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**New Theories / Re: Shooting out of a moving pickup truck.**

« **on:**14/10/2019 21:15:20 »

Muzzle velocity is still finite, so leading the target would still be required.

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Muzzle velocity is still finite, so leading the target would still be required.

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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

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: E

E

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

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?

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But still this equation gives the same number whether the substance is a 150,000g/mole or 150g/mole which is not sensible.

Keep in mind that what you are calculating is kilograms per square meter per second, not molecules per square meter per second. Larger molecules will move slower, but they carry more mass.

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As seen, visual perceiving does not need an exponential function of parameters.

Then it doesn't match experimental results and is therefore falsified.

The subject of Radar: We never see any think simultaneously because of the finite/limited value of light’s velocity theoretically. However we have not any problem for nearby objects. So, negligible effects are mentioned. You are right radar system never gives exact results; but it is functional for our local distances.

A policeman's radar gun doesn't make measurements based on delay time. It detects frequency shifts. So a limited speed of light is a non-issue.

Your additional proofs are not my interest area.

This is a cop-out.

I study astronomy, cosmology, light kinematics.

Gravitational waves, Mercury's orbit and neutron stars are all relevant to astronomy, so you can't use the "not my interest area" excuse for those.

Therefore the root postulate of SR is not consistent.

Non-sequitur. You absolutely

In the case where you aren't moving directly towards or away from the asteroid, but are moving at a right angle to it, you can use the angle that the radio pulse was received from to determine the direction that the asteroid is moving relative to the ship. That information, combined with the other information mentioned above, will still tell you the change of distance to the asteroid over time in your reference frame.

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You already tried that argument and it didn't work. The equation you offered predicted a linear relationship between velocity and red shift, whereas relativity predicts an exponential one. Relativity's prediction is the one with experimental support. You tried to save the equation by claiming that you have to use "universal velocity" instead of relative velocity, but that makes the problem worse. If we did live in a world with an absolute reference frame and red shift was dependent upon absolute velocity, then the radar guns used by police officers would have to be continually calibrated throughout the day and year to reflect that change in velocity over time due to the Earth's rotation and orbit.

If what you are claiming instead is that we have to measure the difference in the universal velocity of the car and the universal velocity of the radar gun, then that simplifies right back to relative velocity again. It doesn't matter if the Milky Way galaxy was sitting still or moving at 50% the speed of light in some absolute frame: the red shift detected by the radar gun would be the exact same because it's measuring relative velocity. That was known long before special relativity was even conceived of.

Time dilation is also far from the only observable prediction of relativity.

An illusion caused by the finite speed of light cannot explain the observation of quadrupolar gravitational waves.

An illusion caused by the finite speed of light cannot explain the observation of mass-energy conversion.

An illusion caused by the finite speed of light cannot explain the precession of mercury's orbit.

An illusion caused by the finite speed of light cannot explain the decay rate of neutron star's orbits.

All I see is a quoting of my posts.

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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)/r

F = ((6.674 x 10

F = (7.926080939013 x 10

F = 3.5418308070902173002872574484922 x 10

Attraction between the Moon and Sun:

F = (GMm)/r

F = ((6.674 x 10

F = (9.7437510158 x 10

F = 4.3315711523075275820924842308266 x 10

Sum of the two forces: 3.5851465186132925761081822908005 x 10

Attraction between the Sun and a hypothetical Earth-Moon combination object:

F = (GMm)/r

F = ((6.674 x 10

F = (8.023518449171 x 10

F = 3.5849742757445814209639045480079 x 10

Those two values for force are different from each other by about 1.722 x 10

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?

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That's definitely not a good start for you. You should shape up before you get banned for threats and insults.

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By inventing methods such as the scientific method.

But if you think the scientific method isn't good enough, then what do you propose to replace it?

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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

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

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

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...

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Philosophy could help prevent the requirement of brute force attempts to get results.

How?

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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.

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1- In universe, everything has a motion. The world has also its motion according to Sun, galaxy, local cluster, ........etc.

We use the earth as a co-reference frame for physical and other events. However , If we want to analyze the motion relation of Earth and a celestial object (e.g. Fornax cluster) we must use an outer/common reference frame how includes both of them. And -as a first step- we have to consider the centre of local cluster as the equivalent partner of Fornax.

And all of that motion is relative (except for acceleration).

2- If claimed deformations are visual; that space-time is also an illusion

As Bored Chemist said, they are physically real.

SR does not impose/effect the life.

That depends on what you mean by "affect".

Even, it cannot transform the astronomical parameters to respected/ useful / simultaneous values.

I don't understand this sentence.

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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

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.

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a search for truth by thinking

How does one test to see if their thinking is accurate?

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but, a similar deformation never be happened in our life because of universal motion of the world.

What is this "universal motion of the world" you speak of? Are you talking about an absolute reference frame? If so, then you are begging the question again: https://en.wikipedia.org/wiki/Begging_the_question

As I say; if the theory SR is consist of this, it will become fairy tale for adults.

Again, why do you conclude this?

I agree, there are visual deformations. The reason of them is the finite/limited value of light's velocity.

You already tried that argument and it didn't work. The equation you offered predicted a linear relationship between velocity and red shift, whereas relativity predicts an exponential one. Relativity's prediction is the one with experimental support. You tried to save the equation by claiming that you have to use "universal velocity" instead of relative velocity, but that makes the problem worse. If we did live in a world with an absolute reference frame and red shift was dependent upon absolute velocity, then the radar guns used by police officers would have to be continually calibrated throughout the day and year to reflect that change in velocity over time due to the Earth's rotation and orbit.

If what you are claiming instead is that we have to measure the difference in the universal velocity of the car and the universal velocity of the radar gun, then that simplifies right back to relative velocity again. It doesn't matter if the Milky Way galaxy was sitting still or moving at 50% the speed of light in some absolute frame: the red shift detected by the radar gun would be the exact same because it's measuring relative velocity. That was known long before special relativity was even conceived of.

Time dilation is also far from the only observable prediction of relativity.

An illusion caused by the finite speed of light cannot explain the observation of quadrupolar gravitational waves.

An illusion caused by the finite speed of light cannot explain the observation of mass-energy conversion.

An illusion caused by the finite speed of light cannot explain the precession of mercury's orbit.

An illusion caused by the finite speed of light cannot explain the decay rate of neutron star's orbits.

Probably it may be also similar to muon experiment.

It can't be. Lithium ions aren't unstable like muons are.

They had neglected to consider the speed of reference muons.

I already pointed out to you that we can make very cold muons for which time dilation is negligible. Even in that particular case, knowing the kinetic energy and mass of the muons allows one to calculate their velocity. You can then calculate from that velocity and the observed decay rate what the resting decay rate should be (as predicted by relativity). If relativity is correct, then the predicted decay rate of those resting muons can be used to accurately predict the decay rate of moving muons.

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Speaking of low-pressure, pure oxygen atmospheres, is it known if such atmospheres have a negative impact on human beings over a very long period of time? Like years? What about other organisms? Plants at least would have an issue with nitrogen fixation if some terraformed planet didn't have any nitrogen in the atmosphere. If we're talking about a sealed colony, then I suppose that could be fixed manually with artificial fertilizer.

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I'd be interested in some numbers from any relativity guru on this forum: what duration of travel might produce a reasonably measurable time difference between two clocks?

It doesn't have to be a long trip at all for the difference to be measurable: https://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment

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This is just going around in circles....

What wasn't clear about what we said?

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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

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The difference between those two values is 1.293786% at apogee.

Moon distance at perigee: 362,600,000 meters

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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)/r

F = ((6.674 x 10

F = (3.985959738 x 10

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)/r

F = ((6.674 x 10

F = (1.992979869 x 10

F = 164.7 newtons

F = (GMm)/r

F = ((6.674 x 10

F = (1.992979869 x 10

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.

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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.