Bar is very important.
There are no connections.  These are not solid objects.  Two smears of dust that intersect do not make physically connected dust.  If that connection is important, then your idea really falls flat, because there's no connection.  Everything in those pictures is free-falling objects interacting only by gravity, much in the same way that the atoms of Earth or a spanner are not.

Quote
Please look at the following image:
http://hubblesite.org/image/1636/news_release/2005-01
Please see the ratio between the bar radius to the arm maximal radius:
Do you agree that the ratio is about 1:3?
About that, yes, but only if you don't include the dimmer outer parts of the arms that they cropped off the image, but you're including those parts when you say the milky way goes all the way to 45 kps:
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In the Milky Way the bar gets to 3 KPC while the spiral arms gets to 45 KPC.
So the ratio in the Milky way (with a ring) is 1:15.
If you look at the Milky Way image in your first link of the prior post, the bar ratio seems to be about 1:5 or 1:6, and a similar ratio for that 'perfect spiral' galaxy you link.
So the ratio is higher, but not a lot higher.

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Do you agree that without a ring (or with minimal ring) there is a maximal load (or maximal spiral arm size/radius) that spiral galaxy can take?
I don't know what 'load' is.  If you're treating the arms or ring as solid objects, then the question is nonsensical.
OK, you say it is a maximal galactic radius, but that is a function of how much mass and angular momentum the galaxy has.  Mass holds it together, and angular momentum flattens and spreads it out.  These numbers are different for different galaxies.  The symmetry of the galaxy has a lot to do with its recent history.  That perfect spiral galaxy has not been disturbed in quite some time, where our galaxy has been continuously eating small snack-size neighbors, and is soon to be eaten itself by a larger fish, all of which will at least temporarily destroy that nice bar/ring/ arm picture of both galaxies.

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Our scientists only try to explain the orbital velocity in the arm, but they ignore completely the Bar. How could it be that they don't see that the bar is directly connected to the arm?
Probably because it isn't.

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3. How could it be that at the ring, the orbital velocity is at its minimal value (190Km/s)?
Perhaps because there is not much additional mass inside the ring to make it go faster.
Maybe the velocity in some places is not tangential.  The graphs do not show velocity, only speed.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/01/2019 06:13:19
If you look at the Milky Way image in your first link of the prior post, the bar ratio seems to be about 1:5 or 1:6, and a similar ratio for that 'perfect spiral' galaxy you link.
So the ratio is higher, but not a lot higher.
Thanks
So, we agree that with a ring the ratio between the bars radius to the maximal arm radius is higher.
If so, why do we ignore the outcome of this evidence?

Quote
Our scientists only try to explain the orbital velocity in the arm, but they ignore completely the Bar. How could it be that they don't see that the bar is directly connected to the arm?
Probably because it isn't.
Why do you claim that it isn't?
Please look again at the following image:
https://en.wikipedia.org/wiki/Spiral_galaxy#/media/File:PIA19341-MilkyWayGalaxy-SpiralArmsData-WISE-20150603.jpg
Please advice if you agree with the following:
1. The starting point of the Soutum-Centaurus arm is located at the same spot where the Bar ends?
 2. Exactly at the other end of the bar we see a symmetrical view. The starting point of Perseus arm is located at that spot.
3. Those two arms seems to be main arms due to their length and width.
4. The ring cross exactly at those  two edges/spots of the bar connections/locations with the those two main arms.
5. Based on this image, it is clear the bar is orbit in clock wise.
6. Near the Soutum - Centaurus arm starting point to the end bar location spot, we also see a starting point of a smaller thinner arm. However, it is located a little bit further from the clock wise orbital direction of the bar.
7. In full symmetrical view, we see another thinner arm which its starting point is located near the other Bar/main arm spot. This arm also starts a little bit further from the clock wise orbital direction of the bar.
8. So do you agree that we see clearly a symmetrical view from both end sides of the Bar?

Don't you think that there is a meaning for what we see?

As you don't like to call this spot as a "connection" point, do you estimate that the bar can move faster/slower than the starting point of the main arms and one day it should be disconnect from the arm?
If so, can you please offer an image of a Barred spiral galaxy with two main arms, while its bar (ends) had been totally disconnected from those two main arms?


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/01/2019 14:47:23
So, we agree that with a ring the ratio between the bars radius to the maximal arm radius is higher.
If so, why do we ignore the outcome of this evidence?
Evidence of what?  A general trend?  No.  The cherry-picked dataset of 3 is not evidence of that.  A trend can be found in a survey of a lot of galaxies.
In that small dataset, I see a tightly wound set of arms, and a loosely wound set.  The one is so tight that it almost looks like a ring, just like toilet paper looks like concentric circles even though it isn't.
I suspect that the loose galaxy had less angular velocity.  Perhaps such slowly rotating barred galaxies generally have larger bar-to-galaxy ratios, but only a larger survey would be evidence of that.

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5. Based on this image, it is clear the bar is orbit in clock wise.
The bar and arms are not clearly in orbit.  The shape of them seems stable, but the image does not show motion, and the ones that do show motion do not show motion that corresponds with that rotation curve.  Thus I do not see the motion of either bar nor arms qualifying as 'orbital motion'.  Yea, they twirl around, but that's just spinning, not necessarily orbiting.

What I see in the arms is that telltale fallen-dominos signature, where material seems grouped in diagonal clumps that overlap. I can really see it in that 'perfect spiral' galaxy you linked, especially in the outer parts of the arms. It is a familiar image:
https://en.wikipedia.org/wiki/Wake#/media/File:Bodensee_at_Lindau_-_DSC06962.JPG
The duck wake appears completely stationary to a camera that is following the duck, and has that overlapping fallen-dominos signature, not just one clean wave that angles out.  The material of the waves does not follow the wave/duck however.  The water is moving in a completely different direction than is the wake.  The wake appears to be a solid object connected to the duck, but it is not connected at all.
The galaxy appear to work exactly that way.  Apparent motion of the arm waves are just wave action, not actual motion of the stars that make them up.  You cannot deduce the arm motion from the motion of the component stars and vice versa.

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8. So do you agree that we see clearly a symmetrical view from both end sides of the Bar?
Almost all galaxies look little different if rotated 180 degrees, so yes, I agree with this one.

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Don't you think that there is a meaning for what we see?
You don't identify what meaning you expect me to get from that, but probably not the sort of things you've been pushing like any of it behaving as a solid.  There are simply no forces that could account for that.  A spanner holds its shape because there are very much forces that connect the components.  Those forces are not gravitational.

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As you don't like to call this spot as a "connection" point, do you estimate that the bar can move faster/slower than the starting point of the main arms and one day it should be disconnect from the arm?
You would have images of galaxies then where that happens.  No, the [main] arms always start at the bar ends, just like the wake always starts at the duck despite not being connected .
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/01/2019 17:36:45
Quote
As you don't like to call this spot as a "connection" point, do you estimate that the bar can move faster/slower than the starting point of the main arms and one day it should be disconnect from the arm?
You would have images of galaxies then where that happens.  No, the [main] arms always start at the bar ends, just like the wake always starts at the duck despite not being connected .

Thanks

So now we agree that the main arms always starts at the bar end!!!
But why is it?
Let's start with the Bar:
What is the Bar?
Based on Wiki:
https://en.wikipedia.org/wiki/Barred_spiral_galaxy
"The creation of the bar is generally thought to be the result of a density wave radiating from the center of the galaxy whose effects reshape the orbits of the inner stars. This effect builds over time to stars orbiting further out, which creates a self-perpetuating bar structure."
As the bar had been set due to "density wave", can we claim that it is an object?
If the ring isn't an object than how could it be that the bar is an object?
If it isn't an object, how can we compare it to duck?

Now, let's look at the spiral arm? What is spiral arm?
https://earthsky.org/space/galaxies-spiral-arm
Astronomers believe that galaxies have spiral arms because galaxies rotate – or spin around a central axis – and because of something called “density waves.
So, the bar is there due to "density wave", while the spiral arms is there due to the same idea of "density wave".
Hence, how could it be that the bar has a totally different shape and structure from the spiral arm while they both had been created based on the same concept of density wave?
How could it be that the "second density wave" (spiral arm) always starts at the end of the "first density wave" (Bar)?
Why just at 3KPC (exactly at the ring) the "first density wave" (bar) had been changed its shape to "second density wave" spiral arm?
If it is all about "density wave", can we calim that the ring is also there due to density wave?
Why the galaxy needs the Bar or the ring?
Why the spiral arms can't start directly from the Bulge or even directly from the SMBH?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/01/2019 21:06:01
As the bar had been set due to "density wave", can we claim that it is an object?
If the ring isn't an object than how could it be that the bar is an object?
Waves are not objects.  A density wave moves through water at about 1500 m/sec despite no bit of water moving at that speed.  Waves don't have mass of their own, and thus are not subject to forces like objects are.  A wave traveling from the depths to the surface for instance is not slowed by gravity at all.  It is a mistake to treat waves as objects like the spanner where the whole spanner reacts if you exert force at only one point of it.

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If it isn't an object, how can we compare it to duck?
I was comparing everything to the wake, not the duck.

I don't know the dynamics of stellar motion within the bar.  I've seen a few pictures of paths of a single star in simulations, and the stars seem to stay within the bar, but move up and down the length of it. That wouldn't happen if it was a solid object.  But still, the star stayed in the bar (while stars in the arms seem not to stay in their arms), so in that way the bar does resemble a sort of liquid object like a blob of water floating in the space station.  The water holds itself together despite minor disturbances, but it also doesn't form a bar shape since the forces are different than the gravitational ones acting on the galactic bar.

Translation, the bar may well be an 'object' of sorts, a bit like the duck, but a liquid one at best, not a solid one. I'm willing to entertain that so long as you don't treat it like a solid spanner. The bar doesn't look like a wave, even if it formed due to density waves from the center.  The arms on the other hand look very much like density waves, like the wake of the duck.

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Astronomers believe that galaxies have spiral arms because galaxies rotate – or spin around a central axis – and because of something called “density waves.
Look up 'density wave theory' on wiki.  There are some really good illustrations that show what they're talking about.
Here's a gif showing how a wave forms moving at one speed when all the material moves at a different speed.
https://imgur.com/gallery/dtb8WrD
Follow any particular star and it doesn't stay with either arm.

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So, the bar is there due to "density wave", while the spiral arms is there due to the same idea of "density wave".
Hence, how could it be that the bar has a totally different shape and structure from the spiral arm while they both had been created based on the same concept of density wave?
I don't understand the dynamics of the bar.  It seems to not be a density wave itself like the arms are, but I could be wrong about that. They give a reference to a paper on the subject, but it is heavy reading. The site says it is formed by density waves from the center, reshaping orbits, and that effect apparently grows. I know that a lot of the 'orbits' is not stars at all, but just dense gas clouds being reshaped.  A lot of stars are born in there from that gas.
I suspect the galaxies with long bars have been undisturbed for a long time, and those with nonexistent bars are in the process of merging, which disrupts all the symmetry.  If this is true, then we should see galaxies that have recently 'eaten' and are forming new bars and reestablishing the symmetry lost during the merger.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/01/2019 16:09:33
I don't understand the dynamics of the bar.  It seems to not be a density wave itself like the arms are, but I could be wrong about that. They give a reference to a paper on the subject, but it is heavy reading.
I assume that even the scientist that wrote the article has no clue how the density wave creates the Bar.
If it was clear to him, he could explain it in few words. There is no need for "heavy reading" to explain something which is quite clear.
Please also be aware that he doesn't even try to explain how the ring had been formed due to density wave.

There are some really good illustrations that show what they're talking about.
Here's a gif showing how a wave forms moving at one speed when all the material moves at a different speed.
https://imgur.com/gallery/dtb8WrD
Follow any particular star and it doesn't stay with either arm.
I wonder how can we get any valid information from this illustration.
There is no ring and no bar in the illustration.
We only see stars that are moving in and out in order to set the spiral arms.
If that was correct, than by definition the 400 Billion stars that move in and out should collide with each other..
How many collisions do we see in the galaxy?
In any case, in this illustration there is no ring no bar no bulge - only spiral arms. So how can we use it as a valid illustration?
Spiral galaxy is not just spiral arms. It's time for our scientists to walk up and see what there is in spiral galaxy.
There is no way that density wave can set all of those different shapes - Bar, ring and spiral arms.

Let's look at your example about the duck wake:

:
https://en.wikipedia.org/wiki/Wake#/media/File:Bodensee_at_Lindau_-_DSC06962.JPG
The duck wake appears completely stationary to a camera that is following the duck, and has that overlapping fallen-dominos signature, not just one clean wave that angles out.  The material of the waves does not follow the wave/duck however.  The water is moving in a completely different direction than is the wake.  The wake appears to be a solid object connected to the duck, but it is not connected at all.
We see the wake - But it has a specific shape.
For example - There is no way to get a wake in a ring shape, zig zag, or that moves in front of the duck.
So, if there is a density wave - it can't technically set all the variety shapes that we need (Bar, Ring, spiral arms, Bulge) and keep each shape at a specific radius and aria while the bar ends exactly at the starting point of the main spiral arms and this point cross the ring.
This is too challenging request from a simple idea of - "Density wave"
Therefore, I would assume that there is a severe mistake with that idea
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/01/2019 00:28:17
I assume that even the scientist that wrote the article has no clue how the density wave creates the Bar.
I don't think any of the articles you've linked have been written by scientists.  The last article claimed no author and was just a seeming effort from magazine staff or something.
I've occasionally linked to articles from actual scientists/physicists, but most of them don't write for general audiences.

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I wonder how can we get any valid information from this illustration.
There is no ring and no bar in the illustration.
It was meant to illustrate a density wave, not be a simulation of a barred galaxy.  I've seen such simulations, but the short videos made from them compress the timespan too much to observe the dynamics of waves vs stars and other material.

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We only see stars that are moving in and out in order to set the spiral arms.
If that was correct, than by definition the 400 Billion stars that move in and out should collide with each other..
By definition of what?  Most stars are pretty much moving the same way in relation to the galaxy as a whole, not in opposite directions.

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Let's look at your example about the duck wake:

We see the wake - But it has a specific shape.
For example - There is no way to get a wake in a ring shape, zig zag, or that moves in front of the duck.
Two ducks circling each other?
The picture shows what a wake arm looks like.  Yes, galactic ones are from rotating motion, not linear motion.  There will be no V wake from that.  The shape I was illustrating was the 'fallen domino' effect, which exactly matches what I see in the arms of galaxies.

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So, if there is a density wave - it can't technically set all the variety shapes that we need (Bar, Ring, spiral arms, Bulge) and keep each shape at a specific radius and aria while the bar ends exactly at the starting point of the main spiral arms and this point cross the ring.
This is too challenging request from a simple idea of - "Density wave"
Not knowing how it works, I am in no position to declare that it cannot work.  They have simulations that reproduce it, and these simulations use only gravity acting on each thing, not physical connections.  That means it is not too challenging of a request.  They needed to posit dark matter to get it to work, but no new laws like a 5th force or something.

You've suggested no model so far, posting only observations and little actual explanation for any of them.

I think a model of a solid spinning ring and bar with solid arms attached to it would result in an asterisk-shaped galaxy with the arms sticking straight out like they would if you spun out in space a ring with attached flexible arms like that .  That is the best description I've seen so far of your model.
There would need to be a rope or some physical structure holding our sun to the arm so it isn't flung away with the spinning, but we see no structure attaching our sun to the arm.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/01/2019 05:47:22
Quote
So, if there is a density wave - it can't technically set all the variety shapes that we need (Bar, Ring, spiral arms, Bulge) and keep each shape at a specific radius and aria while the bar ends exactly at the starting point of the main spiral arms and this point cross the ring.
This is too challenging request from a simple idea of - "Density wave"
Not knowing how it works, I am in no position to declare that it cannot work.  They have simulations that reproduce it, and these simulations use only gravity acting on each thing, not physical connections.  That means it is not too challenging of a request.
In all the illustration that I have found, there is no answer to the real image of Spiral galaxy as the Milky way.
I couldn't find an illustration which shows clearly the two main major arms which start directly from the end of the bar in full symmetrical view including the Bulge, ring and the Bar.
 
They needed to posit dark matter to get it to work, but no new laws like a 5th force or something.
I have no idea what is the meaning of the 5th force, but there is no need for it. There is also no need for dark matter.
Gravity force - That's all we need to set our galaxy.

I think a model of a solid spinning ring and bar with solid arms attached to it would result in an asterisk-shaped galaxy with the arms sticking straight out like they would if you spun out in space a ring with attached flexible arms like that .  That is the best description I've seen so far of your model.
I agree. If you try to set that kind of model without two basic elements, it won't work.

There would need to be a rope or some physical structure holding our sun to the arm so it isn't flung away with the spinning, but we see no structure attaching our sun to the arm.
There is no rope or some physical structure holding our Moon around the Earth. The same force which holds the Moon in its orbital path also holds the sun in the arm.
You've suggested no model so far, posting only observations and little actual explanation for any of them.
Agree.
Now that it is clear that our scientists have a fatal error in their illustration - it's time for me to present my model.

However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
2. All the stars in the galaxy must orbit around some virtual host (I will explain it later on).

Once we know those two key elements – We have actually solved the Spiral galaxy enigma.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/01/2019 19:40:30
Quote from: Halc
They needed to posit dark matter to get it to work, but no new laws like a 5th force or something.
I have no idea what is the meaning of the 5th force, but there is no need for it. There is also no need for dark matter.
Gravity force - That's all we need to set our galaxy.
I meant no new laws.  The 5th force was just an example.  You create new laws, all of physics must be redone from scratch, which makes for a lousy theory when the existing laws suffice.

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It's time for me to present my model.

However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
I count several violations of existing laws of physics here.  Gravitational force is described by GMm/r², and that has no term for friction or for time.  Friction requires physical contact and has an effect on momentum, and has nothing to do with gravitational force.  So you need to explain what you mean by that.
I can go down a greased slide quickly or a sand-paper slide with lots of friction, and the latter will have me going slower and bloodier at the bottom of the slide, but the gravitational force of Earth on me will be completely the same in both cases.
What else?  Oh yea:  Gravity force of an object is increased by its mass, not by its radius, and an object has no gravity force of its own.  Gravitational force is defined between two objects (the M and m above).  One object creates a field with acceleration (not force) that is a function of the object's mass (not radius of the object) and distance (r) from it.  So that's at least 3 things that seem to not agree with existing physical law in your #1.

Maybe you're just not being clear, but that's what I'm reading here.

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2. All the stars in the galaxy must orbit around some virtual host (I will explain it later on).
You've attempted explanation of this at length.  Watch out for Newton's third law, because this particular notion seems to violate it heavily.

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Once we know those two key elements – We have actually solved the Spiral galaxy enigma.
Actually describing your model would also be a plus.  A list of elements is just hand-waving if I cannot predict the Sun's motion from it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/01/2019 16:03:49
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It's time for me to present my model.
However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
Friction requires physical contact and has an effect on momentum, and has nothing to do with gravitational force.  So you need to explain what you mean by that.
That is incorrect.
Based on the current idea of our scientists, there is tidal friction.
In Tidal friction there is no direct contact or friction between the Earth and the moon, however, somehow it cause the Moon to increase its orbital radius.
It shows that even our scientists believe that without direct contact there is a friction.
Therefore in orbital system there is a friction even without a direct contact.

Gravitational force is described by GMm/r², and that has no term for friction or for time.

I agree
At any given moment the formula for gravity force is as follow:
F=GMm/r²
However, Newton didn't take in his account the impact of time frame.
Somehow, our scientists believe that this force can last forever. This is incorrect.
I do believe that even gravity force must be reduced over time (While there is no change in the mass).
You can call it "Tidal friction", "Gravity friction" or "Gravity force reduced over time". The name of that phenomenon is not relevant.
The key idea is that ALL the planets and ALL the moons in the solar system are drifting outwards over time.
You would never ever find even one planet or one moon in the solar system that drifts inwards.
If we could set the measurements - we should find that ALL of them are drifting outwards over time.
Therefore, after each orbit, the orbital object must shifts/drifts outwards. It can be just few Pico millimeter per cycle or few Km per cycle - but it is there and it is real.
So, orbital cycle is spiral by definition.
This is something that our scientists had missed.
We had long discussion on that issue. So, please don't start all over again with the hypothetical idea why it isn't realistic.
I'm ready to accept your claim - If you can prove (only by real measurements) that there is even one moon or planet that drifts inwards. If you can't prove that there is a moon or planet in a solar system that drifts inwards, you can't contradict this key element phenomena in orbital system.
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2. All the stars in the galaxy must orbit around some virtual host (I will explain it later on).
You've attempted explanation of this at length.  Watch out for Newton's third law, because this particular notion seems to violate it heavily.
Please look again at the following diagram.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see clearly that the sun orbits around some virtual host point, while this host point orbits around the galaxy (Gray Line)
We have also discussed this issue. So, I don't see a value to start all over again.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/01/2019 18:33:14
Quote
It's time for me to present my model.
However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
Friction requires physical contact and has an effect on momentum, and has nothing to do with gravitational force.  So you need to explain what you mean by that.
That is incorrect.
Based on the current idea of our scientists, there is tidal friction.
...
It shows that even our scientists believe that without direct contact there is a friction.
The friction is mostly between water and the sea floor, which is direct physical contact, and is where all the heat is generated.
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In Tidal friction there is no direct contact or friction between the Earth and the moon, however, somehow it cause the Moon to increase its orbital radius.
...
Therefore in orbital system there is a friction even without a direct contact.
A fact that you have repeatedly denied in prior posts, but here you are pushing gravitational friction.  What moves the moon to a higher orbit is a positive gravitational force vector.  There is no friction in this.  The moon does not heat up, only the Earth where the friction with the water takes place.

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At any given moment the formula for gravity force is as follow:
F=GMm/r²
However, Newton didn't take in his account the impact of time frame.
Somehow, our scientists believe that this force can last forever. This is incorrect.
If you mean that they believe that the force will not change over time when none of the variables (M,m,r) change, then I agree with them.  If you're saying something else, then I need a reference for this claim.

There are theories that speed of light was different in the past, and this affects a lot of 'constants', G possibly being one of them.
If your nonexistent model involves a significant change in recent time, you are rewriting physics.
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I do believe that even gravity force must be reduced over time (While there is no change in the mass).
Yea, like that.  My point exactly.

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You can call it "Tidal friction", "Gravity friction" or "Gravity force reduced over time". The name of that phenomenon is not relevant.
It is entirely relevant because these are quite different theories and they make different predictions.
Your theory predicts all orbits should increase by the same proportion over time, but the moon orbit increases (as a ratio of its orbital radius) by 1,500,000 times more than does Earth, instead of the same ratio as you predict.
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The key idea is that ALL the planets and ALL the moons in the solar system are drifting outwards over time.
Yes, but they're not.  All of Mars' moons and over half of Jupiter's moons are losing orbit.
If you model depends on this idea, it is trivially falsified.


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You would never ever find even one planet or one moon in the solar system that drifts inwards.
You need to read more.

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This is something that our scientists had missed.
They measured otherwise, failing to wisely realize the truth and discard such obviously biased findings.
Conservation laws were always wrong anyway.

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If you can't prove that there is a moon or planet in a solar system that drifts inwards, you can't contradict this key element phenomena in orbital system.
Orbit of Phobos is changing more than the orbital radius of any other measurable object.  It is measured as losing over 34 cm per Martian year, or 18 cm per Earth year.  You can choose to deny this of course since the scientists failed to discard this finding each and every time they measure it.  You can also choose to deny when the thing rips apart due to getting too close.

Earth is drifting outward (trivially), but due mostly to a change in M, not a change in G.
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 17/01/2019 22:19:06
We have not been understanding gravity completely. I explain that in detail in my paper mentioned below. There-in, the GDE = Gravitation Direction of Evolution and the FDE = Fundamental Direction of Evolution.

This is an excerpt from the paper which I am also discussing in the "Is The Hubble Shift Due To Time Dilation" and "What is Space" threads in New Theories. This is from the section on Galactic Rotation Velocities:

"The masses of flattened spiral galaxy systems and spherical stellar systems have different shapes and, therefore, different shaped time dilation gradients and different effects in the time aspect.

Within a stellar system, where GR works so well, as the dilation gradient deepens more quickly as the center of the dilation "pit" is approached, all events appear to accelerate increasingly in spacetime, appearing to evolve forward faster through its apparently faster velocity “through/in” space. The dilation gradient only equalizes in an infinitesimal focal point at the center of the star, impeding the forward evolution of events in all directions, concentrating energy.

In Einstein’s 1915 paper, substituting X, Y, Z, T for his X1, X2, X3, X4, his Fundamental Metric, which can be considered the basis of the tensors describing a null gravitational field, is:

   X   Y   Z   T
X   -1   0   0   0
Y   0   -1   0   0
Z   0   0   -1   0
T   0   0   0   +1

In flattened spiral galaxies, designating the Y axis as being orthogonal to the flat galactic disk, the dilation gradients along the +Y and -Y axes above and below the flat mass of the disk equalize within the disk and as the dRt → 0 along the Y axes, ∆Y→ 0.

As ∆Y = 0 at Y = 0 in the middle of the plane of the galactic disk, the Galactic Fundamental Metric within the disk of a flattened spiral galaxy is:

   X   Y   Z   T
X   -1   0   0   0
Y   0   0   0   0
Z   0   0   -1   0
T   0   0   0   +1

As all the Y elements go to 0, this metric can be reduced to:

   X   Z   T
X   -1   0   0
Z   0   -1   0
T   0   0   +1

As with Einstein’s Fundamental Metric, this Galactic Fundamental Metric cannot be realized in finite space as it also represents a null gravitational field without time dilation. ∆Y also never actually remains at 0 since particles oscillate above and below the plane of the galactic disk.

However, in this fundamental metric without Y elements, forward evolution can only proceed through the X and Z axes, which share a common plane, and we get circular motion around the center of the galactic mass, orthogonal to the dilation gradients. Note that the orbits in a stellar system are also orthogonal to the dilation field. As above, the GDE can only manifest orthogonal to the FDE.

There is a secondary GDE in along the edges where the Y components lose their dominance and we see an evolution inward and the formation of bars.

As the +1 in Einstein’s g44 element of the Fundamental Metric represents an invariable rate of time for all frames along all axes, the +1 of the g33 element in the Galactic Fundamental Metric represents an invariant rate of time along the X and Z axes. Within the galaxy’s dynamic metric, these time elements then change relative to the mass density along the spiral arms, and the apparent velocity relative to adjacent frames is determined by the relative rates of time along the +Y and -Y axes at Y = 0. In the absence of an accelerating gradient as in a stellar system, the relative rate of evolution within the inertial frames of the disk is primarily determined by the rate of time in the inertial frames. Thus, we see an initial rapid increase in velocities near the center of the galaxy, where the rate of time rapidly increases with distance from the massive central MECO (Magnetic Eternally Collapsing Object - the latest in black hole development).

Velocities appear slower between the arms, despite the faster rate of time, due to the shallower gradient. Within the arms, where densities are concentrated, the gradient is deeper along the Y axes and velocities appear
accelerated more as they do in a deeper gradient in a stellar system.  The dilation gradients along the Y axis decrease in slope as mass densities decrease along the arms, time goes faster at Y = 0, and we see a slightly faster evolution (apparent velocity) of the stellar systems within the continuum with distance from the galactic center.
 
The g33 element also varies slightly relative to the slope and depth of the gradients within the individual stellar systems. This slope effect also manifests the same as we see in a stellar system where relative acceleration increases as the gradient of the slope deepens. A test of this would be that larger masses and groups of masses should therefore appear to be evolving forward faster, and appear to have higher velocities, relative to nearby smaller masses due to their deeper, steeper, individual gradients.

Although the primary dilation gradient is along the Y axes, as the disk flattens there is also a secondary gradient looking in from the edges. The evolution in this directions forms the bars of Sb galaxies."

The full 21 page paper, General Relativity: Effects in Time As Causation, is on vixra.org at http://vixra.org/abs/1804.0109#comment-3850079405 and a final version will be published in some form by a cosmological journal in what I hope is the near future, as discussed in the other threads.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/01/2019 04:33:28
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The key idea is that ALL the planets and ALL the moons in the solar system are drifting outwards over time.
Yes, but they're not.  All of Mars' moons and over half of Jupiter's moons are losing orbit.
If you model depends on this idea, it is trivially falsified.

Can you please specify one moon (around Mars or Jupiter) that we have measured its orbital radius and verify that it is really losing orbit?
You know that we just assume that those moons are drifting outwards, but in reality we didn't confirm our hypothetical idea.
So, please don't say something which is incorrect.
You can't say for sure that those moons drift outwards without real measurements.
It was a big surprise for our scientists when they have discovered that the moon is drifting outwards from Earth. No one at that time had expected to see that increase in the orbital radius.
Therefore, it will also be a big surprise for our scientists when they will discover that ALL moons in the solar system are drifting outwards (Including All of Mars' moons and ALL Jupiter's moons).!!!

The friction is mostly between water and the sea floor, which is direct physical contact, and is where all the heat is generated.
The friction in the Earth is none relevant to the friction between the Earth to the Moon.
Just a simple example -
Let's assume that you sit at a bus while it drive at 120 Km/s
If you set a friction between your hand to the bus body, does it mean that you set a friction between the bus and the road?
In the same token, a friction between water and the sea floor on the Earth can't set a friction between the Earth and the Moon.
Orbital objects as Moon and planets are losing gravity force over time. This is a normal activity in gravity. Therefore, the Moon increases its orbital radius (around the earth) over time.

 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/01/2019 06:16:04
Can you please specify one moon (around Mars or Jupiter) that we have measured its orbital radius and verify that it is really losing orbit?
You know that we just assume that those moons are drifting outwards, but in reality we didn't confirm our hypothetical idea.
See my prior post, giving measured numbers for Phobos, the moon with the largest orbital change per year of anything in the solar system.  It has been repeatedly measured, not just assumed.

Quote from: Halc link=topic=75495.msg565716#msg565716 date=1547749994
The friction is mostly between water and the sea floor, which is direct physical contact, and is where all the heat is generated.
The friction in the Earth is none relevant to the friction between the Earth to the Moon.[/quote]
There is no direct friction between those two.  They don't touch.  The energy added to the moon is pure gravitational.  If there was friction, it would lose kinetic energy and drop down.  Friction reduces total mechanical energy, not increases it.  I can think of no exception to this.

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Just a simple example -
Let's assume that you sit at a bus while it drive at 120 Km/s
If you set a friction between your hand to the bus body, does it mean that you set a friction between the bus and the road?

The friction between the bus and the road was already there.  My hand on the interior doesn't change that.  There is not much friction with the road since little is sliding across it.  Most of the friction would be with the air and the internal bearings of the bus and such.  These are parts that move across each other by physical contact.  Not so much the road since the tires move at the same velocity as the road where they touch, even when accelerating or braking.
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In the same token, a friction between water and the sea floor on the Earth can't set a friction between the Earth and the Moon.
I didn't claim there was friction between Earth and the moon.  They don't touch.  The dynamics between the two is simple gravitational thrust equivalent to a rock accelerating when spun around in a sling shot.

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Orbital objects as Moon and planets are losing gravity force over time. This is a normal activity in gravity. Therefore, the Moon increases its orbital radius (around the earth) over time.
I pointed out several falsification tests for this.  You ignored them, exactly as I assumed you would.  You don't do science.  You ignore any evidence against your view, and cherry pick facts out of context to support a theory that has never been described.  I see no formula for gravitational weakening over time.  You make no explanation as to why galaxies in the past seem to have the same gravity as modern ones.  This is exactly how the scientific method doesn't work.

In addition, your idea (I haven't seen a theory based on the idea) doesn't explain the sun's acceleration about the galaxy.  If gravity is weakening, all objects should be accelerating less than the scientists predict, and even they predicted an acceleration less than what is observed.  Your idea makes that problem even worse, not better.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/01/2019 10:18:42
See my prior post, giving measured numbers for Phobos, the moon with the largest orbital change per year of anything in the solar system.  It has been repeatedly measured, not just assumed.
Please see the following:
https://en.wikipedia.org/wiki/Phobos_(moon)
Phobos has dimensions of 27 km × 22 km × 18 km,[1] and retains too little mass to be rounded under its own gravity.
So, Phobos is very small object (comparing to real moons) and it doesn't have a clear ball shape structure. It has no symmetrical shape.  It is clearly a broken object.
Therefore, why don't we call it asteroid? No more than that.
For example - Vesta — the second-largest asteroid in the solar system at 525 km (326 miles) in size and is located in the asteroid belt between Mars and Jupiter.
So, how could it be that an object at about 500 Km we call Asteroid, while a broken object about 25 Km we call moon...?

"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, closer to its primary body than any other known planetary moon. It is so close that it orbits Mars much faster than Mars rotates, and completes an orbit in just 7 hours and 39 minutes."
So, this asteroid is very close to Mars.
Mars radius about 3400 Km.
So the ratio between mars radius to Phobos orbits from the Martian surface is 1:2.
At this ration it is clear that it is drifting inwards.
Therefore, Phobos is a wrong example.
Please try to offer a real moon (not a small broken object) with a relative longer orbital radius (similar to the Earth moon ratio).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/01/2019 15:34:19
So, how could it be that an object at about 500 Km we call Asteroid, while a broken object about 25 Km we call moon...?
Phobos orbits a planet.  Vesta does not.  That's the difference between moon and asteroid.  Vesta isn't large enough to clean up its orbit, which is why it isn't considered a planet in itself.
The names we give to these types of objects does not have any effect on how gravity acts on them.

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"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, ... "
So, this asteroid is very close to Mars.
Mars radius about 3400 Km.
So the ratio between mars radius to Phobos orbits from the Martian surface is 1:2.
More like 1:2.75.  Orbital radius is measured from the center, not the surface, so it is about 9400 km.  6000 km is its altitude, not its orbital radius.

There are several moons of other planets with tighter ratios than that.  Mimas (around Saturn) has a slightly lower ratio, and is quite a nice large round moon, since you seem to find that important.
Thebe (around Jupiter) is a misshapen lump about 4 times the size of Phobos and orbits at a ratio a little larger (3x) the radius.  Both these moons are increasing their orbits.

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At this ration it is clear that it is drifting inwards.
You spend pages of posts denying inward drift, and suddenly it is clear that it must, totally without explanation.  Does gravity only grow less over time if it is working on large things?  Is there a separate increasing-over-time gravity for the little stuff?
All I see is that you find you cannot continue to deny that Phobos is losing altitude.  You've denied this in several of the recent posts, but all of a sudden "it is clear" that it must.  What is happening is that your idea was falsified.  Weakening gravity would send Phobos to higher orbits if only strength of gravity determined such things.
Earth should be moving away from the sun at 1,500,000 times the value seen to keep up with the decreasing gravity that accounts for the moon movement.  Earth is a bigger round thing, so not under the just-made-up law of non-weakening of gravity for little things.

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Please try to offer a real moon (not a small broken object) with a relative longer orbital radius (similar to the Earth moon ratio).
All the big ones have positive orbital periods and lie outside geosync radius, so they all move outward per tidal forces.    Lack of a large counterexample does not explain why suddenly you find it clear that Phobos orbit must be degrading when all the prior posts asserted otherwise.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/01/2019 16:32:16
Lack of a large counterexample does not explain why suddenly you find it clear that Phobos orbit must be degrading when all the prior posts asserted otherwise.
Phobos is none relevant due to its size, shape and its orbital radius ratio.
However, you have stated that:
All of Mars' moons and over half of Jupiter's moons are losing orbit.
So, would you kindly offer another moon (From Mars or Jupiter) that meets the criteria and is losing orbit?
All the big ones have positive orbital periods and lie outside geosync radius, so they all move outward per tidal forces.
If I understand you correctly, there is no real evidence which can contradicts my statement that all real moons and planets are drifting outwards.
However, you claim that this process is due to "tidal forces" and I claim that it is due to normal gravity force reduction over time!!!
Please look again at the following:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
If we could monitor the orbital motion (gray line) of the Sun, we would clearly find that it drifts outwards from the center.
The assumption that the sun can orbit at the same radius for several billion years is absolutely incorrect.
The Sun' orbital motion must drifts outwards over time.
In the same token, any other star' orbital motion must drifts outwards!!!





Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 19/01/2019 05:09:44
strength of gravity
The trouble is that neither of you has defined gravity, as I do in the paper noted in my previous post. GR describes gravity in terms of the curvature of spacetime. I take it an additional step by pointing out that the curvature also represents the forward evolution of space, and the densities there-in, due to the force of the passage of time.

The passage of time is the fundamental force as it evolves space, and the events (densities) there-in, forward. When a dilation gradient is introduced, we also see an evolution down the gradient, This is the Gravitational Direction of Evolution, or GDE. The curvature of evolution we see in GR is the resultant of the GDE and Fundamental Direction of Evolutuion, or FDE, due to the simple passage of time.

The "strength of gravity" is determined by differences in the rates of time between frames, as per GR and my additional perspective.

I can't even try to read what you guys are writing as it all makes no sense as gravity, as per GR, is not defined as a force other than that due to the differences in the rates of time, Einstein's "energy components", and resultant relative lengths in space, between frames.

In spiral galaxies, this manifests as in my previous post.

Where you appear to be speaking about "friction", you should probably be looking at "frame dragging" which, again, is dependent on relative rates of time, which are also dependent on relative perspectives.....  8)

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/01/2019 05:43:29
How much does gravity degrade over time?
Good question.
There is no current formula for that. I assume that we have to verify the gravity degrade per many orbital objects and try to develop the requested formula for that.
Why does it only affect the moon and not Earth? How does gravity know to degrade only for masses that Dave Lev designates as 'real moons' and not for other masses made of the exact same component materials?
It effects any orbital System.(Asteroid, Moon, Planet, Star, BH, Dust, Gas cloud...)
However, the formula itself should include the orbital drifting direction and amplitude.
It should be based on the current "orbital radius ratio" (Orbital radius to Host radius ratio), Mass ratio (orbital object to host mass), "Friction" (If there is a friction. For example due to atmosphere) and some other factors (for example - Orbital structure or shape and so on).
I'm quite sure that If Newton or Einstein had the current measurements that we have today, they could find the correct formula for orbital drifting amplitude per cycle and its direction.
In any case, all orbital objects must drift in one or the other direction. Nothing can stay forever at the same radius!
Therefore, the assumption that the Sun has to stay at the same radius from day one is absolutely incorrect.
It is quite clear that all the real moons, Planets and stars must drift outwards. Even most of Asteroids drift outwards. The Oort cloud is a perfect example for billions of asteroids that had been drifted outwards from the core of the Solar system while they all orbit around the solar system. Today - all of them are still drifting outwards (even if we see that some of them are moving inwards - There is an explanation for that).
However, few of the Asteroids might drift inwards and eventually collide with their host. It happened on Earth with the big asteroid collision that had killed all the dinosaurs that have lived here (and gave us the opportunity to come). It also should be the final outcome of Phobos with Mars.
On the contrary.  Your theory says it should move immediately quickly from the center since it cannot hold this high acceleration needed to maintain that 8 kpc radius at a whopping 217 km/sec.
It shouldn't drift outward, but rather should never have been this far inward in the first place.  Your idea of degraded gravity makes that problem even worse.
Sorry. This is incorrect.
My theory doesn't say that none realistic idea.

Please, give me the chance to explain the theory.
Please don't continue to argue about those two key elements.
With your permission, please let's assume that those two elements are correct and let's see how they give us the breakthrough knowledge about our Universe.
They should lead us to have excellent understanding of:
- How spiral galaxy had been created?
- Why we see the Bulge, Bar, Ring, spiral arms, and all the other mass around the galaxy.
- What is the answer for the Galaxy rotation curve problem.
- What is the real mass of the SMBH in the Milky way
- Why there is no need for dark matter to set the spiral galaxy
- Why for any star in the galaxy there is at least one outside
- Why there are hydrogen clouds strewn/bridge between Andromeda (M31) and Triangulum (M33).
- Why the far end galaxies are moving away at ultra velocity, while the density of the Universe stay at the same ratio.
- Why the universe has a blackbody radiation
- What is the age of the Universe
- What is the size of the universe
And many more questions...
All of that without any help from dark matter or dark energy.
Just pure gravity force and the two following elements:
1. All stars in the galaxy are increasing their orbital motion radius over time.
2. All stars in the galaxy must orbit around some virtual host..

Please give me the chance to explain the theory.
Let's stop the endless discussion about those two elements.
So, please from now on we will assume that those two elements are correct and try to understand what could be the outcome.
Would you kindly agree for that?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/01/2019 06:14:39
I've agreed to this several times before.  It's not like you need permission from me.
Thanks
So, from now on we will assume that the following two elements are fully correct:
1. All stars in the galaxy must orbit around some virtual host point.
This virtual host point represents the real orbital motion point of a star around the center of the galaxy.
2. The orbital motion radius is increasing over time.
In the same token, Based on the formula (which we still have to find) any orbital object which has a high enough "orbital radius ratio" (and some other factors) is increasing its orbital radius over time. Therefore, all real Moons and planets are increasing their orbital radius over time.
   
I will introduce the theory step by step, based on those two elements and pure gravity force.
In each step I would mostly appreciate to get your feedback (again - based on the idea that those two elements are fully correct.)
In this theory there is no need for dark matter, dark energy, space expansion, density wave and any other none realistic hypothetical idea.
Just after the final step in the theory introduction, we can go back and verify if those two elements are correct or not.
Thanks for the cooperation in advance.


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/01/2019 07:33:39
Let's start the theory by the focusing on the Sun' orbital motion.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see the apparent motion of the solar system (blue dot points) the virtual host point (orange ball) and the real orbital motion of the sun around the galaxy (gray line).
Let's start with the apparent motion of the solar system. (Blue dot points).
We see that it has an elliptical shape. If I remember correctly, the ratio between the maximal length to the minimal lengths is 20 to 7.
This elliptical orbit represents the Sun orbit around its host virtual point.
Based on our understanding, it is claer that the sun must increase its orbital radius over time. Therefore, the length of this elliptical orbital cycle must increase over time.
The sun will never ever come back again to the same point in this elliptical cycle.
It will form some sort of spiral shape around this elliptical cycle.
The virtual host point (orange ball) orbits around the galaxy. This orbit (gray line) represents the real motion of the sun around the galaxy.
Please be aware that also the radius of this orbit must increase over time. Therefore we should see a spiral shape as every movement the sun is increasing its orbital motion radius.
The outcome is quite simple -
The measured sun' orbital speed represents the combination of the Sun orbit around the virtual host + the real Sun orbital motion around the galaxy.
Therefore, if we need to extract the real orbital motion of the sun around the galaxy, we must deduct from the measured sun' orbital speed the impact of the sun orbital speed around its virtual host.
Hence, I can only assume that if the 220 Km/s represents the measured orbital speed of the sun, the real orbital motion of the sun around the galaxy should be in the range of 210km/sec or less.
Please also be aware that the real orbital motion is located at the same distance from the galactic disc.
This is very important outcome from this image. We will use it later on.
Please let me know if you agree with all of that (again - based on the assumptions that the two key elements are fully correct)

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/01/2019 15:21:07
Let's start the theory by the focusing on the Sun' orbital motion.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
You do realize that this diagram you keep referencing is a simplified model based on a series of facts, many of which are incorrect.  It comes from http://www.biocab.org/coplanarity_solar_system_and_galaxy.html
which states that the plane of the solar system is tilted at 90° WRT the galactic plane.  The actual figure is about 63°.
Here is another gem:
Quote
The second movement, which is described in most of astronomy books, is an oscillation of the Solar System from north to south and from south to north with respect to the galactic plane. The oscillation “upwards” and “downwards” is mainly established by the gravitational pull exerted by other bodies of the Solar System on the Sun, i.e. planets, asteroids, etc. The speed of this movement is 7 km/s.[\i]
They apparently seem to claim that our planets and asteroids are capable of dragging the sun back and forth at a pace of 7 km/s, and that "most of anstronomy textbooks" describe this.  Those 'textbooks' are seemingly where they are getting their facts, not from data directly from actual astronomical surveys.

Their model has the axis of rotation for this helical motion (they never call it an orbit) rotating with the position in the galaxy, always staying tangential to the curving dashed grey orbit line (that line they do call an orbit).  Orbits don't change their axis like that, so they rightly don't describe the motion as 'orbital'.  The axis of an orbit tends to be fixed over time, so our solar system has always had this 63° tilt and does not process to a different orientation when we've moved a quarter of the distance around the galaxy.

I say all this because you're seemingly using this one diagram of a poor model as some sort of measured scientific fact instead of just just a diagram of a model that it claims to be.  The site is a biology site, hardly the first place I'd look for accurate information on this subject.

The site also does not explain how the N/S motion (7 km/s) and the in/out (20 km/s) motion just happen to have the same periods.  The model is not described in detail.

Quote
Please let me know if you agree with all of that (again - based on the assumptions that the two key elements are fully correct)
You asked my in my prior post to stop with me pointing out the points with which I disagree.  I'm still doing that, but it doesn't imply that I agree with you.  If you don't want critique, don't ask me if I agree with your posts.  I'm still waiting to see where this goes.

My post here is not so much a critique of your view, but rather a critique of this diagram from which you seem to be drawing your facts.

Quote
I will introduce the theory step by step, based on those two elements and pure gravity force.
In each step I would mostly appreciate to get your feedback (again - based on the idea that those two elements are fully correct.)
I am assuming your two elements are correct.  I am staying quiet about the implications of those two elements since they've not really been spelled out.  I cannot make predictions from them yet, so I cannot falsify them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/01/2019 13:27:47
I am assuming your two elements are correct.  I am staying quiet about the implications of those two elements since they've not really been spelled out.  I cannot make predictions from them yet, so I cannot falsify them.
I do appreciate your good willing to accept the two elements for this discussion.
The idea if the plane of the solar system is tilted at 90° WRT the galactic plane or 63° Isn't relevant to our discussion.
In the same token, it is also not relevant if they call the Sun orbital movement around the virtual host point (orange point) as a swinging motion.
However, the image itself shows clearly that the sun orbits around the virtual host point (orange ball) while this host orbits around the galaxy at the same amplitude (or distance) from the galactic plane.
Please look at the following image (from different article):
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
It is stated that "the solar system moves around the center of the galaxy like planets around the Sun"
"The circular motion around the center is shown by the white dashed line."
This dashed line looks the same as the orbital motion of the virtual sun host point in the other article (which was represented by the gray line).
Hence, in both articles we see a similar view/outcome.

In any case, you didn't reply my questions:
Do you agree that based on the two key elements?
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 03:46:02
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Sorry, neither of these is correct so the resultant conclusion is also wrong.

Motion and velocities are only relative based upon perspective. You are ignoring that. (I would also point out that the disk is not uniform, but consists of conical spiral arms of varying densities. That complication does not need to be considered here.)

For instance, considering the perspective of the orbits of Mercury and Venus relative to the plane of the ecliptic, we assign Mercury a velocity of 47.89 km/s and Venus one of 35.03 km/s, a large difference.

But if we consider the velocity of the Sun and its forward evolution in time relative to the CMB, and the helical distances travelled by the planets we get a much different perspective:

(in the following computations the inclination of the plane of the ecliptic is ignored and:
Planetary orbital lengths and periods are as per NASA.
Orbital periods are related to 1 Earth year.
Orbital lengths are as perceived “around the Sun”.
Helical orbital lengths are computed using the following formula:
(Distance travelled by the Sun)2 + (Orbital length)2 = (Helical length)2
The distance travelled by the Sun is relative to the CMB.
Sun velocity = 368 km/s = 11.60672*109 km/yr.)

Mercury:
Orbital length: 57.909227*106 km
Orbital period = .24 yr
Orbits/yr = 4.1666
Total orbital length = 241.249839*106 km
Helical length = 11.609226961*109 km
Velocity = 368.07948 km/s vs 47.89 km/s

Venus:
Orbital length: 10.8209475*107 km
Orbital period = .62 yr
Orbits/yr = 1.6129
Total orbital length = 17.4531062*107 km
Helical length = 11.608032143*109 km
Velocity = 368.04160 km/s vs 35.03 km/s

Earth:
Orbital length: 14.9598262*107 km
Orbital period = 1 yr
Orbits/yr = 1
Total orbital length = 14.9598262*107  km
Helical length = 11.607684041*109 km
Velocity = 368.03056 km/s vs 29.79

Mars:
Orbital length: 22.7943824*107  km
Orbital period = 1.88 yr
Orbits/yr = .5319
Total orbital length = 121.2467148*106 km
Helical length = 11.607353269*109 km
Velocity = 368.02007 km/s vs 24.13

Jupiter:
Orbital length: 778.340821*106 km
Orbital period = 11.86 yr
Orbits/yr = 0.0843
Total orbital length = 65.6273879*106 km
Helical length = 11.606905535*109 km
Velocity = 368.00588 km/s vs 13.06

Saturn:
Orbital length: 142.6666422*107 km
Orbital period = 29.46 yr
Orbits/yr = 0.0339
Total orbital length = 484.27237*105 km
Helical length = 11.606821027*109 km 12576482920
Velocity = 368.00320 km/s vs 9.64

Uranus:
Orbital length: 287.0658186*107 km
Orbital period = 84.01 yr
Orbits/yr = .0199
Total orbital length = 341.70434*105 km
Helical length = 11.606770299*109 km
Velocity = 368.00159 km/s vs 6.81

Neptune
Orbital length: 449.8396441*107 km
Orbital period = 164.8 yr
Orbits/yr = 0.0060
Total orbital length = 272.96094*105 km
Helical length = 11.606752096*109 km
Velocity = 368.00101 km/s vs 5.43

From this perspective, the velocities, or rate of evolution, of Mercury and Venus are only .038 km/s different. Note also that as we increase distance from the Sun, the velocities decrease until Neptune has a velocity only .001 km/s different from the base velocity of the Sun. Relative velocities equalize with a larger perspective. If we shift out to the local group and its apparent motion relative to the CMB of 627 km/s, the difference between the Sun and Neptune’s velocity is only .00059 km/s. Viewed as a whole, the universal evolutionary rate of evolution, and apparent resultant relative velocity, is 1: all events in space evolving forward at the same rate over a steady rate of time.

This is the evolution of the quantum continuum. This is how I tie quantum and astro- physics together.

In both perspectives, the velocity and acceleration are directly related to the difference in the rate of time (dRt)/distance so are higher in steeper gradients, and this higher apparent acceleration of events in slower time frames maintains their relative positions within the overall continuum as it evolves forward as viewed from both perspectives. 

This means GR is describing the forward evolution of the continuum and the events occurring within it, rather than the evolution of events through pre-existing “curved spacetime”. It is not the masses that determine relative velocities and trajectories, but the dynamics and perspectives in time.

Again, this is extracted from my paper on Relativity found here: http://vixra.org/abs/1804.0109#comment-3850079405.



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/01/2019 13:40:30
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Sorry, neither of these is correct so the resultant conclusion is also wrong.

Motion and velocities are only relative based upon perspective. You are ignoring that.
He is not ignoring it.  There is a 'perspective' (as you put it) for each of the values specified.
Point 2 might be wrong, but not for the reasons you seem to point out.
Quote
For instance, considering the perspective of the orbits of Mercury and Venus relative to the plane of the ecliptic, we assign Mercury a velocity of 47.89 km/s and Venus one of 35.03 km/s, a large difference.
That would be relative to the frame of the solar system, not 'relative to the plane of the ecliptic'.  A plane is not a valid reference frame.
Quote
But if we consider the velocity of the Sun and its forward evolution in time relative to the CMB
The CMB is light, not an object relative to which one might have a velocity.  I presume you mean relative to the frame in which the CMB appears to be isotropic, but you don't say that.
Quote
in the following computations the inclination of the plane of the ecliptic is ignored
But the inclination is not zero, so all of the number you quote are wrong, especially comments like this one:
Quote
Note also that as we increase distance from the Sun, the velocities decrease until Neptune has a velocity only .001 km/s different from the base velocity of the Sun.
I suppose this would be the speed difference in a frame where the inclination put the axis of the solar system parallel with the direction of travel, but this is not the case in the frame you specify, so this comment is totally wrong.  You seem to base some conclusion on this fact, so the conclusion is wrong.  I could not figure out the whole point of your post.  I think it might just have been an excuse to plug the link to your paper, which I have no plans to read.
How is any of this post relevant to the discussion going on in this topic?
Quote
Relative velocities equalize with a larger perspective.
No they don't, at least not until you get to relativistic speeds.  Even if the inclination was flat on like you suggest, the relative velocity between the sun and Neptune would be about 5.4 km/sec, not the .001 km/sec figure you suggest.
The value you quote is a difference in speed, not a difference in velocity.  I can take any two objects with arbitrarily high and random velocity differences and find frames in which they have the same speed.  So what?  The frame you selected is not one of those frames.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/01/2019 15:42:54
Thanks Halc
Now the question is - What is the estimated mass of the Sun' virtual host point?
It shows the vertical motion, same as the other diagram, corresponding to the 7 km/sec component of the helix (the one the bioCab attributes to gravity from our own planets).  The other article didn't give a period to the wiggle, but this one says every 62 million years.  That's about 4 waves per trip around the galaxy, less than the tighter frequency depicted in their image, and far less than the dense helix depicted in the bioCab simulation.
So, we know that the Sun sets one full orbit around this virtual host point in 62 Million years.
However, I have no data about the orbital radius.
I assume that based on the radius and the orbital time we could calculate the requested mass.
Any idea?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/01/2019 15:57:50
Sorry, neither of these is correct so the resultant conclusion is also wrong.
Motion and velocities are only relative based upon perspective. You are ignoring that. (I would also point out that the disk is not uniform, but consists of conical spiral arms of varying densities. That complication does not need to be considered here.)
It seems to me that you have missed the key point of this idea.
Please look again at the following image:
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
Please try to focus on the dashed white line Vs the green line.
Do you agree that the green line is longer in its distance than the dashed White line?
It is clear that in order to set one full orbital cycle around the galaxy, an object at the green line must move faster than an object on the dashed white line.
So, do you understand by now why the virtual host velocity (dashed line) is lower than the sun actual orbital velocity (green line)?
Therefore, if the green line represents the actual orbital velocity of the sun - Let's assume that it is 220 Km/s, than the dashed white line, which represents the orbital motion of the Sun virtual host around the galaxy, must be lower than 220 Km/s.

Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 16:28:19
So, do you understand by now why the virtual host velocity is lower than the sun actual orbital velocity?
Sorry. As I pointed out above, this is non-sensical. You are trying to prove the impossible, i.e., objects evolving against the direction of the time dilation gradient, or backwards gravity. Gravity only has one direction which is down the dilation gradient.

You are trying to imagine a virtual host that does not exist. The white line is meaningless. It only marks the plane of the ecliptic in the pictured galaxy. As Dave pointed out previously the scale is way off in the picture, as well. Even if your virtual host existed the difference in velocity would be meaningless.

You are also not visualizing the evolution of the continuum properly. All events are acted on by two directions of evolution, the Fundamental, in situ, evolution and the Gravitational, down gradient, evolution.   
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/01/2019 16:38:02
Sorry. As I pointed out above, this is non-sensical. You are trying to prove the impossible, i.e., objects evolving against the direction of the time dilation gradient, or backwards gravity. Gravity only has one direction which is down the dilation gradient.
You are trying to imagine a virtual host that does not exist. The white line is meaningless. It only marks the plane of the ecliptic in the pictured galaxy. As Dave pointed out previously the scale is way off in the picture, as well. Even if your virtual host existed the difference in velocity would be meaningless.
Sorry
I'm not going to start the whole discussion about this key idea with you.
In this discussion we assume that:
1. All stars in the galaxy must orbit around some virtual host point.
This virtual host point represents the real orbital motion point of a star around the center of the galaxy (Dashed white line)
2. The orbital motion radius is increasing over time.
Please see reply no. 172
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 19:25:09
Dave, I suggest you just ignore these posts by captcass.
Sorry, I don't think that is wise. From the outset of this thread you all have been using, and trying to modify, Newtonian formulas to come up with your breakthrough. The trouble is, Newtonian formulas only approximate what General Relativity describes, which is the apparent evolution of events in the continuum due to Lorentz contractions in both time and space. People still largely use the Newtonian formulas because the Relativity formulas quickly become very complex when more than 1 body is considered. If you say you can modify the Newtonian formulas, then you are saying you can modify relativity, too. Can you demonstrate how that might work?
(I actually do this by deriving the Hubble Shift from a 2.2686"10-18 s/s acceleration in the rate of proper time that is then added to Einstein's Tensor to eliminate singularities, Big Bangs, and infinitely accelerating expansions of the universe in the "Is the Hubble Shift Due to Time Dilation" thread). .
Any discussion of gravity that does not consider the Lorentz contractions, especially the time dilation, which Einstein calls his "energy components", is off the mark. You are talking about Newtonian "forces" that do not exist. It is all about the evolution of events within the continuum, not the motion of particles "through" a pre-existing space.
That being said, I will leave you all to your debate unless someone wants to ask me a question about anything I've said here. I don't want to hijack a thread, but as I explain galactic rotation velocities in my paper, as copied above, I felt it was on topic. "On topic" doesn't always mean "on the same track" and I can see that.  :) 
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 20:59:11
I see you invoking references to the CMB which is pretty irrelevant to a discussion of the dynamics of a local system.
I did that to illustrate the relative nature of apparent velocity and motion. But what I am trying to contribute does not really fit the thread so I will bow out.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/01/2019 20:12:19
It seems to have maximum acceleration at the peaks of the waves, when it is furthest from this host point, which makes the motion not really 'orbital', where max acceleration is reached when the object is closest to the point around which it orbits.
I wonder why our scientists assume that it has maximum acceleration at the peaks. Why not a sin wave?
Any idea?.
Somehow, it also seems to me that the 62 Million years per cycle it too long.
Just a brief calculation.

The nearest star is about 4 LY away from us.
Its quite clear to me that the maximal orbital radius around the virtual host must be significantly lower than that.
However, just for the calculation, let's assume that the radius is 1LY (9.461e+12 Km)
P = 2 * 3.14 * r = 6.28 * 9.461e+12 Km = 59.425 10^12 Km
In one year there is = 31536000 sec.
In the article it is stated that the orbital speed around the orbital motion (which is the virtual host) is 7 to 20 Km/s.
Just to make it easy, let's use an average speed of 10 Km/s (in a pure cycle orbit).
At a speed of 10 Km/s we can cross in one year:
L = 10 * 31536000 = 315.36 10 ^ 6 Km/Year
T = P/L = 59.425 10^12 Km / 315.36 10 ^ 6 Km/Year = 1.88 10^5 = 188,000 year
So, in order to set a full 1 LY cycle at 10Km/s, 188,000 years are needed.
Hence, how do they have got the idea for 62 Million years per cycle?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/01/2019 11:57:17
There just isn't any significant mass a light year away to give your 'host point' the sort of pull it would need to haul the sun around at 10 km/sec at that tight radius, which would be incredible acceleration.
It seems that you insist to criticize the key elements of my theory.
I didn't call it Host point. The correct name is "virtual host point".
So, it is a virtual point. Therefore, we don't expect to see any real mass at that Sun' "host point".
Please read again the following key elements:
1. All stars in the galaxy must orbit around some virtual host point.
This virtual host point represents the real orbital motion point of a star around the center of the galaxy (Dashed white line)
2. The orbital motion radius is increasing over time.

Why do you keep criticize those two key elements in my theory?
Please - let me introduce the whole theory based on the two key elements and then take a decision if you accept it or reject it. Please, don't criticize it before getting the whole information about the new theory.
Let me give you an example:
Let's assume that you want to teach you kid mathematics.
For one year you tell him that the first two key elements in mathematics are:
1 + 1 = 2
1 - 1 = 0
Unfortunately, your kid doesn't agree to accept those two elements.
So, at some moment you ask him to take it as is.
However, at any step forward he tries to criticize those two basic elements.
So, how can you give him any basic understanding about mathematics, if he doesn't agree with those two key elements?
In the same token.
You have agreed to accept those two key elements (just during the introduction process of the theory)
So, please would you kindly not criticize those elements any more.
Please - try to help me to find what is the outcome due to those two key elements.

If so, there is no real mass at the virtual host point.
It just represents the equivalent mass which is needed to keep the sun in that orbital cycle around that virtual point.
It also doesn't important if the orbital cycle is elliptic or pure cycle.
Actually, in the nature most of the orbital cycles are elliptical.
There is no big difference if we need to use Newton formula or Kepler.
However, during this discussion, if possible - let's consider the orbital cycles as pure cycle ( just to make it more easy to set the calculation).
Do you agree with all of that?


 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/01/2019 13:24:36
Why do you keep criticize those two key elements in my theory?
I didn't criticize the two points.  I commented on the unlikely radius of your orbit about this VHP since you for the first time gave a number for it.  I didn't comment on the second of the two points at all.  I was hoping that by shutting up, you'd actually get to your theory.

Quote
Let me give you an example:
Let's assume that you want to teach you kid mathematics.
For one year you tell him that the first two key elements in mathematics are:
1 + 1 = 2
1 - 1 = 0
Unfortunately, your kid doesn't agree to accept those two elements.
So, at some moment you ask him to take it as is.
However, at any step forward he tries to criticize those two basic elements.
So, how can you give him any basic understanding about mathematics, if he doesn't agree with those two key elements?
And suppose the child agrees to just take those two points as is, and yet the parent keeps harping on those two points instead of moving on.  Eventually the child has no option but to comment more on the two points since that is all that has been presented.
Quote
If so, there is no real mass at the virtual host point.
I never claimed there has to be, but there has to be mass somewhere.  Pluto and Charon orbit a common virtual host point, and there is no mass at that point.  But each constitutes the mass that is responsible for the orbit of the other.  Without either one of those two bodies, there would be no virtual point about which the other could orbit.

I think your problem is that your two points are not premises.  They are proposed effects (not even observations), and you need to come up with viable physics that explain those effects, not come up with implications of those effects.  I'm certain that your ideas might have the implication of a galaxy behaving this or that way, but without the physics to explain those effects, the effects are not grounded in reality.  Hence my questioning them so hard.  You've presented no new physics* to explain these two proposals, and the proposals seem to completely violate existing physics.

You're telling me that 1+1=7 and asking me to take that on faith.  Fine, but eventually you need to justify how 1+1 happens to work out to 7, otherwise the implications of that assumption are grounded in nonsense.

* OK, you've presented one new piece of physics: Selective gravity that decreases over time for certain objects, but remaining constant or increasing for others.  I could build an infinite energy generator with that sort of physics.
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It just represents the equivalent mass which is needed to keep the sun in that orbital cycle around that virtual point.
It also doesn't important if the orbital cycle is elliptic or pure cycle.
Yes and yes.   But it seems there is no mass out that way, so what exactly pulls the sun into that orbital cycle if not mass?
Quote
There is no big difference if we need to use Newton formula or Kepler.
However, during this discussion, if possible - let's consider the orbital cycles as pure cycle ( just to make it more easy to set the calculation).
Do you agree with all of that?
Actually, I do.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/01/2019 15:54:29
Actually, I do.
Many Thanks

Quote
If so, there is no real mass at the virtual host point.
I never claimed there has to be, but there has to be mass somewhere.  Pluto and Charon orbit a common virtual host point, and there is no mass at that point.  But each constitutes the mass that is responsible for the orbit of the other.  Without either one of those two bodies, there would be no virtual point about which the other could orbit.
Yes. You are absolutely correct!
Sorry that I didn't understand correctly your reply.
This is a very important issue. However, in my theory we actually discuss on many objects.
I will explain it.
And suppose the child agrees to just take those two points as is, and yet the parent keeps harping on those two points instead of moving on.
Yes. That is correct.
I will shift gear.
So, as our Sun orbits around a virtual host point, every star must do so.
In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
On the other hand - please see the second element:
2. The orbital motion radius is increasing over time.
Based on that idea, all the stars, gas clouds, dust... are drifting outwards from the center. Nothing can move inwards.
So, how could it be that the center is full with mass, while all the mass is drifting in one way out?
The only logical outcome is that the galaxy must create new mass in order to compensate on all the mass that had been drifted out.
Hence, the accretion disc is actually an excretion disc.
I have already discussed deeply on the activity of new mass creation in the galaxy.
I have poved that an Atom is a cell of energy.
So, the SMBH converts its ultra high gravity force into energy. this energy is locked in new Atoms in the plasma.
It creates new Hydrogen atoms at this axcretion disc. Then due to the collisions between those new born atoms it creats all the variety of atoms and molecular.
The ultra high temp of the plasma in the accretion disc (10 ^9 c) is a clear evidence for the new born atom activity at that disc.
As all the new born molecular drift outwards from the disc - remember - all objects must drift outwards over time including those Atoms), they get to the magnetic shield around the accretion disc.
This Ultra power of magnetic disc, boost the atoms upwards at a speed of almost 0.8 c.
As they move up, they lose the boost of the magnetic force and eventually they fall back at the galactic disc plane.
They set the gas could that we see around the SMBH.
However, In the gas clouds, the molecular orbits around some virtual center of mass (as explained in element 1).
I assume that the orbital velocity in the gas cloud is quite high.
This internal orbit in the gas cloud, pulse the great impact of the nearby SMBH set the Huge New form star activity that we see.
Any new born star must drift outwards while it orbits around some virtual center of mass.
That is all for that step in my theory.
Please let me know if based on the two elements - you agree with the outcome.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/01/2019 00:32:37
So, as our Sun orbits around a virtual host point, every star must do so.
Does  everything orbit a virtual host point, or only stars?  If the latter, what delimits objects with a VHP vs ones that don't?  What about a small black hole, or a cold black dwarf that has less mass than some planets?
For instance, Venus doesn't seem to orbit a VHP.  It goes pretty much in a normal elliptical orbit straight around the sun with no extra wobble of its own.  All the planets actually, Earth and Pluto excepted.  Earth's host point (called its barycenter in normal terminology) has mass present at it, while the barycenter of Pluto is in empty space.  The other planets don't really have a barycenter/virtual-host-point about which they have any sort of motion that could be described as orbital.

Quote
In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
What do you mean by this?  They have an orbit, a wobble (sort of like the green sin wave)?  What do you mean by 'fixed'?  Do all VHP's have this same amplitude, or do you mean each is different but regular, not varying from one peak to the next?

What has this VHP thing got to do with the stuff below where the center of the galaxy creates new matter that is expelled away from there?

Quote
On the other hand - please see the second element:
2. The orbital motion radius is increasing over time.
For some things, but not others you've said.  Do unqualified objects not increase their orbits, such as small rocks or gas clouds and such?

Quote
Based on that idea, all the stars, gas clouds, dust... are drifting outwards from the center. Nothing can move inwards.
Not even irregular 'not real' moons?  You seem to be inconsistent with prior posts, so I'm trying to get it straight.

Quote
So, how could it be that the center is full with mass, while all the mass is drifting in one way out?
The only logical outcome is that the galaxy must create new mass in order to compensate on all the mass that had been drifted out.
That actually sort of follow from your second point, at least when coupled with data with the mass density of the center of the galaxy.

Quote
Hence, the accretion disc is actually an excretion disc.
I have already discussed deeply on the activity of new mass creation in the galaxy.
I have poved that an Atom is a cell of energy.
So, the SMBH converts its ultra high gravity force into energy. this energy is locked in new Atoms in the plasma.
It creates new Hydrogen atoms at this axcretion disc. Then due to the collisions between those new born atoms it creats all the variety of atoms and molecular.
The ultra high temp of the plasma in the accretion disc (10 ^9 c) is a clear evidence for the new born atom activity at that disc.
There is a formula for the temperature of the accretion disk at radius R which is something like
T(R) = √√[3GM⊕/8πσR³ (1−√(Rinner / R)) ]
Where G π and σ are constants, ⊕ is the accretion rate (which is actually an M with a dot over it, but I couldn't figure out how to write that).  Rinner is the Schwarzschild radius.  Plug your idea into that and you get a negative number for the temperature.  This is fine.  You've thrown all of known physics out the window, so there's no reason for that rule to have survived.

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As all the new born molecular drift outwards from the disc - remember - all objects must drift outwards over time including those Atoms), they get to the magnetic shield around the accretion disc.
This Ultra power of magnetic disc, boost the atoms upwards at a speed of almost 0.8 c.R
Sort of like hitting a boost pad in Mario Kart.  How does the magnetic disc know which way to send the material?  I've never heard of a magnetic field with only rotational symmetry, but I could just be ignorant of that.
At what radius does this take place?  That rotation curve has inner things moving pretty slow, and gradually moving up until maxing out at about 1KPC, after which the curve goes down again for a while.

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As they move up, they lose the boost of the magnetic force and eventually they fall back at the galactic disc plane.
They set the gas could that we see around the SMBH.
However, In the gas clouds, the molecular orbits around some virtual center of mass (as explained in element 1).
I assume that the orbital velocity in the gas cloud is quite high.
Each gas molecule has a virtual host point?

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This internal orbit in the gas cloud, pulse the great impact of the nearby SMBH set the Huge New form star activity that we see.
Any new born star must drift outwards while it orbits around some virtual center of mass.
Well, I've seen sites that describe the bar as a stellar nursery, so this seems plausible enough.
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That is all for that step in my theory.
Please let me know if based on the two elements - you agree with the outcome.
The stuff-moving-out element indeed seems to require the generation of mass.  That seems to follow.
The virtual host point idea doesn't seem to contribute anything to the idea.  You haven't used that to explain anything, so I'm not sure why you have to posit this.  You describe the sun moving in a 1LY radius circle, but I see no purpose in proposing that it does that.  What gets explained by that?

Neither of the two elements seems related to this idea of magnetic fields propelling the new mass to, well, nonzero speeds.  Perhaps you need more than just the two elements.  Yes, the angular speed needs something to account for it, else you just get matter moving straight out from the center like lava from a volcano.
It seems that the speed boost from the magnetic field would account for matter moving away from the center, not something like negative gravity repulsing nearby stuff.

Do black holes in general generate new matter like that?  I ask because we don't seem to see this happening around smaller black holes.  What is special at a galaxy center that this process goes on, but we don't see it elsewhere?

Also, what happens to all this new matter after a long time?  Does it fade to nothing after a while, or does each galaxy grow indefinitely?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 06:23:57
Thanks Halc
I really appreciate your excellent reply, and your important questions.
So, let me answer one by one.
 
Does  everything orbit a virtual host point, or only stars?  If the latter, what delimits objects with a VHP vs ones that don't?  What about a small black hole, or a cold black dwarf that has less mass than some planets?
For instance, Venus doesn't seem to orbit a VHP.  It goes pretty much in a normal elliptical orbit straight around the sun with no extra wobble of its own.

All the stars in the galaxy must orbit around some virtual host point. That includes all the stars in the Bulge, Bar, Spiral arms and even outside the disc. I will explain later on how the gravity impact the orbital cycle at each segment.
However, In the Solar system there is only one big star.
The Sun is the leading gravity power in the solar system.
Therefore, there is no need for a virtual host point.
However, it is quite interest to know that the moon orbits around the Earth instead around the sun although the sun gravity force on the moon is more than a twice than the Earth gravity force on the moon.
This is very important evidence.
I wonder why our scientists had neglected this observation.
In order to do so, it is clear to me that in day one of the Sun/Earth/Moon system the gravity force of the Earth/moon was much stronger than the Sun/Moon.
I assume that you head about the word - "Hysteresis".
https://en.wikipedia.org/wiki/Hysteresis
"Hysteresis is the dependence of the state of a system on its history."
That is very clear based on my theory.
All the stars/Planets/Moon... had been formed at those molecular gas clouds at the same moment!!!
Our Sun, Earth and moon had been created at the same day (more or less) and from the same matter.
So, in day one of the solar system there were no rocky planets or moons.
All the objects were very nice round gas objects (yes even our Earth and Moon).
However, it is clear that the gas atoms (especially - Hydrogen) were dominant in all of those new formed objects.
As an example - "The sun is a big ball of gas and plasma. Most of the gas — 91 percent — is hydrogen"
https://www.space.com/17170-what-is-the-sun-made-of.html
If I understand it correctly, less than 2% in the gas clouds are solid molecular (Silicate, Iron... and even water)
Therefore, today the Earth & the Moon has less than 2% from the whole matter which they had in day one.
Hence, the real Earth mass in day one was bigger by 98 from its current mass value. (same issue with all the rocky planets and moons in the solar system).
Therefore, the gravity force between the Earth moon in day one was much higher than the Sun moon.
That is the History!!!
We prove it by - "Hysteresis is the dependence of the state of a system on its history."
In any case, all the rocky planets and moons were too small to hold the gas Atoms.
Due to gravity force, all the heavier molecular moved to the center of the young planets/moons while most/all the gas had been ejected out over time. Therefore, all of them have a very nice ball shape.
We would never ever get a nice ball shape if from day one all we have is a solid matter. The gas is vital for the ball shape of all the planets and moons. In other wards - there is no way that the moon had been created due to some none realistic collision with the Earth
The other gas planets were big enough to hold significant portion of the gas. Therefore, we also call them gas planets.
The Sun is the biggest one. Therefore even after all the hydrogen fusion process it has 91% of Hydrogen atoms.
Even Ceres - was a nice big ball shape planet in early time.
Unfortunately, some other big object had been collided with this planet and broke it to small pieces.
The Earth have got all its water supply from day one. (Therefore, there was no need for secial water delivery as our scientists believed.)
If I understand it correctly, there is no DNA for matter. However, if we could check it, we should find that all the matter in the solar system has the same DNA.
Actually, as all the matter in the Milky way had been created by the same SMBH, they all should carry the same DNA.



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/01/2019 14:08:28
All the stars in the galaxy must orbit around some virtual host point. That includes all the stars in the Bulge, Bar, Spiral arms and even outside the disc. I will explain later on how the gravity impact the orbital cycle at each segment.
However, In the Solar system there is only one big star.
The Sun is the leading gravity power in the solar system.
Therefore, there is no need for a virtual host point.
You seem to contradict yourself.  You say all stars have a virtual host point, but then say now that the sun doesn't have one.  You put it a light-year away in a prior post, and now say it doesn't exist.
It is hard to figure out your VHP idea if the rules change from post to post, or in the same paragraph as has happened just above.

Yes, I agree that binary systems have stars that effectively orbit their barycenter (their mutual center of gravity).  You calling it a virtual host point just changes the terminology.  There's already one word for it.
For systems with 3 or more stars, there is very much a center of gravity which remains relatively fixed no matter the motion of the stars involved, but the motion about that point is no longer strictly orbital, depending on the dynamics of the stars.  This is what they mean by the three-body problem.  Typically pairs of stars orbit each other, and those pairs orbit other stars or other pairs, until there are only 2 bodies left.  3 or more in mutual orbit with no 2 in independent orbit is not stable.

Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length.

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Gravity force on the moon is more than a twice than the Earth gravity force on the moon.
Assuming gravity works like physics says and not like you say, acceleration due to the moon itself at its surface is 0.1630 m/s, 0.0027 due to Earth, and 0.0060 due to sun. That means it is 60 times as much, which indeed is 'more than twice' the acceleration on the moon.  It is about 60x in fact.

I say acceleration because there is no force of gravity on the moon itself.  You would need an object to have a force, and no object was specified.  So if I stand on the moon, there might be a force of 135 N on me, but the 'the Earth gravity force on the moon' is about 2e20 N, which is a lot more than my 135 N.  That huge 2e20 figure is the force that Earth exerts on the moon, and is also exactly the force that the moon exerts on Earth, at least per standard physics.

I'm saying all this because I'm trying to figure out what you meant to say, as opposed to what you actually said.

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This is very important evidence.
I wonder why our scientists had neglected this observation.
I'm sorry, but I didn't get exactly which observation was neglected by scientists.  All the numbers I quoted come from those scientists, and you didn't actually quote any numbers at all, let along ones that are different than what the scientists noticed.

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In order to do so, it is clear to me that in day one of the Sun/Earth/Moon system the gravity force of the Earth/moon was much stronger than the Sun/Moon.
I assume that you head about the word - "Hysteresis".
https://en.wikipedia.org/wiki/Hysteresis
"Hysteresis is the dependence of the state of a system on its history."
The word is used more in molecular changes than macroscopic history.  The orientation of dipoles on the seabed floor is considered hysteresis, while the nice crater in Arizona is not, despite both of them very much being a state that depends on the history of the system.
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That is very clear based on my theory.
All the stars/Planets/Moon... had been formed at those molecular gas clouds at the same moment!!!
Our Sun, Earth and moon had been created at the same day (more or less) and from the same matter.
So, in day one of the solar system there were no rocky planets or moons.
How very biblical.

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All the objects were very nice round gas objects (yes even our Earth and Moon).
However, it is clear that the gas atoms (especially - Hydrogen) were dominant in all of those new formed objects.
As an example - "The sun is a big ball of gas and plasma. Most of the gas — 91 percent — is hydrogen"
https://www.space.com/17170-what-is-the-sun-made-of.html
If I understand it correctly, less than 2% in the gas clouds are solid molecular (Silicate, Iron... and even water)
Therefore, today the Earth & the Moon has less than 2% from the whole matter which they had in day one.
Hence, the real Earth mass in day one was bigger by 98 from its current mass value. (same issue with all the rocky planets and moons in the solar system).
Therefore, the gravity force between the Earth moon in day one was much higher than the Sun moon.
That is the History!!!
We prove it by - "Hysteresis is the dependence of the state of a system on its history."
They actually use analysis of history in doing this sort of work, so yes, it would be interesting to see you use it yourself to demonstrate this new assertion.

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In any case, all the rocky planets and moons were too small to hold the gas Atoms.
If Earth and moon were 50 times larger on that first day, they would touch and be one thing, not two.

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Due to gravity force, all the heavier molecular moved to the center of the young planets/moons while most/all the gas had been ejected out over time. Therefore, all of them have a very nice ball shape.
We would never ever get a nice ball shape if from day one all we have is a solid matter. The gas is vital for the ball shape of all the planets and moons. In other wards - there is no way that the moon had been created due to some none realistic collision with the Earth
The other gas planets were big enough to hold significant portion of the gas. Therefore, we also call them gas planets.
The Sun is the biggest one. Therefore even after all the hydrogen fusion process it has 91% of Hydrogen atoms.

Even Ceres - was a nice big ball shape planet in early time.
Unfortunately, some other big object had been collided with this planet and broke it to small pieces.
It seems to be in one piece to me.  Or are you saying that the belt was once a single planet?  That planet is not Ceres, which is just the name of that one piece.  The hypothetical planet has sometimes been called Phaeton.

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The Earth have got all its water supply from day one. (Therefore, there was no need for secial water delivery as our scientists believed.)
If I understand it correctly, there is no DNA for matter. However, if we could check it, we should find that all the matter in the solar system has the same DNA.
Actually, as all the matter in the Milky way had been created by the same SMBH, they all should carry the same DNA.
If matter doesn't have DNA, then how can it all carry the same DNA?

My comment on all this seems to be that I don't see how any of this follows from or depends on your two primary points about VHP and stuff drifting away from gravitational sources.
You said you are addressing my questions one at a time, and this post started as the answer to my question about which objects have VHP's and which don't.  But none of this really discusses that topic except the first few lines which now says that the sun doesn't need one now, in contrast to your prior posts.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 14:18:42
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In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
What do you mean by this?  They have an orbit, a wobble (sort of like the green sin wave)?  What do you mean by 'fixed'?  Do all VHP's have this same amplitude, or do you mean each is different but regular, not varying from one peak to the next?
All the stars in the galaxy orbit around some virtual host point.
This host point orbits around the galaxy.
So, the stars wobble (sort of like the green sin wave), however, the virtual host point which show the orbital motion, moves at a fixed distance from the galactic disc.
If we will compare the distance between any two nearby virtual points over time (while they are at the same siral arm), we should see that the distance between the two is fixed. Therefore, the relative speed between any two nearby stars should be Zero (or almost zero).
What has this VHP thing got to do with the stuff below where the center of the galaxy creates new matter that is expelled away from there?
I do not understand the question.
Would you kindly explain the question.

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Based on that idea, all the stars, gas clouds, dust... are drifting outwards from the center. Nothing can move inwards.
Not even irregular 'not real' moons?  You seem to be inconsistent with prior posts, so I'm trying to get it straight.
As I have stated, not real moons means a broken objects.
I would assume that a broken object by definition is asteroid (even if we call it moon).
So, if the astroide is close enough to the host, it will move inwards and eventually it should collide with the host.
We must find the correct formula which shows the drifting direction over time.
In any case, as I have stated, all Stars, Planets and real moons must drift outwards.
There is a formula for the temperature of the accretion disk at radius R which is something like
T(R) = √√[3GM⊕/8πσR³ (1−√(Rinner / R)) ]
It is clear to me that in order to prove the none realistic hypothetical ideas, our scientists come with new formula. They think that by mathematics they can force the Universe to act upon their understanding.
Sorry. This is a severe mistake.
If they want to prove something, they have to use real formula by Newton, Kepler or Eisenstein.
 
How does the magnetic disc know which way to send the material?  I've never heard of a magnetic field with only rotational symmetry, but I could just be ignorant of that.
At what radius does this take place?
Good question.
I also think that it must be symmetrical.
So, I assume that the new born molecular are ejected upwards and downward (with regards to the galactic plane) due to the magnetic field.
I have no idea at what radius the magnetic field starts to work.
Each gas molecule has a virtual host point?
Sure.
The atom/molecular must be concentrated in order to set the gas cloud. The only possibility to do so is by orbiting around a virtual host point in the cloud.
So, as the molecular orbit around this virtual host, they start to form all the stars, planets and moons that we have in the galaxy.
However, not all the gas in the cloud is used for the new stars forming activity.
The leftover of the gas cloud is ejected from the disc plane of the galaxy.
We see it clearly in the following example:
https://www.universetoday.com/102011/hydrogen-clouds-discovered-between-andromeda-and-triangulum-galaxies/
Hydrogen Clouds Discovered Between Andromeda And Triangulum Galaxies.
Those are two spiral galaxies.
Andromeda is the mother galaxy of Triangulum.
Triangulum had started as a small black hole at the center of Andromeda.
As it drifts outwards from Andromeda center (with all the other stars) it starts to create its own new mass.
I will explain later on how this activity impacts our observable universe.
In any case, as those two galaxies move away from each other they ejects the leftover of the gas clouds.
Therefore, we see all of those gas clouds between the two galaxies.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 14:33:17
You seem to contradict yourself.  You say all stars have a virtual host point, but then say now that the sun doesn't have one.  You put it a light-year away in a prior post, and now say it doesn't exist.
Sorry if I was not clear.
The Sun orbits around a virtual host point. But there is no need for host point in the solar system.
Therefore, all the planets and moons orbit around real objects.
I have no clue if the orbital radius around the VHP is 1 light year or 100 Light years.
The main idea was that 64 Million years for one orbital cycle of the Sun around that VHP is too long (Based on my understanding).
Yes, I agree that binary systems have stars that effectively orbit their barycenter (their mutual center of gravity).  You calling it a virtual host point just changes the terminology.  There's already one word for it.
For systems with 3 or more stars, there is very much a center of gravity which remains relatively fixed no matter the motion of the stars involved, but the motion about that point is no longer strictly orbital, depending on the dynamics of the stars.  This is what they mean by the three-body problem.  Typically pairs of stars orbit each other, and those pairs orbit other stars or other pairs, until there are only 2 bodies left.  3 or more in mutual orbit with no 2 in independent orbit is not stable.

Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length.
Yes, the whole idea is based on center of mass of multi-star solar systems.
However, there are two main issues to remember:
1. Each star orbits around a virtual host center. This is a gift that each star gets when it had been formed at the gas cloud. So, the center of mass of multi-star solar systems is actually a center of the Virtual host points of all the other solar/stars systems. Therefore, when we set a calculation we actually should ignore the current location of the star itself, but focus on it's virtual host point which is represented by the white dashed line (It is also called the Star motion around the galaxy).
2. This center of mass (of the entire nearby virtual host points) is working under the impact of the galaxy' gravity. It is quite complicate, as at different segments of the galaxy it has different gravity forces. I will explain it later on.

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Gravity force on the moon is more than a twice than the Earth gravity force on the moon.
Assuming gravity works like physics says and not like you say, acceleration due to the moon itself at its surface is 0.1630 m/s, 0.0027 due to Earth, and 0.0060 due to sun. That means it is 60 times as much, which indeed is 'more than twice' the acceleration on the moon.  It is about 60x in fact.

I say acceleration because there is no force of gravity on the moon itself.  You would need an object to have a force, and no object was specified.  So if I stand on the moon, there might be a force of 135 N on me, but the 'the Earth gravity force on the moon' is about 2e20 N, which is a lot more than my 135 N.  That huge 2e20 figure is the force that Earth exerts on the moon, and is also exactly the force that the moon exerts on Earth, at least per standard physics.

I'm saying all this because I'm trying to figure out what you meant to say, as opposed to what you actually said.
You miss the whole point.
The idea is as follow:
let's assume that we break out the Moon/Earth orbital system.
Let's give the moon the possibility to chose around which one it prefers to orbit.
Under the current conditions the gravity force of the Sun/moon is twice stronger than the earth/moon.
Therefore, it is clear to me that it should chose to orbit around the Sun instead of around the moon.
However, the moon orbits around the Earth as in early time when the Moon and the earth were gas objects and the whole solar system were more concentrated, the gravity force between the Earth/moon was higher than the Moon Sun gravity force.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 14:36:04
are you saying that the belt was once a single planet?  That planet is not Ceres, which is just the name of that one piece.  The hypothetical planet has sometimes been called Phaeton.
Yes. That is correct!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 14:58:03
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In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
This host point orbits around the galaxy.
You didn't really answer my question.  OK, the VHP orbits around the galaxy.  Where does 'amplitude' come into that?  An orbit doesn't have an amplitude.  The star might have one, the amplitude being its distance from its host point, but how does the VHP have an amplitude?
Just trying to understand your terminology.

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In any case, as I have stated, all Stars, Planets and real moons must drift outwards.
Why is Earth not drifting measurably outward?

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It is clear to me that in order to prove the none realistic hypothetical ideas, our scientists come with new formula. They think that by mathematics they can force the Universe to act upon their understanding.
No scientist would assert that they are forcing the universe to do anything.  The universe is forcing them to find mathematics that describes it.  I've seen no mathematics from you, so they're doing much better so far.

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Sorry. This is a severe mistake.
If they want to prove something, they have to use real formula by Newton, Kepler or Eisenstein.
But you've discarded the works of all three of these people.  Their mathematics forbid the sort of thing you are describing.  Every one of the conservation laws that derive from their mathematics is violated.  If you're going to going to appeal to these people, your theory needs to conform to the physics that stems from their work.
I've been quiet about the blatant violations because I've assumed that your model would have new mathematics, but here you are suddenly appealing to established views instead of this new one you are pushing.

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Quote from: Halc
How does the magnetic disc know which way to send the material?  I've never heard of a magnetic field with only rotational symmetry, but I could just be ignorant of that.
At what radius does this take place?
Good question.
I also think that it must be symmetrical.
I said it was, but seemingly only rotational symmetry.  If it had any kind of mirror symmetry, material would be accelerated equally in both directions, which would at least conserve angular momentum.  Your idea has no angular momentum conservation.

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I have no idea at what radius the magnetic field starts to work.
So much for the good question...

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Quote from: Halc
Each gas molecule has a virtual host point?
Sure.
The atom/molecular must be concentrated in order to set the gas cloud. The only possibility to do so is by orbiting around a virtual host point in the cloud.
I can concentrate gas by squeezing it in a tube.  I can light a match this way if I squeeze hard on it.  No virtual host point needed, just compressing force.  A virtual host point I suppose would make a gas atom go in a circle, but I don't see how that helps with their being more concentrated since being a circle, the atom will just come back to its original point, or even further out since everything drifts away over time, which results in decreased concentration, not increased concentration.

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So, as the molecular orbit around this virtual host, they start to form all the stars, planets and moons that we have in the galaxy.
No they don't.  They drift away you said.  How can a star form if everything drifts away?

However, not all the gas in the cloud is used for the new stars forming activity.
The leftover of the gas cloud is ejected from the disc plane of the galaxy.
We see it clearly in the following example:
https://www.universetoday.com/102011/hydrogen-clouds-discovered-between-andromeda-and-triangulum-galaxies/
Hydrogen Clouds Discovered Between Andromeda And Triangulum Galaxies.
Those are two spiral galaxies.
Andromeda is the mother galaxy of Triangulum.
Triangulum had started as a small black hole at the center of Andromeda.
[/quote]So black holes can spawn not just new plasma material, but objects like new black holes??
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In any case, as those two galaxies move away from each other they ejects the leftover of the gas clouds.
Therefore, we see all of those gas clouds between the two galaxies.
They're actually on course for collision in 2.5 billion years, but that's according to those scientists who obviously just make stuff up.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/01/2019 15:19:44

SMBH


Do black holes in general generate new matter like that?  I ask because we don't seem to see this happening around smaller black holes.  What is special at a galaxy center that this process goes on, but we don't see it elsewhere?
There are different kinds of black holes.
For quite long time I have thought about that issue.
Let's look at the SMBH at the Milky Way.
We see clearly that there is matter in the accretion disc.
If I understand it correctly, the total mass in the accretion disc is estimated to be in the range of 3 Sun mass.
Actually, based on the orbital velocity of the mass at the accretion disc, and the estimated radius, we could easily calculate the real mass of the SMBH.
Unfortunately, I couldn't find and information about the accretion disc radius.
In any case, it seems to me that the SMBH must increase its mass over time.
However, I wonder how it gets new matter.
There could be two options:
1. Some of the new formed particles drift inwards.
2. In one of the article that I have found it was written that for any new particle that is created, there must be a negative particle. If that is correct, and assuming that all the positive particles are drifting outwards, than the negative must drift inwards. In this case, the SMBH is a concentration of the negative particles. (I have no clue if that is correct...)

Let's look at another kind of SMBH - Quasar:
https://www.forbes.com/sites/startswithabang/2017/07/31/universes-largest-black-hole-may-have-an-explanation-at-last/#52b55966fc55
"The brightest, most luminous objects in the entire Universe are neither stars nor galaxies, but quasars, like S5 0014+81."
So, this SMBH (we call it quasar) has no galaxy.
It creates new matter and ejects it in a stream upwards and downwards as expected.
So it works perfectly as the SMBH of any spiral galaxy should work.
New matter is created at the accretion disc. As the new molecular drifts outwards they get to the magnetic shield around the accretion disc. The ultra high magnetic power blow it upwards and downwards.
It is also clear that those molecular fall back to the disc plane of the Quasar (outwards from the magnetic shield). However, somehow, they don't form the gas cloud. Without the gas cloud there is no new born stars. Without stars there is no galaxy.
So, this is a good example for a SMBH which generates new matter without having a real spiral galaxy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/01/2019 15:50:44
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In any case, as those two galaxies move away from each other they ejects the leftover of the gas clouds.
Therefore, we see all of those gas clouds between the two galaxies.
They're actually on course for collision in 2.5 billion years, but that's according to those scientists who obviously just make stuff up
No, this is incorrect.
They are quite close to each other, but they are drifting away from each other. However, both of them are moving in our direction. So the 2.5 Billion is the requested time for the collision with the Milky way.
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So, as the molecular orbit around this virtual host, they start to form all the stars, planets and moons that we have in the galaxy.
No they don't.  They drift away you said.  How can a star form if everything drifts away?
I don't see any problem with that.
Let's assume that it takes 10 Million years to set a star in a gas cloud, and in this time frame the gas cloud had increased its radius by 10%
In this case, this gas cloud will set new star and planets
Please also remember that the gas cloud itself is drifting away from the center.
So, it is clear that at some point of time the gas cloud (or the leftover from the gas cloud) will not be able to create any more stars and planets.
What is the problem with that ?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 17:46:53
There are different kinds of black holes.
For quite long time I have thought about that issue.
Let's look at the SMBH at the Milky Way.
We see clearly that there is matter in the accretion disc.
If I understand it correctly, the total mass in the accretion disc is estimated to be in the range of 3 Sun mass.
Actually, based on the orbital velocity of the mass at the accretion disc, and the estimated radius, we could easily calculate the real mass of the SMBH.
Yes, that's how it's done in normal physics, but you've denied that when the scientists measure the mass of the galaxy using that exact method and got a mass much larger than all the material they've measured.  The scientists say there must be more mass they can't see, while you say they are using the wrong physics to calculate the mass.

Therefore, you don't know the mass of the SMBH using the speed of the material orbiting it, since you deny the validity of that method.  You need a new way of measuring the mass of an object.

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Unfortunately, I couldn't find and information about the accretion disc radius.
It's not like it stops abruptly.  Technically, the whole galaxy is an accretion disk.  It just changes properties the closer to the gravity source it gets.

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In any case, it seems to me that the SMBH must increase its mass over time.
You have it spewing matter out into the galaxy.  I'd think it that something that does that might be losing mass. I don't see how stuff falls in if the stuff is supposed to tend to drift away, not towards it.

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However, I wonder how it gets new matter.
There could be two options:
1. Some of the new formed particles drift inwards.
2. In one of the article that I have found it was written that for any new particle that is created, there must be a negative particle.

 If that is correct, and assuming that all the positive particles are drifting outwards, than the negative must drift inwards.
The negative particles are antimatter actually, yes.  If the antimatter falls in, the black hole actually gets smaller.  It's how Hawking radiation eventually evaporates any black hole given enough time, but I'm not sure of your idea has this sort of process in it.
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In this case, the SMBH is a concentration of the negative particles. (I have no clue if that is correct...)
That's actually a way to solve it, if the black hole is antimatter, such radiation would increase the mass, not decrease it, assuming it makes sense for the anti-halves to tend to go in and not a random distribution.  Anyway, such radiation (change in mass) is greatest for the smallest black holes and slowest for the big ones.  Sgr-A might lose a few atoms per year via Hawking radiation. Maybe not even that fast.

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Let's look at another kind of SMBH - Quasar:
https://www.forbes.com/sites/startswithabang/2017/07/31/universes-largest-black-hole-may-have-an-explanation-at-last/#52b55966fc55
"The brightest, most luminous objects in the entire Universe are neither stars nor galaxies, but quasars, like S5 0014+81."
They're just young galaxies.  There are not any nearby because nothing nearby is that young.  Only looking waaay into the past do you see the initial formation processes of a galaxy.  Our own galaxy might have been a quasar at some point, but it was probably too small in its youth to be the sort of object they're looking at.  Those are massive objects that have formed what are the proper big galaxies we now see closer by.
This is not an expert statement.  It is just me spouting off at the mouth.  I know they've found black holes that are well of 100x the size of Sgr-A, and the formation of those would have generated the sort of light that you see in a quasar.
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Without the gas cloud there is no new born stars. Without stars there is no galaxy.
You don't know how many stars it has.  Trying to spot one is like spotting a firefly sitting between a car's headlights.  These quasars are also way to far away to see individual stars, except when they go supernova.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 18:01:10
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They're actually on course for collision in 2.5 billion years, but that's according to those scientists who obviously just make stuff up
No, this is incorrect.
They are quite close to each other, but they are drifting away from each other.
But accelerating inward, on schedule for a 2nd collision pass in 2.5 billion years. 

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However, both of them are moving in our direction. So the 2.5 Billion is the requested time for the collision with the Milky way.
Andromeda will takes its first pass in about 4 billion years, well after triangulum galaxy is well within Andromeda, and the 2nd about 2 billion years after that.  After that, you don't really count passes anymore, but it takes even longer for the process to finally finish and our SMBH is swallowed by a much larger one.
Not sure if you support that idea since these things are suppose to drift away, not be pulled in and combined.

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I don't see any problem with that.
Let's assume that it takes 10 Million years to set a star in a gas cloud, and in this time frame the gas cloud had increased its radius by 10%
In this case, this gas cloud will set new star and planets
Your gas cloud just got 33% less dense (33% increase of volume from a 10% increase in radius).  It would seem to require gas to be condensed to form stars, else they would already have been stars when the gas cloud was tighter.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/01/2019 15:11:42
Your gas cloud just got 33% less dense (33% increase of volume from a 10% increase in radius).  It would seem to require gas to be condensed to form stars, else they would already have been stars when the gas cloud was tighter.

Please see the following article about: Star Formation in gas cloud Around Suppermassive Black Holes:

https://arxiv.org/ftp/arxiv/papers/0810/0810.2723.pdf
"The presence of young massive stars orbiting on eccentric rings within a few tenths of a parsec of the suppermassive black hole in the Galactic centre is challenging for theories of star formation"

"The presence of young massive stars in the vicinity of the Galactic centre is difficult to reconcile with current models of star formation where turbulent molecular clouds produce a mostly clustered population of stars with a remarkably constant distribution of stellar masses, covering stars with masses from less than a tenth to greater than 100 times the mass of the sun (6)."

We have to understand the real meaning of: "turbulent molecular clouds".
In the article there is no real explanation. (Why they didn't trace that turbulent activity?)
However, I think that there is an order in that "turbulent".
I assume that it represents several/many internal orbital cycles.
If we could look carefully, we might find that around each orbital cycle there is an activity for new star forming.
So, deep in the gas cloud each star stars to orbit around a virtual center of mass.
All the virtual centers of new born stars are orbiting according to a multi star system under the impact of the nearby SMBH.
Therefore, deep in the gas cloud, each new born star gets a gift of orbiting around a virtual Host point.
That is correct to any star in the galaxy.
Actually, if we will trace S2, we might find that it set a bobbling activity as it orbits around the SMBH. That bobbling proves that even this star orbit around a virtual center of mass.
So, the gas cloud can set new stars as it drifts outwards. It just an issue of verifying the rate of outwards drifting VS the rate of new star forming.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/01/2019 14:47:39
They've tracked it quite closely through more than a full orbit now.  It doesn't have any kind of regular bobbing activity.  S2 does seem to be the sort of object referred to in the article, being a large young star in a highly eccentric orbit about Sgr-A.  That one gets particularly close to it.
Yes it does.
S2 has a clear bobbling orbiting path.
Please look at the following diagram:
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that the measured locations of S2 are not directly at the elliptical orbital cycle.
I do recall that in one of the articles it was stated that in 2002, the light of S2 was exactly at the same location as SMBH.
Our scientists were puzzled from this evidence.
They couldn't explain that verification. Therefore they had an idea of error in their instruments.
This was a big mistake.
There is no error in the data.
S2 orbits around a Virtual host point, while this VHP orbits around the SMBH.
Therefore, we see some Zig Zag in the location of S2 with reference to the expected orbital path.
Hence, S2 can even cross the location of the SMBH, while its VHP is moving exactly at the expected elliptical cycle.
I really don't understand why we have only few measured points on that image.
Technically we should know evry day where it is.
Why this information is not valid for all of us?
Why they hide this important information???

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/02/2019 15:16:09
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S2 has a clear bobbling orbiting path.
Please look at the following diagram:
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that the measured locations of S2 are not directly at the elliptical orbital cycle.
I don't see it.  I see an ellipse (viewed almost edge-on) and every measurement falling on it within the margin of error.  Nothing in the article suggests an irregularity to the orbit, or especially a regular one.
Sorry
This is the biggest mistake of our scientists!
In order to understand this important issue, let's look at the example of the Moon/Sun orbital cycle/
Let's assume that the Earth is invisible from outside. Therefore, it is actually a VHP for the Moon while this one orbits around the Sun.
So, if we try to monitor the Moon/Sun orbit, while we have no clue if the Earth is there, what shall we see?
The Moon/Earth orbital radius is:
R1 (Moon/Earth) = 384,400 Km.
The Earth/Sun orbital radius is
R2 (Earth/Sun) = 149,600,000 Km.
The ration is:
R1/R2 = 384,400 / 149,600,000 = 0.0025
In this example let's assume that the Moon/Earth orbital cycle is located at the same Earth/Sun orbital cycle.
Therefore, if we try to monitor the Moon/Sun orbital cycle, we might see that the moon is moving in almost a perfect cycle, with very minor error.
If we randomly monitor the location of the Moon, it is clear that the error fit is in the range of Zero to 0.0025.
Just if we monitor the moon while it is at its maximal/minimal distance from the sun (R2+R1 or R2-R1) we can get that maximal error fit of 0.0025.
From statistical point of view the chance to get randomly to the maximal 0.0025 Error is very low.
The error should be mainly greater than Zero and lower than 0.0025.
Now, please, let me remind you that in this example we don't know if the Earth (which is the VHP of the Moon) is there.
So, if we give those results to our scientists, what would be the answer?
Based on your answer, our scientists should reply:  "Every measurement falling on it within the margin of error"
So, based on this error margin, the Earth doesn't exist.
Now, let's go back to the S2 orbital cycle.
Please look again at the measured. What is the estimated error margin?
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that many measured points are located outside the Fit orbital cycle.
If we look carefully, we clearly see that the maximal measured error point took place on 1995.36
It seems to me as an error which is significantly bigger than 0.0025.
If you also look carefully, we see that the SMBH is not located at the symmetrical point in that orbital cycle.
It is located too close to the left bottom side.
That is one more evidence that S2 doesn't orbit directly around the SMBH.
What about the orbital velocity? Did we try to verify if there are small variations in the measured velocities?
So, how our scientists could afford themselves to ignore those clear evidences of none fit?
I think that it is a fatal mistake to claim:  "Every measurement falling on it within the margin of error"!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/02/2019 17:23:03
This is the biggest mistake of our scientists!
The scientists know how to do vector arithmetic, and you don't, so the odds of them being the one making the mistakes are pretty low.
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In order to understand this important issue, let's look at the example of the Moon/Sun orbital cycle/
Let's assume that the Earth is invisible from outside. Therefore, it is actually a VHP for the Moon while this one orbits around the Sun.
You just described dark matter, something you deny.

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So, if we try to monitor the Moon/Sun orbit, while we have no clue if the Earth is there, what shall we see?
The Moon/Earth orbital radius is:
R1 (Moon/Earth) = 384,400 Km.
The Earth/Sun orbital radius is
R2 (Earth/Sun) = 149,600,000 Km.
The ration is:
R1/R2 = 384,400 / 149,600,000 = 0.0025
Yes, and if you run a Fourier transform on the data for the position of the moon over time, there would be two huge spikes for the periods of a month and a year.  The virtual host point would be glaringly obvious from that data.

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In this example let's assume that the Moon/Earth orbital cycle is located at the same Earth/Sun orbital cycle.
I don't know what you mean by this.  A cycle doesn't have a location, and I don't think you are proposing that the Earth is where the sun is.  You want the moon to go around the Earth in one sidereal year?  That would put it at one of the Lagrange points, making it a trojan satellite, not really orbiting the Earth.

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Therefore, if we try to monitor the Moon/Sun orbital cycle, we might see that the moon is moving in almost a perfect cycle, with very minor error.
No error at all in fact.

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If we randomly monitor the location of the Moon, it is clear that the error fit is in the range of Zero to 0.0025.
Just if we monitor the moon while it is at its maximal/minimal distance from the sun (R2+R1 or R2-R1) we can get that maximal error fit of 0.0025.
We'd know the moon's position to far greater accuracy than that from the distance you indicate.
If you are appealing to margin of error, then it doesn't support your idea, it just says the data isn't accurate enough to make a determination.  No scientists are making mistakes then.  A regular perpetration would show up in a Fourier transform with enough data points, even with a margin of error far larger than the perpetration.

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From statistical point of view the chance to get randomly to the maximal 0.0025 Error is very low.
OK, your knowledge of statistics is on par with your vector arithmetic prowess I see.  Statistics is exactly how you get an arbitrarily small margin of error from data points each of which has a large margin of error.

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The error should be mainly greater than Zero and lower than 0.0025.
Now, please, let me remind you that in this example we don't know if the Earth (which is the VHP of the Moon) is there.
So, if we give those results to our scientists, what would be the answer?
If it goes around the Earth in the same period that it goes around the sun, then the dark Earth will be undetectable.  I'm not sure why you proposed that period, or even if that is what you meant.

Based on your answer, our scientists should reply:  "Every measurement falling on it within the margin of error"
So, based on this error margin, the Earth doesn't exist.[/quote]
If the moon has a period around Earth that is a month (something less than a year), and the measurements are all a worse margin of error than this 0.0025, then a sufficiently large sample of measurements would suffice to detect Earth, or the lack of it.

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If you also look carefully, we see that the SMBH is not located at the symmetrical point in that orbital cycle.
It is located too close to the left bottom side.
That's because you're looking at an ellipse nearly edge-on which distorts where the focus points are.  Our vantage point is not perpendicular to the plane of S2's orbit, but you seem to assume so.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/02/2019 06:28:50
In this example let's assume that the Moon/Earth orbital cycle is located at the same Earth/Sun orbital cycle.
Sorry if I was not clear.
The idea is:  The Moon/Earth orbital cycle disc plane is located at the same Earth/Sun orbital cycle disc plane.
Again - In this example we want to verify that just by monitoring the locations of the Sun & moon in one cycle (one year) we should find that the Earth is missing (assuming that we don't see it in our verification).
If the moon has a period around Earth that is a month (something less than a year), and the measurements are all a worse margin of error than this 0.0025, then a sufficiently large sample of measurements would suffice to detect Earth, or the lack of it.
O.K.
So, you understand the idea of this example. You also agree that if we see that our measurements of the moon has a small variations from the expected perfect orbital cycle (even if it is less than 0.0025) we should know that the moon must orbit around some other object. We call it Earth, but we can also call it VHP (as we don't see it in our example).
Therefore, I still don't understand why do you still support our scientists while they have clearly neglecting the variations of S2 locations from the expected perfect orbital cycle? They call those variations - Errors. But those variations are very important observations.
Let me use the example at a different perspective:
Let's assume that the  Moon/Earth orbital cycle disc plane is located vertically to the Earth/Sun orbital cycle disc plane and it orbits directly in the Earth/sun orbital cycle.
Let's also assume that our vantage point is perpendicular to the Earth/Sun orbital plane.
Therefore, it is clear that in all the verifications of the Moon location points we should see that it is perfectly located directly at the expected orbital cycle around the Sun.
So, without any variations (or errors) from the expected orbital cycle, we might think there is no need for the earth to explain a perfect orbital fit of the Moon around the Sun.
However, there is another key verification - Velocity.
"KEPLER’S SECOND LAW DESCRIBES THE WAY AN OBJECT’S SPEED VARIES ALONG ITS ORBIT
A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit."
So, in a real elliptical orbit we should find that the velocity of the planet is decreasing smoothly as the farther it is from the Sun.
However, in this example, Due to the Moon/Earth orbital path, sometimes the moon orbits in the orbital Earth/sun direction, and sometimes on the opposite direction (I hope that you understand me correctly).
Therefore, we should find the moon is not smoothly increasing or decreasing its orbital velocity as it orbits around the sun.
So, we have two ways to verify if the Earth (or the VHP) is there.
One - By verify if there is variations (error) in the verified moon location with reference to the expected orbital elliptical cycle.
Two - By verify if there is none smoothly variations in the orbital velocity.
the location of the moon.
With regards to S2 - Even with the assumption that our vantage point is not perpendicular to the plane of S2's orbit, we clearly see that S2 Does not fulfill the two requirements.
Therefore, it shows that S2 must orbit around a VHP while this one might orbit the SMBH.
There is one more key evidence for that:
Please look at the following diagram:
https://en.wikipedia.org/wiki/S2_(star)#/media/File:Galactic_centre_orbits.svg
We see several S stars that orbit around the SMBH.
I assume that our vantage point is not perpendicular to any of those S planes of orbit.
There is good chance that each orbital cycle plane might be different from all the other.
However, if we try to estimate the requested location of the SMBH for each cycle (based on kepler law), we should find that each one needs the SMBH at a different location in space.
So, that also shows that those stars do not orbit directly around the same SMBH.
Each one of them orbits around a VHP while all of those VHP orbits around the SMBH.
Why can't you agree with something which is so clear to us?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/02/2019 14:04:07
The idea is:  The Moon/Earth orbital cycle disc plane is located at the same Earth/Sun orbital cycle disc plane.
Oh, OK.  It already is you know.  There is virtually no tilt to it, something not true in general.
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Again - In this example we want to verify that just by monitoring the locations of the Sun & moon in one cycle (one year) we should find that the Earth is missing (assuming that we don't see it in our verification).
Earth would be detectable in this manner no matter what the tilt of the moon's orbit.
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So, you understand the idea of this example. You also agree that if we see that our measurements of the moon has a small variations from the expected perfect orbital cycle (even if it is less than 0.0025) we should know that the moon must orbit around some other object.
It could have a very large variation (called eccentricity) and we'd still be able to detect the orbit, yes.

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We call it Earth, but we can also call it VHP (as we don't see it in our example).
Yes, we presumably don't know it's there ahead of time, so we certainly don't know its name.

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Therefore, I still don't understand why do you still support our scientists while they have clearly neglecting the variations of S2 locations from the expected perfect orbital cycle? They call those variations - Errors. But those variations are very important observations.
They've probably run a Fourier transform on the data and found no regular orbit to it.  That's why asked if you had done that.  You seem to be getting your data from pictures and conclusions written by somebody else, rather than from the actual measurements taken.  If you claim a pattern to the data, take the data and demonstrate it.

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Let me use the example at a different perspective:
Let's assume that the  Moon/Earth orbital cycle disc plane is located vertically to the Earth/Sun orbital cycle disc plane and it orbits directly in the Earth/sun orbital cycle.
You mean perpendicular I presume.  A plane cannot be vertical to another one.
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Let's also assume that our vantage point is perpendicular to the Earth/Sun orbital plane.
Therefore, it is clear that in all the verifications of the Moon location points we should see that it is perfectly located directly at the expected orbital cycle around the Sun.
No, it would move back and forth in one dimension but not in the other.  So assuming the axis runs right/left from that vantage, the motion you'd see from that vantage is a continuous up/down movement, which is towards and away from the sun when Earth is on the upper or lower side of the orbit, and forwards and backwards in its orbit when Earth is on either side.
If your measurements were good, you could also detect velocity towards and away from your vantage point, which would clue you in to motion in that dimension.  Good measurements are 3D and thus better than 2D pictures.

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So, without any variations (or errors) from the expected orbital cycle, we might think there is no need for the earth to explain a perfect orbital fit of the Moon around the Sun.
I can think of no orientation of axes or vantage point that would make Earth undetectable in this way.

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However, there is another key verification - Velocity.
"KEPLER’S SECOND LAW DESCRIBES THE WAY AN OBJECT’S SPEED VARIES ALONG ITS ORBIT
A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit."
So, in a real elliptical orbit we should find that the velocity of the planet is decreasing smoothly as the farther it is from the Sun.
However, in this example, Due to the Moon/Earth orbital path, sometimes the moon orbits in the orbital Earth/sun direction, and sometimes on the opposite direction (I hope that you understand me correctly).
Therefore, we should find the moon is not smoothly increasing or decreasing its orbital velocity as it orbits around the sun.
So, we have two ways to verify if the Earth (or the VHP) is there.
They're actually the same way.  You variations are detectable despite any orientation of axes or vantage points.  So yes, there would be no way to hide the VHP if there was one.  Earth is not a VHP in that scenario.  It is a dark host mass.  It is a real mass, not a virtual one, and is dark, simply meaning we cannot see it.  For the moon or any other object to orbit an actual VHP would mean acceleration without any of the known forces to account for it.
So if scientists were to detect such regular motion, they'd know.  They detect this sort of motion for many stars, even when the motion is very subtle (less than the radius of the star), and use the motion to detect dark masses like gas giant planets.  Our own star exhibits this sort of motion about a host point, but its motion around it is not elliptical and thus really an orbit.  Sometimes the sun is incredibly close to its host point, and sometimes further away.  It's acceleration is not a function of its distance from that host point, and in the case of Keplerian orbit, it would be a function of that distance, per Kepler's 2nd law you quoted above which only applies to simple 2-body orbits.

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One - By verify if there is variations (error) in the verified moon location with reference to the expected orbital elliptical cycle.
Two - By verify if there is none smoothly variations in the orbital velocity.
Same thing, since any variation in location or velocity would result in a variation of the other.

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With regards to S2 - Even with the assumption that our vantage point is not perpendicular to the plane of S2's orbit, we clearly see that S2 Does not fulfill the two requirements.
You clearly see what you want to.  I don't see it at all, mostly because all I have is a lousy diagram and not actual data, but the perfect orbit path depicted hits every single one of the data points within the margin of error, meaning there many be deviation from that path, but said deviation is anything but clear.

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Therefore, it shows that S2 must orbit around a VHP while this one might orbit the SMBH.
I told you how to clearly show that from the data set, or from a larger one.  You've not done that, so nothing has been shown.

There is one more key evidence for that:
Please look at the following diagram:
https://en.wikipedia.org/wiki/S2_(star)#/media/File:Galactic_centre_orbits.svg
We see several S stars that orbit around the SMBH.
I assume that our vantage point is not perpendicular to any of those S planes of orbit.
There is good chance that each orbital cycle plane might be different from all the other.
However, if we try to estimate the requested location of the SMBH for each cycle (based on kepler law), we should find that each one needs the SMBH at a different location in space.[/quote]
How so?  For it to appear be at one of the focus points of the elliptic paths, you need a perpendicular vantage point, and you admit that our vantage is anything but perpendicular.  The graph does not plot position, but rather ascension and declination, but you're interpreting the plot as position when presuming where the SMBH should be.

The graph is deceptive because they compare the appearance plot of these stars to a position plot of the orbit of several solar-system planets, none of which are presented as they appear to us, which would be nearly straight lines massively outside the narrow range of the diagram.

Anyway, have you ever watched elliptical motion from a non-perpendicular vantage point?  I have, at length even.  I had an entertaining screen saver that did just that, with a dozen objects in random orbits about a central mass, and each object left a trail behind it so the shape of the orbit was easy to see, and it looked exactly like that picture, except that your picture doesn't show where the motion is fast or slow, so you don't get a good feel for the actual point in the path where the approach to the center is closest.  S14 for instance is clearly closest on the far lower right of its path, not at the place where it goes behind the SMBH and appears to be in the same place.  It is a 2D plot of 3D motion, but you're interpreting it as 2D space when making these invalid claims.

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Each one of them orbits around a VHP while all of those VHP orbits around the SMBH.
Why can't you agree with something which is so clear to us?
The moon's path around the sun is a wobbly ellipse.  None of those lines wobble.  Maybe they're simplified, but then they've hidden the evidence you seek.  I see nothing but clean ellipses depicted, so no VHPs.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/02/2019 09:41:28
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Therefore, I still don't understand why do you still support our scientists while they have clearly neglecting the variations of S2 locations from the expected perfect orbital cycle? They call those variations - Errors. But those variations are very important observations.
They've probably run a Fourier transform on the data and found no regular orbit to it.  That's why asked if you had done that.  You seem to be getting your data from pictures and conclusions written by somebody else, rather than from the actual measurements taken.  If you claim a pattern to the data, take the data and demonstrate it.
Our scientists can do it better.
Please look at fig 1.4 in pg. 20
Center of Mass at Offset Position
http://webdoc.sub.gwdg.de/ebook/dissts/Koeln/Mouawad2005.pdf
They claim clearly, that just based on the Red line they could set the Keplerian fit. Based on the Green line there is no fit.
However, they were also forced to shift the major axis in order to get the fit.
Please see pg 78 at Figure 7.6.
They specifically indicate that the shift in major axis was very critical to get the fit.
In the following examples the couldn't get the fit:
"Exemplary three non-fitting orbits. Example of 3 orbits with an error ≥ 5 σ corresponding to: (Left) the case of Western position, a fit with 3.3×106M point mass + 0.8×106M extended component (Middle) the case of Eastern position, a fit with 3.3×106M point mass + 0.4×106M extended component, and (Right) the case of Northern position, a fit with 2.7×106M point mass + 0.3×106M extended component."
Therefore, it is clear that without changing the position of the Major axis and specifically assume that it moves on the red line, there is no fit.
I don't have to prove it. It is written by our scientists.
So, how can we take an elliptical cycle and shift its major axis???
We know that our vantage point is not perpendicular to the plane of S2's orbit. But even if we try to place that elliptical cycle in space, there is no way to shift the major axis while we keep the shape of the elliptical cycle as is.
I have tried to do it without success.
This by itself proves that based on kepler there is no fit between S2 and the SMBH. Just after all those manipulations over manipulations they claim for a fit.
Sorry this is incorrect.
Why they so deeply insist for Fit???
Those none fits are more important than a fit.
The none fit gives us deep understanding about our galaxy.
It proves that S2 doesn't directly orbit around the SMBH.
That proves that there must be something between S2 and the SMBH.
In other words, S2 must orbit around some invisible object while that invisible object might orbit around the SMBH.
I call that invisible object a Virtual Host point - VHP.
  Gravity force is computed by GMm/r² and M is zero for a VHP, so the gravitational force exerted by a VHP is zero, and thus a VHP cannot pull a star into an orbital path by gravity.
I have never claimed that the VHP has Zero mass.
It all based on Multi star system, as you have explained:
Quote from: Halc
Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length.
I will explain it later on how it really works in the galaxy, and how that system can set all the unique features of the spiral galaxy without any need for dark matter.
Anyhow, it seems to me that you support our scientists by all means.
So, what ever I might offer and show, you have already took a discussion to support the current incorrect ideas for good.
Is it correct?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 03/02/2019 14:45:09
Our scientists can do it better.
Please look at fig 1.4 in pg. 20
Center of Mass at Offset Position
http://webdoc.sub.gwdg.de/ebook/dissts/Koeln/Mouawad2005.pdf
They claim clearly, that just based on the Red line they could set the Keplerian fit. Based on the Green line there is no fit.
Careful.  You make it sound like they're getting a Keplerian fit from looking at the red and green lines, but they're getting those fits from the data, and drawing the lines according to the fit.  You are doing it the opposite way.

Anyway, yes, the green line is an apparent better fit, and it plots a non-Keplerian orbit, which is to be expected since the center of the galaxy has far more than two bodies.  Keplerian orbits only occur with two body systems.
The title of the paper indicates that this is a study of mass distribution near the center of the galaxy, so of course it is going to be focusing on the deviations from that Keplerian orbit caused by the distribution of additional mass.

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However, they were also forced to shift the major axis in order to get the fit.
Please see pg 78 at Figure 7.6.
They specifically indicate that the shift in major axis was very critical to get the fit.
In the following examples the couldn't get the fit:
"Exemplary three non-fitting orbits. Example of 3 orbits with an error ≥ 5 σ corresponding to: (Left) the case of Western position, a fit with 3.3×106M point mass + 0.8×106M extended component (Middle) the case of Eastern position, a fit with 3.3×106M point mass + 0.4×106M extended component, and (Right) the case of Northern position, a fit with 2.7×106M point mass + 0.3×106M extended component."
Therefore, it is clear that without changing the position of the Major axis and specifically assume that it moves on the red line, there is no fit.
There is no fit anyway.  I see three different attempts at a fit to the same data, each set apparently fitting a different subset of the measurements.  I don't see any assumption about assuming that a red line moves.

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I don't have to prove it. It is written by our scientists.
So, how can we take an elliptical cycle and shift its major axis???
It isn't elliptical.  It wasn't expected to be, since it isn't a two body system.  The orbit of Uranus isn't elliptical, and measuring the deviations from a pure Keplerian orbit is exactly how they knew where to look for Neptune and eventually Pluto.  That's what they're doing in this paper is learning about the nature and density of other objects using the same techniques used to find additional planets.

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We know that our vantage point is not perpendicular to the plane of S2's orbit. But even if we try to place that elliptical cycle in space, there is no way to shift the major axis while we keep the shape of the elliptical cycle as is.
I have tried to do it without success.
From just 2D point plots, it is actually quite effortless.  You can change the axis to almost anywhere you like.  From a 3D plot (one with temporal data), there is only one fit to the major axis.

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This by itself proves that based on kepler there is no fit between S2 and the SMBH.
There is always a fit.  It is the line that most closely matches the data.  The fit need not match the actual path since that's not what a fit is.
That S2 does not perfectly follow a Keplerian orbit about the SMBH is expected.  It isn't a 2 body system.

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Just after all those manipulations over manipulations they claim for a fit.
Sorry this is incorrect.
Why they so deeply insist for Fit???
It's them doing statistical analysis on the data.  It's what you do.  These guys know their statistics.

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It proves that S2 doesn't directly orbit around the SMBH.
Depends on your definition of directly.  It doesn't orbit only the mass of the SMBH, yes.  There is other mass around which it orbits.  It is a crowded place down there.

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That proves that there must be something between S2 and the SMBH.
Of course.  Why wouldn't there be??  This paper is trying to learn more about how much more something is between S2 and the SMBH, and the distribution of it, and not just prove that something is there.  It would be a far larger claim to say there was not something additional inside that orbit.

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In other words, S2 must orbit around some invisible object while that invisible object might orbit around the SMBH.
That doesn't follow at all.  Just because Venus is inside our orbit doesn't mean we orbit it.  But Venus does make the orbit of the Earth/moon system non-Keplerian, just as we deviate the orbit of Venus despite not being inside its orbit.

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I call that invisible object a Virtual Host point - VHP.
Oh, it's an object now. The story changes. How is it virtual if it's an object?  An invisible object is dark matter.

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I have never claimed that the VHP has Zero mass.
It all based on Multi star system, as you have explained:
Our sun is not a multi-star system.  Such systems are quite common, but we're not one of them.
In a binary system, it is the gravity of the other object that pulls each object into a Keplerian orbit about the other.  The VHP (center of mass of the two) does not exert any force since it is massless.

If you story is now that the VHP has mass, then it is just a dark object and not at all the center of mass of any system.

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I will explain it later on how it really works in the galaxy, and how that system can set all the unique features of the spiral galaxy without any need for dark matter.
You just posited dark matter being the unseen mass around which S2 orbits.  You just don't call it dark matter, but that's what it is: matter you can't see, and if S2 had such a secondary orbit, it would count as having been seen, just like all the other binary systems where only one of the two is luminous.

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Anyhow, it seems to me that you support our scientists by all means.
They know how to do mathematics that is completely beyond my skill set.  You've displayed a need to review high-school mathematics, and sometimes even lower.  Your ideas work in your head because you don't have the mathematics skills to see the trivial violations with your assertions.
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So, what ever I might offer and show, you have already took a discussion to support the current incorrect ideas for good.
Is it correct?
Not at all.  I'm just pointing out places where your idea doesn't hold water.  A new theory that works and makes falsifiable predictions would be warmly received.  But I've seen nothing but bad science from your postings: cherry picking data, drawing conclusions from diagrams that are not to scale instead of actual data, and complete denial of any evidence that counters your ideas.
If you want to get any attention, compute a curve of S2's path based on a VHP that matches the data points better than any the scientists have.  You'd need to find the period of the secondary orbit.  But none of the green non-Keplerian paths match a model with a secondary orbit.  You need to find that line.  Do the work, else you have zero evidence of the VHP thing.

I still have no idea why the VHP idea was necessary for the whole idea of matter being created at the SMBH and moving outward.  It seems that if it worked that way, it would work with or without everything needing a VHP.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/02/2019 16:08:47
Please look at the following article:
http://en.es-static.us/upl/2018/03/S2-black-hole-orbit.png
We see clearly the orbital shape of S2 and some other S stars.
https://earthsky.org/space/star-s2-s0-2-single-milky-way-monster-black-hole
It is stated:
"Until now, it was thought that S0-2 might be a double star. Two stars orbiting each other would have complicated the upcoming gravity test
But a team of astronomers led by Devin Chu of Hilo, Hawaii – an astronomy grad student at UCLA – has found that S2 doesn’t have a companion"
The question is: Why our scientists consider that there are two orbiting stars?
Could it be that there is a difficulty to explain the orbital path of S2 just by one star?
Could it be that they also have thought on a possibility that based on the real verifications, S2 should orbit around some other object as I was expecting?
So, could it be that even our scientists understand that there is a problem with S2 orbital Path?
I still have no idea why the VHP idea was necessary for the whole idea of matter being created at the SMBH and moving outward.
You don't know, but it is clear to me that our scientists know why a companion star is needed!!!
As I was expecting:
There is no need for any real companion star (or object). The VHP by itself can give a perfect explanation for what we see.

It is also stated:
"[S2] orbits Sgr A* on an ellipse that takes about 15 years to complete. The diameter of its orbit is about 300 billion km [200 billion miles], which may sound like a lot, but we’re talking about a suppermassive black hole here! That’s close!
And it gets closer. Because the orbit is an ellipse, the star drops down to a mere 18 billion km [11 billion miles] from the black hole, a positively terrifying close approach. That’s only four times farther from the black hole than Neptune is from our sun."
I wonder if that information represents the S2 elliptical orbit cycle that we see.
It was stated that our vantage point is not perpendicular to the plane of S2's orbit.
So, what we see doesn't represent the real S2 orbital cycle.
Therefore, I would like to understand what we really see and what is the real orbital cycle (based on our understanding).
It was also stated that the major axis had been shifted to the left.
Unfortunately till now I couldn't understand how the Major axis could shift to the left while we see that S2 cycle set a nice Symmetrical elliptical shape.
So, our scientists must give us full information about the real S2 orbital cycle and how it is positioned in space in order to set the expected major axis shift in the nice elliptical cycle that we see.
This is very important information.

 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/02/2019 19:05:51
It is stated:
"Until now, it was thought that S0-2 might be a double star. Two stars orbiting each other would have complicated the upcoming gravity test
But a team of astronomers led by Devin Chu of Hilo, Hawaii – an astronomy grad student at UCLA – has found that S2 doesn’t have a companion"
The question is: Why our scientists consider that there are two orbiting stars?
That's kind of the default state.  Most stars come in pairs or larger groups.  The lone star like our own is in the minority.  So until verified one way or the other, a given star is likely to be in orbit about one or more companions.

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Could it be that they also have thought on a possibility that based on the real verifications, S2 should orbit around some other object as I was expecting?
No.  The verification was finally done, and it turned up no other object.  S2 is by itself, or so says the article you quote.  Such a companion would have interfered with the smooth velocity change being measured, demonstrating relativistic red-shift.

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As I was expecting:
There is no need for any real companion star (or object). The VHP by itself can give a perfect explanation for what we see.
1) We don't see this motion you claim.  2) A VHP by itself has no mass and cannot affect the path of a star.  It is the mass of one or more companion objects that makes a star's path have a semi-regular deviation from a clean path.  S2 does not have a clean path since it passes by many other objects during its 15 year circuit, but none of those objects orbit S2, and S2 does not orbit them.

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It is also stated:
"[S2] orbits Sgr A* on an ellipse that takes about 15 years to complete. The diameter of its orbit is about 300 billion km [200 billion miles], which may sound like a lot, but we’re talking about a suppermassive black hole here! That’s close!
And it gets closer. Because the orbit is an ellipse, the star drops down to a mere 18 billion km [11 billion miles] from the black hole, a positively terrifying close approach. That’s only four times farther from the black hole than Neptune is from our sun."
I wonder if that information represents the S2 elliptical orbit cycle that we see.
If you're asking if that information describes the elliptical path we see from our vantage, then yes.

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It was stated that our vantage point is not perpendicular to the plane of S2's orbit.
I think I said that, yes.  The article didn't bother to mention that.  We're not perpendicular to any of the orbits plotted in that picture, but we seem to be nearly edge-on to at least one of them (S14).

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So, what we see doesn't represent the real S2 orbital cycle.
That we don't have a perpendicular vantage doesn't mean it isn't real.  Venus's orbit (or that of any of the planets for that matter) hardly looks like an ellipse from our vantage, but it is in fact nearly circular.  We very much can see its real orbital cycle without need for a perpendicular vantage.

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It was also stated that the major axis had been shifted to the left.
OK.  To be expected.  It is precessing, even more so due to the relativistic component of its orbit.  Even Earth does this.

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Unfortunately till now I couldn't understand how the Major axis could shift to the left while we see that S2 cycle set a nice Symmetrical elliptical shape.
It isn't a nice symmetrical ellipse.  You saw them trying to fit different curves to the data.  None of them was exact.  It is passing objects that deflect its path, as expected.  But the axis shift is probably more from precession than it is from random deflections from non-orbiting masses.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/02/2019 12:02:45
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So, what we see doesn't represent the real S2 orbital cycle.
That we don't have a perpendicular vantage doesn't mean it isn't real.  Venus's orbit (or that of any of the planets for that matter) hardly looks like an ellipse from our vantage, but it is in fact nearly circular.  We very much can see its real orbital cycle without need for a perpendicular vantage.
You do not answer the question!
What is the real Orbital cycle shape of S2???
What do you mean by: "We very much can see its real orbital cycle without need for a perpendicular vantage."
So, do you mean that what we see is almost the real orbital cycle of S2?
How can you set any sort of calculation on something that is almost correct?
OK.  To be expected.  It is precessing, even more so due to the relativistic component of its orbit.  Even Earth does this.
It isn't a nice symmetrical ellipse.  You saw them trying to fit different curves to the data.  None of them was exact.  It is passing objects that deflect its path, as expected.  But the axis shift is probably more from precession than it is from random deflections from non-orbiting masses.
The major axis shift is very dramatically.
How can you compare it to Earth?
Our scientists have to find the center of mass based on the orbital cycle of S2 (by ignoring the SMBH) and then find how far is it from the SMBH.
Than they have to prove by calculation, that without any need for a perpendicular vantage, we can move the calculated center to the location of the SMBH.
Sorry - you try to offer a solution to a problem without any real calculation.
I would expect from our scientists to offer real solution.


2) A VHP by itself has no mass and cannot affect the path of a star.  It is the mass of one or more companion objects that makes a star's path have a semi-regular deviation from a clean path.  S2 does not have a clean path since it passes by many other objects during its 15 year circuit, but none of those objects orbit S2, and S2 does not orbit them.
After all of our discussion, it seems that you have no clue what is the real meaning of the VHP.
So, let me explain it again for you:
The VHP is the center of mass for a star in a multi mega star system
Each star has a unique VHP.
As you have already explained:
"Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length."
So, the VHP is a virtual point which represents the center of mass of multi star system.
There is no real object there.
But it is a virtual point in space which which is used as the center of mass for a specific star.
Therefore, each star in the galaxy (which is under a Multi star system) must have a unique VHP for itself.
Back to S2:
S2 orbital cycle is directly affected by the nearby multi star system.
Therefore, S2 must orbit around its unique VHP, while this VHP might orbit around the center of the galaxy or the SMBH.
Why is it so difficult for you to understand my simple message?
You can accept it or reject it.
But please, try to understand my explanation.
If you understand the explanation, let's see if it is feasible:
You claim that :
"S2 does not have a clean path since it passes by many other objects during its 15 year circuit, but none of those objects orbit S2, and S2 does not orbit them."
How can we verify if S2 orbits around other objects?
When we normally discuss about a multi star system - we mainly think about two, three, four...or maximal 16 stars.
If that was the case for S2, than we could easily find the other stars in the Multi star system.
However, in the center there are much more than few stars or even few thousands of stars.
Therefore, each S stars that we see must orbit around a unique VHP, while this VHP orbits according to the Multi Mega star system.
So, that multi mega star system is affected by all the nearby stars, by the SMBH and even by the 3KPC RING!!!
Yes, I do believe that this ring also has an important impact on the orbital cycles of the stars in the center.
Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
Each S star orbits perfectly based only on real mass.
It is clear to me that you disagree with my explanation.
So, please - there is no need to argue about it.
The key point for me is to deliver the message about the VHP.
I hope that by now you understand what I mean by that word - VHP.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/02/2019 14:34:45
Halo is a perfect example for the Multi mega star system:
https://scitechdaily.com/astronomers-make-shocking-discovery-about-stars-around-the-milky-way/
"These halo stars are grouped together in giant structures that orbit the center of our galaxy, above and below the flat disk of the Milky Way."
So, what kind of force keeps them together? How many stars there are in those giant structures?
It is clear to me that they all group together due to internal gravity force.
So, based on multi mega star system (as halo stars), each star must orbit around all the other stars, but it is impossible to verify a single orbit around a specific star or group of stars.
Therefore, if we will try to trace one star there, we will not be able to see that it orbits any other star (or group).
In the same token, S2 orbits due to the impact of all the mass in the center - (Multi mega stars system or giant structures), while we will not be able to verify that it actually orbits around any specific stars.

We call those giant star structures as halo stars because they have ejected from the disc of the spiral arm.
Actually, as long as they are in the disc, they must be connected to one of the spiral arms.
Let's look again at the Milky way diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
We see clearly that at the end of each spiral arm there are many points. It looks as a broken chain.
I assume that in each point of the chain there are Millions of stars.
Those millions of stars are grouped together due to gravity.
Each point in this chain set a local giant gravity.
Please look at the last chain in each arm.
This last chain holds itself to the one in front by gravity force.
As long as it is connected to the arm, it is part of the arm, It stays at the galactic disc plane and it follows the orbital path of the spiral arm.
However, it will not have the power to hold itself for long time.
The orbital velocity at that point is too high for the balanced gravity force in that last point of giant star structure.
Sooner or later, this last point in the chain will have to be disconnected from the arm.
However, once it is ejected from the arm, it is also ejected from the disc plane.
So, S2 which had been created at the core of the molecular cloud at the center of the galaxy must drift outwards over time.
One day S2 will arrive to the same radius from the center as the solar system.
At that time, our sun might get to the last point in the arm orbital chain.
Then, we will be ejected from the arm and be part of the Halo stars....

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/02/2019 23:58:32
You do not answer the question!
What is the real Orbital cycle shape of S2???
Something approximated by an ellipse, about 970 AU long and around 450 wide.  That's the actual shape, not what it looks like from here.  Like all orbits, it precesses, so it doesn't trace the exact same path each time around, even without other objects perturbing it.

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What do you mean by: "We very much can see its real orbital cycle without need for a perpendicular vantage."
I was talking about the orbit of Venus, which looks, from our vantage, like an almost flat ellipse with the sun nowhere near either foci.  That orbit is quite real, and yes, we see it, so if you mean something else when you ask if we see the real orbit, I don't get your question.
The image of S2 is delayed by over 25000 years, so in that sense it isn't real.  Maybe the thing has since been eaten and isn't real at all anymore.  Betelgeuse has the same problem.  Never sure if it is really there or we're just seeing an afterimage of a nonexistent star.

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How can you set any sort of calculation on something that is almost correct?
Tell that to the weather prediction guys.  To be completely correct, you need a full description of its state.

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The major axis shift is very dramatically.
How can you compare it to Earth?
Earth orbit is nearly circular.  OK, S2 is not as eccentric as say a comet, but it also has a relativistic orbit, and that might effect major axis shift.  I don't know how to compute that.  I know that relativity very much plays a role because they never were able to predict Mercury's orbit until all GR effects were accounted for.
S2 is also passing random objects, any of which can send it onto a new trajectory, a different ellipse.  That seems to happen multiple times per orbit, as shown by some of the pages you've been linking.

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Our scientists have to find the center of mass based on the orbital cycle of S2 (by ignoring the SMBH) and then find how far is it from the SMBH.
Sorry, but I cannot figure out this statement.  What center of mass are they seeking?  S2?  Sgr-A?  Something else?  S2's CoM is where S2 is.  Ditto for Sgr-A.  Neither seems to be a binary object.

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Sorry - you try to offer a solution to a problem without any real calculation.
Not sure what I was asked to calculate.  I certainly don't see any calculations from you, the guy who expects a different idea to be taken seriously.



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After all of our discussion, it seems that you have no clue what is the real meaning of the VHP.
So, let me explain it again for you:
The VHP is the center of mass for a star in a multi mega star system
Each star has a unique VHP.
Why wouldn't all the stars in said mega-star system have the same VHP?  How does any particular star know which stars to include in the VHP calculation and which to leave off?  Our sun for instance has stars visible in every direction.  So how might I begin to guess which of those contribute to the VHP and which do not?  Most importantly, why would you expect motion about this VHP to be anything resembling something close to a familiar elliptic orbit?

Anyway, I don't deny that our star is potentially part of a group, and that the group sort of travels as a clump in a cleaner path than any individual mass.  I just don't think our motion within that group can be characterized as being in orbit around the group's center of mass or any other virtual point.  That only works for 2 body systems, not larger ones.
For even a simple 3-body system, barring symmetrical arrangements, none of the three biodies will accelerate towards the common center of mass.  In the most simple case of 3 equal masses with unequal spacing in a line, the middle mass will accelerate away from the common center of mass.  You seem to realize none of this.

The only VHP I see is the center of the galaxy.  We also are influenced by the disk, but forces from a disk are not characterized by an orbit.  The force of the disk actually grows with distance (to a point at least), unlike the force from a given point mass.  I think the disk force contributes to that 62 million year cycle thing, but I'm no expert.
The in/out motion you showed in one of your pictures might be due to the solar system having an eccentric orbit about the galactic center. It would have a ~200 million year period if it was that. Few objects seem to have really low eccentricity trajectories.

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So, the VHP is a virtual point which represents the center of mass of multi star system.
I've already accepted this general idea, but in general, all stars in that group have essentially the same VHP then.

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There is no real object there.
But it is a virtual point in space which which is used as the center of mass for a specific star.
It is the center of mass for all the stars in the set.  The center of mass for a specific star is the star itself.
You really need to choose whether you're talking about a virtual center of mass of a set of multiple objects, or the center of mass of a particular object, which tends to be the middle of said object.

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Therefore, each star in the galaxy (which is under a Multi star system) must have a unique VHP for itself.
Yes, called its location.  That point isn't virtual, since the star is there.
Venus's center of mass is at its center, but Venus loosely orbits the center of mass of the solar system, and more accurately orbits the sun itself, but still not a perfect orbit around it.  No object's path can be accurately described by a clean orbit about a VHP which in turn orbits either the solar system or the sun.

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Back to S2:
S2 orbital cycle is directly affected by the nearby multi star system.
What multi-star systems would that be?  There are other stars in that area, but none seem to form a system with S2, and nobody knows how many of these nearby objects are part of multi-object systems (stay with each other in their primary orbit).

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Therefore, S2 must orbit around its unique VHP, while this VHP might orbit around the center of the galaxy or the SMBH.
They've actually eliminated this option as a possiblility.

Your logic is horrible.  S2 is affected by other masses, therefore it must orbit something other than its primary (the SMBH).  That simply doesn't follow.  Venus, for example, is affected by other masses (I cannot think of a mass that doesn't affect it), but it does not orbit any VHP.

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Why is it so difficult for you to understand my simple message?
You can accept it or reject it.
I understand the message, but since it is based on nonsense, I cannot accept it.

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How can we verify if S2 orbits around other objects?
I already told you this.  It would have a period to that orbit.  Orbits have periods.  Do the math that detects regular periods in the data. Making baseless assertions is not the way to go about it.

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When we normally discuss about a multi star system - we mainly think about two, three, four...or maximal 16 stars.
I can think of much larger systems than that.

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If that was the case for S2, than we could easily find the other stars in the Multi star system.
However, in the center there are much more than few stars or even few thousands of stars.
Therefore, each S stars that we see must orbit around a unique VHP, while this VHP orbits according to the Multi Mega star system.
So, that multi mega star system is affected by all the nearby stars, by the SMBH and even by the 3KPC RING!!!
Yes, all of this stuff has an effect, but none of it is a system that orbits as a unit, else all the other objects in the system would have the same 16 year period around Sgr-A, but we don't see that.  We see not one other object travelling with S2, else S2 would have a regular perturbation to its orbit.  You keep asserting a pattern that just isn't being seen.  Not saying it isn't possible, just that it isn't the case with this particular star.

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Yes, I do believe that this ring also has an important impact on the orbital cycles of the stars in the center.
What impact would that be?  I mean, I agree there is one, but I suspect you haven't thought it through.  What impact would a ring have on Earth's orbit for instance?  Suppose it is out at the asteroid belt, but having the same mass in relation to the sun as the galactic ring does to our SMBH.  Our actual asteroid belt has almost no mass compared to the sun, so we'd need to imagine a more massive one.

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Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
But you haven't taken anything into account. You just imagine that if you did, this S2 orbit (and that of our own solar system for that matter) would fit. Imagining it isn't evidence at all.  You need to actually do the work and show how your idea makes the data fit better than the misguided ideas of all the professional astronomers who have to posit objects that cannot be seen in order to get the numbers to work.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/02/2019 00:32:00
Halo is a perfect example for the Multi mega star system:
https://scitechdaily.com/astronomers-make-shocking-discovery-about-stars-around-the-milky-way/
"These halo stars are grouped together in giant structures that orbit the center of our galaxy, above and below the flat disk of the Milky Way."
So, what kind of force keeps them together? How many stars there are in those giant structures?
The article doesn't say if they're kept together or still spreading apart or whatever.  It doesn't give a count either, but I'm sure the count depends heavily on what sort of object counts as a star and what doesn't.  There's not exactly a clean line defining that.

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It is clear to me that they all group together due to internal gravity force.
If they stay together, then yes, gravity seems a lot more plausible than rubber bands or something.

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So, based on multi mega star system (as halo stars), each star must orbit around all the other stars
Yet again, that doesn't follow.  Each star may tend to stay with the group, but it may also be ejected by chance.  The motion within the group is not necessarily anything that resembles the elliptical path of an orbit.

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In the same token, S2 orbits due to the impact of all the mass in the center - (Multi mega stars system or giant structures), while we will not be able to verify that it actually orbits around any specific stars.
They've verified that it doesn't orbit any other specific stars.

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We call those giant star structures as halo stars because they have ejected from the disc of the spiral arm.
Actually, as long as they are in the disc, they must be connected to one of the spiral arms.
Structures between the arms are still in the disk, so this is false.

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Let's look again at the Milky way diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
We see clearly that at the end of each spiral arm there are many points. It looks as a broken chain.
Those broken lines are extrapolated arms, places that we cannot see at all.  They're not broken, but they're guesses as to what's there.  It is reasonable to assume the arms continue through the regions unseen, just like we can see in other galaxies.

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Please look at the last chain in each arm.
This last chain holds itself to the one in front by gravity force.
I'm sorry, but you're asking me to consider gravity of a 'can't see it' designation on a diagram.  This makes no sense.  Those are not real clumps of stars that have been observed like that.  Read the text accompanying the diagram.

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As long as it is connected to the arm, it is part of the arm, It stays at the galactic disc plane and it follows the orbital path of the spiral arm.
You're treating arms as objects, not as what they probably are, which is waves.  The arms do not move with the stars, else they'd have been smeared out a long time ago.  Our sun (and any star) is connected to no arm, and moves from one to the next as it orbits the galaxy at an entirely different rate than the apparent motion of the density waves that form the arms.

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So, S2 which had been created at the core of the molecular cloud at the center of the galaxy must drift outwards over time.
One day S2 will arrive to the same radius from the center as the solar system.
It is a big star and even assuming it will drift like that, it will burn out long before it gets anywhere.  The big stars nearby (e.g. Betelgeuse) similarly could not have come from the center since they're far too young to have made the trip.  Either that or the big stars must drift outward at an incredible pace, making one wonder why S2 isn't getting any further away with each orbit.

Not claiming to know the answer myself.  The bar is a known stellar nursery, but I don't understand the dynamics of the bar.  S2 seems not to be part of that because it doesn't fall in that rotation curve of the galaxy that we were looking at.  I don't know anyone that can explain that curve and how it fits with named objects like S2.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/02/2019 07:19:58
The bar is a known stellar nursery, but I don't understand the dynamics of the bar.
You don't know, but I know it for sure.
The center of the galaxy is the stellar nursery for most (or even all) the stars in the galaxy/Universe!!!
A star can be created ONLY in a Molecular gas cloud which orbiting around/near a massive object as a SMBH.
So, 99.9..9% of the stars in the Milky way galaxy had been created in that stellar nursery.
Each star have got all its Planets and moons in it's first day.
Therefore, the age of the Earth & Moon is identical to the age of our Sun.
All the objects in the Solar system had been created from the same matter at the same day.
The Earth and the Moon had been also created as a compact molecular gas object.
Their mass in the first day was heavier by more than 98 times than it is today.
The planets and moons in the solar system have got all the Water supply in their first day from the molecular gas cloud.
Over time, due to their compact size, they have lost most of the gas as Hydrogen and became rocky objects.
There was a time when water flows at the surface of Mars.
At that time, Mars was orbitting the Sun at the same radius as we are today and the Earth was much closer to the Sun while it's temp was quite higher.
The big stars nearby (e.g. Betelgeuse) similarly could not have come from the center since they're far too young to have made the trip.
As I have stated, all the stars (Yes - including Betelgeuse) had been formed at the center of the Milky Way galaxy.
Betelgeuse isn't a young star. It is as old as our sun.
You have a severe misunderstanding about the age of the stars in the galaxy.
We could set the calculation how long it might take to a star to drift from the center of the galaxy all the way to our current location.
Therefore, all the stars in the center are young stars.
As we move outwards, the stars get older.
So, there is a severe error in the way that our scientists estimate the age of the stars in the galaxy.
If they stay together, then yes, gravity seems a lot more plausible than rubber bands or something.
Thanks
Gravity keeps all the stars in the Halo stellar. In the same token, gravity keeps all the stars in spiral arms.
You're treating arms as objects, not as what they probably are, which is waves.
Yes, spiral arms are objects. All the stars in the arm are connected to each other by gravity force. The same gravity force which keeps the stars at the halo stellar also keeps the stars at the spiral arms.
However, as I have stated before, there are bridges and branches between the Arms.
Therefore, stars can migrate from arm to arm by those bridges and branches.
We can think about an arm as a highway while there are bridges between the arms.
Each star may tend to stay with the group, but it may also be ejected by chance.
That is correct!
As long as the star stay at the arm or cross to the next arm by bridge, it will save its location in the galactic disc.
However, if it will dare to move away from the arm or the bridge, it will be ejected as a rocket from the galactic disc for good.
Nothing can come back to the arm or the galactic disc.
why S2 isn't getting any further away with each orbit.
Yes, it is!
However, it is very difficult to monitor that drifting outwards.
S2 orbits around its unique VHP. (It is a host point - sorry if I used before the idea of center of mass).
So, S2 set a very clean orbit around its VHP while it increases its radius by only few cm or few m. per cycle.
However, this represents the S2 orbit around its orbital motion.
Let's look again at the following diagram of the Sun Motion:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
In the same token, S2 apparent motion (blue dot points) orbits around its orbital motion (gray dashed line).
Please, you don't have to agree with that, just try to understand my message.
So, S2 orbits around a first level of VHP (let's call it VHP1) and set that nice apparent motion (Blue dot point)
Therefore, when we trace S2, we see that it doesn't move exactly on its orbital cycle:
http://www.phy6.org/stargaze/Kep3laws.htm
However, for this big orbital cycle (15 years), there is second level of VHP (Let's call it VHP2).
Therefore, if we assume that In each cycle, VHP1 drifts from VHP2 by few Km per cycle, than VHP2 drifts away from the SMBH by much more than that per cycle.
So far we have mainly discuss on this VHP (VHP2) but we have ignored the first VHP level (VHP1).
So, any star in the galaxy must orbit around a VHP1 which follows with him anywhere from its first day one in the Universe.
Therefore, when we look at the nearby stars, we shouldn't trace the physical location of the stars, but we must monitor the virtual locations of their VHP1.
If we do so, we should see that the distances between all the nearby VHP1' stars in the Orion arm are absolutely fixed!!!
So, all the VHP1 of the stars in the Sun' nearby area are moving exactly at the same velocity!!!
Therefore, no one is moving away from the arm!

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/02/2019 09:13:37
Quote
What is the real Orbital cycle shape of S2???
Something approximated by an ellipse, about 970 AU long and around 450 wide.  That's the actual shape, not what it looks like from here.
Thanks!
So, the ratio between the major axis to the shorter axis is about 1:2
I have just found a fantastic article:
https://www.slideshare.net/NUCLIO-PT/development-and-implementation-of-student-activity-to-find-mass-of-black-hole-by-naoki-matsumoto
Please look at at slide pg. 18
It is stated that the orbital inclination angle as viewed from the Earth of S2 is 45 degree. While after the correction to 90 degree we get the final results at the attached image.
However, it is quite clear to any first year student that the SgrA* (SMBH) can't be used as the host for this orbital cycle. No way!!!
Now, please look at slide in pg 20.
The calculated Periastron date is year 2002.25 + 0.05 = 2002.3
The calculated Apastoron date is year 1994.62
The full one orbital cycle Is:
P = 15.4 Year.
Based on that data, they have found the mass of the host should be (in pg 22)
3.4 * 10^6 Sun Mass.
So far so good.
I fully agree with this calculation!
However, they also claim that after the adjustment (which they do not explain), The estimated value of the SMBH mass is
SMBH = 4.1 * 10 ^6 Sun mass.
That is a severe mistake!!!
As I have stated, S2 does not orbit around the SMBH.
S2 orbits around its VPH1 which orbits around VHP2 which orbits around the SMBH (assuming that there are no more stages in between).
Therefore:
VHP2 (of S2) = 3.4 * 10^6 Sun Mass.
Now, we have to find the exact relationship between VHP2 and SMBH.
First, based on Kepler, we have to find the perfect location of VHP2 in that S2 orbital cycle of 15.4 years.
It is clear to me that it should be located high above the current location of the SMBH.
Than we have to find how long it takes to VHP2 to set one full cycle around the SMBH and the cycle shape (VHP2' Periastron and  Apastoron)
Just after getting this information we can extract the real value of SMBH.

There is a solid prove for my explanation.
Please look at the following article:
https://alchetron.com/S2-(star)#demo
https://alchetron.com/S2-(star)
If we look carefully, we should see that S2 doesn't set a full nice cycle. At the top of this image we clearly see that there is a shift in the orbital cycle (as an open loop).
This shift is a direct outcome of the orbital motion of VHP2 around the SMBH.
Therefore, The SMBH can stay in its place while VHP2 (of S2) orbits around it and set the same famous orbital motion as the Sun does.
Please look again at the following image:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
Hence, S2 will never ever close one full cycle. It moves forwards due to the orbital motion of VHP2 around the SMBH and set some sort of open loop.
Based on this small shift we should find the orbital velocity of VHP2 around the SMBH.
This might help us to estimate the size of this orbital cycle and extract the real value of the SMBH.





Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/02/2019 22:28:30
Please look at at slide pg. 18
It is stated that the orbital inclination angle as viewed from the Earth of S2 is 45 degree. While after the correction to 90 degree we get the final results at the attached image.
However, it is quite clear to any first year student that the SgrA* (SMBH) can't be used as the host for this orbital cycle. No way!!!
I agree.  The rotation was not done on the correct axis, since if it was done correctly, SgrA would have been at the focus. 

Quote
Now, please look at slide in pg 20.
The calculated Periastron date is year 2002.25 + 0.05 = 2002.3
The calculated Apastoron date is year 1994.62
The full one orbital cycle Is:
P = 15.4 Year.
Based on that data, they have found the mass of the host should be (in pg 22)
3.4 * 10^6 Sun Mass.
Mass of host cannot be determined from just orbital period.  Jupiter has a sort of similar orbital period and yet orbits something of far less mass.  Anyway, yes, they measured the time for half an orbit and figured that the other half takes similar time.  Don't need to be a genius to figure that one out.

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I fully agree with this calculation!
They didn't show their work.  There is a reference to a 'equation 4'.  The only math done on that slide is a conversion of units.

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As I have stated, S2 does not orbit around the SMBH.
S2 orbits around its VPH1 which orbits around VHP2 which orbits around the SMBH (assuming that there are no more stages in between).
Therefore:
VHP2 (of S2) = 3.4 * 10^6 Sun Mass.
Quote
Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
You just posited a VHP2 of mass similar to SgrA, which is a serious boat load of dark matter.  You are contradicting yourself.
I think they would notice if S2 was orbiting a second object that massive, but you seem to have other ideas.  How far away is VHP1 from this VHP2?  What period?

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Now, we have to find the exact relationship between VHP2 and SMBH.
They're nearly the same mass and should be orbiting each other.  Other objects should be orbiting the combined mass of the two.

Quote
First, based on Kepler
You can't really quote Kepler.  You've denied his laws completely in the prior post, and Kepler's laws only apply to two-body systems and that isn't the case here.

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we have to find the perfect location of VHP2 in that S2 orbital cycle of 15.4 years.
It is clear to me that it should be located high above the current location of the SMBH.
Which direction is 'high above'?  There's not exactly a clear direction that is 'above' out there.
You're just talking about a distance?  How much is 'high above' then?  How is SgrA not high above this VHP2, since it is so massive?  How far is VHP1 from VHP2?

Quote
There is a solid prove for my explanation.
Do you have even the slightest idea what constitutes evidence, let alone proof?

Quote
Please look at the following article:
https://alchetron.com/S2-(star)#demo
https://alchetron.com/S2-(star)
2nd link is dead.

If we look carefully, we should see that S2 doesn't set a full nice cycle. At the top of this image we clearly see that there is a shift in the orbital cycle (as an open loop).[/quote]
Yes.  Nobody claimed its motion is Keplerian, and this orbit also has a significant relativistic component.

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This shift is a direct outcome of the orbital motion of VHP2 around the SMBH.
So you assert, but where's the proof of that?  You claimed there was a proof somewhere in it.

Quote
Please look again at the following image:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
This image in no way reflects reality.  Nothing is to scale and the periods depicted do not reflect measured values.
Best I can tell from research, the 5-7 km/sec component has that 62 million year cycle (up and down about 3 times per trip around the galaxy, due to disk effects), and the 20 km/sec motion is one cycle per orbit, due to our eccentricity.  The blue dotted line is almost parallel with the grey line if it was drawn more correctly.

If we actually followed the blue dotted line as depicted (which is about 9x the length of the grey dashed line) and the 217 km/sec motion of the VHP was accurate, the sun would be moving at about 2000 km/sec, well above the escape velocity of the galaxy.  We'd just permanently shoot away.

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Hence, S2 will never ever close one full cycle. It moves forwards due to the orbital motion of VHP2 around the SMBH and set some sort of open loop.
I cannot figure out what you mean to convey with this description.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/02/2019 06:11:24
Quote
As I have stated, S2 does not orbit around the SMBH.
S2 orbits around its VPH1 which orbits around VHP2 which orbits around the SMBH (assuming that there are no more stages in between).
Therefore:
VHP2 (of S2) = 3.4 * 10^6 Sun Mass.
Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
You just posited a VHP2 of mass similar to SgrA, which is a serious boat load of dark matter.  You are contradicting yourself.
I think they would notice if S2 was orbiting a second object that massive, but you seem to have other ideas.  How far away is VHP1 from this VHP2?  What period?
Quote
Now, we have to find the exact relationship between VHP2 and SMBH.
They're nearly the same mass and should be orbiting each other.  Other objects should be orbiting the combined mass of the two.
Dear Halc
After all our discussion, it seems to me that you still don't understand the real meaning of VHP.
Please, you don't have to agree with my explanation - but you have to understand my message.
So, let me explain it one more time for you:
1. VHP = Virtual Host Point
2. VHP is not a real object. It is a virtual object.
3. VHP1 - As I have already stated - ALL stars in the galaxy had been formed in a gas cloud near the SMBH. Each new born star must orbit around a virtual host point (VHP1) in the gas cloud in order to crystallize from molecular gas into real Star.
Therefore, the first level of virtual point (VHP1) is some sort of a gift which had been given to any star in the galaxy from the gas cloud in his first day. Hence, once the Star is out from the gas cloud its physical location is not relevant. We have to focus only on the Virtual point of his host (VPH1). Each star in the galaxy has a unique VHP1. This VHP1 goes with it anywhere it goes.
4. VHP can work over VHP. For example - The Moon orbits around the Earth, while The Earth orbits around the Sun. (Technically, the center of the Earth/Moon orbits around the Sun, but for this discussion, we can assume that the Earth orbits around the Sun. So, we see two levels or orbital cycles. In the same token - S2 orbits around VHP1, while this VHP1 orbits around another virtual host which is called VHP2. This VHP2 can technically orbits around other level of virtual host point - VHP3... and so on...
Therefore, if we focus on those Virtual points, we won't find any real object or any dark matter. All of them are virtual points. However, each Virtual Host Point represents mass. We can calculate the estimated mass of each VHP directly from the star mass and its orbital movement.
5. Dark matter - There is no dark matter and there is no need for dark matter. It is a pure fiction due to the simple outcome that our scientists have failed to understand how gravity really works at the galaxy.
6. Planets and Moons - They are all "by products" due to the process of  "new star creation" in the gas cloud. I can explain it later on if you wish. In any case, by definition - Each star in the galaxy gets also planets and moons as another gift from the gas cloud.

Final conclusion:
When we try to look at any nearby star, we shouldn't focus only on its real location and its mass. We have to focus on the location of its VHP1. Hence, by monitoring the movement of each star (physical location) in the galaxy, we can extract the location of its VHP1 and calculate its host mass (based on orbital cycle of the star and its mass).
From now on, this virtual host point and the calculated mass represents the first level of the orbital cycle of any real star in the galaxy. This is the most critical issue in the galaxy!!!
The shape of spiral galaxy is a direct product of VHP1 which is based on real object.

Is that all clear to you?
Please, let me know if you understand the message. Again, you don't have to agree with that. Just let me know if finely you understand the real meaning of VHP.

   
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/02/2019 12:04:18
Take 3 equal masses distributed in a triangle of sides 3, 5, and 7.  From that, you can find the center of mass of the three which is the same for all three of them.  We can call this VHPc.  In addition, each object X Y and Z has a sort of VHP that is halfway between the other two objects

That is good example.
So, it seems that you understand the meaning of VHP.
However, instead of three object, we must think on thousand, Millions or even billions objects.
Not just stars, but also gas clouds, SMBH and any other real object.
There is no dark matter in the galaxy.
This is a pure science fiction.
Our scientists had failed to understand how spiral galaxy really works, therefore, they came with this none realistic idea.
Based on the VHP idea we can get the spiral galaxy shape without any need for dark matter.
The motion of none of the objects (I have not given their velocities) cannot be modeled by any number of tiered VHP's, at least not according to Newtonian physics.
I disagree with that statement.
VHP over VHP is the key element in spiral galaxy which set the structure of spiral arm.
Please look again at the following image:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Please focus on the last chain point in the red spiral Arm (for example)
In this last chain point there might be thousands or millions stars.
As long as it is connected to the chain point in front, it will follow the orbital momentum of the spiral arm.
However, once it disconnected from the point in front, it will soon drift away from the disc plane.
Our scientists, ignore completely the meaning of the disc plane.
They think that this last chain point (with all the millions objects) orbits around the whole mass inwards the orbital cycle of that point. They have found that the real mass in the galaxy can't support its orbital velocity.  Therefore, they have got into conclusion that dark matter is needed.
As explained in the following:
https://www.quora.com/How-is-the-motion-of-stars-similar-to-the-motion-of-the-sun
"The stars don't reduce in orbital velocity as much as they should be doing... unless there's more mass than we can see.  Behold: meet the mysterious dark matter that must be responsible for this.  More mass due to dark matter can explain that."
That of course is a fatal mistake.
They actually ignore completely the disc plane.
Why suddenly at the end of the spiral arm the stars do not stay any more at the galactic disc?
in that article they show the different orbital cycles:
Disc star orbital - Yellow
Halo star orbits - Green
Bulge star orbits - Red.
So, only the Yellow represents the orbital star at the disc plane.
The dark matter can't give a clear explanation for that phenomenon.
Why suddenly at the end of spiral arms, the stars starts to orbits high above and below the disc (higher than the estimated disc width of about 1KLY.)
So, let me explain how it really works.
Let's assume that in this point there are 10,000 stars.
We see clearly that those stars are connected to each other.
Gravity force is the only force that can keep them all together.
In this activity, they all set a VHP which represents the center of mass of all this last chain point.
Let's call it VHP - m1 (m1 represents the total mass in the last chain point).
Now, for each point in this chain there must be also a represented total mass.
So, the one in front will be called - m2, the other one in front will be called m3 and so no.
Hence, we can see a long chain of points that are connected to each other.
m1 is connected to m2.
m2 is connected to m3.
m3 is connected to m4
mn is connected to mn+1.

However, when I say connected I mean gravity force.
In order to set a constant gravity force we must set an orbital cycle.
So, we actually have the following:
VHP m1 orbits around VHP m2.
VHP m2 orbits around VHP m3
VHP mn orbits around VHP mn+1
However, how could it be that we don't see any orbital cycle?
Because, the orbital velocity of VHP m1 around  VHP m2 is (almost) identical to the orbital velocity of VHP m2 around the galaxy.
Therefore, the orbital velocity of VHP mn around  VHP mn+1 is (almost) identical to the orbital velocity of VHP mn+1 around the galaxy.
I specifically say almost identical, as there is a room for this point to drift outwards. (As I have already explained - in real orbital cycles - orbital objects drift outwards from their host)
This is the basic idea of spiral arm and how VHP works over VHP.
However, at some point, the orbital velocity of VHPm1 can't trace any more the orbital velocity of VHP m2 around the galaxy.
Sooner or later, this last chain point should be disconnected from the spiral arm.
So, I give a clear explanation why the last chain point at the end of the spiral arm should be disconnected from the spiral arm and move away from the galactic disc in order to set the halo star.
The dark matter can't explain this phenomenon.
Do you agree that if the dark matter idea was correct, than the disc plane - or the spiral arm had to be continued further more (theoretically - to the infinity.)
I don't see any explanation why there is an end for the spiral arm based on this dark matter fiction.
So, I hope that by now you understand the idea of VHPmn over VHPmn+1 in the spiral arm which is needed to hold the stars in the arm (and on the disc plane).
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/02/2019 17:59:28
Rotation curve problem
Now we need to understand how could it be that the stars are connected to each other while we get the following rotation curve:
https://en.wikipedia.org/wiki/File:M33_rotation_curve_HI.gif
This is the rotation curve of M33 galaxy, but any spiral galaxy should have a similar rotation curve (more or less).
So what do we see.
Please focus only the observation graph.
The expected graph is none relevant as they assume that each star orbits around the galaxy by its own.
However, I claim that the stars are connected to each other.
So, if we think about a rigid disc, it is clear to us that the orbital velocity of a star at 20KLY from the center should be much higher than a star at a distance of 10KLY.
Actually as the radius in 20K is twice longer than in 10K the orbital ring is longer by:
2^2 = 4
Therefore, in a rigid disc, in order to keep itself in the arm, the orbital velocity at 20K should be four time faster than at 10K.
So, how can we explain this problem?
The answer is quite simple.
Spiral galaxy isn't a rigid disc.
The stars in the arm are connected to each other, but they also drift outwards.
Let's monitor few points:
At 5KLY - the orbital velocity is 70 Km/sec
At 10K - 90 Km/sec
At 20K - 110 Km/sec
So, let's assume that at t=0 we trace three stars at those located radius from the center.
If we will try to monitor the location of the stars after T
t=T
We should see that all of them had been drifted outwards.
Let's assume that T represents the time that is needed for a star to drift from 5KLY to 10KLY.
By this time there is high possibility that a star that was at 10KLY had been drifted to 20KLY.
due to the shape of the spiral arm, we should find that the total distance that the first star had to move from 5KLY to 10KLY might be a little bit shorter that the other star which had been drifted from 10KLY to 20KLY.
Therefore, there is no significant change in the orbital velocity as we trace stars that are located further away from the center (although - they are connected by spiral arms).
As I have already explained, new stars are formed in the center.
Those new stars replace the other stars which had been drifted out from in the spiral arms.
Therefore, at any given moment we see that all the arms are full with stars.
Therefore it is clear to me that our Sun is only temporary located at the current location.
As it drifts outwards, sooner or later it will get to the end of the spiral arm and then it will be ejected from the arm and the galactic plane.
So, the spiral galaxy acts as the biggest star sprinkler in the Universe.
Therefore, we see that for any star in the galaxy there is at least one outside.
Hence - all the stars that we see around our galaxy - in the Halo, in the dwarf galaxies... - all of them had been ejected from the galaxy.
One day we will be there too.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/02/2019 05:41:52
Then show me with the above example.  You disagree because you haven't tried it and verified that your idea predicts different motion than does Newton's laws.
With Pleasure:
Please look at the following article:
https://www.livescience.com/63698-milky-way-spiral-wobble.html
It is stated:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern that has only become more tangled over the past eon."
So, this short description fully supports my theory. (Please focus only on what they see and ignore their none realistic explanation for that key evidence!)
It is stated: " these rogue stars also orbit around one another" - So as they orbit around one another they set a center of mass which I call it - VHP.
Hence, the example had been given by a solid verification of our scientists - as you have requested.
However, it is also stated: "these rogue stars also orbit around one another in a wobbly, spiral pattern".
What does it mean "Wobbly"?
Why the stars are wobbling?
This is a solid evidence to my explanation that any star in the galaxy orbits around its unique VHP1.
The wobbling activity is a direct outcome of VHP1!!!
This VHP1 is the first element for Newton activity in the galaxy.
The Sun is wobbling. I have already proved it by an article.
All the nearby stars are wobbling. S2 star in the center of the galaxy and many others are wobbling as I have proved by several articles.

However, it is quite clear to me that you are going to reject also this evidence.
So, what is your mission?
Do you really want to understand my message or do you just want to prove that whatever I say is incorrect?
Why don't you even try to understand the deep explanation which I have offered?
Why do you constantly reject any explanation, example, evidence and article which I have offered so far?
Why don't you answer my questions about the problems in using the dark matter concept?
I have found the explanation for every bit in our spiral galaxy/observable universe/real universe without any need for dark matter or dark energy, as our universe is working according to very simple roles.
So far I have introduced just the first steps from the whole theory.
Unfortunately, you force me to explain the same issue again and again and therefore we can't move on.

Hence - 
Do you have real willing to open your mind and heart to this breakthrough understanding?
Is there any chance for you to focus only on real evidences and verify how my explanation perfectly fits with those verifications/evidences?
Without it, we won't be able to move forward.



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/02/2019 14:01:10
Then show me with the above example.  You disagree because you haven't tried it and verified that your idea predicts different motion than does Newton's laws.
With Pleasure:
Please look at the following article:
A diversion right out of the gate.  Nothing in any article is going to tell me how to determine the location of the VHPs in the simple example I provided, unless the article discusses your idea, which this one doesn't.
The origin is the common center of mass of the three.  I say this because I was doubtful that you are capable of working that out for yourself.
Is that the VHP of all 3 objects?  I ask because you say each object has its own, like they're in different places.
Given the location of the VHP, I can predict motion if I know how to compute the attraction between an object and its VHP.  You've not provided that either.  No website is going to give this answer because your idea is accepted on any website.

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However, it is quite clear to me that you are going to reject also this evidence.
I don't deny that binary systems have stars that orbit each other.  But the site just doesn't answer the question I asked.  You have to answer it, because your physics is not discussed anywhere outside this very long topic which unfortunately has no actual theory described so far.  I am trying to get to the actual theory instead of just a list of anomalies that you claim will be solved.

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So, what is your mission?
I want to know how your theory works.  I gave the most trivial example I can think of to make it easy for you.  But apparently you can't do it.  You have no actual theory.  Every question is met immediately with a diversion.

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Why don't you even try to understand the deep explanation which I have offered?
You've never given a deep explanation.  If you had, I'd have been able to answer my question.
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Why do you constantly reject any explanation, example, evidence and article which I have offered so far?
Those sites are not pushing your idea, so none of them help clarify your idea.

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Why don't you answer my questions about the problems in using the dark matter concept?
You claim a solution to that problem, but without knowing your theory details, I cannot comment on how well it does that. 

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So far I have introduced just the first steps from the whole theory.
You could not answer my question.  That seems to count as having provided a preface to the theory, but not even the first page of the actual theory.

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Unfortunately, you force me to explain the same issue again and again and therefore we can't move on.
No!  Don't move on.  Answer the question!  Where is the VHP of each object in the example, and how does an object's motion relate to that of its VHP?  Newton provided answers like this, where there is no VHP and objects act on each other via force computed by GMm/r².  From that info alone I can compute the Newtonian motion of those 3 objects.  I want to do so with your rules, but I don't know them.  If the paths are different, then your theory is non-Newtonian.  This is OK.  Actual theories are out there (such as MoND, or Modified Newtonian Dynamics) which reject the above formula at certain scales in order to not have to posit dark matter.  The MoND thing looks nothing like what you are describing.

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Hence - 
Do you have real willing to open your mind and heart to this breakthrough understanding?
I'm trying real hard, but you cannot answer the most trivial question, and as you say:
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Without it, we won't be able to move forward.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/02/2019 18:20:57
Where is the VHP of each object in the example, and how does an object's motion relate to that of its VHP? 
Let's focus on VHP1.
Based on Newton - it is not expected to see any wobbling activity during full orbital cycle (it could move in elliptical cycle - but not up and down from the orbital disc).
We must distinguish between "orbit around one another" to "wobbling".
In the example that I have used it was stated:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern.."
So, when we focus on the message: "orbit around one another" this represents a simple Newton solution. In this case we can say that the stars orbit around some VHP. In the same token, in your example - the stars also orbit around each other - so we can also say that the stars orbit around some sort of VHP.
In your example there is no wobbling activity. Newton didn't expect to see any wobbling activity. Newton only focus on planets and moons and there is no wobbling activity in those systems.
However, in the example which I have offered it is specifically stated that the stars are wobbling!!!
This wobbling activity can only be found in Stars!!!
Our scientists don't understand the real impact of this wobbling activity.
I claim that this wobbling activity is the key element activity for our understanding. it highlights how gravity really works in our spiral galaxy.
The wobbling activity is a direct outcome of VHP1.
So, only if we see a wobbling activity, than we know that there is a VHP1.
You won't find it in any Sun/Planet/moon activity. Therefore, you won't expect to find it based on simple Newton law.
However - if you look at any star in the galaxy, you should find that it is wobbling.
Our Sun is wobbling, the nearby stars are wobbling, S2 is wobbling all S stars are wobbling, the stars in that Halo stars are wobbling ... and so on
How can we ignore that important evidence???.
As I have already explained, this VHP1 is a direct outcome from the creation of any new star in the galaxy.
Each star had been created in the core of gas cloud.
The creation of new star from the gas cloud is based on the huge gravity impact of the nearby SMBH.
I'm quite sure that if we set that gas cloud far away from the SMBH (or even far away from the galaxy) - this gas cloud will not be able to set any sort of new star forming activity.
During the crystallize process from dust/gas/molecular into star, I assume that each star must orbit around some virtual center (I call it VHP1). Therefore, this VHP1 is a direct outcome from the nearby SMBH gravity impact during the new star forming activity.
Please be aware, that each star also comes with Planets and moons. However, those objects orbits directly around the star. Therefore, none of them has VHP1.
So, as the star emerges from the gas cloud it also comes with an integrated unique VHP1.
I'm not sure that I have the tools to deeply explain how it really works and how the SMBH gravity force set this process.
However, the wobbling activity is a solid prove for the existence of VHP1.
In your example, there is no wobbling activity. Therefore this example is none relevant to our case.
Please remember - VHP1 only exists in wobbling activity..
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/02/2019 22:34:36
Where is the VHP of each object in the example, and how does an object's motion relate to that of its VHP? 
Let's focus on VHP1.
Based on Newton - it is not expected to see any wobbling activity during full orbital cycle (it could move in elliptical cycle - but not up and down from the orbital disc).
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In your example there is no wobbling activity.
How do you know this?  You've not said where the VHP is for any object.

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Newton didn't expect to see any wobbling activity.
He does if gravity is exerted by any other mass than the primary.

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Newton only focus on planets and moons and there is no wobbling activity in those systems.
This is utterly false.  There is no mention in his formulas for special treatment of planets and moons.  Mass is mass.

If it helps you, you can take the 3 objects in my example and designate them as moons or possibly planets.

I still see nothing in your post that helps me identify where the VHP is for any specific object, or how the motion of that object relates to the position of that VHP.  You continue to evade the simplest question.

You do not have a theory at all.  How is your theory different than the ancient description that God is responsible for the motion of all seen objects?  Whatever you measure, God put it there.  There is no predictions, only declarations that whatever is seen is explained by the 'theory', which is that God is responsible for it.

The only difference between your theory and the God one is that the God one was quite accurate in making predictions.  The heliocentric models (things go around the sun, not the Earth) were resisted at first because their mathematical predictions were not as accurate as those make by a church with centuries of astronomers with tables and fudge factors.

Anyway, I see you taking this path.  "My theory, if it were ever to actually be spelled out, would account for everything we see, as long as it is in hindsight."  It cannot predict new things because without the actual meat of the theory, we can only guess how God wants the new thing to move.  The church for example would not be able to predict how the 3 objects in my example would move since it doesn't know God's intentions for those three objects.  Neither do you because you don't actually have a theory.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/02/2019 07:16:18
I still see nothing in your post that helps me identify where the VHP is for any specific object, or how the motion of that object relates to the position of that VHP.  You continue to evade the simplest question.
As you insist, let's look at a simple orbital motion of three objects by wiki:
https://en.wikipedia.org/wiki/Three-body_problem#/media/File:Three_body_problem_figure-8_orbit_animation.gif
So, we see a very nice motion. But there is no wobbling activity.
In order to understand the wobbling activity lets go back to our Sun.
Please look at the green line of
https://www.quora.com/How-is-the-motion-of-stars-similar-to-the-motion-of-the-sun
It is stated:
"But the stars and sun don't just orbit in a fixed plane like the planets orbit the sun.  They have a slight up and down wavy motion while they move along their trajectory, like so"
We see clearly that the sun is moving up and down in a sine wave while it orbits the center of the galaxy.
So, how can we explain this orbital motion?
Let's look at the other nearby stars.
it is stated:
"Stars in the local solar neighborhood move randomly relative to one another"
We see clearly that some goes up and other goes down.
It they will continue in that momentum, those stars should go upwards or downwards from the disc plane.
There are billions stars in the disc plane, how could it be that none of them continue with its momentum and move away from the disc plane?
Do you agree that this proves that all the nearby stars (and actally, all the stars in the disc plane) also "have a slight up and down wavy motion while they move along their trajectory"?
Please also look at the following diagram:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see clearly that the sun orbits around some sort of center (VHP1) while this center orbits around the galaxy.
Why don't you agree with that?
 
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Newton didn't expect to see wobbling activity.
He does if gravity is exerted by any other mass than the primary.
You claim that Newton should support this wobbling motion.
So would you kindly explain what is the scource for this wobbling motion?
Why do they have a slight up and down wavy motion while they move along their trajectory?
How could it be that this motion is based Newton?
Can you please show me one example for that motion which had been certified by Newton?
Where is the other mass which is needed for that wobbling motion (without VHP1).
Why the stars do not continue with their momentum and move upwards or downwards from the disc plane?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/02/2019 19:33:52
As you insist,
I was insisting you consider my simple example, and not some special case example you found on a website that doesn't support your view.  So you still have given me no clue as to how to figure out where a VHP is.
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let's look at a simple orbital motion of three objects by wiki:
https://en.wikipedia.org/wiki/Three-body_problem#/media/File:Three_body_problem_figure-8_orbit_animation.gif
So, we see a very nice motion. But there is no wobbling activity.
What do you mean by that??  Of course there is 'wobbling' activity.  If one of those objects was luminous, we'd see it wobbling vertically at twice the frequency and about a third the amplitude of the horizontal wobble, which would be a strange sight.  Such an object would not trace a clean ellipse around the galaxy.

That is a weird special case which is barely stable, and my example was not.  But it is a nice example of a wobbling activity without a VHP.  There just isn't one.  There is a center of gravity of course, but none of the objects orbit it, and when one is located at it, acceleration is at zero, not the maximum value as you would expect when an object gets closest to the virtual point about which it orbits.

It seems funny that you would link to an example that refutes what you've been asserting.

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In order to understand the wobbling activity lets go back to our Sun.
Please look at the green line of
https://www.quora.com/How-is-the-motion-of-stars-similar-to-the-motion-of-the-sun
It is stated:
"But the stars and sun don't just orbit in a fixed plane like the planets orbit the sun.  They have a slight up and down wavy motion while they move along their trajectory, like so"
We see clearly that the sun is moving up and down in a sine wave while it orbits the center of the galaxy.
Yes, that near-sine-wave motion is a easily derived from a mass effected by a planar arrangement of mass.  A solar system has almost all its mass at the center leaving little mass to have a wobble like that which has a period of anything on the order of the period of orbit.  Pluto is out of the plane and has such a wobble which manifests itself as precession since the wobble period is far long than its year.

Anyway, we do not 'see clearly' anything, because the green line in that picture is massively out of scale (so that you can distinguish it from a clean elliptical orbit).  It shows a velocity perpendicular to the plane that is the same order as the orbital velocity instead of about 2-3% of it as it should, and it shows a period that is also 4 times the actual measured period.  Your biocab link was even worse in both respects, and was actually factually wrong, not just out of scale.

So yes, this sine-motion is expected from a planar arrangement of mass, which (locally) describes the galatic disk.  The mass certainly isn't at the center like it is with our solar system.  The SMBH masses less than 0.05% of the mass of the galaxy, the vast majority taking a planar arrangement.
There is a VHP (your term) of sorts which is where the sun crosses the plane twice each wave.  It doesn't orbit that point, but just moves back and forth straight across it as would any object pulled by a planar arrangement of mass.  And yes, it is the VHP that moves at perhaps 230 km/sec (per that site) and the solar system itself at maybe 0.04% faster than that as it follows that longer green line.
They show the solar system following a perfectly circular orbit, but our orbit about the galaxy does have some eccentricity, probably around 0.05.  The path around the galaxy certainly isn't circular or even elliptic since it gets massive perturbations from moving through denser concentrations of matter in its travels, so the eccentricity value is fairly meaningless.  I'm just saying it isn't a nice orbit with constant speed like the simplified pictures show.

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Let's look at the other nearby stars.
it is stated:
"Stars in the local solar neighborhood move randomly relative to one another"
We see clearly that some goes up and other goes down.
No velocities are given, so this isn't clear at all.  The arrows are relative to the sun, not relative to the galactic disk, and they're all the same size, meaning it isn't depicting vectors.  It is just a crude picture trying to illustrate those words.  Maybe they're all moving up, just some less than us, and some more.

They might be trivial differences while we (the other stars) move as a group around the galaxy, or they might be significant differences in which case the nearby stars just happen to be nearby now, but we'll never see them close again.

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It they will continue in that momentum, those stars should go upwards or downwards from the disc plane.
Presumably, yes.  Some of them might be stationary relative to the disk, and hence not have that sort of motion.  The site hardly gives any data on this.

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There are billions stars in the disc plane, how could it be that none of them continue with its momentum and move away from the disc plane?
They almost all move relative to the plane in some way, and they (almost) all come back just like we do.  The plane is the average, not a clump of different stuff that doesn't have this motion.  Sure, some stars experience a close encounter with something big and get flung out of the disk, becoming a halo object of sorts.

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Do you agree that this proves that all the nearby stars (and actally, all the stars in the disc plane) also "have a slight up and down wavy motion while they move along their trajectory"?
Statistics says it is highly improbable that a random number is exactly zero, so this would suggest that a given moment, no star has zero motion perpendicular to the plane.  Even if it did, one moment later, gravity of some nearby object would alter this value to something nonzero.  None of the text above proves this, no.  I reached for statistics to agree with the statement.  Point is, some have more than others.  Even the sun has zero motion perpendicular to the plane about 5-6 times per lap around the galaxy, but that zero value lasts but an instant.

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Please also look at the following diagram:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see clearly that the sun orbits around some sort of center (VHP1) while this center orbits around the galaxy.
If that picture was even remotely accurate, then I suppose it would.  These guys are not astronomers or physicists.  They're biologists.  I don't go to astronomers to ask my biology questions either.
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Why don't you agree with that?
The solar system does not move like that.  For one, we'd have to move at about 1500 km/sec to follow that blue line and still keep up with the 217 km/sec grey line.  The blue line is around 7 times longer than the grey line, but labelled with much lower speeds and no mention of periods.

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You claim that Newton should support this wobbling motion.
Of course.  His laws demand it if there is mass acting on the thing that wobbles.  The moon's path wobbles around the sun because Earth is there yanking back and forth once a month, just as Newton describes.

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So would you kindly explain what is the scource for this wobbling motion?
Mass of something other than itself or its primary.  Any third body or cloud or disk or whatever.

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Why do they have a slight up and down wavy motion while they move along their trajectory?
Any object will have this motion relative to a planar arrangement of mass.

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How could it be that this motion is based Newton?
Because his laws describe it perfectly, and he did it first, so his name got attributed to those laws.  A plane of stuff is going to attract.  You can't go around it, so you get pulled straight towards it and carry right through to the other side after which the plane pulls you the other way.  It's pretty easy to work out.

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Can you please show me one example for that motion which had been certified by Newton?
An apple falling to the ground.

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Where is the other mass which is needed for that wobbling motion (without VHP1).
The disk is the other mass.
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Why the stars do not continue with their momentum and move upwards or downwards from the disc plane?
Because the disk pulls them back.  If we're above the disk, then most of the stars are below us, pulling us downward.  Those stars (including us) make up the disk.  We don't hit the disk on the way through since the stars have space between them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/02/2019 16:32:38
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Why the stars do not continue with their momentum and move upwards or downwards from the disc plane?
Because the disk pulls them back.  If we're above the disk, then most of the stars are below us, pulling us downward.  Those stars (including us) make up the disk.  We don't hit the disk on the way through since the stars have space between them.

You claim that: "Those stars (including us) make up the disk.
So, let me start by asking: why all of those stars stay at the disc from the first stage?
What kind of force keeps them all in the disc?
Why just from the starting point of the Bar (at about 1KPC) to the edge of the spiral arms (at about 10KPC) there is a disc?
How could it be that in the bulge (less than 1KPC) stars do not orbit in a disc?
How could it be that all S stars which are located very close to the SMBH do not share the same orbital disc plane?
Why each S star sets its unique orbital disc plane although they are directly affected by the mighty gravity force of the SMBH?
Our scientists claim that the sun orbits around the center of the galaxy due to the dark matter:
http://cdms.berkeley.edu/Education/DMpages/FAQ/question36.html
"The radius of the sun's orbit is about 2.5x1017 km, so the total mass of dark matter within that orbit is 6x1040 kg. This is the mass of 3x1010 (30 billion) stars like the sun! The entire galaxy only contains ~100 billion stars, so the dark matter does have a significant effect on the sun's orbit through the galaxy. For objects farther out near the edge of the galaxy, the dark matter is actually the main thing keeping them in their orbits. This is more or less how dark matter was discovered by astronomer Vera Rubin and others: the orbital speeds of galactic stars and gas clouds don't match our expectations from the visible matter."
Just think about the huge gravity force which is needed to keep the sun in his track around the center of the galaxy. It is also clear that most of the gravity force comes from the dark matter.
Based on our scientists, the dark matter is not located at the disc itself.
So, how could it be that the stars around us with almost neglected gravity force (comparing to the ultra high gravity force due to the dark matter) can hold each other at the same orbital disc plane?
What kind of force hold all the stars in the disc?
The Sun is located at the Orion arm.
The Orion Arm diameter is about 1,000KLY
In one hand our scientists claim that this arm is only a density wave. So, there are no gravity force connections between the stars.
However, now you claim that those stars pull back any star that is moving too high or too low from the disc plane.
How could it be? don't you see a contradiction?
What kind of power/force do they use in order to pull each other back?
Is it gravity force or some other force?
If it is gravity - Can you please explain how the gravity force pulls back any star which goes too high above the disc or too low below the disc??
You claim that it pull back - but in reality it works like a spring. Up and down Up and down in a constant movment.
How could it be?
If I understand it correctly, Newton gave us two options:
1. Direct contact. For example:
An apple falling to the ground.
2. Orbital motion.

However, I have never heard about Newton gravity force which could force a star to move up and down continually without orbital motion.
Based on this theory, how could it be that the local gravity force due to local mass of nearby stars can have any sort of effect comparing to a gravity force due to 30 billion solar mass?
It seems to me as an elephant which is swinging nearby a mosquito, while the mosquito is sure that it is due to is mass. Is it real?

As for example -.
Let's assume that somehow we can force all the stars in 1,000KLY segment in our spiral arm (by glue) at a fixed location in the segment.
Now, let's take a free star and push it to the most upwards side of the arm and release it.
If I understand it correctly, the gravity force of the other stars will pull it back.
But it will have two options:
One - To collide with one of the nearby stars
Two - start orbit around some sort of center of mass or around VHP which is a direct outcome of the stars in that segment.
Can you please explain how it could move up and down in a constant movement?
How could it be that gravity acts like a spring?
Now, let's assume that somehow the gravity force keeps it from moving to high from the most upwards side of the arm or too low from the most downwards side of the arm.
Why the same force can't hold it from moving away from the arm (too left or too right)?
Why only Up/down and not right/left?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/02/2019 19:44:34
You claim that: "Those stars (including us) make up the disk.
Stars, gas, planets, screwdrivers, etc.  Yes.  The disk is made of stuff.

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So, let me start by asking: why all of those stars stay at the disc from the first stage?
Not all do.  Some get thrown out, but that takes effort, some outside force.

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What kind of force keeps them all in the disc?
Gravity.  The list of forces isn't long, even if you are making up new ones.
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Why just from the starting point of the Bar (at about 1KPC) to the edge of the spiral arms (at about 10KPC) there is a disc?
The disk goes further out than that, but sure.  One might consider stuff closer than 1kpc as well.  It doesn't have an abrupt start or end.

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How could it be that in the bulge (less than 1KPC) stars do not orbit in a disc?
Anything that orbits does so in more or less a plane.  The orbit of no single object can be described as a disk.  A disk is a collection of material flattened by mutual angular momentum.

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How could it be that all S stars which are located very close to the SMBH do not share the same orbital disc plane?
I'm no expert, but I imagine that the close objects have a point mass around which they might orbit.  The solar system does not.  That explains an orbital path as opposed to a more characteristic disk path, but not so much the orbital plane.  I suspect that planar objects are in the plane because they've not been knocked out of it.  Those objects or gas clouds that have acquire random trajectories that take some of them by chance very close to the center of the galaxy where they take up a new orbit in the new plane into which they were deflected.

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Why each S star sets its unique orbital disc plane although they are directly affected by the mighty gravity force of the SMBH?
The SMBH is not capable of altering the orientation of the orbital axis of a moving object.

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Our scientists claim that the sun orbits around the center of the galaxy due to the dark matter:
I find that wording misleading in this context, but the context of the page is different.  Keep that in mind. I would say the sun loosely orbits the galaxy, not the center of the galaxy since, unlike any planet, the vast majority of the forces acting on us come from places other than the center.  About 99.9% of the solar system mass is at the center, but less than 1% in the case of the galaxy.  You just can't compare orbital mechanics between the two.

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Just think about the huge gravity force which is needed to keep the sun in his track around the center of the galaxy. It is also clear that most of the gravity force comes from the dark matter.
Based on our scientists, the dark matter is not located at the disc itself.
So, how could it be that the stars around us with almost neglected gravity force (comparing to the ultra high gravity force due to the dark matter) can hold each other at the same orbital disc plane?
Dark matter has no more gravity force than regular matter.  There are not different kinds of gravity.  A KG of dark matter exerts the same gravitational force as a KG rock or a KG black hole.
Dark matter (at least WIMP dark matter, not so much the MACHO parts) is not subject to friction like gas clouds are, so it doesn't tend to flatten into a disk just because it has angular momentum.  So it makes sense that it isn't concentrated in a disk like the interactive matter tends to do.

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The Sun is located at the Orion arm.
The Orion Arm diameter is about 1,000KLY
In one hand our scientists claim that this arm is only a density wave. So, there are no gravity force connections between the stars.
The stars very much attract each other by gravity.  Newton's laws demand this.  If the arm is a density wave, then it attracts stuff more than the less dense places between the arms, but as the density wave moves on, it doesn't carry the solar system or anything else with it.  The density waves pass through us, meaning the sky is more cluttered when we're in a proper arm than when we're in the low-density regions between the arms, just like a leaf in the water rising up and down with the waves without moving along with them.

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However, now you claim that those stars pull back any star that is moving too high or too low from the disc plane.
How could it be? don't you see a contradiction?
If it is gravity - Can you please explain how the gravity force pulls back any star which goes too high above the disc or too low below the disc??
I see no relevance to the arms.  Suddenly you're talking about the disk.  If the disk is below you, wouldn't you expect all that mass to pull you toward it?  You seem to suggest otherwise, which would contradict the laws of gravity.  I have no idea why you find this baffling.

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You claim that it pull back - but in reality it works like a spring. Up and down Up and down in a constant movment.
It is like a spring.  If I extend the spring, it tries to pull me back.  If that shoots me to the other side, the spring pulls me back again.  Hence the green sine wave you see in that one picture.
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How could it be?
How can it not be?  It would seem to violate Newton's laws if all that mass on one side didn't attract us back each time.
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If I understand it correctly, Newton gave us two options:
1. Direct contact. For example:
Quote from: Halc
An apple falling to the ground.
That isn't direct contact.  The apple has all the mass of the Earth on one side of it, so is attracted by gravity.  That's enough to break free of its bond on the tree and enter into orbital free fall around Earth.  I can't help it that the ground gets in the way and causes the apple to stop via forces other than gravity.  The disk isn't like that.  It is unlikely to actually strike anything large enough to prevent us from passing right through.  But we hit small stuff which slows us each time, so eventually the sine wave fades until the next time something big passes close by again to get a new sine wave going.  I think the friction with the disk is pretty minimal, so Earth will burn up before that wave fades noticeably.
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2. Orbital motion.
The apple goes into orbit when it detaches from the tree, yes.  Not a pure orbit since it is not in a vacuum.  There is a lot of friction.  It would never come back to its start point even if the ground didn't get in the way.  The apple is in orbit around a single mass.  The solar system is not in orbit about the disk in this way.

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However, I have never heard about Newton gravity force which could force a star to move up and down continually without orbital motion.
Try dropping an apple from some distance from a dense disk with a hole in it to allow the apple to pass through.  It will go back and forth forever through the hole, or so says Newton's laws.
You might consider that hole to by your VHP since the apple is always drawn to it, but the motion relative to that VHP would not be an orbit.  The closer the apple gets to that point, the smaller the force is acting on it.  With an orbit, the closer you get to the VHP, the greater the force.

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Based on this theory, how could it be that the local gravity force due to local mass of nearby stars can have any sort of effect comparing to a gravity force due to 30 billion solar mass?
No idea what you're talking about here.  What 30 billion masses?  If they're far enough away, the nearby stars have greater effect since forces in inversely proportional to r².  Still, the 30 billion solar masses, while further away, will pull on each of those local stars, accelerating them as a group if the group moves away from the the disk.  The local stars are on all sides, so their pull tends to cancel out.  The disk, when all on one side, does not have its effect cancelled out.

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It seems to me as an elephant which is swinging nearby a mosquito, while the mosquito is sure that it is due to is mass. Is it real?
I don't understand the analogy.  An elephant on a swing moves due to the gravity of Earth which dwarfs the mass of both the mosquito and the elephant.

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As for example -.
Let's assume that somehow we can force all the stars in 1,000KLY segment in our spiral arm (by glue) at a fixed location in the segment.
Glue them to what?  What is a segment?  Are you trying to glue it to the arm?  What if the arm moves at some insane speed?  What if it moves faster than light?  Waves can do that.  No, I don''t suggest arms move that fast, but they probably move much faster than any of the material that lights them up.

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Now, let's take a free star and push it to the most upwards side of the arm and release it.
If I understand it correctly, the gravity force of the other stars will pull it back.
It will probably be moving too fast to do this, and will be flung away to some larger radius from the galaxy.

OK, so let's say we release the star at the mean velocity of all the other material in the arm, but not the speed of the arm itself.  I think that's what you have in mind.  I think the density of the arm will move on and the star will stay at the radius where you put it.  The arms's don't have much pull for very long since they pull one way as they approach and pull the other way after they pass.  The effects cancel.  It is moving out of the disk (in a direction parallel to the galactic axis) that would result in an attraction back to the disk.

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But it will have two options:
One - To collide with one of the nearby stars
That can always happen, but they're small targets.  We've passed through the disk some ~110 times probably and haven't got close to one yet.

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Two - start orbit around some sort of center of mass or around VHP which is a direct outcome of the stars in that segment.
Now you've jumped to your view.  You have refused to define where a VHP is given a certain setup, so this cannot be answered intelligently.
So how about option 3: It will continue its way around the galaxy, except without the green sine wave since you put it at equilibrium at the middle of the disk so it need not move up and down at all until defected by something.

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Can you please explain how it could move up and down in a constant movement?
I think it will not, since you arrested that motion when you glued it in place.

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Now, let's assume that somehow the gravity force keeps it from moving to high from the most upwards side of the arm or too low from the most downwards side of the arm.
That makes no sense.  The arm is not an object.  Nothing can orbit it.  No matter is particularly part of it.  They're just waves that pass through the matter.  Your theory might propose otherwise, but then you have to answer your own questions, like what speed do the arms move, and why they're not completely wound tight by now.

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Why the same force can't hold it from moving away from the arm (too left or too right)?
Why only Up/down and not right/left?
The disk is an object.  The arms are not.  The arms pull in both directions as they pass with a net effect of more or less zero.  The disk never passes away.  It's always there.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/03/2019 08:30:18
Please look again at the following article:
https://www.livescience.com/63698-milky-way-spiral-wobble.html
It is stated clearly:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern that has only become more tangled over the past eon."
Why can't we trust this verification of our scientists?
It is stated clearly: "these rogue stars also orbit around one another in a wobbly, spiral pattern.."
So, do you agree that our scientists see that while the stars in these rogue stars halo orbit around one another, they also move in a wobbly pattern?
What is the meaning of that wobbly pattern?
Do you agree that it means that stars do not orbit exactly at the expected orbital cycle, but they move in a wobbly way around this expected orbital cycle?
We know that the Sun and all the other stars in the disc are wobbling while they orbit around the galaxy.
We caim that it is due to the galactic disc.
But stars also wobble high or below the disc.
I have offered also other evidences for that wobbling outside the disc -
S2:
https://www.researchgate.net/figure/Fit-to-the-orbit-of-the-S2-star-fitted-data-and-relative-errors-are-in-blue-the-red_fig1_272845577
It is stated:
Fit to the orbit of the S2 star: fitted data and relative errors are in blue, the red line is the orbit.
So we see clearly that the measured points are not located directly on the orbital path.
Our scientists call it "relative errors in blue". But those "errors" proves that our scientists see that S2 is wobbling!
Therefore, it should be clear to all of us that our scientists see wobbling path activity of S2.
Please remember that S2 is also not located at the disc.
Hence, it proves that the hypothetical idea about the Up/Down movement or wobbling due to the disc is incorrect as we see the wobbling activity also out the disc.

The disk is an object.  The arms are not.  The arms pull in both directions as they pass with a net effect of more or less zero.  The disk never passes away.  It's always there.
How could it be?
There are stars above and below the disc while there are also stars in between the arms.
Actually, the width of the disc is dictated by the width of the arm.
In all the images of the milky way, we clearly see that most/all of the stars are located in the arms.
As an example:
http://www.astronoo.com/en/articles/galactic-arms.html
What is the estimated ratio between the total stars in the arms comparing to the total stars in the disc outside the arm?
Do you agree that the arm is mainly made out of stars while the disc is mainly made out of arms?
If we take out all stars from all arms, do you think that there will be left something that can still be called disc?
Can you please explain what do you mean by: "The arms pull in both directions as they pass with a net effect of more or less zero."
If I understand it correctly, you claim that stars can cross the arm as they move vertically to the disc, but they can't cross the arm as they move horizontally to the disc.
Hence, a star can't move horizontally and cross the arm as the stars in the Arm pull it back due to gravity force from the stars in the arm.
So, if the stars in the arm prevent from a star to cross the arm vertically (outwards or downwards) due to gravity force, than why the same gravity force (Due to the stars in the arm) can't prevent from a star to cross the arm horizontally?
How could it be that the stars in the arm sets a gravity force which works horizontally, but doesn't work vertically?

With regards to the spiral galaxy structure.
In the bulge - there is no disc. Each star orbits at a different orbital disc plane - (therefore we call it bulge...). This contradicts the solar orbital system, as all the planets orbit at the same disc plane - with the exception of Ploto.
http://planetfacts.org/wp-content/uploads/2013/05/Planet-Orbits.jpg
Why is it? 
If we move further away from the bulge in the direction of the bar we see that stars start to orbit in a disc. (very wide disc close to the bulge, but as we go further away to the edge af the bar (which is also the starting point of the Arms - at 3KPC) we clearly see the disc shape.
Therefore, it is clear that the Arms + Bar set the disc and not vice versa.
Hence, if we call the disc an object than by definition the Arm and the Bar must be considered as an object.
I still don't understand why our scientists just focus on the disc and the arms?
Why they don't explain the Bulge and the Bar?
Why they don't show in their modeling how we could get the full spiral galaxy shape including: Bulge, Bar, Spiral Arms, Disc, Halo stars and so on.
Why they only focus in one segment of the galaxy (spiral arms + disc).
With regards to modeling:
Let's look at the following article:
https://kof.zcu.cz/st/dis/schwarzmeier/galaxy_models.html
A. "5.8  - Spherical galaxies
Our first objective is a simulation of elliptical galaxies with an eccentricity of zero (spherical galaxies). We will perform the simulation of two spherical systems – Plummer’s and Hernquist’s models."
However, the outcome was -
"Figure 5–13: A time sequence of Plummer’s model in the xy plane. Practically no evolution can be seen."
So, if they start with a spherical galaxy - they don't get a disc and they don't get the spiral arms!!!
B. However, if they start the modeling as a disc, they get some spiral arms which have gone after some time:
"5.9        Initial conditions for flattened systems
A spiral galaxy is flat and can be represented by a flattened potential. The disk of the spiral galaxy is generally very flat and we will describe them as infinitely thin. We will suppose that disks are axisymmetrical and therefore, initial conditions will be set up in cylindrical coordinates."
"Figure 5–16: A time sequence for Kepler’s disk model in the xy plane. The disk is ideally flat in the yz plane."
C. Just when they start with a very thin disc, with other parameters including the radial velocity dispersion of the solar neighborhood, they have finelly got the spiral arms.
"We will create a disk with kinematically hot (supported by random motions, rather than the rotational motion) stars. The thickness of the disk is set such that Toomre’s stability criterion is satisfied. The radial velocity dispersion  corresponds to the value of the Solar neighborhood, i.e."
But this is absolutely none realistic.
They have actually proved that when they took a realistic starting point as spherical galaxies, they didn't get by the modeling the expected shape of spiral arms and disc.
D. After the Big bang, we didn't get any disc (especially not a thin disc of stars!!!).
So, they can't just start from the point where the disc is already there with all the requested velocities. This is an error by definition.
They have to show how long it takes from the Big bang moment to set all the requested stars in a spherical galaxies and from that point to the starting point of a very thin disc.
Than they have to add it to the requested time to form a spiral disc.
"Figure 5–20: A time sequence for the hot exponential disk with  in the xy plane evolved with the Barnes-Hut N-body algorithm"
If in the following modeling 1 billion years is needed to set a spiral galaxy from thin disc, they have to add the requested time from the big bang.
So, this modeling is not realistic as they didn't add the requested time from the Big bang to that starting point.
I'm quite sure that they might find that even after 10 billion years no real spiral arm would be formed.
E. As stated, they have only showed spiral arms and disc. Not even one single word about the other sections of spiral galaxy as Bulge & Bar. This is another sever mistake.
F. Actually, this modeling proves that there is no way to get spiral arms galaxy directly from spherical galaxies.
This evidence proves that our scientists have a severe mistake in their understanding how the spiral galaxy had been evolved from time zero - Big bang.
G. please look at:
"Figure 5–21: A time sequence for the hot exponential disk with  in the yz plane evolved with the Barnes-Hut N-body algorithm. The thickening caused by the random motion of stars can be seen"
We see that the starting point is a very thin disc.
After 1.02 Billion years the disc became quite thick.
Why they stopped after this moment?
Why they didn't continue with the modeling?
In one of the modeling they have showed clearly that after getting the spiral shape there is a possibility to lose it again.
Therefore, I'm quite sure that if they will continue with the modeling (for few more billion years) that disc will be so thick that we shouldn't see any spiral structure - maybe only spherical galaxy.

This could prove that spiral galaxy might be converted over time into spherical galaxy but not vice versa.

So, do you agree that based on those key ideas, there must be a severe problem with our current understanding about spiral galaxy and the evolvement activity of this galaxy from the Big Bang starting point?



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/03/2019 18:07:58
It is stated clearly:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern that has only become more tangled over the past eon."
Why can't we trust this verification of our scientists?
It is stated clearly: "these rogue stars also orbit around one another in a wobbly, spiral pattern.."
So, do you agree that our scientists see that while the stars in these rogue stars halo orbit around one another, they also move in a wobbly pattern?
What is the meaning of that wobbly pattern?
It says in the article.  They mapped the position and velocity of a huge number of objects and found this one region that had motion that was anomalous to the typical paths of the rest of them.  It apparently means somebody threw a rock into a calm pond and long after the rock had passed, the disturbance of it is still visible, centered on the passage point via 'wobbling' waves.  They even extrapolated back and found the object (the rock) that is responsible.

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Do you agree that it means that stars do not orbit exactly at the expected orbital cycle, but they move in a wobbly way around this expected orbital cycle?
A 'wobbly' way is the expected cycle.  We're not in empty space so nobody expects our path to be a nice clean circle or something.

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I have offered also other evidences for that wobbling outside the disc -
S2:
https://www.researchgate.net/figure/Fit-to-the-orbit-of-the-S2-star-fitted-data-and-relative-errors-are-in-blue-the-red_fig1_272845577
It is stated:
Fit to the orbit of the S2 star: fitted data and relative errors are in blue, the red line is the orbit.
So we see clearly that the measured points are not located directly on the orbital path.
Interesting, huh?  I also notice that the error bars abruptly get tiny on the right.  Somebody bought a new more accurate telescope during those 18 years.

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Our scientists call it "relative errors in blue". But those "errors" proves that our scientists see that S2 is wobbling!
It is indeed wobbling. It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion.  It has no VHP.

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Hence, it proves that the hypothetical idea about the Up/Down movement or wobbling due to the disc is incorrect as we see the wobbling activity also out the disc.
What happens somewhere not in the disk is no evidence whatsoever about the dynamics of the disk.

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There are stars above and below the disc while there are also stars in between the arms.
Actually, the width of the disc is dictated by the width of the arm.
Disks have width?  You mean thickness?
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In all the images of the milky way, we clearly see that most/all of the stars are located in the arms.
Yes, arms light up because more stars are there, but that doesn't make arms objects unless the motion of the stars is the same as the motion of the arms, and it isn't.

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Do you agree that the arm is mainly made out of stars while the disc is mainly made out of arms?
Disks are made of stars and gas and other material.  That mass moves differently than the arms, so no, I would not say that the disk is made of the arms any more than I say a lake is made out of waves.  It is made out of water, and yes, there is more water where the waves are high than where the waves are low, but the wave moves on and takes none of the water with it.

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Can you please explain what do you mean by: "The arms pull in both directions as they pass with a net effect of more or less zero."
If I understand it correctly, you claim that stars can cross the arm as they move vertically to the disc, but they can't cross the arm as they move horizontally to the disc.
I didn't say that.  Stars cross the disk as they move vertically back and forth across it, and due to the gravity of it.  The arm wave seemingly passes through us as it goes by, but like a leaf on the water, it makes the leaf wobble but doesn't actually change its position in the end.

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Hence, a star can't move horizontally and cross the arm as the stars in the Arm pull it back due to gravity force from the stars in the arm.
I don't know what you mean by this.  I don't know which direction you consider horizontal.  The arm passes through us, so that's pretty much the same as us crossing it.  Yes, I imagine the higher density of the stars causes some alteration to the path we'd otherwise take.

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So, if the stars in the arm prevent from a star to cross the arm vertically (outwards or downwards) due to gravity force, than why the same gravity force (Due to the stars in the arm) can't prevent from a star to cross the arm horizontally?
Nothing is prevented from moving where it is heading.  There is material on all sides, so for the most part our progress in unhampered by that.  Think again of the leaf on the water which is not prevented from crossing the wave, and isn't even really moved by it in the long run.

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With regards to the spiral galaxy structure.
In the bulge - there is no disc. Each star orbits at a different orbital disc plane - (therefore we call it bulge...). This contradicts the solar orbital system, as all the planets orbit at the same disc plane - with the exception of Ploto.
Pluto isn't a planet, and there are plenty of objects that orbit out of the plane.  None of them happen to be a planet, which are planar because they formed out of the same rotating disk of gas as the sun and have had nothing that can divert them out of that original plane.
The solar system is a lousy model of a little galaxy since it is held together by a primary mass and the galaxy isn't.  The dynamics are very different.

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I still don't understand why our scientists just focus on the disc and the arms?
They probably focus more on stars and clouds and their interactions.  No idea what you mean by your statement there.

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Why they don't explain the Bulge and the Bar?
There are models that do.  Your model doesn't because you have no model.  You don't post any simulations of your model.

With regards to modeling:
...
So, if they start with a spherical galaxy - they don't get a disc and they don't get the spiral arms!!!
B. However, if they start the modeling as a disc, they get some spiral arms which have gone after some time:
"5.9        Initial conditions for flattened systems[/quote]
They get arms (at least briefly) with a different model.  It isn't strictly the use of different initial conditions, but the use of a different model.  The arms are very short lived.  The model isn't very predictive.

All the simulations run for only a short time, like a billion years.  Real galaxies run undisturbed for perhaps this long, but in the longer run, they're constantly bombarded by collisions which disrupt the structure.  A good simulation running for the full 13 billion years needs to include such evolution.

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C. Just when they start with a disc it with other parameters including the radial velocity dispersion of the solar neighborhood, they have finelly got the spiral arms.
Briefly, yes, and using a different model.

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They have actually proved that when they took a realistic starting point as spherical galaxies, they didn't get by the modeling the expected shape of spiral arms and disc.
That was a different model, and I don't know what angular momentum they gave it.

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D. After the Big bang, we didn't get any disc.
So, they can't just start from the point where the disc is already there with all the requested velocities. This is an error by definition.
Indeed, so how did it get that way?  Here, the accretion model of solar system formation is somewhat useful since the early stages of a solar system is distributed material with angular momentum but no central mass yet.  Disks do form naturally, at least by standard models.  Your assertions seem to suggest that fully formed solar systems and their defined planes just spring into existence out of black holes, so maybe there is something that similarly spews forth galaxies with their disks already in place.

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They have to show how long it takes from the Big bang moment to set all the requested stars in a spherical galaxies and from that point to the starting point of a very thin disc.
A model of a spherical galaxy doesn't ever become a disk.  There are galaxies that stay spherical.  There are many classes of galaxy types.

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"Figure 5–20: A time sequence for the hot exponential disk with  in the xy plane evolved with the Barnes-Hut N-body algorithm"
That's a third model now, and it still doesn't look like a real galaxy, but at least the arms seem a bit more stable.

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If in the following modeling 1 billion years is needed to set a spiral galaxy from thin disc, they have to add the requested time from the big bang.
The arms are quite distinct after 0.4 billion years.  There were sort of there in the Kepler model as well, but after a billion years, they were gone again.  They were not stable.  Kepler's model I think is one where the arms are objects instead of waves.

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I'm quite sure that they might find that even after 10 billion years no real spiral arm would be formed.
Actually, this modeling proves that there is no way to get spiral arms galaxy directly from spherical galaxies.
A model that doesn't exhibit the desired behavior proves that the model does not correspond entirely to the modelled thing.  It doesn't in any way prove that another model would not fare better.  Your grasp on logic is on the same level as your grasp of physics.

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This is evidence which proves that our scientists have a severe mistake in their understanding how the spiral galaxy had been evolved.
What mistake would that be?  Are they asserting that one of these models is the correct one?  I doubt that seriously.  They're quite aware that they don't have a full understanding of the dynamics yet.

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E. please look at:
"Figure 5–21: A time sequence for the hot exponential disk with  in the yz plane evolved with the Barnes-Hut N-body algorithm. The thickening caused by the random motion of stars can be seen"
We see that the starting point is a very thin disc.
After 1.02 Billion years the disc became quite thick.
The initial conditions are a very thin disk, just to see what thickness would form naturally.  It isn't suggested that the disk was ever that thin.
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Why they stopped after this moment?
Why they didn't continue with the modeling?
Maybe it didn't evolve much after that.  To go longer, it would need to simulate the collisions with objects coming from the outside.  Those happen frequently at first, and still occur more than once every billion years.  I've seen animated simulations that run the whole lifespan of the galaxy and it shows all those outside objects from which most of the mass of the galaxy comes.  We were likely quite small in the beginning.

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I'm quite sure that after 13 billion years that disc will be so thick that we would see just a normal spherical galaxy.
I've not seen any simulation that did that.
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This could prove that spiral galaxy might be converted over time into spherical galaxy but not vice versa.
If there was a model, sure, but you don't have one, so there is no simulation of it.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/03/2019 14:38:57
Dear Halc
You have chosen to reject all the evidences which I have offered.
It seems to me that even if Newton was standing next to me and fully support my message - you would still reject any idea which contradicts the current mean stream which you fully support.
So, it is quite clear to me that whatever I will say you would reject.
Therefore, let me summarize between those none relevant ideas of the mean stream and my real theory.

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In all the images of the milky way, we clearly see that most/all of the stars are located in the arms.
Yes, arms light up because more stars are there, but that doesn't make arms objects unless the motion of the stars is the same as the motion of the arms, and it isn't.
Sorry
Our scientists have no clue how spiral arms really works.
The assumption that - The motion of the stars isn't the same as the motion of the arms, is absolutely incorrect!!!
I have deeply explained how it really works.
Based on my explanation, the arms are made of stars. Any star which will dare to move out of the arm will be ejected from the galaxy (unless it is crossing a bridges or branches between the arms).
So, the arms are real object!
I have introduced again and again the following image, but you have chosen to ignore that real image/evidence.
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
It shows that based on our verifications all/most of the stars in the milky way disc are located at the arms.
However, in the same article it is stated:
https://en.wikipedia.org/wiki/Milky_Way
"Outside the gravitational influence of the Galactic bars, the structure of the interstellar medium and stars in the disk of the Milky Way is organized into four spiral arms.[113] Spiral arms typically contain a higher density of interstellar gas and dust than the Galactic average as well as a greater concentration of star formation, as traced by H II regions[114][115] and molecular clouds.[116]"
What does it mean "higher density of interstellar gas and dust" in the following statement"?
Where are all of those nearby interstellar gas and dust?
If we look nearby, we only see G-type stars.
Why is it?
Why our scientists discuss about interstellar gas and dust while we mainly see G-type stars?
If I understand correctly - there is only one nubila in the whole Orion arm.
Don't forget that in the simulation our scientists are using stars!
So, please don't speak about interstellar gas and dust just to confuse us - speak about G-type stars (as you are using them in the simulation).
Now, let's try to verify the density of those stars.
In the following article it is stated:
http://www.solstation.com/stars3/100-gs.htm
"As many as 512 or more stars of spectral type "G" (not including white dwarf stellar remnants) are currently believed to be located within 100 light-years or (or 30.7 parsecs) of Sol -- including Sol itself. Only around 64 are located within 50 light-years (ly), while some 448 are estimated to lie between 50 and 100 light-years -- a volume of space that is seven times as large as the inner sphere within 50 ly of Sol. A comparison of the density of G-type stars between the two volumes of space indicates that the outer spherical shell has around 100 percent of the spatial density of known G-type stars as the inner spherical volume, which suggests that astronomers have identified the great majority of the G-type stars that are actually located within 100 ly of Sol, assuming the same spatial distribution in the Solar neighborhood".
So we know the density of G-type stars in a 50 LY volume: "Only around 64 are located within 50 light-years".
That fully meets by 100% the G-type stars density expectation at 50 and 100 light-years volume: "while some 448 are estimated to lie between 50 and 100 light-years"
That shows the G-type stars density at the Orion arm (There are 448 + 64 = 512  G -type stars in a 100LY around the Sun)
However, what is the G-type stars density at the same volume (100LY) between the orion arm and the next nearby arm?
Is it only 400, 200, 100 or just Zero?
I assume that they might find an area between the arms with high density of G -type stars (As it is a bridge or branch between the arms)
However, is there any possibility to find in the same volume of 100KL between the arms ZERO G -type stars?
I'm sure by 100% that there are big areas between the arms without even a single star.
That is a direct outcome from my theory.
So, if that is correct, my theory is also correct.
Wobbling
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Our scientists call it "relative errors in blue". But those "errors" proves that our scientists see that S2 is wobbling!
It is indeed wobbling. It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion.  It has no VHP.
Thanks
So you agree that at least S2 is wobbling.
However, why are you so sure that: "It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion?
Once you agree that stars are wobbling - you actually say that you can't explain it by normal orbital cycle.
If we look at our sun, it is not just wobbles up and down. It doesn't move horizontally to the disc. it actually move with some phase to the disc.
If that is correct, than the idea that it moves up and down due to the disc is incorrect by definition.
What is the amplitude of that wobbling activity of the sun?
I'm quite sure that it is less than 100Ly or even 50LY.
However, the diameter of the spiral arm is 1000LY.
There must be stars which are located at the most upwards side or downwards side of the arm. What is their up down wobbling amplitude. If they have the same wobbling amplitude as the sun, than by definition they don't cross the center of the disc.
Hence, if there is only one star that doesn't cross the center of the galaxy, than it is another solid proof that the stars do not wobble due to the disc.
If we look at the solar system, and monitor the moon cycle, while the earth is dark, we should see the it is wobbling as it orbits around the sun.
So, the only real explanation for that wobbling is a Virtual host point - VHP.
Even if you disagree with that - this is real!
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Do you agree that it means that stars do not orbit exactly at the expected orbital cycle, but they move in a wobbly way around this expected orbital cycle?
A 'wobbly' way is the expected cycle.  We're not in empty space so nobody expects our path to be a nice clean circle or something.
Sorry, This is a severe mistake!
Wobbly way is not expected cycle.
Actually, Newton didn't specify "wobbling activity' in his theory.
Each star must orbit at a specific orbital cycle.
Therefore, the only solution for wobbling is orbital cycle around VHP!
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Hence, it proves that the hypothetical idea about the Up/Down movement or wobbling due to the disc is incorrect as we see the wobbling activity also out the disc.
What happens somewhere not in the disk is no evidence whatsoever about the dynamics of the disk.
I also don't agree with this statment.
S2 (and all the other S stars) are wobbling due to the same idea as our Sun (and actually all the stars in the disc).
Therefore the disc can't be the source for the wobbling activity of our sun.
Let's try to have better understanding on Newton gravity force.
The formula is:
F= G M m / R^2
How based on this formula we can justify a stable wobbling activity around the disc plane of the galaxy?
If M represents the mass of the disc, than by definition the Sun must orbit around that mass in order to set a stable wobbling activity.
It cant just move up/down. There is no formula by Newton for up/down.
This is a fatal mistake!
Please also remember that in the modeling our scientists have actually forgot to add the impact of the wobbling.
Dark matter
Dark matter is a solid prove that our scientists have failed to understand how spiral galaxy really works.
As they have no clue how spiral galaxy really works - they have found a wonderful idea - Dark matter.
They don't understand that there is a simple explanation for what we see in the galaxy.
Therefore, instead of looking for real solution which is based only on real matter, they came with none realistic idea that is called: "Dark matter".
In the same token they could call it "Abra Cadabra"
I have an excellent prove for Abra cadabra -
We can claim in one hand that the orbital velocity of the objects around the galaxy is due to "Abra cadabra" while the proof for "Abra cadabra" is the orbital velocity of the objects around the galaxy.
In the old time - (before Newton) people could claim that the Apple is falling down due to "Abra Cadabra", while the proof for "Abra Cadabra" is that the apple is falling down.
Sorry - You can't proof the dark matter by the same idea which it must support.
You have to be more creative!
Based on our scientists understanding, this dark matter gives an explanation for the orbital velocity of all the stars around the galaxy. They claim that this dark matter is the ultimate solution for the orbital velocities that we see.
So, let's try to understand how this "Abra cadabra" or Dark Matter works in order to keep the stars at the same/similar velocities at while the radius is changing:
The formula for velocity is:
V^2 = G M / R
In order to keep the velocity at the same level at different radius, our scientists believe that they can increase the mass in the orbital sphere by dark matter.
The mass in the sphere is represented by its volume
Volume = 4 π R^3 /3
So, if the density of dark matter is ρ, while the impact of the real matter is neglected, than the total mass in the orbital  sphere should be as follow:
M = ρ * Volume
M = ρ * 4 π R^3 /3 = 4/3 * π  * ρ * R^3
So, we find that the mass in a sphere is a function of R^3 while V^2 is a function of 1/R
V^2 = M G / R = (4/3 * π  * ρ * R^3) G / R = (4/3 * π  * G) * ρ * R^3/ R = (4/3 * π  * G) * ρ * R^2
Hence,
The only way to set the velocity at the same/similar value at different R is by setting the following value to constant
ρ * R^2 =  constant
so we need that
ρ = ƒ 1/R^2
So, if we want to see a same velocity at different radius, the dark matter density (ρ) must be function of  1/R^2.
Hence, as we move closer to the center the density should be higher.
But, we know that the spiral arms starts only from 3KPC (which is about 10,000Ly).
Why is it?
Based on that dark matter density, it should be higher as we move inwards to the center.
So, how could it be that suddenly at a distance lower than 10,000 LY there is no spiral arms?
How could it be that at the bulge - a distance lower than 1KPC (3,300LY) there is no orbital disc while stars orbit at a lower velocity although based on this ρ - dark matter density - the density is higher as we move inwards?
Sorry - this is none realistic idea.
Why the galaxy will set that kind of none linear dark matter density just to justify our none realistic ideas?
The universe is not working according to our wishing list.
There is a very simple explanation for that same/similar velocity over different range of radius.
Each star is connected by gravity to its spiral arm.
It is part of the spiral arm and it moves with it in a constant velocity (more or less) as long as it is located at the arm.
However, as the arm orbits around the galaxy, all the stars in the arm drift outwards.
At the same time, new stars joine the arm from the center.
So, the spiral arms set the disc and not vice versa.
However, at the far edge of the arm, the star can't hold itself to the arm any more. Therefore, it is disconnected from the arm. As it is disconnecting from the arm it is also disconnecting from the disc.
The assumption that the sun stay at the same radius from its first day - is a fatal mistake.
The Sun and all its planets and moons had been formed in the molecular gas cloud at the center of the galaxy. All the matter in that molecular gas cloud had been created at the accretion/excretion disc by the SMBH.
Therefore, our Sun is drifting outwards over time.
Sooner or later, it will be disconnected from the Orion arm (with all the nearby stars) and join/set a new globular cluster.
The Bar is the connection media between the bulge (no disc) to the Arm.
It delivers new stars supply from the Bulge to the arms.
It is so simple, it fits Newton gravity by 100% and there is no need for dark matter or "Abra Cadabra".
This is real science!

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Why each S star sets its unique orbital disc plane although they are directly affected by the mighty gravity force of the SMBH?
The SMBH is not capable of altering the orientation of the orbital axis of a moving object.
Good Answer.
But why is it?
Let's look at our Sun.
http://planetfacts.org/wp-content/uploads/2013/05/Planet-Orbits.jpg
All the stars are orbiting at the same orbital disc plane with one clear exception - Ploto.
It is the further most planet/(object - if you wish) from the Sun and we see clearly that it doesn't share the same orbital disc plane as all the other.
However, its Aphelion is 49.305 AU, while it's Perihelion is 29.658 AU. This is the highest Aphelion/Perihelion ratio (for planet) in the solar system. So, it's orbital path is very elliptical (comparing to all the others).Therefore, it actually cross the orbital disc plane of all the others.
In the same token why can't we expect to see similar activity from a star which gets to the edge of the Arm?
If each star is orbiting around the center and facing a direct gravity force from the orbital sphere (including dark matter - as our scientists say), it is expected to see that some of them gets a very elliptical path and cross the orbital disc.
Do we see any star that does so?
Do we see even one single star that in moving in the direction of the disc?
But please - Not a star which is located far above or far below the disc.
Please - only a star that is quite close to the disc and approaching the disc. Not stars in a globular cluster, as due to the fact that they orbit around each other, at any given moment some of them must orbit in our direction.
For example - if the width of the disc/arm near the sun is 1,000KLY, do we see any star at 4000KLY to 2000KLY above or below us that is moving in the direction of the disc?
I'm sure by 100% that we won't find there even a single star which is moving in the direction of the disc.
If there is a star, this star should fly at ultra high speed away from the disc
This is a solid prove that there is a fatal error with the current concept.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/03/2019 19:56:42
Dear Halc
You have chosen to reject all the evidences which I have offered.
It seems to me that even if Newton was standing next to me and fully support my message - you would still reject any idea which contradicts the current mean stream which you fully support.
Exactly so.  Your message makes no predictions, so it isn't a theory.  If Newton supported that, his celebrity status would not prevent the rejection of the message.

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So, it is quite clear to me that whatever I will say you would reject.
Not at all.  Actually pony up the theory and show how it makes better predictions that the main stream views.  You haven't done that.

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The assumption that - The motion of the stars isn't the same as the motion of the arms, is absolutely incorrect!!!
I have deeply explained how it really works.
I see no deep explanation.   No alternate predictions can be made from what you've told me.  Do you have a measurement of the motion of the arms?  The rotation curve reflects average measurements of the material, not the arms, which, not being objects, have no clear way of speed measurement other than watching them for say 25 million years.

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I have introduced again and again the following image, but you have chosen to ignore that real image/evidence.
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
That is a crude diagram naming the arms.  It is evidence of what names were given to the arms.
It also depicts the parts that we can see vs the parts that are guesswork.
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It shows that based on our verifications all/most of the stars in the milky way disc are located at the arms.
That diagram shows no such thing.  Actual data might show that, but I have no way of gleaning that particular data from that diagram.

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"Outside the gravitational influence of the Galactic bars, the structure of the interstellar medium and stars in the disk of the Milky Way is organized into four spiral arms.[113] Spiral arms typically contain a higher density of interstellar gas and dust than the Galactic average as well as a greater concentration of star formation, as traced by H II regions[114][115] and molecular clouds.[116]"
What does it mean "higher density of interstellar gas and dust" in the following statement"?
It means more stars and gas per unit volume than in the same volume elsewhere.

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Where are all of those nearby interstellar gas and dust?
'Those'?  We're not even in one of the four arms mentioned in that statement.  Anyway, the nearby interstellar gas and dust is nearby, as opposed to more distant interstellar gas and dust which is more distant.

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If we look nearby, we only see G-type stars.
Why is it?
Where do you get this information?  It seems quite a claim.
The three nearest stars are a red dwarf, a type G and a type K.  Sound like we see plenty more than just G stars, but that is just a sample of 3.

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If I understand correctly - there is only one nubila in the whole Orion arm.
That would be amazing...

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Don't forget that in the simulation our scientists are using stars!
Depends on what is being simulated.  A simulation of star formation would need to simulate the material from which the stars form.   If it is simulating only the mass of the material, it might sometimes model that mass as stars (point masses) rather than a fluid.  Some simulations do it with pure fluid dynamics and no point masses at all.

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So, please don't speak about interstellar gas and dust just to confuse us - speak about G-type stars (as you are using them in the simulation).
Did I have a simulation?  Did I ever mention G type stars?

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Now, let's try to verify the density of those stars.
In the following article it is stated:
http://www.solstation.com/stars3/100-gs.htm
"As many as 512 or more stars of spectral type "G" (not including white dwarf stellar remnants) are currently believed to be located within 100 light-years or (or 30.7 parsecs) of Sol -- including Sol itself. Only around 64 are located within 50 light-years (ly), while some 448 are estimated to lie between 50 and 100 light-years -- a volume of space that is seven times as large as the inner sphere within 50 ly of Sol. A comparison of the density of G-type stars between the two volumes of space indicates that the outer spherical shell has around 100 percent of the spatial density of known G-type stars as the inner spherical volume, which suggests that astronomers have identified the great majority of the G-type stars that are actually located within 100 ly of Sol, assuming the same spatial distribution in the Solar neighborhood".
So we know the density of G-type stars in a 50 LY volume: "Only around 64 are located within 50 light-years".
That fully meets by 100% the G-type stars density expectation at 50 and 100 light-years volume: "while some 448 are estimated to lie between 50 and 100 light-years"
That shows the G-type stars density at the Orion arm (There are 448 + 64 = 512  G -type stars in a 100LY around the Sun)
What very convenient round numbers.  All powers of 2.
Never mind.  I get the logic of what they're saying here.  They're also not counting other star types.

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However, what is the G-type stars density at the same volume (100LY) between the orion arm and the next nearby arm?
Is it only 400, 200, 100 or just Zero?
The quote you gave didn't convey that information.
It wouldn't be zero.  The space between is part of the disk which has material at varying densities everywhere.

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I'm sure by 100% that there are big areas between the arms without even a single star.
Depends on your definition of 'big area'.  The larger the volume, the lower the odds that there are no stars in that volume.

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That is a direct outcome from my theory.
How can a theory that cannot be used to make predictions have any direct outcomes?

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So, if that is correct, my theory is also correct.
Again, illustrating that logic is not part of your tool belt.  20 very different theories might allow there to be no G class stars in a given volume, and the finding of such a volume would, by your logic, prove them all correct.


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Thanks
So you agree that at least S2 is wobbling.
It doesn't have a clean orbit.  I wouldn't use the word 'wobble' to describe the discrepancies.

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However, why are you so sure that: "It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion?
Because you've linked to several articles that said pretty much exactly that.

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Once you agree that stars are wobbling - you actually say that you can't explain it by normal orbital cycle.
Only a 2 body system has a normal orbital cycle.

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What is the amplitude of that wobbling activity of the sun?
I'm quite sure that it is less than 100Ly or even 50LY.
I love how you make up numbers instead of do the math from the data from your links.

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Hence, if there is only one star that doesn't cross the center of the galaxy, than it is another solid proof that the stars do not wobble due to the disc.
None of the visible stars have ever crossed the center of the galaxy, since Sgr-A would have eaten them if that happened, and they'd no longer be visible.
Some stars have never crossed the plane of the disk either because they're beyond the gravitational influence of the galaxy, or because they're too young to have ever done it yet.  But I agree that all stars that are part of the galaxy will cross the plane of the disk (or any other plane that goes through the center) given enough time.  Similarly, all objects in the solar system will cross any plane that passes through the sun given enough time.  If it doesn't do that, it isn't part of the solar system.  For that matter, I'm trying to think of any object that is not part of the solar system, but is part of the galaxy, which does not ever cross an arbitrary plane through the sun.  Andromeda has never crossed the solar plane, but we've not yet given it enough time.  More distant objects probably never will.

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If we look at the solar system, and monitor the moon cycle, while the earth is dark, we should see the it is wobbling as it orbits around the sun.
With a very regular wobble, yes.  The existence of Earth would be determined from that motion.  It is how they detect a lot of planets around other stars.

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So, the only real explanation for that wobbling is a Virtual host point - VHP.
No, the explanation is Earth.  A VHP has no gravitational pull of its own.  If it does, you've not given any sort of way I might estimate the pull of one.  I've tried several times to ask, using small words and simple cases and such, and you obviously have no answer.

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Let's try to have better understanding on Newton gravity force.
The formula is:
F= G M m / R^2
No mention of a VHP...  You need a different equation for your idea, but you refuse to give it.

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How based on this formula we can justify a stable wobbling activity around the disc plane of the galaxy?
It makes very good predictions concerning that cycle relative to the plane.
You understand no mathematics.  I'm not going into the derivation of what that formula means for a planar distribution of matter.  Short story: If an object has a mass mostly to one side, it will be attracted by gravity in that direction, accelerating it back to the mass.

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If M represents the mass of the disc, than by definition the Sun must orbit around that mass in order to set a stable wobbling activity.
What definition would that be?

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It cant just move up/down. There is no formula by Newton for up/down.
The formula above predicts it.  Your lack of mathematical skills doesn't change that.


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In the same token they could call it "Abra Cadabra"
I have an excellent prove for Abra cadabra -
We can claim in one hand that the orbital velocity of the objects around the galaxy is due to "Abra cadabra" while the proof for "Abra cadabra" is the orbital velocity of the objects around the galaxy.
That's pretty much how it works.  Substitute 'gravity' for Abra Cadabra, and you have a description of how the motion of Earth was explained, proven by the motion of Earth.  The difference is that they gave a formula from which the motion could be compared with reality.  You've not done that, so you don't have any Abra Cadabra.
Newton didn't just throw a tantrum and repeatedly assert to the guys in some pub (which is essentially what this website is) that his idea explained everything.  But that seems to be your approach.  Of course we cannot be expected to accept it.

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In the old time - (before Newton) people could claim that the Apple is falling down due to "Abra Cadabra", while the proof for "Abra Cadabra" is that the apple is falling down.
I remember there being a physics of 'impetus' for ballistic objects.

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So, let's try to understand how this "Abra cadabra" or Dark Matter works in order to keep the stars at the same/similar velocities at while the radius is changing:
The formula for velocity is:
V^2 = G M / R
In order to keep the velocity at the same level at different radius, our scientists believe that they can increase the mass in the orbital sphere by dark matter.
What is an orbital sphere?

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The mass in the sphere is represented by its volume
Volume = 4 π R^3 /3
So, if the density of dark matter is ρ, while the impact of the real matter is neglected, than the total mass in the orbital  sphere should be as follow:
M = ρ * Volume
If it has uniform density, yes.  Uniform mass distribution would have no net gravitational pull at all since it is the same in all directions.  It all cancels out at any given point.

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M = ρ * 4 π R^3 /3 = 4/3 * π  * ρ * R^3
So, we find that the mass in a sphere is a function of R^3 while V^2 is a function of 1/R
V^2 = M G / R = (4/3 * π  * ρ * R^3) G / R = (4/3 * π  * G) * ρ * R^3/ R = (4/3 * π  * G) * ρ * R^2
You can do algebra.  I wasn't sure.
In short, for uniform matter density spheres, V² is proportional to R² (the rest of the values being invariants), so V is proportional to R, which sort of matches the first part of the rotation curve of a typical galaxy.  One can surmise from this that the density of material in the central sphere of the galaxy is fairly uniform.

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Hence,
The only way to set the velocity at the same/similar value at different R is by setting the following value to constant
Except that the matter in the part of the galaxy with somewhat constant speed of contents is not arranged with spherical symmetry, so the equations you quote above do not apply.  Those equations above are both applicable only to spheres, and you're applying them to a non-spherical arrangement of matter.

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So, if we want to see a same velocity at different radius, the dark matter density (ρ) must be function of  1/R^2.
The matter density can be inferred from the speed, yes.  Not at that proportion if it isn't a sphere.  But it doesn't make a difference if the matter is visible or not.  The mathematics says what the density (ρ) must be to account for a given velocity.
Disk dynamics are more complicated than that of a sphere since material outside the 'orbital' radius has an influence on the speed of material at a particular radius.

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Hence, as we move closer to the center the density should be higher.
Probably, yes, but not in the proportion you state.
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But, we know that the spiral arms starts only from 3KPC (which is about 10,000Ly).
Why is it?
Based on that dark matter density, it should be higher as we move inwards to the center.
Dark matter density was not part of any of your equations.  Matter density is presumably higher towards the center, and maybe it is.  I don't know the numbers, and I'm no expert on how the bar and arms form or interact.  The scientists themselves might not agree on any particular model, but at least they don't just assert that some un-backed descriptions serves as the explanation.

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How could it be that at the bulge - a distance lower than 1KPC (3,300LY) there is no orbital disc
There is still a disk that close in, but plenty of material moves outside of that disk.

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Why the galaxy will set that kind of none linear dark matter density just to justify our none realistic ideas?
What is none-linear dark matter?  OK, you probably meant non-linear, but what is that?  You seem to have made up the term.  How is matter (dark or not) linear or non-linear matter?




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It is part of the spiral arm and it moves with it in a constant velocity (more or less) as long as it is located at the arm.
The arms would not have the shape they do if that were true.  The parts further out have longer to travel and hence take longer to do one lap at that constant speed.  The arms would quickly wind so tight that you'd not be able to make them out.  That is the primary evidence that the idea you propose is wrong.  It doesn't predict what we see.
For the arms to hold their shape like that, they'd need to move in proportion to R, not at a rate that is relatively invariant with R as do the stars.

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This is real science!
Cherry picked evidence, no dynamic model, and no equations that make predictions which can be compared to different predictions from other models.  A real scientist would focus on the points where the model fit most poorly since those are the areas most in need of improvement. It is why they need peer review.  But you run from peer review, changing the subject when blatant contradictions are pointed out.
It is pretty much a great example of not even bad science, but not science at all.

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Quote from: Halc
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Why each S star sets its unique orbital disc plane although they are directly affected by the mighty gravity force of the SMBH?
The SMBH is not capable of altering the orientation of the orbital axis of a moving object.
Good Answer.
But why is it?
Let's look at our Sun.

All the stars are orbiting at the same orbital disc plane with one clear exception - Ploto.
Yes.  Pluto isn't a planet, mostly because it wasn't formed in the same was as a planet, as evidence by its orbital inclination 

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It is the further most planet/(object - if you wish) from the Sun and we see clearly that it doesn't share the same orbital disc plane as all the other.
If you count Pluto, there are several more that are further out.
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However, its Aphelion is 49.305 AU, while it's Perihelion is 29.658 AU. This is the highest Aphelion/Perihelion ratio (for planet) in the solar system. So, it's orbital path is very elliptical (comparing to all the others).
Mercury has similar eccentricity, despite the nice clean circular orbit that the picture gives it.  Pluto is 0.25, and Mercury is 0.21, so that isn't a big difference compared to *all* the others.

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Therefore, it actually cross the orbital disc plane of all the others.
All of the planets and rocks and comets do that.  We discussed that above.  There is no way from them not to.  They all cross any plane that goes through the sun.

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If each star is orbiting around the center and facing a direct gravity force from the orbital sphere (including dark matter - as our scientists say),
What is this orbital sphere that you say is putting a force on each star?  Neither the galaxy (about which we loosely orbit) nor the disk (which deflects that path from a simple orbital path) is a spherical shape.

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it is expected to see that some of them gets a very elliptical path and cross the orbital disc.
Do we see any star that does so?
Yes, the ones that orbit a companion.  Most do.

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Do we see even one single star that in moving in the direction of the disc?
I cannot parse this question, but it seems that about 50% of the stars are moving towards the disk and 50% away, some a lot and some negligibly.  Any other number would be a trend.

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But please - Not a star which is located far above or far below the disc.
Even those must be 50/50 in moving closer or further from it at any given moment.

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Please - only a star that is quite close to the disc and approaching the disc.
You've shown diagrams showing the nearby stars moving every which direction.  So stars nearer the middle of the disc have similar random motion, resulting in that 50/50 split.

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For example - if the width of the disc/arm near the sun is 1,000KLY, do we see any star at 4000KLY to 2000KLY above or below us that is moving in the direction of the disc?
Yes, about half of them.

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I'm sure by 100% that we won't find there even a single star which is moving in the direction of the disc.
There you go.  No math, but at least a prediction.  Look up some survey data and you can verify this thing you're so sure about.
Or get a telescope and do it yourself, being careful only to record the stars that are moving away, and discarding the measurements that show motion towards the disk.

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If there is a star, this star should fly at ultra high speed away from the disc
This is a solid prove that there is a fatal error with the current concept.
If they all do that, it would indeed indicate a problem with the current concepts.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/03/2019 07:17:23
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If there is a star, this star should fly at ultra high speed away from the disc
This is a solid prove that there is a fatal error with the current concept.
If they all do that, it would indeed indicate a problem with the current concepts.

Thanks
However, do you really have the willing to consider a "problem with the current concepts?"
If so, I really admire your approach.

So, let me help you as follow:
Please look at the following article:
https://www.space.com/24432-hypervelocity-stars-ejected-from-milky-way.html
It is stated:
"Strange, Hypervelocity Stars Get Ejected from Milky Way"
"A new class of fast-moving stars are on their way out of the Milky Way, scientists say."
"Our new stars are relatively small — about the size of the sun — and the surprising part is that none of them appear to have come from the galactic core."
So, our scientists are surprising to see a group of Sun like stars that are flying away (fast-moving) from the Milky Way, Those stars don't come from the center of the galaxy.
It is also stated:
"What might have provided the needed galaxy-fleeing kick, however, is still a mystery."
"The big question is, what boosted these stars up to such extreme velocities?" Holley-Bockelmann said. "We are working on that now."
So, let me answer: Mr. Holley-Bockelmann:
Dear Mr. - You don't have to work on that. I have already found the solution (If you are willing to listen..)

So, this "galaxy - fleeing kick" is mystery to Mr. Holley-Bockelmann and other scientists.
It is Mystery as they don't have a clue how spiral galaxy really works.
This Mystery by itself should stop all their activity and force them to evaluate the whole current concepts - from step one.
They shouldn't just say: "The big question is, what boosted these stars up to such extreme velocities?" and just look for a new none realistic idea/patch to fix their unrealistic concept.
If they offer a concept/idea/theory and they see clearly that there is a problem/Mystery/surprising part with this concept, than:
Why our scientists don't say:"Huston - We have a problem?"
Why they don't open their mind to other ideas?
How can we consider them as scientists while their main task is to protect/prove their unrealistic concept under any verification???
I have master in electronics and communications.
If my design group team see a none expected phenomenon/mystery/surprising part - they have to start their verification from step one..
However, I'm quite sure that our scientists would never ever ask themselves if there is a problem with their concept.
They will surly come with new (none realistic) idea to close the gap.
Based on my theory - there is no mystery in this verification.
That exactly meets my expectation.
This is one more solid proof that my theory is 100% correct!

So, what is needed for our scientists to understand that there is a problem with their current concept?
Is there any way to open their eyes?
Or is it none realistic request. For ever & never?

Are you really willing to consider an option for problem with the current concept?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/03/2019 05:31:39
I think the article says "New class" which implies these stars are doing something the typical ones don't.
For one moment I was almost convinced that you are willing to consider an option for a problem with the current concept.
In any case, this article is very important.
It contradicts the current concept and it fully meets the expectations based on my Theory.
It is stated clearly that:
"What might have provided the needed galaxy-fleeing kick, however, is still a mystery."
"The big question is, what boosted these stars up to such extreme velocities?" Holley-Bockelmann said. "We are working on that now."
"the surprising part is that none of them appear to have come from the galactic core."
I have expected to see exactly this kind of activity - and we see it clearly.
How do we know if our theory is correct?
How can we distinguish between real theories to science fiction theory?
Don't you think that the only way to prove any theory is to set an expectation and validate those expectations by real measurements?
How many times our scientists were totally "surprise" from new discoveries?
I'm quite sure that If they had "engineering approach", they would surly look for better ideas instead of adding again one more patch for the none realistic theory.
I see so big gap between the engendering approach to our scientists approach.
In our daily working, the design engineers try to be creative and do not hesitate to start from step one if a  "mystery" had been discovered.
How our scientists consider setting any new discovery if they force themselves to stay in one spot forever and ever?
We are so lucky that our scientists do not have to develop our high tech activity.
The current concept had been developed at the same time with the first transistor.
However, the current concept is still with us, while design engineers all over the world have set breakthrough discovering which dramatically improved our daily life.
So let's compare between the current concepts to my theory.
Do you agree that based on the "current concept" this discovery is not expected as it is stated: mystery/surprising/the big question...?
If so, let me ask you the following:
Are you willing to consider an option that there is a problem with the "current concept theory"?
Is this concept is going to stay with us forever and ever under any mystery or surprising discovery?
How could it be that our scientists don't open their mind to all the difficulties in that concept?
What kind of data/evidence would convince you that this current concept theory is incorrect and none relevant?
Why don't you try to have better understanding about my new theory without being so criticize?
Can I ask you for one moment to forget all the "current concept ideas" and validate my theory only based on evidences, real laws and basic common sense?

Are you willing for the following:
If I offer several real evidences/mesurments that contradicts the expectations based on the "current concept" theory - Than, are you willing to accept the idea that this theory is incorrect? (Please advice how many do you need - as I have already offered more than one)?
If you offer one real evidence/mesurment which contradicts the expectation based on my theory  - Than I'm ready to accept the idea that my theory is incorrect. (only one is needed).
But please - not any sort of evidence which is based on the none realistic "current concept".


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/03/2019 15:32:03
I am unaware of any one particular model that is considered the current concept theory.  That there are multiple models, none of which predict what we see exactly, means that there are problems. 
I do not know of this 'current concept' theory, but the various models do not correspond well to reality, so no one theory is billed as the all encompassing correct one. I can agree to that.
Thanks
So you agree that there are problem with any current concept/modeling/theory which we have.


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Can I ask you for one moment to forget all the "current concept ideas" and validate my theory only based on evidences, real laws and basic common sense?
I cannot validate what you don't have.  I tried real hard to make a simple case, but even that was too much.
I can't do that with your idea since there is nothing from which predictions can be made.  I cannot model its behavior.
So, let me highlight the basic elements which are fundamentals for my theory.
This is very normal request.
Our scientists assume that there is dark matter although they couldn't prove it by any real measurements.
So as they expecting you to agree with the dark matter Idea, I would like to ask you to accept the following elements.
Please don't criticize any element, just accept them as is.
Based on those elements, I will explain the theory and then you can try to validate my whole theory and see if the modeling is working.
So, you are requested to accept the following elements as follow:
1. Newton formula for gravity force is: F=GMm/R^2
2. If star orbits around a host (and assuming that there is no direct extra gravity force on that star) - there is no way  for it to increase its gravity force. Therefore, star would never ever drifts inwards. Actually, it should drift outwards. On every orbital cycle the orbital radius -R is increasing by definition (even if this increasing can be represented by less than Pico mm per cycle).
We had long discussion on that and I'm not sure that I can prove it. But as you can't disprove it, let's assume that this statement is correct. This is a key point in my theory and I ask you to accept it as is
3.Virtual Host Point (VPH)  - We had also long discussion on VPH. You don't agree with that idea. it is clear to me. However, this is very important point in my theory.
4. Accretion disc - The accretion disc is actually Excretion disc. The SMBH creates new Atoms and molecular in that disc.The Ultra high gravity force of the SMBH is converted into small cells of energy that we call Atoms and Molecular. Therefore - Atom is a cell of energy by definition
5. Gravity works locally.
6. Spiral arm is an object.
That's all I ask you to accept.
Once you accept those key elements AS IS, I will explain how Spiral galaxy works and how the whole Universe works. I don't need any other none relevant ideas.
No need for dark matter. No need for dark energy. No need for space expansion or inflation. – just based on what we see. You would find that my theory (which is based on those elements) fully meets any verification that we have.
So, do you agree to accept those elements as is?
If even one element is incorrect, than by definition the theory is incorrect.
If the theory is incorrect - we should find severe problem with this theory
However, if my theory (which is based on those elements) meets our observation by 100% than the theory & those elements are fully correct..
Agree?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/03/2019 00:26:51
So you agree that there are problem with any current concept/modeling/theory which we have.
Pretty much every theory, yes.  None are complete.  Hence the search for the illusive 'Theory of Everything'.

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Our scientists assume that there is dark matter although they couldn't prove it by any real measurements.
Measurements of the acceleration of things (the sun say) are very real measurements.  That doesn't prove dark matter, but something needs to explain that acceleration.

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So as they expecting you to agree with the dark matter Idea, I would like to ask you to accept the following elements.
Please don't criticize any element, just accept them as is.
Based on those elements, I will explain the theory and then you can try to validate my whole theory and see if the modeling is working.
I seriously doubt that.
Quote
So, you are requested to accept the following elements as follow:
1. Newton formula for gravity force is: F=GMm/R^2
2. If star orbits around a host (and assuming that there is no direct extra gravity force on that star) - there is no way  for it to increase its gravity force.
For what to increase its gravity force?  The star, or the host?
Not criticizing, but just trying to clarify.
Quote
Therefore, star would never ever drifts inwards. Actually, it should drift outwards. On every orbital cycle the orbital radius -R is increasing by definition (even if this increasing can be represented by less than Pico mm per cycle).
We had long discussion on that and I'm not sure that I can prove it. But as you can't disprove it, let's assume that this statement is correct. This is a key point in my theory and I ask you to accept it as is
3.Virtual Host Point (VPH)  - We had also long discussion on VPH. You don't agree with that idea. it is clear to me. However, this is very important point in my theory.
4. Accretion disc - The accretion disc is actually Excretion disc. The SMBH creates new Atoms and molecular in that disc.The Ultra high gravity force of the SMBH is converted into small cells of energy that we call Atoms and Molecular. Therefore - Atom is a cell of energy by definition
5. Gravity works locally.
6. Spiral arm is an object.
That's all I ask you to accept.
1 and 5 are in direct conflict with each other.  The rest seem to be sufficiently undefined that they're not wrong, but I think you need to reword either 1 or 5 so they're mutually consistent.

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However, if my theory (which is based on those elements) meets our observation by 100% than the theory & those elements are fully correct..
If there is a theory, yes.
A description that just says: "Everything you see, that's the way it is" meets our observation by 100%, but it isn't a correct theory because it isn't a theory.  I can make no predictions from that terse statement.  A theory suggests falsification tests.  If it passes those tests better than other theories, then it is better than those other theories, but still not necessarily correct.  Nothing in science is every declared proven or necessarily correct.  Even a theory of everything will still be just a theory.  It might be better than anything existing today, but still not 'proven'.

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Agree?
I think I pointed out a contradiction in your points.  Perhaps I misunderstood one of them and just need a clarificiation.  Otherwise, I am willing to agree to your 6 points.

OK, point 1 is just a statement about what Newton said. You're perhaps not actually saying that gravity force obeys that function, just that Newton said it does.  In that case, I'm not sure why it needs to be one of your 6 points.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/03/2019 05:19:50
Quote
However, if my theory (which is based on those elements) meets our observation by 100% than the theory & those elements are fully correct.
If there is a theory, yes.
Thanks!!!
Do appreciate your positive approach.
OK, point 1 is just a statement about what Newton said. You're perhaps not actually saying that gravity force obeys that function, just that Newton said it does.  In that case, I'm not sure why it needs to be one of your 6 points.
Agree
Only five points.
1 and 5 are in direct conflict with each other.  The rest seem to be sufficiently undefined that they're not wrong, but I think you need to reword either 1 or 5 so they're mutually consistent.
So, let me start with 5:
Gravity Works Locally
I think that this is the MOST important one.
In order to understand it, we have to look at our solar system.
Based on Newton formula for gravity force: F=GMm/R^2
The gravity force between the Sun/moon is stronger by more than a double than the Earth/moon gravity force.
Even so, the moon orbits around the Earth instead of directly around the Sun.
This is a current status.
Of course, the Sun has a tidal impact, but still with all its higher gravity force, it can't disconnect the moon from the Earth.
When I thought about it, I have asked myself two questions:
1. Why the moon had chosen to orbit around the Earth instead of around the Sun?
2. How could it be that the sun with more than a double gravity force can't disconnect the moon from the Earth?
The answers are quite simple:
1. Based on my theory - (which I will introduce soon) In the first day of the solar system, the gravity force of the Earth/Moon was much stronger than the Sun/Moon.
2. There is an hysteresis phenomenon also in gravity.
So, as the moon is already connected by gravity to the Earth, an increased external gravity force (from the sun) can't directly disconnect that connection.
It should be quite higher to achieve this goal.
I really don't know to which level of hysteresis it must get before breaking down the local gravity connection.
That gravity Hysteresis is a key element in our discussion.
It shows that each object holds its current local gravity bonding and goes with it where ever it goes.
Therefore, if the sun is orbiting around a local host (even if it is only a virtual host) it will keep with this local orbital connection as long as the gravity hysteresis will keep on.
Hence - Gravity works locally.
Is it clear?


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/03/2019 13:26:37
So, let me start with 5:
Gravity Works Locally
I think that this is the MOST important one.
In order to understand it, we have to look at our solar system.
Based on Newton formula for gravity force: F=GMm/R^2
That conflicts with gravity working locally, so are you saying you're discarding this formula?  If not, you've got a direct conflict with gravity working locally since that formula does not yield a zero result no matter how large you make R.

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The gravity force between the Sun/moon is stronger by more than a double than the Earth/moon gravity force.
Even so, the moon orbits around the Earth instead of directly around the Sun.
According to classic physics, this isn't a function of the greater force, and not a function of the mass of the orbiting thing.  It is a function of acceleration.  An object will orbit Earth if the acceleration due to Earth gravity is greater than the difference of acceleration due to the sun between the object and Earth.  So you compute the (mass independent) acceleration due to the sun of Earth, and also the moon (when they're at maximum radius difference), subtract the two, and compare that to the acceleration of the moon due to Earth.  The latter is larger, hence the moon orbits Earth directly.  You seem to be rewriting these rules, so feel free to offer different ones.
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This is a current status.
Of course, the Sun has a tidal impact, but still with all its higher gravity force, it can't disconnect the moon from the Earth.
When I thought about it, I have asked myself two questions:
1. Why the moon had chosen to orbit around the Earth instead of around the Sun?
2. How could it be that the sun with more than a double gravity force can't disconnect the moon from the Earth?
The answers are quite simple:
1. ... In the first day of the solar system, the gravity force of the Earth/Moon was much stronger than the Sun/Moon.
2. There is an hysteresis phenomenon also in gravity.
So, as the moon is already connected by gravity to the Earth, an increased external gravity force (from the sun) can't directly disconnect that connection.
It should be quite higher to achieve this goal.
I really don't know to which level of hysteresis it must get before breaking down the local gravity connection.
That gravity Hysteresis is a key element in our discussion.
It shows that each object holds its current local gravity bonding and goes with it where ever it goes.
Therefore, if the sun is orbiting around a local host (even if it is only a virtual host) it will keep with this local orbital connection as long as the gravity hysteresis will keep on.
Hence - Gravity works locally.
Is it clear?
OK, a different usage of the term 'local'.  Gravity seems to have memory, and play favorites with things to which it attached in the past.   If that is what you mean by 'locally', then I suppose it is clear.  The term usually means that it doesn't work after a certain distance.  In QM, it means any distance.  Nothing that happens here can have an immediate effect elsewhere.  That's the principle of locality, which has never been proven.  The primary evidence against it is 'spooky action at a distance'.  Anyway, that's what 'local' means in physics, but you seem to mean a different thing. I see no contradiction in what you've said.

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Based on my theory - (which I will introduce soon)
You sound like a religious evangelist promising say the rapture 'soon'.  Always soon, but nothing ever produced.
We're up to like 250 posts here.  As I said in the prior post:
Quote from: Halc
I seriously doubt that.
Not saying anything in your post is wrong except that mention of 'soon'.  That is car-salesman speak for 'never', and yes, I put the religious evangelists on the same tier as car salesmen.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/03/2019 15:02:38
I see no contradiction in what you've said.
Thanks
Actually I fully agree with your explanation.
It seems to me that I should call it: "hysteresis phenomenon in gravity". Which means:
Gravity seems to have memory, and play favorites with things to which it attached in the past.
So thanks about your very positive highlight.
Not saying anything in your post is wrong except that mention of 'soon'
Sorry that I don't go directly to the Theory as I need to clarify also the following basic element:
1. Newton formula for gravity force is: F=GMm/R^2
This formula is OK as long as we ignore the time.
However, if we need to calculate the gravity force after some interval time the formula should be as follow:
F=GMm/(R+tΔ)^2
tΔ represents the change in the radius over time.
For example, we know that the Moon drifts outwards by 3.5 cm per year.
Therefore:
Δ (moon) = 3.5 cm /per year.
or
Δ (moon) = 3.5cm/365= 0.00959 cm/per day.
I assume that we might be able to set the formula for Δ as a function of the masses, radius and some constant.
But I didn't consider it deeply yet.
In any case, the updated formula for gravity which also reflects time interval is:
F=GMm/(R+tΔ)^2
Based on this formula, we should be able to estimate the gravity force for any object after any given time. (Assuming that there is no external force/friction).
It also shows that any orbital object has a spiral shape by definition.
So, the updated basic elements are as follow:
1. Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2
2. Virtual Host Point (VPH) 
3. SMBH Excretion disc
4. Hysteresis phenomenon in gravity
5. Spiral arm is an object.
Is it clear so far?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/03/2019 21:17:03
What makes the radius increase?
The radius increases due to the normal activity by gravity.
Any orbital object increases it's orbital radius over time. This radius increasing is not due to any sort of Tidal. It is just normal gravity activity.
Unfortunately, Newton didn't add the impact of time interval in his gravity formula.
Only if tΔ is always positive, and the moon's tΔ is currently negative, making its closes approach on March 19.
Please do not confuse yourself with the elliptical cycle of the orbital object.
The following formula: F=GMm/(R+tΔ)^2 is related to a perfect orbital cycle.
I assume that we can set the adaptation for any elliptical cycle. (based on kepler)
However, the whole idea is that there is a constant increasing in the radius.
Therefore, for elliptical cycle, we need to compare the current orbital cycle to the previous one.
If we do so, we should find that at any given location in the elliptical cycle, there is increasing in the radius.
I hope that it also answer your questions about the Sun/Earth orbital cycle.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 15/03/2019 21:50:54
The radius increases due to the normal activity by gravity.
Any orbital object increases it's orbital radius over time. This radius increasing is not due to any sort of Tidal. It is just normal gravity activity.

Where does the energy come from to raise the object against the gravitational potential?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/03/2019 06:21:48
The Ultimate Theory for our Universe

1. New mass creation
Further the discussion about new matter creation at the excretion disc of the Milky way:
Please see: https://www.thenakedscientists.com/forum/index.php?topic=75261.40
New Atoms and Molecular are created at the accretion/excretion disc.
This matter drifts out from the excretion disc lifted upwards (or downwards) by the mighty magnetic power around the SMBH.
As it falls back near the SMBH, it sets the molecular gas clouds.
2. New Stars/Planets/Moons forming activity
In those clouds the molecular orbits at high velocity around the center (of the cloud) and cristilize into stars/planets/Moons (Due to the gravity impact of the nearby SMBH).
So, by the time that the new born star emerge from the molecular gas cloud, it also carry all of its planets and moons.
However, each object comes as a hot gas ball. Therefore, all/most of the rocky planets and moons are so nicely round balls.
Our Earth and Moon had been formed at the same day as our Sun and also with the same molecular structure as the Sun. Hence, 98% of the new born Earth and moon were mainly gas (mostly Hydrogen - Exactly as the Sun in his first day). However, in the other 2% (or less) it includes all heavy atoms/molecular that we see on Earth today (as Silicate, cupper, Iron, Gold, water and so on).
The big gas planets (as Neptune) could keep most/some of the their molecular gas over time due to their higher gravity mass. However, our relatively small Earth and moon have lost most of the gas into space. We are very lucky that the earth could keep significant portion of its water and atmosphere during all the time from its first day.
So, at the first day, the Earth and the Moon were heavier by at least 98/2 (about 50 times) than their current mass.
Due to element no. 1 (F=GMm/(R+tΔ)^2) it is also clear that all the objects were much closer to each other in the first day.
In any case, at that first day, the Earth/Moon gravity was much stronger than Sun/Moon gravity.
Therefore, the moon had chosen to orbit around the Earth instead of around the Sun. It keeps orbit around the Earth due to element No.4 (Hysteresis phenomenon in gravity).
3. Virtual host Point - VHP
As I have explained, Each new born star had to orbit around the center of the gas cloud in order to crystallize from molecular gas into real star. That orbital cycle sets the first level of Virtual host Point. (as there was no real object at the center of the gas cloud). Therefore, as the star emerge from the gas cloud it also emerge while it orbits around its VHP1.
Please see the following excellent image (It was not set by our scientists - but it is correct by 100%):
http://www.biocab.org/Motions_of_the_Solar_System.jpg
The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).
That VHP1 is valid for the Sun and for any new born star in the galaxy.
So, each star in the galaxy orbits by definition around its unique VHP1 (that was the center of the gas cloud). It will continue to orbit around this VHP1 (due to element No.4 - Hysteresis phenomenon in gravity) and goes where ever the VHP1 goes.
Therefore, with related to gravity force, the nearby aria do not "see" the star itself, but they "see" or set the gravity impact due to its VHP1.
However, there is one more important impact by this VHP1. It doesn't just change the gravity center location of the Sun, but it also increases dramatically the effective mass of the Sun due to VHP1. (I hope that I have used the correct name)
Let's look again in that diagram.
In order to hold the Sun in that blue dots (Apparent motion of the solar system), The orange ball (Virtual point) must represent higher effective mass.
We can extract that effective mass value by simple calculation (based on its elliptical orbit and velocity), as we did for S2. We might find that the effective VHP1' Sun mass is higher by more than 1000 times with regards to the real Sun' mass.
So, the nearby aria gravity "see" the VHP1' Sun location and its effective mass (and ignore the real location of the sun and its real mass value).
Therefore, with regards to the Sun' nearby stars, we have to monitor the location of each VHP1 star and its effective mass.
Based on this theory, we should find that the VHP1 of each star is located almost at the same location in the spiral arm (although each star is moving to different direction), while the effective mass of each VHP1 is much higher than the real mass value of each star.
Therefore, when we set the gravity simulation in the spiral arm, we need to focus on the locations of the VHP1 of each star and its effective mass, (which is estimated to be much higher than the real mass value).
This is a key element why the gravity force of the nearby G stars in the Orion arm can hold/bond them together in the arm, (although by monitoring the real location/mass/density we might think that it's not good/high enough).
I will stop at this point in order to verify that this message (so far) is clear to you.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/03/2019 14:06:57
In any case, at that first day, the Earth/Moon gravity was much stronger than Sun/Moon gravity.
Therefore, the moon had chosen to orbit around the Earth instead of around the Sun. It keeps orbit around the Earth due to element No.4 (Hysteresis phenomenon in gravity).
There is no hysteresis in the function you gave.  It references a current recession rate, nothing from the past.  It reduces force from Earth without reducing the force from the sun, so if the magnitude of the force determines which thing it orbits (it doesn't), then Newton's formula should have Earth hanging onto the moon harder than your formula suggests.

I also can build a fairly trivial perpetual motion machine from your one formula, but you've made it pretty clear that conservation of momentum (both linear and angular), and conservation of energy are all to be discarded, so a perpetual motion machine is not surprising.

The formula you give also conflicts the VHP idea.  The formula has the moon being attracted to Earth (and other objects), not its VHP.  So this discrepancy is very unclear.

3. Virtual host Point - VHP
As I have explained, Each new born star had to orbit around the center of the gas cloud in order to crystallize from molecular gas into real star. That orbital cycle sets the first level of Virtual host Point. (as there was no real object at the center of the gas cloud). Therefore, as the star emerge from the gas cloud it also emerge while it orbits around its VHP1.
Please see the following excellent image (It was not set by our scientists - but it is correct by 100%):
http://www.biocab.org/Motions_of_the_Solar_System.jpg
The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).[/quote]
If that picture were in any way representative of the actual motion, the orange ball moves at 217 km/sec and circles the galaxy in 200 million years.  The blue dot path appears to be about 8 times longer, so if it is supposed to keep pace with this orange VHP, it would need to move about 8 times as fast, or 1700 km/sec.  Instead it labels the speed at 5-20 km/sec, or about 1% of the speed it needs.  The sun's speed is neither, so that picture is not an "excellent image".
I am not commenting on your idea here, just commenting that the solar system cannot follow a path like that around the galaxy.  Get your evidence from real data, not from drawings made by somebody with no credentials.

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That VHP1 is valid for the Sun and for any new born star in the galaxy.
So, each star in the galaxy orbits by definition around its unique VHP1 (that was the center of the gas cloud). It will continue to orbit around this VHP1 (due to element No.4 - Hysteresis phenomenon in gravity) and goes where ever the VHP1 goes.
Not according to that picture you linked.

If the moon is receding from Earth, is it taking its VHP with it?  If so, why does it still orbit Earth and not this moved VHP?

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Therefore, with related to gravity force, the nearby aria do not "see" the star itself, but they "see" or set the gravity impact due to its VHP1.
Your formula references the other mass, not the VHP.  This is the part that is really unclear.  I would expect a formula that says F=some function of the mass of the object and the distance to (and other properties of) the VHP, not the other masses in the system.

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Therefore, when we set the gravity simulation in the spiral arm, we need to focus on the locations of the VHP1 of each star and its effective mass, (which is estimated to be much higher than the real mass value).
This is a key element why the gravity force of the nearby G stars in the Orion arm can hold/bond them together in the arm, (although by monitoring the real location/mass/density we might think that it's not good/high enough).
I will stop at this point in order to verify that this message (so far) is clear to you.
Well, I asked some clarification questions in the last post, which went unanswered, so I don't much see the point in asking them.  Nevertheless I asked a few above.  The fact that they are asked means of course that the message isn't totally clear.  Feel free to clarify those points, or just ignore them and carry on.  It seems to be your style to just evade places where the idea doesn't work.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/03/2019 17:26:55
There is no hysteresis in the function you gave.  It references a current recession rate, nothing from the past.  It reduces force from Earth without reducing the force from the sun, so if the magnitude of the force determines which thing it orbits (it doesn't), then Newton's formula should have Earth hanging onto the moon harder than your formula suggests.
Sorry, I don't understand your question.
In any case, I have introduced five elements as follow:
1. Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2
2. Virtual Host Point (VPH) 
3. SMBH Excretion disc
4. Hysteresis phenomenon in gravity
5. Spiral arm is an object.
Each one stands by its own.
So, I don't understand why do you mix between no. 4. (Hysteresis phenomenon in gravity) to no. 1. (Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2)
Let me start from the molecular gas which falls next to the SMBH, and explain it again:
Due to the high gravity impact of the SMBH, the molecular gas starts to rotate (around the center) at high velocity - as a tornado. That rotation in the molecular gas set the gas cloud.
So, we get a molecular gas cloud which rotate at high velocity around its center while it also orbits around the nearby SMBH.
At the first day of this cloud we might find there only molecular gas. However, due to the high rotation, the molecular gas starts to crystallize into small objects (mainly due to the very high internal rotation velocity and the gravity impact of the nearby SMBH). Later on those small objects merge with each other and set bigger and bigger objects. Eventually, when the star will emerge from the gas cloud it should carry with it all its planets and moons. By that time the gas cloud had been drifted away from the SMBH due to element no. one.
I assume that the crystallize process could take place only if the gas cloud is close enough to the SMBH. If it is too far away, the internal rotation velocity of the gas cloud should decrease and at some point, the whole crystallize process could stop.
In any case, by the time that the star emerge from the gas cloud, each moon orbits directly around its plane, each planet orbits directly around the star while the star orbits around the center of the gas cloud which represents the VHP1 point of that star.
Therefore, those moons and the planets (that orbits around the star) don't have a VHP as they orbit directly around the star.
That should answer your following question:
If the moon is receding from Earth, is it taking its VHP with it?  If so, why does it still orbit Earth and not this moved VHP?
Never the less, there might be some/many other objects in that gas cloud which didn't join that new star system by the time that it emerges from the gas cloud. So, yes, we can claim that there might be moon and planets that orbits directly around the center of the gas cloud - VHP. However, due to their relatively small size, by the time that they emerge from the molecular cloud, they will not be able to hold themselves by gravity around that VHP. Therefore, all of them will be ejected into space.
So, only the star with its very high mass (With all its planets, moons and other objects (as Oort cloud) that it had collected from the gas cloud) will be able to keep on orbiting that VHP1.
Is it clear?

Quote
Please see the following excellent image (It was not set by our scientists - but it is correct by 100%):
http://www.biocab.org/Motions_of_the_Solar_System.jpg
The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).
If that picture were in any way representative of the actual motion, the orange ball moves at 217 km/sec and circles the galaxy in 200 million years.  The blue dot path appears to be about 8 times longer, so if it is supposed to keep pace with this orange VHP, it would need to move about 8 times as fast, or 1700 km/sec.  Instead it labels the speed at 5-20 km/sec, or about 1% of the speed it needs.  The sun's speed is neither, so that picture is not an "excellent image".
I am not commenting on your idea here, just commenting that the solar system cannot follow a path like that around the galaxy.  Get your evidence from real data, not from drawings made by somebody with no credentials.
This image just show the orbital movement of the sun around Its VHP1. You shouldn't monitor the blue dots in order to get an idea about the orbital velocity. We actually know this data. It sets one full cycle in 60 million years..
However, can you please advice what is the amplitude/radius of that orbit which our scientists call "wobbling".
Please - I don't think that we should de- estimate the ability of "somebody with no credentials".
The estimated orbital shape and the radius are very critical for this discussion. Based on this data we can extract the estimated effective mass of the Sun at VHP1. This should help us to verify the gravity forces at the spiral arm,
Title: Re: How gravity works in spiral galaxy?
Post by: syhprum on 17/03/2019 19:45:34
I think that as regards the Suns attraction to our Moon the Earth/Moon system is sufficiently compact with the centre of gravity below the surface of the Earth that it revolves around the Sun as one unit.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/03/2019 00:07:58
There is no hysteresis in the function you gave.  It references a current recession rate, nothing from the past.  It reduces force from Earth without reducing the force from the sun, so if the magnitude of the force determines which thing it orbits (it doesn't), then Newton's formula should have Earth hanging onto the moon harder than your formula suggests.
Sorry, I don't understand your question.
That one wasn't a question.  Just an observation.  For instance, the moon currently has an average recession rate of 3.8 cm/year, but that rate was larger in the past at times, and smaller at other times.  The formula references nothing from these past rates, only the current rate, which isn't hysteresis.

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In any case, I have introduced five elements as follow:
1. Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2

...  no. 1. (Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2)
That is not Newton's formula, and without a way to add two numbers which are in different units, it cannot be used to make any direct calculations, only relative ones, which means I can make a perpetual motion machine, but cannot compute how much energy I can extract from it.  It doesn't explain recession, only makes force a function of that recession, which just changes the orbital speed of something at a particular radius R from what classic physics would compute.

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In any case, by the time that the star emerge from the gas cloud, each moon orbits directly around its plane, each planet orbits directly around the star while the star orbits around the center of the gas cloud which represents the VHP1 point of that star.
Therefore, those moons and the planets (that orbits around the star) don't have a VHP as they orbit directly around the star.
That should answer your following question:
If the moon is receding from Earth, is it taking its VHP with it?  If so, why does it still orbit Earth and not this moved VHP?
You're telling me that the moon orbits Earth directly and doesn't have a VHP of its own.  In fact nothing in our solar system has a VHP since they all directly orbit some mass.  Perhaps the sun doesn't, but you've not described how to compute the location of its VHP or how the motion of an object relates to that VHP.  Does the VHP attract the sun? By how much? If it does, what about the mass of some object that is far closer: Does that divert the sun?  What if the mass is strong enough to pull the sun away from its VHP?
These are really trivial questions with Newtonian physics, and I only need that one formula, plus F=ma to compute the motion of anything.  Force on the moon from the sun is greater than force from Earth, therefore the moon always accelerates towards the sun, even when between the two.  It only accelerates straight towards Earth when Earth is between the two and the sun and Earth are pulling in the same direction.  That's what it means for one force to be greater than the other.  The moon always accelerates primarily towards the greater force.  It does not mean that it directly orbits the thing with the greater force.
Anyway, I cannot make any of these observations with your physics since you don't tell me how to compute the location of a VHP or how to compute the motion in relation to that VHP.  It is why I cannot predict where my 3 unit-masses will go.  Your theory is absent and I can make no predictions.  I tire of this endless repeating of this summary.  Yes, I get the 6 points.  I cannot demonstrate problems with it since they all incompletely defined.

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So, only the star with its very high mass (With all its planets, moons and other objects (as Oort cloud) that it had collected from the gas cloud) will be able to keep on orbiting that VHP1.
Is it clear?
It is clear what you mean by VHP.  It has been for a long time.  I just don't know where to find it or how the motion of something relates to it.  You mentioned it having an elliptic orbit around it sort of like S2 does, except S2 orbits Sgr-A, not a VHP.  Earth orbits the sun, not a VHP.  It seems few things have one.
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Please see the following excellent image (It was not set by our scientists - but it is correct by 100%):
http://www.biocab.org/Motions_of_the_Solar_System.jpg
The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).
This image just show the orbital movement of the sun around Its VHP1. You shouldn't monitor the blue dots in order to get an idea about the orbital velocity. We actually know this data. It sets one full cycle in 60 million years..
If I shouldn't monitor the blue dots, how is that image 100% correct?  The up/down motion in and out of the disk has a 60 million year period.  The in-out period towards and away from the center has a 200 million year period.  The periods are different, so the path isn't a helix at all, and it certainly isn't a tight helix as depicted in that picture.  The picture is wrong in just about every way, but it shows what you want, so you take it for fact instead of actual measurements.  Classic selection bias.

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However, can you please advice what is the amplitude/radius of that orbit which our scientists call "wobbling".
There isn't a circle, so there's no radius.  The sine-wave picture you showed was more representative, even if it also showed many more waves per lap than the 3.5 waves.  Work it out.  If the 5-7km/sec speed is anything near accurate, the amplitude of the sine wave in and out of the disk is about 200 LY.  I think I've read that it is more than that, but that's what I compute from 6km/s and a 60 MYear period sin wave.  If there is a VHP associated with that motion, the sun crosses it every 30 million years, and has maximum acceleration when it is furthest from it, not when closest to it as you would expect if there was an actual object with mass responsible for that motion.  Orbital motion has a fixed axis, and your biocab picture does not depict motion about any axis.

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Please - I don't think that we should de- estimate the ability of "somebody with no credentials".
If they draw that picture and present it as fact, then I can de-estimate them all I want.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/03/2019 15:10:46
For instance, the moon currently has an average recession rate of 3.8 cm/year, but that rate was larger in the past at times, and smaller at other times
How do you know that?
Do you have any valid measurements to prove this idea, or is it based on current concept?
Based on my understanding this recession rate of 3.8 cm/year was never ever larger in the past

The formula references nothing from these past rates, only the current rate, which isn't hysteresis.
Please do not confuse the following formula F=GMm/(R+tΔ)^2 with hysteresis.
This formula gives the data about the radius in the past or into the future.
F=GMm/(R+tΔ)^2
If we know the value Δ we can calculate what was the radius 1 billion years ago, and what it should be in the next one billion.
But that information is not hysteresis.
hysteresis means that if the moons orbit around the Earth than it will continue to do so even if there will be a nearby object with much stronger gravity force (sun).
In fact nothing in our solar system has a VHP since they all directly orbit some mass.
That is correct.
Perhaps the sun doesn't,
That is also correct.
but you've not described how to compute the location of its VHP or how the motion of an object relates to that VHP.
Thanks for the question.
There isn't a circle, so there's no radius.  The sine-wave picture you showed was more representative, even if it also showed many more waves per lap than the 3.5 waves.  Work it out.  If the 5-7km/sec speed is anything near accurate, the amplitude of the sine wave in and out of the disk is about 200 LY.  I think I've read that it is more than that, but that's what I compute from 6km/s and a 60 M Year period sin wave.  If there is a VHP associated with that motion, the sun crosses it every 30 million years, and has maximum acceleration when it is furthest from it, not when closest to it as you would expect if there was an actual object with mass responsible for that motion.  Orbital motion has a fixed axis, and your biocab picture does not depict motion about any axis.
Thanks
If I understand you correctly, the sun is located about 200LY above the disc
Please look at the following diagram:
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
So the sun is located 200 LY above the disc and  it is still moving up.
Based on your explanation in order for the sun to set full cycle it needs to get all the way down, crossing the disc to the same downwards amplitude (200LY) and then get back again to the same distance from the galactic plane.
Therefore, 60 million years is needed for just one cycle.
However - this is a severe mistake.
The Sun has no need to cross the galactic disc plane.
If it is currently high above the disc(200 LY) it will stay above the disc as long as needed.
It will just need to orbit around its VHP1 (as any other nearby star does).
what about the mass of some object that is far closer: Does that divert the sun?  What if the mass is strong enough to pull the sun away from its VHP?
Based on hysteresis it is clear that the sun will continue to orbit around its VHP1 even if there will be a nearby mass with much stronger gravity force. So, the sun is going to orbit its VHP1 as long as it takes.
These are really trivial questions with Newtonian physics, and I only need that one formula, plus F=ma to compute the motion of anything.  Force on the moon from the sun is greater than force from Earth, therefore the moon always accelerates towards the sun, even when between the two.  It only accelerates straight towards Earth when Earth is between the two and the sun and Earth are pulling in the same direction.  That's what it means for one force to be greater than the other.  The moon always accelerates primarily towards the greater force.  It does not mean that it directly orbits the thing with the greater force.
That is clear.
Anyway, I cannot make any of these observations with your physics since you don't tell me how to compute the location of a VHP or how to compute the motion in relation to that VHP.  It is why I cannot predict where my 3 unit-masses will go.  Your theory is absent and I can make no predictions.
My theory is based on the data which is available to all of us.
I would recommend to look at the nearby stars and verify their relatively orbital velocity.
If I understand it correctly, we see that their relatively maximal velocity is about 15 Km/sec in all direction. So, by average they are moving at about 7.5 Km/s around their VHP. That should also be correct for our sun. If that is correct, and assuming that the radius of VHP1 is about 2LY, we can extract the expected cycle time for one full VHP1 cycle. Do you agree?
Yes, I get the 6 points
Thanks
You mentioned it having an elliptic orbit around it sort of like S2 does, except S2 orbits Sgr-A, not a VHP. 
No, I disagree.
S2 orbits around its VHP1. (although it might have much shorter radius to that VHP1 than our sun to its VHP1).
In any case, I'm quite sure that even the VHP1 of S2 does not directly orbit around the SMBH.
It orbits around VHP2 while this one might orbit around the SMBH.
So, S2 orbits around VHP1 that orbits around VHP2 that orbits around SMBH.
However, we will discuss about it once we get to the bulge.
The picture is wrong in just about every way, but it shows what you want, so you take it for fact instead of actual measurements
Thanks
However, if the estimated orbital radius of sun around its VHP1 is about 2LY, with an orbital velocity of about 7.5 Km/s than this picture might be after all fully correct.


.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/03/2019 19:16:18
For instance, the moon currently has an average recession rate of 3.8 cm/year, but that rate was larger in the past at times, and smaller at other times
How do you know that?
Do you have any valid measurements to prove this idea, or is it based on current concept?
Admittedly current understanding.  It assumes Newtonian physics isn't nonsense.
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Based on my understanding this recession rate of 3.8 cm/year was never ever larger in the past
It is definitely measured to be currently slowing, which means it was larger in the recent past.

I don't know how old Earth is in your theory since you deny classical ways (such as nuclear physics) of aging stars and such.  I suppose you would need to measure the recession rate of the sun from the galactic center and extrapolate its age from that.

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This formula gives the data about the radius in the past or into the future.
F=GMm/(R+tΔ)^2
It's a formula for force, not radius.  Kindly show an example of this formula for a real object (say the moon) since, as I said, it is totally unclear how to add two numbers of different units. You can do it for the current force between Earth and moon, since you supposedly know its Δ.

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But that information is not hysteresis.
hysteresis means that if the moons orbit around the Earth than it will continue to do so even if there will be a nearby object with much stronger gravity force (sun).
I see you've ignored everything I've posted on this, and even state that my explanation on it makes things clear for you (quoted below), but it seems it was not clear.

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Quote from: Halc
In fact nothing in our solar system has a VHP since they all directly orbit some mass.
That is correct.
And yet the new moon accelerates away from Earth, something it seemingly should not do if its motion is determined by the Earth and not by Newton's laws.  Or are you denying that the moon accelerates away from us during a new moon?

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Quote from: Halc
but you've not described how to compute the location of its VHP or how the motion of an object relates to that VHP.
Thanks for the question.
I've only asked that one about 20 times
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The location of VHP1 (VHP1 is the first orbital level of any star in the galaxy) is based on the amplitude of the wobbling activity.
So it cannot be predicted at all.   You watch where it goes, and you theory says that's where it should go.  That's making predictions after the things happen.  You don't know.  That's why you could not find the VHP of any of the 3 objects in my simple example. You were not able to watch it for a while and then guess where the 'wobbling' put each of the VHPs.  Guess what, Newtonian physics doesn't need to watch first.  It has a concrete prediction answer as to where those objects will go.  The physics needs some adjustments for relativity to be exactly correct, but for small masses at small speeds, it is impressively accurate.
Your answer is worthless because you have to watch where it goes first, and then declare that your theory predicted that path, and even then cannot extend that pattern into future moments because none of those objects will have a regular 'wobble' about some VHP.  Regular motion only works with objects that can be reduced to pairs like everything in our solar system.  The sun has no regular motion, and requires full information of all mass in the galaxy to predict its path.

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Quote from: Halc
There isn't a circle, so there's no radius.  The sine-wave picture you showed was more representative, even if it also showed many more waves per lap than the 3.5 waves.  Work it out.  If the 5-7km/sec speed is anything near accurate, the amplitude of the sine wave in and out of the disk is about 200 LY.  I think I've read that it is more than that, but that's what I compute from 6km/s and a 60 M Year period sin wave.  If there is a VHP associated with that motion, the sun crosses it every 30 million years, and has maximum acceleration when it is furthest from it, not when closest to it as you would expect if there was an actual object with mass responsible for that motion.  Orbital motion has a fixed axis, and your biocab picture does not depict motion about any axis.
Thanks
If I understand you correctly, the sun is located about 200LY above the disc
I said nothing about where we are now in that cycle.  I said the amplitude of that sine-motion is about 200LY, and that based on one bit of data from a very unreliable site.  I said I think the real amplitude is larger than that, but still of the same order of magnitude.  I did not look up our current location relative to the mean (middle) of the disk.

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Please look at the following diagram:
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
So the sun is located 200 LY above the disc and  it is still moving up.
Moving up yes.  The site says nothing about our current offset or the amplitude of the wave, but if we have 10 million years to go, we're closer to the middle right now than we are to the peak of that wave.

That site shows way too many waves per rotation around the galaxy, but not nearly as many as the biocab site did.

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Based on your explanation in order for the sun to set full cycle it needs to get all the way down, crossing the disc to the same downwards amplitude (200LY) and then get back again to the same distance from the galactic plane.
As the picture shows, yes.  It doesn't show numbers, but it shows the disk (dotted white) being crossed twice per wave like that.
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Therefore, 60 million years is needed for just one cycle.
A figure in that range is consistent across several sites, yes.
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Sorry - this is a severe mistake.
Well, you have to forgive those guys for going with physics that works.  Not sure why they'd want to do that.
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The Sun has no need to cross the disc plane.
Every picture you post shows the sun crossing the plane (frequently), and yet now you say it doesn't do that.
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If it is currently high above the disc(200 LY) it will stay above the disc as long as needed.
Where does this need come from?  How do you explain something not being attracted south when almost all of the mass of the galaxy is in that direction?  I can't say you're wrong since you've totally denied gravitational pull from mass.
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It will just need to orbit around its VHP1 as all the other nearby stars do.
Does the sun bring its VHP up there with it?  The VHP doesn't come down because it has no mass, and the sun is not attracted to the mass below it, but rather to the VHP which 'needs' to be up there for some purpose.

Your story gets more and more entertaining.

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Based on hysteresis it is clear that the sun will continue to orbit around its VHP1 even if there will be a nearby mass with much stronger gravity force. So, the sun is going to orbit its VHP1 as long as it takes.
As long as it takes for what?  Is there some purpose that needs to be completed before it can let go of the VHP?

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Quote from: Halc
These are really trivial questions with Newtonian physics, and I only need that one formula, plus F=ma to compute the motion of anything.  Force on the moon from the sun is greater than force from Earth, therefore the moon always accelerates towards the sun, even when between the two.  It only accelerates straight towards Earth when Earth is between the two and the sun and Earth are pulling in the same direction.  That's what it means for one force to be greater than the other.  The moon always accelerates primarily towards the greater force.  It does not mean that it directly orbits the thing with the greater force.
That is clear.
Apparently not so clear given what you're still posting above.  You still seem to find this hysteresis necessary to explain why the moon orbits Earth.

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Quote from: Halc
It is why I cannot predict where my 3 unit-masses will go.  Your theory is absent and I can make no predictions.
My theory is based on the data which is available to all of us.
That's predicting where it will go after it has already gone there.  A prediction does it without the data being available ahead of time.  I gave the current conditions.  Classic theory can predict the motion from just that.  Your idea cannot.
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Sorry - this is a severe mistake.
Exactly.

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I would recommend to look at the nearby stars and verify their relatively orbital velocity.
If I understand it correctly, we see that their relatively maximal velocity is about 15 Km/sec in all direction. So, by average they are moving at about 7.5 Km/s around their VHP. That should also be correct for our sun. If that is correct, and assuming that the radius of VHP1 is 2LY, we can extract the expected cycle time for one full VHP1 cycle. Do you agree?
If any object moves at 7.5 km/s in a circle of radius 2 LY, yes you can extract the cycle time from that, and fairly trivially. What does that have to do with anything?

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Quote from: Halc
The picture is wrong in just about every way, but it shows what you want, so you take it for fact instead of using actual measurements
Thanks
That wasn't a compliment.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/03/2019 06:00:10
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This formula gives the data about the radius in the past or into the future.
F=GMm/(R+tΔ)^2
It's a formula for force, not radius.  Kindly show an example of this formula for a real object (say the moon) since, as I said, it is totally unclear how to add two numbers of different units. You can do it for the current force between Earth and moon, since you supposedly know its Δ.
I don't understand why you claim that I add two different units.
Let's look at the following at the following value of updated radius:
R(At time t≥0) = R(at time t=0)+tΔ
t = time (let's assume in years)
Δ = drifting outwards speed in Km per year
If t=T = 1 year
than
tΔ = TΔ = drifting distance in 1 year
Therefore,
As R is represented in Km
than
1 year * Δ = drifting distance in Km (in that one year time period)
Hence, with regards to our moon:
We need to present the value of Δ in km instead of by cm.
therefore:
Δ(moon)= 3.8cm/year = (3.8 / 10^5)km/year.
Therefore, the expected Earth/moon radius by next year should be as follow:
R(t=one year from now) = R(t=0, the current radius) km + 1 * (3.8/10^5) Km
What is wrong with that?
In any case, the value of Δ might increase over time but it should never ever decrease over time.
So, for just few years - we can assume that the value of Δ is 3.8 cm/year.
However, it is clear to me that if we will verify the real value in 1 billion years from now, that value should be higher.
In the same token the real value of Δ in 1 billion years ago was smaller than the current value..
I do believe that for any orbital system we can extract the value of Δ without any need for measuring it.
That value must be a function of the masses, current radius and some constant.
However, I'm not sure that I can offer currently the correct formula for Δ.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/03/2019 09:02:09
4. Spiral arm
Does the sun bring its VHP up there with it?  The VHP doesn't come down because it has no mass, and the sun is not attracted to the mass below it, but rather to the VHP which 'needs' to be up there for some purpose.
It seems to me that you still don't understand how VHP works.
So, let me do it for one more time:
The Sun does not bring it's VHP anywhere. It works the opposite way.
The Sun' VHP1 brings the Sun everywhere it goes..
 So, the Sun orbits around its personal VHP1.This VHP1 must obey to the gravity law in the galaxy and in the spiral arms.
In order to understand why VHP1 fully obey to gravity law in the spiral arms we have to monitor the density of G-star in the arm.
Let's look at the following article:
http://www.solstation.com/stars3/100-gs.htm
"As many as 512 or more stars of spectral type "G" (not including white dwarf stellar remnants) are currently believed to be located within 100 light-years or (or 30.7 parsecs) of Sol -- including Sol itself. Only around 64 are located within 50 light-years (ly), while some 448 are estimated to lie between 50 and 100 light-years -- a volume of space that is seven times as large as the inner sphere within 50 ly of Sol. A comparison of the density of G-type stars between the two volumes of space indicates that the outer spherical shell has around 100 percent of the spatial density of known G-type stars as the inner spherical volume, which suggests that astronomers have identified the great majority of the G-type stars that are actually located within 100 ly of Sol, assuming the same spatial distribution in the Solar neighborhood".

So, based on this article, we know that the density of G-type stars in a 50 LY volume: "Only around 64 are located within 50 light-years".
That fully meets by 100% the G-type stars density expectation at 50 and 100 light-years volume: "while some 448 are estimated to lie between 50 and 100 light-years"
64 * 7 = 448
So, in each 50LY sphere in the arm there are about 64 stars.
That is not a density wave and not even close to density wave.
That shows that there is a fixed/constant G stars density in our aria in the arm.
that fixed density set the requested gravity which is needed to hold all the stars in the Orion arm.
In a density wave (or if the stars were wobbling as our scientists estimate) we should see different densities - as the stars are crossing the arm is some sort of a random movement and also moving up and down while they cross the disc)
We can compare it to the flash in our arm. The density of the flash is almost the same. If we try to measure the density of flash outside our arm, we should find that as there is no arm, there is also no flash over there. Therefore, outside our arm the density of the flash is Zero.
In the same token, outside the spiral arm, the star density should drop to ZERO!!!
I don't claim that the arm is a very nice pipe with a diameter of 1,000KL.
In some arias nearby it might be thicker or narrower. However, at any nearby aria it must have the same G-star density!!!
Exactly 64 stars in 50LY sphere!
So, outside the spiral arm there are no stars. (Especially not a single star – as there is no gravity force to hold it in the arm.
There might be some bridges between the arms, but even in the bridge it is expected to find similar star density..
If one star will dare to move outwards the am it would be kicked out from the arm and from the disc at ultra high velocity.
We are so lucky that our sun is located between all of those nearby stars.
They are not just crossing by stars.
They are our Sun' brothers. They all might have been created more or less at the same time.
They are together for very long time.
Each one of those stars orbits around its VHP1. If we are located 200Ly above the galactic disc, than all of us should stay where we are..
Based on 65 Stars per 50 LY sphere, I would assume that the orbital cycle of each star around its personal VHP1 should be in the range of 2LY (or less).
If we could find the exact location of VHP1 for all the nearby stars, we might find that the relative velocity between the VHP1 is almost zero.
There might be a room for small drifting of VHP1 in the arm, but overall, they must keep the same density at this range from the center of the galaxy.
If we move closer to the galaxy center, the density should be higher. If we move further away from the center the density might be lower.
Therefore, we get the following image of the galactic disc.
It is thicker as we move to the center while it is thinner as we move further away.
Please look at the following image of the milky way disc:
https://www.quora.com/How-big-is-the-Milky-Way
So, spiral arm is an object. (A real object)
It is made out of stars, while the disc is made out of arms.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/03/2019 09:42:33
It seems to me that you still don't understand how VHP works.
So, let me do it for one more time:
The Sun does not bring it's VHP anywhere. It works the opposite way.
The Sun' VHP1 brings the Sun everywhere it goes..
Fine.  What brings the VHP of the sun up out of the disk such that the sun will not return to the disk as you describe?
From what you've said, the VHP follows the sun because you need to watch the sun to determine where the VHP is, not watch/compute the VHP to determine where the sun will go.  That's why I'm proposing that the sun takes its VHP with it, not the other way around.  All your descriptions work with the sun determining the VHP instead of the other way around like you're trying to assert here again.  But you obviously have no idea where to put the VHP except to let the sun's motion determine it, which is the VHP following the sun, or its motion being determined by the sun, not the sun's motion being determined by the VHP.
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Each one of those stars orbits around its VHP1. If we are located 200Ly above the galactic disc, than all of us should stay where we are.
If the VHP is not located in the middle of the arm, then that explains why the sun might stay permanently to one side.  But every one of the pictures you link (instead of actual survey data) shows the path of the sun crossing the plane of the disk twice with each oscillation north and south.  You claim these pictures are 100% accurate, and yet here you claim that our motion is local and held for a long time (through many orbits about the VHP) to one side of the disk where our VHP holds us.
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Based on 65 Stars per 50 LY sphere, I would assume that the orbital cycle of each star around its personal VHP1 should be in the range of 2LY (or less).
You suggest here and in a prior post that the VHP is only 2 light years from here (closer than any star) and that we trace a sort of orbital path around that virtual point, which is a complete violation of Newtonian forces.  There is negligible mass inside that small circle (most of it being our own Oort cloud), and thus gravity cannot be the force that accounts for that motion.

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So, the Sun orbits around its personal VHP1.This VHP1 must obey to the gravity law in the galaxy and in the spiral arms.
Gravity law says nothing about a personal VHP, which is not subject to gravity since it has no mass, or if it does, you give no indication on how much or where it comes from, or why that mass is not drawn to the mass of the disk.  Gravity law says an object will go where the forces from gravity take it, and not where the VHP takes it.  If you're going to propose a different law, don't quote the law that you deny.

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We can compare [star density in spiral arms] to the flash in our arm. The density of the flash is almost the same. If we try to measure the density of flash outside our arm, we should find that as there is no arm, there is also no flash over there. Therefore, outside our arm the density of the flash is Zero.
In the same token, outside the spiral arm, the star density should drop to ZERO!!!
It is called flesh.  Anyway, perhaps you could consult actual survey data to verify that the star density between arms is zero, because that would indeed sink the density wave theory, and finding a non-zero density would sink your idea.  Pick a spot that doesn't have one of these inter-arm pipes doing a high-speed transfer.

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Each one of those stars orbits around its VHP1. If we are located 200Ly above the galactic disc, than all of us should stay where we are..
So all those pictures that are 100% accurate are in fact 100% nonsense?  Not somewhere in between?  One of the few things that I agreed with was that each picture shows our motion crossing the plane of the disk periodically.  I just didn't like the period that most of them depicted.

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If we could find the exact location of VHP1 for all the nearby stars, we might find that the relative velocity between the VHP1 is almost zero.
But you cannot find it because no star moves like that.  How does the star determine where its VHP is and thus where it should go? By watching itself and noting the middle of its orbital path?  Do you not see the absurdity of this?  If you cannot find the VHP from a given state, how can the sun?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/03/2019 10:36:58
It is called flesh.  Anyway, perhaps you could consult actual survey data to verify that the star density between arms is zero, because that would indeed sink the density wave theory, and finding a non-zero density would sink your idea.  Pick a spot that doesn't have one of these inter-arm pipes doing a high-speed transfer
Yes!
I fully agree!
If you can find a single star hanging between the arms by itself, than my theory is none relevant by definition.
But please, don't show me an image of far end galaxies. Just a clear image from the nearby aria.
Try to find one single star that is there by itself between the Orion arm and any nearby arm.
However, as I'm sure by 99.9..9% that you won't find it, than this should "indeed sink the density wave theory".
You suggest here and in a prior post that the VHP is only 2 light years from here (closer than any star) and that we trace a sort of orbital path around that virtual point, which is a complete violation of Newtonian forces.  There is negligible mass inside that small circle (most of it being our own Oort cloud), and thus gravity cannot be the force that accounts for that motion.
Why do you insist to ignore my explanation again and again.
Why do you insist to see real object at VHP while I have stated clearly that it is Virtual point.
No real object is there!!!
I wonder what I should say to open your eyes.
So, if I will say that there is a dark matter, would it be more logical for you?
If so, think about virtual dark matter.
You don't see it; you don't smell it and you don't feel it.
It is there becouse I said that it is there.
If our scientists can set a dark matter at the center of the galaxy, why can't I set a dark matter at the VHP?
If Newton can work for the dark matter while it is at the center of the galaxy, why it can't work for the dark matter at the VHP1?
Again I use the dark matter just to show you a possibility that there is an effective mass at the VHP point which is used as a host for our sun. In reality - there is nothing over there (Not even dark matter).
However, please think about dark matter if it is more convenient to you.
I assume that anyone who accepts the idea of dark matter at the center of the galaxy should also accept the idea of dark matter at that VHP1.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/03/2019 12:03:10
If you can find a single star hanging between the arms by itself, than my theory is none relevant by definition.
https://www.astronomynotes.com/ismnotes/s8.htm is a nice note about the problems with the model of arms being objects, and why that theory doesn't produce observed results.  It proposes several alternatives, density wave theory being only one of them.
The reason I linked it is because it also says this near the front:
"There are many stars that are also in-between the spiral arms, but they tend to be the dimmer stars (G, K, M-type stars).".  My simple google search turned up countless sites that verify this, but none that state than the space between arms is devoid of stars.

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But please, don't show me an image of far end galaxies. Just a clear image from the nearby aria.
Any image would be taken from inside the orion arm and shows orion stars nearby.  An image by itself does not show distance, and hence where those stars actually are.  For that you need parallax data or some other method of computing distance.  Hence you consult survey data, not some image.
Of course you will not do that since it is easier to just stand on your box and make up your data.  Real science is work.

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Quote from: Halc
You suggest here and in a prior post that the VHP is only 2 light years from here (closer than any star) and that we trace a sort of orbital path around that virtual point, which is a complete violation of Newtonian forces.  There is negligible mass inside that small circle (most of it being our own Oort cloud), and thus gravity cannot be the force that accounts for that motion.
Why do you insist to ignore my explanation again and again.
Why do you insist to see real object at VHP while I have stated clearly that it is Virtual point.
No real object is there!!!
I didn't claim there was a real object there.  I'm asking why the sun would circle a region of space with nothing in it?  What forces could possibly account for that motion?  Without force, mass tends to move in a straight line, at least according to Newton, but you seem to be discarding Newton's laws without replacing them.  You have the sun pulling this incredibly high acceleration without any valid force to account for that.  Such acceleration would be very measurable, and yet we don't see it.  So it doesn't take this nonsense 2 LY radius path.  That is total fiction.

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I wonder what I should say to open your eyes.
So, if I will say that there is a dark matter, would it be more logical for you?
It wouldn't be virtual then, but it would at least explain a tight orbit like that.  We're a binary star after all orbiting a massive dark companion.  But we don't take a path like that.  You're just making all that up.

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If so, think about virtual dark matter.
You don't see it; you don't smell it and you don't feel it.
It is there becouse I said that it is there.
If our scientists can set a dark matter at the center of the galaxy, why can't I set a dark matter at the VHP?
Then it would be real, not virtual.  The dark matter at the center of the galaxy is a black hole, which is dark by definition.  All you see is the radiation from the material falling into it.  Any such dark object that close to the sun would be very visible by watching the material near it.
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If Newton can work for the dark matter while it is at the center of the galaxy, why it can't work for the dark matter at the VHP1?
So are you changing your VHP idea now to some proposal that there is an actual non-luminous chunk of matter assigned to every star?  That still doesn't explain how we stay above the galactic disk since any VHP with mass would be attracted to the mass of the disk same as a luminous chunk of mass.  What goes up must come down, unless you throw it fast enough, and we have nowhere near the velocity needed to escape the plane of the disk. Something as tiny as Earth has a higher escape velocity than that.

Again I use the dark matter just to show you a possibility that there is an effective mass at the VHP point which is used as a host for our sun. In reality - there is nothing over there (Not even dark matter). [/quote]Then what force accelerates the sun in such a tight orbit?  Can't be the gravity of the VHP since it has no mass.  Can't be the gravity of other objects or the arm since none of that material is so close that the sun passes it during that tight circle.  So where does the force come from to account for that significant acceleration that nobody has ever noticed?

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I assume that anyone who accepts the idea of dark matter at the center of the galaxy should also accept the idea of dark matter at that VHP1.
Sgr-A is not usually listed as dark matter since the thing is quite visible, even if only by virtue of the material falling into it.  So nobody proposes significant dark matter at the center of the galaxy, which would give all objects a much more Keplerian rotation curve than what is observed.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/03/2019 14:48:32
https://www.astronomynotes.com/ismnotes/s8.htm
is a nice note about the problems with the model of arms being objects, and why that theory doesn't produce observed results.  It proposes several alternatives, density wave theory being only one of them.
The reason I linked it is because it also says this near the front:
"There are many stars that are also in-between the spiral arms, but they tend to be the dimmer stars (G, K, M-type stars).".  My simple google search turned up countless sites that verify this, but none that state than the space between arms is devoid of stars.
Sorry.
There is no real data in that article.
It just show that our scientists have no clue about spiral arms and the disc.
In any case, yes - I agree with them that there are many stars in - between the spiral arms - but only as a bridge between the arms and only while they carry the same/similar density as in our aria (64 stars per 50LY sphere)
In this article they don't say even one word about the possibility that there are wide aria without even a single star.
Therefore, this article is none relevant for our discussion.

VHP -
Let me try to explain it for you one more time:
The VHP1 is a direct outcome of the star forming activity in the gas cloud.
I have stated that in order to set a star from a gas cloud, the matter must rotate around the center of the gas cloud (as a tornado). However, as the matter orbits around the center of the gas cloud, in the center itself, there are no real matter. Most of the matter drifts outwards and crystallize into new star. So, during the process of star forming, the star orbits around the center of the gas cloud, while in this gas cloud there is no matter in the center. Therefore, that center of the gas cloud represents the location of VHP1 for any new born star.
However, this gas cloud also set a gravity impact/"connection" with its nearby aria, as the center of the gas cloud represents its center of gravity.
So, the gas cloud is "connected" to its nearby aria by the gravity which is represented by the location of its center.
However, that gravity center also represents the Sun VHP1 as it had been emerged from the gas cloud.
Due to hysteresis from now on the nearby aria "see" that VHP1 as the host center of the Sun.
Is it clear to you?
If not, than please try to accept it as is.
One of the key element in this discussion is VHP.
I have stated that:
3.Virtual Host Point (VPH)  - We had also long discussion on VPH. You don't agree with that idea. it is clear to me. However, this is very important point in my theory.
VHP = Virtual host Point.
There in nothing there. However, the sun directly orbits around that Virtual point. It also represents an effective mass due to the orbital cycle of the sun around that point.
You have agreed to accept this idea.
So please - take it as is and at the end of my introduction you are more than welcome to disagree.
So, VHP1 should represent the following:
1. The sun orbits directly around VHP1.
2. VHP1 represents an effective Sun' mass due to the Sun orbital movement around that point. In order to extract it, we need to monitor the orbital of the sun (around VHP1) and its velocity (as we did for S2 in order to extract the SMBH mass)
Please, would you kindly accept this idea for this discussion?
We can discuss it later on after introducing the full theory.


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/03/2019 17:59:47
Sorry.
There is no real data in that article.
Nor is there any real data in most articles you link. I linked to it for the comment that there were such stars everywhere, and for the bonus explanation of why arms cannot be objects.  I see you could not refute the argument presented.
If I find a star between the arms, you'll claim it is one of these bridges between arms.  So it is quite up to you to find a 100 light year radius space between the arms with no stars since this would actually be heavy evidence against density wave theory.

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In this article they don't say even one word about the possibility that there are wide aria without even a single star.
Yes they do, and I quoted the words.  They say there isn't one.  All the sites I found that mention the subject say this.

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Therefore, this article is none relevant for our discussion.
It doesn't match your fiction, therefore the data is not relevant.  I knew this reply was coming when I posted that link.

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So, during the process of star forming, the star orbits around the center of the gas cloud, while in this gas cloud there is no matter in the center. Therefore, that center of the gas cloud represents the location of VHP1 for any new born star.
However, this gas cloud also set a gravity impact/"connection" with its nearby aria, as the center of the gas cloud represents its center of gravity.
So, the gas cloud is "connected" to its nearby aria by the gravity which is represented by the location of its center.
However, that gravity center also represents the Sun VHP1 as it had been emerged from the gas cloud.
How can the gas all move to one side and become the star and leave the center of gravity somewhere else?  The center of gravity of the sun is in the middle of the sun.  It doesn't orbit itself.
If a VHP is the center of gravity of something, how can the sun orbit a VHP 2 LY away when there is no significant mass in that volume except the sun at the very edge of the circle?  Center of gravity means there is other mass out there, not necessarily any at the center, but it balances.  If you take a sheet of material and put mass everywhere around it to represent several objects, the sheet will balance exactly if held by its center of gravity, and it will balance nowhere else.  But a 2 LY radius sheet with only the sun at one edge and a couple trivial mass planets is going to fall over with all the weight on one side like that.

The center of gravity of any object is itself, and not a VHP.  The center of gravity of a group of objects is somewhere else, and with two (and no more) objects, the two may or may not orbit that center of gravity.  The sun has no binary companion, so it has no orbital relationship with a star or group of stars.

You never brought up center of gravity with my 3-equal-weight object example.  You never suggested that the VHP of any of the objects might be the center of gravity of the group or the center of gravity of the remaining two objects.  Your descriptions imply that a VHP is just some virtual point that is permanently married to a star despite all the mass around that star possibly going its separate ways.  That is not a description of the center of mass of anything.

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Due to hysteresis from now on the nearby aria "see" that VHP1 as the host center of the Sun.
Is it clear to you?
Yes, its quite clear.  It is also completely wrong.  Nothing can 'see' a massless virtual point.  If it is the center of mass of a collection of objects, the what is 'seen' is the collection of objects, not the center of mass of the group or the VHP of each individual object if they all have different VHPs.
Your inability to describe the motion of the most trivial 3-body system shows that the idea is nonsense.

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If not, than please try to accept it as is.
It is clear enough and has been for a while.  I have no intention of accepting something like that.

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VHP = Virtual host Point.
There in nothing there. However, the sun directly orbits around that Virtual point. It also represents an effective mass due to the orbital cycle of the sun around that point.
If the VHP doesn't have mass, then it doesn't have effective mass either.  It cannot attract anything on its own.  Hence it has no effect at all.

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You have agreed to accept this idea.
I agreed to listen, in hopes that the theory would finally get posted, but it is quite clear that there is none, and it never will. So I don't see the point in leading you on. I cannot accept something this so completely wrong from somebody who has demonstrated algebraic manipulation, but not vector arithmetic, which puts you at about late middle-school with your mathematics skills.
I cannot correct anything of yours or be of particular assistance since you seem not to want constructive feedback.  Lack of that desire dooms any hope for your idea.  So no, I can no longer pretend to accept the idea to see where it will go.  I just don't care where it goes if it rests on that foundation.

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So please - take it as is and at the end of my introduction you are more than welcome to disagree.
I'm not sure how much more repetitive introduction I can take.

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We can discuss it later on after introducing the full theory.
You have no such thing.  I don't believe such promises anymore.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/03/2019 04:59:58
If the VHP doesn't have mass, then it doesn't have effective mass either.  It cannot attract anything on its own.  Hence it has no effect at all.
O.K.
As the VHP is very critical for you, than let's discuss this issue and see where the main problem is.

So, let's start from the idea that there is one pure gas cloud that orbits around the SMBH (while the matter in the gas cloud rotates at high velocity around the center of the gas cloud - as a Tornado).
1. Do you agree that in this case the gravity works between the centers of the gas cloud to the center of the SMBH?
if so
2. Let's also assume the following: As the matter in the gas cloud rotate around the center of cloud, and due to the Huge gravity impact of the nearby SMBH, at the first phase it crystallized into small objects that continue to orbit around the center of the gas cloud (all over the gas cloud).
So, how does the gravity between the SMBH work with that gas cloud and all of those small objects there?
Do you agree that the gravity still works based on the center of the gas cloud to the center of the SMBH?
If so;
3. Let's assume that those small objects crystallized/merge with each other and set bigger and bigger objects in the cloud. Eventually we might find few significant objects that orbits around the center of that gas cloud.
Do you agree that the gravity still works based on the center of the gas cloud to the center of the SMBH?
If so;
4.Let's assume that eventually we might find that those big objects have merged with each other and set one main massive object which still orbits around the center of that gas cloud. This object consume significant portion of the matter in the gas cloud, so even the matter that was in the center had drifted outwards and become part of this object.
In this case, how the gravity works?
Do you agree that the gravity still works between the center of the SMBH to the center of the gas cloud?
If so, then you have just confirmed why each new born star orbits around a VHP1 (which is located at the center of the gas cloud).
If no, please advice why and how you think the gravity between that new massive orbital object in the gas cloud (which orbits around the center of the gas cloud) to the SMBH should work.


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/03/2019 12:55:28
So, let's start from the idea that there is one pure gas cloud that orbits around the SMBH (while the matter in the gas cloud rotates at high velocity around the center of the gas cloud).
1. Do you agree that in this case we can set the gravity calculation between the centers of the gas cloud to the center of the SMBH?
That calculation leads one to false motion.  A gas cloud would be attracted more to a primary mass where it is closer.  The non-uniform gravity would pull the cloud apart and also make the inner parts orbit faster than the outer parts.  The cloud is thus pulled into a disk whose center of gravity is eventually the same as that of the black hole.  This happens even to solid objects if the non-uniformity of the gravity exerts greater force that the force of the object's own gravity trying in vain to hold it together.  Hence Saturn having rings right now because a solid object was pulled apart in this manner.  A gas clould is already pulled apart, so it spreads into a ring without resistance.

Yes, a gas cloud at a considerable distance will not be pulled apart like that, but if its components have high relative velocity like you describe, it probably just moves off in separate directions that it is already moving.  Gas clouds that hold together under their own gravity don't have high rotation rates, but they do at least have a meaningful center of gravity.

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2. Let's also assume the following: As the matter in the gas cloud rotate around the center of cloud, and due to the Huge gravity impact of the nearby SMBH, at the first phase it crystallized into small objects that continue to orbit around the center of the gas cloud (all over the gas cloud).
A sufficiently dense gas cloud (which is sort of disk shaped if it is near the black hole) would be capable of forming 'objects', yes.  That's how stars form.  S2 is theorized to have formed from a cloud that came from a considerable distance.  Passing near the black hole strips much of that cloud away, but it also triggers star formation in the cloud which was too uniform to do it without the disturbance.
Your idea seems to have the cloud appear continuously near the black hole.  Continuous influx of material leaves it no choice but to form stars, but it is unclear what forces push these objects away to make room for new ones.  You have matter appearing from nothing, so I suppose energy can also come from nowhere, but it still needs reaction mass to give the angular momentum it needs to escape gravity.  There is a reason rockets always immediately tilt to the East to get away from Earth.  It takes much more energy to go straight up, and when you turn the engines off, the straight up rocket just falls right back down again.  Motion to the side (especially the East) allows it to get to and stay at a desired altitude.  Rockets bring their own reaction mass with them  No idea how your stars do it, or the moon for that matter since you deny classic sources of that lateral force.

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So, how the gravity between the SMBH and that gas cloud with all of those small objects work?
Do you agree that it still works based on the center of the gas cloud to the center of the SMBH?
I never agreed to that, so I don't still agree.  The gas cloud forms a ring or disk whose center of gravity is the black hole itself.  The motion of the gas cloud is not approximated by a point mass at that location.

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3. Let's assume that those small objects crystallized with each and set bigger and bigger objects in the cloud.
Eventually we might find few significant objects that orbits around the center of that gas cloud.
Sure, since the center of the gas cloud is the black hole.
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Do you agree that the gravity still works based between the center of the gas cloud to the center of the SMBH?
They're the same point.  No, gravity does not act on the center of gravity of multiple objects.  It acts on the objects.  Jupiter and Saturn form a center of gravity that is at some virtual location that moves around the solar system.  Nothing is attracted to that virtual point.  Anything that passes very near that point will not be diverted in the least by it.  Virtual points do not have mass, and gravity acts only on mass.

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Let's assume that eventually we might find that those big objects have merged with each other and set one  massive object which still orbits around the center of that gas cloud. This object consume significant portion of the matter in the gas cloud, so even the matter that was in the center had drifted outwards and become part of this object.
In this case, how the gravity works?
If all of the matter of the cloud collects into one object, then the center of mass of the cloud is now the center of mass of the object.  You can't move the matter without moving the center of mass.  If the cloud creates two objects, then the center of mass of that material is somewhere between those two objects, and given where they are (near a black hole), neither will take any notice of this center of mass.  They will each orbit the black hole, and they would need to do so at a sufficiently large distance to form in the first place.  Close to the black hole (say about one AU from Sgr-A), a star would be torn apart by the gravity gradient and reduced to a disk again.

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Do you agree that the gravity still works between the center of the SMBH to the center of the gas cloud even if there is a significant object which orbits around the center of that gas cloud (while the center of the gas cloud has no significant matter or even without any matter?
Why would you suggest I agree to something like this? How can something big orbit a near massless thing?  That makes no sense.  Objects orbit other masses.  Massless things have no significant pull and cannot exert the force necessary to alter the otherwise straight path that said large object would take. The Earth does not orbit Titan, but if Titan were large enough, Earth would indeed orbit it.  And Titan is not exactly a small thing either.  It seems to have more mass than you're giving to this vanishing cloud.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/03/2019 16:11:18
Dear Halc

You don't agree to accept the idea that VHP1 is actually the center of the gas cloud.
However, how do we know that the current hypothetical ideas about gas cloud, star forming activity, gravity forces near the SMBH and many others ideas are correct?
How do we know that the following message is correct?
Anything that passes very near that point will not be diverted in the least by it.  Virtual points do not have mass, and gravity acts only on mass.
Actually, few years ago, our scientists have discovered a star which is moving very close to the SMBH.
They were expecting for fireworks as the SMBH gravity is going to smash that star.
But surprisingly - that star had not been effected by the mighty SMBH gravity force.
So, time after time our scientists set an expectation and surprisingly - it doesn't work according to their expectations.
You also claim that:
The non-uniform gravity would pull the cloud apart and also make the inner parts orbit faster than the outer parts.
Can you prove it?
There are several gas cloud orbiting around the SMBH. Do you see there any cloud that can confirms this theory?
You discuss about rockets:
There is a reason rockets always immediately tilt to the East to get away from Earth.  It takes much more energy to go straight up, and when you turn the engines off, the straight up rocket just falls right back down again.  Motion to the side (especially the East) allows it to get to and stay at a desired altitude.  Rockets bring their own reaction mass with them  No idea how your stars do it, or the moon for that matter since you deny classic sources of that lateral force.
Based on this answer, due to the ultra high gravity force of the galaxy, there is no way for a star to be ejected out without an engine/rocket.
Surprisingly, somehow stars can escape with any rocket:
https://www.sciencealert.com/a-star-has-been-kicked-out-of-the-milky-way-it-knows-what-it-did?perpetual=yes&limitstart=1
"But researchers from the University of Michigan have identified one hypervelocity star that appears to have been ejected from the stellar disk rather than the galactic bulge."
If there was a dark matter, why that dark matter couldn't hold that star in the galaxy/disc?
How could it be ejected without any engin?
I have a solid evidence why the current theory about the the density wave and dark matter are a fatal errors:
Please look at the following image of the galactic disc:
https://www.quora.com/How-big-is-the-Milky-Way
We see that as we move further away from the center of the galaxy, the disc became thinner and thinner.
I have explained why based on my theory we get exactly that disc shape.
However, is it possible to get that shape based on the current theory?
Please see the following explanation about Orbital inclination:
https://en.wikipedia.org/wiki/Orbital_inclination
"Orbital inclination measures the tilt of an object's orbit around a celestial body. It is expressed as the angle between a reference plane and the orbital plane or axis of direction of the orbiting object."
If we look at the solar system, we can find that all the planets/orbital objects has some sort of Orbital inclination:
Mercury: https://en.wikipedia.org/wiki/Mercury_(planet)
inclination - 7.005° to ecliptic
Venus: https://en.wikipedia.org/wiki/Venus
Inclination  3.39458° to ecliptic
Neptune: https://en.wikipedia.org/wiki/Neptune
Inclination   1.767975° to ecliptic
Pluto: https://en.wikipedia.org/wiki/Pluto
Inclination - 17.16°
It is clear that as we move further away from the center, the impact of that Inclination is more visible:
For pluto we see it very clear as it is located far away from all the other planets in the solar system.
https://en.wikipedia.org/wiki/Pluto#/media/File:Plutoorbit1.5sideview.gif
We see clearly that that at that far end location, the orbital cycle Pluto is moving significantly below/above the disc plane.
So, it is clear that any orbital object has some sort of Inclination if it orbits around a central main host.
Therefore, if the stars in the galaxy were orbiting due to the gravity force of the dark matter we would expect to see that as the radius is longer, the disc should be thicker and thicker. As we move further away the Inclination has more significant impact and therefore it is expected to see a thicker disc at the far end. That fully meets our modeling that I have introduced before..
So, by definition - the current view of the galactic disc contradicts the idea that stars in the galaxy orbit around the galaxy due to main gravity force. (As dark matter).
However, as usual, I'm quite sure that you are not going to let one more evidence about the fatal error in the current concept to confuse you.
S2 is theorized to have formed from a cloud that came from a considerable distance. 
So, from where S2' came from?
What does it mean: "considerable distance?"
Where S2' gas cloud was when S2 had been formed?
How far it was from the SMBH at that time?
What is the age of S2?
Have we ever monitored the process of new star formation in the gas cloud?
Have our scientists really monitored the galactic disc?
Did they really try to see if there are arias without stars?
If they do so, I can promise you that they will find that in some places (between the arms) in a sphere of more than 1000LY there is not even a single star.
You have stated that there are many unclear problems with the current theories/concepts.
My theory fully meets all the evidences.
You have accepted the unrealistic idea of dark matter. But you are not willing to accept the idea of VHP that is vital for my theory.
That is OK with me
I really appreciate your great effort so far.
Thank you.




Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/03/2019 21:37:19
You don't agree to accept the idea that VHP1 is actually the center of the gas cloud.
I never said I disagree with that, and I don't think you ever said that a VHP was the center of a cloud.  You sometimes indicated that it might be the center of mass (CoM) of that cloud, and I didn't disagree with that definition.  Any collection of mass has a center of mass, and if you want to label that point a VHP, you are welcome to do so.
What I do not agree with is you giving point-mass properties to the CoM, which is not a valid property of the CoM of a collection.  I also do not agree with you moving all the mass to one side but not having the CoM move with it.  If a VHP is the center of mass, then it moves when the material does.  It follows the material, not the other way around.  The material can go anywhere the forces take it (say in totally different directions for example) and the CoM is ineffectual in preventing that. An object (part of the collection or not) can pass close by to that point and be totally unaffected by it.  So the center of mass is not a center of gravity.  There is no such thing as the latter, which is only a mathematical simplification of a reasonably rigid sphere.

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However, how do we know that the current hypothetical ideas about gas cloud, star forming activity, gravity forces near the SMBH and many others ideas are correct?
It seems not specific to the SMBH.  Sufficiently dense clouds of gas are capable of producing stars, but since the distribution of gas is fairly uniform, gravity pulls it in no particular direction, so the process needs a seed to start, which often takes the form of a passing mass (like the SMBH) or by a shock wave from an explosion somewhere, which is probably what initiated our solar system.

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How do we know that the following message is correct?
Quote from: Halc
Anything that passes very near that point will not be diverted in the least by it.  Virtual points do not have mass, and gravity acts only on mass.
That derives from Newton's equations.  It can be demonstrated with a simple rubber sheet model, which is an illustrative if not exact physical model of Newton's laws.

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Actually, few years ago, our scientists have discovered a star which is moving very close to the SMBH.
They were expecting for fireworks as the SMBH gravity is going to smash that star.
But surprisingly - that star had not been effected by the mighty SMBH gravity force.
The black hole is a tiny target, so it is not likely to hit it.  So parts of the star get stripped away if it gets too close, and I'm not sure how much 'fireworks' they expected from that.  Only the radio and IR light gets through, and most of the gas just joins the accretion disk and doesn't actually fall in all at once.  Maybe they underestimated the distance at which it would pass.

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So, time after time our scientists set an expectation and surprisingly - it doesn't work according to their expectations.
That happens all the time.  Science would make no progress without it.

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Quote from: Halc
The non-uniform gravity would pull the cloud apart and also make the inner parts orbit faster than the outer parts.
Can you prove it?
There are several gas cloud orbiting around the SMBH. Do you see there any cloud that can confirms this theory?
The cloud of which you speak orbit at a considerable distance.  They distort as they get near the black hole, but gravity might be able to pull that distortion back again.  The distortion creates higher and lower density areas and the higher density regions help seed new star formation that would otherwise not have as readily occurred.
As for proving this, it again derives from gravitational law.  I cannot prove it to somebody in denial of such mathematics.

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You discuss about rockets:
Quote from: Halc
There is a reason rockets always immediately tilt to the East to get away from Earth.  It takes much more energy to go straight up, and when you turn the engines off, the straight up rocket just falls right back down again.  Motion to the side (especially the East) allows it to get to and stay at a desired altitude.  Rockets bring their own reaction mass with them  No idea how your stars do it, or the moon for that matter since you deny classic sources of that lateral force.
Based on this answer, due to the ultra high gravity force of the galaxy, there is no way for a star to be ejected out without an engine/rocket.
High force has nothing to do with it, and the gravity force of the galaxy is quite low (many orders of magnitude less than 1g) no matter where you measure it.  Energy is needed to lift an object away from a gravity source, however weak.  The moon cannot drift away on its own.  It needs to acquire energy from somewhere to do that, just like it takes energy to get a vehicle up a hill, which is a direction away from Earth, a local gravity source.

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Surprisingly, somehow stars can escape with any rocket
Trust me, those stars have a rocket.  No object can just start to move without a force being applied to it.

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"But researchers from the University of Michigan have identified one hypervelocity star that appears to have been ejected from the stellar disk rather than the galactic bulge."
If there was a dark matter, why that dark matter couldn't hold that star in the galaxy/disc?
Every mass has an escape velocity, including the galaxy.  If an object moves faster than that, the mass of the galaxy (doesn't matter if its light or dark) is not enough to bring it back.  We are already moving at 40% of escape velocity, so it doesn't take an insane amount of additional speed to exit the galaxy.

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How could it be ejected without any engin?
It wasn't.  It cannot do that without a huge momentum transfer being applied.  It most likely got a boost the same way as the Voyager probes and all other interplanetary objects we send out there.  We don't have vehicles capable of holding the energy required to do that, so we steal momentum from larger objects like moons and planets.  So probably this star was flung out by a close encounter with two larger objects, one of which probably used to be its companion.

I have a random star simulator that regularly ejects the little ones like this, but the simulation has an artificially high density of stars to make such events far more common.

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I have a solid evidence why the current theory about the the density wave and dark matter are a fatal errors:
Please look at the following image of the galactic disc:
https://www.quora.com/How-big-is-the-Milky-Way
We see that as we move further away from the center of the galaxy, the disc became thinner and thinner.
I have explained why based on my theory we get exactly that disc shape.
However, is it possible to get that shape based on the current theory?
I didn't see where you posted the solid evidence of anything.  The actual galaxy doesn't look like that.  It is a diagram making some particular point, but we have views of galaxies edge on and they're not nice and clean angled like that.  It's a diagram, not an artist conception, and neither of those would be an actual view.  Yes, the disk apparently gets thinner further out.  I don't see that as solid evidence against the density wave theory or dark matter theory, neither of which explicitly predicts otherwise.

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It is clear that as we move further away from the center, the impact of that Inclination is more visible
Pluto isn't a planet, and Mercury has the greatest inclination of the actual planets, and it is the one closest in.  So no, the trend doesn't seem to go that way.  One data point is not a trend.  I actually see no trend in the inclination data.

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For pluto we see it very clear as it is located far away from all the other planets in the solar system.
It actually gets nearer than Neptune at Perihelion, so I wouldn't describe that as 'far away'.  It is very close for the Oort object that it is.  The inclination of Oort objects is far more random than that of a planet.

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So, it is clear that any orbital object has some sort of Inclination if it orbits around a central main host.
There needs to be a reference for there to be an inclination.  A sole planet orbiting a star has no inclination since it alone defines the plane.  The inclination of the planets in the solar system are all relative to Earth orbital plane.  It could have been say the average inclination, but they chose that.

Since the solar system doesn't orbit a central main host, it has no meaningful inclination to its motion.  The motion is not Keplerian and hence not the same kind of orbit.

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Therefore, if the stars in the galaxy were orbiting due to the gravity force of the dark matter we would expect to see that as the radius is longer, the disc should be thicker and thicker.
Dark matter would not be a central main host.  Motion in the galaxy is not comparable to the dynamics of a solar system.

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As we move further away the Inclination has more significant impact and therefore it is expected to see a thicker disc at the far end.
You might have a point if the galaxy was a Keplerian system, but it isn't.

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Quote from: Halc
S2 is theorized to have formed from a cloud that came from a considerable distance.
So, from where S2' came from?
What does it mean: "considerable distance?"
Further away than the current furthest distance of S2's orbit.
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Where S2' gas cloud was when S2 had been formed?
Probably quite close to Sgr-A since that disturbance is what seeded the formation of a concentation of matter from a more uniform gas cloud.  S2 didn't form in a day, so the process doubtlessly took it close and away again repeatedly, but the process was likely started by a close encounter.  Other disturbances might have been the trigger, but Sgr-A is the obvious culprit.

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How far it was from the SMBH at that time?
Don't know.  'It' wasn't really a thing with a defined distance when the process started.
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What is the age of S2?
Unknown, but probably on the order of 7 digits of years (millions, but not likely 10 million).  This is according to accepted nuclear physics models, something which you've denied.

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Have we ever monitored the process of new star formation in the gas cloud?
Yes, but there is not one 'the' gas cloud.

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Have our scientists really monitored the galactic disc?
You live in it.
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Did they really try to see if there are arias without stars?
The survey database is always being updated.  Many stars have been charted for where they appear and luminosity, but lack distance, which requires multiple measurements.  Plus, they only see the brighter objects.  Between the arms is quite far away and low luminosity stars take a lot of time to see, so they can't survey the entire sky for them in a short period.  It's a lot of work.

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You have stated that there are many unclear problems with the current theories/concepts.
My theory fully meets all the evidences.
You don't have a theory.
Yes, any hindsight theory always meets all evidence, since that's all it needs to do is copy the evidence and say it predicts it.

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You have accepted the unrealistic idea of dark matter. But you are not willing to accept the idea of VHP that is vital for my theory.
The existence of something that doesn't shine or can pass through a planet without interaction doesn't violate any fundamental laws of physics.  Neutrino's have that property for instance, but they don't mass much.  You seem to be going out of your way to offend every one of those laws, so yes, I am unwilling to accept your ideas.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/03/2019 07:20:35
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Did they really try to see if there are arias without stars?
The survey database is always being updated.  Many stars have been charted for where they appear and luminosity, but lack distance, which requires multiple measurements.  Plus, they only see the brighter objects.  Between the arms is quite far away and low luminosity stars take a lot of time to see, so they can't survey the entire sky for them in a short period.  It's a lot of work.
Sorry - I don't agree with this answer.
We have the technology to see clearly S2 and many other S stars which are located 28,000 LY away - directly Near the SMBH at a very complicated aria.
We also have mapped our nearby stars.
We have found that the density of G star at 50LY sphere is identical to the density at 100LY sphere.
So, how could it be that suddenly we don't have the technology to monitor the location of G stars at a sphere of 1,000LY or 4,000 LY around the solar system?
We are located almost at the edge of the spiral arm.
We could easily monitor the aria between the arms near our location.
So, why our scientists do not present this important verification?
Why they hide that key evidence?

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You have stated that there are many unclear problems with the current theories/concepts.
My theory fully meets all the evidences.
You don't have a theory.
Yes, any hindsight theory always meets all evidence, since that's all it needs to do is copy the evidence and say it predicts it.
That is incorrect
I have full valid theory. You just don't let me go on.
Based on my theory - there are significant aria between the arms without even one single star.
Our scientists don't offer that idea.
Therefore, by definition I offer an outcome which currently no one considers.
Therefore, I don't offer a solution for what we see.
I set the expectation based on my theory, and currently no one from our scientists expects to see that phenomenon.
We have to verify if my expectations are real. If so, my theory is real. If no - We will set it in the garbage.
However, there must be one roll for any theory.
You have to verify my theory at the same base as you verify any theory including the one which you deeply believe on.
If you find one verification that contradicts any outcome from that theory - than this theory is incorrect by definition.
That should apply to any theory.
However, you actually claim that if we see a contradiction in our scientists' theory than this is perfectly Ok, while if we see a contradiction in any other theory - than this isn't Ok.
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So, time after time our scientists set an expectation and surprisingly - it doesn't work according to their expectations.
That happens all the time.  Science would make no progress without it.
Sorry - this isn't science - I would call it science fiction.
It isn't accepted to see any contradiction at any kind of theory - even if 100,001 scientists claim that this is the ultimate theory.
Contradiction is contradiction by definition.
If our scientists verify a contradiction between their expectations to the verified evidences, than - they must set this theory at the garbage of the history and start from point Zero - as we all expected for any theory.
I insist to get the data about G stars between the arms at the nearby aria.
You would find that my expectation is correct by 100%. (Not 90%, not even 99%).
This is real science! That is how our scientists should work.
Set the expectation - and verify if your expectations are correct.
If so - your theory is valid.
If no - please set it at the garbage.
One roll for any theory!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/03/2019 18:12:44
So, how could it be that suddenly we don't have the technology to monitor the location of G stars at a sphere of 1,000LY or 4,000 LY around the solar system?
Technology is there up to a point. A type G star is quite difficult to see at 4000 LY.  It takes more resources to see than a bright star.  Resources, not technology, is what prevents a full mapping beyond a certain point.  I do not know where that point is. For all I know, they've very much done the survey you need. They don't write pop articles about each one, and your research seems confined to pop articles.

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We could easily monitor the aria between the arms near our location.
So, why our scientists do not present this important verification?
They have. Nothing is hidden. Plenty of people have discovered new things by mining the public database. There is far more information in that data than there is manpower to process it.

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[quoter=Halc]You don't have a theory.
That is incorrect
I have full valid theory. You just don't let me go on.[/quote]
Lying to yourself is fairly pathetic.  I've let you go one and all I get is endlessly repeated assertions that would be contradictory if they actually made some predictions, which they don't.

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Therefore, by definition I offer an outcome which currently no one considers.
You've offered no outcome.  I can make no predictions of outcome without a theory.  You cannot predict the most simple model, let alone anything as complex as a galaxy.  Nobody considers the outcome because there isn't one proposed to consider.

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We have to verify if my expectations are real. If so, my theory is real. If no - We will set it in the garbage.
You will do no such thing.  You will discard data that does not meet your expectations. That's how you've worked so far.

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If you find one verification that contradicts any outcome from that theory - than this theory is incorrect by definition.
No theory is correct by that definition.  Your theory, were it to exist, would predict everything, and it would be more correct if it did a better job of predicting everything than any other theory, even if some measurements contradicted it.  One measurement will not do at all.  If it gets one thing right and totally fails everywhere else, it isn't a very good theory.

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It isn't accepted to see any contradiction at any kind of theory - even if 100,001 scientists claim that this is the ultimate theory.
Who has claimed to have the ultimate theory?

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If our scientists verify a contradiction between their expectations to the verified evidences, than - they must set this theory at the garbage of the history and start from point Zero - as we all expected for any theory.
No need to go to point zero.  What is known works very well.  All technology is based on theories that are not perfect. Newton might not have got everything right, but I still duck if a rock comes at my head, because his imperfect theory quantifies how much it will hurt. Going back to point zero, I'd have to let it hit me to test if still hurts as predicted by some totally new idea.

So run your idea past a few simple tests. Your refusal to do that demonstrates your lack of faith in the theory, and more importantly, the lack of the theory. You've made some amazing claims, many of which are far more trivial to verify than searching a petabyte database for a swath of deep space with no stars of a certain class.

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I insist to get the data about G stars between the arms at the nearby aria.
You would find that my expectation is correct by 100%. (Not 90%, not even 99%).
This is real science! That is how our scientists should work.
Actually, it sort of is how it works. Go for it. The data exists. You just need to extract it.
 It will not verify the idea, but it will serve as a falsification test.  No evidence can verify any theory, only falsifications. A good theory is one that isn't easily falsified by any of multiple tests.

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Set the expectation - and verify if your expectations are correct.
Not sure how that works here.  If you find a G-type star in a place you're looking, you figure you're looking in the wrong place and you go elsewhere. How common are these empty regions supposed to be that you'll declare failure if every region turns up with stars in it?  You have no predictions about what percentage of random volumes of space of size X will be empty and what percentage are these bridges.  Without that prediction, I don't see how any finding one way or the other is going to verify or sink the idea.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/03/2019 15:34:20
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So, how could it be that suddenly we don't have the technology to monitor the location of G stars at a sphere of 1,000LY or 4,000 LY around the solar system?
Technology is there up to a point. A type G star is quite difficult to see at 4000 LY.  It takes more resources to see than a bright star.  Resources, not technology, is what prevents a full mapping beyond a certain point.  I do not know where that point is.
Sorry - it shouldn't be so difficult to see them.
In the following article: http://www.solstation.com/stars3/100-gs.htm - it is stated:
"Yellow-orange "G" stars are among the most common stars seen with the naked eye in Earth's night sky due to their relative abundance. "
So, if we can see those nearby G Stars in naked eyes, why can't we see similar stars at 4000LY with simple technology?

Not sure how that works here.  If you find a G-type star in a place you're looking, you figure you're looking in the wrong place and you go elsewhere. How common are these empty regions supposed to be that you'll declare failure if every region turns up with stars in it?  You have no predictions about what percentage of random volumes of space of size X will be empty and what percentage are these bridges.  Without that prediction, I don't see how any finding one way or the other is going to verify or sink the idea.

I have already explained it and I'm ready to do it again:
The theory for Spiral arm is quite simple -
Spiral arm is an object. It is made out of stars. The stars are gravity bonded with each other at any given aria of the Arm. The density of stars is vital element to keep them all in a long arm.
At any distance from the galactic center, we might find different density. As we move inwards - the density should be higher. As we move outwards, the density should be lower.
Between the spiral arms there could gateways (bridges and branches) of stars. However, those getaways are not so common. Therefore, we might find here and there some gateways, but no more than that.
We are located at Orion arm at a radius of 28,000 LY from the center of the galaxy. We know that the density of stars in the nearby aria is 64 stars per 50LY sphere, or 512 per 100LY Sphere.
This fixed density is vital for the gravity bonding of the stars in the arm (at that 28,000 Ly from the center of the galaxy).
Therefore, if there are stars between the arms, those stars should have a similar density as we have in our location and they also must be connected at least to one of the arms or set a bridge between the arms.
The Sun is located at the edge of the Orion arm - facing Sagittarius arm.
As we start to move from our location to that nearby arm, we should see that as long as we are in the Orion arm the density of stars should be: 64 per 50 LY. However, at some point we have to cross the edge of the Orion Arm.
Once we cross that edge, we should get to a zone without stars. (Not even a single star). 
Therefore, the answer to your question is:
Technically it is expected to find wide arias between the arms without even a single star, while it must be more difficult to find those gateways and bridges between the arms.
So, simple and clear.
It is there. I have full confidence about it!
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/03/2019 17:21:24
The empty spaces should outnumber the occupied volumes, but no volume is specified.  If I select random spheres a light hour across within our solar system, I will find only one with a G class star in it.  Your prediction is met trivially right here in the dense arm, and it demonstrates nothing since nowhere are stars that dense.  If I do the same for random spheres of 10000 LY in diameter centerd between arms, I will find zero such volumes devoid of G type stars.  I would have to go between galaxies to find a space that big with nothing of that size in it, and it wouldn't falsifiy your idea because you don't posit empty volumes that large.  So at what size would I still expect to find more empty spaces than occupied ones if I look only between the arms?

Why do you take it to the extreme? (1 light hour or 10,000 LY)
I have specified a volume.
50 Ly or 100 Ly.
We have found 64 stars in a 50 LY sphere and 512 in 100 LY.
So this is the requested volume.
As I have stated, the chance to find a gateway/Bridge (of stars) between the arms is quite low.
If I understand it correctly, the distance between the Orion Arm to Sagittarius is about 1,000 LY - 2000LY.
Therefore, if we set a sphere of about 50 LY - 500 LY just at the center between the arms, (in front of our location) there is high chance that we won't find there even one star.
Why is it so difficult?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/03/2019 06:18:44
It is difficult because you didn't really say that before.  I was pointing out the places that were lacking.  I'm satisfied now.
Thanks
So it is in your hands to prove us all wrong, and that only is evidence for your idea that spiral arms are objects, which was neatly refuted in the article I posted.
Yes, I agree
If they were objects like curved spokes on an alloy car wheel, their speed would be proportional to the distance from the center of rotation (a constant slope rotation curve).  The rotation curve only looks like that up to about 0.5 KPC and the arms don't even reach that far in.
Your unwillingness to face that falsification is a good deal of why nobody can take your ideas seriously.  All reasonable scientific proposals concentrate most on the parts where they fail predictions since those are the areas pointing out where improvement is needed.
Rotation Curve
Actually, our scientists try to give an explanation only for the spiral arms.
They don't even try to explain the activity at the Bulge, Bar and outside the arms.
I can explain it all.
Please look at the following diagram:
https://pages.uoregon.edu/jimbrau/BrauImNew/Chap23/6th/23_21Figure-F.jpg
Our scientists try to explain that rotation cure by dark matter as follow:
http://aerorocket.com/GalaxySpeed.html
"The MathCAD analysis presented below uses Newton's shell theorem to determine the combined effect of ordinary matter and dark matter on the rotation speed of stars as they orbit around the galaxy. Specifically the theorem states that when a star system is at radius r from the center of the galaxy, the portion of the galaxy that lies outside the shell of radius r does not produce a net gravitational force on the star system. In addition, only the portion of the galaxy that is inside a shell of radius r produces a net gravitational force on the star system and that mass is located at the center of the galaxy. Figure-3 presents the results of the analysis to determine the rotation curve of the Milky Way galaxy compared to the rotation curve using a two-body method. Finally, the rotation curve of the solar system is determined using the two-body method and the results are compared to actual planetary orbital speeds. These analyses clearly illustrate that a disk of matter that forms the Milky Way galaxy produces a flattened rotation curve that matches measured orbital speeds of stars that orbit the galaxy"
That is totally incorrect idea as as I will explain.
In order to understand that, we need to look at the following diagram which gives better visibility about the rotation cure:
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
We see that the rotaion cure is not an nice cure as we have seen in the first diagram.
We have the following sections:
1. Very close to the center, (at the bulge) we first see an increased velocity from about 200 Km/s to 250 Km/s.
2. Than it goes down to min level of 190 km/s at about 3Kpcs. (Bar)
during this phase, we see that the measurements fits almost perfectly the Blue line.
3. From 3KPC (Spiral arm) to 4KPC there is increase in the orbital velocity (from 190 Km/s to about 220 Km/s) and than  it starts to slow down at about 9KC to 190 Km/s. In that section we see some variations in the orbital velocities of different stars, but it is no so severe.
4. At 10KPC we see clearly significant variation in the orbital velocities of the stars. It gets from 140 Km/s to 230 Km/s. At 15 KPC we see even greater variations in the velocity (from less than 130 to over than 250 Km/s.
Dark Matter:
If the dark matter was real, than why do we see so significant dispersion in the orbital velocities of the stars at the same radius?
Our scientists need to explain that real rotation curve and not the one which they have used before (In Red).
That real dispersion of the orbital velocities proves that the dark matter in fantasy. It proves that there is no dark matter. Not in the galaxy and not outside the galaxy.
If you still believe in dark matter, would you kindly explain how could it be that we see at the same radius (for example - 10KPC) stars with orbital velocity of 140 Km/s while others at 230Km/s?
Would you kindly also explain why the velocity goes down from 250 Km/s at less than 1KPC to about 190 Km/s at 3KPC?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/03/2019 15:33:15
Rotation Curve
In order to understand how the Rotation Curve works we need to look at the following diagram:
http://www.novacelestia.com/images/galaxy_hubble_classification.html
Let's start with Barred spiral galaxy SBb.
We see that it has a bar which is connected to two symmetrical spiral arms.
So, how does it works?
The Bar is used as a connection media between the Bulge to the Arm.
New matter & stars are formed in the Bulge. Those stars are drifted outwards cross the bar and get to the starting point of the Arm.
Let's assume that this starting point (the connection between the Bar to the Arm) is located at 3KPC from the center (as in the Milky way)
We also know that at this point the orbital velocity is minimal (Let's assume - 190 Km/s).
So, if we could force the star to stay at 3KPC, it's orbital velocity should be 190Km/s - that is quite clear.
However, although the arm is considered as an object, stars do not stay at the same point in the arm during any time interval. They are actually drifting outwards.
New stars emerge from the bar and replace them.
So, at any given moment we might see there stars. But all the stars in that point are new stars which had just joined the arm.
So, let's look again at the SBb diagram.
As the stars drift outwards (In the arm), they increase their orbital cycle and therefore they must also increase their orbital velocity.
If the arm was as a direct line which is going from the center, than as the stars goes outwards, they had to increase their orbital velocity. (higher and higher).
However, the arms are spirals. Therefore, as the stars drift outwards, they also drift backwards (due to the shape of the spiral arm).
Hence, as the stars increase their orbital radius, they also drift backwards and decrease the distance that they cross in a given time. The orbital velocity is a direct outcome of the distance per time interval.
Therefore, the balance between the orbital radiuses increased to the drifting outwards set the stars in almost a constant velocity.
In reality, it is never constant. In reality we might find two stars at two different arms at the same radius but with a slightly different orbital velocity.
This is quite clear as nothing is perfectly symmetric/ideal in the nature.
Let's look at the following diagram:
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
We see that at the bar the dispersion in the orbital velocity of the stars is quite minimal.
That is also correct at the early stage of the arm. However, as we move outwards, the dispersion in the velocity increases.
Stars that cross a bridge between the arms might have different velocity than other group in the same radius in the arm.
In this rotation-curve diagram we have no way to understand where each star is located and at what arm. Therefore, it is very confusing.
However, I'm quite sure that at any given location in any specific arm all the stars must orbit almost in the same orbital velocity.

With regards to mathematics:
Let's start with the assumption that the arm is rigid.
So, each star stay at the same location at the arm during all his life time.
In this case, it is clear that if a star 1 is located at a radius R1, than:
P1 = 2π R1
For a star 2 which is located at a radius 2R1, we get:
P2 = 2π R2 = 2π 2R1
So, it is clear that if the orbital velocity of star 1 is V1
Than the orbital velocity of a star 2 should be:
V2 = 2V1.
This of course is not realistic.
This proves that the arm is not rigid.
The star must drifts outwards and backwards (due to the spiral shape)
So, let's look again on a star 1 which is located at R1 in the arm.
As it orbits around the center it also drifts outwards and backwards.
Now we need to understand the impact of each activity.
Let's start with star 1 that in interval T drifts outwards a distance ΔS in the arm.
That ΔS represents two vectors as follow:
1. ΔR - the value of the increased distance in the radius.
2. ΔP - the distance that it moves backwards.
So, if the orbital velocity of star 1 is V1 than in order to keep the same velocity as it gets to R + ΔR it must drifts backwards at the same distance which is equivalent to the impact of the increasing radius.
Hence we first need to calculate the impact of ΔR.
It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
Therefore:
P1 = 2π (R + ΔR/2) = 2πR + 2πΔR/2 = 2πR + πΔR
However, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):
P1 = 2πR
Hence, in order to keep the orbital velocity we must set that
ΔP = πΔR
In this case, we eliminate the impact of the radius increasing by the ΔP backwards drift.
This actually set the shape of the spiral arm.
It must meet that formula.
So simple and clear.
This actually also set the drifting outwards velocity.
As we are located closer to the center, the drifting velocity is lower. as we move outwards, the drifting velocity must be higher.
That also explains the density of the stars and the thickness of the arm.
Close to the center, the density and the thickness must be higher (about 3KLY at a radius of 3KPC)
As we move outwards, the density and the thickness of the arm get lower (about 1KLY at radius 28,000 LY).
At the edge of the arm the density is minimal while the thickness of the arm might be lower (I assume that it should be lower than 500LY).
So, the lowest thickness of the arm set the ending point of the arm.
However, as the density gets lower, the gravity bonding is lower.
At that point, due to the very low gravity force, the stars can't be hold in the arm any more.
Therefore they must be ejected from the arm.
As they are ejected from the arm they are also ejected from the disc. Therefore, it is expected to see that the disc gets thicker and thicker again as we move further away from the edge of the arm (it should be at about 10-15 KPC from the center)
At that aria we see high dispersion in the orbital velocities of the stars. (As stars are still connected in one arm, while other have already disconected from the other arm)
Some of them might still be connected to the arm, while other had already been disconnected.
All the stars, globular clusters, dwarf galaxies that we see around the Milky Way had been ejected from the galaxy.
So, the Milky Way is a giant star sprinkler.
She doesn't eat any star from outside as all of those stars had been born in the galaxy.
The Milky Way is the mother of all the stars that are still orbiting around the galaxy. As any good mother, she has no intention to eat any one of her children. She let them go and cross the space.
She also ejects the Nubile. If that Nubile has the ability to produce new matter as the Milky way, it will be converted over time to a giant spiral galaxy as her mother.
The Milky way should be very satisfy from all the stars and baby galaxies which she had produced so far.
As Andromeda galaxy should be satisfy from her baby - Triangulum Galaxy.
That new baby galaxies production gives perfect explanation for the Ultra high escape velocities of the far end galaxies which we see.
It will also set the Universe in the same matter density forever and ever while we see so many galaxies that are escaping from our observable Universe.
No need for Inflation or Space Expantion.
Spiral galaxies are the leading force in our Universe.
I will explain later on how it really works.
But first we need to understand how spiral galaxy works.
So let me highlight one more issue with regards to spiral galaxy:
It is very important to have a symmetrical shape in spiral galaxy (especially - at the center).
We would never ever see a spiral galaxy with only one arm.
We can compare it to propeller in an airplane.
If you set only one arm in the propeller is will break down the axis of the motor.
In the same token, there must be some sort of symmetrical shape in any real spiral galaxy in order to let the gravity works symmetrically in the center.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/03/2019 21:52:18
Based on the following formula we can trace a star as it drifts in the spiral arm:
ΔS↑ - Vector of the drifting outwards distance per time interval T
ΔR↑ - Vector of the value of the increased distance in the radius.
ΔP↑ - Vector of the distance that it moves backwards.

ΔS↑ =  ΔR↑ + ΔP↑

We have found that:
P = 2πR + πΔR
Hence, the following ratio set the change in the orbital velocity in the arm
ΔP ≈ πΔR
ΔR ≈ ΔP/π
If: ΔR ≥ ΔP/π
The impact due to the radius increasing is higher than the backwards drifting.
In this case the orbital velocity is increasing
That mainly represents the first segment in the arm (3KPC to 5KPC)
If: ΔR = ΔP/π
The impact due to the radius increasing is equivalent to the backwards drifting.
In this case the orbital velocity is constant
That mainly represents the second segment in the arm (5KPC to 10KPC)
If: ΔR ≤ ΔP/π
The impact due to the radius increasing is lower with regards to the backwards drifting.
In this case the orbital velocity is decreasing.
Once the stars is in the arm, we don't see a significant decreasing in the velocity. In any case, if there is any sort of decreasing - it is due to that ratio.

I have asked myself the following question:
How long the Sun might stay at the arm before it will be ejected from the arm?
My advice is as follow:
If we could eliminate the impact of the VHP1, we technically could measure the increasing in the radius ΔR per time interval T.
Assuming that we keep our current orbital velocity (around the galactic center), we could extract the drifting outwards vector as:
ΔR = ΔP/π
Once we have found the value of  ΔR and  ΔP we can calculate the value of the drifting distance in the arm per time interval T
ΔS↑ =  ΔR↑ + ΔP↑
As we know the length of the Orion arm, we can find how long it might take us to get to the ending point of the arm.
That doesn't mean that we will be automatically ejected.
We might be lucky and find a bridge to one of the nearby arms.
If not, we have to say By By to our protecting arm and the galactic disc and start a voyage to the unknown.
We might join to a Halo globular star (with other ejected stars).
But, that aria is very dangerous. Each star must obey to local gravity forces. So, as the stars orbit around each other, the chance for collision is quite high. Even without direct collision, the Sun might lose some of its planets.
Let's hope that we will survive and stay with our Sun as the globular star drifts away from the Milky Way.
Do you know that for any star in the galaxy there is at least one outside?
One day, our Sun will join those stars.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/04/2019 02:14:48
We see that it has a bar which is connected to two symmetrical spiral arms.
Responses are delayed.  Busy.  You need to state how long it takes the galaxy to make one rotation (for the bar and everything else around once) somewhat back to where it was before, even if the individual stars are elsewhere and replaced by different ones.  If you have a different figure, state it, but otherwise my responses assume something like 100 million years, a figure given by these scientists who you think don't know what they're doing.  It is a measured thing.  It fits in just a little with your idea of stars staying in their arms but drifting towards the ends of them, or for some reason jumping early on these bridges to more outward arms.  If the arm is moving faster than we are, we have no choice but to fall behind, which has us moving outward in the general direction of towards the end of the arm just like you say.  The galaxy form is moving at 480 km/sec at this radius and the solar system at a little under half that, so it would be moving quite quickly in that direction, not a slow 'drift'.

A rotating (not orbiting) galatic form (the bar and the arms and such) forms a straight rotation curve for an object that rotates once per 100 MY.  That line crosses the rotation curve of the material (the stars and gas and such) at only one point, which happens to be about where the bar and the arms meet, which really helps explain the transition that happens there.  If you posit a different rotation speed, it will intersect somewhere else and will be harder to explain.

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Let's assume that this starting point (the connection between the Bar to the Arm) is located at 3KPC from the center (as in the Milky way)
We also know that at this point the orbital velocity is minimal (Let's assume - 190 Km/s).
It gets lower much further in, but yes, it is at a local minimum there.

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New stars emerge from the bar and replace them.
Stars in the bar are moving faster than the bar itself since it is below the point where the two rotation curves meet.  That suggests that stars move quite quickly (up to 260 km/s) in a bar that is moving at say 40 km/s at that point.  It's why I say it isn't exactly a drift outward, but rather a sprint.  At that speed, it should take only a few 10's of millions of years to get to the end of the bar.

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So, let's look again at the SBb diagram.
As the stars drift outwards (In the arm), they increase their orbital cycle and therefore they must also increase their orbital velocity.
They increase yes. This is why the motion is not considered Keplerian.  Orbiting planets move slower, not faster, with greater distance from the sun.
The speed of the stars is nowhere near the rate at which the arms increase their speed with radius.  They're moving at 190 km/s at the bar end but 480 out here at 8kpc and even faster further out.  The speed is cleanly 2πR/100 million years, which is the formula for any rotating object like the galaxy's form.  I call it 'form' because I am reluctant to call any of the form objects.  If they were objects, their motion would be an orbit, not a rotation, and you'd have to explain why one object orbits at a different speed than another at the same radius.  Somehow I suspect you will solve this problem by ignoring it.

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If the arm was as a direct line which is going from the center, than as the stars goes outwards, they had to increase their orbital velocity. (higher and higher).
From whence comes the energy to accelerate them like that?  Somebody else posted that question not too far back and you ignored that post, which again is how you resolve inconsistencies in the idea.

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However, the arms are spirals. Therefore, as the stars drift outwards, they also drift backwards (due to the shape of the spiral arm).
Very much so yes.  So much that in one rotation of the galaxy, the solar system, if it stays in its current arm, will have exited the end of the arm and be outside all the visible material in the galaxy.  That's a real problem for a view that we stay in the arms or worse, drift outward to exit even faster.  Find a picture and follow the arm.  It ends in half a rotation at best from where we are now.  Either the sun is only a couple hundred million years old, or the sun must regularly 'drift' to lower arms (not outward) in order to stay in the galaxy. No, we don't drift inward, but the arms move outward past us as we remain at a relatively fixed radius from the center, so we drift inward relative to the arms but not relative to the center. That puts the path of all stars in the spaces between the arms, making that space quite occupied, if not a quite the density of the arms.

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Hence, as the stars increase their orbital radius, they also drift backwards and decrease the distance that they cross in a given time.
If their speed is going up as you suggest, that would increase the distance that they cross in a given time.  That's what higher speed means.

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The orbital velocity is a direct outcome of the distance per time interval.
It's kind of the definition of speed, not the outcome of it.  Something with higher speed covers more distance in a given time, be it orbital motion or not.

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Let's look at the following diagram:
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
We see that at the bar the dispersion in the orbital velocity of the stars is quite minimal.
Did they plot stars like S2 on there?  It's way off the top of the chart.

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That is also correct at the early stage of the arm. However, as we move outwards, the dispersion in the velocity increases.
Yes it does, especially further out that us.  I notice you hunted down a chart that showed a lot more dispersion closer in than the ones you linked in earlier posts.  You must have a purpose in doing that.

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Stars that cross a bridge between the arms might have different velocity than other group in the same radius in the arm.
The chart shows speed, not velocity and hence loses much measured information.  It would be instructive to get the actual velocity data and see if that was true.

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In this rotation-curve diagram we have no way to understand where each star is located and at what arm. Therefore, it is very confusing.
It isn't a map of location and velocity.  It depicts what it means to and isn't confusing at all in meeting that purpose.  If you want position and velocity, don't look at a rotation curve plot.

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However, I'm quite sure that at any given location in any specific arm all the stars must orbit almost in the same orbital velocity.
Discarding the odd ones out, yes, I agree with that.

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With regards to mathematics:
Let's start with the assumption that the arm is rigid.
It does appear to be.  The galaxy makes a rotation without significant change in form, so in that sense it is rigid. It has a rotation 'curve' that is not curved at all, but a straight diagonal line from the origin on out. The form (appearance of arms and bar) can change if there is a significant collision with another galaxy, but the general stable picture is of the rotating barred galaxy going around once per 100 MY.
In a different sense it is not rigid any more than is my arm.  Stars don't stay put in the arm which is moving at very different speeds than the stars, sort of like the blood in my arm travels up and down my arm.

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So, each star stay at the same location at the arm during all his life time.
Can't.  The arm moves at a massively different speed than the stars, at least around here.  It has to to get all the way around in 100 million years, less than half the time it takes the solar system to do it.  You even assert otherwise, saying stars drift outward from the center, not staying in one place all its life.

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In this case, it is clear that if a star 1 is located at a radius R1, than:
P1 = 2π R1
That's the speed of the arm (which rotates), not the star (which orbits).  The difference is the speed at which the arm moves relative to the star.

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For a star 2 which is located at a radius 2R1, we get:
P2 = 2π R2 = 2π 2R1
So, it is clear that if the orbital velocity of star 1 is V1
Than the orbital velocity of a star 2 should be:
V2 = 2V1.
This of course is not realistic.
For orbiting objects like stars, no. For the rotating arms (and the bar as well), this is exactly what is measured.  Anything else and the arms would quickly smear out and not be visible after a short time.

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This proves that the arm is not rigid.
It shows that it doesn't move at the same speed as what is measured for the material within, yes.  If that's your definition of rigid (rather than 'holds form'), then the arm is not rigid, yes.

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The star must drifts outwards and backwards (due to the spiral shape)
Backwards yes, but if they stay in the arm or move outward from them, they'd reach the ends of the arms in about a hundred million years.  Earth could not be the age it is even if there was something giving it the energy to move like you're suggesting.

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Let's start with star 1 that in interval T drifts outwards a distance ΔS in the arm.
That ΔS represents two vectors as follow:
1. ΔR - the value of the increased distance in the radius.
2. ΔP - the distance that it moves backwards.
Yes.  Giving a figure to the rotation rate of the galaxy lets us put numbers to these symbols.  ΔP is trivial to compute.  The arm moves at roughly 480 km/s at this radius (per the 2πR/ω) and our star at say 220, so ΔP is 260 km/s. This has been measured.

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So, if the orbital velocity of star 1 is V1 than in order to keep the same velocity as it gets to R + ΔR it must drifts backwards at the same distance which is equivalent to the impact of the increasing radius.
If it stays in its arm, yes.  So let's consider 50 million years.  We're at 8 kpc now, and in 50 million years the galaxy rotates about 180° and we move about 80° for a net change of 100°.  Follow our arm out on a map and see where 100° puts you.  We're now just short of twice the distance from the center of galaxy.  In 75 million years we'll have left the galaxy.  ΔR is about 7 kpc in 50 million years or 1/7 kpc per year or about 140 km/sec needed to stay in the Orion arm. That's not what I'd call a mere drift.
If ΔR is less than that (or zero), then we drift 'inward' towards the Carina-Sagitarius arm, or rather that arm moves outward towards us (at a ΔR of over 100 km/s) and we don't drift much at all.

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Hence we first need to calculate the impact of ΔR.
Calculating ΔR first seems a good step to do before working out the impact of it.  I did that above and I don't see you doing it.

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It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
You're guessing at something which is pretty trivial to work out?  ΔR (average) is ΔR/2?  That makes no sense.

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As we are located closer to the center, the drifting velocity is lower. as we move outwards, the drifting velocity must be higher.
That it does.

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So, the lowest thickness of the arm set the ending point of the arm.
What happens to all the young stars spinning of the ends of the arms?  Why hasn't that happened to us already?

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However, as the density gets lower, the gravity bonding is lower.
At that point, due to the very low gravity force, the stars can't be hold in the arm any more.
Therefore they must be ejected from the arm.
OK, that's what happens.  Still, why hasn't this happened to us.  We're supposedly old enough to have gone around the galaxy about 20 times, but the galaxy has done about 45 full rotations in that time, a difference of 25.  The pattern of the galaxy does not have 25 windings.  1.5 at best for the arms.

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She doesn't eat any star from outside as all of those stars had been born in the galaxy.
Tell that to the ones headed our way, or to the one coming our way but is large enough to eat us.

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The Milky Way is the mother of all the stars that are still orbiting around the galaxy. As any good mother, she has no intention to eat any one of her children. She let them go and cross the space.
How very anthropomorphic.

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In the same token, there must be some sort of symmetrical shape in any real spiral galaxy in order to let the gravity works symmetrically in the center.
There are asymmetrical galaxies, but most of them have seen recent violence (collisions) and have yet to reform.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/04/2019 12:35:24
Responses are delayed.  Busy. 
Thanks, I was waiting for you!
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Let's start with star 1 that in interval T drifts outwards distance ΔS in the arm.
That ΔS represents two vectors as follow:
1. ΔR - the value of the increased distance in the radius.
2. ΔP - the distance that it moves backwards.
Yes.  Giving a figure to the rotation rate of the galaxy lets us put numbers to these symbols.  ΔP is trivial to compute.  The arm moves at roughly 480 km/s at this radius (per the 2πR/ω) and our star at say 220, so ΔP is 260 km/s. This has been measured.
Wow!!!
Thanks for this great information/confirmation!
In my best dream I didn't expect that our scientists have measured the orbital velocity of the arm.
It fits perfectly with my theory and it proves that this theory is correct by 100%.
So, Yes - I fully agree. If the orbital velocity of the arm is 480K/s and the orbital velocity of the Sun is 220Km/s than by definition the backwards velocity of the Sun due (ΔP/T) is:
480 - 220 = 260 Km/sec.
Wow.
We are moving very fast backwards!!!

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However, as the density gets lower, the gravity bonding is lower.
At that point, due to the very low gravity force, the stars can't be hold in the arm any more.
Therefore they must be ejected from the arm.
OK, that's what happens.  Still, why hasn't this happened to us.  We're supposedly old enough to have gone around the galaxy about 20 times, but the galaxy has done about 45 full rotations in that time, a difference of 25.  The pattern of the galaxy does not have 25 windings.  1.5 at best for the arms.
Thanks for the confirmation.
However, the Sun must be ejected from the arm and the disc. It is just matter of time.
At that high backwards velocity (260Km/s) we might be ejected from the arm even sooner than my expectations.
The Sun hasn't gone 20 times around the galaxy!
That outcome is based on the idea that we have born in the same radius as we are today.
That is a fatal mistake.
We have been born at the center of the galaxy from a gas cloud that was located very close to the SMBH.
That gas cloud was made out of the molecular jet stream that has been ejected from the SMBH excretion disc.
So, our real age is based on the time that it took the sun from day one to drift outwards from the center of the Bulge, cross the Bar and at 3KPC starts its long journey outwards In the arm to our current location.
I would assume that just at the Bulge we have orbit the galactic center by at least 1 Million times.
As I have stated, the bar is a media connection between the Bulge to the Arm.
It seems to me as a Rolette in the casino.
The Bar split the new born stars between the two symmetrical main arms.
https://phys.org/news/2011-06-distant-arm-milky-galaxy.html
Those two arms are: Perscus and Scutum-Crux
However, some of the stars miss the entrance of each main arm and they join the 3KPC arms.
The stars stay there for only half cycle just in order to join the other Arm.

Stars in the bar are moving faster than the bar itself since it is below the point where the two rotation curves meet.  That suggests that stars move quite quickly (up to 260 km/s) in a bar that is moving at say 40 km/s at that point.  It's why I say it isn't exactly a drift outward, but rather a sprint.  At that speed, it should take only a few 10's of millions of years to get to the end of the bar.
I wounded why do you claim that stars move at 260Km/sec (and up) at the bar.
Please look at the orbital velocity
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
The bar is located between the Bulge (about 1KPC) to the Arm 3KPC.
In that section we see clearly that the average stars orbital velocity get from 250 Km/s to almost 190Km/s at 3KPC.
That shows that the Bar reduce the orbital velocity.
So, why do you claim the oposite?
You even agree with the following:
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Let's assume that this starting point (the connection between the Bar to the Arm) is located at 3KPC from the center (as in the Milky way)
We also know that at this point the orbital velocity is minimal (Let's assume - 190 Km/s).
It gets lower much further in, but yes, it is at a local minimum there.

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It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
You're guessing at something which is pretty trivial to work out?  ΔR (average) is ΔR/2?  That makes no sense.
Let use the following examle:
if we are located at 8KPC and after one cycle we get to 10KPC.
It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is is almost the average lengths of the cycles at 8KPC and at 10KPC?
If you do so, you would find that it is identical to the cycle lengths at 9KPC.
So, as ΔR = 10 - 8 = 2
ΔR/2 = 2/2 = 1
R + ΔR/2 = 8+1 = 9.
That exactly what I did.
Do you still see any problem with that?

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Hence we first need to calculate the impact of ΔR.
Calculating ΔR first seems a good step to do before working out the impact of it.  I did that above and I don't see you doing it.
What do you mean?
If we know the value of ΔP/T we should extract the value of ΔR.
I had the impression that ΔP/T is more difficult to verify. Now we know that value - 260Km/s
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So, if the orbital velocity of star 1 is V1 than in order to keep the same velocity as it gets to R + ΔR it must drifts backwards at the same distance which is equivalent to the impact of the increasing radius.
If it stays in its arm, yes.  So let's consider 50 million years.  We're at 8 kpc now, and in 50 million years the galaxy rotates about 180° and we move about 80° for a net change of 100°.  Follow our arm out on a map and see where 100° puts you.  We're now just short of twice the distance from the center of galaxy.  In 75 million years we'll have left the galaxy.  ΔR is about 7 kpc in 50 million years or 1/7 kpc per year or about 140 km/sec needed to stay in the Orion arm. That's not what I'd call a mere drift.
If ΔR is less than that (or zero), then we drift 'inward' towards the Carina-Sagitarius arm, or rather that arm moves outward towards us (at a ΔR of over 100 km/s) and we don't drift much at all.
I really don't understand your calculation. Please advice.
However, why do you calim that the sun stays at the arm for only 50 Million years. Based on what data?
Please if it is based on the density wave idea - than please forget it.
It is a fatal mistake.
Did they plot stars like S2 on there?  It's way off the top of the chart.
Yes I agree.
They also didn't plot the orbital velocity of the accretion disc (at about 0/.3 the speed of light)
But those orbital velocities are not so critical for our discussion abot the bar and the arm.
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She doesn't eat any star from outside as all of those stars had been born in the galaxy.
Tell that to the ones headed our way, or to the one coming our way but is large enough to eat us.
Please, don't be in panic.
No one is going to eat us or the milky way. The Milky Way is not going to to eat any star or galaxy.
Actually the Milky way is like a giant Sweeper.
It's ultar high gravity force sweeps away any star or galaxy that cross our way. If I understand it correctly, the Milky way cross the space at a speed of 630 Km/s. The space is full with stars. However, all of them sweeped away. Therefore, we don't see even one star that gets into the disc from outside. Never - Ever!
We know that Andromeda is moving directly in our direction. However, I don't think that it will collide with the Milky way.
There are about 400 Billion spiral galaxies in the observable Universe.
How many collisions did we find? Could it be less than 10 or even 5?
so the chance for collisions between giant galaxies is almost 10 to 400 Billion. Similar to lottery win chance.
Therefore, I assume that those two giant spiral galaxies will probably clear the way to each other.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/04/2019 19:56:08
Somehow I suspect you will solve this problem by ignoring it.
...
From whence comes the energy to accelerate them like that?  Somebody else posted that question not too far back and you ignored that post, which again is how you resolve inconsistencies in the idea.
I see you have not responded to places that I predicted would be ignored. My predictions seem more reliable than your own. I point out many more contradictions below, which I also suspect will be ignored.

At that high backwards velocity (260Km/s) we might be ejected from the arm even sooner than my expectations.
The Sun hasn't gone 20 times around the galaxy!
That outcome is based on the idea that we have born in the same radius as we are today.
Yes it is.  If we were closer in and moving at the speeds shown in the rotation curve, it is more like 200 times around the galaxy in the age of the Earth, except not since Earth was created only a couple hundred million years ago (less than 1% of the accepted age).  The galaxy has not rotated that many times, so the sun cannot have stayed withing the bar/arm pattern.  It is either much younger than a billion years or it leaves the bar and arms.

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We have been born at the center of the galaxy from a gas cloud that was located very close to the SMBH.
We must have sprung from the black hole with Dinosaurs and older fossils already in place. Very creationism.
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That is a fatal mistake.
And yet the glaring mistake is ignored.

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So, our real age is based on the time that it took the sun from day one to drift outwards from the center of the Bulge, cross the Bar and at 3KPC starts its long journey outwards In the arm to our current location.
Did you compute that at the speeds we're apparently moving along the arm and bar and such?  It would take only a few hundred million years, a brief journey.  Any other speed would not match the rotation curve, or would take us out of the arm, which you deny.  How do you explain Earth being 5 billion years old?  Just going to deny that as well?

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I would assume that just at the Bulge we have orbit the galactic center by at least 1 Million times.
Then we go outside the bar, which rotates only once every 100 million years.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.  Why is there a bar if all the stars are whizzing around at much higher speeds?  Why is the center of a galaxy not 100 times brighter if it has 99% of the stars? If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?  How do they get out of that bulge and back into the original plane? Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?  Contradictions are piling up.  That was just 7 easy ones I mentioned.

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I wounded why do you claim that stars move at 260Km/sec (and up) at the bar.
See any rotation curve at raidus 0.5 kpc.  The speed peaks at about 260 km/s.  The bar moves at 40 km/s there, which is what you get when plugging 0.5 kpc into R of the rotation speed 2πR/100 million years

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Let use the following examle:
if we are located at 8KPC and after one cycle we get to 10KPC.
Not sure what a 'cycle' is, but one orbit about the galaxy is time for the arms to rotate more than twice, throwing us beyond 40 KPC, not 10.  Follow the arm out for one rotation.  It fades to nothing before that but you can still work out where it would continue.  We exit the galaxy in a very short time.  Remember the ΔR of 140 km/s (which is ever increasing)?  Plot that out over one 'cycle', whatever that is.

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It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is is almost the average lengths of the cycles at 8KPC and at 10KPC?
I just don't know what you think a cycle is.  It seems to be something different at every radius, so it probably isn't the rotation rate of the galaxy.  If we're spiraling out fast like that, we don't even orbit any more than does water spewing from a twirling sprinkler.  There doesn't seem to be anything that cycles.

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So, as ΔR = 10 - 8 = 2
OK, you posit 2 KPC per 'cycle', so a cycle might be about 15 million years?  What is there that has a period of that length?  It takes 13 million years for the sun to move to 10 KPC out from 8 if it is moving down the arm at 260 km/s.  That's how far away the point is in our arm that is that distant from the center.

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That exactly what I did.
Do you still see any problem with that?
You didn't define a 'cycle'.


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If we know the value of ΔP/T we should extract the value of ΔR.
Yes.  All you need to know is the angle of the where we are.  Angle 0 would form a circle.  Angle 90 forms an asterisk shape.  Anything in between forms a spiral.  You can just look at a diagram of the galaxy and measure it.  It's about 20° here but it gets higher further out, and smaller further in.

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However, why do you calim that the sun stays at the arm for only 50 Million years. Based on what data?
The arm, even if extended to the lengths of the larger arms, is only so long.  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.  Further out (say at 17 kpc), the arm is moving at 1000 km/sec but the star only at 230 for a difference of 870, more than trice the current speed we move up the arm.

So I based it on actually working out the time to the end of the arm given your assertions.  I actually calculated it instead of making up a figure that makes me feel nice.

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Please if it is based on the density wave idea - than please forget it.
Not based on that at all.  That idea doesn't have us traveling up the arm like that.


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Please, don't be in panic.
No one is going to eat us or the milky way.
Not in either of our lives, no.  Denying what is coming is something else.  You don't need a telescope to see it, but a nice time exposure shows much better how close the thing is.
It is not a panic situation.  Collision happens, taking a lot of time to do so.  The solar system will likely not be much affected by the process.  Perhaps we are flung out, but more likely we find a new place in the combined thing.

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The Milky Way is not going to to eat any star or galaxy.
It happens regularly.  You don't get big if you don't eat.

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We know that Andromeda is moving directly in our direction. However, I don't think that it will collide with the Milky way.
I actually have little doubt that you hold such beliefs.

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There are about 400 Billion spiral galaxies in the observable Universe.
How many collisions did we find? Could it be less than 10 or even 5?
Trying to find one that hasn't recently is more the challenge.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/04/2019 16:55:57
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At that high backwards velocity (260Km/s) we might be ejected from the arm even sooner than my expectations.
The Sun hasn't gone 20 times around the galaxy!
That outcome is based on the idea that we have born in the same radius as we are today.
Yes it is.  If we were closer in and moving at the speeds shown in the rotation curve, it is more like 200 times around the galaxy in the age of the Earth, except not since Earth was created only a couple hundred million years ago (less than 1% of the accepted age).  The galaxy has not rotated that many times, so the sun cannot have stayed within the bar/arm pattern.  It is either much younger than a billion years or it leaves the bar and arms.
Sorry
I don't understand why you claim that the sun cannot have stay within the arm pattern for long time.
Actually, the Sun had been born in the Center of the Bulge.
So, first we need to verify how long it should take it to drift all the way till the Bar.
Bulge -  what is the radius of the Bulge?
In the following article it is stated 6500LY.
http://astronomy.swin.edu.au/cosmos/B/Bulges
The bulge of the Milky Way appears to be fairly typical – a slightly flattened sphere of radius ~6,500 light years.
This value represents about 2KPC. I had the impression that the radius of the Bulge is only 1KPC.
However, the arms are already there at 3KPC.
So, based on this data, the bar is located between 2KPC to 3KPC.
If so, we need to find how long it took the Sun from its first day in the center of the Bulge to get to our current location at 28,000Ly from the center.
Bulge - We have already found the orbital velocity at the center of the bulge are quite high. (In the range of 260Km/s).
However, the drifting outwards is quite low.
Therefore, it seems to me that the sun could easily spent few Billion years only at the Bulge.
As the Sun gets to the main arm base at 3KPC, its orbital velocity had been decreased to 190Km/s.
At that aria the diameter of the arm is estimated to be 3000 LY.
I estimate that the density of stars there should be about four times than our current location (4 * 512 = about 2000 Stars per a sphere of 100LY).
Based on the following formula we can trace the Sun as it drifts in the spiral arm:
ΔS↑ - Vector of the drifting outwards distance per time interval T
ΔR↑ - Vector of the value of the increased distance in the radius per time interval T.
ΔP↑ - Vector of the distance that it moves backwards per time interval T.
ΔS↑ =  ΔR↑ + ΔP↑
We have found that:
P = 2πR + πΔR
Hence, the following ratio set the change in the orbital velocity in the arm
ΔP ≈ πΔR
ΔR ≈ ΔP/π
Hence, at the base of the arm it is clear that the drifting backwards ΔP is lower than the πΔR:
I estimate that if the ΔP/T at our location is 260 Km/s, the ΔP/T at 3KPC (at the base) should be less than 1Km/s.
We can also say that:
ΔR ≥ ΔP/π
Therefore, from 3KC to about 5KPC the orbital velocity is increasing.
Hence, it is clear to me that it took long time for the Sun to drift outwards while it is in that location.
I expect that it might take the sun few billion years just to drift from 3KPC to 4KPC.
As it gets to 4KPC, the orbital velocity had increased to about 200 Km/s.
That shows again that ΔR is higher than ΔP/π.
This is correct as long as the sun gets to 5KPC (with orbital velocity of about 220 Km/s)
It is clear to me that at 5KPC the diameter of the arm is lower than the diameter at the base (3KPC) and also the density should be lower. So, I expect that the density at that aria should be in the range of two times with regards to our location (2 * 512 = about 1000 stars per 100LY).
With regards to ΔP/T I expect that the value at 5KPC it should be less than 25Km/s
In any case, from 5KPC to about 10KPC, the orbital velocity is quite constant. (220Km/s).
In order to achieve it we have to set that:
ΔR = ΔP/π
The impact due to the radius increasing is equivalent to the backwards drifting.
The outcome is - constant orbital velocity.
There is a possibility that  ΔR ≤ ΔP/π
The outcome is that the orbital velocity is decreasing.
That can explain the dispersion in the orbital velocity of stars mainly from 5KPC to 10KPC.
So, it seems to me that it could easily take the Sun few more Billion years to get from the base of the arm to our current location.

  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.  Further out (say at 17 kpc), the arm is moving at 1000 km/sec but the star only at 230 for a difference of 870, more than twice the current speed we move up the arm.
So I based it on actually working out the time to the end of the arm given your assertions.  I actually calculated it instead of making up a figure that makes me feel nice.
Yes, I agree with you that in that time frame we might be ejected from the Orion Arm.
I wonder what is the backwards velocity at the nearby arms?.
Somehow, it seems to me that the backwards velocity there should be lower than the 260 Km/s (even for the arm that is located further away from the center).
The arm, even if extended to the lengths of the larger arms, is only so long.  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.  Further out (say at 17 kpc), the arm is moving at 1000 km/sec but the star only at 230 for a difference of 870, more than twice the current speed we move up the arm.
So I based it on actually working out the time to the end of the arm given your assertions.  I actually calculated it instead of making up a figure that makes me feel nice.
Yes, I also agree with this answer..
How do you explain Earth being 5 billion years old?  Just going to deny that as well?
I hope that by now you understand that there is no problem with the age of the earth.
Then we go outside the bar, which rotates only once every 100 million years.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.  Why is there a bar if all the stars are whizzing around at much higher speeds?  Why is the center of a galaxy not 100 times brighter if it has 99% of the stars? If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?  How do they get out of that bulge and back into the original plane? Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?  Contradictions are piling up.  That was just 7 easy ones I mentioned.
With regards to your questions:
1.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.
Answer - Stars stay long enough in the arms. So, in total - there is significant portion of the stars in the arms.
2.Why is there a bar if all the stars are whizzing around at much higher speeds? 
Answer - The bar is used as a media connection between the Bulge and the Arm base. It actually reduces the Orbital velocity of the Stars and delivers them from a free orbit in the Bulge to  a disc shape orbit in the arm.
3. Why is the center of a galaxy not 100 times brighter if it has 99% of the stars?
Answer - Please see answer 1. There is a significant portion of the stars in the arm. Therefore, it is not expected to see it brighter in the Bulge.
4. If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?
Answer - The stars had not been form in the excretion disc. However, their molecular/matter had been formed there. The stars had been formed in the gas cloud which is located in the Bulge near the SMBH.
5. How do they get out of that bulge and back into the original plane?
Answer - As I have stated before, all the orbital objects drift outwards. (even if it is just few Pico mm per cycle) Therefore, by definition - any orbital cycle is spiral.
The bar is the media which is taking care of setting the stars in the arms (and therefore in the galactic plane)
6. Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?
Answer -  First, it is incorrect. Significant portion of the stars are located at the arms. Second, the current  understanding about the impact of the mass/stars inside the orbital cycle is also incorrect.

Highlight -
I have introduced a simple idea why the stars can be in the arm while their orbital velocity is constant (or almost constant) at any radius. Therefore - the arm is an object. It is made out of stars, while those stars have a freedon to drift outwards. If any star will dare to move outside the arm (by itself) it will be ejected from the disc as a rocket.
This is the most important issue in this discussion.
Do you accept this idea?
If so, it's time for you to shift gear and understand the great impact of that theory.
If you have any more questions (especially if they are difficult), please do not hesitate to ask.
However, please base your questions on real evidences and not on outcome of wrong understanding from the unproved current theory.
I'm quite sure that one day Students will learn this theory in the University.
The question is - how long it might take? One year from now, or 100 years?
Sooner or later - our scientists will have to accept this theory as it is the only valid theory.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/04/2019 13:14:05
I don't understand why you claim that the sun cannot have stay within the arm pattern for long time.
Calculate the remaining length of the arm and the rate at which we're currently moving up it if we must stay within it.  Also take into account the higher ΔS rate the further out we go.  It is a simple calculation that we'll reach the end in under 100 million years.  Where do all the stars go that have done this before us?  There's a much lower density of stars out that far, and if the arms are continuously spewing their contents out into the cloud, it must be a pretty busy place.

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Actually, the Sun had been born in the Center of the Bulge.
You've said it is excreted by the SMBH into the 'excretion disk', which is not the bulge.

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the bar is located between 2KPC to 3KPC.
The bar goes all the way across, so it goes from -3KPC to +3KPC. 


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We have found that:
P = 2πR + πΔR
P (circumference of a circle of radius R) is 2πR.  If R changes by ΔR, then ΔP = 2πΔR.

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I estimate that if the ΔP at our location is 260 Km/s, the ΔP at 3KPC (at the base) should be less than 1Km/s.
Besides quoting me, how did you arrive at that 260 estimation and especially the 1km/s?  It didn't come from any of the formulas you gave.  I agree that at that point the rotating speed and orbital speed are the same.  They have to cross each other somewhere.  So the stars are making progress up the arm at 1/260th of the rate of here.  Why is the amount of material there not at least 260 times as much?  You say only 4x below.

If stars spend 10 billion years in the bulge and less than a 10th of that time in the arms, all the mass would be centralized and the rotation curve would look nothing like what we see.  Stars, like orbiting planets, would move slower the further out they are.  This has been brought up countless times and you don't begin to address any of it.  Your assertions make this problem worse since you put most of the mass at the center, and then you posit stars not being attracted by gravity but actually accelerating away from it over time. Where does the energy needed to do this come from?  Certainly not gravity, which causes objects to slow down as they move away from the gravity source.
The standard explanations don't have any of these problems, and yet you repeat what a fatal mistake they're all making.  The fatal mistake is this ignoring of all these contradictions.

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We can also say that:
ΔR ≥ ΔP/π
Therefore, from 3KC to about 5KPC the orbital velocity is increasing.
Besides the fact that your formula there is false, how can the velocity conclusion possibly follow from the fact that a circle with a larger circumference happens to also have a larger radius?  It isn't true of Earth/Mars, which have a R of about a 3-5 ratio like that and corresponding larger P, and yet Mars's orbital velocity is not larger then Earth's at that radius.  An increasing ΔR has gravity resisting the motion since the gravity pulls in the opposite direction from the change in radius.  So the object slows down just as does a rock when I throw it upwards in a direction that gives it a positive ΔR.  So quite the opposite.  Any positive ΔR should decrease its orbital velocity.  Look at S2, which is at its fastest when its R is lowest.  When its ΔR is increasing, its speed in decreasing, and v-v.

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As it gets to 4KPC, the orbital velocity had increased to about 200 Km/s.
That shows again that ΔR is higher than ΔP/π.
What has that speed have to do with your incorrect formula?  ΔR is in fact always exactly half that figure, so less than, not higher than. You can't just spout some random formula (especially if its wrong) and then make a random conclusion that does not follow from the thing you made up. Essentially, you are saying that circles are round, therefore skateboards increase speed when rolling up hill.

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With regards to ΔP I expect that the value at 5KPC it should be less than 25Km/s
Is this a guess again?  The value is easily computed, and it isn't 25.

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In any case, from 5KPC to about 10KPC, the orbital velocity is quite constant. (220Km/s).
In order to achieve it we have to set that:
ΔR = ΔP/π
This irrelevant thing again.  It's still wrong.  ΔR = ΔP/2π always.  All this says is that an increase of the radius of a circle will increase its circumference by 2π, which is simple geometry and has nothing to do with the speed of anything. Stating that it might be greater or less than this would mean that the circle does not increase its circumference by that amount, which would be simply wrong, at least in Euclidean geometry.

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Quote from: Halc
  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.
Yes, I agree with you that in that time frame we might be ejected from the Orion Arm.
The Orion arm is really a short fragment and it will take only about 10 million years to exit that.  The 50 million figure was if the arm was as long as the main arms.

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I wonder what is the backwards velocity at the nearby arms?.
Somehow, it seems to me that the backwards velocity there should be lower than the 260 Km/s (even for the arm that is located further away from the center).
So compute it instead of making up the numbers.  It's really easy. Both numbers have been measured, so all you have to do is subtract them.

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I hope that by now you understand that there is no problem with the age of the earth.
I agree, because I don't buy your idea that we were excreted by Sgr-A.

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With regards to your questions:
1.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.
Answer - Stars stay long enough in the arms. So, in total - there is significant portion of the stars in the arms.
Yet you agree that we're about to be flung off the end of the arm in about the time it takes to make ¼ orbit of the galaxy.  This sounds contradictory.

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2.Why is there a bar if all the stars are whizzing around at much higher speeds? 
Answer - The bar is used as a media connection between the Bulge and the Arm base. It actually reduces the Orbital velocity of the Stars and delivers them from a free orbit in the Bulge to  a disc shape orbit in the arm.
The bar does not stop where the bulge starts.  It goes all the way across to the other side.

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3. Why is the center of a galaxy not 100 times brighter if it has 99% of the stars?
Answer - Please see answer 1. There is a significant portion of the stars in the arm. Therefore, it is not expected to see it brighter in the Bulge.
You said above that 10 billion years are spent in the bulge and then this fast ride out through the arms. That contradicts what you just say here.

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4. If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?
Answer - The stars had not been form in the excretion disc. However, their molecular/matter had been formed there. The stars had been formed in the gas cloud which is located in the Bulge near the SMBH.
This is also a change in story from the prior posts.  How do the stars get their velocity to end up in the bulge?  If the SMBH is just producing gas continuously, it should just gather around the thing without moving.  I think you mentioned magnetism, but gasses are not magnetic.  I cannot attract or repel a hydrogen or helium balloon with a magnet of any power.

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5. How do they get out of that bulge and back into the original plane?
Answer - As I have stated before, all the orbital objects drift outwards. (even if it is just few Pico mm per cycle) Therefore, by definition - any orbital cycle is spiral.
Then they would move out in all directions, not in the plane of the disk.  Things in the bulge have random orbital planes.  If they drift outward, they retain those random planes and the bulge grows in all directions.  I was asking how the motion changes from one plane to the other.
Second comment:  Why do stars in the bulge drift by a few picometer per cycle (still unclear what a cycle is) while stars out by us do it at 140 km/sec?  We're maybe 1 order of magnitude further out, so why the 50 orders of magnitude drift difference?

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6. Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?
Answer -  First, it is incorrect. Significant portion of the stars are located at the arms.
Your idea predicts otherwise.  Any star in the arms (especially from 3.5 KPC out) spends only 1% of its life in them since it travels so fast up the arm and is flung out as is about to happen to us. If 1% of stellar life is spent in the arms, the arms should mass 1% of the material in the galaxy.  You're exactly right with your answer.  It is incorrect.  We observe a significant portion of the stars located at the arms. This is a falsification of your idea. We observer stars that are very young.  We observe stars that are just forming now. All this completely falsifies what your idea predicts.

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Second, the current  understanding about the impact of the mass/stars inside the orbital cycle is also incorrect.
You did deny Newton's law of gravity, but without a replacement for it, you don't have a theory.  Current law says that mass concentrated within one sphere will produce Keplerian orbital motion of objects outside that sphere.  The motion of stars outside 3.5 kps is anything but Keplerian, either observed or the strange motion that you assert.

Our measured ΔP (the difference between our orbital speed and the speed of the arm as it rotates by) has indeed been measured at 260 km/sec, but the ΔR of 140 has not.  A motion that fast would have been noticed.  The estimated ΔR may or may not be something more like a value that varies between a small positive and negative number (periodic) that might average perhaps 10 km/sec, mostly due to eccentricity and local perturbations.  That means that our current trajectory has us and the Orion arm parting company in a short time where we join the wealth of stars that fill the spaces between the arms.
The arms are brighter mostly because of the young bright stars that are born with every wave passing and die before they pass to the spaces between the arms.  The long lived stars like our own are nearly as dense between the arms as they are here. You asked about a survey of this. Here is a good place to look for your regions of empty space you're so sure exist: http://gea.esac.esa.int/archive/

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Highlight -
I have introduced a simple idea why the stars can be in the arm while their orbital velocity is constant (or almost constant) at any radius. Therefore - the arm is an object.
A thing is an object if it has material moving at constant velocity through it? That makes no sense.
My definition was that it had inertia (mass).  If the arm moves here at 480 km/sec, there is some mass that has that sort of speed/momentum/kinetic-energy. None of it does, or we'd have to explain why it isn't flung from the galaxy due to moving faster than orbital speed.  The material is all moving at about 220.  With no mass moving at 480, the arm is not an object.  A pressure wave in the ocean moves at 1.5 km/sec even though no molecule of water ever travels anywhere near that speed.  Sound waves are thus not objects.  They have no momentum.

If the arm is an object moving at 480 km/sec here and the sun is an object moving at much slower 220 km/sec, why is it that the slow thing (sun) moves outward and the higher speed thing does not?  Why is the acceleration of the arm 3 times as much if it is in the exact same gravitational field as the sun? Answer: because it isn't an object, has no mass, and is thus not subject to gravity any more than one can measure the weight of a sound wave.

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It is made out of stars, while those stars have a freedon to drift outwards. If any star will dare to move outside the arm (by itself) it will be ejected from the disc as a rocket.
If the star doesn't thus 'dare', how might it prevent this?

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This is the most important issue in this discussion.
Do you accept this idea?
It's self-contradictory from one end to the other. Of course I don't accept it.

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I'm quite sure that one day Students will learn this theory in the University.
Somehow this statement doesn't surprise me.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/04/2019 19:14:35
Dear Halc

Unfortunately, it seems to me that instead of understanding my explanation you focus on highlight why this theory is incorrect.
Therefore, you don't understand my explanation.
Would you kindly set the effort to understand the theory and then claim why it is incorrect?

Let me start by mathematics:
 
P (circumference of a circle of radius R) is 2πR.  If R changes by ΔR, then ΔP = 2πΔR....
It's still wrong.  ΔR = ΔP/2π always.
I have already explained why we need to use ΔR/2 instead of ΔR. (in several answers)
The explanation was as follow:
Let use the following example:
if we are located at 8KPC and after one cycle we get to 10KPC.
It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is almost the average lengths of the cycles at 8KPC and at 10KPC?
If you do so, you would find that it is identical to the cycle lengths at 9KPC.
So, as ΔR = 10 - 8 = 2
ΔR/2 = 2/2 = 1
R + ΔR/2 = 8+1 = 9.
That exactly what I did.
Do you still see any problem with that?
It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
Therefore:
P1 = 2π (R + ΔR/2) = 2πR + 2πΔR/2 = 2πR + πΔR
However, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):
P1 = 2πR
Hence, in order to keep the orbital velocity we must set that
ΔP = πΔR
I really can't understand why do you insist on ΔR?
ΔR is valid only if at T=0 the star jump to the new radius (R+ΔR) and stay there for one full orbital cycle.
This is not our case.
The star is moving over time from R to R+ΔR.
Therefore, the average cycle is R+ΔR/2!!!
Hence, it is a severe mistake to use ΔR (as is) in the formula.
Is it clear to you by now?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/04/2019 00:56:21
Would you kindly set the effort to understand the theory and then claim why it is incorrect?
It is self-inconsistent, and something self-inconsistent isn't correct.  So work on the places where inconsistencies have been shown. That's how science works. You modify the idea until it is consistent with itself, and preferably consistent with empirical evidence.

Let use the following example:
if we are located at 8KPC and after one cycle we get to 10KPC.
You have yet to describe what a 'cycle' is, so I still have no idea what you're talking about with this statement.  This may or may not be consistent with observations, depending on what that cycle is.  How long is a 'cycle'?  Where do you figure 10 KPC?  Is that computed or just another figure you made up?

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It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is almost the average lengths of the cycles at 8KPC and at 10KPC?
If you do so, you would find that it is identical to the cycle lengths at 9KPC.
So, as ΔR = 10 - 8 = 2
ΔR/2 = 2/2 = 1
R + ΔR/2 = 8+1 = 9.
That exactly what I did.
Do you still see any problem with that?
Repeating unclear statements doesn't help.  I asked questions (what is a cycle?) which went unanswered.
ΔR is sometimes a speed and sometimes a distance.  That's inconsistent.  Distance seems more correct since Δ means change-in X and not 'rate of change of' X.
ΔR/2 = 2/2 = 1:  You divide the difference by two, but don't give any reason why.  OK, that's half the radius change.
R + ΔR/2 = 8+1 = 9 :  Yes, 9 is halfway between 8 and 10.  I still have no idea what this means.  We're presumably at radius 9 after half a 'cycle' instead of a whole cycle, whatever that is.  You've managed to find an obscure way to show that stars at radius 9 KPC orbit between stars orbiting at 8 and 10 respectively.  I cannot comment further because I have no idea what you are doing here.

It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
Here you call it an orbital cycle, so a cycle is presumably the time it takes something to orbit.  I am guessing the solar system orbiting the galaxy, except your idea has no such orbit since the solar system follows the arm in a spiral path and not a reasonably round orbital path.  Perhaps you mean the time it takes to orbit its VHP1, but no mention of that has been made. I'm not sure how long you envision a 'cycle' to be.  Perhaps you could use clear terms instead of assuming I can correctly guess at your meaning.  Clearly I'm failing to do that.

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Therefore:
P1 = 2π (R + ΔR/2) = 2πR + 2πΔR/2 = 2πR + πΔR
If P1 is 2πR, then it is the circumference of a circle of radius R.  If you change R by ΔR, then that circumference goes up by 2πΔR or else the circumference at the new R will no longer be 2πR.  As such, the term on the right is just plain wrong.  If you mean totally different things by this equation above, then tell me what you mean by it, because you added no new explanation in this post.  It makes no sense to me as posted.

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However, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):
There seemed to be no mention of velocity or time in the above equation, but then again I seem to be misunderstanding the meaning completely.  As I said, sometimes R is a distance (radius) and sometimes it is a speed (like an outward drift speed or something).  If it is a speed, then ΔR would be a change in speed.
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I really can't understand why do you insist on ΔR?
I'm insisting on anything. I'm not even clear what your ΔR is. I'm not telling you what it is.

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ΔR is valid only if at T=0 the star jump to the new radius (R+ΔR) and stay there for one full orbital cycle.
Here you use R as a distance again, but there is time T involved, but there seem to be no temporal components to your equations.  At time 0 the star jumps to a different radius?  Like instantly?  Takes a left turn and start moving outward for a while?  Not sure what you're trying to convey with this.

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The star is moving over time from R to R+ΔR.
Outward drift means an increase in radius, sure.  R is a distance here.

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Therefore, the average cycle is R+ΔR/2!!!
A 'cycle' is a distance, not a time period?  So in our example of R=8, ΔR=2, the average cycle is 9KPC?  I really don't understand what a cycle is.  9KPC is an average radius between those two radii, like saying Mars sort of has an average orbital radius between Earth and the asteroid belt.  No idea how the word 'cycle' applies to that.  9 KPC is a distance and 'cycle' is usually one iteration of a repetitive process.
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Is it clear to you by now?
I'm clearly baffled by your terminology.  I'm not saying any of it is wrong since I have no idea what you're trying to describe.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/04/2019 04:57:13
Wow
You have totally got lost on a very simple exercise.
Please - just look at that from a mathematical point of view.
Let me start by asking:
What is the circumference of a circle if the radius is R?
You have already gave the answer:
P (circumference of a circle of radius R) is 2πR.
Therefore:
P = 2πR
So, if R1 = 8 cm, the formula is:
P1 = 2πR1 = 2π*8
If R2 = 10cm
P2 =2 πR2 = 2π*10
However, our mission is to verify the circumference of a circle while R increases from R1 (8cm) to R2 (10cm).
We actually try to calculate the circumference of a spiral shape as it moves from radius 8m to radius 10m in one full cycle.
So, please advice how do you calculate that circumference (P) of a spiral shape from 8cm to 10 cm (ΔR = 10-8=2)?
P is the circumference from R=8, t=0 to R=10 t=T
T = the time that it takes the object to move (in spiral shape) from 8 to 10 in one full cycle.
Somehow, you insist on the maximal radius:
P =2πR2 = 2 πR2 = 2 π (R1 + ΔR) = 2π*10
Is it realistic?
Why not the average radius (R1+R2)/2 or (R1+ΔR/2) or just (9)?

In order to help you, please see the following article:
https://sciencing.com/calculate-spiral-6544041.html
How to Calculate a Spiral
"Determine the inner diameter of the spiral. This is the diameter of the circle formed by the innermost ring of the spiral. Call this length "d."

Plug the numbers obtained in the first three steps into the following formula: L = 3.14 x R x (D+d) ÷ 2

For example, if you had a spiral with 10 rings, an outer diameter of 20 and an inner diameter of 5, you would plug these numbers into the formula to get: L = 3.14 x 10 x (20 + 5) ÷ 2.

Solve for "L." The result is the length of the spiral. Using the example from the previous step: L = 3.14 x 10 x (20 + 5) ÷ 2 L = 3.14 x 10 x 25 ÷ 2 L = 3.14 x 250 ÷ 2 L = 3.14 x 125 L = 392.5"

Please be aware that in this example it is stated:
"Determine the number of rings in the spiral. This is the number of times the spiral curve wraps around the center point. Call this number of rings "R."
We try to calculate only one ring!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/04/2019 13:56:17
However, our mission is to verify the circumference of a circle while R increases from R1 (8cm) to R2 (10cm).
OK, I see what you are doing.  You're computing the length of a spiral segment in a 360° arc which you are calling a cycle.  Your method is a close approximation for slow outward spiral just like sinΦ is a close approximation to tanΦ for a small Φ, but it gets pretty inaccurate for a larger Φ, and a glance at any picture of our galaxy shows that you get massively further out than 10 KPC  following any arm for one lap starting at 8 KPC.
So take the outer arm (a particularly long one) which is at radius 8 kpc about a quarter turn back.  It is at 11 kpc after 90° and 16 kpc at 180° after which it sort of ends.  If you extrapolate it all the way around, it is at something like 26 (not 10) kpc after one cycle.  By your approximation, the length of it between those two points is 2π17 or about 107 kpc.  I'm getting about 120 kpc myself, not a whole lot different. The arm ends after only about 50 kpc from that point where it is at the same radius as us.

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We actually try to calculate the circumference of a spiral shape as it moves from radius 8m to radius 10m in one full cycle.
Here you talk about something moving again, so the definition of a cycle changes.  What exactly is moving for one full cycle? Countless times I've asked this and still I'm force to guess by your changing context.
Is it the time it takes the form (bar and arms and such) to make one rotation?  Is it the time it takes a star at a given radius to make one orbit (which can be worked out from the rotation curve) or is it the difference between the two (the time it takes for a star to make one trip around the pattern at a changing ΔP)?  Those are three very different values, and only the first one is any kind of regular cycle per your idea.  The orbit time is a cycle in standard theory, but an orbital motion does not keep a star in any arm, so that isn't a cycle in your view.  A trip with a finite end isn't a cycle, its just something that ends, especially if there's far less than one lap to go.

Anyway, you talk about going from 8 to 10 in this cycle.  What galaxy are you looking at?  The Orion arm reaches 10 kpc after only about a 15th of a cycle, after which that spur ends.

Not sure where we go after that since if there are no stars between the arms, all these stars shooting off the end of the Orion arm need to be destroyed by something.  A bridge to the larger arms on either side would be crowded enough to see.  How did we get in this arm in the first place?  The inner end is about 7kpc out and even closer than the far end.  We can't have been in this arm more than 10 million years at the rate we're traveling through it.

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So, please advice how do you calculate that circumference (P) of a spiral shape from 8cm to 10 cm (ΔR = 10-8=2)?
The 8/10 figures are not representative for our (or any) galaxy, but for those numbers, the figure you're getting suffices as an approximation.  The way to do it proper is to integrate the cos of the spiral angle Φ for one cycle.  For a small angle like you get with 8/10, the difference is insignificant for our purposes.  Φ seems to be a around 20° for our galaxy.

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P is the circumference from R=8, t=0 to R=10 t=T
I'll ask again: Where did you get 10 from?  It certainly isn't the radius of any arm in any barred galaxy after one cycle from some point at least twice the radius of where the arms start.  No galaxy is wound that tight.  Working with 10 means these calculations are meaningless.  The real figure for our galaxy is about 26.  Look at any depiction of it to see that.
This is part of why I didn't know what a cycle is.  It takes perhaps 10 million years for us to reach 10kpc if we stay in the arm, and I was trying to figure out something that had a period of 10 million years. 

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T = the time that it takes the object to move (in spiral shape) from 8 to 10 in one full cycle.
I think you're talking about how long it takes a star to make one trip around the pattern.  Did you work this time out?  You have all the needed numbers.

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Somehow, you insist on the maximal radius:
P =2πR2 = 2 πR2 = 2 π (R1 + ΔR) = 2π*10
Is it realistic?
Why not the average radius (R1+R2)/2 or (R1+ΔR/2) or just (9)?
I didn't know what you were trying to compute with these equations, which was the linear length of a spiral going from 8 to 10 in one lap, which didn't occur to me at all since our galaxy doesn't look like that.  Our arm goes from 8 to 10 in about a 15th of a lap.

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"Determine the number of rings in the spiral. This is the number of times the spiral curve wraps around the center point. Call this number of rings "R."
We try to calculate only one ring!!!
OK, but where did 10 come from then?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/04/2019 15:54:19
Would you kindly focus on my message?
I have stated:
Please - just look at that from a mathematical point of view
So, we ONLY discuss on a pure mathematical issue.
In that example I'm using a radius of 8 cm (centimeter) that increased in spiral way to 10 cm!!!
So we discuss about centimeters, not KPC.
I just want to show you the correct formula to calculate the length of a spiral segment in a 360° arc (which I'm calling a cycle).
I don't claim that in the Milky Way after one full cycle (or 360° arc) the spiral arm will get from 8KPC to 10KPC.
It is just a mathematical example.
Why is it so difficult?
So, If
R1 - the radius at starting point.
R2 - The radius after 360° arc.
ΔR = R2 - R1
Hence - the formula to calculate the length of a spiral segment in a 360° arc (from R1 to R2) is as follow:
P = 2π(R +ΔR)/2 = 2πR +πΔR
Do you agree with that?
Please!
Inner diameter of 5 means they start well away from the center.  Both types A and B can go all the way in, but the mathematics falls apart there for both.
If you think that the mathematics is incorrect, would you kindly advice how to calculate the length of a spiral segment in a 360° arc?


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/04/2019 17:55:24
Would you kindly focus on my message?
I have stated:
Please - just look at that from a mathematical point of view
So, we ONLY discuss on a pure mathematical issue.
I did that in my prior post (294).  Did you read it before posting this?
The formula is only an approximation for a type-A spiral measured at a reasonable distance away from the center.  The site did not present the formula as inappropriate for other situations, nor did it identify it as an approximation.

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In that example I'm using a radius of 8 cm (centimeter) that increased in spiral way to 10 cm!!!
18π cm is a reasonable approximation for that, yes.  It isn't exactly that.  The real figure is larger.

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So we discuss about centimeters, not KPC.
I just want to show you the correct formula to calculate the length of a spiral segment in a 360° arc (which I'm calling a cycle).
It isn't correct, but close enough if that's all you need. Mathematically, close enough is the same as wrong.

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So, If
R1 - the radius at starting point.
R2 - The radius after 360° arc.
ΔR = R2 - R1
Hence - the formula to calculate the length of a spiral segment in a 360° arc (from R1 to R2) is as follow:
P = 2π(R +ΔR)/2 = 2πR +πΔR
Do you agree with that?
I see what is being done, yes.  It is the sum of the tangential component of the line being measured, but discards the radial component, which isn't large for a mathematical spiral going up by 25% in one trip around.  The discarding of the radial component is what makes it wrong though.  It adds about a 28th of a cm to the length of your example.
The formula fails completely for type B (scale-invariant) spirals, and the article doesn't make the distinction, especially since they show a picture of a type B spiral.

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If you think that the mathematics is incorrect, would you kindly advice how to calculate the length of a spiral segment in a 360° arc?
Integrate the distance along its entire length, just like I said in the post you are quoting.  I see you didn't quote that part. Translation: cut it up into little pieces and add up the length of the pieces.  The length is the limit of that process as the size of the pieces approaches zero.
For a type A spiral, a better approximation for one lap (still staying away from the center) is √((2πR)² + ΔR²) where R is the mean radius (9) and ΔR is 2.  This is still an approximation, but at least a better one.

I don't have at my fingertips a good formula for doing the same for a type B spiral.  It would have to be a scale-invariant formula, so there would be a fixed Φ input instead of a fixed ΔR per lap.  We'll leave it as an exercise to the reader.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/04/2019 07:47:36
You insist that there are two spiral types -  Type A and Type B.
With regards to Type A you claim:
For a type A spiral, a better approximation for one lap (still staying away from the center) is √((2πR)² + ΔR²) where R is the mean radius (9) and ΔR is 2.  This is still an approximation, but at least a better one.
With regards to type B
The formula fails completely for type B (scale-invariant) spirals, and the article doesn't make the distinction, especially since they show a picture of a type B spiral.
Can you please direct me to the article which supports your explanation?
You also claim that:
It isn't correct, but close enough if that's all you need.
So, yes, that's all I need.
You have asked me to explain how we can get a constant velocity of stars in spiral galaxy while the arms are objects that based on stars.
In order to prove it, I need to use a simple formula to calculate the length of the spiral.
In all the articles which I have looked at the Web it is clearly show that the simple formula for that length of the spiral is:
P = 2πR +πΔR
I didn't find any relation in any article for Type A or Type B.
So, why do you try to make it so difficult?
Based on this formula I have proved my theory!!!
I have clearly proved that increased drifting velocity of stars in the arm can compensate the extra requested velocity due to increased radius.
If those types of spiral are real in the nature, why our scientists do not offer their solutions for those types of spirals?
Would you kindly show me the article about the modeling for each type of galaxy?
Why do you insist that I should offer a solution for those types of galaxies while our scientists have no obligation for that?
Why do you claim that:
Mathematically, close enough is the same as wrong.
All the modern science is based on "close enough"
In the idea of density wave the whole modeling is based on "close enough".
Our scientists do not offer any direct outcome from the BBT moment to the creation of spiral galaxy.
In their modeling they have used unrealistic idea of a very thin star disc in order to show that it could be evolved into spiral shape, but they don't show how we get this unrealistic thin star disc from the Big bang moment.
They also couldn't show why the disc get's so narrow as the radius increases.
We have solid evidence that at 3KPC the thickness of the disc is 3000LY.
At our location it is 1000LY.
I'm positively sure that if we move further away from the galactic center, the disc should be even less than 500LY. (If not than my theory is incorrect)
Our scientists could not explain this evidence by their theory.
Actually, based on their modeling they have found that the disc should be thicker as we move further away from the center.
But as usual, they have totally ignored this contradiction in the verification.
They also couldn't explain by the modeling the real functionality of the Bar.
Hence, the density wave is not even "close enough". It is just a fatal error.
So, would you kindly explain why "close enough" is not good enough for my theory, but it is perfectly OK for our scientists?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 06/04/2019 15:12:25
You insist that there are two spiral types -  Type A and Type B.
With regards to Type A you claim:
Quote from: Halc
For a type A spiral, a better approximation for one lap (still staying away from the center) is √((2πR)² + ΔR²) where R is the mean radius (9) and ΔR is 2.  This is still an approximation, but at least a better one.
With regards to type B
Quote from: Halc
The formula fails completely for type B (scale-invariant) spirals, and the article doesn't make the distinction, especially since they show a picture of a type B spiral.
Can you please direct me to the article which supports your explanation?
No article.  I worked it out.  There are probably real mathematical names given to those two types of spirals, which I might have learned had I bothered to hunt down an article.  So lacking that, I called them types A and B.  There are other types, some of them probably with names.

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It isn't correct, but close enough if that's all you need.
So, yes, that's all I need.
So there's no problem.

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You have asked me to explain how we can get a constant velocity of stars in spiral galaxy while the arms are objects that based on stars.
This isn't much of a question since you seem to realize that the arm is moving at one speed (480km/s locally) and the stars another (220).  The question was from a long way back when your personal definition of 'object' was unclear to me.  You seem to be talking about form as an object, so this question is answered so long as you don't start giving inertial properties to that form.

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In order to prove it, I need to use a simple formula to calculate the length of the spiral.
You're going to prove a velocity question using a calculation of length?  I can't wait...

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In all the articles which I have looked at the Web it is clearly show that the simple formula for that length of the spiral is:
P = 2πR +πΔR
Simple yes.  Mathematically correct, no, as I have pointed out.  But it works for our purposes here.  If you would like a demonstration of where the formula is wrong by a factor of 2 or more, I can show it.

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I didn't find any relation in any article for Type A or Type B.
If you read my post on it, you'd know I made those terms up. They probably have official names, but I don't know them.

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So, why do you try to make it so difficult?
Based on this formula I have proved my theory!!!
Your theory has yet to see light of day, and never will since it seems to require a rewrite of all physics.  A calculation of a length of a segment of a spiral is not going to prove or disprove 99% of the things you have proposed.

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I have clearly proved that increased drifting velocity of stars in the arm can compensate the extra requested velocity due to increased radius.
Wait, did the proof go by?  What extra velocity requested?  You quoted a length calculation and suddenly you're saying this somehow 'proves' something concerning velocity of stars.  It doesn't mention velocity of anything.

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If those types of spiral are real in the nature, why our scientists do not offer their solutions for those types of spirals?
What kind of solutions do you think are needed?  Their solutions are consistent at least, if not all correct.

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Would you kindly show me the article about the modeling for each type of galaxy?
Not sure what you want.  Look them up yourself if you want to read about the various models.  I recall some good links going by a ways back in this thread.  I am not very familiar with the various models, or if 'articles' have been written about them.

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Why do you insist that I should offer a solution for those types of galaxies while our scientists have no obligation for that?
Not sure what types of galaxies you're taking about.  Type A and B?  No galaxy is either.  I said that galactic spirals are type C in my post about that.  Type C is 'neither A nor B'.  The form is not describable with a 5 character formula.
Type A is: R=Xω where radius R is a linear function of the number of rotations ω.  X is some constant, and defines the constant separation of arms.
Type B is Φ=X where the X is a constant angle Φ of the spiral relative to the line tangential to a circle at any given point.
Those are really simple formulas.  No doubt there are others, but few that short.

Quote from: Halc
I don't have at my fingertips a good formula for doing the same for a type B spiral.  It would have to be a scale-invariant formula, so there would be a fixed Φ input instead of a fixed ΔR per lap.  We'll leave it as an exercise to the reader.
This is clearly beyond the middle school mathematics skills of the reader, so I'll do it myself.
Formula for length (exact, not approximation) of type B spiral between two radius R values is ΔR/sinΦ. My initial difficultly was trying to make it a function of the number of laps (or cycles), but even that can be worked out from the above formula.

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Quote from: Halc
Mathematically, close enough is the same as wrong.
All the modern science is based on "close enough"
That's why I said mathematically, not scientifically, and usually science wants something more accurate than your crude formula for spiral length.  Your formula is wrong because if you integrate with that formula, the value approaches an incorrect figure (in fact the value doesn't change at all if you integrate yours).  If you integrate mine, it approaches the correct figure.  That's why even science would probably reject the P = 2πR +πΔR that you give above.  The fact that you got it from an article doesn't make it right.
I didn't know the correct formula, but I could see right away that the one you use can't be correct.  So I worked out a better one.  I didn't look it up.

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They also couldn't show why the disc get's so narrow as the radius increases.
How so?  Where are these models that don't predict that?

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Actually, based on their modeling they have found that the disc should be thicker as we move further away from the center.
The models do no such thing.

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So, would you kindly explain why "close enough" is not good enough for my theory, but it is perfectly OK for our scientists?
I said 18π was close enough for an estimate of the length of a spiral doing one lap from R 8 to 10.  The formula fails in general, but it works fine for that example.  No idea what that example demonstrates since you seem incapable of getting to the 'therefore' part.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/04/2019 19:52:01
After all your spiral types' formulas, I wonder which kind of formula our scientists have used in order to confirm the density wave theory.
As you claim that:
This is clearly beyond the middle school mathematics skills of the reader, so I'll do it myself.
Formula for length (exact, not approximation) of type B spiral between two radius R values is ΔR/sinΦ. My initial difficultly was trying to make it a function of the number of laps (or cycles), but even that can be worked out from the above formula.
So, would you kindly show me how our scientists have proved any sort of spiral formula by their density wave theory?
Not just nice hand wave and none realistic simulation - but please - based on real middle school mathematics.
This is vital.
Somehow, I have full confidence that they have never ever used any sort of spiral formulas or middle school mathematics which you are offering.
How can you explain that?




Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/04/2019 06:47:54

Nobody claimed that this information was needed for density wave theory.
So, based on the density wave our scientists gave themselves a waiver to skip the need for using any kind of formula for the spiral galaxy shape.
However, when it comes to me, than even the basic spiral formula which is available in the web is not good enough.
I have offered clear explanation about the spiral arms.
I have solved the rotation curve problem based on that formula at the web.
But this isn't good enough.
Actually, based on your reply I clearly see that you didn't even try to understand my message.
You only focus on finding negative aspects before understanding the theory.
This is very clear to me.
Actually, you represents the good old science approach as we had in the past.
Few centuries ago, our scientists were positively sure that we are the center of the Universe.
https://starchild.gsfc.nasa.gov/docs/StarChild/whos_who_level2/galileo.html
"Galileo's observations strengthened his belief in Copernicus' theory that Earth and all other planets revolve around the Sun. Most people in Galileo's time believed that the Earth was the center of the universe and that the Sun and planets revolved around it."
"The Catholic Church, which was very powerful and influential in Galileo's day, strongly supported the theory of a geocentric, or Earth-centered, universe. After Galileo began publishing papers about his astronomy discoveries and his belief in a heliocentric, or Sun-centered, Universe, he was called to Rome to answer charges brought against him by the Inquisition (the legal body of the Catholic Church). "
I'm lucky that no one is going to set me on jail.
However, there is no difference in the science community to new ideas between that old time to our current time.
Somehow - the science community believe that the wisdom is located only at their site.
They surly see all the contradictions in their theory - But they still believe that their theory is here to stay forever and ever.
We have spent so long time - while you don't even try to understand the theory.
So what is the difference between the current science community approach to that approach in the past?
At Galileo's days the science community was controlled by the Catholic Church. At that time no one there has tried to understand what that person (Galileo) has found. They didn't want to hear any other idea.
They were sure that the only real theory is the one that they have found.
So, do we still live at the dark days of the science?
It is clear to me that you have no willing to understand my theory.
However, I do appreciate your knowledge in science and all your effort to convince me that my theory is incorrect.
So, what is needed to convince you that there is a fatal mistake in the current theory?
What kind of evidence will help you to understand that my theory is correct?
Or is it too much to ask?

 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/04/2019 13:44:18
So, based on the density wave our scientists gave themselves a waiver to skip the need for using any kind of formula for the spiral galaxy shape.
However, when it comes to me, than even the basic spiral formula which is available in the web is not good enough.
1. I always said it was good enough for the example you were using.
2. The formula for the shape of a spiral is a different one than the formula to compute the length of a portion of that shape.  The latter is what you found on the web, and that formula is mathematically wrong but close enough for the example you were using.
3. Any formula for the shape of the galactic form itself would apply to all theories, and would not be specific to wave theory, yours, or any other.  The galaxy form has a certain shape which I suppose is mathematically describable if you want, but that shape is an observation and its mathematical description seems irrelevant.  A good model will produce a shape something like what we see, and a bad model produces something else.  The ability to mathematically describe the shape seems to be irrelevant to a model being accurate or not.

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I have offered clear explanation about the spiral arms.
I have solved the rotation curve problem based on that formula at the web.
And since I've said the formula on the web (which gives an approximate length of an arm segment) is good enough, I've been waiting to see how this solves the rotation curve problem, but you just will not move on to that part.
You seem to deny F=ma, so your solution had better find a good replacement for that, because the whole rotation curve problem revolves around it.

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Actually, based on your reply I clearly see that you didn't even try to understand my message.
It took me a long time to see what you were doing.  Most of the confusion was about your refusal to define a 'cycle' even after several posts of asking.

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You only focus on finding negative aspects before understanding the theory.
Kind of hard not to when it is so self inconsistent.

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Actually, you represents the good old science approach as we had in the past.
Thank you.
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Few centuries ago, our scientists were positively sure that we are the center of the Universe.
That model fit well with the empirical evidence at the time. Copernicus posited an alternative view.

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I'm lucky that no one is going to set me on jail.
Try going against a church that still puts people in jail (if they're lucky) for doing so. Still happens in places.

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They surly see all the contradictions in their theory - But they still believe that their theory is here to stay forever and ever.
If they see contradictions, then they believe quite the opposite. It's why their work is never done.

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At Galileo's days the science community was controlled by the Catholic Church.
You think it isn't that way anymore?  I invite you to Kansas schools then.  OK, it may not be specifically Catholics out there, but it is very much the church. They pit themselves against science because its success makes them a threat.  What you are preaching does not, so they'll probably have no problem with your message.  You're safe.

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At that time no one there has tried to understand what that person (Galileo) has found. They didn't want to hear any other idea.
You claiming nobody listened to Galileo?  You're wrong about that.

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It is clear to me that you have no willing to understand my theory.
I've listened all along, and while the necessary details will remain forever hidden from me, I've done my best to understand and glean the implications of what you are saying, such as the fact that all stars this far out have only 50 million years left in the galaxy, and yet this conflicts with the total lack of stars that have been similarly flung out ahead of us.  Sounds like you're the one that doesn't listen to yourself. You seem to agree with these predictions, and yet don't face the evidence against. You describe a furious lawn sprinkler and yet cannot explain why the grass is not wet yet, like all the water evaporates or something just before hitting the ground, leaving only the twirling shape of the water in the air.

You describe a VHP sometimes as the center of mass of something, but then don't find any mass to define it.  That's completely inconsistent.  If it is a center of mass, then what mass is it the center of?  And then, how does the motion of any object relate to that VHP?  You refuse to answer those questions, which means you don't actually have a theory.  It's not that I prevent you from posting it. No amount of questions will get those answers because clearly you don't even have them.  Science isn't done by some guy saying he has all the answers but will not actually reveal any of it. Nobody will publish that and nobody will publish you. Nobody will teach your theory in the universities because you will not reveal it, due to the lack of its existence.

Now you get stuck for 10 posts about the length of a spiral segment going from 8 to 10 in one lap which is approximately 18π, but you refuse to reveal what possible significance this has, but you spend plenty of time asserting how this revelation will explain everything.  So stop with the promises and deliver how this one number explains something.

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However, I do appreciate your knowledge in science and all your effort to convince me that my theory is incorrect.
I don't have that much science behind me.  I need to look up a lot of what I post since there's only so much they teach in schools if you don't major in these things.  The mathematics is more me.  I'm no dunce with numbers or computers.

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So, what is needed to convince you that there is a fatal mistake in the current theory?
Depends on what you mean by fatal.  There's always room for improvement, but if you want to contest F=ma, you need to find an explicit empirical evidence that it's wrong, which would be a pretty fatal blow.  If you don't contest it, then you need to identify the forces and reactions necessary for the sort of motion you describe, because right now everything you describe contradicts F=ma.

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What kind of evidence will help you to understand that my theory is correct?
Or is it too much to ask?
Show me what's wrong with F=ma.  That is not much evidence that your theory is correct, but it shows at least that all your claims about established science being fatally wrong is true.
I will admit that several of the sites you've linked have incorrect science in them, but the writers of these articles are rarely scientific people themselves. They're journalists and such.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/04/2019 16:22:32
And since I've said the formula on the web (which gives an approximate length of an arm segment) is good enough, I've been waiting to see how this solves the rotation curve problem, but you just will not move on to that part.
You seem to deny F=ma, so your solution had better find a good replacement for that, because the whole rotation curve problem revolves around it.
All my theory is based on Newton law including F=ma.
However, if you agree to accept the following formula:
P =  2πR + πΔR
Than let me explain it again:
A star must ALWAYS stay at the arm. (Let's ignore the bridge or any sort of gateway between the arms).
However, as it stay in the arm it also drifts constantly outwards in the spiral arm.
By doing so, it cross a length of P (in the spiral arm) at a given time T
Therefore, if at t=0 the radius is R1, at t=T it will drift outwards to the radius R2.
The total length that it crosses in the spiral arm is represented by the following formula (for one full orbital cycle):
P =  2πR + π(R2-R1) = 2πR + πΔR

If the star was keeping his orbital radius than the expected total lenth that it had to cross at time T (in a cycle) was 2πR.
However, as it drifts outwards in the spiral arms (at the same time T) and based on that formula we have found that it cross 2πR + πΔR.
Therefore, as the star drifts from R1 to R2 (while it stay at the spiral arm) it increases the length that it cross by πΔR comparing to a star that stay at the same radius (R1).
If you understand that:
Let's use the following example:
Let's assume that star A is located at radius 4KPC from the center.
We know that at that distance it is expected to have an orbital velocity of 200Km/s
So, if the star will stay at that radius during time interval T, it will cross one full cycle which is represented by a length of  2πR (R=4KPC).
So, by definition
2πR/T (at R=4KPC) = 200km/s.
Let's look at another star that is located at R=8KPC.
In order to cross the full cycle in the same T it is clear that its orbital velocity should be double = 400Km/s.
Our sun is located at a radius which is greater than 8KPC.
At our location the expected orbital velocity should be 460 Km/s.
That actually represents the orbital velocity of the arm.
So, at our location, the arm orbits at 460Km/s.
At 8KPC it orbits at 400Km/s
At 4KPC it orbits at 200Km/s
But now we can see the great impact of  ΔP = πΔR.
This ΔP represents the length that the star drifts outwards in the arm.
Due to the spiral shape of the arm the star at time interval T increases its radius by ΔR= R2-R1 = 8KPC - 4KPC = 4KPC,
But it also moved backwards.
so
ΔP↑ - Vector of the drifting outwards (length) in the spiral arm per time interval T
ΔR↑ - Vector of the increased radius per time interval T.
ΔB↑ - Vector of the length that the star moves backwards per time interval T.
ΔP↑= ΔR↑ + ΔB↑
So, ΔB/T gives the negative orbital velocity of the star.
Therefore, we know that the orbital velocity of the arm at the Sun location is 460 Km/s. so if the sun had to stay at the same radius and at the arm, it had to increase its velocity to 460Km/s.
However, as it drifts backwards in the arm at ΔB it reduces its velocity to 220Km/s.
Therefore,
ΔB/T = 460-220 = 240Km/s
So ΔB represents the backwards velocity of the sun in the arm, which is 240Km/s.

Now
Our scientists don't think about that ΔB vector.
Therefore, In order to stay at the same radius and keep the velocity at 220Km/s, the Sun MUST exit from the arm.
That is their fatal mistake.
The Sun will stay at the arm as long as it can, but it must drift outwards. That drifting outwards set the spiral shape of the arms in the galaxy.
I hope that by know you understand why the arm is moving at our location at 460Km/s while the orbital velocity of the Sun is only 220Km/s.
Once you understand that key information I will answer other questions.
So please let me know if you understand this explanation.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/04/2019 22:38:29
All my theory is based on Newton law including F=ma.
Oh.  That add another boatload of contradictions then.  I had been refraining since I had assumed your untold theory would provide replacements for all the current physics which I thought you were denying.

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However, if you agree to accept the following formula:
P =  2πR + πΔR
Than let me explain it again:
A star must ALWAYS stay at the arm. (Let's ignore the bridge or any sort of gateway between the arms).
However, as it stay in the arm it also drifts constantly outwards in the spiral arm.
By doing so, it cross a length of P (in the spiral arm) at a given time T
Therefore, if at t=0 the radius is R1, at t=T it will drift outwards to the radius R2.
The total length that it crosses in the spiral arm is represented by the following formula (for one full orbital cycle):
P =  2πR + π(R2-R1) = 2πR + πΔR
Not much use unless you plug some numbers into that.  T=0 for 'now'.  There is no T in that formula, so not sure why it was specified.  R1 is presumably something like 8.2 KPC but you don't give any numbers for the other values, including the time for one 'cycle' which seems to be the time it takes for the form to rotate one more time than how long it takes the sun to go around.  Neither motion is an orbit since orbits don't move like that.
Anyway, the formula computes the length P.  What figure do you get?
I'd plug in numbers for you but I'd rather see you do it.

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If the star was keeping his orbital radius than the expected total lenth that it had to cross at time T (in a cycle) was 2πR.
'At time T' is a moment in time, and things take duration to cross a distance.  So I think you mean interval, but both the sun and the arms are moving, so you can't compute it using the absolute speed of either, but rather the difference between the two.  The length of time it takes the sun to go around the galaxy is different than the time it takes for the Orion arm to be at the same orientation relative to us as it is now.
This is why it would really help if you plug numbers into your formula.  It would help me understand what you're trying to convey here.  All I get so far is repetition from prior posts, and you indicated that this P= formula would prove something.

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However, as it drifts outwards in the spiral arms (at the same time T) and based on that formula we have found that it cross 2πR + πΔR.
Relative to the form, yes.  P is the amount of arm it traverses in that time (not at that time).  I stress that bit for clarity sake.

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Therefore, as the star drifts from R1 to R2 (while it stay at the spiral arm) it increases the length that it cross by πΔR comparing to a star that stay at the same radius (R1).
The one staying at the same radius will exit the arm and the distance it travels relative to that arm isn't really defined any more than is the distance I travel relative to all the cars going by outside. They're not going any particular direction any more than are the arms. So again, not sure what you're trying to say here.  P is a distance relative to the (local) arm, not relative to other parts of the form which are moving with different velocity.

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If you understand that:
Sort of.  I asked for clarifications.

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Let's use the following example:
Let's assume that star A is located at radius 4KPC from the center.
We know that at that distance it is expected to have an orbital velocity of 200Km/s
So, if the star will stay at that radius during time interval T, it will cross one full cycle which is represented by a length of  2πR (R=4KPC).
Interval T is unspecified, but yes, if the motion is a circular orbit, then one orbit (which one of the things you are calling a cycle) covers that distance, presumably relative to the galaxy and not relative to the rotating form as you described above.  This ambiguity of 'speed relative to unspecified X' is quite confusing.

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So, by definition
2πR/T (at R=4KPC) = 200km/s.
Definition of what?  Speed is 200 km/s by observation you mean.  Nothing is defined that way.  You didn't specify orbital period T, but you can work it out from the observed speed.  You seem to be working out the speed from an unspecified T, which doesn't work.  Besides, you used T before to indicate the time to go one 'cycle' relative to the moving form (the arm), not to the stationary galaxy which is what an orbit is.  You need to use different variables for different things.  To use the same ones is again quite confusing.

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Let's look at another star that is located at R=8KPC.
In order to cross the full cycle in the same T it is clear that its orbital velocity should be double = 400Km/s.
In order to orbit in the same time, yes.  But it doesn't, unless you are claiming that it does, in which case you're claiming that the measurements taken are fiction.

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Our sun is located at a radius which is greater than 8KPC.
At our location the expected orbital velocity should be 460 Km/s.
Only if I expect it to orbit in the same time as the stars at 4KPC, and I don't expect that at all.

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That actually represents the orbital velocity of the arm.
So, at our location, the arm orbits at 460Km/s.
The arm goes a bit faster, and it is rotational velocity, not orbital velocity.  But yes, the figure is more or less correct.

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At 8KPC it orbits at 400Km/s
At 4KPC it orbits at 200Km/s
But now we can see the great impact of  ΔP = πΔR.
P was the length of a segment of arm, not a speed.  So this says that a segment of arm (a 10° chunk say) is twice as long at twice the radius.  That's right...  it just doesn't relate to the speeds you worked out above, which are arm speeds relative to the galaxy, not arm speeds relative to the local stars.

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This ΔP represents the length that the star drifts outwards in the arm.
Well, if it is πΔR, then it is the tangential component of that drift down the length of the arm, which is consistent with the way you defined ΔP and ΔR in post 283.  The total motion down the arm is the vector sum of the two.

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Due to the spiral shape of the arm the star at time interval T increases its radius by ΔR= R2-R1 = 8KPC - 4KPC = 4KPC,
You used several different interval T's above.  Which one are we referencing now?  The time it takes for a star to migrate from 4KPC to 8KPC?  You didn't compute that at all.  It isn't one 'cycle' of anything, whatever that is.  You didn't even compute the distance P between those two points since the number of cycles is not known.  It can be done.  The data is there.  But I'm really unclear what you are talking about without seeing what you're actually plugging in to these formulas.

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But it also moved backwards.
so
ΔP↑ - Vector of the drifting outwards (length) in the spiral arm per time interval T
ΔR↑ - Vector of the increased radius per time interval T.
ΔB↑ - Vector of the length that the star moves backwards per time interval T.
ΔP↑= ΔR↑ + ΔB↑
So, ΔB/T gives the negative orbital velocity of the star.
Motion with a nonzero net ΔR is not orbital, but yes, I agree with the calculation.

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Therefore, we know that the orbital velocity of the arm at the Sun location is 460 Km/s. so if the sun had to stay at the same radius and at the arm, it had to increase its velocity to 460Km/s.
However, as it drifts backwards in the arm at ΔB it reduces its velocity to 220Km/s.
Therefore,
ΔB/T = 460-220 = 240Km/s
So ΔB represents the backwards velocity of the sun in the arm, which is 240Km/s.
We had worked out 480-220=260 in a prior post, but close enough.  So far, all this is pretty repetitive of prior posts.  By your theory we should be progressing out the length of the arm at a tangential rate of about 240 km/s.  The actual rate is higher because ΔR has yet to be worked in.

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Now
Our scientists don't think about that ΔB vector.
Therefore, In order to stay at the same radius and keep the velocity at 220Km/s, the Sun MUST exit from the arm.
That is their fatal mistake.
Why?  Yes, they have the arms passing us by one after the other.  You disagree with this, but that disagreement does not demonstrate any inconsistency in the view.  In fact it is sort of a selling point if you ask me.  So what's fatal about their conclusion that the Orion arm will shortly pass us by?  It is moving at over twice our speed after all.

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The Sun will stay at the arm as long as it can, but it must drift outwards.
So you assert despite the lack of force driving it on such a trajectory.  The idea violates this Newtonian physics that you asserted at the top. The acceleration vector you describe does not match the force vector sum on the solar system of all the mass everywhere.  So your story is a blatant contradiction with F=ma.

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That drifting outwards set the spiral shape of the arms in the galaxy.
This is just an assertion. It doesn't follow from anything you've said. In fact, what you said is that the spiral shape sets the outward drift, not the other way around.  You assumed the spiral shape in order to make your calculations of where the star must go to stay within it.

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I hope that by know you understand why the arm is moving at our location at 460Km/s while the orbital velocity of the Sun is only 220Km/s.
I understood that several posts back when I told you the value the first time.  It also necessitates a ΔR rate of about 100 km/s to move at that speed and stay in the arm, yet no such outward motion has been measured.  100km/s would be an impossible thing to miss.  But this, being a trivial falsification of your view, is something that you will simply ignore as you have done all the other times I've pointed it out.

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Once you understand that key information I will answer other questions.
I understood it all back when I first did the math for you in post 285.  You've only since been quoting the numbers back at me.  This entire post has been repetition of prior posts.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/04/2019 06:39:20
I have found the key evidence which proves my theory:
http://galaxymap.org/drupal/node/204
If you look carefully, there are bridges between the arms:
The Sun is located at the Orange arm color.
There are several bridges there including the Perseus bridge.
So, this Proves my expectation for bridges between the arms.
If we could monitor it carefully, we should see that the aria between the arms (without the bridge is totaly empty from any sort of star!!!)
If we focus on the Mapping hydrogen, we get the following:
http://galaxymap.org/drupal/node/202
"the hydrogen emission is more widespread than the visual data and shows regions of the galaxy that at visual frequencies are obscured by dust, as well as bridges or spurs that connect spiral arms.."
So, do you agree that there are bridges between the arms?
It is also stated:
"the hydrogen data shows more detailed structure than the visual data, including many "holes" or bubbles in the hydrogen emission"
So, do you agree that between the arms there are many "holes" or bubbles without even a single star?
This is how real theory works -
You set an expectation - and you prove it by real evidence!!!
Therefore do you agree that this article by itself proves that my theory is correct by 100%?

Motion with a nonzero net ΔR is not orbital, but yes, I agree with the calculation.
Thanks
I fully agree with you as it comes to that specific ΔR.
The Sun doesn't orbit around the center of the galaxy.
Therefore, there is no gravity bonding between the Sun and the center of the galaxy (or the orbital sphere if you wish).
However, the main gravity bonding exists between the Sun and the local nearby stars.
They hold each other by gravity.
So, the arm is there due to the gravity bonding between the stars in the arm.
I will explain later on how the gravity works in the arm.
However, it is important to understand that in any real orbital cycle ΔR is nonzero by definition.
ΔR can be few Pico mm per cycle or few Km per cycle, but it is always there!!!
This is the highlight of my theory!
I think that our scientists have missed that key issue.
Please, you have to accept this idea as is!!!
My whole theory is based on this element and all the other elements including VHP.
We can't argue again and again and... on each element.
Please!!!
So, the sun orbits mainly due to the impact of the gravity force in the local aria. At each orbital cycle  (not around the center of the galaxy - but around the VHP1, it increases it's orbital radius)
If you insist to argue again about those key elements at each discussion, than we clearly wasting our time.
So, please let me know if you are willing to accept them all (Just during the introduction of the theory)
However, after the full introduction of the whole theory - you are more than welcome to argue on all of those elements.
Let me give you an example -
Let's assume that I have to teach you Hebrew or Arabic.
We normally write those languages from right to left.
However, if you insist that we must write it from left to right, how can you open yourself for those kinds of different languages?
In the same token, I ask you to accept the rules of my theory.
At the end, you can decide if you like it or not.
Agree?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/04/2019 13:35:33
The real Impact of VHP

Dear Halc
I know that you don't like the idea of VHP.
However, I think that the VHP is the most important element in the spiral galaxy and it fully meets Newton law.
Our scientists claim that the Sun is bobbling up and down due to the gravity force of the galactic disc.
So, they believe that the Sun has to cross the galactic disc while it bobbles up and down.
However, they ignore the key issue that the sun doesn't go up down, but it has also some inclination with regards to the galactic disc.
That by itself proves that the sun doesn't orbit/bobbling around the galactic disc.
I also couldn't find any bobbling activity under Newton law.
So, there is no gravity force due to galactic disc and the Sun doesn't bobble around the galactic disc while it orbits around the galactic center.
Therefore, this bobbling idea is just a bobbling idea in a science fiction book.
If we could monitor the Sun for few thousand years we would verify that the Sun is not going to cross the galactic disc (Never & ever). We should see that it orbits around some virtual point as it orbits around the galactic center.
This virtual point is called VHP1.
I have already explained the source of this virtual point as follow:
The Sun had been formed is a gas cloud at the center of the galaxy near the SMBH.
The matter in that gas cloud had to orbit around the center in some sort of tornado.
This orbital motion crystallizes the matter into star/planets/moons system.
So, the center of the gas cloud is the basic location of the VHP1.
Hence, as the Sun emerges from the gas cloud it already orbits around its VHP1.
This VHP1 has great impact on gravity force.
In order to understand that let me offer the following example about centrifugal:
https://thewaythetruthandthelife.net/index/2_background/2-1_cosmological/physics/b7.htm
"If the rope in Fig 94 has the length of the seconds pendulum and the body travels around the hand 10 times per second, then the rope is subject to 400 times that load which it carries when the body simply hangs. "
"The centrifugal field, generated under these conditions is therefore 400 times as strong as the gravitational field. People have succeeded in generating such fields in excess of 100 000 times the gravitational field (Ultra-centrifuge) "
So, the centrifugal can increase the gravitational field by even 100 000 times.
In the same to token, the VHP1 can increase significantly the impact of the Sun mass by gravitational field .
The outcome is, that when we set the gravitational calculation, we have to know the real value/impact of the Sun mass at VHP1.
Based on the radius between the Sun to the VHP1 and the orbital velocity we can extract/estimate the Sun mass gravitational field at the VHP1 location.
So, the orbital motion of the Sun around the VHP1, can increase dramatically the impact of the Sun mass.
Hence, the nearby aria is not gravitational effected by the direct mass of the Sun. it is effected by the Sun mass gravitational field at the VHP1.
That could potentially increase the effected sun mass by 1000 times or even 100,000 times.
This is also correct to any star in the galaxy.
So, when we look at our nearby aria we only see about 512 stars per 100LY.
Let's assume that each star has the same Sun mass.
Our scientists might claim that based on this density and the mass per each star , there is not enough gravity force in order to hold them together - and they are fully correct.
However, due to the orbital motion of each star around its unique VHP1, the effected mass of each star at its unique VHP1 could be 1,000 Sun mass (We must verify this number).
So, now we are dealing with 512 stars per 100LY, while the effected mass of each star is 1,000 Sun mass
I'm quite sure that based on this data, our scientists will confirm that now there is enough gravity force that is needed to hold them together.
So, the orbital motion of a star around its VHP1 significantly increases the impact of its mass. (like gyroscope)
This also explains why there is no need for dark matter in the galaxy. (If 512 Stars set a gravity force of 512,000 stars - there is no need for dark matter.)
Hence, that small orbital motion increases the gravity force of the whole galaxy by 1000 times (Or even more)
Therefore - VHP1 is a key element in spiral galaxy as we get so high gravity force with so less of mass.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/04/2019 14:06:44
I will be very slow in replies for about a week.  Just too busy.
I have found the key evidence which proves my theory:
No evidence proves any theory.  Nothing can be proven by inductive evidence, which constitutes almost all empirical measurements.  Some theories just hold up well to falsification tests.
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http://galaxymap.org/drupal/node/204
If you look carefully, there are bridges between the arms:
A triumph of cherry picking.  You found something that labels bridges, even if it is a map of gas instead of stars.
I'd like to see the map without the arms drawn onto it, and without the greyed areas.

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If we could monitor it carefully, we should see that the aria between the arms (without the bridge is totaly empty from any sort of star!!!)
If we focus on the Mapping hydrogen, we get the following:
http://galaxymap.org/drupal/node/202
"the hydrogen emission is more widespread than the visual data and shows regions of the galaxy that at visual frequencies are obscured by dust, as well as bridges or spurs that connect spiral arms.."
That page has a visible image (the stars) and the gas image (orange) and when superimposed, it shows that the stars tend to be most dense between the gas arms.  So the gas is more of a foam structure with bridges connecting it all, and it avoids the stars which blow the gas away via solar wind.  At least, that's what the text says.
The middle orange image shows a lot more connective structure than the Milky Way image which has the disadvantage of not having a good face-on vantage point.

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It is also stated:
"the hydrogen data shows more detailed structure than the visual data, including many "holes" or bubbles in the hydrogen emission"
So, do you agree that between the arms there are many "holes" or bubbles without even a single star?
On the contrary, the text says that it is the stars that clear these bubbles in the gas via their solar wind:
Quote from: GalaxyMap
gas may expand away from the star formation regions, driven by enormous stellar winds. This local movement may explain the bubbles so obvious in most hydrogen maps of other galaxies.

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This is how real theory works -
You set an expectation - and you prove it by real evidence!!!
The evidence you quote talks about star formation in the gas clouds, contradicting your assertion that all stars form near the SMBH and drift outward from there.

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Therefore do you agree that this article by itself proves that my theory is correct by 100%?
Do you have any idea how proof works?  Do you know what the word means?

My alternate theory predicts among other things that there are stars.  I look up and see stars.  Thus my theory must be 100% correct and yours is wrong.  This seems to be the logic you're using, and yours is weaker since you predicted star bridges, not gas bridges.

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The Sun doesn't orbit around the center of the galaxy.
Therefore, there is no gravity bonding between the Sun and the center of the galaxy (or the orbital sphere if you wish).
1) The conclusion doesn't follow from the premise, and 2) the statement is a denial of the Newtonian physics which you say you don't deny.  Newton says that all mass attracts all other mass, a bond which cannot be broken.  Maybe you mean something else by 'gravitational bonding', but your proposed motion is definitely in conflict with Newton's laws.  Just asserting otherwise makes you look the fool.  If the motion of an object is accelerating outward and forward, the forces required for such acceleration need to be accounted for, and you have nothing to account for it.
Even the mainstream science views ran into this problem.  The acceleration vector was noted and it didn't match the one predicted by a computation of the net force on us from all the known mass in the galaxy.  They knew something was wrong.  The vector at least pointed the right way, but the magnitude was wrong.  Yours doesn't even point the right way.

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However, the main gravity bonding exists between the Sun and the local nearby stars.
They hold each other by gravity.
You assert this without showing it.  The local nearby stars are on all sides and pretty much cancel each other out.  Our overall motion seems little affected by them so long as that holds.

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However, it is important to understand that in any real orbital cycle ΔR is nonzero by definition.
Not by Newton/Kepler's definitions.  A nonzero ΔR in a real orbital cycle is a denial of their laws. You said you don't deny them, so I will hold you to it.  Perhaps your problem is that you don't understand them.

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ΔR can be few Pico mm per cycle or few Km per cycle, but it is always there!!!
This is the highlight of my theory!
Please, you have to accept this idea as is!!!
I tried that, but you didn't move on. You reasserted Newton's laws, so I no longer have to accept a blatant contradiction with them.

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My whole theory is based on this element and all the other elements including VHP.
Both contradict Newton's laws.  If you're going to hold to those laws, then your theory is contradicted and wrong.  If you deny those laws, then you need to find replacements for them, which was why I was always complaining about the theory never coming out.  I was looking for those replacements, without which no predictions can be made.

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We can't argue again and again and... on each element.
Please!!!
Yes, it seems you will argue forever despite having your idea trivially falsified.  Besides contradicting Newton's laws, it has our motion needing to be about 100 km/s away from the center of the galaxy, and no such motion has been measured.  That is a direct empirical falsification of your assertions, and you just ignore it.

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If you insist to argue again about those key elements at each discussion, than we clearly wasting our time.
Just now you're figuring that out?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/04/2019 03:29:53
Dear Halc

Thanks again for all your answers.
I do appreciate your great effort!

On the contrary, the text says that it is the stars that clear these bubbles in the gas via their solar wind:
I disagree. You have to read it all and carefully.
The Text says clearly:
"As explained in the section on Velocity, it is possible to measure the velocity of both atomic hydrogen gas and carbon monoxide and in theory this can be used to construct a map of the Milky Way assuming that the velocity of the gas is largely determined by the rotation of the galaxy. This scheme has run into difficulty for several reasons, one of which is that gas moves for other reasons than galactic rotation. For example, gas may expand away from the star formation regions, driven by enormous stellar winds. This local movement may explain the bubbles so obvious in most hydrogen maps of other galaxies."
So, it is highlight clearly that: " it is possible to measure the velocity of both atomic hydrogen gas and carbon monoxide and in theory this can be used to construct a map of the Milky Way assuming that the velocity of the gas is largely determined by the rotation of the galaxy.
Based on that verification they see clearly holes and bubbles in the map. As those holes and bubbles are not expected,
At this point they have to stop and try to understand the great impact of those Bubbles.
But as usual, our scientists don't let the evidence to confuse them.
So, they try to explain the source of the holes and bubbles.
They say "May". It might be "May" or "May not".
So it is just a speculation
"gas may expand away from the star formation regions,"
I can't understand how they can base so important question on speculation.
However, even if the gas expands as they say.
Why do we see the bubbles?
How could it be that this gas expansion could set the holes and bubbles?
On the contrary, if the gas expands – it must expend everywhere. So, we shouldn't see any bubble. Therefore, the idea that the gas expands can't give an answer for the bubbles and holes that we see..
So, why they claim:"This local movement may explain the bubbles"?
Why? How the expended gas could set any sort of bubble or hole?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/04/2019 04:05:13
So, they believe that the Sun has to cross the galactic disc while it bobbles up and down.
However, they ignore the key issue that the sun doesn't go up down, but it has also some inclination with regards to the galactic disc.
What do you consider to be 'inclination'?  The word is usually used to describe the tilt of the axis of rotation relative to the plane in which it orbits.  So the Earth has an inclination of 23° but that has nothing to do with its motion through the solar system.  Similarly the solar system has an inclination of 63° relative to the galactic plane.  So the scientists are not in any sort of denial about our inclination.  The value is pretty trivial to measure.

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I also couldn't find any bobbling activity under Newton law.
You don't understand the implications of the law then.  Work out the force on an object from a planar distribution of material.  It can be derived from Newton's gravitational formula.  The force vector is a function of the thickness/density-curve of the material.

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So, there is no gravity force due to galactic disc and the Sun doesn't bobble around the galactic disc while it orbits around the galactic center.
Your lack of mathematic skills does not constitute evidence against what the mathematics implies..

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The Sun had been formed is a gas cloud at the center of the galaxy near the SMBH.
The matter in that gas cloud had to orbit around the center in some sort of tornado.
This orbital motion crystallizes the matter into star/planets/moons system.
So, the center of the gas cloud is the basic location of the VHP1.
That makes the star its own VHP.  That's not a virtual point.  It is a real mass.  The star isn't going to orbit anything, given that description.

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Hence, as the Sun emerges from the gas cloud it already orbits around its VHP1.
How can the sun orbit itself?

You've basically described an ideal cloud that collapses into a unary solar system like ours. The star is the mass around which everything else orbits, because everything else has mass that is dwarfed by the primary.

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This VHP1 has great impact on gravity force.
If a virtual point (something virtual, with no mass of its own) has any impact on gravity at all, then it doesn't conform to Newton's laws, which describe no such thing.  His law is F=GMm/R², and the M of a VHP is zero, so zero force, and zero impact.

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In order to understand that let me offer the following example about centrifugal:
https://thewaythetruthandthelife.net/index/2_background/2-1_cosmological/physics/b7.htm
You're getting your science from a religious site?

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"If the rope in Fig 94 has the length of the seconds pendulum and the body travels around the hand 10 times per second, then the rope is subject to 400 times that load which it carries when the body simply hangs. "
"The centrifugal field, generated under these conditions is therefore 400 times as strong as the gravitational field. People have succeeded in generating such fields in excess of 100 000 times the gravitational field (Ultra-centrifuge) "
So, the centrifugal can increase the gravitational field by even 100 000 times.
Yes, this is exactly how one might create artificial gravity in a space station or ship, except the article mistakenly misrepresents the effect as a 'field'.  A centrifugal field exists only in a rotating reference frame, which is not what is being depicted.  They're using EM force instead of gravity to apply acceleration to mass.  EM force can take only so much.  It takes an incredibly strong material to achieve 100,000 Gs.  A gas cloud is not that material.

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In the same to token, the VHP1 can increase significantly the impact of the Sun mass by gravitational field .
How so?  Via what force?  Yes, If I tie two rocks together with a cable, I can get them to orbit each other via EM force at a much higher rate than what gravity would produce.  The cable is needed to transfer the force between the objects.  A VHP has no place to attach a cable or anchor any other force.  Not even gravity since it has no mass.

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The outcome is, that when we set the gravitational calculation, we have to know the real value/impact of the Sun mass at VHP1.
If it is going to have any effect on the sun, then you need to instead calculate the real value/impact of the VHP1 on the sun, not the other way around.  Of course under Newton's 3rd law, the two forces are equal and opposite.

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Based on the radius between the Sun to the VHP1 and the orbital velocity we can extract/estimate the Sun mass gravitational field at the VHP1 location.
So you claim, but no way to extract this has ever been provided, so it's fiction until then.

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So, the orbital motion of the Sun around the VHP1, can increase dramatically the impact of the Sun mass.
Nothing in the article you quoted had anything's mass being increased by its motion.  This happens under relativity, but we're not exactly getting into that here.

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Let's assume that each star has the same Sun mass.
Our scientists might claim that based on this density and the mass per each star , there is not enough gravity force in order to hold them together - and they are fully correct.
However, due to the orbital motion of each star around its unique VHP1, the effected mass of each star at its unique VHP1 could be 1,000 Sun mass (We must verify this number).
This is a complete misrepresentation of the article you quoted.  A spinning object has no more mass than the same object stationary (relativity aside).  A centrifuge exerts no more force on objects nearby when it is spinning compared to when it is stopped.  You seem to assert otherwise here.  You also assert that stars orbit a virtual point, which defies all known physics.  If there is orbital motion, it is because there is real mass to orbit, not a virtual nothing.

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So, the orbital motion of a star around its VHP1 significantly increases the impact of its mass. (like gyroscope)
A gyroscope attracts nothing when spinning, nor does its weight or mass change.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/04/2019 06:14:32
How can the sun orbit itself?

You've basically described an ideal cloud that collapses into a unary solar system like ours. The star is the mass around which everything else orbits, because everything else has mass that is dwarfed by the primary.
No
We need to understand how the Sun had been formed in a gas cloud.
You believe that the Sun had been formed as the matter collapses to the center of the gas cloud.
I claim that this is an incorrect scenario.
So how it really works?

Let's start from the excretion disc around the SMBH.
In the excretion disc there is a plasma at a temp of 10^9 c.
This plasma includes a new quarks, Atoms and molecular which had been formed by the Ultra high SMBH gravity force and the ultra velocity ( at about 0.3 speed of light)
As all of this new matter drifts outwards from the excretion disc it gets to the magnetic field around the SMBH.
The Ultra high power of that magnetic field Push that mater upwards (or downwards) with regards to that disc (at almost 0.8 speed of light).
At some point, the Atom/Molecular falls back to the galactic disc (but outwards the SMBH magnetic fields).
As the new born molecular gets to the galactic disc, it starts to orbit around the SMBH. At that point it had also decreased its temp to lower lever (I assume to 10^6 c).
However, the orbital momentum around the SMBH, push the new born molecular to gather in a hot gas clouds.
Each gas cloud set a very strong gravity force with the nearby SMBH.  The gravity force had been set
between the center of the SMBH to the center of that hot gas cloud (Let's call this center - VHP1)
However, that SMBH great gravity force, also force the matter in the gas cloud to orbit around the center of the cloud (VHP1) at a very fast speed.
We can compare it to a tornado storm in the gas cloud, while the center of the gas cloud (VHP1) is connected by gravity to the SMBH.
https://en.wikipedia.org/wiki/Tornado
"Definitions: A tornado is "a violently rotating column of air.."
So, this violently rotating column/cloud push the matter away from the center of the gas cloud (VHP1) to the edge of the gas could.
By doing so, there is high pressure in the matter as it gets to the edge of the cloud. This high pressure, High temp (10^6) and high orbital velocity set the crystallization process of the Sun.
In the first phase, we might find many small hot objects that orbit around VHP1.
However, over time those objects merge with each other and set the Sun and all the other planets and moons,
So, the sun had been formed from the matter which had been pushed away from the center (VHP1). Therefore, its temp at day one is 10^6 c. (Any planet and moon, has the same temp and the same matter as the Sun in its first day.)
Therefore, the Sun (and all its planets & moons) had been crystallized while it orbits around the center of the gas cloud (VHP1) and not IN the center of the gas cloud.
This is very important.
Therefore, when the Sun had been emerged from the gas cloud, it was already orbiting around the center of the gas cloud (VHP1) while this center orbits around the SMBH.
However, at this stage, there is no matter at the center of the gas cloud (VHP1), while the SMBH still holds that VHP1 by gravity.
So, the SMBH holds the center of the gas cloud (VHP1) by gravity while all the matter in the gas cloud had been drifted/pushed outwards and set the Sun.
This activity force the sun to orbit around a virtual host point - VHP1 (that was the center of the gas cloud - but currently without any matter) , while that VHP1 orbits around the SMBH.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/04/2019 12:48:38
What I said was that the star must be at the center of gravity of the gas that comprises it.  Thus an isolated gas cloud (need not be symmetrical) that collapses entirely into a star must put that star exactly at the center of gravity of the original gas cloud.

Please be aware that we discuss about a gas cloud near the SMBH. So, you can't get any conclusion from a gas cloud which is not directly affected by a SMBH.
I say that a gas cloud near a SMBH doesn't set the star forming activity at the center of the gas cloud.
Few questions:
Why are you so sure that a new star must be formed at the center of the gas cloud that comprises it while this gas cloud is located near the SMBH and affected by its high magnetic power?
Can you please prove it by real evidence?
Do we have a solid verification from any gas cloud which is located near the SMBH?
Please let me know if you agree with the following evidences:
1. The Plasma in the accretion disc orbits at 0.3 speed of light, while its temp is 10^9c. Yes or no?
2. There are clear evidences that the matter from the accretion disc drifts outwards. So far our scientists didn't find evidence for any sort of matter (Star, Planet, Moon or even Atom) which had been eaten by the SMBH. Yes or No?
3. There is a very strong magnetic field around the SMBH. Yes or no?
4. We clearly see a molecular jet stream which is blowing upwards/or downwards from the accretion disc at 0.8 speed of light. Yes or no?
5. The molecular falls back to the galactic disc. Yes or no?
6. With regards to S2 - based on our observation,
https://pages.uoregon.edu/imamura/323/lecture-2/sgrAbig.jpg
A. We see clearly that S2 doesn't move exactly at the expected orbital cycle. Yes or No?
B. At 1992.23 the shift is about 0.2 Light days, while at 1998.36 the shift is 0.3 Light day. Yes or no?
C. If S2 is orbiting around a VHP1 with a radius of 0.3 Light day (while this VHP1 orbits around the SMBH), do you agree that this verification is exactly what we should see. Yes or no?

 

 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/04/2019 15:41:01
Quote from: Halc
What I said was that the star must be at the center of gravity of the gas that comprises it.  Thus an isolated gas cloud (need not be symmetrical) that collapses entirely into a star must put that star exactly at the center of gravity of the original gas cloud.
Please be aware that we discuss about a gas cloud near the SMBH. So, you can't get any conclusion from a gas cloud which is not directly affected by a SMBH.
What I said doesn't change if there are external forces moving the gas particles around.  Center of mass is computed by state at a given moment, so if you move some of the particles, the center of mass moves correspondingly.  If you gather all the mass of the material into one object like a star, then the center of mass of that material will be at the center of that star, and not elsewhere.  The presence of the SMBH (besides preventing that formation) does not change that.

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I say that a gas cloud near a SMBH doesn't set the star forming activity at the center of the gas cloud.
I carefull asked what you mean by those words, and as usual you ignored it.  So I don't know what you mean by 'set the star forming activity'.  Perhaps a star might start to form to one side with most of the gas still off to the other side, but if it eventually gathers that remaining gas, the star will be exactly at the center of mass of all that material, and if it doesn't gather in all that gas, then it isn't comprised of all the material of which we're taking the center of mass.

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Why are you so sure that a new star must be formed at the center of the gas cloud that comprises it while this gas cloud is located near the SMBH and affected by its high magnetic power?
External forces are irrelevant.  I am sure of my statement because of the way one computes center of mass.  This is a simple geometric property and not even a law of physics.  How does one find the center of mass of an object?  How does on find the center of mass of an object affected by high magnetic powers of a nearby SMBH?  The answer is the same for both since the SMBH and magnetic powers are not part of the calculation.

BTW, magnetic fields are a form of friction and slow things down, not speed things up.  An SMBH with a strong magnetic field will tend to align the motion of orbiting material to cross as few abstract lines of the field.  This forms a torus shaped cloud in some instances, although this has not been observed in our own galaxy.  Thus:
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The Ultra high power of that magnetic field Push that mater upwards (or downwards) with regards to that disc
Magnetic forces tend to gather material into a disk or a torus if the field is very strong.  The field resists motion in the direction you indicate.

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Can you please prove it by real evidence?
Any computation of the center of mass of the material comprising an object would show it.  I cannot help it if your education didn't include teaching you how to do that.  So I can prove it to the average 7th grader, but I cannot prove it to you.

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1. The Plasma in the accretion disc orbits at 0.3 speed of light, while its temp is 10^9c. Yes or no?
It varies of course, but I don't know the figures myself.  You don't usually call it an accretion disk.  So I'll say yes since I have no reason to contest it.
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2. There are clear evidences that the matter from the accretion disc drifts outwards.
Its very name suggests otherwise.  They wouldn't call it that if matter drifted outward.  There is no clear evidence of this at all.
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So far our scientists didn't find evidence for any sort of matter (Star, Planet, Moon or even Atom) which had been eaten by the SMBH. Yes or No?
No.  All visual measurements of any SMBH are due to the matter being 'eaten' by it.  If it doesn't pull in matter, it is entirely invisible.  Ours is particularly starved for material, so it consumes matter at an unusually low rate, and thus emits far less radiation than a typical one.

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3. There is a very strong magnetic field around the SMBH. Yes or no?
It has one, yes.  It is not particularly strong.  Cygnus A is a nice example of a SMBH with a strong magnetic field.
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4. We clearly see a molecular jet stream which is blowing upwards/or downwards from the accretion disc at 0.8 speed of light. Yes or no?
No.  Nothing like that comes from the accretion disc.  No such jet has ever been observed.
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5. The molecular falls back to the galactic disc. Yes or no?
Material moving away from the galaxy will return to the galaxy if below escape velocity.   The disk has little to do with this, other than to provide most of the mass that determines the escape velocity of the galaxy at a given point.
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6. With regards to S2 - based on our observation,
https://pages.uoregon.edu/imamura/323/lecture-2/sgrAbig.jpg
A. We see clearly that S2 doesn't move exactly at the expected orbital cycle. Yes or No?
There is no plot of expectation vs reality there, so it is not clearly depicted, and thus No.  Since we don't know the distribution of matter near its path, there is no clear expectation of exactly where it will go.

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B. At 1992.23 the shift is about 0.2 Light days, while at 1998.36 the shift is 0.3 Light day. Yes or no?
Don't know what shift is.  You mean the length of the error bars?
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C. If S2 is orbiting around a VHP1 with a radius of 0.3 Light day (while this VHP1 orbits around the SMBH), do you agree that this verification is exactly what we should see. Yes or no?
No.  What one would see is a regular deviation apparent in a Fourier transform of the position data.  That analysis has been done (you did not link to such an analysis) and it has conclusively shows that S2 does not orbit a secondary mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/04/2019 18:35:45
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4. We clearly see a molecular jet stream which is blowing upwards/or downwards from the accretion disc at 0.8 speed of light. Yes or no?
No.  Nothing like that comes from the accretion disc.  No such jet has ever been observed.
https://phys.org/news/2012-05-ghostly-gamma-ray-blast-milky-center.html
Yes, there are clear evidences!
"New evidence of ghostly gamma-ray beams suggests that the Milky Way's central black hole was much more active in the past."
"Those bubbles also stretch 27,000 light-years from the center of the Milky Way. However, where the bubbles are perpendicular to the galactic plane, the gamma-ray jets are tilted at an angle of 15 degrees. This may reflect a tilt of the accretion disk surrounding the supermassive black hole."
Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required.
"Shoving 10,000 suns into the black hole at once would do the trick. Black holes are messy eaters, so some of that material would spew out and power the jets," he said."
Our scientists have totally got lost.
They don't have even a clue about the meaning of that jet.
The total mass in the accretion disc is estimated to be at the range of 3 suns mass.
How can 3 sun mass eject 10,000 sun mass?
The answer is simple -
All of 10,000 suns mass had been ejected from the excretion disc which has only 3 sun mass at any given time.
However - Over time, that 3 Sun mass set that 10,000 sun mass in the jet stream.
Let's assume that the time which is needed to set this jet stream is T.
So, if there are 400 billion stars in the galaxy, than the excretion disc can create them all in the following time frame:
t = (400,000,000 / 10,000) T = 40,000T
So simple!
In the same token - The only power which could boost that jet is the magnetic power.
The image shows how strong is that ultra high magnetic power!!!
Our scientists claim that the image of the "the gamma-ray jets are tilted at an angle of 15 degrees" and therefore, "this may reflect a tilt of the accretion disk surrounding the suppermassive black hole" and I fully agree with that.
So, this jet by itself proves the great functionality of the excretion disc and the magnetic power.
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B. At 1992.23 the shift is about 0.2 Light days, while at 1998.36 the shift is 0.3 Light day. Yes or no?
Don't know what shift is.  You mean the length of the error bars?
Yes
You call it error bar.
I call it "shift".
So, how can you justify that "error bar"?
Since we don't know the distribution of matter near its path, there is no clear expectation of exactly where it will go.
If you "don't know the distribution of matter near its path" how can you claim that this is the ultimate source for the "error bar"?
That analysis has been done (you did not link to such an analysis) and it has conclusively shows that S2 does not orbit a secondary mass.
What kind of analysis had been set to prove that S2 doesn't orbit a secondary mass?
In any case - As I have stated - there is no real mass at VHP1. It is a virtual point of mass.
So, I don't expect to see any real mass there. The analysis is fully correct.
However, VHP1 is the only explanation for that shift ("error bar" if you wish).
So, VHP1 is there because it must be there!
That VHP1 increases the total effective gravity of the galaxy and eliminate the need for the science fiction idea which is called - "Dark matter".
VHP1 is key element in the activity of spiral galaxy.

No.  All visual measurements of any SMBH are due to the matter being 'eaten' by it.  If it doesn't pull in matter, it is entirely invisible.  Ours is particularly starved for material, so it consumes matter at an unusually low rate, and thus emits far less radiation than a typical one.
This is a fatal error.
You want to believe that it eats matter, but you have no prove for that.
Therefore, you claim that this process is invisible.
No. I disagree.
If the SMBH eats a star it shouldn't be invisible. Actually, some time ago our scientists were expecting for fireworks as one of the S stars gets closer to the SMBH, but the star had survive.
So, the SMBH will never ever eat even one atom from the galaxy, as all the matter in the 400 Billion stars had been created by the SMBH!
As I have stated - mothers do not eat their children.
It's time for you to open your mind and understand the real activity in our galaxy!
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I say that a gas cloud near a SMBH doesn't set the star forming activity at the center of the gas cloud.
I carefully asked what you mean by those words, and as usual you ignored it.  So I don't know what you mean by 'set the star forming activity'.
It means - the crystallization process of the molecular in the gas cloud.

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Why are you so sure that a new star must be formed at the center of the gas cloud that comprises it while this gas cloud is located near the SMBH and affected by its high magnetic power?
External forces are irrelevant.  I am sure of my statement because of the way one computes center of mass.  This is a simple geometric property and not even a law of physics.  How does one find the center of mass of an object?  How does one find the center of mass of an object affected by high magnetic powers of a nearby SMBH?  The answer is the same for both since the SMBH and magnetic powers are not part of the calculation.
Sorry - I totally disagree with your following statement: "External forces are irrelevant."
The SMBH has a great impact on the nearby gas cloud! Its ultra high gravity force sets the crystallization process at the gas cloud.
I agree that at the first stage the SMBH sets the gravity bonding with the Gas cloud center of mass.
However, after the bonding, as the matter in the gas orbit around the center of the gas cloud, the SMBH will hold that center by gravity as long as needed.
For example -
If during the crystallization process, there will be a ring of matter/objects which orbits around the center at a radius of R (while the center is totally empty), than I assume that you agree that the SMBH will continue to hold the center (Let's call this center VHP1).
From this moment there is no meaning how that ring will look like.
I claim that even if all the matter in the ring will crystallized to one star, the SMBH will continue to hold this VHP1 while the star orbits around that point with a radius R.
By that activity, the SMBH had increased the gravity force impact of that star by several hundreds or thousands.
So, our sun could set a gravity force which is equivalent to 1000 Sun mass just due to the orbital movement around VHP1. That is correct to S2 and any star in the galaxy.
Hence, 1,000 stars can set a gravity force which is equivalent to one million stars.
Therefore, when we try to monitor the gravity impact of the galaxy, we see that it is much more than the total mass in all the stars in the galaxy.
So, that simple VHP1 eliminates the need for "dark matter"



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/04/2019 19:40:56
Quote from: Halc
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4. We clearly see a molecular jet stream which is blowing upwards/or downwards from the accretion disc at 0.8 speed of light. Yes or no?
No.  Nothing like that comes from the accretion disc.  No such jet has ever been observed.
https://phys.org/news/2012-05-ghostly-gamma-ray-blast-milky-center.html
Yes, there are clear evidences!
Those jets do not come from the accretion disk.  They come from the poles.  That jet always exists, but is very weak for Sgr-A, as this article confirms.  It is very apparent in other black holes (little ones, other galaxies), but it doesn't come from the disk and it is mostly gamma rays, not hydrogen.

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Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required.
"Shoving 10,000 suns into the black hole at once would do the trick. Black holes are messy eaters, so some of that material would spew out and power the jets," he said."
Our scientists have totally got lost.
They don't have even a clue about the meaning of that jet.
The total mass in the accretion disc is estimated to be at the range of 3 suns mass.
How can 3 sun mass eject 10,000 sun mass?
He suggests a mass that large falling in, not coming out.  The current disk has very little mass because so little material is falling in.

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However - Over time, that 3 Sun mass set that 10,000 sun mass in the jet stream.
The jet stream is a gamma ray beam which doesn't even have proper mass.  It is light.

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In the same token - The only power which could boost that jet is the magnetic power.
Magnetic fields have no impact on light.  It carries no charge.

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You call it error bar.
I call it "shift".
So, how can you justify that "error bar"?
It means the measurement has only so much accuracy.  You'll notice that in other plots, the technology has improved and the more recent measurements are more accurate.

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If you "don't know the distribution of matter near its path" how can you claim that this is the ultimate source for the "error bar"?
It is an error in measurement, not a statement of probability of where they think it will be before they measure it.

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Quote from: Halc
What one would see is a regular deviation apparent in a Fourier transform of the position data. That analysis has been done (you did not link to such an analysis) and it has conclusively shows that S2 does not orbit a secondary mass.
What kind of analysis had been set to prove that S2 doesn't orbit a secondary mass?
The one you edited out of my post, and I put back.

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In any case - As I have stated - there is no real mass at VHP1. It is a virtual point of mass.
If it is virtual, it has no gravitational effect.  It takes real mass to do that.

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Quote from: Halc
So I don't know what you mean by 'set the star forming activity'.
It means - the crystallization process of the molecular in the gas cloud.
OK, that works.  The process can begin off-center, but it cannot finish there.


Have to go.  Didn't reply to the whole thing.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/04/2019 04:10:15
Sorry - I totally disagree with your following statement: "External forces are irrelevant.".  The SMBH has a great impact on the nearby gas cloud
Forces (external or otherwise) are irrelevant to a center-of-mass calculation.  Of course gravity has impact on the gas.  It just doesn't come into play with a CoM calculation.  A collection of material cannot be somewhere where it isn't, force or no force.
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Its ultra high gravity force sets the crystallization process at the gas cloud.
That it might do, but more a moderate force.  A high force tends to separate gas and more solid objects.  If you were to orbit close enough to a small black hole in a space suit, you would be pulled to pieces, not compressed.  Crystallization utilizes the gravity of the gas itself, possibly with an external shock wave to seed the process.
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I agree that at the first stage the SMBH sets the gravity bonding with the Gas cloud center of mass.
However, after the bonding, as the matter in the gas orbit around the center of the gas cloud, the SMBH will hold that center by gravity as long as needed.
Gravity exerts no force on a center of mass.  It exerts it on mass.  Newton's formula has nothing about gravity acting on an abstract point.  Such points have no mass and are thus unaffected by any gravitational field.

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For example -
If during the crystallization process, there will be a ring of matter/objects which orbits around the center at a radius of R (while the center is totally empty),
Are you suggesting a ring shaped cloud?  That might happen if there is a gravity source pulling the cloud into that shape, but then the ring isn't exactly empty.  Saturn's rings are a shape like that, but the center has Saturn sitting in it.  Clouds don't form rings by themselves.  They tend to be formless blobs, but if the gravity of the cloud acts on itself, it would be most dense somewhere in the middle, not the edges.  Real clouds tend to be long drawn out stretches of matter which might form several stars along its length.
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than I assume that you agree that the SMBH will continue to hold the center (Let's call this center VHP1).
The SMBH will not hold the center since there is no mass to hold there.  It will attract the nearest part the most and the furthest part the least, and thus pull such a ring apart.  The inner part will orbit faster than the stuff further away, so the ring ends up getting completely dispersed as the various atoms/objects assume different orbits.

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I claim that even if all the matter in the ring will crystallized to one star, the SMBH will continue to hold this VHP1 while the star orbits around that point with a radius R.
Your claim has gravity acting on non-mass, which violates Newton's laws.  Either you rewrite those laws, or your claim is contradictory.
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By that activity, the SMBH had increased the gravity force impact of that star by several hundreds or thousands.
This is also a contradiction.  Newton's laws say no such thing.  The force exerted by gravity is clearly spelled out in the simple formula.

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So, our sun could set a gravity force which is equivalent to 1000 Sun mass just due to the orbital movement around VHP1. That is correct to S2 and any star in the galaxy.
I've already commented on this nonsense above, and you just ignored my comments.  I see little point in repeating it
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/04/2019 14:27:40
Magnetic fields have no impact on light.  It carries no charge.
It is stated:
"The magnetic field embedded in the disk therefore accelerates the jet material along the spin axis of the black hole, which may not be aligned with the Milky Way."
Those jets do not come from the accretion disk.  They come from the poles.  That jet always exists, but is very weak for Sgr-A, as this article confirms.  It is very apparent in other black holes (little ones, other galaxies), but it doesn't come from the disk and it is mostly gamma rays, not hydrogen.
It is also stated:
"The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused. The gamma-ray bubbles likely were created by a "wind" of hot matter blowing outward from the black hole's accretion disk. As a result, they are much broader than the narrow jets."
They specifically claim that the source of the jet is the plasma that squirted out from the galactic center. So, the gamma ray is just an evidence for the "hot matter blowing outward from the black hole's accretion disk."
Therefore, this is a clear indication for the Plasma/hot matter which had been squirted out from the accretion disc.
It is also stated clearly that the "magnetic field kept it tightly focused" and "The magnetic field embedded in the disk therefore accelerates the jet material along the spin axis of the black hole"
So, why do you insist again and again to show that any issue that I highlight is incorrect by definition even if there is clear evidence?
What do you gain with that?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/04/2019 15:39:36
Quote from: Halc
Magnetic fields have no impact on light.  It carries no charge.
It is stated:
"The magnetic field embedded in the disk therefore accelerates the jet material along the spin axis of the black hole, which may not be aligned with the Milky Way."
That makes sense.  The material they're speaking of here is not light (which would not move as slow as 0.8c anyway).  A magnetic field will resist motion across field lines, so motion towards and away from the disk is slowed by friction.  But at the poles, the motion crosses no field lines, so the effect is acceleration much like a linear solenoid.
So the disk (the rotating material outside the black hole) generates the field, but the jets emanate from the material at the poles, not the disk.

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It is also stated:
"The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused. The gamma-ray bubbles likely were created by a "wind" of hot matter blowing outward from the black hole's accretion disk. As a result, they are much broader than the narrow jets."
I actually don't know what they're talking about here.  Gamma rays are light and light emitted a million years ago doesn't hang around to be seen much later.  How can any kind of light form a bubble that persists for an extended time like that?  Ditto for the gamma ray jet that should not be detectable after the jets shut down.  Sure, the galaxy currently has active jets, just not as intense as it seemingly was a million years ago.  I have no idea how they could detect that.  It must be some side effect of the energy that persists and is detectable so much time later.

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They specifically claim that the source of the jet is the plasma that squirted out from the galactic center. So, the gamma ray is just an evidence for the "hot matter blowing outward from the black hole's accretion disk."
Yes, I agree with that.  Some of the matter falling in finds its way to the poles and is squirted out in these plasma jets, powered by the energy of the greater amount of material falling in.  Supernovas do the same thing, where a gravitational collapse of a star powers a fraction of the material to escape at incredible energies.  They have the same characteristic jets at the poles.

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Therefore, this is a clear indication for the Plasma/hot matter which had been squirted out from the accretion disc.
From the part at the poles, which isn't really the disk part, but nevertheless powered by the magnetic field of the rotating disk.  I'm not disagreeing with your comment, just clarifying that no material jets from the sides where the disk is.  It comes from the material at the axis, not the ring of material on the sides.  The inner-most material under the highest gravity might cause some of the material of the disk to find its way around the black hole to the poles where the magnetic field is strong enough to counter the gravity pulling it in.

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It is also stated clearly that the "magnetic field kept it tightly focused" and "The magnetic field embedded in the disk therefore accelerates the jet material along the spin axis of the black hole"
So, why do you insist again and again to show that any issue that I highlight is incorrect by definition even if there is clear evidence?
I don't disagree with the above quote.  Your comments seem to imply that the jets come directly from the disk, and are not just powered by the disk.  If you mean the latter, then we're not in disagreement.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/04/2019 16:18:59
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They specifically claim that the source of the jet is the plasma that squirted out from the galactic center. So, the gamma ray is just an evidence for the "hot matter blowing outward from the black hole's accretion disk."
Yes, I agree with that.
So, do you agree that the hot matter/plasma is blowing outwards from the accretion disc?
Therefore, do you agree that the accretion disc acts as an excretion disc?
Do you agree that this hot matter is boosted upwards & downwards due to the ultra high magnetic field around the SMBH?
If so, why do you still claim:
Your comments seem to imply that the jets come directly from the disk, and are not just powered by the disk.

Don't you see that the hot matter is coming outwards directly from the excretion disc while it powers up/down by the magnetic field and set the jet?
So, the magnetic power boosts that hot matter/plasma/molecular (which had been ejected from the excretion disc) upwards and downwards at almost 0.8 speed of light.
This matter is the source for all the stars in our galaxy.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/04/2019 21:10:43
Quote from: Halc
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They specifically claim that the source of the jet is the plasma that squirted out from the galactic center. So, the gamma ray is just an evidence for the "hot matter blowing outward from the black hole's accretion disk."
Yes, I agree with that.
So, do you agree that the hot matter/plasma is blowing outwards from the accretion disc?
I was not clear about the part with which I am in agreement. I agree with their comment that the jet is plasma squirted out from the galactic center. It says from the center, not from the disk, which is not at the center. I don't agree with you chnaging that to imply that the jet emanates directly from the disk.
The material does come in through the disk, and if it gets 'squirted' to the poles as some of it seems to, then the field there is capable of ejecting it like that.
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Therefore, do you agree that the accretion disc acts as an excretion disc?
What you have described as an excretion disk is a disc created by material excreted from the central object itself.  An accretion disk is material from outside that was drawn in via the gravity of the central mass.  The difference is matter coming from above or below.  No, your excretion description defies gravitational field theory, which is not Newton's work, but rather Einstein's.
Yes, I agree that some of the accreted material is ejected at the poles (not at the disk), rather than completing its fall into the gravity well that drew it in.  There are countless websites that depict the typical disk and jet such as https://i.pinimg.com/originals/25/f0/60/25f060ef8e26c1012d3e137b97a5226d.jpg
These pictures seem to all be artists conceptions and not actual photographs.  The beam of light is quite visible if it gets pointed straight at Earth as it does in pulsars.

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Do you agree that this hot matter is boosted upwards & downwards due to the ultra high magnetic field around the SMBH?
Yes.  Most of that beam is light, but some of it is matter.  The magnetic field only allows it in those two directions.  Any other direction involves crossing the magnetic field lines which reduces motion.

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If so, why do you still claim:
Quote from: Halc
Your comments seem to imply that the jets come directly from the disk, and are not just powered by the disk.
Because that is a different claim than the one above with which I have far less objection.

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Don't you see that the hot matter is coming outwards directly from the excretion disc while it powers up/down by the magnetic field and set the jet?
From the poles.  Not from elsewhere.
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This matter is the source for all the stars in our galaxy.
Well, It's the matter falling in, a small percentage of which is ejected.  There is a net loss, not a net gain.  This is quite apparent in our own galaxy which currently lacks significant material falling in, and thus the jets are all but gone.  Dump new stuff in, and the jets will start up again.
The excretion model would presumably be a more continuous process and not rely on material being dumped in from the outside.  It needs to rewrite relativity, and thus there is no black hole at all, but just some sort of matter/energy generation point, which is completely different physics.  It isn't wrong physics, but it certainly contradicts accepted principles.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/04/2019 13:23:46
Well, It's the matter falling in, a small percentage of which is ejected.  There is a net loss, not a net gain.  This is quite apparent in our own galaxy which currently lacks significant material falling in, and thus the jets are all but gone.  Dump new stuff in, and the jets will start up again.
O.K
Few questions:
In the article it is stated:" The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused"
"The gamma-ray bubbles likely were created by a "wind" of hot matter blowing outward from the black hole's accretion disk."

1. Do you agree that Plasma exsists only at the accertion disc (at 10^9 c)?
2. Do you agree that there is a very strong magnetic field around the accertion disc?
3. Do you agree that this strong magnetic field can boost the Plasma/hot matter (which had been ejected from the accertion disc) upwards or downwards by almost 0.8 speed of light?
4. So, do you agree that we have clear evidences that hot matter/plasma are excerted outwards from the accertion disc?
5. Do you agree that so far we didn't find any evidence for any sort of matter which eccerts from outside directly into the eccertion disc?
6. If we only see matter that get's out from the accertion disc, while we have never ever see any sort of matter that gets in, why do you still believe that the eccertion disc get's its matter from outside?
7. What kind of evidence is needed to convince you that the accertion disc is actually axcertion disc?  That it never ever eats any matter form outside? That the plasma/hot matter is constantly generated by the SMBH gravity force and than ejected outwards from the accertion disc?



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/04/2019 15:41:54
Few questions:
In the article it is stated:" The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused"
"The gamma-ray bubbles likely were created by a "wind" of hot matter blowing outward from the black hole's accretion disk."
I really have no idea what it is trying to convey there.  Gamma rays are not matter, and do not gather anywhere.  They are light and part company with their origin at the speed of light.  So maybe it is some kind of matter that sticks around and visible because it gives off gamma rays of its own.  That sort of fits with that description, and corresponds a little to your description as well.  I don't know what might hold it in those 'clouds' above and below the disk.  If it is matter, gravity would theoretically pull it back in.  Orbiting stuff can't form the shape that they depict in their picture.  If it was moving anywhere, it wouldn't still be there after a million years.

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1. Do you agree that Plasma exsists only at the accertion disc (at 10^9 c)?
Plasma exists in stars and in labs on Earth, although I have no idea of the temperatures in any of those places or near the SMBH.  I had a hard time finding a reference to that one.

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2. Do you agree that there is a very strong magnetic field around the accertion disc?
Most of them have one, yes.  The one at the center of the galaxy has a very weak magnetic field compared to the one in almost any other galaxy like Andromeda for instance, which has almost the same galactic mass as our own, yet a much more well fed central black hole.  You talk about a central black hole, but your theory denies them.  Anything that excretes light or matter or anything is not black.  A black body absorbs everything and puts out nothing in return.  You need a different name for it like super massive matter creator (SMMC) or something.
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3. Do you agree that this strong magnetic field can boost the Plasma/hot matter (which had been ejected from the accertion disc) upwards or downwards by almost 0.8 speed of light?
From the poles, yes.  That's where the magnetic field acts to accelerate mass, not decelerate it like it does near the disk.
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5. Do you agree that so far we didn't find any evidence for any sort of matter which eccerts from outside directly into the eccertion disc?
No excretion disk has ever been observed, so of course there is no evidence of how such a thing might behave.
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6. If we only see matter that get's out from the accertion disc, while we have never ever see any sort of matter that gets in, why do you still believe that the eccertion disc get's its matter from outside?
I don't believe that.  Any image of the disk (they just came out with a famous one) doesn't tag the matter with its origins.  I see Saturn's rings and the material there isn't thus tagged with the address of its origin, but no workable model has it springing from Saturn itself, so it's a fair bet that it came from outside.

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7. What kind of evidence is needed to convince you that the accertion disc is actually axcertion disc?
A replacement for the laws of physics which are contradicted by the idea.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/04/2019 05:10:12
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6. If we only see matter that get's out from the accertion disc, while we have never ever see any sort of matter that gets in, why do you still believe that the eccertion disc get's its matter from outside?
I don't believe that.  Any image of the disk (they just came out with a famous one) doesn't tag the matter with its origins.  I see Saturn's rings and the material there isn't thus tagged with the address of its origin, but no workable model has it springing from Saturn itself, so it's a fair bet that it came from outside.
How can you believe in image???
How can you compare the SMBH to Saturn???
There is no room for "fair bet" in science.
Only clear evidence!!!
So, let me ask you again for the last time:
Do we see any sort of mass/matter that comes from outside into the accretion disc of the SMBH?
How can you ignore the great impact of the ultra power of the magnetic field around the accretion disc?

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3. Do you agree that this strong magnetic field can boost the Plasma/hot matter (which had been ejected from the accertion disc) upwards or downwards by almost 0.8 speed of light?
From the poles, yes.  That's where the magnetic field acts to accelerate mass, not decelerate it like it does near the disk.
Sorry. It seems that your knowledge in electronics and magnetic field is very poor.
Magnetics can't start just from the poles. It covers the whole body!!!
It is similar to the magnetic field around the Earth.
Any solar wind is lifted to the poles.
Therefore, the magnetic field covers the SMBH & the accretion disc. Any molecular/matter that gets closer to that field (from outside the disc, or from the disc) is lifted upwards or downwards (according to the magnetic waves around the SMBH) directly to the Poles. However, due to the Ultra power of that magnetic field the matter is then boosted from the poles upwards/downwards at ultra high speed.
That is the only way how magnetic should work!!!
So, the real "fair bet" is that any atom which will dare to come closer (from outside) to the accretion disc, will never ever cross that ultra power magnetic field. As it will come closer to the impact of the magnetic field it will immediately lifted upwards or downwards to the poles. So there is no chance that any particle/molecular/Moon/Planet/Star will be able to cross that magnetic field in his way to the accretion disc.
That proves that the accretion disc will never ever get any sort of matter from outside.
So, how can you compare it to Saturn???
Therefore, all the matter in the accretion disc had been created in the accretion disc by definition!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/04/2019 05:59:17
Therefore, all the matter in the accretion disc had been created in the accretion disc by definition!!!

Created from what?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/04/2019 11:28:28
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Therefore, all the matter in the accretion disc had been created in the accretion disc by definition!!!
Created from what?
Created from the SMBH' Ultra high gravity force.

Actually, Atom is some sort of energy cell by definition:
If we look at a proton for example -
https://en.wikipedia.org/wiki/Proton
"A proton has a mass of approximately 938 MeV/c2, of which the rest mass of its three valence quarks contributes only about 9.4 MeV/c2; much of the remainder can be attributed to the gluons".
So, the gluons contribute about 99% to the Proton Mass.
However, the galun itself doesn't carry mass. "It is just a type of energy"
https://en.wikipedia.org/wiki/Gluon
"A gluon is an elementary particle that acts as the exchange particle (or gauge boson) for the strong force between quarks"
"Gluons are actually just bosons, since they are the equilibrium force between the two quarks, which together form a triumvirate, and thus the energy force of the boson is in the form of a gluon, and thus the quarks become stable. They cannot separate unless something greater is capable of separating the quarks from each other, and so the gluon appears to hold these forces together. In fact it is just a type of energy while the two smaller forces, the quarks (also forms of energy) can unite under a single force, and this is the gluon's job."
So, the Galun is a type of Energy while the Quarks are also forms of energy.
Therefore, the mass in the proton represents Energy.
So, the Ultra high Energy/Temp/Pressure/Velocity in the accretion disc around the SMBH, create new Protons and Neutrons?
Those new protons/Neutrons will be the basic elements for all atoms in our periodic table (novel atoms) and all the molecular.
With regards to velocities:
Please see the following message from evan_au
Plasma at the inner edge of the black hole has an orbital velocity which is a significant fraction of the speed of light (like 30% of c).
The polar jets are emitted at up to 80% of c...
So, how do we know if the SMBH ejects the matter/plasma in its accretion disc or swallows it?
Our scientists estimates that 99% of the matter ejects out while the SMBH swallow only 1%.
https://edition.cnn.com/2013/08/30/tech/innovation/black-hole-diet/index.html
"Less than 1% of matter will be actually sacrificed for the freedom of 99% of gas," Wang said. "So, 99% of gas can escape from the capture of the black hole."
"According to Wang and colleagues, the black hole needs to throw out more than 99% of the material in order to accomplish this. That ejected 99%, in turn, heats up the environment around it, which affects the evolution of the galaxy as a whole."
Even so, do they have any real evidence for swallowing even that 1%?
The answer is NO!!!
So, by definition all the hot matter/plasma which had been created at the accretion disc is ejected outwards.

I wonder why do we choose to believe in something that we don't see instead of believe in what we really see?
If we see that 99% goes out, and we can't see even 1% goes in, than why don't we understand that 100% goes out?
If we assume that the SMBH eats 1%, how do we know that this 1% comes from outside?
We claim that the SMBH is a bad eater.
"Why our galaxy's black hole is a picky eater"
You might think of black holes as indiscriminate eaters, hungrily gobbling up everything in their vicinity.
But the black hole at the center of our Milky Way galaxy, Sagittarius A*, is not exactly like this, new research suggests. Instead, this black hole -- and likely other black holes in the centers of galaxies -- must spit out a lot in order to swallow a little."
So, let's look at our self.
If we want to eat - we eat. Actually, 99.99..9% of our life time our mouth is empty. - and we are very good eater!
However, 99.99....9% (or 100%) of the SMBH at the center of spiral galaxy has a mouth (accretion disc) full with food.
Why is it?
The total matter in the accretion disc (of the Milky Way) is estimated to be around three Sun mass.
At any similar spiral galaxy that we look we see more or less the same amount of mass. (at any given moment)
None of them has an empty mouth or significantly more mass than that.
How could it be?
The accretion disc is the biggest accelerator in the Universe.
If our scientists at CERN have created the Higgs Boson, why the biggest acceleration in the universe (the accretion disc of the SMBH) can't create new matter?

Few year ago (2014), our scientists thought that the source for the gas cloud is a massive disc stars:
https://www.mpg.de/8777573/gas-cloud-galactic-centre
"A likely source for both G1 and G2 could then be clumps in the wind of one of the massive disk stars, which could have been ejected some 100 years ago close to the apocentre of the G2 orbit. Another possible explanation that has been suggested recently would be a large star, enveloped by an extended gas cloud. Based on the current VLT data, however, this model is highly unlikely.
Astronomers are nevertheless puzzled why they have not yet detected increased Xray emission from the gas cloud near the black hole."
I hope that by now they already know that the new matter is coming from the excretion disc of the galaxy.
Therefore, our scientists shouldn't be so "puzzled".
The black hole has no intention to eat any new matter which it just drifts outwards from its excretion disc.
So, do you agree that the mighty magnetic shield around the excretion disc protects the core from any accretion activity (from outside)?

New star production.
https://en.wikipedia.org/wiki/Baby_Boom_Galaxy
"The Baby Boom Galaxy has been nicknamed "the extreme stellar machine" because it is seen producing stars at a rate of up to 4,000 per year (one star every 2.2 hours). The Milky Way galaxy in which Earth resides turns out an average of just 10 stars per year.[4]"
So, how could it be that our SMBH eats one sun mass per year, produces 10 stars per year, ejects that kind of high-energy particles stream, and - we still see that the galaxy center is still full with mass?
The ultimate answer must be - New mass creation at the accretion disc!!!

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/04/2019 14:28:52
How can you believe in image???
You asked for one.  I didn't state my level of belief in the image.  Do you think it was just made up like every image you've linked which are all artist's impressions?
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How can you compare the SMBH to Saturn???
Saturn's rings is a lovely example of what happens to matter that gets sufficiently close to a gravity source.  The rings did not accrete from bits of material from outside, it accreted from a single object that came from outside and was torn apart by the same forces that would keep a star from forming close to a large gravity well.
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So, let me ask you again for the last time:
Do we see any sort of mass/matter that comes from outside into the accretion disc of the SMBH?
The image they took shows exactly that, so yes, we see it.  It isn't just an artist's conception. It isn't a nice high resolution image like the artist ones, but it is real.
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How can you ignore the great impact of the ultra power of the magnetic field around the accretion disc?
Nobody is ignoring that.  The disk generates most of that field after all.  The jets from the poles become quite visible if by chance they're pointed straight at Earth, and some of them are on a periodic basis.
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Sorry. It seems that your knowledge in electronics and magnetic field is very poor.
It kind of is, yes.  I had to look all of that up.  I'm quite good at mathematics but hold no degree in it.  I'm a computer guy in the end.
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Magnetics can't start just from the poles. It covers the whole body!!!
That it does.  Didn't say otherwise.
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It is similar to the magnetic field around the Earth.
Any solar wind is lifted to the poles.
Where did you find this?  I had to look up the parts I didn't know.  I've never heard this one.  The solar wind is definitely affected by Earth's magnetic fields, but it doesn't accrete, so it isn't a great example of what goes on in an accretion disk.  The rings of Saturn formed from the tearing apart by gravity of a single object that wandered too close.  It isn't a great example of magnetic effects since Saturn's field completely overpowers any magnetic field generated by the spinning rings.

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Therefore, the magnetic field covers the SMBH & the accretion disc. Any molecular/matter that gets closer to that field (from outside the disc, or from the disc) is lifted upwards or downwards (according to the magnetic waves around the SMBH) directly to the Poles.
If that were true, there would be no accretion disk since your description has everything avoiding or leaving it.
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However, due to the Ultra power of that magnetic field the matter is then boosted from the poles upwards/downwards at ultra high speed.
From where comes this energy?  I agree it happens, but if energy is transferred to some matter to eject it at high speed, that energy must be transferred from something else.  What is that, and what effect occurs due to the loss of that energy?

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So there is no chance that any particle/molecular/Moon/Planet/Star will be able to cross that magnetic field in his way to the accretion disc.
That proves that the accretion disc will never ever get any sort of matter from outside.
That asserts that the accretion disc will never ever get any sort of matter from outside.  Your knowledge of magnetics is worse than mine since I at least look up the parts I don't know.

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So, how can you compare it to Saturn???
From a magnetic standpoint, I didn't.  It was simply an illustration of why stars cannot form close to a large mass.
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Therefore, all the matter in the accretion disc had been created in the accretion disc by definition!!!
The definition of accrete is to bring it together from outside under the influence of gravity, so your statement is completely wrong.  If material is created in the disk itself, it would seem to be an entirely new physics where mass just grows from nothing.  Kryptid I see also cannot bear such nonsense in silence.  Good to see there are a few actually reading this thread.

Created from the SMBH' Ultra high gravity force.
Force does not create things.  It accelerates things.

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Actually, Atom is some sort of energy cell by definition:
What definition is this?  Where are you getting all these definitions?  The battery in my phone is some sort of energy cell by definition.  Yes, mass and energy are interchangable, so any mass can be expressed as a quantity of energy.  But that's not the definition of an atom.

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Therefore, the mass in the proton represents Energy.
OK, I actually agree with this.
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So, the Ultra high Energy/Temp/Pressure/Velocity in the accretion disc around the SMBH, create new Protons and Neutrons?
That doesn't follow at all.  The energy/mass is already there.  You are proposing new energy coming from somewhere.  Nothing comes from the SMBH.  That's why it's black.  If something comes from it, then its mass/energy goes down correspondingly until it is gone, and you're describing something other than a black thing.


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With regards to velocities:
Please see the following message from evan_au
Quote from: evan_au
Plasma at the inner edge of the black hole has an orbital velocity which is a significant fraction of the speed of light (like 30% of c).
...
The polar jets are emitted at up to 80% of c...
So, how do we know if the SMBH ejects the matter/plasma in its accretion disc or swallows it?
The matter that gets to the poles might (does?) get ejected.  That takes energy away from the remaining disk that powered it, which means that it will drop down and eventually fall permanently into the gravity well (be it a black hole or pulsar or whatever).  If something shoots away, something else must drop down.  New energy cannot be created.

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Our scientists estimates that 99% of the matter ejects out while the SMBH swallow only 1%.
https://edition.cnn.com/2013/08/30/tech/innovation/black-hole-diet/index.html
OK, That's actually an interesting link.  They say that the gas is too hot in our galaxy so it doesn't form a disk at all.  A higher percentage of cold disk material is swallowed, but hot gas is proposed to have a 99% escape rate.  The 1% that falls in provides the energy to heat the gas even further.  Something like that.  Not sure how much this is accepted or just a new hypothesis.  The article calls the figure 'a suggestion'.

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Even so, do they have any real evidence for swallowing even that 1%?
The answer is NO!!!
The energy needs to come from somewhere, so the answer is quite yes.
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So, by definition all the hot matter/plasma which had been created at the accretion disc is ejected outwards.
Here we go with the 'by definition' thing again.  "I found an article that says most stuff is ejected, so by definition, all of it is."  Logic is not your forte either, as I've noted before.

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I wonder why do we choose to believe in something that we don't see instead of believe in what we really see?
If we see that 99% goes out, and we can't see even 1% goes in, than why don't we understand that 100% goes out?
They're basing it on what they see.  They see the part that goes in.  That's what lights the thing up.  They're wondering why our SMBH consumes so much less matter than comparable galaxies.

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If we assume that the SMBH eats 1%, how do we know that this 1% comes from outside?
That's sort of the definition of 'eats'.  If I eat a pizza, the pizza presumably comes from outside.  I don't eat my heart because it is already inside.

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We claim that the SMBH is a bad eater.
So is the sun.  Here's all these nice meaty planets and asteroids and stuff and it eats none of them.  Gravity is pretty great at moving things in circles but no so good at actually consuming them.  The sun is actually getting lighter since it ejects material faster than it sucks new stuff in.  The ejection is powered by fusion. If it stops burning, it will start gaining weight again.

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If we want to eat - we eat. Actually, 99.99..9% of our life time our mouth is empty. - and we are very good eater!
Do you go out of your way to make up nonsense data?  You claim that a person has food in mouth less than a 10th of a second per day.  Seems suspect.  The black hole always has food in its mouth, but it doesn't swallow most of it.  I tend to swallow the majority of the food that makes it as far as my mouth, but that's just me.

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However, 99.99....9% (or 100%) of the SMBH at the center of spiral galaxy has a mouth (accretion disc) full with food.
Why is it?
The total matter in the accretion disc (of the Milky Way) is estimated to be around three Sun mass.
The article you quoted also says that most of that disk will be swallowed, but most of the matter falling in is hot and not in that cold disk.  It is the cold disk material that is slowly eaten.  The mass estimate might not be agreed upon by the guys that wrote that article.  It is working under different assumptions.

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At any similar spiral galaxy that we look we see more or less the same amount of mass. (at any given moment)
This is false.  The other ones usually have much more mass.  Anromeda is a good example.  Same size galaxy (within 10%), but a far more active/massive black hole.
[/quote]None of them has an empty mouth or significantly more mass than that.[/quote]You're making up fiction.
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The accretion disc is the biggest accelerator in the Universe.
More fiction.
They have something called a blazar that fires plasma blobs with the mass of Jupiter at .999c.  The energy needed to do that boggles the mind.
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If our scientists at CERN have created the Higgs Boson, why the biggest acceleration in the universe (the accretion disc of the SMBH) can't create new matter?
They consumed the power of several cities to create that boson.  They've been created matter in accelerators for a long time, but they do it by consuming energy.  The two are the same thing, remember?  You cannot create new mass/energy, and yet here you are asserting exactly that.


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So, do you agree that the mighty magnetic shield around the excretion disc protects the core from any accretion activity (from outside)?
I know of no excretion disk, so why would I agree to any property it?  Such properties cannot be predicted until you have new physics that allow the existence of an excretion disk.

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So, how could it be that our SMBH eats one sun mass per year,
It doesn't eat one sun mass per year.  It is only 4 million masses, yet 12 billion years old.  That's no more than one sun mass per 3000 years, and the actual figure is much lower now since it was higher in the past.
You're losing your grip on 4th grade mathematics now.  Is your need to make up random numbers greater than your aversion to doing a simple division of two numbers?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/04/2019 17:07:01
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How can you compare the SMBH to Saturn???
Saturn's rings is a lovely example of what happens to matter that gets sufficiently close to a gravity source.  The rings did not accrete from bits of material from outside, it accreted from a single object that came from outside and was torn apart by the same forces that would keep a star from forming close to a large gravity well.
How do you that this info is real?
Had we been hear when it happened?
Do we have any record on that?
Sorry, this is absolutely imagination!
The real story is as follow:
Saturn had been formed from the same matter as the Sun and all the other Planets and moons and at the same day (with all the solar system).
I assume that it had one moon (or more).
Unfortunately, an external object had been collided with this moon (or moons) and break them to those small objects that form the ring.
We have one more evidence for a similar phenomenon - Asteroid belt
I assume that in the early days of the solar system, Cerest was just one big planet in the system.
Unfortunately, it was destroyed by an external object (It could be a comet or any big enough object) and therefore it had been converted to astride belt.
Hence, Saturn didn't accrete any object from outside (As there is no way to accrete objects from outside!). It just lost his moon. (Poor Saturn' moon...).
The definition of accrete is to bring it together from outside under the influence of gravity, so your statement is completely wrong.
The definition of accrete is very clear.
However, the assumption that gravity can accrete/bring objects from outside is just nonsense!!! (unless the object orbits too close to the planet/moon and effected by other factor -  atmosphere for example)

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If our scientists at CERN have created the Higgs Boson, why the biggest acceleration in the universe (the accretion disc of the SMBH) can't create new matter?
They consumed the power of several cities to create that boson.  They've been created matter in accelerators for a long time, but they do it by consuming energy.  The two are the same thing, remember?  You cannot create new mass/energy, and yet here you are asserting exactly that.
You are 100% correct!!!
In CERN we have used electric power to set the acceleration at almost the speed of light.
In the SMBH we get it by the ultra high gravity force!

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Therefore, the mass in the proton represents Energy.
OK, I actually agree with this.
Thanks
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So, the Ultra high Energy/Temp/Pressure/Velocity in the accretion disc around the SMBH, create new Protons and Neutrons?
That doesn't follow at all.  The energy/mass is already there.  You are proposing new energy coming from somewhere.  Nothing comes from the SMBH.  That's why it's black.  If something comes from it, then its mass/energy goes down correspondingly until it is gone, and you're describing something other than a black thing.
Yes it is.
Nothing comes from the SMBH.
However, the ultra high gravity at the accretion disc + the ultra high velocity (0.3 c) + the magnetic/electric field
All of that set excellent conditions for the mighty accelerator in the galaxy which has the ability to create quarks at the first stage.
As those quarks drift outwards in the accretion disc, they gain the requested energy that is needed to form new Atoms and molecular

.
 
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Magnetics can't start just from the poles. It covers the whole body!!!
That it does.  Didn't say otherwise.
Thanks
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So, let me ask you again for the last time:
Do we see any sort of mass/matter that comes from outside into the accretion disc of the SMBH?
The image they took shows exactly that, so yes, we see it.  It isn't just an artist's conception. It isn't a nice high resolution image like the artist ones, but it is real
The image is nonsense. Our scientists have never ever found any evidence for that.
It is just science fiction as nothing can accretes from outwards to the accretion disc.
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It is similar to the magnetic field around the Earth.
Any solar wind is lifted to the poles.
Where did you find this?  I had to look up the parts I didn't know.  I've never heard this one.  The solar wind is definitely affected by Earth's magnetic fields, but it doesn't accrete, so it isn't a great example of what goes on in an accretion disk.
Yes - "The solar wind is definitely affected by Earth's magnetic fields, but it doesn't accrete" - therefore it is an excellent example!!!
However, the magnetic power is quite low. Therefore it only affects the solar wind.
In any case, so far our scientists don't have even one real evidence for that!!!
The article you quoted also says that most of that disk will be swallowed, but most of the matter falling in is hot and not in that cold disk.
Unfortunately, our scientists don't have a clue how the SMBH really works.
Therefore, their hypothetical ideas is just none realistic. They have to offer real evidences - not just some science fiction stories.
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At any similar spiral galaxy that we look we see more or less the same amount of mass. (at any given moment)
This is false.  The other ones usually have much more mass.  Andromeda is a good example.  Same size galaxy (within 10%), but a far more active/massive black hole
That is incorrect
https://www.quora.com/How-big-is-Andromeda-compared-to-the-Milky-Way
"The Andromeda galaxy (also known as Messier 31, M31, or NGC 224) could be considered the big brother of the Milky Way, as it contains over a trillion stars (compared to our 200-400 billion), and is approximately 220,000 light years across to our 100,000."
As it is much bigger, its SMBH is relatively more active.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/04/2019 17:11:46
Created from the SMBH' Ultra high gravity force.

It sounds like you are talking about Hawking radiation. For a black hole of such size, its radiation is far too weak to produce an accretion disk (the radiation would be in the form of radio wave photons anyway, not atoms).
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/04/2019 21:35:08
He says that all material in the whole galaxy has sprung from the central object, making it a white hole of sorts.

If that's his argument, then he is just plain wrong. Matter cannot get out of the event horizon of a black hole (and it isn't a while hole, as the recent photograph of M87* clearly shows).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/04/2019 23:20:46
How do you that this info is real?
Had we been hear when it happened?
Do we have any record on that?
Yes actually.  The breaking up of nearby objects has been witnessed on a number of occasions.  It is happening to Phobos right now (slowly, but the cracks are growing).  Comet Shoemaker–Levy 9 is a more immediately apparent example as it was torn into several large pieces as it approached Jupiter.

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The real story is as follow:
Saturn had been formed from the same matter as the Sun and all the other Planets and moons and at the same day (with all the solar system).
I assume that it had one moon (or more).
It has 53 named moons as of some article I found.  If they were all there in day one, then you assume an awfully low number.
@Kryptid :  See what I mean?


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Quote from: Halc
The definition of accrete is to bring it together from outside under the influence of gravity, so your statement is completely wrong.
The definition of accrete is very clear.
I think I just said that, yes.  But you are defining it otherwise, which is why your comment is wrong.

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In CERN we have used electric power to set the acceleration at almost the speed of light.
In the SMBH we get it by the ultra high gravity force!
Gravity accelerates matter inward (down), not outward (up).  Your ejected material is going the wrong way to be accelerated by gravity.  Gravity resists/slows such motion, but accelerates things falling inward, which is the source of the high temperature.

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Nothing comes from the SMBH.
You say otherwise in a lot of you posts, but I agree that nothing (short of negligible Hawking radiation mentioned by Kryptid) is emitted by a black hole.

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However, the ultra high gravity at the accretion disc + the ultra high velocity (0.3 c) + the magnetic/electric field
All of that set excellent conditions for the mighty accelerator in the galaxy which has the ability to create quarks at the first stage.
That likely actually happens, but creation of matter like that is going to remove the energy used to create it.  The material of the disk slows down when its energy is used up like that, and slower material will fall closer to the black hole.  Material will fall in if some of the energy is being consumed by matter creation.  This results in a net loss of matter outside the black hole.

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As those quarks drift outwards in the accretion disc, they gain the requested energy that is needed to form new Atoms and molecular
If they drift out, they gain gravitational potential energy and lose kinetic energy.  If electromagnetic force propels them out against that resistance (and it does), then that energy is again lost by the disk powering that thrust, and again, a disk that has lost energy will tend to drop some of its mass into the black hole.
At no point is there a net gain of energy outside the black hole.  Some leaks out, some is irretrievably lost to the gravity well, and the balance is unusable entropy.

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The image is nonsense. Our scientists have never ever found any evidence for that.
What, you think they just made it up?  Somebody created it with photoshop or something? That image IS the evidence for that.

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However, the magnetic power [of Earth] is quite low. Therefore it only affects the solar wind.
In any case, so far our scientists don't have even one real evidence for that!!!
Quite the opposite.  They measure it directly.  It has painted quite a good picture of what actually goes on with the solar wind.

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Andromeda is a good example.  Same size galaxy (within 10%), but a far more active/massive black hole
That is incorrect
It seems I am wrong about that one.  Using a poor reference it seems.  The consensus is a larger radius galaxy with considerably more stars (which is not directly a measure of mass).  It still has a more active central black hole, as do the vast majority of galaxies around our size.  My point is that ours is unusually inactive.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/04/2019 14:28:06
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However, the ultra high gravity at the accretion disc + the ultra high velocity (0.3 c) + the magnetic/electric field
All of that set excellent conditions for the mighty accelerator in the galaxy which has the ability to create quarks at the first stage.
That likely actually happens, but creation of matter like that is going to remove the energy used to create it.  The material of the disk slows down when its energy is used up like that, and slower material will fall closer to the black hole.  Material will fall in if some of the energy is being consumed by matter creation.  This results in a net loss of matter outside the black hole.
If they drift out, they gain gravitational potential energy and lose kinetic energy.  If electromagnetic force propels them out against that resistance (and it does), then that energy is again lost by the disk powering that thrust, and again, a disk that has lost energy will tend to drop some of its mass into the black hole.
At no point is there a net gain of energy outside the black hole.  Some leaks out, some is irretrievably lost to the gravity well, and the balance is unusable entropy.
Thanks
Actually, I fully agree with you.
It seems to me that some of the new born quarks should be "eaten" by the SMBH, while the others are drifting outwards.
Those quarks that falls in, will increase the mass of the SMBH, while the others should be converted into new atoms and molecular.
So, the SMBH increases constantly its mass while it generates new mass that excretes from the accretion/excretion disc.
Hence, over time, the SMBH of the Milky Way might be as big and massive as the SMBH in Andromeda galaxy. At that time, its new mass production should be similar to the current production of the Andromeda' SMBH.
More massive SMBH + Higher new star productions mean bigger spiral galaxy.
Therefore, it is quite clear that sometime in the future, our Milky Way galaxy should be as big as the current Andromeda galaxy.

Therefore, once a BH has the ability to create new mass, it increases it's mass by definition (while it also ejects new mass). Over time any compact BH could be converted into a SMBH which is hosting the core of new spiral galaxy.

That actually the highlight of our discussion!
There is no need to create the whole mass of the Universe by one Big Bang.
We only need to have one compact BH in the whole Universe which has the ability to generate new mass based on its sufficient gravity force.
So, with regards to the matter creation in the Universe:
Only one BH is needed to be created by the Big Bang. That single BH has the ability to generate the whole matter in our magnificent universe that we see today (and also the other section that we don't see).

It seems to me that there is high similarity between living creatures and galaxies.
Darwin had claimed that only one living cell was needed to get all the variety of living creatures/animals on Earth.
In the same token, only one BH was needed to set the whole Universe.
However, that first BH must have the ability to generate new mass.
In the process of creation new matter, it increases its mass and sets the first spiral galaxy in the Universe.
This spiral galaxy generates new stars which had been ejected from that first galaxy.
Over time, some of them should be transformed into BH.
Each new BH sets a new baby spiral galaxy and so on...
Therefore, if we give that process the requested time (Infinity/almost infinity) we should get the universe that we see today.
Hence, the BH is the driving power of our Universe!
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/04/2019 15:35:46
The material actually formed from the SMBH (via quantum effects) decreases the mass of the SMBH.  If it falls back in (as most of it does), there is no net change of mass.  The stuff created by the disk is more than balanced by material/energy of the disk falling into the black hole, for a net loss of material/energy to the black hole.  At no point can there be a net gain outside if the black hole is gaining mass.

This is incorrect!!!
The SMBH has the ability to generate new mass based on its pure gravity force.
So, it doesn't consume any matter from its body to create that new matter.
On the contrary, the creation of new matter increases its mass over time
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 19/04/2019 17:21:05
On the contrary, the creation of new matter increases its mass over time

That violates the first law of thermodynamics. Gravity can't create mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/04/2019 05:56:37
That violates the first law of thermodynamics. Gravity can't create mass.

Gravity by itself can't create mass - I agree with that.
However, with regards to BH (or SMBH) due to their Ultra high gravity force and the added unique features, we get the ultimate conditions for mass creation in the accretion disc:

1. Ultra high velocity of the plasma - We have measured 0.3 speed of light. However, I'm quite sure that at the inner most of the disc the velocity should higher (Almost the speed of light).
2. Very strong magnetic/electric field - This magnetic field is so strong enough to boost any particle that gets to its poles (upwards or downwards) at the speed of 0.8 speed of light.
3. Plasma - The plasma temp at the accretion disc is 10^9 c. This temp is high above any fusion activity. Please remember that the internal temp of the Sun is only 10^6. Therefore, this kind of ultra high temp is a clear indication for the creation process at that disc.
4. We see clearly that hot matter is ejected from the accretion disc. I have already introduced several articles which confirm that excretion activity. In one of the article it was stated that 99% of the matter in the accretion disc is ejected outwards.
5. Our scientists want to believe that the matter in the accretion disc is coming from outside the disc. However, so far they couldn't give even one real evidence for that. I have offered several articles about it. In one of the article it was stated that our scientists were very "puzzled" as the SMBH didn't eat the nearby star although it was so close to its location.
6. Magnetic shield - The magnetic field set a very strong magnetic shield around the accretion disc. On Earth we also have a Magnetic shield. That magnetic shield is very low, but even so it is strong enough to shift all the solar winds from the sun to the poles. The magnetic shield around the SMBH should be stronger by several millions (or more) than our local shield. So, if our local (low power) magnetic shield can shift the solar wind to the poles, than the ultra power magnetic field around the SMBH should have the power to break any object to it's basic Atoms and shift it to its poles. From that point we have already know that the matter is boosted upwards/downwards at confirmed speed of 0.8c.
Conclusions - Based on all the above there is no way for the accretion disc to get its matter/Plasma from outside. There is also no justification for the SMBH to orbits its food around it mouth at this ultra high velocity just to eject 99% and eat less than 1% at the end of this activity. Actually, if we could monitor the plasma structure, we should find different matter at different radius. At the inner most side, I'm quite sure that we won't find any atom or molecular. At that radius we should only find basic particles as quarks. Their temp & orbital velocity must be at the higher most level in the accretion disc. All of those quarks are new born particles. Some of those new born quarks falls into the SMBH and increase its mass. Some other are drifting outwards, gets the requested energy and transformed into real Hydrogen Atoms. However, due to the Ultra high orbital velocity - we also should get there all the variety of atoms & molecular that we know about. Therefore, the Jet that is ejected from the Poles is called by our scientists - Molecular Jets. That proves that the particles which are ejected from the outer most radius of the accretion disc are already fully mature & mixed Molecular. As those molecular get to the magnetic shield, they lifted to the poles and set the magnificent Jet stream that our scientists see so clearly! They can't fall back to the accretion disc due to the magnetic shield that keeps them all away from the disc. Those new molecular should fall back to the galactic disc, gather into gas clouds and set new stars. Therefore, most of the new born stars activity in our galaxy takes place near the SMBH. In any case, any new born star/planet/moon gets all its Atoms/molecular in his first day.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/04/2019 06:14:15
mass creation in the accretion disc

Since you agree that gravity can't create mass, then where does the accretion disk come from in the first place? You claim that it doesn't come from outside of the black hole and we know that it can't come from inside of the black hole either. It can't spontaneously spring into existence (that would violate the first law of thermodynamics). So where does it come from?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/04/2019 08:26:46
Since you agree that gravity can't create mass, then where does the accretion disk come from in the first place?
Thanks, that is good question.
If you accept the idea that a SMBH can create new matter, and we only need to deal with the BH first step, then our current situation is very good.
So, I agree that there must be a moment that a compact BH gets its ability to set the accretion disc and start its new matter creation activity.
Once it starts its activity, it will be converted in the future into a SMBH that is hosting a spiral galaxy.
However, the first step is critical. Any first step is critical.
I'm not sure that every BH must have the ability to start this activity, but I assume that most of them will be successful enough and set their first production.
I assume that they need to meet several criteria in order to overcome the first step.
The first matter in the accretion disc could come from a real matter that it accretes constantly in order to increase its mass.
However, even if it sets in the disc real matter, it doesn't mean that it has already the ability to create new matter at the accretion disc. Without enough magnetic field it can't start this activity.
Therefore, I assume that at some point of time (after "eating" enough real matter) that BH should have enough mass to set ultra high orbital velocity of real matter at the accretion disc. That orbital velocity generates the requested magnetic/electric field which is vital for the new matter creation activity at the accretion disc.
In the same token we could ask:
How a star starts it first fusion activity? Are we sure that any new born star has the ability to set this activity?
How the first BH in the Universe had been created? How the first big bang had started?
It is clear that any first step is critical. However, as any child starts at some point of time its first step, most of the BH should start their first step in setting the new mass creation at accretion disc.




Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/04/2019 15:37:49
most of the BH should start their first step in setting the new mass creation at accretion disc.

Technically-speaking mass can't be created. Matter can be created, but not mass. If what you are talking about is energy being converted into mass (which is somewhat redundant, since energy already has relativistic mass associated with it), then where does the energy come from that is required to create the new mass? You can't say it comes from the gravitational force, because force is not energy. They aren't even measured in the same units.

A black hole plus its accretion disk have a finite amount of energy. Some of that energy is in the form of potential energy and some in the form of kinetic energy. Those two forms of energy can be inter-converted, but the total amount must stay the same. So the total energy content (and therefore the total mass) of the black hole/accretion disk system cannot increase over time unless it gets energy or matter from some outside source. To say otherwise would violate the first law of thermodynamics.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/04/2019 16:09:09
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Gravity by itself can't create mass - I agree with that.
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The SMBH has the ability to generate new mass based on its pure gravity force.
These two statements are mutually contradictory. Force is not energy, and energy/mass (same thing) cannot be created or destroyed. It violates thermodynamic law as Krypid points out:
Quote from: Kryptid
It can't spontaneously spring into existence (that would violate the first law of thermodynamics).
Quote from: Kryptid
Since you agree that gravity can't create mass, then where does the accretion disk come from in the first place?
Thanks, that is good question.
If you accept the idea that a SMBH can create new matter,
Nothing can create new mass.  That's what the first thermodynamic law says.  What they do in particle accelerators is transform existing energy into a different kind of energy, but energy/mass is conserved throughout the process.

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So, I agree that there must be a moment that a compact BH gets its ability to set the accretion disc and start its new matter creation activity.
This again violates that first law.

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Once it starts its activity, it will be converted in the future into a SMBH that is hosting a spiral galaxy.
And a third violation.  A compact black hole that excretes mass (via Hawking radiation say) would lose that mass until it is gone.  The only way to make it bigger is to dump mass into it from outside, which reduces the remaining mass/energy outside.

You are describing creationism. Post the idea in a religious forum.

The first matter in the accretion disc could come from a real matter that it accretes constantly in order to increase its mass.[/quote] 'Accretion' means to gather it from elsewhere, something you've denied.  If it is gathering it from elsewhere, the the mass of elsewhere is going down, not up.

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However, even if it sets in the disc real matter, it doesn't mean that it has already the ability to create new matter at the accretion disc. Without enough magnetic field it can't start this activity.
Magnetic fields, just like gravitational fields, are incapable of generating mass.

 
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Therefore, I assume that at some point of time (after "eating" enough real matter) that BH should have enough mass to set ultra high orbital velocity of real matter at the accretion disc.
A tiny black hole can do this.  A strong gravitational field implies high orbital speeds of nearby matter.  But no mass is created by this or any system.

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That orbital velocity generates the requested magnetic/electric field which is vital for the new matter creation activity at the accretion disc.
4th violation. 
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In the same token we could ask:
How a star starts it first fusion activity? Are we sure that any new born star has the ability to set this activity?
The activity is what makes it a star, not the other way around.  Fusion changes matter into thermal energy with a net loss since some of that energy radiates away.  Our sun is losing mass steadily via this process.

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How the first BH in the Universe had been created? How the first big bang had started?
It is clear that any first step is critical.
Different question.  Net sum of mass/energy in the universe seems to be zero, so it doesn't violate thermodynamic law at least.  But it doesn't answer that question either.

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However, as any child starts at some point of time its first step, most of the BH should start their first step in setting the new mass creation at accretion disc.
A black hole is formed by putting any amount of mass within a region defined by the Schwarzschild radius of that mass.  That formation event may or may not involve a situation with an accretion disc already present.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/04/2019 18:44:18
Technically-speaking mass can't be created. Matter can be created, but not mass. If what you are talking about is energy being converted into mass (which is somewhat redundant, since energy already has relativistic mass associated with it), then where does the energy come from that is required to create the new mass? You can't say it comes from the gravitational force, because force is not energy. They aren't even measured in the same units.
A black hole plus its accretion disk have a finite amount of energy. Some of that energy is in the form of potential energy and some in the form of kinetic energy.
Thanks Kryptid for your excellent explanation.
So you claim that:
1. "Matter can be created, but not mass".
2. Gravitational force is not energy. 
3. "A black hole plus its accretion disk have a finite amount of energy."
4. "Some of that energy is in the form of potential energy and some in the form of kinetic energy"
So, you actually prove that my statement about gravity force is fully correct:
A SMBH without accretion disc can't convert any gravity force into energy.
However, A SMBH with accretion disc converts its ultra high gravity force into energy in the accretion disc.
Therefore, I hope that you also agree that based on the energy in the accretion disc it is feasible to create new matter (not mass).
However, in order to set this process we must overcome the first law of thermodynamic.
As you have explained:
So the total energy content (and therefore the total mass) of the black hole/accretion disk system cannot increase over time unless it gets energy or matter from some outside source. To say otherwise would violate the first law of thermodynamics.
We know that the magnetic shield prevents from any matter from outside to get in.
Therefore, let's focus on the first law of thermodynamic and see how it works:
https://en.wikipedia.org/wiki/First_law_of_thermodynamics
"The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed."
I don't see any contradiction with regards to this law.
Actually, I have already proved that Atom is by definition an energy cell.
So, some of the energy in the accretion disc is transformed to the energy cell which we call Atom or new matter if you wish.
Therefore, as long as the total energy in the accretion disc is constant, there is no violation for that first law of thermodynamic.
Hence, it is clear that all the energy that was needed to set the new matter process must be deducted from the energy in the accretion disc.
As "Some of that energy (in the accretion disc) is in the form of potential energy and some in the form of kinetic energy", than some of this energy must go down during that process.
If I understand it correctly, the kinetic energy represents the orbital velocity. Therefore, during the creation process, this orbital velocity should go down. So, we can claim that the measured 0.3 c of the plasma orbital velocity at the accretion disc is direct outcome of that first law of thermodynamic. Without it, the plasma orbital velocity could be higher than that.
Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/04/2019 21:17:03
However, A SMBH with accretion disc converts its ultra high gravity force into energy in the accretion disc.

All it does is convert the gravitational potential energy already in the disc into kinetic energy as it falls through the black hole's gravitational field. The total energy is unchanged.

Therefore, as long as the total energy in the accretion disc is constant, there is no violation for that first law of thermodynamic.

If the black hole gains mass, then the accretion disk must lose mass to compensate. Actually, the black hole-accretion disk system as a whole must lose mass and energy over time since it is constantly radiating energy out into space. That is, if no more mass or energy is being brought in from outside.
Title: Re: How gravity works in spiral galaxy?
Post by: Lance Canham on 20/04/2019 22:01:06
In my black hole thread I conclude that a black is pure matter but maybe its Pure Mass existing outside space time. Matter as you deal with it Or Physics deals with is built of 1.s an 0.s 1 Being mass 0 being space/time.  A black hole is the god particle you all search for It gives mass to Matter as you Know it but the big bang repeats and a Black whole it the most basic of thing. All matter is built on it and space/time.

A galaxy is a More complex particle therefore to figure it out you need to see what dark matter truly is and realize that thing in the middle is more than you think and may have a force none of us are seeing yet.  The energy of the galaxy particle is in its mass - the very slight ration speed difference would give a different wavelength if it were the supper small in another Big bang - its a particle - figure out why it does what it does and you have the connection - the link to gravity and everything else.

Think of the galaxy as looking like it spinning at almost the speed of light to you - Relative to you looking from what above in A universe built on the stuff - Relative to you time is taking a second to it -billions of years

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/04/2019 22:07:43
All it does is convert the gravitational potential energy already in the disc into kinetic energy as it falls through the black hole's gravitational field. The total energy is unchanged
O.K.
You claim that the SMBH "eats" some of the matter from the accretion disc
I actually fully agree with that.
So, it converts some of the gravitational potential energy in the disc into kinetic energy.
Hence, there is extra energy that is needed to set the creation process.

If the black hole gains mass, then the accretion disk must lose mass to compensate.
We have just found that the SMBH converts some of the gravitational potential energy in the disc into kinetic energy.
Why this extra kinetic energy can't be used for the new matter creation at the disc?
Why do you also ignore the following?
As "Some of that energy (in the accretion disc) is in the form of potential energy and some in the form of kinetic energy", than some of this energy must go down during that process.
If I understand it correctly, the kinetic energy represents the orbital velocity. Therefore, during the creation process, this orbital velocity should go down. So, we can claim that the measured 0.3 c of the plasma orbital velocity at the accretion disc is direct outcome of that first law of thermodynamic. Without it, the plasma orbital velocity could be higher than that.
Actually, the black hole-accretion disk system as a whole must lose mass and energy over time since it is constantly radiating energy out into space. That is, if no more mass or energy is being brought in from outside.
That statement proves that there is a fatal error in your theory.
Our scientists couldn't find any indication that matter from outside can cross the magnetic shield around the accretion disc and move inwards. I claim that this functionality is none realistic!
So, if there is no matter from outside? How all the accretion discs around all the SMBH (In any spiral galaxy) are full with matter?

Actually, if we already discuss about the first law of thermodynamic:
I wonder how that law meets the Big bang Theory.
How could it be that there was energy before the Big Bang?
What was the scours of that energy?
Why & how that energy could be transformed into real mass (not just new matter)
How can we believe that the whole mass of the Universe could be created at just one "big bang" while I can't see any feasibility to set even a single Atom by any sort of bang?

There must be several key elements that are vital in the process of new matter creation.
All of those elements are located at the accretion disc that functions as a massive accelerator.
Our scientists at CERN have proved that by using accelerator it is feasible to create new matter.
Therefore, I can't understand how any scientist which in one hand considers about the feasibility of the first law in that ultra strong accelerator (accretion disc), can on the other hand accept the idea of mass creation by big bang???
If he worries about the feasibility of the first law at the accretion disc, why he doesn't worry about its feasibility at the Big bang?   
Title: Re: How gravity works in spiral galaxy?
Post by: Lance Canham on 21/04/2019 01:10:03
There is 2 universe in the same place. Matter is made simply of mass outside of space time.  After asking myself if Mass out side spacetime could be spacetime outside of mass on the other side  it works for me because a black hole is the fundamental particle of everything. A Black hole is 1 space is 0. earth the sun every atom and every part of every atom is made of mass outside of spacetime and spacetime there for if mass outside spacetime is spacetime outside mass both universes are connected at every point we find matter as the fundamental piece is this connection between the 2 is the god particle (black hole).

Space is not stretching its pouring out of the black hole and That is why space compress around matter.  we are cramming it in there its coming out they see it being crammed in we see the compression close and Apparent stretching at a distance.  So how much space does the AGN have to push out to match the effect of in fall to make it look the way it does.
 
Title: Re: How gravity works in spiral galaxy?
Post by: Lance Canham on 21/04/2019 02:15:21
You have to apply space pouring out of everything, Under my view dark matter is everything outside a singularity - in this case outside your view of regular matter as I have the view of multiple Big bangs at ever larger scales over infinite space and time. So space is coming out of literally everything every where not just the AGN. Likely more the dark matter in the earth and moon. the heart of the sun - Have to think that out one day Im thinking when a particle is more than a point in its description its broke down to the realm of gravity within it - Bing works like our universe inside super small - Basically dark matter stars -galaxies gas on a super small scale. So much room for pooling and space poring.

So it comes from everything dependent on amount of matter able to fall into a god particle - hence spew out space.

Im not sold on a 2 universe Idea totally just yet.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/04/2019 02:33:28
Why this extra kinetic energy can't be used for the new matter creation at the disc?

"Mass" and "matter" are non synonyms. Turning kinetic energy into matter won't make the mass of the system go up. It would either stay the same or decrease (due to radiation).

Before I continue, I need to know something: do you claim that the total mass of the black hole-accretion disk system increases over time or not?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/04/2019 05:39:22
Before I continue, I need to know something: do you claim that the total mass of the black hole-accretion disk system increases over time or not?

Yes!
However, before you continue, please let me know if you agree with the following:

1. Around the SMBH there is a Plasma orbiting at ultra high velocity - 0.3 speed of light and at temp of 10^9 c.
2.  This plasma set the accretion disc which generate Ultra high Magnetic field.
3. Our scientists claim that they have evidences that at least 99% from the plasma/hot matter in the accretion disc is ejected outwards.
4. The ultra power magnetic field acts as a magnetic shield. any nearby atom/object will be lifted to its poles. Therefore, any molecular that comes from inside the magnetic shield (from the accretion disc) will be lifted to the poles. In the same token, any Atom /molecular/object that dare to come from outside of the magnetic shield will also be lifted to the poles. Therefore, nothing from outside can cross the magnetic shield and get into the accretion disc.
5. There is a clear evidence that any molecular that get's to the poles of the magnetic shield is boosted upwards/downwards in a molecular jet at 0.8 speed of light.
6. So far our scientists could not find even one real evidence for any matter outside the magnetic shield that can cross it and get directly into the plasma in the accretion disc. If our scientists will verify the Ultra power of that magnetic shield they will find that this Hypothetical idea of accretion matter from the Bulge is just unrealistic.
6. Actuall the Bulge is full with matter. We clearly see there several gas cloud that are orbiting around the SMBH (outside the magnetic shield) while they are forming new stars.
7. There are 10 million stars within one parsec (about 3.26 Light years) of the Galactic Center. https://en.wikipedia.org/wiki/Galactic_Center
This is very high density of stars. Just to remind you that in 50 LY around the Sun there is just 64 stars). If the SMBH was "eating" matter from that aria it was expecting to see there significantly less star density. That proves that the SMBH has no intention to eat even one single atom from the nearby Galactic Center. All of those stars had been created by the new matter which had been ejected from the accretion disc!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/04/2019 05:58:48
Yes!

So I don't need to say anything else. Your idea violates conservation of mass. Therefore it cannot be correct.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/04/2019 06:58:15
So I don't need to say anything else. Your idea violates conservation of mass. Therefore it cannot be correct.
Ok
So, this is what you think.
I respect it, but this is a severe mistake.

Just the evidence that 10 million stars within one parsec (about 3.26 Light years) of the Galactic Center proves that the SMBH doesn't eat even one single Atom from outside.
However, our scientists don't let the evidences to confuse them!
Therefore, you also hold on the unrealistic idea of "eating mass from outside".
You claim that this accretion disc don't meet the first law of thermodynamics while you ignore the idea of energy transformation in that massive accelerator.
In the same token, you fully to accept the unrealistic idea that the Big Bang meets that first law of thermodynamics although it is clear to you that there is no way to set new matter by any sort of bang.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/04/2019 15:01:40
you ignore the idea of energy transformation in that massive accelerator.

No I don't. Energy and mass are equivalent. The total amount of energy and mass before and after a collision in a particle accelerator are the same. If two protons are accelerated up to 1 GeV and slammed into each other, the resulting spray of particles produced will also have a total of 1 GeV of mass/energy. Matter in a high state of potential energy (such as being at high altitude in a strong gravitational field) weighs more than it would in a low state. If that mass is allowed to fall, that potential energy is converted in kinetic energy. However, the mass remains the same because fast-moving matter weighs more than slow-moving matter.

Just the evidence that 10 million stars within one parsec (about 3.26 Light years) of the Galactic Center proves that the SMBH doesn't eat even one single Atom from outside.

That's a non-sequitur.

In the same token, you fully to accept the unrealistic idea that the Big Bang meets that first law of thermodynamics

The Big Bang did not create mass or energy. It represented an extreme expansion of space that mass and energy already existed in.

although it is clear to you that there is no way to set new matter by any sort of bang.

I already explained to you that "matter" and "mass" are not synonyms. Matter can be created. Mass cannot. Light is not matter but it does have relativistic mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/04/2019 15:34:33
The Big Bang did not create mass or energy. It represented an extreme expansion of space that mass and energy already existed in.
If "mass and energy already existed in" before the Big bang, than why do we need the Big bang theory?
How long before the Big Bang that mass and energy existed?
If the whole mass of the Universe was already existed before the Big bang why do we need the energy?
Why do we claim that the age of the Universe is only 13.8 Billion years if the mass of the whole universe was there before that time?
When did we get the first Atom in the Universe?
What is the real age of the Universe?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/04/2019 17:44:54
If "mass and energy already existed in" before the Big bang, than why do we need the Big bang theory?

It explains the relative abundance of the chemical elements in the Universe, the temperature of the microwave background and the expansion of the Universe among other things.

How long before the Big Bang that mass and energy existed?

Nobody knows. Some models posit that we live in a cyclic universe where an infinite number of Big Bangs and Big Crunches happen in sequence.

If the whole mass of the Universe was already existed before the Big bang why do we need the energy?

I'm not sure what this question means. The mass is already there with the energy.

Why do we claim that the age of the Universe is only 13.8 Billion years if the mass of the whole universe was there before that time?

The Big Bang happened 13.8 billion years ago and could be viewed as the starting point of our own Universe. That doesn't rule out the existence of previously-existing universes. We also don't even know if the concept of time is meaningful before the Big Bang.

When did we get the first Atom in the Universe?

About 379,000 years after the Big Bang. That was when the Universe was cool enough to allow electrons to bind to protons to form hydrogen.

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What is the real age of the Universe?

I suppose that depends on how you choose to define the Universe.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/04/2019 06:26:44
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How long before the Big Bang that mass and energy existed?
Nobody knows. Some models posit that we live in a cyclic universe where an infinite number of Big Bangs and Big Crunches happen in sequence.
We also don't even know if the concept of time is meaningful before the Big Bang.
How do you know that "Nobody knows".
Do you mean that "Nobody" from our scientists?
Let me tell you the following:
I'm leading several engineering design teams.
I can accept an answer that the result doesn't meet our expectation.
But I will never ever accept an answer as: "I don't know and "Nobody knows""
If an engineer will dare to give me that kind of answer, he will have to look for new job!
In any case, if there is a severe problem in the design, I normally tell them to start from point Zero or deliver the design to other team.
In engineering there is no room for: "Nobody knows"!!!
However, in science it seems that you live very happily with "Nobody knows"
Somebody must know the answer.
If Nobody from our scientists knows, than why don't they open their mind and listen to other ideas?
I personally don't think that our scientists are so foolish to believe that "Nobody knows".
They must know the answer, at least some of them should know if the BBT is real or unrealistic.
Just to remind you the source for that theory name:
https://www.universetoday.com/54756/what-is-the-big-bang-theory/
"Ironically, it was Hoyle who coined the phrase “Big Bang” during a BBC Radio broadcast in March 1949, which was believed by some to be a pejorative dismissal (which Hoyle denied).
So, that BBT theory which started about 70 years ago must meet all the current discoveries and evidences.
If it meets them all - than it is a valid theory.
If it doesn't, than you shouldn't hide behind the statement: "Nobody Knows"
Yes, some of you should know the answer!
If you claim that "Nobody Knows", than you know that there is a fatal error in your theory.
In order to avoid dealing with this fatal error you prefer to say "Nobody Knows".
So, if our scientists' mission is discovering the real theory of our universe - it is expected that they will immediately start looking for new idea.
I was expecting that you would love to hear my message as it might bring you some good ideas.
However, as I see that you reject any idea, now I understand that you have no willing to discover the real theory about our Universe.
It seems to me that your mission is protecting the BBT under any circumstances, even by presenting yourself so foolish by claiming: "Nobody Knows".
But Why? Why do you reject any new idea while you know that there are severe problems with the BBT and you really don't know all the answers?
What do you gain by protecting that unproved BBT theory? How force you to do so?
Don't you want to understand how our Universe really works?




Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/04/2019 08:17:53
How do you know that "Nobody knows".

I mean it literally. Nobody knows. People have different ideas, but nobody has proof one way or the other. Science isn't even about proof.


Do you mean that "Nobody" from our scientists?

I mean literally nobody.

Let me tell you the following:
I'm leading several engineering design teams.
I can accept an answer that the result doesn't meet our expectation.
But I will never ever accept an answer as: "I don't know and "Nobody knows""
If an engineer will dare to give me that kind of answer, he will have to look for new job!

Too bad, because "I don't know" is the current answer to an awful lot of questions humanity has right now. We don't know what the ultimate unified field theory is. We don't know whether space and time are discrete or continuous. We don't know if humanity will ever achieve interstellar flight. We don't know whether or not there are life forms in Europa's oceans. The list goes on and on.

In engineering there is no room for: "Nobody knows"!!!
However, in science it seems that you live very happily with "Nobody knows"

It isn't a matter of "living happily" with "nobody knows". It's a matter of accepting reality. We can and do try to learn more about nature, but don't know everything. Humanity is not omniscient.

Somebody must know the answer.

Not necessarily.

If Nobody from our scientists knows, than why don't they open their mind and listen to other ideas?

(1) The ideas have to actually be falsifiable.
(2) There has to be a good reason to believe those ideas are correct or at least on the right track to being correct.

I personally don't think that our scientists are so foolish to believe that "Nobody knows".

If there is anyone who knows, they have yet to demonstrate it with good, solid evidence.

If it doesn't, than you shouldn't hide behind the statement: "Nobody Knows"

The Big Bang theory does fit the existing evidence. Nobody is hiding behind anything. The Big Bang theory isn't supposed to explain where mass and energy came from. It only describes the current properties of the Universe based on properties predicted about its distant past.

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If you claim that "Nobody Knows", than you know that there is a fatal error in your theory.
In order to avoid dealing with this fatal error you prefer to say "Nobody Knows".

Except for the fact that the Big Bang theory was never meant to explain where mass and energy came from or whether time itself had a beginning or not. So "I don't know" is not a fatal flaw here.

I was expecting that you would love to hear my message as it might bring you some good ideas.

What message? We were talking about black holes, accretion disks and conservation of mass/energy then you went off on a tangent about the Big Bang.

However, as I see that you reject any idea, now I understand that you have no willing to discover the real theory about our Universe.

I don't "reject any idea", I reject ideas that clash with known science. Violation of conservation of mass/energy clashes with science.

It seems to me that your mission is protecting the BBT under any circumstances

I don't know how you got that impression, especially not from a couple of posts on an Internet forum. If scientists discover something that falsifies the Big Bang theory, then I would have to discard it.

even by presenting yourself so foolish by claiming: "Nobody Knows".

It isn't a claim. It's a fact. No one knows whether this is the only Universe in existence or whether it is cyclical. Anyone who claims otherwise has not yet provided compelling, verifiable evidence to support their stance.

But Why? Why do you reject any new idea

I don't. You are putting words in my mouth.

while you know that there are severe problems with the BBT

Name some.

and you really don't know all the answers?

It's called "being human". Nobody knows "all the answers". If we did, then there would be no point in doing science.

What do you gain by protecting that unproved BBT theory?

Science isn't about proof, it's about evidence. What I gain is some satisfaction in educating others (or at least attempting to).

How force you to do so?

What makes you think that anyone has forced me to do anything? You are starting to sound like a conspiracy theorist (and we have more than enough of those on this board).

Don't you want to understand how our Universe really works?

Yes, and a good place to start is by understanding what we have already learned about nature instead of denying it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/04/2019 19:44:45
Science isn't about proof, it's about evidence. What I gain is some satisfaction in educating others (or at least attempting to).
I do appreciate all your efforts (and Halc efforts) in answering my questions.
With regards to BBT:
You claim that "Science isn't about proof, it's about evidence". The BBT had been set about 70 years ago. During that time how many contradictions did you find between this theory and new evidences?  How many times our scientists had added patch over patch in order to adjust the BBT to new evidences?

It isn't a matter of "living happily" with "nobody knows". It's a matter of accepting reality. We can and do try to learn more about nature, but don't know everything.
About 379,000 years after the Big Bang. That was when the Universe was cool enough to allow electrons to bind to protons to form hydrogen.

There are so many open questions about the BBT that I really can't understand why our scientists still hold that theory. How the BBT goes with reality if we don't know what the reality of our universe is?
1.  Is the universe infinite or finite in its size?
2.  If there was something before the BBT, than why don't we count the time from that something?
3.  Why do we insist to start the BBT theory while all the electrons and protons of the whole universe were already existed in the Universe?
4. Why don't we ask our self how all of those electrons & Protons had been created before the BBT and how long before they had been created?
5. It surly took some time to set all of those electrons & protons which we currently have in whole of our Universe, so why they didn't merge and set the atoms before the BBT?
6. If we start from the Protons and electrons, why don't we start the BBT while all the Atoms and molecular are here and make our life easier? Actually why we don't start it from the moment that we have all the galaxies?
7. Did we try to verify if the first law of thermodynamics meets the electrons & Protons creation process and the creation of the Hydrogen Atoms after the BBT? (or did we gave a waiver for this verification?)
8. Why it took so long time after the Big Bang to set the Hydrogen atoms? Actually, how do we know so well the exact time? From "Nobody knows" you prove that we "perfectly know". How could it be?
9. How electron and proton could merge and set hydrogen Atoms without accelerator? did we try to verify if this merging process is feasible due to big bang?
10. Why they took them so long time (379,000 years) after the bang to set the atoms?
11. What kind of force set that big bang?
The Big Bang theory isn't supposed to explain where mass and energy came from. It only describes the current properties of the Universe based on properties predicted about its distant past.
12. Why not? Why this theory isn't supposed to explain where mass and energy came from?
13. Why do you give that kind of waiver to this theory?

Those are just few questions that I have about the BBT.
There are also big questions about the dark matter and dark energy.
I think that those ideas are absolutely none realistic.
What is the difference between dark matter and magic powder?
Both of them are magic.
We don't see them, we don't feel them, but they are there without any way to confirm their existence.
We use the dark matter in order to prove something that we see and we can't explain.
I consider it as a fatal error.
There must be a better explanation for what we see.
The "dark" matter or energy ideas are clear indication that our scientists have failed in understanding how the Universe really works.
I mean it literally. Nobody knows. People have different ideas, but nobody has proof one way or the other.
Why do you claim that "literally nobody knows"? Don't you think that someone might know the correct theory?
I do believe that I have full explanation on every aspect of our Universe.
I can explain our Universe without any need for dark matter or dark energy
Actually there is high similarity between my theory and Darwin theory.
Darwin couldn't explain how the first cell of life had been created. However, once it had been created – he could explain how the variety of life on Earth had been evolved.
In the same token, I can't explain how the first BH had been created in the whole Universe.
It could had been created due to a Big Bang
 However, once we have only one BH, I can explain how the whole universe had been evolved from it.
Based on my theory, we should know the size of the Universe and its age and some of the answers about the past and the future of our universe.
We can estimate where the Sun might be in the next few millions or billions years from now.
However, if you gave the BBT an option to set key elements in the theory (as dark matter and dark energy) and some waivers, I also ask to get the option to set my key elements in the theory.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/04/2019 21:11:22
The BBT had been set about 70 years ago. During that time how many contradictions did you find between this theory and new evidences?

Given that the Big Bang theory is still going strong, none.

How many times our scientists had added patch over patch in order to adjust the BBT to new evidences?

I don't know, but it's normal for a theory to acquire modifications over time as new evidence comes in.

How the BBT goes with reality if we don't know what the reality of our universe is?

Because it works as best as we can tell.

1.  Is the universe infinite or finite in its size?

Nobody knows.

2.  If there was something before the BBT, than why don't we count the time from that something?

Because we don't know if there was anything before the Big Bang or not.

3.  Why do we insist to start the BBT theory while all the electrons and protons of the whole universe were already existed in the Universe?

They didn't already exist. Their mass and energy were there, but they were not.

4. Why don't we ask our self how all of those electrons & Protons had been created before the BBT and how long before they had been created?

They weren't created before the Big Bang.

5. It surly took some time to set all of those electrons & protons which we currently have in whole of our Universe, so why they didn't merge and set the atoms before the BBT?

First of all, because there weren't any electrons or protons before the Big Bang and secondly atoms can't form unless the conditions are right. If the temperature is too high, electrons can't bind to protons. This is what happens in a plasma.

If we start from the Protons and electrons, why don't we start the BBT while all the Atoms and molecular are here and make our life easier? Actually why we don't start it from the moment that we have all the galaxies?

Because the temperature would have been too hot to allow those things to exist.

7. Did we try to verify if the first law of thermodynamics meets the electrons & Protons creation process and the creation of the Hydrogen Atoms after the BBT? (or did we gave a waiver for this verification?)

There's no reason to assume that it wouldn't.

Why it took so long time after the Big Bang to set the Hydrogen atoms? Actually, how do we know so well the exact time?

Because it was too hot for hydrogen atoms to form. The Universe would have been filled with plasma before then. We know by comparing predictive models with the cosmic microwave background. Before those neutral hydrogen atoms formed, the plasma would have been strongly absorbing to light radiation. Afterwards, light could travel much more freely and therefore would be observable. When that happened has an effect on the overall, observable properties of the Universe. The Wilkinson Microwave Anisotropy Probe showed that these observations agreed with predictions of the Big Bang theory: https://arxiv.org/pdf/astro-ph/0302209.pdf

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From "Nobody knows" you prove that we "perfectly know". How could it be?

You're putting words in my mouth again. I never said that we don't know when the first atoms formed.

9. How electron and proton could merge and set hydrogen Atoms without accelerator? did we try to verify if this merging process is feasible due to big bang?

It's because electrons are negatively-charged and protons are positively-charged. They attract each other naturally to form hydrogen atoms. A hot plasma of hydrogen will automatically reform into neutral atoms if you allow it to cool down. There's nothing unknown about that.

10. Why they took them so long time (379,000 years) after the bang to set the atoms?

Because it was too hot before then.

11. What kind of force set that big bang?

Nobody knows.

12. Why not? Why this theory isn't supposed to explain where mass and energy came from?

Because that isn't its purpose. That's like asking me why the theory of gravity doesn't explain why apples are red.

13. Why do you give that kind of waiver to this theory?

It's not a waiver. I don't ask a theory to explain anything that it isn't supposed to explain.

What is the difference between dark matter and magic powder?
Both of them are magic.
We don't see them, we don't feel them, but they are there without any way to confirm their existence.
We use the dark matter in order to prove something that we see and we can't explain.
I consider it as a fatal error.
There must be a better explanation for what we see.
The "dark" matter or energy ideas are clear indication that our scientists have failed in understanding how the Universe really works.

Nobody said that we understand everything about how the Universe works. Do you really think that means we understand nothing at all? That is a false dichotomy.

Why do you claim that "literally nobody knows"? Don't you think that someone might know the correct theory?

In order for someone to know, they would have to have performed the necessary experiment in order to confirm that it is correct. If anyone has done this, they either haven't yet let anyone else know about it or they haven't given out the evidence for it.

I do believe that I have full explanation on every aspect of our Universe.

Beliefs are not knowledge. Have you performed the necessary experiments to support your explanation?

In the same token, I can't explain how the first BH had been created in the whole Universe.

So you're allowed to say "I don't know" but Big Bang theorists aren't? That sounds like a double standard to me.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/04/2019 06:22:22
Thanks for all your explanations
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In the same token, I can't explain how the first BH had been created in the whole Universe.
So you're allowed to say "I don't know" but Big Bang theorists aren't? That sounds like a double standard to me.

It is not the same.
I agree, any theory should start from a starting point that is ZERO.
I assume that the BBT also starts from Zero.
So, there was a time when the Universe was totally without anything.
Just think about infinite space without even one single particle or energy.
From this point we should start.

However, I would like to ask you to forget all the sympathy that you might have in favor with the BBT and try to evaluate my theory (Let's call it theory-D) Vs the BBT on the same platform.
For one moment try to position yourself as a fair judgment without any special favorite to any side.
Let's assume that you have just arrived to our planet.
You have full knowledge about all the physics law (Newton, thermodynamics...) but you don't know anything about the theories of our Universe.
You have never heard about the BBT or about theory-D.

We will go step by step from that Zero time (while there was nothing in the whole universe) and we will set the BBT Vs theory-D
If you agree, than let's set our first step:

Once upon a time, our universe was totally free from any sort of mass.
It was infinite space without even a single particle or quark.
However, let's assume that something came out of this nothing.
We can't explain you how exactly that happened. (Not based on the BBT and not based on theory-D)
In any case:

Based on BBT everything should come directly from nothing.  So in order to justify the BBT, all the matter in the whole vast Universe had to be created at a single moment. We also need to add a theory of space expansion and inflation. We set full theory just for the first second after the Big bang moment. We also need to add the ideas about dark matter and dark energy.

Based theory-D, only small amount of something is needed out of nothing. So all is needed is a compact BH with small excretion disc. (Just one BH in the whole empty Universe). No need to come with any sort of requirements/theories as inflation, space expansion or the first second, dark matter &dark energy.

My first question is simple:
As a fair judgment - what kind of starting point is more logical:

1. The BBT Theory: With everything out of nothing and all the requested theories?
2. Theory-D: Just Something out of nothing without any other special request. No Inflation, No space expansion, no first second, no dark matter, dark energy or magic powder. What we see in our universe is what we have.?



Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/04/2019 13:10:29
So, there was a time when the Universe was totally without anything.

According to who?

Based on BBT everything should come directly from nothing.

That's not what the Big Bang theory claims.

So all is needed is a compact BH with small excretion disc.

If the total mass/energy of this black hole and excretion disk is smaller than the total mass/energy of the galaxy that comes from it, then you have broken the first law of thermodynamics and therefore falsified your own model.

No need to come with any sort of requirements/theories as inflation, space expansion or the first second, dark matter &dark energy.

How does it get rid of the need for those things?

My first question is simple:
As a fair judgment - what kind of starting point is more logical:

1. The BBT Theory: With everything out of nothing and all the requested theories?
2. Theory-D: Just Something out of nothing without any other special request. No Inflation, No space expansion, no first second, no dark matter, dark energy or magic powder. What we see in our universe is what we have.?

Neither. The first model is a straw-man of the Big Bang theory, not the actual Big Bang theory. The second one violates the first law of thermodynamics.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/04/2019 15:17:03
Covering a few points not mentioned by Kryptid.

I agree, any theory should start from a starting point that is ZERO.
I assume that the BBT also starts from Zero.
It is a convenient place to assign the zero.  It is very difficult to talk about the first second of the big bang if you put the zero say 2019 years ago.  For the same reason, cosmological temperatures are measured in Kelvin scale, not Celcius.  Same scale, but the zero where it really belongs.

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Just think about infinite space without even one single particle or energy.
The big bang theory does not posit a bang happening somewhere in infinite space.  It concerns all of space (everywhere) being not very much and expanding from there, as it is still doing.

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Let's assume that you have just arrived to our planet.
You have full knowledge about all the physics law (Newton, thermodynamics...) but you don't know anything about the theories of our Universe.
You have never heard about the BBT or about theory-D.

We will go step by step from that Zero time (while there was nothing in the whole universe)
You are already assuming a theory, that there is a zero time (which sounds like BBT) and there being nothing (which the BBT does not posit).  You can't assume any of that stuff if you've just arrived at the planet and have never made any observations.  You cannot posit any theories concerning past state at all unless you make some observations and conclude things from them.  So your first step seems to be to just make stuff up, which is a horrible first step.
First step is to make observations.  That's how it's always been done.

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Once upon a time, our universe was totally free from any sort of mass.
You don't know that.  Maybe it still is.  If you add all the positive and negative mass/energy, it seems to add up to zero still.  That would be very much in keeping with thermodynamic law.

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It was infinite space without even a single particle or quark.
No theory claims this.

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However, let's assume that something came out of this nothing.
This is a philosophical assertion which has little to do with science.
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We can't explain you how exactly that happened.
It seems a poor tactical move to posit something that defies explanation.  If it isn't logically possible for something to come from nothing, then it seems illogical to assume that happened.  Your idea is starting off on an unsound foundation.

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My first question is simple:
As a fair judgment - what kind of starting point is more logical:

1. The BBT Theory: With everything out of nothing and all the requested theories?
The BBT does not posit that, as Kryptid has pointed out in prior posts.
This is a terrible starting point.  All we know is some basic laws of motion and thermodynamics, things we might know via local experiments.  We've lived in a cloud and could never see any distance. We've only now just arrived and you want to posit a BBT?  Bad idea.  There is as yet no evidence to support the BBT or any other theory about the distant past.  We need to make observations first.  Get a telescope and tell me what you see.
I look up and see dots of light in the sky.  OK, Newton's laws didn't predict that.  I need a theory about what they are.  Maybe they're bugs.  Do we know about Earth being round and orbiting the sun, or do we need to discover those things still?  How might you go about that?  We're assuming we know nothing except Newton's laws, and maybe some of Einstein's and Bohr's if we get into situations that require it.

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2. Theory-D: Just Something out of nothing without any other special request. No Inflation, No space expansion, no first second, no dark matter, dark energy or magic powder. What we see in our universe is what we have.?
What I see is a lot more than a small mass with an accretion disk, so Theory D is already falsified.  It seems to violate thermodynamic law at every step.  It also violates Newton's laws of motion, so it contradicts the premise you gave that these laws are valid.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/04/2019 16:27:30
You don't know that.  Maybe it still is.  If you add all the positive and negative mass/energy, it seems to add up to zero still.  That would be very much in keeping with thermodynamic law.
Thanks Halc
Few questions:

1. Do you claim that based on the BBT all the mass of the Universe was already there before the Big bang, in order to keep with thermodynamic law?
2. What kind of matter/particles was there in the early Universe? Can we assume: Protons, electrons. Neutrons, quarks...?
3. Can we also assume that the entire early Universe was covered with those particles at the same density?
4. What was the size of that early Universe? Can we assume that it was infinite? So, can we assume that the particles were also infinite?
5. What was the temp of the Universe before the BBT?
6. You have mentioned the idea of: "positive and negative mass/energy". So do you mean that if we add all the positive and negative mass/energy of the early universe we should get zero?
7. If that is the case, why all the negative and positive mass/energy in the early universe didn't transformed into nothing before the big bang?
8. What about our current universe? If we add all the negative and positive would we get zero universe (without any mass) even today?
9. Why do we need the Big bang if all the mass of the Universe was there before the Big bang?
10. why all the particles of the early Universe had not been formed into real Atoms before the Big bang?
11. How could it be that a big bang will be everywhere at the same moment?
12. Is it realistic? Can we set a bang in the infinite early Universe? How can we agree with that none realistic idea?
13. Why did we get that big bang?
14. What was the energy source for the Big bang? How do we know that there is no violation for thermodynamics law just based on that big bang?
15. Why the Big bang had increased the temp of the early Universe? If the Big bang was everywhere at the same moment, than do you agree that after the BBT the temp should go up everywhere? Hence, do you agree that an infinite energy is needed to increase the temp of that unlimited early universe? So what is the source for that energy?
16. If the temp of the Universe went up so high after the big bang, (while there is no limit to the early universe) then theoretically, it shouldn't go down (as there is no limit - infinite?)
17. Did you try to prove the BBT by thermodynamics law/calculations/temp? (Especially, how the BBT have got it's unlimited energy without violating the thermodynamics law.)


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/04/2019 20:33:25
You should answer Kryptid's questions in his latest post.  He asked some relevant stuff, and since they lead to inconsistencies in your view, I notice that you're ignoring them.

1. Do you claim that based on the BBT all the mass of the Universe was already there before the Big bang, in order to keep with thermodynamic law?
The BBT makes no claims about 'before the big bang'.  Kryptid has already stated this, and answers to most of these questions.
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2. What kind of matter/particles was there in the early Universe? Can we assume: Protons, electrons. Neutrons, quarks...?
Again, no.  It was too hot for these things to exist.
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3. Can we also assume that the entire early Universe was covered with those particles at the same density?
Energy density was reasonably uniform, but not perfectly.  Cosmic inflation theory says that quantum fluctuations during the inflation phase, due to the uncertainty principle, were amplified into the large scale density differences that would eventually form the large scale structures like superclusters and such.
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4. What was the size of that early Universe?
Arbitrarily smaller than it is now.
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Can we assume that it was infinite?
Nobody knows this.
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6. You have mentioned the idea of: "positive and negative mass/energy". So do you mean that if we add all the positive and negative mass/energy of the early universe we should get zero?
Possibly yes.  Computing them all is not trivial, but it is suggested, yes.
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7. If that is the case, why all the negative and positive mass/energy in the early universe didn't transformed into nothing before the big bang?
BBT is not about before big bang.  Look to theories whose purpose is to answer such questions.  There are several, and there is no particular consensus.  A unified field theory seems to be required to make meaningful headway.
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8. What about our current universe? If we add all the negative and positive would we get zero universe (without any mass) even today?
If it was ever zero, it is still zero, per thermodynamic law.
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9. Why do we need the Big bang if all the mass of the Universe was there before the Big bang?
Big bang doesn't explain the appearance of nonzero mass, if there is any.  If the total is zero, then it doesn't need to explain why zero is still there.
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10. why all the particles of the early Universe had not been formed into real Atoms before the Big bang?
I'm going to stop answering all the before-the-big-bang questions which are not relevant to the theory.
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11. How could it be that a big bang will be everywhere at the same moment?
If the space of the universe is compressed to an arbitrarily small amount (singularity?), then there is nowhere in that small super-hot space that isn't 'banging'.  The theory does not posit that a bang happened at one point and stuff subesequently expanded into existing space.  The event is indefinite compression of all of space itself.
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12. Is it realistic? Can we set a bang in the infinite early Universe? How can we agree with that none realistic idea?
It is not only realistic, but unrealistic to posit anything else given the observations.  You look around and a steady-state universe is not the thing you see.  Yes, there are those that posit exactly that, but it is a difficult position to defend.
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14. What was the energy source for the Big bang? How do we know that there is no violation for thermodynamics law just based on that big bang?
If total energy is zero, it doesn't need a source.
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15. Why the Big bang had increased the temp of the early Universe?
It didn't.  It posits that the temperature was cooling the whole time.
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If the Big bang was everywhere at the same moment, than do you agree that after the BBT the temp should go up everywhere?
After the BBT?  You mean after the big bang?  It says that the temperature goes down, not up.  You expand a hot thing, you get a cooling effect, just like compression heats things.
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17. Did you try to prove the BBT by thermodynamics law/calculations/temp?
I personally am not involved in the calculations of any scientific theory.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/04/2019 13:59:11
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Can we assume that it was infinite?
Nobody knows this.
I really can't understand how we can set any sort of theory for any kind of object without knowing its shape/size/age
For example -
Let's assume that we give a request to an engineer to design a bridge.
It is clear that the first question will be: what is the size of that bridge.
Is it a toy bridge of 20 Cm or is it a bridge that needs to cross the Atlantic Ocean.
Both are called bridges. But how can we compare a 20 Cm Bridge to 3000 miles bridge.
In the same token - How can you set a theory for a universe without knowing its size?
That by itself shows that our scientists have a fatal error by definition.
A theory for a Universe with a radius of 13.8 BLY can't be the same as for an infinite Universe.
If I recall it correctly, when the BBT Idea came to our life, our scientists were quite sure that what we see is what we have. So, the Universe was quite compact.
Therefore, the idea was that the big bang took place at a singular location. That kind of bang was very logical for a compact Universe at that time..
However,  as we have improved our instruments, our scientists discovered that the Universe should be much bigger.
At the first step they have found that the size of the observable universe is:
https://en.wikipedia.org/wiki/Observable_universe
"The radius of the observable universe is therefore estimated to be about 46.5 billion light-years"
After this discovery they have changed the BBT.  so they didn't use the idea of a singular
However, now some of our scientists start to understand that the Universe might be infinite.
So, how can you adjust a theory for compact Universe to infinite universe?
One option is to claim:
If the space of the universe is compressed to an arbitrarily small amount (singularity?), then there is nowhere in that small super-hot space that isn't 'banging'.  The theory does not posit that a bang happened at one point and stuff subesequently expanded into existing space.  The event is indefinite compression of all of space itself.
However, that answer is based on "If the space of the universe is compressed to an arbitrarily small amount (singularity?)"
But if not? We discuss science not a lottery game.
So, there is a better idea:
The big bang theory does not posit a bang happening somewhere in infinite space.  It concerns all of space (everywhere) being not very much and expanding from there, as it is still doing.
So, now the idea is that the Big Bang took place at the same moment in the whole infinite space.
However, I assume that you understand that this idea could kill down the BBT before we even we start our discussion, as there is no way to cool down the temp of infinite space. When the space was limited (finite) there is some logic in the idea of cooling it down as there is increasing in space. But if the starting space is infinite, how can we cool it down by the expansion? What is the meaning of increasing an infinity space to infinity space? I personally can't see how this kind of universe can cool its temp..
So, I really can't understand how you try to give us an explanation for a universe which you don't know its size.
If you don't know its size - your theory is just irrelevant!!!
 
I'm going to stop answering all the before-the-big-bang questions which are not relevant to the theory
Why it is not relevant?
Why you do not want to discuss about it?
I didn't get an answer for the source of that infinite energy that is requested to that kind of Big Bang.
If our scientists can't give the answer for the time before the BBT than how is responsible for that time?
I also didn't get yet an answer how the thermodynamics law works at the BBT.
You are using this thermodynamics law in order to disqualify the Theory-D, while you give yourself a waiver for the BBT theory.
That's really unfair.
You ask me to look on Evidences
So let's do so:
Magnetic field around the excretion disc: We clearly see that this magnetic field boosts a molecular jet at a speed of 0.8 speed of light. Just based on this data, do you agree that the magnetic power has an ULTRA HIGH energy force?
If you add to that the ultra velocity of the plasma at the excretion disc (0.3 speed of light) and the ultra high temp (10^9c) than don't you think that all that kind of combined energy forces can be transformed into one single quark or one single Atom?
What is the estimated ratio in the energy between one quark to that energy in the excretion disc?
Could it be 1 to 10^100?
Why do you insist that the first thermodynamics law should not work at the excretion disc although it is clear that there is so much extra energy there, while you don't even try to find the ultra high missing energy for the BBT?
Sorry - If our scientists assume that they can give them a waiver for that Ultra high missing energy and bypass the first thermodynamics law for the BBT, how can you claim that an energy ratio of 1 to 10^100 is not good enough to form one single quark.
This is really unrealistic.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/04/2019 17:06:47
So, I really can't understand how you try to give us an explanation for a universe which you don't know its size.
If you don't know its size - your theory is just irrelevant!!!

Then your explanation must be irrelevant too, since you also don't know the size of the Universe. You have no way of measuring its true size.

I didn't get an answer for the source of that infinite energy that is requested to that kind of Big Bang.

Whoever said that the energy was infinite? Your argument is like saying that we can't know that a bomb exploded because we don't know who made the bomb. That's a non-sequitur.

You are using this thermodynamics law in order to disqualify the Theory-D, while you give yourself a waiver for the BBT theory.

The Big Bang theory doesn't require any violation of thermodynamic laws. There is no reason that the energy and mass present at the Big Bang should be any different than the amount present in the modern Universe.

Why do you insist that the first thermodynamics law should not work at the excretion disc although it is clear that there is so much extra energy there

Because "mass" and "matter" are not the same thing. Sure, you can create new subatomic particles, but you can't increase the total mass/energy of the system unless you get some mass/energy from an outside source. The creation of new subatomic particles in a particle accelerator consumes energy from power plants. The total mass and energy before and after the particle collisions is the same. The total mass/energy of the black hole/accretion disk system cannot increase if mass/energy isn't coming in from outside.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/04/2019 17:44:32
Then your explanation must be irrelevant too, since you also don't know the size of the Universe. You have no way of measuring its true size
Yes, I know
The Universe is infinite!!!
I can prove it.
So, theory-D gives an explanation for infinite Universe

Sure, you can create new subatomic particles,

Thanks!!!
So, do you agree that the excretion disc can create subatomic particles as quarks?
If so, that's all I need

The Big Bang theory doesn't require any violation of thermodynamic laws. There is no reason that the energy and mass present at the Big Bang should be any different than the amount present in the modern Universe.
1. If all the energy and mass of the Universe (Finite or infinite) was there before the Big bang, then do you agree that the Universe was already there before the Big Bang? If so, why we don't add that time to the Universe age?
2. Do you agree that the Big bang activity is based on energy, and it also had consumed some portion of the available energy that existed in the Universe before that bang?
As you claim that there is no violation in thermodynamics - so do you agree that the Big bang had reduced the total energy of the Universe? Do you have an idea by what factor?
3. If the Universe was already there, and the BB just reduced the total energy of the Universe, so would you kindly explain the benefit of that bang? Why do we need it?
4. I still don't understand why don't we start the BBT theory when there was no mass/energy in the Universe, or at least when it was with some mass/energy. What is the benefit of theory if it starts when we already have everything?
For example, Theory-D starts when there is only one BH in the whole Universe. This one BH sets the whole Infinite Universe.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/04/2019 20:22:25
A theory for a Universe with a radius of 13.8 BLY can't be the same as for an infinite Universe.
Not true.  It might depend on one's definition of 'universe'.  One definition is everything inside the Hubble sphere, which is to say everything that exists in our inertial reference frame.  That size is known, and yes, it is 13.8 BLY.  Another definition bounds the universe at the event horizon, which is a bit further out, or the theoretically observable universe, which is quite larger.  Still other definitions consider locations beyond that.  All of them are valid definitions under big bang theory, although possibly not under steady state theory.

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If I recall it correctly, when the BBT Idea came to our life, our scientists were quite sure that what we see is what we have. So, the Universe was quite compact.
Even more compact in pre-Copernican days.  Yes, the extent of 'all there is' seems to always be growing.  The universe certainly got bigger if Hilbert space is considered part of it, but the size of Hilbert space isn't measured in light years.

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Therefore, the idea was that the big bang took place at a singular location.
The theory does not posit this.  This is a strawman representation.  Do not presume to state details of a theory you clearly have no intention of understanding.

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At the first step they have found that the size of the observable universe is:
https://en.wikipedia.org/wiki/Observable_universe
"The radius of the observable universe is therefore estimated to be about 46.5 billion light-years"
After this discovery they have changed the BBT.
How so?  The theory did not specify a size of the observable universe before it was computed, and the observable universe is not the universe except when 'universe' is defined that way.  But the theory does say that what is now that observable universe was arbitrarily small once.

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However, now some of our scientists start to understand that the Universe might be infinite.
So, how can you adjust a theory for compact Universe to infinite universe?
The observable part is unchanged by whether or not the universe (obviously defined here as more than the parts with which we can interact) is finite or not.  Even then, it really depends on how one might go about defining said size.  Time seems to have no end (except under big-rip or big-crunch theories, only the former of which might be plausible), and so the universe is infinite if measured in that direction.  Size/distance is not a defined concept outside of special relativity, and our universe does not conform to the special conditions that make SR special.  So one must be more specific about what is being measured.

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One option is to claim:
Quote from: Halc
If the space of the universe is compressed to an arbitrarily small amount (singularity?), then there is nowhere in that small super-hot space that isn't 'banging'.  The theory does not posit that a bang happened at one point and stuff subsequently expanded into existing space.  The event is indefinite compression of all of space itself.
However, that answer is based on "If the space of the universe is compressed to an arbitrarily small amount (singularity?)"
Yes.  The two are different things, like the difference between an arbitrarily large finite number and infinity.

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Quote from: Halc
The big bang theory does not posit a bang happening somewhere in infinite space.  It concerns all of space (everywhere) being not very much and expanding from there, as it is still doing.
So, now the idea is that the Big Bang took place at the same moment in the whole infinite space.
That was not said.

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However, I assume that you understand that this idea could kill down the BBT before we even we start our discussion, as there is no way to cool down the temp of infinite space.
Untrue.  You could double the size of it and reduce the temperature by a factor of 8.  Charles' law.

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What is the meaning of increasing an infinity space to infinity space?
If you double the size, things that were 1 meter apart are now 2 meters apart, exactly as we observe.

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Why it is not relevant?
Read my replies that you edited out.

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Why you do not want to discuss about it?
I didn't say I didn't want to discuss it.  I just said the BBT doesn't discuss it.  The theory concerns other things.  Remember Kryptid talking about why apples are red?  Read that again.

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If our scientists can't give the answer for the time before the BBT than how is responsible for that time?
Nobody said scientists give no answers for that.  But the BBT doesn't, since it is not a theory concerning 'before the big bang'.  BBT is actually a very simple theory in principle, making a simple conclusion from a simple observation.

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I also didn't get yet an answer how the thermodynamics law works at the BBT.
Kryptid has answered this repeatedly.  BBT posits no change in mass energy between the time of the big bang and now.  That means it does not contradict thermodynamic law.  Your theory asserts a direct contradiction with it, and also asserts that the laws are valid, thus your theory is immediately self contradictory.

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Just based on this data, do you agree that the magnetic power has an ULTRA HIGH energy force?
Energy and force (and power for that matter) are all different things, so I cannot agree to a nonsensical statement.  It may not be ultra high force at all.  It takes considerably less force to accelerate a subatomic particle to 0.8c than it takes to lift my pint of beer, something I don't consider to be on the 'ultra-high' end of the force scale.

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Why do you insist that the first thermodynamics law should not work at the excretion disc
Because you asserted that it doesn't.

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although it is clear that there is so much extra energy there
Energy is energy.  There is not a special kind that is 'extra'.  You posit new energy/mass coming from a closed system, which violates thermodynamic law.
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how can you claim that an energy ratio of 1 to 10^100 is not good enough to form one single quark.
Nobody claimed that.  We said it probably happens. The creation of a quark takes one quark of mass away from the everything-else that made it.  There is no net gain by this process.  As long as the theory doesn't posit a net gain, there is no violation.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/04/2019 22:04:03
The Universe is infinite!!!
I can prove it.

Please do so.

So, do you agree that the excretion disc can create subatomic particles as quarks?

I don't know if it can create quarks specifically or not. I know that there are supernova models where electrons and positrons can be created inside of stars at immense temperatures (pair-production supernova). Maybe quarks can be created too, but you'd probably need much higher temperatures (because they are more massive than electrons). If you do end up with quark creation, you must also create antiquarks at the same time in order to satisfy conservation laws. So then you'd end up with the issue of how to keep those quarks and antiquarks from recombining and annihilating.

However, the creation of those particles does not decrease or increase the amount of energy or mass present in the system.

1. If all the energy and mass of the Universe (Finite or infinite) was there before the Big bang, then do you agree that the Universe was already there before the Big Bang? If so, why we don't add that time to the Universe age?

Because we don't know how long that time period is.

2. Do you agree that the Big bang activity is based on energy, and it also had consumed some portion of the available energy that existed in the Universe before that bang?

How can it consume energy? The energy can't disappear. It has to be there at all points in the process. Otherwise, the first law is violated.

As you claim that there is no violation in thermodynamics - so do you agree that the Big bang had reduced the total energy of the Universe? Do you have an idea by what factor?

The Big Bang did not decrease the total energy of the Universe. That itself would violate the first law. The energy has to remain constant. The form that the energy takes can change (from photons into electrons and positrons, for example), but the amount must not change.

3. If the Universe was already there, and the BB just reduced the total energy of the Universe, so would you kindly explain the benefit of that bang? Why do we need it?

The Big Bang didn't reduce the Universe's energy. Whether or not it was "beneficial" or whether it was "needed" for anything is irrelevant to whether or not it happened.

4. I still don't understand why don't we start the BBT theory when there was no mass/energy in the Universe

Whoever said there was ever a point in the past when there was no mass or energy in the Universe? That would violate the first law.

What is the benefit of theory if it starts when we already have everything?

We didn't have everything. We didn't have galaxies, stars, planets or living things back then.

For example, Theory-D starts when there is only one BH in the whole Universe. This one BH sets the whole Infinite Universe.

Then that black hole had better have had all of the mass/energy in the entire Universe or else you're breaking the first law.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/04/2019 23:09:41
Only if the total now is nonzero, a debated thing.

Perhaps the better thing to say would then be, "It isn't any different now than it was then".
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/04/2019 08:09:09
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The Universe is infinite!!!
I can prove it.
Please do so.
With pleasure!.
The evidence is in the CMB:
https://en.wikipedia.org/wiki/Cosmic_microwave_background
"The cosmic microwave background radiation is an emission of uniform, black body thermal energy coming from all parts of the sky."
"The color temperature of the ensemble of decoupled photons has continued to diminish ever since; now down to 2.7260±0.0013 K,[4] it will continue to drop as the universe expands."
So, let's try to understand the real meaning of the CMB on our Universe:
The radiation is coming from "all parts of the sky" -So the same radiation at the same amplitude is coming from any direction.
Let's start by the assumption that the Universe has a ball shape and it is limited in its size (Universe radius - Ru).
So, if the Universe is finite, it is quite clear that in order to get the same radiation from any direction we have to be at the center of this universe, or close enough to the center.
1. Let's verify the statistical chance for that:
In order to do so, let's divide Ru into 1000 segments and see what is the chance that we are located at one of that segments which are located at the center (or close enough to the center) of that Universe ball shape.
So,
R1 = Ru/1000
So, the total volume of the Universe based on R1 is:
Vu= f of Ru^3 = f of (1000R1)^3 = 10^9 R1^3
So the chance to be exactly at the center is 1 to 10^9.
What is the chance that we are located at the most outwards segment (999 to 1000)?
The Volume is:
V(last segment) = V(Ru) - V(Ru -R1) =
f of (1000R1)^3 - ((1000-1)R1)^3= 1000^3 R1^3 - 999^3 R1^3 = 2,997,001 R1^3
So, while the chance to be in the center is 1 to 10^9, the chance to be at the most outwards segment is 2,997,001 to 10^9
Therefore our chance to be at the most outwards envelop of the universe is greater by  2,997,001 than the chance to be at the center.
Let's try to verify what is the chance to be located at the segments between (0 to 500)R1. (Half of the total radius of the Universe)
V(r=500) = f of (500R1)^3 = 0.125 * 10^3 R1^3
Hence, the chance to be at any segment between (0 - 500) R1 is only 0.125 while the chance to be at any segment between (500 -1000)R1 is 0.875
Let's try to find the radius which represents the middle of the volume?
Vu = 10^9 R1^3
Vu/2 = 0.5 10^9 R1^3
If we choose R = 800R1
V = 0.512 10^9 R1^3.
Therefore,
Half of the whole volume is located between (0 to 800) R1
While the other half is located between (800 to 1000) R1.
Conclusion:
Our chance to be at the center or close to the center of the ball shape universe is very minimal.
It is clear that from statistical point of view, we have higher chance to be at the segment from (500 - 1000)R1 comparing to  any other segment at the (0 -500)R1
This proves that if the Universe was finite, we had to receive different radiation amplitude from different directions.
As we get exactly the same amplitude, it proves that the Universe is infinite by definition. Only at infinite universe there is no meaning where we exactly located in the Universe. Any infinite radius from one side is equal to infinite radius from the other side.
2. There is also an issue of the Black body radiation which proves that the Universe is infinite.
However, let's see if it is needed.
3. With regards to the 2.7260±0.0013 K amplitude:
It is stated: "it will continue to drop as the universe expands". That is a fatal error.
First, the Universe is not expanding, only the galaxies.
Second, for any far away galaxies that are leaving our observable Universe, new galaxies will be born. So, the total density of mass in our observable universe (or at any aria in the infinite Universe) will stay the same
Hence, this amplitude is going to stay with us forever and ever as our universe is infinite in its size and infinite in its age.

What kind of engineering are your teams involved in?  I'm an engineer myself, not a scientist.
Electronics (Mainly Communications and computing - I was leading the technical support for Intel & AMD in my aria).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/04/2019 14:37:02
Quote from: Kryptid
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The Universe is infinite!!!
I can prove it.
Please do so.
With pleasure!.
The evidence is in the CMB:
...
The radiation is coming from "all parts of the sky" -So the same radiation at the same amplitude is coming from any direction.
Let's start by the assumption that the Universe has a ball shape and it is limited in its size (Universe radius - Ru).
Your approach is actually pretty good, but an assumption of a 3D ball shape with a center is unrealistic.  It assumes that the universe is expanding from a central location into existing space, not that space itself is expanding.  A finite universe would be a limited size hypersphere which has no edge and no center just like the surface of an expanding balloon has limited area and yet has no edges or center on its surface.
You seem to be describing a situation with an edge where there is stuff only on one side and emptiness on the other.

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So, if the Universe is finite, it is quite clear that in order to get the same radiation from any direction we have to be at the center of this universe, or close enough to the center.
1. Let's verify the statistical chance for that:
In order to do so, let's divide Ru into 1000 segments and see what is the chance that we are located at one of that segments which are located at the center (or close enough to the center) of that Universe ball shape.
So,
R1 = Ru/1000
R1 is about 46 billion light years (size of visible universe).  Yes, if there was an edge and it was closer than that, we'd see the anomaly in the CMB.
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So, the total volume of the Universe based on R1 is:
Vu= f of Ru^3 = f of (1000R1)^3 = 10^9 R1^3
So the chance to be exactly at the center is 1 to 10^9.
What is the chance that we are located at the most outwards segment (999 to 1000)?
The Volume is:
V(last segment) = V(Ru) - V(Ru -R1) =
f of (1000R1)^3 - ((1000-1)R1)^3= 1000^3 R1^3 - 999^3 R1^3 = 2,997,001 R1^3
So, while the chance to be in the center is 1 to 10^9, the chance to be at the most outwards segment is 2,997,001 to 10^9
Therefore our chance to be at the most outwards envelop of the universe is greater by  2,997,001 than the chance to be at the center.
The math is off, but close enough.  The odds of being not at the edge is about 997 to 3 by this calculation.  So most likely we're neither at the edge nor the exact center.  But it seems that only at the edge would the CMB definitely not be isotropic.

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Therefore,
Half of the whole volume is located between (0 to 800) R1
While the other half is located between (800 to 1000) R1.
Conclusion:
Our chance to be at the center or close to the center of the ball shape universe is very minimal.
It is clear that from statistical point of view, we have higher chance to be at the segment from (500 - 1000)R1 comparing to  any other segment at the (0 -500)R1
This proves that if the Universe was finite, we had to receive different radiation amplitude from different directions.
Wait, where did 'different radiation amplitude' suddenly come from?  None of the proof made any mention of why that must be non-isotropic from any of these non-edge volumes.  If I sample a random small spherical volume centered somewhere in a large ball of plastic, I can tell by the flat spot if I managed to sample a location right at the edge, but all the other ones will look identical.

If I sample a random small circle on the surface of a finite area balloon, they'll all look identical, without any probability about it.

You seem to have demonstrated that given
p1) a 3D Euclidean model of the universe (no spacetime) and where
p2) the CMB would taper off uniformly from center to edges and where
p3) the ratio of Ru to R1 is small
c1) that we would probably notice non-isotropy of the CMB.

That is three unlikely premises.  P2 is not unreasonable, but you have proof of none of them, so they're premises, and I agree that the conclusion stands if all three of these premises are true.

P1 corresponds to no accepted cosmological model of the universe.  It involves things moving faster than light through space.
P3 assumes a small Ru.  If it is large but still finite, the CMB variance from various directions might vary (if P2 holds), but below our ability to measure.  As it is, it does vary, being more red shifted in one direction and blue shifted in the other, but that's due to our velocity relative to the mean velocity of everything around us.

Argument by stats is not a proof.  It results in a probability of something being true, but not proof.  I accept such probabilistic evidence, but I don't accept the premises. The accepted models of a finite universe don't look like what you're describing.  They're either non-euclidean, or an Ru in line with eternal inflation theory where there are finite edges because we're using a foliation where the universe is young and still 'banging' presently.

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Second, for any far away galaxies that are leaving our observable Universe, new galaxies will be born. So, the total density of mass in our observable universe (or at any aria in the infinite Universe) will stay the same
This violates thermodynamic law.  A system (OU) that is losing mass should get lower and lower in mass unless there is a balance of mass coming in.

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Quote from: Halc
What kind of engineering are your teams involved in?  I'm an engineer myself, not a scientist.
Electronics (Mainly Communications and computing - I was leading the technical support for Intel & AMD in my aria).
Tech support.  OK.  You don't design the bridges, but you support those that have trouble with their bridges.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/04/2019 20:13:40
R1 is about 46 billion light years (size of visible universe).  Yes, if there was an edge and it was closer than that, we'd see the anomaly in the CMB.

You can use any size of Universe.
If Ru = 46 Billion LY
The total size of the Universe is as big as the Observable Universe
R1 = Ru/1000 = 46 Million LY
This is the Length of each segment.
The odds of being not at the edge is about 997 to 3 by this calculation.  So most likely we're neither at the edge nor the exact center.  But it seems that only at the edge would the CMB definitely not be isotropic.
It seems to me that you have missed the point.
If the radius of the whole Universe is 46 BLY, let's divide it into two sections:
The odds of being in at the range of 0 to 46/2= 32BLY  is only 0.125.
The odds of being in at the range of 32 BLY to 46BLY is 0.875.
Therefore, our chance to be at the radius of 32BLY to 46BLY is higher than the chance to be at the 0 to 32BLY.
So, assuming that we will be located at 32BLY from the center than:
In one direction to the edge of the Universe it will be at a distance of 64 +32 = 96BLY, while in the other side it will only be 32 BLY.
Therefore, the ratio between one side to the other side is 1 to 3 (32 vs 96)
I hope that you agree that in this case, the CMB definitely not be isotropic.
So, we don't need to be exactly at the edge in order to get that none isotropic radiation.

You have selected R1 = 46BLY.
So, in this case
Ru = 46,000BLY
Hence, it this example the real Universe is 1000 time the observable Universe:
Even for this case:
The chance to be at a radius of 0 to 32,000BLY is 0.125
While the chance to be at 32,000BLY to 64,000BLY is 0.8125.
Therefore, even in a really big universe, it is expected to get CMB which is not be isotropic.
Therefore, just if the Universe is infinite we will get a perfect isotropic radiation.
Argument by stats is not a proof.  It results in a probability of something being true, but not proof.  I accept such probabilistic evidence, but I don't accept the premises. The accepted models of a finite universe don't look like what you're describing.  They're either non-euclidean, or an Ru in line with eternal inflation theory where there are finite edges because we're using a foliation where the universe is young and still 'banging' presently.
Sorry - I don't understand why do you reject that kind of statistical calculation.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/04/2019 20:55:04
If Ru = 46 Billion LY
The total size of the Universe is as big as the Observable Universe
R1 = Ru/1000 = 46 Million LY
We can see much further than 46 MLY.  R1 is the CMB, which comes from material that is 'now' close to 45 BLY away.  If the universe is 3D and that size, then we'd need to be at the exact center for there to be a CMB in all directions.

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Quote from: Halc
The odds of being not at the edge is about 997 to 3 by this calculation.  So most likely we're neither at the edge nor the exact center.  But it seems that only at the edge would the CMB definitely not be isotropic.
It seems to me that you have missed the point.
If the radius of the whole Universe is 46 BLY,
Your example had it at 1000 times that figure.  My 997 to 3 calculation assumed that size.

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let's divide it into two sections:
The odds of being in at the range of 0 to 46/2= 32BLY  is only 0.125.
The odds of being in at the range of 32 BLY to 46BLY is 0.875.
Therefore, our chance to be at the radius of 32BLY to 46BLY is higher than the chance to be at the 0 to 32BLY.
If the universe is only 46 BLY in radius, then all those places save just a few very near the center would 'see' the edge.  Dividing it in regions like that doesn't change that.  Yes, this is reasonable evidence that the universe is at least larger than the visible universe.

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So, assuming that we will be located at 32BLY from the center than:
In one direction to the edge of the Universe it will be at a distance of 64 +32 = 96BLY, while in the other side it will only be 32 BLY.
Where did radius suddenly change to 64?  OK, if the radius was that, and we're 32 BLY from the edge, we'd see the edge since we can see that far, so to speak.
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Therefore, the ratio between one side to the other side is 1 to 3 (32 vs 96)
I hope that you agree that in this case, the CMB definitely not be isotropic.
So, we don't need to be exactly at the edge in order to get that none isotropic radiation.
If you're in sight of the edge, sure, the CMB would not be on that side.  For a much larger radius, the odds of being withing 46 BLY is much much smaller.  997 to 3 in your original example, and even lower odds if it's bigger.

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You have selected R1 = 46BLY.
I didn't select that.  The CMB is 'now' ~45 BLY away, but stuff a little further away might have had a causal effect on us, even if not via light.  So 46.  That's what observable universe means.

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So, in this case
Ru = 46,000BLY
Hence, it this example the real Universe is 1000 time the observable Universe:
Yes, that's the example I was working with, and I thought you were using.

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Even for this case:
The chance to be at a radius of 0 to 32,000BLY is 0.125
Who cares?  You can't see the edge from that distance.  I was computing the chance of being within sight of the edge.

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While the chance to be at 32,000BLY to 64,000BLY is 0.8125.
Therefore, even in a really big universe, it is expected to get CMB which is not be isotropic.
So you're asserting that the universe appears isotropic only when less than half way to the edge?  That kind of is grouped under premise 2, and is not really a new premise. How might you back up this claim here? The 'proof' is only as strong as the list of premises.

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Therefore, just if the Universe is infinite we will get a perfect isotropic radiation.
Well it isn't perfect, so by your logic, the universe isn't infinite.

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Quote from: Halc
Argument by stats is not a proof.  It results in a probability of something being true, but not proof.  I accept such probabilistic evidence, but I don't accept the premises.
Sorry - I don't understand why do you reject that kind of statistical calculation.
I just said I accepted that part.  Try to read more carefully.  Such methods are a valid form of argument, but I'm just noting that it doesn't constitute proof.  I've gone through an awful lot of days, each without dying, so does this prove that I will live forever?  No, but it demonstrates that is is dang probable that I'll still be here tomorrow.

If you read further, I said I had issues with your premises, which do not correspond to any accepted model of a finite universe.  That makes the argument a strawman one.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 25/04/2019 21:06:16
Your "proof" isn't proof at all, because other models can also explain isotropy without invoking an infinite Universe: the Big Bang theory itself is one of them. Another model that would explain isotropy would be a hyperspherical universe. As an analogy, imagine a sphere where 2-dimensional creatures live embedded in its surface. They can't perceive the 3-dimensional shape of their world, only the 2-dimensional surface they reside inside of. If this surface was evenly filled with radiation, then that radiation would look isotropic regardless of where the observer stands. All locations inside of the sphere's surface are equal. If we are living in the "surface" of a hypersphere, the same would be true for our own observations even though the hyperspherical universe has a limited size.

Shell theorem should make radiation look isotropic even if your model of a finite, spherical universe is used. Shell theorem is normally used to describe gravity, but it should work here too because radiation intensity falls off at the exact same rate as gravitational strength does (the inverse square law): https://en.wikipedia.org/wiki/Shell_theorem

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html

The edges of this universe would have to be perfectly reflecting (since light simply vanishing at that edge would violate conservation of energy). Let's assume that we are not at the center of the sphere, but closer to one side. Although you would expect that edge to look brighter to us because we are closer to it, there is more total "edge" further away from us on the other side. It all cancels out when you do that math. So the radiation still looks isotropic even if you are far from the center.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/04/2019 06:07:24
Shell theorem should make radiation look isotropic even if your model of a finite, spherical universe is used. Shell theorem is normally used to describe gravity, but it should work here too because radiation intensity falls off at the exact same rate as gravitational strength does (the inverse square law): https://en.wikipedia.org/wiki/Shell_theorem

Thanks for the great example.
Unfortunately, you have missed the key point in this article

https://en.wikipedia.org/wiki/Shell_theorem
"A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the centre, becoming zero by symmetry at the centre of mass."
It is stated clearly that: "the gravitational force within the object varies linearly with distance from the centre,"
So, on the object we get different forces from different directions. However, the net force is zero.
This isn't our case. We do not measure the net radiation. We measure the CMB level from each direction - isotropic:
By Google - isotropic
Having a physical property that has the same value when measured in different directions.
So, If we combined the measured CMB from all directions we might find that the total is Zero (as the idea of the Shell theorem). But this is not our target.
Actually the Shell theorem proves that the gravity forces are not isotropic. as it is stated: "varies linearly with distance from the centre".
Therefore, this theory proves my case.
If the universe is finite (at any size) than by definition we wouldn't get an isotropic CMB. (Unless we will be exactly at the center or close to the center). However, I have proved that the chance for that is very minimal.
Hence, it is so clear to me that as the CMB is isotropic the universe must be infinite.
I really can't understand why do you both reject this simple idea.

If you're in sight of the edge, sure, the CMB would not be on that side.
Thanks
So, you agree that if we were close enough to the edge of a finite Universe the CMB shouldn't be isotropic.
But how close?
So you're asserting that the universe appears isotropic only when less than half way to the edge?

I have found that the chance to be half way to the edge is 0.875.

The CMB value is very accurate: 2.72548±0.00057 K.[4]
Therefore, the universe appears isotropic only if we are at the most center of the finite Universe or close enough to its center.
The chance for that in very minimal.
Hence, our Universe must be infinity.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/04/2019 15:20:15
The Big Bang did not create mass or energy. It represented an extreme expansion of space that mass and energy already existed in

So, the Big Bang doesn't give any answer for the creation of the mass and energy in our Universe.
Not even one gram of mass or extra energy!
Therefore, do you agree that the BBT is only a theory of transformation?
How our scientists could bypass that key element in the theory of creation our Universe.
They have to answer how the total mass and energy had been created in our universe - Finite or Infinite!
They can't say - Sorry it isn't our job!
Who is going to take care on that???

I have tried to find by Google: "How the universe was created?"
There are many explanations about it - all of them are based on the BBT.
For example:
http://www.bbc.com/future/story/20140812-how-was-the-universe-created
"So how was this unimaginably giant Universe created?"
"Around 13.8 billion years ago, all the matter in the Universe emerged from a single, minute point, or singularity, in a violent burst."
This is misleading information.
Our scientists actually mislead all the people in our planet about the BBT, as it does not deal with the most important part of the creation - Mass and Energy.
Hence, how can they call it a theory of "creation?"
They must call the BBT a theory of "transformation"!!!
Do you agree with that?
If so, please advise them to look for a real theory of creation as Theory-D.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/04/2019 21:35:20
Therefore, this theory proves my case.
If the universe is finite (at any size) than by definition we wouldn't get an isotropic CMB.
Do you have any idea what the words 'by definition' mean?  You use the phrase a lot, and always inappropriately, which doesn't help anybody's opinion of your intelligence.  The definition of a finite universe is not 'one without an isotropic CMB'.
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So, you agree that if we were close enough to the edge of a finite Universe the CMB shouldn't be isotropic.
But how close?
I don't even agree with the former.  I can think of models where there is an edge and we are right at it, and the CMB still isotropic.  For instance, suppose the universe is a computer simulation of a finite space, whose primary purpose is to simulate something that is naturally put at the center.  The size need only be a finite size since anything outside that radius has no effect on the preferred location.  Thus we would be part of the more probable 'not at the center places' and would eventually find ourselves at the edge, which we would not detect even when we're 100000 km from it.  The CMB would be isotropic.  The moon might already be gone because it has passed out of the universe, but we would not see its absence since it is impossible to see the present.

So you ask how far we can theoretically see, and there are several answers to that.  I can see zero distance into the present.  The material of the CMB that we see now is currently 45 BLY away, but that doesn't mean I can see 45 BLY.  The CMB material that we measure was perhaps 1.1 million proper light years away from the location that Earth would one day occupy, when it emitted the light we see now.  That's not very far at all.  I see lots of things that were further away from 'here' when they emitted the light we see now, and yet one cannot see beyond the CMB.  Such are the things that need to be clarified when discussing distances in expanding and relativistic space.
So how close you need to be to the edge to see it depends heavily on the model you're using, and also depends on the way you measure distance.
The model you seem to be using is not supported by anybody, so you're disproving a strawman model.  Try arguing against the hypersphere model that both Kryptid and I mentioned.  That one is nice and finite and completely isotropic, and thus illustrates (by counterexample) that the CMB is not evidence against a finite universe at all.

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Quote from: Halc
So you're asserting that the universe appears isotropic only when less than half way to the edge?
I have found that the chance to be half way to the edge is 0.875.

The CMB value is very accurate: 2.72548±0.00057 K.[4]
Therefore, the universe appears isotropic only if we are at the most center of the finite Universe or close enough to its center.
The chance for that in very minimal.
Hence, our Universe must be infinity.
I want to point out here that you repeated a lot of stuff you've said in prior posts, but you totally did not answer the question I asked. That question doesn't matter since it concerns your strawman universe, but I just posted this to show how you evade questions. You should quit the armchair science and go into politics. You have the knack for that.

So, the Big Bang doesn't give any answer for the creation of the mass and energy in our Universe.
Not even one gram of mass or extra energy!
Therefore, do you agree that the BBT is only a theory of transformation?
It describes how the universe got the way it is now from the way it proposes the universe to have been back then.  That's a transformation of sorts I guess, so I don't disagree with this.

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They can't say - Sorry it isn't our job!
Who is going to take care on that???
Philosophers for one.  It's their job after all.

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I have tried to find by Google: "How the universe was created?"
There are many explanations about it - all of them are based on the BBT.
They're not all BBT links.

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This is misleading information.
Google misinterpreted your question and linked sites that say how galaxies and such were created from the early state, not how the universe itself was created.  Go down the list.  There are creation suggestions if you look for them.
I did the search and 5th one down is Stephen Hawking (a scientist) doing philosophy:
http://www.hawking.org.uk/the-origin-of-the-universe.html
That mentions the big bang at some point, but doesn't suggest it as an explanation of how the universe comes to exist.  I have opinions on the topic myself, but I don't think I could explain it to you.  Short story: I favor the relational interpretation of QM which handily solves the problem.

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Our scientists actually mislead all the people in our planet about the BBT, as it does not deal with the most important part of the creation - Mass and Energy.
Hence, how can they call it a theory of "creation?"
I don't think anyone bills it as a theory of creation. Go to church to hear that one.

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If so, please advise them to look for a real theory of creation as Theory-D.
An idea that blatantly say X is true and false at the same time?  That's trivially falsified.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/04/2019 21:51:59
Unfortunately, you have missed the key point in this article

The part that you quoted is about a "solid sphere of uniform density" not a "hollow sphere". Gravity varies linearly from the center of a solid sphere, whereas there is no net force at all inside of a hollow sphere.

So, the Big Bang doesn't give any answer for the creation of the mass and energy in our Universe.

No. It was never supposed to.

Not even one gram of mass or extra energy!

No, because that would violate the first law of thermodynamics.

Therefore, do you agree that the BBT is only a theory of transformation?

Transformation and expansion, yes.

How our scientists could bypass that key element in the theory of creation our Universe.

Every single theory in existence that isn't about the origin of matter and energy are bypassing that. That isn't something unique to the Big Bang theory.

They have to answer how the total mass and energy had been created in our universe - Finite or Infinite!

Nobody has to answer anything, and it's not like you know where mass and energy ultimately came from either.

They can't say - Sorry it isn't our job!

They absolutely can say that. If their profession isn't about deducing the origins of matter and energy, they have no obligation to it.

Who is going to take care on that???

People have been wondering for ages, but until we figure out a way to perform the right kind of experiment, we may never know.

Hence, how can they call it a theory of "creation?"
They must call the BBT a theory of "transformation"!!!
Do you agree with that?

That depends on how you define the word "creation". A piece of art is often called a creation even though the matter and energy that make up the art piece were not created by the artist.

If so, please advise them to look for a real theory of creation as Theory-D.

(1) You can't call your idea a theory until it has been scientifically tested.
(2) How does your idea explain the existence of mass and energy? Be careful not to let your explanation violate the first law of thermodynamics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/04/2019 05:42:17
The part that you quoted is about a "solid sphere of uniform density" not a "hollow sphere". Gravity varies linearly from the center of a solid sphere, whereas there is no net force at all inside of a hollow sphere.
I have just read the article which you have offered:
Shell theorem is normally used to describe gravity, but it should work here too because radiation intensity falls off at the exact same rate as gravitational strength does (the inverse square law): https://en.wikipedia.org/wiki/Shell_theorem
In the article they show clearly that from inside the sphere each point get's different force:
"The magnitude of the gravitational field that would pull a particle at point P in the x direction is the gravitational field multiplied by cos(theta) where theta is the angle adjacent to the x axis.  In this case, cos(theta)=p/sqrt(p^2+R^2)"
If we:
"Integrating the field due to each thin disc from x=-a to x=+a with respect to x, and after careful algebra, beautifully yields Newton's shell theorem."
Hence, by setting the integration on all of those different forces at different locations we get that Newton's shell theorem.
However, we are dealing with the different forces at different locations:
You actually claim that we can compare the gravity forces to radiation:"radiation intensity falls off at the exact same rate as gravitational strength does (the inverse square law)"
So, it is very clear that we should get different radiation if we won't stay at the center of this finite Universe "shell".
If you think differently - please prove it.

I can think of models where there is an edge and we are right at it, and the CMB still isotropic.
Is it real???
If we stay at the far edge of the Universe, while we see half of the sky full with galaxies and the other half without even one star, how can we still get that isotropic radiation?
I really can't understand why do you insist to claim that there is no meaning for that isotropic radiation?
I can also think on a model which proves that 1+1=0.
We can always band the reality.
But, what is our benefit with that?
Why do you totally reject the idea that the Universe is/could be infinite?
Why are you both so afraid from infinite Universe?
What is the big disaster if our Universe is Infinite?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/04/2019 06:00:58
In the article they show clearly that from inside the sphere each point get's different force:
"The magnitude of the gravitational field that would pull a particle at point P in the x direction is the gravitational field multiplied by cos(theta) where theta is the angle adjacent to the x axis.  In this case, cos(theta)=p/sqrt(p^2+R^2)"
If we:
"Integrating the field due to each thin disc from x=-a to x=+a with respect to x, and after careful algebra, beautifully yields Newton's shell theorem."
Hence, by setting the integration on all of those different forces at different locations we get that Newton's shell theorem.
However, we are dealing with the different forces at different locations:
You actually claim that we can compare the gravity forces to radiation:"radiation intensity falls off at the exact same rate as gravitational strength does (the inverse square law)"
So, it is very clear that we should get different radiation if we won't stay at the center of this finite Universe "shell".
If you think differently - please prove it.

You are still making reference to a solid sphere, not a hollow shell. If you go to the part that says "Inside a shell", then you will see what I am talking about. To quote the article, "the net gravitational forces acting on the point mass from the mass elements of the shell, outside the measurement point, cancel out."

This video reveals the same thing. Skip to 1:35 :


To quote the narrator, "So now, we have the second part of Newton's shell theorem. It states that a homogeneous spherically symmetric shell exerts no gravitational force on objects within the shell."
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/04/2019 07:34:28
To quote the narrator, "So now, we have the second part of Newton's shell theorem. It states that a homogeneous spherically symmetric shell exerts no gravitational force on objects within the shell."
Sure, I fully agree with that.
However, all it says that the sum of all the forces from all the directions is zero.
But they don't say that the forces from all the directions are the same.
They say that one direction cancel the force from the opposite direction.
So, if we take one line (left/right) we should find that the force from the left side cancel the force from the right side.
If we take another line (up/down), we should find that the force from upwards cancel the force from downwards.
However, they didn't say that the forces from Left/right are equal with the forces from up/down.
So, do you agree that there is no prove for isotropic forces in this example?
However, we discuss on isotropic radiation/forces from all directions.
Can you please prove that the gravity isn't isotropic?

If no, do you agree that at any size of finite universe with spherically symmetric shell there is no way to get isotropic radiation unless we are located at the center?
Therefore, as the chance to be at the center is virtually zero, do you finally accept the idea that our universe must be infinite in order to get that kind of isotropic CMB radiation?



Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/04/2019 15:27:06
However, all it says that the sum of all the forces from all the directions is zero.
But they don't say that the forces from all the directions are the same.

That is exactly what it means. If the force from any one direction was greater than the force from another direction, then there would be a net force in one direction and thus shell theorem would be wrong.

So, if we take one line (left/right) we should find that the force from the left side cancel the force from the right side.
If we take another line (up/down), we should find that the force from upwards cancel the force from downwards.
However, they didn't say that the forces from Left/right are equal with the forces from up/down.

The forces from the left and right direction are zero and the forces from the up and down direction are zero. Zero is equal to zero, so they are equal.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/04/2019 15:57:35
The forces from the left and right direction are zero and the forces from the up and down direction are zero. Zero is equal to zero, so they are equal.
Sure.
But the forces from left/right are not equal to the up down directions.
For example
If the gravity force form the left is F1 than the force from right must be also F1 (at the opposite direction)
However, the force from up could be F2 which is different from F1. However the force from down must be F2 in order to cancel the force from up.
So, if you add all the forces you get Zero.
But it is clear that from any direction we could get different forces.
Therefore, the forces are not isotropic.
Is it clear to you by now?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/04/2019 19:13:13
I understand what you're asking.  Cancelling forces (net zero) does not imply isotropy.  It wasn't the purpose of the example to demonstrate isotropy.  But the same mathematics of the shell theorem can be used to demonstrate that the force/radiation is isotropic inside a shell, and you'd realize that if you understood how the proof works and didn't just focus on the words used in the conclusion.
So, what do you want to say by that?
I hope that you accept my approval by Newton's shell theorem that our universe must be infinite in order to get that kind of isotropic CMB radiation.

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Why do you totally reject the idea that the Universe is/could be infinite?
Where did I say this?
If you accept the idea that the universe is/could be infinite, then why do you continue to disapprove this idea?
What do you gain with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/04/2019 20:59:49
But the forces from left/right are not equal to the up down directions.

They have to be equal because you already posited a spherical Universe. There is nothing to cause the radiation coming from above you or below you to have a different intensity than the radiation coming at you from the sides.

If they weren't equal, then the microwave background would not look isotropic from any point in the Universe, not even in the middle. If the radiation coming from above you and below you was brighter than the radiation coming from your sides, this would be detectable anywhere. The fact that the radiation is observed to be isotropic means that this is not the case.

A hyperspherical universe or Big Bang universe would also explain an isotropic microwave background without having to rely on anything infinite in size. So your argument still does not work as proof.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/04/2019 05:45:56
Please quote one place where I (or Kryptid) deny that the universe could be infinite.  I'm simply denying that it must be, as you assert.
So, you still do not want to accept the idea that the Universe IS INFINITE.
You prefer to set it under "Could be" infinite, (or: "We don't know"?).
Sorry, you must know the clear answer.
Is it infinite or not?
"Could be" is nice first step but it's not good enough
I have proved that in order to get isotropic CMB radiation the universe MUST BE infinite.
If you are still denying that it "must be" infinite, then please prove it.
Please show why in a finite Universe we can still get isotropic CMB radiation - based on "shell theorem" (and only on shell theorem theory)" .
If you can't do so, would you kindly and finally write clearly that you agree that the Universe IS/MUST BE infinite!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 28/04/2019 06:12:27
So, you still do not want to accept the idea that the Universe IS INFINITE.
You prefer to set it under "Could be" infinite, (or: "We don't know"?).
I have no choice.  There is as of yet no conclusive way to falsify one or the other idea.  The CMB thing doesn't work because there are very viable finite models (the hypersphere one in particular) with perfect isotropy.

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I have proved that in order to get isotropic CMB radiation the universe MUST BE infinite.
You did no such thing.  As I said, the valid finite models predict isotropy, so the observed CMB is exactly as predicted.
Your model on the other hand (besides being a violation of several principles) doesn't predict one.  If the universe started out at one place (instead of everywhere) and spread out from there, it would be finite size after finite time (due to light speed limitation) and probably wouldn't have a CMB at all.

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If you are still denying that it "must be" infinite, then please prove it.
We did. We showed a model that is finite and matches observations.  That counterexample proves that a non-infinite universe is a possibility.

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Please show why in a finite Universe we can still get isotropic CMB radiation - based on "shell theorem" (and only on shell theorem theory)" .
I'll let Kryptid do that.  He brought it up.  The hypersphere model is isotropic due to symmetry and the shell theorem isn't a part of the argument.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/04/2019 07:39:03
Your model on the other hand (besides being a violation of several principles) doesn't predict one.  If the universe started out at one place (instead of everywhere) and spread out from there, it would be finite size after finite time (due to light speed limitation) and probably wouldn't have a CMB at all.
My modeling/theory or any other theory (including the BBT) isn't relevant to the shape of the real Universe.
Somehow, it seems to me that you try to band the evidences in order to meet your BBT theory.
I don't think that I have more wisdom than any average scientist.
Therefore, it is quite clear to me that based on that discovery any scientist will agree that the universe must be infinite.
But you don't want to accept it as it might contradict the BBT theory.
Therefore, you prefer to set it under "could be" and look for some unrealistic modeling which proves the opposite.

Please be aware that the black body radiation in the CMB is also a clear indication for Infinity Universe.
However, I assume that any evidence that contradicts the wishful list should be rejected.
Therefore, I hope that I will not have to discuss about the black body radiation as the isotropic idea will be good enough to prove that our Universe is infinite.

I'll let Kryptid do that.  He brought it up.  The hypersphere model is isotropic due to symmetry and the shell theorem isn't a part of the argument.

Why do you claim that: "the shell theorem isn't a part of the argument."?
How can reject that kind of important law? (Is it just because it doesn't meet the requested wishful list?)

So, yes please.
Although Kryptid is in your side, I have full confidence that he will answer based on "shell theorem" real evidence and not based on wishful list.
So, I'm waiting for Kryptid to let us know if there is a possibility to get isotropic radiation at a finite Universe (at any size and at any location in the Universe) under the "shell theorem" theory
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/04/2019 15:07:59
So, I'm waiting for Kryptid to let us know if there is a possibility to get isotropic radiation at a finite Universe (at any size and at any location in the Universe) under the "shell theorem" theory

I already did:

They have to be equal because you already posited a spherical Universe. There is nothing to cause the radiation coming from above you or below you to have a different intensity than the radiation coming at you from the sides.

If they weren't equal, then the microwave background would not look isotropic from any point in the Universe, not even in the middle. If the radiation coming from above you and below you was brighter than the radiation coming from your sides, this would be detectable anywhere. The fact that the radiation is observed to be isotropic means that this is not the case.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/04/2019 17:32:56
They have to be equal because you already posited a spherical Universe. There is nothing to cause the radiation coming from above you or below you to have a different intensity than the radiation coming at you from the sides.
I disagree with you
We see clearly that the net forces are based on distances
So, if we are located at the center we have exactly the same distance to all directions.
In this case, it is clear that the forces are fully isotropic.
However, once we move from the center, there is no balance between the Up/Down distance to the Left/right distance. Therefore, it is clear that for one we might get F1 force while from the other line we can get F2.
So, I really can't understand why you insist on isotropic forces at the shell theorem.
If they weren't equal, then the microwave background would not look isotropic from any point in the Universe, not even in the middle. If the radiation coming from above you and below you was brighter than the radiation coming from your sides, this would be detectable anywhere. The fact that the radiation is observed to be isotropic means that this is not the case.
I agree with you.
They have to be equal. But that could be only if our universe is infinite or we are located at the center.
 
However, I'm not going to argue about it anymore.
Let's move to the radiation spectrum.
Our scientists have found that the CMB has the most perfect blackbody spectrum in nature ever observed.
They have also calculated a redshift of 1100 in the CMB.
Therefore, I wonder how could it be that we get the same spectrum and the same redshift from all directions while in one side we are closer to the edge (or even right at the edge) of the Universe while in the other side we are far away?
So, how do we get the same CMB spectrum, same black body radiation and the same redshift from all directions as we get closer to one edge of the finite Universe?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/04/2019 17:50:31
We see clearly that the net forces are based on distances

It isn't based only on distance, but also on the amount of mass present at those differing distances. It is the finding of shell theorem that those two factors always perfectly balance each other out inside of a uniform, spherical shell such that the forces in all directions always add up to zero. I don't know why this is so hard for you to understand.

Therefore, it is clear that for one we might get F1 force while from the other line we can get F2.

I addressed that right here:

Quote
If they weren't equal, then the microwave background would not look isotropic from any point in the Universe, not even in the middle. If the radiation coming from above you and below you was brighter than the radiation coming from your sides, this would be detectable anywhere. The fact that the radiation is observed to be isotropic means that this is not the case.

So, how do we get the same CMB spectrum, same black body radiation and the same redshift from all directions as we get closer to one edge of the finite Universe?

Space expanding in all directions would cause an equal redshift at all locations in the Universe.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/04/2019 19:26:51
Space expanding in all directions would cause an equal redshift at all locations in the Universe.
https://en.wikipedia.org/wiki/Expansion_of_the_universe
"the expansion rate of the universe has been measured to be H0 = 73.24 ± 1.74 (km/s)/Mpc.[14] This means that for every million parsecs of distance from the observer, the light received from that distance is cosmologically redshifted by about 73 kilometres per second (160,000 mph)."
We see clearly that the local impact of the expansion is very minimal.
Hence:
1. If our distance to the near edge is R1 and the distance to the far edge is 1000R1, do you agree that the expansion impact of 1000R1 should be higher than R1 by 1000? If so, how could it be that we should get the same redshift in both directions?
2. If we are located only one Mpc from the edge, then the impact of the expansion is only 73Km/s. How that speed can set a redshift of 1100?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 28/04/2019 21:24:16
We see clearly that the local impact of the expansion is very minimal.
Hence:
1. If our distance to the near edge is R1 and the distance to the far edge is 1000R1, do you agree that the expansion impact of 1000R1 should be higher than R1 by 1000? If so, how could it be that we should get the same redshift in both directions?
The CMB is equidistant in all directions.  It is a shell centered on us.  Any CMB light that originated closer by has already passed by us.  Any further out hasn't got here yet.  Therefore the distance to the edge, in a model that has an edge at all, has zero effect on what is seen.

Quote
2. If we are located only one Mpc from the edge, then the impact of the expansion is only 73Km/s. How that speed can set a redshift of 1100?
1100 is approximately the ratio of the temperature at the decoupling event and what we measure from this distance, which has nothing to do with things one Mpc away now.  If we're currently one Mpc from 'the edge', then material at the edge would be moving away from us at 73Km/s.  What we see in such a model depends very much on the model.  The model I described in post 378 would result in no detectable anomaly in any direction regardless of how close we were to the edge.  That's a pretty far fetched model, but so is yours.  The hypersphere model doesn't have an edge.  The eternal inflation model has an edge under certain foliations, but no observer can be near it because the universe is too young there.  The latter two models are both viable finite universe models that predict CMB isotrophy.  The post-378 one predicts it as well, but it isn't a very viable model.  It asserts that there is a preferred location in space and we're not at that location.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/04/2019 23:11:39
1. If our distance to the near edge is R1 and the distance to the far edge is 1000R1, do you agree that the expansion impact of 1000R1 should be higher than R1 by 1000? If so, how could it be that we should get the same redshift in both directions?

Since we can't see any visible boundary to the visible universe, the visible universe must have a radius of at least slightly less than R1, The visible universe represents the limits of what we can see, so anything happening more than a distance of R1 away from us simply would not be visible and therefore could not impact our observations.

2. If we are located only one Mpc from the edge, then the impact of the expansion is only 73Km/s. How that speed can set a redshift of 1100?

We're not a mere 1 mega-parsec from the edge. We are about 46 billion light-years from the edge of the visible universe, so we are at least that far from any hypothetical absolute edge of the Universe as well.

Then there is another possibility I realized. Our visible universe could represent such an incredibly small portion of a gigantic (but finite) total universe that any anisotropies in the CMB that are present are too small for us to detect with current technology. It would be like trying to measure the curvature of the Earth with a school ruler.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/04/2019 06:42:23
Quote
1. If our distance to the near edge is R1 and the distance to the far edge is 1000R1, do you agree that the expansion impact of 1000R1 should be higher than R1 by 1000? If so, how could it be that we should get the same redshift in both directions?

Since we can't see any visible boundary to the visible universe, the visible universe must have a radius of at least slightly less than R1, The visible universe represents the limits of what we can see, so anything happening more than a distance of R1 away from us simply would not be visible and therefore could not impact our observations.
The CMB is equidistant in all directions.  It is a shell centered on us.  Any CMB light that originated closer by has already passed by us.  Any further out hasn't got here yet.  Therefore the distance to the edge, in a model that has an edge at all, has zero effect on what is seen.

Sorry, you both don't answer my question.
Space expanding
You have stated that Space expanding in all directions would cause an equal redshift at all locations in the Universe:
Space expanding in all directions would cause an equal redshift at all locations in the Universe.
I wonder how the Space expanding can set the same redsift to different distances and different velocities due to the impact of space expansion.
1. How could it be that we get exactly the same redshift value from the near edge of the universe which is R1 and from the furthest edge which is higher by 1000 times than R1 (without the impact of space expansion)?
2. If we add the impact of space expansion, that it is clear that the edge which is located 1000 times R1 from us will move even further away at higher velocities. That should even make it much more difficult to get the same redshift. So, how can you claim that the space expansion adjust them all to the same level?
3. https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects
List of the most distant astronomical objects
We see that GN-z11 galaxy is located at a distance of 13.39   billion LY away and its redshift is only z = 11.09.
So, why we get exactly Z = 1400 in the CMB, while for a galaxy that is located at the furthest location (13.39 billion LY)  in our Universe we only get 11.9?
4. If in one side our distance to the edge is only 13.39 billion LY =R1, what do you think should be the CMB redshift?
Do you agree that the radiation amplitude of the furthest galaxy gets to our location at the minimal value (as 1/R^2).
So, if we look directly in this direction we should get radiations from all the galaxies which are located in that line. those galaxies has lower redshift but higher radiation amplitude.
Hence, we should see in the CMB (at the direction of GN-z11 galaxy) the sum of all the radiations from all the galaxies/objects that are located in this direction.
As the furthest galaxy has a minimal radiation amplitude with a redshift of 11.9, while closer galaxies with much more radiation amplitude have a lower redshift (lower even than 1), do you agree that the average redshift should actually be much lower than 11.9. (could it be 2 or even less than 1?).
If so, how could it be that we get a redshift of 1400 in the CMB in all directions? Why not 5 or 11.9?
Do you agree that this redshift value of 1400  is a real enigma for any finite Universe?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/04/2019 07:52:26
Sorry, you both don't answer my question.

Yes I did. Anything further away from us than R1 is too far away to see, so it can't impact our observations.

You have stated that Space expanding in all directions would cause an equal redshift at all locations in the Universe:

Yes, but not all radiation that we see has traveled the same distance to reach our telescopes. Radiation that has traveled further would be more redshifted because it has been moving through an expanding space for a longer period of time.

I wonder how the Space expanding can set the same redsift to different distances and different velocities due to the impact of space expansion.

There is a subtle difference that needs to be pointed out. The rate that space is expanding is the same at any given location in the Universe (as far as we can tell, anyway). So the amount of stretching that any given photon experiences over any given measure of time should be the same for every location in the Universe as well. However, what is different is that different photons are given different amounts of time to stretch because some travel further than others before reaching our telescopes (as I explained before). So the redshift we observe is greater for objects of greater distance from us.

1. How could it be that we get exactly the same redshift value from the near edge of the universe which is R1 and from the furthest edge which is higher by 1000 times than R1 (without the impact of space expansion)?

We can't see past R1 so what is happening 1,000 times further away isn't visible to us. I say that we can't see past R1 because the edge of the total Universe must be somewhere outside of the observable universe (otherwise we could detect that edge). So R1 must be at least slightly beyond the edge of the observable universe, putting R1000 just that much further beyond our observation abilities.

2. If we add the impact of space expansion, that it is clear that the edge which is located 1000 times R1 from us will move even further away at higher velocities. That should even make it much more difficult to get the same redshift. So, how can you claim that the space expansion adjust them all to the same level?

Because we can't see past R1 at all.

3. https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects
List of the most distant astronomical objects
We see that GN-z11 galaxy is located at a distance of 13.39   billion LY away and its redshift is only z = 11.09.
So, why we get exactly Z = 1400 in the CMB, while for a galaxy that is located at the furthest location (13.39 billion LY)  in our Universe we only get 11.9?
4. If in one side our distance to the edge is only 13.39 billion LY =R1, what do you think should be the CMB redshift?
Do you agree that the radiation amplitude of the furthest galaxy gets to our location at the minimal value (as 1/R^2).
So, if we look directly in this direction we should get radiations from all the galaxies which are located in that line. those galaxies has lower redshift but higher radiation amplitude.
Hence, we should see in the CMB (at the direction of GN-z11 galaxy) the sum of all the radiations from all the galaxies/objects that are located in this direction.
As the furthest galaxy has a minimal radiation amplitude with a redshift of 11.9, while closer galaxies with much more radiation amplitude have a lower redshift (lower even than 1), do you agree that the average redshift should actually be much lower than 11.9. (could it be 2 or even less than 1?).
If so, how could it be that we get a redshift of 1400 in the CMB in all directions? Why not 5 or 11.9?
Do you agree that this redshift value of 1400  is a real enigma for any finite Universe?

The reason is because the light from those distant galaxies was emitted when the Universe was relatively old (about 1,000,000,000 years after the Big Bang), whereas the light that would become the CMB was emitted when the Universe first became translucent (only about 379,000 years after the Big Bang). So the CMB has been experiencing redshift for far, far longer than the light from any galaxy has.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/04/2019 12:46:20
Dear Kryptid

I really don't understand how do you set the contrediction:
In one hand you claim that:
Radiation that has traveled further would be more redshifted because it has been moving through an expanding space for a longer period of time.

Based on this answer it is clear to me that the radiation that has traveled further (from 1000 times R1) would be more redshifted than the radiation that traveled only one R1.
So, how could it be that we get the same radiation from a distance of R1 and 1000 times R1.
I read your following explanation and I still don't understand:
You claim that:
We can't see past R1 so what is happening 1,000 times further away isn't visible to us. I say that we can't see past R1 because the edge of the total Universe must be somewhere outside of the observable universe (otherwise we could detect that edge). So R1 must be at least slightly beyond the edge of the observable universe, putting R1000 just that much further beyond our observation abilities.
So, I agree that R1 represents the nearby edge of our universe and must be at least slightly beyond the edge of the observable universe.
However, we know that on the other side of the Universe, our distance to the edge of the Universe is 1000 times R1. So, why in that direction we don't get higher redshift as the distance is longer by 1000 times?
How it could be that we get a redshift of 1400 if the edge is just slightly beyond the edge of the observable universe or if it 1000 times longer?
Actually, if we get a redshift of 1400 from a far end galaxy, can we calculate/extract the estimated distance to this galaxy?
If so, what is the distance that redshift of 1400 represents?
If we get the same redshift from all directions, why we can't assume that we are located just at the center of the Universe?


 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/04/2019 19:59:53
The visible universe represents the limits of what we can see
...

We are about 46 billion light-years from the edge of the visible universe, so we are at least that far from any hypothetical absolute edge of the Universe as well.
I am going to protest the top statement.  The visible universe represents the current proper distance of the furthest material that could ever have had a causal effect on our current location.  That by no means says we can see that far.  The event horizon is only about a third that distance and anything beyond that cannot have an effect here ever, so that's the absolute limit of how far we can see if we're willing to wait forever.

The light from the CMB is the furthest we can see, and it was emitted a scant ~1.3 million light years (proper distance) from the comoving location corresponding to here.  The journey from there to us/here/now took it considerably further away than that, but no more than say a single digit of BLY away (proper distance again).  That's the furthest we can see, which is well inside the Hubble sphere.  If that light's journey took it to the edge of the universe, it would presumably be affected by that.  We'd see it.  We cannot see any further away than that.
So R1 is not very far at all, no more than 20% of that 46 BLY radius of the 'visible universe'.

To illustrate:
3. https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects
List of the most distant astronomical objects
We see that GN-z11 galaxy is located at a distance of 13.39   billion LY away and its redshift is only z = 11.09.
So, why we get exactly Z = 1400 in the CMB, while for a galaxy that is located at the furthest location (13.39 billion LY)  in our Universe we only get 11.9?
You will notice a little mark on the collumn where this 13.39 'location' is listed.  It says there that the number has no direct physical significance.  We can't see that far.  The object isn't that far.  The time it took light to get from there to here is not 13.39 billion years in any meaningful frame.  It isn't proper distance.
The CMB is considerably further away than that, yet is not listed as it isn't a distinct galaxy or other object.  And yet the CMB light was emitted only 1.3 million (not billion) light years away.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/04/2019 20:38:53
Based on this answer it is clear to me that the radiation that has traveled further (from 1000 times R1) would be more redshifted than the radiation that traveled only one R1.

There hasn't been enough time since the Big Bang for any radiation to travel to that far.

ased on this answer it is clear to me that the radiation that has traveled further (from 1000 times R1) would be more redshifted than the radiation that traveled only one R1.
So, how could it be that we get the same radiation from a distance of R1 and 1000 times R1.

We don't because we can't see any radiation from beyond R1.

So, why in that direction we don't get higher redshift as the distance is longer by 1000 times?

Because we can't see any radiation from beyond R1. How do you possibly expect to measure radiation that hasn't even made it to you?



How it could be that we get a redshift of 1400 if the edge is just slightly beyond the edge of the observable universe or if it 1000 times longer?
Actually, if we get a redshift of 1400 from a far end galaxy, can we calculate/extract the estimated distance to this galaxy?
If so, what is the distance that redshift of 1400 represents?

I don't think there are any galaxies in the observable universe with a redshift anywhere near that high, but you can estimate distances to galaxies based on redshift. I don't know what the equation involved is, though.

If we get the same redshift from all directions, why we can't assume that we are located just at the center of the Universe?

We are located in the center of the observable universe. Since the speed of light is the same in all directions, this means that the distance we can see in all directions is equal. Where this observable universe sits inside the total Universe is unknown.

I am going to protest the top statement.  The visible universe represents the current proper distance of the furthest material that could ever have had a causal effect on our current location.  That by no means says we can see that far.  The event horizon is only about a third that distance and anything beyond that cannot have an effect here ever, so that's the absolute limit of how far we can see if we're willing to wait forever.

The light from the CMB is the furthest we can see, and it was emitted a scant ~1.3 million light years (proper distance) from the comoving location corresponding to here.  The journey from there to us/here/now took it considerably further away than that, but no more than say a single digit of BLY away (proper distance again).  That's the furthest we can see, which is well inside the Hubble sphere.  If that light's journey took it to the edge of the universe, it would presumably be affected by that.  We'd see it.  We cannot see any further away than that.
So R1 is not very far at all, no more than 20% of that 46 BLY radius of the 'visible universe'.

Alright.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/04/2019 22:30:08
I don't think there are any galaxies in the observable universe with a redshift anywhere near that high, but you can estimate distances to galaxies based on redshift. I don't know what the equation involved is, though.
I don't know the equation offhand either, but if you look at the redshift figures for the list of 'furthest galaxies' linked, you notice that the whole list has not much distance variance, but the redshift factor number goes up dramatically for the entries at the top of the list.   The number apparently come from the Lambda-CDM model, a plot (from wiki) appearing here:
https://upload.wikimedia.org/wikipedia/commons/thumb/a/a7/Distance_compared_to_z.png/400px-Distance_compared_to_z.png

Notice that a galaxy at 13.4 GLY (that funny figure that corresponds to nothing physical) plots nicely at 11, and the CMB with z=1100 seems to be 46 GLY away by that measure.
That's contradictory to what I've read about most distant objects since they've measured a quazar at something like 22 GLY away, meaning that object is now that far away, but within our event horizon back when the light reaching us now was first emitted.  So I don't know exactly what is being measured on the vertical axis of that graph.
Notice that the graph levels off at 46 GLY, with z approaching indefinite values as distance approaches the 'size of the visible universe'.  If we could see through the CMB barrier, we'd observer redshifts far greater than 1100.
The red line is the Hubble red shift, and that goes to infinite z at the Hubble radius.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/04/2019 14:15:28
I don't know the equation offhand either, but if you look at the redshift figures for the list of 'furthest galaxies' linked, you notice that the whole list has not much distance variance, but the redshift factor number goes up dramatically for the entries at the top of the list.   The number apparently come from the Lambda-CDM model, a plot (from wiki) appearing here:
https://upload.wikimedia.org/wikipedia/commons/thumb/a/a7/Distance_compared_to_z.png/400px-Distance_compared_to_z.png
Notice that a galaxy at 13.4 GLY (that funny figure that corresponds to nothing physical) plots nicely at 11, and the CMB with z=1100 seems to be 46 GLY away by that measure.
Thanks Halc
This is very interesting data.
Hence, the CMB with z=1100 seems to be 46 GLY away by that measure!
So, why do you claim that the CMB was emitted only 1.3 million (not billion) light years away.
The CMB is considerably further away than that, yet is not listed as it isn't a distinct galaxy or other object.  And yet the CMB light was emitted only 1.3 million (not billion) light years away.
I really don't understand that contradiction.
You also see the contradiction, but I couldn't understand what do you really mean:
That's contradictory to what I've read about most distant objects since they've measured a quazar at something like 22 GLY away, meaning that object is now that far away, but within our event horizon back when the light reaching us now was first emitted.  So I don't know exactly what is being measured on the vertical axis of that graph.
Notice that the graph levels off at 46 GLY, with z approaching indefinite values as distance approaches the 'size of the visible universe'.  If we could see through the CMB barrier, we'd observer redshifts far greater than 1100.
The red line is the Hubble red shift, and that goes to infinite z at the Hubble radius.
Actually, our scientists claim that the CMB is evidence for the BBT.
So, if the Big bang had been set 13.8 Billion years ago, how could it be that the CMB which is consider as a product of the BBT comes from a distance of 46GLY?
Actually, if we see today a radiation which had been emitted from a distance of 46GLY, don't you think that it proves that our real universe should be much bigger than the estimated size of the observable Universe?
If so, how can we fit that size of the Universe in only 13.8 BY?
Do you think that it could set a contradiction in the BBT?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/04/2019 15:48:08
Hence, the CMB with z=1100 seems to be 46 GLY away by that measure!
So, why do you claim that the CMB was emitted only 1.3 million (not billion) light years away.
Read the post where I said that. The thing 45 GLY away is not the CMB anymore.  It is a collection of galaxies that it has become after 13.8 billion years just like it did here. We cannot see that. We can only see light that is here, now. The CMB that we see is redshifted radiation that was emitted 1.3 million proper LY away, back when the universe was 379,000 years old. It is a perfect shell of uniform radius at any given moment, and hence appears totally uniform to us due to symmetry. The CMB we see is isotropic because we're at the center of it.

Quote
Quote from: Halc
The CMB is considerably further away than that, yet is not listed as it isn't a distinct galaxy or other object.  And yet the CMB light was emitted only 1.3 million (not billion) light years away.
I really don't understand that contradiction. You also see the contradiction, but I couldn't understand what do you really mean:
I didn't make any contradiction.  The 'distant' galaxies (or at least the material that would become them) were closer than 1.3 million LY away back at the same time that the CMB light we see now was emitted.  Everything was closer back then.  That's what expansion of space means.

Quote
Actually, our scientists claim that the CMB is evidence for the BBT.
Of course.  The BBT predicts it.  Other theories have to find some other way to explain it.

Quote
So, if the Big bang had been set 13.8 Billion years ago, how could it be that the CMB which is consider as a product of the BBT comes from a distance of 46GLY?
It doesn't come from that distance.  It comes from 1.3 million LY away.  45GLY is where that material is now, assuming (possibly incorrectly) that the Lambda-CDM model figures its distances that way. 46GLY is even further, but light from there is blocked by the hot plasma of the early universe. I'm not particularly familiar with the Lambda-CDM model, but it is apparently the computation used when asserting the 46 GLY radius of the visible universe.
Remember, I cannot even see the moon right now.  I see some past image of it.  My ability to see into the present is zero.  Light doesn't travel infinitely fast, so I can see nothing in its current state.

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Actually, if we see today a radiation which had been emitted from a distance of 46GLY,
We can't.  Light from that far away will not get here ever, even after infinite time.  The event horizon is only ~16GLY away.  Anything beyond that can never be seen.

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don't you think that it proves that our real universe should be much bigger than the estimated size of the observable Universe?
It is improbable indeed that the universe is as small as that, but since the furthest we can see (now) is under 10GLY, a hyperspherical universe of radius 1.6GLY would actually look like what we see.  That radius is 30x smaller than 46GLY, but I think perhaps the curvature of such a tight universe might be noticed.

It is an interesting exercise to measure curvature.  Suppose you are on an non-spinning an non-orbiting Earth and you posit Earth is flat and infinite. Your argument seems to be that if it were finite and we were not at the exact center, we'd notice the current of the water as it migrates to the edge and falls off. You do all these calculation showing the improbability of being closer to the center than not. But you never consider a round planet.
How might you measure that? If its a small radius, you might just travel a short ways and notice that the stars have moved, but what if you were confined to one city block?  You can draw a triangle on a parking lot and notice that the angles don't add up to 180 (if you're super accurate with your measurements), but what if the the other angles of the triangle are out of reach?  Such is the difficultly of measuring curvature from one point.

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If so, how can we fit that size of the Universe in only 13.8 BY?
I managed to cram it into a ninth that size.  No, I don't propose the universe is that small, but that's about the limit before we'd see an anomaly in the CMB.
Quote
Do you think that it could set a contradiction in the BBT?
BBT doesn't assert a finite universe that small, so of course that wouldn't contradict it.  It doesn't assert a finite or infinite universe at all.  If it did, I suppose there would be a stronger opinion of which is correct.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/05/2019 06:43:21
The 'distant' galaxies (or at least the material that would become them) were closer than 1.3 million LY away back at the same time that the CMB light we see now was emitted.  Everything was closer back then.  That's what expansion of space means.
Thanks
I don't understand how a redshift which we are using for galaxies can't also be used for the CMB?
A redshift is a redshift. It comes as an information in the radiation. It comes from any kind of matter. We can call this matter: dust, Gas cloud, stars, galaxies... (What about CMB radiation from dark matter or dark energy?)
So, if we claim that a radiation from a galaxy with a redshift of 1100 represents a distance of  46GLY, why a radiation which we call CMB with a redshift of 1100 doesn't represents a radiation which had been emitted from a distance of 46GLY?
Few more questions:
1. Do you mean that the whole mass of the Universe were at some point of time at a maximal distance of only 1.3 Million LY from each other?
2. If so, do you agree that there is no way to set an infinite Universe in only 13.8 BY?
3. In this case, why do you claim that we don't know if the universe is finite or infinite. Do you agree that based on the BBT the Universe must be finite? If we will discover that the Universe is infinite, than what do you prefer: a problem with the discovery or a problem with the BBT?
4. If we could eliminate completely the impact of the BBT on the CMB
What is the expected CMB that we should get with regards to Amplitude, Redshift and isotroic?
5. Do we have any idea what is the estimated amplitude degradation in the CMB per one million year?
6. If we could come back to our Universe in one Billion or 10 Billion years from now, then what kind of CMB we might find?
7. Can we extract from the CMB the total mass of the whole Universe?
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 01/05/2019 11:05:05
Quote from: Dave Lev
I don't understand how a redshift which we are using for galaxies can't also be used for the CMB?
The CMB originated from a soup of hot atomic gas (mostly Hydrogen, some Helium) which had just cooled below 3000K, and become transparent to visible light.

It should not be possible for stars to condense when temperatures are this high, so we don't expect to find galaxies of stars of the same age of the CMB. So the first galaxies (pinwheels of stars) are expected to have a lower redshift than the CMB.

Footnote: Astronomers have seen that the black holes at the core of quasars formed surprisingly early in the early universe, and it is not clear what pulled these supermassive black holes together. It is conceivable that these cores were formed during the Big Bang, started like a giant accretion disk of gas; galaxies of stars later formed around them.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/05/2019 12:56:52
I don't understand how a redshift which we are using for galaxies can't also be used for the CMB?
They don't use a different kind of redshift for different things.  Galaxies are closer, so they're less redshifted, as evan_au points out.  Galaxies are things and have defined distances from us.  The CMB is not a thing or an object and has no distance to us.  It is a single flash of light that occurred everywhere (even here).  It left only the light and no object from which it came.  So there is no meaningful distance to the CMB since there is no CMB to which one might travel.

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A redshift is a redshift. It comes as an information in the radiation. It comes from any kind of matter. We can call this matter: dust, Gas cloud, stars, galaxies... (What about CMB radiation from dark matter or dark energy?)
CMB is radiation from normal plasma/hydrogen, not dark anything.  Yes, it is redshifted like anything else that was emitted at a significant distance.

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So, if we claim that a radiation from a galaxy with a redshift of 1100 represents a distance of  46GLY, why a radiation which we call CMB with a redshift of 1100 doesn't represents a radiation which had been emitted from a distance of 46GLY?
Nothing that distant can be seen.  It is too far away.  Please read what I said in the prior posts about this.

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Few more questions:
1. Do you mean that the whole mass of the Universe were at some point of time at a maximal distance of only 1.3 Million LY from each other?
1.3 MLY is the distance from here that the CMB we see now was emitted.  That was never stated to be the radius of the whole universe at that time.

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2. If so, do you agree that there is no way to set an infinite Universe in only 13.8 BY?
No.

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3. Do you agree that based on the BBT the Universe must be finite?
No.

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4. If we could eliminate completely the impact of the BBT on the CMB
What is the expected CMB that we should get with regards to Amplitude, Redshift and isotroic?
The question makes no sense.  The CMB is the big bang itself we are seeing.  Without the BB, there would be no CMB.

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5. Do we have any idea what is the estimated amplitude degradation in the CMB per one million year?
It isn't a fixed value per million years, but is very predictable. I don't know the current rate.  A million years is about a 14000th the age of the universe, so I imagine it will lower the CMB temperature somewhere around the 5th digit in or thereabouts.

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6. If we could come back to our Universe in one Billion or 10 Billion years from now, then what kind of CMB we might find?
3000K if you go all the way back to when it was emitted.

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7. Can we extract from the CMB the total mass of the whole Universe?
No.  We don't know the size and the CMB is only a temperature of a shell of material at a fixed radius, centered on us.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/05/2019 13:47:31
Here is a good image illustrating all the concepts we're talking about.
(https://i.stack.imgur.com/6yzwk.jpg)

The upper graph shows proper distance vs time, and the 'visible universe' is seen to be where the particle horizon crosses the 'now' line (which is defined as the age of the universe for a comoving ('stationary') object).  A comoving object is one for which the universe appears to be isotropic.  The solar system motion varies in the long term, but we're currently about 0.0013c from being comoving.  It's more than twice that when we get to the other side of the galaxy.
The lower graph shows the same thing in comoving coordinates, and the black dotted worldlines are just vertical lines in that view.

The event horizon (orange) delimits events which can never have a causal effect on us here even given infinite time.  The event horizon would not be there if the expansion of the universe was not accelerating.

The Hubble Sphere (purple) is the line where comoving objects are increasing their proper distance from us at light speed.  It would be straight if the expansion of space were not accelerating.

The red line is the light cone and represents what we can see.  Every event (every star in the sky, the moon, the CMB and the beer you're holding) is on that red line.  It seems to go furthest to the right at about 3.5 BY and reaches the 5.5 BLY mark.  That's as far as we can see.  R1 is 5.5 BLY.  If the universe has an edge and looks different there, the red line needs to cross it for us to notice it.

The CMB is on that red line and so low on the chart that you cannot make out how close it is to here.  I figure it to be around 1.3 MLY away, but I did not actually read that number anywhere.  The best way to figure that is to use the lower graph.

The black dotted lines are worldlines for comoving objects, and those within the particle horizon touch the red line, and those outside it do not.  The red line is 'what we see', so this difference defines 'the observable universe'.

You will notice that they don't even bother drawing a line to represent events simultaneous with us in our inertial frame.  I challenge you to plot that.  The inability to do so illustrates why special relativity is special and does not apply to our universe.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/05/2019 18:25:06
Thanks Halc & Evan_au

I do appreciate the answers.

Let me ask the following:
Is there any way, any evidence, any discovery any issue which could convince you that there is a problem with the BBT?

3000K if you go all the way back to when it was emitted.
If we could travel in time:
If we could verify that at the early days (about 13 Billion years ago) the CMB was exactly as it is today, while also 10 Billion years from now in the future, the CMB is also the same.
What can we learn from that?

The CMB is the big bang itself we are seeing.  Without the BB, there would be no CMB.

Why are you so sure that without the Big bang there is no CMB?
For example:
Thermal emission of dust in the Milky way:
https://irsa.ipac.caltech.edu/applications/DUST/docs/background.html
"The dust temperature varies from 17 K to 21 K, which is modest but does modify the estimate of the dust column by a factor of 5".
So, the Milky Way has a thermal emission.
Let's assume that we could set the whole Milky way in some sort of closed sphere or galactic Oven.
In this case, what would be the thermal radiation amplitude in that galactic oven or closed sphere?
Actually, Oven could be a perfect example:
When the door is closed and we operate the oven, we can get over than 220c in the oven. However, once we open the door, the temp goes immediately down. We can set our hand in the oven and we won't feel that supper high temp.
So, I wonder what might be the "CMB" of the Milky Way if we could set it in some sort of a galactic oven.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/05/2019 19:07:10
Let me ask the following:
Is there any way, any evidence, any discovery any issue which could convince you that there is a problem with the BBT?
...
If we could travel in time:
If we could verify that at the early days (about 13 Billion years ago) the CMB was exactly as it is today, while also 10 Billion years from now in the future, the CMB is also the same.
What can we learn from that?
That would be pretty strong evidence for some sort of steady state theory.  If all the galaxies were younger and closer together 13 BY ago, then the steady state theory would be wrong as well.

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Why are you so sure that without the Big bang there is no CMB?.
There is a CMB.  It is an empirical fact.  So another theory that accounts for it (and everything else) is another contender.

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For example:
Thermal emission of dust in the Milky way:
https://irsa.ipac.caltech.edu/applications/DUST/docs/background.html.
"The dust temperature varies from 17 K to 21 K, which is modest but does modify the estimate of the dust column by a factor of 5".
So, the Milky Way has a thermal emission..
Sure.  So does the sun.  It looks nothing like the CMB.  The picture at the top of that website looks nothing like the CMB picture.

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Let's assume that we could set the whole Milky way in some sort of closed sphere or galactic Oven.
In this case, what would be the thermal radiation amplitude in that galactic oven or closed sphere?.
If you put part of it in a closed sphere like that it would get warm in there and there would be no thermal radiation to the outside because the enclosure would reflect it back in.  Removing the enclosure would be something like opening an oven door, yes.

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So, I wonder what might be the "CMB" of the Milky Way if we could set it in some sort of a galactic oven.
It would be the Milky Way light background (MWLB).  It has no thermostat like an oven so it would just keep getting hotter as energy is converted to heat.  The galaxy would also be less massive because the enclosure would prevent any new fuel from getting in, so the temperature seems to depend partly on how much it has when you seal it off, and how long you let it cook.

Concerning the graph I posted:
Quote from: Dave Lev
So, if we claim that a radiation from a galaxy with a redshift of 1100 represents a distance of  46GLY, why a radiation which we call CMB with a redshift of 1100 doesn't represents a radiation which had been emitted from a distance of 46GLY?
Galaxies are objects, and almost all of them are nearly comoving.  Objects have vertical lines in the 2nd graph.  There is a comoving distance to each of them.
The CMB is not an object, and so does not have a vertical line.  It is a horizontal line (it happened everywhere) in either graph, at the T=379,000 mark, which is at the bottom. It doesn't exist now because now isn't year 379,000. What we see now is anything at (not within) the red line.  We can see a galaxy only if the vertical black worldline of the galaxy object intersects the red line.  We can see now the CMB at exactly the one point where the red line intersects the horizontal CMB line, which in the first picture is at about 1.3 million light years.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/05/2019 22:41:48
Why are you so sure that without the Big bang there is no CMB?

The CMB is extremely uniform (temperature fluctuations from one place to another amount to a mere + 0.00335 kelvins), which means that it isn't radiation emitted by localized sources like stars or galaxies. It must have been emitted by something that once evenly filled all of space.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/05/2019 06:19:50
Black Body Radiation:

The CMB is extremely uniform (temperature fluctuations from one place to another amount to a mere + 0.00335 kelvins), which means that it isn't radiation emitted by localized sources like stars or galaxies. It must have been emitted by something that once evenly filled all of space.
Wow
That exactly the message which I was looking for.
You claim that:"It must have been emitted by something that once evenly filled all of space"
I would like to add that: It is emitted by something that filled all of space.
In order to understand that, let's look at the Black body radiation.
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/01%3A_The_Dawn_of_the_Quantum_Theory/1.1%3A_Blackbody_Radiation_Cannot_Be_Explained_Classically
It is stated clearly:
"A body emits radiation at a given temperature and frequency exactly as well as it absorbs the same radiation"
Therefore:
"Blackbody radiator is any object that is a perfect emitter and a perfect absorber of radiation."
"So how do we construct a perfect absorber in the laboratory? OK, nothing’s perfect, but in 1859 Kirchhoff had a good idea: a small hole in the side of a large box is an excellent absorber, since any radiation that goes through the hole bounces around inside, a lot getting absorbed on each bounce, and has little chance of ever getting out again. So, we can do this in reverse: have an oven with a tiny hole in the side, and presumably the radiation coming out the hole is as good a representation of a perfect emitter as we’re going to find (Figure  1.1.2 )."
They also have used the same idea of Oven as I did: "have an oven with a tiny hole in the side"
So, if I understand it correctly, in order to get from a black body radiation, all/most of the radiation must stay at the object.
However, once we take it out from the "oven", there is no black body radiation any more.
So, Let's go back to the following statement:
It must have been emitted by something that once evenly filled all of space.
Therefore, at the early time the Big bang took all the early available space. Therefore at the first moment the radiation of the Bang was clearly black body.
However, once we start the inflation and the expansion we actually kill any possibility for black body radiation.
As we contradict the basic idea of "perfect absorber of radiation":
"Blackbody radiator is any object that is a perfect emitter and a perfect absorber of radiation"
Therefore I have stated: "It is emitted by something that filled all of space."
So, in order to get the black boday radiation in the CMB, it must also filled all of space today
Therefore, the idea of the inflation and expansion actually contradicts the possibility for black body radiation.
So, the only way to get this black body radiation is by setting the object in an oven (As I have used in my example).
Therefore, let me go back to my example:

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Let's assume that we could set the whole Milky way in some sort of closed sphere or galactic Oven.
In this case, what would be the thermal radiation amplitude in that galactic oven or closed sphere?.
If you put part of it in a closed sphere like that it would get warm in there and there would be no thermal radiation to the outside because the enclosure would reflect it back in.  Removing the enclosure would be something like opening an oven door, yes.
So, you agree that we should get some thermal radiation if we put the Milky way at a galactic oven.
I assume that we will also get a black body radiation - as we fulfill the requirement for:
"Blackbody radiator is any object that is a perfect emitter and a perfect absorber of radiation"
I do not claim that it is feasible to set the Milky way in an oven, but I would like you to look at the impact of this hypothetical activity.
Therefore, do you agree that as long as the Milky Way will be in a galactic oven it will create a "CMB" which carry a black body radiation?
Do you also agree that once we open the oven, the black body radiation will be gone forever?
If so, how could it be that we still get a black body radiation from a CMB while the universe expands?
It seems to me that only if there is some envelope to our universe, than this might keep the black body radiation.
However, if the matter in our universe expands to the open infinity space - I really don't see any possibility to keep the black body radiation in the CMB.
Do you agree with that?
Do you agree that (based on hypothetical idea) if the Milky Way is in a galactic oven it can generate black body radiation?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/05/2019 07:11:55
However, once we start the inflation and the expansion we actually kill any possibility for black body radiation.

The radiation was emitted well after the inflationary epoch ended.

So, in order to get the black boday radiation in the CMB, it must also filled all of space today

The CMB does fill all of space today.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/05/2019 13:41:24
It is stated clearly:
"A body emits radiation at a given temperature and frequency exactly as well as it absorbs the same radiation"
Therefore:
"Blackbody radiator is any object that is a perfect emitter and a perfect absorber of radiation."

So, if I understand it correctly, in order to get from a black body radiation, all/most of the radiation must stay at the object.
I know 'understanding correctly' is not your forte, but you just directly contradicted the quote you gave, which says the black body emits as well as it absorbs.  The radiation does not stay at the object.

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Quote from: Kryptid
It must have been emitted by something that once evenly filled all of space.
Therefore, at the early time the Big bang took all the early available space.
It is space.  It doesn't take available space.
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Therefore at the first moment the radiation of the Bang was clearly black body.
It isn't an object, so isn't a body at all.  There is nowhere else to send the radiation.  There is no oven or barrier which has an inside and an outside.  That's what is meant by 'everywhere'.
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However, once we start the inflation
The start of expansion was 379000 years before, assuming we're talking about the decoupling event that sets up the CMB as we see it.
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"Blackbody radiator is any object that is a perfect emitter and a perfect absorber of radiation"
Therefore I have stated: "It is emitted by something that filled all of space."
So, in order to get the black boday radiation in the CMB, it must also filled all of space today
The CMB does fill all of space today. It isn't blackbody radiation because there isn't a body.  Only the light from the thing that happened everywhere remains.  All we detect is the light.  The plasma from which it was emitted is long gone.
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Quote from: Halc
If you put part of it in a closed sphere like that it would get warm in there and there would be no thermal radiation to the outside because the enclosure would reflect it back in.  Removing the enclosure would be something like opening an oven door, yes.
So, you agree that we should get some thermal radiation if we put the Milky way at a galactic oven.
See the part I bolded. No, I agreed that you would release pent up thermal radiation if you opened the oven.  Putting it in an oven would serve to block that radiation.  Yes, there would be radiation detected from the inside from the walls of the enclosure.  Perhaps this is what you're asking.

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I do not claim that it is feasible to set the Milky way in an oven, but I would like you to look at the impact of this hypothetical activity.
Therefore, do you agree that as long as the Milky Way will be in a galactic oven it will create a "CMB" which carry a black body radiation?
The enclosure is the black body, as viewed from inside?  If it radiates to the outside, is isn't really an enclosure, just a diffuser like frosted glass.  If there was a shell like you describe, it would get hotter in places that receive more radiation than others and the radiation from it would not appear isotropic.  For that, I think the Milky Way would need to be more spherically symmetrical instead of just circularly.
It isn't a model of the CMB because you are modeling a persisting finite object, something that isn't everywhere. Your model has radiation confined to inside an oven, which isn't everywhere.
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Do you also agree that once we open the oven, the black body radiation will be gone forever?
No.  Radiation from it would just keep on going unimpeded into space.  Light doesn't stop just because you turned off the source.  It keeps going until it hits something.
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If so, how could it be that we still get a black body radiation from a CMB while the universe expands?
The CMB is light, not the stuff from which it came.  We see it because that light is just now finding something to hit.  It isn't black body radiation.  It sort of was before year 379000.
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However, if the matter in our universe expands to the open infinity space
No model suggests this.  It is a naive model of the universe as being an object that 'started' at one location instead of everywhere.  Space and time are parts of the universe, not preexisting things in which the universe happened.
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- I really don't see any possibility to keep the black body radiation in the CMB.
Do you agree with that?
I never claimed it was black body radiation.  There's no body.  That would be a property of an object, and the universe is not an object, nor is the CMB.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/05/2019 15:35:34
The CMB does fill all of space today.
How could it be that the CMB fills all space?
If the Universe is finite - that statement might be incorrect as follow:
After any given moment the Universe is bigger. If it is bigger than the CMB covers more space over time. So how can you claim that it fills all space while we know that at any given moment it moves to new space?
As the CMB is radiation - do you agree that it must move at speed of light? Hence, it must move much faster than the expansion in space. While the real matter/galaxies are limited to approximately 70 km/s/Mpc, the CMB is moving at the speed of light.
So, do you agree that the CMB covers much more sphere than our real Universe?
Our scientists only focus on the hypothetical sphere of our real Universe, but what about the Hypothetical sphere of the CMB?
If they are identical - than do you agree that there must be an envelope to our Universe?
If there is envelope, I fully agree that the CMB can carry a black body radiation.
However, what kind of force keeps the CMB in our finite universe (without envelope)?
Why it isn't moving to the infinity in just few moments or years?
What is the possibility for any kind of envelope around our real Universe?
How the CMB could have a black body radiation without an envelope, while it moves to the infinity at the speed of light?

However, if the Universe is infinite - than I fully agree - "the CMB fills all space" to the infinity.
Actually, if the CMB is moving to the infinity at the speed of light, it is still in the infinity.
Therefore, it fills the whole infinite space.
In this case, it can keep its amplitude and it should have black body radiation
Hence, do you agree that the Idea that CMB must fill all of space today, proves that the Universe is infinite?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 03/05/2019 17:31:09
How could it be that the CMB fills all space?

Because it was emitted by a substance that once filled all of space (hydrogen plasma).

After any given moment the Universe is bigger. If it is bigger than the CMB covers more space over time. So how can you claim that it fills all space while we know that at any given moment it moves to new space?

The CMB expands with space.

As the CMB is radiation - do you agree that it must move at speed of light?

Yes.

So, do you agree that the CMB covers much more sphere than our real Universe?

That makes no sense. The Universe in total would represent all of the space there is. If there is a limited amount of space, then the CMB cannot move outside of that limited space (outside of space is an oxymoron).

Our scientists only focus on the hypothetical sphere of our real Universe, but what about the Hypothetical sphere of the CMB?

It's the same, because the CMB fills all of space.

If they are identical - than do you agree that there must be an envelope to our Universe?

That depends on what you mean by an "envelope". There doesn't have to be any kind of solid barrier around the Universe, if that is what you are saying. In a hypersphere universe, for example, space is closed and travel in any one direction will eventually bring you back to your starting point. That represents a limited amount of space but there are no physical barriers present at any point to stop your movement.

However, what kind of force keeps the CMB in our finite universe (without envelope)?

No force, just the fact that you can't move outside of a finite space because there is no space outside of space to move into (that should go without saying).

Why it isn't moving to the infinity in just few moments or years?

If there is no infinity to move to then obviously it can't move there.

What is the possibility for any kind of envelope around our real Universe?

I don't know, but it must be beyond what we can see if it is there.

How the CMB could have a black body radiation without an envelope, while it moves to the infinity at the speed of light?

I don't understand why you think these two points are contradictory.

Actually, if the CMB is moving to the infinity at the speed of light, it is still in the infinity.
Therefore, it fills the whole infinite space.

Unless there is no infinite space in the first place.

Hence, do you agree that the Idea that CMB must fill all of space today, proves that the Universe is infinite?

Absolutely not.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 03/05/2019 19:52:44
A few comments to add to Kryptid's replies, with which I agree.

So how can you claim that it fills all space while we know that at any given moment it moves to new space?
No viable model has anything moving to 'new space'.  The universe does not occupy a fraction of space.  It is the space.  All of it, be it finite or not.  The plasma once filled all of space, meaning there is nowhere where it wasn't.

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So, do you agree that the CMB covers much more sphere than our real Universe?
I want to echo Kryptid that this makes no sense.  There is nowhere that isn't the universe. That would mean there was a place where the plasma wasn't, but the model says it was everywhere.

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Our scientists only focus on the hypothetical sphere of our real Universe, but what about the Hypothetical sphere of the CMB?
You are attributing a naive strawman model to scientists. They focus on no such thing. Finite space is usually modeled as the surface of a hypersphere which has no space that is 'outside'

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However, what kind of force keeps the CMB in our finite universe (without envelope)?
There is nowhere else to go to get out. That's what finite space means.
You keep picturing an enclosure (delimiting 'universe') with space inside and more outside, but the stuff inside is not all of space then.  It isn't a model of finite space at all.

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What is the possibility for any kind of envelope around our real Universe?
Quote from: Kryptid
I don't know, but it must be beyond what we can see if it is there.
I cannot see it being meaningful. If the 'real' universe is confined to space inside the envelope, then the 'real' universe covers a subset of space.  That isn't a universe then, just an object (an isolated system) in a larger space.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/05/2019 06:34:01
The Universe in total would represent all of the space there is. If there is a limited amount of space, then the CMB cannot move outside of that limited space (outside of space is an oxymoron).
No viable model has anything moving to 'new space'.  The universe does not occupy a fraction of space.  It is the space.  All of it, be it finite or not.  The plasma once filled all of space, meaning there is nowhere where it wasn't.
Thanks for the explanation.
So, what is the difference between the Universe and space?
I had the impression that the space represents the infinity, as there is no limit in space, while the Universe represents the size of our universe. (Therefore the Universe can take the whole size of the open infinity space or just part of it).
Now I understand that the universe and the space are identical.
Thanks for the explanation.
.
So, let's assume that our Universe/space is finite and go back to year 400,000 - just after the inflation process while the CMB gets already its Black body radiation.
Let's assume that the radius of the Universe/space at that time was only R1 - which represents a compact universe/space.
All the matter of our current Universe had been there in this compact universe/space.
There was nothing outside of this radius.
Due to the expansion in space of 70 km/s/Mpc, the space of our universe had been increased and therefore after 13.4 Billion year from that moment, our current Universe/space get's to its maximal radius of R2.
(while there is still nothing outside of R2 radius).
However, as our Universe/space had been increased at a rate of 70 km/s/Mpc, the CMB was moving in all directions at the speed of light.
Sooner or later all the energy of the CMB must get to the edge of that compact universe while it increases its size at a relatively low speed of only 70 km/s/Mpc.
What should happen with the CMB when it gets to the edge of the Universe/space?
No force, just the fact that you can't move outside of a finite space because there is no space outside of space to move into (that should go without saying).
As there is no space outside the edge, it is clear that the CMB can't cross the edge.
So, is it going to be reflected back from the edge?
If it is reflected back, what kind of element in the edge of that increasing compact Universe force the CMB to do so?
You claim that there is no force and no element or solid barrier around the Universe.
There doesn't have to be any kind of solid barrier around the Universe, if that is what you are saying. In a hypersphere universe, for example, space is closed and travel in any one direction will eventually bring you back to your starting point. That represents a limited amount of space but there are no physical barriers present at any point to stop your movement.
But, the radiation must move in direct line to all directions.
So, how could it be that a radiation/light which starts at point A moves in a direct line to point B, cross that point while it moves in the same direction and eventually gets back to point A without meeting the edge of the Universe?
If that is correct, do you agree that when we look at the left side of the space/Universe, we could see the light/radiation which is coming from the right side of the finite space/universe?
Is it real?
There is nowhere else to go to get out. That's what finite space means.
You keep picturing an enclosure (delimiting 'universe') with space inside and more outside, but the stuff inside is not all of space then.  It isn't a model of finite space at all.
Yes, now I do understand the meaning of Finite space.
However, any finite space must have some sort of size. Therefore, I don't understand how a light can cross a limited space or Finite space without getting to the edge of the space.
If the Universe has a limited/finite size than somewhere there must be an edge, otherwise it should go to the infinity.
Therefore, I can't understand how a finite Universe with a radius R has no edge.
Once we set a sphere for the finite space, we actually also set the edge of that space/sphere.
So, how could it be that there is a finite space with a finite sphere without edge?
This is a real enigma for me.
Would you kindly explain it.
However, if the radius of the Universe is R and there is an edge, at some point of time the CMB must get to that edge.
So, if there is no solid barrier around the Universe how could it be that the CMB doesn't cross the edge?
Can we assume that it is reflected back?
Let's make a simple calculation:
The speed of light is 299 792.458 kilometers / second, while the speed of the expansion is only 70 km/s/Mpc
The ratio between the two is : 4285 : 1
Hence, it is clear that during the last 13.4 Billion years the CMB had been reflected back from the edge of the Universe/space at least million or billion times.
Even if we set at the edge a perfect mirror, there must be some minor energy reduction at every reflection.
So, after all of those reflections, don't you think that the CMB should be virtually zero by now?
How could it be that we still get any sort of energy in the CMB, while it must be reflected again and again from the edge of limited/finite Universe/Space size?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/05/2019 14:29:23
Quote from: Kryptid
The Universe in total would represent all of the space there is. If there is a limited amount of space, then the CMB cannot move outside of that limited space (outside of space is an oxymoron).
Quote from: Halc
No viable model has anything moving to 'new space'.  The universe does not occupy a fraction of space.  It is the space.  All of it, be it finite or not.
Thanks for the explanation.
So, what is the difference between the Universe and space?
For purposes of the discussion at hand, no difference. But more precisely, the universe is all there is in our spacetime.  Space is 3 dimensional, and spacetime is 4 dimensional.  One can model the universe as 3D with time external to it, but that seems to require the same naive sort of thinking that puts space external to the universe.
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I had the impression that the space represents the infinity, as there is no limit in space
Then you are using a model of infinite space. You need to use a different model when considering a finite size universe. You can't use both models at once. Of course they will contradict each other. Try the surface of hypersphere model that is far more mainstream. There's no infinity in that geometry.
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while the Universe represents the size of our universe. (Therefore the Universe can take the whole size of the open infinity space or just part of it).
If the universe can take just part of something, it is an object within something larger, and the larger thing is the universe and the thing you're thinking of is just some object. There is no valid model that looks like that.
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So, let's assume that our Universe/space is finite and go back to year 400,000 - just after the inflation process while the CMB gets already its Black body radiation.
The CMB is the radiation. There isn't a body that is doing any radiating or absorbing.
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Let's assume that the radius of the Universe/space at that time was only R1 - which represents a compact universe/space.
All the matter of our current Universe had been there in this compact universe/space.
There was nothing outside of this radius.
Nor inside either.  The model I am assuming is the surface of a hypersphere, so space at a given moment covers only the surface, and R1 is not measured in a spatial dimension, but rather the time dimension.  The past lies inside R1 and the future lies outside it.  If the past and future do not exist, then time is external to the universe (making it an object of sorts again) and there is nothing inside or outside of R1.
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Due to the expansion in space of 70 km/s/Mpc, the space of our universe had been increased and therefore after 13.4 Billion year from that moment, our current Universe/space get's to its maximal radius of R2.
So you know, that's the current rate, which reflects a doubling of size in 12-13 billion years (the age of the universe).  It was much younger then and would double in another 400000 years, so the km/s/Mpc figure back then would be about 30000 times that rate.
I really hate the expression of 70 km/s/Mpc because that makes it sound like a constant thing, but it has no dimensions, just a ratio.  That ratio equals the number of seconds in the age of the universe.  The age of the universe was much less back then, so the (unchanged) expansion rate would be that much greater.

I'm not disagreeing with anything you said since you're not computing anything with that 70 figure, but expansion cannot actually have a rate.
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However, as our Universe/space had been increased at a rate of 70 km/s/Mpc, the CMB was moving in all directions at the speed of light.
The CMB would be moving at lightspeed even in non-expanding universe.  The two have little to do with each other.

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Sooner or later all the energy of the CMB must get to the edge
There is no edge.  A circle is not a line segment with endpoints.  You go far enough you just get back to where you started.  That makes it a finite size closed dimension.  A hypersphere is like that, except it has 3 spatial dimensions on its surface and looks exactly like what we see.
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As there is no space outside the edge,
There is no edge.  An edge sort of implies unoccupied space beyond.  Wrong model.  Yes, I agree that any model with an edge is probably implausible.  I do not know of one that is proposed by scientists, so there is little point in talking about it.
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But, the radiation must move in direct line to all directions.
So, how could it be that a radiation/light which starts at point A moves in a direct line to point B, cross that point while it moves in the same direction and eventually gets back to point A without meeting the edge of the Universe?
Yes.  Space on the surface of Earth is finite (else real estate would be less expensive).  Yet there is no direction I can go to get to the edge of it and fall off.  If I go far enough, I get back to where I started.  It isn't any more complicated than that, yet you persist in the flat model with an edge. Yes, such a geometry would be non-Euclidean.
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If that is correct, do you agree that when we look at the left side of the space/Universe, we could see the light/radiation which is coming from the right side of the finite space/universe?
If the size was really small, light may have had time to go around, yes.  Maybe all the CMB radiation (the oldest light there is) has done that.  How would you tell?
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Is it real?
The model works.  So it is possibly real.
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Yes, now I do understand the meaning of Finite space.
However, any finite space must have some sort of size. Therefore, I don't understand how a light can cross a limited space or Finite space without getting to the edge of the space.
If you don't understand the lack of an edge, then you don't understand the meaning of Finite space.
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Once we set a sphere for the finite space
A sphere has and edge.  The surface of a hypersphere does not.  Space is not a sphere.
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The speed of light is 299 792.458 kilometers / second, while the speed of the expansion is only 70 km/s/Mpc
The ratio between the two is : 4285 : 1
You are relating a speed (units km/s) to an expansion rate (km/s/Mpc).  You cannot compare two numbers in different units.  The result is meaningless.
You also give one figure to 9 digits of accuracy and the other to only 2.  That means the 4285 figure is accurate to only 2 digits.

Edit: Not meaningless.  From that you can conclude that the Hubble sphere has a radius of ~4300 Mpc.  The actual figure is more like 4100.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/05/2019 03:37:57
But, the radiation must move in direct line to all directions.
So, how could it be that a radiation/light which starts at point A moves in a direct line to point B, cross that point while it moves in the same direction and eventually gets back to point A without meeting the edge of the Universe?

All "straight" lines in a hypersphere-shaped universe are actually curved. They eventually loop back around on themselves. It's like the lines of longitude and latitude on a globe. Walk far enough on the Earth, and you'll end up back where you started (ignoring the fact that the oceans are in the way, of course). It's a similar idea.

If that is correct, do you agree that when we look at the left side of the space/Universe, we could see the light/radiation which is coming from the right side of the finite space/universe?

If such a universe was small enough, old enough and expanding slowly enough, yes. If not, then the light either hasn't had time to make a complete circuit around the universe yet or the universe is expanding too quickly at those distances for it to ever complete the journey.

However, any finite space must have some sort of size. Therefore, I don't understand how a light can cross a limited space or Finite space without getting to the edge of the space.
If the Universe has a limited/finite size than somewhere there must be an edge, otherwise it should go to the infinity.
Therefore, I can't understand how a finite Universe with a radius R has no edge.

The Earth is finite in size. If you walk far enough, will you reach the edge of the Earth? No, you won't. There is no edge. It's a similar idea to a hypersphere universe.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/05/2019 07:20:22
Nor inside either.  The model I am assuming is the surface of a hypersphere, so space at a given moment covers only the surface, and R1 is not measured in a spatial dimension, but rather the time dimension.  The past lies inside R1 and the future lies outside it.  If the past and future do not exist, then time is external to the universe (making it an object of sorts again) and there is nothing inside or outside of R1.

I'm willing to accept any model that you wish.
However, I can't understand how we can use one Universe modeling for the CMB and other modeling for the expansion.
If we wish to explain the expansion based on three dimensions x,y,z - and one more for -t, than we must also to use the same modeling also for the CMB.
If we wish to explain the CMB based on two diminutions x,y (the surface of a hypersphere) - and one external for -t, than we have to use the same modeling also for the expansion.
We are living in one Universe. This Universe is the home for the expansion and the CMB (and also for us).
Therefore, I assume that we can't use different modeling for each one.
Our scientists have stated that the expansion rate today is 70 km/s/Mpc
If I understand it correctly - The Mpc represents a cube with x = 1Mpc, y = 1Mpc = 1Mpc.
Therefore, they have actually set the Universe in three dimensions x,y z.
If that is correct - they must explain/prove the CMB based on those three dimensions.
If they insist to prove the CMB based on only two dimensions ( the surface of a hypersphere), then please set/prove the expansion based on the same "surface of a hypersphere" modeling.
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If that is correct, do you agree that when we look at the left side of the space/Universe, we could see the light/radiation which is coming from the right side of the finite space/universe?
If the size was really small, light may have had time to go around, yes.  Maybe all the CMB radiation (the oldest light there is) has done that.  How would you tell?
I don't understand how can we force the radiation/light to move around the surface of a hypersphere when we know that it must move in a direct line?
In any case, if there is a size to our Universe/space, do you agree that it also must have a volume? If there is a volume, then there must be three dimensions to this volume.
So, don't you agree that we must explain the CMB radiation based on the real modeling for three dimensions?


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But, the radiation must move in direct line to all directions.
So, how could it be that a radiation/light which starts at point A moves in a direct line to point B, cross that point while it moves in the same direction and eventually gets back to point A without meeting the edge of the Universe?

All "straight" lines in a hypersphere-shaped universe are actually curved. They eventually loop back around on themselves. It's like the lines of longitude and latitude on a globe. Walk far enough on the Earth, and you'll end up back where you started (ignoring the fact that the oceans are in the way, of course). It's a similar idea.
You actually contradicts the Fermat’s principle.
http://scipp.ucsc.edu/~haber/ph5B/fermat09.pdf
Fermat’s principle states that “light travels between two points along the path
that requires the least time, as compared to other nearby paths.
The Earth is finite in size. If you walk far enough, will you reach the edge of the Earth? No, you won't. There is no edge. It's a similar idea to a hypersphere universe.
There is no obligation how do we walk. We can walk in three dimensions, two dimensions or even one dimension.
However, light must move in a direct line and in the least time!
We can't band the light/radiation just because it meet our expectation.
So, if we shot a laser beam in one direction - it won't come back to us unless it meets the law of Reflection and Refraction!
Hence, we can't assume that light/radiation can "walk" on the hypersphere universe and come back without any real impact as Reflection and Refraction.
If such a universe was small enough, old enough and expanding slowly enough, yes.
OK
But, we have to find a way how to overcome on the  Fermat’s Principle and the Laws of Reflection and Refraction.

If not, then the light either hasn't had time to make a complete circuit around the universe yet or the universe is expanding too quickly at those distances for it to ever complete the journey.
How can you keep the energy in the CMB if the Universe expands at the speed of light and the radiation is moving away in all directions?
Don't you agree that in just few moments after the bang - any bang, the radiation should go to zero?
How can you get any sort of radiation after 13.8 Billion years from the bang, while nothing from the radiation is reflected back?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/05/2019 07:34:57
You actually contedicts the Fermat’s principle.
http://scipp.ucsc.edu/~haber/ph5B/fermat09.pdf
Fermat’s principle states that “light travels between two points along the path
that requires the least time, as compared to other nearby paths.

Exactly. The shortest path in a curved space is therefore a curved line.

There is no obligation how do we walk. We can waltk in three dimentions, two dimentions or even one dimention.

I'm obviously talking about walking on the ground. Any path you choose, when walking without veering to the left or right, would eventually put you back where you started.

However, light must move in a direct line and in the least time!

Least time? Yes. Straight line? No. Gravitational lensing bends light so that it doesn't travel in a straight line. The path that will take the least time to traverse in a curved (non-Euclidean) space is a curved line.

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We can't band the light/radiation just becouse it meet our exectation.

I don't understand your use of the word "band" in this sentence.

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So, if we shot a lazer beam in one direction - it woun't come back to us unless it meets the law of Reflection and Refraction!
Hence, we can't assume that light/radiation can "walk" on the hypersphere universe and come back without any real impact as Reflection and Refraction.

The laser beam is traveling through curved space, just like the curved space around massive objects that causes gravitational lensing. That is what would make the beam go all the way around the universe and come right back to its starting point.

Hence, we can't assume that light/radiation can "walk" on the hypersphere universe and come back without any real impact as Reflection and Refraction.

Yes we can assume that. If the space is curved, the light has no choice but to curve along with it.

OK
But, we have to find a way how to overcome on the  Fermat’s Principle and the Laws of Reflection and Refraction.

There is nothing to overcome. Neither of those things are violated.

How can you keep the energy in the CMB if the Universe expands at the speed of light and the radiation is moving away in all directions?

The first law of thermodynamics guarantees that such energy cannot go to zero. It can become more diffuse as the universe expands, but that's all.

Don't you agree that in just few moments after the bang - any bang, the radiation should go to zero?

No, because that would break the first law of thermodynamics.

How can you get any sort of radiation after 13.8 Billion years from the bang, while nothing from the radiation is reflected back?

Why is there any need for reflection? Absolutely all of space is filled with the CMB. In any given volume of space, CMB photons are leaving it, but other CMB photons are entering it.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/05/2019 13:58:20
I'm willing to accept any model that you wish.
Obviously not.
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However, I can't understand how we can use one Universe modeling for the CMB and other modeling for the expansion.
The BBT (a theory of the time evolution of the universe) explains the expansion and the CMB.  That theory does not conflict with various models of the geometry of the universe.
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If we wish to explain the expansion based on three dimensions x,y,z - and one more for -t, than we must also to use the same modeling also for the CMB.
The xyz-t thing has been around for centuries and was probably mainstreamed with Minkowski's and Einstein's work at the beginning of the 20th century.  The BBT theory came later when the expansion was noticed, and the discovery of the CMB later still.  So the 4-D spacetime model had nothing to do with explaining these yet unknown things.

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If we wish to explain the CMB based on two diminutions x,y (the surface of a hypersphere)
A hypersphere is a 4 dimensional object with a 3 dimensional surface.  You're thinking of a surface of a sphere here.

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and one external for -t, than we have to use the same modeling also for the expansion.
Yes, the expansion also would need to be covered by it.  You will note that the 4D model was not falsified by the subsequent discovery of the expansion.  The 4D model also does not assert a finite or infinite space.  Those are variants, and since both are valid, there seems to be no obvious evidence to answer the question of the universe being infinite or not.

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We are living in one Universe. This Universe is the home for the expansion and the CMB (and also for us).
Therefore, I assume that we can't use different modeling for each one.
I use any combination of models that don't contradict each other.  I cannot use a finite and an infinite model at the same time for instance.  But each of these theories might make different assertions, but they don't contradict each other.  They can all be true at once.
BBT says time has a singularity at a known point in the past and has evolved to its current state today.
4D model says time is part of the universe just like space.The two are essentially the same thing and can be measured in common units.
3D model says time is external to the universe, and the universe exists within it.  That means there is a larger container, and the universe is reduced to being more like an object, not really the same definition of 'universe' at all.  3D and 4D models are mutually contradictory, but either works with BBT.

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Our scientists have stated that the expansion rate today is 70 km/s/Mpc
If I understand it correctly - The Mpc represents a cube with x = 1Mpc, y = 1Mpc = 1Mpc.
No.  A Mpc is a linear distance, not a measure of volume.  A liter is a measure of volume, and I'm not sure if they have a unit for cosmological scale volume other than say a Hubble volume with is a function of time, so not really a unit.
The current expansion rate is measured as a percentage increase in linear separation of points in space over time.  Any such specification can be reduced to a statement of the age of universe.  If you halved or doubled the expansion rate, the rate today would still be 70 m/s/Kpc.  So it isn't really a measure of a rate, just a measure of time.

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I don't understand how can we force the radiation/light to move around the surface of a hypersphere when we know that it must move in a direct line?
Please familiarize yourself with non-Euclidean geometry. Don't criticize the mathematics that you clearly don't understand. The lines are very much straight lines.
Kryptid attempts some basics in his reply.

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In any case, if there is a size to our Universe/space, do you agree that it also must have a volume?
Yes.  A finite 3D space has a meaningful finite volume.  An infinite space does not.  Earth has finite radius and thus a finite surface area.
I think the finite spacetime model would have to have a radius computed a way comparable to the Schwarzschild radius where they take the distance required to get back to the staring point and then divide by 2π, which may or may not be the same value as the length in the radial direction.  In the case of Schwarzschild coordinates, the radial distance is not meaningful.
Translation:  The volume of a hyperspherical universe is best expressed as a function of its circumference, not of its radius.

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If there is a volume, then there must be three dimensions to this volume.
Yes. The model would be falsified if it didn't have 3 macro spatial dimension like we see.

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So, don't you agree that we must explain the CMB radiation based on the real modeling for three dimensions?
This has nothing to do with the CMB.  The model was showing a finite volume with no edge. Yes, the CMB needs to be present equally at every point in that volume (in any of the 3D/4D finite/infinite models) at a fixed moment in absolute time.  It is not equal at every point in time, so one can measure a crude age of the universe just by looking at the CMB.

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You actually contradicts the Fermat’s principle.
http://scipp.ucsc.edu/~haber/ph5B/fermat09.pdf
Fermat’s principle states that “light travels between two points along the path
that requires the least time, as compared to other nearby paths.
Non Euclidean space does not violate this principle, as Kryptid points out.


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There is no obligation how do we walk. We can walk in three dimensions, two dimensions or even one dimension.
Yes there is obligation.  You are confined to space, and cannot leave it.  Light is similarly confined.

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How can you keep the energy in the CMB if the Universe expands at the speed of light and the radiation is moving away in all directions?
The energy density goes down as the volume expands, but the total CMB energy only goes down when the radiation is absorbed by objects it encounters.

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How can you get any sort of radiation after 13.8 Billion years from the bang, while nothing from the radiation is reflected back?
How could you not? Suppose I do it in reverse, tracing back light today into a contracting universe in reverse time.  In almost 13.8 billion years (down to year 379000) that light will be somewhere in (much smaller) space, and since the plasma was everywhere in space back then, there will be plasma there.  That's where the CMB we see now comes from.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/05/2019 20:43:22
Thanks Halc & Kryptid
I do appreciate your efforts to explain the issue.
So, yes, now I have better understanding about the difference in the modeling.
Exactly. The shortest path in a curved space is therefore a curved line.

Let's assume that we are located just at the center of the Space/Universe which its radius is 52 BLY.
We see that all the furthest galaxies in all directions are moving away from us at ultra high speed.
Let's draw a line (in one direction) to the infinity.
If I understand it correctly, this line should come back to us, although we have draw a direct line to the infinity.
Let me use the following scenario:
At some point in this line to the infinity, we should get to the furthest galaxy that we still can see - Let's assume that it is located 13 BLY away.
This galaxy (Galaxy A) is moving away from us at almost the speed of light.
Now, let's assume that we can jump and be there in less than one second.
Once we are at galaxy A, do you agree that we should see a similar view as we see from our galaxy?
So, we should see that all the furthest galaxies in all directions are moving away from us at ultra high speed.
If we will continue with the same line that we have set to the infinity, we should get to another furthest galaxy (galaxy B).
So, it is located 13 BLY away from Galaxy A and 26 BLY away from us.
If we continue to jump on the same direction from galaxy to galaxy (while each jump represents 13 BLY)  - after 4 jumps we should get to 52 BLY away from us.
Let's assume that this point represents the last furthest galaxy in this direction.
So, what kind of view shall we see from this last galaxy?
What will happen if we will try to jump in the same direction that we have set?
Based on the curvature, does it mean that we should make a U turn at the edge of the space (however - now I know that there is no edge) and come back to galaxy C then B, A and finely to our galaxy?
All of that while we think that we still move in the same direction to the infinity.
So, do you mean that this scenario sets the shortest path in a curved space?
Hence, even if we are actually always jumping in one direction, eventually we are coming back to the starting point.
If that is correct, I must say that it is very difficult for me to accept this idea.





Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/05/2019 21:31:11
Let's assume that we are located just at the center of the Space/Universe which its radius is 52 BLY.
That's a sphere.  Unrealistic.
If space is the surface of a hypersphere, you can't be at the center since that's not the surface.

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If I understand it correctly, this line should come back to us
You are picturing a sphere with a radial line coming out.  No, that ends when it hits the edge.

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At some point in this line to the infinity
I'm going to assume we're talking about a finite model without infinity.  That means if you go far enough, you get back to the start just like I do when I walk west.
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we should get to the furthest galaxy that we still can see - Let's assume that it is located 13 BLY away.
We can't see that galaxy since we can see zero distance in the present.  But it's there, sure.  What we see is light from events in the past, and if you look beyond 13 billiion years into the past, the galaxies have just not yet had time to form.  That furthest object we see is incredibly young.

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This galaxy (Galaxy A) is moving away from us at almost the speed of light.
Maybe more.  Depends how you measure that 13 BLY.

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Now, let's assume that we can jump and be there in less than one second.
Once we are at galaxy A, do you agree that we should see a similar view as we see from our galaxy?
Yes.  Principle of mediocrity says there is a similar view from anywhere.  An isotropic CMB and uniform sprinkle of galaxies in every direction.

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If we continue to jump on the same direction from galaxy to galaxy (while each jump represents 13 BLY)  - after 4 jumps we should get to 52 BLY away from us.
And more beyond that, yes.

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Let's assume that this point represents the last furthest galaxy in this direction.
No, that violates the principle of mediocrity.  You persist in positing an edge.  There is none.  You posited a radius of 52 BLY, so the circumference would then be ~325 BLY.  If you go that far, you get back here.

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So, what kind of view shall we see from this last galaxy?
What will happen if we will try to jump in the same direction that we have set?
Based on the curvature, does it mean that we should make a U turn at the edge of the space (however - now I know that there is no edge) and come back to galaxy C then B, A and finely to our galaxy?
No edge, last galaxy, nor reflections, u-turns, or any of this stuff.  It looks like it does here from anywhere.  How many times do we have to repeat this?

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All of that while we think that we still move in the same direction to the infinity.
Yes, just like I can walk west forever and never find the end of it.

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So, do you mean that this scenario sets the shortest path in a curved space?
It sets a straight path.  That path isn't always the shortest in non-euclidean space.  I can fly from London to Paris the wrong way around and the path won't be the shortest, but the airplane won't need to turn to do it.  There are only two such paths in this case.  Any other either needs to turn or doesn't end up in Paris.

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Hence, even if we are actually always jumping in one direction, eventually we are coming back to the starting point.
If that is correct, I must say that it is very difficult for me to accept this idea.
As I said before, familiarize yourself with non-euclidean geometry. That or stop being incredulous about its properties. I cannot help it that your education falls short of this point. They teach it in high school.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/05/2019 22:47:54
Based on the curvature, does it mean that we should make a U turn at the edge of the space (however - now I know that there is no edge) and come back to galaxy C then B, A and finely to our galaxy?

No, you would not make a sudden U-turn at any point and there is no edge of space where you could even make such a U-turn. The curvature is not sudden like that. You are traveling through curved space at all times along the journey, just like you are walking over the curved surface of the Earth at all times if you tried to walk around the globe. At no point do you turn around in 3-dimensional space just like at no point do you turn around when you walk around the Earth. Nor do you encounter any edge of space anymore than you encounter an edge to the Earth.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/05/2019 21:23:57
No, you would not make a sudden U-turn at any point and there is no edge of space where you could even make such a U-turn. The curvature is not sudden like that. You are traveling through curved space at all times along the journey, just like you are walking over the curved surface of the Earth at all times if you tried to walk around the globe. At no point do you turn around in 3-dimensional space just like at no point do you turn around when you walk around the Earth. Nor do you encounter any edge of space anymore than you encounter an edge to the Earth.
Thanks
Somehow you insist to explain the CMB only by the idea of hypersphere, while when we discuss on the Expansion and Observable Universe we normally discuss on the normal three dimensions.
So, let's see what is the real meaning of - hypersphere:
https://en.wikipedia.org/wiki/N-sphere
"An n-sphere embedded in an (n + 1)-dimensional Euclidean space is called a hypersphere."
"In mathematics, the n-sphere is the generalization of the ordinary sphere to spaces of arbitrary dimension. It is an n-dimensional manifold that can be embedded in Euclidean (n + 1)-space."
https://en.wikipedia.org/wiki/3-sphere
In mathematics, a 3-sphere, or glome,[1] is a higher-dimensional analogue of a sphere.
It is stated clearly: In mathematics...
So, in the nature there is no real object as hypersphere. It is only exist as a mathematical concept/idea.
Therefore, we can't explain the behavior of the CMB based on that Unreal mathematical model.
We have to explain the CMB on real Universe.
As our real Universe has only three dimensions, then only this real model should be used for our explanation.
In any case, this mathematical idea of the hypersphere confirms your message about the curvature:
"Stereographic projection of the hypersphere's parallels (red), meridians (blue) and hypermeridians (green).
"All curves are circles"
They also show that there is a possibility for a curves with infinite radius:
"the curves that intersect ⟨0,0,0,1⟩ have infinite radius (= straight line).
So, they show that the radius is a function of the fourth dimension in that mathematical concept.
However, as the Fourth dimension is not part of the expansion, it is quite clear to me that there is no way to expand the hypersphere.
So, if we wish to use this unreal mathematical model for our universe in order to prove the CMB, we actually can't use it for the expansion. Therefore, our scientists do not even try to set any connection between the hypersphere to the expansion.
Let me use the earth example:
just like you are walking over the curved surface of the Earth at all times if you tried to walk around the globe.
So, if the Universe is like the surface of the earth, the expansion is like the tectonic plates. Those plates can move or expand, but they can't increase the size of the surface of the earth.
In the same token, the expansion which works perfectly at the real three dimension universe, can't work at that mathematical concept that we call hypersphere (as the expansion has no impact on the fourth dimension)
Do you agree with all of that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/05/2019 22:45:22
Somehow you insist to explain the CMB only by the idea of hypersphere, while when we discuss on the Expansion and Observable Universe we normally discuss on the normal three dimensions.

Three dimensions are already a part of the hypersphere. It's just that there is also an extra dimension that is curved.

So, in the nature there is no real object as hypersphere.

When was that ever demonstrated?

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It is only exist as a mathematical concept/idea.

When was that ever demonstrated?

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Therefore, we can't explain the behavior of the CMB based on that Unreal mathematical model.
We have to explain the CMB on real Universe.

Whoever said it was "unreal"? You obviously can't see a hypersphere with your own, three-dimensional eyes or have one exist in three-dimensional space. It's a four-dimensional object. That's like expecting a two-dimensional being to be able to see a sphere. A hypersphere as a model for the shape of the universe has not been ruled out by any experiments.

As our real Universe has only three dimensions, then only this real model should be used for our explanation.

Three dimensions that you can see with your eyes, anyway. That doesn't rule out extra dimensions.

However, as the Fourth dimension is not part of the expansion, it is quite clear to me that there is no way to expand the hypersphere.

There is no reason why a hypersphere cannot expand, so I don't know what you are talking about. To think of an analogy, consider two-dimensional beings living embedded on the skin of an inflating balloon. The balloon has an extra dimension that those two-dimensional beings cannot observe (a third dimension). The balloon is expanding in three dimensions simultaneously, causing all of the two-dimensional objects embedded in the balloon's surface to become further apart from each other over time. That is an analogy for universal expansion.

So, if we wish to use this unreal mathematical model for our universe in order to prove the CMB, we actually can't use it for the expansion. Therefore, our scientists do not even try to set any connection between the hypersphere to the expansion.

There is no need to "prove" the CMB. It's there for all to observe. I also explained with the balloon analogy why a hypersphere has no trouble expanding and that such an expansion easily explains why we observe objects receding away from us.

So, if the Universe is like the surface of the earth, the expansion is like the tectonic plates. Those plates can move or expand, but they can't increase the size of the surface of the earth.

The proper analogy with an expanding universe would be an expanding Earth. Just pretend the Earth is a giant balloon that is slowly inflating over time. That's more like what we are seeing.

In the same token, the expansion which works perfectly at the real three dimension universe, can't work at that mathematical concept that we call hypersphere (as the expansion has no impact on the fourth dimension)
Do you agree with all of that?

No I don't, because it's wrong. Just like an expanding balloon expands in all three dimensions, an expanding hypersphere expands in all four dimensions.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/05/2019 00:47:11
So, in the nature there is no real object as hypersphere.
Indeed, this has not been demonstrated.  The universe is not an object, and a hypersphere can very much exist in 4 dimensional spacetime.

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Therefore, we can't explain the behavior of the CMB based on that Unreal mathematical model.
The hypersphere model isn't one of the CMB.  It is a model of finite space.  The CMB isotropy is consistent with it due to complete symmetry at any point.

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"Stereographic projection of the hypersphere's parallels (red), meridians (blue) and hypermeridians (green).
"All curves are circles"
They also show that there is a possibility for a curves with infinite radius:
"the curves that intersect ⟨0,0,0,1⟩ have infinite radius (= straight line).
Are you suggesting the universe is a Stereographic projection?  That would render meaningless the classic definition of length.  Two stationary meter sticks would have different lengths depending on their location.

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So, they show that the radius is a function of the fourth dimension in that mathematical concept.
Yes, that would be the time dimension.

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However, as the Fourth dimension is not part of the expansion
It very much is a function of that dimension since expansion happens over time.

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Let me use the earth example:
Quote from: Kryptid
just like you are walking over the curved surface of the Earth at all times if you tried to walk around the globe.
So, if the Universe is like the surface of the earth, the expansion is like the tectonic plates.
The expansion would be more like inflating the Earth to a larger radius, with the continents separating because they're objects that don't particularly expand when the surface space does.  That's why they call it the balloon analogy since it resembles galaxies as stickers or coins on an inflating balloon.

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Those plates can move or expand, but they can't increase the size of the surface of the earth.
Other way around.  Earth expands but the continents stay the same, so the oceans get wider.  Time is the vertical dimension in that analogy, with the big bang event at the center of Earth, and the future is up.

Putting the big bang at the center is considerably further than the balloon analogy is meant to go, but it fits nicely with the small hypersphere model.  The analogy, properly used, considers only a local portion of the surface of a balloon to explain what we observe locally.  The rest (the curvature of the surface) is a possibility but not an empirically measured thing.  Hence no way to demonstrate if the universe is finite or not.

Kryptid also described this.  The analogy is well known.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/05/2019 06:52:26
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So, they show that the radius is a function of the fourth dimension in that mathematical concept.
Yes, that would be the time dimension.
Sorry this is a fatal error.
Where do you see that the fourth dimension can be a time.
Actually it is a severe violation of the whole idea of the n-sphere:
https://en.wikipedia.org/wiki/N-sphere
Based on the following formula:
S^{n}=\left\{x\in \mathbb {R} ^{n+1}:\left\|x\right\|=r\right\}.
"That is, for any natural number n, an n-sphere of radius r may be defined in terms of an embedding in (n + 1)-dimensional Euclidean space as the set of points that are at distance r from a central point, where the radius r may be any positive real number."
Hence, it is clear that all dimensions must be based on the same unites.
If you chose three dimensions of length and one of time - you set a violation in the formula.
How can we add time to length?
If we take two dimensions of length and one dimension of time, shall we get a sphere?
If we add one kilo -gram to one meter  - does it mean that we get a two kilo-gram-meter?
So, do you agree by now that there is no way to fit different unites in the N-sphere formula?
Do you agree that the selection of time in the fourth dimension have set unrealistic N-sphere and therefore - unrealistic universe?
The hypersphere model isn't one of the CMB.  It is a model of finite space.
Sorry
This is also incorrect.
Hypersphere can also be infinite
It is stated clearly:
"the curves that intersect ⟨0,0,0,1⟩ have infinite radius (= straight line)".
So, even if you use in all four dimensions the same unites - we can get an infinite radius.
So, how can you use the hypersphere as an example for finite radius while it is stated clearly that it can also be infinite?
Are you suggesting the universe is a Stereographic projection?  That would render meaningless the classic definition of length.  Two stationary meter sticks would have different lengths depending on their location.
Sure
If you set a severe violation in your calculation/formula - you should get unrealistic outcome.
The universe is not an object, and a hypersphere can very much exist in 4 dimensional spacetime.
Sorry
The Universe is an object as the galaxy is an object and as the the star is an object.
The hypersphere can't exist in 4 dimensional spacetime.
The CMB isotropy is consistent with it due to complete symmetry at any point.
Sorry
The CMB is consistent with unrealistic hypersphere/Universe.
It is clear to me that our Universe is a real sphere with only three dimensions!
Not four dimensions, not five dimensions and not any sort of 3+n dimensions.
We have to prove what we see based on real universe and not on some sort of unrealistic mathematical assumptions/calculations.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/05/2019 08:02:50
The CMB is consistent with unrealistic hypersphere/Universe.

There is nothing about our universe that is incompatible with it being a hypersphere.

It is clear to me that our Universe is a real sphere with only three dimensions!

The universe being a hypersphere is not incompatible with our ability to see only three dimensions. A two-dimensional creature living embedded in the surface of a balloon is only aware of the two dimensional surface that they live in even though the balloon itself is three-dimensional. If the balloon was large enough, the creature wouldn't even be aware of the balloon's curvature and they could just as easily believe that their universe was a completely flat, 2-dimensional space. In the same manner, the universe being a sufficiently-large hypersphere would go unnoticed by us because the curvature would be too gradual to detect.

We have to prove what we see based on real universe and not on some sort of unrealistic mathematical assumptions/calculations.

Science isn't about proof.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/05/2019 14:06:04
Quote from: Halc
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So, they show that the radius is a function of the fourth dimension in that mathematical concept.
Yes, that would be the time dimension.
Sorry this is a fatal error.
Thank you. I'd be less confident about my statement if you considered it otherwise.

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Where do you see that the fourth dimension can be a time.
Actually it is a severe violation of the whole idea of the n-sphere:
https://en.wikipedia.org/wiki/N-sphere
Spacetime modeled as a 4-ball is a simplified model of finite Minkowski spacetime.
The model is far more accurate locally, where space looks exactly like a local portion of a 3-sphere and time the dimension orthogonal to that.  When considered locally, there is no meaningful radius to the thing.  The size is thus not known, but the curvature is close enough to zero to be so far immeasurable.  Not sure how one might go about measuring it, as I described in a prior post.
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Hence, it is clear that all dimensions must be based on the same unites.
In Minkowski spacetime, they are.
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If you chose three dimensions of length and one of time - you set a violation in the formula. How can we add time to length?
Using vector arithmetic for one, but two points one second apart is the same separation as two points about 3e8 meters apart.  That separation is called the interval and is not frame dependent.  It can be computed for any two points in spacetime.
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If we take two dimensions of length and one dimension of time, shall we get a sphere?
If you curve it into a ball, yes.  A sphere is a 2D surface and a ball is a 3D solid. It is the same difference as between a circle and a disk. This is terminology used in the N-sphere wiki site you linked.
Without the curving, you just have flat 3D spacetime, like the one used in a block representation of Conway's game of life.
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Quote from: Halc
The hypersphere model isn't one of the CMB.  It is a model of finite space.
Sorry
This is also incorrect.
Not a fatal mistake this time?  Aww....
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Hypersphere can also be infinite
It is stated clearly:
"the curves that intersect ⟨0,0,0,1⟩ have infinite radius (= straight line)".
That isn't a hypersphere they're talking about.  That is a description of a stereographic projection of a finite 3-sphere onto Euclidean 3-space, which indeed requires infinite space.  Only the projection of a 1-sphere onto Euclidean 1-space can be done in finite space since there are no intersection points to keep perpendicular.

As for an infinite hypersphere, there is indeed no limit to the radius of such a thing.

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So, how an you use the hypersphere as an example for finite radius while it is stated clearly that it can also be infinite?
You taking quotes out of context is not an example of something being clearly stated.
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Sorry
The Universe is an object as the galaxy is an object and as the the star is an object.
There's your problem then. You inherit all the problems of such an assertion such as how distant things are receding from us faster than lightspeed or how it got to infinite size in finite time. I also think the universe is infinite, but that would be completely contradictory if the universe was an object.
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The hypersphere can't exist in 4 dimensional spacetime.
That's like saying a circle can't exist in a 2 dimensional coordinate system.
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It is clear to me that our Universe is a real sphere with only three dimensions!
Maybe you'd make more progress if you would drop the things that are clear to you and open yourself to other possibilities.  Science is not about closed mindedness like this.

A 3D sphere is a finite thing.  Arbitrarily large, sure, but if it is infinite, it is no longer a sphere.  So this contradicts one of the other things that is clear to you.  This illustrates the danger of everything being clear before you actually investigate the consequences.
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Not four dimensions, not five dimensions and not any sort of 3+n dimensions.
A 3D infinite universe is a valid philosophical interpretation of the universe, but not the only valid one.  It becomes an object of sorts because it needs to exist in a container.  So there is a container universe outside the object universe.  That seeming contradiction in terms merely serves to illustrate the need to define your words carefully.

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We have to prove what we see based on real universe and not on some sort of unrealistic mathematical assumptions/calculations.
Something being clear to you is not evidence of anything, yet this seems to be your only evidence offered.  I'm pushing the hypersphere model not because I think it is that way, but because it could be.  The model is of a real thing, not an abstraction, and it yields a universe exactly like the one we see.  If it doesn't, then you have a falsification test.  There is no proof test.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/05/2019 20:49:25
Thanks for the explanation.
Spacetime modeled as a 4-ball is a simplified model of finite Minkowski spacetime.
The model is far more accurate locally, where space looks exactly like a local portion of a 3-sphere and time the dimension orthogonal to that.

Let's try to compare the Newtonian mechanics with Minkowski spacetime:
http://einsteinrelativelyeasy.com/index.php/special-relativity/11-introduction-to-spacetime-diagrams
Newtonian mechanics - "In Newtonian mechanics, events are described using a three-dimensional Euclidean space time plus an independant scale of absolute time."
Minkowski spacetime
"Minkowski spacetime is the most common mathematical structure on which special relativity is formulated."
"Each such observer labels events in space-time by four inertial coordinates t, x, y, z."
" it is described by the spacetime interval ds2 = c2Δt2 - Δx2 - Δy2 - Δz2"
So, Minkowski spacetime is a mathematical structure which labels events in space-time by four inertial coordinates t, x, y, z.
it is described by the spacetime interval ds2 = c2Δt2 - Δx2 - Δy2 - Δz2
It also seems to me that the formula looks different from the four dimensions sphere (but I'm not sure about it.)
However, as it is stated - the Minkowski spacetime is a mathematical structure which set the time as one more dimension in space, therefore - it is unreal universe structure.
Our real universe is represented correctly only by Newtonian mechanics. In Newtonian mechanics there is no curvature in the sphere. Therefore, in our real Universe there is no curvature.
Do you agree with that?
A 3D infinite universe is a valid philosophical interpretation of the universe, but not the only valid one.
I think that A 3D infinite/finite Universe which is based on Newtonian mechanics is the ONLY valid universe.
A 3D sphere is a finite thing.  Arbitrarily large, sure, but if it is infinite, it is no longer a sphere.  So this contradicts one of the other things that is clear to you.

Do you mean that if you take X,Y,Z to the infinity - then we get an infinite 3D universe, (but it isn't a sphere).
If so, I agree with you.
I claim the following:
1. The Universe is infinite 3D
2. There is no curvature in our real universe.
3. The CMB is a direct product of our infinite 3D Universe.
4. Its age is also infinite
So, how can I prove it:
Based on black body radiation. This is the ultimate prove that our universe is infinite.
I claim that if we set our galaxy under insulated enclosure - we should get the following:
All the radiation from the galaxy will stay under this insulated enclosure.
https://en.wikipedia.org/wiki/Black_body
"Suppose the cavity is held at a fixed temperature T and the radiation trapped inside the enclosure is at thermal equilibrium with the enclosure. The hole in the enclosure will allow some radiation to escape. If the hole is small, radiation passing in and out of the hole has negligible effect upon the equilibrium of the radiation inside the cavity. This escaping radiation will approximate black-body radiation that exhibits a distribution in energy characteristic of the temperature T and does not depend upon the properties of the cavity or the hole, at least for wavelengths smaller than the size of the hole"
So, if we will measure the radiation under this insulated enclosure, do you agree that we should get a perfect black body radiation?
Now, let's set 10Mpc of our universe under this kind of insulated enclosure.
If we do so, I'm quite sure that the energy radiation should be in the range of 2.7K  and it should also carry a black body radiation. So, we should get almost the same CMB that we measure (with one exception - redshift)
However, in reality, we can't set the galaxy under insulated enclosure.
So, just think about the following possibility -
Let's assume that our universe is infinity.
Let's divide it to infinite insulated enclosure Cube. The size of each cube will be 10Mpc. The temp in each cube is 2.7K and it has a black body radiation.
So, if we look at X dimension - there will be infinite 10Mpc insulated enclosure cube which are connected to each other.
That will be the case also for Y and z dimensions.
Now, what would be the impact if we eliminate the insulated enclosure between two nearby cubes?
As each one of them has a temp of 2.7K with black body radiation - it is quite clear that the two will keep the same temp and the same black body radiation.
Even if we eliminate the insulated enclosure between the cubes to the 13 BLY sphere the same temp and black body will remain.
However - now as the radiation comes from far locations - we should see the some redshift in the CMB.
Never the less, even if the value of a galaxy which is located at a distance of 13 BLY is 10 (for example), it doesn't mean that we should get all of this redshift in the radiation. It should represent the impact of all the galaxies/matter in that sphere.
So, I can just assume that we might get all range of redshift up to 10, but the average should be much lower.
if we eliminate the insulated enclosure between all the cubes to the infinity - than the temp will stay at 2.7K and also the black body radiation.
With regards to the redshift -
As it comes from a galaxies which are located at the infinity - we should get wide mix of redshift up to the infinity.
However, the average should be exactly 1100.
So, the CMB that we get is the radiation from the real matter in our real infinite Universe which its age is also infinity.
Do you agree with that?


 
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/05/2019 22:00:09
Let's try to compare the Newtonian mechanics with Minkowski spacetime:
http://einsteinrelativelyeasy.com/index.php/special-relativity/11-introduction-to-spacetime-diagrams
Newtonian mechanics - "In Newtonian mechanics, events are described using a three-dimensional Euclidean space time plus an independant scale of absolute time."
It can be done with 4D space/time (as it had been in those days), but it wouldn't be Minkowski spacetime with the Lorentz transformations and all.

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it is described by the spacetime interval ds2 = c2Δt2 - Δx2 - Δy2 - Δz2
It also seems to me that the formula looks different from the four dimensions sphere (but I'm not sure about it.)
Special relativity covers the flat spacetime case, as does the Minkowski spacetime on which it was based.

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However, as it is stated - the Minkowski spacetime is a mathematical structure which set the time as one more dimension in space,
In addition to the 3 spatial dimensions.  It doesn't put time in space, but rather orthogonal to it.
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therefore - it is unreal universe structure.
You seem not to agree with it, so you label it 'unreal'.  How very well argued.

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Our real universe is represented correctly only by Newtonian mechanics.
That has been empirically falsified.

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Do you agree with that?
Stop asking me if I agree with any of your assertions.  You probably know the answer.

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A 3D infinite universe is a valid philosophical interpretation of the universe, but not the only valid one.
I think that A 3D infinite/finite Universe which is based on Newtonian mechanics is the ONLY valid universe.
You're repeating yourself.  You've not falsified the other view, so this wrong.  It might well be the case, but if the other views work, they're equally valid even if not correct.

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A 3D sphere is a finite thing.  Arbitrarily large, sure, but if it is infinite, it is no longer a sphere.  So this contradicts one of the other things that is clear to you.
Do you mean that if you take X,Y,Z to the infinity - then we get an infinite 3D universe, (but it isn't a sphere).
Right (except for terminology).  What you call a sphere (a 3D ball actually) has an edge.  An infinite universe does not.  A sphere may have no edge, but it is only a two dimensional non-Euclidean surface, and hence does not correspond to our universe.

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If so, I agree with you.
I claim the following:
1. The Universe is infinite 3D
2. There is no curvature in our real universe.
3. The CMB is a direct product of our infinite 3D Universe.
4. Its age is also infinite
The last one is on shaky ground since it seems to contradict the 2nd law of thermodynamics, but there are those who hold such a view and attempt ways to get around that.  Just asserting your way past it doesn't work.

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So, how can I prove it:
You can't prove stuff like that.  At best you can demonstrate that the view is consistent, but to prove it you need to falsify not only all other views but also the view that nobody has yet proposed.
We've told you thing countless times, yet you persist in claiming you have proofs of things.

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Based on black body radiation. This is the ultimate prove that our universe is infinite.
I claim that if we set our galaxy under insulated enclosure - we should get the following:
All the radiation from the galaxy will stay under this insulated enclosure.
https://en.wikipedia.org/wiki/Black_body
"Suppose the cavity is held at a fixed temperature T and the radiation trapped inside the enclosure is at thermal equilibrium with the enclosure. The hole in the enclosure will allow some radiation to escape. If the hole is small, radiation passing in and out of the hole has negligible effect upon the equilibrium of the radiation inside the cavity. This escaping radiation will approximate black-body radiation that exhibits a distribution in energy characteristic of the temperature T and does not depend upon the properties of the cavity or the hole, at least for wavelengths smaller than the size of the hole"
So, if we will measure the radiation under this insulated enclosure, do you agree that we should get a perfect black body radiation?
You have a galaxy in there, which is hardly in thermal equilibrium like the quote above describes.  So no.  I've answered this before.  There will be hot spots nearer the points of higher radiation.

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Now, let's set 10Mpc of our universe under this kind of insulated enclosure.
If we do so, I'm quite sure that the energy radiation should be in the range of 2.7K  and it should also carry a black body radiation. So, we should get almost the same CMB that we measure (with one exception - redshift)
The temperature would be more uniform, but still hotter where galaxies are near.  You've not computed the temperature at all, but rather asserted it, which is hardly proof of anything.

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So, if we look at X dimension - there will be infinite 10Mpc insulated enclosure cube which are connected to each other.
That will be the case also for Y and z dimensions.
Now, what would be the impact if we eliminate the insulated enclosure between two nearby cubes?
The galaxies would be able to move between cubes instead of smashing into the sides like they would when you cubed up all of space into small 10Mpc cages like that.  They're not standing still you know.
The distant galaxies are moving away, so the radiation from those distant places is much lower temperature than the nearby places.  If it's 2.7K here, it is far less than that if you look deep into space.  That's what redshift does.  The CMB would be immeasurable since we'd get radiationj only by looking at actual stars and not looking between them.

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As each one of them has a temp of 2.7K with black body radiation - it is quite clear that the two will keep the same temp and the same black body radiation.
I see what you're describing and there would be a background radiation something like that if everything stood essentially still relative to us, but that's not what we see.  Your model does not account for the observed recession of all objects.

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However - now as the radiation comes from far locations - we should see the some redshift in the CMB.
Yes, so that lowers the measured temperature from the 2.7K that it would appear if not redshifted.

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Never the less, even if the value of a galaxy which is located at a distance of 13 BLY is 10 (for example), it doesn't mean that we should get all of this redshift in the radiation. It should represent the impact of all the galaxies/matter in that sphere.
So, I can just assume that we might get all range of redshift up to 10, but the average should be much lower.
Most galaxies are far more distant that a mere 13 BLY in an infinite universe.


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With regards to the redshift -
As it comes from a galaxies which are located at the infinity - we should get wide mix of redshift up to the infinity.
Redshift is an effect, not a thing.  It doesn't come from anywhere.  Light does.  How can distant galaxies move faster than light? Or do they not do this?
Hubble's law says 70 km/s/Mpc, so it doesn't take deep mathematics to realize that anything 14 BLY away is moving faster than light, assuming we're standing still.
For that matter, why are we the only ones standing nearly still?  What would it look like if we were moving at a significant percentage of light speed?  It's your naive belief, so I have no idea how to answer questions like that.
Newtonian mechanics actually allows objects to move at greater than lightspeed, but they would not be able to see anything in most directions just like you can't hear an approaching supersonic jet.
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However, the average should be exactly 1100.
In an infinite universe, the vast majority of galaxies would have a redshift higher than that since only the ones within 45 BLY would exhibit that shift, and that's less than 1% of the universe.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/05/2019 22:05:18
Our real universe is represented correctly only by Newtonian mechanics. In Newtonian mechanics there is no curvature in the sphere. Therefore, in our real Universe there is no curvature.

Newtonian mechanics gives an incorrect prediction for the amount of gravitational lensing of light around the Sun. Einsteinian relativity, which assumes a curved space-time around the Sun, does give the correct prediction for gravitational lensing.

If we do so, I'm quite sure that the energy radiation should be in the range of 2.7K

Show the calculations. We're not going to take your word for it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/05/2019 06:19:51
Thanks
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However, as it is stated - the Minkowski spacetime is a mathematical structure which set the time as one more dimension in space,
In addition to the 3 spatial dimensions.  It doesn't put time in space, but rather orthogonal to it.
That is correct.
Minkowski was a very clever scientist. The idea of spacetime is brilliant with regards to the mathematical concept/structure. However, Minkowski have never ever claimed that in our real universe the time is orthogonal to the other 3 spatial dimensions. So, Minkowski spacetime is a brilliant idea to set a calculation with regards to time/space for special cases.
We can't just take it to the extreme. As we do so, we get unrealistic results which set our universe as unreal universe based on that spacetime. Only Newtonian mechanics gives the correct prediction for our whole universe!
Therefore, there is no curvature in our real Universe, but there is a curvature if we take the Minkowski spacetime to the extreme.

Newtonian mechanics gives an incorrect prediction for the amount of gravitational lensing of light around the Sun. Einsteinian relativity, which assumes a curved space-time around the Sun, does give the correct prediction for gravitational lensing.
There are no black holes under Newtonian mechanics.  That requires curvature of space.
Yes, I fully agree with that.
Newtonian mechanics can't give a perfect prediction to all cases.
Hence, if we look at a very special cases, as the amount of gravitational lensing of light around the Sun or black holes, it is clear that Einsteinian relativity, which assumes a curved space-time is a perfect modeling.
However, in the same token - Einsteinian relativity, which assumes a curved space-time, is a perfect solution for the gravitational lensing of light around the Sun and black holes, but it gives an incorrect prediction if we try to set it at the extreme and verify with it the whole infinite universe.
We need to fit the modeling for each case.
Why don't we use Einsteinian relativity to calculate the gravity forces between stars and planets?
Why do we prefer Newton law?
Therefore:
Einsteinian relativity, which assumes a curved space-time is the ultimate tool for a special cases and it gives the best perfect prediction for those cases. However, if we use that curved space-time as a module for our whole universe we get unrealistic results as it gets to the extreme and therefore, the predictions are incorrect.
Newtonian mechanics is the only ultimate tool for the prediction of our whole universe!
As in our real universe the time isn't orthogonal to the other 3 spatial dimensions, there is no curved space-time in our real universe.
Any radiation that had been emitted from point A should move to the infinity and never ever come back!!!
Therefore, the CMB that we see around us, is coming from all the matter in our real infinite universe.
As our universe is infinite - we get a perfect isotropic radiation from all directions.
So, there is no meaning for our location in this infinite universe, as long as the distance to any "edge" of the universe is still infinite!
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/05/2019 11:31:09
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Now, let's set 10Mpc of our universe under this kind of insulated enclosure.
If we do so, I'm quite sure that the energy radiation should be in the range of 2.7K  and it should also carry a black body radiation. So, we should get almost the same CMB that we measure (with one exception - redshift)
The temperature would be more uniform, but still hotter where galaxies are near.
Thanks.
Yes, I have used the idea of 10Mpc in order to get more uniform temperature. We actually know that Universe is Uniform on Large Scales (10Mpc)
I also agree with you that the temp might be hotter were galaxies are near.
However, we need to think about a galaxy as some sort of heat element in an oven.
Once the oven is closed, and there is no heat loose, then the temp in the oven might go up to almost the temp of the heating elements.
In the same token, if we set the galaxy in insulated enclosure/oven - the temp in that oven should go up and meet almost the temp of the galaxy.
You've not computed the temperature at all, but rather asserted it, which is hardly proof of anything.
I agree.
I didn't set the calculation for the expected temperature in 10Mpc insulated enclosure as I have no clue about the density of galaxies/satrs/matter in that size. However, it is clear to me that if we do so, we should get exactly that 2.7 K temp.


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I claim the following:
1. The Universe is infinite 3D
2. There is no curvature in our real universe.
3. The CMB is a direct product of our infinite 3D Universe.
4. Its age is also infinite
The last one is on shaky ground since it seems to contradict the 2nd law of thermodynamics, but there are those who hold such a view and attempt ways to get around that.
The age of the Universe in infinite. Why do you claim that it contradicts the 2nd law of thermodynamics? Is it because you reject the idea of constant mass creation in the excretion disc around the SMBH. This idea is much more stable and solid than the shaky ground about the curvature universe based on the mathematical structure of spacetime.
I know that it is very difficult for you to accept the idea of new mass creation.
However - this is the driving force of our Universe.
Once you agree with that - you get full answer to the enigma of our universe.
If we could go in back time - we should find that the CMB temp/features and the universe density at 10^100 years ago was exactly as it is today.
So, the same CMB and the same universe density will stay with us forever and ever.

However, the average should be exactly 1100.
In an infinite universe, the vast majority of galaxies would have a redshift higher than that since only the ones within 45 BLY would exhibit that shift, and that's less than 1% of the universe.
I agree
The ultimate answer for that is the Shell_theorem as explained by Kryptid
Shell theorem should make radiation look isotropic even if your model of a finite, spherical universe is used. Shell theorem is normally used to describe gravity, but it should work here too because radiation intensity falls off at the exact same rate as gravitational strength does (the inverse square law): https://en.wikipedia.org/wiki/Shell_theorem
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html
If we set the Shell_theorem on the infinite universe it is clear to me that the average impact of the radiation should be exactly at a redshift of 1100 (although this redshift represents a distance of only 45 BLY). 

 
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As each one of them has a temp of 2.7K with black body radiation - it is quite clear that the two will keep the same temp and the same black body radiation.
I see what you're describing and there would be a background radiation something like that if everything stood essentially still relative to us, but that's not what we see.  Your model does not account for the observed recession of all objects.
Thanks
So we agree that there would be a background radiation something like that, if everything stood essentially still relative to us.
That's all I ask.
The whole idea is that if we can hold the galaxies/matter in each 10Mpc cube, we should get that background radiation.
My modeling gives a perfect solution for the observed recession of all objects/galaxies.
So, how it works?

1. New Born Spiral Galaxy
Each mother spiral galaxy creates new baby spiral galaxies. This Idea perfectly fits and explains the source of the hydrogen "bridge" between Andromeda Galaxy and the Triangulum Galaxy.
http://www.sciencedaily.com/releases/2012/06/120611193632.htm
"The new observations confirm a disputed 2004 discovery of hydrogen gas streaming between the giant Andromeda Galaxy, also known as M31, and the Triangulum Galaxy, or M33."
As both Andromeda and Triangulum are Spiral Galaxies, with rotational suppermassive black hole, they should have the requested power to generate Hydrogen Atoms in their core. Therefore, as they are drifting apart, they are releasing Hydrogen and set this kind of bridge!!! Andromeda has about 1,000 Billion stars. It is the mother spiral galaxy. Triangulum has about 40 billion stars. Therefore, it is a young spiral galaxy. This Hydrogen bridge is like an Umbilical cord which connects the mother galaxy – Andromeda' to her Embryo – Triangulum.
2. "Living" Galaxy
I see high similarity between animals and Galaxies. Let's look at our universe. What do we see? We see more and more galaxies, at different shape and size. So let's compare those galaxies to animals. But first, let's try to understand how animals had been created. It's quite clear that an animal had been born by similar animal. Therefore, life can create new life. Any living animal is a direct product of another living animal. There is no possibility for animal to be created out of a dead body!
Therefore, spiral galaxy should be considered as a "living" galaxy, as it has the ability to create baby spiral galaxy.
In the past, it was believed that the life on Earth had started in some sort of a blast or divine power.
Thanks to Darwin we know how the life on our Earth had been evolved. We don't know how the first living cell had been created. We can set the same concept on spiral galaxies. We don't know how the first "living" spiral galaxy (or BH with excretion disc) had been created. But once it had been created, our universe had been evolved from it. So, the first spiral galaxy in the universe can be considered as the first living galaxy which creates the whole matter in our universe as we see today.
3. Acceleration of far end galaxies
That first "living galaxy" (let's call it - first generation) had created new others living galaxies. Let's assume that each new born galaxy (from the second generation) had been ejected from the first generation galaxy at the orbital speed of the Solar System around the center of the Galaxy - approximately 220 km/s or 0.073% of the speed of light (at a random direction in space). However, those second generation galaxies also have the ability to create other new galaxies. Therefore, they have created the third galaxies generation. As expected the third generation galaxies had also been ejected in all directions at the same 220 Km/s.
Therefore, theoretically, if all the new galaxies will move in one direction than after 1370 generations, the last one will move at a speed of light (1370 x 220 Km/s = 301,400 Km sec) - with regards to the first generation galaxy.
So, there is no request for any sort of space expansion to prove the recession of all objects/galaxies.
All we need is just one first spiral galaxy and infinite time.
This also show why galaxies are moving in all directions while the furthest galaxies have the most recession speed.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/05/2019 12:39:35
Quote from: Halc
It doesn't put time in space, but rather orthogonal to it.
The idea of spacetime is brilliant with regards to the mathematical concept/structure. However, Minkowski have never ever claimed that in our real universe the time is orthogonal to the other 3 spatial dimensions.
Yes he did.  He didn't come up with the idea, which had been around since before Newton's time, but he combined that old idea with Lorentz's transformation equation to show how the time axis is not fixed, but can be arbitrarily oriented, just like there is no obvious X axis to space, leaving one to arbitrarily assign it any direction one finds convenient.

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So, Minkowski spacetime is a brilliant idea to set a calculation with regards to time/space for special cases.
The Euclidean simplicity works locally, but a non-Euclidean extension is needed to account for bent spacetime.
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We can't just take it to the extreme. As we do so, we get unrealistic results which set our universe as unreal universe based on that spacetime.
What unrealistic results are these?

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Quote from: Halc
There are no black holes under Newtonian mechanics.  That requires curvature of space.
Yes, I fully agree with that.
You just contradicted yourself:
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Therefore, there is no curvature in our real Universe

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Einsteinian relativity, which assumes a curved space-time, is a perfect solution for the gravitational lensing of light around the Sun and black holes, but it gives an incorrect prediction if we try to set it at the extreme and verify with it the whole infinite universe.
What prediction would that be?  It contradicting one of your assertions just makes your beliefs incompatible with accepted science.  An assertion is not a prediction.  A prediction is a proposal for an empirical test.
 
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Why don't we use Einsteinian relativity to calculate the gravity forces between stars and planets?
Why do we prefer Newton law?
Because they're not different.  Einstein did not propose a different formula that I know of.

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Einsteinian relativity, which assumes a curved space-time is the ultimate tool for a special cases
Wrong.  Special relativity is the tool for the special case of flat spacetime.  That's why it's called special relativity.  General relativity is for the general case.  That one describes the universe.  SR cannot describe it since it forbids really distant galaxies from increasing their proper distance from us at a rate greater than light speed.  SR can only be used for situations that are local and not in a significant gravitational field.  GR can be used for anything.

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and it gives the best perfect prediction for those cases. However, if we use that curved space-time as a module for our whole universe we get unrealistic results as it gets to the extreme and therefore, the predictions are incorrect.
Name one.

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Newtonian mechanics is the only ultimate tool for the prediction of our whole universe!
Newton envisioned a universe perhaps limited to 10000 light years.  Not sure when the term 'light year' was coined, but they had just learned a figure for it, so I imagine the term wasn't far behind.
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As in our real universe the time isn't orthogonal to the other 3 spatial dimensions
I don't think Newton ever denied the validity of this interpretation of time, but he would have said (as everyone did) that the units are different and that the axis orientation is fixed, not arbitrary.  Minkowski showed that the units are the same and that the orientation is arbitrary, something that was becoming apparent to several people around the end of the 19th century.

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Any radiation that had been emitted from point A should move to the infinity and never ever come back!!!
Even infinite space models do not suggest that.  There is an event horizon, just short of 16 BLY out.  A star there (a very finite distance away) can emit light now and that light will never reach here ever.  So again, you have beliefs incompatible with reality.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 08/05/2019 16:47:51
but it gives an incorrect prediction if we try to set it at the extreme and verify with it the whole infinite universe.

What is that incorrect prediction you speak of?

Why don't we use Einsteinian relativity to calculate the gravity forces between stars and planets?
Why do we prefer Newton law?

They give the same answers.

However, if we use that curved space-time as a module for our whole universe we get unrealistic results as it gets to the extreme and therefore, the predictions are incorrect.

Again, what are those incorrect predictions? What experiment was it that demonstrated this?

there is no curved space-time in our real universe.

Demonstrably wrong, that's exactly what causes gravitational lensing.

Any radiation that had been emitted from point A should move to the infinity and never ever come back!!!

You have yet to demonstrate this.

If we set the Shell_theorem on the infinite universe it is clear to me that the average impact of the radiation should be exactly at a redshift of 1100 (although this redshift represents a distance of only 45 BLY). 

Based on what math? I'm also waiting for you to show the calculations that demonstrate that your model predicts a CMB temperature of 2.7 K.

Therefore, theoretically, if all the new galaxies will move in one direction than after 1370 generations, the last one will move at a speed of light (1370 x 220 Km/s = 301,400 Km sec)

Galaxies can't move at the speed of light, so your model breaks yet another law of physics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/05/2019 07:52:56
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3. Acceleration of far end galaxies
...
Therefore, theoretically, if all the new galaxies will move in one direction than after 1370 generations, the last one will move at a speed of light (1370 x 220 Km/s = 301,400 Km sec)
Speeds don't add that way, at least not under relativity.  Newton would do it that way, but without relativity, you have no black holes, and you seem to require them for your beliefs.
I fully agree. We need the relatively in order to measure the speed between galaxies.
However, I have used Newton in order to prove that based on "galaxies generations idea" we would get the real view that we see from our galaxy (which is similar to the view that we should see from any location in the whole universe).
Based on this idea, the universe is infinite and fix - so there is no need for any sort of space expansion (as we need for the BBT)
In our real Universe nearby galaxies are moving at a relatively low speed with regards to each other. some of them might move further away from each other and some are moving closer to each other. So, at any location in the whole universe, nearby galaxies are moving at a relatively low speed with each other.
However, as we go further away, the galaxies are moving away from each other at higher and higher speed. At the furthermost distance (that we can still see), the galaxies are moving away at ultra speed - almost at the speed of light.
We can't see galaxies at further locations, as they are moving away from us at a speed which is higher than the speed of light.
So, the "galaxies generation idea" can explain all of that without any need for "space expansion".
Quote from: Dave Lev on Today at 11:31:09
Therefore, theoretically, if all the new galaxies will move in one direction than after 1370 generations, the last one will move at a speed of light (1370 x 220 Km/s = 301,400 Km sec)
Galaxies can't move at the speed of light, so your model breaks yet another law of physics.
No, my model doesn't break any law.
Yes, based on Newton - they can move further away from each other at higher speed than light, if the distance is long enough.
All of that while the universe is infinite and fix!

In my explanation I have used 1370 generations of galaxies which are all moving in only one direction.
So, if we are located at the first galaxy (first generation) we should see a long direct line of 1370 galaxies, while each one is moving away from the nearby galaxies at a speed of 220Km/s.
Therefore, the second galaxy will move away from the first galaxy at a speed of 220Km/s. The third one at 440Km/s, the 10th at 2,200Km/s and so on till the last one which moves away at 301,400 Km/s. (based on Newton)
If for example we will stay at galaxy at the middle - (galaxy no 685), the furthest galaxy in one direction is moving at 150,700km/s while on the other side, the furthest galaxy also should move away at the same speed of 15,700 km/s.
However - I fully agree that we need to measure the velocities between the far end galaxies based on relativity. So, we won't get the same velocities as I have stated, but I just gave you the idea why at the far end we should see galaxies which are moving away from us at a very high speed (at least - based on newton).
Our universe is infinite in its size and in its age.
Theoretically, we can extend this direct line of galaxies generation to the infinity.
At any location that we will be (in this line), we should see that most of the nearby galaxies are moving at a relatively low speed with each other. (no more than few times the speed of 220Km/s - as the Milky Way and Andromeda). However, as we look at further away galaxies, the relatively speed gets higher. At some point of distance, the relative velocity is almost the speed of light.
If we monitor the speed between two galaxies with a separation of 100,000 generations, we should find that based on Newton they are moving away from each other at 220,000,000 Km/s.
So far I have only used in my explanation a direct line of "galaxies generations" while they all are moving in one direction.
Let's try now to understand what is the impact for new generation as they move also at the opposite direction.
We know that the second generation moves away from the first generation at a speed of 220Km/s.
However, if the third generation is moving in the direction of the first generation, it is actually moving at a -220K/s (with regards to that first generation). Therefore, the relative velocity between the first generation and that third generation is zero.
Actually Andromeda and triangulum is a perfect example for mother and her baby. Do we know at what speed they are moving away from each other?
Now, think about the possibility that each galaxy creates unlimited number of new galaxies which are moving in all directions.
I'm quite sure that if we will set this modeling in a computer, we should get the same view that we see from our location without any need for none realistic idea of "space expansion".
It is clear to me that all the nearby galaxies around the Milky way had been created by the Milky way.
I wonder how many galaxies had been created so far only by Andromeda and the Milky way.

With regards to the CMB
Not 2.7K, but sure, there would be one.
Thanks
So we agree that infinite Universe with the same density (as we see) should get a CMB.
I hope that we also agree about the black body radiation in the CMB.
The main issue that you have is regards to the redshift and amplitude.
With regards to the amplitude
The temperature of the heating elements is thousands of degrees.  Why are you proposing that the 10 Mpc enclosure only gets to 2.7 then?  To get that, energy would need to leave the box at a greater rate than it comes in from the outside.  Your assertion describes a universe of infinite age, which is a steady-state model of sorts.  In a steady-state model, the box argument works, but it predicts the wrong temperature.  Are you asserting that stars burn at about 2.7K?  If not, where is the heat going?
The heat of the stars and galaxies are going to the infinity, while the heat of the CMB is coming from the infinity of all directions. Therefore, the balance should set the CMB temp at 2.7K.
So, yes - my assertion describes a universe of infinite age and infinite size which is a steady-state model.

I agree with Fred Hoyle concept:
Fred Hoyle - https://en.wikipedia.org/wiki/Fred_Hoyle
"Hoyle was a strong critic of the Big Bang. He is responsible for coining the term "Big Bang" on BBC radio's Third Programme broadcast on 28 March 1949."
"Hoyle, unlike Gold and Bondi, offered an explanation for the appearance of new matter by postulating the existence of what he dubbed the "creation field", or just the "C-field", which had negative pressure in order to be consistent with the conservation of energy and drive the expansion of the universe"
Hence, Fred Hoyle had estimated that galaxies have the ability to produce new matter, but he did not foresee the recent developments and the particle accelerator, and therefore he did not elaborate on how and where the mass could be formed.
He didn't know that the Milky Way's Giant Black Hole Spits Out Its Food:
http://www.space.com/22586-milky-way-giant-black-hole-food.html
"The new findings show definitively that most of the matter in the gas cloud surrounding the black hole is ejected out into space, which explains why it doesn't release light on its way in to be eaten."
This is an indication that new matter is ejected from the Milky Way supper massive black hole.
He also didn't know about the creation of matter and anti matter - the knowledge gained from the accelerator in Europe. Wikipedia: "in November 2010 reported that ATRAP group could develop a new method for producing anti-atoms - hydrogen. Method is based on the slowing down of anti-particles - protons and uniting with slow positrons."
So, as our scientists were able to produce anti-matter in this accelerator, it is clear that the supper massive rotatable black hole is the ultimate natural accelerator which can produce infinite quantity of new matter.

In any case, I agree with you that 10Mpc might be too small for the calculation.
However, as you claim that we can't use the shell theorem in our calculation:.
The shell theorem concerns being off-center in a shell of a uniform field or continuous radiation.
It does not concern anything about more distant things being equally redshifted, something empirically shown to be otherwise.  That difference is how they measure large distances.
So, how can we calculate the CMB in our infinite universe with infinite galaxies which are moving away from each other, while new born galaxies/new matter pop up everywhere in order to compensate (and therefore, the density is kept forever).

With regards to the redshift -

A universe with infinite age predicts no redshift at all.
Can you please explain why you don't see a possibility for redshift in the CMB for infinite Universe with infinite age, with far away galaxies that are moving away from each other at a speed which is much faster than a speed of light (based on Newton)?

Title: Re: How gravity works in spiral galaxy?
Post by: raviraj on 09/05/2019 11:50:17
thanks for this information.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/05/2019 14:29:44
Quote from: Halc
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Therefore, theoretically, if all the new galaxies will move in one direction than after 1370 generations, the last one will move at a speed of light (1370 x 220 Km/s = 301,400 Km sec)
Speeds don't add that way, at least not under relativity.
I fully agree. We need the relatively in order to measure the speed between galaxies.
We don't need relativity to measure them.  We need it to add 220 to 220 like you're doing there.
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I have used Newton in order to prove that based on "galaxies generations idea" we would get the real view that we see from our galaxy
Add 1370 of them together like that and you get about ¾c.  Adding more will not get a recession velocity of more than c.  So your Newtonian conception of space does not correspond to 'the real view that we see from our galaxy'.
That conception of space was already falsified before we had the technology to notice the recession rate being discussed here.
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We can't see galaxies at further locations, as they are moving away from us at a speed which is higher than the speed of light.
That doesn't follow.  We should be able to see all of them by your beliefs.  Light coming from any finite distance will get here in finite time.  High recession speed causes significant redshift but does not itself prevent light from leaving it.  I can hear a supersonic jet that is going away, no matter how fast it goes.  I just can't hear it coming, just like I would not be able to see a galaxy coming at me at greater than c.
And yes, some of those further galaxies are quite visible.
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No, my model doesn't break any law.
We've pointed out it breaking just about every law there is.  You don't have a model, only a belief. The motions predicted by the laws which you claim not to violate differ from the motion you describe with your belief.  That makes for a non-model and a contradictory belief.
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Yes, based on Newton - they can move further away from each other at higher speed than light, if the distance is long enough.
Newton does not have anything in his physics about distance making a difference.
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In my explanation I have used 1370 generations of galaxies which are all moving in only one direction.
Not enough.  That gets only ¾c, and only if each generation emits one galaxy in the same direction and never one in another.
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till the last one which moves away at 301,400 Km/s. (based on Newton)
But you said Newton was wrong in this respect:
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Quote from: Halc
Speeds don't add that way, at least not under relativity.
I fully agree.
So you fully agree at the top of this post that Newton's way is wrong, and yet here you are doing it Newton's way.
You need to pick one.  If Newton's way is wrong, then Newton does not describe how the universe is.
If relativity is wrong, then a thing might move faster than light but would not be able to see anything behind it due to the inability of light to keep up.  Is that how you think the universe is?  You described it looking the same from everywhere, but things would look completely different from one location to the next due to the varying speed of the observer relative to light.  We (Earth, not even the galaxy) would be at the exact center of the universe because only here are we stationary.
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If for example we will stay at galaxy at the middle - (galaxy no 685), the furthest galaxy in one direction is moving at 150,700km/s
Then it should see light moving at half speed in one direction and 1.5c in the other.  Star colors would be vastly shifted from one side of the sky to the other, although since Newton did not support the wave nature of light, he would not have suggested spectrum shift like that.  The night sky on one side would merely have appeared 3x brighter than the other side to an observer moving that fast.
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while on the other side, the furthest galaxy also should move away at the same speed of 15,700 km/s.
No, it would have been stopped on the other side.  Speed is absolute under Newton's universe.  A stopped thing is stopped, and to an observer moving at 150,700 km/s, it is the observer that is moving, not the dim stationary thing he sees when he looks behind him.
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However - I fully agree that we need to measure the velocities between the far end galaxies based on relativity.
Measurement of relative velocity is not different between the two views.  They both measure it as an increase of distance over time.  Both have the same units of say m/sec.
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At some point of distance, the relative velocity is almost the speed of light.
Not almost.
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If we monitor the speed between two galaxies with a separation of 100,000 generations, we should find that based on Newton they are moving away from each other at 220,000,000 Km/s.
What do you mean by 'monitor'?  Anyway, here you correctly state that under Newton, a galaxy (or any object) is not prevented from moving faster than light and thus light cannot catch up to such a thing, just like we observe with sound.
This contradicts your assertion that the view is the same from anywhere in the universe.  It only looks like this if you're nearly stopped, presumably something that happens near the center.
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I'm quite sure that if we will set this modeling in a computer
I have little doubt that you can put such a belief in your head since you have no intention of actually doing it. Whenever you need to push a contradictory assertion, instead of attempting to demonstrate anything, it is always just 'clear to me'.
If you modeled it in a computer, you'd get brownian motion, similar to the expansion of ink when you put a drop of dye in calm water.  If the dye multiplies instead of just spreads, then the calm water will become infinitely dense with dye until it explodes.  If you model your belief on a computer and let it run long enough, there would be no point in space not crammed with matter getting ever more dense.  Light speed is not enough to get rid of it since it is coming in as fast as you can expel it.  Such is a consequence of violation of thermodynamic law.
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It is clear to me that all the nearby galaxies around the Milky way had been created by the Milky way.
Case in point. If you can find a website that depicts a point you want to make, you post it, even if it requires being taken out of context. Points that are too wrong to even do that are just 'clear to me', which means they must be exceptionally wrong.
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So we agree that infinite Universe with the same density (as we see) should get a CMB.
Depends on the model.  You've described several, some with and some without a CMB.
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I hope that we also agree about the black body radiation in the CMB.
The CMB does not come from a body.
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Quote from: Halc
In a steady-state model, the box argument works, but it predicts the wrong temperature.  Are you asserting that stars burn at about 2.7K?  If not, where is the heat going?
The heat of the stars and galaxies are going to the infinity,
And equal amounts are coming from infinity.  That doesn't help.  The heating element of the oven is the temperature of a star, so the CMB would be that temperature.
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while the heat of the CMB is coming from the infinity of all directions.
Yes.  That makes a perfect balance, and the CMB is star hot.
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Therefore, the balance should set the CMB temp at 2.7K.
Stars don't burn at 2.7K.
If the CMB was 2.7K, then our box is radiating more heat than it is taking in from its neighbor, which means the other boxes are different, not uniform.
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I agree with Fred Hoyle concept:
Fred Hoyle - https://en.wikipedia.org/wiki/Fred_Hoyle
"Hoyle was a strong critic of the Big Bang. He is responsible for coining the term "Big Bang" on BBC radio's Third Programme broadcast on 28 March 1949."
"Hoyle, unlike Gold and Bondi, offered an explanation for the appearance of new matter by postulating the existence of what he dubbed the "creation field", or just the "C-field", which had negative pressure in order to be consistent with the conservation of energy and drive the expansion of the universe"
So you're going to jump onto the bandwagon of a theory that has been thoroughly falsified?  At least Hoyle knew something had to be postulated to prevent violation of conservation of energy.
Hoyle still envisioned an expanding universe, not a static Newtonian one.  An expanding steady-state model predicts no CMB, and it was the discovery of the CMB that really put a fatal nail in an alredy dead theory.
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Hence, Fred Hoyle had estimated that galaxies have the ability to produce new matter
He did not.  He postulated new mass, but it didn't come from galaxies.
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http://www.space.com/22586-milky-way-giant-black-hole-food.html
"The new findings show definitively that most of the matter in the gas cloud surrounding the black hole is ejected out into space, which explains why it doesn't release light on its way in to be eaten."
This is an indication that new matter is ejected from the Milky Way supper massive black hole.
There is no mention of 'new matter' in there.
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it is clear that the supper massive rotatable black hole is the ultimate natural accelerator which can produce infinite quantity of new matter.
More for the 'it is clear' heap.
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In any case, I agree with you that 10Mpc might be too small for the calculation.
Pick any size you like and the argument is the same.
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So, how can we calculate the CMB in our infinite universe with infinite galaxies which are moving away from each other, while new born galaxies/new matter pop up everywhere in order to compensate (and therefore, the density is kept forever).
Depends on you model.
If you hold to Newton's non-expanding space, then the CMB should be the temperature of a star since there is nowhere you can look that doesn't hit one if you go far enough.
If space is expanding, then distant things are  redshifted to arbitrarily near 0°K long before their light can reach us.  There would be no CMB unless there was something super bright at a finite distance. Hoyle's steady state idea said it looks the same from anywhere at any time, so no super bright thing at some finite distance. The big bang posits the bright thing (an event more than an object) at a finite distance (due to a finite age of the universe). The temperature of the CMB would be computed as the temperature (3000k) at which hydrogen atoms can form from free protons and electrons, divided by the red-shift computed (1100) for the finite distance to the material which emitted that light.
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Quote from: Halc
A universe with infinite age predicts no redshift at all.
Can you please explain why you don't see a possibility for redshift in the CMB for infinite Universe with infinite age
That was not clearly worded on my part.  A non-expanding universe with infinite age predicts no average redshift.  Objects still move at random velocities in all directions, so some will be redshifted and as many would be blue shifted.  Any expanding universe, finite or infinite age will exhibit a general recession of all objects which will subsequently manifest as a redshift in general accordance with distance.
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with far away galaxies that are moving away from each other at a speed which is much faster than a speed of light (based on Newton)?
Based on Newton, if galaxies moved faster than light, then light would not be able to keep up with them and one would only be able to observe things in one direction but not the other.
Title: Re: How gravity works in spiral galaxy?
Post by: Colin2B on 09/05/2019 15:15:17
thanks for this information.
Did you really only post this so folks would look at your website??
Well, they won’t look anymore as we have changed it to ours, and as a special bonus we have awarded you a ban. Enjoy
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/05/2019 16:42:38
No, my model doesn't break any law.

If you posit that galaxies can move faster than light, then yes, you are breaking a law of physics.

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Yes, based on Newton - they can move further away from each other at higher speed than light, if the distance is long enough.

According to Einstein, they can't. It takes an infinite amount of energy to accelerate an object with mass up to the speed of light. How do you propose to give a galaxy more than infinite energy so that it can go faster than light?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/05/2019 19:19:52
According to Einstein, they can't. It takes an infinite amount of energy to accelerate an object with mass up to the speed of light. How do you propose to give a galaxy more than infinite energy so that it can go faster than light?

Einstein is very clever and he is fully correct.
I fully agree with him that it takes an infinite amount of energy to accelerate an object with mass up to the speed of light.
However, he didn't consider the scenario which I have offered.
Based on my theory, any baby galaxy is ejected from its mother galaxy at only 220Km/s.
This is very normal as most of the stars are ejected from the galaxy at that range of speed.
So, I assume that Einstein has no problem with that.
However, I discuss about ejection over ejection over... ejection.
Each ejection represents a speed of only 220 Km/s.
Let me use the following example-
Let's assume that we can create a rocket with 1370 stages.
So, the main rocket carries 1369 rockets. The second rocket carries 1368 rockets,.... the last one doesn't carry any more rockets.
It works as follow:
We launch the main rocket from a fixed point in space. As it gain a speed of 220Km/s, it launches the second rocket. As the second rocket gain a speed of 220Km/s with regards to the main rocket, it launches the third rocket.
So, let me ask you the following:
Do you agree that the speed of the second rocket (at the moment of launching the third rocket) is 440 Km/s with regards to the fixed starting point?
If so, let's continue with the launching process with the 4th 5th and the other entire rockets till the last one.
So, do you see any violation of law in this process?
What is the relative velocity between the fixed point in space to the last rocket?
What Einstein might say about it?
Why can't we add all the velocities?
I really don't see any violation in law.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/05/2019 20:10:52
Most of your post was just repetition again, but I have a few comments.

Based on my theory, any baby galaxy is ejected from its mother galaxy at only 220Km/s.
How do you compute this figure? Other than pulling it out of, um, somewhere....
It actually can be computed from prior things which 'are clear to you', but I suspect you're not up to the math involved.  It doesn't work out to 220, hence my asking.

It seems to be awful fast.  Hubble's law says things need to be over 3 Mpc away before they recede at that rate.  The average rate of recession of things closer than that is much less.
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So, I assume that Einstein has no problem with that.
Please don't suggest that Einstein would have no problem with anything you've stated.  You contradict him with every post.
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Let me use the following example-
Let's assume that we can create a rocket with 1370 stages.
So, the main rocket carries 1369 rockets. The second rocket carries 1368 rockets,.... the last one doesn't carry any more rockets.
It works as follow:
We launch the main rocket from a fixed point in space. As it gain a speed of 220Km/s, it launches the second rocket. As the second rocket gain a speed of 220Km/s with regards to the main rocket, it launches the third rocket.
So, let me ask you the following:
Do you agree that the speed of the second rocket (at the moment of launching the third rocket) is 440 Km/s with regards to the fixed starting point?
No. That's the whole point. Velocities don't add like that, at least not in the last 100 years. You've displayed no understanding of Newton's physics, so how should I expect you to understand Einstein's?
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What is the relative velocity between the fixed point in space to the last rocket?
¾c (approximately), relative to the frame in which the ship was initially stationary.  There are no fixed points in space, just fixed points in spacetime, and velocity is undefined relative to one of those.
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What Einstein might say about it?
That is what he says.
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Why can't we add all the velocities?
We can, just not the way you're doing it.  You're describing the addition of proper velocity, where each stage accelerates by 220 km/s relative to the frame where it's engine gets lit.  From the pilot's point of view, accelerating at say 1G, each stage burns for just over 6 hours, when the next one kicks in.  After 1370 stages (not quite a year, measured on the pilot's clock), he's moving at ¾c relative to his starting frame.

You can tack on as many stages as you like, but you'll never get to light speed.  The guy riding in front will feel acceleration forever, never dropping off.

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I really don't see any violation in law.
Despite the fact that the violations are point out over and over? Yes, I notice that you just don't see things you don't want to see. Ignoring contradictions because you find them inconvenient is not scientific, nor even rational thought.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/05/2019 21:14:40
Einstein is very clever and he is fully correct.
I fully agree with him that it takes an infinite amount of energy to accelerate an object with mass up to the speed of light.
However, he didn't consider the scenario which I have offered.

As a matter of fact, he did. He deduced that you cannot add velocities linearly when the speed of the objects is relativistic. Here is a link explaining it: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html

Based on my theory, any baby galaxy is ejected from its mother galaxy at only 220Km/s.
This is very normal as most of the stars are ejected from the galaxy at that range of speed.
So, I assume that Einstein has no problem with that.
However, I discuss about ejection over ejection over... ejection.
Each ejection represents a speed of only 220 Km/s.
Let me use the following example-
Let's assume that we can create a rocket with 1370 stages.
So, the main rocket carries 1369 rockets. The second rocket carries 1368 rockets,.... the last one doesn't carry any more rockets.
It works as follow:
We launch the main rocket from a fixed point in space. As it gain a speed of 220Km/s, it launches the second rocket. As the second rocket gain a speed of 220Km/s with regards to the main rocket, it launches the third rocket.
So, let me ask you the following:
Do you agree that the speed of the second rocket (at the moment of launching the third rocket) is 440 Km/s with regards to the fixed starting point?

At such low speeds, linear velocity addition is an okay assumption.



If so, let's continue with the launching process with the 4th 5th and the other entire rockets till the last one.
So, do you see any violation of law in this process?

If you're assuming that the velocities always add together linearly, then yes, there is a problem because that doesn't happen when velocities begin to reach a significant fraction of the speed of light. If one stage of the rocket has reached a speed of 200,000 km/s and then launches another stage at a speed of 220 km/s relative to itself, then the observed velocity of that next stage for an outside observer is not 200,220 km/s, but rather 200,122 km/s. This is less than the sum of the two velocities. That problem gets worse and worse as you get closer and closer to the speed of light. The reason for this seeming discrepancy is length contraction, time dilation and relativistic mass gain. Here is a calculator that will let you do your own calculations for relativistic velocity addition: https://www.omnicalculator.com/physics/velocity-addition

What is the relative velocity between the fixed point in space to the last rocket?

I don't know because I don't feel like using that calculator 1,370 times in order to find the answer. It would definitely be less than the speed of light, though.

What Einstein might say about it?

He would say that you always get a total velocity less than that of light: https://en.wikipedia.org/wiki/Velocity-addition_formula

Why can't we add all the velocities?

Relativistic mass gain. The faster something is moving, the heavier it gets, and so takes even more energy to accelerate it. It's a positive feedback loop that approaches infinity as you near the speed of light. Each stage of the rocket gets heavier and heavier as you try to make it go faster, requiring more and more energy. You can never give that last stage enough energy to push it up to the speed of light.

I really don't see any violation in law.

Then you need to study relativity in greater detail.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/05/2019 05:22:32
He would say that you always get a total velocity less than that of light: https://en.wikipedia.org/wiki/Velocity-addition_formula
Thanks
In the article it is stated:
"According to the theory of special relativity, the frame of the ship has a different clock rate and distance measure, and the notion of simultaneity in the direction of motion is altered, so the addition law for velocities is changed. "
So, they discuss about the "frame of the ship". It is clear that for local aria those formulas are perfectly OK.
However, we discuss now about the infinite open space and therefore It is totally different scale.
Einstein didn't know the total size of our Universe. So, if we just focus on a limited aria - those formulas are perfectly OK.
Let me ask you the following based on the following example:
We stay at galaxy B
If we look to our left side - we see galaxy A at a distance of 13 BLY that is moving away at almost the speed of light.
If we look to exactly to the opposite side (right) - we see galaxy C at a distance of 13 BLY that is moving away at almost the speed of light.
Therefore -
The relative velocity between A to B is almost the speed of light, the relative velocity between B to C is also almost the speed of light - and A B C is located on the same direct line.
Hence, the distance between A to C is 26 BLY.
What is the relative velocity between galaxy A to galaxy C?

Let's make it more difficult:
If we jump to galaxy C, I assume that we should see a similar view.
If we look ahead at the same ABC line we see a galaxy D at a distance of 13 BLY which also moving away at almost the speed of light.
Let's continue to jump from D to E and so on till the 11th galaxy - K.
Hence, there are 10 segments of 13BLY between those 11 galaxies.
The distance between galaxy A to galaxy K is 130 BLY.
In each segment we see that the relative speed between the galaxies (on that segment) is almost the speed of light.
So, what is the relative velocity between A to K?
Is it still " less than that of light"?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/05/2019 05:57:03
Thanks
In the article it is stated:
"According to the theory of special relativity, the frame of the ship has a different clock rate and distance measure, and the notion of simultaneity in the direction of motion is altered, so the addition law for velocities is changed. "
So, they discuss about the "frame of the ship". It is clear that for local aria those formulas are perfectly OK.
However, we discuss now about the infinite open space and therefore It is totally different scale.

Reference frames are about direction, speed or acceleration (or equivalently, in gravitational fields). Distance isn't what it is about. A ship 2 feet away from that first ship that is travelling in a different direction or a different speed or deeper into a gravitational potential is in a different reference frame from the first ship, whereas a ship a billion light-years away travelling in the same direction at the same speed/acceleration is in the same reference frame as the first ship.

Einstein didn't know the total size of our Universe.

He didn't need to because distance doesn't matter.

Let me ask you the following based on the following example:
We stay at galaxy B
If we look to our left side - we see galaxy A at a distance of 13 BLY that is moving away at almost the speed of light.
If we look to exactly to the opposite side (right) - we see galaxy C at a distance of 13 BLY that is moving away at almost the speed of light.
Therefore -
The relative velocity between A to B is almost the speed of light, the relative velocity between B to C is also almost the speed of light - and A B C is located on the same direct line.
Hence, the distance between A to C is 26 BLY.
What is the relative velocity between galaxy A to galaxy C?

If galaxy A is moving away from galaxy B at 90% the speed of light to the left while galaxy C is moving away from galaxy B at 90% the speed of light to the right, then the speed that galaxy C looks like it's moving from the reference frame of galaxy A would be 99.45% the speed of light (based on the calculator I posted earlier).

Let's make it more difficult:
If we jump to galaxy C, I assume that we should see a similar view.
If we look ahead at the same ABC line we see a galaxy D at a distance of 13 BLY which also moving away at almost the speed of light.
Let's continue to jump from D to E and so on till the 11th galaxy - K.
Hence, there are 10 segments of 13BLY between those 11 galaxies.
The distance between galaxy A to galaxy K is 130 BLY.
In each segment we see that the relative speed between the galaxies (on that segment) is almost the speed of light.
So, what is the relative velocity between A to K?
Is it still " less than that of light"?

Yes, it's still less than the speed of light and for the same reason that the speed of C relative to A is below the speed of light. It gets closer and closer with each galaxy, but never quite gets there.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/05/2019 07:38:57
Thanks
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The distance between galaxy A to galaxy K is 130 BLY.
In each segment we see that the relative speed between the galaxies (on that segment) is almost the speed of light.
So, what is the relative velocity between A to K?
Is it still " less than that of light"?
Yes, it's still less than the speed of light and for the same reason that the speed of C relative to A is below the speed of light. It gets closer and closer with each galaxy, but never quite gets there.

If our universe is infinite, there must be infinite segments of galaxies in a row that are moving away from each other (at the same segment) at almost the speed of light.
Therefore, if we take those infinite segments with almost speed of light (in each segment)- you claim that the relatively speed between the first one to the infinity one is still less than the speed of light.
Wow.
How can we believe in this answer?

If I recall it correctly, in one of your answers you have stated that due to the expansion, there is a possibility that far end galaxies are moving away from each other faster than the speed of light.
If so, how could it be that the relative velocity between the galaxies is less than the speed of light, while we know that they are moving away from each other at a speed which is faster than the speed of light?


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/05/2019 13:36:26
Quote from: Kryptid
Yes, it's still less than the speed of light and for the same reason that the speed of C relative to A is below the speed of light. It gets closer and closer with each galaxy, but never quite gets there.
If our universe is infinite, there must be infinite segments of galaxies in a row that are moving away from each other (at the same segment) at almost the speed of light.
This is why I didn't like answering this question with the SR answer.  Yes, in the flat SR universe I described, there are infinite galaxies, the vast majority of which are moving just under light speed from any given point, and hence are way to young to actually be galaxies yet.  You can't talk about them being galaxies because they're not.  They're not even matter yet.  So mathematically, if you view such a universe as populated with mathematical points, there are infinitely many such points and they're almost all moving away at nearly c.  This model doesn't correspond to our universe since we have accelerating expansion, which pushes all but a finite amount of material past the event horizon that doesn't exist in the SR model.  Space is not flat, and these distant galaxies don't exist at all in our reference frame, nor do we exist in theirs.  Mutually nonexistent things don't have relationships, and coordinate velocity is such a relationship.

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Therefore, if we take those infinite segments with almost speed of light (in each segment)- you claim that the relatively speed between the first one to the infinity one is still less than the speed of light.
In the SR universe, yes. There is no infinity one. The most distant point in space is 13.8 BLY from Earth in Earth's reference frame, so these not-galaxies are no further than that, and Hubbles-law says that anything in the Hubble Sphere is moving away at sub-light speed. Again, we don't live in a SR universe. I'm using it to illustrate the concepts.

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If I recall it correctly, in one of your answers you have stated that due to the expansion, there is a possibility that far end galaxies are moving away from each other faster than the speed of light.
This is not the SR universe. So yes, anything outside the Hubble sphere is moving faster than light, as measured in comoving coordinates.
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If so, how could it be that the relative velocity between the galaxies is less than the speed of light, while we know that they are moving away from each other at a speed which is faster than the speed of light?
Things cannot have a coordinate speed faster than c in an inertial reference frame (the coordinate system), but no inertial frame covers all of space, so things outside that frame don't have a defined coordinate speed.  So we use comoving coordinates since that does actually cover all of spacetime. The coordinate system is a little weird since the length of a meter stick changes with time, and there are no meaningful inertial frames.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/05/2019 17:30:59
If our universe is infinite, there must be infinite segments of galaxies in a row that are moving away from each other (at the same segment) at almost the speed of light.
Therefore, if we take those infinite segments with almost speed of light (in each segment)- you claim that the relatively speed between the first one to the infinity one is still less than the speed of light.
Wow.
How can we believe in this answer?

Because it's what the laws of physics say. You can't go faster than light.

If I recall it correctly, in one of your answers you have stated that due to the expansion, there is a possibility that far end galaxies are moving away from each other faster than the speed of light.
If so, how could it be that the relative velocity between the galaxies is less than the speed of light, while we know that they are moving away from each other at a speed which is faster than the speed of light?

The metric expansion of space is not the same thing as moving through space. The galaxies are not actually moving away from each other faster than light, it's that the space between them is expanding faster than light. Actually, we can't even see any galaxies that are receding from us faster than light (obviously). We assume that they are out there because we assume that the universe still continues beyond our ability to see it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/05/2019 22:38:52
Actually, we can't even see any galaxies that are receding from us faster than light (obviously). We assume that they are out there because we assume that the universe still continues beyond our ability to see it.
So, do you mean that you assume that there are galaxies that are moving away from us faster than the light?

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If I recall it correctly, in one of your answers you have stated that due to the expansion, there is a possibility that far end galaxies are moving away from each other faster than the speed of light.
This is not the SR universe. So yes, anything outside the Hubble sphere is moving faster than light, as measured in comoving coordinates.
I have tried to understand the meaning of comoving coordinates:
It is stated:
https://en.wikipedia.org/wiki/Faster-than-light
"Rules that apply to relative velocities in special relativity, such as the rule that relative velocities cannot increase past the speed of light, do not apply to relative velocities in comoving coordinates, which are often described in terms of the "expansion of space" between galaxies. "
"However, because the expansion of the universe is accelerating, it is projected that most galaxies will eventually cross a type of cosmological event horizon where any light they emit past that point will never be able to reach us at any time in the infinite future.."
"There are many galaxies visible in telescopes with red shift numbers of 1.4 or higher. All of these are currently traveling away from us at speeds greater than the speed of light. Because the Hubble parameter is decreasing with time, there can actually be cases where a galaxy that is receding from us faster than light does manage to emit a signal which reaches us eventually"
So, there is a clear message that galaxies can move faster than light.
I didn't ask about Hubble. So, it can be in or out - but it is there!
Because it's what the laws of physics say. You can't go faster than light.
Why do we insist that the same law that works locally should also works at the infinity?
Why don't we adjust our physics law based on what we see (or actually - don't see)?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/05/2019 00:12:21
So, do you mean that you assume that there are galaxies that are moving away from us faster than the light?

Yes, because there's no particular reason that the universe should stop conveniently at the exact the point where those photons from the most red-shifted galaxies came from.

Why do we insist that the same law that works locally should also works at the infinity?

Because we have yet to find a reason to think otherwise. There are no cases that I know of where scientists have detected with high confidence that the laws of physics are different far away from us than they are close to us.

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So, there is a clear message that galaxies can move faster than light.

I already explained to you that moving through space faster than light is not the same thing as receding faster than light due to the metric expansion of space. The first is forbidden by relativity, the second is not. Nothing need be actually moving in the second case.

Why don't we adjust our physics law based on what we see (or actually - don't see)?

We do. There is not yet any need to alter relativity because we have yet to find any cases where the speed of light limit is broken.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/05/2019 15:23:02
Speed is not a property.
A rock cannot go 1 km/sec, but it can go 1 km/sec relative to another rock.
Yes, I fully agree.
Therefore I have asked:
So, what is the relative velocity between A to K?
and...
If so, how could it be that the relative velocity between the galaxies is less than the speed of light, while we know that they are moving away from each other at a speed which is faster than the speed of light?


Quite so.  You can see things that are receding faster than light.  Such objects are the dotted black lines.  If they touch the red line, we can see them today.  Notice the dotted lines are vertical in the comoving coordinates, meaning comoving objects are stationary, and almost all galaxies are within a percent of being comoving.  See the 2nd vertical dotted line at about 21 BLY?  That one is outside the Hubble Sphere and always has been, but the red line touches it at age ~3BY so we can see that galaxy if it had formed before the universe was 3BY old.  It is moving at about 1.2c and is currently outside the event horizon which means we can never see it after it's state when the universe was 9 billion years old (about when life began here) where it touches the orange line.  Light from after that event will never get here.
Thanks for the excellent explanation.
Now I start to understand the answers for some of my questions.
However, it is based on the BBT theory (as the time frame is a direct outcome of the BBT -13.8 BY).
This is quite problematic.
How can we confirm or disconfirm one theory based on another theory which fully contradicts the first one?
In this thread we discuss on new Theory.
As it is new theory - it is forbidden to take your argument from another theory.
You can use any confirmed law. That is perfectly OK. But sorry - you can't reject theory D based on BBT assumption.
There are big gap between the two theories as follow:
1. Universe age: BBT - 13.8BLY, D - Infinite
2. Universe size: BBT - finite. D - Infinite
3. CMB: BBT - Direct product of the bang, D - Direct product of the infinite Universe.
4. Mass creation: BBT - no mass creation, only transformation or evolvement. D - Constantly new mass creation in low quantity at the execration disc of spiral galaxy
5. Receding faster than light: BBT -
receding faster than light due to the metric expansion of space
D - Due to "galaxy generation" or rocket over rocket phenomenon. (no need for metric expansion of space)
and many more.
So, based on theory D there is no limit in time frame as there is in the BBT.
If we discuss just about the time frame (age)
This is my estimation for the age -
The Solar system age -
Based on theory D all the Planets and Moons had been formed at the same day with our Sun. All the rocky planets/moons had been formed as a hot gas objects - with the same matter as the Sun and in the same gas cloud around the SMBH.
With regards to the Earth - its current size is less than 2% from its size in the first day (that represents the solid matter ratio in the gas cloud).  So, in order to understand the minimum age of the Earth, we must verify the requested time for Earth to eject all the gas from the planet (Hydrogen, Hellion...).
Our scientists assume that the earth had got its rock shape about 4 BY. Therefore, I estimate than the minimum age of the earth (or the moon) is 400BY. This is also set the requested time for the solar system to drift outwards from the gas cloud near the SMBH to our current location.
I also estimate that the age of the Milky Way is at least 40,000BY. (From the moment of a compact BH with a small excretion disc till this day). The age of Andromeda must be much more than that.
Therefore, It is clear to me that there is a sever mistake about the "time frame" of the Universe based on the BBT.
Therefore, the following answer is not applicable for theory D
No galaxy existed back then.  The ones we see are much closer, as evan_au points out.
You base your answers on a the assumption that our universe is quite young (13.8BY). That leads to sever misunderstanding about the feasibility of the receding faster than light.
Actually, if you take that diagram to the infinity age & infinity Size - you would find that most of the galaxies are receding faster than light.
I think that there is high similarity between the rocket over rocket theory (based on theory D) to the space expansion (based on BBT). I can show it if you wish.

With regards to the CMB:
The matter that emitted that CMB light we see is now about 45 GLY (proper distance) away.  We don't see stuff like it is now, we see things in the past. That matter has been moving at more than 3x light speed relative to us, so it was much much closer when that light was emitted.  Just divide 45 by 13.8 billion years and multiply it by 379000, and you get something like 1¼ light years.
You also base your understanding about the CMB on the 13.8 BY time frame.
So, I can fully agree and understand your following answer:
In an inertial frame, the relative velocity is a relation between two objects at some point in time.  Those distant galaxies don't exist at any point in time in our reference frame, so there is no defined velocity relation in that framework. 
But again - In theory D the Universe time frame (age is infinite)
That is another error. Remember - In theory D the Universe is infinite in its age and in its size. Therefore, we must extract the projected amplitude/features of the CMB based on the assumptions that the Universe is Infinite in its size and in its age. We also must add to that the idea that galaxies that are located far enough - must move away from us higher than the speed of light. So, we need to understand what is the maximal distance that we still get radiation from galaxies.
I had the impression that if the galaxy is moving faster than the speed of light - we shouldn't get its radiation.
 Would you kindly explain how could it be that there is a possibility to get in a future a radiation from a galaxy which is moving today at the speed faster than light.
 Again - Please do not mix up with the BBT theory and claim that in the past they we were close enough. That idea isn't relevant in this infinite Universe.

It isn't about infinity.
It seems to me that it is all about infinity.
If you knew that our universe is infinite in its size & age - the whole story about the BBT was totally different.
A very good idea.  So why is it you are positing all this nonsense that nobody sees?
Why do you all insist on the BBT
What is needed to convince you that the time frame of our real universe is much bigger than 13.8BLY and our universe is infinite in its size.
What kind of data would convince you that there is a sever mistake with the BBT?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/05/2019 15:33:37
Constantly new mass creation in low quantity at the execration disc of spiral galaxy

As we've said before, we can throw this idea out then because it violates conservation of mass. No known process in the universe creates new mass. It is only transformed from one form to another.

D - Due to "galaxy generation" or rocket over rocket phenomenon. (no need for metric expansion of space)
and many more.

We already explained to you that you can't exceed the speed of light that way. Relativity forbids it. You can't cheat your way past the speed of light like that. So that's two laws of physics that your idea violates. No one has ever seen mass being created or objects with mass moving faster than light.

If I was aboard the first stage of your hypothetical rocket and tried to measure the kinetic energy of the final stage of the rocket (the one that you claim should be going faster than light), what do you think I would measure the kinetic energy as? It would be more than infinite, which makes no sense. Where did that final rocket stage get its more than infinite kinetic energy from? It would have to have come from the rocket fuel carried by the previous stages plus whatever fuel it carried. So explain to me how you can carry more than an infinite amount of rocket fuel on any real rocket.

Why do you all insist on the BBT

It's what the evidence supports.

What is needed to convince you that the time frame of our real universe is much bigger than 13.8BLY and our universe is infinite in its size.
What kind of data would convince you that there is a sever mistake with the BBT?

Any kind of verifiable scientific evidence.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/05/2019 17:51:19
Therefore I have asked:
So, what is the relative velocity between A to K?
Depends on the way you measure velocity.  So in a universe with finite age of 13.8 BY, A and K are moving at about 9c in comoving coordinates.  The two don't exist in any one inertial frame, so they can't have that kind of velocity relative to each other.

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If so, how could it be that the relative velocity between the galaxies is less than the speed of light, while we know that they are moving away from each other at a speed which is faster than the speed of light?
The relative speed between those two galaxies is not sub-light speed, at least not in the finite age universe.  In a universe described by SR, speeds add up to something less than but nearly c, as per the velocity adder.  A and K would be able to see each other in such a universe.

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Quote from: Halc
Quite so.  You can see things that are receding faster than light.  Such objects are the dotted black lines.  If they touch the red line, we can see them today.  Notice the dotted lines are vertical in the comoving coordinates, meaning comoving objects are stationary, and almost all galaxies are within a percent of being comoving.  See the 2nd vertical dotted line at about 21 BLY?  That one is outside the Hubble Sphere and always has been, but the red line touches it at age ~3BY so we can see that galaxy if it had formed before the universe was 3BY old.  It is moving at about 1.2c and is currently outside the event horizon which means we can never see it after it's state when the universe was 9 billion years old (about when life began here) where it touches the orange line.  Light from after that event will never get here.
Thanks for the excellent explanation.
Now I start to understand the answers for some of my questions.
However, it is based on the BBT theory (as the time frame is a direct outcome of the BBT -13.8 BY).
Yes, that picture makes no sense outside the BBT.  You can't use it to describe other theories.

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This is quite problematic.
No it isn't.  Draw a different picture to match a different theory.
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How can we confirm or disconfirm one theory based on another theory which fully contradicts the first one?
Find an empirical difference.  The CMB is such a difference since only BBT predicts it.  The discover of it was a fatal blow to Hoyle's theory which was already falsified in other less fatal ways.

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In this thread we discuss on new Theory.
As it is new theory - it is forbidden to take your argument from another theory.
That's right.  Can't use my picture.  Can't just assert that stuff is the way we see it.  You have to justify your assertions: show that your belief is consistent with the numbers you claim.

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you can't reject theory D based on BBT assumption.
I said your beliefs are wrong because they contradict themselves, not because they contradict BBT.

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There are big gap between the two theories as follow:
4. Mass creation: BBT - no mass creation, only transformation or evolvement.
Lack of mass creation is not a BBT thing.  It is a law of thermodynamics, which is not based at all on BBT.  Take away BBT and this law still stands, as do all the laws of motion that are so heavily violated by your VHP thing.

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5. Receding faster than light: BBT -
D - Due to "galaxy generation" or rocket over rocket phenomenon.
Rocket over rocket will not produce faster than light speeds.

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So, based on theory D there is no limit in time frame as there is in the BBT.
You mean there is not a finite age to the universe relative to now?  Yes, I agree that such a thing would go away if you take away BBT.  There are cyclic models that say time is infinite with repeated bangs and the one we know is just the most recent bang.  That's still BBT, but with serial bangs, and it has a hard time explaining how expansion is accelerating when it predicts a slowing of it.

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You also base your understanding about the CMB on the 13.8 BY time frame.
Yes, the only theory that predicts one. Your belief merely asserts one because you know about it.  Predicting something after it is known is not prediction, just observations.  You cannot explain it, so you resort to assertions.
I showed how your belief predicts the entire sky as bright as the sun, the same temperature that your oven would get to after infinite time baking like that with nowhere for the heat to go.  Yes, you assertions result in a CBR, but not a microwave one at 2.7°.

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So, I can fully agree and understand your following answer:
Quote from: Halc
In an inertial frame, the relative velocity is a relation between two objects at some point in time.  Those distant galaxies don't exist at any point in time in our reference frame, so there is no defined velocity relation in that framework.
But again - In theory D the Universe time frame (age is infinite)
Fine.  Then all those galaxies exist in a nearly inertial frame and exist in such a coordinate system.  Why can't we see them?  Another empirical falsification of this D belief.

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Remember - In theory D the Universe is infinite in its age and in its size. ... We also must add to that the idea that galaxies that are located far enough - must move away from us higher than the speed of light.
No they don't, unless you deny relativity theory, or you take the expansion of space assertion from BBT (which makes it hard to posit infinite age).  If you posit space expansion, then the CBR is wiped out since there is nothing from which it might have been emitted.

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I had the impression that if the galaxy is moving faster than the speed of light - we shouldn't get its radiation.
 Would you kindly explain how could it be that there is a possibility to get in a future a radiation from a galaxy which is moving today at the speed faster than light.
I didn't say that.  In a non-expanding universe, it isn't possible to move at greater than light speed. You need to deny relativity to do that, and I cannot tell you how such a universe would behave since I know of no consistent theory that denies relativity.
I did say that if an object can move faster than light, you would not be able to see it coming, but you would be able to see it receding if you're not outrunning light yourself.  The physics just gets stupid and predicts that most places can see almost nothing since almost nothing is stationary enough to allow light to keep up.

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Quote from: Halc
It isn't about infinity.
It seems to me that it is all about infinity.
I showed an example of an object moving at greater than c only 350000 km from here.  I'm saying you don't need to invoke infinity for an example of it.

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A very good idea.  So why is it you are positing all this nonsense that nobody sees?
Why do you all insist on the BBT
I don't think I mentioned BBT there.  I just noticed that you consistently fail to adjust your physics law based on what is seen.  You commit the very fallacy that you accuse of us.  That's being hypocritical.

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What is needed to convince you that the time frame of our real universe is much bigger than 13.8BLY and our universe is infinite in its size.
A viable alternative model (which would not convince, but at least allow a 2nd interpretation).  To convince, the BBT would also need to be falsified.
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What kind of data would convince you that there is a sever mistake with the BBT?
Different evidence than it predicts.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/05/2019 20:08:58
Thanks Halc & Kryptid

Do appreciate your great support.
I should be in Nederland till the end of the month.
We will hopefully continue our discussion upon my return.
Best Regards
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 13/05/2019 22:13:39
Have fun!
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/05/2019 18:14:43
Dank je wel

We are visiting my parents in law. (My wife is also dutch).
I would love to live there, but my wife prefers to stay at our current location.

Veel succes en geluk
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/06/2019 06:19:12
Hello Allemmal

After a very relaxing vacation I'm back.
I had the time to reconsider few ideas in our discussion.
I would like to start with - hypersphere
There is nothing about our universe that is incompatible with it being a hypersphere.
The universe being a hypersphere is not incompatible with our ability to see only three dimensions. A two-dimensional creature living embedded in the surface of a balloon is only aware of the two dimensional surface that they live in even though the balloon itself is three-dimensional. If the balloon was large enough, the creature wouldn't even be aware of the balloon's curvature and they could just as easily believe that their universe was a completely flat, 2-dimensional space. In the same manner, the universe being a sufficiently-large hypersphere would go unnoticed by us because the curvature would be too gradual to detect.
Based on the hypersphere idea, if one of us could jump all the way to the left, and the other one will jump all the way to the right - at some point they might meet with each other. (As a surface of a balloon).
That meeting point represents a distance from our location. Let's assume that the distance is symmetrical distance from both sides (left & right) and its value is X light year away (from our location).
Let's also assume that a galaxy which is located at that distance must move away from us at a relative velocity of n * c
Hence, a galaxy which is located at a distance of  x to the left is moving away from us at  n * c, while a galaxy which is located at the same distance to the right is also moving at n * c.
Therefore, at the meeting point of that hypersphere, we might have one galaxy from the left which is moving at a velocity of n * c, while from the other side (right) another galaxy is coming at a velocity of n * c.
So, theoretically at that meeting point, we might have two galaxies which are moving head to head at a relative velocity of 2n *c.
Do you agree with that?
If that is correct, do you agree that it contradicts the whole idea of the hypersphere?
In other words -
There is no way that we are living on a "surface of balloon".
Hence, the space of our universe has no limit. It must be infinite.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/06/2019 15:21:45
Do you agree with that?
If that is correct, do you agree that it contradicts the whole idea of the hypersphere?

No, because you're ignoring the fact that the "movement" of the galaxies in this model is actually just the balloon expanding.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/06/2019 19:52:16
Welcome back!  BTW, my wife has only traces of dutch in her, but I'm almost 100%, despite never having lived there.  My mother emigrated as a refugee after the flood of 53, taking her fiancee with her.  All my blood relatives on her side live below sea level.
Thanks
Wow, what a life story...
No.  They're the same galaxy, and it is receding (increasing its comoving proper distance) from us at n*c in any direction.  Two dots on opposite side of a balloon are separated by X in any direction, and are moving apart as the balloon inflates.
Ok
That is clear.
However, do you agree that if the space is represented as a "surface of a balloon" - somehow, we must monitor a curvature in space?
So, why we can't see that curvature in space?
Why it is only a theoretical  idea?
Where is the real evidence for that curvature?
Let me use the surface of the Earth as an example -
"What is the minimum distance over water that, using a laser, the curvature of the earth can be clearly demonstrated?"
https://www.quora.com/What-is-the-minimum-distance-over-water-that-using-a-laser-the-curvature-of-the-earth-can-be-clearly-demonstrated-using-ordinary-household-measuring-devices
"The Earth’s curvature works out to 7.98 inches over one mile*"
7.98 inches = 7.98 *2.54 cm = 20 Cm
1 Mile = 1.6 km
So, in 1Km the curvature is 20/1.6 cm = 12.5 cm.
In 100 Meter the curvature is 1.25 cm.
I assume that based on a laser we should monitor that curvature (1.25 cm) even at 100M = 0.1 km.
Let's assume that below that distance, we couldn't verify the curvature.
The Earth's circumference is 40,075 km.
The ratio between 100 M to the Earth's circumference is:
0.1 km to 40,000 km = 1 to 400,000
In the same token - when we look at a distance of 13 Billion LY we don't see any curvature.
So, the minimal Space circumference should be:
13 BLY * 400,000 = 5,200,000 Billion Ly
Therefore, if you wish to believe in space curvature - the minimal Space circumference must be 5,200,000 Billion Ly.
So how can we create that size of Universe in only 13.8 Billion years?

Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 01/06/2019 22:54:41
Quote from: Dave Lev
I assume that based on a laser we should monitor (the Earth's) curvature (1.25 cm) even at 100M = 0.1 km.
There is an experiment proposed for the 2030s that would search for gravitational waves using lasers bounced of satellites, perhaps 2.5 million km apart.

The goal was to look for tiny variations in the phase of a laser beam over this distance - looking for variations of around 20 picometers with periods ranging from 1 second to 10 hours.

It is outside the mission goals, but the same equipment could be used in the same way as a surveyor's laser to measure the distance between the satellites with an accuracy of perhaps 1nm over a period of a year. That may reveal some deviation from flat space due to the nearby Sun (superimposed on orbital disturbances from Earth, the Moon, Jupiter, Venus and Mars - and the Milky Way itself).

As a bonus, it is expected that this device could get a measure of the expansion of the universe (the Hubble constant) which is independent of the two current (conflicting) answers based on various methods.

However, measuring deviation of intergalactic space from perfectly flat space of the universe would be beyond such a device.
See: https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/06/2019 21:32:59
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However, do you agree that if the space is represented as a "surface of a balloon" - somehow, we must monitor a curvature in space?
Not if the curvature is below the sensitivity of the instruments to measure it.
Thanks
So, you agree that based on our instruments we do not see any curvature in our Universe (as far as we can see and monitor - At least 13 Bly for any direction)
If the surface of Earth was 2D space, how could the 2D creatures on it measure the curvature?  Not by using altitude.  If space went in that direction, it wouldn't be 2D space.  Likewise, there is no direction in the hypersphere that is 'above' the surface that represents a specific moment in comoving time.  You can't see along the surface anyway.  You can only see light that comes from the past.
How can we claim in one hand that based on our mathematical theory the Universe should have a curvature, but on the Other hand we claim that there is no way for you to validate/prove this theory by real verification?
Our scientists must find a way how to verify the curvature in our Universe!
This is a mandatory request, otherwise, the whole idea of the curvature is problematic.
It was s a nice concept to place at the same diagram the space & time.
Based on this 4D concept, we have got an outcome that the space could be considered as "surface of a balloon"
However, if we could find that based on our formula, the Universe acts as a "surface of a balloon", we also must extract the Minimum size of that "Balloon"
In other words - as we see no curvature up to a minimal radius of 13 Billion light year, we must find what is the minimal circumference of  "surface of a balloon" that is needed to support that none curvature at that radius.

I fully accept/agree with the following answer from evan_au:
There is an experiment proposed for the 2030s that would search for gravitational waves using lasers bounced of satellites, perhaps 2.5 million km apart....
If I understand it correctly, our scientists try to find a breakthrough technology/ideas in order to give an answer to some of key questions including the idea of curvature.
Therefore, If for example, we might find in the future that there is some curvature in our "surface of a balloon" space, than based on that curvature verification, we could technically calculate what is the minimal circumference of  "surface of a balloon" that supports that kind of curvature.
Let's use the circumference of the Earth as an example-
At the early time, our scientists didn't really measure the circumference of the Earth.
They just measured the value of the surface Earth curvature, and extract mathematically the circumference of the Earth.
In the same token, we must calculate the value of the minimal circumference of the surface of a balloon/space that can still support a radius of 13 Billion LY universe without any verification for curvature.
That minimal space/universe size is vital information for our theory.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/06/2019 22:09:16
Our scientists must find a way how to verify the curvature in our Universe!
This is a mandatory request, otherwise, the whole idea of the curvature is problematic.

No, it isn't problematic. All it means is that it might be curved or it might not be curved. If it's possible that the Universe is closed, curved and finite in size, then you can't say that you've proven it to be infinite. You would somehow have to rule out even the possibility of a hyperspherical universe of finite size first. No experiment has done that.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/06/2019 23:05:12
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However, do you agree that if the space is represented as a "surface of a balloon" - somehow, we must monitor a curvature in space?
Not if the curvature is below the sensitivity of the instruments to measure it.
So, you agree that based on our instruments we do not see any curvature in our Universe (as far as we can see and monitor - At least 13 Bly for any direction)
I don't see how looking a long distance would help.  You need to propose a way to measure it.  They've done it over a distance of say 50 km, but no curvature was measured at that scale.  Nobody has tried something significant like billions of light years.  We're not capable of putting instruments that far away.

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How can we claim in one hand that based on our mathematical theory the Universe should have a curvature,
Who said that?

Maybe it's finite but not curved.  Read my other post.   Maybe the finite/infinite distinction depends on how the size is measured.

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In other words - as we see no curvature up to a minimal radius of 13 Billion light year, we must find what is the minimal circumference of  "surface of a balloon" that is needed to support that none curvature at that radius.
No measurement at 13 BLY has been taken.

I fully accept/agree with the following answer from evan_au:
There is an experiment proposed for the 2030s that would search for gravitational waves using lasers bounced of satellites, perhaps 2.5 million km apart....
If I understand it correctly, our scientists try to find a breakthrough technology/ideas in order to give an answer to some of key questions including the idea of curvature.[/quote]
That setup would indeed measure curvature of space, but we already know that local space is curved due to there being nearby mass.  A decent measurement of the curvature of the universe probably needs to be done away from local mass.

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Let's use the circumference of the Earth as an example-
At the early time, our scientists didn't really measure the circumference of the Earth.
You're trying to measure the curvature of a 3D ball, not a 2D surface.  Light on a 2D surface follows the surface,  Techniques used to find the size of a solid will not work on a surface geometry.
It can be done, but it requires non-locality.  Even Earth measurement used non-locality.  I can think of some local ways to do it, but they're not obvious, and they won't work for a surface since they require looking in directions other than on the surface.

Sort of a challenge:  How to measure the size of the planet without moving.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/06/2019 06:32:55
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How can we claim in one hand that based on our mathematical theory the Universe should have a curvature,
Who said that?
Maybe it's finite but not curved.  Read my other post.   Maybe the finite/infinite distinction depends on how the size is measured.
Wow
After all of our discussion, do you agree that there is a possibility that there is no curvature in our Universe and maybe it is also infinite?
Please be aware that those ideas are key elements in Theory D.

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Therefore, at the meeting point of that hypersphere, we might have one galaxy from the left which is moving at a velocity of n * c, while from the other side (right) another galaxy is coming at a velocity of n * c.
They're the same galaxy, and receding in all directions.  Nothing that distant is coming at us.  The universe is expanding, remember?
I see a severe contradiction between the idea of expansion and the 4D module.
In order to get the 4D module Minkowski had placed the time as orthogonal to the 3D of space.
So, if we set an expansion in 4D module, we must set an expansion in each dimension of this module.
If we only set an expansion in the 3D space dimensions and ignore the time dimension, than we set a severe violation in Minkowski formula.
So, in order to validate Minkowski concept under the idea of space expansion, we also must set an expansion in the time.
But, if I understand it correctly, there is no way to set an expansion in time.
Therefore, do you agree that Minkowski and expansion can't work together?
This discusses how a 3D creatures would measure the curvature of a 3D ball, not the analogous 2D creatures measuring the curvature of the 2D surface of Earth.
I also do not agree to that analogy.
We are living in a 3D space.
Minkowski had added mathematically one more dimension - time.
If it is correct - than we are living in a 4D Universe.
This should be correct everywhere. So, you can't go back and eliminate one dimension. Hence your analogy to 2D is not realistic.
However, if we are not living in a 4D Universe - than there is no value for Minkowski calculation.
It is just an unrealistic mathematical idea.
My conclusion is as follow:
1. In our space there is only 3D, so we are not living in a 4D Universe. In our real universe the time isn't orthogonal to the 3D space.
2. Hence, Minkowski concept for 4D Universe is just unrealistic mathematical concept of Space-time.
3. As there is no way for setting the expansion in time, Minkowski formula break down under the expansion idea.
4. If the 4D concept of space-time is unrealistic, than its outcome as a curvature in space is also unrealistic.
5. If our universe is infinite (or might be infinite - based on your answer), we must find a theory that gives a clear explanation for infinite Universe!
6. I can't see how the BBT gives explanation for an infinite Universe in a very limited time frame of 13.8 Billion years
7.  Thermodynamics , Atom creation & BBT
Lack of mass creation is not a BBT thing.  It is a law of thermodynamics, which is not based at all on BBT.
Any real theory must show the source for all the matter in our Universe.
If our universe is infinite, than the total matter in the universe is also infinite.
If I understand it correctly - in order to bypass the thermodynamic obstacle, our scientists doesn't give an explanation for the source of particles before the BBT. They just show the transformation process from particles to Atoms after the Big bang: From Particles to Atoms, From Atoms to stars, and from stars to galaxies.
However, we know that particles are not stable over time. So, if there were particles before the big bang - their life time was quite short (few seconds?). Therefore, I can't see any option to accumulate infinite quantity of particles, just in order to wait for the mighty Big bang to come and start the process of converting all of them into real atoms.
By that converting process, our scientists believe that they have overcome the thermodynamics obstacle.
However, thermodynamics is always there. It was there after the Big bang and also before the Big bang.
They can't bypass that thermodynamics law!
They must show how all particles had been created before the Big bang!!!
Without it, the whole BBT is useless.
In any case, if I understand it correctly - it took over 300,000 years after the big bang just to set the first Atom. So, even if there was some quantity of particles before the big bang, they had to stay with us for at least 300,000 years before being converted into real Atom. Is it real? How can we hold a particle for so long time? What is the expected life time of a average particle?
How can we hold those particles there for unlimited time before the Big bang and for the minimum 300,000 years after the big bang just in order to be transformed into Hydrogen Atom which will be used as a basic element for the whole Universe?
I have proved that Atom is a cell of energy.
So, even if there was unlimited quantity of quarks before the big bang, a significant energy must be added in order to set the transformation into Atom. So, how the Big bang could get the energy which was requested for that transformation of Atom creation?
How even one atom can be created without real magnetic/electric field and without acceleration?
How the Big bang could set the requested magnetic/electric field and acceleration in order to accomplish the mission of only atom creation?
What is the source of the energy of the Big bang? Why it had suddenly happened?
How the thermodynamics can supports the Big bang requested power?
If it is so difficult to create/transform even one atom, how the Big bang could set the whole matter in our infinite Universe - if it is infinite)
Sorry - we can't hide behind the idea that it isn't our job to find the source for all the particles before the Big bang and the source of power which is needed for the Big bang to take place. We have to offer a valid theory how all of that particles/matter had been created (before and after the Big bang) - especially if our universe is infinite!!!
We must find a way how the thermodynamics is part of the creation particles/Atoms process - before and/or after the big bang!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/06/2019 14:48:38
Wow
After all of our discussion, do you agree that there is a possibility that there is no curvature in our Universe and maybe it is also infinite?
Since I never said it cannot be infinite, I don't know where 'all this discussion' about it comes from.  My actual personal stance is that the size of the universe is dependent on how you measure it.

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I see a severe contradiction between the idea of expansion and the 4D module.
It's no different than a 2D model (one of space, the other time, so an expanding circle), which is easier to envision if you find 4D too much for you.  The balloon analogy is 2D of space, but even that seems beyond your ability to envision.

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In order to get the 4D module Minkowski had placed the time as orthogonal to the 3D of space.
So, if we set an expansion in 4D module, we must set an expansion in each dimension of this module.
If we only set an expansion in the 3D space dimensions and ignore the time dimension, than we set a severe violation in Minkowski formula.
All the formulas work fine locally.  If you look up the balloon analogy on the web, it is stated that it is only intended to be used locally.  Yes, your point about Minkowski spacetime is relevant on the large scale.  If space was a hypersphere holding to Minkowski's work on the large scale, then the size of the universe would be known.  It would take only about 87 billion light years to get back to here.  The universe is probably not that small.  My point is that in a hypersphere, the 4 dimensions are orthogonal at every point, but they look more like a 4d version of a polar coordinate system, which looks like a normal Cartesian coordinate system locally.

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So, in order to validate Minkowski concept under the idea of space expansion, we also must set an expansion in the time.
But, if I understand it correctly, there is no way to set an expansion in time.
You must not understand it correctly.  If time expanded with space, there would be no apparent expansion.  We'd not see distances to far objects increasing.  At best it would look like everything staying put but getting smaller, with gravity dropping steadily as the mass of everything decreases.  I suppose that is a valid, if not strange, way to describe physics.  All the physical constants would become time dependent.

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Therefore, do you agree that Minkowski and expansion can't work together?
Minkowski's model is a local one, and never was a cosmological scale model. The two models are not contradictory, but Minkowski does not paint a picture of the universe. Mind you it could:

The thread I have going is trying to describe a flat universe according to the Minkowski description.  While not a hypersphere, it turns out that the universe is finite in size and very much has an edge. The universe is still isotropic from any point, despite having an edge. This is the standard model but using Cartesian coordinates, not comoving coordinates.  I said the size depends on how you measure it.  I'm measuring it along straight lines, not bent ones as is usually done.  I'm still working acceleration of expansion into the picture, but that won't change the size of it, just curve a lot of lines that were straight without that acceleration.
I suppose this topic inspired me to explore that view of the universe.

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Quote from: Halc
This discusses how a 3D creatures would measure the curvature of a 3D ball, not the analogous 2D creatures measuring the curvature of the 2D surface of Earth.
I also do not agree to that analogy.
We are living in a 3D space.
So use the hypersphere model then, which is for 3D creatures.  I was trying to simplify it for somebody obviously incapable of envisioning a hypersphere.

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1. In our space there is only 3D, so we are not living in a 4D Universe.
Doesn't follow.  Logic fail.
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2. Hence, Minkowski concept for 4D Universe is just unrealistic mathematical concept of Space-time.
It works, predicting exactly what we see.  If it doesn't, show me how it would make an empirical difference.  Your words are empty without that.
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4. If the 4D concept of space-time is unrealistic, than its outcome as a curvature in space is also unrealistic.
True.  Curvature of space makes no sense in 3D, yet it has very much been measured, so it must be curvature in spacetime.  You just demonstrated the 4D concept by a valid indirect argument.
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5. If our universe is infinite (or might be infinite - based on your answer), we must find a theory that gives a clear explanation for infinite Universe!
Or at least a coordinate system that yields infinite distances.
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6. I can't see how the BBT gives explanation for an infinite Universe in a very limited time frame of 13.8 Billion years
Good argument.  You're using a coordinate system (like Minkowski's) with a finite light speed.  Yes, under that system, the size is finite.  See my other thread.  See the standard picture in post 464 which clearly shows faster than light speed of objects, not possible in the coordinates that Minkowski and special relativity uses.  It is a different coordinate system (I know not the name of it), but the one used when discussing very distant things.
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Any real theory must show the source for all the matter in our Universe.
Any theory of the origin of the universe needs to, yes.  The BBT is not such a theory.  It is only a model of what happened subsequently.
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If our universe is infinite, than the total matter in the universe is also infinite.
You need to add homogeneity postulate to make that statement stick, but that is typically assumed, so yes.
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If I understand it correctly - in order to bypass the thermodynamic obstacle, our scientists doesn't give an explanation for the source of particles before the BBT.
Since the particles formed subsequent to the big bang, I don't see why they would want to do that.  All that matter you talk about formed after the big bang.

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They just show the transformation process from particles to Atoms after the Big bang: From Particles to Atoms, From Atoms to stars, and from stars to galaxies.
You forgot energy to particles.

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However, we know that particles are not stable over time. So, if there were particles before the big bang - their life time was quite short (few seconds?). Therefore, I can't see any option to accumulate infinite quantity of particles, just in order to wait for the mighty Big bang to come and start the process of converting all of them into real atoms.
You're right.  I can't see that either.
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By that converting process, our scientists believe that they have overcome the thermodynamics obstacle.
By attributing the statement above to 'our scientists', you seem to be formulating a strawman argument.

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How can we hold a particle for so long time? What is the expected life time of a average particle?
Some are quite stable.  Some are not and thus come and go. Electrons and protons are quite stable, and neutrons quickly bound with the latter.  Together they could easily last that 379000 years before they turned into the first atoms.

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What is the source of the energy of the Big bang? Why it had suddenly happened?
There are actually hypotheses concerning the dynamics beyond the big bang singularity. Look them up for some interesting reading, but I have little patience for your naive concepts such as talking about time before time.  The total energy of the universe may well be zero, so a theory of the origin of the singularity need not posit any sort of energy anomaly to prevent violation of thermodynamic law.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/06/2019 13:47:31
Thanks Halc

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2. Hence, Minkowski concept for 4D Universe is just unrealistic mathematical concept of Space-time.
It works, predicting exactly what we see.
Why do you claim that it is predicting exactly what we see?
Do we see any special verification/evidence for that confirms space - time concept? Do we really see a curvature in our universe?
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4. If the 4D concept of space-time is unrealistic, than its outcome as a curvature in space is also unrealistic.
True.  Curvature of space makes no sense in 3D, yet it has very much been measured, so it must be curvature in space-time.  You just demonstrated the 4D concept by a valid indirect argument.
Did we really measure that Curvature of space?
It was stated before that so far we couldn't find any curvature in our universe.
Please explain this answer.

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6. I can't see how the BBT gives explanation for an infinite Universe in a very limited time frame of 13.8 Billion years
Good argument.  You're using a coordinate system (like Minkowski's) with a finite light speed.  Yes, under that system, the size is finite.  See my other thread.  See the standard picture in post 464 which clearly shows faster than light speed of objects, not possible in the coordinates that Minkowski and special relativity uses.  It is a different coordinate system (I know not the name of it), but the one used when discussing very distant things.
So do you agree that the BBT can't support an infinite Universe (without curvature)? Only a finite curvature Universe?

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After all of our discussion, do you agree that there is a possibility that there is no curvature in our Universe and maybe it is also infinite?
Since I never said it cannot be infinite, I don't know where 'all this discussion' about it comes from.  My actual personal stance is that the size of the universe is dependent on how you measure it.
Once you accept the idea that the universe is/could be infinite (without curvature), than do you accept the idea that for this infinite Universe the BBT is useless and therefore we must look for more updated theory?
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We are living in a 3D space.
So use the hypersphere model then, which is for 3D creatures.  I was trying to simplify it for somebody obviously incapable of envisioning a hypersphere.
https://en.wikipedia.org/wiki/Hypersphere
"The term hypersphere was introduced by Duncan Sommerville in his discussion of models for non-Euclidean geometry.
The first one mentioned is a 3-sphere in four dimensions."
The main problem with "hypersphere" is that it doesn't represent our real 3-sphere Universe.
Therefore, as our universe isn't hypersphere, there is no room for curvature in our real universe.
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I see a severe contradiction between the idea of expansion and the 4D module.
It's no different than a 2D model (one of space, the other time, so an expanding circle), which is easier to envision if you find 4D too much for you.  The balloon analogy is 2D of space, but even that seems beyond your ability to envision.
I fully understand your explanation about the 2D.
However, I don't agree with this analogy.
It is really strange to me that in one hand you add the time as it an extra dimension to our real 3D space (call it 4D space-time) and then eliminate one real dimension from that 4D space-time just to show that our real 3D space is like a 2D in that unrealistic 4D space-time.
Do you agree that we are still living in a 3D space?
We can add as many other dimensions as we like to our real 3D space.
For example - we could add velocity as another dimension and call it: space -time-velocity space.
However, I don't agree that we can now say that we are living in a 1D of that space -time-velocity Universe.
It is clear to me that in any kind of universe that we might offer - It is forbidden to eliminate one real space dimension (x,y,z). Therefore the analogy of 2D dimension is just not relevant.
I also can't agree with the idea that in one hand we offer an idea of curvature, while on the other hand we show that we won't be able to verify if this curvature idea is correct or incorrect (as our space is considered as a surface of a balloon in an unrealistic space-time concept) .
If there was a curvature in our Universe - why don't we see it?
I have full confidence that even if we could install our measurements tools at 13 Billion years from each other we won't find even one centimeter of curvature in our Universe.
Therefore, it seems to me that the main success of the Curvature idea is just an extra living time for the BBT.
Therefore, I still wonder why our scientists do whatever it takes to keep the BBT in life?
Why is it so important for them?
Why they don't start looking on all the evidences without the BBT glass and try to find a more accurate theory to our real Universe?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 14/06/2019 13:58:54
Sorry - If there was a curvature in our Universe we should see it!

Not if it's too subtle for us to measure with current technology.

I have full confidence that even if we could install our measurements tools at 13 Billion years from each other we won't find even one centimeter of curvature in our Universe.

Your confidence is not evidence. Even if we didn't observe curvature under such circumstances, that still wouldn't eliminate the possibility of curvature. The curvature could still be so subtle that it cannot be detected with any possible technology.

Therefore, I still wonder why our scientists do whatever it takes to keep the BBT in life?

Because it matches the evidence.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 14/06/2019 16:08:27
Why do you claim that it is predicting exactly what we see?
To be more precise, it doesn't introduce any predictions that contradict empirical measurements.

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Do we see any special verification/evidence for that confirms space - time concept? Do we really see a curvature in our universe?
The theory does not predict that these things can be distinguished.

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Did we really measure that Curvature of space?
Sure.  In principle, you can measure the circumference C of Earth with a tape measure and also its diameter D through a hole passing through.  In flat spaceitme, the ratio would be π.  In reality, the diameter would be larger than C/π.  No, they've not put a tape measure through the center of Earth, but there any theory that predicts a ratio of π has been falsified.

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It was stated before that so far we couldn't find any curvature in our universe.
Please explain this answer.
I meant on a large scale, not a small scale like a planet or star.  Black holes cannot exist without curved space.  If your idea denies curved space, then it cannot have black holes.  They're space bent to the point of breaking.

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So do you agree that the BBT can't support an infinite Universe (without curvature)? Only a finite curvature Universe?
It can.  It just depends on how you measure distance. I got a finite universe by using Minkowski coordinates (a defined inertial reference frame, constant light speed, etc).  The other coordinates (the standard one you see in all the books) do away with all that.  Distances are not measured along inertial lines.  Objects move faster than light due to this different way of measuring their distance.  A completely different set of events is labeled as 'now' and this events are not simultaneous with ours in any inertial sense.

So same universe, different picture of it.  One is finite, and one is not.

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Once you accept the idea that the universe is/could be infinite (without curvature),
Depends what you mean by curvature.  I said the standard picture measures the size of the universe along a line that would be curved in a Minkowski style coordinate system.  The straightness of the lines all depends on how you map the events.
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than do you accept the idea that for this infinite Universe the BBT is useless and therefore we must look for more updated theory?
The BBT is not inconsistent with either view.  In the standard coordinates used, things move arbitrarily faster than light, so a limit to the size is not concluded.  In the Minkowski coordinates, light speed is fixed in any frame and the size of the universe is frame dependent, but always finite.

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The main problem with "hypersphere" is that it doesn't represent our real 3-sphere Universe.
How do you know that? 

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Do you agree that we are still living in a 3D space?
I was planning to move next Tuesday, but so far I'm still here.

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If there was a curvature in our Universe - why don't we see it?
I have full confidence that even if we could install our measurements tools at 13 Billion years from each other we won't find even one centimeter of curvature in our Universe.
There you go.  You know the answer, so why bother putting the instruments out there?

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Therefore, it seems to me that the main success of the Curvature idea is just an extra living time for the BBT.
Who said the hypersphere model was a particularly successful one?  It's just a possibility that hasn't been falsified.  My Minkowski model is another finite one that is not curved. It has an edge like the hypersphere doesn't.
I drew a picture of it yesterday here:
https://www.thenakedscientists.com/forum/index.php?topic=76976.msg577095#msg577095
The purple line is the edge of the universe, growing further away from 'here' as time progresses.  There's another one on the other side not depicted.

The upper dark blue line represents comoving 'now', and distance is measured along that line.  It is curved, meaning the distance is measured with a bunch of meter sticks all moving at different velocities relative to each other.
The horizontal lighter blue line is the inertial now, with all the meter sticks having the same velocity.  They are stationary in the selected frame.

The light blue line intersects the purple line. Thus the univserse is finite in size when measured that way.  The dark blue line never hits the purple line, and thus the universe is infinite when measured that way.
Same (standard) model, but different ways of measuring things.

I had not added dark energy to the picture.  Still working on that.  Adding it will not change what I say here in this thread, but it very much affects that red light cone line.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/06/2019 15:12:53
Thanks Halc & Kryptid

1. With regards to curvature:
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Sorry - If there was a curvature in our Universe we should see it!
Not if it's too subtle for us to measure with current technology.
Yes, I agree
With the current technology we can't see any curvature.
In the future, we might use LISA as advised by Evan_au:
There is an experiment proposed for the 2030s that would search for gravitational waves using lasers bounced of satellites, perhaps 2.5 million km apart.
The goal was to look for tiny variations in the phase of a laser beam over this distance - looking for variations of around 20 picometers with periods ranging from 1 second to 10 hours.
It is outside the mission goals, but the same equipment could be used in the same way as a surveyor's laser to measure the distance between the satellites with an accuracy of perhaps 1nm over a period of a year. That may reveal some deviation from flat space due to the nearby Sun (superimposed on orbital disturbances from Earth, the Moon, Jupiter, Venus and Mars - and the Milky Way itself).

As a bonus, it is expected that this device could get a measure of the expansion of the universe (the Hubble constant) which is independent of the two current (conflicting) answers based on various methods.
However, measuring deviation of intergalactic space from perfectly flat space of the universe would be beyond such a device.
See: https://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna
However, Even LISA might not discover any curvature.
Let's assume that in 2200 we will develop a supper advanced technology that sets the satellites at 2.5 Billion Km apart (instead of only 2.5Million Km in Lisa). However,  even in that range we won't find any curvature.
So, what is needed to us to understand that there is no curvature in our Universe?
In any case, we know the accuracy of our current technology. Therefore, why can't we say that based on the current accuracy there is no curvature in space?
Why instead of assuming that one day we should find a curvature in our Universe, we don't say today that up to the current level of accuracy our universe is flat and we don't see any curvature?
Don't you agree that based on the statement of zero curvature (at the current accuracy level) we must solve the enigma of our Universe?
How long do we have to wait in order to understand that there is no curvature in our Universe?
Just an example -
If I will tell you that an elephant is hiding behind the tree, would you believe that there is an elephant there?
If you will be very positive, how long are you going to wait for that elephant to move away from the tree in order to show himself?
In the same token:
Our scientists want to believe that there is a curvature in our Universe but it is hiding and therefore we can see it with the current technology. That is perfectly clear to me.
However, I would like to ask - How long do we have to wait until we all understand that the curvature is just in our imagination?


2. With Regards to Hypersphere:
Who said the hypersphere model was a particularly successful one?  It's just a possibility that hasn't been falsified. 
Thanks
So, the hypersphere model is a possibility that hasn't been falsified yet.
Therefore, do you agree that as long as there is no solid prove for that hypersphere possibility - we can't really use it as an evidence for the curvature in the Universe?

3. With regards to Minkowski model:
My Minkowski model is another finite one that is not curved. It has an edge like the hypersphere doesn't.
I drew a picture of it yesterday here:
https://www.thenakedscientists.com/forum/index.php?topic=76976.msg577095#msg577095
The purple line is the edge of the universe, growing further away from 'here' as time progresses.  There's another one on the other side not depicted.
I have looked at your modeling but I'm not sure that I fully understand how it really works.
In one hand you claim that: "Minkowski model is another finite one that is not curved"
On the other hand you claim that: "The purple line is the edge of the universe, growing further away from 'here' as time progresses"
So, if the purple line is the edge of the universe, then how could it be that there is no curvature in the Universe?
How could it be that in a finite Universe with an edge there is no curvature?
Don't you agree that a finite space/universe must have a curvature?
In any case, please advice if I understand it correctly as follow:
The main idea with the 4D space-time of Minkowski is that even if we see that the Universe is completely flat (without any sort of evidence for curvature) the Universe/space must be a finite universe.
However, what is the size of that finite Universe?
As you claim that Minkowski model has an edge like the hypersphere doesn't, then would you kindly calculate the size/edge of our current space/universe based on this model?
If the 4D space-time of Minkowski model is correct, than why based on the flat Universe that we see (and measured), and the Picture that you drew, we can't extract the total size of our real finite Universe?
Why don't we try to calculate the minimum size of the finite Universe as a direct outcome of Minkowski 4D space-time model?
How can we claim that: The edge of the universe is growing further away from 'here' as time progresses, while we have no clue about the where is 'here' and what is the edge/size of the Universe?
You also add:
The light blue line intersects the purple line. Thus the univserse is finite in size when measured that way.  The dark blue line never hits the purple line, and thus the universe is infinite when measured that way.
Same (standard) model, but different ways of measuring things.
So, we can get different outcomes as we use the Minkowski model differently.
Therefore, do you agree that theoretically if we move along the opposite direction of the lines, than we might find that the Universe is shrinking?
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The main problem with "hypersphere" is that it doesn't represent our real 3-sphere Universe.
How do you know that? 
In our real 3D space all the three dimensions are orthogonal to each other.
That is very clear to all of us.
However, I still don't see any prove for the idea that the time is orthogonal for those 3 Dimension in space.
It might be a great mathematical concept. But if in our real life, the time isn't orthogonal to the space, than the outcome of Minkowski space-time model is an imaginary mathematical Universe.
In this imaginary Universe, there might be a curvature/edge in the space or even in the time.
Therefore, we get that magneficent image as you have drawn.
However, in order to use it in our real Universe, we must first prove that the time is orthogonal to the 3D space.
So, can you please prove that our time is orthogonal to the 3D space?
I wonder what Einstein would say about that modeling...
In any case, I see it as follow:
If we can prove that the time is orthogonal to the 3D space, than your image represents our real Universe and it proves that it can be finite with edge - as a direct outcome from the image.
However, if we can't prove it - than why don't we set this model as a mathematical concept?
Why do we use it as a real model for our universe?


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Therefore, I still wonder why our scientists do whatever it takes to keep the BBT in life?
Because it matches the evidence.
Which evidence?
Do we see any curvature in our Universe?
Can we prove that our Universe is Hypersphere?
Can we prove that in our real universe the time is orthogonal to the 3D space?
If we can't prove the above and our universe is infinite, how the BBT could fit in this Universe?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 15/06/2019 20:49:18
Which evidence?

The abundance of the chemical elements matches Big Bang predictions, the metric expansion of space, the existence of the cosmic microwave background, the morphology and distribution of the galaxies and quasars, the fact that the Big Bang model agrees with the age of some of the oldest stars in the universe as measured using other methods, the fact that we now know that the CMBR temperature was higher in the past among other things.

Do we see any curvature in our Universe?

Not so far. But we don't need to.

Can we prove that our Universe is Hypersphere?

It might not be a hypersphere, so we don't know.

Can we prove that in our real universe the time is orthogonal to the 3D space?

I don't think that's necessary.

If we can't prove the above and our universe is infinite, how the BBT could fit in this Universe?

The inability to prove any of the above things does not rule out the Big Bang. Science is not about proof.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/06/2019 14:05:41
Therefore, why can't we say that based on the current accuracy there is no curvature in space?
If current technology lacks the precision to distinguish a thing from the absence of the thing, no conclusion about the thing can be drawn from the technology in question.

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Just an example -
If I will tell you that an elephant is hiding behind the tree, would you believe that there is an elephant there?
That's a positive claim being compared to one that is not.  We're only claiming the tree is large enough to hide an elephant (by like 20 orders of magnitude), so we can't be sure.
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Our scientists want to believe that there is a curvature in our Universe but it is hiding and therefore we can see it with the current technology. That is perfectly clear to me.
Got any evidence for this claim, or it just another item on the long list of things that are clear to you?


Therefore, do you agree that as long as there is no solid prove for that hypersphere possibility - we can't really use it as an evidence for the curvature in the Universe?

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3. With regards to Minkowski model:
Quote from: Halc
My Minkowski model is another finite one that is not curved. It has an edge like the hypersphere doesn't.
I drew a picture of it yesterday here:
https://www.thenakedscientists.com/forum/index.php?topic=76976.msg577095#msg577095
The purple line is the edge of the universe, growing further away from 'here' as time progresses.  There's another one on the other side not depicted.
I have looked at your modeling but I'm not sure that I fully understand how it really works.
It's the standard model (the accepted one  * ), so it works like that one.  All I did was draw a picture of it using a standard inertial coordinate system where measurements are taken with meter sticks that are all moving at the same speed instead of different speeds like they usually depict.  Minkowski space is inertial, so he uses a coordinate system like I drew there, but rarely uses it at that scale.  I figured out that it is quite possible to scale it up all the way.

* The picture differs from the actual accepted model in that I didn't include effects of dark energy into it.  Haven't got that far yet.

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In one hand you claim that: "Minkowski model is another finite one that is not curved"
On the other hand you claim that: "The purple line is the edge of the universe, growing further away from 'here' as time progresses"
So, if the purple line is the edge of the universe, then how could it be that there is no curvature in the Universe?
By not curved, I mean it is Euclidean space.  Big triangles drawn anywhere have 3 angles that add up to 180°, which is not true of the curved hypersphere model.

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How could it be that in a finite Universe with an edge there is no curvature?
Kind of like a Euclidean ball is finite in size and has an edge.
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Don't you agree that a finite space/universe must have a curvature?
Locally it does. No large scale curvature has ever been measured.  You just posted all this stuff about denial of curvature, and here you are insisting on it.  My coordinate system doesn't demand flat space, but it works with flat space, so that's what I drew.

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In any case, please advice if I understand it correctly as follow:
The main idea with the 4D space-time of Minkowski is that even if we see that the Universe is completely flat (without any sort of evidence for curvature) the Universe/space must be a finite universe.
Measured that way, it would need infinite time to grow to infinite size.  The model has finite time (it is a big bang model), so it hasn't had time to get bigger than I've drawn.  Nothing can move faster than light in Minkowski spacetime.
It is actually an infinite model.  There is no upper edge (time goes up), so after arbitrary time X years, the universe will grow to X light years in radius.

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However, what is the size of that finite Universe?
Currently about 1.09e31 cubic light years, assuming a comoving frame.  It is only slightly larger as measured in Earth's frame.
The thread topic was about the limit of the size of a stationary object.  I had to draw a picture of the universe to see it.  The max size of the object is about 2.75e10 light years if it is stationary relative to Earth, but it can be arbitrarily larger if it is moving fast enough.  See below for more on that.

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As you claim that Minkowski model has an edge like the hypersphere doesn't, then would you kindly calculate the size/edge of our current space/universe based on this model?
13.75 BLY away of course.  The universe is that old and nothing can move faster than light in that coordinate system, so that's how large it is.
Again, this isn't a different model, just the same model in Minkowski coordinates.

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If the 4D space-time of Minkowski model is correct, than why based on the flat Universe that we see (and measured), and the Picture that you drew, we can't extract the total size of our real finite Universe?
Why don't we try to calculate the minimum size of the finite Universe as a direct outcome of Minkowski 4D space-time model?
We can.  I did it above.  The size they typically quote measures distance a different way (with the meter sticks all moving at different speeds).

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How can we claim that: The edge of the universe is growing further away from 'here' as time progresses, while we have no clue about the where is 'here' and what is the edge/size of the Universe?
Here is here.  Relativity says there are no absolute coordinates.  The Minkowski coordinates do not contradict that.  In drawing the picture, I selected a frame where Earth is at the exact center of the universe.  See the dark blue line labeled 'now'?  I can choose a frame that puts us anywhere on that line.  By doing so, the universe gets larger, which is why the selection of reference frame allows me to posit larger objects than the one you get in Earth's frame.  The choice of frame similarly gets us closer to one of the purple lines, so we can be arbitrarily close to the edge if you like.

You also add:
The light blue line intersects the purple line. Thus the universe is finite in size when measured that way.  The dark blue line never hits the purple line, and thus the universe is infinite when measured that way.
Same (standard) model, but different ways of measuring things.
So, we can get different outcomes as we use the Minkowski model differently. [/quote]It's not a different model, just different coordinates.  We get different answers by measuring via a different method.

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Therefore, do you agree that theoretically if we move along the opposite direction of the lines, than we might find that the Universe is shrinking?
One cannot move along any line except a worldline.  The brown lines show some worldlines, but they can angle upwards at any angle up to 45°.


Quote from: Halc
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The main problem with "hypersphere" is that it doesn't represent our real 3-sphere Universe.
How do you know that? 
In our real 3D space all the three dimensions are orthogonal to each other.
That is very clear to all of us.[/quote]Oohh  you can see them?  No wonder stuff is clear to you and nobody else.  Anyway, coordinate lines are orthogonal on a hypersphere as well.

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However, I still don't see any prove for the idea that the time is orthogonal for those 3 Dimension in space.
Newton assumed that, and worked out a speed of light that was measurably different depending on your speed. If you moved the time axis, the space axes should have stayed the same. It wasn't, so time is orthogonal, per relativity theory.  If you assign a different direction to the time axis, the space dimensions must move with it, which is mathematically what relativity of simultaneity is.  The theory followed from constant light speed and showed that the two are the same thing (spacetime).  Relativity of simultaneity (four orthogonal axes) has been repeatedly demonstrated.  Some deny it and assume a 3D universe and no time dimension at all, but their inability to orient the x,y,z axes without assuming an orientation of the nonexistent t axis seems kind of hypocritical.  The math is much more complex if done in a 3D universe.

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It might be a great mathematical concept. But if in our real life, the time isn't orthogonal to the space, than the outcome of Minkowski space-time model is an imaginary mathematical Universe.
It isn't a universe at all.  It's a model, and one that corresponds exactly to what we see.  If you disagree, find the empirical contradiction, but please stop it with the 'I don't like it, so it is clear that it is wrong' nonsense.  You not liking something is probably good evidence that something is correct.
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In this imaginary Universe, there might be a curvature/edge in the space or even in the time.
Therefore, we get that magneficent image as you have drawn.
What I drew is Euclidean everywhere, even at the edges.

However, in order to use it in our real Universe, we must first prove that the time is orthogonal to the 3D space.
So, can you please prove that our time is orthogonal to the 3D space?
I wonder what Einstein would say about that modeling...[/quote]Einstein (together with the other contributers like Minkowski) showed it as described above.  If you don't understand that theory, then you can't assert that it doesn't show this.  I can't help you.  You imagine your idea being taught in textbooks someday, and yet you're incapable of understanding a theory where somebody else has done all the work, let along working out your own.

In any case, I see it as follow:
If we can prove that the time is orthogonal to the 3D space, than your image represents our real Universe and it proves that it can be finite with edge - as a direct outcome from the image.
However, if we can't prove it - than why don't we set this model as a mathematical concept?
Why do we use it as a real model for our universe?

Quote from: Kryptid
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Can we prove that in our real universe the time is orthogonal to the 3D space?
I don't think that's necessary.
Einstein showed it.  You get a Newtonian universe without it, and that has been falsified.  Even those that deny the existence of the time dimension (presentists) find that they need to posit a fixed yet unmeasurable orientation for it.  The relativity people are free to move it around of course.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/06/2019 15:43:42
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However, I still don't see any prove for the idea that the time is orthogonal for those 3 Dimension in space.
Newton assumed that, and worked out a speed of light that was measurably different depending on your speed.
How can we call it a proof?
Actually, Newton sees it quite differently:
https://astarmathsandphysics.com/a-level-physics-notes/special-and-general-relativity/3013-newton-s-views-on-space-and-time.html
"Newton's Views on Space and Time"
"Newton founded classical mechanics on the view that space is something distinct from matter and that time passes uniformly at every point in space, without regard to whatever happens in the world. For this reason he spoke of absolute space and absolute time, so as to distinguish these entities from the various ways by which we measure them.
It is stated clearly:
"Space had three dimensions – the normal directions of three dimensional space, and time threaded the whole of space and moved uniformly at the same rate throughout space."
So, where do you see Newton approval for the idea that the time should be orthogonal to space?
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It might be a great mathematical concept. But if in our real life, the time isn't orthogonal to the space, than the outcome of Minkowski space-time model is an imaginary mathematical Universe.
It isn't a universe at all.  It's a model, and one that corresponds exactly to what we see.
On the contrary.
Based on the 4D module, the math shows that the Universe has a curvature.
Do we really see ant curvature in our Universe?
Why can't we assume that if so far we don't see any curvature in the Universe, than there is high chance that there is no curvature in the Universe?
So, why do you claim that the model corresponds exactly to what we see?
Where is the curvature?
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Can we prove that in our real universe the time is orthogonal to the 3D space?
I don't think that's necessary. Einstein showed it. 
Is it?
Please see the following info about:
Einstein's Spacetime
https://einstein.stanford.edu/SPACETIME/spacetime2.html
By 1905 he had shown that FitzGerald and Lorentz's results followed from one simple but radical assumption: the laws of physics and the speed of light must be the same for all uniformly moving observers, regardless of their state of relative motion. For this to be true, space and time can no longer be independent. Rather, they are "converted" into each other in such a way as to keep the speed of light constant for all observers. (This is why moving objects appear to shrink, as suspected by FitzGerald and Lorentz, and why moving observers may measure time differently, as speculated by Poincaré.) Space and time are relative (i.e., they depend on the motion of the observer who measures them) — and light is more fundamental than either. This is the basis of Einstein's theory of special relativity ("special" refers to the restriction to uniform motion)
It is also stated:
"Einstein did not quite finish the job, however. Contrary to popular belief, he did not draw the conclusion that space and time could be seen as components of a single four-dimensional spacetime fabric."
So, Einstein proved that "Space and time are relative" but he didn't claim that they are orthogonal to each other.
The Fourth Dimension is Minkowski idea:
That insight came from Hermann Minkowski (1864-1909), who announced it in a 1908 colloquium with the dramatic words: "Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality".
Please be aware that Einstein lived long after Hermann Minkowski.
Therefore, he was fully aware about Minkowski 4D dimension.
He could accept that idea. But based on the above explanation it seems clearly that Einstein didn't agree with the idea that the Time is orthogonal to space.
He claimed that "Space and time are relative".
So, If Einstein didn't accept the idea of 4D Dimension, who am I to disagree with him?
Speaking about Einstein:
If I understand it correctly - He actually disagree with the idea of BBT.
https://guardianlv.com/2014/03/albert-einstein-debunked-the-big-bang-theory/
"Albert Einstein’s theory of relativity walked hand in hand with the Big Bang theory, but recently resurfaced manuscripts show that the physicist debunked this idea and believed that the universe expanded steadily and eternally."
He also disagree with his personal idea about the cosmological constant:
https://www.space.com/9593-einstein-biggest-blunder-turns.html
"In 1917, Albert Einstein inserted a term called the cosmological constant into his theory of general relativity to force the equations  to predict a stationary universe in keeping with physicists' thinking at the time. When it became clear that the universe wasn't actually static, but was expanding instead, Einstein abandoned the constant, calling it the '"biggest blunder" of his life"
So, Einstein called the "cosmological constant"  as the '"biggest blunder" of his life.
"But lately scientists have revived Einstein's cosmological constant (denoted by the Greek capital letter lambda) to explain a mysterious force called dark energy that seems to be counteracting gravity -- causing the universe to expand at an accelerating pace."
Therefore, we must respect this scientist and NOT use his "biggest blunder" to prove the unrealistic idea of "dark energy", as there is no dark energy in our Universe.
How our scientists dare to use the "cosmological constant" in Einstein formula against his clear wish and then claim that it is based on Einstein?
Don't you agree that once we add this "cosmological constant" we clearly contradict Einstein formula?
So, our scientists use in one hand the 4D as Einstein Idea - which is incorrect, while on the other hand the also use the  "cosmological constant" in Einstein formula against his clear wish just to prove the BBT which Einstein did not accept.
So, how our scientists could use "Einstein" name in order to prove the BBT which he didn't agree with?


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 21/06/2019 18:27:42
Actually, Newton sees it quite differently:
https://astarmathsandphysics.com/a-level-physics-notes/special-and-general-relativity/3013-newton-s-views-on-space-and-time.html
"Newton's Views on Space and Time"
"Newton founded classical mechanics on the view that space is something distinct from matter and that time passes uniformly at every point in space, without regard to whatever happens in the world. For this reason he spoke of absolute space and absolute time, so as to distinguish these entities from the various ways by which we measure them.
It is stated clearly:
"Space had three dimensions – the normal directions of three dimensional space, and time threaded the whole of space and moved uniformly at the same rate throughout space."
Yes, Newton viewed it as 3D.  Time in that interpretation (presentism) is something that flows at a uniform rate everywhere.  If that model is done in 4D, there is one direction that time can be oriented that is orthogonal.  Newton didn't see things in 4D like that, but some did.  The block universe concept is quite old actually, but spacetime (them being the same thing in the same units) is something different, introduced in the 20th century.

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So, where do you see Newton approval for the idea that the time should be orthogonal to space?
If you're going to have a coordinate system, the coordinates (like x y z) need not be orthogonal to work, but things sure work a lot easier if you do it.  They just can't be parallel.  Then it ceases to function as a coordianate system.
Since Newton posited absolute space and time, there is really only one way to orient the time axis orthogonally if you're going to represent it as an axis.  It works at other orientations, but only one is orthogonal.

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Quote from: Halc
It isn't a universe at all.  It's a model, and one that corresponds exactly to what we see.
On the contrary.
Based on the 4D module, the math shows that the Universe has a curvature.
Not sure which model you are referring to.  The 4D thing is an interpretation of time (block universe) and it doesn't assert curvature or lack of it, or even relativity.  The Newtonian model was falsified whether it is interpreted as 3D or 4D.  It predicts frame dependent measurement of light speed.

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So, why do you claim that the model corresponds exactly to what we see?
Where is the curvature?
The Minkowski model is a local model, and if expanded beyond locality, corresponds to a non-curved space.

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Please see the following info about:
Einstein's Spacetime
https://einstein.stanford.edu/SPACETIME/spacetime2.html
By 1905 he had shown that FitzGerald and Lorentz's results followed from one simple but radical assumption: the laws of physics and the speed of light must be the same for all uniformly moving observers, regardless of their state of relative motion. For this to be true, space and time can no longer be independent. Rather, they are "converted" into each other in such a way as to keep the speed of light constant for all observers. (This is why moving objects appear to shrink, as suspected by FitzGerald and Lorentz, and why moving observers may measure time differently, as speculated by Poincaré.) Space and time are relative (i.e., they depend on the motion of the observer who measures them) — and light is more fundamental than either. This is the basis of Einstein's theory of special relativity ("special" refers to the restriction to uniform motion)
It is also stated:
"Einstein did not quite finish the job, however. Contrary to popular belief, he did not draw the conclusion that space and time could be seen as components of a single four-dimensional spacetime fabric."
So, Einstein proved that "Space and time are relative" but he didn't claim that they are orthogonal to each other.
The stuff you quote does not imply that.
As I said, a coordinate system is abstract, so if you don't mind totally complicating the math, you can put the axes in any non-orthogonal orientation that turns you on.

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The Fourth Dimension is Minkowski idea:
It was proposed centuries earlier than that.  A 4 dimensional universe was suggested for instance by Anselm of Canterbury somewhere around year 1100.  It is probably older than that.  Minkowski may have first suggested that time and space are ontologically the same thing.


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If I understand it correctly - He actually disagree with the idea of BBT.
https://guardianlv.com/2014/03/albert-einstein-debunked-the-big-bang-theory/
"Albert Einstein’s theory of relativity walked hand in hand with the Big Bang theory, but recently resurfaced manuscripts show that the physicist debunked this idea and believed that the universe expanded steadily and eternally."
Expansion and BBT was unknown at the time when relativity theory was published. 

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He also disagree with his personal idea about the cosmological constant:
https://www.space.com/9593-einstein-biggest-blunder-turns.html
"In 1917, Albert Einstein inserted a term called the cosmological constant into his theory of general relativity to force the equations  to predict a stationary universe in keeping with physicists' thinking at the time. When it became clear that the universe wasn't actually static, but was expanding instead, Einstein abandoned the constant, calling it the '"biggest blunder" of his life"
Just so.  That was him accepting the BBT as a better explanation of things than his cosmological constant.

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So, how our scientists could use "Einstein" name in order to prove the BBT which he didn't agree with?
I don't think Einstein's name comes up much in any evidence of BBT.  It isn't his theory.  Try Hubble and others.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/06/2019 05:28:28
That was him accepting the BBT as a better explanation of things than his cosmological constant.
It seems to me that you have missed the key point about Einstein formula.
So the question is as follow:
In order to prove the BBT, is it correct that our scientists have used Einstein formula?
If so, could they prove the BBT without adding the cosmological constant to Einstein formula?
Do you agree that Einstein have stated that adding this cosmological constant is the biggest mistake of his life?
Therefore, Do you agree that Einstein will not accept the idea that our scientists have used that cosmological constant in his formula?
Hence, do you agree that it is forbidden to add that cosmological constant to Einstein formula and still call it Einstein formula?
In other words do you agree that the statement that the BBT had been mathematically proved based on Einstein formula - is totally incorrect?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/06/2019 14:41:15
Which formula is this?
And I don't think any cosmological theory has ever been claimed as proved.
In the following article it is stated:
https://en.wikipedia.org/wiki/Cosmological_constant
From the 1930s until the late 1990s, most physicists assumed the "cosmological constant to be equal to zero.[4] That changed with the surprising discovery in 1998 that the expansion of the universe is accelerating, implying the possibility of a positive nonzero value for the cosmological constant.[5]"
The question is as follow:
Do we need to use the cosmological constant in Einstein formula in order to find an explanation for "the discovery in 1998 that the expansion of the universe is accelerating"?
Can we get an explanation for that discovery without the cosmological constant?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/06/2019 05:39:03
OK. That quote suggests that the cosmological constant and dark energy are the same explanation, not alternate explanations for the same effect.
In order to prove the idea of the dark energy, our scientists were obliged to set a positive cosmological constant value in Einstein formula.
Therefore, do you agree that without this positive value, they would not be able to add the dark energy into Einstein formula?
Yet your statement here seems to suggest that this one explanation is unrealistic, leaving no alternative.  So your statement certainly doesn't follow from the quote you gave
As I have proved, Einstein had claimed that it is forbidden to use this cosmological constant in his formula.
Hence, without using the cosmological constant, how our scientists could prove the existence of that dark energy in Einstein formula?
If they can't use the cosmological constant, than they have a severe problem with the idea of dark energy.
Without the dark energy, they can't explain the acceleration in the expansion process.
Without this explanation, they have a severe problem with the BBT.
So, do you see the contradiction?


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/06/2019 14:06:51
You claim to have proved that Einstein made a claim about the forbidden usage of some constant in some unspecified formula.  That sounds pretty inconsistent if the formula makes any mention of this constant.
I'm not sure that I fully understand the meaning of this answer.
However:
1. Do you agree that in order to prove mathematically the BBT, our scientists are Using Friedmann equations?
2. Do you agree that Friedmann equations are based on Einstein' general relativity formulas?
https://en.wikipedia.org/wiki/Friedmann_equations
"The Friedmann equations are a set of equations in physical cosmology that govern the expansion of space in homogeneous and isotropic models of the universe within the context of general relativity."
"They were first derived by Alexander Friedmann in 1922 from Einstein's field equations of gravitation for the Friedmann–Lemaître–Robertson–Walker metric and a perfect fluid with a given mass density {\displaystyle \rho } \rho  and pressure {\displaystyle p} p"
3. Do you agree that in that Einstein general relativity formula the cosmological constant is a key element that is needed to validate the existence of the dark matter? So it must have a positive value?

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In order to prove the idea of the dark energy, our scientists were obliged to set a positive cosmological constant value in Einstein formula.
Therefore, do you agree that without this positive value, they would not be able to add the dark energy into Einstein formula?
See post 497 for my reply to this repeated question.
Your message at 497 was:
Which formula is this?
And I don't think any cosmological theory has ever been claimed as proved.
and my reply at 498 was:
In the following article it is stated:
https://en.wikipedia.org/wiki/Cosmological_constant
From the 1930s until the late 1990s, most physicists assumed the "cosmological constant to be equal to zero.[4] That changed with the surprising discovery in 1998 that the expansion of the universe is accelerating, implying the possibility of a positive nonzero value for the cosmological constant.[5]"
In that article it is stated clearly:
Since the 1990s, studies have shown that around 68% of the mass–energy density of the universe can be attributed to so-called dark energy.[6] The cosmological constant Λ is the simplest possible explanation for dark energy, and is used in the current standard model of cosmology known as the ΛCDM model. While dark energy is poorly understood at a fundamental level, the main required properties of dark energy are that it functions as a type of anti-gravity, it dilutes much more slowly than matter as the universe expands, and it clusters much more weakly than matter, or perhaps not at all."
So, do you agree that dark matter is a key element in our understanding of the Universe?
It is stated clearly: "The cosmological constant Λ is the simplest possible explanation for dark energy, and is used in the current standard model of cosmology known as the ΛCDM model"
Therefore, the cosmological constant Λ in Einstein formula is used as a possible explanation for the existence of dark matter.
4. Do you agree that Einstein had rejected the idea of using the cosmological constant in his formula?
5. Therefore, do you agree that without that cosmological constant we can't confirm mathematically the existence of dark matter?
6. If so, do you agree that without the dark matter we can't explain the acceleration in the expansion?
7. If we can't explain that acceleration in the expansion, how can we prove the BBT?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/06/2019 17:22:12
1. Do you agree that in order to prove mathematically the BBT, our scientists are Using Friedmann equations?
Nobody proves anything, and I've said that repeatedly.  So I will not agree to such a statement.
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2. Do you agree that Friedmann equations are based on Einstein' general relativity formulas?
Quote from: wiki
wiki Friedmann_equations
"The Friedmann equations are a set of equations in physical cosmology that govern the expansion of space in homogeneous and isotropic models of the universe within the context of general relativity."
"They were first derived by Alexander Friedmann in 1922 from Einstein's field equations of gravitation for the Friedmann–Lemaître–Robertson–Walker metric and a perfect fluid with a given mass density and pressure"
It says "within the context of GR" and "derived ... from [GR equations for this specific metric]", so yes, the expansion of the universe was first worked out directly from GR theory, even before it was first observed.  Einstein's equations predict expansion.
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3. Do you agree that in that Einstein general relativity formula the cosmological constant is a key element that is needed to validate the existence of the dark matter? So it must have a positive value?
The GR formulas that Friedmann used did not contain a cosmological constant.  Einstein added it to attempt a model of a static universe, and abandoned it when Hubble falsified the static model.
None of this has anything to do with dark matter.  I think you mean dark energy, and none of this has to do with that either.  All the above discussion concerns work done in the 20's and 30's.  We're still discussing an early model with steady expansion.

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So, do you agree that dark matter is a key element in our understanding of the Universe?
It is stated clearly: "The cosmological constant Λ is the simplest possible explanation for dark energy, and is used in the current standard model of cosmology known as the ΛCDM model"
Therefore, the cosmological constant Λ in Einstein formula is used as a possible explanation for the existence of dark matter.
I assume you mean dark energy, and the universe does not behave in a way a model without it (or an equivalent) predicts.
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4. Do you agree that Einstein had rejected the idea of using the cosmological constant in his formula?
History records him abandoning it, yes, since he was using it for a static model (a non-big-bang one).  Nobody is resurrecting such a static model.
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5. Therefore, do you agree that without that cosmological constant we can't confirm mathematically the existence of dark matter?
This statement makes no sense.  If you mean dark energy, the two are the same thing.  If you're asking if it needs to be a constant, I don't think that has been demonstrated.
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6. If so, do you agree that without the dark matter we can't explain the acceleration in the expansion?
I am unfamiliar with other suggested models (than say λCDM) and the terminologies they use, so I don't know if the term 'dark energy' is used in them.  You really need to stop saying dark-matter in all these statements. 
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7. If we can't explain that acceleration in the expansion, how can we prove the BBT?
Yet again, science isn't involved in proofs.  They have explained it, so the if-we-don't part of your question is irrelevant, like asking why apples fall if we've no explanation for the effect?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/06/2019 17:39:45
Thanks Halc
None of this has anything to do with dark matter.  I think you mean dark energy,
Yes, it is all about dark Energy.
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1. Do you agree that in order to prove mathematically the BBT, our scientists are Using Friedmann equations?
Nobody proves anything, and I've said that repeatedly.  So I will not agree to such a statement.
So, what is the added value of Friedmann equations?
Do you mean that there is no mathematical prove for the BBT?
If so, why do you so positively support this theory?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 25/06/2019 21:21:08
Do you mean that there is no mathematical prove for the BBT?

There is no mathematical proof for any theory. That isn't how theories work.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/06/2019 07:52:38
Do you mean that there is no mathematical prove for the BBT?
here is no mathematical proof for any theory. That isn't how theories work.
If there is no mathematical proof for the BBT, than do you agree that there is a chance that it is incorrect?
During this discussion, I have set several obstacles for the feasibility of the BBT.
One of the main obstacles is time frame.
I really can't see how can we fit all of this magnificent Universe in only 13.8 Billion years.
If the Universe is infinite in its size and there is no curvature - than why do you insisted that the 13.8 BY is still ok to set it?
Few days ago I have watched two amazing programs of "How the universe works" at the national geographic channel TV.
In one program they have discussed about the SMBH.
They have asked how the SMBH had been created.
They have set a calculation that if we take a BH and give it the time to eat as much as it wants, it can't technically be converted to SMBH after 13.8 BY.
Much more time is needed.
However, we see massive spiral galaxies (with embedded SMBH) with estimated age of about 12 BY.
So, they have wondered how could it be that the SMBH had been created in so short time after the BBT and could also add around it all the Billion stars in order to set the mighty spiral galaxy?
They have tried to come with some ideas, but they were not sure about those hypothetical ideas.
They also wondered why we see mainly find BH and SMBH.
In SMBH there are over than Billion BH.
Why don't we see as many mid size BH with only few Millions of BH?
In any case, they were positively sure that the SMBH can't evolve from a BH due to time limitation.
 
In the other program they have discussed about the heavy metal as Iridium and gold.
They have found that those metals could not been created by a supernova. The maximum power of that supernova is Setting Iron.
They have stated that one hypothetical idea for the creation of those two heavy metals is an explosion in twin neutron stars.
So, in order to calculate the requested time for the gold in our Planet, we need to add all the requested time as follow:
1. Time for the first Hydrogen
2. Time for the first Hydrogen cloud.
3. Time to set the first star. However, we have already found that in order to set a star in Hydrogen cloud there must be a nearby SMBH). so if there is no SMBH after the Big bang, how the gas cloud could be set any sort of star?
In any case, let's assume that somehow the hydrogen cloud had set the first Hydrogen star.
3. Time for the first hydrogen star to set the first Supernova explosion. Due to this explosion we get the iron. However, it is very rare and it is stes the iron as dust in space.
4. So, how long it might take to gather all the dust in a first iron dust cloud.
5. How long it might take to this first iron dust could to set the first twin neutron star systems.
6. How long it might take to this first twin Neutrom star system to set the explosion in order to create the first iridium and Gold in the whole Universe?
7. How long it might take to those heavy metal to arrive to our planet?
If they are too far away, do you agree that due to the expansion in space, they have no chance to get by Asteroids to our planet?
So, how do we have got Gold and iridium in only 13.8 Billion years?

In this program they didn't even try to explain how heavier Atoms as plutonium had been created.
It seems to me that even an explosion in a twin Neutron star is not good enough.
So, how do the the plutonium had been created and how it had been arrived to our planet in only 13.8 BY?

Based on theory D, all the atoms had been created at the excretion disc of a SMBH (Yes, including the plutonium).
There is no need to any chain of explosions to set any sort of heavy Atom.
Our planet had got all its matter in his first day. No needed for any special delivery ( as water, gold, plutonium...) from any kind of asteroid.
So, if the BBT has no real approval, why don't you open your mind to a breakthrough theory?
A theory which is so simple and gives a perfect explanation to any activity in our infinite universe?
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 29/06/2019 10:50:17
Quote from: Dave Lev
in order to set a star in Hydrogen cloud there must be a nearby SMBH
I don't see this dependency.

You need a hydrogen cloud, and it must be fairly dense to collapse under it's own gravity. But the early universe was fairly dense.

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Time for the first hydrogen star to set the first Supernova explosion.
This is thought to be fairly quick (in astronomical terms): probably less than a million years.

The reason is that the early stars were formed from Hydrogen and Helium, and these have difficulty radiating away energy. To collapse under their own gravity, the first "Population III" stars must have been very massive, perhaps 100 times the mass of the Sun. Since the rate of burning their nuclear fuel increases as something like the 4th power of the mass, these stars would have had short lives.

When these first stars exploded, they would have distributed a lot of elements up to iron, which made the formation of the next generation of "Population II" stars much easier.
See: https://en.wikipedia.org/wiki/Stellar_population#Population_III_stars

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one hypothetical idea for the creation of those two heavy metals is an explosion in twin neutron stars.
I think the phrase you want is "merger of twin neutron stars"; but I agree that certainly produces a dramatic explosion!

It is no longer hypothetical - in 2017 such an event was observed in gamma rays and visible light; the imminent event was heralded by gravitational waves arriving more than a minute before the gamma rays.

Spectroscopic analysis suggested that gold (and other heavy elements) were formed in this event. It is likely that Plutonium and heavier nuclei were also formed, but these radioactive elements have fairly short half-lives compared to Uranium, Thorium and Potassium-40 that we do find in reasonable concentrations on Earth.

Conceivably, a similar mechanism could occur if a neutron star were disrupted by a black hole - some of the neutron-rich material may be sprayed out into space, beyond the escape velocity of the black hole.
See: https://en.wikipedia.org/wiki/Neutron_star_merger#Observed_mergers

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how do the the plutonium had been created and how it had been arrived to our planet in only 13.8 BY?
There is no measurable amount of natural plutonium that arrived on Earth during its formation almost 5 billion years ago. The longest-lived isotope of Plutonium (Pu-244) has a half-life of 80 million years, so it will have decayed by now.

There is some natural Plutonium generated continually on Earth, as a neutron occasionally strikes a Uranium atom in Uranium ores; this unstable Uranium nucleus decays into Plutonium (and then into Americium). The balance between creation of Plutonium and decay of Plutonium is biased heavily on the decay side, so the long-term average level of Plutonium in Uranium ores is extremely low.

A considerable amount of Plutonium is generated in nuclear reactors; the operation of some nuclear reactors is optimised to maximise production of Plutonium.
See: https://en.wikipedia.org/wiki/Isotopes_of_plutonium

There are signs that elements with half-lives of millions of years have arrived on Earth - sedimentary bands with traces of Fe-60 (half-life 2.7 million years) have been found in ocean basins. But as you say, Iron is likely to be produced in supernova explosions, which are expected to be far more frequent than neutron star mergers.

After all, any isolated massive star can go supernova - but it takes two very closely orbiting neutron stars to merge (and this implies two closely orbiting massive stars that had already gone supernova).
 
See: https://en.wikipedia.org/wiki/Near-Earth_supernova#Past_events
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/06/2019 15:04:02
If there is no mathematical proof for the BBT, than do you agree that there is a chance that it is incorrect?
Any theory might be falsified some day.  If it was mathematically proved, it would be a theorem, not a theory.
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I have set several obstacles for the feasibility of the BBT.
One of the main obstacles is time frame.
I really can't see how can we fit all of this magnificent Universe in only 13.8 Billion years.
How is you unwillingness to see something an obstacle to the theory?  It seems only an obstacle to you.
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If the Universe is infinite in its size and there is no curvature - than why do you insisted that the 13.8 BY is still ok to set it?
Question lacks syntactic sense.  Set what exactly?  The age?  The size?  The former is finite, and the latter depends on the coordinate system under which it is measured.
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In one program they have discussed about the SMBH.
They have asked how the SMBH had been created.
They have set a calculation that if we take a BH and give it the time to eat as much as it wants, it can't technically be converted to SMBH after 13.8 BY.
Much more time is needed.
If you put enough stuff nearby, it can be done in a few minutes, so this obviously doesn't follow.  Sgr-A is fairly small.  There are galaxies with black holes 100 or 1000 times as massive, meaning they achieved SMBH status in far less than 13 billion years.
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However, we see massive spiral galaxies (with embedded SMBH) with estimated age of about 12 BY.
You have a link for that, or it it just another assertion?  I agree that galaxies with SMBH of age 12 BY are probably out there, but I don't necessarily agree that we see it. 
Edit:  Kryptid's link posted just below seems to satisfy this request.
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In SMBH there are over than Billion BH.
No, there is just the one.  They don't consist of a collection of objects.
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Why don't we see as many mid size BH with only few Millions of BH?
A BH is not a unit of mass.  Sgr-A is 4 million solar masses, so does that count?  A solar mass IS a unit of mass.
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In any case, they were positively sure that the SMBH can't evolve from a BH due to time limitation.
What alternative do they propose?  I don't see any other path than being small at first, and then growing, even if it takes a short time to do the growing part.  I also don't know a specific designation distinguishing a BH from a SMBH.  What mass separates the two?  Is there a standard one?
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In the other program they have discussed about the heavy metal as Iridium and gold.
They have found that those metals could not been created by a supernova. The maximum power of that supernova is Setting Iron.
Iron is the lowest energy form, but creation of the higher elements has not been shown to be impossible.
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They have stated that one hypothetical idea for the creation of those two heavy metals is an explosion in twin neutron stars.
Out of curiosity, how does this TV program justify that a supernova lacks the ability to create these higher elements and yet a neutron star merger can do it?  Is the supernova insufficiently violent?  I mean, the scientists can create these elements in a lab without involving the sort of degenerate matter than exists in a neutron star.
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3. Time to set the first star. However, we have already found that in order to set a star in Hydrogen cloud there must be a nearby SMBH).
The show found this?  Or is this just you?  None of us has found this at all, as Evan points out.
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3. Time for the first hydrogen star to set the first Supernova explosion. Due to this explosion we get the iron. However, it is very rare and it is stes the iron as dust in space.
This is again pointed out above:  The time for this is quite short and the event is quite common.
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5. How long it might take to this first iron dust could to set the first twin neutron star systems.
Neutron stars tend to form from first generation stars (the mostly hydrogen ones), and not from second generation stars like our own.  They can form from 2nd generation stars.
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7. How long it might take to those heavy metal to arrive to our planet?
They were always here.  If they were not (just passing through say), then a solar system would not form here.
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If they are too far away, do you agree that due to the expansion in space, they have no chance to get by Asteroids to our planet?
Space expansion has nothing to do with it unless you propose that the elements needed come from other galaxies and for some reason stop here.
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So, how do we have got Gold and iridium in only 13.8 Billion years?
By not watching National Geographic TV shows that deny the accepted dynamics of a supernova explosion.
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In this program they didn't even try to explain how heavier Atoms as plutonium had been created.
Plutonium doesn't naturally occur except as a short lived byproduct of the decay of more stable things like Uranium.
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So, how do the the plutonium had been created and how it had been arrived to our planet in only 13.8 BY?
It doesn't.  Virtually all plutonium on Earth was manufactured here.
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So, if the BBT has no real approval, why don't you open your mind to a breakthrough theory?
Yours isn't a theory, it is just an idea, and one that is trivially falsified. One need not employ a closed mind to discount it.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/06/2019 15:39:05
Of relevance to this thread's discussion on supermassive black holes: https://mediarelations.uwo.ca/2019/06/28/black-hole-formation/

Paper is here: https://academic.oup.com/mnras/article/370/1/289/1026607
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/07/2019 15:49:46
Thanks
When these first stars exploded, they would have distributed a lot of elements up to iron, which made the formation of the next generation of "Population II" stars much easier.
See: https://en.wikipedia.org/wiki/Stellar_population#Population_III_stars
I have read carefully the explanation about the Stellar_population.
The reason is that the early stars were formed from Hydrogen and Helium, and these have difficulty radiating away energy.
In one hand we claim that after the BBT we have mainly got Hydrogen. So how the Helium had been pop up immediately after the Big bang in order to create the first Population_III_stars?
https://en.wikipedia.org/wiki/Big_Bang
"As the universe cooled, the rest mass energy density of matter came to gravitationally dominate that of the photon radiation. After about 379,000 years, the electrons and nuclei combined into atoms (mostly hydrogen);"
In this explanation about the Big bang they don't say even one word about helium.
However, in the following artical it is stated that the Helium had been created by the Big bang:
https://stardate.org/astro-guide/hydrogen-and-helium
"Hydrogen is an element, usually in the form of a gas, that consists of one proton and one electron. Hydrogen is the most abundant element in the universe, accounting for about 75 percent of its normal matter, and was created in the Big Bang. Helium is an element, usually in the form of a gas, that consists of a nucleus of two protons and two neutrons surrounded by two electrons. Helium is the second-most abundant element in the universe, after hydrogen, and accounts for about 25 percent of the atoms in the universe. Most of the helium in the universe was created in the Big Bang, but it also is the product of hydrogen fusion in stars."
So how could it be that the Big bang could form so high percentage of Helium (25% of all the matters in the Universe are Helium)?
Why there is no information about this massive Helium creation in the Big Bang activity?
If the Big bang could create Helium, why not more heavier elements?
Why not iron or even gold?
Why are we so sure that it is only Hydrogen and Helium?
To collapse under their own gravity, the first "Population III" stars must have been very massive, perhaps 100 times the mass of the Sun. Since the rate of burning their nuclear fuel increases as something like the 4th power of the mass, these stars would have had short lives.
Why we can't find even one "Population III" star in the whole Universe?
What is the chance that 100% of the "Population III" stars got a mass of at least 100 times the mass of the Sun?
I fully understand why a star with 100 times the mass of the Sun had a very sort lives. However, why there were no compact "Population III" stars (at the Sun size)? Don't you agree that at this compact size they could stay with us even today?
As we can't find even one Hypothetical "Population III" star, could it be that all the stars in the Universe had been formed from day one from more than just Hydrogen and Helium? Therefore, could it be that this hypothetical idea about "Population III" star  is not realistic.
There is some natural Plutonium generated continually on Earth, as a neutron occasionally strikes a Uranium atom in Uranium ores; this unstable Uranium nucleus decays into Plutonium (and then into Americium). The balance between creation of Plutonium and decay of Plutonium is biased heavily on the decay side, so the long-term average level of Plutonium in Uranium ores is extremely low.
Actually, I have to ask about the Uranium.
Iron is the lowest energy form, but creation of the higher elements has not been shown to be impossible.
So, how the Uranium had been formed?
If it is so difficult to creat Gold atom, what kind of explosion or activity could create Uranium atom?
Of relevance to this thread's discussion on suppermassive black holes: https://mediarelations.uwo.ca/2019/06/28/black-hole-formation/
It is stated:
"During the last decade, many suppermassive black holes that are a billion times more massive than the Sun have been discovered at high ‘redshifts,’ meaning they were in place in our universe within 800 million years after the Big Bang. The presence of these young and very massive black holes question our understanding of black hole formation and growth."
So, there is real question how those SMBH had been created in less than 800 Million years after the Big bang.
In the article it is stated:
"The model is based on a very simple assumption: suppermassive black holes form very, very quickly over very, very short periods of time and then suddenly, they stop. This explanation contrasts with the current understanding of how stellar-mass black holes are formed, which is they emerge when the centre of a very massive star collapses in upon itself."
I really don't know if our scientists accept this assumption.
In any case, if the SMBH can't get to that size at a very short period, Do you agree that the age of the SMBH (the very massive one) should be longer than just 800 Million years?
If that is corret, do you agree that the age of the Universe should be longer than just 13.8 billion years?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/07/2019 00:51:48
I really don't know if our scientists accept this assumption.

It's a brand new idea. What do you expect?

In any case, if the SMBH can't get to that size at a very short period, Do you agree that the age of the SMBH (the very massive one) should be longer than just 800 Million years?

Only if it can be demonstrated that it must take longer than 800 million years for it to form and reach its measured mass.

If that is corret, do you agree that the age of the Universe should be longer than just 13.8 billion years?

See my answer above.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/07/2019 06:31:21
Only if it can be demonstrated that it must take longer than 800 million years for it to form and reach its measured mass.
In the article that you have offered it was stated clearly:
The presence of these young and very massive black holes question our understanding of black hole formation and growth."
Also in the TV program it was stated clearly that there is a limitation for the BH size increase over time.
So, there is a big question about the minimal requested time to form a SMBH.
Let's look at our SMBH.
If I recall it correctly, our scientists verified that it consumes about 3 Sun mass per year.
So, in one billion year is will add about 0.75 to its current mass.
There is unlimited number of black holes in our Universe.
We must trace the BH activities and verify how long it might take them to increase their mass over time.
If we will discover that on average it takes about one Billion year for a black hole to double its mass, we could set the clock time for our Universe.
Is there any way to set a SMBH without crossing the stage of compact BH?
Is there any possibility to form a man without crossing the stages of boyhood, childhood or babyhood?
Why do we assume that a SMBH can be formed without crossing the stages of compact BH, Mid size BH and Massive BH?
Our scientists know/should know the maximal consumption rate of a BH.
Based on that, they could easily calculate how long is needed to transform a compact BH into a mid size BH and how long it might take to a mid size BH to be transformed into a SMBH.
We know that the SMBH at the core of the Andromeda galaxy is much more massive than the SMBH in our galaxy.
So, we can also calculate the requested time to form that kind of Ultra SMBH out of a  (Milky Way)' SMBH.
Hence, we must look for the biggest SMBH in the Universe and try to calculate how long it might take it to be formed out of a compact BH.
That time frame sets the minimal age of the Universe.
So, why can't we use the SMBH to adjust the time clock for our Universe?

What about the Helium?
So how could it be that the Big bang could form so high percentage of Helium (25% of all the matters in the Universe are Helium)?
How long it took the BBT to set all the Helium mass in the Universe?
Is there a possibility to set the first Population_III_stars without Helium?
How long it took to set the first Population_III_stars with Helium?

What about the Uranium?
How long it took to set the first Uranium Atom in the Universe?
How long it took to get the Uranium delivery to Mother Earth?

Why do we always insist to force the age of 13.8 BY to everything we see?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/07/2019 07:18:47
3billion is not 0.75 of 4 million.  It is 750 times its current mass.
Thanks
I agree
No, Sgr-A consumes almost no mass per year, far less than the average it would need (a star every 3000 years) to have grown to its present mass.
We need to verify the meaning of:
 
Sgr-A consumes almost no mass per year,
Is it 0.001 Sun mass per year?
If so, do you agree that in order to double the size about one billion years is needed?
If it is less than that, do you agree that we might get to a conclusion that 10 Billion or even 100 Billion years is needed to double a SMBH size?
If you feed it, it will eat it.  It isn't going to say "No-thanks, that's enough for today".  There is no max rate.
We know that the SMBH is picky-eater
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"When astronomers used Chandra to study Sgr A*, in one of its longest ever observations, they found that more than 99% of the infalling material was ejected long before reaching the event horizon "
So, there must be a Max rate.

In any case - what is the real meaning of: "more than 99%"
Why they don't say: 100%???
Do they see even 0.0...1% that is reaching the event horizon?
Do they see any infalling material?
Do you agree that even if we only see more than 99%, and we don't see any star or gas cloud that is infalling in - than we have to call the accretion disc as excretion disc?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/07/2019 17:41:17
Only if it can be demonstrated that it must take longer than 800 million years for it to form and reach its measured mass.
In the article that you have offered it was stated clearly:
The presence of these young and very massive black holes question our understanding of black hole formation and growth."

If you had read the whole article, you'd see that the scientists were offering a solution to that problem.

Quote
Also in the TV program it was stated clearly that there is a limitation for the BH size increase over time.
So, there is a big question about the minimal requested time to form a SMBH.

A time that depends on the model you use to calculate it.

Quote
Let's look at our SMBH.
If I recall it correctly, our scientists verified that it consumes about 3 Sun mass per year.
So, in one billion year is will add about 0.75 to its current mass.

The amount that a black hole consumes now is not necessarily how much it consumed in the past. Nor does that take into account what the black hole's mass was when it was initially formed.

Quote
There is unlimited number of black holes in our Universe.

Based on what evidence?

Quote
We must trace the BH activities and verify how long it might take them to increase their mass over time.
If we will discover that on average it takes about one Billion year for a black hole to double its mass, we could set the clock time for our Universe.

Averages can be deceiving. Humans increase in mass quickly when they are infants, but very slowly once they become adults. The average rate of a black hole's mass increase today may not be what it was in the distant past.

Quote
Is there any way to set a SMBH without crossing the stage of compact BH?
Is there any possibility to form a man without crossing the stages of boyhood, childhood or babyhood?
Why do we assume that a SMBH can be formed without crossing the stages of compact BH, Mid size BH and Massive BH?

If you had read the article, you would know.

Quote
Our scientists know/should know the maximal consumption rate of a BH.
Based on that, they could easily calculate how long is needed to transform a compact BH into a mid size BH and how long it might take to a mid size BH to be transformed into a SMBH.
We know that the SMBH at the core of the Andromeda galaxy is much more massive than the SMBH in our galaxy.
So, we can also calculate the requested time to form that kind of Ultra SMBH out of a  (Milky Way)' SMBH.
Hence, we must look for the biggest SMBH in the Universe and try to calculate how long it might take it to be formed out of a compact BH.
That time frame sets the minimal age of the Universe.
So, why can't we use the SMBH to adjust the time clock for our Universe?

That all depends on what model you are using to calculate the growth rate. The direct collapse model doesn't require a small black hole to grow into a large one. The gas cloud collapses directly into a heavy black hole.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/07/2019 17:34:42
Quote
Do they see any infalling material?
Yes. Sgr-A would not be visible in certain frequencies if no material fell in.
How the Sgr-A is visible?
Do you mean that we see in "certain frequencies" the matter/plasma in the accretion disc around the Sgr-A?
If so, how the visible plasma can give any sort of indication about the infalling matter?
So, let me ask again:
Do we see any direct gas cloud, star, planet, stone or even one Atom/molecular that are falling in from outside the disc?
If we don't see, than why can't we say clearly that we have never seen any direct infalling matter?
There is a clear evidence that over than 99% of the matter in the accretion disc are ejected out from the accretion disc:
"more than 99% of the infalling material was ejected long before reaching the event horizon".
However, it seems to me that our scientists want to believe that all the matter in the accretion disc is due to infalling matter.
Therefore they claim: "more than 99% of the infalling material" instead of "more than 99% of the matter/plasma in the accretion disc", although they don't have even one single evidence for direct infalling matter.
It is clear to me that they didn't find infalling matter and they we will never ever find.
There is a reason for that
 Magnetic Field around the SMBH -
This magnetic field is so powerful that it boosts any atom or molecular upwards/downwards (relative to the galactic disc) at a verified velocity of 0.8 the speed of light. How any infalling matter can cross this ultra power magnetic filed?
Around our planet there is just a friction of a friction of that magnetic field. Even so, it has the power to shift the solar wind to the poles.
Do you agree that if we could set around our planet a magnetic filed in the same amplitude as the one around the SMBH, we would discover that nothing can cross that magnetic line?
What kind of magnetic power is needed in order to boost any nearby atom and molecular high above the galactic disc at that ultra velocity of 0.8 Speed of light?
Why our scientists didn't try to verify the amplitude of that magnetic power?
Why they have decided to neglect the great impact of this power?
Why they constantly insist that matter is falling in, while they didn't find any evidence to prove it?
How could they believe that the Sgr-A is just a messy eater:
Milky Way's Giant Black Hole Spits Out Its Food
https://news.yahoo.com/milky-ways-giant-black-hole-spits-food-180338156.html;_ylt=AwrCxGHe.SVdeSoA6x8PxQt.;_ylu=X3oDMTByMHZ0NG9yBGNvbG8DYmYxBHBvcwM3BHZ0aWQDBHNlYwNzcg--
"The colossal black hole at the heart of the Milky Way galaxy is a messy eater. "
How could it be?
If the mouth of that colossal black hole is full with food, why does it eject over than 99% from its food?
Can we prove that it eats 0.1% or even 0.0...01% from that food?
How many animals do we know that eject over than 99% from the food in their mouth?
Why our scientists ignore impact of the magnetic field around the SMBH, while there is a solid prove that it is so powerful that any nearby molecular/atom is ejected at 0.8 speed of light.
Why do they insist to believe that the matter in the accretion disc is in falling matter, while there is no direct evidence for that and they should know that nothing can cross the mighty magnetic power around the SMBH?
Why do they insist to believe that the SMBH eats some of the matter in the accretion disc, while there is no evidence for that?
Could it be that our scientists don't let the evidences to confuse them?

I still don't understand why we insist to call it accretion disc.
If more than 99% is ejected out from the disc, why don't we call it excretion disc?
So, let me know if I understand it correctly (based on our scientists):
Accretion disc means that 99.99% ejects out from the disc while there is no evidence for any infalling matter to the SMBH.
while,
Excretion disc means that 99.99999999…9999% ejects out while there is no evidence for any infalling matter to the SMBH.
Is it correct?

That all depends on what model you are using to calculate the growth rate. The direct collapse model doesn't require a small black hole to grow into a large one. The gas cloud collapses directly into a heavy black hole.
Did we ever find evidence in the whole universe for a gas cloud that collapses directly into a SMBH?
If our scientists have no clue how the SMBH works today, how can they set any sort of modeling for its growth rate at the early time?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/07/2019 19:37:17
"The colossal black hole at the heart of the Milky Way galaxy is a messy eater. "
How could it be?
If the mouth of that colossal black hole is full with food, why does it eject over than 99% from its food?
Can we prove that it eats 0.1% or even 0.0...01% from that food?
How many animals do we know that eject over than 99% from the food in their mouth?
(https://media.giphy.com/media/HGe4zsOVo7Jvy/giphy.gif)
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/07/2019 21:37:38
Did we ever find evidence in the whole universe for a gas cloud that collapses directly into a SMBH?

Please take note that no one claims to have proven that this is how super-massive black holes formed. The study simply showed that it is plausible. This means that we now know of a way that super-massive black holes could have formed within the known age of the Universe.

Quote
If our scientists have no clue how the SMBH works today, how can they set any sort of modeling for its growth rate at the early time?

They don't have "no clue". There are a lot of good ideas about how they work. Testing those ideas is difficult, because of how far away these black holes are.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/07/2019 06:07:35
Quote
How many animals do we know that eject over than 99% from the food in their mouth?
(https://media.giphy.com/media/HGe4zsOVo7Jvy/giphy.gif)
Lovely.
Similar but with one small difference -
We clearly see the cookie as it moves directly to the monster' mouth.
Why don't we see the requested cookie (gas cloud) as it moves directly to the mouth of our colossal SMBH monster?
There are SMBH monsters at any spiral galaxy.
Our scientists claim that there are more than 100 Billion spiral galaxies in our visible Universe.
Some of them are clearly visible.
They also claim that as the cookie moves to the colossal SMBH it should set magnificent fireworks.
So, how could it be that we have never ever seen any fireworks that highlight the gas cloud as it moves to the mouth of the colossal SMBH monster?
Not even one in the whole Universe!!!
Just an example -
Let's assume that we stay at the center of the Highway for one full year and we only see that all cars are moving in one direction.
Why can't we assume that we are watching a one direction high way?
How long do we have to wait in order to understand that simple meaning?
In the same token -
Our scientists want to believe that the colossal SMBH eats gas cloud.
They are looking for this evidence in the whole universe, but they can't find it.
So how long do we have to wait till our scientists will get the unbelievable understanding that our colossal SMBH monster has no willing to eat any gas could.
One year? 10 year? 1000 years or one trillion years?



Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/07/2019 19:52:59
Are you asserting a 1-way highway is seen?  The computed percentage of in/out would be considerably higher than 100% then.
So, if we clearly see that more than 99% of the matter in the disc around the SMBH is ejected out, while NOTHING is coming in - Than, why our scientists still insist that it is accretion? Why not excretion?
How more time do they need in order to open their eyes and finely see that simple outcome?

They don't have "no clue". There are a lot of good ideas about how they work. Testing those ideas is difficult, because of how far away these black holes are.

Unfortunately, our Universe is not so cooperative with the wishful thinking of our scientists.
However, they don't give up. They bring more and more good ideas - but all of those ideas must obey to the BBT.

It is like a person how is looking his lost key under the light while he lost it at totally different location.
Therefore, as long as our scientists lock themselves at the BBT black box, they will continue to complain about this uncooperative Universe and will never find the answer for the big enigma.

Hence, we have two options -
Replace our universe in order to meet the good ideas of our scientists.
or
Replace our scientists or take them out from the BBT black box. Once they are free from the BBT limitation, they will surly find the ultimate idea..
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/07/2019 21:36:28
It is like a person how is looking his lost key under the light while he lost it at totally different location.
Therefore, as long as our scientists lock themselves at the BBT black box, they will continue to complain about this uncooperative Universe and will never find the answer for the big enigma.

You're assuming that the "key" is in fact lost in a place away from the Big Bang theory.

Hence, we have two options -
Replace our universe in order to meet the good ideas of our scientists.
or
Replace our scientists or take them out from the BBT black box. Once they are free from the BBT limitation, they will surly find the ultimate idea..

You forgot the third option: the Big Bang theory is on the right track and that is where we will find "the ultimate idea".
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/07/2019 11:55:26
It is like a person how is looking his lost key under the light while he lost it at totally different location.
Therefore, as long as our scientists lock themselves at the BBT black box, they will continue to complain about this uncooperative Universe and will never find the answer for the big enigma.
You're assuming that the "key" is in fact lost in a place away from the Big Bang theory.
You forgot the third option: the Big Bang theory is on the right track and that is where we will find "the ultimate idea".
Dear Kryptid
I would like to take you with me to a voyage outside the BBT black box.
A voyage to far end galaxy that is called M.W galaxy.
No one there knows anything about the BBT.
Once we start our voyage we get a reset to all our knowledge about the BBT (and only the BBT).
So
Those people at that M.W galaxy tell us about their following evidences/discoveries:
1. In the center of their M.W. galaxy there is a very massive object. They call it Abra Cadabra.
2. Around this Abra Cadabra they clearly see a disc with very hot matter which is orbiting at a velocity of 0.3 speed of light. They call this hot matter -Plasma. Its temp is 10^9 c.
At the inner most of the disc the plasma is mainly made out of particles.
At the outer most of the disc there are mostly Atoms and Molecular.
3. More than 99% of the matter in that plasma disc is ejected out and there is no indication that even one particle (0.0...01 from the total plasma disc) is falling directly to the abra cadabra.
4. A mighty magnetic field covers the Abra cadabra with its disc.
5. As those ejected Molecular gets closer to the impact of the mighty magnetic field, they are boosted upwards/downwards (with related to the disc) at a velocity of 0.8 Speed of light and set a very high/long stream jet above/below the disc.
6. The estimated total molecular mass in that jet stream above the disc is 10,000 solar mass (Our solar mass).
7. They also see that the molecular fall back to the galactic disc (outside the impact of the magnetic field)
8. At the galactic disc, those molecular gathers together and form a gas cloud.
9. As the gas cloud orbits around the Abra Cadbra (outside the magnetic field) it starts to form new stars.
10. They claim that this aria around the Abra cadabra is the most reach with Hydrogen atoms in the whole galaxy and most of the star forming activity is taking there.
11.  They have never seen any evidence for any molecular/atom or even particle that drifts directly inwards to the Abra Cadabra mouth.
12. They have also never ever seen any gas could or star that drifts inwards to the disc and try to cross the great impact of the magnetic field.

So, the questions are as follow (please remember - We are outside the BBT black box!):

A. If we compare this Abra Cadabra to our SMBH, do we see any difference in the evidences/discoveries?
If so, please let us know which evidence is incorrect/difference.
B. Is there any possibility for any Atom/Rock/Star/Gas cloud from outside to cross the mighty magnetic field and get into the plasma disc?
If positive - please explain.
C. If nothing can get into the plasma disc, how it gets its mass and constantly ejects hot matter outside. (There is a clear evidence for about 10,000 solar mass above the galactic disc.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 12/07/2019 21:37:46
B. Is there any possibility for any Atom/Rock/Star/Gas cloud from outside to cross the mighty magnetic field and get into the plasma disc?

Obviously it can, otherwise the disc would disappear over time as its mass was lost to space. When you take mass away from something without adding any back in, the mass eventually disappears. The first law of thermodynamics guarantees it. You can't expect that magnetic field to be absolutely perfect at preventing plasma from entering the hole, especially since that field is generated by the plasma itself.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/07/2019 06:09:33
B. Is there any possibility for any Atom/Rock/Star/Gas cloud from outside to cross the mighty magnetic field and get into the plasma disc?

Obviously it can, otherwise the disc would disappear over time as its mass was lost to space. When you take mass away from something without adding any back in, the mass eventually disappears. The first law of thermodynamics guarantees it. You can't expect that magnetic field to be absolutely perfect at preventing plasma from entering the hole, especially since that field is generated by the plasma itself.
How long we will continue to say "in falling matter" without any proof for that?
How long we can we reject the evidences which are located infront of our eyes?
How long we would continue to believe that the SMBH eats a nearby matter while we see clearly that nothing is falling in and everything (more than 99%) is ejected out (without even a clear evidence for 0.0…01 falling in).
Our scientists have clearly verified the mighty power of the magnetic field around the plasma disc.
Nothing... Absolutely nothing can cross this magnetic shield.
I'm positively sure that our scientists are fully aware about that obstacle.
If even one Atom can bypass it - Please prove it.
If we can't, then the only possibility to see a constant ejected steam of hot molecular from the plasma disc is by the activity of new mass creation at the SMBH.
So easy and so simple.
How long can we refuse to accept the real knowledge of our Universe?
How long are we going to lock ourselves in the BBT black box?.

 
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 13/07/2019 06:25:03
If we can't, then the only possibility to see a constant ejected steam of hot molecular from the plasma disc is by the activity of new mass creation at the SMBH.

As has been pointed out to you many times before, that would break the laws of physics. Mass/energy cannot be created or destroyed. So we know that's not what's going on. If it isn't coming from the black hole, then it's coming from an outside source. Magnetic fields are not like those force fields you see in science fiction. They aren't impassable barriers. Besides, the outer, cooler area of the accretion disk wouldn't be hot enough to be a plasma and as such wouldn't have an intrinsic magnetic field.

Quote
How long can we refuse to accept the real knowledge of our Universe?

You're doing exactly that by claiming that black holes can create mass/energy. And no, don't say anything about particle accelerators creating mass/energy, because they absolutely do not do that.

Quote
How long are we going to lock ourselves in the BBT black box?

The Big Bang theory has nothing at all to do with this. Black holes simply don't work the way you claim them to.
Title: Re: How gravity works in spiral galaxy?
Post by: pensador on 13/07/2019 10:41:01
Mass/energy cannot be created or destroyed
That is the generally accepted idea.
except
Big Bang theory suggests otherwise. Inflation > Baryogenesis > nucleosynthesis etc
Hoyle theorised mass was constantly appearing in space, a cold slower form of particle creation.
Hawking theorised virtual particles can be converted to real particles via black holes gravity (converting gravitational energy to particles, energy is conserved here)

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 13/07/2019 15:36:55
Big Bang theory suggests otherwise. Inflation > Baryogenesis > nucleosynthesis etc

No it doesn't. All of those particles were created using energy that already existed. That doesn't violate conservation of mass/energy.

Hoyle theorised mass was constantly appearing in space, a cold slower form of particle creation.

This is not the accepted model of today.

Hawking theorised virtual particles can be converted to real particles via black holes gravity (converting gravitational energy to particles, energy is conserved here)

This doesn't violate conservation of mass/energy either. When the black hole emits Hawking radiation, it loses an amount of mass equal to what the radiation carried away.

I am not saying that particles cannot be created or destroyed. I'm talking about mass and energy.
Title: Re: How gravity works in spiral galaxy?
Post by: pensador on 14/07/2019 10:52:48
No it doesn't. All of those particles were created using energy that already existed. That doesn't violate conservation of mass/energy.

What energy, where did it come from.?

The influence of dark energy appears to be increasing the accelerating expansion of the universe. Are you perhaps referring to dark energy. ?

If Dark energy has its origins in the HUP/zero point energy of the vacuum/quantum vacuum, is it real energy or is it just a temporary violation, because the energy is only momentarily borrowed and payed back to the vacuum. ?

Did the energy really exist prior to the big bang, or was it just a temporary violation of the laws of thermodynamics?




This is not the accepted model of today.
That is not the most widely accepted model of today, it is still a model.

This doesn't violate conservation of mass/energy either. When the black hole emits Hawking radiation, it loses an amount of mass equal to what the radiation carried away.

I am not saying that particles cannot be created or destroyed. I'm talking about mass and energy.

Yes I know, but the interesting point with Hawking radiation is that virtual particles are separated and become real, in a rapidly  inflating universe due to dark energy could virtual particles not be separated in a similar way prior to the hot big bang and what follows.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 14/07/2019 15:03:40
What energy, where did it come from.?

The energy present in the Big Bang singularity.

The influence of dark energy appears to be increasing the accelerating expansion of the universe. Are you perhaps referring to dark energy. ?

No, I'm talking about "regular" energy.

If Dark energy has its origins in the HUP/zero point energy of the vacuum/quantum vacuum, is it real energy or is it just a temporary violation, because the energy is only momentarily borrowed and payed back to the vacuum. ?

Nobody knows what dark energy is.

Did the energy really exist prior to the big bang, or was it just a temporary violation of the laws of thermodynamics?

As far as we can tell, the total energy in the Universe has been constant since the beginning of time. So there was no violation.

That is not the most widely accepted model of today, it is still a model.

Just like the flat Earth model is a model.

Yes I know, but the interesting point with Hawking radiation is that virtual particles are separated and become real, in a rapidly  inflating universe due to dark energy could virtual particles not be separated in a similar way prior to the hot big bang and what follows.

I don't know. I've wondered about this myself. I presume that this would still not represent a violation of the first law, as the energy supplied by the expansion is being converted into particles.
Title: Re: How gravity works in spiral galaxy?
Post by: pensador on 14/07/2019 16:12:00
The energy present in the Big Bang singularity.
The inflationary model has no big bang singularity. It happened over a region of inflating space.

I don't know. I've wondered about this myself. I presume that this would still not represent a violation of the first law, as the energy supplied by the expansion is being converted into particles.

This link might interest you https://arxiv.org/pdf/1903.05523.pdf Particle pair creation by inflation of quantum vacuum inn an ion trap.

Nobody knows what dark energy is.

I can think of several theoretical physicists that have reasonably plausible ideas on what dark energy is, and how gravity works. They might be correct or not as the case may be :) A popular theory is Eric Verlindes entropic gravity.?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/07/2019 20:55:00
Quote
Mass/energy cannot be created or destroyed
That is the generally accepted idea.
except
Big Bang theory suggests otherwise. Inflation > Baryogenesis > nucleosynthesis etc
Hoyle theorised mass was constantly appearing in space, a cold slower form of particle creation.
Hawking theorised virtual particles can be converted to real particles via black holes gravity (converting gravitational energy to particles, energy is conserved here)
Thanks for your great support.
Quote
If we can't, then the only possibility to see a constant ejected steam of hot molecular from the plasma disc is by the activity of new mass creation at the SMBH.
As has been pointed out to you many times before, that would break the laws of physics. Mass/energy cannot be created or destroyed. So we know that's not what's going on. If it isn't coming from the black hole, then it's coming from an outside source. Magnetic fields are not like those force fields you see in science fiction. They aren't impassable barriers. Besides, the outer, cooler area of the accretion disk wouldn't be hot enough to be a plasma and as such wouldn't have an intrinsic magnetic field.
How can we explain the molecular jet stream that is moving Upwards/downwards at ultra velocity of 0.8 speed of light?
What kind of force can set that jet stream which is moving high above the accretion disc plane and includes about 10,000 Sun mass?
Don't you agree that the only power that can set this molecular jet stream is the mighty magnetic filed?
How can you de estimate this mighty power?
I still don't understand why the magnetic field  can trap any molecular that is ejected out from the accretion disc (more than 99% of the matter in the disc), but it can't trap any molecular that try to cross it?

I am not saying that particles cannot be created or destroyed. I'm talking about mass and energy.
So, do you agree that particles can be created or destroyed?
If so, that's all we need for Atom creation.
I have already proved that Atom is a cell of energy:
https://www.thenakedscientists.com/forum/index.php?topic=75261.40
Let's focus on a proton:
https://en.wikipedia.org/wiki/Quark
"A proton is composed of two up quarks, one down quark, and the gluons that mediate the forces "binding" them together."
The total mass of those three quarks is:  9 MeV/c^2.
Therefore, if those quarks can be created at the accretion disc around the SMBH, they can be the seed for the Proton.
The ultra high energy/temp/ eclectic/magnetic field at the accretion disc, can add the requested energy for setting the gluons.
This gluons contribute 99% of the total mass in the Proton (As the total mass of a Proton is over than 900 MeV/c^2)
Therefore, in the accretion disc we can see particles at the most inwards side of the disc - At this aria - quarks are created, while as they drifts outwards, they gain the energy of the gluons which converts them into Protons and Neutrons.
Those elements set the Hydrogen Atoms.
However, as all the plasma orbits at Ultra high velocity (0.3 c), Hydrogen Atoms collide with each other and sets Helium and the whole list of Atoms and molecular that we know including water, gold and even Uranium.
As this mass ejected from the accretion disc, they are trapped by the mighty magnetic field around the accretion disc and boosted upwards/downwards at 0.8 speed of light.

P.S. - Unfortunately, I couldn't reply as I had to go to Nederland due to a very sad circumstances.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/07/2019 21:06:51
How can we explain the molecular jet stream that is moving Upwards/downwards at ultra velocity of 0.8 speed of light?
What kind of force can set that jet stream which is moving high above the accretion disc plane and includes about 10,000 Sun mass?
Don't you agree that the only power that can set this molecular jet stream is the mighty magnetic filed?
How can you de estimate this mighty power?
I still don't understand why the magnetic field  can trap any molecular that is ejected out from the accretion disc (more than 99% of the matter in the disc), but it can't trap any molecular that try to cross it?

First of all, none of that requires energy/mass to be created out of nowhere and as such is not evidence for you idea. Second of all, "over 99%" is not 100%. You wouldn't expect the matter on the very innermost of the disk to be stopped by the magnetic field anyway since the magnetic field is generated by the disk itself and not by the black hole.

So, do you agree that particles can be created or destroyed?

Yes, it happens all the time. A muon decays into an electron, a neutrino and anti-neutrino. The muon is destroyed and three new particles are created in its place. However, and this is the important part, the total amount of mass and energy remains exactly the same.

If so, that's all we need for Atom creation.

Atoms can be created, yes. It happens all the time in nuclear fission. The total mass/energy does not change at all, though.

The ultra high energy/temp/ eclectic/magnetic field at the accretion disc, can add the requested energy for setting the gluons.

The total mass/energy before and after the creation of the protons would be the same. If the mass/energy to create it came from the disk, then the disk now has less mass/energy than it had before.

However, as all the plasma orbits at Ultra high velocity, Hydrogen Atoms collide with each other and sets Helium and the whole list of Atoms and molecular that we know including water, gold and even Uranium.
As this mass ejected from the accretion disc, they are trapped by the mighty magnetic field around the accretion disc and boosted upwards/downwards at 0.8 speed of light.

That's nice and all, but what happens once all of the mass/energy of the disk is depleted? Then the accretion disk should disappear.
Title: Re: How gravity works in spiral galaxy?
Post by: pensador on 24/07/2019 12:22:35
On a small scale the dynamic Casimir effect produces particles.

On a larger scale Fast spinning Neutron Stars/Pulsars have huge magnetic fields, that are also thought to be able to produce particles from the vacuum of space. The surfaces of some neutron stars are apparently moving at 0.25c, mind bogglingly fast. 8) I guess that this form of particle production will take some energy from the neutron star and slow its spin.  :-\


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/07/2019 18:57:34
On a small scale the dynamic Casimir effect produces particles.

On a larger scale Fast spinning Neutron Stars/Pulsars have huge magnetic fields, that are also thought to be able to produce particles from the vacuum of space. The surfaces of some neutron stars are apparently moving at 0.25c, mind bogglingly fast.  I guess that this form of particle production will take some energy from the neutron star and slow its spin. 
Wow!
That is really important information.
Neutron Stars/Pulsars have a mass size that is a friction of a friction of the SMBH.
We know that just in the accretion disc there is about three Sun mass.
So, if "Neutron Stars/Pulsars have huge magnetic fields" what kind of magnetic field could be evolve around the SMBH accretion disc?
If the magnetic field around a Neutron star/Pulsars are thought to be able to produce particles from the vacuum of space, than the magnetic filed around the SMBH should be able to produce much more particles.
But, particles are not Atoms.
Atoms can be created, yes.
In order to set an Atom we first must set Proton & Neutron
https://en.wikipedia.org/wiki/Quark
"A proton is composed of two up quarks, one down quark, and the gluons that mediate the forces "binding" them together."???
Hence, if we break down the proton - what do we get?
Do you agree that we should get Quarks and Gluons?
Quarks are very clear particles, but what about gluons?
Is it a particle?
Is there any gluons particles?
Have we ever found one gluons particle in the whole Universe?
Do you agree that gluons represents a pure energy?
If the gluons is an Energy, than do you agree that the Atom represents a cell of Energy?
Hence, if you wish to creat a new proton you must have the requested quarks + the requested energy to set the gluons.
Please remember - the mass contribution of the gluons to the proton is more than 900 MeV/c^2. while all the quarks is a proton contribute about 9 MeV/c^2. (less than 1% from the proton).
Therefore, if a gluons isn't a particle and before the BBT there were only particles, how can we dream that the BBT can create even one atom without the gluons?
Is there any possibility for a bang (even if we call it big bang) to create gluons?
If it can't create gluons, how it can create an Atom?
If the BBT doesn't add energy to the existed articles (that were there before the bang), how it could create Atoms?
Don't you agree that the accretion disc around the SMBH is the only place in the whole Universe where the energy of the magnetic/electric power + Temp (10^9) + Velocity (0.3c) + pressure is high enough to add the gluons to the quarks in order to form new proton?
Do you believe that the Big Bang can create those kinds of energy/power/conditions (especially the magnetic/electric power)?
How the BBT can add that gluons without having those key requested elements?
How the BBT can't set even one proton if the gluons is not a particle?
Do you agree that if "Neutron Stars/Pulsars have huge magnetic fields, that are also thought to be able to produce particles from the vacuum of space" than the SMBH should also be able to produce articles (especially quarks) at the vacuum of space in the innermost of the accretion disc?
Therefore, we clearly see mainly particles at the innermost ring of the accretion disc, while at the outermost side of the accretion ring we clearly see molecular.
That proves that the new born particles are drifted outwards as they orbit at ultra high velocity (of over than 0.3 c).
Do you agree that as they drift outwards they become part of the plasma in the accretion disc, and directly affected by the ultra magnetic/electric power?
So, do you agree that the power/temp (10^9)/density/peruse at the accretion disc is the only place in the whole universe that can add the gluons to the quarks and form new proton and new atom?

Quote from: Dave Lev
How can we explain the molecular jet stream that is moving Upwards/downwards at ultra velocity of 0.8 speed of light?
What kind of force can set that jet stream which is moving high above the accretion disc plane and includes about 10,000 Sun mass?
There is a 10k sun-mass amount of material moving at 0.8c?  How does it not exit the galaxy?  That's well above escape velocity from 'high above the disc plane'.
Ghostly Gamma-ray Beams Blast from Milky Way's Center
https://www.cfa.harvard.edu/news/2012-16
The two beams, or jets, were revealed by NASA's Fermi space telescope. They extend from the galactic center to a distance of 27,000 light-years above and below the galactic plane.
The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused.
It would take a tremendous influx of matter for the galactic core to fire up again. Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required.
In this article they try to explain this jet as: "a corkscrew-like magnetic field that kept it tightly focused"
I think that magnetic field is the only source of power that can boost 10,000 Sun mass in this kind of molecular jet steam which had been ejected from the accretion disc.
So, just try to figure what kind of magnetic field is needed to do that kind of job.
It is clear to me that all the molecular that ejected from the accretion disc are boosted upwards/downwards due to the ultra magnetic field.
Therefore, if a molecular from outside the magnetic field will dare to come closer it will also will boosted upwards/downwards with that jet stream.
Hence, nothing can theoretically cross the impact of that ultra magnetic shield.
Not even one molecular from outside.
Therefore, I totally disagree that the source for that mass is:
"Shoving 10,000 suns into the black hole at once would do the trick. Black holes are messy eaters, so some of that material would spew out and power the jets,"
The SMBH does not shove even one Atom/molecular from outside.
We have never found any evidence for that and we will never find!
It is also not logical that while more than 99% of the mass in the plasma is ejected out from the accretion disc, a mass from outside will try to get in. The same power that ejects all of that mass (over than 99%) from the accretion disc should prevent from any molecular to get in!!!
You wouldn't expect the matter on the very innermost of the disk to be stopped by the magnetic field anyway since the magnetic field is generated by the disk itself and not by the black hole.
Why the SMBH can't generate magnetic field?
Do we have prove for that?
Do we know the real process at the SMBH?


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/07/2019 19:55:03
Therefore, if a gluons isn't a particle and before the BBT there were only particles
BBT says all the 'particles' existing today were created after the big bang.

Quote
Do you agree that if "Neutron Stars/Pulsars have huge magnetic fields, that are also thought to be able to produce particles from the vacuum of space"
They're not produced from the vacuum of space.  They're produced from the magnetic field of the Neutron star according to the statement you quote.  If just the vacuum was enough, you'd see such particle creation all the time everywhere.

Quote
than the SMBH should also be able to produce articles (especially quarks) at the vacuum of space in the innermost of the accretion disc?
The accretion disk is not a vacuum.

Quote
Quote from: Halc
There is a 10k sun-mass amount of material moving at 0.8c?  How does it not exit the galaxy?  That's well above escape velocity from 'high above the disc plane'.
Ghostly Gamma-ray Beams Blast from Milky Way's Center
https://www.cfa.harvard.edu/news/2012-16
Quote from: harvard
The two beams, or jets, were revealed by NASA's Fermi space telescope. They extend from the galactic center to a distance of 27,000 light-years above and below the galactic plane.
The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused.
It would take a tremendous influx of matter for the galactic core to fire up again. Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required.
OK, that mentions a 10k solar mass cloud being required to get back into 'gulp' mode from its current 'sip' mode.  No mention of anything moving at 0.8c mentioned.  The detected bubble is faint (not massive), and is an effect from a 'gulp' perhaps a million years ago.  The bubble extends to 27000 LY, so it's moving at best at 27 ly per 1000 years, hardly 0.8c.

The jets, if fast moving material, are the produce of the current 'sip' mode.  If they're a remnant of the big mass from a million years ago, the jets must be slow indeed to still be there.

Quote
Therefore, if a molecular from outside the magnetic field will dare to come closer it will also will boosted upwards/downwards with that jet stream.
That cloud is the source of the energy for the jet.  It is the gun, not the bullet.

Quote
Therefore, I totally disagree that the source for that mass is:
"Shoving 10,000 suns into the black hole at once would do the trick. Black holes are messy eaters, so some of that material would spew out and power the jets,"
OK, so you quote this article to support your assertions, but reject what it says.  Classic.

Quote
Why the SMBH can't generate magnetic field?
Do we have prove for that?
A reasonable question.  Black holes preserve charge and angular momentum.  Seems plausible.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/07/2019 21:06:49
Hence, if we break down the proton - what do we get?
Do you agree that we should get Quarks and Gluons?

You can't have free quarks due to color confinement, except, debatably, in a quark-gluon plasma (but that automatically involves more than a single proton).

Quarks are very clear particles, but what about gluons?
Is it a particle?
Is there any gluons particles?

Yes, gluons are particles too.

Have we ever found one gluons particle in the whole Universe?

Yes, we've seen their signatures in particle accelerators.

Do you agree that gluons represents a pure energy?

There's no such thing as "pure energy". Energy is a property of physical entities,, not a physical entity unto itself.

If the gluons is an Energy, than do you agree that the Atom represents a cell of Energy?

Atoms are more than just energy.

Hence, if you wish to creat a new proton you must have the requested quarks + the requested energy to set the gluons.
Please remember - the mass contribution of the gluons to the proton is more than 900 MeV/c^2. while all the quarks is a proton contribute about 9 MeV/c^2. (less than 1% from the proton).
Therefore, if a gluons isn't a particle and before the BBT there were only particles, how can we dream that the BBT can create even one atom without the gluons?
Is there any possibility for a bang (even if we call it big bang) to create gluons?
If it can't create gluons, how it can create an Atom?

Protons can be created as a product of pair production in extreme gravitational environments like those at a black hole's event horizon. An antiproton has to be created at the same time due to conservation laws. By the way, gluons are automatically created by quarks as a part of their strong nuclear field. If you have quarks, you already have gluons.

If the BBT doesn't add energy to the existed articles (that were there before the bang), how it could create Atoms?

Pair production, as I pointed out before. That gives you both protons and electrons to work with.

Don't you agree that the accretion disc around the SMBH is the only place in the whole Universe where the energy of the magnetic/electric power + Temp (10^9) + Velocity (0.3c) + pressure is high enough to add the gluons to the quarks in order to form new proton?

No, because quarks already have gluons.

Do you believe that the Big Bang can create those kinds of energy/power/conditions (especially the magnetic/electric power)?

The Big Bang was orders of magnitude hotter and more energetic than any other event in the universe.

Do you agree that if "Neutron Stars/Pulsars have huge magnetic fields, that are also thought to be able to produce particles from the vacuum of space" than the SMBH should also be able to produce articles (especially quarks) at the vacuum of space in the innermost of the accretion disc?

I'm not sure that neutron stars can actually do that, but if so, then part of the neutron star's mass/energy is depleted in the process of creating those particles.

How the BBT can add that gluons without having those key requested elements?
How the BBT can't set even one proton if the gluons is not a particle?

I've already pointed out why both of these assumptions are wrong.

Therefore, we clearly see mainly particles at the innermost ring of the accretion disc, while at the outermost side of the accretion ring we clearly see molecular.
That proves that the new born particles are drifted outwards as they orbit at ultra high velocity (of over than 0.3 c).

No, it does not.

Do you agree that as they drift outwards they become part of the plasma in the accretion disc, and directly affected by the ultra magnetic/electric power?

No, because they don't drift outward.

So, do you agree that the power/temp (10^9)/density/peruse at the accretion disc is the only place in the whole universe that can add the gluons to the quarks and form new proton and new atom?

No, because quarks already have gluons regardless of where they are or what environment they are in. You can't add gluons to quarks any more than you can add electric fields to electrons: they already have them.

It is also not logical that while more than 99% of the mass in the plasma is ejected out from the accretion disc, a mass from outside will try to get in. The same power that ejects all of that mass (over than 99%) from the accretion disc should prevent from any molecular to get in!!!

So what happens when all of the mass/energy of the disk is depleted? The disk would disappear and the magnetic field would vanish.

Why the SMBH can't generate magnetic field?
Do we have prove for that?
Do we know the real process at the SMBH?

In order for a black hole to have a magnetic field, it must be both electrically-charged and spinning. The black hole should indeed spin, but it should not be electrically-charged. The reason for this is two-fold. Firstly, the very nature of the electromagnetic force means that positive and negative charges seek each other out and form neutral bodies. So almost all of the matter around a black hole would already be neutral. Secondly, an electrically-charged black hole would preferentially attracted oppositely-charged matter that would neutralize it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/07/2019 06:44:00
Protons can be created as a product of pair production in extreme gravitational environments like those at a black hole's event horizon. An antiproton has to be created at the same time due to conservation laws. By the way, gluons are automatically created by quarks as a part of their strong nuclear field. If you have quarks, you already have gluons.
Thanks Kryptid
I couldn't ask for a better answer!!!
You have just solved several main questions:
1. How the SMBH increases its mass?
2. Why all the matter from the accretion disc is ejected outwards?
3. Why the SMBH doesn't need to eat any matter from the nearby aria?

Based on your answer:
"Protons can be created as a product of pair production in extreme gravitational environments like those at a black hole's event horizon"
So, based on your reply, Proton and Antiproton could be created at the same moment at the SMBH's event horizon.
I would like to compare those two new born particles to a twin star that orbit around a BH.
We know that if one star is falling in, the other one is ejected outwards.
In the same token, if the Antiproton will fell into the SMBH, the Proton should be drifted outwards.
The "falling in" activity of the Antiproton, have solved the problems.
In one hand it increases the mass of the SMBH, while it the other hand it push the Proton outwards.
Therefore, we clearly see that more than 99% of the mass in the plasma in the accretion disc is ejected outwards.
Hence, there is no need for any matter from outside the disc to drift inwards.
If that scenario is correct, it might also answer your following remark about the magnetic field:

In order for a black hole to have a magnetic field, it must be both electrically-charged and spinning. The black hole should indeed spin, but it should not be electrically-charged. The reason for this is two-fold. Firstly, the very nature of the electromagnetic force means that positive and negative charges seek each other out and form neutral bodies. So almost all of the matter around a black hole would already be neutral. Secondly, an electrically-charged black hole would preferentially attracted oppositely-charged matter that would neutralize it.
So you claim that:
"an electrically-charged black hole would preferentially attracted oppositely-charged matter that would neutralize it"
So, as the SMBH is made out of Antipositron, we need to call it "A NEGATIVLY electrically-charged SMBH".
You claim that: "an electrically-charged black hole would preferentially attracted oppositely-charged matter that would neutralize it"
In one of the articles about the BBT that I have read it was stated that about 99% of the Protons/Antiprotons that had been created by the BBT have probably eliminated each other at the same moment of their creation.
However, based on the NEGATIVLY electrically-charged SMBH, the story is totally different:
At the same moment that the new born Antipositron & Proton is created the "NEGATIVLY electrically-charged SMBH" would preferentially attracted NEGATIVLY-charged matter that would neutralize it (which is the Antiproton), while the Proton will be ejected outwards.
Therefore, there is no time for the Antiproton & proton to eliminate each other at the moment of their creation.
At the same moment that they are created they also are separated by the mighty power of the "NEGATIVLY electrically-charged SMBH".
Hence, the SMBH ejects all protons that it had created to our galaxy/universe, while all Antiprotons is used as its real food.

So, the SMBH sets all the activity.
It forms new Antiproton and New Proton at the same moment.
It eats the Antiproton and therefore it increases its mass constantly during the mass creation activity.
Actually, on any new born Proton that it contributes to our Universe, it eats one equivalent Antiproton.
Hence, the mass of the SMBH indicates the total Antiproton mass that it have consumed during its life time frame.
In the same token that mass also represents the total proton production that it contributes to the Galaxy/Universe.
Therefore, we can easily calculate the total Sun mass that it had been created during his life time.

Do you agree with that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/07/2019 07:05:48
Based on your answer:
"Protons can be created as a product of pair production in extreme gravitational environments like those at a black hole's event horizon"
So, based on your reply, Proton and Antiproton could be created at the same moment at the SMBH's event horizon.
I would like to compare those two new born particles to a twin star that orbit around a BH.
We know that if one star is falling in, while the other one is ejected outwards.
In the same token, if the Antiproton will fell into the SMBH, the Proton should be drifted outwards.
The "falling in" activity of the Antiproton, have solved the problems.
In one hand it increases the mass of the SMBH, while it the other hand it push the Proton outwards.
Therefore, we clearly see that more than 99% of the mass in the plasma in the accretion disc is ejected outwards.
Hence, there is no need for any matter from outside the disc to drift inwards.
If that scenario is correct, it might also answer your following remark about the magnetic field:

You have it all wrong.

(1) Pair production at a black hole's even horizon causes the black hole to lose mass. This is Hawking radiation: https://en.wikipedia.org/wiki/Hawking_radiation

(2) Supermassive black holes don't even generate proton-antiproton pairs. The space-time curvature at the event horizon is far too gentle for that. Counterintuitively, smaller black holes release more energetic radiation than larger ones. Protons are too heavy to be created by a supermassive black hole. Instead, those black holes can only create very, very weak photons (with a temperature much lower than that of the cosmic microwave background). Only very small black holes could create proton-antiproton pairs.

At the same moment that the new born Antipositron & Proton is created the "NEGATIVLY electrically-charged SMBH" would preferentially attracted NEGATIVLY-charged matter that would neutralize it (which is the Antiproton), while the Proton will be ejected outwards.

You have it backwards. Negative charges attract positive charges. If the black hole was negative, it would preferentially attract protons, not antiprotons. I already pointed out to you why an electrically-charged black hole is very unlikely to occur in nature and why it would quickly neutralize itself if it did occur.

It eats the Antiproton and therefore it increases its mass constantly during the mass creation activity.

No it doesn't. You're trying to break the law of conservation of mass again. I already pointed out to you that Hawking radiation decreases a black hole's mass, not increase it.

Do you agree with that?

No. Please fix your misconceptions before continuing.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/07/2019 21:40:53
(1) Pair production at a black hole's even horizon causes the black hole to lose mass. This is Hawking radiation: https://en.wikipedia.org/wiki/Hawking_radiation
Thanks for the great article
It is stated:
"Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.[9] As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole.[10]
Let's start with the following message:
"This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles."
Do you agree that it is stated that the gravitational power of the black hole is boosted virtual particles into real particles?
Therefore, the gravity power of the BH itself has the ability to create new real particles?
This actually proves your following statement:
Protons can be created as a product of pair production in extreme gravitational environments like those at a black hole's event horizon.
So, you agree that BH can generate new particles as Proton and Antiproton due to its simple gravity force.
However, why do you claim that the space-time curvature at the event horizon of the SMBH is much more gentle than the one at a BH?
How could it be that Protons are too heavy to be created by a suppermassive black hole but they are not too heavy for the BH?
Would you kindly prove it?
Would you also kindly explain what is so unique in those special very small black holes that only them could create proton-antiproton pairs.
Hawking doesn't give any special information about small BH or SMBH.

In any case, most of us consider that gravity comes for free.
Therefore, if the gravity of a BH could create proton-antiproton pairs, than why it doesn't come for free?
Based on Hawking radiation "the escape of one of the particles lowers the mass of the black hole."
If that is correct, than as the BH decreases its mass due to the creation activity, it actually decreases also its gravity.
Hence, please let me know if you agree with the following:
1. BH gravity creates new particle. Therefore:
2. BH mass is reduces. Therefore:
3. Gravity is reduces.
So, Hawking proved/assumed that gravity doesn't come for free.
If it doesn't come for free for mass creation, than why it comes for free for Sun/Earth orbit system.
Why the Sun doesn't need to lose mass (or gravity) due to this orbital activity?
Why the gravity reduction is only applicable for BH during new particle creation activity?
Please remember - Those new particles had been created directly from the gravity that boosted virtual particles.
So, the BH doesn't take any mass from itself to create those new particles
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/07/2019 01:59:27
Do you agree that it is stated that the gravitational power of the black hole is boosted virtual particles into real particles?

I've recently come to the knowledge that virtual particles aren't an actual thing, but are just modeling tools that are useful in the mathematics of quantum mechanics. Regardless, the idea of virtual particles being changed into real particles is still probably good enough to visualize the process, so I'll tentatively agree.

Therefore, the gravity power of the BH itself has the ability to create new real particles?

It isn't just gravity that is doing it, it is specifically the tidal forces caused by gravity. To be more specific, it is the difference between the force of gravity at different distances from the event horizon that allow the process to work. The tidal forces "pull" the particle-antiparticle pairs apart. The stronger the tidal forces, the faster particle-antiparticle pairs can be separated and the faster the hole shrinks.

So, you agree that BH can generate new particles as Proton and Antiproton due to its simple gravity force.

Only if the tidal forces are strong enough at the event horizon. In other words, only if the black hole is small enough.

However, why do you claim that the space-time curvature at the event horizon of the SMBH is much more gentle than the one at a BH?
How could it be that Protons are too heavy to be created by a suppermassive black hole but they are not too heavy for the BH?
Would you kindly prove it?
Would you also kindly explain what is so unique in those special very small black holes that only them could create proton-antiproton pairs.

The tidal forces at the event horizon of a super-massive black hole are lower because the radius of that black hole is larger. Let's use two examples to illustrate this point. We start off with a very small black hole with an event horizon radius of 1 centimeter. How does the gravitational force right at the event horizon compare to the gravitational force 1 centimeter away from the event horizon? Right at the event horizon, the distance to the singularity is 1 centimeter, whereas 1 centimeter away from from the event horizon, the singularity is 2 centimeters away. Since gravity follows the inverse square law, we know that a doubling of distance causes the acceleration due to gravity to be divided by 4. So over that tiny distance of 1 centimeter, the gravitational force is different by a factor of 400%.

Let's compare this to a much larger black hole that is 1 kilometer in radius. What is the difference in gravitational strength right at its event horizon and 1 centimeter away from the horizon? Since 1 centimeter is 10,000 times shorter than a kilometer, then we are comparing total distances from the singularity of 1 kilometer and 1.00001 kilometers. What is the difference in gravitational acceleration between these two distances? It would be 1.00001 squared, or 1.00002. That means the gravitational acceleration right at the event horizon is only 0.002% stronger than what it is 1 centimeter away from the horizon.

So the small black hole has tidal forces that are much, much stronger at the horizon than the large black hole.

The more massive the subatomic particles, the more energy (and therefore tidal force) is needed to "boost" them into existence. Photons can have arbitrarily low mass/energy, so black holes can always create photons as Hawking radiation regardless of how weak the tidal forces are. Protons, on the other hand, are pretty heavy (almost 2,000 times heavier than electrons) and so require large amounts of energy (or very strong tidal forces) to generate. Only small black holes have the needed tidal forces to create them.

In any case, most of us consider that gravity comes for free.
Therefore, if the gravity of a BH could create proton-antiproton pairs, than why it doesn't come for free?
Based on Hawking radiation "the escape of one of the particles lowers the mass of the black hole."
If that is correct, than as the BH decreases its mass due to the creation activity, it actually decreases also its gravity.
Hence, please let me know if you agree with the following:
1. BH gravity creates new particle. Therefore:
2. BH mass is reduces. Therefore:
3. Gravity is reduces.
So, Hawking proved/assumed that gravity doesn't come for free.
If it doesn't come for free for mass creation, than why it comes for free for Sun/Earth orbit system.
Why the Sun doesn't need to lose mass (or gravity) due to this orbital activity?
Why the gravity reduction is only applicable for BH during new particle creation activity?
Please remember - Those new particles had been created directly from the gravity that boosted virtual particles.
So, the BH doesn't take any mass from itself to create those new particles

Gravity does come for "free" for all objects with mass, even black holes. Remember, mass is not vanishing into nothingness when a black hole evaporates due to Hawking radiation. What is occurring is that mass is moving out of the black hole and being sent into the universe at large in the form of radiation. That radiation has mass/energy that is equal to the mass/energy lost by the black hole and as such also has gravity equal to the amount of gravity lost by the hole.

The reason that the Sun and Earth don't emit Hawking radiation is because they don't have event horizons that can capture one member of the particle pair. So any virtual particle-antiparticle pairs popping up around them can annihilate and return to the vacuum.

The reason that the black hole's mass decreases when Hawking radiation is a bit harder to understand. It is often said that the member of the particle pair that escapes into the universe has positive mass/energy, while the particle that is consumed by the black hole has negative mass/energy. When that negative mass is added to the positive mass of the hole, it causes the hole to lose total mass. It isn't quite that simple, though, as the following description of the process explains: https://physics.stackexchange.com/questions/30597/black-holes-and-positive-negative-energy-particles/30601#30601

Halc has a better handle on relativity than I do, so perhaps he can chime in on how the reference frames can make one member of the particle pair look like it has negative energy. Alternatively, maybe I'll go look at my black hole book later and find relevant quotes about this matter.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/07/2019 03:25:07
Therefore, the gravity power of the BH itself has the ability to create new real particles?
It isn't just gravity that is doing it, it is specifically the tidal forces caused by gravity. To be more specific, it is the difference between the force of gravity at different distances from the event horizon that allow the process to work. The tidal forces "pull" the particle-antiparticle pairs apart. The stronger the tidal forces, the faster particle-antiparticle pairs can be separated and the faster the hole shrinks.
Like to add that tidal forces near a small mass is greater than that near a large mass, so the tidal forces near a neutron star will kill you, despite the gravitational force on you being a fraction of that near the event horizon of a black hole.  The tidal force of a small black hole (like Jupiter mass) will be far worse, and that near the largest black holes not particularly bothersome.

I imagine a black hole the mass of say an elephant might be able to produce a proton.  I don't know.  I'm guessing.  I know it's small.

Quote
1. BH gravity creates new particle. Therefore:
2. BH mass is reduces. Therefore:
3. Gravity is reduces.
Only if the particle radiates away does the mass reduce.  The vast majority of them fall immediately back in, resulting in no mass change.

Quote
If it doesn't come for free for mass creation, than why it comes for free for Sun/Earth orbit system.
Why the Sun doesn't need to lose mass (or gravity) due to this orbital activity?
Hawking radiation has nothing to do with orbital activity.  The sun very much does radiate away mass at billions the rate that the SMBH does, and yes, it loses mass steadily as it does so.  Most of the energy on Earth comes from a tiny fraction of that radiated energy that manages to hit us and stick for a while.  Yes, the sun's gravity reduces accordingly as this goes on.

Quote from: Kryptid
The reason that the Sun and Earth don't emit Hawking radiation is because they don't have event horizons that can capture one member of the particle pair.
Or because they don't have significant tidal fields anywhere.

Quote
Halc has a better handle on relativity than I do, so perhaps he can chime in on how the reference frames can make one member of the particle pair look like it has negative energy.
Sorry, I think one needs to be more of a particle physicist than a relativity guy to describe clearly how all that works.  I'd muck it up.  The stack exchange guys often put out some top rate answers.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/07/2019 21:29:55
Thanks
It isn't just gravity that is doing it, it is specifically the tidal forces caused by gravity. To be more specific, it is the difference between the force of gravity at different distances from the event horizon that allow the process to work.
In all the articles that I have found, it is specifically stated about the gravity that is needed to boost the virtual particle into real particle.
I couldn't find even one word about the tidal with related to the creation on new particles around a BH or SMBH.
So, why do you advice that it is tidal?
Actually, it seems to me that you discuss about simple gravity forces at different distances: "it is the difference between the force of gravity at different distances from the event horizon".

https://en.wikipedia.org/wiki/Tidal_force
"In celestial mechanics, the expression tidal force can refer to a situation in which a body or material (for example, tidal water) is mainly under the gravitational influence of a second body (for example, the Earth), but is also perturbed by the gravitational effects of a third body (for example, the Moon)."
In our example there are no second & Third bodies. It is just the gravity impact of the BH on the virtual particles.
So, why do you call it Tidal? Is there a third body?
Why not a simple gravity force.
Don't you agree that the meaning of gravity force is different force at different distance?
So, I don't understand why you insist on tidal.

The reason that the black hole's mass decreases when Hawking radiation is a bit harder to understand. It is often said that the member of the particle pair that escapes into the universe has positive mass/energy, while the particle that is consumed by the black hole has negative mass/energy. When that negative mass is added to the positive mass of the hole, it causes the hole to lose total mass. It isn't quite that simple, though, as the following description of the process explains: https://physics.stackexchange.com/questions/30597/black-holes-and-positive-negative-energy-particles/30601#30601
So we agree by now that due to the BH gravity force (you can call it tidal if you wish) two identical particles with negative polarities to each other (one negative and one positive) are popped up.
As the negative particle is falling into the BH, the positive is ejected outwards.
If the BH is made out of positive particles than the negative particles that fall in actually decrease the total mass of the BH. That is very clear.
However, if the BH is made out of negative particle, than the negative particle that falls in must increase the BH mass.
Do you agree with that?
Therefore, it is a severe mistake to assume that under any condition the BH should disappear.
We have to say that the BH might disappear just if the polarity of the falling particle is the opposite to the polarity of that BH.

The tidal forces "pull" the particle-antiparticle pairs apart. The stronger the tidal forces, the faster particle-antiparticle pairs can be separated and the faster the hole shrinks.
Can you please explain how the Tidal can "pull" the particle-antiparticle pairs apart?
Tidal is gravity force.
Why the gravity will have different effect on particle with different polarity.
Don't you agree that it should affect both partials at the same impact?
Do we know about gravity with negative polarity?
We know what is the meaning of negative or positive voltage, while there is no meaning of negative or positive gravity.
Therefore, if we wish to pull apart two particles (one negative and one positive) don't you agree that we MUST set them under electric/magnetic field?
So, do you agree that Tidal (or any sort of gravity force) can convert the virtual particles into real particles, but it can't separate them.
Do you agree that without a direct impact of electric/magnetic field they might immediately merge again and disappear forever?
Therefore, how we can discuss about particle creation while magnetic/electric field doesn't involve in the process?




The tidal forces at the event horizon of a super-massive black hole are lower because the radius of that black hole is larger. Let's use two examples to illustrate this point. We start off with a very small black hole with an event horizon radius of 1 centimeter. How does the gravitational force right at the event horizon compare to the gravitational force 1 centimeter away from the event horizon? Right at the event horizon, the distance to the singularity is 1 centimeter, whereas 1 centimeter away from from the event horizon, the singularity is 2 centimeters away. Since gravity follows the inverse square law, we know that a doubling of distance causes the acceleration due to gravity to be divided by 4. So over that tiny distance of 1 centimeter, the gravitational force is different by a factor of 400%.
I imagine a black hole the mass of say an elephant might be able to produce a proton.  I don't know.  I'm guessing.  I know it's small.
Thanks for the explanation. However, I really don't understand how could it be that a SMBH with estimated mass of Million BH, has less power than a BH with a size of an elephant???
Let me give you other example:
If I will tell you that the thrust of the jet 747 is less than a bicycle- would you believe me?
In order to prove it, I will place one person on the bicycle while other person will stay at a distance of 1 km from the 747.
Which one will get higher trust?
Is that a correct way to verify the trust?
In the same way it seems to me that you do not evaluate correctly the impact of the SMBH vs a BH.
The other issue is orbital velocity.
It seems to me that inorder to convert a virtual particles into real particle, you need to set them at an Ultra high velocity.
you even claim that:
The more massive the subatomic particles, the more energy (and therefore tidal force) is needed to "boost" them into existence
Hence, we must boost the particles in order to create them.
The meaning of boost is to give them a possibility to run at ultra high speed for a long enough distance. (I call that distance - Runway distance)
If you try to set any object at a very small orbital cycle, and try to run it at ultra velocity, the force outwards should be stronger as the cycle is shorter.
Therefore, the virtual particle will be ejected outwards from the event of horizon cycle before setting the whole Runway that is needed to form a real particle.

One last question:
Why are we so sure that the new particle creation is taking care only at the event horizon?
Why it couldn't be created even inwards to the event horizon and then imidiatly be separated by the electric/magnetic field.
If the field is strong enough - why it can't eject the new born particle even if the particle had been created deep into the event horizon radius?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/07/2019 22:25:39
"In celestial mechanics, the expression tidal force can refer to a situation in which a body or material (for example, tidal water) is mainly under the gravitational influence of a second body (for example, the Earth), but is also perturbed by the gravitational effects of a third body (for example, the Moon)."
In our example there are no second & Third bodies. It is just the gravity impact of the BH on the virtual particles.
So, why do you call it Tidal? Is there a third body?

You missed the very first sentence of that article: "The tidal force is a force that stretches a body towards and away from the center of mass of another body due to a gradient (difference in strength) in gravitational field from the other body". That gradient in gravitational force is what I'm talking about.

In all the articles that I have found, it is specifically stated about the gravity that is needed to boost the virtual particle into real particle.
I couldn't find even one word about the tidal with related to the creation on new particles around a BH or SMBH.
So, why do you advice that it is tidal?
Actually, it seems to me that you discuss about simple gravity forces at different distances: "it is the difference between the force of gravity at different distances from the event horizon".

Perhaps you haven't found any, but I have a book that mentions it. Here is a quote from Black Holes & Time Warps: Einstein's Outrageous Legacy by Kip S. Thorne:

Quote
Tidal gravity near the horizon is very strong; it pulls the virtual photons apart with a huge force, thereby feeding great energy into them, as seen by the infalling observer who is halfway between the photons. The increase in photon energy is sufficient, by the time the photons are a quarter of a horizon circumference apart, to convert them into real long-lived photons (right half of Figure 12.2), and have enough energy left over to give back to the neighboring, negative-energy regions of space. The photons, now real, are liberated from each other. One is inside the horizon and lost forever from the external Universe. The other escape from the hole, carrying away the energy (that is, the mass) that the hole's tidal gravity gave to it. The hole, with its mass reduced, shrinks a bit.

I've uploaded figured 12.2 from the book to illustrate this:


* BlackHoleParticles.jpg (97.02 kB . 573x434 - viewed 3854 times)

So we agree by now that due to the BH gravity force (you can call it tidal if you wish) two identical particles with negative polarities to each other (one negative and one positive) are popped up.
As the negative particle is falling into the BH, the positive is ejected outwards.
If the BH is made out of positive particles than the negative particles that fall in actually decrease the total mass of the BH. That is very clear.
However, if the BH is made out of negative particle, than the negative particle that falls in must increase the BH mass.
Do you agree with that?
Therefore, it is a severe mistake to assume that under any condition the BH should disappear.
We have to say that the BH might disappear just if the polarity of the falling particle is the opposite to the polarity of that BH.

Black holes are always made from matter with positive mass, like stars and clouds of gas.

Can you please explain how the Tidal can "pull" the particle-antiparticle pairs apart?

The same way the tidal forces cause spaghettification: https://en.wikipedia.org/wiki/Spaghettification

Why the gravity will have different effect on particle with different polarity.

It doesn't. The difference is caused by the particles being different distances from the singularity.

Don't you agree that it should affect both partials at the same impact?

It affects them by different amounts because they are different distances from the horizon.

Do we know about gravity with negative polarity?
We know what is the meaning of negative or positive voltage, while there is no meaning of negative or positive gravity.

No, but negative gravity is neither required nor even mentioned in this process.

Therefore, if we wish to pull apart two particles (one negative and one positive) don't you agree that we MUST set them under electric/magnetic field?

No. Particles that don't even respond to electromagnetic fields can be pulled apart by these tidal forces, so its irrelevant.

So, do you agree that Tidal (or any sort of gravity force) can convert the virtual particles into real particles, but it can't separate them.

No, I don't agree with this. We've seen objects separated by tidal forces before. Jupiter's tidal gravity pulled apart comet Shoemaker-Levy 9 back in 1994: https://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9

Thanks for the explanation. However, I really don't understand how could it be that a SMBH with estimated mass of Million BH, has less power than a BH with a size of an elephant???
Let me give you other example:
If I will tell you that the thrust of the jet 747 is less than a bicycle- would you believe me?
In order to prove it, I will place one person on the bicycle while other person will stay at a distance of 1 km from the 747.
Which one will get higher trust?
Is that a correct way to verify the trust?

You are equivocating two different concepts: energy and power. Just because an entity has more energy does not automatically mean that it has more power. The tidal forces at a black hole's event horizon are what determines the rate that it radiates particles. For a large black hole, those tidal forces are smaller, so the hole radiates more weakly.

Hence, we must boost the particles in order to create them.
The meaning of boost is to give them a possibility to run at ultra high speed for a long enough distance. (I call that distance - Runway distance)
If you try to set any object at a very small orbital cycle, and try to run it at ultra velocity, the force outwards should be stronger as the cycle is shorter.
Therefore, the virtual particle will be ejected outwards from the event of horizon cycle before setting the whole Runway that is needed to form a real particle.

You're taking the concept of a virtual particle too literally. They are convenient modelling tools. They don't actually exist and as such can't actually travel. It may be better to think of the tidal forces acting on the various fields around the black hole to create real particles from those fields.

Why are we so sure that the new particle creation is taking care only at the event horizon?
Why it couldn't be created even inwards to the event horizon and then imidiatly be separated by the electric/magnetic field.

I suppose it could happen inside of the event horizon, but it would have no effect on the black hole's mass. If it eats both the positive mass and the negative mass particle, then the total mass hasn't changed.

If the field is strong enough - why it can't eject the new born particle even if the particle had been created deep into the event horizon radius?

A particle would have to travel faster than light to escape an event horizon. Since relativity forbids that, no field can be strong enough to push it that fast.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/07/2019 17:17:08
1. SMBH - matter or Antimatter
Black holes are always made from matter with positive mass, like stars and clouds of gas.
Why are you so sure about it?
Is it just a wishful thinking or is it based on real verifications and evidences?
Let me ask the following:
Let's assume that our SMBH is made out of Antimatter.
Do you think that it would effect differently the galaxy with regards to a SMBH which is made out of matter?
Please explain the differences in the effects?

2. Different Tidal forces on new born particle (Negative & Positive)
Quote
Why the gravity will have different effect on particle with different polarity.
Don't you agree that it should affect both partials at the same impact?
It doesn't. The difference is caused by the particles being different distances from the singularity.
It affects them by different amounts because they are different distances from the horizon.
How could it be that the Negative particle will have a different distance from the Horizon comparing the positive particle on their moment of birth?
Do you agree that at this moment of birth, the distance between the Negative particle to the positive is less than a Pico mm?
So, how that very short distance could set a different tidal forces?
What is the minimum requested distance between the two in order for the tidal to take its impact?
How the BH split them to that minimum distance request?
Why they don't eliminate each other at the moment of birth?
How do we know that always the negative particle is inwards to the Horizon, while the positive is outwards - on their birth moment?
Do you agree that based on statistics, the positive particle could be sometimes inwards and sometimes outwards?
If the positive particle is inwards (while the negative is outwards), do you estimate that the BH will be able to change their locations? How?
What might be the outcome if Antiproton is ejected outwards, while the proton is falling inwards?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/07/2019 20:16:47
Thanks Halc
Antimatter still constitutes positive mass/energy.  A black hole is made of mass, period.  It isn't matter or antimatter anymore.
So, the gravity impact of a BH with only Antimatter is identical to a BH with only matter (as long as they have the same mass).
Therefore, do you agree that we can't know if our SMBH is made out of Matter or Antimatter?
The mathematics is pretty trivial.  A=Gm/r², so the question is the same as asking what is the minimum change in r to get a different value for A:  Anything nonzero.  Keep in mind that r here is the distance to the event horizon, not including the Schwarzschild radius, which isn't a real distance.
So, do do agree that the chance for in falling antimatter to the BH is identical for the chance of in falling matter?
Therefore, do you agree that there is a possibility that the BH could be made out of antimatter or matter?
Only small black holes create such things as particles with proper mass.  I imagine in such cases they might create as much antimatter as matter.
Please let me know if you agree with the following scenario:
If the BH is very small and is made out of antimatter, while the most in falling matter is antimatter, than -
Do you agree that this BH should increase its mass over time?
If so, don't you agree that over time this small BH might become SMBH?
If not - Why not?
I'm unaware of there being a conservation law that applies, especially given the obvious imbalance we see in the universe.
Do you mean the imbalance that we see between the matter and Antimatter in our Universe, as we actually only see matter, but we don't see any antimatter?
Therefore, in any particle creation activity - the total antimatter in the Universe should be identical to the total matter.
So, how can we explain that we only see matter while we don't see any antimatter?
Let's assume that you have to solve this enigma. What are you going to do?
I have an advice for this problem:
Let's set all the antimatter in big unseen barrels, while all the matter will be free outwards.
Don't you agree that there is a possibility that this is exactly the solution that our Universe had found?
If all the antimatter in the whole Universe is saved in the Billion over millions BH or SMBH barrels, than all the matter will be free to form stars planets and moons that we see.
Therefore - do you agree that Antimatter SMBH/BH can solve the enigma of our Universe?
It eats Antimatter and ejects matter.

Therefore - after all, our SMBH isn't so bad eater.
For any proton that it ejects - it eats one antiproton.
Hence, as it ejects over than 99% of the matter in the accretion disc - it eats over than 99% of the antimatter that were a byproduct of those new particles creation in that accretion disc.

So simple solution – for so difficult question.



Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 31/07/2019 21:35:49
Why are you so sure about it?
Is it just a wishful thinking or is it based on real verifications and evidences?

Because all matter that has ever been observed in the Universe is made of matter with a positive mass. The particle falling into the black hole ends up with negative momentum, and therefore negative energy, because of the weirdness of space and time within the black hole. I don't fully understand it, but Stephen Hawking (and the majority of physicists who have analyzed his work) agree that it is sensible in the framework of relativity.

Let's assume that our SMBH is made out of Antimatter.
Do you think that it would effect differently the galaxy with regards to a SMBH which is made out of matter?
Please explain the differences in the effects?

A black hole made from antimatter would be identical to one made out of matter. This is called the "no-hair theorem", which proposes that black holes are extremely simple objects which possess only a handful of distinct properties, such as electric charge, mass, angular momentum and linear momentum: https://en.wikipedia.org/wiki/No-hair_theorem

How could it be that the Negative particle will have a different distance from the Horizon comparing the positive particle on their moment of birth?

It wouldn't always, but on those particular occasions where it does, it can result in Hawking radiation. In other cases, the particle pair can self-annihilate and disappear into the vacuum again.

Do you agree that at this moment of birth, the distance between the Negative particle to the positive is less than a Pico mm?

The distance probably varies a lot, in part depending on the wavelengths of the particles in question.

So, how that very short distance could set a different tidal forces?

Any distance has tidal forces involved. It's just weaker when those distances are shorter. Hence why very small black holes with very strong tidal forces are needed to produce short-wave length (and therefore high energy) particles.

What is the minimum requested distance between the two in order for the tidal to take its impact?

Any distance will have tidal forces of some amount. I don't know the specific numbers for given particles, though.

How the BH split them to that minimum distance request?

With tidal forces, the same as I've been saying this whole time.

Why they don't eliminate each other at the moment of birth?

Many of them probably do.

How do we know that always the negative particle is inwards to the Horizon, while the positive is outwards - on their birth moment?
Do you agree that based on statistics, the positive particle could be sometimes inwards and sometimes outwards?
If the positive particle is inwards (while the negative is outwards), do you estimate that the BH will be able to change their locations? How?

The interior geometry of the black hole is what makes that particular particle negative in the first place, so you can't have the opposite scenario.

What might be the outcome if Antiproton is ejected outwards, while the proton is falling inwards?

You would indeed expect antiprotons to be emitted at the same rate as protons, but nothing of particular note would happen to the black hole because black holes have no "hair".

Quote
Therefore, do you agree that we can't know if our SMBH is made out of Matter or Antimatter?

Black holes are identical whether they formed from an object made of matter or antimatter.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 31/07/2019 21:51:34
So, the gravity impact of a BH with only Antimatter is identical to a BH with only matter (as long as they have the same mass).
Therefore, do you agree that we can't know if our SMBH is made out of Matter or Antimatter?
It isn't a meaningful distinction to describe the mass of a black hole being matter or antimatter.  The mass isn't in the form of material that has properties like that.
So, yes, black holes created by dumping matter into a place, or antimatter, or a mixture, all behave identically.

Quote
So, do do agree that the chance for in falling antimatter to the BH is identical for the chance of in falling matter?
Not necessarily.  In-falling antimatter is likely to hit matter in the accretion disk along the way and get annihilated.

Quote
Therefore, do you agree that there is a possibility that the BH could be made out of antimatter or matter?
They're 'made out of' neither.  A neutron star can be made out of one or the other, but not a black hole.  The matter might be degenerate, but no so degenerate that it isn't still matter.

Quote
Please let me know if you agree with the following scenario:
If the BH is very small and is made out of antimatter, while the most in falling matter is antimatter, than -
Do you agree that this BH should increase its mass over time?
I don't agree that it is made out of anything.  Yes, if you dump mass in, its mass increases.

Quote
If so, don't you agree that over time this small BH might become SMBH?
If not - Why not?
If you dump enough mass in, it will cross whatever you think is the threshold for being 'super massive'.

Quote
Quote from: Halc
I'm unaware of there being a conservation law that applies, especially given the obvious imbalance we see in the universe.
Do you mean the imbalance that we see between the matter and Antimatter in our Universe, as we actually only see matter, but we don't see any antimatter?
Yes
Quote
Therefore, in any particle creation activity - the total antimatter in the Universe should be identical to the total matter.
No.  I said there seems to be no such law.

Quote
I have an advice for this problem:
Let's set all the antimatter in big unseen barrels, while all the matter will be free outwards.
Don't you agree that there is a possibility that this is exactly the solution that our Universe had found?
How did it get sorted like that?  Sounds like antimatter behaves differently than matter, which contradicts the symmetry of it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/08/2019 05:41:38
Quote
I have an advice for this problem:
Let's set all the antimatter in big unseen barrels, while all the matter will be free outwards.
Don't you agree that there is a possibility that this is exactly the solution that our Universe had found?
How did it get sorted like that?  Sounds like antimatter behaves differently than matter, which contradicts the symmetry of it.
Yes, that is an important question.
However, let's start with the following simple question - How did a new born pair particle get sorted like that?
In order to understand that let's focus on the meaning/creation of pair particle:
https://en.wikipedia.org/wiki/Pair_production
"Pair production is the creation of a subatomic particle and its antiparticle from a neutral boson. Examples include creating an electron and a positron, a muon and an antimuon, or a proton and an antiproton. Pair production often refers specifically to a photon creating an electron–positron pair near a nucleus. For pair production to occur, the incoming energy of the interaction must be above a threshold of at least the total rest mass energy of the two particles, and the situation must conserve both energy and momentum.[1] "
So, let's focus on electron and a positron:
For the creation we "MUST conserve both energy and momentum."- That is clear.
So, a BH should be able to create those pair.
What is the difference between the two:
https://www.answers.com/Q/What_is_the_difference_between_a_positron_and_an_electron
"The fact that the electron and positron are matter and anti-matter, and that they have a charge of -1 and +1 respectively are the major differences. A positron is an electron's anti-particle, and when the electron and positron come in contact with each other to combine, they annihilate each other in a process called electron-positron annihilation."
So, "they have a charge of -1 and +1 respectively."
However, if they stay together they will annihilate each other in a process called electron-positron annihilation"
By now we already know that tidal can't do the job of separation as on the moment of creation they are located close to each other.
We must have a real external force to set the separation.
They only real power is - Magnetic Field!!!
If those electron and positron had been created under the impact of strong magnetic/electric field, the outcome of their charge of -1 and +1 must push them away from each other. (I assume that it is due to the Lorenz law trust).
therefore - if we see any activity of particle creation - it is clear that magnetic field must be there.
Without it - the new born pair will "annihilate each other in a process called electron-positron annihilation" on the same moment of the creation!
Do you agree with that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/08/2019 05:59:07
By now we already know that tidal can't do the job of separation as on the moment of creation they are located close to each other.

It can if the tidal forces are strong enough. I just did calculations showing that a small enough black hole (one about 10 nanometers in radius, in my particular calculations) produce tidal forces many orders of magnitude above what is needed to separate at least some positron-electron pairs. The calculations are a bit involved, so I'll post them up tomorrow.

We must have a real external force to set the separation.
They only real power is - Magnetic Field!!!

Naturally existing black holes either don't have magnetic fields or have extremely weak ones because they would be very close to electrically neutral.

therefore - if we see any activity of particle creation - it is clear that magnetic field must be there.

Or the tidal force of gravity, as Stephen Hawking and Kip S. Thorne agree.

Without it - the new born pair will "annihilate each other in a process called electron-positron annihilation" on the same moment of the creation!
Do you agree with that?

No, such will not necessarily occur.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/08/2019 16:51:53
It can if the tidal forces are strong enough. I just did calculations showing that a small enough black hole (one about 10 nanometers in radius, in my particular calculations) produce tidal forces many orders of magnitude above what is needed to separate at least some positron-electron pairs.
Why do you select a BH with so tiny radius?
Actually, if I understand it correctly, based on Hawking radiation we must focus on the event of horizon when we discuss about new particle creation.
I have found the event of horizon for a BH with a Sun mass is 3 Km.
So, at that radius, we expect to get the creation of a particle pair.

Let's assume that you can calculate the exact radius where tidal takes care.
Why do you think that all the new born particle pair will be created exactly at that radius?
Do you agree that if the pair is created inwards - both of them will be pulled inwards, while if both are created outwards - both of them will be pushed outwards?
What is the chance that they will be created exactly at that radius?
Even if they have been created there, how the tidal can distinguish between less than one Pico meter?
Don't you agree that at the moment of creation that is the distance between the pair (positron to electron)?
Do you also agree that there is no confidence if the positron will be located inwards or outwards in the moment of its creation?
How any sort of BH tidal will be able to split between the pair while their distance between each other in the first moment is less so small?
How could it be that the tidal will be able to pull ALWAYS inwards one of the pair (Positron?) while the other one will be pushed ALWAYS outwards (or vice versa)?
Naturally existing black holes either don't have magnetic fields or have extremely weak ones because they would be very close to electrically neutral.
Can you prove it?
Did we monitor BH/SMBH and verify that they don't have magnetic field?
Do we really know all the aspects about BH/SMBH?
Do we know for sure if a BH/SMBH rotates or not?
If I understand it correctly we even don't know for sure if they made out of matter or Antimatter as the gravity impact of antimatter is actually identical to matter.
So why are you so sure that the BH is made out of matter and not antimatter?
Did we ever send a monitor into a BH/SMBH?
How can you claim that the matter in the BH is close to electrically neutral while we don't have deep understanding/verification about a BH/SMBH and we even don't know for sure that it isn't antimatter?


Close is still nonzero, which might mean a lot of tidal difference if the gravitational gradient is steep, which it isn't near a SMBH.  Electrons and positrons are not created by such massive objects.
Why are you so sure about it.
So far our scientists couldn't find any evidence for even one Atom that is falling inwards to the accretion disc of our SMBH.
So, please - would you kindly backup the hypothetical ideas/assumptions by real evidences?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/08/2019 17:43:12
Why do you select a BH with so tiny radius?

Because the strong tidal forces of small black holes are needed to separate electron-positron pairs. I just went with a very small radius because it was more likely to work with my calculations.

Actually, if I understand it correctly, based on Hawking radiation we must focus on the event of horizon when we discuss about new particle creation.
I have found the event of horizon for a BH with a Sun mass is 3 Km.
So, at that radius, we expect to get the creation of a particle pair.

Yes, the creation of photon pairs (and maybe neutrinos and gravitons) in particular. There was an analysis done in 1976 that showed that black holes with a mass much above 1017 grams don't radiate electron-positron pairs. The Sun, however, is far more massive than that: https://en.wikipedia.org/wiki/Hawking_radiation#1976_Page_numerical_analysis

Let's assume that you can calculate the exact radius where tidal takes care.
Why do you think that all the new born particle pair will be created exactly at that radius?

Because the event horizon is a necessary component of the proces. Once one particle crosses the horizon, it is causally separated from the other because no signal can get out of the horizon.

Do you agree that if the pair is created inwards - both of them will be pulled inwards, while if both are created outwards - both of them will be pushed outwards?

You can't create both of them outside of the horizon.

What is the chance that they will be created exactly at that radius?

100%, because the event horizon is necessary for the process to work.

Even if they have been created there, how the tidal can distinguish between less than one Pico meter?
Don't you agree that at the moment of creation that is the distance between the pair (positron to electron)?

No. The distance between the particles can be significantly larger than that (above a nanometer). My calculations showed that a separation that large is enough for them to be pulled apart by a small enough black hole. I'll post those calculations when I get back from work.

Do you also agree that there is no confidence if the positron will be located inwards or outwards in the moment of its creation?

That's correct. The positron could be the one going in or it could be the one going out.

How any sort of BH tidal will be able to split between the pair while their distance between each other in the first moment is less so small?

If the tidal forces are very strong, it can be done.

How could it be that the tidal will be able to pull ALWAYS inwards one of the pair (Positron?) while the other one will be pushed ALWAYS outwards (or vice versa)?

If they were both pulled in, then they would be able to annihilate each other and return to the vacuum. If they were both pushed out, then they would be able to annihilate each other and return to the vacuum. One going in and the other going out is the only way that both can become real particles.

Can you prove it?

It's the nature of matter. In its natural state, it strives to be electrically neutral because opposites attract.

Did we monitor BH/SMBH and verify that they don't have magnetic field?

We wouldn't be able to tell from this distance, but we know enough about physics to know that naturally-formed ones wouldn't have much of a field (for reasons that I have discussed ad nauseum by now).

Do we really know all the aspects about BH/SMBH?

Black holes are simple as per the "no hair" theorem. There isn't much to know about them.

Do we know for sure if a BH/SMBH rotates or not?

Yes: https://resonance.is/super-massive-black-holes-spin-near-the-speed-of-light/

If I understand it correctly we even don't know for sure if they made out of matter or Antimatter as the gravity impact of antimatter is actually identical to matter.
So why are you so sure that the BH is made out of matter and not antimatter?

They aren't made of matter or antimatter. Both Halc and I have already said this.

How can you claim that the matter in the BH is close to electrically neutral while we don't have deep understanding/verification about a BH/SMBH and we even don't know for sure that it isn't antimatter?

Because natural matter like gas clouds or stars don't have net electrical charge.

So far our scientists couldn't find any evidence for even one Atom that is falling inwards to the accretion disc of our SMBH.
So, please - would you kindly backup your theory by real evidence?

That doesn't have anything to do with Hawking radiation.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/08/2019 21:25:41
Here are the calculations:

The first thing I want to calculate is the strength of electromagnetic attraction between the electron and positron. So I need two values: the magnitude of electric charge on each particle and their distance from each other. The electric charge on the electron is known to be -1.6021766208 x 10-19 Coulombs, whereas that of the positron is of equal magnitude and opposite in sign.

The distance between the two particles will depend upon their wavelength, as stated by Kip S. Thorne:

Quote
The virtual photons can separate from each other easily, so long as they both remain in a region where the electromagnetic field has momentarily acquired positive energy. That region can have any size from tiny to huge, since vacuum fluctuations occur on all length scales; however, the region’s size will always be about the same as the wavelength of its fluctuating electromagnetic wave, so the virtual photons can move apart by only about one wavelength.

So now we need to find the wavelengths of the electron and positron.The DeBroglie wavelength of a particle can be determined using the following website: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/debrog2.html

When we input a rest mass of 0.511 electronvolts (that of an electron) and a kinetic energy of 100 electronvolts, we end up with a wavelength of 1.22637367 x 10-10 meters. Remember, the kinetic energy chosen is arbitrary, as particles on pretty much all kinetic energy scales would be pulled from the vacuum.

Now that we know the charges and distance involved, we can calculate the magnitude of the electrical attraction between the particles using Coulomb’s law. The following online calculator allows us to calculate that attraction: https://www.omnicalculator.com/physics/coulombs-law

Inputting the values gives us an electrical attractive force of 7.6691 x 10-13 newtons.

Now we need to determine the tidal forces on these particles. We can use the following calculator to determine the radius of a black hole from its mass: https://www.omnicalculator.com/physics/schwarzschild-radius I will go with a mass of 1016 kilograms, since the analysis by Page suggested that such a mass can produce electron-positron pairs. The resulting radius is 1.4852 x 10-10 meters.

So we will place one of the particles right at the horizon (1.4852 x 10-10 meters from the singularity) and the other particle one wavelength away from the event horizon (2.71157367 x 10-10 meters from the singularity).

We can calculate the force of gravitational attraction using the following online calculator: https://www.ajdesigner.com/phpgravity/newtons_law_gravity_equation_force.php The mass of the electron is equal to the mass of the positron, which is 9.10938356 x 10-31 kilograms.

For the particle right at the horizon, the resulting gravitational attractive force between the particle and the black hole is 2.7555874548284 x 10-4 newtons. For the particle one wavelength away from the horizon, the resulting gravitational attractive force between the particle and the black hole is 8.2668729594675 x 10-5 newtons.

Subtracting these two values gives us the tidal force, which is 1.9289001588165 x 10-4 newtons.

This tidal force is about 2.5 x 108 times greater than the electrical attractive force between the particles.

So yes, black holes can, under the right circumstances, pull electron-positron pairs apart with tidal force.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/08/2019 05:14:03
Here are the calculations:
Thanks for the great explanation.
You have calculated the magnitude of the electrical attraction between the particles using Coulomb’s law while the particles are at a distance of one wavelength away from each other.
Based on that calculation you have proved that the tidal can separate the particles.
That is very clear and I fully agree!

However, why you didn't try to calculate that magnitude of the electrical attraction between the particles at the moment of their birth?
Actually, what is the expected distance between the particles on that specific moment of birth?
Do you agree that it should be much less than one wavelength?
If one wavelength is 1.22637367 x 10-10 meters, Could it be that the distance at the moment of birth is less than 1x10^-20m, 1x10^-50m or even less than 1x10^-1000 m.?
Let's use a distance of 1x10^-51 and calculate the electrical attractive force:
Inputting the values gives us an electrical attractive force of 6.6726E+29 Newton instead of 7.6691 x 10-13 Newton for one wavelength.
Don't you agree that this electrical attractive force of 6.6726E+29 Newton is very strong?
Even if we assume that the distance is longer than 1x10^-51m, as long as it is still significantly shorter than one wavelength, it is clear that the electrical attractive force is quite strong.
Sorry, but tidal is not good enough for that.
So why Mr. Kip S. claims that:
The virtual photons can separate from each other easily....
The separation at the moment of birth It is not easy at all.
Therefore it was clear to me that we must use other force for the first separation.
I have stated several times that the only force that can do that job is magnetic/electric field.
Mr. Kip S. fully supports my understanding as he also claims:
... so long as they both remain in a region where the electromagnetic field has momentarily acquired positive energy.
So, based on Mr. Kip S, the electromagnetic field is vital for the first separation process.

If there is electromagnetic field, the separation is quite easy and I also fully agree that once we get to a distance of one wavelength, the tidal can continue with the separation process.

With regards to the electromagnetic field around our SMBH.
In one hand you claim that the SMBH can't generate that field, but on the other hand you have stated that the accretion disc can do it.
So, I really don't care what is the source for the electromagnetic field around our SMBH.
As long as it is there it can set the job of the first separation process as Mr. Kip S have stated.

There are also two more vital effects of the electromagnetic field around our SMBH.

1. The creation process for new particles can take at a wide range.
There is no need to be right at the horizon.
If the electromagnetic field can overcome a force of 6.6726E+29 Newton, it can easily separate particles deep into the horizon and pull one out.
You have stated that the SMBH can't produce particles at its horizon, but I assume that you have no obligation that it can create new particles deep into the Horizon.
So, as long as the electromagnetic field is strong enough, it can easily separate between the new born particles - deep into the horizon, and pull them into the accretion disc.

2. That electromagnetic field also should sort between the Positive to Negative particles. It will push the positive to one side and pull the negative to the other side, regardless of their location at the moment of birth.
Hence, that electromagnetic field keeps all the antimatter in the SMBH while it brings the Universe new life with the matter that is ejected outwards from the accretion disc. Some of that matter will be converted into new formed stars.
And many thanks to Mr Kip S.

Do you agree with all of that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 03/08/2019 05:38:18
However, why you didn't try to calculate that magnitude of the electrical attraction between the particles at the moment of their birth?
Actually, what is the expected distance between the particles on that specific moment of birth?
Do you agree that it should be much less than one wavelength?
If one wavelength is 1.22637367 x 10-10 meters, Could it be that the distance at the moment of birth is less than 1x10^-20m, 1x10^-50m or even less than 1x10^-1000 m.?
Let's use a distance of 1x10^-51 and calculate the electrical attractive force:
Inputting the values gives us an electrical attractive force of 6.6726E+29 Newton instead of 7.6691 x 10-13 Newton for one wavelength.
Don't you agree that this electrical attractive force of 6.6726E+29 Newton is very strong?
Even if we assume that the distance is longer than 1x10^-51m, as long as it is still significantly shorter than one wavelength, it is clear that the electrical attractive force is quite strong.
Sorry, but tidal is not good enough for that.

The particles are not "born" until they become real. Like I said before, you are taking the concept of a virtual particle too literally. They are a modeling tool. Virtual particles are not real and as such don't have defined locations. What is being described is the behavior of the local electromagnetic field (which is present everywhere in space). What the tidal forces are doing is adding energy to that field until it forms local excitations in the form of real particles.

EDIT: Actually, I don't know whether to say that virtual particles are real or not. I keep seeming to read contradictory responses from actual physicists. Some say they are real, others say they are just a model. In the case of a "real" quantum vacuum, you could say that it has the ability to form any and all possible virtual particle pairs. This would include those particle pairs that are too close together to separate (like the ones you mention) as well as those that are far enough apart to be separated (and those become real particles). The vacuum state would represent a superposition of these different particles. Of course, only those particles that just so happen to be far enough apart to separate are the only ones that are important to the evaporation of a black hole.

Let's use a distance of 1x10^-51 and calculate the electrical attractive force:

That's less than the Planck length (~10-35 meters), where our understanding of physics breaks down. Our equations don't give meaningful answers at that level.

Sorry, the separation at the moment of birth It is not easy at all.

So you know theoretical physics better than a Nobel Prize-winning theoretical physicist does?

So, based on Mr. Kip S, the electromagnetic field is vital for the first separation process.

He's talking about the electromagnetic field that exists literally everywhere in space, not some electromagnetic field possessed by the black hole. There is a field for every single particle. There is an electron field, a proton field, a neutron field, and so on. The local value of those fields at most regions of space is zero because that space is empty. This is where the virtual particle picture comes in. Virtual particles are mathematical tools to represent the behavior of those fields.

But the electromagnetic field is only necessary when speaking of the formation of photons by tidal forces. For other particles, you have to speak of their own, individual fields. For the formation of a pair of gravitons, for example, you would only speak of the gravitational field, not the electromagnetic field.

You have stated that the SMBH can't produce particles at its horizon

I never said that. Super-massive black holes do produce particles (long wavelength photons). What they don't produce is electron-positron pairs.

but I assume that you have no obligation that it can create new particles deep into the Horizon.
I hope that you agree that the SMBH can create new particle deep into the horizon.
So, as long as the electromagnetic field is strong enough, it can easily separate between the new born particles - deep into the horizon, and push them into the accretion disc.

No they can't. I already told you this before: particles cannot get out of a black hole's event horizon. That would require them to travel faster than the speed of light.

That electromagnetic field also will take care on that the all the Positive particle will move to one side while all the negative will pushed to the other side.
So, that electromagnetic field keeps all the antimatter in the SMBH while it brings us only the matter.

Again, no. The black hole itself doesn't have an intrinsic electric field. Even if it did, it would be quickly neutralized by the preferential attraction and absorption of those particles that are attracted to it (and therefore opposite in charge). So if it has the negative charge needed to attract only positrons, then those positrons are going to cancel out the negative charge once they get eaten by the hole and the electric field goes away.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/08/2019 08:26:44
Thanks!

Would you kindly answer the following key question:
what is the expected distance between the particles on that specific moment of birth?
What do you mean by the following Planck length of 10^-35 meters?
That's less than the Planck length (~10-35 meters), where our understanding of physics breaks down. Our equations don't give meaningful answers at that level.
Do you mean that the distance can't be closer than that?
Do you mean that if our scientists have set that Planck length of 10^-35 meters, than the distance in the moment of Birth can't be shorter than that?
So please - would you kindly answer this important question?
You also claim that:
So you know theoretical physics better than a Nobel Prize-winning theoretical physicist does?
My knowledge is neglected to this Nobel Prize-winning theoretical physicist, but how he can set any sort of theory without calculate the real magnitude of the electrical attractive force at the moment of birth?
He's talking about the electromagnetic field that exists literally everywhere in space, not some electromagnetic field possessed by the black hole. There is a field for every single particle. There is an electron field, a proton field, a neutron field, and so on.
If he assume that the electromagnetic field that exists literally everywhere in space, can do the job, then don't you agree that he must also prove that assumption?
Don't you agree that he must first evaluate the distance between the particles in the moment of Birth, extract the magnitude of the electrical attractive force between the particles and than prove that the "electromagnetic field that exists literally everywhere in space" can overcome on that force?
How do you really work in science?
Do you mean that just because he is a Nobel Prize-winning theoretical physicist than whatever he says is 100% correct while whatever I say is 100% incorrect?
Please prove the assumption that  the "electromagnetic field that exists literally everywhere in space" can separate the two particles from their location at the moment of birth.

Super-massive black holes do produce particles (long wavelength photons). What they don't produce is electron-positron pairs.
Thanks
That is perfect for me.
So you agree that the SMBH can produce particles!
With regards to the electromagnetic field around the SMBH:

Would you kindly advice if the accretion disc generates electromagnetic field around the SMBH?
Is it correct that our scientists have monitor an electromagnetic field around the SMBH?
https://iopscience.iop.org/article/10.1088/0004-637X/758/2/103/pdf
MAGNETICALLY LEVITATING ACCRETION DISKS AROUND SUPERMASSIVE BLACK HOLES
"In this paper, we report on the formation of magnetically levitating accretion disks around suppermassive black
holes (SMBHs). The structure of these disks is calculated by numerically modeling tidal disruption of magnetized
interstellar gas clouds. We find that the resulting disks are entirely supported by the pressure of the magnetic
fields against the component of gravitational force directed perpendicular to the disks. The magnetic field shows
ordered large-scale geometry
that remains stable for the duration of our numerical experiments extending over 10% of the disk lifetime. Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation. This in combination with the repeated feeding of magnetized molecular clouds to an SMBH yields a possible solution to the long-standing puzzle of black hole growth in the centers of galaxies."
It is stated clearly that there is a magnetic field around the SMBH.
So why do you insist that there is no magnetic field?


No they can't. I already told you this before: particles cannot get out of a black hole's event horizon. That would require them to travel faster than the speed of light.
If you ignore the impact of the magnetic field, than yes this answer is correct.
However, if you add the trust of the magnetic field to the particle, then don't you agree that it can easily get out of a black hole's event horizon without a need to break the speed of light?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 03/08/2019 15:36:52
Do you mean that the distance can't be closer than that?
Do you mean that if our scientists have set that Planck length of 10^-35 meters, than the distance in the moment of Birth can't be shorter than that?
So please - would you kindly answer this important question?

It's more a matter that we don't know how physics works on scales that small: https://en.wikipedia.org/wiki/Planck_length#Theoretical_significance

My knowledge is neglected to this Nobel Prize-winning theoretical physicist, but how he can set any sort of theory without calculate the real magnitude of the electrical attractive force at the moment of birth?

Because, as I said before:

In the case of a "real" quantum vacuum, you could say that it has the ability to form any and all possible virtual particle pairs. This would include those particle pairs that are too close together to separate (like the ones you mention) as well as those that are far enough apart to be separated (and those become real particles).

If he assume that the electromagnetic field that exists literally everywhere in space, can do the job, then don't you agree that he must also prove that assumption?

It's already been done on mathematical grounds by several people.

Don't you agree that he must first evaluate the distance between the particles in the moment of Birth, extract the magnitude of the electrical attractive force between the particles and than prove that the "electromagnetic field that exists literally everywhere in space" can overcome on that force?

Virtual particles are constantly forming and moving around in the vacuum. Some are too close together to be separated, but others are sufficiently far apart. As demonstrated by my calculations, those that are far enough apart can be separated by tidal forces. Recall that these are quantum objects that are represented by a probability distribution. There is a certain probability that the particles will be close together or far apart.

The hydrogen atom itself is a good example of this. From classical mechanics, you would expect the negatively-charged electron to move towards the positively-charged proton until the distance between them was zero. This, however, does not happen. Instead, the electron has a finite average distance from the proton (called the Bohr radius, which is about 5.3 x 10-11 meters). Sometimes the electron moves closer, sometimes further away. This is the same thing that would happen with an electron-positron pair (also called positronium: https://en.wikipedia.org/wiki/Positronium) The Bohr radius of a positronium atom in its ground state is about twice that of a hydrogen atom. This is very close to the distance I used in my calculations.

How do you really work in science?

Hawking radiation is a theoretical prediction based on the known laws of physics.

Do you mean that just because he is a Nobel Prize-winning theoretical physicist than whatever he says is 100% correct while whatever I say is 100% incorrect?

No, but it does mean that he is more likely to know what he is talking about than you are. You have a track record of misconceptions about black holes.

Please prove the assumption that  the "electromagnetic field that exists literally everywhere in space" can separate the two particles from their location at the moment of birth.

The math has been done that shows it can happen.

Would you kindly advice if the accretion disc generates electromagnetic field around the SMBH?

Yes, because the disk is an electrically-conducting fluid (plasma) that is in motion. That generates a magnetic field.

It is stated clearly that there is a magnetic field around the SMBH.
So why do you insist that there is no magnetic field?

Because the magnetic field is generated by the disk, not the black hole. Again, for reasons that I have repeated over and over already.

If you ignore the impact of the magnetic field, than yes this answer is correct.

No, the assertion is correct whether there is a magnetic field present or not.

However, if you add the trust of the magnetic field to the particle, then don't you agree that it can easily get out of a black hole's event horizon without a need to break the speed of light?

I absolutely do not agree. You can't get out of a black hole using force alone.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/08/2019 07:05:10
Magnetic Field
Quote
It is stated clearly that there is a magnetic field around the SMBH.
So why do you insist that there is no magnetic field?
Because the magnetic field is generated by the disk, not the black hole. Again, for reasons that I have repeated over and over already.
Why is it important to understand the source of the magnetic field around the SMBH?
My main task in this discussion is to prove that there is Magnetic field around the SMBH (I really don't care about its source).
However, how can I prove that there is a strong magnetic field around the SMBH?
In the Article it is stated clearly:
https://iopscience.iop.org/article/10.1088/0004-637X/758/2/103/pdf
"MAGNETICALLY LEVITATING ACCRETION DISKS AROUND SUPERMASSIVE BLACK HOLES.
"The magnetic field shows ordered large-scale geometry..."
"Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation."
So, there is a strong magnetic field in the accretion disc.
However, Magnetic field should also works around the whole SMBH.
If the body is SMBH, than that magnetic field must work around it and we should see its great impact in every aspect around that SMBH.
We should compare it to the magnetic field around the Earth.
Can we claim that the magnetic filed works only at a narrow disc around the Earth?
Can we claim that it works only at the poles?
The Answer is quite clear. Magnetics works around the whole body.
The magnetic field around the Earth pulls the solar wind to the poles and protects our life.
In the same token, the magnetic files abound the SMBH covers the whole SMBH.
We clearly see its impact in the following elements:
1. In the accretion disc - "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation"
2. Outwards from the accretion disc - It Push more than 99% of the Atom and molecular in the accretion disc outwards. Actually, few years ago, our scientists were very sure that all the matter in the accretion disc is falling into the SMBH. They have estimated that the gravity force of the SMBH at the accretion disc is so strong that it should grab all the matter inwards. They were sure that the matter in the accretion disc is the food of the SMBH. That assumption could be correct if we eliminate the impact of the magnetic field and we only count the mighty gravity force of the SMBH at the accretion disc.
However, now they clearly see that more than 99% of the matter in the accretion disc is ejected outwards. But they still hope that at least one present is falling in. Do they see any evidence for that? Do they see even one atom or one molecular from the accretion disc that is falling in? The answer is clear - NO NO NO.
But they still have a great wish/hope that somehow something from the accretion disc will fall into the SMBH. They refuse to accept the evidences as those evidences contradicts their wishing list. They don't want to understand that the mighty magnetic field around the SMBH will also prevent from any atom in the accretion disc to fall into the SMBH. The polarity of the magnetic field force 100% of the matter in the accretion disc to be ejected outwards. So, not even one Atom will fall into the SMBH. But of course - they will continue to hope...
3. Molecular jet Stream. We clearly see molecular jet stream above and below the accretion disc. What kind of force can set that kind of jet stream which is estimated to have about 10,000 solar mass? It is clear to me that the only power that can do it is - magnetic field. It actually grab all (over 99%) of the molecular that had been ejected from the accretion disc bring them all to the poles (as the earth with the solar wind) and then push them (from the poles of the magnetic fields) upwards/downwards at ultra velocity.
4. No matter is falling into the accretion disc - It is clear that no star, no planet, no moon, no rock not even atom is falling into the accretion disc from outwards. Our scientists have never ever found any sort of evidence that confirms that wrong assumption. However, they still hope that one day something will fall in. They can't accept the simple understanding that the magnetic field/shield around the SMBH will prevent from any object to fall in. If something will dare to come closer, the magnetic field should boost it upwards/downwards into that molecular jet stream with all the other molecular that had been ejected from the accretion disc.
So, if you still believe that the magnetic field around the SMBH is neglected -
Would you kindly show one solid evidence for even one atom that is falling inwards from the accretion disc into the SMBH and/or from outside into the accretion disc?.
Science is not a wishful list.
If you think something - you have to prove it by solid evidence.
If you think that the SMBH eats the mass from outside - please prove it by solid verification.
How long do we still have to wait until our scientists will finally understand that NOTHING - NOTHING at all, is falling in?
How long do we still have to wait until our scientists will finally understand the great impact of the mighty Magnetic field around the SMBH?
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 05/08/2019 13:49:23
You quote as evidence:
Quote
Strong magnetic pressure allows high accretion rate

And then you deduce:
Quote
4. No matter is falling into the accretion disc - It is clear that no star, no planet, no moon, no rock not even atom is falling into the accretion disc
This clearly does not follow from the evidence that you cite.

Quote
Would you kindly show one solid evidence for even one atom that is falling inwards from the accretion disc into the SMBH and/or from outside into the accretion disc?
Only this year has a blurry image been released of the M87 accretion disk, which is much wider than our solar system.
- On this scale, monitoring the movement of individual atoms is not possible, so don't demand it.
- However, astronomers estimate that on average, about 90 Earth masses per day is falling into this accretion disk.
See: https://en.wikipedia.org/wiki/Messier_87#Supermassive_black_hole

However, the strong magnetic field does need to be generated from some source of energy, and that is the flow of matter through the accretion disk and into the black hole.
- What do you propose powers the magnetic field? (even though you don't care where it comes from?)
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/08/2019 19:01:42
If you claim that matter cannot enter the accretion disk from the outside, then your model invariably predicts that the magnetic field becomes weaker over time. Mass from the accretion disk is constantly being thrust away from the poles of the black hole, so its mass is constantly decreasing. The emission of radiation by the hot plasma means that the energy is constantly decreasing in the disk over time as well. The disk becomes smaller, less massive and less energetic if mass/energy is not replenished from an outside source. The magnetic field must therefore become weaker and weaker until matter can indeed pass it and enter the black hole. Eventually, the accretion disk will disappear because you don't allow any mass/energy to be added to it.

By the way, the electromagnetic radiation, gravitational radiation and neutrinos emitted by the disk are not stopped by magnetic fields and would therefore easily enter the black hole.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/08/2019 05:49:51
Only this year has a blurry image been released of the M87 accretion disk, which is much wider than our solar system.
- On this scale, monitoring the movement of individual atoms is not possible, so don't demand it.
- However, astronomers estimate that on average, about 90 Earth masses per day is falling into this accretion disk.
See: https://en.wikipedia.org/wiki/Messier_87#Supermassive_black_hole
In that Article it is stated:
"The galaxy experiences an infall of gas at the rate of two to three solar masses per year, most of which may be accreted onto the core region.[52] "
However, if you go to 52 you get an article from Aug 1981
In that article it is stated:
"Resent X-ray spectroscopic results suggest that 2-3 M0 yr^-1 of cooling gas may be falling …M87 ...in reach cluster).
What do they mean by word "may"?
Is it "may yes" or "may no"?
Do you see any word about accretion disc?
How can they base this understanding on that unclear article from 1981?
Please be aware that at that time our scientists were positively sure that all the matter in our accretion disc is falling into the SMBH.
Therefore, this article doesn't give any updated evidence for in falling matter into the accretion disc or from the accretion disc into the SMBH.
Today our scientists have very advanced monitoring tools. Based on those tools they have found that more than 99% of the matter in the accretion disc of our SMBH is ejected outwards.
If something was falling in, why they didn't declare about it?
As they didn't say that even 0.0..1% is falling in, why our scientists are so sure that something is falling in?
I wonder why they have used the message "over than 99%" instead of 100%?
Could it be that it was a good willing for all those scientists who still hope that one day something might fall in?

If you claim that matter cannot enter the accretion disk from the outside, then your model invariably predicts that the magnetic field becomes weaker over time.

Why don't we focus on the evidences before offering the solution?
In any new discovery - our scientists ALWAYS set the explanation with or in front of that discovery.
Why is it so important to offer a solution before deeply understanding the discovery?
This is a severe mistake.
I'm working in electronics/communications engineering.
If we have a problem - We always focus on the evidences.
We think that understanding the evidences is 50% of the solution.
Therefore, we always try to understand all the evidences and discoveries and just then we look for a solution.
This question shows that before understanding clearly the evidences, our scientists wish to find the solution.
Why is it?
It seems to me that our scientists try to fit the evidences into the theory instead of the other way
Our theories could be correct or incorrect.
However, the evidences and discoveries are solid.
Don't you agree that first we must focus on those evidences/discoveries and just after having/agree on all the evidences/discoveries, we can look for a solution?

Therefore, the key question is as follow:
Do you agree that so far our scientists found solid evidences that over than 99% of the matter in the accertion disc are ejected outwards, while we have no real prove that something is falling in?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/08/2019 07:59:18
In any new discovery - our scientists ALWAYS set the explanation with or in front of that discovery.

Nonsense. If that was true, then we would already have an explanation for everything, even (somehow) before we discovered it.

Why is it so important to offer a solution before deeply understanding the discovery?

You have done exactly that. You are trying to explain galaxy formation using black holes even though you have demonstrated a severe lack of understanding of how black holes work.

Don't you agree that first we must focus on those evidences/discoveries and just after having/agree on all the evidences/discoveries, we can look for a solution?

Yes, which is why you should realize that your model won't work. If a system is constantly losing mass and energy (by spitting it out as galactic jets) without replenishing it, then it eventually winds down and stops working. That's the first law of thermodynamics at work (which is very strongly supported by the existing evidence).

Do you agree that so far our scientists found solid evidences that over than 99% of the matter in the accertion disc are ejected outwards, while we have no real prove that something is falling in?

Firstly, science isn't about proof. Secondly, let's assume that you are right and all of the matter is thrown off by the jets. What happens when all of the mass and energy in the accretion disk are gone?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/08/2019 09:11:26
Firstly, science isn't about proof. Secondly, let's assume that you are right and all of the matter is thrown off by the jets.
I think differently – "Science is all about proof".
However, why do we need to assume that I'm right?
Would you kindly advice if we have ever found any real evidence for in falling matter to the accretion from outside or from the accretion into the SMBH?

With the regards to its name - accretion disc:
Normally, we have to call an object based on its major section.
So, do you agree that even if only 51% was ejected outwards - we had to call it - excretion?
However, now that we know that over than 99% is ejected outwards and there is no proof for any in falling matter, why do we still insist to call it accretion disc instead of excretion disc?

What happens when all of the mass and energy in the accretion disk are gone?
Once we know for sure that I'm right - everyone can easily understand the answer for that question.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/08/2019 14:44:27
I think differently – "Science is all about proof".

That's not true. You can't prove a scientific theory. You can only gain evidence for or against it. As new data comes about, a theory that was formerly supported by the data may fail to align with the new data. Newtonian mechanics is one such example of this. In the past, the existing data supported its accuracy. As new data came along, it was found to be inaccurate in some cases and was replaced by relativity and quantum mechanics.

However, why do we need to assume that I'm right?

It's to explore the consequences of that assumption being true.

Would you kindly advice if we have ever found any real evidence for in falling matter to the accretion from outside or from the accretion into the SMBH?

Yes. If you go look at the Wikipedia article that evan_au linked and go down to reference 72 (which is the one that evan_au was referencing), then you will see that it says in the opening of that article, "We measure the X-ray
gas temperature and density profiles and calculate the Bondi accretion rate, M_ Bondi  0:1 M yr-11" https://iopscience.iop.org/article/10.1086/344504/pdf

Another quote of importance from the article:

Quote
Thanks to its high spatial resolution and sensitivity, the Chandra X-Ray Observatory is able to provide some of the most stringent constraints on the properties of low-luminosity black holes. In particular, at the distance of M87 (18 Mpc), the spatial resolution of Chandra corresponds to a radius of less than 100 pc or, equivalently, a few 105 Schwarzschild radii. For M87, this allows us to measure, for the first time, fundamental properties of the ISM at the accretion radius of the black hole and thereby estimate the mass supply into the accretion flow.

So actual data is, in fact, being used in models to estimate the rate of accreting matter. That is evidence.

So, do you agree that even if only 51% was ejected outwards - we had to call it - excretion?

No, because the only way that the disk could have formed in the first place is by matter coming in from outside. A supermassive black hole does not produce matter, it only produces extremely weak Hawking radiation. So that cannot be the source of the matter in the disk. If the black hole started off with no accretion disk, then gravity would have eventually gathered gas and other matter together into a disk around the hole. To start off with, the disk would have been cool and lacked a significant magnetic field. As the disk grew and became hotter over time, then the magnetic field would have increased in strength until it became powerful enough to produce jets.

However, now that we know that over than 99% is ejected outwards and there is no proof for any in falling matter, why do we still insist to call it accretion disc instead of excretion disc?

Because the mass in the disc must have originally come from the outside. That is true whether or not any more matter is being added to it now or not.

Once we know for sure that I'm right - everyone can easily understand the answer for that question.

And what do you propose that answer to be?
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 06/08/2019 17:08:08
Quote from: Dave Lev
we know that over than 99% is ejected outwards
You keep repeating this 99% figure...
- Where do you get the 99% figure? It seems excessively high.

Have a look at this article:
http://www.mpia.de/homes/fendt/Lehre/Lecture_OUT/pudritz.pdf

Section 2.2 states that:
- the Mass of the wind (Mw, mass flowing into the jet) is about 10% of the Mass flowing into the accretion disk (Ma).   (10% is a lot lower than 99%!)
- The angular momentum carried away by the jet can be 60% to 100% of the angular momentum of the accretion disk (maybe this is where you got the 99%? But it's not 99% of the mass!)
- If you rob angular momentum from the accretion disk to feed the jet, matter in the accretion disk will slow down and fall into the black hole much faster.
- This is backed up by observations of DG Tau (and over 300 other jets)
- They also observe that jets can occur during the formation of new stars - these objects can even have much lower mass than the Sun. So it doesn't need to rely on any mysterious behavior of super-massive black holes; it also applies to accretion disks without a black hole.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/08/2019 21:34:28
You keep repeating this 99% figure...
- Where do you get the 99% figure? It seems excessively high.
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"One such telescope is the Chandra Xray Telescope, which detects the Xrays emitted by superheated matter and other sources.  When astronomers used Chandra to study Sgr A*, in one of its longest ever observations, they found that more than 99% of the infalling material was ejected long before reaching the event horizon...
The main question is - do we have any evidence for in falling matter into the accretion?
How could it be that we see clearly that more than 99% of the matter in the accretion disc is ejected outwards, while there is no confirmation for any sort of matter that is falling into the accretion disc.
I have found a very interesting article:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"First detection of matter falling into a black hole at 30 percent of the speed of light"
That article proof my statement that so far our scientists didn't find any evidence for in falling matter.
So, in all the articles that had been published till 2018 there is no evidence for in falling matter.
Not around our SMBH and not around M87 SMBH
Not in any galaxy in the whole Universe and especially not in the Milky way.
However, let's see what is stated in that article:
"A UK team of astronomers report the first detection of matter falling into a black hole at 30 percent of the speed of light, located in the centre of the billion-light year distant galaxy PG1211+143."
So, only at a galaxy which is located at a distance of one billion light year - our scientists have found for the first time ever that matter is falling in.
But, is it really falling in?
Let's see the explanation:
"The researchers found the spectra to be strongly red-shifted, showing the observed matter to be falling into the black hole at the enormous speed of 30 per cent of the speed of light, or around 100,000 kilometers per second. The gas has almost no rotation around the hole, and is detected extremely close to it in astronomical terms, at a distance of only 20 times the hole's size (its event horizon, the boundary of the region where escape is no longer possible)."
If I understand it correctly, they say that they see gas which is orbiting round the event of horizon of that SMBH. Therefore, they are sure that this matter can't escape. Hence, they have got the conclusion that this matter must fall in. So, they didn't really see that the gas is falling into the SMBH. They assume that based on its current location it must fall in.
They also admit that they trace this gas for only one day.
However, in the following message, they say clearly that the gas had been swallowed by the SMBH:
"He continues: "We were able to follow an Earth-sized clump of matter for about a day, as it was pulled towards the black hole, accelerating to a third of the velocity of light before being swallowed up by the hole."
I wonder what the correct information is:
Did they really see a matter that is falling into the SMBH and disappear for ever, or they just found that it is at the event of horizon and therefore they have assumed that it should be eaten by the SMBH?
In any case, my understanding from this article is as follow:
1. Our scientists claim that till 2018 they have never ever found any evidence for in falling matter to the SMBH.
2. The idea that they have found it for the first time in a galaxy at a distance of one billion light years from us proves that they have not find any evidence for in falling matter - not in the milky way and not in any other galaxy up to a distance of one billion year.
3. How can they get real measurements from a galaxy that is located at a distance of one billion light years away?
How could they calculate and monitor the exact location of the event of horizon for a galaxy which is so far away?
Could it be that they have an error in their verifications?
4. How they can see a gas cloud with a size of an earth at a distance of one billion light year away?
Our scientists struggle to see an earth size which is located just 300 Light years away.
So, how can they see an earth gas cloud at one billion light year away? is there any possibility to see an earth size at one billion light year away from us?
5. Do they really see that it swallowed by the SMBH or they assume that it should be swallowed due to its location - event of horizon?
Do you agree with the following concussion:?
1. Our scientists have no evidence for in falling matter. Not to our SMBH and not to any SMBH at a distance of up to one billion light year away. So I was fully correct when I have stated that there is no proof for in falling matter.
2. How can we accept this understanding for earth size gas which is falling in at a distance of one billion light year away?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/08/2019 22:08:01
"First detection of matter falling into a black hole at 30 percent of the speed of light"
That article proof my statement that so far our scientists didn't find any evidence for in falling matter.

How is it that you quote a sentence from the article that contradicts your position and then somehow claim that it supports it? How much of that article did you even read?

Not around our SMBH and not around M87 SMBH

I already posted a link to observations that support matter entering the accretion disk of M87.

If I understand it correctly, they say that they see gas which is orbiting round the event of horizon of that SMBH.

You are not understanding it correctly. The very section you quote says that it has "almost no rotation", which means it isn't orbiting the event horizon.

So, they didn't really see that the gas is falling into the SMBH.

Yes they did. It says so right here: "We were able to follow an Earth-sized clump of matter for about a day, as it was pulled towards the black hole, accelerating to a third of the velocity of light before being swallowed up by the hole." If it fell into the black hole, then the radiation that it was emitting would disappear. The disappearance of that radiation would be detectable.

They also admit that they trace this gas for only one day.

Because it only took a day for it to reach the black hole and get eaten. Anyone with English as a first language could figure that out from what was said (although, admittedly, I'm not sure what your first language is. This may be why you misunderstood it).



1. Our scientists claim that till 2018 they have never ever found any evidence for in falling matter to the SMBH.

No, that is not what the article says. This is the first direct detection, not the first evidence.

2. The idea that they have found it for the first time in a galaxy at a distance of one billion light years from us proves that they have not find any evidence for in falling matter - not in the milky way and not in any other galaxy up to a distance of one billion year.

Again, this article is about a direct observation. Evidence of other types have been known before then.

3. How can they get real measurements from a galaxy that is located at a distance of one billion light years away?
How could they calculate and monitor the exact location of the event of horizon for a galaxy which is so far away?
Could it be that they have an error in their

Because you underestimate what information can be extracted from spectra with modern technology. Unless, of course, you want to posit some kind of bizarre conspiracy where they falsified their data.

4. How they can see a gas cloud with a size of an earth at a distance of one billion light year away?
Our scientists struggle to see an earth size which is located just 300 Light years away.
So, how can they see an earth gas cloud at one billion light year away? is there any possibility to see an earth size at one billion light year away from us?

The methodology is important. I went out and found the article in question: https://academic.oup.com/mnras/article/481/2/1832/5090165

When he says "Earth-sized", I'm guessing he meant "Earth-mass", as the following quote from the article details how the mass of the gas cloud was measured:

Quote
We use the higher resolution flow data of rev2659 (Table 2) to estimate the peak mass flow rate, where – as a stream of matter plunges towards the black hole – its increasing compactness ensures the third segment represents a maximum fraction passing through the line of sight. We assume a cylindrical inflow element at a radial distance 20Rg, with length 5Rg constrained by the well-defined velocity, and diameter 2Rg to allow a reasonable chance of detection along a line of sight to the hard X-ray source. The mean particle density is then NH/5Rg, and the observed mass element min=(volume×density×protonmass)∼15R3g×(3.6×1023/Rg)×1.7×10−24 g. For a black hole mass of 4 × 107 M⊙, Rg ∼ 6 × 1012 cm and min ∼ 3.3 × 1026 g. Since the observed element will cross the line of sight in ∼3000 s, the instantaneous observed inflow mass rate is ∼1023gs−1⁠.

It wasn't like taking a picture of something with a conventional camera. It was a combination of observing the characteristics of the spectra and the times scales over which it varied.

5. Do they really see that it swallowed by the SMBH or they assume that it should be swallowed due to its location - event of horizon?

If they can see the X-ray signature of the cloud, then they can also see it disappear. We know from conservation of energy that it didn't vanish into nothingness, so such a disappearance must necessarily mean that it was consumed by the black hole.

Do you agree with the following concussion:?

I disagree with both of your conclusions (as usual) because they are a blatant rejection and/or misunderstanding of the evidence.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/08/2019 06:44:06
This is the first direct detection, not the first evidence.
Again, this article is about a direct observation. Evidence of other types have been known before then.
Would you kindly explain the difference between Direct detection/observation to evidence?
You have stated that in order to detect this infilling matter they have used X-ray.
The methodology is important. They are detecting X-rays, which have a much, much smaller wavelength than visible light and therefore allow (1) much greater resolution and (2) are much more energetic than visible light and therefore are much easier to detect.
So, if X-ray is considered as direct observation, than what kind of verification/tools they have used for "evidence"?
How can we distinguish between direct observation which is based on X-ray, to evidence which is based on other tool?
I already posted a link to observations that support matter entering the accretion disk of M87.
Why do you call it "Observation" and not evidence?
Do you mean that they have used also X-ray in 1981?
I have already replied to  this observation which is based on the assumption that took place in Aug 1981.
Please see thread 568:
In that Article it is stated:
"The galaxy experiences an infall of gas at the rate of two to three solar masses per year, most of which may be accreted onto the core region.[52] "
However, if you go to 52 you get an article from Aug 1981
In that article it is stated:
"Resent X-ray spectroscopic results suggest that 2-3 M0 yr^-1 of cooling gas may be falling …M87 ...in reach cluster).
What do they mean by word "may"?
Is it "may yes" or "may no"?
.

So, do you mean that the meaning of "May" in English is "Yes" - 100% confirmation?
If that is correct, than do you mean that this "solid observation" which took place in Aug 1981 to confirm the in falling matter into the M87 was the second one, while this observation in 2018 is the first one?
If so,  why that observation in 2018 is considered as the" first" while we have already "observed" the M87 in aug 1981?
"A UK team of astronomers report the first detection of matter falling into a black hole at 30 percent of the speed of light, located in the centre of the billion-light year distant galaxy PG1211+143."
So do you mean that in English we count backwards?
In any case, if you still believe that there was a solid observation or evidence for in falling matter (before 2018) into our SMBH or to any SMBH at  any nearby galaxy (which is less than 1 billion light year away) would you kindly offer that observation/evidence (Please add a link)?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/08/2019 07:12:23
Would you kindly explain the difference between Direct detection/observation to evidence?

Models and mathematical calculations are evidence when they are based on the known laws of physics. They allow us to make inferences about phenomenon that we have not yet directly observed.

So, if X-ray is considered as direct observation, than what kind of verification/tools they have used for "evidence"?
How can we distinguish between direct observation which is based on X-ray, to evidence which is based on other tool?

See my previous response.

Why do you call it "Observation" and not evidence?

Observations are evidence, but they are not the only form of evidence. Hawking radiation itself has never been observed, but it's based on sound math and the known laws of gravity and quantum physics. That supports it as a form of evidence.

Do you mean that they have used also X-ray in 1981?

X-ray telescopes have been used since at least 1963.

I have already replied to  this observation which is based on the assumption that took place in Aug 1981.
In that article, it is clearly stated that the matter "may" fall in.
So, do you mean that the meaning of "May" in English is 100% confirmation?

Having not read that 1981 article, I cannot comment on it.

If that is correct, than do you mean that this "solid observation" which took place in Aug 1981 to confirm the in falling matter into the M87 was the second one, while this observation in 2018 is the first one?
If so,  why that observation in 2018 is considered as the" first" while we have already "observed" the M87 in aug 1981?

These are different kinds of observations. Matter accreting onto the accretion disk is not the same as matter falling into a black hole.

So do you mean that in English we count backwards?
In any case, if you still believe that there was a solid observation or evidence for in falling matter (before 2018) into our SMBH or to any SMBH at  any nearby galaxy (which is less than 1 billion light year away) would you kindly offer that observation/evidence (Please add a link)?

Again, matter accreting onto the accretion disk is not the same as it falling into the black hole. Those are two different observations.

On another note, it's a shame that you quoted me when you did, as you didn't catch the edit that I was making to my post. I provided a link to the scientific paper that describes the matter falling into a super-massive black hole at 30% the speed of light. Go back and read it if you want clarification on particular matters, such as how they made their observations.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 07/08/2019 12:57:51
Quote from: Dave Lev
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
...they found that more than 99% of the infalling material was ejected long before reaching the event horizon...
Thanks for the reference.
- Now I can see why you were confused.
- This is talking about the black hole at the center of our Milky Way galaxy
- This is surrounded by hot gas which produces some X-Rays, which is why it showed up in early X-Ray telescopes
- However, our SMBH does not have a very active accretion disk (as we observe it, at present)
- This hot gas is (currently) slowing down the rate at which gas can approach the SMBH in out galaxy
- So our galaxy is not (currently) producing a strong relativistic jet

However, the behavior of SMBHs does change over time
- If there is another SMBH in the vicinity, they will work together to eject stars and gas clouds away from the center.
- The activity of the accretion disk depends on how much matter is pouring in
- At present, not much is pouring in

So I think that you are better start using the 10% figure for the amount of mass falling into the accretion disk that ends up in a relativistic jet
- Not the 99% figure you have been quoting, as this is for a SMBH that does not have an accretion disk or a relativistic jet

We can hope to see some new data in the next year. The Event Horizon Telescope took data on our SMBH. They are currently trying to process this data to see if they can see any evidence of an accretion disk at microwave wavelengths.
- The Max Planck Institute continues to monitor the orbits of some extremely bright stars passing close to the location of the central black hole. But these orbits do not pass close enough to the central black hole to be tidally disrupted. This may take decades to reveal new information
- Progress with High speed X-Ray observations may reveal more information about the innermost stable orbit of a black hole. See: https://en.wikipedia.org/wiki/Quasi-periodic_oscillation
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/08/2019 18:08:10
Models and mathematical calculations are evidence when they are based on the known laws of physics. They allow us to make inferences about phenomenon that we have not yet directly observed.

Are you sure that Models and mathematical calculations are evidence?
Do you agree that when our scientists have used Models and mathematical calculations (till the last 10 years ago?), they were positively sure that ALL the matter in the accretion disc is falling into the black hole?
Now, we clearly have direct observation that over than 99% is ejected outwards.
Do you think that they would dare to call it "accretion" if they knew on day one that 99% of the matter is ejected outwards?
So if our Models and mathematical calculations didn't give us any indication for the 99% ejected matter, than how can we call it - evidence?
Based on Google -
Evidence - the available body of facts or information indicating whether a belief or proposition is true or valid.
How can we call those Models and mathematical calculations - "evidences" if  we have found that our belief (based on those models) isn't true?
Do you agree that the evidence (which is based on Models and mathematical calculations) is representing the current belief of our scientists? However, in order to verify if the evidence is true or false - Direct observation is needed?

I provided a link to the scientific paper that describes the matter falling into a super-massive black hole at 30% the speed of light.
Do you mean the link which I have offered:
I have found a very interesting article:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"First detection of matter falling into a black hole at 30 percent of the speed of light"
In this article they claim for direct observation of in falling matter into a super-massive black hole at 30% the speed of light.
However, that article is dated for 2018.
In that Article it is stated that it is the first time that our scientists discovered a direct observation:
"A UK team of astronomers report the first detection of matter falling into a black hole at 30 percent of the speed of light, located in the centre of the billion-light year distant galaxy PG1211+143."
However, if we ignore that direct observation:
Do you agree that till this first detection in 2018:
1. There was no direct observation (by X-ray) for any in falling matter. Not to the accretion disc and not to the SMBH. Not in our Galaxy and not in any galaxy in the whole Universe?
2. There is a direct observation that over than 99% of the matter in the accretion disc is ejected outwards.
3. There is direct observation for Jet molecular stream that is boosted upwards and downwards from the accretion disc plane (That jet steam is reaching 27,000 Light year above and below the disc with estimated 10,000 solar mass)

Please answer the following:
1. Severe mistake in the Models and mathematical calculations
How could it be that our evidences (Models and mathematical calculations) have missed the true? (Over than 99% of the matter that is ejected outwards from the accretion disc)?
What is the source for our severe mistake in our Models and mathematical calculations which we have set just 10 years ago?
2. How our updated Models and mathematical calculations can explain this new discovery of outwards ejected matter?
3. Why tidal/gravity force of the mighty SMBH can't pull all/most/some/more than 1% of that matter inwards?
4. How our updated Models and mathematical calculations can explain the discovery of the Molecular jet stream?
5. What is the source for that molecular jet stream?
6. Why tidal/gravity force of the SMBH doesn't pull the molecular jet stream inwards?
7. Why the jet moves at the opposite direction from the SMBH? (Directly upwards/downwards to the accretion disc)
8. Why the upwards/downwards velocity of the jet stream is so high (I assume that it is almost 0.8 speed of light)?
9. Do you agree that our scientists are mainly using Models and mathematical calculations to prove that their theory is correct? In order to do so they set the parameters that meets their needs. Therefore, theoretically, they could prove the opposite, if they would use different parameters/setup.
10. If the direct observation proves that there is a severe mistake in their Models and mathematical calculations, why they are not brave enough to say - sorry, we have failed in our "evidences".


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 08/08/2019 22:09:48
Are you sure that Models and mathematical calculations are evidence?

When they are based on the known laws of physics, yes, they are. Gravitational waves are one good example of this. Their existence and characteristics were described by relativity long before we ever detected them. When we finally did detect them, they had exactly the properties that they were predicted to have. So calculations and predictions based on successful models are good evidence for the phenomena that they predict.

Do you agree that when our scientists have used Models and mathematical calculations (till the last 10 years ago?), they were positively sure that ALL the matter in the accretion disc is falling into the black hole?

I'm extremely doubtful of your claim. Astrophysical jets have been known to exist since at least 1918, and we knew that black holes couldn't emit those jets directly. We must have known long before "10 years ago" that the jets had to originate from matter in the accretion disk.

Now, we clearly have direct observation that over than 99% is ejected outwards.

Perhaps in the Milky Way specifically, but that would not necessarily follow for other galaxies where their super-massive black holes have different masses, different spins and accretion disks of different masses, temperatures, composition, rotation rates, etc.

Do you think that they would dare to call it "accretion" if they knew on day one that 99% of the matter is ejected outwards?

Yes, because the matter was originally accreted from an outside source. That's how you get an accretion disk around a black hole in the first place, since such black holes don't emit matter. The fact that the matter in the accretion disk is blasted off in the form of jets is beside the point, since that step comes after the accretion process.

So if our Models and mathematical calculations didn't give us any indication for the 99% ejected matter, than how can we call it - evidence?

Do you have a link showing that any particular models describing the Milky Way's super-massive black hole can't account for that?

Evidence - the available body of facts or information indicating whether a belief or proposition is true or valid.
How can we call those Models and mathematical calculations - "evidences" if  we have found that our belief (based on those models) isn't true?

Again, how has that 99% figure falsified all of our models of accretion disks and super-massive black holes? Remember, you are talking about one particular case which may not apply to other galaxies.

Do you agree that the evidence (which is based on Models and mathematical calculations) is representing the current belief of our scientists?

You have it the wrong way around. The belief is what it is because of the mathematical evidence.

However, in order to verify if the evidence is true or false - Direct observation is needed?

Yes, I never denied that. I never said that evidence is the same as proof. But evidence is still evidence. And if you're going to be picky about this and demand direct observations, then you should realize that your own assertions lack observational evidence as well (we have never seen a black hole create matter, for example).

In that Article it is stated that it is the first time that our scientists discovered a direct observation:

To finish your sentence, it says that this is the first observation of matter falling into a black hole at 30% the speed of light. It doesn't say that this is the first observation of matter falling into a black hole ever.

1. There was no direct observation (by X-ray) for any in falling matter. Not to the accretion disc and not to the SMBH. Not in our Galaxy and not in any galaxy in the whole Universe?

No, the article doesn't say that this was the first ever detection of any form of matter falling into a black hole, so we cannot draw that conclusion.

2. There is a direct observation that over than 99% of the matter in the accretion disc is ejected outwards.

Only in our galaxy specifically. Evan_au pointed out a case where it was only 10%.

3. There is direct observation for Jet molecular stream that is boosted upwards and downwards from the accretion disc plane (That jet steam is reaching 27,000 Light year above and below the disc with estimated 10,000 solar mass)

Where did you get those numbers from? Even if those numbers are right, it doesn't matter. Those are no problem for our models.

1. Severe mistake in the Models and mathematical calculations
How could it be that our evidences (Models and mathematical calculations) have missed the true? (Over than 99% of the matter that is ejected outwards from the accretion disc)?

Again, give me a link where it was found that this figure violated our models.

2. How our updated Models and mathematical calculations can explain this new discovery of outwards ejected matter?

That was answered in the very article that you got the 99% figure from:

Quote
Thanks to the conservation of angular momentum, anything falling in towards a black hole has to lose some of its energy if it is not to miss the black hole.  Particles within dense clouds lose kinetic energy easily, due to friction within the cloud.  Similarly, the particles within cool clouds have little individual energy to start with, so are more easily captured by the black hole.  But the Chandra observation, collected over five weeks in 2012, revealed that the material in the vicinity of Sgr A* is not only very thin and diffuse, but is also extremely hot.  As a consequence, it escapes the black hole while it is still far away, and before it has time to form a superheated accretion disk, which explains the coolness of the ejected material.

3. Why tidal/gravity force of the mighty SMBH can't pull all/most/some/more than 1% of that matter inwards?

Apparently, the jets are formed far enough away from the black hole that over 99% of the matter reaches escape velocity (which is below the speed of light outside of the event horizon). The prior quote from the article says as much. You are very bad about quote-mining.

4. How our updated Models and mathematical calculations can explain the discovery of the Molecular jet stream?

You mean the astrophysical jet? Those have been easily explained for a long time. The magnetic field generated by the disk propels them away from the black hole.

5. What is the source for that molecular jet stream?

If you're talking about the astrophysical jets, then that would be the accretion disk. As much as you talk about this, I figure you would have known that already.

6. Why tidal/gravity force of the SMBH doesn't pull the molecular jet stream inwards?

It's moving too fast.

7. Why the jet moves at the opposite direction from the SMBH? (Directly upwards/downwards to the accretion disc)

It's because of the alignment of the generated magnetic field lines relative to the disk. You end up with a north pole and a south pole in the direction of the jets.

8. Why the upwards/downwards velocity of the jet stream is so high (I assume that it is almost 0.8 speed of light)?

The matter is hot, which corresponds to a high particle velocity.

9. Do you agree that our scientists are mainly using Models and mathematical calculations to prove that their theory is correct?

No, scientific theories are. Not. Proven.

In order to do so they set the parameters that meets their needs. Therefore, theoretically, they could prove the opposite, if they would use different parameters/setup.

No, you can't modify the laws of physics to suit your desires and come up with a realistic model.

10. If the direct observation proves that there is a severe mistake in their Models and mathematical calculations, why they are not brave enough to say - sorry, we have failed in our "evidences".

Because the observations don't contradict our models.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 09/08/2019 08:55:54
Quote from: Dave Lev
Now, we clearly have direct observation that over than 99% is ejected outwards.
You continue to quote this 99% expulsion as characteristic of a SMBH with an accretion disk. But this figure is actually for a SMBH without an accretion disk (as we see it at present).

Quote
How our updated Models and mathematical calculations can explain this new discovery of outwards ejected matter?
The radio lobes around galaxies is not a new discovery. They were described and categorised in the 1970s.
See: https://en.wikipedia.org/wiki/Radio_galaxy#Radio_structures

Models explain it by twisted magnetic fields embedded in the swirling plasma of the accretion disk, which funnelout matter  along the polar axis of the central object.About 10% of the mass falling into the accretion disk ends up in the jets, and 90% falls onto the central object (SMBH, stellar-mass black hole, neutron star, white dwarf, protostar, etc).

Quote
6. Why tidal/gravity force of the SMBH doesn't pull the molecular jet stream inwards?
Because around 90% of it does get pulled inwards.
The models show that the magnetic fields in the rotating plasma can propel around 10% of the accretion disk flow up to twice the escape velocity of the central object (SMBH, stellar-mass black hole, neutron star, white dwarf, protostar, etc).

Depending on the radial distance of the source material, that escape velocity can be fairly low (for a small protostar), but approaching the speed of light for a black hole.

Quote
Do you agree that till this first detection in 2018:
1. There was no direct observation (by X-ray) for any in falling matter.
The first millisecond pulsar was discovered in 1982. We now know of several hundred of these objects, and many of them emit X-Rays.

The majority of these are believed to be neutron stars that have absorbed matter from a companion star, forming an accretion disk. Most of the matter in the accretion disk rains down on the neutron star along the magnetic field lines.
- This produces the "hot spots" detected as pulsations in light output
- This additional angular momentum causes the pulsar to spin faster & faster - the fastest known example is spinning 716 times per second, with an equatorial speed of 24% of the speed of light.
See: https://en.wikipedia.org/wiki/Millisecond_pulsar

From the viewpoint of the donor star, there is no difference between the gravitational field of a neutron star & a black hole of similar mass. It only makes a difference when you get within about 20km.
- However, for astronomers, the pulsar has a significant benefit that we can measure the orbital period of its companion star by the emitted pulses from the central object. This companion star is the source of the accretion disk (which probably only operates in bursts as it periodically sucks matter off the companion star).
- Black holes are much harder to observe, because no matter or light has ever been observed being emitted from a black hole.
See: https://en.wikipedia.org/wiki/PSR_J1748%E2%88%922446ad
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/08/2019 17:40:09
Quote
7. Why the jet moves at the opposite direction from the SMBH? (Directly upwards/downwards to the accretion disc)
It's because of the alignment of the generated magnetic field lines relative to the disk. You end up with a north pole and a south pole in the direction of the jets.
Wow!!!
Thanks for your great answer!
You have just confirmed the great impact of the magnetic field around the SMBH.
That magnetic field is a key element in the galaxy.
We clearly see that it boosts the Molecular jet stream at ultra speed of almost 0.8c upwards/downwards.
It takes its matter from the accretion disc:
Quote
5. What is the source for that molecular jet stream?
If you're talking about the astrophysical jets, then that would be the accretion disk. As much as you talk about this, I figure you would have known that already.
If it is so powerful, and if you agree that it takes the matter from the accretion disc, than don't you agree that this is the answer for why 99% of the matter in the accretion disc is ejected outwards?
In other words - the mighty gravity force, force the matter in the accretion disc to be ejected outwards.
Therefore, do you agree by now that the same ultra magnetic force that push the matter in the accretion disc outwards and boosts the jet stream at 0.8c - must also prevent from any matter from outside to fall into the accretion disc. If something will come in - it will be boosted upwards/downwards.

That is also the answer for that gas which our scientists thought that the SMBH eats:
https://phys.org/news/2018-09-falling-black-hole-percent.html
The researchers found the spectra to be strongly red-shifted, showing the observed matter to be falling into the black hole at the enormous speed of 30 per cent of the speed of light, or around 100,000 kilometers per second. The gas has almost no rotation around the hole, and is detected extremely close to it in astronomical terms, at a distance of only 20 times the hole's size (its event horizon, the boundary of the region where escape is no longer possible).
Look carefully what is written:
"The gas has almost no rotation around the hole"
So this gas cloud does not orbit around the super massive black hole. Therefore, it can't be in the accretion disc as the plasma must orbit around the SMBH. Hence, it must be outside the accretion.
However, it is also stated that it is "falling into the black hole at the enormous speed of 30 per cent of the speed of light".
So, our scientists assume that now it is moving to the center of the Black hole.
That is a severe mistake.
What they really see is the great impact of the magnetic field.
As that gas cloud had been ejected outwards from the accretion disc, it had been traped by the mighty power of the magnetic filed.
Let's look again on the following answer:
It's because of the alignment of the generated magnetic field lines relative to the disk. You end up with a north pole and a south pole in the direction of the jets.
That magnetic field pulls the gas cloud directly the north/south pole and then boosts it at ultra velocity upwards or downwards as a molecular jet stream.
So, as the gas cloud gets to the pole, it is actually located above or below the SMBH (depending on the galaxy view from our location. Therefore, our scientists thought that this gas cloud is falling into the SMBH as they have no real technology to verify its upwards/downwards location with regards the SMBH. Once it gets to the pole, the magnetic fields break it down to a molecular jet that is boosted upwards/downwards.
This jet stream is very difficult to verify. Therefore, they thought that the SMBH had swallowed the gas cloud:
He continues: "We were able to follow an Earth-sized clump of matter for about a day, as it was pulled towards the black hole, accelerating to a third of the velocity of light before being swallowed up by the hole."
Sorry dear scientist let me tell you the following:
You have a fatal error.
This gas cloud had been boosted into the molecular jet stream. The SMBH didn't eat it.
Remember the fireworks???
That fireworks that we are expecting to see if matter is falling into the SMBH?.
Where is that fireworks?
There is no fireworks as the SMBH didn't eat that gas cloud.

Quote
Now, we clearly have direct observation that over than 99% is ejected outwards.
There is no direct observation of this.  We do not directly observe material moving in and out of Sgr-A's accretion disk.  At best we have a view of the radiation signatures surrounding it,
In the article it is stated:
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"When astronomers used Chandra to study Sgr A*, in one of its longest ever observations, they found that more than 99% of the infalling material was ejected long before reaching the event horizon (the point of no return around a black hole, from which not even light can escape) and was unusually cool, and therefore quite dim in the Xray spectrum."
So, it is stated that based on X-ray (real direct observation) our scientists observered that more than 99% of the matter in our accertion disc (which they consider as infalling matter), had been ejected outwards.
We have already agreed that X-ray means -- Observation.
So, why don't you accept this clear data?
Quote
Now, we clearly have direct observation that over than 99% is ejected outwards.
You continue to quote this 99% expulsion as characteristic of a SMBH with an accretion disk. But this figure is actually for a SMBH without an accretion disk (as we see it at present).
What do you mean by:"SMBH without an accretion disk"
Don't you agree that our SMBH has an accretion disk?

but our edge-on view is far too obscured to see things smaller than say S2 which doesn't get close enough to count as material moving in or out of the disk.
The accretion disc is located at about 28,000 LY from us, while the galaxy in which we have observed (also by X-ray) in falling matter for the first time is located at a distance of 1,000,000,000 LY (One billion light year away)
Hence - If our scientists wants to prove something - than they have the technology to observe an Earth size cloud at a  distance of One billion light year away. However, If they don't see the expected in falling matter in a distance of only 28,000 light year, than suddenly our technology is very poor.
Is it real?
Quote
How our updated Models and mathematical calculations can explain this new discovery of outwards ejected matter?
The radio lobes around galaxies is not a new discovery. They were described and categorized in the 1970s.
See: https://en.wikipedia.org/wiki/Radio_galaxy#Radio_structures

Models explain it by twisted magnetic fields embedded in the swirling plasma of the accretion disk, which funnelout matter  along the polar axis of the central object.About 10% of the mass falling into the accretion disk ends up in the jets, and 90% falls onto the central object (SMBH, stellar-mass black hole, neutron star, white dwarf, protostar, etc).
So you claim that our modeling is an evidence for 90% in falling matter onto the central object.
However, we have already agreed that modeling is just an "evidence". Evidence is not a proof. Therefore, we still must prove this modeling. So, far we didn't find/observe any in falling matter into our SMBH or any other SMBH which is located at a distance of up to one billion light year away.
Actually, only in 2018 we have observed for the first time -Ever - the in falling matter in a galaxy which is located at one billion light year away.

From the viewpoint of the donor star, there is no difference between the gravitational field of a neutron star & a black hole of similar mass. It only makes a difference when you get within about 20km.
So, you claim that there is no difference between the gravitational field of a neutron star & a black hole of similar mass.
While Kryid claims that even SMBH might be different from each other:
Quote
Now, we clearly have direct observation that over than 99% is ejected outwards.
Perhaps in the Milky Way specifically, but that would not necessarily follow for other galaxies where their super-massive black holes have different masses, different spins and accretion disks of different masses, temperatures, composition, rotation rates, etc.
So, do you mean that if we want to prove that our SMBH eats the mass in its accretion disc, than we can base our understanding even on the activity of  neutron star, while in the same token we can claim that our SMBH can't be used as a study case for other SMBH because it doesn't fulfill our expectations?
Why our SMBH is so naughty? Why does it eject 99% of the food in the accretion buffet?   
In any case, do you mean that if we find a SMBH (even if it is located at a distance of a billion years away and even if we only find one in the whole Universe)  which eats his food (as expected by our scientists) - than this unique SMBH can be used as a key model for all the SMBH in the Universe. However, all the other Millions and billions SMBH (including our local SMBH) that insist to eject there food can't be used as a model as they do not fulfill our expectations?
So the only one exception SMBH in the whole Universe (which at least fulfill our expectation and show some signs of eating - at a distance of one billion LY away) is used as the key model for all the other Billions SMBH (that eject their food and don't show any eating activity) - Do you agree with that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/08/2019 22:01:22
Wow!!!
Thanks for your great answer!
You have just confirmed the great impact of the magnetic field around the SMBH.
That magnetic field is a key element in the galaxy.
We clearly see that it boosts the Molecular jet stream at ultra speed of almost 0.8c upwards/downwards.
It takes its matter from the accretion disc:

I don't know why you're so surprised. I never denied the existence of the field.

If it is so powerful, and if you agree that it takes the matter from the accretion disc, than don't you agree that this is the answer for why 99% of the matter in the accretion disc is ejected outwards?

That's far from the whole story. The temperature of the matter is also important because hotter gases have faster-moving particles. The faster the particles move, the more easily they can reach escape velocity. The article you linked said the same thing.



Therefore, do you agree by now that the same ultra magnetic force that push the matter in the accretion disc outwards and boosts the jet stream at 0.8c - must also prevent from any matter from outside to fall into the accretion disc. If something will come in - it will be boosted upwards/downwards.

No, because the magnetic field strength is not uniform throughout the disk. It would be stronger the closer you get to the black hole because it spins faster and is more strongly ionized there. The inner region can exceed a million kelvins, but outer, cooler regions can be a mere 10,000 kelvins: https://www.sciencenews.org/article/accretion-disk-milky-way-galaxy-black-hole There is nothing keeping gases from being added to these outer regions.

So this gas cloud does not orbit around the super massive black hole. Therefore, it can't be in the accretion disc as the plasma must orbit around the SMBH. Hence, it must be outside the accretion.

It originally was in the accretion disk. Read the following excerpt from the article:

Quote
The observation agrees closely with recent theoretical work, also at Leicester and using the UK's Dirac supercomputer facility simulating the 'tearing' of misaligned accretion discs. This work has shown that rings of gas can break off and collide with each other, cancelling out their rotation and leaving gas to fall directly towards the black hole.

As that gas cloud had been ejected outwards from the accretion disc, it had been traped by the mighty power of the magnetic filed.

That magnetic field pulls the gas cloud directly the north/south pole and then boosts it at ultra velocity upwards or downwards as a molecular jet stream.
So, as the gas cloud gets to the pole, it is actually located above or below the SMBH (depending on the galaxy view from our location. Therefore, our scientists thought that this gas cloud is falling into the SMBH as they have no real technology to verify its upwards/downwards location with regards the SMBH. Once it gets to the pole, the magnetic fields break it down to a molecular jet that is boosted upwards/downwards.

So you really think that astrophysicists are so stupid that they can't tell whether a gas cloud is moving towards or away from the black hole? For one, the gas cloud was observed to become increasingly dense as it accelerated, which is what you would expect if the gas cloud was actually approaching the hole (it would be compressed by the hole's gravity). If it instead owed its great velocity to being accelerated into the jets, the gas cloud would become much less dense as it was stretched out. Also, if the gas cloud was converted into a jet, its X-ray signature would not have disappeared because there is nothing to keep the gas from continuing to emit X-rays (it would still be hot). If it fell into the black hole, then you would indeed expect it the X-ray signature to disappear (and it did).

The accretion disc is located at about 28,000 LY from us, while the galaxy in which we have observed (also by X-ray) in falling matter for the first time is located at a distance of 1,000,000,000 LY (One billion light year away)
Hence - If our scientists wants to prove something - than they have the technology to observe an Earth size cloud at a  distance of One billion light year away. However, If they don't see the expected in falling matter in a distance of only 28,000 light year, than suddenly our technology is very poor.
Is it real?

The paper about infalling matter moving at 30% the speed of light has the answer to that:

Quote
The transient nature of the rev2659 inflow largely explains why a compelling detection has not been reported before. Isolated claims are not uncommon, however, with several single-line detections noted in the ‘Introduction’ section, and both XMM–Newton and Suzaku archival searches finding – but not discussing – transient absorption lines in the region occupied by redshifted Fe K lines.

In order to see a gas cloud get eaten by a black hole, you have to be looking in the right place at the right time (this event happened over the scale of hours). It's easy to miss.

Therefore, we still must prove this modeling.

How many times do I have to tell you that you don't prove scientific models or theories?

So, you claim that there is no difference between the gravitational field of a neutron star & a black hole of similar mass.
While Kryid claims that even SMBH might be different from each other:

You are taking my statements out of context. What evan_au is saying is absolutely true: due to shell theorem, a distant body will react to gravity in exactly the same manner whether the object it is orbiting is a normal star, a neutron star or a black hole (if they all have identical mass). My statements are different because, for one, I am talking about black holes of different masses. My statements aren't only about gravity, either, but are about differences in spin, temperature and gas densities as well. You're comparing apples with oranges.

So, do you mean that if we want to prove that our SMBH eats the mass in its accretion disc, than we can base our understanding even on the activity of  neutron star, while in the same token we can claim that our SMBH can't be used as a study case for other SMBH because it doesn't fulfill our expectations?

It depends on the circumstances. For a stellar mass black hole, yes, you'd expect it to behave in a similar manner to a neutron star put in its place when it accretes matter. A super-massive black hole, however, is many, many times more massive than a neutron star and (potentially) has an accretion disk many, many times more massive. You would still expect the super-massive black hole to pull the matter in just as the neutron star does, but the rate may be very different.

Why our SMBH is so naughty? Why does it eject 99% of the food in the accretion buffet?   

Would you please quit asking the same questions over and over again when they've already been answered? I've already given you the answer from that same article that you seem to love so much:

Quote
But the Chandra observation, collected over five weeks in 2012, revealed that the material in the vicinity of Sgr A* is not only very thin and diffuse, but is also extremely hot.  As a consequence, it escapes the black hole while it is still far away, and before it has time to form a superheated accretion disk, which explains the coolness of the ejected material.

In any case, do you mean that if we find a SMBH (even if it is located at a distance of a billion years away and even if we only find one in the whole Universe)  which eats his food (as expected by our scientists) - than this unique SMBH can be used as a key model for all the SMBH in the Universe. However, all the other Millions and billions SMBH (including our local SMBH) that insist to eject there food can't be used as a model as they do not fulfill our expectations?

No, you can't use any one super-massive black hole as a fool-proof model for all others. Each one is a unique entity with its own properties. Some may eat most of their accretion disk, others may expel most of it, while yet others may eat half and expel half. You have to look at it on a case-by-case basis.

So the only one exception SMBH in the whole Universe (which at least fulfill our expectation and show some signs of eating - at a distance of one billion LY away) is used as the key model for...

No, there is no good reason to believe that it is unique. As the article pointed out before, eating gas clouds is a transient phenomenon so it would be rarely observed. Assuming that this galaxy is somehow so magically different from all the others that it is the only one where the black hole consumes matter would be a ridiculous conclusion.

Quote
all the other Billions SMBH (that eject their food and don't show any eating activity)

Since when were "all the other billions" of super-massive black holes demonstrated to "eject their food"? You are making that up. Do you honestly think that we have closely investigated the accretion disk behavior of "billions" of super-massive black holes?

Quote
- Do you agree with that?

No, because, as usual, you don't know what you are talking about.

Sorry dear scientist let me tell you the following:
You have a fatal error.
This gas cloud had been boosted into the molecular jet stream. The SMBH didn't eat it.

Yes it did. The data is consistent with it having fallen into the hole, not with it having been converted into a jet.

Remember the fireworks???
That fireworks that we are expecting to see if matter is falling into the SMBH?.
Where is that fireworks?
There is no fireworks as the SMBH didn't eat that gas cloud.

What fireworks?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/08/2019 02:01:03
Quote from: Halc
Quote
Now, we clearly have direct observation that over than 99% is ejected outwards.
There is no direct observation of this.  We do not directly observe material moving in and out of Sgr-A's accretion disk.  At best we have a view of the radiation signatures surrounding it,
In the article it is stated:
"When astronomers used Chandra to study Sgr A*, in one of its longest ever observations, they found that more than 99% of the infalling material was ejected long before reaching the event horizon (the point of no return around a black hole, from which not even light can escape) and was unusually cool, and therefore quite dim in the Xray spectrum."
So, it is stated that based on X-ray (real direct observation) our scientists observered that more than 99% of the matter in our accertion disc (which they consider as infalling matter), had been ejected outwards.
Yes, the X-rays were directly observed.  The matter falling in was not.  We cannot see the matter, only a general dim X-ray signature.  The one observation actually saw a clould move and accelerate over the course of hours, but that wasn't Sgr-A.  A direct observation of a mass moving in and/or out like that has not been made for Sgr-A.

Just because I see red light doesn't mean I've observed an apple.  I might deduce that it's an apple causing the red light, but that might not be a direct observation.  In the one distant galaxy, they actually saw the apple, and not just a signature glow from a steady process.

Quote
What do you mean by:"SMBH without an accretion disk"
Don't you agree that our SMBH has an accretion disk?
Only a negligible one, barely active at all.

Quote
The accretion disc is located at about 28,000 LY from us, while the galaxy in which we have observed (also by X-ray) in falling matter for the first time is located at a distance of 1,000,000,000 LY (One billion light year away)
Hence - If our scientists wants to prove something - than they have the technology to observe an Earth size cloud at a  distance of One billion light year away. However, If they don't see the expected in falling matter in a distance of only 28,000 light year, than suddenly our technology is very poor.
Is it real?
With high technology, we can see a magnesium fire in the dark from space, but that's looking down at the fire.  Looking at Sgr-A is like spotting a firefly from the vantage point on the ground with a forest between us and the firefly.  Sure, its a lot closer, but much less bright and completely obscured by the forest.

Quote
So, you claim that there is no difference between the gravitational field of a neutron star & a black hole of similar mass.
At a given radius, yes, there is no difference.  If the black hole had the mass of Earth and you put it in the middle of a hollow massless shell of Earth radius, we on the surface would feel the same gravity.  This should be obvious from Newton's formula which you've quoted.  There are not different kinds of gravity that depend on the nature of the mass involved.

Quote
In any case, do you mean that if we find a SMBH (even if it is located at a distance of a billion years away and even if we only find one in the whole Universe)  which eats his food (as expected by our scientists)
The Andromeda one is very similar to the one where they witnessed the cloud falling in.  About the same mass (10x that of Sgr-A), and also being fed far more than is ours.  The distant one is apparently accreting at an even greater rate than Andromeda.  Our galaxy is sort of the anomaly here.  Very little mass is falling into what's left of the disk.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/08/2019 08:39:53
Quote
In the article it is stated:
"When astronomers used Chandra to study Sgr A*, in one of its longest ever observations, they found that more than 99% of the infalling material was ejected long before reaching the event horizon (the point of no return around a black hole, from which not even light can escape) and was unusually cool, and therefore quite dim in the Xray spectrum."
So, it is stated that based on X-ray (real direct observation) our scientists observed that more than 99% of the matter in our accretion disc (which they consider as in falling matter), had been ejected outwards.
Yes, the X-rays were directly observed.  The matter falling in was not.  We cannot see the matter, only a general dim X-ray signature.
Thanks Halc
If I understand you correctly, the X-rays of the ejected matter were directly & clearly observed.
However, there is no indication at all for any in falling matter.
Therefore, our scientists have stated that 99% of the matter is ejected outwards.
That is clear to me by now.

With regards to that gas cloud at earth size which had been discovered at the one billion light year away in 2018:
It originally was in the accretion disk. Read the following excerpt from the article:
"The observation agrees closely with recent theoretical work, also at Leicester and using the UK's Dirac supercomputer facility simulating the 'tearing' of misaligned accretion discs. This work has shown that rings of gas can break off and collide with each other, cancelling out their rotation and leaving gas to fall directly towards the black hole".
So, our scientists claim that " rings of gas can break off and collide with each other, cancelling out their rotation and leaving gas to fall directly towards the black hole"
Let see if this is feasible:
1. Orbital velocity of the matter/plasma in the "inner edge of the accretion disk"
Based on the article which Kryptid had offered:
Do we know for sure if a BH/SMBH rotates or not?
Yes: https://resonance.is/super-massive-black-holes-spin-near-the-speed-of-light/
"This research project utilized x-ray emission fluorescence produced by the reflection of hard X-rays off the inner edge of the accretion disk. The x-rays are produced in the region outside of the black hole as iron molecules are excited in the chaotic region outside of the event horizon."
So, they specifically measure the orbital velocity the inner side of the accretion disc (they even call it:  "the chaotic region outside of the event horizon").
In that region the have found that the orbital velocity is 0.85c:
"the suppermassive black hole called NG1365 is spinning at an extreme speed of 85% of the speed of light or 670 million miles per hour."

2. The impact of the magnetic field on the Accretion rings:
https://iopscience.iop.org/article/10.1088/0004-637X/758/2/103/pdf
MAGNETICALLY LEVITATING ACCRETION DISKS AROUND SUPERMASSIVE BLACK HOLES
"In this paper, we report on the formation of magnetically levitating accretion disks around suppermassive black
holes (SMBHs). The structure of these disks is calculated by numerically modeling tidal disruption of magnetized
interstellar gas clouds. We find that the resulting disks are entirely supported by the pressure of the magnetic
fields against the component of gravitational force directed perpendicular to the disks. The magnetic field shows
ordered large-scale geometry that remains stable for the duration of our numerical experiments extending over 10% of the disk lifetime. Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation. This in combination with the repeated feeding of magnetized molecular clouds to an SMBH yields a possible solution to the long-standing puzzle of black hole growth in the centers of galaxies."
It is stated clearly: "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation."
Hence, the ultra magnetic force "inhibits disc fragmentation".
Therefore, there is no possibility for: "rings of gas can break off and collide with each other" in the accretion disc.
We have already found that the matter in the "inner edge of the accretion disk" orbits in one direction at ultra high velocity of 0.85c.

So, even if there is a collision, while all the matter orbits at the same direction at ultra velocity - how can they suddenly stop the orbital velocity of plasma?
How our scientists believe that somehow in the "inner edge of the accretion disk" there is "disk fragmentation" while it is stated that -  "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation", and than somehow two plasma sections that orbit at the same direction and at the same ultra high velocity of 0.85C can collide and stop at their orbital momentum?
How could it be that after this imaginary stop the gas cloud move so nicely in the direction of the black hole and cross the event of horizon? No more orbital velocity of 0.85c, no more "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation" In that  "inner edge of the accretion disk".
How a plasma can be converted back into gas cloud?
Wow!
So, let's summarize:
A gas cloud from outside is falling into the accretion disc.
As it falls in it gain ultra high orbital velocity and ultra high temp, converted into plasma under the ultra gravity force of the SMBH and under the Ultra magnetic force.
At the "inner edge of the accretion disk" which is also called: "the chaotic region outside of the event horizon" the orbital velocity of that plasma is almost 0.85c.
Than suddenly, the plasma stops its orbital velocity, (from 0.85c to zero!) converted back into a nice gas cloud at the size of the Earth and move nicely and directly to the event of horizon of the SMBH (at 0.3c) without any orbital momentum.
If this is not science fiction - than I clearly don't know the meaning of science fiction..


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/08/2019 13:45:35
However, there is no indication at all for any in falling matter.
...
That is clear to me by now.
There is indication.  My point was that it isn't directly seen.  Have you seen actual pictures of Sgr-A?  It looks like the forest I mentioned.  I'd personally never have guessed there was something special there, but then it's not my job.  Here's an image from a prior post in this thread when we were talking about S2:
(https://en.es-static.us/upl/2018/03/S2-milky-way-black-hole-2-e1520510847503.jpg)
Sgr-A isn't even centered on one of the bright spots.
The X-ray shots allow one to actually see it, if only an image of a fuzzy spot that isn't nearly as bright as many other nearby X-ray sources.

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So, they specifically measure the orbital velocity the inner side of the accretion disc (they even call it:  "the chaotic region outside of the event horizon").
In that region the have found that the orbital velocity is 0.85c:
"the suppermassive black hole called NG1365 is spinning at an extreme speed of 85% of the speed of light or 670 million miles per hour."
That comment is obviously wrong.  Black holes don't have a spin rate, they have angular momentum. Your post says it is a measurement taken from the inner edge of the ring, not the black hole itself. So the article contradicts itself, which is typical of an article written by some guy instead of the scientist himself.  Yes, black holes spin, but it isn't measured in units of linear speed.
Similarly, the one article was similarly wrong when it said 'a cloud the size of Earth'.  There is no way an Earth-size object could be seen at that distance.  I suspect they meant mass-of-Earth.  I'm surprised you didn't latch onto that error and forever quote that 'clearly' Earth-size objects can be seen from a billion light years away.

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It is stated clearly: "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation."
Hence, the ultra magnetic force "inhibits disc fragmentation".
For the most part, a single disk prevents itself from breaking up on its own.  It doesn't really cover the case of multiple accretion disks working on different planes, as is the case of PG1211+143, which had at least 7 disks according to the article.

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Therefore, there is no possibility for: "rings of gas can break off and collide with each other" in the accretion disc.
Contradicted by findings you linked.  Again, I notice one quote is 'clearly' and the other dismissed.  Science cannot be done with that sort of blatant selection bias.

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We have already found that the matter in the "inner edge of the accretion disk" orbits in one direction at ultra high velocity of 0.85c.

So, even if there is a collision, while all the matter orbits at the same direction at ultra velocity - how can they suddenly stop the orbital velocity of plasma?
It wasn't that black hole where they measured that speed.  That speed was also the inner edge of one exceptional disk.  In the PG1211+143 case, it was a interaction of a pair of slow-moving non-inner rings that was in what they call the 'tearing zone', somewhat between the inner and outermost rings where the conflicting forces can tear chunks off the rings and allow them to hit each other.

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How our scientists believe that somehow in the "inner edge of the accretion disk" there is "disk fragmentation" and somehow two plasma sections that orbit at the same direction and at the same ultra high velocity of 0.85C can collide and stop at their orbital momentum?
Nobody said that except you.

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How could it be that after this imaginary stop the gas cloud move so nicely in the direction of the black hole and cross the event of horizon? No more orbital velocity of 0.85c, no more "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation" In that  "inner edge of the accretion disk".
How a plasma can be converted back into gas cloud?
Nobody said it was plasma.  It probably became so just before entering the black hole, but not while they were watching it.

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Wow!
So, let's summarize:
A gas cloud from outside is falling into the accretion disc.
As it falls in it gain ultra high orbital velocity and ultra high temp, converted into plasma under the ultra gravity force of the SMBH and under the Ultra magnetic force.
At the "inner edge of the accretion disk" which is also called: "the chaotic region outside of the event horizon" the orbital velocity of that plasma is almost 0.85c.
This is a terrible summary.  You're mixing descriptions of different things and different scenarios.  No real scenario is described by this summary.

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If this is not science fiction - than I clearly don't know the meaning of science fiction.
I actually agree with this statement.  The above description is fiction.  For once, you didn't preface it with "Do you agree with that?".
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 10/08/2019 14:27:01
Oops! overlap with Halc...
Quote from: Kryptid
the gas cloud was observed to become increasingly dense as it accelerated, which is what you would expect if the gas cloud was actually approaching the hole
Just to clarify this: A gas cloud that is "stationary" relative to the black hole will fall towards the black hole.
- Because it needs to maintain a minimum orbital velocity to remain in orbit
- Different parts of the gas cloud will fall towards the black hole
- Parts of the cloud that are closer to the black hole will accelerate away from more distant parts (because gravity is stronger when you are closer to the black hole): The "spaghetti effect"
- Parts of the cloud that are equidistant from the black hole will accelerate towards each other (because they are all aiming for the somewhat small black hole)

So whether the gas cloud is overall more or less dense depends on the relative sizes of the gas cloud and the black hole, and their relative velocities.

If the "stationary" gas cloud was formed by the collision of two gas clouds orbiting on different orbital planes, you could expect the collision to be violent, and to produce a debris cloud with a high temperature and a wide spread of velocities.   

Quote from: DaveLev
even if there is a collision, while all the matter orbits at the same direction at ultra velocity - how can they suddenly stop the orbital velocity of plasma?
If you look at the orbits of the stars around Sgr A*, you will see that they have very different orbital planes and eccentricities - much like comets orbiting the Sun.

If two of these stars or dust clouds formed an accretion disk around SGR A*, they would be orbiting on different orbital planes - and perhaps even in opposite directions. These accretion disks would collide at very high speeds (we are talking of orbital velocities which are a fair fraction of c).
- This would produce heating and material that collapses into the black hole

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If this is not science fiction - than I clearly don't know the meaning of science fiction..
You are just confusing several different scenarios:
- A single active accretion disk, with a dominant orbital plane (such as M87, imaged with the Even Horizon Telescope). At present we see mass steadily falling into the black hole, with around 10% siphoned off in jets.
- A very short-lived accretion disk, containing matter in different orbital planes. The angular momentum is not enough to sustain an accretion disk, and it quickly disappears into the black hole.
- A minimal accretion disk, such as we see in our galaxy at present = 26,000 years ago. The high temperatures around the black hole (perhaps from a previous accretion disk?) are preventing a new accretion disk from forming, at present. (After it cools down, if something massive comes close enough in a decade or so, probably an accretion disk will resume in our galaxy.)

Note that cosmologists expect that the growth rate of black holes is somewhat self-limiting.
- A dietary binge is likely to heat up the environment enough so that it slows down subsequent consumption.
- This may be what we see in our own galaxy at present
- Cosmologists are trying to refine these dietary limits.
- In particular, it is unclear how SMBH which form the cores of galaxies formed so early in the universe.
See: https://en.wikipedia.org/wiki/Supermassive_black_hole#Formation
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/08/2019 14:58:22
However, there is no indication at all for any in falling matter.

I already explained why there is. Objects don't disappear into nothingness, so the disappearance of the X-ray signature indicates that the gas cloud was consumed by the hole.

Let see if this is feasible:

Yes, it's feasible. For reasons the others have described.

How a plasma can be converted back into gas cloud?

Plasma is already a gas. The fact that the cloud was emitting X-rays clearly showed that it was very hot.
It is stated clearly: "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation."
Hence, the ultra magnetic force "inhibits disc fragmentation".
Therefore, there is no possibility for: "rings of gas can break off and collide with each other" in the accretion disc.

The article is talking specifically about gravitational fragmentation, not fragmentation due to different parts of different disks at different angles to each other colliding. The Lense-Thirring effect, which is responsible for these multiple, out-of-plane disks, is not even mentioned in that particular article.

By the way, that very same study you cite also says, "We find that the resulting disks are completely dominated by the magnetic field pressure and display high accretion rates due to the Maxwell stress associated with the large-scale magnetic field the structure of which remains stable over the duration of the simulation." So you are willing to accept what that paper says (which is talking about a numerical simulation, which you normally seem to be suspicious of) when it agrees with your ideas, but when it contradicts them (like saying that black holes accrete matter) you ignore it?

So are calculations published by astrophysicists trustworthy evidence or are they not?

So whether the gas cloud is overall more or less dense depends on the relative sizes of the gas cloud and the black hole, and their relative velocities.

That's true, although the paper did say that the density increased in this particular case.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/08/2019 05:21:08
" So you are willing to accept what that paper says (which is talking about a numerical simulation, which you normally seem to be suspicious of) when it agrees with your ideas, but when it contradicts them (like saying that black holes accrete matter) you ignore it?
May I ask you the same question?
Are you willing to accept the data which had been given in this article?
If so, why do you take one and reject the other?
In the article it is stated very clearly:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"First detection of matter falling into a black hole at 30 percent of the speed of light"
So, do you accept the statement that this is the FIRST detection/observation?
Do you agree that before 2018 we didn't find any observation for in falling matter?
Do you agree that after this discovery we have also didn't find any observation for in falling matter?
Do you agree that this discovery is the only real observation that we have ever found?
We didn't find any observation for in falling matter. Not in M87 in 1987, not in Andromeda galaxy and not at any SMBH at any location in the whole Universe. Not before and not after this unique discovery
Please let me know if you agree to accept this clear statement from the article.

With regards to the technology:
Please remember that we have the technology to monitor a gas cloud at a size of Earth from a distance of one billion light year away.
if we could observe this in falling matter, don't you agree that we should have the technology to see in falling matter at any other galaxy in the Universe (up to one Billion Light year away)?

If so, do you agree that the following answer for our limitation to see the in falling matter by X-ray signature is not relevant?
Yes, the X-rays were directly observed.  The matter falling in was not.  We cannot see the matter, only a general dim X-ray signature.
So, can we agree that we have never ever see any in falling matter (except of this last verification of Gas cloud in the size of the Earth which had been observed at a distance of ONE BILLION LIght year away) although we have the most advanced technology to see any in falling matter For up to one Billion light year away?
Why do you claim:
With high technology, we can see a magnesium fire in the dark from space, but that's looking down at the fire.  Looking at Sgr-A is like spotting a firefly from the vantage point on the ground with a forest between us and the firefly.  Sure, its a lot closer, but much less bright and completely obscured by the forest.
Please be aware that the Gas cloud was observed by our scientists for one full day (24 H?).
Please see the image (red arrow):
https://phys.org/news/2018-09-falling-black-hole-percent.html
It came from outside, cross several accretion discs and than even cross the alighted accretion disc (that disc surely has the Plasma - at temp of 10^9c and minimal orbital velocity of 0.3 speed of light) and then it had been swallowed by the SMBH event of horizon.
Our scientists follow the whole rote. Therefore, if we can trace a gas cloud at Earth size from a distance of one billion light year away, crossing the ultra high fire of the accretion discs (one by one) and even the most biggest fire in the inner most disc (aligned accretion disc), than you should agree that we have the technology to see also any in falling matter at any nearby galaxy including our galaxy?
Therefore the conclusion is very simple:
We didn't see any in falling matter in our galaxy - not because our technology limitation, but because there is no in falling matter.
In the same token -
We have never ever found any in falling matter in any galaxy in the Universe (except that one at one billion years away) not because our technology limitation but because there is no in falling matter into the accretion disc or from the accretion directly into the SMBH!!!

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/08/2019 06:00:31
May I ask you the same question?

I will answer that question if you answer it first.

We didn't see any in falling matter in our galaxy - not because our technology limitation, but because there is no in falling matter.
In the same token -
We have never ever found any in falling matter in any galaxy in the Universe (except that one at one billion years away) not because our technology limitation but because there is no in falling matter into the accretion disc or from the accretion directly into the SMBH!!!

I don't know if you missed it or intentionally ignored it, but the paper already addressed this:

Quote
The transient nature of the rev2659 inflow largely explains why a compelling detection has not been reported before. Isolated claims are not uncommon, however, with several single-line detections noted in the ‘Introduction’ section, and both XMM–Newton and Suzaku archival searches finding – but not discussing – transient absorption lines in the region occupied by redshifted Fe K lines.

Bring this up again and I'm going to post this quote again. The answer won't change just because you don't like it.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/08/2019 13:13:33
With regards to the technology:
Please remember that we have the technology to monitor a gas cloud at a size of Earth from a distance of one billion light year away.
We probably don't.  Yes, the article uses those words.

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if we could observe this in falling matter, don't you agree that we should have the technology to see in falling matter at any other galaxy in the Universe (up to one Billion Light year away)?
Sure, if it's an isolated 'object' like that cloud.  Usually it's more of a bleeding of material in a steady rain, like all the local black holes they see siphoning material off a companion star.  No individual 'object' is seen moving.

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If so, do you agree that the following answer for our limitation to see the in falling matter by X-ray signature is not relevant?
No.  There are no 'objects' falling into Sgr-A.  It is not well fed.  Instead of watching it eat a burger, we're watching it take in a thin stream of atoms. We can't see those individual atoms since groups of them don't move as a unit.

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Please be aware that the Gas cloud was observed by our scientists for one full day (24 H?).
Please see the image (red arrow):
The image is not a picture of PG1211+143, but PG1211+143 has a configuration something like that.

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It came from outside, cross several accretion discs and than even cross the alighted accretion disc
The article doesn't say that.  Point is that the rings are misaligned, so infalling material does not cross the more inner rings.
Quote
(that disc surely has the Plasma - at temp of 10^9c and minimal orbital velocity of 0.3 speed of light)
Perhaps, but the article doesn't say this, so you can't draw a conclusion from asserting it.

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Therefore, if we can trace a gas cloud at Earth size from a distance of one billion light year away, crossing the ultra high fire of the accretion discs (one by one) and even the most biggest fire in the inner most disc (aligned accretion disc)
Article doesn't say this.  These are your words.

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than you should agree that we have the technology to see also any in falling matter at any nearby galaxy including our galaxy?
Nearby, if the galaxy is not seen edge-on, sure.  Ours, perhaps not, because of our view being obscured.  If S2 fell in, we'd see that because suddenly it would blink out.  It apparently is bright enough to see despite all the clutter between us and it.

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Therefore the conclusion is very simple:
We didn't see any in falling matter in our galaxy - not because our technology limitation, but because there is no in falling matter.
There are no 'objects' falling in, sure.  There is a steady, albeit thin stream of matter falling in, and we 'see' the signature from that.  So in that sense, we directly see material falling into that black hole just like any other.  If we don't see that, we can't tell the black hole is there except perhaps due to lensing if it passes in front of something.

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In the same token -
We have never ever found any in falling matter in any galaxy in the Universe (except that one at one billion years away) not because our technology limitation but because there is no in falling matter into the accretion disc or from the accretion directly into the SMBH!!!
No, we see it all the time.  The stuff falling in is just not 'objects', so we can't track individual bits of it.  That's the new thing that was observed with PG1211+143, and even then, we could only track the displaced cloud due to its spectral lines being different than all the other material around it as it 'stopped' and then was pulled straight in, accelerating on the way down.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/08/2019 08:32:54
Yes, it's feasible. For reasons the others have described.
As we discuss on a gas cloud at the size of the Earth, let's verify the confidence in this feasibility:
1. "Gas can break off and collide with each other":
"The observation agrees closely with recent theoretical work, also at Leicester and using the UK's Dirac supercomputer facility simulating the 'tearing' of misaligned accretion discs. This work has shown that rings of gas can break off and collide with each other, cancelling out their rotation and leaving gas to fall directly towards the black hole."
When they say - "rings of gas can break off and collide with each other" what do they mean?
2. Same size?
If the gas cloud from one ring has different size, than it is clear that it will win and grab the other one with it.
So, what is the chance for the same size?
3. Face to face?
As each ring orbits at a different radius, what is the chance for face to face collision?
What is the chance that the inner ring will send a gas outwards, while the outer ring will send a gas cloud inwards - both at the same size and face to face?
4. Structure of the gas cloud:
If I understand it correctly a cloud is not a It is not a dense star planet moon or even rock. So cloud is based on Atoms Molecular & particles that are not too dense together (Atom to Atom). There must be a gap between the Atoms/Molecular/Particles.
Now, let's try to collide two Hydrogen gas cloud (at the same size of 1/2 Earth size) at the same size while each one is moving at a velocity of 0.3c.
What might be the outcome?
As there is a gap between the particles, what is the chance that one particle from one cloud will collide with other particle from other cloud?
Do you agree the chance for that is very low? So, there is good chance that the clouds can cross through each other without any severe impact to each other.
On the other hand. Let's assume that some particles would collide with particles from the other cloud:
What is the chance that one particle from one cloud will collide with other particle with the same mass and also face to face?
Do you agree that in order to stop at the spot of the collision, a particle from one cloud must collide with the same particle (Same size same mass) from the other cloud and the collision mast be face to face by 100%?
If they are not in the same size, or if they are not directly face to face than they should be ejected into different directions.
So, what is the chance that a particle from one cloud will collide face to face with the same particle from the other cloud?
Somehow, it seems to me that the highest chance is that they should cross each other without any significant impacts to each other.
On the other hand assuming that the gas could have the same size, it's made out of one sort of particle, and it is very dense, than don't you agree that the outcome should be like a bomb that explode to all directions.
So, if we add all the chances - it is very clear to me that the chance to have only one nice gas cloud moving directly into the center of the SMBH out of a collision - it is less than one to one million of a trillion.
This is just one section.
I would like to get your feedback before continue with the other one.

Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 12/08/2019 12:14:14
Quote
If the gas cloud from one ring has different size, than it is clear that it will win and grab the other one with it.
Not quite.

If they had exactly equal angular momentum, both would fall straight into the black hole.
If they had a 50% difference in angular momentum (with opposite sign), about 2/3 would fall straight into the black hole.

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What is the chance that the inner ring will send a gas outwards, while the outer ring will send a gas cloud inwards - both at the same size and face to face?
It is quite possible that gas clouds arrive with a highly elliptical orbit. This means that the gas moves inwards and outwards on every orbit. Due to friction within the cloud, the cloud will spread out radially, and due to gravitational dispersion it will also spread out along its orbit, starting to form an accretion disk in the orbital plane of the incoming gas cloud.

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As there is a gap between the particles, what is the chance that one particle from one cloud will collide with other particle from other cloud?
Almost none.
Although there is a gap between atoms in a gas, there are a lot of atoms in a gas cloud with half the mass of the Earth..

Once the gas turns into a plasma, we are not talking about collision of electrically neutral atoms.
We are talking about electrically charged electrons and nuclei diverting other electrically charged particles.
Effectively all the spaces disappear.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 12/08/2019 19:15:23
When they say - "rings of gas can break off and collide with each other" what do they mean?

The "breaking-off" is referring to the breaking of an accretion disk into multiple, smaller rings that are oriented at different angles to each other (caused by the Lense-Thirring effect). Gas clouds are turbulent things and they don't have well-defined boundaries. Right at the interface where two rings of gas cross, friction between the clouds would be expected to cause some of the gas to be ripped off, slow down, and falling into the hole.

Since the rest of your post has been addressed by Halc and evan_au, I will now await your answer to my earlier question: can calculations done by astrophysicists be considered trustworthy sources of evidence? You said:

Quote
It is stated clearly: "Strong magnetic pressure allows high accretion rate and inhibits disk fragmentation."
Hence, the ultra magnetic force "inhibits disc fragmentation".
Therefore, there is no possibility for: "rings of gas can break off and collide with each other" in the accretion disc.

You are citing the results of calculations as if they were absolutely factually. Your use of the phrase "no possibility" clearly demonstrates the weight of authority you give to these calculations. So are you suddenly saying that calculations are facts?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/08/2019 06:31:53
The "breaking-off" is referring to the breaking of an accretion disk into multiple, smaller rings that are oriented at different angles to each other (caused by the Lense-Thirring effect). Gas clouds are turbulent things and they don't have well-defined boundaries. Right at the interface where two rings of gas cross, friction between the clouds would be expected to cause some of the gas to be ripped off, slow down, and falling into the hole.
Let's look at the following image of the accretion disc of that far end galaxy:
https://phys.org/news/2018-09-falling-black-hole-percent.html
In the article it is stated:
"The orbit of the gas around the black hole is often assumed to be aligned with the rotation of the black hole, but there is no compelling reason for this to be the case. In fact, the reason we have summer and winter is that the Earth's daily rotation does not line up with its yearly orbit around the Sun."
So, they claim that in the accretion disc there are un aligned rings in the accretion disc and they also orbit in opposite directions.
I wonder if they really see those rings in the accretion disc, or is it one more hypothetical idea.
In order to give a confidence to the opposite orbital directions of the rings in the accretion disc they claim:
"This is particularly relevant to the feeding of suppermassive black holes since matter (interstellar gas clouds or even isolated stars) can fall in from any direction."
Hence, as stars/gas cloud might fall in different directions, they theoretically could set different orbital directions in the accretion Rings.
Based on this answer, it seems to me that they have no real prove to those Unaligned rings.
They just want to prove something, so they invent something.
Sorry - If they want to prove that there is "different angles" between the rings and thery orbit at opposite directions - they must prove it based on real rings in Astronomy.
For example let's look at Saturn's Rings:
https://en.wikipedia.org/wiki/Rings_of_Saturn 
https://en.wikipedia.org/wiki/Rings_of_Saturn#/media/File:Unraveling_Saturn's_Rings.jpg
It has several rings. However, all of them orbit at the same plane and at the same direction.
I positively sure that all the rings around all the other planets – must be fully aligned and orbit at the same direction.
The spiral disc/ring in our galaxy can also be used as a perfect example
We call it disc, but in reality it is ring as it starts from about 3KPC to about 12KPC.
There are billion of stars there.
All of them are located at the same plane and all of them orbit at the same direction.
You won't find even one star that dare to orbit at the opposite direction in that spiral disc/ring.
Why is it? How could it be that all the stars that are falling inwards to the spiral disc/ring must orbit in only one direction?
How can they assume that star "can fall in from any direction" while we see clearly that this isn't the case in the spiral disc/ring?
So, in all the available rings in astronomy, those rings are fully aligned and orbit in one direction.
We actually can also confirm the aligned idea with our own accretion disc.
We see it from the side. So, if there were unaligned rings, we could easily see them.
Do we see any unaligned rings in our accretion disc or in any nearby accretion disc?
Is it science or wishful list?
You can't assume something and we all have to accept this unreal assumption just because you are professor in a famous University. (The team, led by Professor Ken Pounds of the University of Leicester)
If you want to prove your assumption - you need to offer information/example about similar rings in Astronomy.
Can you please show even one real example for unassigned rings around any Star or SMBH?
How can you compare a ring to a solar system?
Can we assume that planets around solar system are ring???
If he can't prove the basic idea about unassigned rings in accretion disc, how can we trust the theory of collision between clouds?

As evan says, almost zero chance that any two particles (improbably small targets) will hit.
Do you mean that they can't hit each other? (That exactly what I Claim).
Let's look at the radius of hydrogen Atom:
https://en.wikipedia.org/wiki/Bohr_radius
"The Bohr radius (a0 or rBohr) is a physical constant, exactly equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It is named after Niels Bohr, due to its role in the Bohr model of an atom. Its value is 5.29177210903(80)×10−11 m.[1][note 1]"
Let me ask again:
What is the chance that one Hydrogen from one cloud will hit that Atom directly at the center (face to face)?
If they hit each other even at only 0.29177210903(80)×10−11 m from their center, than they will both ejected to different directions.
So, it is not just about the collision chance between each two atoms, but it is also about the exact face to face collision point between the atoms.

Consider a small 'cloud' of blue gas to the left and another red one to the right, about a meter apart.  Put a little portable fan in the red cloud, pointed at the blue, and turn it on.  Does the red cloud pass through the blue leaving most of the blue cloud behind before it starts to move? 
The Answer is YES.
A cloud in the Space is not as a cloud in the Earth.
It has a density. So, there are limited no. of atoms in a limited space. So there is an empty space between the atoms.
I didn't set the calculation yet, but I assume that if the radius of Hydrogen atom is 0.29177210903(80)×10−11, than in a real gas cloud the distance between two nearby atoms cloud be one million or even one billion bigger than this no.
Therefore, even if billions of atom are crossing by, there is no real confidence that they will hit that specific atom.
If nothing will hit it, it will continue with its momentum and cross the other cloud.
If something will hit it - and not at the exact center, they will be ejected to different directions.
So, I wonder what is the chance to hit an atom exactly at its center of mass.
Please also take in account that gas cloud has different particles. It might be Hydrogen Atom but also it might be different kind of atom or molecular.
Therefore, even if two different mass atoms/molecular will hit at their center (face to face), due to the mass change, one atom could grab the other one.
We need to compare it to car accident in a highway.
If two cars exactly at the same size collide with each other exactly at their center of mass, than, yes - theoretically they could stop at their current accident spot.
However, how many times we have really found that kind of scenario?
Do you agree that it should be less than one to one million?
If one car has a little difference in its mass, don't you agree that  the heavier must grab the other one with it?
If it is not directly face to face, don't you agree that they will continue to move at different directions.
So, the chance for an Atom to collide and stay at its current location, is clearly less than this kind of unique collision between same cars in highways.
It seems to me as one will claim that a collision between two cars can create instantly new truck, without any garbage around the collision point.
Let's assume that somehow due the collision we have set a new truck out of two collided cars without any garbage.
Why that new truck will move now at 90 degrees from the highway direction?
In the same token?
Let's assume that all the atoms from the two gas clouds have been collided with the same atom mass and also face to face.
How could it be that they can cross the accretion ring without any difficulty?
Did you ever try to cross a highway?
Why the aligned accretion ring does not grab that gas cloud with its plasma orbital stream?
Therefore, I still consider that the idea of forming that single gas cloud out of a single collision which is moving directly to the SMBH while crossing the rings one by one,  is very problematic.
It is clear to me that the only possibility for a gas cloud to cross the highway accretion disc is by flying above those rings.
Therefore, it isn't falling into the SMBH, but it is moving to the pole of the SMBH. Magnetic force is the only power that can set this activity. It is moving directly under the magnetic field line directly to the pole.
Hence, this gas cloud will be ejected as a molecular jet stream above the SMBH.
I really sorry that the team didn't try to trace that earth size gas cloud as it was getting closer to the SMBH pole.
They could see some sort of X-ray as the gas cloud was boosted upwards.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 13/08/2019 08:09:46
Hence, as stars/gas cloud might fall in different directions, they theoretically could set different orbital directions in the accretion Rings.
Based on this answer, it seems to me that they have no real prove to those Unaligned rings.
They just want to prove something, so they invent something.
Sorry - If they want to prove that there is "different angles" between the rings and thery orbit at opposite directions - they must prove it based on real rings in Astronomy.

There is nothing made up about this. The Lense-Thirring effect is a prediction of relativity and has been observed to be consistent with the behavior of a jet emitted by V404 Cygni: https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession#Experimental_verification

It has several rings. However, all of them orbit at the same plane and at the same direction.
I positively sure that all the rings around all the other planets – must be fully aligned and orbit at the same direction.

If you actually understood what Lense-Thirring precession was, you'd know why all of these rings and orbits don't exhibit much distortion. The Lense-Thirring effect is normally extremely weak and only manifests to a significant degree when the object in question is very massive and spinning very quickly. Those are exactly the attributes that a super-massive black hole has.

We actually can also confirm the aligned idea with our own accretion disc.
We see it from the side. So, if there were unaligned rings, we could easily see them.
Do we see any unaligned rings in our accretion disc or in any nearby accretion disc?

Whether or not this even happens with the Milky Way's SMBH depends on the nature of the accretion disk. The Lense-Thirring precession would only break the disk into multiple rings if it is orbiting closely enough to the hole (the effect is stronger the closer you get). If it's too far away, there will be no breaking in the disk. The fact that the earlier article about 99% of material getting thrown away from Sagittarius A* stated that the material in the accretion disk gets thrown off before it can get close to the hole would support that idea. Then, on top of that, these out-of-plane disks would have to be bright enough for us to directly image them. Our disk is much dimmer than many other black hole disks.

Not that any of this matters. Even if the idea of multiple rings is wrong, that doesn't invalidate our observations. We saw a gas cloud get eaten by a black hole. It happened. Complaining about it won't make it go away.

I really sorry that the team didn't try to trace that earth size gas cloud as it was getting closer to the SMBH pole.

They traced the cloud until it was gone. Did you not even read the paper?

They could see some sort of X-ray as the gas cloud was boosted upwards.

Exactly. This is how we know that it didn't get blasted out as a jet. They followed the cloud until it disappeared. So it did go into the hole.

By the way, quit ignoring this:

Quote
You are citing the results of calculations as if they were absolutely factually. Your use of the phrase "no possibility" clearly demonstrates the weight of authority you give to these calculations. So are you suddenly saying that calculations are facts?
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 13/08/2019 13:56:01
Quote
There are billion of stars there.
All of them are located at the same plane and all of them orbit at the same direction.
The spiral arms of our galaxy is like a flat disk.
We can't see the core of our galaxy clearly (due to dust), but spiral galaxies like our own have a galactic bulge which is like a swarm of angry bees around a central black hole, all moving in different directions, in different planes. The same goes for all of an elliptical galaxy (and globular clusters).

The black hole develops an accretion disk when it gets fed, and with matter arriving from different directions at different times from within the galactic bulge, a temporary accretion disk could form at any angle.

If this accretion disk runs into an accretion disk at a different angle, both will end up with less angular momentum, and both will move closer to entering the event horizon.

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Therefore, even if two different mass atoms/molecular will hit at their center (face to face), due to the mass change, one atom could grab the other one.
The velocity of these stars and gas clouds would be quite high when they approach the black hole on an elliptical orbit.

That means that any collision with a gas cloud will be quite energetic, so they are unlikely to form molecules during the collision - they are more likely to form a plasma.

A single star would probably bore a hole through a dust cloud, leaving swirls behind it.
- Provided it is well outside the Roche limit, the star will maintain its integrity
- If it is within the Roche limit, the star will be tidally disrupted, spreading gas along its orbit
- If the star is near the Roche limit, it may lose some of its outer atmosphere, but continue for another encounter on the next orbit.
https://en.wikipedia.org/wiki/Roche_limit

However, a dense gas cloud meeting another dense gas cloud will have a very energetic interaction, spread over a large volume of space.

You can see this when a SMBH jet interacts with the very thin intergalactic medium. For example, see the end of the M87 jet.
https://en.wikipedia.org/wiki/Astrophysical_jet#Relativistic_jets
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 13/08/2019 15:11:51
I feel like all of this recent talk is a distraction from the main thrust of this thread. Dave_Lev, you have complained that the scientific models we speak of are "unreal" and not "proven", yet your model has these exact problems. You have claimed that super-massive black holes can conjure mass and energy from nothingness such that they both increase their own mass and create an accretion disk with new mass at the same time. We have repeatedly told you that this violates the first law of thermodynamics and is therefore very much "unreal". You seem to want proof for everything we claim, so why is it that you do not hold yourself to the same standard and offer proof of super-massive black holes creating mass and energy from nothingness? When was a black hole ever observed to create net mass or energy? They never have been and you know it. So why is it that we have to provide proof but you don't?

You have also cited the results of calculations from a scientific paper as being absolutely factual when they agreed with your model, yet you reject the results of calculations from other scientific papers when they disagree with your model. Do you know what that is called? A double standard.

Even if every model and theory about black holes and accretion disks was wrong, it wouldn't make your model correct. One model being wrong is not evidence that a competing model is right. Each model has to stand on its own individual merits, which your model does not have because it breaks the laws of physics.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/08/2019 15:23:18
Let's look at the following image of the accretion disc of that far end galaxy:
https://phys.org/news/2018-09-falling-black-hole-percent.html
That is not an image of that 'far end galaxy'.  It's an image from a computer simulation run 15 years ago, not meant specifically to model any particular galaxy.
There are images of that galaxy, but that article didn't include one.

Quote
In the article it is stated:
"The orbit of the gas around the black hole is often assumed to be aligned with the rotation of the black hole, but there is no compelling reason for this to be the case. In fact, the reason we have summer and winter is that the Earth's daily rotation does not line up with its yearly orbit around the Sun."
So, they claim that in the accretion disc there are un aligned rings in the accretion disc and they also orbit in opposite directions.
It doesn't say that.  There is no 'the disk' if there are multiple ones.  If they're misaligned, they can't orbit in opposite directions.  East is not opposite of North.

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For example let's look at Saturn's Rings:
https://en.wikipedia.org/wiki/Rings_of_Saturn 
https://en.wikipedia.org/wiki/Rings_of_Saturn#/media/File:Unraveling_Saturn's_Rings.jpg
It has several rings. However, all of them orbit at the same plane and at the same direction.
Because it all came from the same object that was torn apart.  Mars will soon have rings of its own, and all in one plane like that.

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All of them are located at the same plane and all of them orbit at the same direction.
You won't find even one star that dare to orbit at the opposite direction in that spiral disc/ring.
Not so.  The halo objects have fairly random orbits, so about half of them orbit the opposite way as the galaxy.  It is less frequent in the galactic plane, but there are exceptions like 'complex H' which is a group of stars/objects orbiting at around 33 Kpc and going the opposite way.

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Why is it? How could it be that all the stars that are falling inwards to the spiral disc/ring must orbit in only one direction?
Or maybe that all these things of which you're so sure is just made-up assertions with no basis.

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How can they assume that star "can fall in from any direction" while we see clearly that this isn't the case in the spiral disc/ring?
Things out that far are fairly stable and don't fall in any more than Earth is expected so suddenly make a turn and drop into the sun.  If a close encounter with another mass disrupts that stable orbit, then its new trajectory will be a random one, and if it drops near Sgr-A, it might come from any random direction.  Only disrupted objects get close to it.  If their normal orbit took it close, it would already have been absorbed long ago.

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Can we assume that planets around solar system are ring???
Not even Mercury is anywhere close enough to become one.


Quote from: Halc
As evan says, almost zero chance that any two particles (improbably small targets) will hit.
Do you mean that they can't hit each other? (That exactly what I Claim).[/quote]The can and do, but the probability of a collision between two specific particles is small.

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Let me ask again:
What is the chance that one Hydrogen from one cloud will hit that Atom directly at the center (face to face)?
Let me answer again:  A small number.

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If they hit each other even at only 0.29177210903(80)×10−11 m from their center, than they will both ejected to different directions.
That's still a hit.  Momentum has been exchanged.

Quote
So, it is not just about the collision chance between each two atoms, but it is also about the exact face to face collision point between the atoms.
Are you expecting them to fuse or something?  If not, this statement is on par with your other assertions.  Fusion requires a pretty direct hit.  Not much of that takes place in the cloud collisions at the lower speeds we're considering.

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A cloud in the Space is not as a cloud in the Earth.  It has a density.
Clouds on Earth have no density.  Or maybe you meant the other way around.  Either way, that's an interesting assertion.  I see little point in continuation of this discussion for fear that others might not be able to tell the difference, as the saying goes.

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If two cars exactly at the same size collide with each other exactly at their center of mass, than, yes - theoretically they could stop at their current accident spot.
Atoms have no requirement to all stop. That would make their temperature 0, and nobody is claiming the process cools the clouds. A cloud is not one atom. If a pair of gas clouds collide, there will remain a remnant of both that has little momentum, and there will also be ejecta in all directions. The total energy can't go up or down.

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It seems to me as one will claim that a collision between two cars can create instantly new truck, without any garbage around the collision point.
Such a collision will result in slow garbage in the middle and yes, plenty more faster garbage to the sides.  It's a poor analogy because cars absorb the energy of the impacts and atoms do not.  You can't make dents in them.

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Let's assume that somehow due the collision we have set a new truck out of two collided cars without any garbage.
Why that new truck will move now at 90 degrees from the highway direction?
Because gravity pulls it that way.  These cars were not on a road, but rather in orbit.  If they hit head on in orbit and create a stopped truck, that truck will drop straight down, 90° from the direction they were moving before.  Gas doesn't behave like that.  More like a pair of water balloons colliding in low orbit creating not a bigger water balloon 'truck', but just a big splash.  Result is that more than half the water hits the Earth.  At a certain orbital altitude there is a point that water balloon collision results in less than half the water reaching Earth.

The computer simulation you link simulates exactly such a collision and demonstrates the incorrectness of many of the ideas you're asserting here.  You've obviously no background in fluid dynamics either.  You can't write such a simulation with flawed fluid dynamics.  Trivial empirical cases would falsify the validity of the simulation.

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How could it be that they can cross the accretion ring without any difficulty?
Nobody said they do.  The rings are misaligned, so they're going around the rings, not through them.

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It is clear to me that the only possibility for a gas cloud to cross the highway accretion disc is by flying above those rings.
Since it is clear to you, my instincts are to say no, but in this case I agree.  What is wrong with the world when you can't count on a rule of thumb anymore?  The cloud goes around it, and they watched it drop in, not move to the poles.  I can cross a busy highway by using an overpass (or any route that does not intersect the highway).
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/08/2019 07:12:56
You have also cited the results of calculations from a scientific paper as being absolutely factual when they agreed with your model, yet you reject the results of calculations from other scientific papers when they disagree with your model. Do you know what that is called? A double standard.
You complain about double standard, but unfortunate, that exactly what you do.
In the article which I have offered, it is stated clearly in the title:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"First detection of matter falling into a black hole at 30 percent of the speed of light"
You fully and happily accept the idea of "detection of matter falling into a black hole at 30 percent of the speed of light", but you reject the meaning of "First".
Although English is not my first language, I will try to help you by Google:
"coming before all others in time or order; earliest; 1st. his first wife."
Do you understand by now the meaning of the word "first"?
When I have asked you about it, your answer was:
I will answer that question if you answer it first.
Are we in the kindergartner?
Somehow, it seems to me that you do understand the meaning of the word "First"
However. you don't want to discuss about it as it contradicts your wishful list that we have already got many other examples for in falling matter long before this discovery. (Examples for in falling matter in M87, Andromeda galaxy, Milky way and many other long before 2018).
If this is not "double standard", than would you kindly explain to me the real meaning of this message in English?
I have no "double standard". I accept by 100% all the observations and all the laws in physics.
However, I reject the outcome of those observations.
You have stated that our science community call "modeling" - Evidence.
That by itself a fatal error.
If they are using wrong assumptions and setups in their modeling they will surly get wrong outcome.
As our scientists insist to look at all observations and discoveries through the BBT filter.
They also are using the "correct" setups and assumptions to fit the modeling into the BBT direction.
Therefore, any modeling which is based on a wishful list of setups and assumptions can't prove anything!
It seems to me that any new scientist get its diploma without the function of "eliminate BBT filter"
Hence, it is impossible mission to any scientist to look on any discovers and observations as I do.
I have no obligation to BBT or to any other theory.
If one day you will be able to eliminate the "BBT filter" (even momentarily), you would find why our Universe is so simple.
So, if you claim that you don't have a double standard - than you must except the very sad outcome that this article is the ONLY real observation for "detection of matter falling into a black hole"
In any case, I fully accept the idea that this is the first detection, but I don't agree with the message of "swallowing" due to the following:
It is also stated:
"The researchers found the spectra to be strongly red-shifted, showing the observed matter to be falling into the black hole at the enormous speed of 30 per cent of the speed of light, or around 100,000 kilometers per second."
Can you please explain the meaning of:
"showing the observed matter to be falling into the black hole"
Why not - "We have observed clearly that the matter is falling into the Black hole? "or "We saw clearly the impact of that falling in activity."?
The answer for that is given in the following:
"The gas has almost no rotation around the hole, and is detected extremely close to it in astronomical terms, at a distance of only 20 times the hole's size (its event horizon, the boundary of the region where escape is no longer possible)."
So, they didn't see it really falling into the Black hole.
They just saw that the gas cloud came very close to the BH "at a distance of only 20 times the hole's size - event horizon".
As the "event horizon" represents "the boundary of the region where escape is no longer possible", they were very sure that it must fall in.
So, they didn't see it falling in, they just assume that as it get so close the the event horizon, which represents the boundary of the region where escape is no longer possible, that gas must fall in.
I claim that this is a fatal understanding.
That gas cloud was not even close to the event of horizon.
As they see the galaxy from the top at a distance of 1 Billion light year away, they can't really monitor the height of the gas cloud from the SMBH.
If this gas cloud was just flying above the SMBH, how can the monitor few million Km, from a top view of 1 Billion light year away?
They have totally neglected the great impact of the:
"The gas has almost no rotation around the hole"
In all the modeling that I have ever seen for in falling matter, our scientists have claimed that the in falling gas must orbit around the SMBH before falling in.
So, you want us to believe in the following process:
1. There was a gas cloud outside the accretion disc
2. This gas cloud had been accreted into one of the rings in the accretion disc.
3. As it falls in, it gain high orbital velocity of 0.3 c and above. Please be aware that I have offered clear observation also for 0.85c at the innermost side of the accretion disc. It was stated that this aria is the most chaotic aria in the whole galaxy.
4. It gain high temp - of 10^9 c and be converted into Plasma.
5. It works under the ultra power of the Magnetic field (around the SMBH)
6. Then suddenly there is a collision between two nearby rings in the accretion disc.
Based on the observations, as we move closer to the SMBH, the orbital velocity of the plasma is higher.
However, our scientists want us to believe those nearby rings are moving at opposite directions, and at the same orbital velocity.
7. Somehow, due to this unreal collision - our scientists want to believe that we get a very nice gas cloud that lost completely its orbital momentum.
Is it real? Why our scientists don't offer a modeling for that?
I Have tried to find at the web a modeling for a collision. I thought that a collision between two drops could do the Job
There is no way for the drops to stop at the spot of the collision:
http://cdn.intechopen.com/pdfs/23094/InTech-3d_coalescence_collision_of_liquid_drops_using_smoothed_particle_hydrodynamics.pdf
"These authors identified two types of collisions leading to drops separation"
If you still believe that it is feasible - please show me the modeling for that.
8. Gas cloud temp - There is a big difference from gas cloud to plasma in many aspects especially in their temp. If I understand it correctly, a gas cloud has an estimated temp of less than 10^6 c. Plasma has a temp of 10^9c. So, our scientists want us to believe that the collision had also decrease dramatically the temp of the plasma in order to create a gas cloud.
9. The location of the gas cloud - Let's assume that somehow, we have got this unreal gas cloud of the collision. Don't you agree that it must be created just between the two rings? If it moves just a little bit inwards, why the inwards accretion ring will not grab it?
10. Separated rings in the accretion disc:
Quote
Let's look at the following image of the accretion disc of that far end galaxy:
https://phys.org/news/2018-09-falling-black-hole-percent.html
That is not an image of that 'far end galaxy'.  It's an image from a computer simulation run 15 years ago, not meant specifically to model any particular galaxy.
There are images of that galaxy, but that article didn't include one.
This modeling had not been confirmed by any observation. It was just a fantasy of our scientists (15 years ago) and long before understanding how real Accretion disc works.
Therefore, it was forbidden to use that fantasy.
11. Observation
Even if the idea of multiple rings is wrong, that doesn't invalidate our observations. We saw a gas cloud get eaten by a black hole. It happened. Complaining about it won't make it go away.
So, no.
The gas cloud is not an evidence for in falling matter.
We see a gas cloud that moves directly to the pole of the SMBH.
This is not a typical falling in activity as there is no orbital movment in the process of that falling in.
It is very unrealistic to belive that after all the efforts in setting the accertion disc, most of the matter in that accertion disc will be ejected outwards, while the only real observation that we see, will bypass the accretion ring/disc and fall directly into the SMBH.
This is unreal.
The answer is very simple:
As the gas cloud gets close enough to the pole it is ejected outwards - NOT inwards to the SMBH.
You can see this when a SMBH jet interacts with the very thin intergalactic medium. For example, see the end of the M87 jet.
https://en.wikipedia.org/wiki/Astrophysical_jet#Relativistic_jets
"An astrophysical jet is an astronomical phenomenon where outflows of ionised matter are emitted as an extended beam along the axis of rotation."
"One explanation is that tangled magnetic fields[2] are organised to aim two diametrically opposing beams away from the central source by angles only several degrees wide (c. > 1%)."
So it is stated that "tangled magnetic fields[2] are organized to aim two diametrically opposing beams."
Therefore, the magnetic field is a key element in astrophysical jet stream.
Our SMBH also ejects that kind of molecular jet stream.
Therefore, it should be clear to all of us that the magnetic field has a key functionality in our SMBH and at any SMBH in the Universe. I really don't understand why our scientists ignore its great impact.
This is the only explanation for a direct movement of the Gas cloud to the SMBH pole.
Not Gravity force, but Magnetic force!!!
As it gets there, it is boosted upwards.
So simple and clear.


After all of this information, let's assume that you insist to avoid "double standard" from both of us.
Do you understand that out of the whole proves for the in falling matter, we end up with only one gas cloud at a size of the Earth which is located at a distance of one Billion Light year away.

So, assuming that you also respect your request about "double standard", do you agree that we see clearly that 99% (or at least 90%) from the mass in the accretion disc is ejected outwards and we actually don't have any real observation (by X-ray) for any in falling matter in any SMBH at the whole Universe (except of that unreal falling in gas cloud)?






Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/08/2019 14:00:43
Quote from: Kryptid
You have also cited the results of calculations from a scientific paper as being absolutely factual when they agreed with your model, yet you reject the results of calculations from other scientific papers when they disagree with your model. Do you know what that is called? A double standard.
You complain about double standard, but unfortunate, that exactly what you do.
In the article which I have offered, it is stated clearly in the title:
"First detection of matter falling into a black hole at 30 percent of the speed of light"
You fully and happily accept the idea of "detection of matter falling into a black hole at 30 percent of the speed of light", but you reject the meaning of "First".
Titles are not written by scientists and often not even by the author of the article.  Example: Einstein's Theory of Special Relativity.
That said, the observation was a first, but not necessarily the first that you make it out to be:
Quote
you must except the very sad outcome that this article is the ONLY real observation for "detection of matter falling into a black hole"
There you go, bending an article (mostly from just the title) into what you want it to be.
Matter being pulled into black holes has been observed for a long time now, and all of it must move at 30% light speed at some point along the way.  Neither of those two points is a first, and you know that, but you cherry pick the assertion you want to make, from the title no less.
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I have no "double standard". I accept by 100% all the observations and all the laws in physics.
Clearly you do not.  The whole thread started off with your rejection of tidal forces, and moved on to assertion of the creation of mass in the same breath where you state that you accept the law of physics saying mass cannot be created or destroyed.  No, you reject what you don't like, despite countless posts and sites explaining it in simple terms, but you unquestioningly embrace poorly worded quotes from fringe sources if it seems to support something you're pushing.  That's the double standard.  If you accept the laws of physics, read the books about them and get educated instead of approaching the subject from deliberate ignorance.
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If they are using wrong assumptions and setups in their modeling they will surly get wrong outcome.
That's right. The wrong outcome would be a falsification of the model. It goes on all the time.
Quote
As our scientists insist to look at all observations and discoveries through the BBT filter.
There were plenty of non-BBT models. Pretty much all of them 'got the wrong outcome' you speak of above.
Many models (like the multi-ring black hole one you link) are not based on BBT since it isn't a model of the early universe.
Quote
Therefore, any modeling which is based on a wishful list of setups and assumptions can't prove anything!
You keep implying that science or models produce proof of anything.  They don't.
Quote
It is also stated:
"The researchers found the spectra to be strongly red-shifted, showing the observed matter to be falling into the black hole at the enormous speed of 30 per cent of the speed of light, or around 100,000 kilometers per second."
Can you please explain the meaning of:
"showing the observed matter to be falling into the black hole"
Why not - "We have observed clearly that the matter is falling into the Black hole? "or "We saw clearly the impact of that falling in activity."?
1: Most writers don't litter their sentences with the word 'clearly' all the time. The word means 'I unquestioningly accept this statement' as distinct from statements to be rejected (labeled 'unreal').  Usage of the words just illustrates the selection bias being employed.
2. They report on the actual observation (the red shift), not the conclusion of that observation. Your wording does not.  As for seeing the impact of that falling activity, they said they observed red shift, which could be considered the impact.  If I moved away at 0.3c, the impact on somebody's observation of me would be a red shift.
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The answer for that is given in the following:
"The gas has almost no rotation around the hole, and is detected extremely close to it in astronomical terms, at a distance of only 20 times the hole's size (its event horizon, the boundary of the region where escape is no longer possible)."
So, they didn't see it really falling into the Black hole.
They just saw that the gas cloud came very close to the BH "at a distance of only 20 times the hole's size - event horizon".
You omit the part where they observe it accelerate in.
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As the "event horizon" represents "the boundary of the region where escape is no longer possible", they were very sure that it must fall in.
So, they didn't see it falling in, they just assume that as it get so close the the event horizon, which represents the boundary of the region where escape is no longer possible, that gas must fall in. I claim that this is a fatal understanding.
The article suggests none of this. You deliberately omitted the part where they saw it accelerate in, which indeed is fatal misunderstanding on your part.
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That gas cloud was not even close to the event of horizon.
Then it could not have accelerated the way it  was observed.
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If this gas cloud was just flying above the SMBH, how can the monitor few million Km, from a top view of 1 Billion light year away?
The article explains this.  Read it.
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They have totally neglected the great impact of the:
"The gas has almost no rotation around the hole"
In all the modeling that I have ever seen for in falling matter, our scientists have claimed that the in falling gas must orbit around the SMBH before falling in.
The impact of a lack of orbital speed is that it will fall right in. Again, you misrepresent what scientists say.  Matter already orbiting a mass (like Earth) tends to continue that orbit.  Matter not orbiting will fall in, like a meteor, which almost never comes from orbit.
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So, you want us to believe in the following process:
Nobody wants you to believe anything.  They're just reporting what was observed, and its similarity to a simulation that had already run.
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1. There was a gas cloud outside the accretion disc
2. This gas cloud had been accreted into one of the rings in the accretion disc.
There is no 'the accretion disk'.  Each ring is a different accretion disk.  The cloud in question did not accrete into any of them. With every statement you demonstrate that you don't care what the article actually says.
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3. As it falls in, it gain high orbital velocity of 0.3 c and above. Please be aware that I have offered clear observation also for 0.85c at the innermost side of the accretion disc. It was stated that this aria is the most chaotic aria in the whole galaxy.
The article makes no mention of the observed speed of the any of the disk materials.  You're referencing an observation of a different object with that figure.
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4. It gain high temp - of 10^9 c and be converted into Plasma.
5. It works under the ultra power of the Magnetic field (around the SMBH)
None of this was stated. The strength of the magnetic fields in this situation is not stated.
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6. Then suddenly there is a collision between two nearby rings in the accretion disc.
Based on the observations, as we move closer to the SMBH, the orbital velocity of the plasma is higher.
However, our scientists want us to believe those nearby rings are moving at opposite directions, and at the same orbital velocity.
None of this was stated or required.  Interaction between rings (which cannot have the same orbital velocity) can occasionally generate clumps of material that is relatively stationary.  The simulation demonstrated this, meaning the mathematics of the dynamics involved is sound.
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7. Somehow, due to this unreal collision - our scientists want to believe that we get a very nice gas cloud that lost completely its orbital momentum. Is it real?
The collision and resulting cloud was observed, so it isn't unreal.  They really don't care if you believe it or not.  You're not their audience.
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Why our scientists don't offer a modeling for that?
The model is referenced at the top (and bottom) of the article.
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I Have tried to find at the web a modeling for a collision. I thought that a collision between two drops could do the Job
There is no way for the drops to stop at the spot of the collision:
Water has surface tension, so it can be done.  Ever watch them play with water drops on the ISS?
Anyway, with the gas cloud thing, energy cannot be destroyed, so much of the gas from the collision is ejected at higher speeds just like the splat of water drops hitting.  The simulation they ran took into account all the forces involved, which is a lot more complex than simulating a pair of water drops.
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8. Gas cloud temp - There is a big difference from gas cloud to plasma in many aspects especially in their temp. If I understand it correctly, a gas cloud has an estimated temp of less than 10^6 c. Plasma has a temp of 10^9c. So, our scientists want us to believe that the collision had also decrease dramatically the temp of the plasma in order to create a gas cloud.
The article did not suggest this.
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9. The location of the gas cloud - Let's assume that somehow, we have got this unreal gas cloud of the collision. Don't you agree that it must be created just between the two rings?
No.  The simulation has arrows indicating the cloud formed well away from the rings supplying the material for it.
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If it moves just a little bit inwards, why the inwards accretion ring will not grab it?
It isn't in the plane of that ring, as I stated in the prior post.
10. Separated rings in the accretion disc:
Quote from: Halc
That is not an image of that 'far end galaxy'.  It's an image from a computer simulation, not meant specifically to model any particular galaxy.
There are images of that galaxy, but that article didn't include one.
This modeling had not been confirmed by any observation.[/quote]The article reports on the actual observation that confirmed parts of the model used in the simulation. Suddenly you deny the words in the article in claiming lack of observation.  What happened to 'clearly'?
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We see a gas cloud that moves directly to the pole of the SMBH.
No such thing was observed.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 15/08/2019 21:51:08
You complain about double standard, but unfortunate, that exactly what you do.
In the article which I have offered, it is stated clearly in the title:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"First detection of matter falling into a black hole at 30 percent of the speed of light"
You fully and happily accept the idea of "detection of matter falling into a black hole at 30 percent of the speed of light", but you reject the meaning of "First".
Although English is not my first language, I will try to help you by Google:
"coming before all others in time or order; earliest; 1st. his first wife."
Do you understand by now the meaning of the word "first"?

But I do agree with the title of the article:

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First detection of matter falling into a black hole at 30 percent of the speed of light

That is the title of the article. I agree with this. On the other hand...

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First detection of matter falling into a black hole

This is not the title of the article. I do not agree with it.

However. you don't want to discuss about it as it contradicts your wishful list that we have already got many other examples for in falling matter long before this discovery. (Examples for in falling matter in M87, Andromeda galaxy, Milky way and many other long before 2018).

But the article doesn't contradict those other discoveries. Nowhere does the article say that this was the first evidence we have ever had of matter falling into a black hole. That's the important distinction.

I have no "double standard". I accept by 100% all the observations and all the laws in physics.

So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.

Which is it? You can't have it both ways.

You have stated that our science community call "modeling" - Evidence.
That by itself a fatal error.

Then you can no longer cite that article about "magnetically levitating accretion disks" because it is a model.

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If they are using wrong assumptions and setups in their modeling they will surly get wrong outcome.

Then by your own standards, your own model is wrong because because you use a "wrong assumption" (i.e. that black holes can magically introduce new mass and energy into the Universe).

As our scientists insist to look at all observations and discoveries through the BBT filter.

The Big Bang Theory has literally nothing to do with the behavior of accretion disks around super-massive black holes.

We see a gas cloud that moves directly to the pole of the SMBH.

Please quote the part of the scientific paper that states the gas cloud was observed moving towards the pole of the black hole. You actually contradict your own claim when you say:

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As they see the galaxy from the top at a distance of 1 Billion light year away, they can't really monitor the height of the gas cloud from the SMBH.

Okay, so you claim that we saw the gas cloud approach the pole of the black hole and somehow at the same time we can't see that it is approaching the pole. Right...

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This modeling had not been confirmed by any observation.

Neither has yours. Unless you know of some observation where a black hole creating new mass and energy was seen.

The gas cloud is not an evidence for in falling matter.

You are contradicting yourself again. You can't say on one hand "the gas cloud didn't fall into the hole" and on the other hand "this is the first time we have seen a gas cloud fall into a black hole".

As the gas cloud gets close enough to the pole it is ejected outwards - NOT inwards to the SMBH.

That is your hypothesis, and you already provided an experiment to test it. You stated yourself that, if it had been ejected outwards, we would have been able to continue to detect its X-ray emission. As I have stated many times before, they did follow the gas cloud and it disappeared. That means there were no longer any X-rays to detect and that the gas cloud was not ejected. So the observations have falsified your hypothesis.

EDIT: Actually, I am going to have to correct myself here. After giving the scientific paper another read, it seems that the gas cloud itself was not actually seen to disappear. The observations done by the spacecraft were done in sections. The period of observation that found the 0.3c cloud was labelled "rev2659". The cloud was seen until the end of rev2659. After that, there was a gap in the observations for a period of 4 days. When the observations were finally resumed (called "rev2661"), the gas cloud was no longer there to be observed. This could obviously imply that the cloud was consumed by the black hole, but the event itself was not actually seen firsthand. This would also explain why there were no "fireworks" seen because the actual consumption of the cloud itself was not witnessed.

8. Gas cloud temp - There is a big difference from gas cloud to plasma in many aspects especially in their temp. If I understand it correctly, a gas cloud has an estimated temp of less than 10^6 c. Plasma has a temp of 10^9c. So, our scientists want us to believe that the collision had also decrease dramatically the temp of the plasma in order to create a gas cloud.

What part of "plasma is already a gas" did you not understand the first time I told you? The Sun is mostly plasma, yet its surface temperature is only about 5,500 oC. So you don't "understand correctly". Your number is way off.

So, assuming that you also respect your request about "double standard", do you agree that we see clearly that 99% (or at least 90%) from the mass in the accretion disc is ejected outwards

Only for the Milky Way's black hole and others similar to it (for reasons that I have already brought up ad nauseum).

If you are going to claim that all matter is ejected away from black holes, you are going to have to provide the calculations that show this is what is to be expected (astrophysicists have already done their own calculations and they don't agree with your conclusion). If you had read that paper that evan_au posted a link to, you will see some of those scientist's calculations about the properties of accretion disks around black holes and they came to the conclusion that matter can enter the hole. If you think their numbers are wrong, then you need to point out the problems with their calculations. But you can't stop there. You need to actually show us what the correct calculations are and that those calculations demonstrate that the magnetic field will prevent any and all matter from entering the hole. Get to it.

and we actually don't have any real observation (by X-ray) for any in falling matter in any SMBH at the whole Universe (except of that unreal falling in gas cloud)?

There are varying degrees of observational evidence for matter falling into black holes, so to say that there is only one "real" observation depends on your definition of "real".

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I really don't understand why our scientists ignore its great impact.

They don't ignore it. They have already done the calculations and it doesn't keep all matter out of the black hole. They have calculations to support their claims. You don't.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/08/2019 02:20:05
Since I'm having trouble keeping track of all these different articles and papers, I'm going to try to compile them here. I hope this proves helpful to others as well:

Magnetically Levitating Accretion Disks around Supermassive Black Holes: https://iopscience.iop.org/article/10.1088/0004-637X/758/2/103/pdf

Accretion onto the Supermassive Black Hole in M87: https://iopscience.iop.org/article/10.1086/344504/pdf

Disk Winds, Jets, and Outflows: Theoretical and Computational Foundations: http://www.mpia.de/homes/fendt/Lehre/Lecture_OUT/pudritz.pdf

Milky Way's Black Hole a Picky Eater: https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/

The paper that the above news article is about: Dissecting X-ray-Emitting Gas Around the Center of our Galaxy: https://arxiv.org/pdf/1307.5845.pdf

First Detection of Matter falling into a Black Hole at 30 Percent of the Speed of Light: https://phys.org/news/2018-09-falling-black-hole-percent.html

The paper that the above news article is about: An ultrafast inflow in the luminous Seyfort PG1211+143: https://watermark.silverchair.com/sty2359.pdf?token=AQECAHi208BE49Ooan9kkhW_Ercy7Dm3ZL_9Cf3qfKAc485ysgAAAkUwggJBBgkqhkiG9w0BBwagggIyMIICLgIBADCCAicGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMqnabWV9m-0kKhJ7oAgEQgIIB-C_O1niF1ETvTDVF6omnqQfqvYQpfwyJhmugfTUacCtwOfhRq1QD4mN9P7aCWCRUiw-3lWDoRemwXgEdsGJAU4ziqnDh327JK1i9DwR6l-_hUz6FOnk6USyHcK8E1ZfecTtMuyzMq0d2TKyBP5tUa_HQotJnvcBRYLqextviCD55SuU__Je5qcPjZlYpBFsqoc69merRkOOKfuFzh7QuuKrT-C7WO0K-ISVcViUvRG3AHM5gg44T_WM2y64MAAfBlVgvFJPnbIthaXXFBTRgX3jfA4ivEEg248PWbsgT0_7TWNUPqSfrP2AqahQ89bVhZmGK-9j46pi04txtNHvp1IGXlmzVrZIUeQekoMs8tlgFuyVJmN4XJxIcEjI9dL8oYWEyKvpw4nYHQp6Hidw7pO5vi1NL6NUwVAMBpmDYjV0giwbf96_W7A9G7Zxhqe9m9EUN0qmrpEbrLT1nTTD7XurOmQitythuhCnTQ-rwa3phRdjNgIfTWOb2EcMylLRu6IT0nbPK-Hv9RZ0p40e8O6pYkjEaZs4dT1PcGXgUj_XcDHErOwu9ir6q3QMQ3bRRVYam2Y_STdv089AWhzICYWm-AbEo2N1beNlt5DcOGXUeMLUiS4wXQpAGNxXI40imInlsR7s0Qr4P9HPossOf31oduYbQieeh1g
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/08/2019 05:18:39
Just a brief reply about the meaning of the word "First"
Titles are not written by scientists and often not even by the author of the article.
Sorry, I don't agree with this answer.
Title is a key part of the article.
However, I fully agree with the following answer.
I do agree with the title of the article:
"First detection of matter falling into a black hole at 30 percent of the speed of light"
On the other hand...
"First detection of matter falling into a black hole
This is not the title of the article. I do not agree with it.
You actually claim that the word first is related to the velocity (0.3c) and not to the "detection of matter falling into a black hole".
I'm not expert in English, but in my language I will set it as follow (after the translation):
"Detection of matter falling into a black hole for the first time at 30 percent of the speed of light"
With regards to the speed of "30 percent of the speed of light".
Don't you agree that any in falling matter should fall in that that range of speed?
Is there any possibility to fall in to a SMBH at a velocity of 1km per hour?
However, as I have stated, your answer is still perfectly OK and unfortunately, I'm obliged to agree with you.
In any case, would you kindly summarize all the articles that confirm real observation or detection of matter falling into a black hole by X-ray at any galaxy at any speed in the whole Universe (Please - no simulation or modeling)?


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/08/2019 15:14:36
Magnetic field around the SMBH:

Disk Winds, Jets, and Outflows: Theoretical and Computational Foundations: http://www.mpia.de/homes/fendt/Lehre/Lecture_OUT/pudritz.pdf
Thanks Kryptid for this great article.
"The general stability of disk–outflow solutions is still being debated and the result may depend on the detailed assumptions about the model. In the solutions discussed here the accretion stress comes entirely from the large scale magnetic field rather than some small scale turbulence"
So, it is stated clearly that "the accretion stress comes entirely from the large scale magnetic field"
That proves that there is high magnetic force around the accretion disc.
They have even set a calculation with the impact of the external magnetic torque:
"How much angular momentum can such a wind extract from the disk? The angular momentum equation for the accretion disk undergoing an external magnetic torque may"
I have found the word "magn" (for: magnetic, magnetized, hydromagnetic,magnetosphere) 150 times!!!
In several threads I have highlighted the great impact of the magnetic filed around the SMBH.
I have also offered several articles that link the Molecular jet stream directly to the impact of the Magnetic field.
So, why our scientists insist to ignore the great impact of the magnetic field around the SMBH?
I wonder why our scientists are so surprised:
Milky Way's Black Hole a Picky Eater: https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"astronomers studying Sgr A* (the supermassive black hole at the centre of the Milky Way Galaxy) were surprised to notice that less than 1% of the gas and dust drawn into its gravitational field ever get consumed – almost everything else gets ejected."
"they found that more than 99% of the infalling material was ejected long before reaching the event horizon"
Could it be that they didn't read the article about the magnetic field around the SMBH, or could it be that in 2014 our scientists were not fully aware about this great power?
It is also stated that "less than 1%" drawn into its gravitational field.
Do you agree that:
1. Zero is also less than 1%?
2. Less than 1% is not equal to 1%?
3. More than 99% could be also 100%?
Hence, they don't have a solid proof for even 1% of in falling matter and there is a possibility that 100% from the matter in the accretion disc is ejected outwards.

In any case, I hope that by now we all agree that there is an Ultra high gravity force around the accretion disc.
It pushes the matter from the accretion disc outwards (More than 99%?).
It Boosts the molecular jet stream at almost 0.8c u upwards/Downwards (up to 27,000 light year).
If we agree for all of that, why the magnetic field can't work also below the event of horizon of the SMBH?
We have already agreed that at this aria the SMBH creates new particles. (Positive and negative).
The magnetic field has a polarity.
As it pushes the atoms (Positive) out of the accretion disc, why it can't also pull the new created positive particle deep from the event of horizon?
Is there any barrier at the event of horizon that prevent the magnetic field from do that kind of job?
Our scientists were very surprised that the magnetic field can push outwards 99% of the matter in the accretion disc.
As they didn't have a clue about the outwards flow from the accretion disc, they also have no clue about the matter flow from the SMBH. 
Now they should be one more time surprise to know that all the matter in the accretion disc is coming deep from the SMBH event of horizon.
This process doesn't contradict any law of physics
where you state that you accept the law of physics saying mass cannot be created or destroyed.
Do you still see any conterediction with law of physics?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/08/2019 23:44:09
Sorry, I don't agree with this answer.
Title is a key part of the article.

News articles are often written with simplistic or hyperbolic headlines. The more authoritative source would be the very scientific paper that the article is reporting on. Nowhere in the paper does it state that this is the first detection (be it direct or indirect) of matter falling into a black hole ever.

Don't you agree that any in falling matter should fall in that that range of speed?

No, because the velocity would be dependent on (1) how far away the gas cloud is from the event horizon (if it starts further away, it has more time to accelerate), and (2) how much angular momentum it has (if it is circling the hole just barely slower than the required velocity to keep it in orbit, it will fall much more slowly than if it had no angular momentum at all and thus fell straight in).

Is there any possibility to fall in to a SMBH at a velocity of 1km per hour?

If it was in a very slowly decaying orbit, I don't see why not.

In any case, would you kindly summarize all the articles that confirm real observation or detection of matter falling into a black hole by X-ray at any galaxy at any speed in the whole Universe (Please - no simulation or modeling)?

How do you define a "real observation"?

So, why our scientists insist to ignore the great impact of the magnetic field around the SMBH?

I already told you before that they don't ignore it. How many of my replies do you even read?

Could it be that they didn't read the article about the magnetic field around the SMBH, or could it be that in 2014 our scientists were not fully aware about this great power?

The magnetic field isn't the only component responsible for this. The very high temperature of the gas is another one (as I posted before).

Do you agree that:
1. Zero is also less than 1%?
2. Less than 1% is not equal to 1%?
3. More than 99% could be also 100%?
Hence, they don't have a solid proof for even 1% of in falling matter and there is a possibility that 100% from the matter in the accretion disc is ejected outwards.

(1) You are talking about the Milky Way's black hole specifically, so don't forget that.
(2) Of course we don't have "solid proof" that matter falls into a black hole (just as you don't have solid proof that all matter is ejected away from a black hole). What we have is evidence.

In any case, I hope that by now we all agree that there is an Ultra high gravity force around the accretion disc.
It pushes the matter from the accretion disc outwards (More than 99%?).

No, gravity does not push matter. It pulls it.

It Boosts the molecular jet stream at almost 0.8c u upwards/Downwards (up to 27,000 light year).

No it doesn't. Gravity pulls, it doesn't push.

If we agree for all of that, why the magnetic field can't work also below the event of horizon of the SMBH?

It's not that it doesn't work below the event horizon, it's that it doesn't matter because any matter inside of the horizon can't get out.

The magnetic field has a polarity.
As it pushes the atoms (Positive) out of the accretion disc, why it can't also pull the new created positive particle deep from the event of horizon?

How many times do we have to repeat our explanations over and over and over again? You can't get them out of the event horizon because that would require them to be accelerated beyond the speed of light. That cannot be done with a finite amount of energy. No magnetic field of finite strength can provide infinite energy.

Is there any barrier at the event of horizon that prevent the magnetic field from do that kind of job?

Yes, it's called the escape velocity. But there is a second component as well. The space and time inside of a black hole is so distorted that anything falling in must inevitably travel towards the singularity regardless of how much force is applied to it. I've once heard this likened to space now behaving like time. As we inevitably travel forward in time, matter inside the horizon must inevitably travel towards the singularity.

Now they should be one more time surprise to know that all the matter in the accretion disc is coming deep from the SMBH event of horizon.

No, that's no possible. Matter cannot get out of the event horizon for reasons I just stated.

Do you still see any conterediction with law of physics?

That depends. Are you still claiming that black holes create mass and energy? If your answer is still yes, then there is your contradiction. As is your claim that particles can get out of the event horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/08/2019 02:46:44
In any case, I hope that by now we all agree that there is an Ultra high gravity force around the accretion disc.
No mention of that is made in the article.  In fact, the word 'force' does not appear anywhere in it.
The black hole by itself does not generate any force at all.  A second mass is needed, and an underfed accretion disk amounts to very little mass, thus force well outside the range of 'ultra high gravity force' as you word it.
As for jets at the poles, it takes less than a Newton of force to accelerate a particle to something 0.8c over a short distance.
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It pushes the matter from the accretion disc outwards (More than 99%?).
As Kryptid points out, you have this completely backwards.  Gravity pulls matter in, just like the sun does (which also consumes less than 1% of the matter that has been drawn into its gravitational field).  The sun ejects more mass than it takes in, but gravity is the one force not particularly involved in this process.

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It Boosts the molecular jet stream at almost 0.8c u upwards/Downwards (up to 27,000 light year).
The article you quote makes no mention of that.  A different article suggested that figure for millions of years ago when much more material was falling in and a far higher percentage of matter falls to the event horizon.  You're mixing your scenarios.  If something moves at 0.8c out of the plane of the galaxy, it isn't going to stop by itself at 27,000 light years like you imply with your statement. 

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If we agree for all of that, why the magnetic field can't work also below the event of horizon of the SMBH?
We have already agreed that at this aria the SMBH creates new particles. (Positive and negative).
New particles may or may not be created below the event horizon.  It is irrelevant since they can't get out.

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Is there any barrier at the event of horizon that prevent the magnetic field from do that kind of job?
It is literally like trying to push a particle from here to 2018.  Matter just can't be made to move in that direction.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/08/2019 06:01:14
Quote
In any case, I hope that by now we all agree that there is an Ultra high gravity force around the accretion disc.
It pushes the matter from the accretion disc outwards (More than 99%?).
No, gravity does not push matter. It pulls it.
Yes, I fully agree with this answer.
Gravity does not push matter. It pulls it.
However, It was typo error.
My intention was about Magnetic force.
The correct question should be:
"I hope that by now we all agree that there is an Ultra high Gravity Magnetic force around the accretion disc."
In one hand you claim that our scientists do not ignore the magnetic force:
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So, why our scientists insist to ignore the great impact of the magnetic field around the SMBH?
I already told you before that they don't ignore it. How many of my replies do you even read?
On the other hand, why they don't consider the great impact of that magnetic force on the activity of the SMBH (Around the SMBH and especially on the Molecular jet stream and the matter ejection from the accretion disc)?
1. Molecular Jet stream
Please look again on the following image of the molecular jet stream:
https://www.thunderbolts.info/wp/wp-content/uploads/2012/08/Image169.gif
In the article it is stated:
http://www.thunderbolts.info/wp/2013/02/21/jet-streams-2/
"Recently, astronomers from the Harvard-Smithsonian Center for Astrophysics announced that our own Milky Way galaxy is expelling enormous jets of gamma rays from a putative suppermassive black hole residing in its nucleus. In 2010, twin funnels of gamma ray emissions were detected above and below the galactic plane, measuring 65,000 light-years in diameter."
It is also stated:
"The prevailing theory of “compacted gravitational point sources” exciting gas and dust as they orbit does not address the existence of collimated jets. There is only one force that can hold such a matter stream together over those distances: magnetism."
So, starting from 2010 our scientists know that there is a massive molecular jet stream above and below the accretion disc.
They also know for sure that: Magnatism is only one force that can hold such a matter stream together over those distances.
So, what they did with this critical information?
Why our scientists didn't try to calculate/evaluate the magnitude of the Magnetic field?
Why they didn't try to verify how that ultra magnetic force might impact the SMBH?
2. Outflow ejection from the accretion disc
Quote
Do you agree that:
1. Zero is also less than 1%?
2. Less than 1% is not equal to 1%?
3. More than 99% could be also 100%?
Hence, they don't have a solid proof for even 1% of in falling matter and there is a possibility that 100% from the matter in the accretion disc is ejected outwards.
(1) You are talking about the Milky Way's black hole specifically, so don't forget that.
Yes, we are talking about the Milky Way's black hole specifically - Why not?
I would like to understand how the Milky way galaxy really works.

(2) Of course we don't have "solid proof" that matter falls into a black hole (just as you don't have solid proof that all matter is ejected away from a black hole). What we have is evidence.
Of course we don't have "solid proof" (X-ray observation) that ANY matter falls into a black hole, But we have "solid proof" (X-ray observation) that more than 99% is ejected away from a black hole.
We really don't need the 100%. Don't you agree that more than 99% is high enough?
We also have a clear indication that the only force that can set this activity is Magnetic force:

Disk Winds, Jets, and Outflows: Theoretical and Computational Foundations: http://www.mpia.de/homes/fendt/Lehre/Lecture_OUT/pudritz.pdf
"The general stability of disk–outflow solutions is still being debated and the result may depend on the detailed assumptions about the model. In the solutions discussed here the accretion stress comes entirely from the large scale magnetic field rather than some small scale turbulence"
So, it is stated clearly that "the accretion stress comes entirely from the large scale magnetic field"
That proves that there is high magnetic force around the accretion disc.
Again, our scientists have found that Magnetic force is the only force that can set this stress on the accretion disc.
So, why do you insist to minimize the great impact of the magnetic field?
The magnetic field isn't the only component responsible for this. The very high temperature of the gas is another one (as I posted before).
Do you really belive that the "very high temperature of the gas" can set the outflow of more than 99% from the accretion disc?
If so, please prove it?
3. Magnetic force VS Gravity force:
We have already agreed that:
"Gravity does not push matter. It pulls it."
However with regards to magnetic force:
Do you agree by now that the magnetic force is the only force (or the main force) that can boosts that molecular jet stream up to 27,000 Light year and set the outflow of 99% of the matter from the accretion disc (All of that - against the gravity force)?
So, do you agree that the magnetic force works on the opposite direction from the gravity force and therefore: "Magnetic does not pull matter. It pushes it."
Hence, there are two main forces that works on the opposite directions:
Gravity force pulls the matter inwards to the SMBH, while Magnetic force pushes the matter away from the SMBH.
Who wins the game?
If the magnetic force pushes more than 99% from the matter in the accretion disc outwards - than it is clear that the Magnetic force is the champion!!!
If we could shut down the magnetic force, it is clear that no molecular jet stream could evolve above and below the accretion disc. It is also clear that ALL the matter in the accretion disc should fall into the SMBH.
That statment is correct all around the SMBH, in the accertion disk, below the accertion disc and even below the event horizon.
That is the most important issue in our SMBH.
4. Polarity of the magnetic field
Gravity force has no polarity. Therefore, it pulls inwards matter and antimatter.
However, Magnetic force has a polarity.
Therefore Magnetic force pushes matter (positive) outwards, and pulls antimatter (negative) inwards.
Hence, Antimatter is pulled inwards due to the magnetic force + gravity force, while matter is pushed outwards as the magnetic force is stronger than the gravity force.
5. Magnetic force impact deeper into the event of horizon:
Quote
Is there any barrier at the event of horizon that prevent the magnetic field from do that kind of job?
Yes, it's called the escape velocity. But there is a second component as well. The space and time inside of a black hole is so distorted that anything falling in must inevitably travel towards the singularity regardless of how much force is applied to it. I've once heard this likened to space now behaving like time. As we inevitably travel forward in time, matter inside the horizon must inevitably travel towards the singularity.
Yes, this answer is correct if we ignore the magnetic force.
But we can't shut down the Magnetic field.
Therefore, the magnetic force around the SMBH is very valid and we have two main proves that the magnetic force is much stronger than the gravity force of the SMBH!!!
So, if the magnetic force is so powerful, why it can't work deeper into the event of horizon?
If there is a particle exactly at the event of horizon (or below), don't you agree that it should also feel the great impact of the magnetic force?
If you still assume that the magnetic force can't work below the event of horizon - than please prove it.
6. New matter creation
New particles may or may not be created below the event horizon.  It is irrelevant since they can't get out.
So, we all agree that New particles may be created below the event horizon.
We have also found that the magnetic force is stronger than the gravity force and works on the opposite direction (based on the polarity of the particle)
So, as new particle pair is created (positive + negative) the positive is ejected outwards while the Negative is pulled inwards to the SMBH.
Therefore - This new particles creation is very important.
The new created particle is not eliminated on the moment of creation.
The ultra high magnetic force pulls the positive particle inwards and pushes the negative particle outwards at the same moment of creation.
Therefore:
This is the ultimate answer for:
why we do not see any Antimatter in our Universe.
Any antimatter/negative particle will be pulled inwards to the SMBH by the gravity force + the magnetic force.
However, any matter/positive particle will be pushed outwards as Magnetic force - Gravity force is bigger than zero.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/08/2019 10:11:19
Why our scientists didn't try to calculate/evaluate the magnitude of the Magnetic field?

Who said they didn't?

Yes, we are talking about the Milky Way's black hole specifically - Why not?
I would like to understand how the Milky way galaxy really works.

Fine. Just don't pretend that any conclusions we form about it can be applied universally to all super-massive black holes.

Don't you agree that more than 99% is high enough?

High enough for what? It certainly isn't high enough to conclude that it never eats any matter at all.

So, why do you insist to minimize the great impact of the magnetic field?

I don't know where you keep getting this impression. I have repeatedly agreed that there is a strong magnetic field present.

Do you really belive that the "very high temperature of the gas" can set the outflow of more than 99% from the accretion disc?
If so, please prove it?

It was stated in the very same article that you got those numbers from in the first place. If the material is hot enough, the average particle velocity will be higher than the local escape velocity. I also never said that it was the only contributing factor. I agree that the magnetic field takes part in it.

Who wins the game?

That depends on your distance from the hole. This is partly because gravity obeys an inverse square law while magnetism obeys an inverse cube law and partly because the fields are generated in different locations (gravity coming from the black hole and magnetism coming from the disk).

It is also clear that ALL the matter in the accretion disc should fall into the SMBH.

Except for those particles that are travelling above the local escape velocity away from the black hole.

However, Magnetic force has a polarity.
Therefore Magnetic force pushes matter (positive) outwards, and pulls antimatter (negative) inwards.
Hence, Antimatter is pulled inwards due to the magnetic force + gravity force, while matter is pushed outwards as the magnetic force is stronger than the gravity force.

These statements are filled with misconceptions:

(1) The polarity of a magnetic field is north and south, not positive and negative (this is important to distinguish it from an electric field).
(2) Magnetic fields neither attract nor repel electric charges. Magnetic fields have no effect at all on stationary electric charges. All magnetic fields do is deflect the path of a moving electric charge at an angle dependent upon the direction of the field lines and the polarity of the electric charge: https://en.wikipedia.org/wiki/Lorentz_force#Trajectories_of_particles_due_to_the_Lorentz_force
(3) Matter is not "positive" nor is antimatter "negative". It depends on the particle in question. Electrons have a negative electric charge while their antimatter counterparts have a positive electric charge. The opposite is true for protons, which are positively-charged whereas antiprotons are negatively-charged. Neutrons and anti-neutrons are both neutral.

Yes, this answer is correct if we ignore the magnetic force.

It's correct even when we consider the magnetic force. You can't take things out of the event horizon. The extreme curvature of space-time won't allow it. No amount of force will do it.

Therefore, the magnetic force around the SMBH is very valid and we have two main proves that the magnetic force is much stronger than the gravity force of the SMBH!!!

Again, that depends on where you are and it also doesn't matter how strong the magnetic field is anyway.

So, if the magnetic force is so powerful, why it can't work deeper into the event of horizon?
If there is a particle exactly at the event of horizon (or below), don't you agree that it should also feel the great impact of the magnetic force?
If you still assume that the magnetic force can't work below the event of horizon - than please prove it.

It isn't that the magnetic field is no longer present there, it is that space and time have changed in such a way that movement towards the outside of the horizon is no longer possible: http://www.jimhaldenwang.com/black_hole.htm

Quote
Notice how the minus sign has moved from the t coordinate to the r coordinate.  This means that inside the event horizon, r is the timelike coordinate, not t.  In relativity, the paths of material particles are restricted to timelike world lines.  Recall the discussion of timelike separation earlier in this paper (2).  It is the coordinate with the minus sign that determines the meaning of "timelike."  According to relativity, inside a black hole time is defined by the r coordinate, not the t coordinate.  It follows that the inevitability of moving forward in time becomes, inside a black hole, the inevitability of moving toward r = 0.  This swapping of space and time occurs at r = 2M.  Thus, r = 2M marks a boundary, the point where space and time change roles.  For the observer inside this boundary, the inevitability of moving forward in time means that he must always move inward toward the center of the black hole at r = 0.  All timelike and lightlike world lines inside r = 2M lead inevitably to r = 0 (the end of time!)  Because it is not possible for any particle or photon inside r = 2M to take a path where r remains constant or increases, the boundary r = 2M is called the event horizon of the black hole.  No observer inside the event horizon can communicate with any observer outside the event horizon.  The event horizon can be thought of as a one-way boundary.

This is the ultimate answer for:
why we do not see any Antimatter in our Universe.

Except that it isn't, because black holes don't treat matter differently from antimatter. An antimatter particle is just as likely to move away from the black hole as a matter particle.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 17/08/2019 10:25:55
A paper published last week tracks the orbit of a superluminous star S2 (https://en.wikipedia.org/wiki/S2_(star)) around the Milky Way's black hole.
By using the red shift of spectral lines from this star, they have been able to:
- Plot its orbit in 3 dimensions
- Determine that at closest approach it travels at almost 2% of c (even though it's distance is still more than 1000x the black hole radius
- Confirm Einstein's prediction of gravitational redshift in the vicinity of a black hole

With further observations, they hope to estimate the density of matter in the vicinity of the Milky Way's black hole
- Astronomers estimate that there are hundreds or thousands of stars and gas clouds orbiting in this region of space
- These stars are not bright enough to be visually identified through the dust near the center of the galaxy
- They could add mass to an accretion disk at any time (provided the temperature is not too hot, which it is at present)
- This hidden mass should perturb the orbit of S2, which should (eventually) be measurable. 

Quote
In any case, I hope that by now we all agree that there is an Ultra high gravity force around the accretion disc.
It pushes the matter from the accretion disc outward
I see that you have subsequently said that you made an error here. But just to clarify...
I agree that the gravitational field around the SMBH is strong.
But look at the orbits of the stars around the black hole - the stars go faster as they get closer, and slow down as they move away.
- The SMBH gravity is attractive, not repulsive
- And it is attractive all the way down to 1000 x event horizon radius
- General Relativity states that it will continue to be attractive all the way down to the event horizon
- No-one really knows what happens inside the event horizon, but General Relativity is interpreted to mean that it continues to be attracted all the way down to the central singularity

Quote
There are billion of stars there.
All of them are located at the same plane and all of them orbit at the same direction.
Look at the plotted orbits of bright stars in the Youtube video above.
- They don't orbit in nearly the same plane - it looks pretty random.
- You can't tell the actual direction without actually knowing the red shift of each star
- The average angular velocity of all stars is much less than you get by adding up the magnitude of the angular momentum of each star
- That means if these stars interact via some dissipative effect (eg an accretion disk, rather than gravity), then they will move closer to the black hole (not farther away).

Quote
and we actually don't have any real observation (by X-ray) for any in falling matter in any SMBH at the whole Universe (except of that unreal falling in gas cloud)?
Here is yet another case where matter fell into a massive black hole, from a tidally disrupted star.
- A supernova survey picked up the tidally disrupted star
- Follow-up observations found a white dwarf orbiting the black hole, near the Innermost Circular Stable Orbit (ISCO)
- The white dwarf star will follow the tidally disrupted star into the event horizon, probably within a century or so.
See: https://www.sciencedaily.com/releases/2019/01/190109184731.htm

The ISCO is about 3 x the event horizon radius - any matter within this radius will end up falling into the event horizon.
- At about 1.5 x the event horizon radius is the photon sphere - any light within this radius will end up falling into the event horizon
- So high-speed X-Ray telescopes often detect blobs of matter falling into black holes
- Where the following article talks about "unstable" orbits, it means that the matter will fall into the black hole under the black hole's gravity (although it omits to mention that - they thought it would be obvious)
- Where it mentions "unstable" orbits, it does not mean that the matter will be blown away from the black hole by the black hole's gravity
https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit

Quote
If the magnetic force pushes more than 99% from the matter in the accretion disc outwards - than it is clear that the Magnetic force is the champion!!!
This fallacy is called a "false dichotomy".

In fact, temperature is another factor, and increased temperature always produces increased pressure, and it is high temperatures that are quoted as pushing away the 99% of matter.

Quote
However, Magnetic force has a polarity.
Therefore Magnetic force pushes matter (positive) outwards, and pulls antimatter (negative) inwards.
Matter does not work the way you imagine.
- Electrons (negative) are matter
- Protons/nuclei (positive) are matter
- Antimatter makes up a very small fraction of the mass of our galaxy

Magnetic forces do not work the way you imagine.
- Magnetic fields don't actually attract or repel charged objects (matter or anti-matter): Charged particles circle around the magnetic field
- The "north" and "south" beams are both made of balanced positive and negative particles, so it will be an electrically neutral plasma
- Gamma rays are electrically neutral, and cannot be accelerated by a magnetic field. They jets emit electromagnetic energy at all wavelengths (including gamma rays), but you would not describe them a gamma ray jets.

What repels matter is when charged particles are trapped in the magnetic field
- the magnetic field is twisted by the magnetic field of the rotating accretion disk
- The magnetic fields stretch and then "reconnect", releasing a burst of particles at high speed. This could explain pulsed beams. We can study this effect close-up in Coronal Mass Ejections from the Sun.
- Computer modelling also show that magnetic fields can collimate about 10% of the infalling accretion disk in a continual stream to form a jet.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/08/2019 13:06:26
1. Molecular Jet stream
Please look again on the following image of the molecular jet stream:
https://www.thunderbolts.info/wp/wp-content/uploads/2012/08/Image169.gif
In the article it is stated:
http://www.thunderbolts.info/wp/2013/02/21/jet-streams-2/
"Recently, astronomers from the Harvard-Smithsonian Center for Astrophysics announced that our own Milky Way galaxy is expelling enormous jets of gamma rays from a putative suppermassive black hole residing in its nucleus. In 2010, twin funnels of gamma ray emissions were detected above and below the galactic plane, measuring 65,000 light-years in diameter."
The article seems to have been written by somebody like you with a personal agenda. This one is obviously in denial of the existence of black holes. It calls the black hole 'putative' and the result of irrational manipulation of the equations. Interesting choice of authors for getting your info about black holes. It makes zero reference to 'molecular jet stream'. It reports that the jets are gamma rays. Gamma rays are light, and light beams are not held together with magnetism.

The article is pushing the concept of electrical currents between linearly arranged galaxies, with streams of electrons (held together with magnetism) passing from one galaxy to the next to form a circuit.

The image (an artist image, not an empirical one) comes from https://news.harvard.edu/gazette/story/2012/06/a-milky-way-cooling-its-jets/
The actual finding come from The Astrophysical Journal, (Finkbeiner, Su) but I have no link.
This article is closer to what was actually seen and posited concerning the clouds and jets and such. Again, no mention of molecular streams. This article estimates about 10% of the matter pulled in gets flung back out, and 90% falls into the black hole.

Quote
So, starting from 2010 our scientists know that there is a massive molecular jet stream above and below the accretion disk
As for that, or the 'Ultra high Gravity Magnetic force around the accretion disc' that you speak of, the article quotes the scientist concerning the power of the jets:
Quote from: Finkbeiner
As far as jets go, this is a faint jet. This is a pathetic jet.
This is a quote from the actual scientist, not the article author.

Quote
So, do you agree that the magnetic force works on the opposite direction from the gravity force and therefore: "Magnetic does not pull matter. It pushes it."
Then why do my magnets hold my kids' artwork to the fridge instead of being repelled by it.
Quote
However, Magnetic force has a polarity.
Therefore Magnetic force pushes matter (positive) outwards, and pulls antimatter (negative) inwards.
Ah, so it works because my fridge is antimatter.
Do you know nothing of electromagnetism?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/08/2019 19:26:49
Quote
Why our scientists didn't try to calculate/evaluate the magnitude of the Magnetic field?
Who said they didn't?
So can you please offer the calculation /estimation for the magnetic force?
What kind of magnetic force can boost that kind of molecular jet stream?
Please remember there are about 10,000 solar mass in that beam, the molecular is boosted at a speed of 0.8c to a high of 27,000 light years away from the accretion disc.
If we know the level of the magnetic field around the SMBH, would you kindly offer it?

(gravity coming from the black hole and magnetism coming from the disk).
Why are you so sure that magnetism is coming from the disk?
Do we really know how SMBH works?
Do we know if the matter in the that SMBH rotates or not?
It seems to me that the accretion disc by itself can't produce the requested magnetism level.
In order to set magnetism, we normally need a rotational core. For example, the Earth produce magnetism due to its core rotation. Therefore, I'm not sure that an orbital rotation disc by itself can create any significant magnetism or even any sort of magnetism. If you have an example that supports your claim - please offer it.
In any case, even if it the accretion disc can produce some sort of magnetism it surly might be too weak to support that kind of molecular jet stream of 10,000 solar mass up to 27,000 light year and that is boosted at  0.8c.
Don't forget that the total mass in the accretion disc is only three sun mass.
I see it differently.
The 4 millions solar mass at the core of the SMBH sets the magnetic force.
It rotates (as the core of the earth) and sets the ultra magnetic force around the SMBH.
Therefore, A SMBH with an accretion disc will generate the same magnetism amplitude as identical SMBH without any accretion disc.
However, the accretion disc has a key function. It acts as a rotor under magnetic field (generator).
As it orbits under the magnetic field of the SMBH, an Ultra high current is flowing in its matter.
This increases it's temperature and converts it into plasma.
Therefore, we call the matter in the accretion disc - plasma (it gets to ultra high temp of 10^9). The orbital velocity by itself can't generate that kind of temp.
Without the magnetic field of the SMBH, there will be no current in the accretion disc.
Without the current flow, it might be an impossible mission to achieve that high level of temperature
Actually, we can compare it to plasma that we create in order to cut metal
https://www.thenakedscientists.com/forum/index.php?topic=75495.600#quickreply
"Plasma cutting is a process that cuts through electrically conductive materials by means of an accelerated jet of hot plasma."
"OSHA recommends a shade 8 for arc current less than 300 A"
So, we are using very high current to set the plasma. In the same token, the ultra high current in the accretion disc converts the matter in the accretion disc into plasma.
I wonder why our scientists didn't even thought about it.
However, this high current sets an Electric field.
So, the contribution of the accretion disc is that it creates an Electromagnetic field around the SMBH.
(1) The polarity of a magnetic field is north and south, not positive and negative (this is important to distinguish it from an electric field).
As I have stated, around the SMBH there is electromagnetic force. Therefore it is a combination of magnetic and electric fields.
Magnetic fields have no effect at all on stationary electric charges.
A new created pair of particle must move at ultra high speed.
Therefore, we can't consider them as stationary electric charges.
We must evaluate them based on their speed in the first moment of creation under the electromagnetic fields and Lorentz force.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 18/08/2019 00:25:29
Quote from: Dave Lev
it seems to me that the accretion disc by itself can't produce the requested magnetism level.
Computer modeling shows that it can.
Evidently, the physicists' computer knows more about magnetic fields than you do.

Quote
The 4 millions solar mass at the core of the SMBH sets the magnetic force.
In saying that the SMBH generates a magnetic field, you may be confusing the "No Hair Theorem".
This states that the only parameters of a black hole that are externally visible are:
- Mass (already measured for the Milky Way SMBH)
- Angular momentum (not yet measured for Milky Way SMBH, but they have hopes)
- Electrical charge (not yet measured for Milky Way SMBH, but expected to be very close to 0)

Electrical charge is a monopole, and so can come from a singularity.
- As far as we know, magnetic fields do not have a monopole component, and so cannot come from a singularity. There has been no verifiable detection of magnetic monopoles to date.
See: https://en.wikipedia.org/wiki/No-hair_theorem

Quote
In order to set magnetism, we normally need a rotational core. For example, the Earth produce magnetism due to its core rotation.
Although something similar to Earth's oscillating magnetic field has been recreated in the lab (using liquid sodium), the exact mechanism for the Earth is still under investigation.

Earth has a solid inner core, and a liquid outer core, made of metallic nickel/iron. The magnetic field originates in the liquid outer core, where physical movement and circulation is possible, creating a dynamo. The central solid core does not create a dynamo because movement is not possible in a solid.

Similarly, the magnetic field of a galactic SMBH originates in the plasma of the surrounding accretion disk where physical rotation movement and circulation is present (due to Kepler's laws), creating a dynamo. The central black hole does not create a dynamo as matter is on a 1-way path to the singularity, and circulation is impossible.
 
See: https://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Physical_origin

Quote
As (the accretion disk) orbits under the magnetic field of the SMBH, an Ultra high current is flowing in its matter.
This increases it's temperature and converts it into plasma.
I agree that electrical/resistive heating will occur in the accretion disk.
- But magnetic fields only induce electrical currents in conductors
- And electrical currents only flow in conductors - and cool molecular gas clouds are not electrical conductors.
- You need a plasma for magnetic heating.
- More effective mechanisms for heating the accretion disk up to plasma temperatures are:
       + matter coming into the accretion disk at different speeds and different angles. Collisions/Friction causes heating.
       + Even when the accretion disk has settled into a single plane, there is differential rotation of the accretion disk at different distances from the black hole. Friction causes heating

Quote
and we actually don't have any real observation (by X-ray) for any in falling matter in any SMBH at the whole Universe
Listen to an interview with one of the leaders of the Event Horizon Telescope.

In the 1990s (when he was a Phd candidate) he was involved in radiotelescope observations of Sagittarius A*.
- From those observations, they were able to deduce that the Milky Way SMBH is currently consuming about 10-8 solar masses per year (compared to 1 solar mass per year for quasars). At present, our SMBH is on a starvation diet.
- With further observations, they were able to determine that microwave intensity from the accretion disk drops off at shorter than 1mm wavelength - and these wavelengths are generated by matter very close to the event horizon.
- It is only the magnifying effect of the black hole's gravity that allows us to observe the black hole's "shadow", using a millimeter-wave telescope the width of the Earth.
Listen, starting at 32 minutes, for 5 minutes: https://omegataupodcast.net/320-the-event-horizon-telescope/
...but the whole podcast is interesting...
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/08/2019 01:52:36
So can you please offer the calculation /estimation for the magnetic force?
What kind of magnetic force can boost that kind of molecular jet stream?

From "Disk Winds, Jets, and Outflows: Theoretical and Computational Foundations", it's at least 1 kilogauss:

Quote
The recent direct detection of a rather strong of a true disk field of strength 1 kG at 0.05 AU in FU Ori, provides new and strong support for the disk wind mechanism (Donati et al. 2005).

Why are you so sure that magnetism is coming from the disk?
Do we really know how SMBH works?
Do we know if the matter in the that SMBH rotates or not?

I know you are not going back over this again after I have already explained it to you so many times already...

Yes, the black hole spins. No, the black hole does not have significant charge. Don't you ask me how we know again, because I have already answered that several times now. I'm not going to give you a new answer. Go look up the old one if you've forgotten already.

In order to set magnetism, we normally need a rotational core. For example, the Earth produce magnetism due to its core rotation. Therefore, I'm not sure that an orbital rotation disc by itself can create any significant magnetism or even any sort of magnetism. If you have an example that supports your claim - please offer it.

For someone who seems to like magnetism so much, you sure have a poor understanding of how it works. Magnetism can be generated by spinning, electrically-conducting fluids. The liquid metal in the Earth's core is rotating around the inner core. This is what generates the magnetism: https://en.wikipedia.org/wiki/Dynamo_theory The exact same kind of thing happens in an accretion disk because it too is a spinning, electrically-conducting fluid (a plasma, in this case).

The 4 millions solar mass at the core of the SMBH sets the magnetic force.
It rotates (as the core of the earth) and sets the ultra magnetic force around the SMBH.

No.
It.
Won't.

The orbital velocity by itself can't generate that kind of temp.

No one said that the orbital velocity alone is what did it. Friction and compression have a lot to do with it too.

Without the magnetic field of the SMBH, there will be no current in the accretion disc.

Wrong. There is no magnetic field of the black hole. Ask me how I know and I'll tell you to go back and read the numerous previous instances of where I explained it.

As I have stated, around the SMBH there is electromagnetic force. Therefore it is a combination of magnetic and electric fields.

It is still incorrect to say that it attracts or repels electric charges. Magnetic fields don't do that.

We must evaluate them based on their speed in the first moment of creation under the electromagnetic fields and Lorentz force.

Evaluate all you want to. That still won't make matter come out of the black hole with antimatter falling in instead. That isn't how electromagnetic fields affect them.

By the way, I'm still waiting for you to address these:

Are you still claiming that black holes create mass and energy? If your answer is still yes, then there is your contradiction.
So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.

Which is it? You can't have it both ways.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/08/2019 16:40:37
The figures you are quoting are wrong.  Yes, there was an article mentioning a figure of 10000 solar masses, but that wasn't a reference to said jet stream nor any current state of Sgr-A.  A jet stream going at 0.8c isn't going to stop at 27k light years.  The jet stream of our galaxy has instead been described as faint and pathetic with negligible matter.
Said 10000 solar masses has never been measured.
I have already introduce this article.
However, let me do it again for you.
https://www.cfa.harvard.edu/news/2012-16
"The newfound jets may be related to mysterious gamma-ray bubbles that Fermi detected in 2010. Those bubbles also stretch 27,000 light-years from the center of the Milky Way. However, where the bubbles are perpendicular to the galactic plane, the gamma-ray jets are tilted at an angle of 15 degrees. This may reflect a tilt of the accretion disk surrounding the supermassive black hole.
"Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required"
Do you still think that this 10,000 solar mass is negligible matter?
Our scientists have stated clearly in one of the article that I have offered that magnetism is the only force that can boost that kind of molecular jet.
So, why we do not find what kind of magnetism force is needed for that jet stream?
From "Disk Winds, Jets, and Outflows: Theoretical and Computational Foundations", it's at least 1 kilogauss:
Based on this data, the requested magnetism for the accretion disc is 1 kilogauss.
Do we really think that this magnetism can set that kind of molecular jet stream?
With regards to the accretion disc magnetism:
Quote
it seems to me that the accretion disc by itself can't produce the requested magnetism level.
Computer modeling shows that it can.
Evidently, the physicists' computer knows more about magnetic fields than you do.
So, you claim that the accretion disc can produce the magnetism which is needed to its function.
This is really big enigma for me.
Let me tell you a story from the time that I was teaching Electricity.
One of my student came with a brilliant idea to invent a gen-motor that should work without any electricity source.
The idea was that the rotor of the Motor will also be used to generate current which will be used for the motor.
Based on his simulation, if we ignore the friction, it should work.
I have told him that it won't work as the his Gen-motor will never be able to produce enough current that is requested to keep its rotation.
In the same token, it is quite clear to me that the accretion disc can't produce the magnetism that is need for its requirement.
The magnetism must come from other source.
That is my point of view.
I really don't understand how a thin disc which orbits around a SMBH can produce any sort of magnetism.
Would you kindly share with me those Simulations/Computer modeling?
In any case, even if it produces some magnetism, it seems to me that it surly can't be enough to boost the 10,000 solar mass at ultra high velocity into those molecular jet stream.
As the magnetism field that the accretion disc might generate is very weak (especially - for the jet stream) it is clear that the main source for the magnetism must come for the SMBH itself.
Although something similar to Earth's oscillating magnetic field has been recreated in the lab (using liquid sodium), the exact mechanism for the Earth is still under investigation.
So, we don't know for sure how the magnetism works for Earth, while we are living on the surface of the Earth.
In our vision we see hot layers deep into the center of the Earth that orbits around a metal core .
That might be correct or incorrect.
In the same token, we really don't know the structure of the matter deep into the SMBH.
There is a possibility, that it also has layers that rotates around some core.
That theoretically can generate Magnetic field as the Earth does but much more stronger.

There has been no verifiable detection of magnetic monopoles to date.
See: https://en.wikipedia.org/wiki/No-hair_theorem
"The no-hair theorem states that all black hole solutions of the Einstein–Maxwell equations of gravitation and electromagnetism in general relativity can be completely characterized by only three externally observable classical parameters: mass, electric charge, and angular momentum."
In this article they discuss about a BH.
BH isn't a SMBH. It is totally different.
If you try to gain data from BH and use it for SMBH than it is a sever mistake.
Do you agree that only SMBH can hold Galaxy while there is no galaxy around any BH?
We call them both Black Holes. However, the characteristics of SMBH might be totally different form those of BH.
As I have stated, there is good chance that deep into the SMBH there are layers that orbit around a core and generate the requested ultra high magnetic field.
This magnetic field affects the accretion disc and the molecular jet stream.
So please, try to offer dedicated articles about a SMBH and not about BH.
They are totally different!

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/08/2019 18:42:07
Do we really think that this magnetism can set that kind of molecular jet stream?

You sound doubtful. What are those doubts based on, exactly? Don't say something generic like, "It doesn't seem like enough". Have you actually done some math that shows that measurement to be wrong? By the way, where did you get that number of 0.8c for the relativistic jet from a super-massive black hole?

But another point is that the measurement of 1,000 Gauss may not apply to all accretion disks (or even to all areas in that particular example). Closer in, the magnetic field could be much stronger (the measurement was made in a part of the disk about 7.5 million kilometers from the center of the disk). This article estimates field strengths of up to 10,000 Gauss for NGC 4258 right at the event horizon: https://arxiv.org/pdf/0909.1207.pdf

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It is shown that the second model is preferable for estimating the magnetic field in NGC 4258. For estimations we used the standard accretion disk model assuming that the same power-law dependence of the magnetic field follows from the range of the optical emission down to the horizon. The observed optical polarization from NGC 4258 allowed us to find the values 103 − 10[/sup]4[/sup] Gauss at the horizon, depending on the particular choice of the model parameters.

The Large Hadron Collider has magnets with a strength of 80,000 Gauss and they are capable of accelerating protons to more than 99.9% the speed of light. So if we are being generous, perhaps we can say that the field in the accretion disk of some of the more powerful active galactic nuclei have a strength somewhere in the 10,000 to 80,000 Gauss range. That should give you what you need for accelerating protons (which are the major mass component of a cloud of hydrogen plasma anyway) very close to the speed of light. Now that you have these numbers, what do you propose to do with them?

So, you claim that the accretion disc can produce the magnetism which is needed to its function.
This is really big enigma for me.
Let me tell you a story from the time that I was teaching Electricity.
One of my student came with a brilliant idea to invent a gen-motor that should work without any electricity source.
The idea was that the rotor of the Motor will also be used to generate current which will be used for the motor.
Based on his simulation, if we ignore the friction, it should work.
I have told him that it won't work as the his Gen-motor will never be able to produce enough current that is requested to keep its rotation.
In the same token, it is quite clear to me that the accretion disc can't produce the magnetism that is need for its requirement.

This a false analogy. What your student was talking about was a perpetual motion machine of the third kind (the kind that generates enough of its own power to power itself). Accretion disks do not "power" themselves: they are "powered" by gravity. The fact that they generate a magnetic field is just a consequence of the fact that they are both very hot and spinning. A plasma cloud does not require magnetism in order to enter a decaying orbit around a black hole. Anything with both mass and angular momentum can do that.

I really don't understand how a thin disc which orbits around a SMBH can produce any sort of magnetism.

Maybe if you actually read the replies sent to you, you'd understand. Go look at the dynamo thing I talked about earlier.

In any case, even if it produces some magnetism, it seems to me that it surly can't be enough to boost the 10,000 solar mass at ultra high velocity into those molecular jet stream.

80,000 Gauss can accelerate protons to almost the speed of light in the LHC. Would that not be enough for your tastes?

As the magnetism field that the accretion disc might generate is very weak (especially - for the jet stream) it is clear that the main source for the magnetism must come for the SMBH itself.

No matter how many times you say it, super-massive black holes are not going to magically sprout magnetic fields because you want them to have them. The fact that you have consistently refused to accept this clearly demonstrates that you do not accept all of the laws of physics. Electrically-neutral black holes cannot have magnetic fields. The no-hair theorem forbids this.

There is a possibility, that it also has layers that rotates around some core.
That theoretically can generate Magnetic field as the Earth does but much more stronger.

The only way that rotation in a black hole can generate a magnetic field is if the hole has a net electric charge. The no-hair theorem isn't invalid just because you don't like or understand it. In order for there to be some kind of complex internal structure inside of black holes that can produce a dynamo, whatever it is made of would have to be capable of withstanding the enormous crushing forces present there. No known form of matter can endure it. So what is your evidence for such a form of super-strong matter?

But let's say that you are right and super-massive black holes do, in fact, have very powerful intrinsic magnetic fields. If it was generated by a mechanism similar to that of the Earth, then you would expect the field lines to look similar as well. That is, the field lines would converge at the poles of the black hole just as they do on Earth, which draws charged particles towards the poles, not away from the poles. That's why we have auroras near the Earth's poles. If the magnetic field lines pushed those particles away in the form of jets, they wouldn't come down to Earth and make auroras. They would move away from the Earth instead. So you would have the same thing happen with a black hole. Particles are focused towards the poles, where they are consumed as soon as they follow those field lines down through the event horizon.

BH isn't a SMBH. It is totally different.

Super-massive black holes are black holes. That's like saying that large birds are not birds. The only thing that makes them what they are is their great mass. They obey all of the same physics as black holes because they are black holes. The no-hair theorem applies. You can't dance around that because you don't like it. The fact that thought that super-massive black holes were "something different" from black holes just further demonstrates your stubborn unwillingness to learn about the very subject matter that you claim to know more about than leading scientists. Until you cure that ignorance problem, you will never develop a working model.

And quit ignoring these questions. You expect me to answer your questions, so why won't you answer mine?

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So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.

Which is it? You can't have it both ways.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/08/2019 20:51:38
The figures you are quoting are wrong.  Yes, there was an article mentioning a figure of 10000 solar masses, but that wasn't a reference to said jet stream nor any current state of Sgr-A.  A jet stream going at 0.8c isn't going to stop at 27k light years.  The jet stream of our galaxy has instead been described as faint and pathetic with negligible matter.
Said 10000 solar masses has never been measured.
I have already introduce this article.
However, let me do it again for you.
"Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required"
Do you still think that this 10,000 solar mass is negligible matter?
I never said 10,000 solar masses was negligible matter.  I left my quote up there.  I quoted a scientist's description (yes, the scientist from that very article) of the jet from Sgr-A as being pathethitic and negligible, and I happen to agree with that assessment. The large mass figure is what one might have to drop into the black hole to fuel a more active jet. A proper wording would have been a rate of consumption instead of a total mass,. Is it 10,000 solar masses since the birth of the galaxy?  That would still be pathetic since one would have to question how it got to a mass of 4 million if you only fed it 1/400th of that.  So 10,000 per day?  Per second?  The figure Evan quotes is over 14 orders of magnitude lower than those.  Hence my saying that your figures are wrong, apparently by at least 14 zeros.
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it seems to me that it surly can't be enough to boost the 10,000 solar mass at ultra high velocity into those molecular jet stream.
Did you actually read the article you linked?
Quote from: Finkbeiner
"Shoving 10,000 suns into the black hole at once would do the trick"
He's suggesting dropping the large mass mass into the black hole (rate not specified), not jetting that kind of mass outward.  No estimate of the mass of the jet is given in the article.

You claim to have been a teacher, but lack any reading comprehension skills.
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In our vision we see hot layers deep into the center of the Earth that orbits around a metal core.
For instance, do you know what the word 'orbit' means?
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In this article they discuss about a BH.
BH isn't a SMBH. It is totally different.
Further embarrassment.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 19/08/2019 13:52:08
Some more information about magnetic fields:

You can calculate the force that a magnetic field has on a charged particle using the following equation:

F = qvBsinθ, where

“F” is the resulting force in newtons
“q” is the charge in coulombs
“v” is the velocity in meters per second
“B” is the magnetic field strength in teslas (1 tesla = 10,000 gauss)
“sinθ” is the angle between the direction of the particle’s motion and the direction of the magnetic field lines

When we consider moving a proton across a magnetic field at an angle of 90 degrees to the field line orientation (in order to maximize the force), two things stand out. (1) A strong magnetic field can apply a large force to a slow-moving proton (doubling the field strength doubles the force), but just as importantly (2) A weak magnetic field can apply a large force to a fast-moving proton (doubling the speed doubles the force). So you don’t need an especially powerful field if the proton is already moving very quickly. The particles in an accretion disk near the event horizon are indeed moving very quickly because the disk is very hot.

For hydrogen plasma at a temperature of 1 million degrees Celsius, the average speed of the protons is about 146,000 m/s. For 10 million degrees, it’s around 460,000 m/s. 100 million is 1,455,000 m/s and 1 billion is 4,600,000 m/s. These numbers were calculated with this: https://www.omnicalculator.com/physics/particles-velocity#average-velocity-of-gas-particles

Just as a caution for the above calculator, it assumes classical physics. That is, the resulting numbers will become inaccurate when the particle velocity approaches a significant fraction of the speed of light. So the velocity it outputs for temperatures of many billions or trillions of degrees may be inaccurate (the resulting velocity will be an over-estimate).

So for a field strength of 1 tesla (10,000 gauss), the resulting force on a proton moving at 146,000 m/s is:

F = qvBsinθ
F = (+1.6021766208 x 10-19 C)(146,000 m/s)(1 T)(sin90o)
F = (2.3391779 x 10-14)((sin90o)
F = 2.0912172 x 10-14 newtons

This amounts to an acceleration of about 1.25 x 1013 meters per second squared (or about 1.276 trillion times the acceleration due to Earth’s surface gravity). Since the force increases linearly with particle velocity, the force at 10 million degrees is about 4 trillion times Earth’s gravity, 100 million degrees is 12.7 trillion times Earth’s gravity and 1 billion degrees is 40 trillion.

So super-strong fields aren’t strictly necessary for very high particle accelerations.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/08/2019 14:38:57
This article estimates field strengths of up to 10,000 Gauss for NGC 4258 right at the event horizon: https://arxiv.org/pdf/0909.1207.pdf
Wow
That's really good news.
Closer in, the magnetic field could be much stronger (the measurement was made in a part of the disk about 7.5 million kilometers from the center of the disk)
So, as we move closer to the center of the SMBH, the magnetic field is stronger.
Therefore, if new pair of particles had been created at the event of horizon or very close it:
https://en.wikipedia.org/wiki/Pair_production
"Pair production is the creation of a subatomic particle and its antiparticle from a neutral boson."
"if one particle has electric charge of +1 the other must have electric charge of −1, or if one particle has strangeness of +1 then another one must have strangeness of −1"
Do you agree that at the moment of creation, one particle has electric charge of +1 while the other must have electric charge of −1 and they are probably moving very close to each other at high orbital velocity.
Do you agree that Lorentz force due to that magnetic power of 10,000 Gauss could affect their movement directions?
If one will be pushed inwards to the center of the SMBH, the other one will be pulled outwards from the event horizon (directly into the accretion disc) at the same moment of creation due to Lorentz force under opposite electric charges.
Why not?
If you agree with that, than the accretion disc gets its matter from the event of horizon (or deeper) without any violation of any law of physics.
By the way, where did you get that number of 0.8c for the relativistic jet from a super-massive black hole?
Halc have asked the same question and even had set the calculation:
There is a 10k sun-mass amount of material moving at 0.8c?  How does it not exit the galaxy?  That's well above escape velocity from 'high above the disc plane'.
Ghostly Gamma-ray Beams Blast from Milky Way's Center
https://www.cfa.harvard.edu/news/2012-16
The two beams, or jets, were revealed by NASA's Fermi space telescope. They extend from the galactic center to a distance of 27,000 light-years above and below the galactic plane.
The jets were produced when plasma squirted out from the galactic center, following a corkscrew-like magnetic field that kept it tightly focused.
It would take a tremendous influx of matter for the galactic core to fire up again. Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required.
OK, that mentions a 10k solar mass cloud being required to get back into 'gulp' mode from its current 'sip' mode.  No mention of anything moving at 0.8c mentioned.  The detected bubble is faint (not massive), and is an effect from a 'gulp' perhaps a million years ago.  The bubble extends to 27000 LY, so it's moving at best at 27 ly per 1000 years, hardly 0.8c.
The jets, if fast moving material, are the produce of the current 'sip' mode.  If they're a remnant of the big mass from a million years ago, the jets must be slow indeed to still be there.
That brings me to the following critical issue with regards to The Virial Theorem and Dark Matter:
Our scientists claim that our Sun orbits around the center of the galaxy due to the impact of The Virial Theorem and Dark Matter.
It is also stated:
https://web.stanford.edu/~ajlucas/The%20Virial%20Theorem%20and%20Dark%20Matter.pdf
"Modern estimates put the percentage of the universe made up of dark matter at about 80-90%. We still
have very little idea as to what makes up dark matter."
Based on the Virial Theorem, do you agree that same total mass that holds by gravity any star (including the Sun) in the spiral disc must also apply to that molecular Jet stream (for the same radius)?
If so, how could it be that in the spiral arms we see clearly that the all stars obey to that Virial Theorem and orbit around the center, while the molecular jet stream ignore it completely?
It might contradicts the basic idea of  Virial Theorem and Dark Matter, unless we set the followig calculation:
Have you actually done some math that shows that measurement to be wrong?
OK.
The orbital velocity for any star in the spiral disc is:
V^2 = G M /r
M total (due to viral + Dark matter) = V^2 r /G
As most of the stars in the spiral disc orbits at 220 K/s
Than the total center mass that with regards to the radius is:
M total = 220 K/s ^2 * r /G
This formula must work also for the jet stream due to the Virial theorem (based on the same radius).
So, do you agree that if a molecular gets to a distance of 20,000 Ly above the center, it should feel the impact of a total center mass of:
M total = 220 K/s ^2 * 10,000LY /G
In order for the molecular jet stream to fly direct upwards (without any orbital movement) it is clear that the upwards force must be significantly stronger than the impact of the gravity force (due to the M total mass by Virial theorem).
So, which kind of force can do it?
Our scientists say that it must be magnetism.
So, the total gravity force on a molecular at radius r is:
F = G M (total) m(molecular)/ r^2
After placing the value of M total in the formula:
F = G V^2 * r m /G r^2 = m V^2 /r
The gravity force on any molecular in that molecular jet stream and at any radius from the center should be:
F gravity = m (molecular) 220km/s ^2 /r
Hence, In order to set that kind of a direct flow of the molecular jet stream, F magnetism must be significantly higher than F gravity.
Any idea about how much stronger it should be?

You can calculate the force that a magnetic field has on a charged particle using the following equation:
F = qvBsinθ, where
“F” is the resulting force in newtons
“q” is the charge in coulombs
“v” is the velocity in meters per second
“B” is the magnetic field strength in teslas (1 tesla = 10,000 gauss)
“sinθ” is the angle between the direction of the particle’s motion and the direction of the magnetic field lines
Thanks for this explanation.
Although, we might not need it. As the gravity force can give us fairly good indication for the requested magnetism force.
In any case, I still wonder why the molecular jet stream doesn't orbit around the center of the galaxy as any other star in the galaxy?
Could it be that they are not effected by the Virial theorem and dark matter?
If so, why is it?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/08/2019 18:37:06
That brings me to the following critical issue with regards to The Virial Theorem and Dark Matter:
Our scientists claim that our Sun orbits around the center of the galaxy due to the impact of The Virial Theorem and Dark Matter.
I don't think any scientist claims that.  Theorems don't have any impact for one thing.
Yes, the sun orbits the galaxy at a rate that is a function of the mass distribution, including dark matter, which I see you're accepting now instead of being in total denial in prior posts.

Quote
It is also stated:
https://web.stanford.edu/~ajlucas/The%20Virial%20Theorem%20and%20Dark%20Matter.pdf
"Modern estimates put the percentage of the universe made up of dark matter at about 80-90%."
The figure is more like 27%, but maybe it's not including all mass/energy.  The statement doesn't specify 80-90% of what, so it is kind of vague.

Quote
Based on the Virial Theorem, do you agree that same total mass that holds by gravity any star (including the Sun) in the spiral disc must also apply to that molecular Jet stream (for the same radius)?
The theorem is irrelevant to the question.  The theorem only applies to systems in thermal equilibrium, and thus not to a jet stream.
Secondly, no.  The total mass around which the sun orbits is orders of magnitude larger than the total mass acting on your jet of molecules.

Quote
The orbital velocity for any star in the spiral disc is:
V^2 = G M /r
That formula is for a spherical distribution of mass.  A disk is not a sphere.  The calculation is more complicated than this simple case.  Yes, you can compute Earth's orbital speed around the sun with that formula.

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M total (due to viral + Dark matter) = V^2 r /G
Virial theorem is not a mass.  It can be used to compute limits on energy.  I think you mean
M total (Ordinary matter + Dark matter) = V^2 r /G, which is still wrong because the mass is not spherically distributed.

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As most of the stars in the spiral disc orbits at 220 K/s
Than the total center mass that with regards to the radius is:
M total = 220 K/s ^2 * r /G
This formula must work also for the jet stream due to the Virial theorem (based on the same radius).
It does not work for the jet stream which is not in a stable orbit.

Quote
So, do you agree that if a molecular gets to a distance of 20,000 Ly above the center, it should feel the impact of a total center mass of:
M total = 220 K/s ^2 * 10,000LY /G
No.  It isn't orbiting, so the calculation is meaningless.

Quote
In order for the molecular jet stream to fly direct upwards (without any orbital movement) it is clear that the upwards force must be significantly stronger than the impact of the gravity force (due to the M total mass by Virial theorem).
You haven't invoked the virial theorem in any of this, other than to drop its name now and then.  You seem to have no idea what it means.

Quote
So, the total gravity force on a molecular at radius r is:
F = G M (total) m(molecular)/ r^2
'Molecular' is an adjective, not a noun. A molecular has no more mass than does 'daft'.  Perhaps you mean the force on a given molecule.

Quote
After placing the value of M total in the formula:
F = G V^2 * r m /G r^2 = m V^2 /r
The gravity force on any molecular in that molecular jet stream and at any radius from the center should be:
F gravity = m (molecular) 220km/s ^2 /r
The formula is I suppose meaningful, but your lack of ability in applied mathematics means you've no idea what you've computed.  Want a clue?

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Hence, In order to set that kind of a direct flow of the molecular jet stream, F magnetism must be significantly higher than F gravity.
What you computed was obviously independent of gravity or the mass of the galaxy or a black hole, none of which appear in your equation.  So your 'Hence' is completely mistaken.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 19/08/2019 21:57:15
So, as we move closer to the center of the SMBH, the magnetic field is stronger.
Therefore, if new pair of particles had been created at the event of horizon or very close it:
https://en.wikipedia.org/wiki/Pair_production
"Pair production is the creation of a subatomic particle and its antiparticle from a neutral boson."
"if one particle has electric charge of +1 the other must have electric charge of −1, or if one particle has strangeness of +1 then another one must have strangeness of −1"
Do you agree that at the moment of creation, one particle has electric charge of +1 while the other must have electric charge of −1 and they are probably moving very close to each other at high orbital velocity.
Do you agree that Lorentz force due to that magnetic power of 10,000 Gauss could affect their movement directions?

This would only be true for black holes that are small enough to create positron-electron pairs or heavier particles. Super-massive black holes can only create neutral particle pairs such as photons or gravitons.

If one will be pushed inwards to the center of the SMBH, the other one will be pulled outwards from the event horizon (directly into the accretion disc) at the same moment of creation due to Lorentz force under opposite electric charges.
Why not?

Magnetic fields neither attract nor repel electric charges. All they do is deflect their path (assuming they weren't traveling exactly parallel to the field lines, that is). The result is that the charged particles spiral around the lines (either clockwise or counterclockwise, depending on their charge). So it isn't a matter of one kind of charge being attracted towards the source of the magnetic field while the other is repelled. So you wouldn't get any preferential treatment of one kind of charge moving away from the hole with the other moving towards it.

If you agree with that, than the accretion disc gets its matter from the event of horizon (or deeper) without any violation of any law of physics

Well, not from super-massive black holes. All they do is emit radio waves. They don't have the tidal forces needed to produce matter like electrons or protons.

Halc have asked the same question

So where is the answer? Where did you get the measurement of 0.8c for a jet from a super-massive black hole?

Based on the Virial Theorem, do you agree that same total mass that holds by gravity any star (including the Sun) in the spiral disc must also apply to that molecular Jet stream (for the same radius)?
If so, how could it be that in the spiral arms we see clearly that the all stars obey to that Virial Theorem and orbit around the center, while the molecular jet stream ignore it completely?

Astrophysical jets travel far, far above the escape velocity of galaxies. Most stars do not.

Hence, In order to set that kind of a direct flow of the molecular jet stream, F magnetism must be significantly higher than F gravity.
Any idea about how much stronger it should be?

For any particular particle being send out into the jets, I agree that the total force sending it away from the hole must be much stronger than any pulling it inward. I would like to do some calculations in order to find that, but I'm missing some information. For example, I don't know the particle's acceleration. We have measured the final velocity of the jets (close to light speed in some cases), but how long did it take to reach those speeds?

I did do a very rough calculation that suggested a proton could be accelerated to 80% the speed of light by a 10,000 gauss field in something like 0.02 seconds, but the problem with that math is that it assumes that the field strength is constant throughout those 0.02 seconds. Since the proton travels over 2,000 kilometers in that time frame, that is almost certainly not true. Magnetic fields quickly weaken the further you get from their source. I would probably need calculus to get a good answer, and I'm no good at using that form of math.

In any case, I still wonder why the molecular jet stream doesn't orbit around the center of the galaxy as any other star in the galaxy?
Could it be that they are not effected by the Virial theorem and dark matter?
If so, why is it?

Escape velocity.

I'm still waiting for you to address this:

So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.

Which is it? You can't have it both ways.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 20/08/2019 00:14:52
Quote from: DaveLev
Hence, In order to set that kind of a direct flow of the molecular jet stream, F magnetism must be significantly higher than F gravity.
Any idea about how much stronger it should be?
This information was provided in a (much) earlier reference.
- The jet can be ejected from the accretion disk at up to 2x escape velocity
- For locations near the event horizon, the escape velocity exceeds 0.5c, so relativistic speeds for the jet is understandable
- But this mechanism also applies to jets from stellar-mass black holes, neutron stars, protoplanetary disks and even brown dwarf stars. So it doesn't need Hawking radiation  or pair production from a micro black hole to produce a jet.

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So, as we move closer to the center of the SMBH, the magnetic field is stronger.
A few corrections:
1. As you move closer to the event horizon of a black hole, the magnetic field is stronger. (Reason: nothing inside the event horizon will have an effect outside the event horizon)
2. As you move closer to the innermost edge of an accretion disk, the magnetic field is stronger. (Reason: the magnetic field originates in the accretion disk.)
3. The jet does not originate from the black hole, it originates from matter already in the accretion disk, fed from matter already outside the black hole.
3.1 The jet does not originate from within the photon sphere (1.5x event horizon radius), as only electromagnetism can easily escape from there
3.2 The jet probably does not originate from within the LSCO (3x event horizon radius), as matter has great difficulty escaping from there
3.3 We need to look at the MHD (Magneto-Hydro Dynamic) simulations to see exactly where the matter and magnetic fields come from - but it's not from within the event horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/08/2019 19:20:10
Quote
The orbital velocity for any star in the spiral disc is:
V^2 = G M /r
That formula is for a spherical distribution of mass.  A disk is not a sphere.  The calculation is more complicated than this simple case.  Yes, you can compute Earth's orbital speed around the sun with that formula.
Thanks
So, let's look on Newton Shell theorem
https://en.wikipedia.org/wiki/Shell_theorem
"the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy."
Outside a shell"
"A solid, spherically symmetric body can be modelled as an infinite number of concentric, infinitesimally thin spherical shells. If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass"
So, this gives a clear indication how we should use that Newton Shell theorem.
Based on this explanation, there is no restriction for any orbital movement.
Therefore, if we could hold the Sun at it's current location - (without any orbital movement, it shouldn't change it's gravity force - as long as it there and there is no change in the radius.
Dark matter -
We have already discussed deeply the great impact of the dark matter.
It is clear to all of us the the density of the dark matter is increasing as we move in.
There is a spicial formula for the dark matter in the Milky way.
Our scientists claim that the real matter can't hold the Sun in its orbital path.
Therefore, the real matter is almost neglected with regards to the dark matter (which has the greatest impact on the Sun gravity force - based on our scientists).

Newton Shell theorem works the same for any direction.
Therefore, if the Sun is on the spiral disc or high above the accretion disc - The gravity force should be the same for the same radius.
So, if we will place the Sun directly above the accretion disc at 27,000 Ly it will face exactly the same gravity force as it gets while it is on the spiral disc at that radius.
In the same token, if the Sun is orbiting around the SMBH, falling in directly to the SMBH or ejected  directly outwards from the SMBH - the gravity force at 27,000 LY will be absolutely the same.
In the same token, there is no difference if at 27,000 Ly we place the Sun, the moon or just single water' molecular.
They all should feel the same impact from the same center of mass (As all the mass in the shell is concentrated at the center of the Shell.
 
It does not work for the jet stream which is not in a stable orbit.
I don't understand on which law do you base this statement?
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1-anim.gif
In this diagram we see clearly that m is located at a fixed distance/radius from the shell' center.
There is no requirement for any stable orbit in the Newton Shell Theorem.
So, would you kindly explain why do you add that unexppected request?
In
No.  It isn't orbiting, so the calculation is meaningless
If there is no request for orbiting - then my calculations are correct.
This would only be true for black holes that are small enough to create positron-electron pairs or heavier particles.
Thanks
Super-massive black holes can only create neutral particle pairs such as photons or gravitons.
In one hand you claim that a SMBH is the same as BH.
On the other hand based on this answer, they are different.
Is it just the size that set this impact?
Why at different radius of the SMBH we can't get different kind of new partials?
Why are you so sure about the activities in the SMBH while our real data about it is so limited?
Do we really know if there are layers in the SMBH or not? If yes, how many? If no - Please prove it.
Magnetic fields neither attract nor repel electric charges. All they do is deflect their path (assuming they weren't traveling exactly parallel to the field lines, that is).
So, the magnetic field deflects the new born particles path.
Each charge will be deflected to different direction.
That's all we need!
Hence, assuming that particles pair orbit the SMBH/BH at ultra high velocity on their first moment of birth, than if based on Lorentz force one charge will be deflected outwards, than the other one should be deflected inwards.
I don't understand why do you claim:
So it isn't a matter of one kind of charge being attracted towards the source of the magnetic field while the other is repelled.
I don't care about the source of the magnetic field. As long as it can deflected the path - that is perfectly Ok with me.
Astrophysical jets travel far, far above the escape velocity of galaxies. Most stars do not.
That is 100% correct.
The question is: What can we learn from that? Remember, Newton Shell theorem works on m if it is Hydrogen Atom or if it is the Sun.
We can't just claim that Newton Shell theorem is working according to our wishful list. It must work everywhere and anywhere exactly the same.
Therefore, we need to explain that verification based on real law.
Escape velocity.
I'm still waiting for you to address this:
Once you agree that Atom and stars are working based on the same law and the same Newton Shell theorem, you would see how easy it is.
- The jet can be ejected from the accretion disk at up to 2x escape velocity
- For locations near the event horizon, the escape velocity exceeds 0.5c, so relativistic speeds for the jet is understandable
- But this mechanism also applies to jets from stellar-mass black holes, neutron stars, protoplanetary disks and even brown dwarf stars.
Do you really see in your vision that all those stars are working together just to eject this jet stream?
Don't you think that they also need to complete their orbital cycle?
So, at any given moment the forces from those stars are changing, while the jet stream is moving constantly in the amplitude and at the same direction.
Hence, do you still consider that those stars can set that kind of jet stream?
A few corrections:
1. As you move closer to the event horizon of a black hole, the magnetic field is stronger.
Agree

2. As you move closer to the innermost edge of an accretion disk, the magnetic field is stronger.
Agree
(Reason: the magnetic field originates in the accretion disk.)
I don't agree.
3. The jet does not originate from the black hole, it originates from matter already in the accretion disk,
Agree, but not directly.
fed from matter already outside the black hole.
disagree
3.1 The jet does not originate from within the photon sphere (1.5x event horizon radius), as only electromagnetism can easily escape from there
We have already agreed that the jet originates from matter already in the accretion disk.
However, from this zone, new particles escape directly to the accretion disc. Yes - electromagnetism is responsible for that activity.
3.3 We need to look at the MHD (Magneto-Hydro Dynamic) simulations to see exactly where the matter and magnetic fields come from - but it's not from within the event horizon.
I'm not sure about the simulation.
However, I'm quite sure that one day we will discover the real main source for that ultra requested magnetic field (and it isn't the accretion disc).
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/08/2019 20:30:56
No more running, Dave Lev. I'm going to keep posting this question until you answer it.

So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.

Which is it? You can't have it both ways.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/08/2019 21:01:52
Quote from: Halc
Quote
The orbital velocity for any star in the spiral disc is:
V^2 = G M /r
That formula is for a spherical distribution of mass.  A disk is not a sphere.  The calculation is more complicated than this simple case.  Yes, you can compute Earth's orbital speed around the sun with that formula.
So, let's look on Newton Shell theorem
https://en.wikipedia.org/wiki/Shell_theorem
"the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy."
"A solid, spherically symmetric body can be modelled as an infinite number of concentric, infinitesimally thin spherical shells. If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass"
So, this gives a clear indication how we should use that Newton Shell theorem.
That clear indication is that you can't use it since matter both inside and outside the shell of the sun's orbit is not spherically arranged.  That's why I said your formula above doesn't work for a galaxy.
Again, your lack of reading comprehension skills is showing.  You quote the very parts that invalidate the use of the shell theorem in a model of a disk-arrangement of matter like a galaxy.

Quote
Dark matter -
We have already discussed deeply the great impact of the dark matter.
You denied its existence actually. So you both deny it and find it having a great impact.  You're at least consistent in your inconsistency.

Quote
Newton Shell theorem works the same for any direction.
Therefore, if the Sun is on the spiral disc or high above the accretion disc
The theorem doesn't apply to the motion of the sun, so there is no 'therefore' about it.
Quote
So, if we will place the Sun directly above the accretion disc at 27,000 Ly it will face exactly the same gravity force as it gets while it is on the spiral disc at that radius.
Exactly wrong.
Quote
Quote from: Halc
M total = 220 K/s ^2 * r /G
This formula
must work also for the jet stream due to the Virial theorem
It does not work for the jet stream which is not in a stable orbit.[/quote]
I don't understand on which law do you base this statement?[/quote]If you actually read about the virial theorem, it only concerns stable systems like regular orbits.  The jet is not such a thing, so the theorem (which you never invoked except by name) is inapplicable. The formula you quote does not come from the virial theorem. It is an invalid application of one of Newton's laws, essentially giving a spherically distributed mass necessary for our orbit.  Since the mass is not spherically distributed, it inaccurately computes the higher actual mass of all the material within our orbital radius.
Quote
There is no requirement for any stable orbit in the Newton Shell Theorem
So, would you kindly explain why do you add that unexppected request?.
Never said there was.  The shell theorem concerns spherical shells, not orbits.  The virial theorem on the other hand concerns stable systems.
Matter exiting a system at escape velocity is not a stable system.
Try actually reading about both and understanding them before using both incorrectly.
Quote
In
No.  It isn't orbiting, so the calculation is meaningless
If there is no request for orbiting - then my calculations are correct.
The formula you quoted is only meaningful to a description of an orbit.
Quote
In one hand you claim that a SMBH is the same as BH.
On the other hand based on this answer, they are different.
Is it just the size that set this impact?
Yes.  Tidal forces near the event horizon are inversely proportional to the mass of the object.
Quote
Why at different radius of the SMBH we can't get different kind of new partials?
Tidal forces get even weaker with distance.
Quote
Do we really know if there are layers in the SMBH or not? If yes, how many? If no - Please prove it
Doesn't matter, per no-hair theorem.
Quote
We can't just claim that Newton Shell theorem is working according to our wishful list. It must work everywhere and anywhere exactly the same.
It must work only where there are shells.  A galaxy is lacking in them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/08/2019 20:47:26
The shell theorem concerns spherical shells, not orbits.  The Virial theorem on the other hand concerns stable systems.
Matter exiting a system at escape velocity is not a stable system.
Try actually reading about both and understanding them before using both incorrectly.
Ok
Let's verify the Virial Theorem
http://hosting.astro.cornell.edu/academics/courses/astro201/vt.htm
"We can estimate the Virial Mass of a system if we can observe:
The true overall extent of the system Rtot
The mean square of the velocities of the individual objects that comprise the system"
" in a spiral galaxy, the dominant motion of the stars in the disk is circular rotation in the plane of the disk. The variation in the orbital velocities with radius V(r) is called the rotation curve."
With regards to spiral galaxy-
We have good verification about all stars in the spiral disc including their mass and orbital velocities.
However:
1. Dark matter
It is stated clearly - "We can estimate the Virial Mass of a system if we can observe"
Do we observe the dark matter?
What do we really know about the dark matter? How does it work in the galaxy?
Does it stay in some sort of cloud? Is it moving as a wind in a turbulent or is it orbit around something?
Could it be that it orbits around the real matter and not around the galaxy center and at what velocity?
Do we know for sure that the dark matter meets the following two main requirements for using the Virial theorem:
A. The true overall extent of the system Rtot - for Dark matter?
B. The mean square of the velocities of the individual dark matter objects that comprise the system?
How can we still use the Virial Theorem if we can't observe the the dark matter and if we have no idea how and if it orbits?
Could it be that our scientists are using a none relevant theorem for something that they don't know how it works?

2. Stable system:
The Virial theorem on the other hand concerns stable systems.
Matter exiting a system at escape velocity is not a stable system.

Do you really believe that the Milky way is a stable system?
https://curiosity.com/topics/this-hypervelocity-star-was-ejected-from-the-milky-way-curiosity
This Hypervelocity Star Was Ejected from the Milky Way.
Hypervelocity stars are rather rare in our galaxy. The first one was discovered in 2005, and so far researchers have discovered fewer than 30 of them. They travel at more than 1 million miles per hour, or 500 kilometers per second, twice as fast as other stars, and it takes an enormous amount of energy to propel them to that velocity."
We have found 30 Hypervelocity stars that had been ejected from the galaxy.
How the galaxy can be considered as "stable system" while we clearly see that at least 30 Hypervelocity stars had been ejected from it.
It is also stated:
"it takes an enormous amount of energy to propel them to that velocity."
As the Virial theorem is all about energy, how that "enormous amount of energy" could impact our calculation?
Yes, we only see 30 Stars, but it gives an indication for none stable system
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/08/2019 21:15:29
Why are you afraid of this question?

So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.

Which is it? You can't have it both ways.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/08/2019 12:39:45
So that gives us two options. Either:

(1) You agree with the laws of physics and therefore agree that your own model (which posits a violation of the first law of thermodynamics) is wrong.
(2) You agree with your model and therefore do not agree with the first law of thermodynamics.
There is no contradiction between my model to the First law of thermodynamics.
I have already explained it.
However, I'm not going to ask you to look for that answer as I will do it one more time for you and with pleasure.

1. New created particles - New pair of particles are created constantly around the SMBH (at the event of horizon or below). If one particle carry a positive charge, the other one gets a negative charge.
2. Magnetics field - Around the SMBH there is magnetic field. This magnetic field is quite strong at the event of horizon (or deeper?...)
As you move closer to the event horizon of a black hole, the magnetic field is stronger.
3. Lorentz force - Based on Lorentz force, the magnetic fields deflects differently the path of the orbital new born particles pair:
Magnetic fields neither attract nor repel electric charges. All they do is deflect their path (assuming they weren't traveling exactly parallel to the field lines, that is).
Hence, if the positive charged particle will be deflected outwards, the negative charged particle will be deflected inwards.
Therefore, while the negative is pushed inwards into the center of the SMBH, the positive is pulled outwards and get's eventually into the accretion disc.

Would you kindly explain where is the contradiction?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/08/2019 16:00:19
1. New created particles - New pair of particles are created constantly around the SMBH (at the event of horizon or below).

Right.

Quote
If one particle carry a positive charge, the other one gets a negative charge.

For very small black holes, this would be true. This would not be true for super-massive black holes as they have insufficient tidal forces to produce anything other than photons, gravitons and maybe neutrinos. Charged particles like electrons, protons or muons have too much mass-energy to be generated by the (relatively) weak tidal forces present.

2. Magnetics field - Around the SMBH there is magnetic field. This magnetic field is quite strong at the event of horizon (or deeper?...)

Although true black holes cannot have such a magnetic field, I will submit to the possibility that something like MECOs (magnetospheric eternally collapsing objects) could, maybe, be what "black holes" actually are. So I will tentatively agree that "black holes" could be MECOs and as such could have magnetic fields. I'll consider this plausible for the sake of discussion: https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object

3. Lorentz force - Based on Lorentz force, the magnetic fields deflects differently the path of the orbital new born particles pair:

Hence, if the positive charged particle will be deflected outwards, the negative charged particle will be deflected inwards.
Therefore, while the negative is pushed inwards into the center of the SMBH, the positive is pulled outwards and get's eventually into the accretion disc.

The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in". The Lorentz force would deflect the path of particles at a right angle to the field lines (assuming that they were already on a path perpendicular to the field lines. If they are parallel to the field lines, there is no force).

Would you kindly explain where is the contradiction?

The contradiction is your claim that this process causes the black hole to grow in mass. The mass of the black hole has to shrink, not grow, as the negative mass (not negative charge, an important difference) particle is invariably the one that passes into the hole (because the swapping of time and space coordinates inside of the event horizon is what makes that particular particle have a negative mass in the first place). That negative mass subtracts from the positive overall mass of the hole, causing it to become smaller. If you are willing to accept this point, then I will agree that your model no longer violates the first law of thermodynamics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/08/2019 20:13:31
The contradiction is your claim that this process causes the black hole to grow in mass. The mass of the black hole has to shrink, not grow, as the negative mass (not negative charge, an important difference) particle is invariably the one that passes into the hole (because the swapping of time and space coordinates inside of the event horizon is what makes that particular particle have a negative mass in the first place). That negative mass subtracts from the positive overall mass of the hole, causing it to become smaller. If you are willing to accept this point, then I will agree that your model no longer violates the first law of thermodynamics.

Thanks Kryptid for this great answer.
So, you agree that new particles could be ejected outwards from the SMBH without any violation of first law of thermodynamics. However, you need my confirmation for your point of view:
If you are willing to accept this point, then I will agree that your model no longer violates the first law of thermodynamics.
So, YES!!! I fully agree with your point!
If the matter in the SMBH is "positive overall mass", than "That negative mass subtracts from the positive overall mass of the hole, causing it to become smaller.
That is 100% correct.
However, I claim that the matter in the SMBH isn't "Positive overall mass" but "Negative overall mass".
Therefore, a new falling in negative particle should increase the total  "Negative overall mass".
As I have stated before, the SMBH acts as a "Negative overall mass barrel" that collects all new born negative particles.
So, we get into the accretion Disc the "Positive overall mass" while the SMBH gets all the "negative overall mass".
This activity is a Win-Win situation for our accretion disc and for the SMBH.
Both are increasing their mass constantly from this new particle pair creation.

Therefore, our SMBH isn't so bad eater. It actually has great appetite.
For any positive particle that it offers the accretion disc, it eats one negative particle.
Hence, it doesn't need to eat any particle from the accretion disc.
This is the solution for the following enigmas -
1. Why more than 99% of the matter in the accertion disc is ejected outwards?
2. Why we couldn't see any in falling matter - not from the accretion disc into the SMBH and not from outside into the accretion disc?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/08/2019 07:21:44
However, I claim that the matter in the SMBH isn't "Positive overall mass" but "Negative overall mass".

So you've gone from violating the first law of thermodynamics to violating the law of gravity.

If a black hole had negative mass, it wouldn't be a black hole because it would repel light and matter, not attract it (just as Halc said). Nothing could ever be in orbit around a black hole because it wouldn't be attracting anything. There would be no accretion disk, no orbiting stars, nothing of the sort.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/08/2019 14:21:47
Then a black hole would repel everything, since it would exert a negative gravitational force.
F=GMm/r².  So S2 would not orbit Sgr-A if the latter had negative mass.
No real object has negative mass.  The concept is strictly a mathematical one concerning non-real virtual particles, as discussed in prior posts.
On what kind of "evidence" do you base this understanding?
I do recall that you have stated that matter and Antimatter has the same gravity impact.
We can also see a confirmation for that in the following article:
https://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter
"The gravitational interaction of antimatter with matter or antimatter has not been conclusively observed by physicists. While the consensus among physicists is that gravity will attract both matter and antimatter at the same rate that matter attracts matter, there is a strong desire to confirm this experimentally."
If you still not sure about it, I would like to offer the following article from:
Sabine Hossenfelder - Sabine is a theoretical physicist specialized in quantum gravity and high energy physics. She also freelance writes about science.
https://www.forbes.com/sites/startswithabang/2017/03/29/why-doesnt-antimatter-anti-gravitate/#67be1a5419e2
"The standard model also includes all anti-particles, which are identical to their partner-particles except for having opposite electric charge. Is it possible that the anti-particles also anti-gravitate?
Theory clearly answer this question with a resounding, “No.” From the standard model, we can derive how antimatter gravitates – it gravitates exactly the same way as normal matter."
"Those with little faith in theoretical arguments might want to argue that maybe it’s possible to find a way to make antimatter anti-gravitate only sometimes. I am not aware of any theorem which strictly proves this to be impossible, but neither is there – to my best knowledge – any example of a consistent theory in which this has been shown to work."
"And if that still wasn’t enough to convince you, the ALPHA experiment at CERN has not only created neutral anti-hydrogen, made of an anti-proton and a positron (an anti-electron), but has taken great strides towards measuring exactly how anti-hydrogen behaves in Earth’s gravitation field. Guess what? So far there is no evidence that anti-hydrogen falls upwards – though the present measurement precision only rules out that the anti-hydrogen’s gravitational mass is not larger than (minus!) 65 times its inertial mass."
"Both matter and antimatter particles hold together the quarks that make up neutrons and protons. Indeed, the anti-particles’ energy makes a pretty large contribution to the total mass of neutrons and protons, and hence to the total mass of everything around us. This means if antimatter had a negative gravitational mass, the equivalence principle would be badly violated. It isn’t, and so we already know antimatter doesn’t anti-gravitate."

However, if you still have little faith in theoretical arguments and you disagree with Sabine Hossenfelder, than you are more than welcome to offer any article which supports your point of view.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/08/2019 16:39:27
You suggested a negative mass object, not antimatter.
A negative mass black hole would very very much be distinct from a positive mass one due to the entirely empirical effect as described above.  The object would repel the entire galaxy and effectively squirt out one side somewhere.
Yes, I agree.
However, this was not my intention.
Sorry for that.
In the article it was stated:
"The standard model also includes all anti-particles, which are identical to their partner-particles except for having opposite electric charge. Is it possible that the anti-particles also anti-gravitate?
Theory clearly answer this question with a resounding, “No.” From the standard model, we can derive how antimatter gravitates – it gravitates exactly the same way as normal matter."

In this discussion we have focused on a pair of particles
One is partner-particle and the other one is anti particle.
They are fully identical except for having opposite electric charge.
As Kryptid had used the "Positive overall mass" for the positive electrical charged particles, I have used the "Negative overall mass" for the Negative electric charged particles.
So, it is not about negative or positive mass, it is all about positive or negative electric charged particles.
Therefore, as the positive electric charged particle represents the partner-particle, the Negative electric charged particle represents the Anti -Particle.
In any case, as you agree that:
Antimatter still has positive mass.
An antimatter black hole is not distinct from a regular matter black hole, per no-hair theorem.
It is clear by now that antimatter or antiparticle has a positive mass
Therefore antimatter or antiparticle must have the same gravitational force as matter or positive particle.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/08/2019 17:31:05
As Kryptid had used the "Positive overall mass" for the positive electrical charged particles

At no point did I equate positive mass with positive charge. They are distinct concepts. If the electron is the particle that escapes, then that would be a case of a negatively-charged particle with positive mass. Alternatively, if the positron is what falls in, then that is a positively-charged particle with negative mass. For a neutral black hole, the probability for an electron escaping is equal to the probability of a positron escaping because there is no preferential mechanism at work (and no, magnetic fields won't change this).

So, it is not about negative or positive mass, it is all about positive or negative electric charged particles.

It absolutely is about negative and positive mass, since we are talking about super-massive black holes (which cannot produce charged particles because those particles are too massive).

Therefore, as the positive electric charged particle represents the partner-particle, the Negative electric charged particle represents the Anti -Particle.

Positrons are antimatter. They are positively-charged.

To sum things up, your idea doesn't work because:

(1) Super-massive black holes have positive mass. We know this because things orbit them, and
(2) Super-massive black holes cannot produce electrically-charged particles.

It would still be a black hole of sorts because it would attract itself. It wouldn't self-explode or anything.

Negative mass actually repels other negative mass. When you plug in the numbers for a pair of negative mass objects, you would initially expect the two to attract each other because two negative numbers multiplied by each other are positive. However, a negative inertial mass in addition to a negative gravitational mass means that it responds in the opposite way to a force acting on it. So an attraction is experienced as a repulsion: https://en.wikipedia.org/wiki/Negative_mass#Runaway_motion
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/08/2019 19:13:54
Negative mass actually repels other negative mass. When you plug in the numbers for a pair of negative mass objects, you would initially expect the two to attract each other because two negative numbers multiplied by each other are positive. However, a negative inertial mass in addition to a negative gravitational mass means that it responds in the opposite way to a force acting on it. So an attraction is experienced as a repulsion: https://en.wikipedia.org/wiki/Negative_mass#Runaway_motion
I had actually worked that out myself, but hadn't posted any of it yet.  There is a positive attraction between pairs of negative-mass objects, but that positive force results in negative acceleration due to F=ma, hence they act as if repelled.  Hence I figured out the funny positive and negative planets chasing each other, a sort of reactionless force.  It doesn't violate conservation of energy or Newton's 3rd law of motion since combined kinetic energy and momentum remain zero.  I was trying to drive it to inconsistency and have yet to actually do it.
I didn't know wiki had a site on the physics of macro negative mass.
The site calls the runaway motion 'preposterous', but what's actually wrong with it?

My idea of the black hole squirting out of the galaxy seems incorrect.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/08/2019 20:15:00
To sum things up, your idea doesn't work because:
(1) Super-massive black holes have positive mass. We know this because things orbit them, and
(2) Super-massive black holes cannot produce electrically-charged particles.
I disagree with both statement:
"(1) Super-massive black holes have positive mass. We know this because things orbit them"
There is no such thing as "Negative mass". This is a fiction. We only discuss on matter/Antimatter, or Positive/Negative electric charged particles.
So, by now it is clear that the gravity force for matter/particle is identical to antimatter/antiparticle
"(2) Super-massive black holes cannot produce electrically-charged particles"
I disagree with that statement, mainly because our knowledge about BH and SMBH is very poor.
You give that kind of statement which is full with confidence, while we have no real verification. We don't know how the BH/SMBH really works and from what kind of matter/antimatter they have made of.
We don't know if they have a core and how many layers covers this core.
You even don't believe that the SMBH can generate magnetic fields.
Therefore, as long as our knowledge about the BH/SMBH is so poor, it's better to avoid from that kind of statement.
In any case, based on your theory a BH can produce electrically-charged particles.
So what is the difference between the BH to the SMBH?
The main difference is Mass.
That mass affects the gravity force.
So, do we know the requested gravity force that is needed to produce those electrically-charged particles in a BH?
Let's assume that we can sent a probe into the BH and monitor exactly the radius that is needed for the production of those electrically-charged particles.
Based on that radius, we can extract the gravity force.
However, we can get the same gravity force also at the SMBH.
We only need to get to the relevant radius.
For example -
F=GMm/r²
Let's assume that:
If BH mass = M
SMBH mass = M * 10^6
Therefore
If
r1 -  represents the requested radius in a BH for the production of electrically-charged particles.
Than:
F (BH)=GMm/r1²
r2 represents the requested radius in the SMBH which has the same gravity force as r1 in a BH
so the gravity force for the SMBH is:
F (SMBH) = GM* 10^6 m/r2²
F (BH)= F (SMBH)
GMm/r1² = GM* 10^6 m/r2² 
1/r1² = 10^6 /r2²
r2² = 10^6 * r1²
r2 = 10^3 * r1
So, if the mass of the SMBH is higher by 10^6 than the mass of a BH, at a radius which is bigger by 10^3 from the radius of the BH we will get the same gravity force.
That shows that the total mass of the SMBH is not relevant for production of electrically-charged particles.
It's all about gravity force which is a direct outcome of the radius.
If the BH can create electrically-charged particles at radius r1 the SMBH can do the same at radius r2.

In any case:
I have proved that whole your idea about the Virial theorem is totally wrong.
In that theory we must OBSERVED the orbital objects. As we can't observe the dark matter we can't know its orbital velocity (and if it has any sort of velocity)
Therefore our understanding of how the galaxy works must also be updated.
As we have based the gravity force that holds the Sun around the galaxy on the Virial theorem - than we must look for better theory.
How are you going to address this key issue?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/08/2019 21:35:17
I disagree with both statement:
"(1) Super-massive black holes have positive mass. We know this because things orbit them"
There is no such thing as "Negative mass". This is a fiction. We only discuss on matter/Antimatter, or Positive/Negative electric charged particles.
So, by now it is clear that the gravity force for matter/particle is identical to antimatter/antiparticle

You're contradicting yourself. If I say that black holes have positive mass and you also say that black holes have positive mass, then we are agreeing, not disagreeing. If you do believe that black holes have positive mass, then why did you make this statement earlier?

However, I claim that the matter in the SMBH isn't "Positive overall mass" but "Negative overall mass".

You can't say in one post that black holes have negative mass and in another post that they have positive mass. Make up your mind.

I disagree with that statement, mainly because our knowledge about BH and SMBH is very poor.

We wouldn't need to know anything about black holes at all in order to know that some certain minimum tidal force is necessary to separate a particle pair of a given energy. It's the basic laws of physics at work. Conservation of energy will not allow particle pairs to pop into existence without being given sufficient energy. Electrons and positrons, unlike photons, have a finite rest mass. This means you cannot have an electron or positron with arbitrarily low energy. Even in the rest frame of those particles, that minimum energy is 0.511 MeV per particle (or slightly less if they are bound in the form of a positronium atom). If the tidal forces cannot supply that needed level of energy, then you don't get particles. If you claim otherwise, then you are arguing for a violation of the first law of thermodynamics.

We don't know how the BH/SMBH really works and from what kind of matter/antimatter they have made of.

Irrelevant. Objects inside of the event horizon cannot have any effect on those outside the horizon (except from gravity). This is a widely-known and well-understood property of black holes. If information could get out, then it wouldn't be an event horizon.

We don't know if they have a core and how many layers covers this core.

Again, irrelevant as per my prior statement.

You even don't believe that the SMBH can generate magnetic fields.
Therefore, as long as our knowledge about the BH/SMBH is so poor, it's better to avoid from that kind of statement.

"Belief" is irrelevant. It is absolutely true that a neutral, classical black hole with a singularity and an event horizon cannot have an intrinsic magnetic field. This is why the no hair theorem is called a theorem. A theorem cannot be proven wrong because it has already been proven true: https://en.wikipedia.org/wiki/Theorem

The only way around that would be if black holes have a different structure than predicted. I will concede that this is possible, and have offered the MECO as an alternative example which could have an intrinsic magnetic field.

In any case, based on your theory a BH can produce electrically-charged particles.

Don't call it "my" theory. It isn't. It is the currently-accepted theory. The law of conservation of energy will only allow charged particles to be produced when the tidal forces are strong enough. Mathematical analysis by physicists have already been done that exclude super-massive black holes from producing charged particles. The needed tidal force is not present.

So what is the difference between the BH to the SMBH?
The main difference is Mass.
That mass affects the gravity force.

This is like asking what the difference is between a cat and a mammal. A cat is a type of mammal. Perhaps what you meant to ask is "what is the difference between a stellar-mass black hole and a super-massive black hole?" That question would make more syntactic sense.

So, do we know the requested gravity force that is needed to produce those electrically-charged particles in a BH?

Yes. I don't know what it is off the top of my head, but the calculations have already been done by physicists (Don N. Page, for example). I do know that, according to those calculations, a black hole with a mass much larger than 1014 kilograms (about one-tenth the mass of Mar's satellite Deimos) doesn't produce electron-positron pairs. Super-massive black holes are many, many orders of magnitude more massive than that. So you don't get charged particles from them. Such calculations are based purely on the known laws of physics and conservation of energy.

Come to think of it, since Hawking radiation resembles black-body radiation, I might actually be able to calculate the required minimum mass of a black hole where the majority of photons produced are at the needed energy level to form positron-electron pairs. I'll get on that.

The main difference is Mass.
That mass affects the gravity force.
So, do we know the requested gravity force that is needed to produce those electrically-charged particles in a BH?
Let's assume that we can sent a probe into the BH and monitor exactly the radius that is needed for the production of those electrically-charged particles.
Based on that radius, we can extract the gravity force.
However, we can get the same gravity force also at the SMBH.
We only need to get to the relevant radius.
For example -
F=GMm/r²
Let's assume that:
If BH mass = M
SMBH mass = M * 10^6
Therefore
If
r1 -  represents the requested radius in a BH for the production of electrically-charged particles.
Than:
F (BH)=GMm/r1²
r2 represents the requested radius in the SMBH which has the same gravity force as r1 in a BH
so the gravity force for the SMBH is:
F (SMBH) = GM* 10^6 m/r2²
F (BH)= F (SMBH)
GMm/r1² = GM* 10^6 m/r2² 
1/r1² = 10^6 /r2²
r2² = 10^6 * r1²
r2 = 10^3 * r1
So, if the mass of the SMBH is higher by 10^6 than the mass of a BH, at a radius which is bigger by 10^3 from the radius of the BH we will get the same gravity force.
That shows that the total mass of the SMBH is not relevant for production of electrically-charged particles.
It's all about gravity force which is a direct outcome of the radius.
If the BH can create electrically-charged particles at radius r1 the SMBH can do the same at radius r2.

The fatal flaw in your calculations is that you are considering the overall gravitational force, not the tidal force. Redo the calculations, but for tidal forces, and you'll be on the right track.

In any case:
I have proved that whole your idea about the Virial theorem is totally wrong.

Since when have I ever said anything about virial theorem, let alone say that it was "my idea"? Or are you talking to Halc here?

In that theory we must OBSERVED the orbital objects. As we can't observe the dark matter we can't know its orbital velocity (and if it has any sort of velocity)
Therefore our understanding of how the galaxy works must also be updated.
As we have based the gravity force that holds the Sun around the galaxy on the Virial theorem - than we must look for better theory.
How are you going to address this key issue?

What does any of that have to do with the behavior of super-massive black holes?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/08/2019 22:43:07
Okay, now for some calculations.

Since the total mass of an electron-positron pair is 1.022 MeV, a photon must also have at least this much energy in order to be converted into such a pair (by interacting with an atomic nucleus, for example).

Hawking provided us with equations that allow us to predict the properties of Hawking radiation based on a black hole’s mass. The following calculator does this: https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator

When we put in a peak photon energy of 1.022 MeV, we get an associated black hole mass of around 4 x 1013 kilograms. This is somewhat smaller than the previous number I gave of 1014 kilograms, but the mechanism of producing the electron-positron pairs is also different. The electrons and positrons are not being produced directly as a part of Hawking radiation, but rather as the interaction of photon-based Hawking radiation interacting with some form of matter surrounding the black hole (either hydrogen atoms in the vacuum or perhaps accretion disk material).

Theoretically, electron-positron pairs can be produced from photon-photon interactions. This would require half the photon energy, since it involves two photons instead of one (with a resulting black hole twice as heavy as the previous calculation). The following paper discusses this phenomenon around black holes: https://pdfs.semanticscholar.org/f4bc/24655948d538fdca3804a41cbf812e39cf34.pdf

Of interest, the results of calculations done by Don Page can be seen in the image here: https://i.stack.imgur.com/vmSpB.jpg

To summarize the image, a black hole with a mass above of 1017 grams emits Hawking radiation with a consistency of 81.4% neutrinos and antineutrinos, 16.7% photons and 1.9 % gravitons, whereas one with a mass between 1017 grams and 1014 grams emits 45% positrons and electrons, 45% neutrinos and antineutrinos, 9% photons and 1% gravitons. Between 1014 grams and 3.16 x 1013 grams, the hole emits 12% neutrons and antineutrons, 28% electrons and positrons, 48% neutrinos and antineutrinos, 11% photons and 1% gravitons.

The image was found here: https://physics.stackexchange.com/questions/89983/how-many-of-which-particles-are-in-hawking-radiation
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/08/2019 06:22:18
With regards to the Tidal VS Lorentz force
The fatal flaw in your calculations is that you are considering the overall gravitational force, not the tidal force. Redo the calculations, but for tidal forces, and you'll be on the right track.
We have already agreed that Lorentz force under the magnetic field can do that job. Therefore, there is no need for tidal.
The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in". The Lorentz force would deflect the path of particles at a right angle to the field lines (assuming that they were already on a path perpendicular to the field lines. If they are parallel to the field lines, there is no force).
Therefore, the only open question is about the feasibility of a SMBH to create new pair of charged particles as a BH can do.
Let's start with a BH:
Yes. I don't know what it is off the top of my head, but the calculations have already been done by physicists (Don N. Page, for example). I do know that, according to those calculations, a black hole with a mass much larger than 1014 kilograms (about one-tenth the mass of Mar's satellite Deimos) doesn't produce electron-positron pairs.
However, in all the articles that you have offered they specifically focus on the black body radiation:
https://physics.stackexchange.com/questions/89983/how-many-of-which-particles-are-in-hawking-radiation
"The black body radiation (semiclassical form) description of Hawking radiation is realistic for the low temperatures of black holes acting as black bodies, because the energy needed to create a detectable particle from the vacuum fluctuation framework is large, 1 MeV for e+and e- , one of them falling back."
So, If I understand it correctly, they have calculated the requested energy that is needed to create those charged particles, extract the requested mass (or gravity) and then verify if the BH will can set the radiation.
So, if we will try to work with the same formulas of energy/gravity at SMBH, we can easily find that SMBH can also create those kinds of particles (at the relevant radius/energy). However, due to it's massive size, those particles would not be ejected and therefore, there will be no radiation. That is very clear to me.
However, in all of those articles, I couldn't find even one word about magnetism.
So, they have totally ignored the great impact of Lorenz force.
That was a severe mistake.
If our scientists will add the impact of the Lorentz force (especially at the SMBH) they should find how easy it is for the SMBH to eject those new born positive charged particles as a BH can do without the need for Lorentz force.
Therefore, the following statement is fully correct:
Super-massive black holes are many, many orders of magnitude more massive than that. So you don't get charged particles from them.
However, it is correct as long as we ignore the magnetism and Lorentz force.
Never the less, once we add Lorentz force, we should find that it can easily extract the radiation of the Positive charged particles from the SMBH.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/08/2019 06:29:33
We have already agreed that Lorentz force under the magnetic field can do that job. Therefore, there is no need for tidal.

Who is "we"? I never agreed that Hawking radiation can be created by a magnetic field.

However, in all of those articles, I couldn't find even one word about magnetism.
So, they have totally ignored the great impact of Lorenz force.
That was a severe mistake.
If our scientists will add the impact of the Lorentz force (especially at the SMBH) they should find how easy it is for the SMBH to eject those new born positive charged particles as a BH can do without the need for Lorentz force.

Absolutely not! I have explained this to you several times already: you can't use magnetism to pull particles out of an event horizon! The swapping of space-time coordinates inside of a black hole do not allow matter that is moving below the speed of light to travel away from the singularity. It's a consequence of the geometry of space inside of the horizon. It doesn't work. It can't be done. Period. The one and only way to escape from a black hole is by traveling faster than light, which is not a velocity that a magnetic field can provide.

Again, here is the link that speaks of space and time "swapping" such that escape is impossible for any particles travelling slower than light: http://www.jimhaldenwang.com/black_hole.htm

Quote
Notice how the minus sign has moved from the t coordinate to the r coordinate.  This means that inside the event horizon, r is the timelike coordinate, not t.  In relativity, the paths of material particles are restricted to timelike world lines.  Recall the discussion of timelike separation earlier in this paper (2).  It is the coordinate with the minus sign that determines the meaning of "timelike."  According to relativity, inside a black hole time is defined by the r coordinate, not the t coordinate.  It follows that the inevitability of moving forward in time becomes, inside a black hole, the inevitability of moving toward r = 0.  This swapping of space and time occurs at r = 2M.  Thus, r = 2M marks a boundary, the point where space and time change roles.  For the observer inside this boundary, the inevitability of moving forward in time means that he must always move inward toward the center of the black hole at r = 0.  All timelike and lightlike world lines inside r = 2M lead inevitably to r = 0 (the end of time!)  Because it is not possible for any particle or photon inside r = 2M to take a path where r remains constant or increases, the boundary r = 2M is called the event horizon of the black hole.  No observer inside the event horizon can communicate with any observer outside the event horizon.  The event horizon can be thought of as a one-way boundary.

Here is a video explaining it:


Another analogy for this is to visualize space as "flowing" into the black hole. Outside of the event horizon, this flow is moving below the speed of light. This allows particles that are moving quickly enough to escape. Inside the event horizon, however, space is flowing faster than light. Since no material object can move faster than light, it is impossible for anything to move against this flow. It must inevitably be pulled into the singularity. Even infinite force would be insufficient (as infinite force can only get you up to the speed of light, not beyond it).

Here is yet another explanation of space and time swapping roles (I want to be thorough with this because it's a strange concept that can be hard to understand): http://www.einstein-online.info/spotlights/changing_places.html

By the way, your agreement that black holes have positive mass means that the infalling negative mass particles formed from the Hawking process decrease the mass of the hole (exactly as modern physics says). Give it long enough and the hole eventually disappears.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 24/08/2019 12:34:06
Some news: A recent outburst was seen in Sgr A*: It was suddenly 75x brighter than normal, when observed over a 2-hour period.
- It was observed in the infra-red, by the Keck telescope
- Normally, the black hole is hardly visible in the infra-red, but on this occasion, Sgr A* was brighter than S2, confusing the astronomers somewhat.
- During the observation period, Sgr A* was declining in brightness, so it was probably even brighter before they observed it.

It sounds like a big clump of matter got added to the accretion disk and/or swallowed by the black hole, causing a burst of infra-red.
See: https://www.universetoday.com/143150/milky-ways-black-hole-just-flared-growing-75-times-as-bright-for-a-few-hours/
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/08/2019 12:33:46
Some news: A recent outburst was seen in Sgr A*: It was suddenly 75x brighter than normal, when observed over a 2-hour period.
- It was observed in the infra-red, by the Keck telescope
- Normally, the black hole is hardly visible in the infra-red, but on this occasion, Sgr A* was brighter than S2, confusing the astronomers somewhat.
- During the observation period, Sgr A* was declining in brightness, so it was probably even brighter before they observed it.
See: https://www.universetoday.com/143150/milky-ways-black-hole-just-flared-growing-75-times-as-bright-for-a-few-hours/
The orbital period cycle of S2 is about 15.2 Years
https://www-istp.gsfc.nasa.gov/stargaze/Kep3laws.htm
In 2002 and in 2017 it was very close to SMBH.
The next time should be in 2032.
So, how could it be that in 2019 it comes back again to the SMBH?
Is it just to make the impossible – possible??
I have explained this to you several times already: you can't use magnetism to pull particles out of an event horizon!
The magnetism can pull particles out of an event horizon.
There is no need for a unrealistic high velocity. Lorentz force can do it!!!
This is very clear to me and I thought that it was also clear for you.
However, are you the same person that sent me the following reply #635?
Dave - 1. New created particles - New pair of particles are created constantly around the SMBH (at the event of horizon or below).
Kryptid - Right.
Dave- If one particle carry a positive charge, the other one gets a negative charge.
Kryptid - For very small black holes, this would be true. This would not be true for super-massive black holes as they have insufficient tidal forces to produce anything other than photons, gravitons and maybe neutrinos. Charged particles like electrons, protons or muons have too much mass-energy to be generated by the (relatively) weak tidal forces present.
Dave - 2. Magnetics field - Around the SMBH there is magnetic field. This magnetic field is quite strong at the event of horizon (or deeper?...)
Kryptid- Although true black holes cannot have such a magnetic field, I will submit to the possibility that something like MECOs (magnetospheric eternally collapsing objects) could, maybe, be what "black holes" actually are. So I will tentatively agree that "black holes" could be MECOs and as such could have magnetic fields. I'll consider this plausible for the sake of discussion: https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object
Dave - 3. Lorentz force - Based on Lorentz force, the magnetic fields deflects differently the path of the orbital new born particles pair:
Hence, if the positive charged particle will be deflected outwards, the negative charged particle will be deflected inwards.
Therefore, while the negative is pushed inwards into the center of the SMBH, the positive is pulled outwards and get's eventually into the accretion disc.
Kryptid - The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in". The Lorentz force would deflect the path of particles at a right angle to the field lines (assuming that they were already on a path perpendicular to the field lines. If they are parallel to the field lines, there is no force).
So, you had agreed that the positive particles can be ejected outwards due to Lorentz force.
However, you were sure that the negative particles that fall in could decrease the size of the BH:
The contradiction is your claim that this process causes the black hole to grow in mass. The mass of the black hole has to shrink, not grow, as the negative mass (not negative charge, an important difference) particle is invariably the one that passes into the hole (because the swapping of time and space coordinates inside of the event horizon is what makes that particular particle have a negative mass in the first place). That negative mass subtracts from the positive overall mass of the hole, causing it to become smaller. If you are willing to accept this point, then I will agree that your model no longer violates the first law of thermodynamics.
I wonder why did you agree in reply 635 to accept the impact of the magnetism and Lorentz force, while now you have decided to withdraw from your willing to accept it.
I'm not going to force you to do it and I have no more questions about magnetism at this phase.

So, would you kindly answer the following question?
I have proved that whole your idea about the Virial theorem is totally wrong.
In that theory we must OBSERVED the orbital objects. As we can't observe the dark matter we can't know its orbital velocity (and if it has any sort of velocity)
Therefore our understanding of how the galaxy works must also be updated.
As we have based the gravity force that holds the Sun around the galaxy on the Virial theorem - than we must look for better theory.
How are you going to address this key issue?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/08/2019 20:58:57
The orbital period cycle of S2 is about 15.2 Years
https://www-istp.gsfc.nasa.gov/stargaze/Kep3laws.htm
In 2002 and in 2017 it was very close to SMBH.
The next time should be in 2032.
So, how could it be that in 2019 it comes back again to the SMBH?
Is it just to make the impossible – possible??

Nobody said that it did. Its closest approach was "in the middle of 2018" according to the article. The hypothesis that the article is talking about is the following:

Quote
It’s possible that SO-2’s close approach disrupted the way that material flows into Sgr. A*. That would generate the kind of variability and bright flaring that astronomers saw in May, about one year after the star’s close approach.

Since the star's closest approach was 17 light hours from the hole (18.7 billion kilometers), any disruption of the material orbiting at that distance would not have had any kind of immediate impact on the flow of material into the hole. It would have taken time for such a disruption in an accretion disk to propagate to the hole.

The magnetism can pull particles out of an event horizon.

No it can't. That would require the laws of physics to be broken. Magnetism could only move particles away from the hole if they are already outside of the horizon.

There is no need for a unrealistic high velocity. Lorentz force can do it!!!
This is very clear to me and I thought that it was also clear for you.

I feel as if you completely and utterly ignored everything in my last post in regards to space and time swapping inside of an event horizon. Almost the entirety of post 650 was about that subject, so it isn't possible that you merely missed it. I posted a video from PBS about it. I posted an analogy about space flowing into the hole faster than light. I posted two links explaining it in text form (one of which used visual figures to aid in the explanation). Go back and actually look at them. Nothing can get out of an event horizon because the inevitable flow of space into the singularity will not let it. It isn't the same as pushing yourself away from a conventional gravity field with rocket engines (or, in this case, with a magnetic field).

Right now, you are unavoidably travelling into the future. You can't stop that. It's the same with traveling towards the singularity inside of a black hole because space has taken on the unstoppable movement property of time. You can't stop yourself from going into the future outside the hole, and you can't stop yourself from moving towards the singularity inside the hole. Magnetism, even if it is literally infinitely strong, will make no difference.

So, you had agreed that the positive particles can be ejected outwards due to Lorentz force.

You are lying. What is truly bizarre about this is that you even quoted where I stated, "The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in"." So I stated the opposite of what you claim. What were you trying to gain by lying about what I said? I am genuinely beginning to wonder if you are a troll. I don't see how a serious person could continuously cite one thing and then say the exact opposite in the same post the way you do. I refuse to believe that your reading comprehension skills are really that bad (unless, perhaps, you have some kind of language disability akin to dyslexia that I have never heard of before). If you did, then I would be more understanding.

Quote
However, you were sure that the negative particles that fall in could decrease the size of the BH:

Negative mass particles, not negatively-charged particles. Don't make that mistake again.

I wonder why did you agree in reply 635 to accept the impact of the magnetism and Lorentz force, while now you have decided to withdraw from your willing to accept it.

You are being deceptive. I agreed that magnetism would have an effect on electrically-charged particles generated as Hawking radiation (which are outside of the horizon), but at no point did I agree that magnetism can pull them out of an event horizon. It can't.

So, would you kindly answer the following question?

I already said, "Since when have I ever said anything about virial theorem, let alone say that it was "my idea"? Or are you talking to Halc here?"

I don't know much about virial theorem or what it has to do with super-massive black holes, so I don't even know how to answer your question.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/08/2019 00:25:24
At unequal velocities, one has momentum that is not the negative of the other. The sum of the momentum of the system is not zero.

If they started at unequal velocities, then shouldn't the system have started with a momentum sum that wasn't zero anyway?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/08/2019 18:49:07
Dear Kryptid
Sorry as I didn't understand you correctly.
Magnetism could only move particles away from the hole if they are already outside of the horizon.
I agreed that magnetism would have an effect on electrically-charged particles generated as Hawking radiation (which are outside of the horizon), but at no point did I agree that magnetism can pull them out of an event horizon. It can't.
So, do you mean that if there is a positive particle out of the event of horizon of a SMBH (But very close to it), based on magnetism and Lorentz force this particle could be ejected outwards. However, if it is located inwards into the event of horizon, than there is no way to push it outwards even if the magnetism is infinity?
you can't stop yourself from moving towards the singularity inside the hole. Magnetism, even if it is literally infinitely strong, will make no difference.
In order to get better understanding:
Let assume that one positive particle is located one meter inwards from the Event of horizon and the other one is located one meter outwards from the event of horizon
Both particles orbits almost at the same velocity around the SMBH.
Therefore:
Do you agree that the gravity force on both two positive particles is finite and almost identical?
If the magnetism is infinite do you agree that Lorentz force should also be infinite?
If so, I can't understand why infinite Lorentz force can only extract the particle that orbits outwards from the event of horizon but can't do it with the one that orbits two meter inwards?
Let me offer a solution:
1. A particle that orbits inside the event of horizon (even one meter inwards) gets infinite gravity force in order to overcome the infinite Lorentz force. Is it feasible?
2.  A particle that orbits inside the event of horizon doesn't get infinite gravity force. However, the Lorentz force there is zero. So, could it be that once we cross the event of horizon the magnetic force goes to zero?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/08/2019 22:01:59
So, do you mean that if there is a positive particle out of the event of horizon of a SMBH (But very close to it), based on magnetism and Lorentz force this particle could be ejected outwards.

Possibly. It depends on the direction the particle is travelling in and the relative orientation of the magnetic field.

However, if it is located inwards into the event of horizon, than there is no way to push it outwards even if the magnetism is infinity?

Yes, that's exactly what I'm saying. In order to escape an event horizon, you have to travel faster than the speed of light. Infinite force, whether it is produced by a magnetic field or otherwise, can only get you up to the speed of light. That isn't enough.

In order to get better understanding:
Let assume that one positive particle is located one meter inwards from the Event of horizon and the other one is located one meter outwards from the event of horizon
Both particles orbits almost at the same velocity around the SMBH.

There are no stable orbits inside of an event horizon. In fact, there are no stable orbits within a radius of 1.5 times that of the event horizon (called the photon sphere). At this distance, only objects moving at the speed of light can remain in orbit. Inside of this radius, not even light can stay in orbit.

Do you agree that the gravity force on both two positive particles is finite and almost identical?

The gravitational force is finite, but that is not what inevitably draws particles towards the singularity once they get inside the horizon. The particles are driven by the swapping of space and time. Time travels forwards outside the horizon whether you want it to or not. Space travels inwards towards the singularity whether you want it to or not. It's like trying to swim upstream when the current is moving too fast for even an infinite force to swim against.

If the magnetism is infinite do you agree that Lorentz force should also be infinite?

Yes, but infinite force won't help. Infinite force can only get you up to the speed of light, not beyond it.

If so, I can't understand why infinite Lorentz force can only extract the particle that orbits outwards from the event of horizon but can't do it with the one that orbits two meter inwards?

It's for the same reason that you can't stop getting older. Space behaves like time inside the horizon.

1. A particle that orbits inside the event of horizon (even one meter inwards) gets infinite gravity force in order to overcome the infinite Lorentz force. Is it feasible?

Orbits don't exist inside the horizon nor is the gravity there infinite in strength.

2.  A particle that orbits inside the event of horizon doesn't get infinite gravity force. However, the Lorentz force there is zero. So, could it be that once we cross the event of horizon the magnetic force goes to zero?

Magnetism probably doesn't go to zero. I think I've seen papers about magnetic field lines (from an outside source) threading the event horizons of spinning black holes as a way to draw angular momentum away from them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/08/2019 04:56:32
Thanks Kryptid
So, the info is as follow:
1. A positive particle out of the event of horizon will be ejected outwards due to magnetism and Lorentz force
2. A positive particle in the event of horizon won't be ejected outwards due to space time.
3. Magnetism in the event of horizon doesn't go to zero.

If that is correct, let me ask the following:

1. Space time.
If I understand it correctly, Minkowski had developed the Space time module for the Universe.
Based on this module our scientists have found that there must be a curvature in our universe.
However, so far they didn't find any curvature in our universe.
Therefore, my personal understanding is that the space time is just a module which doesn't necessarily represents the real Universe  - especially, once we get into the edge of the module.
Now our scientists have decided to use this model also for the aria inwards the event horizon.
Why is it? Based on what data?
What is the chance that this model is not relevant for the event horizon?
Can you please prove that space time module works also at the event of horizon?
How the extreme conditions at that aria could affect the space time?
If we use this module for the event of horizon, why we can't use it for outside that aria?
Why not using it for the center Bulge or even for the whole galaxy?
Why do we insist to fix it only for that limited aria - Event horizon?


2. Speed limit -
We all know that the maximal speed limit is the speed of light.
We also know that if we go in one direct line, we will never come back to the starting point.
However, based on the space time module, we have discovered that if we go in one direct line, we might come back to the same starting point.
Therefore in the same token - if the space time can set a curvature in space, why it can't eliminate the speed limit?
Why under the space time module, the speed can't get to infinite.
Did we try to verify this issue in our space time modeling?

3. Creation of new particle
You have already confirmed that new particle can be created below the event of horizon in a BH.
This particle can even be ejected from the event horizon if the BH is small enough.
That proves that particle can orbits below the event horizon.
Now you claim that:
Orbits don't exist inside the horizon nor is the gravity there infinite in strength.
How could it be?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 28/08/2019 05:52:57
Thanks Kryptid
So, the info is as follow:
1. A positive particle out of the event of horizon will be ejected outwards due to magnetism and Lorentz force
Even a regular stellar black hole is far too large to create charged particles.  Photons are unaffected by magnetism and such.
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1. Space time.
If I understand it correctly, Minkowski had developed the Space time module for the Universe.
Based on this module our scientists have found that there must be a curvature in our universe.
However, so far they didn't find any curvature in our universe.
It is found wherever mass is nearby, like here on Earth.
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We also know that if we go in one direct line, we will never come back to the starting point.
True.  One cannot cross the event horizon, and you would need to cross it to go all the way around if there was an 'around' to go.
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However, based on the space time module, we have discovered that if we go in one direct line, we might come back to the same starting point.
I don't know what 'the space time module' is.  Minkowski didn't develop a module.  A model perhaps, but that model is a local model, not one of the universe.
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Therefore in the same token - if the space time can set a curvature in space, why it can't eliminate the speed limit?
Why under the space time module, the speed can't get to infinite.
Maximum speed of c follows from frame independent speed of light, not from curved spacetime.
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You have already confirmed that new particle can be created below the event of horizon in a BH.
We have not. We're just not in denial of the possibility.
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This particle can even be ejected from the event horizon if the BH is small enough.
No it cannot.
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That proves that particle can orbits below the event horizon.
Again, no it cannot. Asserting nonsense is not proof of anything. You can't even orbit anywhere close to the event horizon on the outside.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/08/2019 06:02:40
1. A positive particle out of the event of horizon will be ejected outwards due to magnetism and Lorentz force

This is no more likely than a negatively-charged particle being ejected.

1. Space time.
If I understand it correctly, Minkowski had developed the Space time module for the Universe.
Based on this module our scientists have found that there must be a curvature in our universe.
However, so far they didn't find any curvature in our universe.
Therefore, my personal understanding is that the space time is just a module which doesn't necessarily represents the real Universe  - especially, once we get into the edge of the module.
Now our scientists have decided to use this model also for the aria inwards the event horizon.
Why is it? Based on what data?

You are talking about a universe with overall curvature. That is a different matter than the curvature of space-time around a massive body due to gravity. A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat. Don't confuse local curvature with universal curvature.

What is the chance that this model is not relevant for the event horizon?
Can you please prove that space time module works also at the event of horizon?
How the extreme conditions at that aria could affect the space time?
If we use this module for the event of horizon, why we can't use it for outside that aria?
Why not using it for the center Bulge or even for the whole galaxy?
Why do we insist to fix it only for that limited aria - Event horizon?

It's only a matter of degree. Space-time mixing happens in every day life as well, but the effects are too subtle to notice unless you have sufficiently-sensitive equipment (the phenomena of length contraction and time dilation are manifestations of this mixing). Here is an excerpt from "Black Holes & Time Warps" about this:


* spacetimemix.jpg (622.89 kB . 711x1198 - viewed 3668 times)

It's just that, at the event horizon, the conditions become so extreme that the mixing becomes a complete swapping so that space becomes fully time-like. This is what defines the event horizon and what makes it different from other locations in a galaxy.

Therefore in the same token - if the space time can set a curvature in space, why it can't eliminate the speed limit?
Why under the space time module, the speed can't get to infinite.
Did we try to verify this issue in our space time modeling?

There is no reason for this to be the case. The speed of light in a vacuum has consistently been measured as finite and constant regardless of reference frame.

This particle can even be ejected from the event horizon if the BH is small enough.

Only if it is outside the horizon.

That proves that particle can orbits below the event horizon.

I have no idea how you came to that conclusion, but it's wrong. The closer you get to a gravitational source, the faster your orbit has to be. At the photon sphere, that required orbital speed is the speed of light. Inside of that radius, you would have to move faster than light to remain in orbit, which cannot be done with either light or material objects. So merely orbiting is not an option to prevent you from falling in. Please keep in mind that the inability to orbit is not the same as the inability to move away from the black hole. You can do that for so long as you are still outside the horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/08/2019 07:35:51
You are talking about a universe with overall curvature. That is a different matter than the curvature of space-time around a massive body due to gravity. A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat. Don't confuse local curvature with universal curvature.
How do we know that  "A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat?
spacetimemix.jpg (622.89 kB . 711x1198 - viewed 3 times)
It's just that, at the event horizon, the conditions become so extreme that the mixing becomes a complete swapping so that space becomes fully time-like. This is what defines the event horizon and what makes it different from other locations in a galaxy.
In the articale which you have offered there is not even one word about event horizon.
So, why did you decide to set the space time only below the event horizon?
There must be a solid proof for that. Would you kindly offer it?
However, if there is no mathematical roof for that, why the accretion disc can't be also part of the space time?
It is also around a massive Body.  The conditions there are also extreme. The orbital velocity could be above 0.3c.
If we can claim that the accretion disc could also be part of the space time, than nothing should also be ejected from it.
Why can't we look at the whole central bulge as space time?
Can you please prove it?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/08/2019 16:37:14
How do we know that  "A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat?

The consequences of that curvature (time dilation, gravitational lensing, relativistic orbital precession, frame dragging and the geodetic effect) have been measured.

In the articale which you have offered there is not even one word about event horizon.
So, why did you decide to set the space time only below the event horizon?

What do you mean I decided to "set the space time"? Space-time is everywhere, not only below the horizon.

There must be a solid proof for that. Would you kindly offer it?
However, if there is no mathematical roof for that,

Go back and watch that PBS video I posted. They go into the math of the behavior of space-time below the horizon.

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why the accretion disc can't be also part of the space time?

It is a part of space-time. Everything we know of is. Why wouldn't it be? The gravity in that area isn't strong enough to swap space and time, though.

It is also around a massive Body.  The conditions there are also extreme. The orbital velocity could be above 0.3c.

Not extreme enough. The escape velocity has to exceed the speed of light for this space-time swapping to take place.

If we can claim that the accretion disc could also be part of the space time, than nothing should also be ejected from it.

It's not inside of an event horizon.

Why can't we look at the whole central bulge as space time?
Can you please prove it?

Again, space-time is everywhere so I don't know what you are talking about.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/08/2019 19:10:35
Thanks Halc and Kryptid
where the accretion disc is is part of spacetime, as is Earth.
What do you mean I decided to "set the space time"? Space-time is everywhere, not only below the horizon.
space-time is everywhere so I don't know what you are talking about.
So now you both agree that even the accretion disc is under the space time.
However - we clearly see that 99% of the matter in the accretion disc is ejected outwards.
Hence
How could it be that the accretion disc which is fully under the impact of space time can eject most of its mass, while the event of horizon can't do so just because of the same space time.
You have stated that based on space-time, nothing could be ejected from inside the horizon:
Please see the following:
The gravitational force is finite, but that is not what inevitably draws particles towards the singularity once they get inside the horizon. The particles are driven by the swapping of space and time. Time travels forwards outside the horizon whether you want it to or not. Space travels inwards towards the singularity whether you want it to or not. It's like trying to swim upstream when the current is moving too fast for even an infinite force to swim against.

Why space time that works so nicely at the event of horizon and prevents from any particle to be ejected outwards, works so badly at the accretion disc and has no resistance for the massive ejection at that aria?
In other words -
If the space time has no objection that most of the particle can be ejected from the accretion disc, could it be that it also has no objection that particles can be ejected also below the horizon?
Why it prevents from particles to be ejected from the horizon, while it has no resistance that most of the particles can be ejected from the accretion disc?
Do you have better idea than space-time to show the difference between the horizon and the accretion?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 28/08/2019 20:20:00
So now you both agree that even the accretion disc is under the space time.
Spacetime is simply where and when events occur.
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the accretion disc which is fully under the impact of space time
Spacetime has no impact.  As I said, it is simply where and when events occur.  It is a geometric 4D space in which a coordinate system can be mapped.
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You have stated that based on space-time, nothing could be ejected from inside the horizon:
For the same reason I cannot give an apple to a hungry mammoth, no matter what force I put on the apple.  I am in the future of said mammoth and cannot push the apple back in time.  It isn't a matter of lacking the energy or force to do it.
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Why space time that works so nicely at the event of horizon
Spacetime does not work so nicely at the event horizon.  There is a singularity there, meaning all the laws of physics break down.  Lacking a unified theory, we cannot meaningfully describe what goes on at the event horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/08/2019 21:48:24
So now you both agree that even the accretion disc is under the space time.

"Under the space time" makes no sense. How can you be "under" space-time?

How could it be that the accretion disc which is fully under the impact of space time can eject most of its mass, while the event of horizon can't do so just because of the same space time.
You have stated that based on space-time, nothing could be ejected from inside the horizon:
Please see the following:

What you are missing is that space-time behaves differently inside of the horizon than it does outside of the horizon.

Why space time that works so nicely at the event of horizon and prevents from any particle to be ejected outwards, works so badly at the accretion disc and has no resistance for the massive ejection at that aria?
In other words -
If the space time has no objection that most of the particle can be ejected from the accretion disc, could it be that it also has no objection that particles can be ejected also below the horizon?
Why it prevents from particles to be ejected from the horizon, while it has no resistance that most of the particles can be ejected from the accretion disc?

Because space is fully time-like inside the horizon but it isn't outside the horizon.

Do you have better idea than space-time to show the difference between the horizon and the accretion?

The differing behavior of space-time at those two locations is what makes all of the difference. Space-time behaves differently in different reference frames. It behaves differently in strong gravitational fields than it does in weak ones. It behaves differently at high accelerations than at low ones. This "all or nothing" thinking process of yours is wrong.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/08/2019 04:58:13
What you are missing is that space-time behaves differently inside of the horizon than it does outside of the horizon.
Because space is fully time-like inside the horizon but it isn't outside the horizon.
Thanks
So, there are two types of space-time
A. Fully time-like Space-time which works exactly up to the event Horizon - Nothing can escape from it.
B. Not fully time-like space time which works in the accretion disc - Everything can escape from it.
Questions:
1. How could it be that the two types of space-time behave so differently?
2. What is so unique in the "fully time-like" that prevents from any particle to escape, while his brother - the "None fully time-like" has no problem that everything can escape? Can you please show the different formulas for each one?
3. How the space-time knows exactly where is the border of the "fully time-like"?
4. Why the "fully time-like" is exactly located at the event horizon?
5. Why not 5% or 50% outwards or inwards from the horizon.
6. Would you kindly offer real proof for the border of the "fully time-like" space-time version (If possible -mathematical calculation)



Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/08/2019 06:16:16
So, there are two types of space-time
A. Fully time-like Space-time which works exactly up to the event Horizon - Nothing can escape from it.
B. Not fully time-like space time which works in the accretion disc - Everything can escape from it.

This is overly simplistic. There is actually a gradual transition from one to the other. Any time there is a gravitational field present, it causes some degree of space-time mixing. This causes objects in space to move in a preferred direction: towards the source of the field. In weak fields like on Earth, this direction is merely preferred but not mandatory. As the gravitational field becomes stronger, space becomes increasingly time-like and the preference for movement into the field becomes stronger. Once the field becomes so strong that the escape velocity reaches the speed of light space has become fully time-like and what was merely a preferred direction becomes a mandatory direction instead. It's like travelling down a river that goes faster and faster until it eventually becomes impossible to resist its flow.

Questions:
1. How the space-time knows exactly where is the border of the "fully time-like"?

Space-time doesn't "know" anything. It simply "is" fully time-like at the event horizon. This border automatically exists at any location where the escape velocity reaches the speed of light. So it is defined by gravity.

2. Why the "fully time-like" is exactly located at the event horizon?

Because that is the very thing that defines it. An event horizon is what it is specifically because nothing can get out of it and the reason nothing can get out is because space is fully time-like there.

3. Why not 5% or 50% outwards or inwards from the horizon.

Because then the event horizon wouldn't be an event horizon.

4. Would you kindly offer real proof for the border of the "fully time-like" space-time version (If possible -mathematical calculation)

The math is discussed in the PBS video, as well as here: http://www.jimhaldenwang.com/black_hole.htm

This page explains it using light cones: https://www.quora.com/If-the-pulling-force-inside-the-event-horizon-is-not-infinite-why-cant-I-push-myself-out-with-a-greater-force

For more information on light cones: https://en.wikipedia.org/wiki/Light_cone
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/08/2019 16:48:12
Don't understand this.  There are 4 dimensions, and they don't seem to 'mix'.  Nothing moves through spacetime. They move through space, but have worldlines in spacetime. No worldline can be angled so much that its events become separated in a space-like manner.  That's the speed of light restriction. It's the same as saying that my worldline must be entirely contained in my own causal cones.
All this is no different inside the event horizon.  There's still 1 time and 3 spatial dimensions, but the event horizon is in the past light cone of any event in there.  None of it is in the future light cone.

I'm just using the terminology that Kip Thorne uses in his book (see the page I scanned above). The main thrust of what I'm saying is that the light cone tilting is not something that happens suddenly at the event horizon. It becomes more and more tilted as you approach it.

Not necessarily true inside the event horizon.  There are still the 3+1 dimensions with no preference for direction of movement through space. In theory, masses could form with stuff orbit them and such. There would be no obvious tidal force tearing such systems apart. The spacetime is reasonably normal and not different in a way that one is termed 'time-like' and the other not.

Perhaps my terminology is poor then. I figured there were degrees of "time-like"-ness.

I find the river a poor analogy because sufficient force would allow one to resist any arbitrarily large flow.  The river is still space. Time is not something that flows in a spacetime model.

Perhaps you are right. Perhaps I could instead say that the river flows at the speed of light. Then it becomes more apparent that force alone won't allow you to resist it.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/08/2019 18:52:20
I'm just using the terminology that Kip Thorne uses in his book (see the page I scanned above).
OK.  As for 'mixing', it just means that space and time are the same thing that can be measured in the same units if you wish. That's what spacetime is, and one does not move through spacetime.
I don't see references to 'time-like' spacetime, like it's something different.  The whole scanned page just seems to run through the basics of relativity of simultaneity.  That's the terminology that prompted my post. Maybe a different page uses such terms in a discussion involving an event horizon. I've not seen the book.

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The main thrust of what I'm saying is that the light cone tilting is not something that happens suddenly at the event horizon. It becomes more and more tilted as you approach it.
Agree. I've seen pictures where the tilt is abrupt at the event horizon, but it doesn't work that way.  Picture an object somewhat near a black hole.  Time moves upward in the picture, with the black hole spatially off to the right. The light cones of any event along that worldline point more or less up and down for future and past respectively. The past cone never touches the event horizon, even in the original picture where it grows straight down.  So the presence of the black hole already deforms the past light cone into a shape that isn't a cone.
As the worldline of the object approaches the event horizon, the light cone from any event along that line begins to tilt to the right so the future part of the cone begins to point rightward and the past cone away from it. As you get close, very little of the future light cone fails to intersect the event horizon.
At the event horizon is a singularity and there are not defined cones there.  In this sense, the transition is not a clean one.
For events inside, the event horizon does not exist anywhere in the future light cone, but physics is not particularly abnormal. You could live there. In my picture, the light cones now point left and right instead of up and down.  Vertical in the picture (what used to be the time dimension) is now just one of the 3 spatial dimensions.

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Perhaps you are right. Perhaps I could instead say that the river flows at the speed of light. Then it becomes more apparent that force alone won't allow you to resist it.
Force (alone or with help) will not allow you to push an object into the past. Speed of light restriction makes it sound like it's a spatially separated place you're trying to get to and you just can't go fast enough.  No.  It's a temporal separation, and you can't travel back in time. That's what prevents inside events from having a causal effect on the outside.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/08/2019 21:54:17
Halc, I am thankful that you are actively participating in this thread. Space-time swapping in a black hole is not something I have ever delved into in depth before this debate started. I can use all the correction I can get.

For others reading this thread: if in doubt, refer to Halc.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/08/2019 05:26:10
An event horizon is what it is specifically because nothing can get out of it and the reason nothing can get out is because space is fully time-like there.
Let's see if I understand it correctly:
We all know the meaning of Event horizon:
https://en.wikipedia.org/wiki/Event_horizon
In astrophysics, an event horizon is a boundary beyond which events cannot affect an observer on the opposite side of it. An event horizon is most commonly associated with black holes, where gravitational forces are so strong that light cannot escape.
This is very clear.
Therefore, based on that and our understanding of Space time - we have decided that the border of the "fully time-like" space time is at the event horizon:
An event horizon is what it is specifically because nothing can get out of it and the reason nothing can get out is because space is fully time-like there.
So far so good.
However, now we use the space time to prove that nothing can escape from this point.
An event horizon is what it is specifically because nothing can get out of it and the reason nothing can get out is because space is fully time-like there.
There must be an error in this logic.
How can we say?
The boarder of the" fully time-like" space time is at the event of horizon because nothing can escape from that point and than in order to proof that nothing can escape from there we use the space-time formula.
Sorry -
Nothing can escape from the event horizon based on its meaning and not because of space-time.
Therefore, space time can't be used as a formula or evidence to proof that nothing can escape from the horizon.
This approach has a built in logical error in my point of view.
Hence, the following statement isn't logical:
"An event horizon is what it is specifically because space is fully time-like there".




Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/08/2019 06:20:09
The boarder of the" fully time-like" space time is at the event of horizon because nothing can escape from that point and than in order to proof that nothing can escape from there we use the space-time formula.

You have it backwards. The reason nothing can escape is because space is time-like within the horizon. This, in turn, is what causes it to be an event horizon.

There must be an error in this logic.

Then show where the math is wrong.

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Nothing can escape from the event horizon based on its meaning and not because of space-time.

You are confusing the definition of an event horizon with what causes it to be an event horizon. This is like arguing that, "water is wet based on its meaning and not because of its molecular properties".
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/08/2019 12:24:21
Space-time:
https://en.wikipedia.org/wiki/Spacetime
Space-time interval = (Δs)
(Δs)^2 = (Δx)^2 - (Δt)^2
As: t represents the time, and x represents the space.
I was under the impression that in space-time the space is orthogonal to time.
Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions
"The basis vectors of the four dimensional Minkowski space (which is used in relativity) are orthogonal. But, you need to remember that a visualization of these four dimensions works different than just adding one extra dimension to our common three dimensional Euclidean space.
The time dimension is something special. Distances between two points in our three dimensional Euclidean space are always real positive numbers (the distance from New York to Chicago is 790 miles). Minskowski space allows for imaginary distances that need to be interpreted as timelike distances. These are events that can have a causal relation, they are in our future or past. Minskowski allows also for zero distances between two points that are spatially separated. These need to be interpreted as the paths of light signals send between two points. Positive distances in a Minkowski space are interpreted as events that don't have a causal relationship. Some observers may see these events happening at the same time, some don't. But everyone agrees they happen at different locations.
So yes, time is "perpendicular" to the other spacial dimensions in Minkowski space.  No, there is an important twist in the mathematics, interpretation, making the time dimension not the same as the spatial dimension."

So, "Minkowski space allows for imaginary distances that need to be interpreted as timelike distances."
Hence, this imaginary distances (or Imaginary time), sets the Minus sign.
However, once we start with imaginary distance, the time could also be imaginary and the whole outcome of the space time formula could also be imaginary.
Let's assume that Minkowski formula for the two dimensions was as follow:
(Δc)^2 = (Δa)^2 - (Δb)^2
So, does it mean that this formula is correct?
What Mr. Pythagoras would say about that?
In the same token, I wonder what Mr. Einstein (who was the student of Minkowski) had stated about that imaginary formula.
I really can't understand how our scientists can get any real information from that imaginary formula..
In my opinion, the outcome of imaginary formula of Minkowski must also be imaginary/unreal.
Therefore, this imaginary formula sets the imaginary/unreal curvature in our Universe and imaginary/unreal curvature/singularity/outcome in a BH.
Hence, If Mr. Einstein didn't approve the use of that formula by Minkowski, I will also not going to accept it and we all should reject it.
Any idea about Mr. Einstein approach to this formula?

Let's continue our discussion about
You have it backwards. The reason nothing can escape is because space is time-like within the horizon. This, in turn, is what causes it to be an event horizon.
So, what comes first?
As we know, Space time is not about a BH or a Universe.
However, due to the Imaginary space, there is a situation where
(Δs)^2 = (Δx)^2 - (Δt)^2 = 0
I assume that this set the border of the "fully time-like" space time.
But there is no real data about the exact radius in this space-time formula.
So, how do we know what is the value of the radius?  Where is the boarder?

Space-time doesn't "know" anything. It simply "is" fully time-like at the event horizon. This border automatically exists at any location where the escape velocity reaches the speed of light. So it is defined by gravity.
I agree it's defined by gravity.
Space time doesn't give any information about Gravity at the BH.
If we could extract the gravity and the radius from the space-time, than we could clearly say that it is he reason for the event horizon.
However, we must use different formula to find the boarders.
So, the Space time can't have any influence on the formula which sets the event horizon formula.
Therefore - the "fully time-like" of the space time can't be used to set the event horizon.
This Event Horizon is set by the following (Schwarzschild) Radius Calculation:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
So, Schwarzschild tells Minkowski where are the  boarders and not the other way.
Therefore, Minkowski space time has no impact on those boarder and therefore, it doesn't prove anything about those boarders.
Hence, I really don't understand why you insist that:
The reason nothing can escape is because space is time-like within the horizon
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/08/2019 14:42:53
Space-time:
https://en.wikipedia.org/wiki/Spacetime
Space-time interval = (Δs)
(Δs)^2 = (Δx)^2 - (Δt)^2
As: t represents the time, and x represents the space.
I was under the impression that in space-time the space is orthogonal to time.
Yes, it's like the Pythagorean theorem that works only for triangles with two orthogonal sides.  The formula is more complicated if non-orthogonal axes are chosen.

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Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions
Be very cautious about anything you get from quora. The don't have a mechanism to eliminate the crap answers, and there are a lot of them.

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"The basis vectors of the four dimensional Minkowski space (which is used in relativity) are orthogonal. But, you need to remember that a visualization of these four dimensions works different than just adding one extra dimension to our common three dimensional Euclidean space.
The time dimension is something special. Distances between two points in our three dimensional Euclidean space are always real positive numbers (the distance from New York to Chicago is 790 miles). Minskowski space allows for imaginary distances that need to be interpreted as timelike distances.
This seems nonsense.  Distance along the time dimension is measured in real numbers as well.  It doesn't take me 2i-.5 hours to do something.

Quote from: quora
Minskowski allows also for zero distances between two points that are spatially separated. These need to be interpreted as the paths of light signals send between two points.
Blatantly wrong.  The statement is self contradictory. Light cannot travel between two spatially separated spacetime points, nor can it travel between temporally separated points (at least not in a vacuum).  Only light-like separated points (events) can have light travel between them.
The quora answer also speaks of points, not events, a terminology error.

Quote from: quora
Positive distances in a Minkowski space are interpreted as events that don't have a causal relationship.
Here at least they use 'events' correctly, but it should read 'Positive intervals', not 'Positive distances' for the statement to be correct.  Intervals are not distances.  You might observe a star exploding, and the event of the explosion and the event of you seeing it might be separated by a (frame dependent) distance of 500 light years, but a frame independent interval of zero.  Distance and interval are very different things.

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So, "Minkowski space allows for imaginary distances that need to be interpreted as timelike distances."
Hence, this imaginary distances (or Imaginary time), sets the Minus sign.
You have a troll's talent for spotting the crap in a web site and quoting it as gospel, but questioning all the correct sites.
Imaginary numbers don't set a minus sign.  They set an imaginary component i.  If you want to make sense of time being measured as 5i or something, be my guest.  Last I looked, my clock didn't have an i on it.

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However, once we start with imaginary distance, the time could also be imaginary and the whole outcome of the space time formula could also be imaginary.
Let's assume that Minkowski formula for the two dimensions was as follow:
(Δc)^2 = (Δa)^2 - (Δb)^2
So, does it mean that this formula is correct?
What Mr. Pythagoras would say about that?
The interval is not a measurement of a hypotenuse, something that would be frame dependent.  Pythagoras was describing a spatial relation, not a spacetime one.

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Therefore, this imaginary formula sets the imaginary/unreal curvature in our Universe and imaginary/unreal curvature/singularity/outcome in a BH.
The formula is not imaginary and does not describe curvature.  It works in Euclidean space, and is thus more of an SR thing than a GR thing.

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Hence, If Mr. Einstein didn't approve the use of that formula by Minkowski,
Where do you get this assertion?

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As we know, Space time is not about a BH or a Universe.
However, due to the Imaginary space, there is a situation where
(Δs)^2 = (Δx)^2 - (Δt)^2 = 0
I assume that this set the border of the "fully time-like" space time. But there is no real data about the exact radius in this space-time formula.
The above is word salad and has nothing to do with event horizons.  Read my prior posts for description of how the orientation of causal light cones relates to the Schwarzschild radius and hence an event horizon (a singularity) must exist there.

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So, how do we know what is the value of the radius?
Approximately 2GM/c².  This isn't a real radius (an actual length), but rather a measure of the circumference divided by 2π.
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Where is the boarder?
In the spare room.

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This Event Horizon is set by the following (Schwarzschild) Radius Calculation:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
There, see?  You answered your own question.
The site is a calculator and lets you specify a gravitational constant that is in units of m/sec², which is wrong.  Mistakes (including grammatical ones) on the site aside, the quote above is correct.

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So, Schwarzschild tells Minkowski where are the  boarders and not the other way.
Therefore, Minkowski space time has no impact on those boarder and therefore, it doesn't prove anything about those boarders.
And we're back to word salad. Minkowski never published anything about black holes.  Schwarzschild didn't tell Minkowski anything since the latter died before making that contribution.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/08/2019 21:56:46
In the same token, I wonder what Mr. Einstein (who was the student of Minkowski) had stated about that imaginary formula.
I really can't understand how our scientists can get any real information from that imaginary formula..
In my opinion, the outcome of imaginary formula of Minkowski must also be imaginary/unreal.

Don't mistake the word "imaginary" for "fictional" in the context of mathematics.

Any idea about Mr. Einstein approach to this formula?

He was initially unenthusiastic, but later realized its great importance. To quote from "Black Holes & Time Warps":

Quote from: Page 93
When Einstein learned of Minkowski's discovery, he was not impressed. Minkowski was merely rewriting the laws of special relativity in a new, more mathematical language; and, to Einstein, the mathematics obscured the physical ideas that underlie the laws. As Minkowski continued to extol the beauties of his spacetime viewpoint, Einstein began to make jokes about Gottingen mathematicians describing relativity in such complicated language that physicists wouldn't be able to understand it.

The joke, in fact, was on Einstein. Four years later, in 1912, he would realize that Minkowski's absolute spacetime is an essential foundation for incorporating gravity into special relativity.

Quote from: Page 106
The idea of warpage of both time and space was rather daunting. Since the Universe admits an infinite number of different reference frames, each moving with a different velocity, there would have to be an infinity of warped times and an infinity of warped spaces! Fortunately, Einstein realized, Hermann Minkowski had provided a powerful tool for simplifying such complexity: "Henceforth, space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality." There is just one, unique, absolute, four-dimensional spacetime in our Universe; and a warpage of everyone's time and everyone's space must show up as a warpage of Minkowski's single, unique, absolute spacetime.

This was the conclusion to which Einstein was driven in the summer of 1912 (though he preferred to use the word "curvature" rather than "warpage"). After four years of ridiculing Minkowski's idea of absolute spacetime, Einstein had finally been driven to embrace is, and warp it.

So yes, Einstein did accept the Minkowski metric.

So, what comes first?

This is like asking, "which comes first, the water or the wetness?" Neither "comes first".

However, due to the Imaginary space, there is a situation where
(Δs)^2 = (Δx)^2 - (Δt)^2 = 0
I assume that this set the border of the "fully time-like" space time.
But there is no real data about the exact radius in this space-time formula.
So, how do we know what is the value of the radius?  Where is the boarder?

That's easy. Anywhere that (Δx)^2 equals (Δt)^2, you will get zero as your answer. That's basic algebra.

I agree it's defined by gravity.
Space time doesn't give any information about Gravity at the BH.
If we could extract the gravity and the radius from the space-time, than we could clearly say that it is he reason for the event horizon.
However, we must use different formula to find the boarders.
So, the Space time can't have any influence on the formula which sets the event horizon formula.
Therefore - the "fully time-like" of the space time can't be used to set the event horizon.
This Event Horizon is set by the following (Schwarzschild) Radius Calculation:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
So, Schwarzschild tells Minkowski where are the  boarders and not the other way.
Therefore, Minkowski space time has no impact on those boarder and therefore, it doesn't prove anything about those boarders.
Hence, I really don't understand why you insist that:
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The reason nothing can escape is because space is time-like within the horizon

This is a bunch of nonsense. You know that gravity is a distortion in space-time, right? So obviously, the local curvature of space-time is going to tell you everything about the gravity there.

Also, no one is pretending that a single equation is going to tell you everything you want to know about something. You can't use one equation to tell you both what the surface gravity on Earth is and what its orbital period around the Sun is. Those two equations, however, are closely related and rely on the same basic physics of gravity to work. They are consistent with each other. Same thing for Minkowski's equations and those of Schwarzschild.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/08/2019 05:54:32
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Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions
Be very cautious about anything you get from quora. The don't have a mechanism to eliminate the crap answers, and there are a lot of them.
Thanks Halc
So, you  don't like the explanation by " quora", however, you didn't offer any real explanation for the following formula by Minkowski:
(Δs)^2 =  (Δx)^2 - (Δt)^2
How could it be that the (Δt)^2 comes with Minus sign???
Don't forget that even if the original sign of (Δt) was minus, than after the following calculation:
(-Δt)^2 = (Δt)^2
We should get it as a positive value.
So, even if the (-Δt) represents a time in the past, the outcome of the (-Δt)^2 calculation MUST  be positive.
I have never ever found any possibility that
(-Δt)^2 = -(Δt)^2
Don't mistake the word "imaginary" for "fictional" in the context of mathematics.
I'm not sure that I understand this answer.
Do you agree that the Δt is "imaginary" as stated in " quora"?
Would you kindly explain how could it be that:
(+/-Δt)^2 = -(Δt)^2
I really want to understand this calculation.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 31/08/2019 06:04:09
How could it be that the (Δt)^2 comes with Minus sign???

It doesn't. What the equation is doing is subtracting (Δt)2 from (Δx)2.

I have never ever found any possibility that
(-Δt)^2 = -(Δt)^2

Good, because that's not what the equation says.

I'm not sure that I understand this answer.
Do you agree that the Δt is "imaginary" as stated in " quora"?

https://en.wikipedia.org/wiki/Imaginary_time

Would you kindly explain how could it be that:
(+/-Δt)^2 = -(Δt)^2

No one is saying that is the case.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 31/08/2019 06:07:12
Quote
Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions
Be very cautious about anything you get from quora. The don't have a mechanism to eliminate the crap answers, and there are a lot of them.
Thanks Halc
So, you  don't like the explanation by " quora", however, you didn't offer any real explanation for the following formula by Minkowski:
(Δs)^2 =  (Δx)^2 - (Δt)^2
How could it be that the (Δt)^2 comes with Minus sign???
Because it only works that way.  There would be no Lortentz transform if it were a plus sign.  Light speed would be frame dependent.  It would not be the universe we observe.
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Don't forget that even if the original sign of (Δt) was minus, than after the following calculation:
(-Δt)^2 = (Δt)^2
We should get it as a positive value.
So, even if the (-Δt) represents a time in the past, the outcome of the (-Δt)^2 calculation MUST  be positive.
Correct again.  That means that the spacetime interval is imaginary for any pair of events separated in a time-like manner.  That's how you know the separation is time-like.  Real for space-like (causally separated), and zero for light-like.

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Do you agree that the Δt is "imaginary" as stated in " quora"?
Quora is wrong more than it is right.  Don't get your science from there.
More specifically, if Δt was imaginary, then squaring it would yield a potentially negative number and the interval calculation would not be frame independent. That makes it wrong if he said that, but he didn't say time was imaginary. He called the interval an 'imaginary distance', but the word 'distance' is misleading.  An interval is not a distance.
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I really want to understand this calculation.
Why are you trying to understand the concept of spacetime interval when the discussion was concerning event horizons?  There isn't a defined interval between events on opposite sides of a gravitational event horizon.

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I'm not sure that I understand this answer.
Do you agree that the Δt is "imaginary" as stated in " quora"?
https://en.wikipedia.org/wiki/Imaginary_time
Oh that's interesting.  It is a different interpretation that assumes t itself is imaginary in an effort to make the interval a real number. I don't think the quora answer is assuming this interpretation. It requires a change of Minkowski's formula to: d²=x²+y²+z²+(it)² which always yields a real distance (Δd) instead of an imaginary interval Δs for causally connected events.  But can time have a nonzero real component then?  If not, the formula above seems incorrect in the general case.  Does it make sense to say time imaginary like that?  Kinetic energy would then always be negative, which seems to make no sense.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/08/2019 12:10:14
Why are you trying to understand the concept of spacetime interval when the discussion was concerning event horizons?
Because the Space-time is used as a proof for the idea that nothing can escape from the radius below the event horizon.
An event horizon is what it is specifically because nothing can get out of it and the reason nothing can get out is because space is fully time-like there.
If you give up on that, I won't ask any more questions about the space-time.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/08/2019 14:14:01
1) Spacetime is not used as such a proof,
Perfect.
So, particle pair that had just been created below the event of horizon, are not forced to stay there due to space-time.
However, the event horizon radius is very clear and it is calculated by the following (Schwarzschild) Radius formula:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the Schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
We all know that the maximal velocity of particle is the speed of light.
Therefore, even  if a particle orbits at the maximal speed (speed of light) and it is located below that radius, it won't be able to be ejected from that radius due to gravity force.
Therefore, the simple conclusion is that nothing can be ejected from the event of horizon.
So far so good.
However, if we add the impact of the Lorentz force due to magnetic field, than now the outcome is quite different.
Let's assume that one of the new particle pair carry positive charge, while the other one carry negative charge.
So, please don't say the following message again:
It absolutely is about negative and positive mass, since we are talking about super-massive black holes (which cannot produce charged particles because those particles are too massive).
Let's assume that one has a negative charge while the other carry a positive charge.
Let's also assume that there is a magnetic field around that BH (let's assume that it is very high (infinite).
Kryptid- Although true black holes cannot have such a magnetic field, I will submit to the possibility that something like MECOs (magnetospheric eternally collapsing objects) could, maybe, be what "black holes" actually are. So I will tentatively agree that "black holes" could be MECOs and as such could have magnetic fields. I'll consider this plausible for the sake of discussion: https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object
Let's also assume that the positive charged particle is deflected outwards due to Lorentz force (assuming that it is already on a path perpendicular to the field lines):
Dave - 3. Lorentz force - Based on Lorentz force, the magnetic fields deflects differently the path of the orbital new born particles pair:
Hence, if the positive charged particle will be deflected outwards, the negative charged particle will be deflected inwards.
Therefore, while the negative is pushed inwards into the center of the SMBH, the positive is pulled outwards and get's eventually into the accretion disc.
Kryptid - The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in". The Lorentz force would deflect the path of particles at a right angle to the field lines (assuming that they were already on a path perpendicular to the field lines. If they are parallel to the field lines, there is no force).
Therefore, as the Magnetic field is almost infinite, let's assume that the Lorentz force on the positive charged particle might is also infinite.
So, do you agree that this infinite Lorentz force should deflect the path of the new born positive particle?
Actually, this particle will face two forces (assuming that it orbits at ultra high velocity - almost the speed of light).
One - Gravity force - that pulls it inwards.
Two - Lorentz force - That pushes it outwards
So based on all above assumptions, do you agree that If Lorentz force is stronger than the gravity force, the positive particle should eventually be ejected outwards from the event horizon even while its velocity is still below the speed of light?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 31/08/2019 14:46:58
Perfect.
So, particle pair that had just been created below the event of horizon, are not forced to stay there due to space-time.

It's forced to move inward because space is time-like inside the horizon.

However, if we add the impact of the Lorentz force due to magnetic field, than now the outcome is quite different.

No it isn't, because space is still time-like below the horizon.

Let's assume that one of the new particle pair carry positive charge, while the other one carry negative charge.
So, please don't say the following message again:

Sorry, but if you are going to insist that super-massive black holes can produce charged particles, then I am going to have to keep repeating myself: they can't.

Let's also assume that the positive charged particle is deflected outwards due to Lorentz force (assuming that it is already on a path perpendicular to the field lines):

That's not how the Lorentz force works.

So, do you agree that this infinite Lorentz force should deflect the path of the new born positive particle?

Only if it is already outside of the horizon and even then it won't deflect it away from the black hole. It will deflect it at a right angle to its path instead: https://www.khanacademy.org/test-prep/mcat/physical-processes/magnetism-mcat/a/using-the-right-hand-rule

Let's say you are sitting at the north pole of the black hole and you are watching positively-charged particles stream away from the equator. As they move away, they encounter magnetic field lines running perpendicular to their path (the field lines running downwards relative to you). You can use the right-hand rule to tell you what reaction these charged particles will have to the field lines. If the particle is moving away from the equator while the direction of the magnetic field line is downward, then the force acting on a positively-charged particle is to the left.

However, since this left-moving particle is still interacting with magnetic field lines, it will want to turn left (relative to its current direction) yet again. This results in the particle travelling into a counter-clockwise circle (from your vantage point atop of the black hole). So the magnetic field neither pushes the particle away from the horizon nor pulls it in. The opposite is true for a negatively-charged particle: it will be deflected right and end up circling around the field lines in a clockwise circle.

Actually, this particle will face two forces (assuming that it orbits at ultra high velocity - almost the speed of light).
One - Gravity force - that pulls it inwards.
Two - Lorentz force - That pushes it outwards
So based on all above assumptions, do you agree that If Lorentz force is stronger than the gravity force, the positive particle should eventually be ejected outwards from the event horizon even while its velocity is still below the speed of light?

No, for the reasons we have repeated over and over and over...
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/09/2019 21:21:20
Thanks Halc
It is clear to me that you don't agree with (almost) any idea that I offer.
Let me understand if you agree with the observation about the Milky Way
1. More than 99% of the matter in the accretion disc is ejected outwards. So out of 100 particles - more than 99 are ejected outwards.
2. So far our scientists couldn't find even one particle that is falling into the accretion disc from outside, or from the accretion disc into the SMBH, although they monitor the SMBH almost constantly with several X-ray detectors.
3. There is a magnetic field around the accretion disc. This magnetic field sets the Molecular jet steam above and below the accretion disc. We have already found that the jet stream velocity is almost 0.8 c while it moves up to 27,000 LY above and below the disc.
I hope that at least you agree with those observations.

If so, let's assume that there is a particle outside the magnetic field which is bagging to be eaten by the SMBH.
So, first it must cross the magnetic field.
Can you please explain how any particle can cross that magnetic field? Why it won't be boosted by the magnetic field and join all the other particles in that molecular jet stream? What is the chance that it can do it?  Is it one to 1 Million or 1 to 1 Billion?
Let's assume that somehow one particle was very lucky and could cross the magnetic field.
So, now this particle is there trying to cross the last gap to the accretion disc, while it is facing all of that 99% stream of matter that are ejected outwards.
If those 99% are ejected outwards, don't you agree that there must be some sort of force that drives them outwards?
I assume that you call this force - tidal.
Why that tidal don't push outwards also this particle?
How could it be that the tidal ejects those 99% of the matter in the accretion disc, while it lets that particle to get inwards to the accretion disc?
Do you think that there is a policeman that knows where the particle is coming from?
How could it be that out of 100 particles in the accretion disc, more than 99 are ejected outwards, while the way for our particle will be clear to fall into the accretion disc?
Let's assume that against all odds it was so lucky and finely arrived to the accretion disc.
However, after all the difficulties in its way to the accretion disc, now it has a chance of less than 1% to fulfill its goal and fall finely into the SMBH. Actually it has a chance of more than 99% to be ejected outwards.
Did we try to find how many times that poor particle must come back again to the accretion disc in order to be eaten by the mighty SMBH? Do you agree on at least 99 times?
I wonder how you still believe that a particle which had been ejected at least 99 times from the accretion disc will come back again with great hope that this is the last time.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/09/2019 01:50:09
1. More than 99% of the matter in the accretion disc is ejected outwards. So out of 100 particles - more than 99 are ejected outwards.

That appears to be the case with Sagittarius A*, yes.

2. So far our scientists couldn't find even one particle that is falling into the accretion disc from outside, or from the accretion disc into the SMBH, although they monitor the SMBH almost constantly with several X-ray detectors.

I'm not so sure that is true. Indirect evidence of matter accreting onto the disk or falling into the hole would come from detected X-ray emissions.

3. There is a magnetic field around the accretion disc. This magnetic field sets the Molecular jet steam above and below the accretion disc. We have already found that the jet stream velocity is almost 0.8 c while it moves up to 27,000 LY above and below the disc.

I don't remember if you ever answered this question, but where did you get the 0.8c figure from?

Can you please explain how any particle can cross that magnetic field? Why it won't be boosted by the magnetic field and join all the other particles in that molecular jet stream? What is the chance that it can do it?  Is it one to 1 Million or 1 to 1 Billion?

That strongly depends on the trajectory of the particle. Remember, charged particles move in spirals around magnetic field lines. If the particles are already moving in an upward or downward direction, then they could spiral up the field lines and into the jets. If they hit the field lines at a right angle, however, then the particles would be stuck moving in circles without significant movement up or down. This spiraling also causes the charged particles to release electromagnetic radiation, which makes them lose energy and slow down. Eventually, they should slow down enough to fall into the hole because the magnetic force on a charged particle becomes weaker as the particle slows down.

Alternatively, the particle that is stuck moving in circles could be pushed in by other particles hitting it from behind.

So, now this particle is there trying to cross the last gap to the accretion disc, while it is facing all of that 99% stream of matter that are ejected outwards.
If those 99% are ejected outwards, don't you agree that there must be some sort of force that drives them outwards?
I assume that you call this force - tidal.
Why that tidal don't push outwards also this particle?

The magnetic field is what does this, not tidal forces. As I said before, the trajectory of the particle is what determines whether the magnetic field can move it towards the poles or not.

Do you think that there is a policeman that knows where the particle is coming from?

Was that a joke?

How could it be that out of 100 particles in the accretion disc, more than 99 are ejected outwards, while the way for our particle will be clear to fall into the accretion disc?

Not all particles are moving at the same velocity or in the same direction. You wouldn't expect all of them to react to the magnetic field in exactly the same way.

Let's assume that against all odds it was so lucky and finely arrived to the accretion disc.

I presumed we were talking about a particle that was already in the accretion disk. It's not hard for particles to add themselves to the outer regions of an accretion disk because the outer regions are cool and slow-moving. That would make any magnetic fields present there weak. At least, that is the case with sufficiently large disks.

However, after all the difficulties in its way to the accretion disc, now it has a chance of less than 1% to fulfill its goal and fall finely into the SMBH. Actually it has a chance of more than 99% to be ejected outwards.
Did we try to find how many times that poor particle must come back again to the accretion disc in order to be eaten by the mighty SMBH? Do you agree on at least 99 times?

Who said anything about the particle coming back into the accretion disk? How could it do so if it was already being ejected by the jet? I think you are also underestimating the sheer number of particles that exist in clouds of gas. Even a single gram of hydrogen plasma contains 5.975 x 1023 electrons and 5.975 x 1023 protons. Even if 99.999% of those particles were ejected, that's still 1.195 x 1019 particles that fall in. There are lots and lots of grams in an accretion disk.

I also came to another realization. If you propose that the magnetic field is too strong to allow charged particles from the accretion disk to fall into the black hole, then it must also prevent movement in the opposite direction. That is, any charged particles generated around the black hole must not be able to get out into the accretion disk. So your idea that the accretion disk is actually an excretion disk is unworkable if you consider the magnetic field to be impassable.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/09/2019 02:36:40
Thanks Halc
It is clear to me that you don't agree with (almost) any idea that I offer.
You do seem to be going out of your way to misrepresent or ignore anything you read here or on sites, so yes, what do you expect from us?

Quote
Let me understand if you agree with the observation about the Milky Way
1. More than 99% of the matter in the accretion disc is ejected outwards.
 So out of 100 particles - more than 99 are ejected outwards.
One site said something like that, yes.  It doesn't say how the figure is measured.  Sgr-A masses less than 0.000005 the mass of the galaxy, so obviously less than 1% (0.01) has fallen in.  How is the 99% figure computed?  If 99% of the accretion disk is ejected outward, why is there still an accretion disk sitting there?  Just saying the article wasn't very specific about how it got those numbers.

Quote
2. So far our scientists couldn't find even one particle that is falling into the accretion disc from outside, or from the accretion disc into the SMBH, although they monitor the SMBH almost constantly with several X-ray detectors.
This is blatantly wrong. This was stated on no site. In fact, the opposite was recently observed (see posts 597 and 651) , totally countering the 99% thing you state in point 1.  Yes, when the disk is not fed, almost nothing falls further in, just like in the solar system where far less than 1% of the matter orbiting the sun is seen to fall in.
The article you reference in point 3 talked about 10,000 solar masses falling in, which is quite a bit more than 'not even one particle'.

Quote
3. There is a magnetic field around the accretion disc. This magnetic field sets the Molecular jet steam above and below the accretion disc. We have already found that the jet stream velocity is almost 0.8 c while it moves up to 27,000 LY above and below the disc.
This is a misrepresentation of what the article says.

Quote
If so, let's assume that there is a particle outside the magnetic field which is bagging to be eaten by the SMBH.
So, first it must cross the magnetic field.
Can you please explain how any particle can cross that magnetic field? Why it won't be boosted by the magnetic field and join all the other particles in that molecular jet stream? What is the chance that it can do it?  Is it one to 1 Million or 1 to 1 Billion?
I have posted little opinions on how the magnetic fields work. All I know is that if the energy is boosted, it needs to be taken from somewhere else.  If it falls in slow and is ejected at 0.8c, then it took energy away from the system, and that loss of energy must be reflected in either a weaker disk energy or lower potential energy (the mass of the disk drops towards/into the black hole). Conservation of mass will not allow otherwise.
Only charged particles will be affected by a magnetic field, no? That excludes anything that the Hawking radiation from the black hole itself produces.

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Let's assume that somehow one particle was very lucky and could cross the magnetic field.
So, now this particle is there trying to cross the last gap to the accretion disc, while it is facing all of that 99% stream of matter that are ejected outwards.
If it is coming from anywhere except the plane of the disk or the poles, it meets none of this.  The main reason it it ejected is perhaps because its aim was poor and it misses the black hole. That's the same reason almost all comets (at least 99% of them in fact) are ejected by the sun after they fall in.  Most of them simply are not aimed close enough to the sun, just like Earth isn't.

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If those 99% are ejected outwards, don't you agree that there must be some sort of force that drives them outwards?
Inertia isn't a force.  If there's a force involved, there needs to be energy expended by something to apply that force.  If that comes from the magnetic field, then the field is weakened by this expenditure.  That's why we don't get free electricity just by spinning magnets.  The spinning slows down if they are used to exert a force on anything.  If the field was generated by an orbit of particles around something, that energy loss would cause them to drop further towards the thing orbited.  Reduction of kinetic energy like that is how they take the space shuttle out of orbit.

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I assume that you call this force - tidal.
No, of course not.  Tidal forces are not involved in the ejection of material. Tidal forces act to separate connected objects. Such forces are greatest where the gravitational gradient is greatest, and an SMBH has a low gravitational gradient.

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Why that tidal don't push outwards also this particle?
How could it be that the tidal ejects those 99% of the matter in the accretion disc, while it lets that particle to get inwards to the accretion disc?
No such thing happens. Tides play no role. I do recall you saying that magnetic force was involved. The articles I read say the physics is not fully known, which is why I'm not claiming to know the correct answers.

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Let's assume that against all odds it was so lucky and finely arrived to the accretion disc.
However, after all the difficulties in its way to the accretion disc, now it has a chance of less than 1% to fulfill its goal and fall finely into the SMBH. Actually it has a chance of more than 99% to be ejected outwards.
Did we try to find how many times that poor particle must come back again to the accretion disc in order to be eaten by the mighty SMBH? Do you agree on at least 99 times?
I don't know what those numbers represented, or how accurate they are, especially when negligible material is falling in most of the time.  I agree that it is long odds that a random particle on a random trajectory will find itself anywhere near our black hole.  It's a dang insignificant target.  Earth is in danger of falling in, but the odds are still incredibly small.  If it were to be deflected straight in, nowhere near 99% of it would be ejected, but it also wouldn't all go in no matter how well aimed.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/09/2019 05:01:18
I don't know what those numbers represented, or how accurate they are, especially when negligible material is falling in most of the time. 

Although that 99%+ figure was obtained using observational data, the irony of it is that assumptions and models were used in order to get to that number (which Dave Lev seems to be leery of). Here are some quotes from the paper:

Quote
The flat density profile of the flow (i.e., s ∼ 1) suggests the presence of an outflow that nearly balances the inflow (26). As a result, <∼ 1% of the matter initially captured by the SMBH reaches the innermost region around Sgr A*, limiting the accretion power to <∼ 1039 erg s−1.

And about that "flat density profile":

Quote
The RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ2/n.d.f. = 187/218) and in terms of matching individual lines (Fig. 2b). The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed [e.g., (14,17)], this suggests a very flat density profile of the flow (i.e., s ∼ 1), indicating an outflow mass-loss rate that nearly balances the inflow.

So it's not like they directly measured the flow. They had to model it based on spectral data.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/09/2019 16:41:53
Quote
The RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ2/n.d.f. = 187/218) and in terms of matching individual lines (Fig. 2b). The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed [e.g., (14,17)], this suggests a very flat density profile of the flow (i.e., s ∼ 1), indicating an outflow mass-loss rate that nearly balances the inflow.
So it's not like they directly measured the flow. They had to model it based on spectral data.
Thanks Krypid
So you agree that they don't measured directly the inflow.
This article is actually based on assumptions:
"If θ ∼ 1 as typically assumed"
"assuming that the magnetic field is near equipartition,"
"With this upper limit, assuming that ri ∼ 102 rs we can infer s > 0.6"
"where PBondi is assumed to have a 10% efficiency of the Bondi mass accretion rate"
"The dashed circle around Sgr A* marks its Bondi capture radius (assumed to be 400)"
"..assuming collisional ionization equilibrium"
"..because the measurement then depends sensitively on the assumed thermal plasma model"
" The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed.

If we will assume that (θ ∼ 1) is incorrect, does it mean that our assumption is correct or incorrect?
Sorry - is it about science, or assumption game?
I can't understand how those scientists from arxiv do whatever it takes to show that somehow there must be in falling matter.
Are they looking to find the real understanding about our galaxy or they just wish to fit the current mainstream to the observations against all odes.
So, would you agree to say: "They had to model it based on spectral assumption"?

Let me offer the following articles from the same arxiv:
https://arxiv.org/pdf/1104.5443.pdf
"In regions close to the black hole, the AGN outflows are revealed through blueshifted absorption lines in X–ray emission (Pounds et al. 2003a, b; King 2010a). Tombesi et al. (2010a, b) show that they are present in more than 35 percent of a sample of over 50 local AGN, and deduce that their solid angles are large (certainly > 0.6 × 2π, and probably greater). The observed absorption columns imply that in many cases the outflows are quite recent (few years), suggesting that outflows are an almost ubiquitous feature of central black hole activity (King 2010b)"

Therefore, they clearly say that there is an outflow from the SMBH –" the AGN outflows are revealed through blueshifted".
In any case, so far I couldn't find any real X-Ray observation for any in falling matter into the SMBH or from outside into the accretion disc.
We only get clear observations for OUTFLOWS.

In the following article (from arxiv):
https://arxiv.org/ftp/arxiv/papers/1501/1501.07664.pdf
"Wind from the black-hole accretion disk driving a molecular outflow in an active galaxy"
"Recent observations of large-scale molecular outflows3,4,5,6,7,8 in ultraluminous infrared galaxies (ULIRGs) have provided the evidence to support these studies, as they directly trace the gas out of which stars form.
So, now they claim that the outflow wind from the BH (or SMBH) drives the star formation activity.
This fits perfectly with Theory D.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/09/2019 19:47:14
To measure it directly, you'd have to be there and be able to capture everything going in and measure the mass before letting it go on, and even then it can be argued to be not direct.  So any measurement has levels of indirection.
How could it be that we see so clearly the outflow from the SMBH or from the accretion disc?
Do you agree that we see the outflow directly by X-ray and by the blueshift: " the AGN outflows are revealed through blueshifted".
So, do you agree that we don't need to be there in order to see the outflow?
However, when it comes to the Inflow - Why we need modeling?
Why we can't see directly the inflow as we clearly see the outflow?
If the total quantity of outflow is almost identical to the Inflow than why we can't see both of them in the same observation tools?

Don't think matter outside the accretion disk is likely to radiate X-rays.
It seems to me that as the matter radiates when it is ejected outwards, it should also radiates when it is falling inwards.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/09/2019 22:08:56
Thanks Krypid
So you agree that they don't measured directly the inflow.
This article is actually based on assumptions:
"If θ ∼ 1 as typically assumed"
"assuming that the magnetic field is near equipartition,"
"With this upper limit, assuming that ri ∼ 102 rs we can infer s > 0.6"
"where PBondi is assumed to have a 10% efficiency of the Bondi mass accretion rate"
"The dashed circle around Sgr A* marks its Bondi capture radius (assumed to be 400)"
"..assuming collisional ionization equilibrium"
"..because the measurement then depends sensitively on the assumed thermal plasma model"
" The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed.

You know what you're doing, don't you? You're attacking the very study that you have continually cited as saying that over 99% of the matter in the Milky Way's accretion disk is ejected outward. If you are skeptical of assumptions and models, then you must also be skeptical of the very study that yielded that 99%+ figure. Modelling and assumptions were involved in the acquiring of that figure. I hope you understand that now.

If we will assume that (θ ∼ 1) is incorrect, does it mean that our assumption is correct or incorrect?
Sorry - is it about science, or assumption game?

When it comes to studying phenomena that are very difficult to observe (such as if they are very, very far away or very, very small), assumptions are necessary in order to come to conclusions. As I said before, science is about evidence, not proof. The question then becomes whether the assumptions are reasonable or not. Predictions about the properties of an exoplanet that we have not seen visually are much more likely to be correct if we assume that it is sphere-shaped instead of cube-shaped. Assumptions are practically unavoidable.

Are they looking to find the real understanding about our galaxy or they just wish to fit the current mainstream to the observations against all odes.
So, would you agree to say: "They had to model it based on spectral assumption"?

Like I said, assumptions are often unavoidable if we cannot directly observe something. The most we can hope for is to make the assumptions as reasonable as possible. Either that, or wait until we have technology that is good enough to make direct observations. That's no reason not to try to figure out what is happening the in mean time. Can these assumptions be wrong? Yes. Are they likely to be wildly wrong? That's a different matter.

Let me offer the following articles from the same arxiv:
https://arxiv.org/pdf/1104.5443.pdf
"In regions close to the black hole, the AGN outflows are revealed through blueshifted absorption lines in X–ray emission (Pounds et al. 2003a, b; King 2010a). Tombesi et al. (2010a, b) show that they are present in more than 35 percent of a sample of over 50 local AGN, and deduce that their solid angles are large (certainly > 0.6 × 2π, and probably greater). The observed absorption columns imply that in many cases the outflows are quite recent (few years), suggesting that outflows are an almost ubiquitous feature of central black hole activity (King 2010b)"

That paper you linked also contains assumptions and models:

Quote
We assume that Sgr A* outflow either carries with it CR protons created near the black hole, or that the CR protons are accelerated on shock fronts where the outflow runs into the interstellar medium.
Quote
If instead we assume that the gamma–ray lobes were produced in this event, we must conclude that the energy–driven outflow is still proceeding, with a mean velocity hvi ≃1000 km s−1 over its lifetime.
Quote
Our model is similar to that of Crocker & Aharonian (2011) in terms of assuming the CR protons (rather than electrons) produce the observed gamma-ray emission.
Quote
They further consider a quasi–steady state model in which the CR protons are continuously injected by supernova explosions.
Quote
Within this model, then, in galaxies with large σ>∼ 150 km s−1, Eddington outflows tend to sweep the vicinity of the hole clear of gas of density (2) and prevent further accretion and growth, establishing the M−σ relation for the black hole mass (King 2003; 2005).
Quote
For the present day Milky Way and directions well out of the Galactic plane, we expect fg to be significantly less than fc, so we parametrize fg as fg = 1.6 · 10−3 f0.01, where f0.01 ∼ 1 is a dimensionless free parameter of the model.

In the following article (from arxiv):
https://arxiv.org/ftp/arxiv/papers/1501/1501.07664.pdf
"Wind from the black-hole accretion disk driving a molecular outflow in an active galaxy"
"Recent observations of large-scale molecular outflows3,4,5,6,7,8 in ultraluminous infrared galaxies (ULIRGs) have provided the evidence to support these studies, as they directly trace the gas out of which stars form.
So, now they claim that the outflow wind from the BH (or SMBH) drives the star formation activity.

This also contains assumptions and models:

Quote
The efficiency f = 0.2 is assumed.
Quote
We model the broad absorption at E ~ 9 keV with the XSTAR31 code v. 2.2.1bn. We consider a Γ = 2 power-law continuum, consistent with the observed value (see SI), and standard Solar abundances. A turbulent velocity of 30,000 km s−1 is assumed for the fast wind model.
Quote
This equation assumes a spherical, but clumpy, geometry.
Quote
We derive a physical characterization of this outflow using a dedicated photoionization absorption model (see Methods).

To say that you don't like models or assumptions, you sure do cite a lot of studies that use them...
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/09/2019 05:51:37
Like I said, assumptions are often unavoidable if we cannot directly observe something
Yes, I fully agree. However, the only question is: "what do we really assume?".
Please look at the following article: "The Suzaku view of highly ionized outflows in AGN"
https://academic.oup.com/mnras/article/430/1/60/983995
"Measured outflow velocities span a continuous range from <1500 km s−1 up to ∼100 000 km s−1, with mean and median values of ∼0.1 c and ∼0.056 c,"
"Observational evidence for outflows and winds in active galactic nuclei (AGN) is seen in multiple energy regimes, "
So, yes, they have also set several assumptions.
However, they have only found OUTFLOWS. Not even a single word about inflow.
I have also found a very interesting article:
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf
Please look at the page before the last one
It is stated:
"Innermost   highly   ionized   wind launched   from   within 100   Rg (1016 cm)   of
black   hole   – ultra   fast   iron   K   absorption   (0.3c)"
We see clearly the location of the BH and the accretion disc.
However, the outflow wind comes directly from the BH while the accretion disc is located far away.
This proves my understanding that the matter is originally ejected from the BH.
In that picture we see that the outflow wind is banding upwards from the accretion disc.
Somehow, they don't show the outflow wind that is banding downwards.
The even ask:
"What   is   the   driving   mechanism   for   the   disk   wind?   Radiation,   
Magnetic   driving?"
I think that there is only one explanation for that outflow wind - Lorenz force Due to the strong magnetic field.
Actually, if we could trace the outflow wind upwards we should see the great impact of the magnetic field.
I assume that some of this outflow wind also gets into the accretion disc (Yes, I also assume).
In this article they discuss about AGN which must be a SMBH.
So, we see clearly that the SMBH ejects matter outwards.
Some of that matter might get into the accretion disc. If so, this is the ultimate matter source for the accretion disc - As I have offered in my theory.
They also ask:
"What   is   the   role   of   the   clumpy   gas,   is   it   
needed   to   accelerate   the   gas?"
NO!!!
There is no need for in falling matter/gas cloud/stars as the matter is ejected outwards directly from the SMBH.
In any case in both articles they only discuss about outflows and this is a solid proof that SMBH is ejecting matter outwards.
You have tried to explain why a SMBH can't eject any matter.
Now you need to adjust the theory in order to explain how could it be that a SMBH ejects matter (or outflow wind) directly from its core.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/09/2019 06:35:53
However, they have only found OUTFLOWS. Not even a single word about inflow.

So now you think a single study contains all of the observations that we have ever made of inflows and outflows? Outflows are easier to observe because they cover a much larger volume of space and are further away from the hottest, inner regions of the accretion disk than material flowing into the black hole would be.

I have also found a very interesting article:
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf
Please look at the page before the last one
It is stated:
"Innermost   highly   ionized   wind launched   from   within 100   Rg (1016 cm)   of
black   hole   – ultra   fast   iron   K   absorption   (0.3c)"

1016 centimeters is equal to 1011 kilometers. The black hole in PG1211+143 has a mass of 4 x 107 solar masses. That corresponds to an event horizon radius of about 1.18 x 108 kilometers. That, in turn, means that the outflows are originating about 9.99 x 1010 kilometers out from the event horizon, which is a distance almost 17 times greater than the average distance between the Sun and Pluto. To put that into perspective, you could shrink that black hole down to the size of a small marble (1 centimeter across) and that innermost highly ionized wind boundary would be almost the full length of an A-37 Dragonfly attack aircraft (8.62 meters long) away from it. So the following statement of yours...

Quote
This proves my understanding that the matter is originally ejected from the BH.

...is not at all supported by this data. In fact, it is contradicted by the data. We don't see the outflow originating right at the event horizon. Instead, we see it starting much, much further out.

What's more is that you are misinterpreting the diagram. The accretion disk is represented by the two wedge shapes on either side of the diagram. Those shapes extend all the way inward to the black hole in the diagram. So this claim...

Quote
the accretion disc is located far away.

...is also wrong.

Now you need to adjust the theory in order to explain how could it be that a SMBH ejects matter (or outflow wind) directly from its core.

There is no need to because the data doesn't support that conclusion at all. 99 billion kilometers is far, far beyond the event horizon.

As I have offered in my theory.

You don't have a theory. You have a hypothesis that has already been thoroughly falsified.

Quote
Some of that matter might get into the accretion disc. If so, this is the ultimate matter source for the accretion disc - As I have offered in my theory.

As I pointed out before, if the magnetic field is too strong to let any charged matter into the black hole, then it will also be too strong to let charged matter out of the black hole. So take your pick. Is the field too strong or is it not too strong?

In any case in both articles they only discuss about outflows and this is a solid proof that SMBH is ejecting matter outwards.

Woah, woah, wait a minute. Now you are saying that studies using assumptions and models (which both of those articles used) are proof of something? I thought your current stance was that assumptions and models can't prove anything? Have you changed your mind all of a sudden?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/09/2019 09:03:20
016 centimeters is equal to 1011 kilometers. The black hole in PG1211+143 has a mass of 4 x 107 solar masses. That corresponds to an event horizon radius of about 1.18 x 108 kilometers. That, in turn, means that the outflows are originating about 9.99 x 1010 kilometers out from the event horizon, which is a distance almost 17 times greater than the average distance between the Sun and Pluto.
Thanks
Do you have any idea about the radius of the accretion disc in that AGN?
The inwards radius and the outwards radius?

In any case, do you agree that we clearly see outflow from that AGN?
Do you also agree that our scientists have found a clear outflow by the Suzaku detector/observation?
"The Suzaku view of highly ionized outflows in AGN"
https://academic.oup.com/mnras/article/430/1/60/983995
"Measured outflow velocities span a continuous range from <1500 km s−1 up to ∼100 000 km s−1, with mean and median values of ∼0.1 c and ∼0.056 c,"
"Observational evidence for outflows and winds in active galactic nuclei (AGN) is seen in multiple energy regimes, "
Actually, at any detector that I have looked it was stated that they have only found outflow.
This is correct even for NASA's Chandra detector:
NASA's Chandra Helps Confirm Evidence of Jet in Milky Way's Black Hole
https://www.nasa.gov/press/2013/november/nasas-chandra-helps-confirm-evidence-of-jet-in-milky-ways-black-hole/#.XXNdUi5vbIU
"Astronomers have long sought strong evidence that Sagittarius A* (Sgr A*), the supermassive black hole at the center of the Milky Way, is producing a jet of high-energy particles. Finally they have found it, in new results from NASA's Chandra X-ray Observatory and the National Science Foundation's Very Large Array (VLA) radio telescope."
" Additionally, the energy signature, or spectrum, in X-rays of Sgr A* resembles that of jets coming from supermassive black holes in other galaxies."
Now, let's go back to that unrealistic article from arxiv which tries to confirm an inflow:
https://arxiv.org/pdf/1307.5845.pdf
It is stated clearly that they took the data from NASA Chandra observation:
" Our own Galaxy’s SMBH provides a uniquely instructive exception, and we present a close-up view of its quiescent X-ray emission based on 3 mega-second of Chandra observations"
So, how could it be that NASA Chandra observations see only OUTFLOW jets (in our SMBH and in many others), while arxiv (which is based on the same data from NASA Chandra observations) see suddenly Inflow?
Is it real???
Can you please show one real detector/observation that see inflow?
Even only one real inflow please.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/09/2019 13:34:02
Quote from: Kryptid
1016 centimeters is equal to 1011 kilometers. The black hole in PG1211+143 has a mass of 4 x 107 solar masses. That corresponds to an event horizon radius of about 1.18 x 108 kilometers. That, in turn, means that the outflows are originating about 9.99 x 1010 kilometers out from the event horizon, which is a distance almost 17 times greater than the average distance between the Sun and Pluto.
Do you have any idea about the radius of the accretion disc in that AGN?
The inwards radius and the outwards radius?
The diagram stopped labeling the outward radius at 1018 cm which is over a light year.  It doesn't say it stops there, it's just where the labeling stops.  The diagram actually goes out to about 20 light years.
As for inner radius, the size of the black hole limits it to about 3.5e8 km, which is a little outside the orbital radius of the asteroid belt.  The diagram doesn't show this because the labeling also stops in that direction.

Quote
In any case, do you agree that we clearly see outflow from that AGN?
This black hole is about the size of the one in Andromeda, although perhaps even more active since the study seems to seek particularly active ones.  Note that our galaxy does not have an AGN.

Quote
Now, let's go back to that unrealistic article from arxiv which tries to confirm an inflow:
Why is it unrealistic?  Does it say something that you've decided ahead of time is wrong?

Quote
So, how could it be that NASA Chandra observations see only OUTFLOW jets (in our SMBH and in many others), while arxiv (which is based on the same data from NASA Chandra observations) see suddenly Inflow?
Nowhere does it say that the observations saw only outflow.  The outflow is what they were concentrating on.  The inflow has a different signature and that signature was always there.  This new study seems to be more concentrated on that portion of the data, and still not on the myriad of other things shown during the observation periods.

Quote
Is it real???
Can you please show one real detector/observation that see inflow?
Even only one real inflow please.
You just quoted one.

Evan and I both pointed to articles about the event last May where a larger than usual inflow was seen.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/09/2019 15:16:41
Nowhere does it say that the observations saw only outflow.  The outflow is what they were concentrating on.  The inflow has a different signature and that signature was always there
Sorry
In both articles it is stated clearly that they have found an outflow jet.
For example - X-rays detected by Chandra:
"Astronomers have long sought strong evidence that Sagittarius A* (Sgr A*), the suppermassive black hole at the center of the Milky Way, is producing a jet of high-energy particles."
"Jets of high-energy particles are found throughout the universe, on large and small scales. They are produced by young stars and by black holes a thousand times larger than the Milky Way's black hole. They play important roles in transporting energy away from the central object and, on a galactic scale, in regulating the rate of formation of new stars."
"We were very eager to find a jet from Sgr A* "
"The jet appears to be running into gas near Sgr A*, producing X-rays detected by Chandra and radio emission observed by the VLA."
So they clearly gives an outflow jet that is running outwards from the Sgr A* directly into the gas around it.
Therefore, they have observed an outflow jet and not an inflow jet.
However, they have a very interesting explanation for that:
"Scientists think jets are produced when some material falling toward the black hole is redirected outward. Since Sgr A* is presently known to be consuming very little material, it is not surprising the jet appears weak. A jet in the opposite direction is not seen, possibly because of gas or dust blocking the line of sight from Earth or a lack of material to fuel the jet."
So, they claim that this outflow jet is an outcome from the in falling material. They even say claerly that "A jet in the opposite direction is not seen". So there is no observation for any in falling (inflow) material into the SMBH.
Evan and I both pointed to articles about the event last May where a larger than usual inflow was seen.
The one that you and Evan pointed was not an observation for in falling matter.
Our scientists have discovered a flare.
This flare could be an outcome of magnetic field.
So, our scientists have never ever discovered any sort of clear X-Ray/observation of inflow matter.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/09/2019 18:11:19
Thanks
Do you have any idea about the radius of the accretion disc in that AGN?
The inwards radius and the outwards radius?

I don't know. If the inner edge of the accretion disk did extend all the way to the event horizon, then I don't think we would be able to see any defined inner edge. As for the outer radius, I don't know if that has been measured or not.

In any case, do you agree that we clearly see outflow from that AGN?
Do you also agree that our scientists have found a clear outflow by the Suzaku detector/observation?

Yes. The next question then becomes, are you willing to say that conclusions based on those models and assumptions are trustworthy?

Actually, at any detector that I have looked it was stated that they have only found outflow.

You must have a poor memory, since we extensively discussed that paper which stated that an inflow moving at 30% the speed of light was detected.

So, how could it be that NASA Chandra observations see only OUTFLOW jets (in our SMBH and in many others), while arxiv (which is based on the same data from NASA Chandra observations) see suddenly Inflow?

Your assumption here is that, since the particular NASA article you looked at doesn't mention inflows, they must not have detected any. Have you actually looked at the scientific paper that the article was based on? I strongly doubt you have, since the link to it from that particular website doesn't seem to allow public access to the file. Even if the paper doesn't mention inflows, that doesn't mean that none were seen. So there is no contradiction here.

Can you please show one real detector/observation that see inflow?
Even only one real inflow please.

https://arxiv.org/pdf/1808.09373.pdf

They even say claerly that "A jet in the opposite direction is not seen".

That doesn't have anything to do with inflows. Active galaxies typically have two jets moving in opposite directions, one from the north pole and one from the south pole. Apparently, they only saw one of them.

By the way, you didn't answer these questions:

Woah, woah, wait a minute. Now you are saying that studies using assumptions and models (which both of those articles used) are proof of something? I thought your current stance was that assumptions and models can't prove anything? Have you changed your mind all of a sudden?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/09/2019 20:01:48
You must have a poor memory, since we extensively discussed that paper which stated that an inflow moving at 30% the speed of light was detected.
I do remember that.
However, you might have missed the key point  that I have offered this article.
It was about in falling gas cloud in the size the Earth into a galaxy which is located at a distance of almost one billion light years away.
I have told you that our scientists have seen this gas cloud as it was moving directly in the direction of the SMBH, but it was not falling in. If it was falling in due to gravity, it had to set some sort of orbit.
That gas cloud cross the accretion disc and moved directly to the pole of the SMBH.
It is not gravity. It is magnetic field. From the pole it is boosted upwards as we clearly see the outwards jet stream in our galaxy.
If it was falling in, we had to see a fireworks.
So, that article doesn't proof any in falling matter.
Please look again at the following image at the pg before the last one.
However, they have only found OUTFLOWS. Not even a single word about inflow.
I have also found a very interesting article:
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf
Please look at the page before the last one
We see the Wind Structure as it moves away from the accretion disc.
We clearly see that it starts to band upwards.
If we could trace the wind, we should see that it should move upwards in the direction of the pole due to the magnetic field.
That wind which is driven by the magnetic fields sets the jet stream.
So, the Earth size gas cloud is actually the same wind that we see here.

Quote
Can you please show one real detector/observation that sees inflow?
Even only one real inflow please.
https://arxiv.org/pdf/1808.09373.pdf
This is about the same in falling gas cloud at the same galaxy.
As I have stated, this gas cloud is not falling inwards into the SMBH.
In any case,  you have only one discovery of in falling gas cloud in the whole Universe. This galaxy is located at a distance of 1 billion LY away.
That's all you have for in falling matter.
Wake up please!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/09/2019 20:24:30
If it was falling in due to gravity, it had to set some sort of orbit.

Not if it was falling directly towards the black hole it wouldn't. Orbits are not an automatic result of gravitational attraction (case in point, meteorites).

That gas cloud cross the accretion disc and moved directly to the pole of the SMBH.

Where was it ever stated that there was evidence that the gas cloud was moving towards the black hole's pole?

It is not gravity. It is magnetic field. From the pole it is boosted upwards as we clearly see the outwards jet stream in our galaxy.

Where was it ever stated that the gas cloud was boosted upwards?

If it was falling in, we had to see a fireworks.

The detector wasn't looking at the black hole during the time period the cloud would have been consumed, so that's the obvious reason why no "fireworks" were observed.

So, that article doesn't proof any in falling matter.

It isn't proof, but it is good evidence.

If we could trace the wind, we should see that it should move upwards in the direction of the pole due to the magnetic field.

You don't know that.

So, the Earth size gas cloud is actually the same wind that we see here.

It can't be. The cloud was moving inward, not outward.

As I have stated, this gas cloud is not falling inwards into the SMBH.

You never provided any evidence to support that claim and it goes against the findings of the scientists. So who should I believe, physicists who are experts in astrophysics or some random person on the Internet who has a long history of misunderstanding black holes?

In any case,  you have only one discovery of in falling gas cloud in the whole Universe. This galaxy is located at a distance of 1 billion LY away.
That's all you have for in falling matter.

One is all I need, but there are others: https://iopscience.iop.org/article/10.3847/0004-637X/829/2/96

Wake up please!

Wake up to what? Your consistent inability to understand physics? I'm quite awake to that, thank you.

By the way, I'm still waiting for you to answer my questions:

Quote
Woah, woah, wait a minute. Now you are saying that studies using assumptions and models (which both of those articles used) are proof of something? I thought your current stance was that assumptions and models can't prove anything? Have you changed your mind all of a sudden?
Quote
The next question then becomes, are you willing to say that conclusions based on those models and assumptions are trustworthy?
Quote
As I pointed out before, if the magnetic field is too strong to let any charged matter into the black hole, then it will also be too strong to let charged matter out of the black hole. So take your pick. Is the field too strong or is it not too strong?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/09/2019 14:55:10
Quote
So, how could it be that NASA Chandra observations see only OUTFLOW jets (in our SMBH and in many others), while arxiv (which is based on the same data from NASA Chandra observations) see suddenly Inflow?

Your assumption here is that, since the particular NASA article you looked at doesn't mention inflows, they must not have detected any. Have you actually looked at the scientific paper that the article was based on? I strongly doubt you have, since the link to it from that particular website doesn't seem to allow public access to the file.
Would you kindly offer the NASA Chandra observation which arxiv have used to prove the inflow accretion in their article?
How could it be that NASA Chandra had detected any sort of inflow without giving any report on that?
If NASA Chandra had detected an inflow - they must highlight this inflow observation!!!
 
Even if the paper doesn't mention inflows, that doesn't mean that none were seen. So there is no contradiction here.
How could it be that there is no contradiction between NASA Chandra and arxiv?
NASA Chandra is the observation site.
If I understand it correctly, NASA Chandra have never ever observed any inflow into the Milky way SMBH.
So, how could it be that arxiv which have used the data from NASA Chandra claims that suddenly they have discovered an inflow?
Woah, woah, wait a minute. Now you are saying that studies using assumptions and models (which both of those articles used) are proof of something? I thought your current stance was that assumptions and models can't prove anything? Have you changed your mind all of a sudden?
The question is as follow:
Could it be that the unrealistic set ups/assumptions that arxiv have used, converted the Outflow observation by NASA Chandra into inflow?
Please remember - arxiv has no observation station/detector. They have stated clearly that they took the data from NASA Chandra
So, would you kindly explain what kind of manipulation arxiv have set in NASA Chandra data/assumptions in order to convert the outflow observation that NASA Chandra had verified into an inflow?
As I pointed out before, if the magnetic field is too strong to let any charged matter into the black hole, then it will also be too strong to let charged matter out of the black hole. So take your pick. Is the field too strong or is it not too strong?
Let me explain it again for you.
We all agree that there is quite strong magnetic field around the accretion disc.
Our scientists see the impact of the magnetic field on the plasma in the accretion disc.
They also claim that the Molecular stream is a direct product of the magnetic field.
In the following presentation that I had offered we see clearly the impact of the magnetic field.
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf
Please look at the page before the last one
If we ignore the Magnetic field, the outflow had to be ejected at the accretion disc plane.
However, we clearly see that the outflow is banding upwards.
There is only one force that can do it - Magnetic filed.
It shows that the matter which had been ejected from the accretion disc is lifted upwards.
That is the source for the molecular jet steam that we see above and below our accretion disc.
Therefore, the scenario is as follow:
New born particles are ejected deep from the SMBH event horizon into the accretion disc.
Those particles under the Ultra high temp (10^9c) high orbital velocity (0.3c) + electromagnetic and Fusion activity is converted into real Atoms and molecular.
As the atoms and molecular are ejected outwards, they are trapped by the magnetic field and banded directly backwards and upwards. at that stage there is no orbital velocity any more. (As the magnetic field is much stronger at that aria than the gravity of the SMBH). When the Magnetic field takes control it boosts the molecular upwards (and in the direction of the pole).
This also explains the "Earth size" gas cloud that our scientists had verified at that far away galaxy.
The gas cloud was originally a plasma in the accretion disc. As it was ejected outwards it was trapped by the magnetic field and lost the orbital momentum.
The magnetic field takes this gas cloud and boosts it upwards at a velocity of 0.8c.
However, this galaxy is located at direct face on with regards to our location. Therefore, we clearly see the accretion disc, but we can't see the height of the molecular jet stream.
Hence, as the gas cloud is boosted upwards (in the normal direction of any Molecular Jet stream) it is actually moving directly upwards to us.
Unfortunately, we can't see that galaxy in 3D. we only see it in 2D.
Therefore, instead of 3D movement of 0.8c upwards/and inwards to the pole, we only monitor the 2D inwards movement to the pole direction.
Hence, our scientists have only got the 0.3c instead of the real 0.8c. Please remember, that the confirmed velocity of the molecular jet stream in our galaxy is 0.8c. Therefore, I assume that also in this galaxy the velocity of the molecular jet stream should be similar.
In any case, as the molecular jet stream in the Milky way is boosted upwards without any impact of the SMBH gravity force, it also does not have any orbital momentum.
Therefore, that Earth size gas cloud is moving directly upwards while we only see that it moves closer to the pole (without any orbital momentum).
Actually, we think that it is moving closer to the SMBH (or even falling into it) but in reality it is moving far above the SMBH).
In our galaxy the molecular jet stream gets almost to 27,000 LY above the SMBH. That is the real destiny of that Earth gas cloud.
As it moves upwards, the gas cloud brakes down and therefore we can't see it any more. But all the molecular are there - directly high above the pole of the SMBH.
Later on, as the molecular are far enough from the impact of the magnetic field, they should lose their upwards momentum and fall back to the disc plane of the galaxy (Not far away from the SMBH, but outside the impact of the magnetic field/shield)
Their mass will be used to create new gas clouds that orbit around the SMBH.
In those gas clouds new stars will be formed. Each star will get planets and moons from the same matter in the gas cloud and at the same day.
Those stars will be drifted outwards and join all the other stars in the Bulge and later on in the galactic disc.
Why is it so difficult for you to see our real galaxy structure?
Why do you desperately insist for Inflow?
There is no inflow as there is no need for Inflow!

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/09/2019 01:40:59
Would you kindly offer the NASA Chandra observation which arxiv have used to prove the inflow accretion in their article?

You sure do like to misuse the words "prove" and "proof", since neither are offered by the article in question. I already told you before that the results of this study (and many, many others) were obtained by putting the spectral data into a model and making conclusions based on what that model showed.

How could it be that NASA Chandra had detected any sort of inflow without giving any report on that?
If NASA Chandra had detected an inflow - they must highlight this inflow observation!!!

Says who? They might not have even known that they had such an observation. Data has to be analyzed in order to figure out what it means. They may simply have not done the necessary analysis required to find an inflow.

How could it be that there is no contradiction between NASA Chandra and arxiv?

Because NASA never said that there wasn't an inflow. They may not have even been looking for any and didn't scan the data for it.

If I understand it correctly, NASA Chandra have never ever observed any inflow into the Milky way SMBH.
So, how could it be that arxiv which have used the data from NASA Chandra claims that suddenly they have discovered an inflow?

Because what Chandra acquired was data. That data has to be analyzed in order to learn things about what produced the data. That almost certainly requires inputting the data into particular models and finding out what the model predicts. Interestingly enough, the NASA article which you claim doesn't show an inflow does have this to say:

Quote
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.

Could it be that the unrealistic set ups/assumptions that arxiv have used, converted the Outflow observation by NASA Chandra into inflow?

Who said the assumptions were "unrealistic"? The answer, of course, is no. The arXiv article already concludes that over 99% of the matter is being ejected, so they obviously already think that outflow represents the major phenomenon there. If what they were doing was merely converting an outflow into an inflow, then they should have concluded that over 99% of the matter was flowing in instead of out. That's not what they did.

So, would you kindly explain what kind of manipulation arxiv have set in NASA Chandra data/assumptions in order to convert the outflow observation that NASA Chandra had verified into an inflow?

They didn't manipulate an outflow into an inflow so your question is nonsensical.

New born particles are ejected deep from the SMBH event horizon into the accretion disc.

No, quit saying this. You can't eject particles from within the super massive black hole's event horizon. We have explained this to you a multitude of times. Do you think you live in some kind of magical world where repeatedly stating false information eventually makes it correct? You can repeat this a trillion times, but it will still violate the laws of physics. The swapping of space and time inside of the horizon won't let it happen. For objects inside the horizon, the horizon is in the object's past and the singularity is in its future. So please do explain how a magnetic field can push an object into the past. If it can't do that, then it can't get particles out of the horizon.

The second half of your question also ignores another problem that I pointed out. If the magnetic field is so strong that it won't let matter from the accretion disk into the black hole, then it also won't let matter from the black hole into the accretion disk. Is the magnetic field too strong for matter to cross it or is it not? You can't have it both ways. Pick one and stick with it.

Those particles under the Ultra high temp (10^9c) high orbital velocity (0.3c) + electromagnetic and Fusion activity is converted into real Atoms and molecular.

The mass/energy of the black hole/accretion disk system can't increase this way, though. The jets themselves are removing mass and energy from the system, so it is losing mass and energy as time goes on. Do you agree with this? If you don't agree with this, then you also don't agree with the first law of thermodynamics.

Therefore, instead of 3D movement of 0.8c upwards/and inwards to the pole, we only monitor the 2D inwards movement to the pole direction.

I'm still waiting for you to provide evidence that the gas cloud was moving towards either of the poles.

Actually, we think that it is moving closer to the SMBH (or even falling into it) but in reality it is moving far above the SMBH).

Don't just say it, give evidence for it.

Their mass will be used to create new gas clouds that orbit around the SMBH.
In those gas clouds new stars will be formed. Each star will get planets and moons from the same matter in the gas cloud and at the same day.

And in the process, the black hole/accretion disk system must lose mass and energy because it is giving that mass and energy to something else (stars, planets, etc.).

Those stars will be drifted outwards and join all the other stars in the Bulge and later on in the galactic disc.

Where does the energy to make the stars "drift outward" come from? Remember, you're raising them against a gravitational potential and that requires an input of energy.

Why is it so difficult for you to see our real galaxy structure?

Your model contradicts the laws of physics. No more need be said. It has already been falsified. You just stubbornly refuse to accept that.

Why do you desperately insist for Inflow?

That's what all of the relevant physics points too.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/09/2019 03:56:05
The articles focus on trying to spot the outflow against the bright background of the inflow radiation.  It's hard to spot the jets when looking at all the X-rays due to the inflow heating the disk.  The disk isn't going to stay hot by itself without energy input from stuff falling in.  So it isn't a matter of trying to find the inflow (weak though it is in our galaxy).  Trick is to spot the evidence of the jets despite it.  And the jet has a different signature, so yes, they've been able to pick it out much of the time.

That is true, although I'm sure Dav Lev is going to try to argue that just because we see a lot of heat and X-rays, that doesn't mean it's being caused by an inflow. I imagine what he is looking for is some kind of direct detection where we can visually watch matter move unambiguously into the black hole. We obviously don't have that.

Sometimes I feel like this is a pointless battle. Despite how broken his model is shown to be, he keeps insisting on it...

(1) He says that particles can be pulled out of the horizon by a magnetic field. This doesn't work because magnetic fields can't move particles into the past. He has consistently failed to understand how event horizons work.
(2) He says that those particles can cross the magnetic field into the accretion disk but not the other way around. This doesn't work because magnetic fields are not one-way doors. He has consistently failed to understand how magnetism works.
(3) He says that the particles that move out eventually end up in stars and planets. Yet at the same time, he keeps insisting that the black hole doesn't lose mass. He has consistently failed to understand the first law of thermodynamics.
(4) He says that planets and stars spiral outwards while ignoring that energy is required to make this happen. He denied that tidal acceleration can do this and seems to assume that it just happens naturally. Another case of him failing to understand the first law of thermodynamics.

I can't help but wonder at times if he is a troll.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/09/2019 07:12:17
Quote
How could it be that there is no contradiction between NASA Chandra and arxiv?
Because NASA never said that there wasn't an inflow. They may not have even been looking for any and didn't scan the data for it.
Yes, I fully agree.
However, the past is a past.
We can gess, we can believe on, but we were not there 10 billion years ago.
The present is real.
Let's read again that statement by NASA:
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.
Do you see any conclusion that they see today any sort of inflow?
They say that: "the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way".
So, does it mean that they see today any sort of inflow???
The answer is very clear.
It is NO NO NO.
They say that it gives an indication to our astronomers that in the PAST (not today) "gas and dust have migrated steadily into Sgr A* over the past 10 billion years"
But again - they don't see any sort of inflow TODAY!!!
Actually, the following statement  by NASA: "which indicates to astronomers" proves that there is no current inflow.
If there was any current inflow, they had to say it clearly as they say clearly that there is outflow.
Please read again the article from NASA site.
They don't mention any observation for current inflow.
I also wonder why the use the word "astronomers" instead of "scientists".
Could it be that the leading team in NASA are scientists.
They surly have the ultimate knowledge to analyze the data that they fetch from their observation site.
However, astronomers (that believe in the BBT) don't like that real verified current outflow observation by NASA
Therefore, NASA have added this remark in order to give them the possibility to continue with their dream.
If they have discoed any sort of inflow in their data, they were surly announce it - clearly and loudly.
Actually, in NASA article it is stated that they see outflow from the accretion disc at many others SMBH. Not even one word about any current indication for inflow.
Why NASA don't say:
The observation data "indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years."
This is one more evidence that NASA have full confidence that the data observation as it only indicates for OUTFLOW.
Therefore, my conclusion is very clear.
NASA didn't find TODAY any sort of inflow in the observation data. Not even a one to one million chance for inflow.
However, when you read the article from arxiv it shows that they have found in the same data from NASA an evidence for CURRENT inflow. So they see TODAY an inflow indication in the same NASA data.
Sorry - this is manipulation!!!
I have full trust in NASA' scientists knowledge.
Unless, you can show that NASA scientists are so foolish that they can't read correctly the data which they have extracted from their observation site. While Astronomers at arxiv are much more cleaver.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/09/2019 16:49:14
They say that it gives an indication to our astronomers that in the PAST (not today) "gas and dust have migrated steadily into Sgr A* over the past 10 billion years"

I know you are not a native English speaker, so perhaps I should explain this. When someone says "over the past 10 billion years", they are speaking of something that is on-going. It is something that is still happening now and has been happening for 10 billion years. If someone says "It's been raining over the past three hours", they mean it is still raining.

Actually, the way you phrase this has its own implications. Does this mean you agree that there was an inflow in the past and therefore black holes can experience inflows? That certainly seems to be what you are saying.

However, astronomers (that believe in the BBT) don't like that real verified current outflow observation by NASA

Why wouldn't they? Outflows don't contradict the Big Bang theory.

I have full trust in NASA' scientists knowledge.

If that's true, then that means you must agree with them that there was inflow in the past and therefore black holes can experience inflows.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/09/2019 04:20:46
I know you are not a native English speaker, so perhaps I should explain this. When someone says "over the past 10 billion years", they are speaking of something that is on-going. It is something that is still happening now and has been happening for 10 billion years. If someone says "It's been raining over the past three hours", they mean it is still raining.
Ok
Let's try to have better understanding:
https://www.nasa.gov/press/2013/november/nasas-chandra-helps-confirm-evidence-of-jet-in-milky-ways-black-hole/#.XXmqVC5vbIV
1. clear outflow observation of a jet of high-energy particles by NASA:
"Astronomers have long sought strong evidence that Sagittarius A* (Sgr A*), the supermassive black hole at the center of the Milky Way, is producing a jet of high-energy particles. Finally they have found it, in new results from NASA's Chandra X-ray Observatory and the National Science Foundation's Very Large Array (VLA) radio telescope."
So, NASA observation have found that " the suppermassive black hole at the center of the Milky Way, is producing a jet of high-energy particles"
2. Explanation about the impact of that outflow jet on the gas around the SMBH:
"The jet appears to be running into gas near Sgr A*, producing X-rays detected by Chandra and radio emission observed by the VLA. The two key pieces of evidence for the jet are a straight line of X-ray emitting gas that points toward Sgr A* and a shock front -- similar to a sonic boom -- seen in radio data, where the jet appears to be striking the gas. Additionally, the energy signature, or spectrum, in X-rays of Sgr A* resembles that of jets coming from supermassive black holes in other galaxies."
3. Outflow Jets throughout the universe at other galaxies:
"Jets of high-energy particles are found throughout the universe, on large and small scales. They are produced by young stars and by black holes a thousand times larger than the Milky Way's black hole."
4. The Molecular jet stream had been set by the outflow jet.
"Astronomers have suggested the giant bubbles of high-energy particles extending out from the Milky Way and detected by NASA's Fermi Gamma Ray Telescope in 2008 are caused by jets from Sgr A* that are aligned with the rotation axis of the galaxy. The latest results from Chandra support this explanation."
5. The outflow jet stream tells the direction of the black hole's spin axis
"We were very eager to find a jet from Sgr A* because it tells us the direction of the black hole's spin axis. This gives us important clues about the growth history of the black hole," said Mark Morris of the University of California at Los Angeles, a co-author of the study. So, that confirmed outflow get shows the direction of the black hole's spin axis.
6. Mr. Mark Morris study:
That direction of the black hole's spin axis is important for another study which is totally not connected to NASA observation. This information was important to Mr. Mark Morris study at the University of California at Los Angeles.
"The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years."
So, based on the study of Mr. Mark Morris the University of California at Los Angeles, it was found that the direction of the spin axis of Sgr A* indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.
7. Other scientists' wishful thinking:
"Scientists think jets are produced when some material falling toward the black hole is redirected outward. Since Sgr A* is presently known to be consuming very little material, it is not surprising the jet appears weak. A jet in the opposite direction is not seen, possibly because of gas or dust blocking the line of sight from Earth or a lack of material to fuel the jet".
So, Scientists think that jets are produced when some material falling toward the black hole is redirected outward.
However, do we see any evidence for inflow jet by NASA observation?
The answer is very clear: " A jet in the opposite direction is not seen"
So, NASA didn't find any observation for inflow jet (Not even for native English speaker).
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 12/09/2019 06:22:46
. Mr. Mark Morris study:
That direction of the black hole's spin axis is important for another study which is totally not connected to NASA observation.

Where do you get that it was a "different study"? Mark Morris is explicitly mentioned as a co-author of the study that the news report is about (with Zhiyuan Li being another co-author). There is another news report by Scientific American that says the same thing: https://www.scientificamerican.com/article/milky-way-black-hole-jet/ And another: https://chandra.harvard.edu/blog/node/467 And another: http://www.sci-news.com/astronomy/science-chandra-jet-supermassive-black-hole-01564.html

1. clear outflow observation of a jet of high-energy particles by NASA:
"Astronomers have long sought strong evidence that Sagittarius A* (Sgr A*), the supermassive black hole at the center of the Milky Way, is producing a jet of high-energy particles. Finally they have found it, in new results from NASA's Chandra X-ray Observatory and the National Science Foundation's Very Large Array (VLA) radio telescope."
So, NASA observation have found that " the suppermassive black hole at the center of the Milky Way, is producing a jet of high-energy particles"
2. Explanation about the impact of that outflow jet on the gas around the SMBH:
"The jet appears to be running into gas near Sgr A*, producing X-rays detected by Chandra and radio emission observed by the VLA. The two key pieces of evidence for the jet are a straight line of X-ray emitting gas that points toward Sgr A* and a shock front -- similar to a sonic boom -- seen in radio data, where the jet appears to be striking the gas. Additionally, the energy signature, or spectrum, in X-rays of Sgr A* resembles that of jets coming from supermassive black holes in other galaxies."
3. Outflow Jets throughout the universe at other galaxies:
"Jets of high-energy particles are found throughout the universe, on large and small scales. They are produced by young stars and by black holes a thousand times larger than the Milky Way's black hole."
4. The Molecular jet stream had been set by the outflow jet.
"Astronomers have suggested the giant bubbles of high-energy particles extending out from the Milky Way and detected by NASA's Fermi Gamma Ray Telescope in 2008 are caused by jets from Sgr A* that are aligned with the rotation axis of the galaxy. The latest results from Chandra support this explanation."

This is like trying to prove there are no white crows by counting black crows.

So, Scientists think that jets are produced when some material falling toward the black hole is redirected outward.
However, do we see any evidence for inflow jet by NASA observation?

Yes, since that article says that matter has been migrating into Sagittarius* over the past 10 billion years.

The answer is very clear: " A jet in the opposite direction is not seen"
So, NASA didn't find any observation for inflow jet (Not even for native English speaker).

If you think that, then you clearly don't understand English as well as you think you do.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/09/2019 18:14:25
Quote
The answer is very clear: " A jet in the opposite direction is not seen"
So, NASA didn't find any observation for inflow jet (Not even for native English speaker).
If you think that, then you clearly don't understand English as well as you think you do.
I really can't understand your conclusions
Let's try to verify the following step by step:
"Scientists think jets are produced when some material falling toward the black hole is redirected outward. Since Sgr A* is presently known to be consuming very little material, it is not surprising the jet appears weak. A jet in the opposite direction is not seen, possibly because of gas or dust blocking the line of sight from Earth or a lack of material to fuel the jet".
1. "Scientists think jets are produced when some material falling toward the black hole is redirected outward."
So,  "Scientists think". It is very nice that they think. However, does it mean that they really see? Don't you agree that thinking is one issue and observation is totally another issue?
So, do you agree that this message doesn't gives any real observation for "material falling toward the black hole"?
2. " Since Sgr A* is presently known to be consuming very little material, it is not surprising the jet appears weak"
Which jet? Is it the Inflow or outflow?
I was quite sure that the following message: "it is not surprising the jet appears weak" is about outflow. Remember, was quite difficult to observed the outflow jet.
In the other article it is stated that:
"For example, scientists have detected a weak jet emanating from the black hole at the center of the nearby galaxy M81, but models suggest that if the same jet projected from the Milky Way’s core, we wouldn’t be able to see it. "
So, do you agree that the "a weak jet emanating from the black hole" is the outflow jet?
3. "A jet in the opposite direction is not seen,". If the "jet appears weak" is about outflow, than why the "A jet in the opposite direction" could not represents the inflow?
If it is inflow - than it is stated clearly - "is not seen".
Can you please explain what is the meaning in English for "is not seen"
Could it be that NASA see inflow but can't tell us what they see?
Or, NASA  just don't see any inflow?
If you still assume that NASA see inflow, would you kindly explain how do you read/understand that message?

Yes, since that article says that matter has been migrating into Sagittarius* over the past 10 billion years.
In the article it is stated:
 
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.
So, What do you understand from:
"The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way"
Do you see there any word about the Observation data?
Don't you agree that the Observation data is only used to find the pointing direction of "the spin axis of Sgr A*" 
So, do you agree that the "study" is not the "Observation data"?
Hence, the "study" indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years and not the "Observation data".
However, if you still sure that the "Study" means "NASA observation data" then would you kindly explain this assumption to a person that isn't a "native English speaker".
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 12/09/2019 21:45:18
1. "Scientists think jets are produced when some material falling toward the black hole is redirected outward."
So,  "Scientists think". It is very nice that they think. However, does it mean that they really see? Don't you agree that thinking is one issue and observation is totally another issue?

Yes, but all of the evidence points to this being the source of power for the jets. It's the only source of matter available. Super-massive black holes can't produce matter.

So, do you agree that this message doesn't gives any real observation for "material falling toward the black hole"?

They did when they said this:

Quote
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.

2. " Since Sgr A* is presently known to be consuming very little material, it is not surprising the jet appears weak"
Which jet? Is it the Inflow or outflow?

Outflow, obviously, since inflow isn't a jet.

So, do you agree that the "a weak jet emanating from the black hole" is the outflow jet?

Yes.

3. "A jet in the opposite direction is not seen,". If the "jet appears weak" is about outflow, than why the "A jet in the opposite direction" could not represents the inflow?
If it is inflow - than it is stated clearly - "is not seen".
Can you please explain what is the meaning in English for "is not seen"

I already corrected you on this. Here's a hint: inflows aren't jets.

Could it be that NASA see inflow but can't tell us what they see?
Or, NASA  just don't see any inflow?
If you still assume that NASA see inflow, would you kindly explain how do you read/understand that message?

Like this:

Quote
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.

So, What do you understand from:
"The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way"
Do you see there any word about the Observation data?
Don't you agree that the Observation data is only used to find the pointing direction of "the spin axis of Sgr A*"
So, do you agree that the "study" is not the "Observation data"?
Hence, the "study" indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years and not the "Observation data".
However, if you still sure that the "Study" means "NASA observation data" then would you kindly explain this assumption to a person that isn't a "native English speaker".

The study was based on the observational data. The conclusion that matter has been moving inwards for 10 billion years is obviously a result of their analysis of the data. That seems painfully obvious.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/09/2019 06:43:46
Quote
3. "A jet in the opposite direction is not seen,". If the "jet appears weak" is about outflow, than why the "A jet in the opposite direction" could not represents the inflow?
If it is inflow - than it is stated clearly - "is not seen".
Can you please explain what is the meaning in English for "is not seen"
Outflow, obviously, since inflow isn't a jet.
I already corrected you on this. Here's a hint: inflows aren't jets.
Ok
So, do you fully agree that there is no observation for inflow jet?.
http://www.sci-news.com/astronomy/science-chandra-jet-supermassive-black-hole-01564.html
"Jets of high-energy particles are found throughout the Universe, on large and small scales. They are produced by young stars and by black holes a thousand times larger than the Milky Way’s black hole."
So, all the jet that we see in any SMBH is always outflow JET.
We have never ever observed any inflow jet.
However, you still want to believe that somehow we should observe inflow.
If that is correct, why you couldn't say it long time before that you agree that we have never ever found an inflow jet, but you believe that just an inflow is feasible.
So, can you please explain how could it be that all the outflows are coming in jet forms, while the inflows must be just an inflow without jet?
Why there is no clear observation for inflow jet?
If an object will fall into the accretion disc or the SMBH, why it can't create some sort of a jet stream as it falls in?
In any case, I still claim that NASA didn't see any sort of inflow.
Not with a jet and not without a jet.
You claim the opposite.
So let's look again on the following statement:
Quote
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.
Your explanation was:
The study was based on the observational data. The conclusion that matter has been moving inwards for 10 billion years is obviously a result of their analysis of the data. That seems painfully obvious.
What is the meaning of "Their" in the following: "their analysis of the data"?
Do you mean NASA analysis?
If so, I totally disagree.
Who set that Study? Do you see any involvement in the Study by NASA?
Where do you get that it was a "different study"? Mark Morris is explicitly mentioned as a co-author of the study that the news report is about (with Zhiyuan Li being another co-author). There is another news report by Scientific American that says the same thing: https://www.scientificamerican.com/article/milky-way-black-hole-jet/ And another: https://chandra.harvard.edu/blog/node/467 And another: http://www.sci-news.com/astronomy/science-chandra-jet-supermassive-black-hole-01564.html
In all of those articles which you have offered, you can't find even a single word about the involvement of NASA team in that "Study".
The study was based on the observational data
This is an error.
In that study, they are using NASA observation to find "the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way"
So, they don't extract any inflow (Jet or none jet) from NASA data observation, but the observed outflow jet (which had been discovered by NASA), had been used to show the direction of that "spin axis of Sgr A*"
Why do you insist to mislead yourself and still hope that NASA observation data itself is used to show the inflow?
As it is stated that - "the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way", it proves that in that "study" they have used the verified "spin axis pointing direction" and not the observation data itself.
Therefore, it is clear to me that in this "study" they didn't try to find any sort of inflow directly from NASA observed data.
Hence, in that "Study" they didn't try to set any sort of manipulation in the data itself in order to prove any sort of inflow.
However, in that article from arxiv, they have clearly stated that they have used NASA data observation to find inflow.
Therefore, this is a clear manipulation in my point of view.
In any case, it is quite clear to me that you will continue to see it differently.
You wish to show that in NASA observation data there is a prove for inflow - and I fully disagree with that.
Super-massive black holes can't produce matter.
This is a wrong assumption of the science community.
They just disagree with what they see.
Based on their theory- the SMBH must eat matter from outside.
So, they ignore all the clear observations (from any sort of detector) and try to show that somehow something must fall in.
Do you really believe that we have the power to tell our SMBH if it must eat something or not?
As NASA claims clearly and loudly that they only see outflow jets from any SMBH that they have observed, why is it so difficult for our science community to accept this message as is.
Why can't they accept this clear observation by NASA?
Why can't they go back to the design table and restart their theory?
This is what any design engineer will do once he discovers a contradiction between his theory and the observed data.
Therefore, we live in a world with magnificent breakthrough discoveries and development in electronics and computing.
Each one of us carry a modern I-phone based on the Engineering approach that we must accept the observed data as is and change our theory accordingly.
However, in Astronomy - they don't agree with NASA observed data as is.
Our ASTRONOMERS at arxiv insist to manipulate NASA observation data in order to find a fit to their wishful thinking of INFLOW (any sort of inflow is good for them).
Why is it?
Don't you agree that it is much more logical to update your theory instead of the manipulation in NASA observation data?

In any case.
I really don't want to upset you.
I do appreciate your great support!!!
Let's agree that we disagree on this "inflow" and go on to the next issue.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 14/09/2019 14:59:34
Ok
So, do you fully agree that there is no observation for inflow jet?.

Yes, because inflows aren't jets.

So, can you please explain how could it be that all the outflows are coming in jet forms, while the inflows must be just an inflow without jet?

I never said that all outflows are in the form of jets. The outflow at the poles of the black hole obviously are, though. There is no reason that an inflow should be a jet.

Why there is no clear observation for inflow jet?

Because inflows aren't jets. But I am also confused. You were saying earlier that there was inflow in the past, which means that you agreed that inflow can happen. Are you changing your mind all of a sudden?

f an object will fall into the accretion disc or the SMBH, why it can't create some sort of a jet stream as it falls in?

Because that's not what a jet is. That's just a falling gas cloud.

What is the meaning of "Their" in the following: "their analysis of the data"?
Do you mean NASA analysis?
If so, I totally disagree.
Who set that Study? Do you see any involvement in the Study by NASA?

Obviously, I'm talking about Zhiyuan Li, Mark Morris and whoever else worked on that study. As long as NASA supplied the data, that's the only involvement on their part that was needed.

In all of those articles which you have offered, you can't find even a single word about the involvement of NASA team in that "Study".

They were obviously involved in obtaining the data.

This is an error.
In that study, they are using NASA observation to find "the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way"
So, they don't extract any inflow (Jet or none jet) from NASA data observation, but the observed outflow jet (which had been discovered by NASA), had been used to show the direction of that "spin axis of Sgr A*"
Why do you insist to mislead yourself and still hope that NASA observation data itself is used to show the inflow?

Because they explicitly said so when they said:

Quote
The study shows the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way, which indicates to astronomers that gas and dust have migrated steadily into Sgr A* over the past 10 billion years.

As it is stated that - "the spin axis of Sgr A* is pointing in one direction, parallel to the rotation axis of the Milky Way", it proves that in that "study" they have used the verified "spin axis pointing direction" and not the observation data itself.

The observational data is what they used to obtain the spin axis. Otherwise, why bother mentioning Chandra at all?

Therefore, it is clear to me that in this "study" they didn't try to find any sort of inflow directly from NASA observed data.

Inflow can't be observed directly. It must always be inferred from data. This would be true for many forms out outflow as well.

Hence, in that "Study" they didn't try to set any sort of manipulation in the data itself in order to prove any sort of inflow.

Once again, studies like this never "prove" anything...

You wish to show that in NASA observation data there is a prove for inflow - and I fully disagree with that.

No, no I don't. Quit using the word "prove".

This is a wrong assumption of the science community.

It's not an assumption. It's a consequence of the laws of physics. Things cannot get out of an event horizon.

They just disagree with what they see.

We have never observed a black hole creating matter.

Based on their theory- the SMBH must eat matter from outside.
So, they ignore all the clear observations (from any sort of detector) and try to show that somehow something must fall in.

They have observed things falling in (as pointed out numerous times in the past).

Do you really believe that we have the power to tell our SMBH if it must eat something or not?

They aren't "telling" it anything.

As NASA claims clearly and loudly that they only see outflow jets from any SMBH that they have observed, why is it so difficult for our science community to accept this message as is.

NASA never said that they have only ever observed outflow.

Why can't they accept this clear observation by NASA?

Because it isn't what NASA claims.

Why can't they go back to the design table and restart their theory?

Because no compelling evidence against the theory has yet to surface.

This is what any design engineer will do once he discovers a contradiction between his theory and the observed data.

There is no contradiction.

However, in Astronomy - they don't agree with NASA observed data as is.

They obviously do, since they explicitly used that data to come to their conclusions.

Our ASTRONOMERS at arxiv insist to manipulate NASA observation data in order to find a fit to their wishful thinking of INFLOW (any sort of inflow is good for them).
Why is it?
Don't you agree that it is much more logical to update your theory instead of the manipulation in NASA observation data?

You have yet to demonstrate that there was any such manipulation.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 15/09/2019 05:14:08
Quote from: Dave Lev
Do you really believe that we have the power to tell our SMBH if it must eat something or not?

A study released this past week shows some bubbles of hot gas around the Milky Way's SMBH.
- One possible interpretation of this observation is that the SMBH had a feeding frenzy a couple of million years ago.
- This hot gas would tend to discourage more gas from flowing into the SMBH
- But it won't stop stars orbiting the SMBH

See: https://news.northwestern.edu/stories/2019/09/milky-way/

In future, we can expect far more data, with much higher resolution, as the MeerKAT pathfinder telescope gets expanded into the Square Kilometer Array radio telescope.
See: https://en.wikipedia.org/wiki/MeerKAT

We are not telling the SMBH when to eat, but we are trying to see when it had breakfast (and what it ate for breakfast)...
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/09/2019 20:29:48
The outflow at the poles of the black hole obviously are, though. There is no reason that an inflow should be a jet.
Dear Kryptid
There is no outflow from the pole of the black hole.
This is a severe mistake.
The outflow is from the accretion disc itself.

Please look again at the Wind structure outflow from the accretion disc at the page before the last one:
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf
So, the matter is first ejected outwards from the accretion disc and at the plane of the accretion disc.
In the following article it is stated that it is due to magnetic field:
 https://www.scientificamerican.com/article/milky-way-black-hole-jet/
"Jets arise because the black hole is spinning. As matter falls into the black hole, the matter’s magnetic field gets twisted and amplified by the black hole’s spin, and this pumped-up magnetic field launches material outward in the form of jets."
Therefore, the outflow from the SMBH is always from the accretion disc due to magnetic field.
However, due to the magnetic field lines, the matter is lifted upwards.
Again - we can see it in the presentation which I have offered.

So, let me ask you if you agree with the following:
1.The magnetic field launches material outward from the accretion disc and set the first stage of outflow?
2. The same magnetic field also Pushes this outflow in the direction of upwards/downwards with regards to the accretion disc plane?
3. As it boosted outwards (at almost 0.8c) due to the magnetic field lines it set the molecular jet stream that we see above and below the galactic disc. (this jet stream move directly above the pole)?
4. So, the jet stream outflow that we see above and below the accretion disc plane is a direct outcome from the impact of the magnetic field on the matter in the accretion disc?

If you agree with all the above;
I really can't understand why you constantly refuse to accept the following direct outcome:
This magnetic field is so strong that it can pull matter (including particles, atoms, molecular) from the accretion disc, than band them outwards and shoot them at a speed of 0.8 directly upwards above (or below) the pole of the SMBH.
If so,
1. How any sort of matter can fall into the accretion disc? Why the same mighty magnetic field that sets the outflow jets can't push outwards any matter that just think to fall into the accretion disc?
2. Why the same magnetic field can't pull particles from the event horizon of the SMBH?
Once you agree with that we have solved the enigma of the whole Universe.

Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 15/09/2019 22:47:06
Quote from: Dave Lev
Please look again at the Wind structure outflow from the accretion disc at the page before the last one:
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf
So, the matter is first ejected outwards from the accretion disc and at the plane of the accretion disc.
In the following article it is stated that it is due to magnetic field:
 https://www.scientificamerican.com/article/milky-way-black-hole-jet/
"Jets arise because the black hole is spinning. As matter falls into the black hole, the matter’s magnetic field gets twisted and amplified by the black hole’s spin, and this pumped-up magnetic field launches material outward in the form of jets."
Therefore, the outflow from the SMBH is always from the accretion disc due to magnetic field.
However, due to the magnetic field lines, the matter is lifted upwards.
Again - we can see it in the presentation which I have offered.

So, let me ask you if you agree with the following:
1.The magnetic field launches material outward from the accretion disc and set the first stage of outflow?
2. The same magnetic field also Pushes this outflow in the direction of upwards/downwards with regards to the accretion disc plane?
3. As it boosted outwards (at almost 0.8c) due to the magnetic field lines it set the molecular jet stream that we see above and below the galactic disc. (this jet stream move directly above the pole)?
4. So, the jet stream outflow that we see above and below the accretion disc plane is a direct outcome from the impact of the magnetic field on the matter in the accretion disc?
Fantastic description! Bravo!
- You have just described the operation of an accretion disk, its magnetic fields and generation of polar jets, as understood by current theories.
- It's a pity that you didn't post this 15 pages ago...

Now given that agreement, there is no need for the rest of your post, which presumes that matter can be extracted from beneath the event horizon of a black hole using magnetic fields.
- According to current theories, this can't be done
- According to current theories, the only thing that can come out of a SMBH event horizon is a tiny wisp of Hawking radiation, which is immeasurably small, even without the energetic maelstrom which is the nearby accretion disk.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 15/09/2019 23:19:14
Dear Kryptid
There is no outflow from the pole of the black hole.
This is a severe mistake.
The outflow is from the accretion disc itself.

Please look again at the Wind structure outflow from the accretion disc at the page before the last one:
http://phsites.technion.ac.il/talks/agn2017/Reeves-J.pdf

That particular website is about “disk winds”. It’s true that they originate from the accretion disk, but they are a different phenomenon from the polar jets. The word “jet” isn’t even in that document.

In the following article it is stated that it is due to magnetic field:
 https://www.scientificamerican.com/article/milky-way-black-hole-jet/
"Jets arise because the black hole is spinning. As matter falls into the black hole, the matter’s magnetic field gets twisted and amplified by the black hole’s spin, and this pumped-up magnetic field launches material outward in the form of jets."
Therefore, the outflow from the SMBH is always from the accretion disc due to magnetic field.
However, due to the magnetic field lines, the matter is lifted upwards.
Again - we can see it in the presentation which I have offered.

Yes, all of that is true, but it isn’t related to disk winds. The polar jets are a different phenomenon. These are the jets from the poles that I am talking about: http://astronomy.swin.edu.au/cosmos/G/Galactic+Jets There is one on each side of the galaxy. The jets don’t literally come from the poles of the black hole, the accretion flow is just lifted along those particular axes.

So, let me ask you if you agree with the following:
1.The magnetic field launches material outward from the accretion disc and set the first stage of outflow?
2. The same magnetic field also Pushes this outflow in the direction of upwards/downwards with regards to the accretion disc plane?

For disk winds, yes.


3. As it boosted outwards (at almost 0.8c) due to the magnetic field lines it set the molecular jet stream that we see above and below the galactic disc. (this jet stream move directly above the pole)?

No, the polar jets are a different phenomenon (even though they also originate from the accretion disk and are propelled by magnetic fields).


4. So, the jet stream outflow that we see above and below the accretion disc plane is a direct outcome from the impact of the magnetic field on the matter in the accretion disc?

Yes.

1. How any sort of matter can fall into the accretion disc? Why the same mighty magnetic field that sets the outflow jets can't push outwards any matter that just think to fall into the accretion disc?

Because magnetic fields neither attract nor repel charged particles. The way that the particles interact with the field is strongly dependent upon their path relative to the field lines. Those moving up or down can continue to move up or down, but those moving directly at right angles to the field lines will only spin around the field lines without moving up or down significantly. This spin will cause the particle to lose energy, weakening the effect of the magnetic field on it and allowing it to fall further towards the black hole.

2. Why the same magnetic field can't pull particles from the event horizon of the SMBH?

Because magnetic fields can’t move things into the past. This has been explained to you many times before.

Quote
Once you agree with that we have solved the enigma of the whole Universe.

No we won’t, because your model violates conservation of mass/energy (a black hole cannot send mass/energy out in the Universe without losing mass/energy itself).
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/09/2019 01:20:19
The jets are going to be aligned with the axis of rotation of the accretion disk, not the axis of the black hole itself.

Yes, that's true. I was making a simplified assumption.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/09/2019 15:57:53
The polar jets are a different phenomenon. These are the jets from the poles that I am talking about: http://astronomy.swin.edu.au/cosmos/G/Galactic+Jets There is one on each side of the galaxy. The jets don’t literally come from the poles of the black hole, the accretion flow is just lifted along those particular axes.
Thanks Kryptid
Do appreciate this great article.
It is stated:
"Another question troubling astronomers is what exactly are these jets made of? The modern consensus is that the main constituents are electrons and their anti-particle equivalent, positrons. These are thought to be produced by ‘pair-production’ close to the event horizon of the central black hole, and squirted out at almost light speed by the action of the combined magnetic fields."
So, our scientists estimates that there is a possibility for ‘pair-production’ close to the event horizon of the central black hole, and squirted out at almost light speed by the action of the combined magnetic fields.
That meets my expectations form the mighty magnetic fields.
So, why our scientists can justify those pair-production near the event of horizon that are squirted out by magnetic force, while it is forbidden for me to use the same identical idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/09/2019 18:16:59
So, why our scientists can justify those pair-production near the event of horizon that are squirted out by magnetic force, while it is forbidden for me to use the same identical idea?

Your idea wasn't identical. You were proposing that they come from within the event horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/09/2019 20:49:43
Your idea wasn't identical. You were proposing that they come from within the event horizon.
Well, I really don't care about the distance between those new created pair production to the event horizon.
As long as they are squirted out at almost light speed by the action of the combined magnetic fields from the SMBH into the accretion disc - that is perfectly OK with me.
Not sure how that sources material for the jets since the positrons typically find another electron somewhere in the accretion disk and annihilate it, for a net zero gain of charged particles
It seems to me that our knowledge about the functionality of the accretion disc is quite poor.
In the accretion disc new arrival created partials which had been squirted out from the SMBH are converted into real Atoms and molecular.
Therefore, at the innermost accretion ring, we should see mainly those kind of new born particles.
However, as the particles are drifted outwards in the accretion disc, they are converted into atoms and molecular.
Finally, they are ejected outwards from the accretion disc and lifted upwards (or downwards) by the mighty magnetic fields and set the molecular jet stream.
There is an excellent reason for why the galaxy needs to create that jet stream.
Not all the particles have been converted into real atoms and molecular at the moment that they are ejected from the accretion disc.
Therefore, in the outflow from the accretion disc we should find particles, broken atoms, atoms and molecular.
If those  particles will be part of a new star it might create a very negative impact.
Somehow, those immature particles must be pushed away.
Therefore, the accretion outflow is boosted upwards into a molecular jet stream by the magnetic field.
The relatively heavy atoms and molecular are falling back to the galactic disc plane and are gathered into gas cloud in order to start the new star forming activity, while the immature particles are pushed away from the galaxy.
So, the molecular jet stream set the cleaning activity and push away the immature particles from the matter which should be used as the building blocks for any new star.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/09/2019 22:49:29
Well, I really don't care about the distance between those new created pair production to the event horizon.
As long as they are squirted out at almost light speed by the action of the combined magnetic fields from the SMBH into the accretion disc - that is perfectly OK with me.

The pair production happens in the accretion disk near the event horizon, where the temperatures are very, very high. Another example of such pair production is in "pair-instability supernova": https://en.wikipedia.org/wiki/Pair-instability_supernova

Since the temperature of the material itself is what drives this, the energy and mass of the accretion disk is reduced by it when the pairs are blasted off into the jets. So you are still left with the conundrum of where the matter in the accretion disk comes from, since you don't allow it to come in from outside nor can it come out of the black hole because of the event horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/09/2019 07:58:24
The pair production happens in the accretion disk near the event horizon, where the temperatures are very, very high
Dear Kryptid
The modern consensus is that the main constituents are electrons and their anti-particle equivalent, positrons. These are thought to be produced by ‘pair-production’ close to the event horizon of the central black hole, and squirted out at almost light speed by the action of the combined magnetic fields."
Our scientists specifically claim that the ‘pair-production’ are produced close to the event horizon of the central black hole.
Therefore, it is clear that the ‘pair-production’ have been produced outside the accretion disc.
Hence the pair- production activity itself has no effect on the energy and mass in the accretion disk.
However, as the pair- production are squirted out at almost light speed by the action of the combined magnetic fields and drift into the innermost ring of the accretion disc - the mass of the accretion disc is increasing.
In the accertion disc, those particales are transformed into real Atoms and Molecular. The main energy for that activity is coming from the same magnetic field. Therefore, in total the mass of the accretion disc is increasing due to the new arrival of the  pair- production and due to the transformation activity.
You claim:
Since the temperature of the material itself is what drives this, the energy and mass of the accretion disk is reduced by it when the pairs are blasted off into the jets
The pairs are blasted off into the jets by the power of the magnetic fields.
Hence, the energy and mass in the accretion disc is quite stable in the long run due to the balance between the inflow into the accretion disc of the new born pair- production  from the event horizon (or close to it) and their transformation activity into new atoms and molecular with regards to the total jet outflow from the accretion disc.
Don't you see that the game is over?
You have offered solid evidence to that theory.
Fron now on we must call it - Excretion disc.
Why don't we share our efforts in bringing the breakthrough information to the science community and mankind?
We need to get a reward for our discovery!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/09/2019 14:11:55
Therefore, it is clear that the ‘pair-production’ have been produced outside the accretion disc.

Not at all. Pair production is modeled as happening inside the accretion disk. There are other sources that state this:
https://iopscience.iop.org/article/10.1088/0004-637X/735/1/9/meta

Quote
Here, we use a self-consistent dynamical and radiative model to investigate pair production by γγ collisions in weakly radiative accretion flows around a black hole of mass M and accretion rate M.

http://cxc.harvard.edu/cdo/accr10/pres/moscibrodzka_monika.pdf:

Quote
For first time we compute non-equilibrium electron-positron pair production rates by γγ from turbulent accretion disk around spinning black hole


https://www.sciencedirect.com/science/article/pii/S0273117797000458:

Quote
The high pair creation rate during this state is indicated by the presence of a broad e−e+ annihilation line-like feature and needs the presence of a high temperature radiation field. We put forward a scenario in which the observed spectrum originates from the inner region of an accretion disk around a rapidly rotating black hole.
Hence the pair- production activity itself has no effect on the energy and mass in the accretion disk.

Yes it does, since the energy of the accretion disk is what allows them to be formed in the first place.

the mass of the accretion disc is increasing.

So where do you claim that this extra mass comes from? If the pairs aren't produced by the accretion disk, what are they produced by? Where does the energy necessary to create those pairs come from? Do you recognize that whatever source you choose must lose energy in the process of creating those pairs?

Don't you see that the game is over?

It was over as soon as you made the first post in this thread, because your model was wrong since the first post.

You have offered solid evidence to that theory.

I have offered nothing that shows the pairs are produced anywhere outside of the plasma around the black hole. Energy is required to produce particle pairs. That energy has to come from somewhere.

Fron now on we must call it - Excretion disc.

Non-sequitur.

Why don't we share our efforts in bringing the breakthrough information to the science community and mankind?
We need to get a reward for our discovery!!!

You haven't made any discoveries.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/09/2019 17:33:49
Quote
Therefore, it is clear that the ‘pair-production’ have been produced outside the accretion disc.
Not at all. Pair production is modeled as happening inside the accretion disk. There are other sources that state this:
https://iopscience.iop.org/article/10.1088/0004-637X/735/1/9/meta
In this article it is stated clearly that pair production is concentrated close to the event horizon, but not in the accretion disc:
"Electron-positron pairs may be produced near accreting black holes by a variety of physical processes"
"We have studied electron–positron pair production in black hole magnetospheres by γγ collisions."
"The pair production rates are calculated nearly ab initio within 40GM/c2 of the event horizon, using Monte Carlo methods."
"The main results of this work are the fitting formulae for the rate and spatial distribution of pair production in terms of m8 and $\dot{m}$ (Equation (26)) and in terms of m8, LX, and α (Equation (30)). These indicate that γγ pair production is concentrated close to the event horizon, and is sensitive to model parameters such as $\dot{m}$."
So, where do you see that Pair production is happening inside the accretion disk?
http://cxc.harvard.edu/cdo/accr10/pres/moscibrodzka_monika.pdf:
"For first time we compute non-equilibrium electron-positron pair production rates by γγ from turbulent accretion disk around spinning black hole"
In this model they discuss about "turbulent accretion disk around spinning black hole".
I don't think that they aim for the accretion disc.
Please look at the following article:
https://iopscience.iop.org/article/10.1088/0004-637X/735/1/9/meta
It is stated:
Our pair production rate simulations are based on a GRMHD time-dependent model of a magnetized disk around a spinning black hole.
Therefore, the real meaning of "turbulent accretion disk around spinning black hole" is "magnetized disk around a spinning black hole."

https://www.sciencedirect.com/science/article/pii/S0273117797000458:

Quote
The high pair creation rate during this state is indicated by the presence of a broad e−e+ annihilation line-like feature and needs the presence of a high temperature radiation field. We put forward a scenario in which the observed spectrum originates from the inner region of an accretion disk around a rapidly rotating black hole.
In this article they don't discuss about the location of the pair production. They just speak about the the observed spectrum originates from the inner region of an accretion disk around a rapidly rotating black hole."
This is very interesting.
Actually as the pair production is drifted outwards, it must meet the inner region of the accretion disc.
Therefore, it is very interesting to understand what kind of spectrum they have observed and what can we learn from that.
Quote
Hence the pair- production activity itself has no effect on the energy and mass in the accretion disk.
Yes it does, since the energy of the accretion disk is what allows them to be formed in the first place.

Sorry, in most of the articles it is stated that the pair - production activity is NEAR the event horizon (and not in the accretion disc itself)

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/09/2019 21:41:07
So, where do you see that Pair production is happening inside the accretion disk?

Here:

Quote
For first time we compute non-equilibrium electron-positron pair production rates by γγ from turbulent accretion disk around spinning black hole.

and here:

Quote
We put forward a scenario in which the observed spectrum originates from the inner region of an accretion disk around a rapidly rotating black hole.

I don't think that they aim for the accretion disc.

What do you mean when you say "aim" for the accretion disk?

Therefore, the real meaning of "turbulent accretion disk around spinning black hole" is "magnetized disk around a spinning black hole."

Which, in other words, is the accretion disk.

Sorry, in most of the articles it is stated that the pair - production activity is NEAR the event horizon (and not in the accretion disc itself)

You realize that "near the event horizon" and "in the accretion disk" are not mutually-exclusive locations, don't you?
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 17/09/2019 22:41:10
Quote from: DaveLev, with original emphasis
In this article it is stated clearly that pair production is concentrated close to the event horizon, but not in the accretion disc:
"Electron-positron pairs may be produced near accreting black holes by a variety of physical processes"
These indicate that γγ pair production is concentrated close to the event horizon
So, where do you see that Pair production is happening inside the accretion disk?
You do realise that as the matter in the accretion disk spirals closer to the black hole, most of it actually spirals right into the event horizon?
So there is no actual gap between the event horizon and the inner edge of the accretion disk?
So every time an article says "near a black hole event horizon", they actually mean that it happens within, or on the edges of, the accretion disk.

Quote from: Dave Lev
Fron now on we must call it - Excretion disc.
According to current computer models, about 90% of the matter in the accretion disk falls into the black hole, and only around 10% goes into the polar jets. So retaining the term "accretion disk" is 900% more accurate than changing it to "excretion disk".
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/09/2019 19:56:00
You realize that "near the event horizon" and "in the accretion disk" are not mutually-exclusive locations, don't you?
So every time an article says "near a black hole event horizon", they actually mean that it happens within, or on the edges of, the accretion disk.
Are you both sure about it?
If I recall correctly, you (or Halc) have stated that there is no possibly that the pair-production will take place at the accretion disc, as special conditions are needed for the particles creation. I have got the impression from you that the Pair-production activity must take place even below the event horizon.
Now, suddenly, you claim that the pair-production can be created in the accretion disc, high above the event horizon.
Thanks for that Great news.
If new particles can be created at the accretion disc, than why do I need to find a solution how to squirt out the new particles from the event horizon (or close to it)?
I don't need any more to call the magnetic field for help. Therefore, the following explanation is not needed:
"The modern consensus is that the main constituents are electrons and their anti-particle equivalent, positrons. These are thought to be produced by ‘pair-production’ close to the event horizon of the central black hole, and squirted out at almost light speed by the action of the combined magnetic fields."
Now you agree that new particle are created at the accretion disc. Therefore, the solution is as follow:
New particle are created at the innermost accretion ring.
Due to their ultra high velocity (at the moment of creation), high gravity force and mighty magnetic field those particles are transformed into real atoms as they are drifted outwards.
The requested energy for that transformation is contributed by the gravity force + magnetic energy.
Therefore, the outflow jet from the accretion disc is a direct outcome from that process.
I hope that this explanation meets your expectations.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/09/2019 21:31:01
The requested energy for that transformation is contributed by the gravity force + magnetic energy.

The only way to get energy out of a gravitational field is by converting gravitational potential energy into kinetic energy. This, in turn, can only be done by lowering a mass into the gravitational field. You, however, are taking a mass and pushing it out and far away from the black hole (in the form of jets). That doesn't give you energy, it requires it instead. So gravity is not a source of energy you can use for this.

Magnetism can indeed transfer energy (electric motors use this principle all the time), but that energy doesn't come from nowhere. The source of that magnetic field must lose energy in the process. If the source of the field is the accretion disk, then the disk must lose energy. If the source of the field is the black hole (in your hypothetical case), then the black hole must lose energy (likely by slowing down its spin). So you still have a system that loses energy and mass over time.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/09/2019 15:11:26
The only way to get energy out of a gravitational field is by converting gravitational potential energy into kinetic energy. This, in turn, can only be done by lowering a mass into the gravitational field. You, however, are taking a mass and pushing it out and far away from the black hole (in the form of jets). That doesn't give you energy, it requires it instead. So gravity is not a source of energy you can use for this.
Yes, I agree.
However, without the ultra high gravity, no new particle would be created.
Magnetism can indeed transfer energy (electric motors use this principle all the time), but that energy doesn't come from nowhere.
That is fully correct.
So the energy should come from the magnetic field.
The source of that magnetic field must lose energy in the process.If the source of the field is the accretion disk, then the disk must lose energy.
That is also correct.
However, SMBH is the main source for the magnetic field and not the accretion disc itself.
In the article it is stated:
Our pair production rate simulations are based on a GRMHD time-dependent model of a magnetized disk around a spinning black hole.
The spinning black hole/SMBH sets the magnetic fields.
The amplitude/energy of the magnetic filed is directly effected by the total mass of the SMBH.
If the source of the field is the black hole (in your hypothetical case), then the black hole must lose energy (likely by slowing down its spin).
That is also correct if we isolate the SMBH.
However, our SMBH is not working for free.
Some of the new created particles must fall into the SMBH from the Event Horizon or close to it.
Therefore, as new particles are squirted outwards into the accretion disc, some of them are falling directly into the SMBH.
Those particles increase the total mass of the SMBH and therefore, there is increasing in the magnetic fields.
Hence, this increasing in the magnetic field overcomes the energy lost due to the creation activity of new atoms and molecular in the accretion disc and due to the requested energy that drifts them all outwards into that jet outflow.
So you still have a system that loses energy and mass over time.
No, there is a win-win situation.
Both - the accretion disc and the SMBH get new particles.
The accretion disc converts them into Atoms and molecular that will be used to form new stars, while the SMBH increases its mass during that activity.
Therefore, we get so massive objects as SMBH.
All the matter in the Milky way had been created by the SMBH (including the nearby dwarf galaxies and any other object that orbits around the galaxy)
Our galaxy will not accept any matter from outside.
As it crosses the space at a speed of over than 600 Km/sec its incredible gravity force swifts away any star and any smaller galaxy.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/09/2019 17:18:22
Those particles increase the total mass of the SMBH and therefore, there is increasing in the magnetic fields.
Hence, this increasing in the magnetic field overcomes the energy lost due to the creation activity of new atoms and molecular in the accretion disc and due to the requested energy that drifts them all outwards into that jet outflow.

That violates the first law of thermodynamics. You can't get more energy out than you put in.

Both - the accretion disc and the SMBH get new particles.

"New particles" does not equal "new mass/energy". The total mass/energy of the system cannot increase over time without violating the first law of thermodynamics.

Quote
Some of the new created particles must fall into the SMBH from the Event Horizon or close to it.

I thought you said that nothing can fall into the black hole because the magnetic field won't let it?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/09/2019 19:12:50
Quote
Some of the new created particles must fall into the SMBH from the Event Horizon or close to it.
I thought you said that nothing can fall into the black hole because the magnetic field won't let it?
I have stated clearly - Nothing outside the accretion disc can fall into the accretion disc or the SMBH.
However, new created particles close to the event of horizon can fall into the accretion disc or the SMBH.
That violates the first law of thermodynamics. You can't get more energy out than you put in.
"New particles" does not equal "new mass/energy". The total mass/energy of the system cannot increase over time without violating the first law of thermodynamics.
No, there is no violation of the first law of thermodynamics.
Please remember, Atom is a cell of energy.
I have discussed about it very deeply.
So, the energy in the atom is achieved by the Supper high gravity force pulse the mighty magnetic field.
Therefore, the creation of new atom should be represented as an energy transformation.
Remember the famous formula by Einstein
E = M c^2
In Atomic bomb the mass is converted into pure energy without any violation of the first law of thermodynamics.

In the accretion disc the opposite activity takes place.
Energy is transformed into new atom.
Therefore, we can claim that the energy E is contributed by the mighty force (gravity + magnetic) of the spinning black hole.
As an outcome we get:
M = E / c^2
That represents the Energy that is requested to create new atom without any violation of thermodynamics law.
Some of that mass falls inwards (into the SMBH) and the other part is ejected outwards as outflow jet.

Therefore, when we see an atomic bomb in action, we actually see the Energy of the SMBH that was kept in the Atom cell.
In the same token - Our Sun lights our life with the Energy that it have got from the SMBH (As it converts mass to Energy by fusion activity).
Even our body holds the Energy from the SMBH.
The whole galaxy is there due to the Energy from the SMBH.
Therefore, Energy can be transformed into mass as mass can be transformed into Energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/09/2019 20:46:27
No, there is no violation of the first law of thermodynamics.

If you propose that a black hole can become heavier by eating some of the mass that was taken out of it, you are absolutely are proposing such a violation. Let's say that the magnetic field transfers 1.022 keV of mass-energy out of the black hole into order to form an electron-positron pair. The black hole must now weigh 1.022 keV less than it did before. Now, one of those particles (0.511 keV) is thrown either into the accretion disk or into the jet, while the other 0.511 keV particle falls back into the black hole. The black hole lost 1.022 keV by forming the particle pair and only got 0.511 keV back by eating one member of the pair. That's still a net loss of 0.511 keV. Elementary school arithmetic demands that the black hole has lost mass, not gained it.

Remember the famous formula by Einstein
E = M c^2
In Atomic bomb the mass is converted into pure energy without any violation of the first law of thermodynamics.

The total mass-energy before and after the explosion is identical, unlike what you are proposing.

Therefore, Energy can be transformed into mass as mass can be transformed into Energy.

Mass and energy are technically already equivalent. all they are doing is changing form from potential energy into kinetic energy or vice-versa. That still does not explain how you think a black hole can get heavier by eating mass that has already been taken out of it.

How do you think that a magnetic field and gravity can form new particles anyway? What mechanism is behind it? You can't be talking about Hawking radiation, since that causes a black hole to lose mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/09/2019 04:34:27
Let's say that the magnetic field transfers 1.022 keV of mass-energy out of the black hole into order to form an electron-positron pair. The black hole must now weigh 1.022 keV less than it did before.
Yes, that could be correct if the the creation of new partial/atom was only based on Magnetic force.
However, we already know that the magnetic force by itself can't create any new particle.
The combination of ultra high gravity of the SMBH + the mighty magnetic fields sets the creation of new particle/atom.
Therefore, the gravity has an important job in this new creation activity.
However, the gravity comes almost for free.
I say "almost" as our scientists consider that it is totally free.
We have already discussed deeply about the impact on the gravity force in the long run.
Let's look again on the Sun/Earth gravity system.
If I recall it correctly, you assume that the gravity stays at the same amplitude over time (assuming that there is no reduction in the mass).
So, you don't see any reduction in the gravity force while the sun holds the Earth in its orbital momentum by gravity.
You have stated that the Earth is drifting outwards not due to gravity force reduction but due to tidal.
So, you actually claim that the gravity is there for free for ever.
I have stated that there must be a "friction" or reduction also in gravity over time.
So, the Earth is drifting outwards over time as the gravity is reducing due to the orbital activity.
Therefore, I would like to understand from you:
Why when it comes to pair production there is a reduction in the gravity force, while when it comes to any gravity system (as the orbital path of the Sun around the galaxy) there is no gravity force reduction?
Our scientists assume that the Sun stays exactly at the same radius from its first moment.
Therefore, they claim that the Sun orbits around the galaxy for the last 6 billion years exactly at the same radius.
I don't agree with that as there must be also a reduction in gravity over time (even if it is a very low reduction).
In any case, if you assume that the sun is not losing gravity force due to its orbital momentum around the galaxy, than I can claim that a new created particle is also do not set any reduction in the gravity force of the SMBH (close to the Event Horizon).
Now, let's go back to your question:
I claim that the SMBH is losing much less than 1.022 keV in order to form an electron-positron pair as the activity of that creation is based on Gravity + Magnetic Energy.
As the Gravity comes almost for free, than the total reduction in the Gravity + Magnetic Energy must be less than 1.022 keV/2
For this explanation let's assume that the total reduction in the Gravity +Magnetic Energy of the SMBH that is needed to create electron-positron pair is 1.022 keV/4.
Now, if we assume that one particle is falling in the SMBH it gets a mass of 1.022 keV/2.
Therefore, it gains 1.022 keV/2 - 1.022 keV/4 in that activity.
Hence, while the accretion disc gets 1.022 keV/2 for free, the SMBH gains 1.022 keV/4.
This is a win win situation.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/09/2019 05:49:43
The combination of ultra high gravity of the SMBH + the mighty magnetic fields sets the creation of new particle/atom.

How do you propose that works, exactly? What is the precise mechanism?

I say "almost" as our scientists consider that it is totally free.
We have already discussed deeply about the impact on the gravity force in the long run.
Let's look again on the Sun/Earth gravity system.
If I recall it correctly, you assume that the gravity stays at the same amplitude over time (assuming that there is no reduction in the mass).
So, you don't see any reduction in the gravity force while the sun holds the Earth in its orbital momentum by gravity.
You have stated that the Earth is drifting outwards not due to gravity force reduction but due to tidal.
So, you actually claim that the gravity is there for free for ever.
I have stated that there must be a "friction" or reduction also in gravity over time.
So, the Earth is drifting outwards over time as the gravity is reducing due to the orbital activity.

And why do you propose that gravity gets weaker over time? Why should that be the case? What causes it? More importantly, what evidence do you have for this?

Why when it comes to pair production there is a reduction in the gravity force, while when it comes to any gravity system (as the orbital path of the Sun around the galaxy) there is no gravity force reduction?

There isn't. The same total amount of mass is present and therefore the same total amount of gravity is also present. The only thing that has changed is the location of the mass. When a particle pair forms, the mass that formed the pair was taken out of the black hole. The total (black hole + particle pair) has the same mass as the original black hole before the formation of the pair.

I don't agree with that as there must be also a reduction in gravity over time (even if it is a very low reduction).

There would be a reduction of gravity from the Sun over time because the Sun is continually shooting particles, gas, and radiation out into the Universe. That process carries mass away from the Sun and therefore also decreases its gravity over time. There is no need for new physics to explain that.

Our scientists assume that the Sun stays exactly at the same radius from its first moment.
Therefore, they claim that the Sun orbits around the galaxy for the last 6 billion years exactly at the same radius.

I'm rather doubtful that this is the consensus view. Stars are constantly on the move, with some coming closer to the Sun and others moving further away. You would expect those gravitational interactions to change the Sun's position at least a little over time. Again, no need for new physics.

In any case, if you assume that the sun is not losing gravity force due to its orbital momentum around the galaxy, than I can claim that a new created particle is also do not set any reduction in the gravity force of the SMBH (close to the Event Horizon).

The Sun is losing gravity because it is slowly losing mass. Gravity is directly linked to mass. Take away some mass and you take away some gravity. If you take particles out of the Sun, the Sun loses a bit of mass and gravity. This is happening constantly. The exact same thing happens if you remove mass from a black hole in order to form particles.

I claim that the SMBH is losing much less than 1.022 keV in order to form an electron-positron pair as the activity of that creation is based on Gravity + Magnetic Energy.

Then you are explicitly breaking the first law of thermodynamics.

As the Gravity comes almost for free,

Gravity is "free" in the sense that a certain amount of gravity is always associated with a certain amount of mass. That's the gravitational constant.

Quote
than the total reduction in the Gravity + Magnetic Energy must be less than 1.022 keV/2

Non-sequitur.

For this explanation let's assume that the total reduction in the Gravity +Magnetic Energy of the SMBH that is needed to create electron-positron pair is 1.022 keV/4.

If you "assume" that, then you "assume" that there is a violation of the first law of thermodynamics.

This is a win win situation.

It is beyond my understanding how you think that a system which creates more mass-energy than it uses doesn't violate the first law of thermodynamics. That's the very definition of what the law is.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/09/2019 06:47:27
The total (black hole + particle pair) has the same mass as the original black hole before the formation of the pair.
That is absolutely correct.
Please look at the following article:
https://en.wikipedia.org/wiki/Pair_production
"the created particles shall have opposite values of each other. For instance, if one particle has electric charge of +1 the other must have electric charge of −1, or if one particle has strangeness of +1 then another one must have strangeness of −1."
So, we get the pair-production without any effect on the SMBH mass as the total mass of the Pair is zero. (One is positive while the other is negative).
When a particle pair forms, the mass that formed the pair was taken out of the black hole.
That is totally incorrect
As I have proved the total mass of the pair is Zero, therefore, the pair production doesn't take any mass out of the black hole.
Therefore, the pair-production is for free.
The SMBH is not losing any mass due to this production process.
Now, let's assume that one particle is falling inwards into the SMBH, while the other one is squirted outwards to the accretion disc.
If a particle that falls inwards into the SMBH has the same polarity as the SMBH than it should increase the total mass of the SMBH. Otherwise, it should decrease the total mass.
The same idea is correct also for the particle which is squirted outwards to the accretion disc.
I claim that as the SMBH is made out of negative polarity mass, the in falling negative particle must increase its mass, while the other positive particle is squirted outwards into the accretion disc.
Our scientists consider that the SMBH is made out of positive polarity mass.
Therefore, they assume that the in falling negative particle should decrease its total mass.
This is their biggest mistake.
As I have already explained, the SMBH is a huge barrel for negative particles.
The gravity effects of Negative mass are identical to a positive mass.
Therefore, while the SMBH increases its negative mass, the accretion disc gets for free the positive particle.
Therefore, both are increasing their total mass without violation the first low of thermodynamics.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/09/2019 07:00:26
That is absolutely correct.
Please look at the following article:
https://en.wikipedia.org/wiki/Pair_production
"the created particles shall have opposite values of each other. For instance, if one particle has electric charge of +1 the other must have electric charge of −1, or if one particle has strangeness of +1 then another one must have strangeness of −1."
So, we get the pair-production without any effect on the SMBH mass as the total mass of the Pair is zero. (One is positive while the other is negative.

That isn't talking about mass (unless you are speaking of Hawking radiation, in which case the particle falling into the hole does indeed have a negative mass).

That is totally incorrect
As I have proved the total mass of the pair is Zero, therefore, the pair production doesn't take any mass out of the black hole.

Alright, let's assume that you are creating a negative-mass and positive-mass particle pair, for the sake of argument...

The SMBH is not losing any mass due to this production process.

If it's eating the negative-mass particle, it is.

I claim that as the SMBH is made out of negative polarity mass, the in falling negative particle must increase its mass, while the other positive particle is squirted outwards into the accretion disc.

That's not how math works. Making a negative number more negative is a decrease, not an increase. Going from -10 to -20 is a decrease. So the black hole is still losing mass if it's eating negative-mass particles.

The gravity effects of Negative mass are identical to a positive mass.

Anyone who can do basic arithmetic can prove this wrong by putting a negative number for mass into the gravitational force equation. You end up with a repulsion, not an attraction. A negative mass producing an attractive gravitational field indistinguishable from a positive mass would also lead to a violation of the first law of thermodynamics. It would allow you to create increasingly strong gravitational potentials out of nothing, which in turn creates extra gravitational potential energy of any orbiting objects out of nothing. That's a violation of conservation of energy.

Also, when are you going to explain what the mechanism is that allows gravity and magnetism alone to create charged particle pairs? The Hawking process isn't going to do it for a SMBH (only neutral particles can be created by them), so how does your idea work?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/09/2019 07:18:16
That's not how math works. Making a negative number more negative is a decrease, not an increase. Going from -10 to -20 is a decrease. So the black hole is still losing mass if it's eating negative-mass particles.
Sorry, that is a fatal mistake
-10-1 = -11
So, as the SMBH is made out of negative polarity matter/mass, by adding a negative particle we do increase its total mass.
Why you don't agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/09/2019 14:31:00
You didn't learn in 3rd grade that -11 is less than -10?  Adding a negative number to any number (positive or negative) decreases the value.
Also,  other masses (both positive and negative) will accelerate away from negative mass and thus cannot form gravitationally bound objects like a planet or black hole.  If you actually worked through the trivial equations of Newton's gravitational force and resulting acceleration (F=GMm/r² and F=ma, or a=GM/r²), this would be apparent.
Let's look at the following:
https://en.wikipedia.org/wiki/Negative_mass
"In theoretical physics, negative mass is matter whose mass is of opposite sign to the mass of normal matter, e.g. −1 kg."
In our case, we do not discuss on a Negative mass.
We actually discuss on a negative charged mass.
Let's look on a pair particles (Positron/Electron)
https://en.wikipedia.org/wiki/Positron
"The positron or antielectron is the antiparticle or the antimatter counterpart of the electron. The positron has an electric charge of +1 e, a spin of 1/2 (same as electron), and has the same mass as an electron. "
So, the mass of the Positron is equal to the mass of the electron.
We don't have a negative mass, but we have a negative charged mass.
Therefore, the mass is there even for electron.
In gravity, we do not count the polarity of the mass charged.
We only monitor the total mass.
So, the total mass of one billion positrons is equal to the total mass of one billion electrons.
It is a mistake to assume that the mass of electron is –M Kg while the mass of the positron is +M kg.
Both have a positive mass, while their charged polarity is different.
So, the total mass of an object with only one billion positrons will be M, while the total mass with one billion electrons will also be M.
Again, both will be represented by real positive mass (and not negative mass).
Therefore, the gravity force of an object which is fully made with positrons should be identical to an object which is full made with the same numbers of electrons.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/09/2019 14:36:24
Why you don't agree with that?

Because -11 is smaller than -10...

Let's look at the following:
https://en.wikipedia.org/wiki/Negative_mass
"In theoretical physics, negative mass is matter whose mass is of opposite sign to the mass of normal matter, e.g. −1 kg."
In our case, we do not discuss on a Negative mass.
We actually discuss on a negative charged mass.
Let's look on a pair particles (Positron/Electron)
https://en.wikipedia.org/wiki/Positron
"The positron or antielectron is the antiparticle or the antimatter counterpart of the electron. The positron has an electric charge of +1 e, a spin of 1/2 (same as electron), and has the same mass as an electron. "
So, the mass of the Positron is equal to the mass of the electron.
We don't have a negative mass, but we have a negative charged mass.
Therefore, the mass is there even for electron.
In gravity, we do not count the polarity of the mass charged.
We only monitor the total mass.
So, the total mass of one billion positrons is equal to the total mass of one billion electrons.
It is a mistake to assume that the mass of electron is –M Kg while the mass of the positron is +M kg.
Both have a positive mass, while their charged polarity is different.
So, the total mass of an object with only one billion electrons will be M, while the total mass with one billion electrons will also be M.
Again, both will be represented by real positive mass (and not negative mass).
Therefore, the gravity force of an object which is fully made with positrons should be identical to an object which is full made with the same numbers of electrons.

Now you are contradicting yourself, because you explicitly said:

The gravity effects of Negative mass are identical to a positive mass.
Therefore, while the SMBH increases its negative mass, the accretion disc gets for free the positive particle.

If we are still talking about positive mass, then what I said here still applies:

Quote
If you propose that a black hole can become heavier by eating some of the mass that was taken out of it, you are absolutely are proposing such a violation. Let's say that the magnetic field transfers 1.022 keV of mass-energy out of the black hole into order to form an electron-positron pair. The black hole must now weigh 1.022 keV less than it did before. Now, one of those particles (0.511 keV) is thrown either into the accretion disk or into the jet, while the other 0.511 keV particle falls back into the black hole. The black hole lost 1.022 keV by forming the particle pair and only got 0.511 keV back by eating one member of the pair. That's still a net loss of 0.511 keV. Elementary school arithmetic demands that the black hole has lost mass, not gained it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/09/2019 15:05:21
Now you are contradicting yourself, because you explicitly said:
Quote
The gravity effects of Negative mass are identical to a positive mass.
Therefore, while the SMBH increases its negative mass, the accretion disc gets for free the positive particle.
Sorry for that.
My intention was negative charged mass.
In any case, from the SMBH point of view, those new created particles has an opposite charged and therefore, they do not have any effect on its mass at the moment of their creation.
Actually if one second later on they will meet with each other, they will eliminate each other:
"When a positron collides with an electron, annihilation occurs."
So, the creation of the pair-production does not change the total mass of the SMBH as the annihilation does not change its mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/09/2019 20:05:06
In any case, from the SMBH point of view, those new created particles has an opposite charged and therefore, they do not have any effect on its mass at the moment of their creation.

If the needed mass-energy to create the electron-positron pair didn't come from the black hole, then where did it come from? It has to come from somewhere. Whatever that source of mass-energy may be, the source must lose mass-energy in the process of creating that particle pair because that particle pair has a positive net mass-energy. That mass-energy cannot come from the gravitational or magnetic field themselves, as fields can only transform or transfer mass-energy, not create it.

Actually if one second later on they will meet with each other, they will eliminate each other:
"When a positron collides with an electron, annihilation occurs."
So, the creation of the pair-production does not change the total mass of the SMBH as the annihilation does not change its mass.

If the particles annihilate, then you don't have a scenario where one falls into the black hole and the other moves out into the accretion disk. But how can it move out into the accretion disk anyway? You claim that the magnetic field is an impassable barrier. A charged particle shouldn't be able to get through it.

This is the key question to know whether your model breaks the first law of thermodynamics or not:

Does the black hole-accretion disk-jet system increase the total amount of mass-energy in the Universe over time?

If your answer to this question is "no", then you have obeyed the first law of thermodynamics.

If your answer to this question is "yes", then your model violates the first law of thermodynamics and has therefore falsified itself.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/09/2019 05:37:15
If the needed mass-energy to create the electron-positron pair didn't come from the black hole, then where did it come from? It has to come from somewhere. Whatever that source of mass-energy may be, the source must lose mass-energy in the process of creating that particle pair because that particle pair has a positive net mass-energy. That mass-energy cannot come from the gravitational or magnetic field themselves, as fields can only transform or transfer mass-energy, not create it.
Based on what data do you set this assumption?
https://en.wikipedia.org/wiki/Hawking_radiation
In the following article it is stated:
"Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles."
"An alternative view of the process is that vacuum fluctuations cause a particle–antiparticle pair to appear close to the event horizon of a black hole."
It is stated clearly that the particle–antiparticle radiation/creation does not come directly from the black hole itself.
Therefore, the BH doesn't lose any mass during this creation process.
In the article it is stated also:
"As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole"
So, "the particle–antiparticle pair was produced by the black hole's gravitational energy"
Hence. the creation itself has no effect on the BH mass. As the positive charged particle is ejected outwards, the Negative charged particle must fall in. Our scientists estimate that the BH is made out of positive charged mass; Therefore, this in falling Negative charged particle should reduce its total mass.
So, again - the creation of the pair particles do not have any impact on the total mass of the BH.
Therefore, how could it be that you both are so sure that the creation of the Pair- production must decrease the BH mass while in this article it is stated clearly that it does not come directly from the mass of the black hole itself?
If you still believe that the creation of the pair-production must decrease the mass of the BH than please prove it by real article.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/09/2019 07:19:35
Based on what data do you set this assumption?
https://en.wikipedia.org/wiki/Hawking_radiation
In the following article it is stated:
"Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles."
"An alternative view of the process is that vacuum fluctuations cause a particle–antiparticle pair to appear close to the event horizon of a black hole."
It is stated clearly that the particle–antiparticle radiation/creation does not come directly from the black hole itself.
Therefore, the BH doesn't lose any mass during this creation process.
In the article it is stated also:
"As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole"
So, the creation itself has no effect on the BH mass. As the positive charged particle is ejected outwards, the Negative charged particle must fall in. Our scientists estimate that the BH is made out of positive charged mass; Therefore, this in falling Negative charged particle should reduce its total mass.
So, again - the creation of the pair particles do not have any impact on the total mass of the BH.
Therefore, how could it be that you both are so sure that the creation of the Pair- production must decrease the BH mass while in this article it is stated clearly that it does not come directly from the mass of the black hole itself?
If you still believe that the creation of the pair-production must decrease the mass of the BH than please prove it by real article.

The particle pair created from the Hawking process does not come directly from the black hole, no, but the net result of the process is the removal of mass from the hole. Adding a negative mass particle (take careful note that I said negative mass and not negative charge, because charge is not what this is about) to a positive mass black hole must necessarily result in the reduction of the black hole's mass.

Hawking radiation involves the creation of a positive mass particle and a negative mass particle (take note again that I am talking about mass here, not positive and negative charge). The negative mass particle falls in and reduces the black hole's mass while the positive mass particle carries the equivalent amount of mass lost by the black hole into the outside Universe.

This, however, does not apply to your model since you insist that negative mass is not involved with whatever process it is that you posit is taking place. Hence why I say that you need to explain the mechanism by which this occurs. Where does the energy come from to create the particle pair in your model? If negative mass is not involved, then each particle must have a positive mass in your model. As such, a source of energy is needed to create them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/09/2019 13:35:51
The particle pair created from the Hawking process does not come directly from the black hole, no, but the net result of the process is the removal of mass from the hole. Adding a negative mass particle (take careful note that I said negative mass and not negative charge, because charge is not what this is about) to a positive mass black hole must necessarily result in the reduction of the black hole's mass.
Where do you get the idea of negative mass and not negative charge???
Did you read the article?
It is stated clearly:
"As the particle–antiparticle pair was produced by the black hole's gravitational energy"
https://en.wikipedia.org/wiki/Antiparticle
"In particle physics, every type of particle has an associated antiparticle with the same mass but with opposite physical charges (such as electric charge). For example, the antiparticle of the electron is the antielectron (which is often referred to as positron). While the electron has a negative electric charge, the positron has a positive electric charge, "
So it is stated clearly that the antimatter is a negative charged mass.
What is your source for negative mass?
Why do you insist on this none realistic idea?
Title: Re: How gravity works in spiral galaxy?
Post by: jeffreyH on 22/09/2019 14:36:06
"As the particle–antiparticle pair was produced by the black hole's gravitational energy"

The gravitational energy extends beyond the event horizon and into the external universe.

Some particles are theorised to be their own antiparticle. So opposite charge would not apply.
https://en.m.wikipedia.org/wiki/Majorana_fermion (https://en.m.wikipedia.org/wiki/Majorana_fermion)
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/09/2019 15:02:16
Where do you get the idea of negative mass and not negative charge???
Did you read the article?
It is stated clearly:
"As the particle–antiparticle pair was produced by the black hole's gravitational energy"
https://en.wikipedia.org/wiki/Antiparticle
"In particle physics, every type of particle has an associated antiparticle with the same mass but with opposite physical charges (such as electric charge). For example, the antiparticle of the electron is the antielectron (which is often referred to as positron). While the electron has a negative electric charge, the positron has a positive electric charge, "
So it is stated clearly that the antimatter is a negative charged mass.

I see your reading comprehension went out the window again. Electrons, which are matter particles, are the negatively-charged particles. The positrons, which are antimatter, are positively-charged.

In the case of a positron-electron pair, yes, one is negatively-charged and one is positively-charged. But this need not be the case. A pair of photons, which are neutral, can be formed instead. We've been through all of this before...

What is your source for negative mass?
Why do you insist on this none realistic idea?

Stephen Hawking himself, for one. He mentions it in his book A Brief History of Time. The reason one particle has positive mass and the other negative mass is specifically because the first law of thermodynamics has to be obeyed. The particles were pulled out of a zero-energy state, so their total mass has to add up to zero. By one being positive and the other negative, this is accomplished.

We've already discussed all of this before, don't you remember? We even discussed why the negative mass particle has to be the one that falls into the hole.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/09/2019 15:20:27
In the case of a positron-electron pair, yes, one is negatively-charged and one is positively-charged.
OK
So do you finally agree that we discuss on Negative charged mass and not about Negative mass?

Stephen Hawking himself, for one. He mentions it in his book A Brief History of Time. The reason one particle has positive mass and the other negative mass is specifically because the first law of thermodynamics has to be obeyed. The particles were pulled out of a zero-energy state, so their total mass has to add up to zero. By one being positive and the other negative, this is accomplished.
I really don't understand why do you keep pushing this none realistic idea of Negative mass.
If you still want to hope that Stephen Hawking himself mentioned in his book A Brief History of Time that it is about negative mass (and not about negative charged mass), than please offer a direct web link to his statement.

We've already discussed all of this before, don't you remember? We even discussed why the negative mass particle has to be the one that falls into the hole.
As I have already stated, my intention was about Negative chraged mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/09/2019 15:28:08
So do you finally agree that we discuss on Negative charged mass and not about Negative mass?

In the case of a positron-electron pair formed during the Hawking process, one is negatively-charged and the other is positively-charged. One, however, must also have a negative mass while the other has a positive mass.

If you still want to hope that Stephen Hawking himself mentioned in his book A Brief History of Time that it is about negative mass (and not about negative charged mass), than please offer a direct web link to his statement.

I own A Brief History of Time and I will post the quote from the book once I get back to my house (that's where it is right now). If I'm not mistaken, Black Holes & Time Warps by physicist Kip S. Thorne says the same thing and I'm pretty sure I quoted it earlier in this thread (if I can find it...).
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/09/2019 16:27:08
In the case of a positron-electron pair formed during the Hawking process, one is negatively-charged and the other is positively-charged.
That is correct
One, however, must also have a negative mass while the other has a positive mass.
This is imagination!!!
In the article it is stated:
https://en.wikipedia.org/wiki/Antiparticle
"particle and antiparticle must have
the same mass m
the same spin state J
opposite electric charges q and -q."
Therefore, both have a positive mass.
Hence, would you kindly stop that none realistic idea of Negative mass, or offer a real article which supports this imagination...
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/09/2019 18:00:24
This is imagination!!!
In the article it is stated:
https://en.wikipedia.org/wiki/Antiparticle
"particle and antiparticle must have
the same mass m
the same spin state J
opposite electric charges q and -q."
Therefore, both have a positive mass.

Under normal circumstances, you are correct. All normal matter has positive mass and pair production that occurs in places like particle accelerators has positive mass. However, the circumstances of the Hawking process are different because of the extreme nature of black holes. It's also only a relative thing. The particle only looks negative to a distant observer (anyone outside the event horizon).

Quote
Hence, would you kindly stop that none realistic idea of Negative mass, or offer a real article which supports this imagination...

Later tonight, when I have access to Hawking's book, I'll quote it.

This article from LiveScience does mention it, though: https://www.livescience.com/65683-sonic-black-hole-spews-hawking-radiation.html

Quote
Normally, after a pair of virtual particles appears, they immediately annihilate each other. Next to a black hole, however, the extreme forces of gravity instead pull the particles apart, with one particle absorbed by the black hole as the other shoots off into space. The absorbed particle has negative energy, which reduces the black hole's energy and mass. Swallow enough of these virtual particles, and the black hole eventually evaporates. The escaping particle becomes known as Hawking radiation.

Here is another web page describing the process: https://www.asc.ohio-state.edu/mathur.16/infopublic/info2.3.html

There is also something else I want to emphasize. The idea that virtual particle pairs pop out of the vacuum and are then split by tidal forces at the black hole's event horizon is only an analogy that is used by books in an attempt to illustrate the process to average people who are not experts in the physics involved. This page explains: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

Quote
How does this work?  Well, you'll find Hawking radiation explained this way in a lot of "pop-science" treatments:

Virtual particle pairs are constantly being created near the horizon of the black hole, as they are everywhere.  Normally, they are created as a particle-antiparticle pair and they quickly annihilate each other.  But near the horizon of a black hole, it's possible for one to fall in before the annihilation can happen, in which case the other one escapes as Hawking radiation.

In fact this argument also does not correspond in any clear way to the actual computation.  Or at least I've never seen how the standard computation can be transmuted into one involving virtual particles sneaking over the horizon, and in the last talk I was at on this it was emphasized that nobody has ever worked out a "local" description of Hawking radiation in terms of stuff like this happening at the horizon.  I'd gladly be corrected by any experts out there...  Note: I wouldn't be surprised if this heuristic picture turned out to be accurate, but I don't see how you get that picture from the usual computation.

Honestly, I'm struggling to understand how the Hawking process actually works myself. Virtual particles don't seem to be a necessary component to the calculations at all. Instead, it seems to have something to do with the local positive and negative frequencies present in the vacuum around the event horizon: https://en.wikipedia.org/wiki/Unruh_effect

Regardless of the specifics of how it all works, there is one thing that is certain: the net result is that conservation of mass and energy are not violated. That is the most important thing to take home from all of this.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/09/2019 03:33:38
The absorbed particle has negative energy,
Negative Energy means Negative energy charge.
It doesn't mean negative mass as you might hope for.
Any Energy - positive or Negative - must have real positive mass.
The idea of negative mass is none realistic in my point of view.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/09/2019 04:06:36
Negative Energy means Negative energy charge.
It doesn't mean negative mass as you might hope for.
Any Energy - positive or Negative - must have real positive mass.
The idea of negative mass is none realistic in my point of view.

It does mean negative mass/energy. Here is the relevant quote from Stephen Hawking's book about it:

Quote
Because energy cannot be created out of nothing, one of the partners in a particle/antiparticle pair will have positive energy, and the other partner negative energy. The one with negative energy is condemned to be a short-lived virtual particle because real particles always have positive energy in normal situations. It must therefore seek out its partner and annihilate with it. However, a real particle close to a massive body has less energy than if it were far away, because it would take less energy to lift it far away against the gravitational attraction of the body.

Normally, the energy of the particle is still positive, but the gravitational field inside a black hole is so strong that even a real particle can have negative energy there. It is therefore possible, if a black hole is present, for the virtual particle with negative energy to fall into the black hole and become a real particle or antiparticle. In this case it no longer has to annihilate with its partner. Its forsaken partner may fall into the black hole as well. Or, having positive energy, it might also escape from the vicinity of the black hole as a real particle or antiparticle (Fig. 7.8 ). To an observer at a distance, it will appear to have been emitted from the black hole. The smaller the black hole, the shorter the distance the particle with negative energy will have to go before it becomes a real particle, and thus the greater rate of emission, and the apparent temperature, of the black hole.

The positive energy of the outgoing radiation would be balanced by a flow of negative energy particles into the black hole. By Einstein's equation E = mc2 (where E is energy, m is mass, and c is the speed of light), energy is proportional to mass. A flow of negative energy into the black hole therefore reduces its mass.

That last sentence makes it unambiguously clear that Hawking is talking about negative energy and not negative charge. This is only further reinforced by the fact that Hawking radiation can be composed of neutral particles like photons, neutrinos and gravitons that don't have any charge at all.
Title: Re: How gravity works in spiral galaxy?
Post by: Bogie_smiles on 23/09/2019 13:13:38
As a member who knows nothing "for sure", I would submit that gravity expresses itself in waves traveling through space, and  they affect the speed of light based on the density of those waves in space. LIGO evidence shows that massive objects emit gravitational waves, and logic tells me that massive objects absorb gravitational waves. Any elucidation would be appreciated.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/09/2019 13:30:11
No it doesn't.  If you mean negative charge, then say negative charge.  Charge and energy are different things.
Gravitational potential energy is the obvious example of negative energy/mass, so it is quite real.
I have found myself very confused with the meaning of Negative energy/mass.
Based on the following formula:
E = mc^2
A negative Energy means a negative mass (as c^2 is always positive).
However, in all/most of the articles that I have read - The moment of the pair creation is described as pair particles
One with Positive electrical charge while the other one has a Negative electrical charge.
So, I was wondering what the source is for that Negative Energy and what is the real difference between Negative electrical charge to Negative Energy.
In order to answer this question I have looked again on the following article:
This article from LiveScience does mention it, though: https://www.livescience.com/65683-sonic-black-hole-spews-hawking-radiation.html
In this article it is stated:
"(Antimatter particles have the same mass as their matter counterparts, but opposite electrical charge.)"
So, it is clear to me that when we discuss on Matter and Antimatter, or Particle and antiparticle we actually discuss on:
"the same mass as their matter counterparts, but opposite electrical charge"
However, if during the moment of creation we get Particle and Antiparticle (both with positive mass) then how suddenly we get the Negative energy out that?
This article gives the answer for this question:
"Normally, after a pair of virtual particles appears, they immediately annihilate each other. Next to a black hole, however, the extreme forces of gravity instead pull the particles apart, with one particle absorbed by the black hole as the other shoots off into space. The absorbed particle has negative energy, which reduces the black hole's energy and mass. Swallow enough of these virtual particles, and the black hole eventually evaporates. The escaping particle becomes known as Hawking radiation."
So, if I understand it correctly:
We consider the Negative energy only when the Negative electrical charge is falling into the BH:
"The absorbed particle has negative energy, which reduces the black hole's energy and mass."
Hence, as long as the antiparticle is out of the BH, It is considered as a negative electrical charged particle
However, when this antiparticle (with its negative electrical charge) falls into the BH it actually reduces the total energy/mass of the BH. Therefore our scientists consider the in falling Antiparticle as it has a Negative energy or Negative mass.
Do you agree with that?
Do you have better explanation?
Why in all/most of the articles when it comes to moment of the creation of pair production they clearly discuss on Positive/Negative electrical charged and not Positive/Negative Energy?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/09/2019 14:42:07
However, in all/most of the articles that I have read - The moment of the pair creation is described as pair particles
One with Positive electrical charge while the other one has a Negative electrical charge.
You misinterpret almost every sentence you read then.  Most created pairs are not charged at all.
Sgr-A (or any stellar black hole for that matter) is incapable of producing charged particles via Hawking radiation.  The energy required to create such massive particles is not there.
What happens more often is that the hot accretion disk (not the black hole) radiates light (reducing the mass/energy of the disk), and those photons gain energy (blue shift) as they get close to the black hole.  A pair of photons with sufficient positive mass/energy (over 1 MeV) might interact, resulting in this electron/positron pair, which has the same positive mass as the photons.  Yes, that's a combined positive energy like you say, but the energy was already there, having been removed from the accretion ring.  No new mass was created or destroyed.  It's just being moved around.
One of the pair might escape the black hole, and on average, as many positrons as electrons escape.  The positrons will interact with a different electron in the disk where they annihilate,  becoming photons again, for a net loss of mass to the accretion disk.

Quote
So, I was wondering what the source is for that Negative Energy and what is the real difference between Negative electrical charge to Negative Energy.
Charge is an EM property of any particle/system.  Most particles don't have charge, but all have mass/energy.  The two are completely different things.

Quote
"(Antimatter particles have the same mass as their matter counterparts, but opposite electrical charge.)"
So, it is clear to me that when we discuss on Matter and Antimatter, or Particle and antiparticle we actually discuss on:
"the same mass as their matter counterparts, but opposite electrical charge"
And since most/all the particles from Hawking radiation have zero charge, the opposite charge is also zero.  You're mistakenly taking a statement that the particles have opposite charge to mean that they have charge at all.  Hawking radiation from any known black hole is confined to photons and gravitons and such.  It involves virtual particles, and I am no expert on the mathematics involved, but there are good books/articles that explain it.  I don't think you are up to understanding those articles, given your consistent misrepresentation of even the most trivial physics.  I don't understand the articles, and they are not backed by a unified field theory, so their conclusions are in question.

Quote
However, if during the moment of creation we get Particle and Antiparticle (both with positive mass) then how suddenly we get the Negative energy out that?
You don't. Antimatter reactions go kablooey.  Lots of positive energy out of that.  No net change.

Quote
This article gives the answer for this question:
"Normally, after a pair of virtual particles appears, they immediately annihilate each other. Next to a black hole, however, the extreme forces of gravity instead pull the particles apart, with one particle absorbed by the black hole as the other shoots off into space. The absorbed particle has negative energy, which reduces the black hole's energy and mass. Swallow enough of these virtual particles, and the black hole eventually evaporates. The escaping particle becomes known as Hawking radiation."
So, if I understand it correctly:
We consider the Negative energy only when the Negative electrical charge is falling into the BH:
Energy and charge are completly different things.  The bit you quote above doesn't even mention charge.
It is in fact talking about virtual particles, not physical particles.  The former might have a combined mass of zero, which is why they can appear out of a zero energy state.

Quote
"The absorbed particle has negative energy, which reduces the black hole's energy and mass."
Hence, as long as the antiparticle is out of the BH, It is considered as a negative electrical charged particle
This is what I mean by you not being able to read any sentence without twisting it.  It says the negative energy particle falls in, not the anti-particle.  The positive mass particle that escapes might be matter or antimatter.  There should be no predominance of one over the other.


Quote
However, when this antiparticle (with its negative electrical charge)
There's no mention of an antiparticle or one with charge.  It says the negative energy particle falls in.  Remember: Energy and charge are completely different things.

Quote
Therefore our scientists consider the in falling Antiparticle as it has a Negative energy or Negative mass.
Do you agree with that?
About as completely wrong as possible.

Quote
Do you have better explanation?
Yes.  The quotes you gave is the better explanation.  Read them instead of replacing all the words with ones of your choice.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/09/2019 19:30:58
I have found myself very confused with the meaning of Negative energy/mass.

Think of units of energy like electron-volts, joules or calories. Instead of, say, 10 joules you would have -10 joules.

So, if I understand it correctly:
We consider the Negative energy only when the Negative electrical charge is falling into the BH:

You don't understand correctly. The particle that falls into the black hole could have positive charge, negative charge or no charge at all. The energy is what is negative, not the electric charge. The absorption of electric charge wouldn't reduce something's mass because mass and charge are different things.

Hence, as long as the antiparticle is out of the BH, It is considered as a negative electrical charged particle

How many times do I have to tell you that antimatter isn't always negatively-charge? Positrons are positively-charged.

Why in all/most of the articles when it comes to moment of the creation of pair production they clearly discuss on Positive/Negative electrical charged and not Positive/Negative Energy?

Because most articles talking about pair production talk about electron-positron pairs. Conservation of charge is just as important as conservation of energy.

In this article it is stated:
"(Antimatter particles have the same mass as their matter counterparts, but opposite electrical charge.)"
So, it is clear to me that when we discuss on Matter and Antimatter, or Particle and antiparticle we actually discuss on:
"the same mass as their matter counterparts, but opposite electrical charge" However, if during the moment of creation we get Particle and Antiparticle (both with positive mass) then how suddenly we get the Negative energy out that?

They are talking about "normal" circumstances. Both matter and antimatter have positive mass/energy under normal circumstances. Stephen Hawking already explained one circumstance that can make that normally positive mass/energy be negative:

Quote
Normally, the energy of the particle is still positive, but the gravitational field inside a black hole is so strong that even a real particle can have negative energy there.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/09/2019 21:03:23
Quote
In this article it is stated:
"(Antimatter particles have the same mass as their matter counterparts, but opposite electrical charge.)"
So, it is clear to me that when we discuss on Matter and Antimatter, or Particle and antiparticle we actually discuss on:
"the same mass as their matter counterparts, but opposite electrical charge" However, if during the moment of creation we get Particle and Antiparticle (both with positive mass) then how suddenly we get the Negative energy out that?

So, those scientists are talking about "normal" circumstances. In this case, both matter and antimatter have positive mass/energy under normal circumstances.
As they are talking about "normal" circumstances, why I also can't talk about "normal" circumstances?
Why do you push that discussion into the direction of none normal circumstances?
I would like to remind you that in my theory both matter and antimatter have to have positive mass/energy under normal circumstances.
So, why those scientists can claim that Both matter and antimatter have positive mass/energy under normal circumstances while you both insist that it is forbidden for me to use the same "normal circumstances"?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/09/2019 22:28:37
Why do you push that discussion into the direction of none normal circumstances?

The conditions inside of a black hole's event horizon are not normal.

So, why those scientists can claim that Both matter and antimatter have positive mass/energy under normal circumstances while you both insist that it is forbidden for me to use the same "normal circumstances"?

It's not forbidden so long as you are actually talking about normal physics conditions. By merely discussing black holes, you are entering "abnormal" territory.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/09/2019 13:59:20
The conditions inside of a black hole's event horizon are not normal.

OK
Based on that, the conditions outside of a black hole's event horizon are normal.
Is it correct?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/09/2019 14:44:31
OK
Based on that, the conditions outside of a black hole's event horizon are normal.
Is it correct?

It depends on how near the horizon you are. In Black Holes & Time Warps, it is stated:

Quote
Then, in 1974, came a great surprise: Hawking inferred as a by-product of his discovery of black-hole evaporation (Chapter 12) that vacuum fluctuations near a hole's horizon are exotic: They have negative average energy density as seen by outgoing light beams near the hole's horizon. In fact, it is this exotic property of the vacuum fluctuations that permits the hole's horizon to shrink as the hole evaporates, in violation of Hawking's area-increase theorem. Because exotic material is so important for physics, I shall explain this in greater detail.

Recall the origin and nature of vacuum fluctuations, as discussed in Box 12.4: When one tries to remove all electric and magnetic fields from some region of space, that is, when one tries to create a perfect vacuum, there always remain a plethora of random, unpredictable electromagnetic oscillations- oscillations caused by a tug-of-war between the fields in adjacent regions of space. The fields "here" borrow energy from fields "there," leaving the fields there with a deficit of energy, that is, leaving them momentarily with negative energy. The fields there then quickly grab the energy back and with it a little excess, driving their energy momentarily positive, and so it goes, onward and onward.

Under normal circumstances on Earth, the average energy of these vacuum fluctuations is zero. They spend equal amounts of time with energy deficits and energy excesses, and the average of deficit and excess vanishes. Not so near the horizon of an evaporating black hole, Hawking's 1974 calculations suggested. Near a horizon the average energy must be negative, at least as measured by light beams, which means that the vacuum fluctuations are exotic.

I'm not sure where the cut-off point is. That is, I'm not sure how far away you have to be from the horizon before the physics goes back to "normal". The fact of the matter is this, however: Hawking radiation must obviously be formed in a region of space close enough to the horizon where negative energy can exist. That's the only way that the first law of thermodynamics can be preserved. The energy of both particles must add up to zero. The only way to do that if one particle has positive energy would be if the other has an equal amount of negative energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/09/2019 09:37:57
I'm quite confused:
In one hand you claim that
Hawking radiation must obviously be formed in a region of space close enough to the horizon where negative energy can exist. That's the only way that the first law of thermodynamics can be preserved. The energy of both particles must add up to zero. The only way to do that if one particle has positive energy would be if the other has an equal amount of negative energy.
Hence, if "The energy of both particles must add up to zero. The only way to do that if one particle has positive energy would be if the other has an equal amount of negative energy" than you actually claim that at any pair production there must be a negative energy.
On the other hand you claim that the Normal condition depends on:
It depends on how near the horizon you are.
Would you kindly advice clearly when can we get the Normal condition?
Do you agree that in order to get the Normal conditions it must be close enough to the innermost side of the accretion disc (As further away there is no feasibility for the pair production activity)
If you don't agree with that, than please specify exactly the location for Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge.


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/09/2019 15:03:23
For one, the normal condition requires the energy to already be there.
Thanks
So, you agree that if there is energy, we will get Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge.
That is perfect
The question is: What is the source for this energy?
If there is no energy than it is quite clear that it must come from something.
However, If there is an energy, than we don't need to deduct any mass from any nearby object.
The gravitational Energy is an excellent energy source for the new creation:
https://en.wikipedia.org/wiki/Hawking_radiation
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
Hence, the Gravitational energy is already there.
So, why can't we assume a Normal circumstances outside the event horizon while the energy is coming from the gravitational energy?
Please be aware that it is also stated:
https://en.wikipedia.org/wiki/Pair_production
"the created particles shall have opposite values of each other. For instance, if one particle has electric charge of +1 the other must have electric charge of −1, or if one particle has strangeness of +1 then another one must have strangeness of −1."
https://en.wikipedia.org/wiki/Positron
"The positron or antielectron is the antiparticle or the antimatter counterpart of the electron. The positron has an electric charge of +1 e, a spin of 1/2 (same as electron), and has the same mass as an electron. "
So, there is a possibility for the  pair-production with the same positive mass at each particle, but with opposite electrical charge.
They don't say even one word about negative mass or deduct the mass/energy from the BH.
In ALL the articles that I have found (except of the one that you have offered) our scientists do not claim for Negative Energy/mass.
So, why do you consider that the scientists which wrote about the negative energy/mass are much more cleaver than all the others?
With regards to Negative mass:
https://en.wikipedia.org/wiki/Negative_mass
"In theoretical physics, negative mass is matter whose mass is of opposite sign to the mass of normal matter, e.g. −1 kg."
It is stated: "In theoretical physics..."
So does it mean that we have never verified a Negative mass?
Do we have any real observation for Negative mass?
Actually, it seems to me that even negative energy should have some sort of mass.
If it has mass, it must be a positive mass.
Did we ever found a Negative mass that sets a negative gravity?
If we can't see that Negative mass, why do you push in that direction?



Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 25/09/2019 16:34:09
Thanks
So, you agree that if there is energy, we will get Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge.
That is perfect
The question is: What is the source for this energy?

Under normal pair production circumstances, that energy would come from the photons that they were created from.

 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
Hence, the Gravitational energy is already there.
So, why can't we assume a Normal circumstances outside the event horizon while the energy is coming from the gravitational energy?

Because this isn't a case of gravitational field energy being converted into a particle pair. The tidal forces of the gravitational field allowed the pair to come into existence, but you can't extract energy from a gravitational field itself. Gravitational fields can transform gravitational potential energy into kinetic energy (such as when an object falls into them), but that's all. Gravitational fields are not some kind of reservoir of energy that can be tapped and drained. Hence why the particle pair has to have a total energy of zero.

So, why do you consider that the scientists which wrote about the negative energy/mass are much more cleaver than all the others?

I don't. The articles you are referencing are talking about "normal" pair production (like the kind that happens inside of particle accelerators). Those create particles with net positive energy because net positive energy was already present that they could be created from. You are comparing apples with oranges.

"In theoretical physics, negative mass is matter whose mass is of opposite sign to the mass of normal matter, e.g. −1 kg."
It is stated: "In theoretical physics..."
So does it mean that we have never verified a Negative mass?
Do we have any real observation for Negative mass?
Actually, it seems to me that even negative energy should have some sort of mass.
If it has mass, it must be a positive mass.
Did we ever found a Negative mass that sets a negative gravity?
If we can't see that Negative mass, why do you push in that direction?

Those are different circumstances. It may well be impossible to have a container of "negative water" in your refrigerator. However, we are not talking about the normal space-time here on Earth. We are talking about the extremely warped space-time of a black hole. It is that which allows particles with negative energy to exist for prolonged periods of time.

If it did turn out that negative energy particles of the type that Hawking theorizes cannot exist either, then that would mean that Hawking radiation wouldn't exist and tidal force-induced pair production cannot happen at all. The first law of thermodynamics demands that energy not be created out of nowhere, so if you cannot have a negative and positive particle pair to balance that energy budget, then you get no particle pairs.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/09/2019 05:16:11
1. Gravitational energy
Strictly speaking, gravity is not energy.  It is an acceleration field.  Earth gravity is expressed as acceleration (m/sec²), not in joules.
Why do you insist that Gravity is not energy?
In the article it is stated clearly about " black hole's gravitational energy":
https://en.wikipedia.org/wiki/Hawking_radiation
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
So our scientists are fully aware about the energy that BH can create with its Gravitational force.
Do you think that they have no knowledge about that BH, gravity or Energy?
As an example, our scientists are using that gravitational energy to boost space probe on its journey to the moon.
https://en.wikipedia.org/wiki/Planetary_flyby
"Flybys commonly use gravity assists to "slingshot" a space probe on its journey to its primary objective, but may themselves be used as primary means."
2. BH
An article about pair production isn't even talking about black holes.
Why?
In that article it is stated:
"the particle–antiparticle pair was produced by the black hole's gravitational energy"
So, BH is there and also gravitational energy is fully there!
3. Negative mass
Quote
Do we have any real observation for Negative mass?
Not real, no.  No observation of Hawking radiation either.
We've never found a real object with negative rest mass, no.  Virtual particles are not real objects, but the mathematics of virtual particles has been verified.
Thanks
So you confirm that we have never found Negative mass.
Therefore, let's take out this unrealistic assumption from our discussion.
There is no way for the pair production activity to generate one particle in negative mass.
Negative mass is imagination - let's keep it there.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/09/2019 05:46:30
Why do you insist that Gravity is not energy?

Gravitational energy is a thing, but gravity itself is not energy. It is a field of force.

"the particle–antiparticle pair was produced by the black hole's gravitational energy"
So our scientists are fully aware about the energy that BH can create with its Gravitational force.

If you consider that Wikipedia article accurate, then you need to consider what else was written there:

Quote
As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole.
Quote
One of the pair falls into the black hole while the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). This causes the black hole to lose mass, and, to an outside observer, it would appear that the black hole has just emitted a particle.

Quote
Do you think that they have no knowledge about that BH, gravity or Energy?

You seem to think so, as you disregard what Stephen Hawking has to say about it.

Quote
As an example, our scientists are using that gravitational energy to boost space probe on its journey to the moon.
https://en.wikipedia.org/wiki/Planetary_flyby
"Flybys commonly use gravity assists to "slingshot" a space probe on its journey to its primary objective, but may themselves be used as primary means."

That process causes a transfer of energy from the planetary body to the spacecraft, lowering the total orbital energy of the planet afterwards. If anything, that only confirms that energy is being transferred away from the black hole itself.

Why?
In that article it is stated:
"the particle–antiparticle pair was produced by the black hole's gravitational energy"
So, BH is there and also gravitational energy is fully there!

Did you read the rest of that sentence? You can't just quote the part you like and ignore the rest of it:

Quote
As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole.

Thanks
So you confirm that we have never found Negative mass.
Therefore, let's take out this unrealistic assumption from our discussion.

Then you must also take pair production due to gravity out of the discussion, since it requires negative mass in order to balance the energy budget.

Quote
There is no way for the pair production activity to generate one particle in negative mass.
Negative mass is imagination - let's keep it there.

You're just demonstrating your ignorance of the physics involved by saying that, but let's say for the sake of argument you are right. We can cut to the chase a lot faster if you go ahead and answer the question that I posted earlier:

Does the black hole-accretion disk-jet system increase the total amount of mass-energy in the Universe over time?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/09/2019 15:45:12
Gravitational energy is a thing, but gravity itself is not energy. It is a field of force.
That is OK, as long as Gravitational energy represents the requested Energy which is needed to create the "Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge" (as stated in the article).
If you consider that Wikipedia article accurate, then you need to consider what else was written there:
"As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole".
Let's read it all:
"Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.[citation needed] As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole."
So, it is stated clearly that this particle–antiparticle radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.
Therefore, The creation of the particle–antiparticle does not come from the mass of the BH itself, therefore, there is no mass lost during to that creation process.
However, what does it mean: "the escape of one of the particles lowers the mass of the black hole"?
The answer is given:
"One of the pair falls into the black hole while the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). This causes the black hole to lose mass, and, to an outside observer, it would appear that the black hole has just emitted a particle."
So, the BH is losing mass due to the in falling particle and not due to the creation process.
The next question is:
Why the in falling particle creates a negative energy?
Let's look again on the starting point of the pair creation. It is stated:
"Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge"
Do you see any Negative energy in that description?
They only claim that both particles have the same mass as their matter counterparts, but opposite electrical charge.
So, there are two particles with real mass. Not even one word about negative energy.
Now let's try to verify what could happen when each one is falling in:
1. If a particle with a positive electrical charge will fall in, do you see any negative energy?
2.  If a particle with a Negative electrical charge will fall in, do you see any negative energy?
Could it be that they actually was aiming for Negative electrical charge, but by a typo error they have written it as negative energy?
Do you have an idea how any particle (With Negative electrical charge or with positive electrical charge) can set a negative energy as it falls in?
This is real enigma for me.
In the article it is also stated as follow:
"while Hawking radiation seems to contain no such information, and depends only on the mass, angular momentum, and charge of the black hole (the no-hair theorem)."
So, could it be that the BH has a charge?
If it has a charge, than now it is fully clear to me.
If the Negative electrical charge particle is falling into a BH with positive charge, than by definition there must be a mass lost.
However, if a positive electrical charge particle is falling into a BH with positive charge, than by definition there must be mass increase.
What is your advice about all of that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/09/2019 16:44:44
That is OK, as long as Gravitational energy represents the requested Energy which is needed to create the "Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge" (as stated in the article).

Too bad, because it doesn't. That isn't how the Hawking process works.

So, it is stated clearly that this particle–antiparticle radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.
Therefore, The creation of the particle–antiparticle does not come from the mass of the BH itself, therefore, there is no mass lost during to that creation process.

No mass is lost during the creation process itself because one of the particles has negative energy.

Why the in falling particle creates a negative energy?

Because the black hole's gravitational field is so strong that it makes the energy of the particle negative as seen in the reference frame of a distant observer.

Let's look again on the starting point of the pair creation. It is stated:
"Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge"
Do you see any Negative energy in that description?

You have been corrected on this before. That is talking about normal circumstances.

1. If a particle with a positive electrical charge will fall in, do you see any negative energy?
2.  If a particle with a Negative electrical charge will fall in, do you see any negative energy?

In the reference frame of a distant observer, yes, they will see negative energy.

Could it be that they actually was aiming for Negative electrical charge, but by a typo error they have written it as negative energy?

Absolutely not. Stephen Hawking's book makes that abundantly clear:

Quote
Because energy cannot be created out of nothing, one of the partners in a particle/antiparticle pair will have positive energy, and the other partner negative energy. The one with negative energy is condemned to be a short-lived virtual particle because real particles always have positive energy in normal situations. It must therefore seek out its partner and annihilate with it. However, a real particle close to a massive body has less energy than if it were far away, because it would take less energy to lift it far away against the gravitational attraction of the body.

Normally, the energy of the particle is still positive, but the gravitational field inside a black hole is so strong that even a real particle can have negative energy there. It is therefore possible, if a black hole is present, for the virtual particle with negative energy to fall into the black hole and become a real particle or antiparticle. In this case it no longer has to annihilate with its partner. Its forsaken partner may fall into the black hole as well. Or, having positive energy, it might also escape from the vicinity of the black hole as a real particle or antiparticle (Fig. 7.8 ). To an observer at a distance, it will appear to have been emitted from the black hole. The smaller the black hole, the shorter the distance the particle with negative energy will have to go before it becomes a real particle, and thus the greater rate of emission, and the apparent temperature, of the black hole.

The positive energy of the outgoing radiation would be balanced by a flow of negative energy particles into the black hole. By Einstein's equation E = mc2 (where E is energy, m is mass, and c is the speed of light), energy is proportional to mass. A flow of negative energy into the black hole therefore reduces its mass.

If the Negative electrical charge particle is falling into a BH with positive charge, than by definition there must be a mass lost.

Are you serious? Now you think mass and charge are the same thing?

What is your advice about all of that?

My advice to you would be to learn the difference between charge and mass/energy. I'm also still waiting for your answer to this question:

Does the black hole-accretion disk-jet system increase the total amount of mass-energy in the Universe over time?

Stop ignoring it.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/09/2019 01:50:22
Disagree.

It depends on how you define it. Gravitational energy in the form of the potential energy of a mass in a gravitational field is a thing.
Title: Re: How gravity works in spiral galaxy?
Post by: Hayseed on 27/09/2019 04:55:17
I believe the velocity gradient of a spiral galaxy is due to structure.  I think the spirals are helices.  And all gravity spirals are slowly releasing.

Therefore the mass(gravity)density is not centered.  Think of earth's internal gravity gradient.

I believe Sol is orbiting something much closer than the galactic center.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/09/2019 06:02:27
I believe the velocity gradient of a spiral galaxy is due to structure.  I think the spirals are helices.  And all gravity spirals are slowly releasing.
Therefore the mass(gravity)density is not centered.  Think of earth's internal gravity gradient.
I believe Sol is orbiting something much closer than the galactic center.
Wow!!!
It seems to me that your ideas are quite correlated with my theory.
Therefore,
Would you kindly answer the following?
1. "spirals are helices". "Therefore the mass(gravity)density is not centered" - Do you mean that all the stars in each spiral arm are connected to each other as in helices?
2. "I believe Sol is orbiting something much closer than the galactic center." - Why do you think so? What could be this "something"?
3. "And all gravity spirals are slowly releasing." - Do you mean that the Sun (and all the other stars in the spiral arms) are drifting outwards over time?

Dear Kryptid
I promise to answer your question ASAP.

Title: Re: How gravity works in spiral galaxy?
Post by: Hayseed on 27/09/2019 07:01:42
Dave Lev,  I have not read this thread.  Was answering the Title of thread.  I disagree with most modern theory, especially astronomy.   My understanding of gravity comes from charge study, not stars.

1. Yes, also i think that G has a perpendicular component.  The G forces are exchanged thru perpendiculars.   The center of galaxy is not the densest, just like the earth.  Gravity does not come from a point.  No black holes.  No singularities.  No probabilities.  No randomness.

2.  There are too many G sources between here and the center.  The center probably has little effect on us........except thru a chain of G.

3.  Yes.  No BB.  A decay of G.  At first a very fast decay(inflation), now and for eternity.....very slow.

Once G falls, it can not stop.

Please forgive my rudeness for not reading thread.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/09/2019 15:17:58
Quote
As an example, our scientists are using that gravitational energy to boost space probe on its journey to the moon.
https://en.wikipedia.org/wiki/Planetary_flyby
"Flybys commonly use gravity assists to "slingshot" a space probe on its journey to its primary objective, but may themselves be used as primary means."
That process causes a transfer of energy from the planetary body to the spacecraft, lowering the total orbital energy of the planet afterwards. If anything, that only confirms that energy is being transferred away from the black hole itself.
Dear Kryptid
I fully agree that the spacecraft is lowering the total orbital energy of the planet afterwards.
However, that is normal process for any orbital system.
After any orbital cycle, the total orbital energy is decreasing.
Therefore, in real life, any orbital cycle has a shape of spiral (or helices?)
It might drift one Pico millimeter per cycle, but it is there for any orbital cycle.
The Moon is drifting outwards from the Earth and the earth is drifting outwards from the Sun.
So, this spiral orbital shape is real for any orbital system.
Unless, the objects are too close to each other and therefore the in falling momentum/gravity is faster/stronger than the drifting outwards due to the lowering total orbital energy.

Does the black hole-accretion disk-jet system increase the total amount of mass-energy in the Universe over time?

As I have already stated:
The black hole's gravitational energy creates New pair- particles.
Both particle and antiparticle have the same real mass.
The energy for that creation is taken from the black hole's gravitational energy.
https://en.wikipedia.org/wiki/Hawking_radiation
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
So, yes, the SMBH might loose gravitational energy due to this process, but it doesn't lose any mass.
As the SMBH is full with Antimatter, the in falling antimatter actually increases its total mass.
That SMBH mass increase compensates the decrease in the gravitational energy due to the pair production.
Therefore, if we could count the total mass + gravitational energy of the SMBH before the pair production, we should see that this total mass + gravitational energy is increasing after the SMBH gets the in falling antimatter particle.
Hence, the antimatter particle is used to increase the total mass of the SMBH and overcome the decreasing in the gravitational energy due to the pair production, while the other particle is ejected into the accretion disc.
Therefore, we actually gain new particles in the accretion disc (almost for free).
Over time, those new particles are transformed into new atoms and molecular while they are drifting from the innermost accretion ring/disc to the outermost.
Eventually, they are ejected outwards from the accretion disc and boosted upwards as a molecular jet stream by the power of the magnetic force

Therefore, the new creation of the pair-particles by the black hole's gravitational energy increases the total amount of mass-energy in the Universe over time!

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/09/2019 17:16:29
drifting outwards due to the lowering total orbital energy.

You have it backwards. A lowering of orbital energy results in a decrease in the size of the orbit, not an increase. Moving away from a source of gravity requires an input of energy because you are moving against a gravitational potential. It's the same reason that energy input is required to lift something off of the ground, but not required to drop something.

Therefore, the new creation of the pair-particles by the black hole's gravitational energy increases the total amount of mass-energy in the Universe over time!

Then we can end the thread right here, as that violates the first law of thermodynamics. Your model has falsified itself.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/09/2019 05:44:04
You have it backwards. A lowering of orbital energy results in a decrease in the size of the orbit, not an increase.
I don't agree with that.
Lowering of orbital energy means lowering the orbital velocity and lowering the gravity force!
The formula for gravity is:
F = G m1 m2 / r^2
As there is no change in the mass or the G, than the only way to decrease the gravity is by increasing the radius.
Moving away from a source of gravity requires an input of energy because you are moving against a gravitational potential. It's the same reason that energy input is required to lift something off of the ground, but not required to drop something.
The orbital energy is a direct outcome of the orbital velocity which means - kinetic energy and not potential energy.
As the radius of the orbital object is smaller, its orbital velocity is faster.
If we set the moon at a distance of 1000 Ly away from the Earth, does it mean that its orbital energy is stronger?
What will be its orbital velocity?
At this distance, the gravity force between the two objects will be virtually zero.
Therefore, loosing orbital energy means increasing the orbital radius and decreasing the orbital velocity.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/09/2019 05:55:38
I don't agree with that.

Too bad, because you're objectively wrong.

Lowering of orbital energy means lowering the orbital velocity and lowering the gravity force!

The orbital velocity actually isn't lowered: objects in closer orbits move faster. You also aren't lowering the gravitational force, since velocity has no impact on that.

The formula for gravity is:
F = G m1 m2 / r^2

That formula only confirms what I've said. The force of gravity depends only upon the masses and their distance, not their velocity relative to each other.

As there is no change in the mass or the G, than the only way to decrease the gravity is by increasing the radius.

A decrease in orbital energy is not the same thing as a decrease in the gravitational force felt by an object. Orbital energy is the sum total of gravitational potential energy and kinetic energy that the orbiting object has.

The gravity energy is a direct outcome of the orbital velocity which means - kinetic energy and not potential energy.

It's both, actually. You can't ignore potential energy.

As the radius of the orbital object is smaller, its orbital velocity is faster.

Yes, but its potential energy is lower.

If we set the moon at a distance of 1000 Ly away from the Earth, does it mean that the gravity energy is stronger?
What will be its orbital velocity?
At this distance, the gravity force between the two objects will be virtually zero.

You need to learn the difference between gravitational force and gravitational potential energy.

Therefore, loosing orbital energy means increasing the orbital radius and decreasing the orbital velocity.

By that reasoning, airplanes should fly up into outer space when they run out of fuel instead of crashing to Earth.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/09/2019 06:26:42
Quote
Therefore, the new creation of the pair-particles by the black hole's gravitational energy increases the total amount of mass-energy in the Universe over time!
Then we can end the thread right here, as that violates the first law of thermodynamics. Your model has falsified itself.
There is no violation for the first law of thermodynamics!
The black hole's gravitational energy is for free and It is the ultimate energy source for the pair production.
In the article they claim about virtual particles that orbit around the BH.
Their orbital velocity might be faster than the speed of light due to the ultra black hole's gravitational energy.
Those virtual particles are transformed into real particles. However, in this process the orbital energy between the virtual particle (which had just been transformed into real particle) to the BH must decrease.
The outcome is a reduction in the orbital energy/velocity of the new born particle with reference to the Virtual particles.
Therefore, the orbital velocity of the new born particle must be equal or lower than the speed of light.
In other words, the transformation of the virtual particle into real particle takes some energy from the black hole's gravitational energy
Therefore, the new pair is created by using real energy (black hole's gravitational energy) that some of it must be lost during this creation process.
Hence, as some of the black hole's gravitational energy is converted into real particles, there is no violation for the first law of thermodynamics!
New particles are created by using real energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/09/2019 06:36:09
Therefore, the new creation of the pair-particles by the black hole's gravitational energy increases the total amount of mass-energy in the Universe over time!
There is no violation for the first law of thermodynamics!

Do you know what the first law of thermodynamics even is? Right now, you sound like someone who is saying, "I robbed a store, but I didn't steal anything".

The black hole's gravitational energy is for free

So now you think that gravitational fields contain an infinite amount of energy? Seriously? Do I really need to explain why that isn't the case?

In case you need a mathematical demonstration that objects in higher orbits have more energy, you can look at the equations on this website: https://deutsch.physics.ucsc.edu/6A/book/gravity/node15.html

The gravitational potential energy of an orbiting body can be calculated with U = -(GMm)/R, where

“U” is the gravitational potential energy in joules
“G” is the gravitational constant (6.674 x 10−11 m3⋅kg−1⋅s−2)
“M” is the mass of the larger body in kilograms
“m” is the mass of the smaller body in kilograms, and
“R” is the distance between the two in meters.

So for the Moon in its current orbit:

U = -(GMm)/R
U = -((6.674 x 10-11)(5.97237 x 1023)(7.342 x 1022))/384,399,000
U = -7.61316 x 1027 joules

The kinetic energy for an orbiting body can be calculated with K = (1/2)(GMm/R), where the variables have the same meaning as in the prior equation (except for “K”, which is kinetic energy instead of potential energy). So the Moon’s orbital kinetic energy is:

K = (1/2)(GMm/R)
K = (½)(((6.674 x 10-11)(5.97237 x 1023)(7.342 x 1022))/384,399,000)
K = 3.80658 x 1027 joules

Add them together and we get the total energy:

E = U + K
E = (-7.61316 x 1027) + (3.80658 x 1027)
E = -3.80658 x 1027 joules

So now I’ll consider an extreme situation like the one you described (where the Moon is 1,000 light-years from Earth).

U = -(GMm)/R
U = -((6.674 x 10-11)(5.97237 x 1023)(7.342 x 1022))/(9.46 x 1018)
U = -3.09373 x 1017 joules

K = (1/2)(GMm/R)
K = (½)((6.674 x 10-11)(5.97237 x 1023)(7.342 x 1022))/(9.46 x 1018)
K = 1.5468665 x 1017 joules

E = U + K
E = (-3.09373 x 1017) + 1.5468665 x 1017 joules
E = -1.5468635 x 1017 joules

Since -1.5468635 x 1017 joules is larger than -3.80658 x 1027 joules, the Moon would have more total energy if it was 1,000 light-years from Earth than if it were in its current orbit. So energy has to be put into the system in order to increase the distance between two gravitating bodies.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/09/2019 16:43:54
Since -1.5468635 x 1017 joules is larger than -3.80658 x 1027 joules, the Moon would have more total energy if it was 1,000 light-years from Earth than if it were in its current orbit. So energy has to be put into the system in order to increase the distance between two gravitating bodies.
Thanks for your effort in setting the calculation.
I do appreciate that.
Quote
The gravity energy is a direct outcome of the orbital velocity which means - kinetic energy and not potential energy.
It's both, actually. You can't ignore potential energy.
Let's look at the meaning of Gravitational energy:
https://en.wikipedia.org/wiki/Gravitational_energy
"Gravitational energy (GPE) is the potential energy of a physical object with mass has in relation to another massive object due to gravity. It is potential energy associated with the gravitational field. Gravitational energy is dependent on the masses of two bodies, their distance apart and the gravitational constant (G)."
Therefore, when we discuss about BH's Gravitational energy (GPE) we actually have to focus on the BH's potential energy with related to the new born Pair particles. So, we can't ignore it as you have stated.
Let's go back to that article:
https://en.wikipedia.org/wiki/Hawking_radiation
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
So, real particle and antiparticle pair was produced by the black hole's gravitational energy or actually BH's potential energy.
Let's stop at that moment of creation.
Before any particle is ejected outwards or inwards.
Before we set any change in the mass of the BH or the accretion disc due to in falling particle.
Do you agree that the creation of those particles must come from the BH's potential energy?
As it is stated:
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
If so, how that BH's Potential Energy is converted into real pair of particle?
Can you please explain it?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/09/2019 20:41:40
Do you agree that the creation of those particles must come from the BH's potential energy?
As it is stated:
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
If so, how that BH's Potential Energy is converted into real pair of particle?
Can you please explain it?

The black hole's potential energy isn't converted into a particle pair. No energy is consumed in the creation of the particle pair because the pair's net energy is zero. You are getting too hung up on the "particle pair created by gravity" analogy. It isn't literally true. It's just a metaphor used by scientists in an attempt to make the process easier for laymen to understand.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/09/2019 15:43:12
Quote
Do you agree that the creation of those particles must come from the BH's potential energy?
As it is stated:
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
If so, how that BH's Potential Energy is converted into real pair of particle?
Can you please explain it?
The black hole's potential energy isn't converted into a particle pair. No energy is consumed in the creation of the particle pair because the pair's net energy is zero. You are getting too hung up on the "particle pair created by gravity" analogy. It isn't literally true. It's just a metaphor used by scientists in an attempt to make the process easier for laymen to understand.

Sorry
It isn't metaphor. It is a clear explanation about the creation process/energy for new particles.
They didn't try to support any sort of theory.
Therefore, I have full trust in those scientists and it is clear to me that they are fully correct.
Not just in that message but also in the following one:
"Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge"
Those two key messages perfectly fit to my theory and proves that new real particles are created due to the black hole's gravitational energy!
Those new particles increase the total mass in the Universe as one is ejected outwards into the accretion disc while the other is falling into the SMBH.
If you still think that our scientists have an error in their understanding, than you have to argue with them about it.
So far our scientists have never observed any matter falling into the accretion disc. (For sure - Not NASA)
They will never ever find any sort of in falling matter into the accretion disc even if they will stand near the accretion disc with the best telescope.
It's a time to change a disc in the theory of our Universe.
The SMBH is the source for all the matter in the galaxy/Universe.
Nothing moves in.
Everything is moving outwards.
All the stars in the galaxy including our Sun with all its planets and moons have been created by our SMBH and are drifting outwards.
Once you agree with that - you will understand how our Galaxy/Universe really works.
We see today that all the far away galaxies are moving away from us at almost the speed of light.
It is clear that without new mass creation, the density of our (observable) universe should decrease.
However, the creation of new mass by all the SMBH in the universe are fully balanced with the matter that escapes from our observable universe.
If we could come back again to our Universe in 100 Billion years from now, we should find exactly the same observable Universe density, the same cosmic radiation and almost the same space view.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/09/2019 18:08:35
Sorry
It isn't metaphor. It is a clear explanation about the creation process/energy for new particles.

If it really was that clear, then you would stop misinterpreting it to mean what you want it to mean instead of what it actually means. I already posted a link that discusses how the particle pair analogy is not representative of the actual math behind the Hawking process.

Therefore, I have full trust in those scientists and it is clear to me that they are fully correct.

If you had "full trust in those scientists", then you would trust them when they say that the particle pair has net zero energy and would therefore stop this nonsense about gravitational fields creating net positive energy. Gravity can't break the law of conservation of energy. In practice, your actions say "I have full trust in those scientists when what they say agrees with what I believe".

Not just in that message but also in the following one:
"Antimatter particles that have the same mass as their matter counterparts, but opposite electrical charge"

This has been explained to you already. Go back and read our past explanations if your memory is really that bad.
 
Those two key messages perfectly fit to my theory and proves that new real particles are created due to the black hole's gravitational energy!

First of all, learn what it means to prove something. Second of all, you can't drain energy from a gravitational field. Gravitational fields are an intrinsic attribute of the mass that produces them. There isn't any usable energy in a gravitational field. It would be like trying to drain the energy from the charge of an electron. It doesn't work. If it was possible to draw unlimited energy from a gravitational field, then we could have perpetual motion machines on Earth that worked by converting the local gravitational field into energy. But we can't because that isn't how physics works. Again, gravity can't break the first law of thermodynamics.

If you still think that our scientists have an error in their understanding, than you have to argue with them about it.

The error is with your understanding, not that of scientists. Scientists aren't the ones arguing that the created particle pairs have a net positive energy: you are.

Those new particles increase the total mass in the Universe

Again, I have to ask you, do you know what the first law of thermodynamics is?

Everything is moving outwards.
All the stars in the galaxy including our Sun with all its planets and moons have been created by our SMBH and are drifting outwards.

Is your memory really that bad? Have you already forgotten the math I just did that shows that it takes an energy input to increase orbital distances? Or did you just ignore it because it shows that you are wrong?

However, the creation of new mass by all the SMBH in the universe are fully balanced with the matter that escapes from our observable universe.

When was it ever discovered that there is a net loss of matter from our observable universe?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/09/2019 06:10:29
If you had "full trust in those scientists", then you would trust them when they say that the particle pair has net zero energy and would therefore stop this nonsense about gravitational fields creating net positive energy. Gravity can't break the law of conservation of energy. In practice, your actions say "I have full trust in those scientists when what they say agrees with what I believe".
OK
There are two different issues
1. New real particle creation near the BH event horizon
2. The energy source for that creation
I hope that by now we all agree that new real particle and real antiparticle are created near the event horizon.
So, let's focus on no. 2 - the energy source for that creation.
In the article it is stated:
 "the particle–antiparticle pair was produced by the black hole's gravitational energy"
But you don't like it.
You claim that gravity  by itself has no energy and therefore due to the first law of thermodynamics, there is no way to extract new energy from gravity in order to generate new particles.
Surprisingly - I agree with your explanation - as long as we only focus on gravity.
However, BH is not just about gravity, it is much more than that.
In the following article it is stated:
https://en.wikipedia.org/wiki/Black_hole_thermodynamics
"In physics, black hole thermodynamics[1] is the area of study that seeks to reconcile the laws of thermodynamics with the existence of black-hole event horizons. As the study of the statistical mechanics of black-body radiation led to the advent of the theory of quantum mechanics, the effort to understand the statistical mechanics of black holes has had a deep impact upon the understanding of quantum gravity, leading to the formulation of the holographic principle.[2]"
That shows that our scientists are not fully sure how thermodynamics works at BH.
In order to overcome the First low of thermodynamics it is stated:
The first law
"The left side, {\displaystyle dE}dE, is the change in energy (proportional to mass). Although the first term does not have an immediately obvious physical interpretation, the second and third terms on the right side represent changes in energy due to rotation and electromagnetism. Analogously, the first law of thermodynamics is a statement of energy conservation, which contains on its right side the term {\displaystyle TdS}{\displaystyle TdS}."
It is stated: "changes in energy due to rotation and electromagnetism"
Therefore, it seems to me that when we add to BH gravity the impact of the orbital rotation plus the electromagnetism, than we can gain the extra energy which is needed to create new real particles near the event horizon.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/09/2019 07:37:43
You claim that gravity  by itself has no energy and therefore due to the first law of thermodynamics, there is no way to extract new energy from gravity in order to generate new particles.

I'm not so sure that I would say that a gravitational field has no energy. In quantum theories, gravitational fields are modeled as being composed of virtual gravitons. Each of those gravitons should carry some amount of energy. However, this is not energy that can be accessed because those are virtual particles that must quickly vanish back into the vacuum and carry their energy with them in the process. That is the important distinction I want to make. It is very possible to have energy around that you can't use to perform work. The zero-point energy of a vacuum is one such example.

Therefore, it seems to me that when we add to BH gravity the impact of the orbital rotation plus the electromagnetism, than we can gain the extra energy which is needed to create new real particles near the event horizon.

Rotation is indeed an accessible source of energy. It, however, still follows the first law of thermodynamics. As energy is drained from the black hole's spin, the black hole's spin slows down. Once all of the black hole's spin energy is drained, it is no longer spinning and can therefore no longer donate energy via that method.

Electromagnetism is like gravity in the sense that you can't suck energy out of an electromagnetic field because it is an intrinsic property of electric charge. It doesn't contain usable energy in itself, but it can transfer energy from one source to another (which is the basis for our electrical technology). As such, a magnetic field being dragged around by a rotating black hole can transfer energy from the black hole's spin to magnetized objects in that same field. Once all of that spin is gone, however, there is nothing left to drain energy from.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/09/2019 11:36:17
Thanks Kryptid for your great explanation

So, now we agree that the BH's gravity + BH's spinning + BH's Electromagnetism can set the requested energy which is needed to create new pair particles.
However - There is a cost for that.
It should reduce the spinning energy of the BH.
However, for any new pair particles creation, one particle is ejected outwards into the accretion disc and the other Antiparticle is falling into the BH.
Assuming that the BH is full with antiparticles - the in falling new antiparticle is actually increases the BH mass.
More mass in the BH means more spinning power/energy.
Therefore, the mass of the antiparticle that falls into the BH overcomes the energy spinning reduction due to the pair creation.
What do you think about this idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/09/2019 13:51:05
Therefore, it seems to me that when we add to BH gravity the impact of the orbital rotation plus the electromagnetism, than we can gain the extra energy which is needed to create new real particles near the event horizon.
Rotation is indeed an accessible source of energy.
Just a nit.  Dave didn't mention spin, he spoke of orbital rotation, and while some black holes have this, a SMBH does not unless it is merging with another.  So Sgr-A does have orbital energy relative to the SMBH in Andromeda, and that energy will indeed be used to propel random stars to escape velocities, thus draining Sgr-A of its 'orbital energy' and causing it to quickly (in only 10ish billion years) spiral in (gaining negative gravitational potential energy) until the end.

Quote
As such, a magnetic field being dragged around by a rotating black hole can transfer energy from the black hole's spin to magnetized objects in that same field.
I'm no expert in EM fields, but does spinning what seems to be a monopole charged object produce a magnetic field?  The no-hair theorem lists charge but not magnetic field, but perhaps the latter can be computed given the angular momentum figures.

Wiki does say this line: "Magnetic charge, if detected as predicted by some theories, would form the fourth parameter possessed by a classical black hole."

I don't even know what 'magnetic charge' is, as opposed to magnetic field.  I also don't know what they mean by 4th parameter since location, mass, momentum, spin, and charge seems already to be 5 parameters.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/09/2019 20:51:33
Assuming that the BH is full with antiparticles

That's not a good assumption. Particles are as likely to fall in as antiparticles.

Therefore, the mass of the antiparticle that falls into the BH overcomes the energy spinning reduction due to the pair creation.

No, no it doesn't. Basic math will tell you that much. If 1.022 MeV of rotational kinetic energy is extracted from the hole in order to produce a positron-electron pair, then the black hole can only get 0.511 MeV of that energy back by consuming one of the particles. It would only get back half of the energy that it expended.

Any method you use to try to get a system to create net energy is a violation of the first law of thermodynamics. You might as well stop trying.

Just a nit.  Dave didn't mention spin

He did in reply #802.

I'm no expert in EM fields, but does spinning what seems to be a monopole charged object produce a magnetic field?

If by "monopole charge" you mean electric charge, yes. Black Holes & Time Warps states that a spinning black hole with a net electric charge will have a magnetic field. However, I have mostly given up on trying to teach Dave Lev that a neutral black hole can't have a magnetic field. I just try to go with it and demonstrate that, even if such was the case, it still won't give him the result he wants.

I don't even know what 'magnetic charge' is

Magnetic charge is a single north or south pole, which could only exist if magnetic monopoles exist. If they do, then a black hole that consumes magnetic monopoles could have a net magnetic charge.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/10/2019 06:18:14
Any method you use to try to get a system to create net energy is a violation of the first law of thermodynamics.
Dear Kryptid
I was quite sure that we have already agreed that there is no violation of the first law of thermodynamics.
You have already confirmed it:
Rotation is indeed an accessible source of energy. It, however, still follows the first law of thermodynamics. As energy is drained from the black hole's spin, the black hole's spin slows down.
Electromagnetism is like gravity in the sense that you can't suck energy out of an electromagnetic field because it is an intrinsic property of electric charge. It doesn't contain usable energy in itself, but it can transfer energy from one source to another (which is the basis for our electrical technology). As such, a magnetic field being dragged around by a rotating black hole can transfer energy from the black hole's spin to magnetized objects in that same field.
So Please - It is not a tango - two step forwards, one step backwards.
You have to respect your confirmation that the BH has the ability/energy to create new real particles pair!!!
However, we also agree that there is a cost for that new creation. It is the energy reduction at the BH.
So, please - we have already found a solution for the first law of thermodynamics:
BH's gravity + BH's Spinning/rotation + BH's electromagnetic force - all of that are transformed into the requested energy that sets the creation of new real particles.
So, please don't go back again to that thermodynamic issue any more.
I fully agree that we still have to overcome few milestones, but would you kindly confirm the above?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/10/2019 06:55:17
Dear Kryptid
I was quite sure that we have already agreed that there is no violation of the first law of thermodynamics.
You have already confirmed it:

As long as new mass-energy isn't being created, then the first law of thermodynamics is being obeyed.

So Please - It is not a tango - two step forwards, one step backwards.
You have to respect your confirmation that the BH has the ability/energy to create new real particles pair!!!

What I confirmed is that energy can be drained from a black hole's rotation. Whether that rotation can naturally result in the creation of new particles is a different matter. I'm mostly just going with it for the sake of argument because the main thrust of what I'm trying to say is that a black hole contains a finite amount of mass-energy and therefore can only send a finite amount of mass-energy out into the Universe.

However, we also agree that there is a cost for that new creation. It is the energy reduction at the BH.
So, please - we have already found a solution for the first law of thermodynamics:
BH's gravity + BH's Spinning/rotation + BH's electromagnetic force - all of that are transformed into the requested energy that sets the creation of new real particles.
So, please don't go back again to that thermodynamic issue any more.
I fully agree that we still have to overcome on few milestones, but would you kindly confirm the above?

The gravity and electromagnetism don't contribute to the black hole's expendable energy, but the rotation does. Otherwise, I'll say that maybe rotation can be used to create particle pairs. It's not that the rotation doesn't have energy that could theoretically be converted into particles. The energy is certainly there. It's just that the right mechanism to bring about this conversion would also have to be present.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/10/2019 07:40:18
I'll say that maybe rotation can be used to create particle pairs.
I really don't understand why suddenly you are so afraid that BH's rotation can be used to create new particle pairs?
Why can't you just say?
Yes, the rotation of the BH can be used to create particle pairs, as long as there is a balance between the energy that are drained from a black hole's rotation to the energy that is needed for the new creation particle pairs?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/10/2019 07:49:19
I really don't understand why suddenly you are so afraid that BH's rotation can be used to create new particle pairs?
Why can't you just say?
Yes, the rotation of the BH can be used to create particle pairs, as long as there is a balance between the energy that are drained from a black hole's rotation to the energy that is needed for the new creation particle pairs?

Actually, I just remembered that Chapter 12 of Black Holes & Time Warps does indeed mention that a black hole's rotation can produce radiation. So I will agree that such particles can be produced.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/10/2019 10:25:37
Actually, I just remembered that Chapter 12 of Black Holes & Time Warps does indeed mention that a black hole's rotation can produce radiation. So I will agree that such particles can be produced.
Thanks
I really appreciate this answer.

So, let's start to deal with the milestones:
Quote
Assuming that the BH is full with antiparticles
That's not a good assumption. Particles are as likely to fall in as antiparticles.
We have already agreed that  the BH's electromagnetic force transformed the requested energy for the pair creation.
Therefore, It must be there
In the same token, new created particles should carry opposite electrical charge.
https://en.wikipedia.org/wiki/Pair_production
"Pair production is the creation of a subatomic particle and its antiparticle from a neutral boson. Examples include creating an electron and a positron, a muon and an antimuon, or a proton and an antiproton."
" if one particle has electric charge of +1 the other must have electric charge of −1, or if one particle has strangeness of +1 then another one must have strangeness of −1."
Assuming that the new created particles are fully affected by electromagnetic force, don't you agree that one should pushed outwards, while the other one must pulled inwards due to Lorentz force? 

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/10/2019 14:09:28
Assuming that the new created particles are fully affected by electromagnetic force, don't you agree that one should pushed outwards, while the other one must pulled inwards due to Lorentz force? 

Only if the black hole has net electric charge.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/10/2019 15:42:58
Assuming that the new created particles are fully affected by electromagnetic force, don't you agree that one should pushed outwards, while the other one must pulled inwards due to Lorentz force?
Only if the black hole has net electric charge.
Thanks
So by assuming that BH has a net electrical charge, Antiparticles will fall into the BH, while the accretion disc will get particles.
That is perfect solution.
It shows that the BH increases its mass due to the in falling new born Antiparticles. However - we still need to set the energy issue.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/10/2019 16:46:30
So by assuming that BH has a net electrical charge, Antiparticles will fall into the BH, while the accretion disc will get particles.
That is perfect solution.

That depends upon the charge. If it's negative, positrons will preferentially fall in. If it's positive, electrons will preferentially fall in. This, however, is a problem. As more charged particles fall in, their opposite charge starts canceling out the net charge of the black hole. Once the charge is all gone, then no preference for negatively or positively-charged particles being pulled in will exist.

It shows that the BH increases its mass due to the in falling new born Antiparticles.

No, it won't. The black hole is sending more mass-energy out than is coming back in, remember? Two particles out, one particle back in.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/10/2019 17:29:22
That depends upon the charge. If it's negative, positrons will preferentially fall in. If it's positive, electrons will preferentially fall in. This, however, is a problem. As more charged particles fall in, their opposite charge starts canceling out the net charge of the black hole. Once the charge is all gone, then no preference for negatively or positively-charged particles being pulled in will exist.
I really don't understand why the BH charge is needed?
Why the magnetic force by itself is not good enough?
Lorentz force sets the force direction based on the charge of the particle
If the particle is positive - it will be ejected in one direction, if it is negative - it will be ejected to the other direction.
What is the problem with that?

Energy issue - soon to come.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/10/2019 17:53:32
Let's try to find a solution for the Energy:
Basic math will tell you that much. If 1.022 MeV of rotational kinetic energy is extracted from the hole in order to produce a positron-electron pair, then the black hole can only get 0.511 MeV of that energy back by consuming one of the particles. It would only get back half of the energy that it expended.

There is a solution for that.
It is called - Tidal
Please look at the following article:
https://www.space.com/31385-saturn-moon-enceladus-geysers-losing-steam.html
"The 330-mile-wide (530 kilometers) Enceladus hosts a global ocean of salty liquid water beneath its icy shell. This ocean stays liquid because Saturn's powerful gravity twists and stretches Enceladus, generating internal heat through tidal forces. (This tidal heating also provides the energy that powers the jets.)"

Hence, Tidal heating provides the energy that powers the jets.
Therefore, as long as the BH is under tidal forces it could potentially get a compensation for the energy lost due to the pair production activity.
It seems to me that the tidal heating can generate much more energy than the energy lost due to that pair production.
Even if it can only add the energy for one extra gram per year - this new mass is added to the total energy/mass of the whole universe.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 01/10/2019 21:40:03
What is the problem with that?

It's wrong. I have told you over and over and over and over and over again that magnetic fields don't attract or repel electric charges.

Repeat after me:

Magnetic fields neither attract nor repel electric charges.
Magnetic fields neither attract nor repel electric charges.
Magnetic fields neither attract nor repel electric charges.

Even if it can only add the energy for one extra gram per year - this new mass is added to the total energy/mass of the whole universe.

Let me ask you for the third time: do you know what the first law of thermodynamics is?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/10/2019 04:05:26
It's wrong. I have told you over and over and over and over and over again that magnetic fields don't attract or repel electric charges.
Repeat after me:
Magnetic fields neither attract nor repel electric charges.
Magnetic fields neither attract nor repel electric charges.
Magnetic fields neither attract nor repel electric charges.
Are you positively sure about it?
https://en.wikipedia.org/wiki/Lorentz_force
"Lorentz force F on a charged particle (of charge q) in motion (instantaneous velocity v). The E field and B field vary in space and time."
"Charged particles experiencing the Lorentz force"
https://en.wikipedia.org/wiki/Lorentz_force#/media/File:Lorentz_force.svg
"Trajectory of a particle with a positive or negative charge q under the influence of a magnetic field B, which is directed perpendicularly out of the screen."
Any updated understanding from that?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/10/2019 04:14:27
Are you positively sure about it?

Yes, I am positively sure of it.

https://en.wikipedia.org/wiki/Lorentz_force
"Lorentz force F on a charged particle (of charge q) in motion (instantaneous velocity v). The E field and B field vary in space and time."
"Charged particles experiencing the Lorentz force"
https://en.wikipedia.org/wiki/Lorentz_force#/media/File:Lorentz_force.svg
"Trajectory of a particle with a positive or negative charge q under the influence of a magnetic field B, which is directed perpendicularly out of the screen."
Any updated understanding from that?

The force they experience is neither an attraction or repulsion. The direction of the force depends on the direction of the particle's path and the relative orientation of the magnetic field lines. I already told you this. Watch this video:
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/10/2019 05:24:51
The direction of the force depends on the direction of the particle's path and the relative orientation of the magnetic field lines
What about the particle charge?
Even in this video they clearly show a different impact due to the particle charge.
So, how can you constantly claim that the magnetic field has no influence on the direction of a moving charged particle?
You can see it very clearly in the following image:
https://en.wikipedia.org/wiki/Lorentz_force#/media/File:Lorentz_force.svg
"Lorentz force. This diagram illustrates the behavior of a charged particle q moving with velocity v in a magnetic field B. The latter is oriented out of the page, as indicated by the dot in the circle. The particle is accelerated by the Lorentz force, with its behavior determined by the formula: where F is the force vector, E is the electrical field, and c is the speed of light. The × symbol represents the mathematical curl operator. The direction of the electron's curved path is determined by the sign of its charge, which modifies the sign of the v × B in the formula above."
Let's assume that we just stand above the accretion disc (Near the event horizon)

S and N - represent the Poles of the magnetic fields (Below and above the accretion disc)
B - represents the magnetic fields. "The latter (B) is oriented out of the page (directly in our point of view), as indicated by the dot in the circle.
V - represents the orbital velocity/direction of the new pair particles which have just been created.
At the same moment of creation, they also cross the magnetic field.
As one has a positive charge (+q) and the other had a negative charge (-q), one of them will be pulled inwards while the other will be pushed outwards.
Why is it so difficult for you to understand that simple activity of Lorentz force?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/10/2019 05:39:41
So, how can you constantly claim that the magnetic field has no influence on the direction of a moving charged particle?

Stop misrepresenting me. I never said that magnetic fields don't influence the direction of a moving charged particle. I have said the exact opposite. What I said is that the force is not in the form of an attraction or a repulsion. Charged particles are not drawn towards the field nor pushed away from the field. Instead, their path is deflected by the field.

As one has a positive charge (+q) and the other had a negative charge (-q), one of them will be pulled inwards while the other will be pushed outwards.

No they won't. Did you even look at the animation? When the charged particle entered the field at a 90 degree angle to the field lines, it was not pushed away from the field nor drawn into it. Instead, it moved around in a circle. An oppositely-charged particle will move in a circle as well, just in the opposite rotational direction (clockwise vs. counter-clockwise).

Why is it so difficult for you to understand that simple activity of Lorentz force?

I do understand it. Nothing about the Lorentz force causes charged particles to be drawn towards or pushed away from the source of the field. You need to learn the difference between attraction/repulsion and a deflected path.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/10/2019 06:11:26
Did you even look at the animation? When the charged particle entered the field at a 90 degree angle to the field lines, it was not pushed away from the field nor drawn into it. Instead, it moved around in a circle. An oppositely-charged particle will move in a circle as well, just in the opposite rotational direction (clockwise vs. counter-clockwise).

It seems to me that you have missed the key point in my explanation.
So, let's set it again step by step
1. Magnetic Field- B
S and N - represent the Poles of the magnetic fields (Below and above the accretion disc)
B - represents the magnetic fields. "The latter (B) is oriented out of the page (directly in our point of view), as indicated by the dot in the circle.
Do you understand that diagram and the direction of the magnetic field?
If we stand just above the accretion disc we should see that the magnetic fields (B) is pointed directly to us.

2. Direction of the new charged Particles
V - represents the orbital velocity/direction of the new pair particles which have just been created.
At the same moment of creation, they also cross the magnetic field.
As one has a positive charge (+q) and the other had a negative charge (-q), one of them will be pulled inwards while the other will be pushed outwards.
Hence, the particales are pushed/pulled in "90 degree angle to the field lines". (Not to the direction of the B itself).
The one which is pulled in should move in the direction of the BH, while the other must move in the direction of the accretion disc.
3. "moved around in a circle"
Instead, it moved around in a circle. An oppositely-charged particle will move in a circle as well, just in the opposite rotational direction (clockwise vs. counter-clockwise).
Please look at the following diagram:
https://en.wikipedia.org/wiki/Lorentz_force#/media/File:Lorentz_force.svg
Actually, if we ignore the impact of the BH gravity than yes, the Particles should "moved around in a circle".
One moves inwards in a circle, while the other one move outwards in a circle (as we can see in the video which you have offered). So, they could stay close to each other while they "moved around in a circle".
4. Impact of the BH gravity 
However, the BH has also gravity force.
As one particle starts to move inwards, the impact of the BH's gravity on this particle is increasing. therefore it might continue its  "moved around in a circle" but on any given moment it will come closer and closer to the BH and faster - Due to increased gravity force. Eventually it must fall in.
The other one will face less gravity force at the moment that it starts to move outwards. Therefore it might ""moved around in a circle" while its gravity force gets weaker and weaker. Eventually it will get into the inwards side of the accretion disc.

Is it clear by now?
If not, please specify what is not clear for you.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/10/2019 06:18:38
Hence, the articales are pushed/pulled in "90 degree angle to the field lines". (Not to the direction of the B itself).
The one which is pulled in should move in the direction of the BH, while the other must move in the direction of the accretion disc.

They are indeed deflected 90 degrees, but not in the directions you think they are. If we are looking down on the accretion disk from above, we will see the charged particles deflected to the left or the right, not towards the black hole and away from the black hole. Here is yet another video that shows this:


The negatively-charged electron beam is not attracted towards the magnetic field nor is it repelled away from the field. Instead it bends downward or upward (depending on which pole is pointed towards the beam).

4. Impact of the BH gravity
However, the BH has a also gravity force.
As one particle starts to move inwards, the impact of the BH on this particle is increasing. therefore it might continue its  "moved around in a circle" but on any given moment it will come closer and closer to the BH (and faster). Eventually it must fall in.
The other one will face less gravity force and therefore it will ""moved around in a circle" while its gravity force gets weaker and weaker. Eventually it will get into the inwards side of the accretion disc.

That's not how that works. If there are two particles at an equal distance from the black hole, one moving in a clockwise circle and the other moving in a counter-clockwise circle, each will still experience identical gravitational forces. Gravity doesn't care whether the circling motion is clockwise or counter-clockwise.

There is a thing called "right-hand rule" which allows you to deduce the motion of a charged particle in a magnetic field:

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/10/2019 10:56:51
They are indeed deflected 90 degrees, but not in the directions you think they are. If we are looking down on the accretion disk from above, we will see the charged particles deflected to the left or the right, not towards the black hole and away from the black hole.
OK
So, you agree that the charged particles are deflected to the left or the right.
Once you agree that the charged particles are deflected (to any direction) than we have already overcome 90% of the problem.
However, you assume that the deflection is to the left or the right, not towards the black hole and away from the black hole.
This is something that is not clear to me.

If there are two particles at an equal distance from the black hole, one moving in a clockwise circle and the other moving in a counter-clockwise circle, each will still experience identical gravitational forces. Gravity doesn't care whether the circling motion is clockwise or counter-clockwise.
Yes, that could be correct, if the particles are moving up and down. So, the radius to the BH is fixed.
This is not our case.
Particles can't move up/down as that represents the magnetic filed (B)
Lorentz force do not direct the particles in the direction of the magnetic fields.

So let's understand our looking point location:

we are looking down on the accretion disk from above,
So, we are located just above the accretion disc or actually - just above the point of the new created particles (which should be near the event horizon)
In order to get better understanding about this location.
Let's say that if we get down to the that specific creation point near the event horizon (at the galactic/accretion disc plane), the BH will be located at a radius r to the left, while the distance to the closest point in the accretion disc is x.
Therefore, the radius of the inner most side of the accretion disc is
r + x
the radius of the particle creation point is r.
If we stay above that point, to our left we see the BH and to our right we see the inner most side of the accretion disc.
Now, at the moment of creation, the new pair particles are located at radius r from the BH.
Without the impact of the magnetic field (assuming they do not cancel each other) they had to orbit around the BH at the same radius r. (remember - on the left side we have the BH on the right side we have the inner side of the accretion disc.)
However, the magnetic field is there. Lorentz force starts to works at the same moment that the particles had been created.
With regards to particle velocity/direction:
At the moment of creation V is directly vertical to r.
Therefore, as one particle will be move to the left (direction to the BH) the other one will move to the right (direction to the accretion disc)
Please be aware that due to Lorentz force, the particles can't move up or down to the disc plane as that is the direction of the magnetic field.
So, one will move into the direction of the BH while the other will move in the direction of the accretion disc.
Hence, one particle will get to r-Δr while the other one will get to r+Δr.
I only focus on the first moment due to the impact of Lorentz force. (before setting the whole loop cycle)
Therefore, I don't understand why do you insist that there is no change in the radius.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/10/2019 16:20:46
I only focus on the first moment due to the impact of Lorentz force. (before setting the whole loop cycle)

So what do you think happens when you do consider the whole loop?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/10/2019 20:00:51
So what do you think happens when you do consider the whole loop?
Let's assume that the orbital velocity of the new particles should be at almost the speed of light (c) at the first moment of creation while the radius to the BH is r (Near the event horizon).
If there was no magnetic field the particle could orbit at that radius without any difficulties (so there is a balance between the orbital velocity to the radius r)
However, due to the magnetic field one particle will be pulled inwards while the other will be pushed outwards.
Let's assume that due to Lorentz force the time that it takes to the particle to set 45 degree in the loop is t1. (Which represents a movement of Δr). 
That time is a direct outcome of the magnitude of magnetic field
So, after t1 (from the moment of creation), one particle will be located at r-Δr and the other at r+Δr.
However, Lorentz force doesn't change the orbital velocity. it just changes the radius of the particle and its direction.
So, now we have one particle at r-Δr that is also moving inwards at 45 degrees.
That changes the balance between the orbital velocity to the radius.
Therefore, the ultra high gravity at that radius will force the particle inwards without setting any sort of loop.
Similar issue with the other particle.
As it gets to r+Δr its direction is 45 degree outwards from the orbital path. therefore, it will be pushed strongly outwards without any ability to come back. Therefore it should be ejected outwards to the accretion disc.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/10/2019 21:32:43
Let's assume that the orbital velocity of the new particles should be at almost the speed of light (c) at the first moment of creation while the radius to the BH is r (Near the event horizon).

How close are we talking? No stable orbits can exist within 1.5 times the radius of the black hole.

However, Lorentz force doesn't change the orbital velocity.

The overall velocity is unchanged, but the orbital velocity (the speed at which it travels around the black hole) is indeed changed. If a third, neutral particle was traveling with them, it would see the two charged particles lagging behind as it kept going forward. Since an orbit is just a form of "falling and missing", then this relative slow-down will draw both particles in closer to the black hole due to the hole's gravity.

therefore, it will be pushed strongly outwards without any ability to come back.

Pushed strongly by what? And where are your calculations showing that this mysterious force is stronger than the black hole's gravity at that distance? How does the particle get out of the magnetic field so that the field doesn't change its trajectory back towards the hole? I just calculated the radius of curvature for an electron/positron traveling at 99% the speed of light through a 1 Tesla field at about 500 kilometers. The magnetic field, however, extends much, much further out than that.

There is also still the problem of the build up of electric charge in the hole. If the black hole did manage to gain a net electric charge through this process, it is going to preferentially attract opposite-charged particles towards it while preferentially repelling similarly-charged particles. As the hole becomes more and more charged, it becomes increasingly difficult for oppositely-charged particles to escape and similarly-charged particles to enter. So this obviously isn't something that can go on forever. There comes a time where no further charge can be added to the hole.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/10/2019 16:41:14
Pushed strongly by what? And where are your calculations showing that this mysterious force is stronger than the black hole's gravity at that distance? How does the particle get out of the magnetic field so that the field doesn't change its trajectory back towards the hole? I just calculated the radius of curvature for an electron/positron traveling at 99% the speed of light through a 1 Tesla field at about 500 kilometers. The magnetic field, however, extends much, much further out than that.
All your questions are valid.
The main problem is that even our scientists don't have a real knowledge how the black holes really works from inside.
Let's start with:
What Is At The Center Of A Black Hole?
https://interestingengineering.com/what-happens-inside-a-black-hole
"At the center of a black hole is something called a gravitational singularity, or singularity for short. This is where gravity and density are infinite and space-time extends into infinity."
How a BH with a gravitational singularity can generate any sort of electromagnetism?
Why the BH spin/rotates?
In any case, the spin/rotate by itself can't generate any magnetic force.
We can get some ideas from the our planet:
https://cosmosmagazine.com/geoscience/what-creates-earth-s-magnetic-field
"The Earth's core works like a giant bicycle dynamo in reverse."
So, does it mean that the BH works according to similar concept?
If so, it must have a core and several layers around it.
Do we have any clue about it?
Do we know how to calculate the magnetic field magnitude around the black hole?
What Is A Black Hole Made Of?
"Put simply we cannot really be sure. Black holes are by definition regions of space time where extreme gravitational forces prevent anything, including light from escaping."
"Thanks to General Relativity, we think we understand what happens in this extreme gravity and, with the help of Quantum Mechanics, we can make an intelligent estimate as to what happens at smaller, microscopic scales. But if the two theories are combined – like they would be at the center of a black hole – they break down, leaving us with no idea as to what’s going on!" - spaceanswers."
In other words - Our scientists don't know how BH really works from inside.
It they don't know, do you really expect me to know?
It seems to me that understanding how the Universe works is a piece of a cake comparing to the same question about BH/SMBH.
In any case:
There is also still the problem of the build up of electric charge in the hole.
In one of the articles that I have found it was stated that in the nature there are no charged BH.
So, somehow, the matter that falls in must lose its electric charge.
We can also claim that charged BH acts as a battery. However, battery can't be function as a dynamo. So, if the BH generates electromagnetic it can't be in the same time a charged BH.

The overall velocity is unchanged, but the orbital velocity (the speed at which it travels around the black hole) is indeed changed. If a third, neutral particle was traveling with them, it would see the two charged particles lagging behind as it kept going forward.
I agree with that.
Since an orbit is just a form of "falling and missing", then this relative slow-down will draw both particles in closer to the black hole due to the hole's gravity.
Yes and no.
Yes - for the Antiparticle that is drifted inwards due to Lorentz force. That one should fall in.
No-  for the particle that is drifted outwards.
In order to understand that - we must look again at the accretion disc.
The average orbital velocity of the particles there is about 0.3c. However, I assume that the orbital velocity of particles at the inner most ring is much faster from the one at the outermost ring.

As the accretion disc is still under direct impact of the BH magnetic field, all particles/Atoms are drifted outwards due to Lorentz force. They actually have no other alternative. Lorentz force pushes them all outwards. So, even the mighty gravity force of the BH can't pull inwards even one particle from the accretion disc!

I don't know if our scientists have any clue how long each particle must stay at the accretion disc.
However, based on the orbital velocity pulse the drifting time, we might be able to extract the Lorentz force/magnetic field.
In any case, that shows that the drifting is quite minimal.
So, the new born particles are not drifting sharply inwards/outwards. They actually are drifted at a quite low magnitude.
Therefore, there is no possibility for the one that is drifted outwards to fall back into the BH.
As it gets into the inner most accretion ring its orbital velocity must be above the 0.3c. However, over time it is  transformed into Atom and molecular, reduces its velocity and at the end it will be ejected outwards from the accretion disc (thanks to Lorentz force) with all the other new Atom and molecular.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 04/10/2019 17:30:34
How a BH with a gravitational singularity can generate any sort of electromagnetism?

It wouldn't normally. It could only do so if it has consumed matter with a net electric charge. Conservation of electric charge doesn't allow electric charge to be destroyed, so the black hole must necessarily take on the charge of anything that falls in.

Why the BH spin/rotates?

Conservation of momentum. If the collapsing star or gas cloud that formed it was rotating, then the hole must as well.

So, does it mean that the BH works according to similar concept?
If so, it must have a core and several layers around it.
Do we have any clue about it?

An Einsteinian black hole can't be structured like that and therefore can't have a magnetic field unless it is both charged and rotating. Something like a MECO can have such a field, though.

Do we know how to calculate the magnetic field magnitude around the black hole?

It can be measured: https://scitechdaily.com/researchers-measure-magnetic-fields-in-the-vicinity-of-a-black-hole/ Although conventional models propose that the field is generated by the accretion disk and not the hole itself.

What Is A Black Hole Made Of?

This is unknown, but an Einsteinian black hole isn't made of any particular type of matter. It's just mass concentrated into a singularity.

It they don't know, do you really expect me to know?

This is your model. You are the one required to make the testable predictions.

In one of the articles that I have found it was stated that in the nature there are no charged BH.

That's because most matter is neutral or close to neutral.

So, somehow, the matter that falls in must lose its electric charge.

That would violate conservation of charge, so that is absolutely incorrect.

We can also claim that charged BH acts as a battery.

Batteries don't work that way.

However, battery can't be function as a dynamo.

Charged black holes aren't batteries anyway.

So, if the BH generates electromagnetic it can't be in the same time a charged BH.

That's a huge non-sequitur.

As the accretion disc is still under direct impact of the BH magnetic field, all particles/Atoms are drifted outwards due to Lorentz force. They actually have no other alternative. Lorentz force pushes them all outwards.

The Lorentz force doesn't do anything to neutral matter like plasma, so this is wrong.

So, even the mighty gravity force of the BH can't pull inwards even one particle from the accretion disc!

Demonstrate this with math please. I'm not going to accept it just because you say so.

However, based on the orbital velocity pulse the drifting time, we might be able to extract the Lorentz force/magnetic field.

You want to know the Lorentz force? Here is the equation: F = qvBsinθ

"q" is the value for charge. What happens when that charge is zero? The force is zero as well.

In any case, that shows that the drifting is quite minimal.

What shows that it is "quite minimal"? You haven't provided any numbers.

So, the new born particles are not drifting sharply inwards/outwards. They actually are drifted at a quite low magnitude.

Show me the math to support these claims.

Therefore, there is no possibility for the one that is drifted outwards to fall back into the BH.

Another non-sequitur.

and at the end it will be ejected outwards from the accretion disc (thanks to Lorentz force)

The Lorentz force can't eject neutral matter.

All of this is ultimately irrelevant, though. Even if you do have a valid way of getting a black hole to produce matter and eject it the way you want it to, that still ignores the fact that a black hole cannot generate unlimited mass-energy. The mechanisms are irrelevant. The specifics are unimportant. The first law of thermodynamics simply won't let your model work. No amount of figuring will allow you to get more mass-energy out of the black hole than was there to begin with. Doing so would violate the first law by definition.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/10/2019 04:56:37
Quote
Do we know how to calculate the magnetic field magnitude around the black hole?
It can be measured: https://scitechdaily.com/researchers-measure-magnetic-fields-in-the-vicinity-of-a-black-hole/ Although conventional models propose that the field is generated by the accretion disk and not the hole itself.
Thanks Kryptid
You have just offered clear observation for my theory.
Our scientists have measured the magnetic field that the BH generates:
"Two particle jets shoot out from the heart of active galaxy NGC 1052 at the speed of light, apparently originating in the vicinity of a massive black hole. A team of researchers headed by Anne-Kathrin Baczko from the Max Planck Institute for Radio Astronomy Bonn have now measured the magnetic fields in this area. They observed the bright, very compact structure of just two light days in size using a global ensemble of millimeter-wavelength telescopes. The magnetic field value recorded at the event horizon of the black hole was between 0.02 and 8.3 tesla. The team concludes that the magnetic fields provide enough magnetic energy to power the twin jets."
They are stated clearly: "Two particle jets shoot out from the heart of active galaxy NGC 1052 at the speed of light"
That is almost identical to the molecular jet stream that we see at our galaxy. However, in our galaxy, the molecular jet stream is boosted at only 0.8c instead of the speed of light.
In any case, that proves that the magnetic field has direct impact on any particle/atom/molecular.
If the magnetic field can trap any particles, Atoms, molecular outside the accretion disc, it surly can affect the same matter while they are in the accretion disc.
Therefore, the following statement is totally incorrect:
The Lorentz force can't eject neutral matter.
The matter in the accretion disc might not be so neutral (but that is not the issue).
If the magnetic field can collect the particles/molecular (after been ejected outside from the accretion disc) and boost them upwards/downward at almost the speed of light, it can also surly effect them while they are still in the accretion disc.
One of the side effect of magnetic force is - Lorentz force.
Therefore, as long as the particles/molecular are in the accretion disc they are fully affected by the magnetic force including Lorentz force.
Our scientists are also fully aware about the great impact of the magnetic force on the accretion disc.
However, somehow you insist to believe that the source for the magnetic field in the accretion disc is - the accretion disc itself:
Although conventional models propose that the field is generated by the accretion disk and not the hole itself.
This is a fantasy.
In order to understand that, let's go back to the unrealistic story of in falling matter into the accretion disc.
Let's monitor the temperature of an average atom outside in the galaxy.
Normally, the temp of a rock or asteroid should be much less than 0 c.
At the surface of the sun the temp is 5,000 c
At the core of the sun the temp is close to 10^6c
So, how could it be that suddenly at the accretion disc the temp is rising to 10^9c?
Somehow, new energy should come in.
We already know that gravity by itself can't create new energy.
Therefore, if we take a particle at 20 c and set it in an orbital path of 0.3 speed of light, would it increase its temp to 10^9 c?
They answer is quite clear - NO!
Therefore, the energy that is needed to heat up the accretion disc must come from somewhere.
The magnetic field is an excellent source for the new requested energy.
So, we can claim that the extra energy is coming from the impact of the magnetic fields on every particle/atom/molecular in the accretion disc.
It feeds them with the requested energy and therefore they all gain the super high temp of 10^9.
However, we have already agreed that the magnetic field is just a transformation tool.
So, the energy must come from a specific source of energy.
How can we assume that the source of the magnetic is the accretion disc.
The accretion disc can't use its own energy to heat itself
This is a clear violation of the first thermodynamic law.
Therefore, the source of the energy must come from an external source as the BH itself
.Hence, the energy in the BH is transformed into accretion disc by the magnetic field and have a direct impact on every particle there. This energy is used to transform the new born particles into atom and molecular. The 10^9 c is the outcome product of that activity.
Even if you do have a valid way of getting a black hole to produce matter and eject it the way you want it to, that still ignores the fact that a black hole cannot generate unlimited mass-energy. The mechanisms are irrelevant.
Yes it can.
The mechanisms is very important.
It works as follow:
Gravity set the tidal forces on the BH.
Tidal forces increases the heat/energy in the BH.
Magnetic field transformes some of that extra energy into the creation of new particles.
Lorentz force is using the same magnetic field to split between the new born particle to antiparticle.
One will get into the accretion disc while the other will fall into the BH.
The same magnetic field is also used to convert the new born particles into real atoms and molecular in the accretion disc.
Lorentz force is responsible to drift the particles outwards while they are still at the accretion disc and also to eject them all outside at the end of the new Atom/molecular creation process.
The first law of thermodynamics simply won't let your model work.
This law won't let your unrealistic model (that the accretion disc is the source of the magnetic field) to work!
No amount of figuring will allow you to get more mass-energy out of the black hole than was there to begin with. Doing so would violate the first law by definition.
Tidal is responsible to generate the extra requested energy in the BH.
Therefore, my explanation fully meets the first law by definition.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/10/2019 05:30:33
In any case, that proves that the magnetic field has direct impact on any particle/atom/molecular.

No it doesn't. The beams are made of charged particles like electrons and positrons. It isn't made of neutral matter like atoms.

Therefore, the following statement is totally incorrect:
Quote
The Lorentz force can't eject neutral matter.

So now you're telling me that you can multiply zero by some number and the answer isn't also zero. This demonstrates that you don't understand basic algebra either.

The matter in the accretion disc might not be so neutral (but that is not the issue).
If the magnetic field can collect the particles/molecular (after been ejected outside from the accretion disc) and boost them upwards/downward at almost the speed of light, it can also surly effect them while they are still in the accretion disc.
One of the side effect of magnetic force is - Lorentz force.
Therefore, as long as the particles/molecular are in the accretion disc they are fully affected by the magnetic force including Lorentz force.
Our scientists are also fully aware about the great impact of the magnetic force on the accretion disc.

Magnetic fields do affect neutral matter, but not in the same way that it affects charged particles. You can't take a piece of neutral matter and pretend that a magnetic field will do the exact same thing to it that it will to a charged particle. The Lorentz force equation doesn't work for neutral particles. If you don't believe me, do that math yourself. You'll get a force of zero as your answer every time.

However, somehow you insist to believe that the source for the magnetic field in the accretion disc is - the accretion disc itself:
Quote
Although conventional models propose that the field is generated by the accretion disk and not the hole itself.
This is a fantasy.

My, isn't this an interesting double standard. You cite the dynamo model that allows the Earth to generate a magnetic field on one hand, but on the other you deny the existence of the dynamo effect when it comes to accretion disks. Dynamo-induced magnetic fields are generated by electrically-conducting fluids in motion. The liquid metal in the Earth's outer core is a rotating, electrically-conducting fluid. The plasma in an accretion disk is a rotating, electrically-conducting fluid. It's the same thing. Both of them generate magnetic fields.

In order to understand that, let's go back to the unrealistic story of in falling matter into the accretion disc.
Let's monitor the temperature of an average atom outside in the galaxy.
Normally, the temp of a rock or asteroid should be much less than 0 c.
At the surface of the sun the temp is 5,000 c
At the core of the sun the temp is close to 10^6c
So, how could it be that suddenly at the accretion disc the temp is rising to 10^9c?
Somehow, new energy should come in.
We already know that gravity by itself can't create new energy.
Therefore, if we take a particle at 20 c and set it in an orbital path of 0.3 speed of light, would it increase its temp to 10^9 c?
They answer is quite clear - NO!
Therefore, the energy that is needed to heat up the accretion disc must come from somewhere.
The magnetic field is an excellent source for the new requested energy.
So, we can claim that the extra energy is coming from the impact of the magnetic fields on every particle/atom/molecular in the accretion disc.
It feeds them with the requested energy and therefore they all gain the super high temp of 10^9.
However, we have already agreed that the magnetic field is just a transformation tool.
So, the energy must come from a specific source of energy.
How can we assume that the source of the magnetic is the accretion disc.
The accretion disc can't use its own energy to heat itself
This is a clear violation of the first thermodynamic law.
Therefore, the source of the energy must come from an external source as the BH itself
.Hence, the energy in the BH is transformed into accretion disc by the magnetic field and have a direct impact on every particle there. This energy is used to transform the new born particles into atom and molecular. The 10^9 c is the outcome product of that activity.

You are missing the point. It doesn't matter where the heat came from to heat the accretion disk. The accretion disk is a rotating, electrically-conducting fluid. That makes it generate a magnetic field. That doesn't violate the first law of thermodynamics. Even in a scenario where your hypothetical black hole does have a magnetic field and that field did heat the accretion disk, that would simply mean that both the black hole and the accretion disk have fields.

Quote
So, the energy must come from a specific source of energy.

Conventional models say that it comes from a transformation of gravitational potential energy into kinetic energy. No violation of the first law there. The total amount of energy doesn't change (unlike in your physics-defying model).

Quote
The accretion disc can't use its own energy to heat itself

Why not? That's exactly what the Sun is doing right now. Sticks of dynamite do that too.

Quote
This is a clear violation of the first thermodynamic law.

The more you type, the more you demonstrate that you don't know what the first law of thermodynamics is. There are numerous examples of objects becoming hot because of the energy contained within themselves. How do you think explosives work? Why do your think your palms get hot when you rub your hands together quickly? What you are doing is transforming potential energy into heat energy. That doesn't break any laws of physics because the total amount of energy is the same before an after the process. All it did was change form.

Yes it can.
The mechanisms is very important.

Then you deny the first law of thermodynamics. The first law won't let you create energy. You don't know what the first law of thermodynamics even is. If you did, we wouldn't be having this discussion right now.

Tidal forces increases the heat/energy in the BH.

Energy can't be created, so no it doesn't. The first law has killed your idea stone dead.

This law won't let your unrealistic model (that the accretion disc is the source of the magnetic field) to work!

More ignorance. The first law of thermodynamics doesn't prevent potential energy from being transformed into kinetic energy. What it prevents is the creation of new energy.

Tidal is responsible to generate the extra requested energy in the BH.
Therefore, my explanation fully meets the first law by definition.

Tidal forces can't create energy.

The first law of thermodynamics states that energy cannot be created.
You claim that your mechanism can create energy.
How do you not see the blaring contradiction there? Both cannot be true simultaneously. Either energy can be created or it can't. If energy can be created, then the first law has been violated. If energy can't be created, then your model won't work. Which is it? You can't have it both ways.

Quote
my explanation fully meets the first law by definition.

You don't know the definition of the first law.

Just so you know about conservation of energy: https://www.grc.nasa.gov/WWW/K-12/airplane/thermo1f.html

Quote
The conservation of energy is a fundamental concept of physics along with the conservation of mass and the conservation of momentum. Within some problem domain, the amount of energy remains constant and energy is neither created nor destroyed. Energy can be converted from one form to another (potential energy can be converted to kinetic energy) but the total energy within the domain remains fixed.

https://opentextbc.ca/physicstestbook2/chapter/conservation-of-energy/

Quote
Total energy is constant in any process. It may change in form or be transferred from one system to another, but the total remains the same.

https://www.sciencedirect.com/topics/engineering/conservation-of-energy

Quote
conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/10/2019 05:14:57
Quote
Our scientists are also fully aware about the great impact of the magnetic force on the accretion disc.
Magnetic fields do affect neutral matter, but not in the same way that it affects charged particles. You can't take a piece of neutral matter and pretend that a magnetic field will do the exact same thing to it that it will to a charged particle. The Lorentz force equation doesn't work for neutral particles. If you don't believe me, do that math yourself. You'll get a force of zero as your answer every time.
How can you contradict yourself?
In one hand you claim that the matter in the accretion disc is "electrically-conducting fluid":
The accretion disk is a rotating, electrically-conducting fluid. That makes it generate a magnetic field.
You even want to believe that it generates magnetic field.
But somehow, you insist that Lorentz force has no impact on that "electrically-conducting fluid" matter/plasma in the accretion disc.
Sorry if it is "electrically-conducting fluid" it must obey to Lorentz force and be pushed outwards!!!
Therefore, all the matter in the accretion disc MUST be drifted outwards - to the last particle or atom.
As they are pushed outwards - they are all still "electrically-conducting fluid".
Therefore, the BH's Magnetic field can easily boost them all upwards/downwards at almost the speed of light.

Tidal forces can't create energy.
The first law of thermodynamics states that energy cannot be created.
You claim that your mechanism can create energy.
How do you not see the blaring contradiction there? Both cannot be true simultaneously. Either energy can be created or it can't. If energy can be created, then the first law has been violated. If energy can't be created, then your model won't work. Which is it? You can't have it both ways.
Sorry
You miss the key functionality of "TIDAL".
Please look again on the following article:
https://www.space.com/31385-saturn-moon-enceladus-geysers-losing-steam.html
"The 330-mile-wide (530 kilometers) Enceladus hosts a global ocean of salty liquid water beneath its icy shell. This ocean stays liquid because Saturn's powerful gravity twists and stretches Enceladus, generating internal heat through tidal forces. (This tidal heating also provides the energy that powers the jets.)"
What is the meaning of:
"Saturn's powerful gravity twists and stretches Enceladus, generating internal heat through tidal forces. "
Don't you see that it is a clear indication of transforming gravity force into internal heat?
That internal heat is used as a source of energy that powers the jets
"This tidal heating also provides the energy that powers the jets."
So, we have simple explanation how Gravity is transformed into new Heat/energy.
That exactly the scenario in our planet:
The moon's tidal force twists and stretches the surface of the earth and increases the internal heat in our planet.
Thanks to that Tidal the core in our planet still rotates and generate the magnetic field that protects our life.
Look at Mars. Similar planet without tidal forces - lost the energy that was needed to create magnetic field and therefore lost the protection against the solar wind. From a planet full with water and potentially could support life - it became a frozen planet.
The same issue with our moon.
Its face is looked up with our planet. No tidal forces. No internal heat.
The outcome - Frozen moon.
So, tidal is key element that is needed to transform gravity forces into internal heat/Energy.
How can you reject that clear observation?
It does not contradict the first law of thermodynamics.
As the energy doesn't come from nothing. it comes from Gravity.
However - Gravity comes for free.
So, free gravity generates tidal forces that generate new internal heat/energy inside moons/planets/stars/BH/SMBH. That new Heat/energy is added to the total heat/energy in the Universe.
Conventional models say that it comes from a transformation of gravitational potential energy into kinetic energy. No violation of the first law there. The total amount of energy doesn't change (unlike in your physics-defying model).
Gravitational potential energy + kinetic energy represent the total orbital energy.
With or without tidal that orbital energy is fixed.
So, tidal force has no impact on that total orbital energy.
However, without tidal - it has no effect on the internal heat/energy inside the orbital object.
If you wish to believe that Tidal decreases that total orbital energy - than please show me the formula by Newton or Einstein about it.





Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/10/2019 07:04:23
I'm putting a hold on all of the other matters for the moment and focusing on conservation of energy because that is the crux of the problem right now. I'm not moving on until that issue is solved first.

So, free gravity generates tidal forces that generate new internal heat/energy inside moons/planets/stars/BH/SMBH.

No, it doesn't. Tidal forces transform existing orbital or rotational energy into heat energy. It does not create new energy.  Gravity is the medium by which that energy is transferred. Nothing more. Since the total amount of orbital/rotational energy in a planet-satellite system is limited, tidal heating is not a process that can go on forever. Eventually, sufficient rotational energy from the planet is lost in the form of heat so that it takes the same amount of time to spin on its axis as it does for the satellite to orbit it. This is known as tidal locking (which is what Pluto and Charon have). Since the planet and satellite are no longer changing position relative to each other, they no longer tidally flex one another and therefore tidal heating ceases. So tidal force is by no means an infinite energy source.

Quote
It does not contradict the first law of thermodynamics.
As the energy doesn't come from nothing. it comes from Gravity.

Gravity doesn't have energy that it can give. Don't you remember when I told you that? You can't suck energy out of a gravitational field. Energy conservation isn't about energy coming from "nothing". It's about the total amount of energy in a system remaining constant over time. If new energy could be created from a gravitational field, that would violate conservation of energy because the energy is no longer constant. I provided you with three different links stating that energy cannot be created. If energy cannot be created, then gravity cannot create energy either.

To summarize:

(1) Conservation of energy states that the energy in a system must remain constant over time (unless already-existing energy comes in from an outside source or if the system transfers some of its own energy to a different system).
(2) If gravity could create energy, then the energy of a system could increase over time all on its own.
(3) Since conservation of energy forbids that, gravity cannot increase the energy content of a system over time.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/10/2019 15:12:58
I'm putting a hold on all of the other matters for the moment and focusing on conservation of energy because that is the crux of the problem right now. I'm not moving on until that issue is solved first.
Yes, fully agree.
If new energy could be created from a gravitational field, that would violate conservation of energy because the energy is no longer constant.
There is no violation!
Tidal force creates new energy out of gravitational field as it has no negative impact on the gravity force.
Tidal forces transform existing orbital or rotational energy into heat energy.
This is your biggest mistake.
The formula of gravity is:
F=G m1 x m2 / r^2
This formula is correct for each and every particle in an orbital object.
The formula is fixed under any tidal force.
Hence, Tidal does not change the gravity forces and therefore tidal energy is coming for free.
If you wish to believe that tidal energy changes the gravity forces, than please offer your updated Newton formula for gravity under Tidal forces.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/10/2019 15:21:48
The formula is fixed under any tidal forces.

The formula you provided measures force for a fixed distance. Almost all natural orbits are eccentric. They change their distances over time. This change in force is part of what causes the generation of heat. If two orbiting objects are at a constant distance from each other and tidally-locked to each other, there is no change in force over time and as such there is no generation of heat.

tidal energy is coming for free.

Did you even read this?

Quote
conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/10/2019 16:05:14
The formula you provided measures force for a fixed distance. Almost all natural orbits are eccentric. They change their distances over time. This change in force is part of what causes the generation of heat. If two orbiting objects are at a constant distance from each other and tidally-locked to each other, there is no change in force over time and as such there is no generation of heat.
You only offer words over words.
Where is the formula that proves your understanding that tidal decreases the gravity force???

Let me offer an exaple why you are totally wrong:
https://en.wikipedia.org/wiki/Two-body_problem#/media/File:Orbit5.gif
"Two bodies with similar mass orbiting a common barycenter external to both bodies, with elliptic orbits—typical of binary stars"
So we see clearly their common barycenter (the sign + in red)
Now, from any direction that we will look at this system, the center of mass will always be the common barycenter.
The same issue works for tidal.
As long as the common barycenter of an object under Tidal force is fixed, than there is no negative impact on the gravity force.
So, please, if you think that tidal can change the gravity force or the common barycenter - than please show the formula for that!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/10/2019 17:49:31
Where is the formula that proves your understanding that tidal decreases the gravity force???

I never said that it did. What decreases is the rotational or orbital energy possessed by the planet-satellite system. Gravity is constant with distance (as indicated by the "r" in the equation you had in your earlier post). Changing distance (a change in "r") changes the gravitational force. That is what happens in an eccentric orbit: the distance changes over time. That change in gravitational force over time results in a change in tidal force over time. That is what generates the heat.

Force is "free", but energy isn't. Force is not energy. Force cannot be turned into energy. Force only allows one form of energy to be turned into another.

As long as the common barycenter of an object under Tidal force is fixed, than there is no negative impact on the gravity force.

Correct, but under those circumstances, we are no longer talking about an eccentric orbit. So orbital eccentricity is not generating any heating effects here. If the planet is rotating, then there is a change in distance (and therefore a change in force) between different parts of the planet and the satellite. As a beach on Earth moves closer to the Moon due to the Earth's spin, the gravitational force becomes stronger and the sea level rises. If the Earth and Moon were tidally-locked (no longer rotating relative to each other) and it their orbits had an eccentricity of zero (the distance between them is constant), then there would no longer be any form of tidal heating at all.

So, please, if you think that tidal can change the gravity force or the common barycenter - than please show the formula for that!!!

You have it backwards. It is a changing gravitational field strength or direction over time that creates tides. The distance of the Moon from some beach on Earth, for example, is changing over time. Since gravitational field strength is directly tied to distance, that causes the sea level at the beach to change in accordance with the changing distance (tides).

Tidal force creates new energy

ScienceDirect disagrees: https://www.sciencedirect.com/topics/engineering/conservation-of-energy

Quote
energy can neither be created nor destroyed
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/10/2019 20:01:39
If the planet is rotating, then there is a change in distance (and therefore a change in force) between different parts of the planet and the satellite
Dear Kryptid
Let's agree on the following:
A rotating planet by itself, has no effect on Newton gravity formula!
However, you claim that if the planet is rotating, then there is a change in distance.
A change in distance means a change in the center of mass.
So, would you kindly show the formula for that?
Is it one more wishful thinking?
If a planet is rotating, how it could affect its center of mass?
Did Newton mention that a rotating object changes the location of its center of mass or distance?
Actually this statement contradicts Newton Shell theorem:
https://en.wikipedia.org/wiki/Shell_theorem
"Isaac Newton proved the shell theorem and stated that:
1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Therefore, it can rotate as fast as you want.
As long as all of it is a spherically symmetric body there is no change in its center of mass and therefore there is no change in gravity force.
It is also stated:
"2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell."
Therefore, any object inside the shell can move at any direction - As Long as the whole body is still spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside,
Hence, again, there is no change in the center of mass for a rotating object and no net gravitational force is exerted by the shell on any object inside!
If you still believe in that unrealistic Idea of rotating body that can change the center of mass, than this time you must show the formula which links between the rotating activity to distance?

There is very much an impact on the gravity force as the radius changes periodically, and the tidal forces acting on the moon thus vary over time, and heat it, despite it being tide-locked.
Now, let's assume that instead of a spherically symmetric shell (hollow ball), we have a spherically symmetric shell (in American football shape, or tidal shape) that always point to the other body.
Do you see any change in the center of mass?
If there is no change in the center of mass, there is no change in the distance and no change in gravity force.

So, would you kindly once and for all offer the formula that supports your idea (and in the same time contradicts Newton law).

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/10/2019 21:18:53
A football shape is not spherically symmetric.
That is correct.
However, the extra symetrical matter infront and at the back don't really change the center of mass point.
Therefore, even if the tidal shape is not a spherically symmetric, as long as it face directly the other object - there is no change in the center of mass.

You haven't described its motion
If you mean the motion inside the body - than Newton didn't describe any motion.
He specifically claims that "no net gravitational force is exerted by the shell on any object inside".
Therefore, any object inside the shell can set any sort of motion without getting any net gravity force by the shell.
Therefore, its motion is none relevant to the gravity force.

If you mean the motions between the Earth/Moon:
If it represents a hollow moon that is more or less in a similar orbit about our Earth in an isolated 2-body system, they yes, it's center of mass is continuously changing since Earth exerts force on it, and force on a mass results in its acceleration, which moves its center of mass.  The moon's center of mass orbits Earth, and does not stay put.
Than this is correct with or without tidal.
Therefore, tidal has no effect on the gravity force between the two or on their orbital cycle.
So far you couldn't offer any formula that links the tidal impact on gravity force.
You couldn't show the formula that decreases the gravity force due to tidal.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/10/2019 22:59:02
It isn't correct.  1, there is no barycenter of an object.  But there is one for a system of two objects like say Earth/moon.

That was a mis-reading on my part. I thought he was talking about a pair of objects the whole time.

2 That barycenter is fixed (a non-accelerating center of gravity of the two body system) despite the fact that the moon's orbit is eccentric. 

I suppose that depends on how you define "fixed". As the Moon moves further away from the Earth in its orbit, the distance to the barycenter from both the Earth's center and the Moon's center must increase very slightly.

Dear Kryptid
Let's agree on the following:
A rotating planet by itself, has no effect on Newton gravity formula!
However, you claim that if the planet is rotating, then there is a change in distance.
A change in distance means a change in the center of mass.
So, would you kindly show the formula for that?
Is it one more wishful thinking?
If a planet is rotating, how it could affect its center of mass?
Did Newton mention that a rotating object changes the location of its center of mass or distance?
Actually this statement contradicts Newton Shell theorem:
https://en.wikipedia.org/wiki/Shell_theorem
"Isaac Newton proved the shell theorem and stated that:
1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Therefore, it can rotate as fast as you want.
As long as all of it is a spherically symmetric body there is no change in its center of mass and therefore there is no change in gravity force.
It is also stated:
"2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell."
Therefore, any object inside the shell can move at any direction - As Long as the whole body is still spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside,
Hence, again, there is no change in the center of mass for a rotating object and no net gravitational force is exerted by the shell on any object inside!
If you still believe in that unrealistic Idea of rotating body that can change the center of mass, than this time you must show the formula which links between the rotating activity to distance?

You are completely misunderstanding what I am saying. I'm not saying that rotation causes a change in distance between a planet and its satellite as a whole. I said that it causes a change in distance of different parts of the planet. The distance between Orlando, Florida and the Moon's center is changing over time because the Earth is rotating. When Orlando, Florida is directly under the Moon, the Moon is closer to it than when it is on the opposite side of the Earth from the Moon. Thus, the tidal forces acting on Orlando, Florida due to the Moon is changing over time.

But this is all beside the point. The law of conservation of energy won't allow you to create energy. Period: https://www.sciencedirect.com/topics/engineering/conservation-of-energy

Quote
conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics.

I see that you have conveniently avoided this quote in all of your posts so far.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 06/10/2019 23:17:31
Quote from: Halc
2 That barycenter is fixed (a non-accelerating center of gravity of the two body system) despite the fact that the moon's orbit is eccentric.
I suppose that depends on how you define "fixed". As the Moon moves further away from the Earth in its orbit, the distance to the barycenter from both the Earth's center and the Moon's center must increase very slightly.
Yea, but that's the Earth and Moon both moving away from the barycenter, not any motion of the barycenter itself.  Acceleration is absolute, and the barycenter does not accelerate due to the varying distance between a pair of objects with an eccentric orbit, or even if they're connected with springs and hinged counterweights and whatnot.  It can't move while it's a contained system.  That's just simple conservation of momentum.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/10/2019 01:21:45
Yea, but that's the Earth and Moon both moving away from the barycenter, not any motion of the barycenter itself.  Acceleration is absolute, and the barycenter does not accelerate due to the varying distance between a pair of objects with an eccentric orbit, or even if they're connected with springs and hinged counterweights and whatnot.  It can't move while it's a contained system.  That's just simple conservation of momentum.

Duly noted.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/10/2019 05:23:49
That was a mis-reading on my part. I thought he was talking about a pair of objects the whole time.

You have understood me correctly and perfectly.
Halc didn't understand my intention.
Those two orbiting objects represent one Body.
In that body there could be much more than just two objects.
For example - Globular cluster:
https://en.wikipedia.org/wiki/Globular_cluster
"A globular cluster is a spherical collection of stars that orbits a galactic core. Globular clusters are very tightly bound by gravity, which gives them their spherical shapes, and relatively high stellar densities toward their centers."
Their galactic core represents their center mass. with regards to all the orbiting objects inside this cluster, this galactic core represents a "fixed point".
I suppose that depends on how you define "fixed". As the Moon moves further away from the Earth in its orbit, the distance to the barycenter from both the Earth's center and the Moon's center must increase very slightly.
So, the "fixed point" is the barycenter or galactic center to the objects in one body.
The Earth for example has finite no. of atoms.
All those atoms set the Earth center of mass. That center of mass is fixed/same with tidal or without tidal
Therfore – Tidal does not change the location of the Eath center of mass.
As gravity is all about center of mass – Tidal has no impact on the gravity forces or orbital energy

But this is all beside the point. The law of conservation of energy won't allow you to create energy. Period: https://www.sciencedirect.com/topics/engineering/conservation-of-energy
"conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics."
OK
I understand by now that you can't offer any formula that show that Tidal energy can reduce the gravity force.
However, You hope that the "conservation of energy Physics" will support this kind of wrong understanding.
This is a BIGGEST mistake
Gravity is force.
However, it can be represented by Orbital Energy:
https://en.wikipedia.org/wiki/Specific_orbital_energy
"In the gravitational two-body problem, the specific orbital energy {\displaystyle \epsilon }\epsilon  (or vis-viva energy) of two orbiting bodies is the constant sum of their mutual potential energy ({\displaystyle \epsilon _{p}}\epsilon_p) and their total kinetic energy ({\displaystyle \epsilon _{k}}\epsilon _{k}), divided by the reduced mass. "
So, the total orbital energy is the sum of Kinetic Energy + potential energy.
That Orbital energy is fully under conservation of energy Physics.
However, this total orbital energy isn't affected by tidal energy that represents extra heat in any orbital body.
The extra heat due to tidal is a side effect of the "fixed" orbital energy.
Hence, Tidal energy has no impact on that total orbital energy.
Again - If you still think differently - than please would you kindly offer the matematics how the gravity force (or the total orbital energy) is reduced/increased by the Tidal extra energy/heat.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/10/2019 08:11:48
I understand by now that you can't offer any formula that show that Tidal energy can reduce the gravity force.

Because I never said that. Tidal forces don't reduce the force of gravity.

That Orbital energy is fully under conservation of energy Physics.

So is tidal energy. All energy is.

However, this total orbital energy isn't affected by tidal energy that represents extra heat in any orbital body.
The extra heat due to tidal is a side effect of the "fixed" orbital energy.
Hence, Tidal energy has no impact on that total orbital energy.

Heat is a form of energy. As such, it cannot be created. Heat can only come into being if other forms of energy are transformed into heat.

Again - If you still think differently - than please would you kindly offer the matematics how the gravity force (or the total orbital energy) is reduced/increased by the Tidal extra energy/heat.

Why would I offer mathematics for a position that I don't even support? You are building a straw-man argument.

Gravity alone doesn't cause tidal heating. Tidal forces alone don't cause tidal heating either. Your house right now is under tidal forces because the floor is closer to the Earth's core than its roof is. This means that the Earth is pulling very slightly more strongly on the floor than on the roof. That is the definition of a tidal force. Force doesn't generate heat, though. If you want to generate heat using tidal forces, you have to change those tidal forces over time. One way to accomplish that is to change the distance of an object from a source of gravity over time (which is equivalent to an object taking on an eccentric orbit).

To show this, consider the equation to calculate tidal heating: https://en.wikipedia.org/wiki/Tidal_heating

Etidal = -Im(k2)(21/2)((R5n5e2)/G), where

"Etidal" is the rate of tidal heating in watts
"-Im(k2)" is the efficiency of body dissipation (a dimensionless parameter)
"R" is the radius of the body in meters
"n" is the body's mean orbital motion in radians per second
"e" is the orbital eccentricity, and
"G" is the gravitational constant

If I calculate this heating rate for something like Io:

Etidal = -Im(k2)(21/2)((R5n5e2)/G)
Etidal = -(0.005)(10.5)(((1,822,000)5)((4.1 x 10-5)5)((0.0041)2))/(6.674 x 10-11)
Etidal = (-10.5)((2 x 1031)(1.159 x 10-22)(1.681 x 10-5))/6.674 x 10-11)
Etidal = -6.13 x 1015 watts

But what happens if we modify the scenario where the tidal forces are constant? That is, what if we take away the orbital eccentricity?

Etidal = -Im([k2)(21/2)((R5n5e2)/G)
Etidal = -(0.005)(10.5)(((1,822,000)5)((4.1 x 10-5)5)((0.0)2))/(6.674 x 10-11)
Etidal = (-10.5)((2 x 1031)(1.159 x 10-22)(0))/6.674 x 10-11)
Etidal = 0 watts

The power is zero watts. No heat is generated at all.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/10/2019 13:38:18
To show this, consider the equation to calculate tidal heating: https://en.wikipedia.org/wiki/Tidal_heating
Thanks for the explanation about the Tidal heat calculation.
However, you agree that the Tidal doesn't reduce the force of gravity.
Tidal forces don't reduce the force of gravity.
Without changing the gravity force, there is no change in the total orbital energy.
Therefore, with or without Tidal - the gravity force or the total rotation energy stay the same.
Hence, tidal is only a side effect of the gravity force.
The extra tidal heat has no impact on the gravity force or the total rotation energy.
In order to get better understanding for that let me go back to the - Globular cluster:
https://en.wikipedia.org/wiki/Globular_cluster
"A globular cluster is a spherical collection of stars that orbits a galactic core. Globular clusters are very tightly bound by gravity, which gives them their spherical shapes, and relatively high stellar densities toward their centers."
Let assume that in the whole universe there are only one Globular Cluster that orbits around one main and massive body.
In this example - instead of stars we will set several millions or billions rigid balls. they collide with each other, there is also a friction between the balls, but there is no mass lost in the total globular cluster.
Therefore - the internal movement of the rigid balls and the internal friction/collision increases the internal heat in that globular cluster.
However, as there is no change in the gravity force between the globular cluster and the main body, that extra heat don't affect the total orbital energy.
Therefore, the internal heat in the globular cluster is for free!!!
By doing that new extra heat is created without decreasing other energy source or without any transformation of energy from other party..
So is tidal energy. All energy is.
Tidal energy is totally different from all the other energies.
Heat is a form of energy. As such, it cannot be created. Heat can only come into being if other forms of energy are transformed into heat.
That is correct for any kind of energy except - Tidal heat/energy
I agree, any Heat/energy can only come into being if other forms of energy are transformed into heat.
For example - Electromagnetic energy represents a transformation of energy from one source to other.
However, Tidal heat/energy it totally different as there is no transformation of energy in that activity.
Tidal is a side effect of gravity activity and not part of the total orbital energy.
Tidal don't have any impact on Newton gravity force formula or the total orbital energy.
So, there is no transformation of energy from the total orbital energy into tidal.
Hence, tidal heat/energy is only a side effect of the orbital gravity forces.
Tidal has no impact on the gravity force or the total rotation energy
Tidal heat heat/energy doesn't use any energy from the orbital system.
Therefore, as Tidal doesn't consume energy from the orbital system, and there is no need for energy transformation - Tidal heat/energy comes for free.



Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/10/2019 20:02:24
Without changing the gravity force, there is no change in the total orbital energy.
Therefore, with or without Tidal - the gravity force or the total rotation energy stay the same.

The strength of the gravitational field produced by each body remains the same strength, but the force felt by each body changes over time due to the eccentricity of the orbit. It becomes higher as the satellite approaches the planet and then becomes lower when it moves away in its orbit. This change in force stretches the satellite, then relaxes the stretching, then stretches it again, then relaxes it again. It is that stretching that produces tidal heating. A constant force doesn't do that. I already demonstrated that with the equation.

A good explanation for why tidal forces slow down a satellite (and therefore reduce its total orbital energy) can be found here: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833

To sum it up, internal friction due to the stretching of the satellite causes Jupiter to pull on it slightly harder when it tries to move away from the planet. This extra pull has a braking effect, slowing it down and lowering its orbital energy. Thus, less kinetic energy is now available to raise the satellite out and away from the planet on its way to apogee. This loss of kinetic energy means that the satellite cannot travel as far away from the planet as it could before, which means that its apogee has become slightly smaller. Since the difference between apogee and perigee has become smaller, the orbit has become less eccentric. Every time the satellite completes an orbit, the difference between apogee and perigee thus becomes smaller and smaller. This is known as tidal circularization: https://en.wikipedia.org/wiki/Tidal_circularization

As the orbit becomes less and less eccentric, the tidal heating becomes weaker and weaker. Once the orbit becomes completely circular, tidal heating ceases (as per my calculation). So tidal heating is not "for free". It comes at the cost of orbital energy, just like I said before.

This video explains that the energy for tidal heating does come from orbital energy. It states, "orbital energy is dissipated as heat":


Wikipedia says the same thing: https://en.wikipedia.org/wiki/Tidal_heating_of_Io

Quote
Orbital and rotational energy are dissipated as heat in the crust of the moon.

This website agrees: https://astrobites.org/2012/08/30/astrophysical-classics-predicting-tidal-heating-of-io/

Quote
This is because as tides distort a moon, rock is stretched and compressed. As you can imagine, there’s friction when you stretch rock, and friction releases some of the satellite’s orbital energy as heat.

Therefore - the internal movement of the rigid balls and the internal friction/collision increases the internal heat in that globular cluster.
However, as there is no change in the gravity force between the globular cluster and the main body, that extra heat don't affect the total orbital energy.

The friction between the balls that generates the heat will cause them to slow down. You should be quite aware that friction does exactly that: it slows things down. It is a conversion of the ball's energy of motion into heat energy. Since the balls are slowing down, they have less total orbital energy.

Tidal energy is totally different from all the other energies.

Please back that statement up with an authoritative source. If this was true, it would be extremely noteworthy. Yet I have never seen a single source state that "energy cannot be created or destroyed unless it's tidal energy." Every source I have ever seen simply says "energy cannot be created or destroyed". It doesn't make exceptions for any kind of energy. Energy is energy. No energy is free.

That is correct for any kind of energy except - Tidal heat/energy

So then you finally admit that you think energy can be created and therefore you disagree with the law of conservation of energy. So stop claiming that you agree with it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/10/2019 14:50:53
The friction between the balls that generates the heat will cause them to slow down. You should be quite aware that friction does exactly that: it slows things down. It is a conversion of the ball's energy of motion into heat energy. Since the balls are slowing down, they have less total orbital energy.
Thanks
So, your message is very clear:
There is no change in gravity force, however, the extra tidal heating/energy must be deducted from the total orbiting energy.
Let's see if it is freezable by using the example of that globular clusters balls and convert words into real formulas.
In that Globular clusters balls there are billions of balls.
All of them orbit around the center of mass point while they create tidal heating due to friction with each other.

Hence
Their total mass is estimated as m2.
The main body mass is m1.
The radius between the two centers of mass is r

Kinetic Energy = Ek = 1/2 m2 v^2
Potential Energy = Ep = m2 G h = m2 G r
Tidal Energy = E(tidal)
The total rotation energy (without tidal) = Er = Ek +Ep =1/2 m2 v^2 + m2 G r
However if we wish to deduct the E(tidal) from that Er we must set a change in one of the following argument
G, m2, v, or r
G = no option to change it
m2 = it is fixed, so there is no way to change.
Based on gravity formula
v^2 = G (m1 + m2) / r
So, if we wish to change v we must change r
Therefore, in order to decrease Er by the value of Etidal we must change r
However, by changing r we actually change the gravity force, as the formula for gravity is:
Gravity force = F = G m1 m2 /r^2

Therefore, the statement that the Tidal energy decreases the total rotation energy without changing the gravity force is technically not realistic.
Any comment?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 08/10/2019 15:54:29
Tidal Energy = E(tidal)

If you want to calculate the rate of tidal heating, you need to use the formula I posted earlier: Etidal = -Im(k2)(21/2)((R5n5e2)/G)

Therefore, the statement that the Tidal energy decreases the total rotation energy without changing the gravity force is technically not realistic.
Any comment?

I see this as a possible source of confusion. The strength of the overall gravitational field of a planet is unchanging. That is what I mean by the force of gravity being unchanging. However, I need to distinguish this from the force felt by the satellite. In the case of an elliptical orbit, the force experienced by the satellite does change over time because the distance changes over time.

Gravity force = F = G m1 m2 /r^2

The critical part of this formula is the "r". In an elliptical orbit, the value of "r" changes over time. This makes the gravitational force experienced by the satellite change over time as well. If you want specific numbers, I will calculate them for our Moon. When the Moon is at apogee, it is at a maximum distance from the Earth (405,400 kilometers). When it is at perigee, it is at a minimum distance (363,600 kilometers).

For apogee:

F = G((m1m2)/r2)
F = (6.674 x 10−11)((5.97237 x 1024)(7.342 x 1022)/((405,400,000)2)
F = 2.92649 x 1037/1.64349 x 1017
F = 1.78 x 1020 newtons

For perigee:

F = G((m1m2)/r2)
F = (6.674 x 10−11)((5.97237 x 1024)(7.342 x 1022)/((363,600,000)2)
F = 2.92649 x 1037/1.322 x 1017
F = 2.21 x 1020 newtons

The gravitational force experienced by the Moon is 24% stronger when it is at perigee than when it is at apogee. This is because the orbit is eccentric. If the orbit was not eccentric, perigee and apogee would be the same and there would be no change of force over time. No change of force means no tidal heating. Tidal heating causes an orbit to become less eccentric over time because the energy drained from the orbit does not allow the satellite to move quite as far away from the planet as it did on each prior orbit. This slowly decreases the eccentricity and as such slowly decreases the amount of tidal heating over time.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/10/2019 17:02:45
If the orbit was not eccentric, perigee and apogee would be the same and there would be no change of force over time. No change of force means no tidal heating.
Thanks
So,  in case of a pure cycle orbit (If the orbit not eccentric, perigee and apogee would be the same) there is no way to get tidal forces.
Do you agree with that?
However, I assume that in the nature all the orbit cycles must have some eccentric.
 
The critical part of this formula is the "r". In an elliptical orbit, the value of "r" changes over time. This makes the gravitational force experienced by the satellite change over time as well. If you want specific numbers, I will calculate them for our Moon. When the Moon is at apogee, it is at a maximum distance from the Earth (405,400 kilometers). When it is at perigee, it is at a minimum distance (363,600 kilometers).

For apogee:

F = G((m1m2)/r2)
F = (6.674 x 10−11)((5.97237 x 1024)(7.342 x 1022)/((405,400,000)2)
F = 2.92649 x 1037/1.64349 x 1017
F = 1.78 x 1020 newtons

For perigee:

F = G((m1m2)/r2)
F = (6.674 x 10−11)((5.97237 x 1024)(7.342 x 1022)/((363,600,000)2)
F = 2.92649 x 1037/1.322 x 1017
F = 2.21 x 1020 newtons

The gravitational force experienced by the Moon is 24% stronger when it is at perigee than when it is at apogee. This is because the orbit is eccentric.
This explanation is fully clear to me.
We see the difference between the perigee to apogee
That activity represents only one full cycle.
However, do you agree that in the next cycle - we should get exactly the same output?
So, what can we learn from that?
Can we assume that there is no change in the gravity force between one full orbital cycle to the next full cycle?
I really don't understand how could it be that the total Rotation energy per cycle should be decreased due to tidal energy dissipation, while the gravity force between one full orbital cycle to the next one must stay unchanged.
It is clear to me that if we reduce the total rotation energy per cycle, r must be changed from one full cycle to the next cycle, otherwise, the total rotation energy per cycle is unchanged.
So, would you kindly set the formula for the total rotation energy per one full cycle, and show why this total rotation energy must go down due to tidal energy in the next full orbital cycle, while there is no change in gravity force between one full orbital cycle to the next full cycle.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 08/10/2019 22:04:41
However, do you agree that in the next cycle - we should get exactly the same output?

Nope. The satellite's apogee gets very slightly smaller each time an orbit is completed. This makes the difference between apogee and perigee smaller. That smaller difference results in a smaller difference between the gravitational forces, which consequently makes the tidal heat generated also less. It becomes smaller and smaller with each completed orbit.

I really don't understand how could it be that the total Rotation energy per cycle should be decreased due to tidal energy dissipation, while the gravity force between one full orbital cycle to the next one must stay unchanged.

That's the issue: the gravitational force felt by the satellite isn't the same on each orbit because the distances involved are becoming smaller.

So, would you kindly set the formula for the total rotation energy per one full cycle,

The link I provided earlier gives the formula to calculate the energy loss per orbit and even gives an example calculation: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833

Quote
and show why this total rotation energy must go down due to tidal energy in the next full orbital cycle

I posted a link to the explanation earlier. Start off with the assumption that Io is composed of a frictionless fluid. It obviously isn't, but this is to demonstrate a point. As Io nears perigee, the increase in tidal forces felt by it causes it to stretch in the direction of Jupiter. Since there is no friction, no heat is generated by the stretching. At the same time, the lack of friction means there is no lag between the time that Io feels Jupiter's increased gravity and its response to that gravity. So as Io nears Jupiter, its tidal bulge becomes larger and larger, reaching a maximum height exactly where Jupiter's gravity is the strongest (perigee). Then, at an equal rate, the tidal bulge becomes smaller and smaller as it moves away from Jupiter.

This perfect symmetry preserves the total orbital energy of Io. That is, Jupiter is pulling equally hard on Io 1 minute before it reaches perigee and 1 minute after it passes perigee. So the acceleration that Io feels as it nears Jupiter is exactly equal to the deceleration it feels as it moves away from Jupiter.

This symmetry is broken once friction is added. Now there is a lag time between when Io feels Jupiter's gravity and when Io responds to that gravity. So once Io reaches perigee, its tidal bulge has not quite reached its peak height because the stretching is being slowed down by internal friction. So Io's tidal bulge actually reaches its peak height after it has already passed perigee. This makes Io slightly more stretched 1 minute after it has passed perigee than it was 1 minute before perigee. This causes it to experience and unequal pull from Jupiter's gravity. Since the bulge is larger as Io leaves than as it approached, Jupiter decelerates Io faster than it accelerated it. This is a braking effect that robs Io of orbital energy and prevents it from traveling all the way to its previous apogee. The internal friction that caused this is simultaneously responsible for generating the tidal heat.

So the mechanism that links orbital energy loss to tidal heating is friction. Get rid of the friction and you get rid of the loss of orbital energy (while also getting rid of tidal heating).

Quote
while there is no change in gravity force between one full orbital cycle to the next full cycle.

The gravitational force does change, and that is due to the changing distances.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/10/2019 04:16:12
The link I provided earlier gives the formula to calculate the energy loss per orbit and even gives an example calculation: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833
In this article it is stated:
"When a moon stretches, because of the tidal effect, it's gravitational attraction to the planet will become bigger by
M(moon) * M(planet) * G * d * r / R^4
Where d is the height of the bulge, r is the radius of the moon and R the distance to the planet. (this is actually a simplification, assuming that the mass of the moon is divided in two halves, at a distance R+r+d and R-r-d from the planet. The real effect will be smaller, but proportional to this).
Source https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833
So, the extra gravitation due to tidal bulge = M(moon) * M(planet) * G * d * r / R^4"

This idea totally contradicts the shell theorem by Newton
https://en.wikipedia.org/wiki/Shell_theorem
"Isaac Newton proved the shell theorem[1] and stated that:
A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Let's look at the following image:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1-anim.gif
m represents the planet, while the shell represents the moon.
In this example we see a pure cycle.
However, we can divide it to two halves and claim that the half at the front has more impact than the one at the back.
Newton have proved (after long calculation) that as long as the half at the front in symmetrical to the half at the back the total mass of that cycle or moon is:
"which suggests that the gravity of a solid spherical ball to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass."
Now, if we add a symmetrical bulge at the front and at the back of this cycle, don't you agree that as the half at the front is fully symmetrical with the half at the back, than the outcome should be:
That the gravity of a solid spherical ball (with symmetrical bulge) to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass.?

Actually, we could split this pure cycle two halves and set the total mass of each cycle in one new cycle.
So, we could have two balls connecting one to each while they are directly in line with the planet.
Don't you agree that even in this case, their gravity impact to an exterior object can be simplified as point of mass in the meeting point between those two balls while the total mass is the sum of the two balls?

Therefore, the assumption that a symmetrical bulge (in front and in the back) can change the center of mass of the moon is a fatal mistake.

If you still think differently, would you kindly explain why Newton proved that although the first half is closer to the planet, the total two halves set the center of the mass exactly at the center of the ball?
What is the difference between this scenario to symmetrical bulge that one is added in the front and one in the back?

Gravity is all about mass and center of mass.
The bulge doesn't change the total mass of the moon.
However, if you can prove that the Symmetrical bulge can change the Moon's center of mass than there is a change in the gravity force between the planet and the moon.


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/10/2019 05:26:20
The link I provided earlier gives the formula to calculate the energy loss per orbit and even gives an example calculation: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833
In this article it is stated:
"When a moon stretches, because of the tidal effect, it's gravitational attraction to the planet will become bigger by
M(moon) * M(planet) * G * d * r / R^4
Where d is the height of the bulge, r is the radius of the moon and R the distance to the planet. (this is actually a simplification, assuming that the mass of the moon is divided in two halves, at a distance R+r+d and R-r-d from the planet. The real effect will be smaller, but proportional to this)."
That's correct.  The F=GMm/r² only works exactly for spherically symmetric bodies.

Quote
This idea totally contradicts the shell theorem by Newton
No it doesn't.
Quote
"Isaac Newton proved the shell theorem[1] and stated that:
A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Let's look at the following image:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1-anim.gif
m represents the planet, while the shell represents the moon.
It does not represent a tidally distorted moon.  Only a spherical approximation.

Quote
However, we can divide it to two halves and claim that the half at the front has more impact than the one at the back.
Newton have proved (after long calculation) that as long as the half at the front in symmetrical to the half at the back the total mass of that cycle or moon is:
"which suggests that the gravity of a solid spherical ball to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass."
Now, if we add a symmetrical bulge at the front and at the back of this cycle, don't you agree that as the half at the front is fully symmetrical with the half at the back
Per the article you first linked, the two do not cancel out.  The two masses are not spherically distributed.  The part I bolded above is something you made up.  Newton could trivially demonstrate that this isn't the case.

Quote
Therefore, the assumption that a symmetrical bulge (in front and in the back) can change the center of mass of the moon is a fatal mistake.
A pair of bulges like that indeed has no effect on the center of mass of the moon.  Nobody claimed otherwise.

Quote
Gravity is all about mass and center of mass.
Obviously not, since those bulges did not change the total mass or the center of mass, yet per your first link, the gravitational force is different.  So gravity is not just about center of mass.
Given the right distribution of mass, I can have gravity actually repel a pair of objects (as measured from their respective centers of gravity).
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/10/2019 05:38:53
Halc has it right. Shell theorem only applies to spherically-symmetrical objects. A tidally-distorted satellite is not spherically symmetrical.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/10/2019 05:58:42
Halc has it right. Shell theorem only applies to spherically-symmetrical objects. A tidally-distorted satellite is not spherically symmetrical.
Ok
Let me offer the following:
You agree that it works perfectly in a spherically symmetric cycle.
Now, let's look again on the following diagram:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1.svg
1. If we divide the spherically symmetric cycle into two halves -One in the front and one in the back - Do you agree that the center of mass should stay at the same point?
2. If we push away the two halves from each other on the same radius line from the planet - So one will be pushed back by r Km from the center of mass, while the one in the front will be pushed forward also by r km) - do you agree that the center of mass should also stay at the same point as before?
3. If so, let's use this spited spherically symmetric cycle to represent the symmetric bulge (one in front and one in the back)
So, one spherically symmetric cycle will represent the moon minus the mass of the bulge, while the bulge will be represented by splited spherically symmetric cycle - one in the front and one in the back.
Why the two spherically symmetric cycles can't fully represents the moon + the bulge with the center of mass at the same point as before?



Quote
Therefore, the assumption that a symmetrical bulge (in front and in the back) can change the center of mass of the moon is a fatal mistake.
A pair of bulges like that indeed has no effect on the center of mass of the moon.  Nobody claimed otherwise.
So, if you agree that the Bulge do not change the total mass of the moon and it also do not change the center of mass of the moon, than how can you claim that there is a change in the total gravity force between the planet and the moon?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/10/2019 06:10:06
The bulges are not symmetrical: https://physics.mercer.edu/hpage/tidal%20asymmetry/asymmetry.html
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/10/2019 06:29:42
The bulges are not symmetrical: https://physics.mercer.edu/hpage/tidal%20asymmetry/asymmetry.html
This calculation might not be fully correct.
The gravity force is transformed to Taylor's series:
"Next, expand F(r) in a Taylor's series as follows:"
In order to get closer to the reality we must take this Taylor's series to the infinity.
If we don't take it to infinity, than by definition there is a room for error.
Any none perfectly assumption or set up could also cause some minor error.
They even claim:
"The gradient expression for the tidal force is a poor approximation in the Fig. 3 case,"
So, I really don't know what might be the magnitude of the error in this kind of calculation but it is there for sure.
That error could be reflected as the unrealistic change in the center of mass or in the gravity force.
Newton has used different approach.
I prefer to use Newton solution.
If the splited spherically symmetric cycle can represent the symmetric bulge, than there shouldn't be any change in the center of mass point.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/10/2019 06:36:10
In order to get closer to the reality we must take this Taylor's series to the infinity.

What is that even supposed to mean?

If we don't take it to infinity, than by definition there is a room for error.

What do you mean by "take it to infinity"? How can you take anything to infinity?

Any none perfectly assumption or set up could also cause some minor error.

Then show where the error in the math is.

Newton has used different approach.
I prefer to use Newton solution.

When did Newton show anything that was counter to what is said on this page? The method mentioned on the page used Newtonian math to arrive at the result:

Quote
The conventional treatment of the tidal field of the Moon is an approximation that uses calculus differentials, As the distance of a moon from its parent decreases progressively, the assumptions of this first-order approximation become ever more unacceptable. It is then necessary to apply Newton's universal law of gravity to virtually all components of the body.

If the splited spherically symmetric cycle can represent the symmetric bulge

That's just it, though. The bulges aren't symmetrical. Using the equations on that page, you can find that the difference between nadir and zenith (the far bulge and the near bulge, respectively) is around 5% for the Earth, about 1.4% for the Moon and 1.3% for Io.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/10/2019 08:16:38
Quote
In order to get closer to the reality we must take this Taylor's series to the infinity.
If we don't take it to infinity, than by definition there is a room for error.
What do you mean by "take it to infinity"? How can you take anything to infinity?
If you convert something (gravity) into Taylor's series than you have to use infinity no. in this series in order to get high accuracy.
When did Newton show anything that was counter to what is said on this page? The method mentioned on the page used Newtonian math to arrive at the result:
They are using the same picture as in Newton's shell theorem:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1.svg

That's just it, though. The bulges aren't symmetrical. Using the equations on that page, you can find that the difference between nadir and zenith (the far bulge and the near bulge, respectively) is around 5% for the Earth, about 1.4% for the Moon and 1.3% for Io.

Ok
I was not aware about that difference in size between the front bulge to the rear bulge.
However - Do you agree that if they were exactly at the same size - the center of mass had to be located exactly at the center of the cycle as explained by Newton's shell theorem?
If so, let's look at the Earth.
It's front bulge is bigger by 5% from the rear one.
That for sure will set the center of mass closer to the moon (with regards to spherically symmetric bulge)
That will increase the gravity force between the Moon -Earth comparing to spherically symmetric bulge.
So, now we have stronger gravity force.
However, as there is no change in this ratio (assuming that it is constantly at 5%), there is also no change in the location of center of mass and therefore - there is no change in gravity force.
Hence, as long as the 5% stay, there is no change in the center of mass and no change is the gravity force.
Without a change in those two segments, there will be no change in the total rotation energy.
Therefore, the bulge by itself doesn't change the center of mass point, the gravity force or the total rotation energy. Only a change in the bulge ratio can change those values.
However, they can go up or down.
If the ratio be get to 4% the center of mass will be shifted backwards and therefore les gravity force and less total rotation energy.
If the ratio will get to 6%, the center of mass will be shifted inwards and therefore more gravity force and more total rotation energy.
Therefore - the bulge does not consume energy from the total rotation energy - it just change the location of center of mass and therefore, it sets the amplitude of that gravity force or the total energy based on the bulge rear/front ratio.
.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 09/10/2019 17:11:21
Ok
I was not aware about that difference in size between the front bulge to the rear bulge.
However - Do you agree that if they were exactly at the same size - the center of mass had to be located exactly at the center of the cycle as explained by Newton's shell theorem?
If so, let's look at the Earth.
It's front bulge is bigger by 5% from the rear one.
That for sure will set the center of mass closer to the moon (with regards to spherically symmetric bulge)
That will increase the gravity force between the Moon -Earth comparing to spherically symmetric bulge.
So, now we have stronger gravity force.
However, as there is no change in this ratio (assuming that it is constantly at 5%), there is also no change in the location of center of mass and therefore - there is no change in gravity force.
Hence, as long as the 5% stay, there is no change in the center of mass and no change is the gravity force.
Without a change in those two segments, there will be no change in the total rotation energy.
Therefore, the bulge by itself doesn't change the center of mass point, the gravity force or the total rotation energy. Only a change in the bulge ratio can change those values.
However, they can go up or down.
If the ratio be get to 4% the center of mass will be shifted backwards and therefore les gravity force and less total rotation energy.
If the ratio will get to 6%, the center of mass will be shifted inwards and therefore more gravity force and more total rotation energy.
Therefore - the bulge does not consume energy from the total rotation energy - it just change the location of center of mass and therefore, it sets the amplitude of that gravity force or the total energy based on the bulge rear/front ratio.

Actually, the 5% thing is an average value based on the semi-major axis. The difference in size of the bulges actually does change over time because of the eccentricity of the orbit. The difference will be larger at perigee and smaller at apogee. To do some example calculations for the size difference of the bulges on the Moon:

Moon distance at apogee: 405,400,000 meters

az = (2RGm)/(r3)(1+(3R/2r)
az = (2(1,737,100)(6.674 x 10−11)(7.342 x 1022)/((1,737,100)3)(1+(3(1,737,100)/2(405,400,000,000)
az = (1.702375648936 x 1019)/(5.241727755811 x 1018)(1.006427355698)
az = (3.24773763)(1.006427355698)
az = 3.2686119969964766

an = -(2RGm)/(r3)(1-(3R/2r)
an = -(2(1,737,100)(6.674 x 10−11)(7.342 x 1022)/((1,737,100)3)(1-(3(1,737,100)/2(405,400,000,000)
an = -(1.702375648936 x 1019)/(5.241727755811 x 1018)(0.9935726443)
an = -(3.24773763)(0.9935726443)
an = -3.226863265

The difference between those two values is 1.293786% at apogee.

Moon distance at perigee: 362,600,000 meters

az = (2RGm)/(r3)(1+(3R/2r)
az = (2(1,737,100)(6.674 x 10−11)(7.342 x 1022)/((1,737,100)3)(1+(3(1,737,100)/2(362,600,000,000)
az = (1.702375648936 x 1019)/(5.241727755811 x 1018)(1.007186)
az = (3.24773763)(1.007186)
az = 3.2710759299

an = -(2RGm)/(r3)(1-(3R/2r)
an = -(2(1,737,100)(6.674 x 10−11)(7.342 x 1022)/((1,737,100)3)(1-(3(1,737,100)/2(362,600,000,000)
an = -(1.702375648936 x 1019)/(5.241727755811 x 1018)(0.992813982)
an = -(3.24773763)(0.992813982)
an = -3.22439933

The difference between those two values is 1.4476% at perigee. So the difference does change over time.

Because it does.  Do the math.  It is trivial.  Compute the force on a 2m mass all at one point, and then a pair of 1m masses that are connected but separated by some distance, but have the same center of mass.  The combined force on the two 1m masses will be larger than that of the 2m mass. All you need is F=GMm/r².

As a matter of fact, I'll do exactly that. For the case of a 100 kilogram sphere orbiting 10,000,000 meters above Earth's surface:

F = (GMm)/r²
F = ((6.674 x 10−11(5.97237 x 1024)(100))(10,000,000)2
F = (3.985959738 x 1016)/1014
F = 398.6 newtons

Now I'll consider a scenario where there are two spheres, each with a mass of 50 kilograms. One is orbiting at 11,000,000 meters and the other is orbiting at 9,000,000 meters:

F = (GMm)/r²
F = ((6.674 x 10−11)(5.97237 x 1024)(50))(11,000,000)2
F = (1.992979869 x 1016)/(1.21 x 1014)
F = 164.7 newtons

F = (GMm)/r²
F = ((6.674 x 10−11)(5.97237 x 1024)(50))(9,000,000)2
F = (1.992979869 x 1016)/(8.1 x 1013)
F = 246 newtons

Add those two forces together and you get a total force of 410.7 newtons, which is larger than the 398;6 newtons of the single sphere scenario. So asymmetrical tidal bulges aren't even necessary to get an increase in total force.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/10/2019 14:40:53
an = -(2RGm)/(r3)(1-(3R/2r)
As I have already explained this formula is INCORRECT.
Gravity force should only be represented by the following formula:
F = (GMm)/r²
Gravity is all about mass and center of mass.
The tidal bulge only changes the location of the center of mass.
As the front bulge is bigger by 5% from the rear one - than it should shift the center of mass inwards.
Let's look at our moon.
Its face is locked with the Earth.
What is the reason for that?
I think that the answer is as follow:
The Moon's center of mass had shifted inwards (to the Earth) due to Tidal.
Therefore, the moon had lost its spherical Symmetrical shape.
It is similar to an extra weight in a cube.
That extra weight should force the cube to fall always at the same side - as the center of its mass had shifted to that side in the cube.
In the same token; the tidal bulge shifts the location of the moon's center of mass and now it can't continue to rotate.
We can also think about a Gyro or wheel.
The rotation momentum will keep the spin (assuming that there is no friction).
However, if we will set a bulge in that wheel - that is always pointed to one direction, that bulge should decrease the rotation momentum over time
Therefore, the tidal bulge acts as some sort of a friction in the rotation Momentum.
So, I agree with your idea that the tidal energy should be taken from something, but it has to be taken from the rotation momentum and not from the total orbital energy.
There is another issue with gravity.
Gravity works locally.
In other words - The impact of local gravity force is much higher than the impact of far away gravity.
The moon prefers to orbit around the Earth, even as the gravity force from the sun is stronger by more than twice than the gravity from the earth.
So we are lucky that we orbit at the first gravity layer around the moon.
As the moon's tidal energy dispassion on the Earth is relatively quite low (the mass of the Earth is much higher than the mass of the moon), the earth is still keeping its rotation momentum.
However, As the Earth's tidal energy dispassion on the moon is relatively quite high (again - the mass of the Earth is much higher than the mass of the moon), the moon has totally lost its rotation momentum.
Other planets without moons - were not so lucky as the earth. They have lost their rotation momentum as their first gravity layer is based on the Sun. As the Sun mass is much bigger than their own mass, they have totally lost all their rotation momentum.
So, Tidal energy works on the Rotation momentum/energy and not on Total gravity energy!!!

However, I have a question with regards to none perfect orbit (eccentricity less than 1).
You claim that:
The difference in size of the bulges actually does change over time because of the eccentricity of the orbit. The difference will be larger at perigee and smaller at apogee.
Let me focus on:
"The difference will be larger at perigee and smaller at apogee" - "Over time"
Do you agree that as there are changes in the gravity force between perigee to apogee, with or without tidal bulge the gravity should be go down over time?
If you agree with that - than you have solved one more enigma that I'm trying to solve.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/10/2019 17:14:32
As I have already explained this formula is INCORRECT.

Your "explanation" is what was incorrect.

Gravity force should only be represented by the following formula:

Those two formulas are telling us different things. The formula I posted tells what the force at the tidal bulges is, whereas the one you posted tells what the overall force is between two spherical objects. Each formula is giving information that the other does not provide.

But even if you don't like the formula I provided, the one you posted still allows for an increase in net gravitational force without a change in the center of gravity. I posted such an example calculation at the end of post #862. That increase in force, caused by Io's distorted shape as it moves away from Jupiter, provides a braking effect that lowers its total orbital energy.

but it has to be taken from the rotation momentum and not from the total orbital energy.

It can be taken from rotation as well, but it can also definitely be taken from orbital energy. I already explained in a prior post how a braking effect from Jupiter lowers the total orbital energy of Io.

So, Tidal energy works on the Rotation momentum/energy and not on Total gravity energy!!!

It's both, actually.

So, I agree with your idea that the tidal energy should be taken from something

Then we can end this thread here and now, since that was the point all along. You can't make an infinite energy factory using tidal forces.

Do you agree that as there are changes in the gravity force between perigee to apogee, with or without tidal bulge the gravity should be go down over time?

No, the force of gravity as felt by a satellite should actually go up over time because the loss of eccentricity will bring it closer to the planet.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/10/2019 17:50:37
Quote from: Dave Lev 3
but it has to be taken from the rotation momentum and not from the total orbital energy.
It can be taken from rotation as well, but it can also definitely be taken from orbital energy. I already explained in a prior post how a braking effect from Jupiter lowers the total orbital energy of Io.
Rotation momentum, if transferred to the satellite, increases its total orbital angular momentum, and thus energy.
This is the case with Io, and our moon: the orbital radius is increasing as the rotational momentum of Jupiter is transferred to the orbital momentum of Io, an angular acceleration effect, not a braking effect.
This is because Jupiter rotates about 4x the speed of Io's orbit. Phobos is losing altitude because its orbit has a shorter period than the rotation rate of Mars.

This was discussed heavily in the early posts of this thread, and Dave was in denial of all of it, and now I see he's pushing different (but still wrong) ideas.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/10/2019 20:51:02
Rotation momentum, if transferred to the satellite, increases its total orbital angular momentum, and thus energy.
This is the case with Io, and our moon: the orbital radius is increasing as the rotational momentum of Jupiter is transferred to the orbital momentum of Io, an angular acceleration effect, not a braking effect.

That is technically true, but both of those effects need to be taken into account. If Jupiter is transferring rotational energy to Io faster than it is losing orbital energy due to a braking effect, then it is true that its semi-major axis will increase over time. However, The orbital eccentricity should still decrease over time anyway, thus lowering the rate of tidal heating.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/10/2019 22:03:40
Quote
As I have already explained this formula is INCORRECT.
Your "explanation" is what was incorrect.
My explanation is fully correct.
It was a severe mistake to use Taylor_series
https://en.wikipedia.org/wiki/Taylor_series
"In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point"
It is some sort of mathematical fiction.
We take something - gravity force, convert it to infinite something by Taylor series and than use finit something just in order to prove other something - Tidal negative impact on total rotation energy.

But even if you don't like the formula I provided,
It's not an issue that I don't like it.
It is just a totally incorrect formula.
We shouldn't use it  - Never and ever.

I posted such an example calculation at the end of post #862. That increase in force
As a matter of fact, I'll do exactly that. For the case of a 100 kilogram sphere orbiting 10,000,000 meters above Earth's surface:
F = (GMm)/r²
F = ((6.674 x 10−11(5.97237 x 1024)(100))(10,000,000)2
F = (3.985959738 x 1016)/1014
F = 398.6 newtons

Now I'll consider a scenario where there are two spheres, each with a mass of 50 kilograms. One is orbiting at 11,000,000 meters and the other is orbiting at 9,000,000 meters:

F = (GMm)/r²
F = ((6.674 x 10−11)(5.97237 x 1024)(50))(11,000,000)2
F = (1.992979869 x 1016)/(1.21 x 1014)
F = 164.7 newtons

F = (GMm)/r²
F = ((6.674 x 10−11)(5.97237 x 1024)(50))(9,000,000)2
F = (1.992979869 x 1016)/(8.1 x 1013)
F = 246 newtons

Add those two forces together and you get a total force of 410.7 newtons, which is larger than the 398;6 newtons of the single sphere scenario. So asymmetrical tidal bulges aren't even necessary to get an increase in total force.
You have a severe mistake in this calculation!
As I have stated -
Gravity is all about mass and center of mass.
So, it was a mistake to set two diffrent calculations - One for the object at 11,000,000 meters and other one for the object at 9,000,000 meters.
You had to find the center of mass of those two objects and than find the gravity force to the main body.
As an example -
Look at the Moon/Earth/Sun system.
We do not set a separate gravity force calculation for the Moon/Sun and Earth/Sun
We set a calculation for the Earth/moon center of mass and just than set the gravity calculation for that center of mass while it orbits around the Sun.

quote]but it has to be taken from the rotation momentum and not from the total orbital energy./quote]
It can be taken from rotation as well, but it can also definitely be taken from orbital energy.
No.
It can be taken ONLY from the rotation energy.
It is a severe mistake to assume that it take anything from the orbital energy!

Quote
So, I agree with your idea that the tidal energy should be taken from something
Then we can end this thread here and now, since that was the point all along. You can't make an infinite energy factory using tidal force
Yes, I was expecting for this answer from you.  I thought about it before.
However, I'm going with my discovery even if it contradicts key idea in my theory.
That shows you that the true is more important for me than to prove a key section in my theory.
In any case, I still don't think that it contradicts the theory.
The main reason for that is that we don't know how the BH or the SMBH really works from inside.
I must say that I still need to think about it.
So, let's set the activity and verify if we can overcome that milestone:

Gravity from nearby objects set tidal forces on the SMBH.
Those Tidal forces increase the internal heat energy.
However, they also decrease the rotation momentum of the SMBH.
So, how can we overcome that issue?
My idea is based on the following explanation from Halc with minor adaptation:
the rotational momentum of Jupiter is transferred to the orbital momentum of Io
The Accretion disc orbits at 0.3c. That must have an impact of the SMBH orbital rotation momentum.
Actually, at the moment of creation, the pair particles are moving at almost the speed of light.
We know that as they get to the outer side of the accretion disc, their orbital velocity is decreasing to 0.3c
So, we can claim that some of the decreasing orbital energy is transformed back to the SMBH and maintain its orbital momentum.
You might claim that somehow new energy must be added in order to maintain this activity. Otherwise, it won't work.
My answer for that will be - "Gravity force".
We have already agreed that the new pair particles are created due to the Magnetic force + the gravity force.
However, some of the energy for that creation is taking also from the gravity force.
So, the SMBH's gravity force contributes some new energy to our Universe.
In any case, new energy must be created by the SMBH – For sure
That SMBH's extra new energy is used to create the mass in the accretion disc and then the SMBH is using some of this new mass to form stars planets and moons.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/10/2019 22:31:31
It is some sort of mathematical fiction.

So where in the Wikipedia article you cited does it say that it's "fiction"?

We take something - gravity force, convert it to infinite something by Taylor series and than use finit something just in order to prove other something - Tidal negative impact on total rotation energy.

Then you must think calculus itself is wrong, since it also involves infinite sums.

It's not an issue that I don't like it.
It is just a totally incorrect formula.
We shouldn't use it  - Never and ever.

Why do I get the feeling that your understanding of mathematics doesn't go beyond simple algebra?

So, it was a mistake to set two diffrent calculations - One for the object at 11,000,000 meters and other one for the object at 9,000,000 meters.
You had to find the center of mass of those two objects and than find the gravity force to the main body.

Then let's start off by considering each sphere in isolation. Now we have two separate objects that are not connected to each other. The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 246 newtons.

The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 164.7 newtons. The sum total of the forces acting on both spheres is therefore 410.7 newtons.

Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters). If your reasoning is correct, this now-singular object is suddenly and magically feeling 12.1 newtons of force less than it did when the objects were separate (because you claim that these two tethered spheres should feel the same force as a single sphere of identical mass and center of mass). How can the mere addition of a string cause the Earth to pull on the spheres less than it did before?

The answer is that it can't: strings don't have magical antigravity properties. We can therefore safely conclude that your reasoning is incorrect.

Tidally-distorted objects don't satisfy shell theorem. Your equation is only valid for points and spheres. The Wikipedia article on shell theorem even shows that different equations are used for rings and disks, despite the fact that they can just easily have the same center of mass and total mass as a sphere, Different shapes require different equations: https://en.wikipedia.org/wiki/Shell_theorem#Derivation_of_gravitational_field_outside_of_a_solid_sphere

Nothing on a tidally-distorted object is sphere-shaped or even hemisphere-shaped. The whole object is shaped like an ellipsoid, basically a chicken egg. So the standard gravitational equation is insufficient to accurately calculate gravitational forces acting on it.

That shows you that the true is more important for me than to prove a key section in my theory.

Then stop trying to violate the law of conservation of energy.

So, how can we overcome that issue?

You can't. The law of conservation of energy won't let you.

So, we can claim that some of the decreasing orbital energy is transformed back to the SMBH and maintain its orbital momentum.

Do you seriously not know how to do simple addition and subtraction? If an energy value of "1" was taken from the black hole to create the particles, then those particles can only give at most an energy unit of "1" back to the black hole. If it gives less than that back (which it would have to in order to continue existing), then the energy received back by the black hole must be less than it expended creating the particles. You're not going to cheat the law of conservation of energy.

My answer for that will be - "Gravity force".

Force is not energy.

However, some of the energy for that creation is taking also from the gravity force.

No, it isn't. Gravity does not create net energy.

So, the SMBH's gravity force contributes some new energy to our Universe.

No, it doesn't.

In any case, new energy must be created by the SMBH – For sure

Not if the law of conservation of energy has anything to say about it.

That SMBH's extra new energy is used to create the mass in the accretion disc and then the SMBH is using some of this new mass to form stars planets and moons.

Except that it doesn't. Again, conservation of energy...
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/10/2019 05:21:46
Then let's start off by considering each sphere in isolation. Now we have two separate objects that are not connected to each other. The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 246 newtons.

The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 164.7 newtons. The sum total of the forces acting on both spheres is therefore 410.7 newtons.

Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters). If your reasoning is correct, this now-singular object is suddenly and magically feeling 12.1 newtons of force less than it did when the objects were separate (because you claim that these two tethered spheres should feel the same force as a single sphere of identical mass and center of mass). How can the mere addition of a string cause the Earth to pull on the spheres less than it did before?

The answer is that it can't: strings don't have magical antigravity properties. We can therefore safely conclude that your reasoning is incorrect.
Thanks for that explanation.
Your calculation is correct by 100%.
In order to prove it, please use the Moon/Earth/Sun system.
Please calculate the gravity force between the Moon/Sun and Earth/Sun systems.
Please use your following idea to set one common center of mass to the Earth moon system:
"Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres."
Make the gravity forces calculations for the two scenarios. I promise you that you would be surprised to find that the results are almost identical to the systems that you have offered!!!
Therefore, your final outcome is incorrect.

Quote
We take something - gravity force, convert it to infinite something by Taylor series and then use finite something just in order to prove other something - Tidal negative impact on total rotation energy.
Then you must think calculus itself is wrong, since it also involves infinite sums.
Well, the issue is very simple.
Taylor series is an excellent tool to get close to the solution.
Theoretically, we could take a square wave and convert it to infinite no of sign waves.
However, as we can't use in reality infinite no of waves, somehow there must be a small error in our calculation.
Let's assume that this Taylor series represents the reality by 99%.
That by itself is an excellent estimation. However, there is still a delta or error of 1%.
In that formula, our scientists actually are focusing on that delta.
So, they take the delta and set a new formula for gravity which shows the connection between the tidal to the delta.
This was their severe mistake.



Quote
However, some of the energy for that creation is taking also from the gravity force.
No, it isn't. Gravity does not create net energy.

Well, I thought about it in the last few hours and I have found a solution.
 I Fully agree that the new created pair particles energy must be taken from the SMBH's energy by magnetic field
All the 100% energy!.
That is clear to all of us.
However, Gravity force still set its contribution.
The SMBH's mighty gravity force accelerates those new pair creation into the speed of light.
So, that acceleration comes for free due to the SMBH's mighty gravity force.
That acceleration sets the inwards particles rings in the accretion disc at an orbital velocity of almost the speed of light.
That ultra high orbital velocity of the particles/plasma in the accretion disc is transformed back to the SMBH's and increases its internal rotation.
That activity compensates the degradation of its rotation due to the tidal forces.
So, now we have found an excellent solution for the new energy cycle as follow:
Gravity set the Tidal force in the SMBH.
The Tidal force increases the Energy in the SMBH, and decreases its rotation velocity.
Some of that Tidal heat/energy is transformed by the magnetic field to create new pair particles.

However, each new particle get a present from the SMBH - an orbital velocity at the speed of light.
All the antiparticles fall into the SMBH, increasing its mass and its orbital velocity.
The other particles are drifted outwards into the accretion disc.
Their ultra high velocity in the accretion disc - is also transformed back into the SMBH and sets some more compensation in its rotation velocity.

In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy/mass in our Universe.
 
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 12/10/2019 06:24:18
Make the gravity forces calculations for the two scenarios. I promise you that you would be surprised to find that the results are almost identical to the systems that you have offered!!!

You realize that would make me right, not wrong, don't you?

Attraction between the Earth and Sun:

F = (GMm)/r2
F = ((6.674 x 10−11(5.97237 x 1024)(1.9885 x 1030))(149,598,023,000)2)
F = (7.926080939013 x 1044)/(2.2379568485508529 x 1022)
F = 3.5418308070902173002872574484922 x 1022 newtons

Attraction between the Moon and Sun:

F = (GMm)/r2
F = ((6.674 x 10−11(7.342 x 1022)(1.9885 x 1030))(149,982,422,000)2)
F = (9.7437510158 x 1042)/(2.2494726908986084 x 1022
F = 4.3315711523075275820924842308266 x 1020

Sum of the two forces: 3.5851465186132925761081822908005 x 1022 newtons

Attraction between the Sun and a hypothetical Earth-Moon combination object:

F = (GMm)/r2
F = ((6.674 x 10−11(6.04579 x 1024)(1.9885 x 1030))(149,602,691,137)2)
F = (8.023518449171 x 1044)/(2.2380965195362677778850495607477 x 1022)
F = 3.5849742757445814209639045480079 x 1022 newtons.

Those two values for force are different from each other by about 1.722 x 1018 newtons. This, of course, means that I was right and you don't get the right answer when you try to calculate the gravitational force of a non-spherical object with an equation that was made for spherical and point-like objects.

Taylor series is an excellent tool to get close to the solution.

Then what was that nonsense you were saying when you called it "mathematical fiction"?

Let's assume that this Taylor series represents the reality by 99%.
That by itself is an excellent estimation. However, there is still a delta or error of 1%.
In that formula, our scientists actually are focusing on that delta.
So, they take the delta and set a new formula for gravity which shows the connection between the tidal to the delta.
This was their severe mistake.

If the calculation for zenith was off by about 1%, then the calculation for nadir would also be off by about 1%. So a difference between the forces at nadir and zenith would still show up (especially if the calculated difference was much higher than 1%, such as is the case for the Earth, which has a tidal bulge difference of about 5%).

The SMBH's mighty gravity force accelerates those new pair creation into the speed of light.
So, that acceleration comes for free due to the SMBH's mighty gravity force.

There are two scenarios where you would get gravitational acceleration: (1) where both particles are falling into the black hole, or (2) where the particles are using a kind of gravitational slingshot effect to boost their speed. In the first scenario, obviously the particles aren't leaving the black hole. In the second scenario, the energy gained by the particles from the gravitational slingshot is subtracted from the hole's total energy. So the hole still loses energy and therefore mass.

That ultra high orbital velocity of the particles/plasma in the accretion disc is transformed back to the SMBH's and increases its internal rotation.

If the rotation of the disk is spinning up the black hole, then the rotational energy of the disk must decrease as a result. Energy is being transferred, not created.

The ultra high velocity of the new created particles in the accretion disc - is transformed back into the SMBH and sets the compensation in its rotation velocity.

No. It. Doesn't.

Do the math. The black hole can't get back more energy back than it gave out in the first place. Again, do you know what simple addition and subtraction are?

In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.

No it doesn't. Conservation of energy won't let you. Give it up already. What part of "energy cannot be created or destroyed" do you not understand?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/10/2019 05:56:01
Quote
In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.

No it doesn't. Conservation of energy won't let you. Give it up already. What part of "energy cannot be created or destroyed" do you not understand?
OK
My job is to prove that the SMBH's gravity force contributes more energy than the energy that is needed to create those new particle pairs.

Let's set the calculation and verify the real contribution of the mighty SMBH's Gravity force to the total energy.

Let's start with the Virtual particle pair that orbit at almost the speed of light.
Before the creation they have no mass and therefore they have no Kinetic Energy.
At the moment of the creation, they actually get all their mass energy from the SMBH (by the magnetic field).
However, they also get kinetic energy from the SMBH's gravity force.
In other words-
The SMBH's internal energy is transformed into real particle and antiparticle, while the Kinetic energy for those particles are contributed by the SMBH's gravity force.
So, the calculation should be as follow:
M - Represents the mass of the particle or Antiparticle
Eout - The total energy that the SMBH have lost in order to generate those particle and Antiparticle air
Eout = 2M
Ek - The kinetic energy of each particle at the moment of creation. As their orbital velocity is almost the speed of light then:
Ek= 1/2 M c^2
Ep - The potential energy of a particle at the moment of creation
Er = The Total rotation energy of the particle at the moment of creation:
Er = Ek + Ep
The Ep is transformed into kinetic energy as the antiparticle is falling into the SMBH.
Therefore, if the orbital velocity of the antiparticle at the moment of creation is the speed of light, than as it drifts inwards to the SMBH than theoretically, its orbital velocity should be higher than the speed of light.
However, as it gets below the event horizon, even at that Ultra high velocity, it can't go out any more.
However, at the moment of the collision between the antiparticle to the SMBH, all the total rotation energy is transformed into new Energy in the SMBH.

The SMBH creates two particles (Particle + Antiparticle). However, only the antiparticle is falling in.
So, we need to verify if its total rotation energy is greater that the energy that was needed to create the pair.
Therefore, we need to prove that:
Er is greater than Eout
Er = Ek + Ep
For this simple calculation lets ignore the impact of the Ep
So, let's assume that:
Er = Ek = 1/2 M c^2 
Eout = 2M
Er = n * Eout
n = represents how bigger is Er from Eout
Er = 1/2 M c^2 = n 2M
n = 1/4 c^2
Conclusions:
For any energy that the SMBH contributes to create the Pair particle, it gets back at least 1/4c^2 times the "invested" energy.
This does not include:
1. The extra mass of the in falling antiparticle
2. It's Potential energy
3. The particle that is drifted outwards to the accretion disc
4. The transformation of the energy from the accretion disc back to the SMBH.
Therefore, I have proved that only the Kinetic energy of the in falling antiparticle contributes much more energy that was needed to set the whole creation activity.


You realize that would make me right, not wrong, don't you?

Attraction between the Earth and Sun:

F = (GMm)/r2
F = ((6.674 x 10−11(5.97237 x 1024)(1.9885 x 1030))(149,598,023,000)2)
F = (7.926080939013 x 1044)/(2.2379568485508529 x 1022)
F = 3.5418308070902173002872574484922 x 1022 newtons

Attraction between the Moon and Sun:

F = (GMm)/r2
F = ((6.674 x 10−11(7.342 x 1022)(1.9885 x 1030))(149,982,422,000)2)
F = (9.7437510158 x 1042)/(2.2494726908986084 x 1022
F = 4.3315711523075275820924842308266 x 1020

Sum of the two forces: 3.5851465186132925761081822908005 x 1022 newtons

Attraction between the Sun and a hypothetical Earth-Moon combination object:

F = (GMm)/r2
F = ((6.674 x 10−11(6.04579 x 1024)(1.9885 x 1030))(149,602,691,137)2)
F = (8.023518449171 x 1044)/(2.2380965195362677778850495607477 x 1022)
F = 3.5849742757445814209639045480079 x 1022 newtons.
It is totally a diffrent senario from the one that you have used.
In this case, you have set all the objects at the same radius of 149,602,691,137.
However, in  your first example you have set one object at minimal distance of 9,000,000 m
The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms.
While the other at 11,000,000 m:
The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms.
After merging the two objects, their center of mass had been set at 10,000,000 m:
Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters).
I can promise you that if in your example you have used the same radius for all objects - you would probably get fully balanced forces.
Therefore, please use the minimal and maximal distance/radius from the Earth Moon to the Sun (maximal tidal) and then set your recalculation.


Quote
Taylor series is an excellent tool to get close to the solution.
Then what was that nonsense you were saying when you called it "mathematical fiction"?
Mathematical fiction is a direct outcome if we use the error/delta in the Taylor series to prove something which is totally based on that error/delta.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 14/10/2019 07:42:24
My job is to prove that the SMBH's gravity force contributes more energy than the energy that is needed to create those new particle pairs.

Gravity doesn't contribute any energy. All it can do is transfer or transform energy that already exists.

However, they also get kinetic energy from the SMBH's gravity force.

Every single bit of that kinetic energy will come from that black hole's rotation, reducing the black hole's rotation (and therefore its total energy) by the same amount.

Therefore, we need to prove that:
Er is greater than Eout

So you're trying to prove one of the laws of physics wrong despite the fact that you have said many times earlier that you accept the laws of physics...

For this simple calculation lets ignore the impact of the Ep

If you do that, you'll get the wrong answer. The gravitational potential energy of a particle near a super massive black hole is nowhere close to zero.

Er = Ek = 1/2 M c^2

This is wrong for two reasons. Firstly, you can't use the Newtonian kinetic energy equation when considering relativistic velocities. Secondly, particles with a non-zero rest mass like electrons and positrons can't move at the speed of light no matter how much kinetic energy you give them. If you are talking about electrons and positrons moving near the speed of light, then you have to use the relativistic kinetic energy equation: Ek = (mc2)/√(1-(v2/c2)), where,

Ek is the energy in joules
m is the mass in kilograms
v is the velocity of the particle in meters per second, and
c is the velocity of light in meters per second

So your math is going to be wrong until you fix this problem.

Eout = 2M

This will have to include the particles' potential energy, which means you can't ignore it in the calculation.

Therefore, I have proved that only the Kinetic energy of the in falling antiparticle contributes much more energy that was needed to set the whole creation activity.

All you have proven is that you don't know how to use the correct equations. Gravitational potential energy contributes to the total energy. All the infalling particle is doing is converting potential energy that it already had into kinetic energy as it falls into the hole. It isn't giving the hole any more energy than it had to begin with. The particles don't get kinetic energy out of nowhere. They are draining it from the black hole's spin. Any kinetic energy gained by the outgoing particle is lost by the black hole. Again, do you not know what "energy cannot be created or destroyed" means?

In this case, you have set all the objects at the same radius of 149,602,691,137.

Look again. I have the Earth set at a distance of 149,598,023,000 meters from the Sun, the Moon is set at a distance of 149,982,422,000 meters from the Sun and the Earth-Moon barycenter at 149,602,691,137 meters from the Sun. All three of those are different numbers.

I can promise you that if in your example you have used the same radius for all objects - you would probably get fully balanced forces.
Therefore, please use the minimal and maximal distance/radius from the Earth Moon to the Sun (maximal tidal) and then set your recalculation.

I already did put the proper distances into the equations. You probably just saw the "149" and assumed that the rest of the number was the same in all three cases, didn't you?

Mathematical fiction is a direct outcome if we use the error/delta in the Taylor series to prove something which is totally based on that error/delta.

When was the Taylor series ever proving anything by using an error? Can you find me a source that states this?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/10/2019 03:52:59
Gravity doesn't contribute any energy. All it can do is transfer or transform energy that already exists.
Every single bit of that kinetic energy will come from that black hole's rotation, reducing the black hole's rotation (and therefore its total energy) by the same amount.
Do you mean that if the black hole's rotation is zero, than its gravity force should also be zero?
Do you really mean the gravity force is only some sort of energy transformation?
If so, why Newton didn't call it Gravity Field instead of Gravity Force?
You have to agree that once an object have lost its rotation energy, than it also should lose its gravity force.
How could it be?
It is a severe contradiction to the Newton Gravity Law.
Newton didn't set any connection between the gravity force to the rotation energy.
If there is a formula that connects the rotation energy to gravity force - Would you kindly offer it?
We all know that a function of energy is force times distance.
Therefore, Gravity force by itself can generate energy.
That energy is a direct outcome of the force. Therefore, the force is not a transformation. It generates energy based on Newton gravity force/Law.
Let's take the moon as an example.
Its face is looked with the earth.
So, it lost its ability to rotate.
Can we claim that the moon has no ability to converts its gravity force into energy just because it does not rotate any more?
If an object will come closer to the moon, its movement will not be affected by the Moon's gravity force?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/10/2019 06:02:22
Do you mean that if the black hole's rotation is zero, than its gravity force should also be zero?

No. The fact of the matter is that you've been arguing that a black hole's rotation is what is creating the particles since at least post #809 and I agreed that such a thing can work and that it was even mentioned in theory a long, long time ago. Rotating holes radiate particles. Tidal forces and gravity are not the source of energy, rotation is.

Do you really mean the gravity force is only some sort of energy transformation?

No, what I'm saying is that energy transformation is all that gravity can do. It can't create energy.

If so, why Newton didn't call it Gravity Field instead of Gravity Force?

Gravitational fields can produce a force. They are not mutually exclusive.

You have to agree that once an object have lost its rotation energy, than it also should lose its gravity force.
How could it be?

It will not lose all of its gravity. The rotation of an object contains kinetic energy. Due to mass-energy equivalence, that energy also has a mass. That resulting mass increases the strength of the gravitational field. This additional gravity will be extremely tiny compared to the gravity produced by the object's rest mass in most cases (like for planets and stars). In the case of a black hole with the maximum possible rotation rate, however, 29% of its mass-energy will be contained in that rotation (which is fairly significant). So if some of the rotation (and therefore energy) of the hole is extracted, the overall energy content (and therefore mass) of the hole is reduced.

It is a severe contradiction to the Newton Gravity Law.
Newton didn't set any connection between the gravity force to the rotation energy.
If there is a formula that connects the rotation energy to gravity force - Would you kindly offer it?

I never said that an object's entire gravitational field is generated by rotation, so this is a straw-man.

We all know that a function of energy is force times distance.
Therefore, Gravity force by itself can generate energy.

No it can't. Energy can't be created or destroyed. All "force times distance" is telling you in a gravitational field is how much of that gravitational potential energy has been transformed into kinetic energy (if the object is falling) or how much kinetic energy is being transformed into gravitational potential energy (if it is being lifted against gravity). The total energy is unchanged.

That energy is a direct outcome of the force.

Nonsense. Force cannot create energy.

Therefore, the force is not a transformation.

Force is not a transformation, but it causes transformations of energy.

It generates energy based on Newton gravity force/Law.

If that was true, then we would have recognized long ago that conservation of energy was wrong. But that's not true. Force doesn't generate energy. It only changes its form.

Can we claim that the moon has no ability to converts its gravity force into energy just because it does not rotate any more?

It never had that ability in the first place. All the Moon's gravitational field can do is transform existing energy from one form to another.

If an object will come closer to the moon, its movement will not be affected by the Moon's gravity force?

It will be affected, but the object's total energy will be unchanged. All that is happening is that some of the object's gravitational potential energy is being converted into kinetic energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/10/2019 20:30:07
Tidal forces and gravity are not the source of energy, rotation is.
Sorry
I see it differently.
Gravity force and tidal creates energy.
Rotation is a direct outcome of gravity force.
Newton has set the gravity force as:
F=GMm/r^2
The Kinetic energy is directly affected by the orbital velocity.
While that orbital velocity is fully affected by the gravity force.
F = M V^2 / r
So, Newton gravity force which had been set ONLY by mass and radius (and the constant G), has a direct impact on the orbital velocity and that orbital velocity sets the Kinetic orbital energy.
Therefore, The kinetic energy is a direct outcome of the gravity force and not the other way.
Same issue with the Potential kinetic
Ep = m G h
So, even the potential energy is a direct outcome of mass and distance (or radius). As the gravity force is a direct outcome of mass and radius - the potential energy is also an outcome of gravity force.

No, what I'm saying is that energy transformation is all that gravity can do. It can't create energy.
I disagree
Gravity force can create new energy
Quote
You have to agree that once an object has lost its rotation energy, than it also should lose its gravity force.
How could it be?
It will not lose all of its gravity. The rotation of an object contains kinetic energy. Due to mass-energy equivalence, that energy also has a mass. That resulting mass increases the strength of the gravitational field. This additional gravity will be extremely tiny compared to the gravity produced by the object's rest mass in most cases (like for planets and stars). In the case of a black hole with the maximum possible rotation rate, however, 29% of its mass-energy will be contained in that rotation (which is fairly significant). So if some of the rotation (and therefore energy) of the hole is extracted, the overall energy content (and therefore mass) of the hole is reduced.
How did you get the 29% for the BH?
What should be the gravity lost in the Moon which have already lost its rotate energy?
Please show the formula which links the rotate energy to gravity force.
I never said that an object's entire gravitational field is generated by rotation,
Again - Please show the formula which links between the two.
If that was true, then we would have recognized long ago that conservation of energy was wrong. But that's not true. Force doesn't generate energy. It only changes its form.
It seems to me that you are using the conservation of energy at the wrong place.
I have already discussed about Newton formula that shows that Gravity generates energy.

 
It will be affected, but the object's total energy will be unchanged. All that is happening is that some of the object's gravitational potential energy is being converted into kinetic energy.
If I understand you correctly, you claim that the total rotation energy (Kinetic + potential) is fixed.
So, the gravity just transfers the energy from one to the other.
This is incorrect.
Gravity works locally.
So, if we take an object to the infinity - we can also claim that its potential energy is infinite.
Just based on this idea we can claim that the energy in our Universe is infinity.
In reality - if it is far enough, it will not be affected by the gravity force any more.
So, although the potential energy should be infinite, the gravity force between those two objects at a distance of infinite is virtually zero and therefore there is no meaning for that Potential energy.
On the other way, I could claim that the virtual pair are coming from the infinity.
So, at the moment of creation, their infinite potential energy had been transformed into Kinetic energy.
I assume that if you give the possibility to transform energy between kinetic and potential than you should also agree with that sort of idea.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/10/2019 22:29:09
Sorry
I see it differently.

Then you disagree with the laws of physics.

Gravity force and tidal creates energy.

Not according to the law of conservation of energy, it doesn't. Do you think you are some kind of higher authority than the laws of physics themselves?

Rotation is a direct outcome of gravity force.

Gravity isn't needed at all for rotation. Take a CD player, for example.

The Kinetic energy is directly affected by the orbital velocity.

Kinetic energy isn't merely "affected" by velocity, it is what causes the velocity to be what it is. Movement of any kind requires kinetic energy.

So, Newton gravity force which had been set ONLY by mass and radius (and the constant G), has a direct impact on the orbital velocity and that orbital velocity sets the Kinetic orbital energy.

And that kinetic energy had to come from somewhere. In the case of an asteroid captured in orbit around a planet, that kinetic energy came in part from the gravitational potential energy of the asteroid and partly from the kinetic energy that the asteroid already had.

Therefore, The kinetic energy is a direct outcome of the gravity force and not the other way.

You are right in the sense that gravity is converting gravitational potential energy into kinetic energy.

So, even the potential energy is a direct outcome of mass and distance (or radius). As the gravity force is a direct outcome of mass and radius - the potential energy is also an outcome of gravity force.

Yes, but gravity is not a source of unlimited potential energy. The amount of potential energy that an object has in a gravitational field is finite and any attempt to increase that potential energy requires an input of energy from an outside source. Raising an object against gravity, for example, increases its gravitational potential energy. But that rise in potential energy had to come from somewhere. In the case of a rocket, that energy came from the rocket engines.

I disagree
Gravity force can create new energy

Then you disagree with the law of conservation of energy. I'm still waiting for you to cite some authoritative source that agrees with you that gravity does not obey conservation of energy. The only evidence you have provided so far is  your own misunderstanding of how gravity works. That, of course, is not evidence at all.

How did you get the 29% for the BH?

Look at page 29: http://www.phys.unm.edu/~gbtaylor/astr421/lectures/24_A421_BlackHolesnew.pdf

What should be the gravity lost in the Moon which have already lost its rotate energy?

That would depend upon how fast it was rotating in the past.

Please show the formula which links the rotate energy to gravity force.

First, you calculate the mass equivalent of the rotational kinetic energy using E=mc2 (or rather, the rearranged version: m = E/c2). Then you put that resulting mass into the standard gravitational attraction equation if what you have is a spherical body (as well as the mass and distance of whatever object is being affected by the field): F = G((m1m2)/r2). That will tell you how much gravitational force is contributed by the rotational energy.

Again - Please show the formula which links between the two.

See above.

It seems to me that you are using the conservation of energy at the wrong place.

There is no such thing as "the wrong place" for conservation of energy.

I have already discussed about Newton formula that shows that Gravity generates energy.

If you think that's correct, then give a link to an authoritative source that agrees with you that gravity is an exception to the conservation of energy. If you are correct, then that shouldn't be difficult, given how incredibly important such a discovery would be.

If I understand you correctly, you claim that the total rotation energy (Kinetic + potential) is fixed.

You don't understand correctly. Rotation can slow down (but only if that rotational energy is transferred elsewhere. It can't disappear into nothingness due to conservation of both energy and momentum).

So, if we take an object to the infinity - we can also claim that its potential energy is infinite.

No you can't. The potential energy at infinity is limited. This is because gravity becomes weaker and weaker as you get further away from its source. In calculus, this is called approaching a limit. The fact that gravitational potential energy is finite at infinity is the basis for the concept of the escape velocity. An object traveling beyond escape velocity has more kinetic energy than the same object's gravitational potential energy at infinity, which means that it can keep moving away from a source of gravity forever without ever being slowed down to zero speed.

The energy required to move an object against a gravitational field out to infinity is (GMm)/r, where G is the gravitational constant, M is the mass of the planet, m is the mass of the object and r is the radius of the planet. For the Earth, moving a 1 kilogram mass out to infinity would require ((6.674 x 10−11)(5.97237 x 1024)(1))/(6,371,000) = 62,564,114.55 joules of energy. That is very much a finite number: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/gravpe.html

Just based on this idea we can claim that the energy in our Universe is infinity.

No, we can't.

In reality - if it is far enough, it will not be affected by the gravity force any more.

This is also wrong. Gravity is a force of infinite range. It becomes increasingly weak with distance, but the force never falls all the way to zero.

So, although the potential energy should be infinite

It isn't.

the gravity force between those two objects at a distance of infinite is virtually zero

At infinity, the force would be literally zero. But that's only a mathematical consequence, since it's impossible to move any object an infinite distance away from any other object in a finite span of time.

and therefore there is no meaning for that Potential energy.

Since the potential energy actually has a finite value, it still does have meaning.

On the other way, I could claim that the virtual pair are coming from the infinity.

No you can't. That wouldn't make any sense.

So, at the moment of creation, their infinite potential energy had been transformed into Kinetic energy.

They don't have infinite potential energy.

I assume that if you give the possibility to transform energy between kinetic and potential than you should also agree with that sort of idea.

Absolutely not. Potential energy never becomes infinite, no matter what the distance.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/10/2019 20:08:18
Not according to the law of conservation of energy, it doesn't. Do you think you are some kind of higher authority than the laws of physics themselves?
You are right in the sense that gravity is converting gravitational potential energy into kinetic energy.
OK

Let's verify the idea of conservation of energy:
Ek = 1/2 m V^2 
V^2 = G M /r
Ek = 1/2 G M m /r
Ep = M G r
Et = Ek + Ep = 1/2 G M m / r + M G r
If r = 2R
Et = 1/2 G M m / 2R + M G 2 R
In other words by increasing the radius by twice, we decrease the Ek by twice and increase the Ep by twice.
Therefore, as Ek is not equal to Ep any change in the radius also set a change in Et.
If the total rotation energy is changing due to radius change - where do you get the idea of law of conservation of energy in gravity force???
Do you see any error in this simple example?

In any case, let me remind you that we have started this discussion as I have stated that there is no energy transformation in Gravity.
I have used the moon example -
It had totally lost its rotate ability as it is fully locked with the Earth.
Even so, any object that has a wish to orbit around the moon must fully obey to the same Ep, Ek and Et as I have used.
So, with the ability to rotate/spin or without it, the outcome due to gravity force is the same.
Therefore, there is no need for transformation of energy from the rotate/spin energy into the orbital rotation energy and vice versa.
This idea is a fiction and I have just proved it!
How those scientists that set laws of physics can't get that simple outcome?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/10/2019 22:28:20
Therefore, as Ek is not equal to Ep any change in the radius also set a change in Et.

Right, but the radius doesn't change spontaneously. Energy either has to be added to the orbit or lost from it in order for such a radius change to occur. So it still obeys conservation of energy.

If the total rotation energy is changing due to radius change - where do you get the idea of law of conservation of energy in gravity force???

If you are referring to the rotational energy of the black hole being reduced by raising the particles into a higher orbit, there is no challenge to the law of conservation of energy there. An increase in the orbital radius requires a net input of energy, and that required amount of energy comes from the black hole's spin, which is reduced in the process. The total energy remains the same both before and after the orbit change.

Do you see any error in this simple example?

Yes, your error is in thinking that orbits can change spontaneously without any input of energy from an outside source or loss of energy to an outside source.

In any case, let me remind you that we have started this discussion as I have stated that there is no energy transformation in Gravity.

Gravity changes potential energy into kinetic energy, so this argument is objectively wrong. All you have to do is drop something to test that idea.

I have used the moon example -
It had totally lost its rotate ability as it is fully locked with the Earth.

The Moon is still rotating, it just takes the same amount of time to complete one rotation as it takes to complete one orbit around the Earth.

Even so, any object that has a wish to orbit around the moon must fully obey to the same Ep, Ek and Et as I have used.
So, with the ability to rotate/spin or without it, the outcome due to gravity force is the same.

Technically, it isn't, since the value of "m" changes very slightly over time when a satellite's rotation slows down. The value of "m" also changes very slightly when its orbit changes, since orbital energy also has an associated mass due to E=mc2. In most cases, that extra mass is negligible when compared to the total mass. According to this site, the rotational kinetic energy of the Earth is 2.138 x 1029: https://courses.lumenlearning.com/boundless-physics/chapter/rotational-kinetic-energy/

Dividing that by the speed of light squared (299,792,4582 = 89,875,517,873,681,764) yields an equivalent mass of 2.3788 x 1012 kilograms. That sounds like a lot, but the Earth's total mass is 5.97237 x 1024 kilograms. So the mass of the Earth's rotational kinetic energy accounts for a mere 0.0000000000419% of the Earth's total mass. You'd never notice it was gone if the Earth stopped spinning.

Therefore, there is no need for transformation of energy from the rotate/spin energy into the orbital rotation energy and vice versa.

Non-sequitur.

This idea is a fiction and I have just proved it!

All you have proven is that you don't know how gravity, orbits and energy works.

How those scientists that set laws of physics can't get that simple outcome?

First of all, scientists don't "set" the laws of physics: they discover them. Secondly, they didn't get your outcome because they actually know what they are doing instead of making fallacious, misinformed assumptions.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/10/2019 16:11:51
Right, but the radius doesn't change spontaneously. Energy either has to be added to the orbit or lost from it in order for such a radius change to occur.
Thanks
So, you agree that as the object is changing its orbital radius, "Energy either has to be added to the orbit or lost from it"
That is a key element. It proves that there is no conservation of energy in the total rotation energy
So, how can you claim that:
So it still obeys conservation of energy.
If you mean that there is a conservation of energy after adding new energy - that is fully clear.
But again - There is no conservation of energy in the total rotation energy!
An increase in the orbital radius requires a net input of energy,
So, we fully agree that once you change the radius you need to add energy or decrease energy.
The total energy remains the same both before and after the orbit change.
If you add or decrease an energy to or from the total energy - than how can you still think that the total energy remains the same?
Sorry - this is a fatal Mistake!
and that required amount of energy comes from the black hole's spin
So, you think that the required amount of energy comes from the black hole's spin.
Do you agree that based on this idea, in order to decrease an energy of the total rotation energy, the black hole's spin rotation should be increased?
In any case, you claim for a direct linkage between the Total rotation energy of the orbital object around the main body, to the spin of the main body.
However, I have also proved that the Moon has no ability to transfer its spin energy.
The Moon is still rotating, it just takes the same amount of time to complete one rotation as it takes to complete one orbit around the Earth.
That is correct.
The moon is still rotating and it just takes the same amount of time to complete one rotation as it takes to complete one orbit around the Earth.
HOWEVER - that spin velocity is FIXED!!!
If you set there one billion objects that orbits clockwise or the other side - do you agree that all of them will fully obey to Newton law whithout any ability for the moon to transfer even one bit of its spin energy (especially not to the one of the orbital direction). So the moon's spin would continue to rotate at the same fixed velocity while all the Billion objects will fully obey to Newton law and orbital velocity whithout getting any extra energy from the Moon's spin.
Hence, there is virtually no ability for any real energy transformation from the Moon spin to the requested orbital objects.
Therefore, If the moon doesn't transfer any energy from its spin velocity to any orbital object (while it's Kinetic energy + potential energy will fully obey to Newton law) - could it be that even the SMBH won't need to transfer any energy from its spin velocity?
Please - So far you couldn't offer any real formula (By Newton or Einstein) that directly links the spin energy/velocity of the main body to the orbital energy of the orbital object.
If you have any idea about a linkage between the two, would you kindly show in that formula why it doesn't work for the Moon?


I have a question for you:
Why are you so sure that "an increase in the orbital radius requires a net input of energy"?
Actually the Kinetic Energy is decreasing as we increase the Radius:
Ek = 1/2 m V^2 = 1/2 G M m /r
In the same token, the gravity force is also decreasing (even more dramatically):
F = G M m / r^2
So, less Kinetic energy and less Gravity force is needed to maintain the orbital momentum as the radius is increasing.
Therefore, an increase in the orbital radius requires less and less input of energy.
On the other hand, we can claim that as we increase the radius, we also increase the potential energy.
Ep = m G r
However, it seems to me that there is a rang limit for that potential energy.
If the objects are quite close to each other, than the gravity force is strong enough and therefore - the Potential energy has a great impact.
In this case we can clearly claim that:
"an increase in the orbital radius requires a net input of energy"
For example - at the surface of the Earth - we clearly know that " an increase in the orbital radius requires a net input of energy"
However, if the object is too far away from the Earth or even from the sun, the gravity force is decreasing dramatically and I wonder what is the real meaning of the potential energy while the gravity force is almost gone to zero.
For Example - The Oort cloud around the solar system.
I'm not fully sure about it, but it seems to me that for any orbital system up to a certain radius - we can claim for sure that:  "an increase in the orbital radius requires a net input of energy.
However, from that certain radius - "an increase in the orbital radius does not require any input of energy".
We know that in the past that matter in the oort cloud was part of the mass that have set the Solar system.
Now, we know for sure that the Oort cloud is drifting outwards constantly.
So, without any need for external energy - those objects at Oort cloud are increasing their radius over time.
Hence, the main questions are?
1. Is it a Normal activity at gravity that too far away orbital objects are drifting outwards over time?
2. Could it be that the gravity force is decreasing over time for orbital objects that are located too far away?
3. How can we find that radius range that converts the " an increase in the orbital radius requires a net input of energy" to  "an increase in the orbital radius over time requires no net input of energy - It is a normal activity of Gravity force"?
4. Could it be that this range is set by finding the radius when The Kinetic energy is equal to potential energy?
In other words - if Kinetic energy is greater than Potential energy -  an increase in the orbital radius requires a net input of energy, however if the potential energy is greater that the kinetic energy than " an increase in the orbital radius over time requires no net input of energy"?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/10/2019 17:24:01
Thanks
So, you agree that as the object is changing its orbital radius, "Energy either has to be added to the orbit or lost from it"
That is a key element. It proves that there is no conservation of energy in the total rotation energy

You have it backwards. An object won't change its orbital radius unless it is gaining energy from an outside source or losing it to an outside source. A spacecraft won't spontaneously increase its orbital radius from the Earth, for example. If the astronauts want to increase the orbital radius, they will have to add energy by firing the engines.

If you mean that there is a conservation of energy after adding new energy - that is fully clear.
But again - There is no conservation of energy in the total rotation energy!

Since energy is neither created nor destroyed, yes there is.

So, we fully agree that once you change the radius you need to add energy or decrease energy.

Again, you have it backwards. Adding energy from an outside source will increase the orbital radius. You can't say, "I want the  orbital radius to increase, so the energy required to do that pops up out of nowhere". The radius won't increase unless the object in orbit is supplied with energy from another source that already existed. Look at my example with the spaceship.

If you add or decrease an energy to or from the total energy - than how can you still think that the total energy remains the same?

Because the energy lost by the black hole's spin is now present in the particles it created. The total energy (black hole + particles) is the same.

Do you agree that based on this idea, in order to decrease an energy of the total rotation energy, the black hole's spin rotation should be increased?

That is the exact opposite of what would happen. Decreasing the black hole's rotational kinetic energy causes the spin to be slowed down.

In any case, you claim for a direct linkage between the Total rotation energy of the orbital object around the main body, to the spin of the main body.

That depends on what you're talking about. When you say "total rotation energy" are you talking about the total orbital energy or what?

If you set there one billion objects that orbits clockwise or the other side - do you agree that all of them will fully obey to Newton law whithout any ability for the moon to transfer even one bit of its spin energy (especially not to the one of the orbital direction). So the moon's spin would continue to rotate at the same fixed velocity while all the Billion objects will fully obey to Newton law and orbital velocity whithout getting any extra energy from the Moon's spin.

You would actually expect the opposite to occur. Instead of the objects in orbit being given energy by the Moon's spin, you'd expect them to transfer energy to the Moon and spin it up instead. That's because they would be orbiting the Moon faster than the Moon rotates. If the opposite was true and the Moon was spinning faster than the orbit, then you would expect the Moon the slow down while raising the orbital radius of the object

Quote
Hence, there is virtually no ability for any real energy transformation from the Moon spin to the requested orbital objects.
Therefore, If the moon doesn't transfer any energy from its spin velocity to any orbital object (while it's Kinetic energy + potential energy will fully obey to Newton law) - could it be that even the SMBH won't need to transfer any energy from its spin velocity?

This is a completely ridiculous conclusion. You are basically arguing, "Objects that don't spin can't transfer spin energy, therefore spinning objects don't transfer spin energy either." That's like arguing that a car without gas in its tank can't run therefore a car with gas in its tank can't run either. Spin is the very thing that makes these two scenarios different.

Please - So far you couldn't offer any real formula (By Newton or Einstein) that directly links the spin energy/velocity of the main body to the orbital energy of the orbital object.

If you are talking about finding the rate of energy transfer over time, you would need to use multiple equations to figure that out. They are listed on this page: https://en.wikipedia.org/wiki/Tidal_acceleration#Theory

It's a much simpler matter if you are asking how much spin energy is required to increase the orbital radius by a given amount. First you would calculate the increase in total orbital energy when going from the old radius to the new radius (using the orbital kinetic and potential energy equations). Then you subtract that resulting energy difference from the total rotational kinetic energy of the primary body. You can then use the rotational kinetic energy equation to calculate how much the primary body's spin should have slowed down.

I have a question for you:
Why are you so sure that "an increase in the orbital radius requires a net input of energy"?

Because the law of conservation of energy demands it. Energy doesn't pop up out of nowhere.

Therefore, an increase in the orbital radius requires less and less input of energy.

Yes, but that input of energy is still non-zero at all distances. That's the important part.

However, it seems to me that there is a rang limit for that potential energy.

Yes, potential energy reaches its maximum (and finite) value at infinity.

However, if the object is too far away from the Earth or even from the sun, the gravity force is decreasing dramatically and I wonder what is the real meaning of the potential energy while the gravity force is almost gone to zero.

It still has the same meaning. Just because the gravity is weak doesn't mean it isn't there. The math still works.

I'm not fully sure about it, but it seems to me that for any orbital system up to a certain radius - we can claim for sure that:  "an increase in the orbital radius requires a net input of energy.

It works for all radii because gravity has infinite range.

Now, we know for sure that the Oort cloud is drifting outwards constantly.
So, without any need for external energy - those objects at Oort cloud are increasing their radius over time.

Given that the Oort cloud is theoretical at this point, I'm going to have to ask you to cite an authoritative source that supports your assertion that the cloud as a whole is drifting away from the Sun (if it exists).

1. Is it a Normal activity at gravity that too far away orbital objects are drifting outwards over time?

No, although it is easier for such objects to be dislodged from their orbits by even tiny amounts of energy contributed from other sources. Impacts from asteroids or gravitational tugging from other objects in the vicinity are possible sources of such energy.

2. Could it be that the gravity force is decreasing over time for orbital objects that are located too far away?

Due to the gravitational constant, no.

3. How can we find that radius range that converts the " an increase in the orbital radius requires a net input of energy" to  "an increase in the orbital radius over time requires no net input of energy - It is a normal activity of Gravity force"?

You can't, because it doesn't exist.

4. Could it be that this range is set by finding the radius when The Kinetic energy is equal to potential energy?
In other words - if Kinetic energy is greater than Potential energy -  an increase in the orbital radius requires a net input of energy, however if the potential energy is greater that the kinetic energy than " an increase in the orbital radius over time requires no net input of energy"?

Nope.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/10/2019 16:13:02
potential energy reaches its maximum (and finite) value at infinity.
Sorry, I totally disagree with that.
As we increase the radius, we decrease the gravity force by r^2.
However the potential energy is increasing only by r.
So, the decreasing in the gravity force is dramatically higher than the increasing of the potential energy.
If the radius is long enough, the gravity force should go almost to Zero.
At the infinity, the gravity force is virtually ZERO. 
So, what is the meaning of Potential energy at a distance with zero (or virtually zero) gravity force?
Therefore, there is a meaning for potential energy ONLY as long as the gravity force is high enough.
You have set the calculation of the potential energy of the Moon/Earth while the radius is 1000 LY and the outcome was quite impressive Potential energy.
However, at that distance - the gravity force between the Moon/Earth should go down to virtually zero.
So, do you really think that the moon will be affected by the earth gravity (which is virtually zero)?
Therefore, that high potential energy at a distance of 1000 LY is useless in the Erath moon system - as the gravity force goes to almost zero at relatively high radius.
Why our scientists do not add the potential energy when they look at the spiral arm?
If you think that the Earth/moon potential energy has still an impact at long distance of 1000Ly, than why the potential energy of nearby stars (at a range of less than 100 Ly - there are 512 stars at that range) have no impact on the Sun orbital movement?
Please be aware that due to their massive mass and fairly shorter distances - the gravity forces between the nearby stars in the spiral arms are not so neglected.
So, the potential energy should have an impact on nearby stars at the spiral arms.
Why our scientists totally ignore the potential energy when it comes to fairly short distances at the spiral arms (only 100Ly- when gravity force is still quite high) while we claim/hope that potential energy still has an impact even at the infinity (when gravity force is virtually zero)?
Why do you think the moon orbits around the earth instead of the Sun.
We have already found the gravity force of the Sun/moom is more than twice than the Moon/Earth.
Therefore, the only answer for that is:  Gravity Works Locally!

An object won't change its orbital radius unless it is gaining energy from an outside source or losing it to an outside source. A spacecraft won't spontaneously increase its orbital radius from the Earth, for example. If the astronauts want to increase the orbital radius, they will have to add energy by firing the engines.
That's not fully correct
You miss the impact of time.
A spacecraft changes its orbital radius from the Earth over time (without operate the internal rockets).
Normally it falls in. So it reduces its orbital radius over time.
However, that is correct as long as the spacecraft is close enough to the Earth.
Try to set a spacecraft at a significantly further from the Earth - and you would surly find that it should increase its orbital radius over time.
I wonder why our scientists do not add the "over time" impact to the gravity formula.
You would probably claim that it isn't due to due to tidal or spin energy.
I think differently.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 19/10/2019 21:24:20
Sorry, I totally disagree with that.

Then you disagree with the basic mathematics that are universally-accepted enough to appear in school textbooks.

Sorry, I totally disagree with that.
As we increase the radius, we decrease the gravity force by r^2.
However the potential energy is increasing only by r.
So, the decreasing in the gravity force is dramatically higher than the increasing of the potential energy.
If the radius is long enough, the gravity force should go almost to Zero.
At the infinity, the gravity force is virtually ZERO.
So, what is the meaning of Potential energy at a distance with zero (or virtually zero) gravity force?

It has the same meaning as it does at any other distance, only a different value. Remember, infinity is conceptual. It can only be approached, never actually reached.

Therefore, there is a meaning for potential energy ONLY as long as the gravity force is high enough.

Define "high enough".

You have set the calculation of the potential energy of the Moon/Earth while the radius is 1000 LY and the outcome was quite impressive Potential energy.
However, at that distance - the gravity force between the Moon/Earth should go down to virtually zero.
So, do you really think that the moon will be affected by the earth gravity (which is virtually zero)?

Virtually zero is not literally zero, so yes, it will be affected.

Therefore, that high potential energy at a distance of 1000 LY is useless in the Erath moon system - as the gravity force goes to almost zero at relatively high radius.

There is nothing "useless" about it. The problem is that 1,000 light-years is well outside of Earth's Hill sphere, so the Moon could not actually orbit at that distance because there are so many other stars and planets around that would be pulling on the Moon much more strongly than the Earth. If the Universe was empty of all matter except for the Earth and Moon, however, such an orbit could indeed exist because there would no longer be such a Hill sphere.

Why our scientists do not add the potential energy when they look at the spiral arm?

Who said they don't?

If you think that the Earth/moon potential energy has still an impact at long distance of 1000Ly, than why the potential energy of nearby stars (at a range of less than 100 Ly - there are 512 stars at that range) have no impact on the Sun orbital movement?

They do. The Sun is far from sitting still, you know.

Please be aware that due to their massive mass and fairly shorter distances - the gravity forces between the nearby stars in the spiral arms are not so neglected.
So, the potential energy should have an impact on nearby stars at the spiral arms.

Nobody ever said that they should be neglected.

Why our scientists totally ignore the potential energy when it comes to fairly short distances at the spiral arms (only 100Ly- when gravity force is still quite high) while we claim/hope that potential energy still has an impact even at the infinity (when gravity force is virtually zero)?

You're going to have to find some kind of authoritative source supporting your claim that they ignore the potential energy of stars in the spiral arm.

Why do you think the moon orbits around the earth instead of the Sun.

Because the strength of Earth's gravity as felt by the Moon is higher than what it feels from the Sun. That's because distance matters as much as mass does. If the Sun was heavy enough, the Moon would feel it pulling on it more strongly than the Earth does. In that case, the Moon would indeed orbit the Sun.

We have already found the gravity force of the Sun/moom is more than twice than the Moon/Earth.

No, no we have not. What I calculated was gravitational potential energy, not gravitational force. Those are two very different quantities. The force goes down with distance whereas the potential energy goes up with distance.

Therefore, the only answer for that is:  Gravity Works Locally!

No, it doesn't. There is no known limit to gravity's range. The inverse-square law tells you that much. Do you know what the inverse-square law is?

A spacecraft changes its orbital radius from the Earth over time (without operate the internal rockets).
Normally it falls in. So it reduces its orbital radius over time.

That is due to atmospheric drag. If the Earth didn't have an atmosphere, that wouldn't happen.

However, that is correct as long as the spacecraft is close enough to the Earth.

Because of atmospheric drag, right.

Try to set a spacecraft at a significantly further from the Earth - and you would surly find that it should increase its orbital radius over time.

This would absolutely not happen, because it would violate conservation of energy. The gravitational potential energy of the spacecraft cannot spontaneously increase. This is something I have repeated ad nauseum. The only way to increase the gravitational potential energy is via energy input. I can't help but wonder what part of "energy cannot be created or destroyed" you don't understand. It isn't a difficult concept to grasp, so why are you struggling with it?

I wonder why our scientists do not add the "over time" impact to the gravity formula.

You can if you want to, but no amount of waiting will change the total orbital energy until energy is lost or gained by the object in orbit. Again, energy cannot be created or destroyed. You can't win an argument against nature.

You would probably claim that it isn't due to due to tidal or spin energy.

I will argue no such thing, since there is nothing stopping you from adding a time component to the equation. It just won't tell you anything unless energy is being brought into or lost from the system.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/10/2019 12:54:04
Dear Kryptid
Thanks again for your great support!

1. Potential energy/gravity force for nearby stars
Quote
If you think that the Earth/moon potential energy has still an impact at long distance of 1000Ly, than why the potential energy of nearby stars (at a range of less than 100 Ly - there are 512 stars at that range) have no impact on the Sun orbital movement?
They do. The Sun is far from sitting still, you know.
Would you kindly advice what our scientists really understand from that nearby Potential energies/gravity forces?
Let's verify the observations:
At a radius of 100Ly around the sun there are 512 Stars.
That volume represents 8 times the volume of 50Ly.
So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:
512 / 8 = 64 stars per 50Ly.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)
This is not a random activity!!!
A density wave can't generate that kind of fixed density.
The whole idea of density wave is that in some arias there is higher density of stars while in other arias there are less density of stars.
This is not the case in the Orion arm and not in any other spiral arm.
I would love to know the real 50 Ly radius volume densities of stars per distance from the galactic center - in the arm.
Based on my theory, we should find that as we go closer to the galactic center the density is higher.
As we go further away from the galactic center - the density is lower.
However - at any spiral arm - at any distance from the galactic center, the density of stars is FIXED.

Outside the arm - the density MUST be ZERO!
However, the arm might have any kind of shape and there are bridges between the arms.
I can promise you that as we cross the edge of the arm - there will be ZERO stars there!!!
So, at the nearby aria around the sun - the density could be: or 64 per 50Ly radius or ZERO.

There is another issue which is a direct outcome of theory D.
As we go outwards from the galactic center the arm gets thinner and thinner.
If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.
That sets the end point of the spiral arms.
Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.
If they would try to understand the real impact of the Potential energy/gravity force in short range - less than 500 or 1000Ly, they should get to the same conclusions which I have got.
If we could go back in time - let's say 100 M years ago - we would find that all the nearby stars are still there.
So, any star that is moving away from us today - should come back in the future.
While any star that is moving in our direction, should move away in the Future.
How is it possible?
The density wave doesn't give an answer for all of that.
However, the Potential energies and the gravity forces between nearby stars give a perfect solution.
They are all gravity bonded IN THE ARM itself!!!
So please - show me how our scientists translate that nearby potential energy/gravity force into real understanding about the structure of spiral arms.

2. Rotational energy conservation/transformation after the BBT
You claim that stars can't just decrease or increase their orbital radius without external energy.
If that is correct - than how the whole galaxies had been formed after the Big bang while there were no rotational energy at all?
At the early days - 380M years after the Big Bang we have only got matter.
No stars, No galaxies, No orbital objects no main body, no kinetic Energy No rotation energy - only matter everywhere.
So, if we take a cube of one billion LY - in that cube there was no Kinetic energy, no spin energy, no rotation energy - Nothing)
However, Today at the same volume cube there is so high rotational energy. It is everywhere. Try to take the same cube today and calculate the total rotation energy (of all the stars, galaxies) in that cube.
From Zero rotation energy in age of 380 M years we have got so high total rotation energy today.
So, how your idea of rotational energy conservation could be correct if in the early days there was no rotation energy at all???
Please - If you offer a random activity, than why the same idea for random gravity force/Energy that creates the whole stars, planets, moons, BH, SMBH, and even unlimited no of galaxies without any request for early rotational energy transformation/conservation in the Early days, can't work today even for just one pair of particle???
How can anyone believe that somehow in the early days of our universe the whole galaxies have been formed without any need for rotational energy transformation but today even for one pair of particle you insist for rotational energy transformation?
So please - would you kindly show how the whole orbital activities in the early galaxy had been formed out of nothing (no orbital activities after the matter creation - age of 380 MY after the Big bang), while we have to assume that the same rotational energy conservation should works also for those early days.

3. Moon/sun gravity force
Because the strength of Earth's gravity as felt by the Moon is higher than what it feels from the Sun. That's because distance matters as much as mass does. If the Sun was heavy enough, the Moon would feel it pulling on it more strongly than the Earth does. In that case, the Moon would indeed orbit the Sun.
So, you don't agree that the Sun/Moon gravity force is stronger than the earth/moon force?
https://www.universetoday.com/116158/why-doesnt-the-sun-steal-the-moon/
"If you’re up for some napkin calculations, you little mathlete, by using Newton’s law of gravity, you find that even with its greater distance, the Sun pulls on the Moon about twice as hard as the Earth does.
So why can’t the Moon escape the Earth?"
The answer is:
"So, yes, the Sun is totally trying to rip the Moon away from the Earth, but the Earth is super clingy.
The speed of the Moon around the Earth is about 1 km/s. At the Moon’s distance from the Earth, the escape velocity is about 1.2 km/s. The Moon simply isn’t moving fast enough to escape the Earth."
However, I have asked why the Moon had started to orbit around the Earth at its first day?
Our scientists claim:
https://www.space.com/29047-how-moon-formed-earth-collision-theory.html
"The formation of the moon has long remained a mystery, but new studies support the theory that the moon was formed from debris left from a collision between the newborn Earth and a Mars-size rock, with a veneer of meteorites coating both afterward."
I claim that this story is incorrect because:
a. Meteorites can't form new Moon. They only can set some sort of a ring, as we see around Saturn.
b. If one of the meteorite was as big as the moon, it had to follow the gravity force of the Sun at the moment of the impact (as at that moment the gravity force of the sun was stronger than the earth).


4. Could it be that the "Hill sphere" is the answer for why gravity works locally?
There is nothing "useless" about it. The problem is that 1,000 light-years is well outside of Earth's Hill sphere, so the Moon could not actually orbit at that distance because there are so many other stars and planets around that would be pulling on the Moon much more strongly than the Earth. If the Universe was empty of all matter except for the Earth and Moon, however, such an orbit could indeed exist because there would no longer be such a Hill sphere.
Wow
Thanks for that information about the Hill sphere:
https://space.stackexchange.com/questions/3015/how-large-is-the-earths-gravitational-sphere-of-influence-and-how-can-it-be-cal
"The Hill sphere (aka the Roche sphere) looks at things from the perspective of energy rather than force.
The Hill sphere asks a rather different question: Given a smaller body orbiting a larger body, can an even smaller body orbit the small body? "
Hence, as we monitor the rotational energy rather than gravity force we might find the answerfor why gravity works locally and why the moon orbits around the earth instead of orbiting directly around the Sun.

5. "Over time"
I wonder why our scientists do not add the "over time" impact to the gravity formula.
You can if you want to, but no amount of waiting will change the total orbital energy until energy is lost or gained by the object in orbit. Again, energy cannot be created or destroyed. You can't win an argument against nature.
Well, I think differently
Every orbit has a shape of spiral.
In the nature nothing comes back exactly to the same point (same orbital radius).
It might fall in or it might drift out.
Unfortunately, our scientists have missed that key element of "Over time".
I can promise you that by 100% all the stars in the spiral arms are drifting outwards over time.
After every cycle around the galactic center - the sun is drifting outwards.
If we could come back in one billion years from now we might find that our sun had been ejected from the arm and from the galactic disc.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/10/2019 14:50:11
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
There are 133 stars visible to the naked eye within that radius, which is less than 10% of the stars in total.  Map of the 133 here: http://www.icc.dur.ac.uk/~tt/Lectures/Galaxies/LocalGroup/Back/50lys.html
It says there are an estimated 1800 stars in 1300 systems (so much for binary being prevalent).

Your problem is that you get your data from hand-selected google pages instead of where scientists get it: From looking at the stars.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/10/2019 15:22:12

Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
There are 133 stars visible to the naked eye within that radius, which is less than 10% of the stars in total.  Map of the 133 here: http://www.icc.dur.ac.uk/~tt/Lectures/Galaxies/LocalGroup/Back/50lys.html
It says there are an estimated 1800 stars in 1300 systems (so much for binary being prevalent).

Your problem is that you get your data from hand-selected google pages instead of where scientists get it: From looking at the stars.
Sorry if I was not clear enough.
I'm speaking about G stars
http://solstation.com/stars3/100-gs.htm
As many as 512 or more stars of spectral type "G" (not including white dwarf stellar remnants) are currently believed to be located within 100 light-years or (or 30.7 parsecs) of Sol -- including Sol itself. Only around 64 are located within 50 light-years (ly),
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 20/10/2019 18:21:03
Would you kindly advice what our scientists really understand from that nearby Potential energies/gravity forces?

That there are certain limitations placed on how the stars can move in the galaxy. Stars can’t randomly fly out of the galaxy without sling-shotting off of other stars (and thereby sending them deeper into the galaxy), for example.

Let's verify the observations:
At a radius of 100Ly around the sun there are 512 Stars.
That volume represents 8 times the volume of 50Ly.
So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:
512 / 8 = 64 stars per 50Ly.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)
This is not a random activity!!!
A density wave can't generate that kind of fixed density.

That density isn't fixed. The stars are moving relative to each other. It’s also a very bad assumption to calculate the density of stars within 100 light-years of the Sun and extrapolate that to the density of all of the other stars in the spiral arm. What does that have to do with anything, anyway?

Let's verify the observations:
At a radius of 100Ly around the sun there are 512 Stars.
That volume represents 8 times the volume of 50Ly.
So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:
512 / 8 = 64 stars per 50Ly.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)
This is not a random activity!!!
A density wave can't generate that kind of fixed density.
The whole idea of density wave is that in some arias there is higher density of stars while in other arias there are less density of stars.
This is not the case in the Orion arm and not in any other spiral arm.
I would love to know the real 50 Ly radius volume densities of stars per distance from the galactic center - in the arm.
Based on my theory, we should find that as we go closer to the galactic center the density is higher.
As we go further away from the galactic center - the density is lower.
However - at any spiral arm - at any distance from the galactic center, the density of stars is FIXED.

Outside the arm - the density MUST be ZERO!
However, the arm might have any kind of shape and there are bridges between the arms.
I can promise you that as we cross the edge of the arm - there will be ZERO stars there!!!
So, at the nearby aria around the sun - the density could be: or 64 per 50Ly radius or ZERO.

There is another issue which is a direct outcome of theory D.
As we go outwards from the galactic center the arm gets thinner and thinner.
If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.
That sets the end point of the spiral arms.
Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.
If they would try to understand the real impact of the Potential energy/gravity force in short range - less than 500 or 1000Ly, they should get to the same conclusions which I have got.
If we could go back in time - let's say 100 M years ago - we would find that all the nearby stars are still there.
So, any star that is moving away from us today - should come back in the future.
While any star that is moving in our direction, should move away in the Future.
How is it possible?
The density wave doesn't give an answer for all of that.
However, the Potential energies and the gravity forces between nearby stars give a perfect solution.
They are all gravity bonded IN THE ARM itself!!!

We are getting away from the issue at hand. What does any of this have to do with whether gravity obeys conservation of energy or not?

Quote
So please - show me how our scientists translate that nearby potential energy/gravity force into real understanding about the structure of spiral arms.

We aren’t talking about the structure of the galaxy right now. We are still talking about the issue of the energy source of your model.

2. Rotational energy conservation/transformation after the BBT
You claim that stars can't just decrease or increase their orbital radius without external energy.

No, that is not what I claim. An increase in orbital radius requires input energy, whereas a decrease in orbital radius requires energy to be removed.

Quote
If that is correct - than how the whole galaxies had been formed after the Big bang while there were no rotational energy at all?

Energy can be lost in other ways, such as gas clouds contracting by radiating heat.

no kinetic Energy

This is absolutely wrong. Kinetic energy would have been present in every single moving particle in the Universe.

At the early days - 380M years after the Big Bang we have only got matter.
No stars, No galaxies, No orbital objects no main body, no kinetic Energy No rotation energy - only matter everywhere.
So, if we take a cube of one billion LY - in that cube there was no Kinetic energy, no spin energy, no rotation energy - Nothing)
However, Today at the same volume cube there is so high rotational energy. It is everywhere. Try to take the same cube today and calculate the total rotation energy (of all the stars, galaxies) in that cube.
From Zero rotation energy in age of 380 M years we have got so high total rotation energy today.
So, how your idea of rotational energy conservation could be correct if in the early days there was no rotation energy at all???

The rotational axes of different galaxies are in random directions relative to each other, so there is no evidence that the Universe at large has any net angular momentum. So there is no evidence that the Universe has gained net angular momentum over time: the total should still be zero.

Please - If you offer a random activity

What is a "random activity"?

Please - If you offer a random activity, than why the same idea for random gravity force/Energy that creates the whole stars, planets, moons, BH, SMBH, and even unlimited no of galaxies without any request for early rotational energy transformation/conservation in the Early days, can't work today even for just one pair of particle???

It didn't work in the "early days" and doesn't work now. Gravity doesn’t create energy.

How can anyone believe that somehow in the early days of our universe the whole galaxies have been formed without any need for rotational energy transformation but today even for one pair of particle you insist for rotational energy transformation?

Galaxies were formed from gravitational collapse, which releases energy. No energy creation of any kind is necessary.

So please - would you kindly show how the whole orbital activities in the early galaxy had been formed out of nothing (no orbital activities after the matter creation - age of 380 MY after the Big bang), while we have to assume that the same rotational energy conservation should works also for those early days.

You are talking about opposite processes. Galaxy formation does not require any net energy input because the matter involved is entering a lower energy state. This is the opposite of increasing an orbital radius, which requires an input of energy.

So, you don't agree that the Sun/Moon gravity force is stronger than the earth/moon force?
https://www.universetoday.com/116158/why-doesnt-the-sun-steal-the-moon/
"If you’re up for some napkin calculations, you little mathlete, by using Newton’s law of gravity, you find that even with its greater distance, the Sun pulls on the Moon about twice as hard as the Earth does.

So it appears that I was wrong about that particular issue, but it is interesting to know that the Moon technically is already orbiting the Sun.

However, I have asked why the Moon had started to orbit around the Earth at its first day?
Our scientists claim:
https://www.space.com/29047-how-moon-formed-earth-collision-theory.html
"The formation of the moon has long remained a mystery, but new studies support the theory that the moon was formed from debris left from a collision between the newborn Earth and a Mars-size rock, with a veneer of meteorites coating both afterward."
I claim that this story is incorrect because:
a. Meteorites can't form new Moon. They only can set some sort of a ring, as we see around Saturn.
b. If one of the meteorite was as big as the moon, it had to follow the gravity force of the Sun at the moment of the impact (as at that moment the gravity force of the sun was stronger than the earth).

This is beginning to step away from the main issue again. How the Moon formed is not relevant.

Hence, as we monitor the rotational energy rather than gravity force we might find the answerfor why gravity works locally and why the moon orbits around the earth instead of orbiting directly around the Sun.

You already have the answer for why the Moon orbits the Earth in that link you posted...

Well, I think differently

The law of conservation of energy doesn’t care what you think.

Quote
Every orbit has a shape of spiral.
In the nature nothing comes back exactly to the same point (same orbital radius).
It might fall in or it might drift out.

The reason for that is that there are many different factors at play. Asteroid impacts, gravitational interactions with other objects, solar wind, mass loss of the Sun over time, drag from the thin gas in space, the release of gravitational radiation, etc. all have some impact on orbits. Those are sources of energy loss and energy gain.

Quote
Unfortunately, our scientists have missed that key element of "Over time".

No, they haven’t. They are well aware that orbits can change over time due to the factors I just mentioned.

Quote
I can promise you that by 100% all the stars in the spiral arms are drifting outwards over time.
After every cycle around the galactic center - the sun is drifting outwards.

That is actually quite possible, given that the stars are continually losing mass over time. That wouldn't represent a violation of energy conservation, however, as the mass lost by the stars would still be out there in the form of solar wind and radiation.

Quote
If we could come back in one billion years from now we might find that our sun had been ejected from the arm and from the galactic disc.

That’s probably far too soon, but it is technically possible due to a gravitational slingshot off of another star.

The fact of the matter is that energy cannot be created or destroyed. Gravity is no exception to this rule: https://wtamu.edu/~cbaird/sq/2014/01/08/since-gravity-is-unlimited-can-we-use-it-as-an-infinite-energy-source/

You may not accept the law of conservation of energy, but nature doesn't care what you accept.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/10/2019 06:35:19
Quote
Let's verify the observations:
At a radius of 100Ly around the sun there are 512 Stars.
That volume represents 8 times the volume of 50Ly.
So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:
512 / 8 = 64 stars per 50Ly.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)
This is not a random activity!!!
A density wave can't generate that kind of fixed density.

That density isn't fixed. The stars are moving relative to each other. It’s also a very bad assumption to calculate the density of stars within 100 light-years of the Sun and extrapolate that to the density of all of the other stars in the spiral arm. What does that have to do with anything, anyway?
Dear Kryptid

I would like to remind you that the title of this discussion is:
"How gravity works in spiral galaxy?"
We will continue our discussion about the Energy.
However, the density of G stars in spiral arm and the thickness of the galactic disc are very important for our understanding - "How gravity works in spiral galaxy?".

So, do you agree that there is clear observation for 64 Stars per radius of 50 Ly in the radius of 100 Ly around the sun?
If we will discover that this is correct for any location in the 1000 Ly segment around the sun (as long as you are in the Orion arm), If we also discover that outside the arm the density drop to Zero - What can we learn from that?

Quote
There is another issue which is a direct outcome of theory D.
As we go outwards from the galactic center the arm gets thinner and thinner.
If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.
That sets the end point of the spiral arms.
Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.

Actually our scientists have discovered that:
Close to the Ring (3KPC) - the thickness of the arms is 3000Ly
At our current location  (8KPC) - the thickness of the arms is 1000Ly
At the end of the spiral arms (12KPC?) - the thickness of the arms is less than 500Ly

So how the density wave theory can explain all of those observations?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/10/2019 06:45:10
We will continue our discussion about the Energy.

We are still focusing on energy. Jumping around to different topics at the same time is counterproductive. You even agreed with me that we are to focus on the energy issue until that is resolved:

I'm putting a hold on all of the other matters for the moment and focusing on conservation of energy because that is the crux of the problem right now. I'm not moving on until that issue is solved first.
Yes, fully agree.

So we'll talk about the other aspects of your model if and when the energy issue is properly dealt with.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/10/2019 06:04:48
Dear krptid

Our scientists have developed some ideas/assumptions about gravity and energy.
For example you have stated:
An object won't change its orbital radius unless it is gaining energy from an outside source or losing it to an outside source. A spacecraft won't spontaneously increase its orbital radius from the Earth, for example. If the astronauts want to increase the orbital radius, they will have to add energy by firing the engines.
Again, you have it backwards. Adding energy from an outside source will increase the orbital radius. You can't say, "I want the  orbital radius to increase, so the energy required to do that pops up out of nowhere". The radius won't increase unless the object in orbit is supplied with energy from another source that already existed. Look at my example with the spaceship.
However, spacecraft is not a star.
We need to verify the stars activity in the galaxy and see if those assumptions are correct.
The only way to verify that is by our observation on the galaxy.
The thickness of the spiral arms and the density of the stars at any section give us a perfect understanding.
So, if you look at our Milky Way spiral arms you would see immediately that stars can't migrate inwards.
Actually, the stars in the spiral arms MUST drift outwards.
We all see many stars that are ejected at ultra velocity from the galactic disc.
https://curiosity.com/topics/this-hypervelocity-star-was-ejected-from-the-milky-way-curiosity
"Every once in a while, the Milky Way ejects a star. The evicted star is typically ejected from the chaotic area at the center of the galaxy, where our Super Massive Black Hole (SMBH) lives. But at least one of them was ejected from the comparatively calm galactic disk, a discovery that has astronomers rethinking this whole star ejection phenomenon."
In order to explain this phenomenon our scientists claim that it is due to Binary star:
"When a star is kicked out of the galaxy, it's usually one star from a binary pair. Scientists think that as a binary pair get too close to the SMBH and its overwhelming gravity, the hole captures one of the stars. The other star is shot out into space in a "gravitational slingshot." The black hole has to be a supermassive one because only they have powerful enough gravity to accelerate these run-away stars to such high velocities."
The specifically claim: "the hole captures one of the stars"
So, please show me those stars that were captured by the SMBH?
If we see that stars are ejected outwards than we also must see stars that pushed inwards.
We see quite many of those Hypervelocity Star that were Ejected from the Milky Way.
Did we ever see one single Hypervelocity Star that was pushed inwards to the SMBH in the Milky Way?
The answer is very clear - No No and No.
Those Hypervelocity Star were not ejected due to the Binary affect.
They were ejected from the galaxy as they have moved to far away from the requested density in the arm/galaxy
Those 64 stars per 50 Ly is a living factor for our solar system in the galaxy.
Our sun is located near the edge of the Orion arm. If we will dare to move even 30 LY away from the edge of the arm, we will lose the gravity bonding of all our nearby stars.
At that moment we will be boosted outwards from the galaxy and will never come back again.
The galaxy doesn't take back any matter or stars from outside.
All the stars and matter that we see in the galaxy had been created by our mighty SMBH.
So, please - would you kindly answer the questions about the thickness and G stars density in the spiral arms?
How the "density wave theory" can still be valid under those observations?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/10/2019 15:00:08
Our scientists have developed some ideas/assumptions about gravity and energy.

Conservation of energy isn't an assumption. It's an observable fact.

However, spacecraft is not a star.

Gravity can't tell the difference. Mass is mass.

We need to verify the stars activity in the galaxy and see if those assumptions are correct.
The only way to verify that is by our observation on the galaxy.

Stars obey the laws of physics, including conservation of energy.

So, please show me those stars that were captured by the SMBH?

You know there are many stars observed to be in orbit around our super-massive black hole, don't you?

Did we ever see one single Hypervelocity Star that was pushed inwards to the SMBH in the Milky Way?

Of course not, because that isn't how it works. Gravitational sling-shotting speeds one body up at the expense of slowing another body down. If you knew how conservation of energy worked (which you have abundantly demonstrated that you don't), you would have known that. Both stars can't make each other go faster as that would violate conservation of energy. If one speeds up, that is because it has extracted energy from the other star via gravitational interactions, which makes the other one slow down. You really don't know what you are talking about.

Those Hypervelocity Star were not ejected due to the Binary affect.

You're in no position to make such a claim when you don't even know how gravitational slingshotting works.

They were ejected from the galaxy as they have moved to far away from the requested density in the arm/galaxy

You really don't know how gravity works. If a star is gravitationally-bound to other stars, it can't move too far away from those other stars in the first place unless the energy needed to move away from those stars is imparted to it somehow. If it happened spontaneously, that would violate conservation of energy.

Our sun is located near the edge of the Orion arm. If we will dare to move even 30 LY away from the edge of the arm, we will lose the gravity bonding of all our nearby stars.

That is as ridiculous as claiming that "If I dared to move even a hundred million miles off of the Earth, I would lose my gravitational bond with Earth". Can I spontaneously float away from the Earth? No, I can't. Gravity is holding me here. If I want to move a hundred million miles from Earth, I have to be given enough energy to move that far away. So a star can't spontaneously drift away from other stars either. It has to receive enough energy from some source in order to drift that required distance because it is moving against a force that is trying to pull it back. Moving against a force requires energy. Period.

At that moment we will be boosted outwards from the galaxy and will never come back again.

Boosted by some magical thing you made up, it seems.

All the stars and matter that we see in the galaxy had been created by our mighty SMBH.

No matter how many times you say this, the law of conservation of energy won't let it be true. You might hate that law, but that doesn't make it go away.

So, please - would you kindly answer the questions about the thickness and G stars density in the spiral arms?
How the "density wave theory" can still be valid under those observations?

Even if "density wave theory" was wrong (which I don't know enough about to make a judgment call on anyway), it wouldn't matter because your explanation is also wrong. Models that violate the laws of physics are automatically wrong. Your model is automatically wrong because it violates conservation of energy. Is gravitational energy a form of energy? Yes. Therefore, it must obey the law of conservation of energy.

What makes you think you can break the laws of physics?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/10/2019 06:48:13
Gravitational sling-shotting speeds one body up at the expense of slowing another body down. If you knew how conservation of energy worked (which you have abundantly demonstrated that you don't), you would have known that. Both stars can't make each other go faster as that would violate conservation of energy. If one speeds up, that is because it has extracted energy from the other star via gravitational interactions, which makes the other one slow down
So you claim that: "Both stars can't make each other go faster as that would violate conservation of energy."
We know that the average orbital velocity of the stars around the galaxy is about 200 Km/s
I have found a  runaway star at 1200Km/s:
https://www.space.com/28737-fastest-star-galaxy-strange-origin.html
The runaway star, US 708, is traveling at 745 miles per second (1200 km/s) — that's  26 million miles per hour (43 million km/h) —making it the fastest star in the Milky Way ever clocked by astronomers, according to the new research"
It is also stated:
"The monster black hole at the center of the Milky Way has the gravitational muscle to fling a star on a one-way-track out of the neighborhood, and many other hypervelocity stars are thought to originate from there. But US 708 didn't start its journey near the galactic center, the new research shows."
Based on your advice it must come from a binary star in the galactic disc (as it is far away from the SMBH).
However, you have stated that: : "Both stars can't make each other go faster as that would violate conservation of energy."
So, how can we convert the energy of 200Km/s to 1200Km/s?

Our scientists must find a real explanation for all the observations!!!
If we can't explain the observations by the theory that we believe - than this theory is not relevant.
Therefore, If we can't explain the G star's density and thickness - than we have to look for better theory.
Let's assume that 100,000 professors and scientists believe in the same assumption/theory.
Does it mean that this assumption is correct?
If you all believe in that assumption - than you have to explain the observation.
So, a theory can't be correct just because all of you believe in that theory.
There must be full correlation between any theory to all observations.
Theory D is the only valid theory that gives a perfect explanation for everything we see.
It meets all the observations by -100%.
Actually, there is a simple way to verify which theory is real.
You can tell me what is your expectation based on your theory/understanding, and I will tell you the expectation by theory D.
Let's start by the nearby G stars density.
I can tell you by 100% that at any 50LY radius in the nearby aria in the Orion arm - the density MUST be 64 G stars.
If you go out the arm - the density must drop to zero.
So, there are only two possibilities for star density - or 64 per 50Ly or Zero. Nothing in between (unless we are located exactly at the edge of the arm or the bridge).
Would you kindly advice what would be your expectation for the G star density around us (take radius of 2000 Ly?
If we will cross the Orion arm, what might be the star density?
Why don't we ask NASA to verify my expectation?
If we will discover that theory D is correct, do you agree to reconsider your objection against this theory?

Your model is automatically wrong because it violates conservation of energy. Is gravitational energy a form of energy? Yes. Therefore, it must obey the law of conservation of energy.

Gravity force is a FORCE!!!
Newton didn't call it Gravity energy
Gravity is not just an energy transformation.
A force can generate new energy.
Evey orbit of a star consume energy.
If I will give you a rope with a ball at the end, don't you need to generate new force/energy in order to keep it orbiting around you?
In the same token -  Every orbit of a star consumes Force and energy.
However, as the gravity force comes for free (Due to Newton , there is no need to fuel the gravity force or set any energy transformation), the energy due to this force also comes for free.
So, far you couldn't show any formula which links the gravity force into energy transformation.
In electromagnetic - energy transformation is must.
In gravity force - As Newton and Einstein didn't specify any request for gravity energy transformation - your assumption is totally incorrect.
Therefore, the statement that gravity force can't generate new energy is a severe violation of Newton law


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/10/2019 07:28:26
So you claim that: "Both stars can't make each other go faster as that would violate conservation of energy."
We know that the average orbital velocity of the stars around the galaxy is about 200 Km/s
I have found a  runaway star at 1200Km/s:
https://www.space.com/28737-fastest-star-galaxy-strange-origin.html
The runaway star, US 708, is traveling at 745 miles per second (1200 km/s) — that's  26 million miles per hour (43 million km/h) —making it the fastest star in the Milky Way ever clocked by astronomers, according to the new research"
It is also stated:
"The monster black hole at the center of the Milky Way has the gravitational muscle to fling a star on a one-way-track out of the neighborhood, and many other hypervelocity stars are thought to originate from there. But US 708 didn't start its journey near the galactic center, the new research shows."
Based on your advice it must come from a binary star in the galactic disc (as it is far away from the SMBH).
However, you have stated that: : "Both stars can't make each other go faster as that would violate conservation of energy."
So, how can we convert the energy of 200Km/s to 1200Km/s?

The answer is in that very link:

Quote
The fastest-known star in the Milky Way is on a path out of the galaxy, and new research suggests it was a supernova that gave it the boot.

Did you even bother to read that before posting the link here? Not all fast-moving stars must have the same cause.

Theory D is the only valid theory that gives a perfect explanation for everything we see.
It meets all the observations by -100%.

No it doesn't, as It violates the law of conservation of energy. We observe that energy is conserved. Hypothesis D (I won't call it a theory, because it isn't) goes against that observation.

Actually, there is a simple way to verify which theory is real.
You can tell me what is your expectation based on your theory/understanding, and I will tell you the expectation by theory D.
Let's start by the nearby G stars density.
I can tell you by 100% that at any 50LY radius in the nearby aria in the Orion arm - the density MUST be 64 G stars.
If you go out the arm - the density must drop to zero.
So, there are only two possibilities for star density - or 64 per 50Ly or Zero. Nothing in between (unless we are located exactly at the edge of the arm or the bridge).
Would you kindly advice what would be your expectation for the G star density around us (take radius of 2000 Ly?
If we will cross the Orion arm, what might be the star density?
Why don't we ask NASA to verify my expectation?
If we will discover that theory D is correct, do you agree to reconsider your objection against this theory?

No, because Hypothesis D violates conservation of energy. It cannot be correct for that reason.

A force can generate new energy.

Not according to the law of conservation of energy.

Evey orbit of a star consume energy.

No, no it doesn't.

If I will give you a rope with a ball at the end, don't you need to generate new force/energy in order to keep it orbiting around you?

Only because the drag of the air is slowing it down. If I was floating in space, I would have no trouble keeping the ball going without adding any extra force or energy.

In the same token -  Every orbit of a star consumes Force and energy.

No it doesn't. You just don't understand how force works. Your false analogy just displayed that.

However, as the gravity force comes for free (Due to Newton , there is no need to fuel the gravity force or set any energy transformation), the energy due to this force also comes for free.

More proof that you don't understand how force relates to energy. Force is not energy. The law of conservation of energy won't allow force to create new energy. A table is under a constant force from gravity but no new energy is being generated by the force.

So, far you couldn't show any formula which links the gravity force into energy transformation.

I did give you such an equation. The gravitational potential energy equation shows exactly that: how much energy is released by allowing an object to fall into a gravitational field or how much energy is consumed moving an object up against a gravitational field.

In gravity force - As Newton and Einstein didn't specify any request for gravity energy transformation - your assumption is totally incorrect.

Both Newtonian and relativistic physics allow energy to be converted from potential to kinetic by a gravitational field, so what you are saying is false: https://en.wikipedia.org/wiki/Gravitational_energy#Newtonian_mechanics

Therefore, the statement that gravity force can't generate new energy is a severe violation of Newton law

Nowhere do any of Newton's laws state that gravitational fields can generate new energy. Supply a link from an authoritative source if you want me to take you seriously.

Again, what makes you think you can break the laws of physics?
Title: Re: How gravity works in spiral galaxy?
Post by: syhprum on 26/10/2019 08:42:22
When you are in space twirling your test mass around on a string both your hand and the test mass will revolve around their mutual centre of gravity if you make an effort to maintain your hand away from the centre of gravity it will absorb energy from the system there will also be a small amount of energy radiated away in the form of gravitational waves.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/10/2019 17:56:16
The answer is in that very link:
"The fastest-known star in the Milky Way is on a path out of the galaxy, and new research suggests it was a supernova that gave it the boot".
This is fantasy.
If we set a ball on atomic bomb, is there a chance that the atomic bang will rock the ball to the moon?
Sorry, a supernova near a star should break it to pieces.
In any case, we discuss on orbital energy transformation.
Our scientists were positively sure that all of those hyper star had been ejected from the galaxy due to binary star.
They didn't mention any word about supernova.
Now that they find a star at 1200Km/s than suddenly they look for better alternative.
That proves that they really don't have a basic clue how gravity really works in the galaxy.
If they have stated that it is due to binary star and the observation contradicts this statement - It is expected that they should set this idea in the garbage.
You can't just invent new idea every time that new observation contradicts your current assumption.
If our scientists give an explanation for something and this explanation can't meet the observation - they should say: Sorry we have an error in our assumption. Clear and loud!
Once they find that a hyper star at that ultra velocity - they have to say, sorry our binary star idea can't meet this observation.
We see again and again that our scientists have a fatal error in understanding how gravity really works in the galaxy.
So far they couldn't offer any real theory that covers everything we see.
They could not even give a valid explanation for the star density and thickness of the spiral arms.
I have proved that the density wave is a pure fiction.
So, how can you claim that your current understanding in gravity is the ultimate one?
Sorry - if you wish to show that your understanding in relevant, than please give full explanation for whatever we see.
You can't just disqualify my understanding without offering real alternative.

Let me ask you again:
Please show how your current theory about gravity meets the density and thickness of the spiral arms

I did give you such an equation. The gravitational potential energy equation shows exactly that: how much energy is released by allowing an object to fall into a gravitational field or how much energy is consumed moving an object up against a gravitational field.
Yes and no.
As long as we convert the potential energy into falling kinetic energy - there is energy conservation.
However, once we deal with orbital kinetic energy - there is no energy conservation between the potential energy to the orbital kinetic energy.
I have already proved that:
If you decrease the distance/radius between the objects by twice you actually increase the Potential energy by twice and get only half of the orbital kinetic energy.
So, we see clearly that the sum of the total orbital energy changing as we change the radius.
You actually have agreed with that.
Therefore, theoretically, we can agree that as long as the radius is fixed, the gravitational energy is fixed.
However, as I have already claimed, this statement is not fully correct.
Based on my understanding, the orbital kinetic energy is decreasing over time.
Therefore, the radius/distance between the orbital objects is increasing over time.
So, the increased radius between the earth/moon systems is not due to tidal but it is due to that "over time" phenomenon.
I wonder how Einstein and Newton didn't discover that phenomenon.
I know that by now you don't agree with that, but sooner or later our scientists would find that this is fully correct.

Both Newtonian and relativistic physics allow energy to be converted from potential to kinetic by a gravitational field, so what you are saying is false: https://en.wikipedia.org/wiki/Gravitational_energy#Newtonian_mechanics
Again - this doesn't cover the orbital kinetic energy.
Nowhere do any of Newton's laws state that gravitational fields can generate new energy.
He also didn't say that the orbital kinetic energy needs a transformation of energy.
He had offered the formula for gravity force.
Based on that gravity force we can extract the orbital velocity and the orbital kinetic energy.
So, for any orbital radius, there is a specific orbital velocity.
That orbital velocity is a direct outcome of the gravity force.
Hence, the orbital kinetic energy is also a direct outcome of that orbital velocity.
If the orbital is new, than the orbital kinetic energy is new!!!
Newton didn't ask for any transformation in orbital kinetic energy.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/10/2019 20:04:23
When you are in space twirling your test mass around on a string both your hand and the test mass will revolve around their mutual centre of gravity if you make an effort to maintain your hand away from the centre of gravity it will absorb energy from the system there will also be a small amount of energy radiated away in the form of gravitational waves.
At the moment that you change the location of your hand - you actually change the energy.
As you change the energy you get different orbital velocity.
Our scientists think that somehow a matter can fall in and set a new star or even a new galaxy.
This is a fiction.
If you have a floating matter in space, it will never ever merge and form a new star.
The only way to set a star is by a gas cloud that orbits near a SMBH.
Even so, the matter isn't falling inwards to the center of the gas cloud in order to form a single new born star.
In reality, the matter orbits around the center of the gas cloud.
As it orbits (under the impact of a nearby SMBH's gravity force), it crystallized and theoretically could form several stars per gas cloud (that orbit around the center of the cloud).
However, as they emerge from the gas cloud, they are actually orbiting around their common center of mass, (which is probably the center of the gas cloud).
This center represents the VHP (virtual host point) which I have described in the past.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/10/2019 23:52:32
If we set a ball on atomic bomb, is there a chance that the atomic bang will rock the ball to the moon?

As a matter of fact, it is possible to accelerate objects to extremely high velocities (and even into space) using nuclear explosions: https://www.businessinsider.com/fastest-object-robert-brownlee-2016-2

Quote
Sorry, a supernova near a star should break it to pieces.

Stars are much, much larger than a ball and have a binding energy that is many orders of magnitude larger. The mass and diameter of the star isn't mentioned, but typical subdwarf stars have a mass up to about half that of the Sun with a diameter one-tenth that of the Sun. That would be a mass of about 9.9425 x 1029 kilograms and a diameter of about 139,140 kilometers. The results in a surface gravity of about 1,396 times that of Earth. The calculated binding energy of such a star would be 5.6741 x 1041 joules: http://www.stardestroyer.net/Resources/Calculators/PlanetaryParameter.html

The question then becomes: can enough kinetic energy be imparted to the star to speed it up to 1,200 kilometers per second from 200 kilometers per second without giving it enough energy to destroy it in the process? The mass of the star is estimated at 9.9425 x 1029 kilograms, which we put into the kinetic energy equation Ek = 0.5mv2. So we get Ek = (0.5)(9.9425 x 1029)(1,000,000)2 = 4.97125 x 1041 joules. It's close, but this demonstrates that the star can indeed acquire the needed energy from the supernova to speed up to 1,200 kilometers per second without being destroyed. Much of the star probably was blown off in the process, actually, since O-type subdwarfs are thought to be the result of red giants having their outer layers removed by some powerful process.

Our scientists were positively sure that all of those hyper star had been ejected from the galaxy due to binary star.

According to what source?

They didn't mention any word about supernova.

The article that you linked did.

That proves that they really don't have a basic clue how gravity really works in the galaxy.

They know more than you do, since at least they recognize that gravity doesn't violate conservation of energy.

If they have stated that it is due to binary star and the observation contradicts this statement - It is expected that they should set this idea in the garbage.

The observation doesn't contradict that. I just demonstrated mathematically that a binary star system where one of the stars went supernova is a plausible mechanism for accelerating the remaining star to the observed velocity.

You can't just invent new idea every time that new observation contradicts your current assumption.

It would be stupid not to do that. You don't hold on to ideas that are shown to be wrong, so you have to invent new ideas to explain new data.

If our scientists give an explanation for something and this explanation can't meet the observation - they should say: Sorry we have an error in our assumption. Clear and loud!

They do. Our explanations for hypervelocity stars have not been falsified, however. The supernova idea works.

We see again and again that our scientists have a fatal error in understanding how gravity really works in the galaxy.

You have yet to demonstrate this. All you have demonstrated is that your mind is filled with misconceptions about physics.

I have proved that the density wave is a pure fiction.

You have done no such thing.

So, how can you claim that your current understanding in gravity is the ultimate one?

Don't put words in my mouth. I never did.

You can't just disqualify my understanding without offering real alternative.

I absolutely can do that. If you're wrong, then you're wrong. It is possible for every single idea used to explain a phenomenon to be wrong. One idea being wrong is not evidence for an alternative idea being correct. That is known as the "argument from ignorance fallacy". We know for a fact that your idea is wrong because it violates conservation of energy.

Please show how your current theory about gravity meets the density and thickness of the spiral arms

I don't need to. All I need to do is show that your explanation is wrong. It is wrong because it breaks the law of conservation of energy.

However, once we deal with orbital kinetic energy - there is no energy conservation between the potential energy to the orbital kinetic energy.

There absolutely is conservation of energy in that case. Potential energy and orbital kinetic energy are forms of energy. Therefore, they fall under energy conservation.

I have already proved that:
If you decrease the distance/radius between the objects by twice you actually increase the Potential energy by twice and get only half of the orbital kinetic energy.
So, we see clearly that the sum of the total orbital energy changing as we change the radius.
You actually have agreed with that.

Of course the energy values are different at different radii. That doesn't violate conservation of energy because energy either has to be added to the orbit in order to make it larger to taken away from it to make it smaller. The orbit won't change its radius on its own and therefore the energy won't change on its own either. We've been through this before...

Based on my understanding, the orbital kinetic energy is decreasing over time.
Therefore, the radius/distance between the orbital objects is increasing over time.

The only way this can happen is if energy is being put into the system from some outside source (since a larger orbital radius has a larger total energy).

So, the increased radius between the earth/moon systems is not due to tidal but it is due to that "over time" phenomenon.

If not from tidal energy transfer, then where do you propose the energy is coming from in order to increase the total orbital energy of the Earth-Moon system? Energy that is already in existence has to transferred to the system somehow. The law of conservation of energy demands it. Remember, orbital energy is a form of energy and therefore must be conserved.

Again - this doesn't cover the orbital kinetic energy.

Orbital kinetic energy is a form of energy and therefore falls under conservation of energy.

He also didn't say that the orbital kinetic energy needs a transformation of energy.

It's a conversion of gravitational potential energy into kinetic energy, so that's exactly what Newtonian physics says. An orbit is the exact same thing as falling. It's just "missing" instead of hitting the object it's orbiting because it's travelling on a curved trajectory.

If the orbital is new, than the orbital kinetic energy is new!!!

No, it isn't. The energy was already there in the form of gravitational potential energy. Orbital kinetic energy is a form of energy and therefore must be conserved.

Newton didn't ask for any transformation in orbital kinetic energy.

Newtonian physics absolutely does recognize a transformation of potential energy into kinetic energy.

I'm still waiting for you to give some kind of authoritative source that supports your claim that gravity can create new energy. The law of conservation of energy says that it can't. All of this other stuff is irrelevant, honestly. We know that your model is wrong because gravity can't create energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/10/2019 06:38:23
Thanks Kryptid
The question then becomes: can enough kinetic energy be imparted to the star to speed it up to 1,200 kilometers per second from 200 kilometers per second without giving it enough energy to destroy it in the process? The mass of the star is estimated at 9.9425 x 1029 kilograms, which we put into the kinetic energy equation Ek = 0.5mv2. So we get Ek = (0.5)(9.9425 x 1029)(1,000,000)2 = 4.97125 x 1041 joules. It's close, but this demonstrates that the star can indeed acquire the needed energy from the supernova to speed up to 1,200 kilometers per second without being destroyed. Much of the star probably was blown off in the process, actually, since O-type subdwarfs are thought to be the result of red giants having their outer layers removed by some powerful process.
Sorry. I totally disagree
A star isn't an elastic ball that can observe the kick energy and restore it for its kinetic movement.
It is a solid object that keeps its ball shape due to gravity force.
Did you try to calculate the energy that holds any atoms in the star?
If an atom at the surface of the star will get that kind of energy, how it should react?
It is quite clear to me that if you kick a star with an energy of 4.97125 x 10^41 joules, that energy is much stronger than the total gravity energy that keeps the atoms together in order to set the ball shape.
However, Supernova is not just a pure energy. It comes with big broken objects.
We all know the outcome of one big object collision with the earth 65 million year ago (and it was not due to supernova, just a free falling object).
If some of the big broken objects from the supernova hit a nearby star at that ultra energy, they should cross though the star even before the star can open his eyes.
So, in a few moments the whole star should be broken to pieces by that ultra energy that comes with big broken objects.
It is similar to shooting a ballet into a water melon.
Try to do it and verify the outcome.
Therefore, supernova can't just kick a nearby star in order to gain the 1200Km/s without breaking it to pieces.


There is another issue - Star density.
We have a solid observation that in our aria the star density is 512 stars per 100 Ly.
So, if the supernova took place in the same radius as we are, it should affect several hundreds of stars.
Therefore, even if we accept the impossible mission and somehow it could deliver the requested energy, than why do we see only one lonely star at 1200Km/s?
Where are all the other nearby stars that were affected by that mighty supernova energy?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/10/2019 14:05:59
A star isn't an elastic ball that can observe the kick energy and restore it for its kinetic movement.

Stars are very elastic because they are composed entirely of fluid.

It is a solid object that keeps its ball shape due to gravity force.

Stars are "ball-shaped" and are held together by gravity despite the fact that they aren't solid.

Did you try to calculate the energy that holds any atoms in the star?

That's exactly what the gravitational binding energy is.

If an atom at the surface of the star will get that kind of energy, how it should react?

The outer layers of the star would indeed be removed by the explosion (explaining why the star is an O-type subdwarf), but the total binding energy is too high to destroy the entire star.

It is quite clear to me that if you kick a star with an energy of 4.97125 x 10^41 joules, that energy is much stronger than the total gravity energy that keeps the atoms together in order to set the ball shape.

4.97125 x 1041 joules is smaller than 5.6741 x 1041 joules, so you are objectively wrong.

However, Supernova is not just a pure energy. It comes with big broken objects.
We all know the outcome of one big object collision with the earth 65 million year ago (and it was not due to supernova, just a free falling object).
If some of the big broken objects from the supernova hit a nearby star at that ultra energy, they should cross though the star even before the star can open his eyes.
So, in a few moments the whole star should be broken to pieces by that ultra energy that comes with big broken objects.
It is similar to shooting a ballet into a water melon.

Supernovas don't have have anything solid in them. They are too hot.

Try to do it and verify the outcome.

I don't need to because I already have. The amount of energy needed to destroy the star is known. If the energy applied is less than that, the star won't be destroyed.

Therefore, supernova can't just kick a nearby star in order to gain the 1200Km/s without breaking it to pieces.

The math says otherwise.

There is another issue - Star density.
We have a solid observation that in our aria the star density is 512 stars per 100 Ly.
So, if the supernova took place in the same radius as we are, it should affect several hundreds of stars.
Therefore, even if we accept the impossible mission and somehow it could deliver the requested energy, than why do we see only one lonely star at 1200Km/s?
Where are all the other nearby stars that were affected by that mighty supernova energy?

Because of the inverse-square law. The other stars are much, much further away from the supernova than the hypervelocity star would have been. That system would have been what is known as a cataclysmic variable, where two stars are so close together that the outer layers of the larger star are slowly pulled off by the gravity of the white dwarf. Such systems can be so compact that it only takes hours for them to orbit each other. That would make them far closer to each other than even Mercury is to the Sun.

I'm finished talking about this hypervelocity star issue. It doesn't have anything to do with conservation of energy.

The logic is very simple.
Does your model create energy? Yes.
Does the law of conservation of energy forbid the creation of energy? Yes.
Is the law of conservation of energy a law of physics? Yes.
Therefore, your model violates a law of physics and must be wrong.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/10/2019 05:19:06
I'm finished talking about this hypervelocity star issue. It doesn't have anything to do with conservation of energy.

The logic is very simple.
Does your model create energy? Yes.
Does the law of conservation of energy forbid the creation of energy? Yes.
Is the law of conservation of energy a law of physics? Yes.
Yes it is.
You discuss on the name of the gravitational conservation of energy in the galaxy.
However, in order to discuss about that key issue, do you really understand how gravity works in the galaxy?
So, we already know the conservation of energy between the potential energy to kinetic energy.
We all know that if we want to lift an object from the surface of the earth, we need to set a kinetic energy.
Therefore, if we shoot a bullet up to the air at very high velocity, that velocity represents a kinetic energy.
That kinetic energy is transformed into potential energy.
Therefore, as the bullet goes higher its velocity is decreasing.
At some point, (if we ignore the friction) when the potential energy is equal to the First moment kinetic energy, the bullet should stop completely and than fall back.
That is how gravity works on earth.
So, let's go back to our galaxy.
As this hyper star is climbing above the center of the galaxy, it increases its potential energy.
Therefore, its kinetic energy should go down as it goes higher and higher.
As it is moving away from the center of the galaxy, every moment it must gain higher potential energy and therefore, it also must decrease its kinetic energy due to the conservation of energy.
Therefore if we want to verify the first moment kinetic energy, we must add to that  4.97125 x 10^41 joules, all the kinetic energy that had been transformed into potential energy.
If you set this calculation you would surly find that the requested current potential energy represents 10^n times the current kinetic energy. 
Therefore, the Kinetic energy at the first moment due to the supernova, should be in the range of
4.97125 x 10^(41+n) joules
That first kick energy must be significantly higher than the 5.6741 x 10^41 joules
4.97125 x 1041 joules is smaller than 5.6741 x 1041 joules, so you are objectively wrong.
Therefore, by using your idea of the conservation of energy, I have proved that the requested first moment kinetic energy that is needed for this Hyper star, must break it to it's atoms.
We can ask also, why none of those Hyper stars don't fall back to the center of the galaxy?
Actually, the galaxy is crossing the space at a speed of 600Km/s.
It radius is more than 50,000 Ly.
We already know that there are more stars in the open space outside the galaxies than the total number in all the galaxies.
So, any star outside the galaxy had quite high potential energy (with regards to the galaxy center)
Therefore, it is expected that all the stars in the path of the milky way should fall in.
Similar to the falling meteorites on Earth.
How many fall in stars do we see?
One Million? One Thousand? or none?
How could it be that not even a single star can fall into the galaxy?
Don't they have any basic knowledge about the conservation of energy?
How could it be that they refuse to convert their high potential energy to Kinetic energy and fall in? 
Sorry, the outcome is very clear - Our scientists don't have a clue how gravity really works in the galaxy.
Once you understand how gravity works in the galaxy, then let's discuss about the gravitational conservation of energy in the galaxy.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/10/2019 06:40:43
If you set this calculation you would surly find that the requested current potential energy represents 10^n times the current kinetic energy.
Therefore, the Kinetic energy at the first moment due to the supernova, should be in the range of
4.97125 x 10^(41+n) joules
That first kick energy must be significantly higher than the 5.6741 x 10^41 joules

Then do the math and prove it.

Therefore, by using your idea of the conservation of energy, I have proved that the requested first moment kinetic energy that is needed for this Hyper star, must break it to it's atoms.

You haven't proven anything until you've done that math. You're simply assuming that it's higher than the star's binding energy. Even if you did prove it to be the case, it wouldn't matter because it has nothing to do with the fact that your model violates conservation of energy.

Therefore, by using your idea of the conservation of energy, I have proved that the requested first moment kinetic energy that is needed for this Hyper star, must break it to it's atoms.
We can ask also, why none of those Hyper stars don't fall back to the center of the galaxy?
Actually, the galaxy is crossing the space at a speed of 600Km/s.
It radius is more than 50,000 Ly.
We already know that there are more stars in the open space outside the galaxies than the total number in all the galaxies.
So, any star outside the galaxy had quite high potential energy (with regards to the galaxy center)
Therefore, it is expected that all the stars in the path of the milky way should fall in.
Similar to the falling meteorites on Earth.
How many fall in stars do we see?
One Million? One Thousand? or none?
How could it be that not even a single star can fall into the galaxy?
Don't they have any basic knowledge about the conservation of energy?
How could it be that they refuse to convert their high potential energy to Kinetic energy and fall in?
Sorry, the outcome is very clear - Our scientists don't have a clue how gravity really works in the galaxy.
Once you understand how gravity works in the galaxy, then let's discuss about the gravitational conservation of energy in the galaxy.

This is nothing more than another set of unevidenced assumptions and misundestandings that is irrelevant to the fact that the law of conservation of energy won't let your model work. It's just another attempt to distract me. We wouldn't even need to know how gravity worked (even though we do understand it much better than you do) in order to know that the law of conservation of energy does work. That alone would tell us all we need to know about the energetics associated with gravity.

This first statement is true: your model proposes the creation of energy.
This second statement is also true: the law of conservation of energy doesn't allow energy to be created.
The inevitable consequence of those two statements being true should be obvious. Do you want to take a guess as to what that consequence is? Hint: it's something I have repeated innumerable times.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/10/2019 04:57:03
You haven't proven anything until you've done that math. You're simply assuming that it's higher than the star's binding energy. Even if you did prove it to be the case, it wouldn't matter because it has nothing to do with the fact that your model violates conservation of energy.
Let's try to understand the real meaning of Orbital energy conservation especially with regards to potential energy.
Let's assume that in the whole Universe there are only two objects.
Due to the potential energy, the gravity force will convert the potential energy into falling kinetic energy.
Hence, they will move directly to each other until the final collision. The Vector Vp (velocity due potential energy) will be in the direct line between the two objects.
We all know that in orbital kinetic energy, the Velocity Vk (Velocity due to orbital kinetic energy) must be vertically to Vp.
Therefore, do you agree that the potential energy can ONLY be converted into falling kinetic energy (which is represented by Vp).
We know that in the open space objects maintain their momentum.
So, if Vp represents a velocity vector between the two objects, Do you see any possibility to convert that Vp to Vk?
If we can't convert Vp to Vk how could it be that Potential energy is part of the total orbital energy?
How can we discuss on conservation of orbital energy, if in the reality, a potential energy can't be transformed into orbital kinetic energy?

In the same token, let me ask how the accretion disc had been formed?
Let's assume that once upon a time there was a BH or a SMBH without any accretion disc.
Do you agree that any in falling atom or star should fall directly to that BH/SMBH?
So, how could it be that suddenly the Vp (in falling velocity vector due to potential energy) which was in a direct line between the Atom/star to the BH/SMBH is shifted by 90 degrees in order to be converted into a Vk (orbital velocity vector) at almost the speed of light?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/10/2019 06:15:15
Hence, they will move directly to each other until the final collision.

If you're talking about two objects moving in a straight line towards each other then this isn't even an orbit. The concept of "orbital" energy therefore does not apply. Try to keep that in mind.

Therefore, do you agree that the potential energy can ONLY be converted into falling kinetic energy (which is represented by Vp).
We know that in the open space objects maintain their momentum.
So, if Vp represents a velocity vector between the two objects, Do you see any possibility to convert that Vp to Vk?
If we can't convert Vp to Vk how could it be that Potential energy is part of the total orbital energy?

I don't understand what you are trying to say. On the one hand, you are saying that potential energy can only be converted into kinetic energy and at the same time imply that there is no possibility for converting potential energy into kinetic energy. Can you clear this up?

If we can't convert Vp to Vk how could it be that Potential energy is part of the total orbital energy?
How can we discuss on conservation of orbital energy, if in the reality, a potential energy can't be transformed into orbital kinetic energy?

It can be converted. It's happening constantly in eccentric orbits. As a planet approaches perihelion (the closest point to the Sun in its orbit), it speeds up. This speed increase comes from a conversion of the planet's gravitational potential energy into kinetic energy. Once it passes perihelion, it starts moving further away from the Sun and begins to convert its kinetic energy back into potential energy, slowing it down.

In the same token, let me ask how the accretion disc had been formed?
Let's assume that once upon a time there was a BH or a SMBH without any accretion disc.
Do you agree that any in falling atom or star should fall directly to that BH/SMBH?
So, how could it be that suddenly the Vp (in falling velocity vector due to potential energy) which was in a direct line between the Atom/star to the BH/SMBH is shifted by 90 degrees in order to be converted into a Vk (orbital velocity vector) at almost the speed of light?

This is starting to get away from the energy conservation discussion again, but I will say this: it's unwise to assume that all stars and particles will be traveling in a straight line towards the black hole. Conservation of momentum and the initial trajectories and velocities are relevant.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/10/2019 01:40:09
Quote
If we can't convert Vp to Vk how could it be that Potential energy is part of the total orbital energy?
How can we discuss on conservation of orbital energy, if in the reality, a potential energy can't be transformed into orbital kinetic energy?
It can be converted. It's happening constantly in eccentric orbits. As a planet approaches perihelion (the closest point to the Sun in its orbit), it speeds up. This speed increase comes from a conversion of the planet's gravitational potential energy into kinetic energy. Once it passes perihelion, it starts moving further away from the Sun and begins to convert its kinetic energy back into potential energy, slowing it down.
Dear Kryptid
It seems to me that you didn't understand me correctly.
You discuss on a stable orbital cycle - with eccentric orbit. (Every full cycle, the distance between the objects is constant).
I don't discuss on this case.
Let's start with the meaning of Gravitational Field
https://en.wikipedia.org/wiki/Gravitational_energy#/media/File:Gravitational_field_Earth_lines_equipotentials.svg
"Image depicting Earth's gravitational field. Objects accelerate towards the Earth, thus losing their gravitational energy and transforming it into kinetic energy."
As you can see from any direction, the vector is pointing to the center of mass.
It is also stated:
https://en.wikipedia.org/wiki/Potential_energy
Common types of potential energy include the gravitational potential energy of an object that depends on its mass and its distance from the center of mass of another object,"
Therefore, Vp represents a Velocity vector between the center of mass of the two objects.
Hence,at any distance (even if we call it - perihelion), the Vp vector will be on the direct line between the two center of mass.
If they don't have any orbital velocity, is there any possibility for the Vp to be shifted by 90 degrees in order to be transformed into new orbital velocity Vk?
I assume that if we could add wings to the object than this might help.
So, if one of the objects is an airplane or shuttle, it could convert its Vp into Vk.
Another possibility is by adding external force as magnetic field or rocket.

t's unwise to assume that all stars and particles will be traveling in a straight line towards the black hole. Conservation of momentum and the initial trajectories and velocities are relevant.
Why not?
Let's assume that the objects have already some low orbital kinetic energy (too low for a stable orbital cycle)
So, due to the Conservation of momentum they will keep that orbital velocity.
However, as the Vp (Falling velocity due to potential energy) is forcing the objects to get closer every cycle, eventually - they must collide with each other.

If you have an idea how to convert the Vp by 90 degrees to - new Vk (orbital velocity), without external force or wings - than please show it mathematically.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 31/10/2019 06:04:50
Therefore, Vp represents a Velocity vector between the center of mass of the two objects.

No, it doesn't. Potential energy is energy. It isn't a velocity. It isn't a vector.

If they don't have any orbital velocity, is there any possibility for the Vp to be shifted by 90 degrees in order to be transformed into new orbital velocity Vk?

If we're talking about two objects falling directly towards each other in a straight line with no other forces at work, then no, you won't get any orbital energy because what you have isn't an orbit.

I assume that if we could add wings to the object than this might help.
So, if one of the objects is an airplane or shuttle, it could convert its Vp into Vk.

Which, of course, has nothing to do with orbits (since they happen in a vacuum where wings don't work).

Another possibility is by adding external force as magnetic field or rocket.

If two objects are falling towards each other in a straight line, yes, some other force would have to be added in order to get an orbit out of it. I don't know what this has to do with anything, though. It certainly doesn't have anything to do with the fact that gravity doesn't violate conservation of energy.

Why not?

Because stars and particles will have their own random trajectories and velocities before getting anywhere near the black hole. Probability alone strongly suggests that any given star or particle will not be traveling directly towards the black hole on a straight path. Conservation of momentum won't let the particle or star suddenly stop its current trajectory and fall in a straight line towards the black hole. It will follow a curved path instead.

Let's assume that the objects have already some low orbital kinetic energy (too low for a stable orbital cycle)
So, due to the Conservation of momentum they will keep that orbital velocity.

No they won't. The black hole's gravity will accelerate them. Momentum will be conserved because some of the momentum will be transferred to the black hole via gravitational forces.

However, as the Vp (Falling velocity due to potential energy) is forcing the objects to get closer every cycle, eventually - they must collide with each other.

It won't necessarily force them to get closer each cycle. It all depends on the specifics.

If you have an idea how to convert the Vp by 90 degrees to - new Vk (orbital velocity), without external force or wings - than please show it mathematically.

Why would I do that when that's not even what I'm arguing?

Enough of this. No more irrelevant talk. You have consistently avoided the key issue. So answer this question: Is orbital energy a form of energy?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/10/2019 07:21:38
No, it doesn't. Potential energy is energy. It isn't a velocity. It isn't a vector.
Potential energy is converted into falling kinetic energy that creats a velocity vector between the two objects.
If two objects are falling towards each other in a straight line, yes, some other force would have to be added in order to get an orbit out of it.
Thanks
So, you agree that Vp represents the direct falling velocity between the two objects.
Because stars and particles will have their own random trajectories and velocities before getting anywhere near the black hole
Actually, I agree with you.
Therefore, I have stated that there might be some initial sort of orbital kinetic energy.
Let's assume that the objects have already some low orbital kinetic energy (too low for a stable orbital cycle)
So, due to the Conservation of momentum they will keep that orbital velocity.
However, even if there is some sort of random orbital velocity - Vk, and that velocity can't hold a stable orbital momentum, than sooner or later the object should fall in (if it is too low) or just cross by and go away if it is too fast.
No they won't. The black hole's gravity will accelerate them. Momentum will be conserved because some of the momentum will be transferred to the black hole via gravitational force
There is no orbital acceleration.
The only acceleration is due to potential energy.
As we all agree that potential energy set the falling in kinetic energy and that energy represents a velocity Vp of a falling object.
So, there is acceleration in the Vp vector as
Ep = MGR
But there is no acceleration in the orbital velocity due to that potential energy.
It won't necessarily force them to get closer each cycle. It all depends on the specifics.
As long as there is a stable orbital momentum with regards to the falling vector (due to potential energy) than the orbital object will get back to the same distance every full orbital cycle.

Probability alone strongly suggests that any given star or particle will not be traveling directly towards the black hole on a straight path. Conservation of momentum won't let the particle or star suddenly stop its current trajectory and fall in a straight line towards the black hole. It will follow a curved path instead.
This is incorrect assumption.
We can see it from our location on Earth
There are many objects/asteroids that orbit around the sun.
From time to time the earth cross their orbital path.
In this case, some of the objects are colliding/falling directly to the earth. We see them as meteorites.
However, if I understand it correctly, none really is attracted by the earth and starts its stable orbital cycle around the earth instead of the sun.
Hence, the Conservation of momentum won't let any object/asteroid/particle suddenly to change its current trajectory/velocity and starts its orbital cycle around the Earth or any other main object including the BH/SMBH.

Enough of this. No more irrelevant talk. You have consistently avoided the key issue. So answer this question: Is orbital energy a form of energy?
This is the most important issue in all our discussion so far!!!
I have proved that random trajectories or velocities around the object can't hold a stable orbital cycle by itself.
This proves that our assumption that somehow objects start to orbit around each other is totally INCORRECT.
Our scientists must look for better explanation why there are so many objects that orbits around each other in a stable orbital cycle
Again – you can't convert any falling in kinetic energy into orbital kinetic energy!!!
However, If you are a satellite and you have full control on the power, than there is a possibility to approach the main object at the correct velocity and phase and find the correct window to get a stable orbital cycle, but this can't be done randomly (you can ask NASA about it)
Therefore, the assumption that somehow by random falling/crossing by activity, objects should orbit around each other - is the most fatal mistake of our scientists.
Hence, this issue is much more important than any other open issue including the conservation of energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 31/10/2019 17:39:03
Hence, this issue is much more important than any other open issue including the conservation of energy.

It doesn't matter how important you think it is, we both agreed that we will focus on conservation of energy until that particular issue is resolved. So answer my question: is orbital energy a form of energy?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/11/2019 08:03:58
It doesn't matter how important you think it is, we both agreed that we will focus on conservation of energy until that particular issue is resolved. So answer my question: is orbital energy a form of energy?

Yes it is.
In order to get better understanding about the gravitational conservation of energy let me use the following example for Newton’s orbital cannon:
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon
"Newton reasoned that, if the cannon ball was fired with exactly the right velocity, the ball would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. It would be placed in orbit around the Earth."
What do we understand from that:
1. Potential energy: it is stated clearly that the ball is "always falling in the gravitational field" to the Earth (actually to the center of the Earth.
That proves that the potential energy of the ball is converted constantly to a falling kinetic energy.
2. Orbital kinetic energy - It is also stated that "if the cannon ball was fired with exactly the right velocity" "It would be placed in orbit around the Earth"
So, if its orbital kinetic energy can exactly compensate its free falling energy (due to its potential energy) "It would be placed in orbit around the Earth".
In other words:
If we shoot that ball at the correct velocity (with correct orbital kinetic energy) high enough from the earth surface than although the ball is always converts its potential energy to in falling kinetic energy "It would be placed in orbit around the Earth" - forever (assuming that there is no friction as air),.
So, the ball keeps its orbital velocity/Orbital kinetic energy.
The Potential energy has one mission. To force that ball to meet the earth, while the orbital kinetic energy keeps that ball in the same shooting velocity.
Therefore, when we discuss on conservation of energy in orbital cycle, we actually discuss on conservation energy of orbital kinetic energy.
That proves that the potential energy can't be converted into orbital kinetic energy.
On the contrary. the potential energy is constantly forcing the ball to fall directly to earth while the kinetic energy keeps it in a constant orbital cycle.
Therefore, I wonder what is the added value of Total Kinetic Energy?
Our scientists claim that
Total Orbital Kinetic energy = Orbital kinetic energy + Potential energy
However, based on Newton’s orbital cannon it seems to me that we should claim that:
Total Orbital Kinetic energy = Only the Orbital kinetic energy
Total energy = Orbital kinetic energy + Potential energy
Hence, due to the Momentum conservation in space (no friction) there is a conservation of energy only  in the Orbital Kinetic energy.
Now, let's assume that we gave that Newton's cannon ball the correct orbital velocity in order to keep it at constant orbital cycle around the earth.
As there is no friction, the orbital velocity should stay at the same orbital radius forever.
The value of the requested velocity is:
V^2 = MG/r
That velocity is called the "magic velocity"
https://www.scienceabc.com/nature/universe/what-is-orbital-velocity.html
"Newton realized that when the ball is propelled at a certain, magic velocity, it will never fall."
It is also stated:
"One can infer from the expression that first, the velocity decreases with r, the orbit’s distance from the center of Earth. This means that satellites orbiting closer to Earth’s surface must travel faster than satellites orbiting further away. The moon, which lies almost 385,000 km away, races around us at 1.002 km/s, while the International Space Station (ISS), merely 400 km away, completes a lap every 1.5 hours, racing at 7.67 km/s. Secondly, the velocity is independent of the mass m of the orbiter, which means that a multimillion-ton moon or a single-gram nail must travel at the same velocity to achieve orbit around Earth (at the same distance, r, that is)."
It is clear that as the ball will be placed higher from the Earth the requested orbital velocity should be lower and vice versa.
Hence, if the ball or Moon is already in a stable orbital cycle and we push it inwards (to the earth) its current orbital velocity would be too low to keep it in a stable orbital velocity (at a shorter radius) and therefore it should eventually fall in.
In the same token, if the ball or Moon is in a stable orbital cycle and we pull it outwards (from the earth) it's current orbital velocity would be too high too keep it in a stable orbital velocity (at a higher radius) and it should eventually escape from the gravitational Earth.
https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg

Conclusion -
I have proved that the potential energy can't increase the orbital kinetic energy.
At the maximum, due to the momentum conservation or energy conservation, that energy will stay as is forever.
Therefore, the assumption that a practical can increase its orbital velocity to almost the speed of light as it falls into the SMBH is a pure imagination.

 
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/11/2019 14:04:09
Yes it is.

Okay, good. Now what does the law of conservation of energy say?

In order to get better understanding about the gravitational conservation of energy let me use the following example for Newton’s orbital cannon:
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon
"Newton reasoned that, if the cannon ball was fired with exactly the right velocity, the ball would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. It would be placed in orbit around the Earth."
What do we understand from that:
1. Potential energy: it is stated clearly that the ball is "always falling in the gravitational field" to the Earth (actually to the center of the Earth.
That proves that the potential energy of the ball is converted constantly to a falling kinetic energy.
2. Orbital kinetic energy - It is also stated that "if the cannon ball was fired with exactly the right velocity" "It would be placed in orbit around the Earth"
So, if its orbital kinetic energy can exactly compensate its free falling energy (due to its potential energy) "It would be placed in orbit around the Earth".
In other words:
If we shoot that ball at the correct velocity (with correct orbital kinetic energy) high enough from the earth surface than although the ball is always converts its potential energy to in falling kinetic energy "It would be placed in orbit around the Earth" - forever (assuming that there is no friction as air),.
So, the ball keeps its orbital velocity/Orbital kinetic energy.
The Potential energy has one mission. To force that ball to meet the earth, while the orbital kinetic energy keeps that ball in the same shooting velocity.
Therefore, when we discuss on conservation of energy in orbital cycle, we actually discuss on conservation energy of orbital kinetic energy.
That proves that the potential energy can't be converted into orbital kinetic energy.
On the contrary. the potential energy is constantly forcing the ball to fall directly to earth while the kinetic energy keeps it in a constant orbital cycle.
Therefore, I wonder what is the added value of Total Kinetic Energy?
Our scientists claim that
Total Orbital Kinetic energy = Orbital kinetic energy + Potential energy
However, based on Newton’s orbital cannon it seems to me that we should claim that:
Total Orbital Kinetic energy = Only the Orbital kinetic energy
Total energy = Orbital kinetic energy + Potential energy
Hence, due to the Momentum conservation in space (no friction) there is a conservation of energy only  in the Orbital Kinetic energy.
Now, let's assume that we gave that Newton's cannon ball the correct orbital velocity in order to keep it at constant orbital cycle around the earth.
As there is no friction, the orbital velocity should stay at the same orbital radius forever.
The value of the requested velocity is:
V^2 = MG/r
That velocity is called the "magic velocity"
https://www.scienceabc.com/nature/universe/what-is-orbital-velocity.html
"Newton realized that when the ball is propelled at a certain, magic velocity, it will never fall."
It is also stated:
"One can infer from the expression that first, the velocity decreases with r, the orbit’s distance from the center of Earth. This means that satellites orbiting closer to Earth’s surface must travel faster than satellites orbiting further away. The moon, which lies almost 385,000 km away, races around us at 1.002 km/s, while the International Space Station (ISS), merely 400 km away, completes a lap every 1.5 hours, racing at 7.67 km/s. Secondly, the velocity is independent of the mass m of the orbiter, which means that a multimillion-ton moon or a single-gram nail must travel at the same velocity to achieve orbit around Earth (at the same distance, r, that is)."
It is clear that as the ball will be placed higher from the Earth the requested orbital velocity should be lower and vice versa.
Hence, if the ball or Moon is already in a stable orbital cycle and we push it inwards (to the earth) its current orbital velocity would be too low to keep it in a stable orbital velocity (at a shorter radius) and therefore it should eventually fall in.
In the same token, if the ball or Moon is in a stable orbital cycle and we pull it outwards (from the earth) it's current orbital velocity would be too high too keep it in a stable orbital velocity (at a higher radius) and it should eventually escape from the gravitational Earth.
https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg

Conclusion -
I have proved that the potential energy can't increase the orbital kinetic energy.
At the maximum, due to the momentum conservation or energy conservation, that energy will stay as is forever.
Therefore, the assumption that a practical can increase its orbital velocity to almost the speed of light as it falls into the SMBH is a pure imagination.

This sounds like you are attacking existing orbital models instead of defending your own model. As I said before, evidence against an existing model is not evidence in favor of your model: https://en.wikipedia.org/wiki/False_dilemma

Right now, we are to focus our discussion on the energetic mechanism behind your model as per our mutual agreement.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/11/2019 15:35:53
This sounds like you are attacking existing orbital models instead of defending your own model. As I said before, evidence against an existing model is not evidence in favor of your model: https://en.wikipedia.org/wiki/False_dilemma

Right now, we are to focus our discussion on the energetic mechanism behind your model as per our mutual agreement.

Dear Kryptid
Newton have proved by it's "orbital cannon" that Potential energy can't be transformed into orbital kinetic energy.
Do you agree with that?
Please answer: Yes or No
This is the most important issue in our discussion on the energetic mechanism for orbital systems.
How could it be that there are so many orbital systems in our galaxy/universe while there is no way to increase any orbital kinetic energy by potential energy.
If the potential energy can't be used as a source of orbital kinetic energy, than what could be the source for that?
Newton have found that "if the cannon ball was fired with exactly the right velocity", and gaining the "magic orbital velocity" "it would be placed in orbit around the Earth".
So, we all understand that a canon can set the requested energy that is needed for a stable orbital system.
What kind of "canon" the Universe is using in order to set all of those orbital systems in the galaxy?
How that "canon" could set the magic velocity that perfectly fits any orbital system?
What is the real source for the orbital kinetic energy in our universe?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 02/11/2019 19:34:17
Dear Kryptid
Newton have proved by it's "orbital cannon" that Potential energy can't be transformed into orbital kinetic energy.
Do you agree with that?
Please answer: Yes or No
This is the most important issue in our discussion on the energetic mechanism for orbital systems.
How could it be that there are so many orbital systems in our galaxy/universe while there is no way to increase any orbital kinetic energy by potential energy.
If the potential energy can't be used as a source of orbital kinetic energy, than what could be the source for that?
Newton have found that "if the cannon ball was fired with exactly the right velocity", and gaining the "magic orbital velocity" "it would be placed in orbit around the Earth".
So, we all understand that a canon can set the requested energy that is needed for a stable orbital system.
What kind of "canon" the Universe is using in order to set all of those orbital systems in the galaxy?
How that "canon" could set the magic velocity that perfectly fits any orbital system?
What is the real source for the orbital kinetic energy in our universe?

Again, you are trying to point out what you consider to be a short-coming in contemporary cosmological models. We are not focusing on that right now as per our prior agreement to focus purely on how your model deals with conservation of energy. You can keep these questions in mind for future use, if you wish, but I am going to ignore all irrelevant issues at the moment. So please don't try to move the discussion to other areas, as I will simply ignore anything that doesn't have to do with how your model is supposed to get around the conservation of energy issue.

You agreed that orbital energy is a form of energy. Next, what does the law of conservation of energy say?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/11/2019 04:00:18
We are not focusing on that right now as per our prior agreement to focus purely on how your model deals with conservation of energy.
If you wish to focus on my model or any sort of model, don't you think that first you must know how the gravitational conservation of energy really works?
Newton have proved with his cannon example that there is no energy transformation between potential energy to orbital kinetic energy. Therefore, that kind of assumption is a fatal error.
He also proved that in order to set a stable orbital bonding you must first set the orbital object at that "magic velocity".
Actually, we have to ask how could it be that there is any sort of change in the orbital radius.
It is a severe contradiction to the orbital magic velocity.
We all know that the moon is drifting outwards.
Due to the conservation of energy, (and as there is no transformation of energy between the orbital kinetic energy to potential energy), it shouldn't change its current kinetic orbital energy.
Therefore, theoretically, it can't increase or decrease its orbital velocity.
So, once it's drifted inwards, the current orbital velocity will be too low to keep it in the orbital path and therefore, it MUST fall in.
In the same token, once it is drifted outwards, it should eventually be ejected from the Earth.
That shows that your understanding about conservation of energy is very poor.
So how can you speak in the name of the conservation of energy in my model while your current understanding/assumption in this key issue is totally incorrect?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/11/2019 06:25:42
If you wish to focus on my model or any sort of model, don't you think that first you must know how the gravitational conservation of energy really works?

How it works is already known. So what does the law of conservation of energy say?

That shows that your understanding about conservation of energy is very poor.

Except that it doesn't. I've provided links to authoritative sources clearly stating what conservation of energy is. Are you saying those links give the incorrect definition? Then what is the correct definition?

So how can you speak in the name of the conservation of energy in my model while your current understanding/assumption in this key issue is totally incorrect?

It seems like you are now claiming that "energy cannot be created or destroyed" is not the definition of the law of conservation of energy. Is this what you are saying?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/11/2019 16:58:59
It seems like you are now claiming that "energy cannot be created or destroyed" is not the definition of the law of conservation of energy. Is this what you are saying?
The question is: What Newton is claiming by its Cannon ball example?
Please look again at that image:
https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg
Newton shows that any orbital object must be boosted horizontally at the magic velocity in order to keep it in an orbital path.
Nothing is random in orbital system. You can't just hope to boost an orbital object at a random velocity and wish that that somehow those objects will set an orbital system
This is IMAGINATION
If that velocity is too low it will fall directly to the main object, if it is too high it will be ejected outwards.
Newton gave us the clear understanding that potential energy can't be transformed into Kinetic energy.
So, the outcome is quite simple.
1. There is no way to increase the Orbital kinetic energy by decreasing the potential energy.
2. Therefore, there is no way to increase the orbital velocity of the object by decreasing the potential energy. This is the most important issue in all our discussions so far!!!
3. Hence, if an object orbits at the magic velocity which keeps it in the orbital path, it can't increase that velocity (even if it falls in). In other words - Orbital objects couldn't increase their orbital velocity as they decrease their radius to the main object
4. Therefore, due to the conservation energy (kinetic orbital energy) if the orbital object is decreasing its radius while it can't increase its orbital velocity, it breaks the magic orbital velocity and therefore it must collide with the main object.
5. In the same token - if the orbital object is increasing its radius while it can't decrease its orbital velocity, it breaks the magic orbital velocity and therefore it must ejected outwards.
Do you agree with all of that?
If you don't agree than you don't accept Newton's cannon ball explanation for gravity
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 07/11/2019 17:04:24
The question is: What Newton is claiming by its Cannon ball example?

No, the question is about conservation of energy. This is a question that you seem to be afraid of answering because you know it kills your model.

Newton shows that any orbital object must be boosted horizontally at the magic velocity in order to keep it in an orbital path.
Nothing is random in orbital system. You can't just hope to boost an orbital object at a random velocity and wish that that somehow those objects will set an orbital system
This is IMAGINATION
If that velocity is too law it will fall directly to the main object, if it is too high it will be ejected outwards.
Newton gave us the clear understanding that potential energy can't be transformed into Kinetic energy.
So, the outcome is quite simple.
1. There is no way to increase the Orbital kinetic energy by decreasing the potential energy.
2. Therefore, there is no way to increase the orbital velocity of the object by decreasing the potential energy. This is the most important issue in all our discussions so far!!!
3. Hence, if an object orbits at the magic velocity which keeps it in the orbital path, it can't increase that velocity (even if it falls in). In other words - Orbital objects couldn't increase their orbital velocity as they decrease their radius to the main object
4. Therefore, due to the conservation energy (kinetic orbital energy) if the orbital object is decreasing its radius while it can't increase its orbital velocity, it breaks the magic orbital velocity and therefore it must collide with the main object.
5. In the same token - if the orbital object is increasing its radius while it can't decrease its orbital velocity, it breaks the magic orbital velocity and therefore it must ejected outwards.
Do you agree with all of that?
If you don't agree than you don't accept Newton's cannon ball explanation for gravity

These are just more misunderstandings that aren't relevant to the definition of conservation of energy. We aren't currently discussing how orbits come to be. One thing is for sure, though: the scenarios you discuss actually violate conservation of energy. Objects can't spontaneously move away from the object they orbit. That would make extra gravitational potential energy pop up out of nowhere. Therefore, conservation of energy in itself automatically demonstrates that your understanding of orbits is wrong.

However, I will address this. In return, I expect you to give me the courtesy of finally answering my question.

If you start with some simplified assumptions (such as making the orbit perfectly circular, removing all other interference such as the gravitational pulls of other objects and make the Earth and Moon perfectly-rigid, perfectly-uniform, perfect spheres) then the scenario you envision where the Moon remains at the "magic velocity" will be true. Under such circumstances, the gravitational force and the centrifugal force will perfectly balance each other and the Moon will continue to move around the Earth with a constant kinetic energy and therefore a constant velocity. In this case, it is true that gravitational potential energy cannot be transformed into orbital kinetic energy or vice-versa because there are no extraneous factors that can cause such a transformation.

As a consequence of all of these things, the orbital distance of the Moon from the Earth must also remained fixed over time. There are two different ways of looking at this: in terms of net forces and in terms of net energy,

Net forces: In order for the Moon to drift away from or towards the Earth, something would have to occur to disrupt the balance between the gravitational force and the centrifugal force. If the centrifugal force became stronger than the gravitational force, the Moon would move outward. If the gravitational force became stronger than the centrifugal force, the Moon would move inward. However, since the centrifugal force is determined by the Moon's velocity, the centrifugal force cannot change because the Moon's velocity is unchanging. The gravitational force is determined by the combined mass of the Earth and Moon. Since that mass isn't changing over time either, then the gravitational force cannot change. So the Moon, under these circumstances, is trapped at its orbital distance for all time.

Net energy: We are considering a system in isolation that has no means of transforming gravitational potential energy and orbital kinetic energy into each other. For this reason, both the gravitational potential energy and the orbital kinetic energy are constant and unchanging. This means it is impossible for the Moon to drift away from the Earth because that would result in an increase of gravitational potential energy. However, there is nowhere for this new gravitational potential energy to come from, since all forms of energy transformation are impossible in this isolated system. Likewise, the Moon can't drift towards the Earth because it has no means of shedding extra gravitational potential energy. The law of conservation of energy won't allow the Moon to spontaneously create or destroy gravitational potential energy.

So now that I've addressed that, it's time for you to do your part and finally answer my question:

What is the definition of the law of conservation of energy?

If you respond with an answer that isn't the conventional one (and you know what it is), then I will ask for you to supply an authoritative, scientific source to back up your definition. I will not respond to any further arguments from you until you have finally answered this question in a non-dodging manner.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/11/2019 10:00:49
No, the question is about conservation of energy. This is a question that you seem to be afraid of answering because you know it kills your model.
OK
Let me start with the conservation of energy in my model.
Please look again at the following image by Newton:
https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg
Let's assume that in the center we have a SMBH and there is a virtual canon up above.
That virtual cannon fire a virtual pair of particles at the exact magic velocity.
I hope that you agree that so far there is no violation in the energy conservation.
The Ultra gravity force of the SMBH keeps that virtual pair in the orbital path with about the speed of light.
We all know that around the SMBH there is a very strong electromagnetic field.
At some point of time that virtual pair is converted by the electromagnetic field/energy into real particle pair.
So, due to energy conservation, some energy from the electromagnetic had been transformed into this new born particle pair.
Now we need to find the source for that electromagnetic field:
The SMBH has a very strong gravity force that holds million (or even billions) of objects in orbital cycles around it.
Those objects set strong tidal force on the SMBH that is converted to extra heat and rotates its core/or even layers in the core.
This activity increases the electromagnetic field/energy of the SMBH.
Therefore, the conservation of energy works as follow:
The SMBH's ultra high gravity force sets a tidal impact on the SMBH itself.
That tidal force generates extra heat/extra core rotatation that increases the ultra high electromagnetic field.
Some of that electromagnetic energy is used to transformed the virtual pair particle into real pair particle.
Any objection?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 08/11/2019 16:49:39
I will wait until you have answered this:

So now that I've addressed that, it's time for you to do your part and finally answer my question:

What is the definition of the law of conservation of energy?

If you respond with an answer that isn't the conventional one (and you know what it is), then I will ask for you to supply an authoritative, scientific source to back up your definition. I will not respond to any further arguments from you until you have finally answered this question in a non-dodging manner.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/11/2019 17:06:10
I will wait until you have answered this:
What is the definition of the law of conservation of energy?.
The Law of Conservation of Energy Defined
https://www.thoughtco.com/law-of-conservation-of-energy-605849
"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."
So, where is the problem with my model?
The SMBH's ultra high gravity force sets a tidal impact on the SMBH itself.
That tidal force generates extra heat/extra core rotation that increases the ultra high electromagnetic field.
Some of that electromagnetic energy is used to transformed the virtual pair particle into real pair particle.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/11/2019 17:26:38
"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."

Very good.

So, where is the problem with my model?

Your model posits the creation of energy, violating the "energy cannot be created" part of conservation of energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/11/2019 20:14:55
Your model posits the creation of energy, violating the "energy cannot be created" part of conservation of energy.
My model is based on Tidal renewable Energy:
https://energyfive.net/2017/10/07/tidal-energy/
"Tidal energy sources come from the gravitational movements of sun and moon as these creates high and low tides. When you think of it, it means that the ocean wave energy sources are renewable as compare to fossil fuels and nuclear reserves, moon and sum gravitational fields is not though of to be ceased in near future"
So, Tidal is based on Gravitational fields, it is renewable and it is not though of to be ceased in near future.
Therefore:
As long as there is gravitational field - there is Tidal.
As long as there is Tidal - there is renewable energy source.
https://ui.adsabs.harvard.edu/abs/2010Natur.468..952B/abstract
"Tidally driven flow in the Earth's liquid core develops internal shear layers, which distort the internal magnetic field and generate electric currents."
As long as there is Tidal renewable energy source - There is renewable Magnetic Field
As long as there is renewable Magnetic Field - There is renewable electromagnetic energy
So what is the problem with that?
Why can't we use that renewable energy source for the pair particle creation?
Do you still see any contradiction with conservation of energy?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 10/11/2019 23:09:53
So what is the problem with that?

If you posit that it creates new energy, then this is the problem:

"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."

Renewable energy is not the creation of new energy nor is it unlimited.

So, Tidal is based on Gravitational fields, it is renewable and it is not though of to be ceased in near future.

"Near future" being the operative word there. It will run out eventually.

As long as there is gravitational field - there is Tidal.
As long as there is Tidal - there is renewable energy source.

Nope. Go back and look at the tidal energy equation. More than a source of gravity is needed.

Do you still see any contradiction with conservation of energy?

You can answer that question yourself. Just ask yourself "does this create new energy?" If the answer is "yes", then you've contradicted the law of conservation of energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/11/2019 23:35:10
Quote from: Dave Lev
So, Tidal is based on Gravitational fields, it is renewable and it is not though of to be ceased in near future.
"Near future" being the operative word there. It will run out eventually.
In fact, over 80% of it has been used up already.  It will run completely out in 7.5 billion years, more due to the breakdown of the mechanism than to the tank running dry.  In one billion years, the energy will become very difficult to access. The energy source for tidal energy is kinetic energy, not gravity at all.  Gravity is a field or a force, not energy.

Quote
Quote from: Dave Lev
Do you still see any contradiction with conservation of energy?
You can answer that question yourself. Just ask yourself "does this create new energy?" If the answer is "yes", then you've contradicted the law of conservation of energy.
The whole discussion is summed up here .https://m.xkcd.com/2217/
Note the hover text, which is almost exactly your response Kryptid. :)
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/11/2019 16:11:42
You can answer that question yourself. Just ask yourself "does this create new energy?" If the answer is "yes", then you've contradicted the law of conservation of energy.
Well, the answer is clearly YES, while there is no contradiction with the law of conservation of energy .
In order to understand it, let's start with the following:
http://www.isustainableearth.com/energyefficiency/tidal-power-facts-understanding-how-tidal-energy-works
Quick Fact on How Tidal Energy Slows Earth Rotation
"During our research, we found an interesting detail surrounding tidal energy and the effect its generation could have on the Earth.  Naturally the movement of tides causes a loss of mechanical energy that is associated with the Earth’s rotation.  According to some studies, our planet has actually lost 17 percent of its rotational energy in the past 620 million years."
So, our scientists assume that the Tidal energy is taken from the rotational energy of the objects.
This is incorrect assumption.
I can prove that there could be a tidal energy without any need for any rotation.
Let's look at the Moon.
Its face is locked with the Earth.
Therefore, The tidal bulge doesn't cross its surface. It is a frozen Tidal Bulge. Hence, there is no new energy due to Tidal and therefore, the moon is a frozen object.
That is correct, as long as the object is only affected by one main orbital object.
However, let's assume that around the moon we will set four objects. Each object will be in the size of 1/1000 Moon mass and each one orbits at a different radius (R, 1.2R 1.4R, 1.6R) and different direction.
Let's assume that the Moon and all of those four objects have totally lost their rotational energy.
So, the question is:
Could it be that a tidal energy can be created at this moon although it has totally lost its rotational energy?
The answer should be -YES!!!
If you need an explanation - I will be happy to give.
So, we can see that without any need for rotation, tidal forces can set new energy.
That is exactly the case with our SMBH.
Let's assume that it doesn't rotate at all.
Let's even start with the assumption that it is a totally frozen object.
However, there are so many objects that are orbiting around that SMBH (Lets also assume that none has a rotation energy).
So, each object orbits at a different radius and even at a different orbital direction.
Therefore, Each one of them sets a small moving tidal bulge on the SMBH.
Each tidal bulge must set some small tidal energy as it cross the SMBH surface.
The Impact of millions tidal's bulge that are crossing that surface at different directions and amplitude must cause Ultra high tidal energy at the SMBH.
Therefore, even if the SMBH was frozen, after some time it Must be melted from the Ultra high energy of that tidal activity.
Therefore, Tidal energy is a NEW energy.
It is a direct outcome from gravity force.
So, as long as the gravity is there, and there are more than just one orbital object - we should get new tidal energy even if the rotation of all the objects is Zero!.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 11/11/2019 22:07:29
Well, the answer is clearly YES, while there is no contradiction with the law of conservation of energy .

You said this yourself:

"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."

I bolded four words in that sentence. Can you read them? What do they say?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/11/2019 15:25:27
"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."
I bolded four words in that sentence. Can you read them? What do they say?

There is no contradiction.
Force can generate new energy:
http://energywavetheory.com/forces/force-equation/
"Force is energy over distance (F=E/r)."
In the same token we can claim that
Energy is as follow:
E = F * r
Gravity force is also a force.
Therefore, Gravity force can create new energy by tidal.
In the articale about the tidal it is stated clearly that the Tidal is a direct outcome of a gravity force.
Therefore:

1. Do you agree that Tidal energy is a direct outcome of gravity force?
2. Do you agree that by using orbital system with one main object (BH/SMBH) and several orbital objects around it (that are orbiting at different radius/directions/velocity) they MUST set tidal energy on the main object, (even if all the objects do not rotate)?

If so, do you agree that the SMBH can gain new tidal energy from all the objects that are orbiting around it (without any need to transfer the rotation energy into tidal energy).
That new energy is the source for the ultra high magnetic field around the SMBH and also for all the new particles in the accretion disc.

Let's look again at your following answer:
Net forces: In order for the Moon to drift away from or towards the Earth, something would have to occur to disrupt the balance between the gravitational force and the centrifugal force. If the centrifugal force became stronger than the gravitational force, the Moon would move outward. If the gravitational force became stronger than the centrifugal force, the Moon would move inward. However, since the centrifugal force is determined by the Moon's velocity, the centrifugal force cannot change because the Moon's velocity is unchanging. The gravitational force is determined by the combined mass of the Earth and Moon. Since that mass isn't changing over time either, then the gravitational force cannot change. So the Moon, under these circumstances, is trapped at its orbital distance for all time.
Yes, I fully agree with your explanation that the Moon (or any orbital object) can't increase its orbital kinetic energy (with or without changing the potential energy).
We all know by now that the potential energy can't be converted into orbital kinetic energy.
So how the moon has got its orbital velocity for the first time?
How all the orbital objects in the galaxy have got their magic orbital velocity?
What was the source for the first orbital kinetic energy???
Your answer was - By random.
Sorry - there is no way to convert potential energy to orbital kinetic energy and it is very clear that there is no "random" in that Magic orbital velocity.
The answer is very simple:
All the objects in the galaxy got their first orbital energy at the moment of their creation.
Each particale in the galaxy was born around the SMBH, while it orbits at a magic velocity which is as high as the speed of light.
Those new particles had been converted into real atoms and molecular in the accretion disc and later on they were used to form new stars/planets/moons in the gas cloud around the SMBH.
From that moment, each star/planet/moon can only lose some of its magic orbital velocity (or orbital kinetic energy) over time.
That is a key element in any orbital system.
They can lose orbital kinetic energy over time but they can't gain it.
Therefore, there must be a "friction" in real orbital kinetic energy.
That "friction" force the orbital object to reduce its velocity and be drifted outwards.
Hence, the moon is drifting away from the earth over time and the Earth is drifting away from the Sun over time.
In the same token, the Sun is drifting away from the center of the galaxy over time.
As they are drifting away, they are also decreasing their orbital velocity.
Eventually, they must be ejected from the spiral Arm/disc.
Therefore, there are so many stars outside the galaxy.
Actually, for any star in the spiral galaxy, there is at least one outside.
All of those stars have been ejected from their spiral galaxy home.
None of them can come back to any spiral galaxy.
They have all lost forever.
One day, our solar system will be ejected from the Milky Way galaxy without return.


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 12/11/2019 15:30:17
I will wait until you answer these questions.

I bolded four words in that sentence. Can you read them? What do they say?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/11/2019 05:16:08
I will wait until you answer these questions.
"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."

Dear Kryptid
This statement is fully correct, but it is totally irrelevant for our discussion.
I have clearly proved that tidal gravity energy is/could be a new energy.
How long are you going to hide behind this irrelevant statement
Why are you so afraid to deal with my simple explanation???
Please read it again:
Let's look at the Moon.
Its face is locked with the Earth.
Therefore, The tidal bulge doesn't cross its surface. It is a frozen Tidal Bulge. Hence, there is no new energy due to Tidal and therefore, the moon is a frozen object.
That is correct, as long as the object is only affected by one main orbital object.
However, let's assume that around the moon we will set four objects. Each object will be in the size of 1/1000 Moon mass and each one orbits at a different radius (R, 1.2R 1.4R, 1.6R) and different direction.
Let's assume that the Moon and all of those four objects have totally lost their rotational energy.
So, the question is:
Could it be that a tidal energy can be created at this moon although it has totally lost its rotational energy?
The answer should be -YES!!!
If you need an explanation - I will be happy to give.
So, we can see that without any need for rotation, tidal forces can set new energy.
That is exactly the case with our SMBH.
Let's assume that it doesn't rotate at all.
Let's even start with the assumption that it is a totally frozen object.
However, there are so many objects that are orbiting around that SMBH (Lets also assume that none has a rotation energy).
So, each object orbits at a different radius and even at a different orbital direction.
Therefore, Each one of them sets a small moving tidal bulge on the SMBH.
Each tidal bulge must set some small tidal energy as it cross the SMBH surface.
The Impact of millions tidal's bulge that are crossing that surface at different directions and amplitude must cause Ultra high tidal energy at the SMBH.
Therefore, even if the SMBH was frozen, after some time it Must be melted from the Ultra high energy of that tidal activity.
Therefore, Tidal energy is a NEW energy.
It is a direct outcome from gravity force.
So, as long as the gravity is there, and there are more than just one orbital object - we should get new tidal energy even if the rotation of all the objects is Zero!.

If you think that tidal energy can't be a new energy - than please answer the following questions:
1. Do you agree that tidal gravity can set energy?
2. If yes, do you agree that our scientists think that this energy is only a transformation from the rotational activity of the main object? In other words, they think that the earth gain tidal energy from the Moon, but that energy must reduce the rotational velocity of the earth.
3. If Yes, do you agree that by using two moons (at a different radius/orbital velocity), than even if all the three objects have totally lost their rotational ability, and the Earth is fully faced locked with one of the moons, the other moon can easily set a Tidal energy on the earth?
4. Do you agree that without the ability to rotate, than this tidal energy must be a new energy by definition???
5. So, do you agree that even if our SMBH has totally lost its rotational ability, all the orbital objects around it must create new tidal energy on that SMBH?
Please don't be afraid and stop hiding behind your irrelevant statement. Please face directly those key questions!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 16/11/2019 05:47:38
This statement is fully correct

If the statement is fully correct, then that means it is correct when it says that energy cannot be created. So why is there even any debate here? Your black hole can't create energy. You said so yourself.

but it is totally irrelevant for our discussion.

Now you believe the definition of conservation of energy is irrelevant to a discussion about conservation of energy. Right...

I have clearly proved that tidal gravity energy is/could be a new energy.

That contradicts what you said here:

energy cannot be created

1. Do you agree that tidal gravity can set energy?

Depends on what you mean by "set".

2. If yes, do you agree that our scientists think that this energy is only a transformation from the rotational activity of the main object? In other words, they think that the earth gain tidal energy from the Moon, but that energy must reduce the rotational velocity of the earth.

It depends on the specifics, but it all boils to down to this: something must be moving relative to something else.

3. If Yes, do you agree that by using two moons (at a different radius/orbital velocity), than even if all the three objects have totally lost their rotational ability, and the Earth is fully faced locked with one of the moons, the other moon can easily set a Tidal energy on the earth?

Again, that depends on what you mean by "set".

4. Do you agree that without the ability to rotate, than this tidal energy must be a new energy by definition???

You answered your own question here:

energy cannot be created

For that reason, any energy present isn't "new" energy. It has always existed in one form or another.

5. So, do you agree that even if our SMBH has totally lost its rotational ability, all the orbital objects around it must create new tidal energy on that SMBH?

You answered your own question here:

energy cannot be created

For that reason, any energy present isn't "new" energy. It has always existed in one form or another.

Now, do you understand that "energy cannot be created" and "energy can be created" are opposite, mutually-exclusive statements?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/11/2019 03:19:17
For that reason, any energy present isn't "new" energy. It has always existed in one form or another.
Now, do you understand that "energy cannot be created" and "energy can be created" are opposite, mutually-exclusive statements?

Thanks Kryptid
It seems that I don't understand correctly the real meaning of the following law:
"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."

The main question is: Do we have to count in this law also the gravity forces, especially - tidal forces (or tidal gravity energy)?
1. If the tidal force is part of the " law of conservation of energy" than any energy that it generates can't be considered as new energy. Therefore, any energy that tidal gravity force generates does not violate that law.
2.If the tidal force (or tidal energy) isn't part of the " law of conservation of energy" than any energy that it generates should be considered as new energy. In ant case, as it was not part of the law at the first stage, than it also can't violate that law.
So please - which one is correct?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/11/2019 05:13:53
It seems that I don't understand correctly the real meaning of the following law:

It's pretty self-explanatory. It says that energy can't be created. Plain and simple.

The main question is: Do we have to count in this law also the gravity forces, especially - tidal forces (or tidal gravity energy)?

Yes, because "tidal gravity energy" is energy and "energy cannot be created".

1. If the tidal force is part of the " law of conservation of energy" than any energy that it generates can't be considered as new energy. Therefore, any energy that tidal gravity force generates does not violate that law.

This only works if you mean "changes previously-existing energy from one form into another" when you say "generate". If you mean "create energy" when you say "generate", then that still violates conservation of energy because "energy cannot be created". This should not be so difficult to grasp.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/11/2019 07:03:29
Quote
The main question is: Do we have to count in this law also the gravity forces, especially - tidal forces (or tidal gravity energy)?
Yes, because "tidal gravity energy" is energy and "energy cannot be created".

Thanks!

So, any tidal energy that we see in our universe is already part of the " law of conservation of energy".
However, we know that the source for that energy is Gravity.
Therefore, one of the impacts of the gravity - which is Tidal energy -is clearly part of that law.
Hence, in each gravity system there must be a "potential tidal energy".
It is actually similar to potential energy that can be converted to "falling" kinetic energy.
Therefore, when we look at any orbital system, we have to count the following energies:
1. Orbital kinetic energy (current orbital kinetic energy due to gravity force)
2. Potential energy (Future energy that can be converted into "falling" kinetic energy due to gravity force)
3. Current Tidal energy (current heat due to tidal gravity force)
4. Potential tidal energy (Future heat due to tidal gravity force)

If all of that is part of " law of conservation of energy" than the issue is quite clear.
As the potential energy can be converted in the future into "falling" kinetic energy, the "potential tidal energy" can also be converted (in the future) into heat energy without any need to be taken from any other source.
Once we agree on that we have actually solved the main problem with the " law of conservation of energy".
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/11/2019 14:51:32
Once we agree on that we have actually solved the main problem with the " law of conservation of energy".

Only so long as you recognize that the total energy is finite and does not increase over time because energy cannot be created.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/11/2019 17:12:41
Quote
Once we agree on that we have actually solved the main problem with the " law of conservation of energy".
Only so long as you recognize that the total energy is finite and does not increase over time because energy cannot be created.
Wow!

Thanks for this confirmation!!!
So we agree that:

when we look at any orbital system, we have to count the following energies:
1. Orbital kinetic energy (current orbital kinetic energy due to gravity force)
2. Potential energy (Future energy that can be converted into "falling" kinetic energy due to gravity force)
3. Current Tidal energy (current heat due to tidal gravity force)
4. Potential tidal energy (Future heat due to tidal gravity force)

We see clearly that the total energy in any orbital system is based on (1+2+3+4).
However, all of those energies are fully based on gravity force.
Therefore, if the gravity force is finite, than the energy must also be finite
However, as long as we have gravity force in the system - than by definition all the four energies are still there.
So, the main question is as follow: Could it be that the gravity force in the galaxy (for example) stays forever?
If gravity forces stay forever in the Milky Way than we can clearly claim that the total energy of the Milky Way is infinite.
If the gravity forces in the milky way is limited (finite), than by definition the energy in our galaxy must also be finite.

In order to verify this issue let's assume that in the whole Universe there is only one galaxy.
This galaxy is the Milky Way as we see it today.
So, the question is as follow:
What our scientists expect to see in the universe if we will come back 10^Billion years from now (or infinite time from now)?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 17/11/2019 17:38:26
If gravity forces stay forever in the Milky Way than we can clearly claim that the total energy of the Milky Way is infinite.

No. If the galaxy contained an infinite amount of energy, it would collapse into a black hole. The amount of gravity generated by a given amount of mass is finite and it doesn't change over time. Likewise, the gravitational potential energy is finite and (if distance and mass is held constant) doesn't change over time either.

What our scientists expect to see in the universe if we will come back 10^Billion years from now (or infinite time from now)?

The galaxy would certainly be much different after such an enormous time (all the stars would have long since burned out and even all of the black holes would have evaporated due to Hawking radiation). Even all of the chemical elements may have decayed (if proton decay is a real phenomenon). If so, then it may be that the entire mass of the galaxy had been converted into subatomic particles and radiated out into space. I don't think the galaxy would even exist anymore. However, the gravity created by each subatomic particle would still remain unchanged. Gravity doesn't get weaker over time. If it did, that in itself would violate conservation of energy because it would make the total gravitational potential energy of a gravitationally-bound system decrease over time even if it wasn't converted into other forms. But that can't happen because energy can't be created or destroyed.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/11/2019 06:31:49
No. If the galaxy contained an infinite amount of energy, it would collapse into a black hole. The amount of gravity generated by a given amount of mass is finite and it doesn't change over time. Likewise, the gravitational potential energy is finite and (if distance and mass is held constant) doesn't change over time either.
Yes, I agree with your key message.
At any given moment the energy is finite.
We can clearly see it in the following total 1-4 energies:
1. Orbital kinetic energy (current orbital kinetic energy due to gravity force)
2. Potential energy (Future energy that can be converted into "falling" kinetic energy due to gravity force)
3. Current Tidal energy (current heat due to tidal gravity force)
4. Potential tidal energy (Future heat due to tidal gravity force)

Energies - 1 and 3 represents current real kinetic energy.
Those energies must be finite.
Energies 2 and 4 represent potential energies.
Potential Energy 2 must be finite.
However, potential energy no 4 is there as long as there is orbital system.
You calim that:
Gravity doesn't get weaker over time.
So, let's read it again:
4. Potential tidal energy (Future heat due to tidal gravity force)
So, as long as there is a gravity force, and as long as the main orbital system has at least two orbital objects, it must gain tidal heat energy.
That energy is very low..
Therefore, in one hand it doesn't increase at any given moment the current energy to infinity.
However, on the other hand, as long as the orbital system works - the potential tidal energy is converted to real heat.
Therefore, if the orbital system works forever (infinite) the accumulated Tidal heat is also infinite.
So simple and clear.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/11/2019 08:08:48
However, potential energy no 4 is there as long as there is orbital system.

No.

(1) That energy is finite (if it was infinite, it would be a black hole of infinite size).
(2) That energy is converted into other forms over time, eventually resulting in no more tidal heating. Any finite amount of potential energy being changed into other forms over time must mean that the potential energy eventually goes to zero. If you pour water out of a 10 liter bucket, then the bucket will eventually be empty because it has a limited capacity. The bucket cannot create new water. It's the same way with energy.

So, as long as there is a gravity force, and as long as the main orbital system has at least two orbital objects, it must gain tidal heat energy.

No. That would violate conservation of energy. We already know that the energy is finite, so that tidal heating converts that potential energy into other forms until it is all gone. Then no more tidal heating occurs. In order for tidal heating to continue forever, new energy would have to be created. Of course, as we know, "energy cannot be created". Gravity alone doesn't cause tidal heating, which is something I've told you many times.

Therefore, in one hand it doesn't increase at any given moment the current energy to infinity.

It doesn't increase at all. Ever. Waiting a long time won't change that. That would violate the "energy cannot be created" part of conservation of energy.

However, on the other hand, as long as the orbital system works - the potential tidal energy is converted to real heat.

No, see above.

Therefore, if the orbital system works forever (infinite) the accumulated Tidal heat is also infinite.

"Energy cannot be created". You seem to keep forgetting that. If you propose that the energy in the future is higher than the energy now, then you have violated conservation of energy.

So simple and clear.

So simple and wrong. Why do you struggle so much with "energy cannot be created"? If you were isolated with 78 rubber balls and there was a physical law that stated, "rubber balls cannot be created or destroyed", then you can never have more than 78 rubber balls. You can rearrange them all you want to, wait as long as you want to, throw and catch them as much as you want to, but you'll never have more than 78 of them. It's exactly the same way with energy. Your black hole can never make energy.

So let me ask you this again: do you realize that "energy cannot be created" and "energy can be created" are opposite, mutually-exclusive statements? Which one of those statements is in accordance with the law of conservation of energy?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/11/2019 09:18:25
Quote
Quote from: Dave Lev on Today at 06:31:49
However, potential energy no 4 is there as long as there is orbital system.
No.

(1) That energy is finite (if it was infinite, it would be a black hole of infinite size).
(2) That energy is converted into other forms over time, eventually resulting in no more tidal heating. Any finite amount of potential energy being changed into other forms over time must mean that the potential energy eventually goes to zero. If you pour water out of a 10 liter bucket, then the bucket will eventually be empty because it has a limited capacity. The bucket cannot create new water. It's the same way with energy.

Let me ask you the following:
Do you agree that the potential tidal gravitational energy is based GRAVITY force?
If so, you have to agree that as long as there is a gravity force - there also must be Potential tidal energy.
How can you compare any sort of "bucket" to gravity force.
If gravity force works forever, than the "bucket" must be unlimited.
Please be aware that in tidal energy we actually use the gravity force to create Energy.
Therefore - why can't you agree with the simple idea that the "bucket" is full as long as there is gravity force?

No. That would violate conservation of energy. We already know that the energy is finite, so that tidal heating converts that potential energy into other forms until it is all gone. Then no more tidal heating occurs. In order for tidal heating to continue forever, new energy would have to be created. Of course, as we know, "energy cannot be created". Gravity alone doesn't cause tidal heating, which is something I've told you many times.
You insist that the potential Tidal energy must be limited.
Let's assume that this is the case.
So, you agree that at some point of time the tidal energy should be zero.
Therefore, do you agree that even if there is a gravity force and the orbital system is still working, the tidal heat must go to zero?
However, I have proved that any main host with at least two orbital objects MUST gain tidal energy as a direct outcome of the gravity force (even it it had already created infinite tidal energy over infinite time).
So, if you wish to show that the Potential tidal energy is limited, you have to show how it could be that a host with more than two main orbital objects (at different radius) won't create tidal energy any MORE.


"Energy cannot be created". You seem to keep forgetting that. If you propose that the energy in the future is higher than the energy now, then you have violated conservation of energy.
Again - you miss the whole idea of Tidal gravitational energy.
That energy is a direct outcome of gravity force.
Therefore, as long as there is gravity force there is a potential room for Tidal energy.
I don't see any contradiction between that explanation to the law of energy conservation.
We have already agreed that the potential tidal energy is an integrated part of this law.
Therefore, as long as the gravity can converts potential tidal energy to real heat - it will continue to do so.



Why do you struggle so much with "energy cannot be created"? If you were isolated with 78 rubber balls and there was a physical law that stated, "rubber balls cannot be created or destroyed", then you can never have more than 78 rubber balls. You can rearrange them all you want to, wait as long as you want to, throw and catch them as much as you want to, but you'll never have more than 78 of them. It's exactly the same way with energy. Your black hole can never make energy.
I really don't need 78 rubber balls.
Two gravity balls are good enough.
As long as they keep their gravity force my SMBH will gain more heat energy.
Yes - so simple and clear.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 18/11/2019 22:27:12
I will wait until you have addressed this:

So let me ask you this again: do you realize that "energy cannot be created" and "energy can be created" are opposite, mutually-exclusive statements? Which one of those statements is in accordance with the law of conservation of energy?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/11/2019 07:53:45
I will wait until you have addressed this:

So let me ask you this again: do you realize that "energy cannot be created" and "energy can be created" are opposite, mutually-exclusive statements? Which one of those statements is in accordance with the law of conservation of energy?
Dear Kryptid
 
Energy is not created - it is transformed.
The energy is locked in the potential tidal gravitational energy (remember no 4?).
However, as this potential energy is fully based on gravitational force, than as long as gravitational force is working - then the transformation of that potential energy to heat is also working.
Therefore, the tidal heat energy is a direct outcome of the current tidal potential energy and orbital cycles due to gravitational force.

With regards to energy transformation:
If you don't accept the idea of direct tidal heat energy transformation from the potential tidal energy, I'm ready to consider other source of energy transformation.
Our scientists think that the Tidal energy dissipation reduces the rotational orbit of the object.
I have already proved that this idea is totally incorrect.

However, I start to think that there is a possibility that the tidal heat energy is reducing the orbital kinetic energy.
Therefore, as the tidal heat is accumulated over time, the orbital kinetic energy is reducing over time.
The orbital kinetic energy sets the orbital velocity.
Hence, the orbital velocity is reducing over time.
In order to keep the object at the reduced orbital velocity it must increase its radius.
Otherwise it must fall in.
Therefore, if the Tidal energy is taken from the Orbital kinetic energy, then this activity must increase the radius over time.
This could be the ultimate answer why the moon is receding from the earth and the Earth is receding from the Sun.
If that is correct- than by definition every orbital object is receding from its main central host object over time.
What do you think about it?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 21/11/2019 17:06:27
Energy is not created - it is transformed.

Good. Then you must realize the consequences of this.

(1) The amount of energy contained within any system at any given time is finite (including all forms of potential energy).
(2) Since energy cannot be created, the amount of energy now must be the same as the amount of energy later.
(3) This precludes the idea of unlimited heat because that would mean more energy is present later than is present now.

The energy is locked in the potential tidal gravitational energy (remember no 4?).

Which is finite.

However, as this potential energy is fully based on gravitational force, than as long as gravitational force is working - then the transformation of that potential energy to heat is also working.

No. Particular circumstances are needed for tidal heating to occur. Gravity alone is not enough. The orbit must be eccentric, for one. I demonstrated this earlier in the thread with the tidal heating equation. Here is what I said:

To show this, consider the equation to calculate tidal heating: https://en.wikipedia.org/wiki/Tidal_heating

Etidal = -Im(k2)(21/2)((R5n5e2)/G), where

"Etidal" is the rate of tidal heating in watts
"-Im(k2)" is the efficiency of body dissipation (a dimensionless parameter)
"R" is the radius of the body in meters
"n" is the body's mean orbital motion in radians per second
"e" is the orbital eccentricity, and
"G" is the gravitational constant

If I calculate this heating rate for something like Io:

Etidal = -Im(k2)(21/2)((R5n5e2)/G)
Etidal = -(0.005)(10.5)(((1,822,000)5)((4.1 x 10-5)5)((0.0041)2))/(6.674 x 10-11)
Etidal = (-10.5)((2 x 1031)(1.159 x 10-22)(1.681 x 10-5))/6.674 x 10-11)
Etidal = -6.13 x 1015 watts

But what happens if we modify the scenario where the tidal forces are constant? That is, what if we take away the orbital eccentricity?

Etidal = -Im([k2)(21/2)((R5n5e2)/G)
Etidal = -(0.005)(10.5)(((1,822,000)5)((4.1 x 10-5)5)((0.0)2))/(6.674 x 10-11)
Etidal = (-10.5)((2 x 1031)(1.159 x 10-22)(0))/6.674 x 10-11)
Etidal = 0 watts

The power is zero watts. No heat is generated at all.

When the orbital eccentricity is zero, you don't get any tidal heating. Do the math for yourself if you don't believe me. Since tidal heating causes an orbit to become increasingly circular over time (closer to an eccentricity of zero), the heat isn't unlimited. It stops when the orbit is a circle.

Our scientists think that the Tidal energy dissipation reduces the rotational orbit of the object.
I have already proved that this idea is totally incorrect.

No. No you have not.

However, I start to think that there is a possibility that the tidal heat energy is reducing the orbital kinetic energy.
Therefore, as the tidal heat is accumulated over time, the orbital kinetic energy is reducing over time.
The orbital kinetic energy sets the orbital velocity.
Hence, the orbital velocity is reducing over time.
In order to keep the object at the reduced orbital velocity it must increase its radius.
Otherwise it must fall in.

You have it backwards. The orbiting object gets closer, not further away. Remember, objects that are more distant from a source of gravity have more total energy (potential plus kinetic) than those that are closer. So if the total energy of the object is being reduced, it must be getting closer over time instead of further away. If it was getting further away instead, then that would mean that energy is being created because the total amount of energy the orbiting object possesses is increasing despite the fact that you claim it is also being drained of energy. Conservation of energy won't let energy be created.

If that is correct- than by definition every orbital object is receding from its main central host object over time.

Phobos and Deimos are not. They are getting closer to Mars over time. This is because they orbit Mars faster than Mars rotates. The tidal acceleration drains the orbital energy from those two satellites, causing their orbital radius to decrease.

What do you think about it?

I think that, despite everything I've tried to teach you, you still don't understand how orbits work.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/11/2019 17:37:02
(1) The amount of energy contained within any system at any given time is finite (including all forms of potential energy).
Agree - as long as we agree for - "at any given time"
2) Since energy cannot be created, the amount of energy now must be the same as the amount of energy later.
Again - the energy is not created.
It is transformed from the potential tidal energy.
In the article it is stated clearly:
"This energy gained by the object comes from its gravitational energy,"
So, the potential tidal energy - is the source for the tidal heat.
However, that tidal potential energy represents a gravitational energy.
Therefore - as long as the gravity works, there is a possibility to gain tidal heat energy.
Hence - the tidal energy is a direct outcome from Tidal potential gravitational energy.
There is no way to gain infinite energy at any given time frame, however, as long as the gravity is working, than the tidal energy is available.
In any case, if you wish, I agree that some of it might also be transformed from the orbital kinetic energy (not from the rotational energy).
It is stated:
https://en.wikipedia.org/wiki/Tidal_heating
"Tidal heating (also known as tidal working or tidal flexing) occurs through the tidal friction processes: orbital energy is dissipated as heat in either the surface ocean or interior of a planet or satellite. "
But in this case it must come from the kinetic orbital energy and not from the rotational energy.
Actually you didn't answer this issue.
Do you agree that the Orbital kinetic energy could also be the one that the tidal heat is used for it's energy transformation.
I really can't understand why our scientists have decided that the rotational energy must be the only source for the tidal heat.
Quote
Quote from: Dave Lev on Yesterday at 07:53:45
Our scientists think that the Tidal energy dissipation reduces the rotational orbit of the object.
I have already proved that this idea is totally incorrect.
No. No you have not.
Yes, I did.
 
Quote from: Dave Lev on 11/11/2019 16:11:42
Let's look at the Moon.
Its face is locked with the Earth.
Therefore, The tidal bulge doesn't cross its surface. It is a frozen Tidal Bulge. Hence, there is no new energy due to Tidal and therefore, the moon is a frozen object.
That is correct, as long as the object is only affected by one main orbital object.
However, let's assume that around the moon we will set four objects. Each object will be in the size of 1/1000 Moon mass and each one orbits at a different radius (R, 1.2R 1.4R, 1.6R) and different direction.
Let's assume that the Moon and all of those four objects have totally lost their rotational energy.
So, the question is:
Could it be that a tidal energy can be created at this moon although it has totally lost its rotational energy?
The answer should be -YES!!!
If you need an explanation - I will be happy to give.
So, we can see that without any need for rotation, tidal forces can set new energy.
That is exactly the case with our SMBH.
Let's assume that it doesn't rotate at all.
Let's even start with the assumption that it is a totally frozen object.
However, there are so many objects that are orbiting around that SMBH (Lets also assume that none has a rotation energy).
So, each object orbits at a different radius and even at a different orbital direction.
Therefore, Each one of them sets a small moving tidal bulge on the SMBH.
Each tidal bulge must set some small tidal energy as it cross the SMBH surface.
The Impact of millions tidal bulge that are crossing that surface at different directions and amplitude must cause Ultra high tidal energy at the SMBH.
Therefore, even if the SMBH was frozen, after some time it Must be melted from the Ultra high energy of that tidal activity.
Therefore, Tidal energy is a NEW energy.
It is a direct outcome from gravity force.
So, as long as the gravity is there, and there are more than just one orbital object - we should get new tidal energy even if the rotation of all the objects is Zero!.
If you still disagree, than would you kindly show why a tidal energy can't be transformed for any system with two orbital objects.
With regards to eccentricity
When the orbital eccentricity is zero, you don't get any tidal heating. Do the math for yourself if you don't believe me. Since tidal heating causes an orbit to become increasingly circular over time (closer to an eccentricity of zero), the heat isn't unlimited. It stops when the orbit is a circle.
The formula is clear.
However, it seems to me that our scientists have a severe error in there assumption about the eccentricity.
The main question is - how do we get the tidal heat?
One option is as stated in the article:
https://en.wikipedia.org/wiki/Tidal_heating
"When an object is in an elliptical orbit, the tidal forces acting on it are stronger near periapsis than near apoapsis. Thus the deformation of the body due to tidal forces (i.e. the tidal bulge) varies over the course of its orbit, generating internal friction which heats its interior. This energy gained by the object comes from its gravitational energy,"
However, let's assume that the object is in a pure orbital cycle (eccentricity - 0)
Let's assume that the main host object had lost its rotation. Same issue with the orbital object.
So, as the orbital object orbits around the host, it must set a tidal bulge at the surface of the host that run forward with its orbital cycle.
Why that bulge (which run on the surface of the host) can't generate the requested tidal heat?
So, in this case - even without any rotational energy from both objects and while the orbital eccentricity is Zero, a tidal heat is generated on the surface of the host.
You have it backwards. The orbiting object gets closer, not further away. Remember, objects that are more distant from a source of gravity have more total energy (potential plus kinetic) than those that are closer. So if the total energy of the object is being reduced, it must be getting closer over time instead of further away. If it was getting further away instead, then that would mean that energy is being created because the total amount of energy the orbiting object possesses is increasing despite the fact that you claim it is also being drained of energy. Conservation of energy won't let energy be created.
Do you remember Newton?
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon
"Newton reasoned that, if the cannon ball was fired with exactly the right velocity, the ball would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. It would be placed in orbit around the Earth."
So, let's assume that the orbital object is a cannon ball.
Hence, if the orbital velocity is too low (from the requested magic velocity) - it must getting closer.

However, it seems to me that there must be some sort of mechanism in any orbital system that decreases the orbital velocity and push the orbital object outwards over time.
I'm not sure how this mechanism really works, but it is surly there (although there might be some exception for that mechanism).
So, if 99% of all orbital Moons are drifting outwards, only 1% are drifting inwards.
Phobos and Deimos are excellent examples for the exceptions.
They clearly represent the exceptions of that mechanism.
Phobos and Deimos are not. They are getting closer to Mars over time. This is because they orbit Mars faster than Mars rotates. The tidal acceleration drains the orbital energy from those two satellites, causing their orbital radius to decrease.
Yes, they are getting closer.
However, I totally disagree that "This is because they orbit Mars faster than Mars rotates"
This assumption is totally wrong!
There must be other explanation for those exceptions.
I still need to verify the source for this issue (Mars atmosphere Friction- as they are too close?).
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 22/11/2019 18:05:53
Agree - as long as we agree for - "at any given time"

It has to be for all time. If the total energy later is higher than the total energy now, then energy must have been created. Conservation of energy won’t let that happen.

Again - the energy is not created.
It is transformed from the potential tidal energy.

That “potential tidal energy” is finite. If it was infinite, then the system would collapse into a black hole of infinite mass.

In the article it is stated clearly:
"This energy gained by the object comes from its gravitational energy,"
So, the potential tidal energy - is the source for the tidal heat.
However, that tidal potential energy represents a gravitational energy.
Therefore - as long as the gravity works, there is a possibility to gain tidal heat energy.

No. Once the eccentricity goes to zero, the tidal heating is gone. I showed this with the math.

There is no way to gain infinite energy at any given time frame, however, as long as the gravity is working, than the tidal energy is available.

You can’t have infinite energy at all. Back to my bucket analogy: if you claim that the bucket is always full regardless of how much water is poured out of it, then that means that bucket must be creating water to replace the water that has been poured out. If I have a one gallon bucket that is magically full all the time no matter how much water has been poured out of it (which is what you seem to be arguing, except with energy), then I can start off with one gallon. Then, I take an empty bucket and fill it up with water from the magic bucket until it is full as well. Now I have two full buckets and two gallons of water. Since I now have two gallons of water but started off with only one gallon, that means that one of those gallons wasn't there before. If it wasn't there before, then that water was created.

That's what you are trying to do with energy. But you can’t do that. Energy cannot be created.

Actually you didn't answer this issue.
Do you agree that the Orbital kinetic energy could also be the one that the tidal heat is used for it's energy transformation.

Only if the bodies are not tidally locked with each other.

Yes, I did.

Then you are claiming a violation of conservation of energy. An increase in tidal heat must be compensated for by a decrease in energy elsewhere. If tidal heating occurs without such a decrease elsewhere, then the total energy of the system has increased over time. That would mean that energy has been created. That is not allowed.

Let's look at the Moon.
Its face is locked with the Earth.
Therefore, The tidal bulge doesn't cross its surface. It is a frozen Tidal Bulge. Hence, there is no new energy due to Tidal and therefore, the moon is a frozen object.
That is correct, as long as the object is only affected by one main orbital object.
However, let's assume that around the moon we will set four objects. Each object will be in the size of 1/1000 Moon mass and each one orbits at a different radius (R, 1.2R 1.4R, 1.6R) and different direction.
Let's assume that the Moon and all of those four objects have totally lost their rotational energy.
So, the question is:
Could it be that a tidal energy can be created at this moon although it has totally lost its rotational energy?
The answer should be -YES!!!
If you need an explanation - I will be happy to give.
So, we can see that without any need for rotation, tidal forces can set new energy.
That is exactly the case with our SMBH.
Let's assume that it doesn't rotate at all.
Let's even start with the assumption that it is a totally frozen object.
However, there are so many objects that are orbiting around that SMBH (Lets also assume that none has a rotation energy).
So, each object orbits at a different radius and even at a different orbital direction.
Therefore, Each one of them sets a small moving tidal bulge on the SMBH.
Each tidal bulge must set some small tidal energy as it cross the SMBH surface.
The Impact of millions tidal bulge that are crossing that surface at different directions and amplitude must cause Ultra high tidal energy at the SMBH.
Therefore, even if the SMBH was frozen, after some time it Must be melted from the Ultra high energy of that tidal activity.
Therefore, Tidal energy is a NEW energy.
It is a direct outcome from gravity force.
So, as long as the gravity is there, and there are more than just one orbital object - we should get new tidal energy even if the rotation of all the objects is Zero!.

Let's say that we do indeed wait until the Earth-Moon system is both tidally-locked and the orbit is fully circular. All tidal heating has ceased. It is still possible for an asteroid to swing by and be captured by the Moon's gravity (or brought there by an advanced spaceship, if you prefer). It is true that the addition of that asteroid will cause new tidal heating. However, that asteroid would have brought its own gravitational potential energy with it when you set it in orbit around the Moon. That potential energy is finite, however. Since it would be orbiting the Moon faster than the Moon rotates, its orbit would decay over time and it would eventually collide with the Moon. That collision will obviously mark an end to any tidal heating it could cause. So you still do not have a perpetual motion machine.

If you still disagree, than would you kindly show why a tidal energy can't be transformed for any system with two orbital objects.

Tidal locking plus zero eccentricity equals no tidal heating because there would no longer be any flexing due to local changes in force.

However, let's assume that the object is in a pure orbital cycle (eccentricity - 0)
Let's assume that the main host object had lost its rotation. Same issue with the orbital object.
So, as the orbital object orbits around the host, it must set a tidal bulge at the surface of the host that run forward with its orbital cycle.
Why that bulge (which run on the surface of the host) can't generate the requested tidal heat?
So, in this case - even without any rotational energy from both objects and while the orbital eccentricity is Zero, a tidal heat is generated on the surface of the host.

Orbiting objects become tidally-locked long before their orbits become circular (Pluto and Charon are tidally-locked even though their orbit is still slightly eccentric). That tidal locking means that the tidal bulges will be at the same place on their surfaces at all times. No movement of the tidal bulge equals no tidal heating.

Hence, if the orbital velocity is too low (from the requested magic velocity) - it must getting closer.

That’s exactly what happens.

However, it seems to me that there must be some sort of mechanism in any orbital system that decreases the orbital velocity and push the orbital object outwards over time.

Too bad, because that's wrong. That would violate conservation of energy because it would mean that objects are moving from a state of lower total energy to a state of higher total energy without any energy input. That would mean that energy is being created. That is not allowed.

So, if 99% of all orbital Moons are drifting outwards, only 1% are drifting inwards.

Because they actually have a source of energy to lift them into a higher energy state. It’s called tidal acceleration.

Yes, they are getting closer.
However, I totally disagree that "This is because they orbit Mars faster than Mars rotates"
This assumption is totally wrong!

That “assumption” is based on the known physics of gravity and Newton’s laws. Just because you don’t understand it, doesn’t mean it’s wrong.

Quote
I still need to verify the source for this issue (Mars atmosphere Friction- as they are too close?).

They are about 6,000 and 23,000 kilometers from Mars, and Mars’ atmosphere already has a surface pressure a mere 0.6% that of Earth’s. Atmospheric drag would be a non-factor.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/11/2019 06:08:30
Thanks Kryptid
I do appreciate all your effort in answering my questions.
I will reply back on all the open issues.
However, I prefer to deal with them one by one
Quote
Quote from: Dave Lev on Yesterday at 17:37:02
Hence, if the orbital velocity is too low (from the requested magic velocity) - it must getting closer.
That’s exactly what happens.
Sorry, I wasn't clear in this message.
I had to say: "Hence, if the orbital velocity is too low (from the requested magic velocity) - it must "falling in".
So, if the orbital object is decreasing its orbital velocity than it must "falling in" - but not "getting in".
This is not the same.
"Falling in" means that it breaks the requested "magic velocity".
"Getting in" means that as it moves in it is increasing its orbital velocity in order to meet the "magic velocity" at any given moment.
Therefore if  Phobos is decreasing its orbital velocity than after just few orbital cycles it should collide with Mars.
So, we have to distinguish between "falling in" to "getting in".
In other words - It is increasing its orbital velocity as it is reducing its radius.
So, let's look again at your explanation:
Quote
Quote from: Dave Lev on 21/11/2019 07:53:45
If that is correct- than by definition every orbital object is receding from its main central host object over time.
Phobos and Deimos are not. They are getting closer to Mars over time. This is because they orbit Mars faster than Mars rotates. The tidal acceleration drains the orbital energy from those two satellites, causing their orbital radius to decrease.
So, you offer an examples of Phobos and Deimos.
You claim that their orbital radius is decreasing.
However, based on Newton - as their orbital radius is decreasing they must increase their orbital velocity.
This was very clear by Newron cannon ball explanation.
At any given radius - there must be a "magic velocity"
So, if 10,000 years ago - Phobos radius was R and its "magic orbital velocity" was V;
Than today - as its radius is R-r its magic orbital velocity MUST be V+v.
So, we clearly see that if we wish to keep the orbital activity of the object, than as the radius is decreasing - the orbital velocity is increasing. That is the simple outcome from the "magic orbital velocity" request.
Therefore, as Phobos is decreasing its radius, it also must increase its velocity - (if it still wish to keep itself in a clear orbital cycle.)
However - if Phobos is just decreasing its orbital velocity without increasing its radius - than, it breaks the key request for "magic orbital velocity. Therefore it MUST fall in and after few cycles it has to collide with Mars.
However, we know for sure that currently Phobos is orbiting at the "magic orbital velocity"
That is very clear to all of us.
Therefore, 10,000 years ago when its radius was higher, its orbital velocity was lower.
So do you understand by now that if the object is willing to keep its orbital activity while it is "getting in", it must increase its velocity, while at any given moment it's velocity must meet the "magic orbital velocity"?


 
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/11/2019 06:31:43
So do you understand by now that if the object is willing to keep its orbital activity while it is "getting in", it must increase its velocity, while at any given moment it's velocity must meet the "magic orbital velocity"?

Yes, the net result of a decrease in orbital radius is an increase in orbital velocity. Although the kinetic energy of Phobos has increased, its overall energy (potential plus kinetic) has decreased. This is why a loss of total energy must result in the orbit becoming smaller, not larger.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/11/2019 08:59:51
Yes, the net result of a decrease in orbital radius is an increase in orbital velocity. Although the kinetic energy of Phobos has increased, its overall energy (potential plus kinetic) has decreased. This is why a loss of total energy must result in the orbit becoming smaller, not larger.
So, do you mean that the extra kinetic energy is getting from decreasing the potential energy?
If so, you have a severe error.
Let me explain:
Let's converts the energies into velocities
The potential energy represents a falling in velocity vector.
Actually at any given moment - the orbital object is falling in at a vector velocity - Vf (falling in velocity).
The kinetic energy represents a forward kinetic velocity vector - Vk (kinetic velocity).
If we will shut down the potential energy - the object will keep its Vk momentum and will be ejected from the main host.
If we will shut down the kinetic energy - the object will fall in directly to the center of the main host.
So, Vk is fully orthogonal to Vf, therefore, each one of them is working on a totally different dimension.
They can't have any effect on each other (as they are working on a different dimension).
If we increase or decrease the Vf it can't set any impact on Vk.
The orbital system works as long as the falling in velocity fully match the forwards kinetic velocity vector in order to keep a fixed radius (let's focus only on a pure cycle orbit).
So, it is clear that the potential energy sets only the falling in kinetic energy (which is based on Vf).
That falling in kinetic energy can't increase or decrease the orbital kinetic energy (which is based on Vk).
Therefore, I have just proved that there is no transformation between the potential energy to the orbital kinetic energy.
As you claim that energy can't be created, than how the extra orbital kinetic energy which is needed to increase the Vk (orbital velocity vector) is created?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/11/2019 15:14:51
So, do you mean that the extra kinetic energy is getting from decreasing the potential energy?

Yes, that's how falling works.

Let's converts the energies into velocities
The potential energy represents a falling in velocity vector.

That's nonsense. Potential energy is neither a vector nor a velocity.

If we will shut down the potential energy - the object will keep its Vk momentum and will be ejected from the main host.

No it won't. That would violate conservation of energy.

So, Vk is fully orthogonal to Vf, therefore, each one of them is working on a totally different dimension.

Potential energy doesn't have a direction or a speed. It changes with a change of distance from the source of gravity, but that's it.

They can't have any effect on each other (as they are working on a different dimension).

They can and do. If the orbiting object is traveling at a given speed at a given orbital radius, then it is traveling sufficiently quickly to complete a circle (or, more likely, an ellipse) of a given size. If something occurs that slows the object down slightly, then that means that it will fall towards the object it is orbiting more than it normally would on its orbit because the distance it covers in an given time is smaller now. It must complete its orbit in a shorter time span. This is equivalent to a smaller overall orbit radius. This extra falling is exactly what converts gravitational potential energy into orbital kinetic energy, causing the object to speed up once more.

Therefore, I have just proved that there is no transformation between the potential energy to the orbital kinetic energy.

All you have proven is that you don't understand how orbits work.

As you claim that energy can't be created

I don't merely "claim" it. It's objectively true. See conservation of energy.

Quote
than how the extra orbital kinetic energy which is needed to increase the Vk (orbital velocity vector) is created?

It comes from gravitational potential energy, regardless of your complaints. The mechanism is well-understood.

You do understand that "energy cannot be created" means that the total energy now must be the same as the total energy later, don't you?
Title: Re: How gravity works in spiral galaxy?
Post by: puppypower on 23/11/2019 17:42:35
I wonder why any new particle/Atom/Molecular is drifting outwards from the excretion disc of the milky way.
Let's look at the following formula for gravity force:
F = G * M * m / R^2
M = The mass of the SMBH
m = The mass of the particle/Atom/Molecular.

There are two additional variables. The mass of the galaxy is decreasing with time, due to the mass burn that occurs within nuclear fusion; E=MC2.  Secondly, although this mass burn energy and other energy is conserved, the second law says that the entropy of the universe increase with time.

Since entropy needs energy to increase, the second law implies that the useable energy of the universe is decreasing with time, because it gets tied up as entropy that is always increasing. Entropy is consistent with energy conservation, but the energy within entropy, is no longer useable in any net way. The net result is the net mass and net unstable energy is decreasing.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/11/2019 18:59:39
Quote
Quote from: Dave Lev on Today at 08:59:51
Let's converts the energies into velocities
The potential energy represents a falling in velocity vector.
That's nonsense. Potential energy is neither a vector nor a velocity.
It seems that you have missed the key idea
Please look at the following diagram
https://www.quora.com/Why-doesnt-Mercury-fall-to-the-Sun
The Sun is in yellow and Mercury is in green
The blue line represents the gravitational force on Mercury.
However, that line also represents the falling in velocity due to the potential energy.
As it is stated: "Mercury is in a constant state of falling into the Sun"
"The force of gravity from the Sun acts to pull Mercury towards it. If Mercury were stationary, it would fall straight towards the sun (just like what happens when you drop an apple on Earth). "
So, this blue line represents the falling in velocity due to gravity - or due to potential energy.
Therefore we clearly see that the potential energy sets a falling in velocity vector.
Hence, your following statement is technically incorrect:
 
Potential energy doesn't have a direction or a speed. It changes with a change of distance from the source of gravity, but that's it.
Sorry, Potential energy due to gravity sets a clear falling in velocity vector.
Therefore, potential energy (or gravity) can ONLY control on that blue line (falling in vector), as it was stated: "Mercury is in a constant state of falling into the Sun"
The red vector represents the orbital kinetic velocity vector.
We clearly see that the red vector is orthogonal to the blue vector.
Therefore, if we will decrease or increase the blue vector (falling in velocity due to gravity or potential energy) we can't set any change of effect on the Blue vector (orbital kinetic velocity vector).
Therefore
If the orbiting object is traveling at a given speed at a given orbital radius, then it is traveling sufficiently quickly to complete a circle (or, more likely, an ellipse) of a given size
That is fully correct.
If something occurs that slows the object down slightly, then that means that it will fall towards the object it is orbiting more than it normally would on its orbit because the distance it covers in an given time is smaller now. It must complete its orbit in a shorter time span. This is equivalent to a smaller overall orbit radius. This extra falling is exactly what converts gravitational potential energy into orbital kinetic energy, causing the object to speed up once more.
I'm not sure that I understand this explanation:
1. "If something occurs that slows the object down slightly" - do you mean that the orbital velocity is decreasing (the red vector)?
2."then that means that it will fall towards the object it is orbiting more than it normally would on its orbit because the distance it covers in an given time is smaller now".
Yes, I also agree with that.
3. "This is equivalent to a smaller overall orbit radius."
I disagree with that. As you decrease the orbital velocity (red), without changing the falling in velocity vector (blue), than you actually break the "requested magic velocity" - (Please try to remember that idea from Newton cannon ball).
Therefore, you break the following key point in orbital system as stated in the article:
"The key point here is Mercury has some velocity. So, although the Sun pulls Mercury straight towards it, Mercury’s velocity keeps it moving, resulting in a circular path."
So, if you only decrease the orbital velocity of Mercury (without changing the gravity force or the falling in velocity), then it must getting closer to the Sun.
4." This extra falling is exactly what converts gravitational potential energy into orbital kinetic energy, causing the object to speed up once more"
Sorry - this is totally incorrect!!!
The falling in vector (blue) is orthogonal to the orbital velocity vector (red) there is no way for transformation of energy (or velocity) between the potential energy (or gravity -blue) to the orbital kinetic energy (red).
5. "So, the objects has to keep its orbital velocity while it get closer and closer to the Sun"
Incorrect.
As the orbital velocity is fixed, then as it get closer to the Sun, that orbital velocity is too low to keep it in a stable orbital cycle.
Therefore, it will move in a spiral way till final collision after few cycles.

So, the potential energy sets the falling in velocity vector.
This vector (in blue) is orthogonal to the orbital kinetic velocity (red)
Therefore - there is no way to transfer velocity from one to other!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 23/11/2019 21:34:50
I will wait until you have addressed this:

You do understand that "energy cannot be created" means that the total energy now must be the same as the total energy later, don't you?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/11/2019 05:27:20
I will wait until you have addressed this:

Quote from: Kryptid on Yesterday at 15:14:51
You do understand that "energy cannot be created" means that the total energy now must be the same as the total energy later, don't you?
Sorry.
Now it is your mission.
I have clearly proved that potential energy can't be transformed into orbital kinetic energy.
It can only be transformed into falling in kinetic energy.
Therefore, potential energy can't have any impact on the orbital kinetic energy.
Please let me know if you still don't understand why the potential energy can't contribute any orbital kinetic energy.
Hence, the total orbital kinetic energy is actually only the current kinetic orbital energy.
Do you have any idea how to "create new orbital kinetic energy" as the orbital object gets closer to the main mass?
Therefore, do you understand that "energy cannot be created" means that the total orbital Kinetic energy now can't be less than the total Orbital kinetic energy later?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/11/2019 06:06:28
Sorry.
Now it is your mission.

I asked that question in reply #945, which is before reply #947 (which was your last reply). Since my post came first, it is only proper that my question be answered first. Once that is done, I will address your other arguments.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 24/11/2019 06:40:25
Quote from: Kryptid
If something occurs that slows the object down slightly ....This is equivalent to a smaller overall orbit radius."
Quote from: Dave Lev
I disagree with that.
It doesn't happen as much today, but something that could slow down the speed of an orbiting body like Mercury would be a head-on collision with another smaller astronomical body.

This would reduce it's tangential velocity, and change the orbit from circular to elliptical, dropping it closer to the Sun at its point of closest approach to the Sun (perihelion).
- But this would not affect the distance when farthest from the Sun (aphelion), which would remain at the original radius of the circular orbit
- The effective radius of elliptical orbits is called the "semi-major axis", which is the average of aphelion and perihelion distances (for circular orbits, aphelion distance=perihelion distance=semi-major axis).
- Such a collision would reduce the semi-major axis, and would result in a shorter orbital period than the original circular orbit.

Less dramatic, but more common in our Solar System today: the orbits of planets are continually affected by tiny periodic tugs from Jupiter and nearby planets, causing the orbits to slowly change over time from circular to more elliptical and back again; this is measured by the "eccentricity" of the orbit.
- Effectively this trades angular momentum between the different planets; it mostly it balances out over time, but in extreme cases it could result in planets getting ejected from the Solar System, or dropped into the Sun.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/11/2019 06:59:08
Quote from: Kryptid
If something occurs that slows the object down slightly ....This is equivalent to a smaller overall orbit radius."
Quote from: Dave Lev
I disagree with that.
It doesn't happen as much today, but something that could slow down the speed of an orbiting body like Mercury would be a head-on collision with another smaller astronomical body.

This would reduce it's tangential velocity, and change the orbit from circular to elliptical, dropping it closer to the Sun at its point of closest approach to the Sun (perihelion).
- But this would not affect the distance when farthest from the Sun (aphelion), which would remain at the original radius of the circular orbit
- The effective radius of elliptical orbits is called the "semi-major axis", which is the average of aphelion and perihelion distances (for circular orbits, aphelion distance=perihelion distance=semi-major axis).
- Such a collision would reduce the semi-major axis, and would result in a shorter orbital period than the original circular orbit.

Less dramatic, but more common in our Solar System today: the orbits of planets are continually affected by tiny periodic tugs from Jupiter and nearby planets, causing the orbits to slowly change over time from circular to more elliptical and back again; this is measured by the "eccentricity" of the orbit.
- Effectively this trades angular momentum between the different planets; it mostly it balances out over time, but in extreme cases it could result in planets getting ejected from the Solar System, or dropped into the Sun.

Thanks
You discuss on the "eccentricity" of the orbit.
This is an important issue by itself. However, this is not the main issue in the current discussion.
We currently discuss on the real meaning of Total orbital kinetic energy.
Our scientists claim that:
Total orbital kinetic energy = current orbital kinetic energy + potential energy.
However, I have proved that the potential energy can only be transformed to in falling kinetic energy. Therefore, it can't be transformed into Orbital kinetic energy.
Hence
Potential energy = Only Falling in kinetic energy
Total  kinetic energy = Current kinetic energy + potential energy (or falling in kinetic energy)
However
Total orbital kinetic energy = Only the current orbital kinetic energy.

My simple question is: Do we all agree with that?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/11/2019 07:01:49
My simple question is: Do we all agree with that?

I'll answer that question once you've answered mine.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/11/2019 14:47:22
This would reduce it's tangential velocity, and change the orbit from circular to elliptical
Not disagreeing with your points, but Mercury's orbit is already quite elliptical and in fact the current tidal locking harmonic of Mercury (exactly 3 rotations every 2 orbits) depends on that eccentricity. Hence unlike other planets, the day on Mercury is essentially fixed, not slowing over time. The quora page Dave linked assumed a massively simplified circular orbit in order to illustrate the most basic concepts.
Typical of quora, the diagram is inconsistent with itself.  If mass is not taken into consideration, the red and blue vectors should be labeled velocity and acceleration respectively.  Any object in that orbit will have the same measurements.  If mass is taken into consideration, then the two arrows should be labeled momentum and force respectively. Acceleration causes a change in velocity. Force causes a change in momentum. I find it to be poorly expressed that they're taking one term from each way of looking at it.

My simple question is: Do we all agree with that?
We agree on almost nothing you post.  Let's keep score.

Quote
We currently discuss on the real meaning of Total orbital kinetic energy.
Our scientists claim that:
Total orbital kinetic energy = current orbital kinetic energy + potential energy.
1) They do not claim this.  That seems to be a description of total orbital energy.  2) The total orbital kinetic energy cannot be kinetic energy plus something not kinetic.  So the total orbital kinetic energy of a system of n bodies is the sum of the kinetic energy of each of the n bodies, each of which is computed as mv²/2, plus the rotational energy of each of the bodies: Iω²/2.

Quote
However, I have proved that the potential energy can only be transformed to in falling kinetic energy.
Don't agree with that. 3) There are obvious counterexamples like potential energy being transformed into electricity at hydroelectric facilities.  4) The fact that you think all such nonsense assertions constitutes a 'proof' is evidence that you're deliberately trolling us.

Quote
Therefore, it can't be transformed into Orbital kinetic energy.
5) Wrong again.  Mercury (since this seems to be our example of the moment) regularly transforms potential energy into orbital kinetic energy.  Quite a bit of it as a matter of fact.

Quote
Potential energy = Only Falling in kinetic energy
Nonsense term made up.  All bodies in an orbiting system are always in free fall by definition, so calling them 'falling' is redundant.  6) Potential energy is not a form of kinetic energy, so your statement is wrong.

Quote
Total  kinetic energy = Current kinetic energy + potential energy (or falling in kinetic energy)
7

Quote
However Total orbital kinetic energy = Only the current orbital kinetic energy.
This contradicts your (wrong) statement above.  I will refrain from counting this last statement as incorrect.

Quote
My simple question is: Do we all agree with that?
7 wrong things, and only one correct one (seemingly by accident).  And you didn't even assert creation of new energy in this post.

Your more regular mistakes are equivocating force, energy, and acceleration.  A gravitational field is an acceleration field.  A body by itself (a black hole say) has a massive gravitational field, but is by itself a source of neither force nor energy. Most of your attempts to posit energy from nowhere seem to revolve around some sort of assertion that this field is energy, mostly by introducing nonsense terms like gravitational energy, or tidal energy.  Gravity isn't energy, and neither is force.  A large rock in my front yard exerts considerable force continuously on my property, and yet I cannot harness that force as an energy source.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/11/2019 15:06:51
Quote
Quote from: Dave Lev on Today at 05:27:20
Sorry.
Now it is your mission.
I asked that question in reply #945, which is before reply #947 (which was your last reply). Since my post came first, it is only proper that my question be answered first. Once that is done, I will address your other arguments.

Dear Kryptid

I'm not sure that I fully understand what are you looking for
If you focus on energy creation, than Potential energy is a key element in this discussion.
In reply #945 you claim the following:
That's nonsense. Potential energy is neither a vector nor a velocity.
Potential energy doesn't have a direction or a speed. It changes with a change of distance from the source of gravity, but that's it.
This extra falling is exactly what converts gravitational potential energy into orbital kinetic energy, causing the object to speed up once more.
Quote
Quote
than how the extra orbital kinetic energy which is needed to increase the Vk (orbital velocity vector) is created?
It comes from gravitational potential energy, regardless of your complaints. The mechanism is well-understood.

So, how can we discuss on energy creation or any sort of energy if we don't really understand the real meaning of potential energy.
Please, Potential energy is a key element in the energy.
Why are you so afraid to deal with that potential energy?
Would you kindly give your confirmation for the clear understanding that Potential energy can ONLY be transformed to falling in kinetic energy and let me know what is your question.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/11/2019 15:21:11
Quote
Quote
Therefore, it can't be transformed into Orbital kinetic energy.
5) Wrong again.  Mercury (since this seems to be our example of the moment) regularly transforms potential energy into orbital kinetic energy.  Quite a bit of it as a matter of fact.
If you estimate that potential energy can be transformed into orbital kinetic energy, than please use the following diagram and show how it works.
https://www.quora.com/Why-doesnt-Mercury-fall-to-the-Sun
Do you agree that the Blue line represents the falling in velocity due to gravity force /potential energy?
If yes, do you agree that it is orthogonal to the orbital velocity (red)
So, how could it be that the blue vector (falling in velocity due to gravity, or potential energy) has any impact on the red one (orbital kinetic energy) if they are orthogonal to each other?
If no, how the potential energy is converted to orbital kinetic energy?

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/11/2019 17:39:18
So, how can we discuss on energy creation or any sort of energy if we don't really understand the real meaning of potential energy.

Physicists already know what potential energy is: https://www.livescience.com/65548-potential-energy.html

But the fact of the matter is that potential energy is a kind of energy and therefore must be subject to conservation of energy.

Why are you so afraid to deal with that potential energy?

I'll deal with it, but only once you've answered my question.

Would you kindly give your confirmation for the clear understanding that Potential energy can ONLY be transformed to falling in kinetic energy

I'll tell you what I think once you've answered my question.

Quote
and let me know what is your question.

I've already told you my question:

You do understand that "energy cannot be created" means that the total energy now must be the same as the total energy later, don't you?

Or, if you want me to rephrase it: do you agree that the energy in a system must remain constant if energy cannot be created or destroyed?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/11/2019 06:19:35
That diagram depicts Mercury with a circular orbit, which is very wrong.  I said that in my post above.

Current orbital kinetic energy of Mercury is something like 3.1e32 joules due to a velocity of around 47 km/sec.  Potential energy is about -6.86e32 .
By mid-February, Mercury's orbital kinetic energy will rise to about 4.5e32 joules with potential energy falling to about -8.26e32 joules.  It does this by converting potential energy to kinetic energy as it's distance from the sun falls from ~70 million km to about 46 million and its speed increases to over 56 km/sec. Your incorrect (#5) statement above says this transformation cannot happen, and thus contradicts empirical observations.
Total orbital energy of Mercury is fairly stable at about -3.76e32 joules, consistent with energy not be created or destroyed.
Dear Halc
You miss the whole point.
You discuss on none circular orbit.
In that kind of orbit, every full cycle, the object gets to the same point.
So, in total, the average radius is fixed, the average potential energy is fixed and even the average orbital velocity is fixed.
This isn't the case in our discussion.
I have already informed evan_au about it:
You discuss on the "eccentricity" of the orbit.
This is an important issue by itself. However, this is not the main issue in the current discussion.
We currently discuss on the real meaning of Total orbital kinetic energy.
So, please answer the following:
1. Do you agree that the potential energy is a direct outcome of Gravity?
2. If so, what Is Gravity?
https://spaceplace.nasa.gov/what-is-gravity/en/
So do you agree that "Gravity is the force by which a planet or other body draws objects toward its center?"
Therefore in that article it is stated:
"Mercury is in a constant state of falling into the Sun"
" If Mercury were stationary, it would fall straight towards the sun (just like what happens when you drop an apple on Earth). "
Therefore, gravity by itself represents a direct falling in velocity vector.
By itself, it can't generate any sort of orbital velocity.
Unless - we discuss on airplane (an object with a wings) that is flying in an atmosphere.
So, if gravity force represents a direct falling in velocity vector, (the blue line in the diagram), than that must be the real meaning of potential energy.
Therefore, the potential energy can ONLY be converted to direct falling in velocity vector, or falling in kinetic energy as it changes the average radius of the object.
Hence, the potential energy can't contribute even 0.000...1 Joule to the Orbital kinetic energy (again - assuming that it isn't an airplane) even if the average radius is falling from R to 1/10...0 R.
Is it clear?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/11/2019 06:54:28
Dear Halc
As you might know, we discuss on orbital energy.
1. What do you mean by "other kinds of potential energy":
No. Gravitational potential energy is related to gravity, but there are other kinds of potential energy.
Lets look at total orbital energy:
http://bogan.ca/orbits/kepler.html
Don't you agree that our scientists assume that:
"The Total energy of an object in orbit is the sum of kinetic energy (KE) and gravitational potential energy (PE).
KE = 1/2 mv2
PE = - GMm/r
r = the distance of the orbiting body from the central object and
v = the velocity of the orbiting body
E = 1/2 mv2 - GMm/r"

2. kinetic energy (KE).
KE = 1/2 mv2
Do you agree that the following formula represents the ORBITAL kinetic energy?
KE = 1/2 mv2
If so, Let's call it
KE(orbital) = 1/2 mv2
Do you also agree that v (velocity vector) is represented as a red arrow in the diagram that we have discussed?

3. Gravitational potential energy:
Do you agree that the following formula represents the gravitational potential energy (PE):
PE = - GMm/r
Based on your explanation the meaning of gravitational potential energy (PE) is:
Gravitational potential energy is arguably a direct outcome of the gravity of a system. The value is the energy yielded by bringing all matter in the system to zero potential (out of each other's gravity wells) from a stationary state (so not including kinetic energy).
PE = - GMm/r
Don't you agree that by your own explanation the direct outcome of the PE is the blue arrow in the diagram.
You can call it acceleration, velocity, force, momentum or "impact"  - as you wish.
Quote
Quote
Therefore, gravity by itself represents a direct falling in velocity vector.
11: "Falling in vector" is undefined.  Use accepted terms such as acceleration, velocity, force, or momentum, all of which are vector quantities. Your undefined nonstandard terms serve only to obfuscate (which I realize is a goal of yours). If you mean force, then say force.
Quote
Quote
So, if gravity force represents a direct falling in velocity vector, (the blue line in the diagram),
13: The blue vector is not labeled as a kind of velocity. The red line is. Your reading skills are lacking. 14: You seem to be equivocating force and velocity here, making this statement utterly wrong.
So, do you agree by now that the Blue arrow represents the "impact" of the gravitational potential energy?

4. Energy transformation
The key issue in this discussion is the energy transformation.
Especially the transformation from Gravitational Potential energy PE to ORBITAL kinetic energy KE(orbit)
Do you think that by changing the average radius of the orbital object (From r to r2), it can transform the change in the PE to added Orbital kinetic energy?
If so, would you kindly explain how that Potential energy can be transformed into added orbital kinetic energy?
Let's look at the following example:
Starting point:
KE = 1/2 mv2
PE = - GMm/r
If the average radius is changed to r2 (while there is no change in v)
The updated Potential energy will be
PE2 = -GMm/r2
So we get a change in the PE
Let's call that change PE'
As the Gravity works only on the Blue vector, would you kindly advice how that change in gravitational potential energy PE' can set any effect on orbital velocity vector -v?


Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 26/11/2019 23:19:01
I realize that my earlier explanation may have been difficult to follow without some kind of visual aid. Note: this is a simplification that uses circular orbits. The real orbits would be eccentric and thus would pass through regions of different gravitational field strengths. In particular, a decrease in orbital radius caused by a collision would probably result in an orbit even more eccentric than the original. In the case of tidal deceleration, the opposite is true: the orbit becomes less eccentric than the original:

 [ Invalid Attachment ]

(1) The satellite in orbit remains in a region around the planet of constant gravitational field strength. Its velocity is constant.

(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat. Since the satellite is now going slower than it was before, it can't follow its original path around the planet and instead begins to fall closer to the planet. Since it is traveling closer to the planet, it is passing through regions where the gravitational field strength is higher than it was in the original orbital radius. A stronger field means more force means more acceleration. This is what causes the satellite to accelerate to an average velocity in excess of its original orbital velocity. This is how gravitational potential energy gets converted into orbital kinetic energy when some event causes the satellite's original velocity to slow down (whether it is a collision, tidal deceleration, gravitational wave emission, or whatever else).

(3) The new orbit is closer to the planet than before, with the satellite's velocity having increased because gravity is stronger at this radius than it was at the original orbital radius.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/11/2019 00:15:37
The new orbit would still come back to the collision point every time unless a 2nd application of force (another collision or millions of years of tidal correction) circularizes the orbit like that.  The new orbit would remain like the 2nd picture, not the third.

I figured someone would bring that up (hence why I said I was simplifying it by using only circular orbits). However, for the sake of making it more correct, I'll add in this image (which would go between (2) and (3):


* Orbit2.jpg (30.47 kB . 390x390 - viewed 3938 times)

Image (3) comes, as you say, after millions of years of tidal effects.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/11/2019 06:29:24
Ow ow No!
Elliptical orbits put the planet at one of the foci, not in the middle like that.

You're right. I went back and edited it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/11/2019 18:31:43
(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat. Since the satellite is now going slower than it was before, it can't follow its original path around the planet and instead begins to fall closer to the planet. Since it is traveling closer to the planet, it is passing through regions where the gravitational field strength is higher than it was in the original orbital radius. A stronger field means more force means more acceleration
Yes, I agree with all of that.
Let's assume that due to the collision, the orbital velocity had been dropped from v to 3/4v (while it was at radius r).
At that moment, it had been lost its requested "magic velocity"
Remember:
v^2 = G M /r
Due to the collision, the velocity had dropped to 3/4v.
The outcome due to Newton is very clear
Let's look at Newton cannon
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon
"Newton reasoned that, if the cannon ball was fired with exactly the right velocity, the ball would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. It would be placed in orbit around the Earth."
As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.
There is NO WAY for it to increase its orbital velocity due to increased gravity force!!!
There is another explanation:
"A stronger GRAVITATIONAL field means more force means more acceleration"
That force is working between the satellite to the center of the main object.
Therefore, the acceleration that we get represents a direct falling in acceleration.
As an example:
"https://sciencebasedlife.wordpress.com/2011/08/15/free-fall-on-the-moon/
Galileo conducted several experiments and concluded that the effect of gravity on earthly objects is the same, regardless of the mass of those objects."
So, if it is a satellite, elephant or just 1 Kg of mass, all of them must fall at the same acceleration, while they keep the same orbital velocity - 3/4v.
Therefore, it is quite clear that there is no way to increase the orbital velocity due to this falling in acceleration.Hence, the following explanation is a total MISTAKE!!!
This is what causes the satellite to accelerate to an average velocity in excess of its original orbital velocity. This is how gravitational potential energy gets converted into orbital kinetic energy when some event causes the satellite's original velocity to slow down (whether it is a collision, tidal deceleration, gravitational wave emission, or whatever else).
A falling in acceleration can't increase any current Orbital velocity.
It must be clear to all of us.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 27/11/2019 18:51:48
As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.

The velocity doesn't stay that way, though. It accelerates again once it passes into the stronger regions of the gravitational field. This is exactly the same thing that happens in an eccentric orbit: an orbiting object speeds up when it approaches perihelion because the force of gravity is stronger closer to the main object.

There is NO WAY for it to increase its orbital velocity due to increased gravity force!!!

And yet that's exactly what happens in an eccentric orbit. Or do you deny the empirical data that shows planets accelerate when they move closer to the Sun in their orbit? If not gravity, then what force is it that increases the planet's orbital velocity as it moves towards perihelion?

A falling in acceleration can't increase any current Orbital velocity.

It does in an eccentric orbit.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/11/2019 06:13:03
Quote
Quote from: Dave Lev on Yesterday at 18:31:43
As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.
The velocity doesn't stay that way, though. It accelerates again once it passes into the stronger regions of the gravitational field. This is exactly the same thing that happens in an eccentric orbit: an orbiting object speeds up when it approaches perihelion because the force of gravity is stronger closer to the main object.

Let me explain:
In order to gain a stable orbital cycle - (with eccentricity = 0), there must be a full match between the radius to the orbital velocity. The requested velocity is called by Newton - "Magic velocity"
If you decrease the radius - then you actually MUST increase the orbital velocity.
Gravity can't increase the orbital velocity.
1. If the orbital velocity is too low it must collide with the main mass:
Please look at the following experiment for Newton ball:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is low, it will simply fall back on Earth."
So, just by decreasing the velocity of the orbital object - you won't get any sort of stable orbital cycle.

2. If the Orbital velocity it too high it should be escape from the main mass.
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=10000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. if the speed is very high, it will indeed leave Earth"

3. In order to gain the eccentric orbit you actually must increase the orbital velocity (by only some percentage - I assume)
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=8000.gif
In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is higher than the orbital velocity, but not high enough to leave Earth altogether (lower than the escape velocity) it will continue revolving around Earth along an elliptical orbit."

So, yes there is a possibility to get an elliptical orbit - but at this case it seems to me that you need to INCREASE the orbital velocity by some low percentage. If it is too high - it will be ejected from the main mass.
However, based on Newton experiment it is clear that if the orbital velocity is too low, the ball/orbital object must fall in and collide with the main mass.

Actually, you could ask NASA how do they set a satellite in a stable orbital cycle around the Earth.
It is quite clear that they have to bring the satellite to the requested radius and give it the requested orbital velocity.
If they won't do so, the satellite could fall in or be ejected to the open space.

Therefore, it is impossible mission to get a stable eccentric orbit by just random falling in.
You must have full control on the engine in order to set a satellite in a stable eccentric orbit.

Asteroids can be used as excellent example
The Erath is crossing their path from time to time.
We see them as meteorites while they fall in to our planet.
So, in total Million over billions of asteroids have cross the Earth path.
How many of them had been trapped in a stable orbital cycle around the Earth.
Do you agree that the number is ZERO???
So, just falling in by random velocity won't set any sort of stable orbital cycle.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/11/2019 06:48:07
Gravity can't increase the orbital velocity.

Then you need to answer this question:

If not gravity, then what force is it that increases the planet's orbital velocity as it moves towards perihelion?

How many of them had been trapped in a stable orbital cycle around the Earth.
Do you agree that the number is ZERO???

The average asteroid travels at more than twice Earth's surface escape velocity, so of course you'd expect the vast majority of them to evade capture. Then doesn't mean all of them have gotten away, though:  https://www.cnn.com/2016/06/16/us/nasa-asteroid-circles-earth/index.html
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/11/2019 08:38:41
Then you need to answer this question:
If not gravity, then what force is it that increases the planet's orbital velocity as it moves towards perihelion?

Newton has already answered this question.
Please read again my following reply:

3. In order to gain the eccentric orbit you actually must increase the orbital velocity (by only some percentage - I assume)
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=8000.gif
In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is higher than the orbital velocity, but not high enough to leave Earth altogether (lower than the escape velocity) it will continue revolving around Earth along an elliptical orbit."

So, yes there is a possibility to get an elliptical orbit - but at this case it seems to me that you need to INCREASE the orbital velocity by some low percentage. If it is too high - it will be ejected from the main mass.

Again - In order to get the elliptical orbit - you must INCREASE the orbital velocity or the orbital kinetic energy (at a given moment).
Therefore, by the decreasing of the orbital velocity – or the kinetic orbital energy, we should get a clear collision.

Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 28/11/2019 21:19:19
Newton has already answered this question.

Yes he did. His answer is gravity. But what is your answer? You already said that it wasn't gravity. What force do you say causes a planet's orbital velocity to increase as it approaches perihelion?

We are getting away from conservation of energy. If energy cannot be created, then the energy later must be the same as the energy now. Recall my bucket analogy. This is why your black hole cannot increase the total amount of energy in the Universe over time. Increasing the total amount of energy requires new energy to be created. Conservation of energy doesn't allow that.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/11/2019 13:50:19
So instead let's consider a rock at the summit of Mount Chimborazo.  No cannon.  Suddenly all of Earth is compressed into a Neutron planet and has a radius of only 10km.  The only thing left behind is this one rock with no change in its initial velocity sitting up there where the mountain used to be. It begins to fall.  It has a speed well below this magic speed at which it would have a circular orbit.  So it goes into an elliptical orbit and misses the surface of Earth (not by much) and come right back to where it started in about a half hour.
No, this is a fatal Error!!!
Alternatively, if there were a tower built on that mountain 16,500 km tall, a rock dropped from it would be in a stable orbit without the need for Earth to shrink to get out of the way.  Venus would need a considerably taller tower.
No No!
If that rock is falling down, (without any starting orbital velocity) it must directly collide at  the center with that 10Km Neutron planet.

Quote
Quote from: Dave Lev
There is NO WAY for it to increase its orbital velocity due to increased gravity force!!!
This is clearly false, and violates conservation of energy, so you know it's false.
If I drop a rock off a bridge, it gains speed from zero to enough speed to splash into the water.  That is empirical evidence refuting your statement.  Yes, the rock is in orbit, at least until the water gets in the way.  It should continue its orbit because you said no friction.

Secondly, the radius has dropped to a lower value, so by the formula for PE, the PE is less. You assert the velocity cannot increase, so this asserts that somehow the kinetic energy cannot increase.  That would be a net loss of energy, violating conservation of energy, contradicting your own statement. Your assertion is wrong.
You have to distinguish between falling in acceleration to orbital velocity.
As the radius has dropped to a lower value, the falling kinetic energy is increasing. But that velocity or acceleration is in a direct line between the center of the object mass to the center of the main mass. - Please look again at the Blue arrow in the diagram.
That falling in kinetic energy (or acceleration) must set a clear collision between the two objects.
As I have already stated - Potential energy can't generate any sort of Orbital energy or orbital velocity!!!
Please let me know if you still don't understand that kind of basic falling in acceleration due to gravity.

Newton has already answered this question.
Yes he did.
As stated by Newton:
"If the speed is low, it will simply fall back on Earth."
Hence, there is no way to increase the orbital velocity or the orbital kinetic energy over time. So, the potential energy CAN'T increase the orbital energy at all.
If energy cannot be created, then the energy later must be the same as the energy now.
There is no way for an object to get inwards and increase its orbital velocity over time.
That must be clear for all of us!
His answer is gravity. But what is your answer? You already said that it wasn't gravity. What force do you say causes a planet's orbital velocity to increase as it approaches perihelion?

Sorry.
My answer is fully correlated with Newton.
YES - it is all about gravity!!!
However, in our discussion we only focus on real changes in the orbital radius or in the average orbital kinetic energy & average potential energy.
Therefore, If the orbital path is a pure cycle or elliptical, there is no change in the average radius over time.
In the same token there is no change in the average orbital velocity or the average Orbital kinetic energy and the potential energy.
If you insist to understand how the energy is working in elliptical orbit, than you have to ask Kepler about it. 
In any case, we focus only on real changes in the average orbital radius over time.
Newton has proved that by decreasing the potential energy as the object is falling in, there is no way to gain extra orbital velocity.
Therefore: "If the speed (magic velocity) is low, it will simply fall back on Earth."
In other words : By decreasing the potential energy we gain extra kinetic falling in energy (or falling in acceleration). However, that falling in kinetic energy doesn't contribute even 0.0..01 j to the current orbital kinetic energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/11/2019 17:30:05
Hence, there is no way to increase the orbital velocity or the orbital kinetic energy over time. So, the potential energy CAN'T increase the orbital energy at all.

And yet that's exactly what happens in an eccentric orbit. A planet's orbital velocity increases as it approaches perihelion. If the orbital velocity has increased, then its orbital energy has increased as well.

There is no way for an object to get inwards and increase its orbital velocity over time.

This is what happens in an eccentric orbit.
This is what is happening to Phobos and Deimos.
This is what is happening to the Hulse-Taylor binary neutron star system: https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary

That also avoids the issue I was bringing up. Even if you were correct and gravitational potential energy could not be converted into orbital kinetic energy, that still would not change the fact that the total energy in the system cannot increase over time. As such, your black hole cannot create new energy.

Sorry.
My answer is fully correlated with Newton.
YES - it is all about gravity!!!

So now you are contradicting yourself. You said earlier that gravity cannot increase the orbital velocity. Now you are saying that it can. Which is it?

However, in our discussion we only focus on real changes in the orbital radius

An eccentric orbit does have a real change in orbital radius. For Pluto, that orbital radius change is particularly extreme (perihelion is 4.4 billion kilometers and aphelion is 7.4 billion kilometers).

Therefore, If the orbital path is a pure cycle or elliptical, there is no change in the average radius over time.

If you think that a planet in an elliptical orbit doesn't change its orbital radius over time, then you don't know what "elliptical" means.

If you insist to understand how the energy is working in elliptical orbit, than you have to ask Kepler about it. 

Yes, and Kepler says that the orbital velocity in an eccentric orbit increases when the planet approaches perihelion.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/11/2019 04:47:28
Quote
Therefore, If the orbital path is a pure cycle or elliptical, there is no change in the average radius over time.
If you think that a planet in an elliptical orbit doesn't change its orbital radius over time, then you don't know what "elliptical" means
At any point in the elliptical orbit cycle the radius is the same after full cycle.
So, there is no change in the average radius from one full cycle to the next one.
Therefore there is no change in the average radius per cycle over time.
In the same token there is no change in the average orbital kinetic energy from one full cycle to the next one.
Therefore, there is also no change in the average orbital kinetic energy per cycle over time.
In our discussion we focus only on real changes in the average radius or in the average energies (from one full cycle to the next one).
Please see your following example:
(1) The satellite in orbit remains in a region around the planet of constant gravitational field strength. Its velocity is constant.

(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat. Since the satellite is now going slower than it was before, it can't follow its original path around the planet and instead begins to fall closer to the planet. Since it is traveling closer to the planet, it is passing through regions where the gravitational field strength is higher than it was in the original orbital radius. A stronger field means more force means more acceleration. This is what causes the satellite to accelerate to an average velocity in excess of its original orbital velocity. This is how gravitational potential energy gets converted into orbital kinetic energy when some event causes the satellite's original velocity to slow down (whether it is a collision, tidal deceleration, gravitational wave emission, or whatever else).

(3) The new orbit is closer to the planet than before, with the satellite's velocity having increased because gravity is stronger at this radius than it was at the original orbital radius.

In this example you start with a satellite which is in orbit around the planet of constant gravitational field strength. Its velocity is constant.
Hence, it's average radius is constant from cycle to cycle and also its average velocity (or orbital kinetic energy) is constant from cycle to cycle.
However, due to the collision with asteroid the satellite had lost some of its orbital velocity.
"(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat.
Let's stop here and try to understand the scenario:
1. The satellite is still located at the same average orbital radius - r
2. Due to the collision, the average velocity (or orbital kinetic energy) had been dropped.
In order to understand the outcome - we only need to ask Newton about it:
Please look at the following experiment for Newton ball:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is low, it will simply fall back on Earth."
It is stated clearly:
"If the speed is low, it will simply fall back on Earth"
So, that satellite Must fall in and collide with the Earth as Newton had clearly stated.
Quote
Please look again at the Blue arrow in the diagram.
That falling in kinetic energy (or acceleration)
30: equivocation of energy and force. The blue arrow is force in that picture.  Kinetic energy is not a vector, but is a function of velocity, which is represented by the red arrow, not the blue one.
Somehow you still insist that the potential energy must be transformed to Orbital kinetic energy.
This is a severe mistake.
Newton has told us that if the current orbital velocity (red) it too low, there is no way to increase it by any sort of potential energy:
again:
"If the speed is low (red), it will simply fall back on Earth"
If you both still don't agree with Newton and you still hope that the somehow the potential energy can increase the orbital velocity, than would you kindly show Newton why his message is incorrect..




Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/11/2019 06:27:07
So now you are contradicting yourself. You said earlier that gravity cannot increase the orbital velocity. Now you are saying that it can. Which is it?

I'm still waiting for an answer to this question.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/11/2019 08:29:33
Dear Halc
It seems that you have totally got lost.
You're switching from average to momentary radius and kinetic energy to average and back again.
1. Momentary radius - the current radius in the orbital cycle. In elliptical orbit it can change from maximal to minimal. However, in a stable elliptical orbit the momentary radius doesn't change the average radius or the average orbital velocity per cycle. Therefore it is none relevant to our discussion.
2. Average radius - It means the average radius PER cycle. Therefore, if there is no change in the average radius per cycle there is also no change in the average orbital velocity per cycle or orbital kinetic energy per cycle. We only focus on that radius in our discussion. Therefore, we only verify a real change in the average velocity from one full orbital cycle to the other one.
3. In order to get better understanding, let's assume that we only focus on a pure orbital cycle (eccentric =0). Therefore, please ignore the issue of Momentary radius or average radius. Just radius in a pure orbital cycle & pure orbital velocity
Quote
Quote
In this example you start with a satellite which is in orbit around the planet of constant gravitational field strength. Its velocity is constant.
Hence, it's average radius is constant from cycle to cycle and also its average velocity (or orbital kinetic energy) is constant from cycle to cycle.
However, due to the collision with asteroid the satellite had lost some of its orbital velocity.
See what I mean?  You're talking about it losing its orbital velocity, but above you said you wanted to concentrate on the average velocity and radius.  It has not done this.  Lower orbits have higher average velocities, so this new orbit is going to have higher average speeds than before.  Look at the average velocities of each of the planets: (km/s) 47 35 30 24 13 10 7 5
Smaller average radius means higher velocity, yet you continue to assert this:
4. In that example by Kryptid, we discuss on a pure orbital cycle. Kryptid have stated that the satellite had only lost some of its orbital velocity due to collision with asteroid. However, it is also clear that at the same moment of the impact there was no change in the radius.
Therefore: the radius is R before the collision and stay the same just immediately after the collision. However, the orbital velocity that was V before the collision, had been dropped to V' immediately after the collision.
The question was - how that decrease in the Orbital velocity could affect the satellite, while the radius had not been change - again, immediately after the collision?
4. Kryptid had estimated that as the average orbital velocity had been dropped, than the satellite can't keep on with its current radius. therefore, he assumed that the satellite must fall inwards. We can see it in the following diagram.

 Orbit2.jpg (30.47 kB . 390x390 - viewed 218 times)
A question to Kryptid - Did I understand your example correctly?
If so,
That actually is a perfect match to Newton explanation.
"Please look at the following experiment for Newton ball:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is low, it will simply fall back on Earth."
In this Newton example the cannon ball is fired vertically to the ground. Therefore, it's first moment speed represents an orbital velocity.
Newton's ball example is fired into the ground, violating the condition that we're ignoring friction.  No satellite orbits right at the surface of a star or planet.  The same under-speed cannon shot would be in orbit if the friction was ignored.
Therefore, you clearly don't understand how Newton ball really works.
The meaning of the first moment speed is the first moment orbital velocity!!!
Therefore, When Newton claim for : "If the speed is low, it will simply fall back on Earth"
He means : If at the first moment the orbital velocity is too low (comparing to the requested "magic orbital velocity"), the cannon ball must fall to earth.
Please see again the following image:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
Therefore, the Satellite in Krypid example MUST fall in to Earth.
It can't convert its potential energy to ORBITAL kinetic energy.
His message is entirely correct.  Nowhere does it say that velocity will not increase as it falls to Earth.
Yes, as it falls in it must gain higher total velocity.
However, the potential energy is converted kinetic energy, but this energy creates higher falling in velocity vector. Let's call it Vf. That falling in velocity is vertically to the ground. I hope that we all agree that when something is falling (due to potential energy) it must fall vertically to the ground. I'm not aware about any object that can fall horizontally to the ground.
Therefore, the total velocity vector is the sum of the current orbital velocity V' vector -horizontally to the ground, and Vf vector - vertically to the ground (due to the potential energy that is converted to kinetic energy).
So, I agree that as the potential energy is converted to kinetic energy it increases the total velocity, however it doesn't increase the orbital velocity V' which is horizontal velocity vector - in red)).
Therefore, the total velocity could be higher than the requested "magic orbital velocity", but as it is not horizontally to the earth, it must lead the satellite to a direct collision with the earth - As was expected by Newton explanation.
I hope that by now it is fully clear.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/11/2019 14:49:18
A question to Kryptid - Did I understand your example correctly?

I'll answer that once you've clearly answered this:

So now you are contradicting yourself. You said earlier that gravity cannot increase the orbital velocity. Now you are saying that it can. Which is it?
Title: Re: How gravity works in spiral galaxy?
Post by: puppypower on 30/11/2019 16:47:45
The mass of spiral galaxies decrease with time due to the mass burn within stars. It is estimated our sun burns the mass equivalent of the earth every 70,000 years. This mass burn times hundreds of millions of stars per galaxy results in the gravitational force decreasing with time.

The fusion in stars creates energy, while the second law states that the entropy of the universe needs to increase, with entropy increasing by absorbing energy. The entropy increase from fusion  is expressed in the complexity of higher atoms. It is also expressed as solar flares where confined and compressed materials lower pressure and temperature and have more freedom of expression; radical and stable chemical states.

The solar wind then goes into space where temperatures can approach absolute zero causing the chemical entropy to fall. Since the second and first law is still in effect, energy is released from entropy and recycled into other forms of entropy with a net increase in universal entropy; diffusion and super conductivity. and magnetism.

Our sun is a second generation star, formed from chilled gases and dust compressed by gravity. This lowered the entropy since it reverses super conducting states. This gives off energy trapped as entropy. This is expressed as a very low energy background universal background signal.that randomizes the galaxy.

If gravity was a force it would give off energy as the potential lowers. This exothermic output is related to the lowering of material entropy which is proportional to the gravitational work. The summation of all the forming stars adds energy into entropy resulting in layers of rotations.
. .

.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/12/2019 04:41:07
If you're saying it cannot increase the orthogonal velocity, you're wrong.  Look at Kryptid's elliptical orbit in picture 2.5.  The orbital velocity is initially V' at the top.  At the bottom of the orbit it is much closer to the primary (lower potential energy).  If it was still moving at the same speed as at the top, its kinetic energy would be the same, so the total energy would have gone down, violating energy conservation.  Vf is zero at both points since motion is horizontal at both of them.  So you're asserting that total energy has gone down, a violation of energy conservation.  The orthogonal velocity vector (which is the same as the total velocity at those two points) is much larger in magnitude at the bottom to account for the gained kinetic energy needed to balance the loss of the potential energy.
Simple question:
Please focus on Kryptid example.
Let's assume that due to the collision between the satellite and the asteroid, the Satellite had totally lost its orbital velocity (orthogonal velocity = 0)..
So, just one moment after the collision, that satellite is still located at a radius r from the planet and its orthogonal velocity is Zero.
Hence, as the Satellite must fall in due to gravity, do you see any possibility that it should gain any orthogonal/orbital velocity and restart to orbit the planet at a lower radius?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/12/2019 18:06:56
We're assuming  a small object orbiting a large one, so yes, in the situation you describe, the impact will probably take place before the orthogonal component of its velocity regains the magnitude of the original orbital velocity.
Thanks
So you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.
But why do you claim: "before the orthogonal component of its velocity regains the magnitude of the original orbital velocity" and why do you highlight the issue that it is just because the orbital object is very small?

If the orbital object was a moon that falls in to Earth, do you see any possibility that it will regain the magnitude of the original orbital velocity?
Please look at the following diagram:
https://www.quora.com/What-is-the-work-done-by-the-force-of-gravity-on-a-satellite-moving-around-the-earth
We see that the orbital velocity vector V (or orthogonal velocity) is in green is zero.
The gravity force (vector Fg) is in red.
Let assume that at one moment the moon had suddenly lost completely its orbital velocity.
So, V =0.
In this case, do you agree that the moon will fall directly to Earth - in the green vector of Fg?
So, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?
Even if we "they were both point masses (or sufficiently small)":
If they were both point masses (or sufficiently small), then no impact would result, and yes, there would be a point where the orthogonal component of the new orbit would vastly exceed the original orbital speed, as it must as it must convert potential energy into kinetic energy.

I really don't understand why do you see any possibility to convert the in falling velocity (or falling kinetic energy) to orthogonal velocity (or orbital velocity) at any sort of orbital objects.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/12/2019 03:57:23
Thanks Halc
Quote
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I really don't understand why do you see any possibility to convert the in falling velocity (or falling kinetic energy) to orthogonal velocity (or orbital velocity) at any sort of orbital objects.
Look at comets, which have almost the exact sort of orbit you're describing here.  They start out coming almost straight in (nearly pure 'falling' velocity as you put it), and suddenly as they pass the sun, the high velocity vector doesn't change much, but the sun is suddenly off to the side and that high velocity vector is now completely orthogonal to the force vector.  The force vector (representing 'down') rotates around quickly but the velocity vector doesn't so much.
Comets are irrelevant to our discussion as they are constantly orbiting the sun:
https://solarsystem.nasa.gov/asteroids-comets-and-meteors/comets/overview/?page=0&per_page=40&order=name+asc&search=&condition_1=102%3Aparent_id&condition_2=comet%3Abody_type%3Ailike
"Comets are cosmic snowballs of frozen gases, rock and dust that orbit the Sun. When frozen, they are the size of a small town... There are likely billions of comets orbiting our Sun in the Kuiper Belt and even more distant Oort Cloud."
Therefore, we can't use comets as example for falling in objects even if their orbital elliptic shape is very sharp.
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So, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?
By rotation of the vectors.  It is falling fast, and suddenly 'down' is a different direction as it passes by Earth.  The direction of 'down' is changing all the time.  Takes month to go all the way around, but less time in this new orbit you've given it.
I really don't understand that explanation.
Please look at the following image for:
1. Free Falling Object Dropped From a Known Height
https://owlcation.com/stem/Solving-Projectile-Motion-Problems-Applying-Newtons-Equations-of-Motion-to-Ballistics
We see that the final velocity (at the collision point) is v = √(2gh)
It is also very clear that this velocity vector is horizontal to the Earth.
So, how this Horizontal falling in velocity vector could be transformed to orthogonal velocity vector (orbital velocity)?
Would you kindly use Newton formula & mathematics to prove that unbelievable idea? 
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/12/2019 06:29:47
They are completely relevant since the orbits of the objects you describe (the ones that don't involve impacts) will orbit forever in their new highly eccentric orbits.
The ones we see happen just like you describe: A reasonably circular orbit is suddenly altered, stopping the comet in place. It thus begins to fall into the inner solar system, achieving far higher velocities than it ever had before.  This is exactly what I've been describing.
Well, if we can show or prove that the comet had completely lost its orbital velocity due to collision, but now it regain it as it falls in, than, yes that could be a good example.
So, do we have any sort of evidence for current immediate outcome due to comet collision?
Why do you claim: "A reasonably circular orbit is suddenly altered, stopping the comet in place."
Why the comet had stopped in place?
Do you see any sort of collision or sudden orbital lost?
Don't you agree that we actually see a comet in a normal orbital cycle that just got to its maximal radius and then comes back?
That activity doesn't give any indication for sudden orbital velocity lost.
This is a very normal activity for any orbital cycle as the orbital object gets to its maximal radius.
So, if you don't have a clear evidence for a comet collision (and the direct outcome from that collision) than the comet is just irrelevant for our discussion.
In the article it is stated clearly:
"Comets are cosmic snowballs of frozen gases, rock and dust that orbit the Sun".
So, it is just in a constant and stable orbital cycle around the Sun, while its eccentric could even be close to one.
Therefore, the comet should come again and again to the same minimal radius and the same maximal radius after every full orbital cycle.
Hence, there is no change in the potential energy or orbital kinetic energy due to a direct collision.
Again - we try to understand the impact of collision that force the orbital object to lose suddenly its orbital velocity.
As the comet is already in a constant orbital status, it can't give any indication for the outcome due to our main focus on suddenly orbital velocity lost due to collision.
In any stable orbital system, with any sort of eccentric (zero to almost one) the orbital object gets to the same point (radius) after every full orbital cycle.
Therefore, do you agree by now that the eccentric can't give any indication for orbital lost energy due to collision?

The picture does not depict an orbit.  Find a picture of a comet orbit to see a real example of this so called unbelievable idea.  Here's one:
http://www.khadley.com/Courses/Astronomy/ph_205/topics/pluto/images/cometorbit.jpg
The comet moves clockwise.  Draw a velocity vector at say the 1985 mark and notice it is pointed almost the same direction as the force vector, not perpendicular at all.  So the comet is gaining speed as it fall nearly directly towards the sun.  It reaches its highest speed as it swings behind the sun near where the word 'Mars' is written.  At that perihelion point the velocity is entirely perpendicular to the force vector.  It is no longer 'falling in' at all, but is moving faster than any other point in the orbit. That is because the velocity vector has rotated only about 80° from the 1985 point to the perihelion point, but the force vector has rotated around 170°, pointing nearly the opposite direction as it was in 1985.  That's the vector rotation I'm talking about.
That could be a normal explanation for a any orbital cycle with eccentric close to one.
So, we only see a comet in a stable orbital cycle around the sun.
Therefore, that Vector rotation is a very normal outcome from any orbital system at any sort of eccentric
Again - this doesn't give any indication for a sudden orbital lost due to collision.

What velocity vector?  The picture you asked me to look at shows a guy dropping a rock from a small height.  The (implied, not depicted) velocity is totally vertical in the picture.
So, do you agree that  the velocity of an object that falls in (to a planet) must be represented by a totally vertical velocity vector (with reference to the planet)?
If so, again - how that "totally vertical" velocity vector can be transformed to orthogonal velocity vector (orbital velocity vector)?

We see that the final velocity (at the collision point) is v = √(2gh)
Has nothing to do with collision point.  That formula assumes a uniform gravitational field g, which means it only works for short distances.  Gravity force on our moon varies by its distance from the planet, so that formula doesn't work.  Use the PE formula, and convert that figure into KE.  That gives an accurate final velocity for dropping something from a large height.
Yes, that is correct.
However, the exact formula for the vertical falling in velocity is not so important.
It is very important that any falling in must be vertical to the planet.

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Would you kindly use Newton formula & mathematics to prove that unbelievable idea?
Use the formulas I indicate above, which give accurate velocity values at any point in the orbit.  Those formulas do not give eccentricity for a given scenario, but they give what you asked: velocity resulting from dropping an object from altitude X to altitude Y.
This answer isn't clear.
How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 05/12/2019 06:43:49
Hence, there is no change in the potential energy or orbital kinetic energy due to a direct collision.

But there is a change of potential and orbital kinetic energy due to the immense change in orbital radius of the comet over time.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 06/12/2019 16:21:42
Quote
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This answer isn't clear.  How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?
The answer I gave yields orbital speed S, not component speeds. Orbital velocity V is the velocity with speed S tangential to the orbital path. If you want to break that velocity into your two components at any point in the orbit, find the angle between the acceleration (or force) vector and the velocity (or momentum) vector.  What you call the 'falling in vector' Vf is Scos(θ) in the direction of the primary.  The vector orthogonal to Vf (Vo) is Ssin(θ) in the direction perpendicular to Vf.  It's that easy.
Dear Halc
So you claim that:
S = Orbital velocity
Vf (falling in velocity vector)= Scos(θ)
Vo (Ortogonal velocity vector) = Ssin(θ)
That could be correct as long as we monitor the velocity vectors.
However, in reality Vf is a direct outcome from gravity force, while Vo is the first moment orthogonal velocity of the object (as was explained by Newton cannon ball.
So, just if there is a full match between the Vf and Vo (orthogonal) we can get the magic velocity that we can call Orbital velocity.
Therefore Vf in reality has no impact on Vo (orthogonal) and vice versa.
However, both of then set the orbital velocity.
Therefore, the formula should be as follow:
S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)
So, the orbital velocity is the outcome between the Vf and Vo.

Let's go back to Krypptid example.
Please remember that we have stated the due to the collision the orbital object had totally lost its orbital velocity.
At that moment, as there is no orbital velocity, R is actually represents H.
So, just after the collision, the object is located at H, its orbital velocity (Vo) is zero, its falling in velocity vector Vf is zero and also its orthogonal velocity is zero.
So, We have the main object (let's assume that it is a planet), and there is also the object.
You have stated that if the orbital object is small enough, most of the time it should collide with the planet.
Quote
So you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.
Most of the time, yes.
I need to understand why do you claim "only most of the time" and how it could gain any orthogonal velocity as it falls in if its orbital velocity just after the collision is zero.
In order to understand that, let's assume that we have an object with mass of 1,000Km.
Let's drop it at a high of 10,000Km above the planet.
So at that moment, H = 10,000Km, Vo =0 , Vf =0.
I assume that we all agree that (if we ignore the air) it should fall horizontally to the planet.
Therefore, as it falls in it accelerates, its falling velocity - Vf (horizontally to the planet) is increasing over time, while its Vo should be zero.
So, even if Vf will get to it's maximal velocity, Vo must be zero.
Now, if we take much heavier object and set it at a higher distance from the planet,
Do you see any possibility that it will increase its Vo as it falls in?

But there is a change of potential and orbital kinetic energy due to the immense change in orbital radius of the comet over time.
In a stable orbital cycle as the eccentric is greater than zero and less than one, that change in the energy (potential/kinetic) is  actually a temporary change.
Every full orbital cycle it should get to the same minimal and maximal radius.
Therefore, the potential energy or the kinetic energy at the minimal or the maximal radius will always be the same after one full orbital cycle.
We discuss on totally different issue.
We want to understand what will be the real impact due to a collision with the orbital object.
First you have stated:
(3) The new orbit is closer to the planet than before, with the satellite's velocity having increased because gravity is stronger at this radius than it was at the original orbital radius.
Than Halc has claimed that:
The new orbit would still come back to the collision point every time unless a 2nd application of force (another collision or millions of years of tidal correction) circularizes the orbit like that.  The new orbit would remain like the 2nd picture, not the third.
Than you have accepted this explanation.
I don't agree with none of you.
I claim that
S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)

Let's look at Newton cannon ball explanation:
http://www.hscstudyguides.com.au/hsc-physics-course-summary/hsc-physics-course-summary-space/rocket-launches-orbital-motion/
"Newton envisaged a cannon firing a projectile horizontally from the Earth’s surface. Ignoring air resistance, the projectile would prescribe a parabola, eventually falling back to Earth. However, as the speed of the projectile is increased, the projectile will take progressively longer to hit the ground, because although gravity is pulling towards the centre of the field, the Earth’s surface is falling away from the projectile at the same time due to its horizontal motion. Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth."
It is stated clearly:
"gravity is pulling towards the centre of the field"
The Vf is a direct outcome due to gravity. therefore, Vf is in a directly towards the centre of the field (or the center of the planet if you wish)
The "horizontal motion" is represented the Orthogonal velocity Vo
It is stated clearly that in order to "travelling in a circle around the Earth" there must be a full match between Vf and Vo.

With regards to ellipse orbit, It is stated clearly:
"As the velocity increases even more, the circle becomes an ellipse"
So, when you think about that ellipse orbital cycle (with eccentric greater than zero, less than one), the orbital object must get a boost in its orthogonal velocity.
It is also stated:
" if the speed is increased enough, the trajectory becomes hyperbolic. At this point, the projectile has enough velocity to leave the gravitational field."
At that moment the eccentric is greater than one.
So, your both assumption that the object can set even one full orbital cycle after loosing significantly it orthogonal velocity (or even zero) is totally wrong.
From now on you have to argue with Newton.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 06/12/2019 16:27:02
In a stable orbital cycle as the eccentric is greater than zero and less than one, that change in the energy (potential/kinetic) is  actually a temporary change.

Temporary or not, it is still a demonstration of an increase in orbital kinetic energy caused by gravity. Something which you claimed doesn't happen.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/12/2019 11:50:22
But the moon is massive, so Earth will be moving relatively quickly in its own orbit around the moon, so if the moon is halted, there's a chance the Earth gets out of the way in time.  Probably not.  As I said, only Charon has a chance of doing this. I haven't worked out the numbers.
If the planet is moving fast enough, the moon will miss it as it falls and go into that eccentric new orbit.
This is incorrect.
You are missing a key element in gravity.
Almost Every object in the galaxy orbits around something due to gravity force.
This something could be a planet, Star, a center mass of several objects, the center of the galaxy and so on.
The moon for example is orbiting around the Earth, while it is totally ignore the Sun Gravity.
Please be aware that the Moon/sun gravity force is more than twice with regards to the Moon/Earth gravity force.
The Earth orbits around the Sun, while it is totally ignore the gravity force of the center of the galaxy, while the Sun orbits around the center of the galaxy at a speed of over than 600 Km/s.
In the same token, none of the star in the galaxy care about the idea that the galaxy is also crossing the space at over than 600 km/s.
So, each object in the galaxy is actually "gravity bonded" in its current orbital system and almost totally ignore the impact of other gravity forces (with the exception of tidal forces).
So, with regards to the moon - it is "gravity bonded" to Earth. Hence, wherever the Earth goes, the moon goes with it.
The mass of the Moon is totally none relevant.
Therefore, it was a mistake to claim that:
"the moon is massive, so Earth will be moving relatively quickly in its own orbit around the moon, so if the moon is halted, there's a chance the Earth gets out of the way in time."
Please be aware that in our theoretical discussion, we only have reduced the current orbital velocity of the object to Zero.
In that theoretical concept we also didn't break of the orbital object from the current gravity force.
So, we didn't break the gravity force between the Moon/Earth.
We have just reduced the moon orbital velocity to zero. (due to some imaginary collision or any idea which you like).
Therefore, It should be clear to all of us that as long as the moon is still "gravity bonded" with the Earth, it must fall directly to the center of the Earth. It can't miss the Earth as it falls in. Therefore, there is no way for the moon to gain any sort of orbital velocity as it falls in.
Most of the time the planet is barely moving.  So you stop any typical satellite and it's going to drop pretty much straight down.
The Earth can move at the speed of light. As long as the satellite is "gravity bonded" with the earth, it should fall in directly to the center of the earth without any ability to gain any sort of orbital velocity.

In my discussion, I assume it was halted in place in the frame in which it had a circular orbit (the frame of the two-body system).  In any other frame, the moon's path was not a circle, but more like a helix.
So, please let's focus only on circular orbit and a frame of two-body system.

Quote
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The formula should be as follow:
S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)
So, the orbital velocity is the outcome between the Vf and Vo.
We called the orbital velocity V, not S. S is a speed, not a vector..With that terminology exception, your statement is actually correct, and I never said otherwise.
Thanks
So let's agree
V (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)

But keep in mind that these vectors are constantly changing over time for an object in orbit.  They're not fixed values, and your argument seems to hinge on them being fixed. There is force on the orbiting thing, so there is always acceleration, and acceleration means all vectors (V, Vf, and Vo) are constantly changing.
As we discuss on "circular orbit and a frame of two-body system" do you agree that the V, Vf and Vo are fixed?

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So, even if Vf will get to it's maximal velocity, Vo must be zero.
If it falls straight down, they yes.  It isn't in orbit in any scenario with an impact like that.
I hope that you also agree by now that if the moon had totally lost its orbital velocity, while it is still in a two body system with the earth, it must fall in directly to the center of the earth.
Therefore, Its Vf at the moment of the collision with Earth must be maximal, while its orbital velocity should be zero.

48: If the orbit is circular, Vf is zero, which hardly sounds like a 'full match' with Vo.
This is incorrect.
Every orbital object is actually falling in constantly.
https://www.quora.com/Why-do-things-orbit-the-earth-instead-of-falling-towards-its-gravity
"objects in orbit around the earth are in a perpetual state of “falling” toward the earth. If the object's velocity were to change (in this case, decrease), then the force of gravity would cause the “falling” to have a greater influence, resulting in the object's eventual plummet into the atmosphere and toward the surface."
I know that you really dislike those kind of answers from Quora, but I think that they are perfectly correct.
Let's look at the Moon.
This moon is actually falling constantly directly to the center of the earth, therefore its Vf is not zero.
However, as it also move forward at the orthogonal velocity - Vo, that movement compensate the falling in velocity.
Therefore, it keeps itself with the same radius (again, let's assume that the orbital is a cycle with eccentric =0).
So, even as Vf is greater than Zero, due to the full magic fit with Vo we get that magic orbital velocity that keeps it in the orbital cycle (while we might  think that Vf is zero).
47: V is orbital velocity whether or not there is this 'full match' between these two component vectors.
If there is no full match between the Vo and the Vf there will be no orbital system.
Please - we focus only on a pure orbital cycle with eccentric = 0.
That brings us to Newton cannon ball:
As was clearly stated by Newton, the magic orbital velocity is
"Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth."
Hence, If we can boost the (Vo -orthogonal velocity) of an object to the magic velocity that perfectly fit with its Vf, than this object will set a pure orbital cycles around the main mass.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/12/2019 07:24:07
I don't.  My mailbox doesn't. Newton's slow cannon ball doesn't. These objects are subject to gravity force, and yet they don't orbit.  Yes, all objects orbit if you restrict 'objects' to things that are in orbit, which is like saying every red object is red.
I mean free objects in space. We are all part of Earth mass.

51 Our own moon is currently stopped in its own frame (by definition), and in that frame, Earth is moving fast enough that the moon misses it and instead goes into an eccentric orbit.  Thus my statement is exactly correct.
It seems that you didn't understand my message:
Let's look at the following two options:
1. The moon is in the Earth frame
You have already agreed that:
54 Our moon is gravity bonded with Earth...
In this case, if you stop it, you stop it with reference to earth.
Therefore, it should be clear that it is similar to the idea that we just drop the Moon at its current location.
In that frame, the planet is always moving, so when you stop the moon, the planet keeps on going.
I disagree. We stop the moon with reference to Earth.
In this case, it should collide directly with the earth.
2. The moon is in another orbital frame.
Let's assume that the moon orbits around the Sun instead around the Earth ("Gravity bonded" with the Sun).
So, yes in this case it could miss the Earth.
However, as it is "gravity bonded" with the Sun, it will continue to orbit around the Sun.
Therefore, there is no way for the Moon to start orbiting around the earth unless it really collide with the Earth and break itself to several objects.
There is no magic 'match' required for Vo and Vf.  If you take a random asteroid from the asteroid belt and give it a random Vo and Vf with no regard to 'matching' them, so long as the sum of them (V) yields a kinetic energy that when added to its potential energy is a negative figure, the object will probably orbit the sun.
This is totally incorrect.
The Earth is crossing again and again the orbital path of the asteroids that actually orbit the Sun.
So, far we can clearly assume that it had crossed the orbital path of Million over billions asteroids.
How many of them had been really captured by the earth gravity and set a circular orbital path around the earth?
You have told me that we have discovered one object and even this one is not orbiting in a circular orbit.
So, do you agree that the chance to set any sort of orbital path even with eccentric that close to 1 (Very sharp elliptical orbit) is less than one to million over billions.
However, the chance to set a circular orbital cycle is virtually ZERO!!!
I still claim that all the comets were orbiting the Sun from day one.
However, the shape of their orbital cycle (eccentric close to one) shows that they are in their way to totally disconnect from the Solar system.
So, they can't be used as a prove or evidence for setting an orbital cycle but on the contrary - they are a clear indication that the orbital system is losing "gravity bonded" over time.
Therefore, it is quite clear to me that those comets should eventually be ejected from the Solar frame.
I wonder if we can find a proof for that.
In other words, I claim that an eccentric orbital system can be converted to circular orbit!!!

That brings me again to Newton
It is quite clear that you refuse to accept Newton explanation.
http://www.hscstudyguides.com.au/hsc-physics-course-summary/hsc-physics-course-summary-space/rocket-launches-orbital-motion/
So, Please let me know which one from the following explanations you reject:
http://www.hscstudyguides.com.au/wp-content/uploads/2016/06/figure-5.jpg

Do you agree that for a Circular Orbit - A magic orbital velocity is needed?

For example, Mercury's velocity is currently decreasing, and yet the force of gravity is decreasing because of this, or "causing the 'falling' to have a lesser influence" as that quote words it.
Somehow you insist to take an eccentric orbital system and try to understand how the forces works based on location in the orbital cycle.
This is not relevant to our discussion.
We want to understand how external forces can affect the orbital system.
Actually we wish to understand how orbital system can be created from the first stage.
Especially with regards to the Accretion disc around the SMBH.
Do you agree that all the atoms/particles in that disc/ring are actually orbiting at a perfect circular orbit at almost the speed of light?
If so, what is the chance that a falling star from outside could be break down to its atoms/particles and start orbiting in that velocity in a pure circular orbit?
I claim that the chance for that is less than a ZERO!!!
The accretion disc can't be created based on falling in matter.
This MUST be a direct outcome from Newton!



Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 12/12/2019 20:26:14
The most powerful gamma-ray burst yet was seen this year.

GRB are thought to originate when:
- A very massive star collapses to a black hole or neutron star
- A burst of matter is ejected from its poles, at very close to the speed of light
- This matter runs into nearby gas, emitting a narrow beam of high-energy gamma rays
- We happen to be in line with this narrow beam of gamma rays

In this case, the gamma rays had energy up to 1 TeV (the LHC only operates up to 14 TeV)

Of course, the small amount of matter ejected along the poles is more than compensated by the large amount of matter falling into the black hole/neutron star.
See, for example: https://www.nasa.gov/feature/goddard/2019/hubble-studies-gamma-ray-burst-with-highest-energy-ever-seen
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2019 08:52:34
Please look at the following article:
Interstellar Misfit ‘Oumuamua Could Be Dormant Comet
http://astrobob.areavoices.com/2017/12/19/interstellar-misfit-oumuamua-could-be-dormant-comet/
"This diagram shows the orbit of the interstellar asteroid / comet ‘Oumuamua as it passes through the solar system. Unlike all other asteroids and comets observed before, this body is not bound by gravity to the Sun. It has come from interstellar space and will return there after its brief encounter with our star system. Its hyperbolic orbit is highly inclined and it does not appear to have come close to any other solar system body on its way in"

We have excellent example for a "body that is not bound by gravity to the Sun" which comes very close to the sun but eventually that body is ejected from the solar system without setting even one full orbital cycle around the sun.

That is a key indication that object that is not bounded by gravity to the main mass might not be able to set even one full orbital cycle.
Somehow you wish to believe that this is feasible.
However even if I agree with this none realistic assumption, you and Kryptid have confirmed that  the orbital cycle must be very elliptical at the first falling/orbital stage.

In the following message you claim that the time that it takes to convert an elliptical orbit to circular orbit is almost infinite:
Most orbits tend to circularize over time via tidal effects, but it takes infinite time to finish the job in an ideal case, but in fact a perfect circular orbit can only be achieved with a 2-body system, and there is no 2-body system isolated from all other bodies.

You also add one more limitation: " in fact a perfect circular orbit can only be achieved with a 2-body system,"
When we look at ANY accretion disc around the SMBH at the center of spiral galaxy we clearly see that ALL of them have a perfect CIRCULAR orbit.
None of the accretion disc in the whole UNIVERSE shows any signs of eccentricity.

Even if we believe to our scientists that all the matter in the accretion disc is coming from outside, than somehow we must see around any accretion disc eccentric orbit for the matter that had fallen inwards in less than infinity.
I still don't understand how any falling in star or gas cloud can set a perfect circular orbit around the SMBH at a velocity of 0.3c, even after the infinity.

However, what is your intention for "infinity"?
Is it more than 13.8 Billion years or more than 10 billion years?
What is the requested Minimal/maximal age of the spiral galaxies in order to get that pure circular orbit of the plasma in the accretion disc?

If nothing was falling inwards during the last 10 billion years, while all of the accretion discs are clearly ejecting matter, how could it be that they are still full with matter/plasma (at orbital velocity of 0.3c)?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2019 21:21:04
Disks are not objects, and ring-objects cannot have a stable orbit.  If they were objects, they would fall in.  The rings are never perfectly circular due to the presence of other objects perturbing the motion.  Saturn's rings are the best nearby example.

How can you compare Saturn's rings to accretion ring?
Can you prove that the Saturn's rings are coming/falling in from outside?
Why do you claim that "Disks are not objects".
In each disc there are many objects. Every stone, rock and even Atom is the disc is an object by itself.

So what is the source for all the Atoms in the accretion disc?
1.Do you agree that the matter/plasma/Atoms in the accretion disc is coming from outside?
2. If so, than why the orbital cycle of that matter is so circular orbit?
Why it isn't elliptical as you have explained:
The new orbit would still come back to the collision point every time unless a 2nd application of force (another collision or millions of years of tidal correction) circularizes the orbit like that.  The new orbit would remain like the 2nd picture, not the third.
How could it be that something is falling in and set a circular/ or almost circular orbit?
3.How could it be that the orbital velocity of the plasma is 0.3c?

Let's look again on Oumuamua:
Oumuamua isn't in orbit since it has positive energy.  It is moving at greater than escape velocity at all times.
Oumuamua is a perfect example for an object that is coming/falling in from outside into the center of the main mass. We clearly see that it comes close to the sun, but then it is ejected outwards for never return.
You claim:
Oumuamua isn't in orbit since it has positive energy.  It is moving at greater than escape velocity at all times.
I agree with that.
However, if Oumuamua isn't in orbit since it has positive energy than any falling in object should behave in a similar way. Therefore, it proves that there is no way to set any orbital path for an object that falls in.
Hence, it proves that even a gas cloud or a star that is falling into the direction of a SMBH should have a positive energy and should be ejected immediately outwards.
Even if you wish to belive that a falling object can set an orbital path, it is clear that the eccentricity of that orbit should be very high.
You have offered a comet as an example. We clearly see that the eccentricity of the comet is almost close to one.
However, when we discuss about the matter in the accretion disc (that our scientists claim for coming from outside) you suddenly forget that in this case the matter must also set an orbital path with very high eccentric.

Conclusions

1. If Oumuamua can't set an orbital path around the Sun, than any falling in matter to the direction of the SMBH can't  also set an orbital path around the SMBH.
2. If you still hope that somehow a falling matter can set an orbital path, than based on your explanation about the comet the falling matter must set high eccentricity orbital path.
3. Hence, if the matter in the accretion disc was coming from outside than their orbital eccentricity had to be very high. However, as the orbital eccentricity of the matter in the accretion disc is close to zero (almost a perfect circular orbit), than it proves that this matter can't technically comes from a matter that is falling from outside the SMBH.


Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/12/2019 04:12:00
Why do you claim that "Disks are not objects".
In each disc there are many objects. Every stone, rock and even Atom is the disc is an object by itself.
I can loop a strap around a hula-hoop (or a rubber ring for that matter) and drag it away by pulling on the strap.  Try that on the rings of Saturn and all you get is a mild disturbance of a couple bits of gravel.  The hula hoop, if spun, all rotates at the same angular rate.  The angular rate of the material in Saturn's rings moves at a rate that is a function of each bit's distance from Saturn.

These distinctions are true because the hula hoop and rubber ring are objects and the rings are not.  For this reason, the rings of Saturn are in a stable orbit and a ring object cannot be.  A ring object cannot be in a stable orbit about a mass at its center regardless of any spin (or lack of it).

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So what is the source for all the Atoms in the accretion disc?
1.Do you agree that the matter/plasma/Atoms in the accretion disc is coming from outside?
2. If so, than why the orbital cycle of that matter is so circular orbit?
Fluid friction.  They hit each other if in eccentric orbits.  Energy is lost to friction until the motion is nearly circular.

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Why it isn't elliptical as you have explained:
Anything infalling obviously was elliptical at some time.

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3.How could it be that the orbital velocity of the plasma is 0.3c?
Apparently the temperature at that radius is enough to form plasma.

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Let's look again on Oumuamua:

Oumuamua is a perfect example for an object that is coming/falling in from outside into the center of the main mass.
It is departing, not falling in.  It already passed said center of the main mass a couple years ago, and obviously missed it, so your comment shows a lack of reading comprehension.

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However, if Oumuamua isn't in orbit since it has positive energy than any falling in object should behave in a similar way.
64 Not the ones with negative energy, like Earth.

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Therefore, it proves that there is no way to set any orbital path for an object that falls in.
65 Moon is currently falling in (getting closer), but it is in orbit, so obviously wrong.

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Hence, it proves that even a gas cloud or a star that is falling into the direction of a SMBH should have a positive energy and should be ejected immediately outwards.
66, mostly for the 'Hence'. It may or may not have positive energy, and it may or may not keep it.

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Even if you wish to belive that a falling object can set an orbital path, it is clear that the eccentricity of that orbit should be very high.
Something can transition from positive to negative energy by losing some of that energy, and thus fall into orbit from a non-orbiting trajectory.

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You have offered a comet as an example. We clearly see that the eccentricity of the comet is almost close to one.
0.6 might be more typical.

However, when we discuss about the matter in the accretion disc (that our scientists claim for coming from outside) you suddenly forget that in this case the matter must also set an orbital path with very high eccentric.[/quote]Not so.  Look at Saturn's rings.  That material never had a very eccentric orbit that I know of, but it also never had positive energy.  Accretion material rarely forms from positive energy material, but it can.

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1. If Oumuamua can't set an orbital path around the Sun, than any falling in matter to the direction of the SMBH can't  also set an orbital path around the SMBH.
But Oumuamua could have fallen into an orbit had a sufficient force acted on it to slow it down.  It just so happens that nothing ever did in this case.

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2. If you still hope that somehow a falling matter can set an orbital path, than based on your explanation about the comet the falling matter must set high eccentricity orbital path.
The comet was always in orbit, but a recent outside force altered it to a far more eccentric one, much like Kryptid's example.  The moon is currently falling in, yet its orbit isn't highly eccentric, so that part is wrong as well.  All these assertions, and every one of them has a counterexample.

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3. Hence, if the matter in the accretion disc was coming from outside than their orbital eccentricity had to be very high. However, as the orbital eccentricity of the matter in the accretion disc is close to zero (almost a perfect circular orbit), than it proves that this matter can't technically comes from a matter that is falling from outside the SMBH.
All wrong.  Pretty much saying the words 'had to be', 'must', 'cannot', 'proves', etc. tends to generate false statements since there are usually exceptions.  Not always, but as a rule of thumb.

Most material that gets near the accretion disk collides with it, and that slows the material down, circularizing its path.  All those eccentric orbit bits of sand and such that hit Earth?  Notice that the collision immediately nearly circularizes their subsequent motion.  The excess energy due to the orbit change is discarded as heat, which is why the accretion disk glows.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/12/2019 07:56:53
1. Collision
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3. Hence, if the matter in the accretion disc was coming from outside than their orbital eccentricity had to be very high. However, as the orbital eccentricity of the matter in the accretion disc is close to zero (almost a perfect circular orbit), than it proves that this matter can't technically comes from a matter that is falling from outside the SMBH.
All wrong.  Pretty much saying the words 'had to be', 'must', 'cannot', 'proves', etc. tends to generate false statements since there are usually exceptions.  Not always, but as a rule of thumb.
Most material that gets near the accretion disk collides with it, and that slows the material down, circularizing its path.  All those eccentric orbit bits of sand and such that hit Earth?  Notice that the collision immediately nearly circularizes their subsequent motion.  The excess energy due to the orbit change is discarded as heat, which is why the accretion disk glows.
How do you know that: "Most material that gets near the accretion disk collides with it, and that slows the material down, circularizing its path"
We can clearly see billions of Spiral galaxies. In each galaxy there is an accretion disc.
Can you please give us one single example to this imagination???
Few years ago, our scientists were positively sure that they are going to see fireworks as S2 is going to collide with the SMBH.
They even verify that S2 and the SMBH were in the same direct view line from us.
Unfortunately for them and for you, there were no fireworks and no collision.
S2 had passed very close to the SMBH without setting any sort of effect on the Accretion disc or on itself.
There is an explanation for that, but as expected, our scientists had claimed that there was an error in their instruments.
They didn't want to deal with the impact of their verification; therefore they have solved it by the "error factor"
In any case, let's assume that in the next time it would collide with the accretion disc.
What would be the impact of that collision?
Don't forget that the total mass in the accretion disc is about three sun mass. Therefore, the mass dissipation in the whole accretion disc is very low.
Did you try to set a simulation for that kind of collision?
Why the low dissipation mass in the accretion disc shouldn't be ejected in all directions due to a collision with the massive object as S2 that comes in with great energy?
Actually, don't you think that even if S2 comes too close to the accretion disc (without touching it), it should generate severe interruption in that disc?
How can you believe that due to that collision: " Notice that the collision immediately nearly circularizes their subsequent motion.  The excess energy due to the orbit change is discarded as heat, which is why the accretion disk glows."
Actually, this is a clear contradiction to Kepler law.
S2 has elliptical orbit. Its orbital path is quite similar to a comet path.
You claim that the eccentricity of the comet is 0.6. so what is the eccentricity of S2?
Based on Kepler, the same energy that brings S2 or comet near the main mass should take it outwards.
So, now you claim that this energy due to kepler law can be converted into heat?
Sorry, this isn't science this is science fiction.
Due to kepler law, that energy should blow out the whole accretion disc.
It is similar to truck that collide with a bicycle.
How can anyone believe that a bicycle can drain the energy from a truck due to collision?
If you still believe in your imagination, then please prove it.
Let's assume that we accept this kind of imagination.
Let's assume that we even agree to ignore kepler law and somehow S2 is acting according to our wishful list.
Let's assume that it collides with the accretion disc and coverts its "kepler" energy to heat as you wish to believe..
Even in this case, there must be severe interruption in the accretion disc due to the collision.
Don't forget that the orbital velocity of S2 might not be fully aligned with the orbital velocity of the accretion disc.
So, we must have a severe interruption in the accretion disc.
Where are the fireworks?
What could be the eccentricity of the accretion disc due to that collision?
So, if for example the eccentricity of the objects/atoms in that accretion disc is less than 0.01, it is quite clear that due to that collision, the objects/atoms in the accretion disc should get high eccentricity.
But in reality - we don't see that.
So, out of the Billion accretion discs that we monitor, why we do not see any signs of falling in stars?
Do you agree that we have never ever seen any signs for this kind of interruption?
If we don't see and it and we will never see, what should be the outcome?
Actually, you hope that somehow a star can come closer and stay closer.
You have already rejected the explanation by Newton about the cannon ball.
Now you reject the impact of the orbital energy due to kepler law. You hope that somehow this outwards energy can be converted to heat.
In other words, you hope that orbital objects can reduce their average orbital radius.
Let's verify the chance for that:

2. Statistical chance to reduce the average orbital radius
We clearly know that all the planets in the solar system and almost all moons (with one or two exceptions) are increasing their average orbital radius per cycle.
The Moon is drifting outwards from the Earth over time. The Earth is drifting outwards from the Sun.
So, they are both increasing their average orbital radius over time.
The total orbital objects (Moons + Planets) mass in the solar system that are increasing their average orbital radius, is significantly higher than the total moons mass that are reducing their average orbital radius.
I hope that you agree that the ratio between the total orbital mass in the solar system that are decreasing their average orbital radius to the total mass that are increasing  their average orbital radius is less than 1 to 1/10^20
If so, the chance for any orbital mass/matter to decrease its average orbital radius is less than 1/10^20
However, you WANT to believe that somehow all the matter in the accretion disc must come from outside.
Therefore, you hope that the matter from outside the accretion disc (It can be a star, gas cloud or even a rock) is reducing its average orbital radius around the SMBH and finely gets into the accretion disc.
You don't care about the simple evidence that the chance to decrease the average radius is less than 1/10^20.
You are using that very low chance as a leading concept for your unrealistic idea.
You hope that somehow the accretion disc eats its food from outside without any evidence for that and with a clear contradiction with Newton and Kepler law.

3. Mass ratio between the accretion disc to the molecular jet stream.
We all know that the total estimated mass in the molecular jet stream above and below the accretion disc is about 10,000 sun mass.
All of that mass is coming from the accretion disc.
So, how could it be that a total mass of only three sun mass in the accretion disc can supply 10,000 sun mass to the molecular jet stream?
Our scientists hope that this mass is a direct outcome from eating process in the past.
That of course is imagination, as the accretion disc can't eat and it will never ever eat even one atom from outside.
All the matter in the accretion disc is coming from inside!!!
I have many other examples that shows that you have a severe misunderstanding about the accretion disc
However, I'm quite sure that you won't let them to change your idea.
It is clear to me that you don't let the evidences to confuse you.
Why you are not using your deep knowledge in science to see the reality?
How can you still believe it that science fiction?



Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 21/12/2019 15:42:04
Few years ago, our scientists were positively sure that they are going to see fireworks as S2 is going to collide with the SMBH.
They even verify that S2 and the SMBH were in the same direct view line from us.
Unfortunately for them and for you, there were no fireworks and no collision.
S2 had passed very close to the SMBH without setting any sort of effect on the Accretion disc or on itself.
Link please, because I've heard no such thing.  Cannot comment on things made up.
For S2 and Sgr-A to line up in our view, we'd have to cross the plane of S2's orbit, and that only happens every 100M years or so.  It's called an eclipse when it happens and nobody expects collisions from an eclipse.

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let's assume that in the next time it would collide with the accretion disc.
What would be the impact of that collision?
S2 would be torn apart by tidal forces and some of it would form a new disk oriented with S2's orbit.  The old disk would probably be absorbed by the new one.

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Actually, don't you think that even if S2 comes too close to the accretion disc (without touching it), it should generate severe interruption in that disc?
I suppose it would.
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How can you believe that due to that collision: " Notice that the collision immediately nearly circularizes their subsequent motion.  The excess energy due to the orbit change is discarded as heat, which is why the accretion disk glows."
Actually, this is a clear contradiction to Kepler law.
There are 3 laws, and I don't see a violation of any of them, especially since none of the laws deal with colliding material.

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You claim that the eccentricity of the comet is 0.6.
There is no 'the comet', so no, I don't claim that.

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so what is the eccentricity of S2?
.885  Look it up.  Not hard to do.

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So, now you claim that this energy due to kepler law can be converted into heat?
No, I didn't claim that.

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Due to kepler law, that energy should blow out the whole accretion disc.
It seems you're the one claiming Kepler's involvement here.  Kindly show which of his laws says this.

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It is similar to truck that collide with a bicycle.
Agree, but which of Kepler's laws describe a truck-bicycle collision?

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How can anyone believe that a bicycle can drain the energy from a truck due to collision?
Oh, Newton's laws have something to say about that.  Try the 3rd law.  Once again you seem to be positing energy from nowhere when the bicycle gains energy but the truck doesn't lose any.

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Don't forget that the orbital velocity of S2 might not be fully aligned with the orbital velocity of the accretion disc.
So, we must have a severe interruption in the accretion disc.
Where are the fireworks?
No fireworks because S2 does not collide with anything.  It gets no closer than about 30 au.  I'm sure there would be a light show if S2 got that close.

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2. Statistical chance to reduce the average orbital radius
We clearly know that all the planets in the solar system and almost all moons (with one or two exceptions) are increasing their average orbital radius per cycle.
Statistics has nothing to do with that.  A great number of moons are drifting inward.  More than half of Jupiter's moons are moving inward. Depends how many of them you count.

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The Moon is drifting outwards from the Earth over time. The Earth is drifting outwards from the Sun.
So, they are both increasing their average orbital radius over time.
The total orbital objects (Moons + Planets) mass in the solar system that are increasing their average orbital radius, is significantly higher than the total moons mass that are reducing their average orbital radius.
I hope that you agree that the ratio between the total orbital mass in the solar system that are decreasing their average orbital radius to the total mass that are increasing  their average orbital radius is less than 1 to 1/10^20
No.  Unrealistic number you made up. Venus probably isn't drifting inward, and it masses well over that unreasonably small percentage of the total.  The moon and every other moon and planet will eventually drift inward (if not destroyed before then), so statistically over time, they spend more time going inward than outward.  I'd say 9/10, not 1/1020.

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3. Mass ratio between the accretion disc to the molecular jet stream.
We all know that the total estimated mass in the molecular jet stream above and below the accretion disc is about 10,000 sun mass.
67.  This is you heavily twisting out of context a number you saw in a article.  The article did not say any such thing.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/12/2019 17:22:39
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Due to kepler law, that energy should blow out the whole accretion disc.
It seems you're the one claiming Kepler's involvement here.  Kindly show which of his laws says this.
https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion
Based on Second law of Kepler:
https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#/media/File:Kepler-second-law.gif
"The same (blue) area is swept out in a fixed time period. The green arrow is velocity. The purple arrow directed towards the Sun is the acceleration. The other two purple arrows are acceleration components parallel and perpendicular to the velocity."
S2 has an elliptical orbit with eccentricity of 0.885:
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So what is the eccentricity of S2?
.885  Look it up.  Not hard to do.
Therefore, due to kepler, the same energy that brings S2 with eccentricity of 0.885 MUST also take it out.
Please look at the (blue) area!!!
Every atom in S2 must obey to the Second law of Kepler
Therefore, if S2 would be collided with the accretion disc, it is not going to lose its outwards energy due to Kepler's Second law even if it would it be torn apart. Each Atom from S2 would have to obey to the Kepler's law and must continue the momentum outwards.
Therefore, due to the collision of S2 with the accretion disc, it would take significantly portion of the mass from the accretion disc outwards.
Try to set a simulation and verify that this is correct by 100%.

S2 would be torn apart by tidal forces and some of it would form a new disk oriented with S2's orbit.  The old disk would probably be absorbed by the new one

Ok.
Let's agree that S2 would be torn apart.
However, due to kepler law the matter in the accretion disc should be blow away with the outwards momentum of all the atoms in S2.
In any case, as the eccentricity of S2 is 0.885 and the eccentricity of the accretion disc is 0.01, than if S2 will be part of the accretion disc then the eccentricity of the accretion disc after the collision should be high above 0.01 (it might be lower than 0.885).
So, in order to verify a star collision with the accretion disc, we only need to find an accretion disc with high eccentricity.
Our scientists could find an Earth size gas cloud falling into the SMBH at a galaxy of 1 Billion LY away.
Based on that high verification, we should have the tools to see any severe interrupt in the accretion disc eccentricity of billions spiral galaxies.
Do we find any interruption in the accretion disc out of the billions spiral galaxies that we can monitor?
If there was a collision, we should also see fireworks.
Do we see any fireworks at any galaxy in the whole universe?
If we don't see any fireworks and any interruption in the eccentricity of the accretion disc than it proves that stars have never ever collide with any accretion disc in the whole Universe.

For S2 and Sgr-A to line up in our view, we'd have to cross the plane of S2's orbit, and that only happens every 100M years or so.  It's called an eclipse when it happens and nobody expects collisions from an eclipse.
Why can't we expect for collision at the eclipse?
Our scientists assume that S2 orbits around the SMBH.
In this case, they are both located in a 2D.
If S2 set an eclipse with the SMBH is MUST collide with it.
Unless, the SMBH is not located at the orbital 2D of S2.
Only in this case there is a possibility for eclipse without a collision.

In any case, with regards to the Molecular jet stream:
Do you agree that the total mass in that jet stream is 10,000 solar mass which had been formed from a constant supply stream from the accretion disc?
However, the total mass in the accretion disc is only 3 sun mass?
If so, how could it be that accretion of only 3 sun mass can supply a constant stream of 10,000 sun mass.
In one of the article that I have found it was stated that
"Shoving 10,000 suns into the black hole at once would do the trick"
https://phys.org/news/2012-05-ghostly-gamma-ray-blast-milky-center.html
However, they didn't give an answer if the accretion disc can hold 10,000 sun mass at once and why the jet stream is nicely and constant (without any interruption) as we can see in the following image:
https://phys.org/news/2012-05-ghostly-gamma-ray-blast-milky-center.html
If the accretion disc had Shoved in the past 10,000 suns at once, then at that moment the outwards stream from the accretion disc must be significantly higher than today, while there is only three sun mass in that accretion disc.
We had to see a severe difference in the thickness of the jet stream as the mass in the accretion disc is reducing.
However, the image of the jet proves that the molecular jet stream is quite constant.
Therefore, it isn't possible that the Milky way' accretion disc had Shoved 10,000 suns at once!!!
There is one more issue: If that was correct, than how could it be that out of the billions spiral galaxies (that we can monitor and which also must supply over 10,000 sun mass to their molecular jet stream), we can't see even one accretion disc with fireworks (due to star collision), with severe interruption in the eccentricity of the accretion disc or with a total of 10,000 sun mass in the accretion disc???

A great number of moons are drifting inward.  More than half of Jupiter's moons are moving inward.
How do we know that?
Did we really measure the drifting direction?
Would you kindly show one (only one moon in the whole universe, except our moon) that we really measured its inwards drifting direction.
Is it one more wrong assumption which is based on our scientists' wishful list (as usual)?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/12/2019 01:01:17
Quote from: Halc
Quote from: Dave
Due to kepler law, that energy should blow out the whole accretion disc.
It seems you're the one claiming Kepler's involvement here.  Kindly show which of his laws says this.
Based on Second law of Kepler:
"The same (blue) area is swept out in a fixed time period.
OK, you've chosen Kepler's 2nd law.  How does your statement above followed from that law?

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Therefore, due to kepler, the same energy that brings S2 with eccentricity of 0.885 MUST also take it out.
Kepler's 2nd law makes no mention of energy.  It also doesn't mention collision with another object.  Again, you draw a conclusion from a random unrelated fact.  Not saying the conclusion is wrong, just that it doesn't follow from the law you've seemingly randomly selected.

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Please look at the (blue) area!!!
Every atom in S2 must obey to the Second law of Kepler
68: Not true.  It certainly isn't true of every atom on Earth for instance.  The law isn't about atoms or other components of an object in orbit.  My mailbox does not obey Kepler's 2nd law.

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Therefore, if S2 would be collided with the accretion disc, it is not going to lose its outwards energy due to Kepler's Second law even if it would it be torn apart.
69: There you go with the 'Therefore' again.  Your statement doesn't follow from any of the above.  Not the 2nd law, and not from the other statements.  You have zero grasp on logic.  It's like saying 2+2=4 (no incorrect), therefore grass is green.  It just doesn't follow from 2+2=4 even if yes, grass is green.

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Therefore, due to the collision of S2 with the accretion disc, it would take significantly portion of the mass from the accretion disc outwards.
70: Again, the 'therefore' is misplaced.  It doesn't follow from the preceding statements.

Kepler's 2nd law does not predict that a bicyclist would likely get severely injured by collision with a fast moving bus.

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Let's agree that S2 would be torn apart.
However, due to kepler law the matter in the accretion disc should be blow away with the outwards momentum of all the atoms in S2.
71: Kepler's 2nd law (nor the 1st or 3rd) does not say this.  It doesn't follow from that law.  You have zero grasp of logic.

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If there was a collision, we should also see fireworks.
Do we see any fireworks at any galaxy in the whole universe?
Indeed we do, almost everywhere except our own galaxy, which currently has a notoriously inactive rate of new matter falling in.

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Why can't we expect for collision at the eclipse?
I've seen many eclipses, and never seen anyone worried that the sun and moon were colliding.  No, nobody has seen S2 line up with Sgr-A. You made that up.

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In this case, they are both located in a 2D.
If S2 set an eclipse with the SMBH is MUST collide with it.
Unless, the SMBH is not located at the orbital 2D of S2.
Only in this case there is a possibility for eclipse without a collision.
Word salad.  'located in a 2D' and 'orbital 2D' are meaningless phrases.  I have no idea what you think you are claiming.

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In any case, with regards to the Molecular jet stream:
Do you agree that the total mass in that jet stream is 10,000 solar mass which had been formed from a constant supply stream from the accretion disc?
No.  'Pathetically weak' was the term I believe was quoted.  It's another thing you've made up.

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If so, how could it be that accretion of only 3 sun mass can supply a constant stream of 10,000 sun mass.
That's right.  It can't, so the 10,000 figure you made up is obviously wrong.

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In one of the article that I have found it was stated that
"Shoving 10,000 suns into the black hole at once would do the trick"
Yes.  They're positing 10,000 suns falling in, not jetting out.  That much falling mass might be enough to power the more significant jets posited in the past.

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If the accretion disc had Shoved in the past 10,000 suns at once
This is mass falling from the outside into the black hole.  The accretion disk is not particularly involved in this, although certainly there would be a disk of some proportion during this kind of activity.  But nobody is positing a disk of that mass.

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Quote from: Halc
A great number of moons are drifting inward.  More than half of Jupiter's moons are moving inward.
How do we know that?
Because there is no known force adding to their energy, while there is very much a known force subtracting from it.

Planets like Jupiter are moving outward mostly because the gravity of the sun is weakening as its mass is radiated away.

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Did we really measure the drifting direction?
Would you kindly show one (only one moon in the whole universe, except our moon) that we really measured its inwards drifting direction.
Moon is heading outward on average.  There is a force adding to its total energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 24/12/2019 03:53:43
Moon is heading outward on average.  There is a force adding to its total energy.

And, by contrast, we know that Phobos is drifting towards Mars. It's been measured. Dave seems to be ignoring that.

Both of the neutron stars in the Hulse-Taylor binary are drifting inward towards each other as well: https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary

These two white dwarfs are drifting towards each other: https://iopscience.iop.org/article/10.1088/2041-8205/757/2/L21/meta

As are these white dwarfs: https://arxiv.org/abs/1910.11389

The extrasolar planet WASP-12b has been measured to have a decaying orbit: https://arxiv.org/pdf/1812.02438.pdf
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/12/2019 16:33:06
Dear Halc
In that article it is stated:
"These jets probably flickered on and off as the supermassive black hole alternately gulped and sipped material," said Finkbeiner. It would take a tremendous influx of matter for the galactic core to fire up again. Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required."

Do you agree that the estimated mass of the molecular jet stream is 10,000 Sun mass?
However, can you please explain how the suppermassive black hole can gulped and sipped material of 10,000 Sun mass outwards from its core below its event of horizon?
Do you see any possibility for that activity?
You have already claimed that this is impossible mission.
How can you now agree for that Idea?
What kind of force is needed to sipped material outwards from the SMBH?
In any case, if matter is ejected outwards from the SMBH, than why do we even think that the accretion disc must get its mass from outside?
Why the same SMBH that can "gulped and sipped material" to the molecular jet stream, can't also gulped and sipped material directly to the accretion disc?
Actually, based on the diagram the molecular jet stream has a stable density.
Let's set a simple calculation:
the total mass in the jet stream is 10,000 sun mass.
The size of that jet stream is 27,000 LY
So, in average density of each one LY segment of that jet stream is:
10,000 / 27,000 = 0.27 sun mass per 1 LY
We have already discussed the speed of that jet stream.
You have stated that the estimated speed is 0.8 c
So, in only one year the total mass that is requested to be supplied to the jet stream is:
0.27 * 0.8 = 0.216 sun mass per year.
I wonder what is the estimated mass ejection per year from the accretion disc.
Do you agree that it should be 0.216 sun mass per year?
Title: Re: How gravity works in spiral galaxy?
Post by: Janus on 28/12/2019 17:45:35
I assume that by now we all understand how the tidal friction set the thrust which is needed to push away the Moon from the Earth.
However, the water bulges in oceans are very unique for the Earth/Moon system.

We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
They don't have water. They don't have relatively big moon around them. They don't form those water bulges.
So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?


Actually not all the moon's move away.  Phobos orbits close enough to Mars to be spiraling in,  There are outer moons of all the gas giants that orbit retrograde and thus spiral in.
You don't need oceans for there to be tidal drag.   The Moon also produces tidal bulges in the Earth itself, it just isn't as large as the Ocean bulges.   These bulges are capable of producing a tidal drag just like the ocean ones do.   
Having oceans increases the drag, but not having them would not eliminate it entirely. 

The degree of tidal drag is determined by a factor known as the Love number ( Named after Augustus Love),  which in turn depends on the density and rigidity of the bodies involved.   The only way you could avoid tidal drag entirely is if you only used perfectly rigid bodies.  Since no astronomical bodies are perfectly rigid, you are always going to have some degree of tidal drag.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/12/2019 23:07:20
Quote
Quote from: Dave Lev on Today at 16:33:06
Do you agree that the estimated mass of the molecular jet stream is 10,000 Sun mass?...
I agree to none of this.  It's all nonsense you're making up sans evidence.
It can't, so the 10,000 figure you made up is obviously wrong.
Yes.  They're positing 10,000 suns falling in, not jetting out.  That much falling mass might be enough to power the more significant jets posited in the past.
So, would you kindly advice why they have used the size of 10,000 sun mass? Why not 1,000 or 100,000 sun mass?
As it is stated: "Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required"
What is the meaning of "required"?
In the following article it is stated:
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"astronomers studying Sgr A* (the supermassive black hole at the centre of the Milky Way Galaxy) were surprised to notice that less than 1% of the gas and dust drawn into its gravitational field ever get consumed – almost everything else gets ejected."
So, if less than 1% of the falling gas is drawn into the SMBH, while 10,000 sun mass are required to fall in, than it is a clear indication that at least 9,900 sun mass is used for the molecular jet stream.
What is the estimated mass in the molecular jet stream?
Do we have any clue about the total mass ejection from the accretion disc per year?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/12/2019 10:14:46
And, by contrast, we know that Phobos is drifting towards Mars. It's been measured. Dave seems to be ignoring that.
Would you kindly direct me to that article?
I would like to understand how our scientists have measured that Phobos is drifting towards Mars.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/12/2019 14:53:35
Would you kindly direct me to that article?
I would like to understand how our scientists have measured that Phobos is drifting towards Mars.

https://archive.briankoberlein.com/2015/11/12/phobos-is-running-out-of-time/index.html

Quote
Measurements of its orbit since the 1950s have found its orbit is decaying at a rate of about 1.8 centimeters per year.

https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2004JE002376

Quote
[2] The Mars Orbiter Laser Altimeter (MOLA) instrument on the Mars Global Surveyor spacecraft has observed 15 transits of the shadow of Phobos across the surface of Mars, and has directly measured the range to Phobos on one occasion. The observed positions of Phobos and its shadow are in good agreement with predictions from orbital motion models derived from observations made prior to 1990, with the notable exception that Phobos is gradually getting ahead of its predicted location. This effect makes the shadow appear at a given location earlier than predicted, and the discrepancy is growing by an amount which averages 0.8 s/yr.

This is a direct observation that Phobos' orbital period is decreasing over time, meaning that its orbit is getting closer to Mars over time as well.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/12/2019 17:27:11
Quote
Measurements of its orbit since the 1950s have found its orbit is decaying at a rate of about 1.8 centimeters per year.
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2004JE002376
[2] The Mars Orbiter Laser Altimeter (MOLA) instrument on the Mars Global Surveyor spacecraft has observed 15 transits of the shadow of Phobos across the surface of Mars, and has directly measured the range to Phobos on one occasion. The observed positions of Phobos and its shadow are in good agreement with predictions from orbital motion models derived from observations made prior to 1990, with the notable exception that Phobos is gradually getting ahead of its predicted location. This effect makes the shadow appear at a given location earlier than predicted, and the discrepancy is growing by an amount which averages 0.8 s/yr.
This is a direct observation that Phobos' orbital period is decreasing over time, meaning that its orbit is getting closer to Mars over time as well.
No, it is not based on real measurements.
It is based on Tidal modeling.
It is stated: ".... which averages 0.8 s/yr. We model this effect, and interpret the required modification in the orbital model as implying a revision to the rate of tidal dissipation within Mars"
Please read it again:
" The observed positions of Phobos and its shadow are in good agreement with predictions from orbital motion models derived from observations made prior to 1990, with the notable exception that Phobos is gradually getting ahead of its predicted location. This effect makes the shadow appear at a given location earlier than predicted, and the discrepancy is growing by an amount which averages 0.8 s/yr. We model this effect, and interpret the required modification in the orbital model as implying a revision to the rate of tidal dissipation within Mars. It has long been understood that tides can be effective in transferring angular momentum from the spin of a planet to the orbit of a satellite, or vice versa, depending on whether the satellite is above or below the synchronous elevation, at which the satellite orbital period matches the planetary rotation period"
In the article they do not give any calculation for their REAL measurements.
They claerly say that they have used Tidal process as was first examined in detail by Darwin:
"The process was first examined in detail by Darwin [1911] and in the case of the Earth‐Moon system there is a very clear signal, with the Earth's spin slowing down, such that the length of the day is increasing by (2.3 ± 0.1) ms per century [Stephenson and Morrison, 1995] and the size of the lunar orbit is increasing at a rate of (3.84 ± 0.07) cm/yr [Dickey et al., 1994]."
They clearly claim that there is a gap between the observation and the model (but it can be minimized):
"The residual between observations and model can be minimized by adjusting the along‐track position of Phobos with respect to the nominal ephemeris. The position in a direction perpendicular to the line of sight to the Sun, and to the along‐track direction, may also be adjusted, while variations in the line of sight direction have virtually no effect.
They even discuss about a misfit between MOLA observation and Phobos predicted position:
"Misfit between MOLA observation and Phobos predicted position.."
They discuss about a misfit of 4km:
"In all but two cases the minimum misfit position lies within 4 km of the nominal Phobos orbital plane. We do not consider the minimization in the cross‐track direction to be sufficiently informative, pending a better understanding of the possible atmospheric effects on the shadow."
How can they get a perfect measurements of few centimeters while this is the shape of phobos?
https://archive.briankoberlein.com/2015/11/12/phobos-is-running-out-of-time/index.html
They try to overcome that problem and set some Error estimations:
"Therefore we apply only the along‐track perturbation in our analysis. Where the surface is reasonably uniform and the light curve is well‐modeled, the precision of the estimate approaches 0.2 s, whereas in other cases the time and duration of the transit is less well resolved. Error estimates are therefore based on the discrepancy between the modeled perturbation with across‐track shift constrained and unconstrained."
The only real calculation that they did is clearly based on tidal:
"A remaining problem in estimating the tidal phase lag is that the higher degree Love numbers are completely unconstrained by direct observations. If we take the uniform density, elastic body values listed above as estimates of the relevant Love numbers, and assume that the phase lag is the same for all harmonic degrees, we can write..."
They are also using tidal quality factor in their calculation:
"and the corresponding tidal quality factor is..."
The Summary tells us that it is all about estimation: "Our new estimate for that parameter..."
"5. Summary
[70] We report on new observations of the orbital position of Phobos, which require an adjustment to previous estimates of secular acceleration in longitude. Our new estimate for that parameter is slightly higher, and much better constrained, than the several estimates which were published circa 1990."
Are you sure that after all of this you still want to call it "measurements"?
How can they really verify real measurements of 1.8 centimeters per year under all of those estimations/errors?
It is clear that the 1.8 centimeters per year is a direct outcome of tidal concept + Error + Estimations.
Sorry, Tidal calculation (with or without error + estimations) isn't a real measurement!
This is just misleading information which is based on one more wishful thinking...
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 29/12/2019 17:50:06
Did you completely miss this quote of mine?

Quote
[2] The Mars Orbiter Laser Altimeter (MOLA) instrument on the Mars Global Surveyor spacecraft has observed 15 transits of the shadow of Phobos across the surface of Mars, and has directly measured the range to Phobos on one occasion. The observed positions of Phobos and its shadow are in good agreement with predictions from orbital motion models derived from observations made prior to 1990, with the notable exception that Phobos is gradually getting ahead of its predicted location. This effect makes the shadow appear at a given location earlier than predicted, and the discrepancy is growing by an amount which averages 0.8 s/yr.

Phobos' orbital period is getting shorter every year. We know this because the shadow has been observed to move further and further ahead of schedule. It takes less time for the shadow to reach the same spot every year. That is a direct observation.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/12/2019 12:30:51
Did you completely miss this quote of mine?
Quote
[2] The Mars Orbiter Laser Altimeter (MOLA) instrument on the Mars Global Surveyor spacecraft has observed 15 transits of the shadow of Phobos across the surface of Mars, and has directly measured the range to Phobos on one occasion. The observed positions of Phobos and its shadow are in good agreement with predictions from orbital motion models derived from observations made prior to 1990, with the notable exception that Phobos is gradually getting ahead of its predicted location. This effect makes the shadow appear at a given location earlier than predicted, and the discrepancy is growing by an amount which averages 0.8 s/yr.

Phobos' orbital period is getting shorter every year. We know this because the shadow has been observed to move further and further ahead of schedule. It takes less time for the shadow to reach the same spot every year. That is a direct observation.
Yes, that message is only valid at the "Introduction".
However, in the article it is stated totally differently:
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2004JE002376
[34] However, as the corrections required to fit the new observations are very small, and appear to be almost entirely along‐track perturbations, we can use a very simple linear perturbation analysis and adjust only 3 parameters. If the adjustments in time of along‐track position were large, compared to the shortest forcing periods, this simple linearization would not be sufficient. The orbital model has secular effects, long period effects, and short period effects, and the forcing periods are mainly at harmonics of the Mars heliocentric orbital period, and the Phobos mean orbital period. As these periods are incommensurate, the orbital motion is quasi‐periodic and a shift in time by a significant fraction of the shortest period will alter the structure of the beat patterns between the input periods. However, the rate of accumulation of along‐track position error is equivalent to 0.8 s/yr, or a few parts in 10^7, and the linear analysis is warranted."

So, we see clearly that they had to make several assumptions:
1.  Adjustments : "If the adjustments in time of along‐track position ..."
2. Linearization: :"this simple linearization would not be sufficient."
3. orbital model: "orbital model has secular effects, long period effects, and short period effects..."
4. shift in time - "As these periods are incommensurate, the orbital motion is quasi‐periodic and a shift in time by a significant fraction of the shortest period will alter the structure of the beat patterns between the input periods"

After all of this, how could it be that they can really measure the difference of 0.8sec per year???
Actually they do not claim for that.
They only say: "the rate of accumulation of along‐track position error is equivalent to 0.8 s/yr...
So, the 0.8 s/yr represents a position error, so I really can't understand how a position error had been suddenly converted to real position location.
In the article they don't show any real calculation for that  0.8 s/yr and how all the assumptions and Adjustments set the outcome.
However, they deeply discuss and set calculation based on tidal as follow:
"[40] We will consider four elements of the tidal process. The response of an ideal elastic body to tidal perturbations sets much of the required background. We then consider the influence on the orbit of a delay in tidal deformation, which causes the tidal bulge to be misaligned with the tide raising body. Next we consider the behavior under tidal forcing of a particular model of viscoelastic deformation. Finally, we consider the long term orbital evolution of a pair of tidally interacting bodies."
So, its quite clear that Tidal formulas has a severe impact on our scientists understanding.
If our scientists have really set the measurements for that result, why most of the calculations in the article are based on tidal formulas???

In any case, let's try to see what kind of sensitivity we really need to measure that 0.8 s /year/
In a year there are 31,556,926 Seconds.
0.8/31,556,926 = 2.5 * 10^-8
Do you really think that it is possible to achieve that kind of sensitivity for a none circular object (Phobos) which orbits with some low eccentricity around Mars, by using the measurements from another object that orbits in space (MOLA)?
It really seems to me as an impossible mission.
But as usual, our scientists make the impossible - possible.

Why our scientists didn't try to find further information?
1. Eccentricity - it is stated that Phobos has low eccentricity, but how low it is?
"Both orbits (MGS and Phobos) have low eccentricities, and thus nearly uniform speed in the along‐track direction."
How that low eccentricity could affect the outcome of their calculation.
I'm not sure if they really took that impact in their calculation and if they did, don't you think that this issue by itself could set an error which is bigger than just 0.8s/year?
and the MOST important issue:
2. Phobos orbital velocity -
Did they measure the orbital velocity of Phobos?
This is the most important issue in my point of view.
If phobos is drifting inwards without increasing its average orbital velocity than it proves that there is no possibility to increase the orbital energy by reducing the potential energy.
In any case, the Eccentricity is quite important factor for the orbital velocity.
So, as our scientists have totally ignored the Eccentricity, the orbital velocity, add to that all the assumptions and tidal effect/calculations, it is quite clear to me that they can't get that 0.8 s/y accuracy.
Therefore, their conclusion is totally useless.

If they wish to set real measurements, than they have to set a laser beam between Phobos to Mars.
Title: Re: How gravity works in spiral galaxy?
Post by: Kryptid on 30/12/2019 19:39:56
You are unteachable. I give up.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/12/2019 21:49:04

You are unteachable. I give up.
I have just proved that Phobos isn't necessarily moving in
If you still don't agree with my explanation, then please let me know why you don't.
If you see that I'm correct, then why don't you accept the advantages of my explanation?
Actually, if you really monitor all the planets and moons in the whole Universe (Billion over billion moons & planets), you won't find even one that is drifting inwards while it is also increasing its orbital velocity in that process.
Title: Re: How gravity works in spiral galaxy?
Post by: jeffreyH on 31/12/2019 01:15:15
The velocity of an object in a decaying orbit will increase. It will spiral into the central mass. Please stop making yourself look foolish.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/12/2019 17:13:54
The velocity of an object in a decaying orbit will increase. It will spiral into the central mass. Please stop making yourself look foolish.
What do you mean by "Velocity"
Based on Google: "Velocity = the speed of something in a given direction".
So, I fully agree that "the velocity speed of an object in a decaying orbit will increase."
However, what about the direction of that speed?
We should distinguish between Orbital velocity (or Horizontal speed direction) to "falling in" velocity (or vertical speed direction)?
You discuss on an object in a decaying orbit.
In order to get better understanding, let's agree that we start from a pure orbital circular.
So, as the orbital object is reducing its radius and spiraling in its vertical speed direction is increasing. That is very clear.
However, that vertical speed direction has no affect on the horizontal speed direction.
So, let's assume that at t=T1 the radius = r1 therefore:
The orbital velocity (or horizontal speed direction) is Vo1
The "falling in" velocity (or vertical speed direction) is 0 (zero)
At t=T2 the radius = r-h = r2
So, that decrease in the radius "h" is actually set a transformation in the Potential energy at  r1 to the kinetic energy at r2.
So, we can claim that the decreasing in the radius in a value "h" is similar to "falling in" for the object from a high of "h"
That "falling in" from that high "h" set a vertical speed direction (or falling in velocity) due to the transformation from potential energy to falling in kinetic energy.
So, as the object gets to radius r2 at T2 it velocity is a combination of the following two vectors:
Its orbital velocity at T1 (which represents the horizontal speed direction) Vo1 + Its new Falling in velocity at T2 (which represents the vertical speed direction) Vf2.
It is quite clear that the horizontal seed direction (orbital velocity = Vo1 ) is orthogonal to the vertical speed direction (falling in velocity = Vf2)
So,
With regards to your statement:
"The velocity of an object in a decaying orbit will increase. It will spiral into the central mass."
It should be as follow:
The Speed of an object in a decaying orbit will be increased.
However, its horizontal speed direction (orbital velocity=Vo1) will stay as it was at T1, while its vertical speed direction at T2 (falling in velocity) will be increased due to the reduce in the radius (or H (high)= h) to Vf2.
Therefore, It will spiral into the central mass
In order to get better understating, I highly advice to read the following explanation by Newton:
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon.
With regards to my statement:
Actually, if you really monitor all the planets and moons in the whole Universe (Billion over billion moons & planets), you won't find even one that is drifting inwards while it is also increasing its orbital velocity in that process.
It should be:
Actually, if you really monitor all the planets and moons in the whole Universe (Billion over billion moons & planets), you won't find even one that is drifting inwards while it is also increasing its orbital velocity in that process, but it can increase the falling in velocity.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/01/2020 11:00:11
Actually not all the moon's move away.  Phobos orbits close enough to Mars to be spiraling in,  There are outer moons of all the gas giants that orbit retrograde and thus spiral in.
I would like to add that retrograde orbit can't be an indication for drifting inwards or spiraling in.
The only way to prove that an object is "spiraling in" is by real laser measurements as we have alreay found the discovery about the earth/moon "spiraling out".
In any case, we all agree that in order to be claim that an object is in orbital path around any main mass; its velocity must meet the following "magic" formula:
V^2 = G M / R
That orbital velocity is also called: "the magic orbital velocity"
Let's assume that at T1 (while the radius is r1) the magic orbital velocity is: Vo1
Hence, by decreasing the radius (assuming that it is in a pure circular orbit) the orbital velocity must be increased.
That orbital velocity represents the horizontal speed direction.
As I have stated, by decreasing the radius we only increase the falling in velocity vector or the vertical speed direction (without any impact on the Orbital magic velocity)
Therefore, at T2 (while the radius had been reduced to r2) the orbital velocity is also Vo1
That orbital velocity is too low to keep it in the orbital path.
So, the object or the moon must now spiral in at acceleration "falling in" velocity.

With regards to Phobos:
If I understand it correctly, we monitor that moon for almost 20 Years.
During this time we didn't see any real change in the orbital path (even if we agree for 0.8s per year it is really neglected verification).
That proves that Phobos is orbiting at the requested magic orbital velocity for the last 20 years.
Hence - it is surly not spiraling in.

Title: Re: How gravity works in spiral galaxy?
Post by: jeffreyH on 01/01/2020 16:58:32
You take the 'instantaneous' velocity at tangents along the orbital path. These velocities when plotted against the time axis will show an increase. Or do you not understand physics?

NOTE: The magnitude of the velocity is equivalent to a speed without  a defined direction. Geometrically, speed has no meaning. Since it is a scalar quantity.
Title: Re: How gravity works in spiral galaxy?
Post by: jeffreyH on 01/01/2020 17:27:38
This may help to ease your confusion.
Title: Re: How gravity works in spiral galaxy?
Post by: evan_au on 02/01/2020 10:42:40
Quote from: Dave Lev
Actually, if you really monitor all the planets and moons in the whole Universe (Billion over billion moons & planets), you won't find even one that is drifting inwards
These days, astronomers are happy to merely detect that there is a planet around a star.
- It is beyond our current technical ability to measure the orbital period to millisecond accuracy

That means it is also beyond your ability. So your statement is merely a wild guess driven by wishful thinking

However, there was one unusual case where astronomers were able to measure the orbital period with millisecond accuracy
- And they found that there was an inwards spiral
- This discovery was awarded a Nobel Prize in 1993

See: https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary

PS: I just saw that this case had been mentioned in a previous post...
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/01/2020 20:14:25
However, there was one unusual case where astronomers were able to measure the orbital period with millisecond accuracy
- And they found that there was an inwards spiral
- This discovery was awarded a Nobel Prize in 1993
See: https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary
Thanks
However, in that article it is stated:
"The pulsar and its neutron star companion both follow elliptical orbits around their common center of mass. The period of the orbital motion is 7.75 hours, and the two neutron stars are believed to be nearly equal in mass, about 1.4 solar masses."

This can't be used as an example for our case.
We only discuss on an object with relatively low mass orbiting around main mass.
For example: A moon that orbits around planet, planet around a star or even Atom around a SMBH.
It is quite clear to me that in all of those cases, the low mass orbital object can't spiral inwards while it also increases its orbital velocity in order to meet the requested "magic" orbital velocity.
Therefore, that discovery is none relevant to our discussion.

In any case, I would like to ask how can we distinguish between real modeling/theory to absolutely none relevant one.
As an Engineer, my answer for that is very simple:
You set a theory based on that set expectations and then verify the results.
If the results/verifications meet the expectations than the theory is correct.
If not, than we normally set that theory in the garbage and start from zero.
However, somehow, it seems to me that our scientists have never stopped for one moment and ask themselves if their current mainstream/theory is correct or not.
If they predict something and it doesn't work, this is still perfectly ok for them.
I have seen in several articles that our scientists were "surprised" to discover that their expectations didn't materialized.
But they don't stop. They keep on with their current assumptions and hope to find a positive discovery in the future.

For quite long time our scientists were quite sure that if you dump in a huge gas, significant portion of that must falls into the SMBH.
Therefore, they were absolutely surprised when they have discovered that only less than 1% of the gas and dust drawn into the SMBH/accretion disc. (and even this one is incorrect - as so far they couldn't find any real object that falls into the accretion disc)
In the following article it is stated:
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"astronomers studying Sgr A* (the supermassive black hole at the centre of the Milky Way Galaxy) were surprised to notice that less than 1% of the gas and dust drawn into its gravitational field ever get consumed – almost everything else gets ejected."
However, even after that discovery Halc still believes that somehow much more than 1% must fall in, without any real indication/observation for that.
That 1%, if accurate, refers to its current dormant state of barely sipping any new mass in at all.  You dump a huge cloud in like that, a lot more than 1% falls into the black hole.

Unfortunately our universe is not so cooperative and refuses to act according to our scientists wishing list.
I would expect that after the first "surprised" discovery, our scientists had to clear the table and look for better theory.
Why they do not wish to understand that if the Universe disagrees with something, they will not find it even if they climb to the top of the Everest...
Our scientists are 100% sure that matter must fall in to the accretion disc.
But so far they couldn't find any real object as star or gas cloud that falls into the accretion disc.
From time to time they see some flares coming from the accretion disc without any real observation for any falling in object.
Those flares could be simple outcome from the activity at the accretion disc.
Even so, they are using those flares as an indication for falling in objects.
So, how long they would still hope that their unrealistic wishing list will materialized?


Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/01/2020 09:55:29
"The large-scale magnetic field of a thin accretion disk with outflows"
https://arxiv.org/pdf/1901.10103.pdf
In this article our scientists would like to assume that the magnetic field is generated due to the rotation of the accretion disc.
So, out of the blue, they have stated that above the disk a potential field is assumed:
"Above the disk, a potential field is assumed, which is a good approximation for a tenuous outflow, and is widely adopted in most previous modeling of magnetically driven outflows"
Based on that assumption and after long calculation they have concluded that:
"The disk is compressed in the vertical direction both by the magnetic stress (which acts like a negative pressure) and the vertical component of the gravity"
This actually explains why the accretion disc is so thin and compressed.
Now, let's try to understand the impact of that "compressed thin accretion disc"
1. It is clear that nothing can be above or below the disc (as it is compressed and thin)
2. The whole disc MUST rotate in one direction. That must be key assumption in this article. If the accretion disc was based on several rings that orbits at the opposite directions, we could assume that each one generates magnetic field at a different direction. This could eliminate completely the combined magnetic field. So, it is clear that the whole accretion disc must rotate in one direction.
I hope that you agree with all the above.

Now, based on those key understandings, let's go back to the following discovery that we have already discussed on:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"We were able to follow an Earth-sized clump of matter for about a day, as it was pulled towards the black hole, accelerating to a third of the velocity of light before being swallowed up by the hole"
My questions are as follow:
1. How could it be that an Earth-sized clump of matter could exists above or below of a thin and compressed accretion disc???
If the accretion disc is compressed and thin, than this Earth-sized clump of matter must also be compressed to the disc!!!
2. How this Earth-sized clump of matter could even be created?
In this article our scientists assume that accretion disc is based on several rings that orbit at opposite directions:
https://phys.org/news/2018-09-falling-black-hole-percent.html
"This work has shown that rings of gas can break off and collide with each other, cancelling out their rotation and leaving gas to fall directly towards the black hole."
However, we have already found that the entire accretion disc must orbit in one direction. So there is no room for this "Rings" assumption.
Therefore, the idea that that Earth-sized clump of matter could be created due to a collision between two nearby rings (in the accretion disc) that rotate in different directions is totally incorrect.
Therefore, Our scientists must find better explanation for that Earth-sized clump of matter.
I clearly know the answer for that.
That Earth-sized clump of matter is not located near the accretion disc and not falling into the SMBH.
It is just a clump of matter that was already ejected from the accretion disc and now due to the magnetic field it is boosted upwards to the molecular jet stream that we see above and below any SMBH.
That clump of matter is moving at 0.8 c. However, as we see that clump from above (directly face on to the galaxy spiral disc), we only monitor 0.3c.
So, it is not falling in but actually moving far above the SMBH.
I also think that our scientists are not so foolish and they also must KNOW that.
However, as they are bagging for something that falls into the SMBH, they are using that discovery to prove their unrealistic assumption.

 
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/01/2020 16:28:35
Quote
Quote from: Dave Lev on 21/12/2019 07:56:53
Few years ago, our scientists were positively sure that they are going to see fireworks as S2 is going to collide with the SMBH.
They even verify that S2 and the SMBH were in the same direct view line from us.
Unfortunately for them and for you, there were no fireworks and no collision.
S2 had passed very close to the SMBH without setting any sort of effect on the Accretion disc or on itself.
Link please, because I've heard no such thing.  Cannot comment on things made up.
For S2 and Sgr-A to line up in our view, we'd have to cross the plane of S2's orbit, and that only happens every 100M years or so.  It's called an eclipse when it happens and nobody expects collisions from an eclipse.

Dear Halc
As requested, please see the following:
https://ui.adsabs.harvard.edu/abs/2005sao..conf..286C/abstract
"We present the first NACO thermal infrared observations of the Galactic Center. L'-band imaging has been performed during NACO Science Verification nights in 2002. During these observations, the separation between S2, the closest star to Sgr A*, and the black hole was too small to directly see a counterpart of the latter."
So, it is clearly stated that they couldn't separate between S2 to Sgr A* in 2002.
Therefore, S2 and Sgr-A* lined up in our view
I have found several articles that try to explain that phenomenon.
Mostly, they try to claim that there was some error in the observation.
My answer is quite simple:
No, there is no error in this observation.
It is possible to see Sgr A* and S2 in the same line view and we even can see Sgr A* outside the S2 orbital cycle as S2 doesn't orbit around Sgr A**.
Somehow our scientists do not wish to accept that simple outcome from their clear observation.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/01/2020 06:03:42
Dear Friends

In this thread I have proved that the accretion disc is actually excretion disc.
The SMBH doesn't "eat" even one atom from that disc.
On the contrary;
The SMBH generates/creates all the particles/atoms that we see in the excretion disc.
New particles pair is created by the SMBH near the event horizon.
As one is drifted outwards to the innermost side of the disc, the other one is used as the main food for the SMBH.
Hence, for any new particle that the SMBH contributes to our universe it eats one.
Therefore the SMBH is so massive object and has the ability to generate an ultra electromagnetic filed.
That electromagnetic field is acting as the biggest "CERN" in the Universe.
It transforms the requested energy to the plasma in the disc and set ultra pressure that is needed to form all new Atoms/molecular that we know in our Universe including water!!!
All the 100% mass in that disc is drifted outwards, than boosted upwards/downwards by the SMBH' mighty electromagnetic fields (up to 27,000 Ly above/below the SMBH' poles).
After losing the energy and falling back to the galactic plane near the SMBH it is used to form new stars in a gas cloud.
This activity is the "life" creation for our whole Universe.
I will open a new thread and explain how our whole magnificent Universe had been formed by the ability of the SMBH' excretion disc to create new atoms.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/03/2020 14:34:00
New mass creation around the SMBH:

Let's start with magnetic field that is generated by the spinning SMBH:

1. Magnetic Field
Black Holes & Time Warps states that a spinning black hole with a net electric charge will have a magnetic field.

Galactic nucleus - the nucleus of the Spiral galaxy is supper massive black hole – Wikipedia: "A supper massive black hole defined mass ranges from100 thousand to 10 billion solar masses. Scientists tend to assume that such a black hole exists at the center of most galaxies in the universe, including the Milky Way."   It holds around hundreds of billions of stars. So clearly, the nucleus creates tremendous power and energy. 
The spin of the SMBH generates ultra powerful magnetic field. 
"A team of researchers has measured the magnetic fields in the vicinity of the suppermassive black hole at the center of NGC 1052."
https://scitechdaily.com/researchers-measure-magnetic-fields-in-the-vicinity-of-a-black-hole/
Two particle jets shoot out from the heart of active galaxy NGC 1052 at the speed of light, apparently originating in the vicinity of a massive black hole.
The team concludes that the magnetic fields provide enough magnetic energy to power the twin jets.
Similar particle jet stream stretches 27,000 light-years from the center of the Milky Way galaxy:
https://www.cfa.harvard.edu/news/2012-16
"The newfound jets may be related to mysterious gamma-ray bubbles that Fermi detected in 2010. Those bubbles also stretch 27,000 light-years from the center of the Milky Way. However, where the bubbles are perpendicular to the galactic plane, the gamma-ray jets are tilted at an angle of 15 degrees. This may reflect a tilt of the accretion disk surrounding the suppermassive black hole.
"Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required"
In order to blow those kind of particles jet stream to that distance of 27,000 LY  it is clear that an Ultra Magnetic field is needed.
ONLY a spinning SMBH can generate that kind of magnetic field!!!

2. New mass creation process:
The gravity and electromagnetism don't contribute to the black hole's expendable energy, but the rotation does.
Chapter 12 of Black Holes & Time Warps does indeed mention that a black hole's rotation can produce radiation. So, new pair of particles can be created around a BH or SMBH.
In order to produce a positron-electron pair, 1.022 MeV of rotational kinetic energy is extracted from the BH
Let's assume that we are looking down on the most inwards side of the accretion disk (or even below) from above.
Let's also assume that electron and positron had been created at some radius below the inmost accretion ring. At the moment of creation they will probably orbit at almost the speed of light.
Please remember that at the moment of creation, the new created particles pair must fully meet the orbital speed for the attitude (or radius) from the SMBH.  It must fully obey to Newton orbital law.
We can get better understanding by look at the following Newton Cannon Ball explanation:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=7300.gif
If the speed is the orbital speed at that altitude it will go on circling around the Earth along a fixed circular orbit just like the moon.
How Lorentz force works on those new particles pair?
In order to get better understanding let's look at the following video:
https://www.youtube.com/watch?time_continue=135&v=RqSode4HZrE&feature=emb_title
The North/South Poles of the SMBH is up/down with reference to their orbital direction. Therefore, based on that video, one charged particle should be deflected to the left while the other one would be deflected right. Hence, one particle should be deflected inwards to the SMBH direction, while the other one would be deflected outwards to the direction of the accretion disc.
The deflection inwards would decrease its altitude or radius from the SMBH. Therefore, it will face stronger gravity force from the SMBH.
That radius change will force it to fall in as its current orbital velocity would be too low. As it is stated in the following video:
"If the speed is low, it will simply fall back on Earth" (or to the SMBH in our case)
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif

On the other hand, the other particles must be deflected outwards from the SMBH. Therefore, its speed would be too high with reference to its current radius. Even a small deflection should bring it under the influence of the inwards side of the accretion disc. At that aria it would have to obey to the magnetic forces/pressures that are generated by the accretion disc itself. We know that the average orbital velocity at the accretion disc is about 0.3c. So, the new arrival particle might bang with the other particles already orbiting at the inwards side of the accretion ring and reduces its velocity from almost the speed of light to about 0.3c. At that moment it would become a new member at the plasma.
With regards to temperature – A new created particle must come with Ultra high temp. Adding to that the ultra high pressures, forces, Electric current flow and fusion activity in the plasma would increase the temp to almost 10^9 c at the accretion disc.   
This separation deflection process is vital. Without it, any new created particle pair would be eliminated at the same moment of their creation as each particle carry a negative charged with reference to the other.

3. Energy transformations
The requested energy for electron-positron pair is 1.022 MeV. That energy had been taken from the energy of the SMBH by the transformation of the magnetic field.
So, theoretically, the SMBH had lost 1.022Mev (due to the creation of the particle pair) and gain only half of that as the mass of a falling in particle as stated by kryptid:
If 1.022 MeV of rotational kinetic energy is extracted from the hole in order to produce a positron-electron pair, then the black hole can only get 0.511 MeV of that energy back by consuming one of the particles. It would only get back half of the energy that it expended.

Any method you use to try to get a system to create net energy is a violation of the first law of thermodynamics. You might as well stop trying.

However, at the moment of the creation the orbital velocity is almost at the speed of light. That speed is given for free from the Ultra gravity force of the SMBH.
Hence, the Kinetic orbital velocity of each particle -with mass m at the moment of creation (assuming that its velocity is the speed of light) is as follow:
Ek = 1/2 m v^2 = 1/2 m c^2
Each falling in particle (as electron for example) is increasing the total mass of the BH by only 0.511 MeV.

However, It also increases the spin of the SMBH due to Conservation of momentum and Tidal. We only discuss on a tiny particle. However, unlimited number of falling in particles can have a similar impact as a falling star with the same total mass.
So the SMBH gravity force had contributed Ultra rotational energy to the created particle pair for free. Some of that rotational energy is transformed back to the SMBH due to Conservation of momentum and due to Tidal energy transformation.
Please remember that Tidal forces transform existing orbital or rotational energy into heat energy.
Therefore, this process doesn't contradict the first law of thermodynamics as was assumed by Kryptid:
Even if you do have a valid way of getting a black hole to produce matter and eject it the way you want it to, that still ignores the fact that a black hole cannot generate unlimited mass-energy. The mechanisms are irrelevant. The specifics are unimportant. The first law of thermodynamics simply won't let your model work. No amount of figuring will allow you to get more mass-energy out of the black hole than was there to begin with. Doing so would violate the first law by definition.
Since the total amount of orbital/rotational energy in a New particle pair around the SMBH is ultra high (and it is for free due to the SMBH mighty gravity force), Conservation of momentum, tidal heating process, SMBH Spin, Transformation of energy by magnetic force to new creation particles pair cycle can go on forever.

Conclusions:
As the universe age is infinite, than unlimited number of falling in particles should increase dramatically the total Energy & mass of the BH and converts it over time to a SMBH without violating the first law of thermodynamics.