Naked Science Forum

On the Lighter Side => New Theories => Topic started by: guest39538 on 14/03/2018 11:38:39

Title: Black hole equations.
Post by: guest39538 on 14/03/2018 11:38:39
Hello, I can not access the closed thread to re-post the provided BH equations, it is showing white screen.

Could somebody please re-post them in this thread so I can discuss them and try to understand them.
Title: Re: Black hole equations.
Post by: Kryptid on 14/03/2018 15:22:10
Here is my post from before:

You can verify the calculation for yourself. The event horizon is simply the location where the black hole's escape velocity equals the speed of light. The equation used to calculate the escape velocity of an object is:

ve = √((2GM)/r)

Where ve is the escape velocity in meters per second, G is the gravitational constant (6.67 x 10-11 m3kg-1s-2), M is the mass in kilograms and r is the radius from the center of the object in meters.

Plugging in the numbers for the Earth, we get:

ve = √((2 x (6.67 x 10-11 m3kg-1s-2) x (5.972 x 1024 kg))/(6,371,000 meters))

ve = 11,182 meters per second (11.182 kilometers per second)

Given how many times we've sent spacecraft into orbit, we've have plenty of occasion to thoroughly test the validity of this equation.

Now let's rearrange the equation so that what we are looking to find is not the escape velocity, but the radius at which the escape velocity takes on a particular value:

r = √((2GM)/(ve2))

Put in the relevant data for the Earth and you can verify that this rearranged equation accurately predicts the radius of the Earth based on its mass and escape velocity.

Now we can enter the speed of light as the escape velocity (299,792,458 meters per second) in order to find the distance from the center of a black hole at which the event horizon must exist for a given mass. Well enter the measured mass of the Cygnus X-1 black hole of 14.8 solar masses:

r = (2 x (6.67 x 10-11 m3kg-1s-2) x (2.943 x 1031 kg))/(299,792,458 meters per second)2

r = 43,682 meters (43.682 kilometers)

So there you have it, a step-by-step explanation on how to calculate the radius of a black holes event horizon based on an experimentally-verified equation.

Calculations aside, it should be pretty obvious that the orbiting star is outside of the black hole's event horizon because we can see it.
Title: Re: Black hole equations.
Post by: evan_au on 15/03/2018 19:52:56
Quote from: Kryptid
r = √((2GM)/(ve2))
I plugged this equation into a spreadsheet, and got a different answer from what you got.
It looks like the "√" is missing when you do the calculation?
Title: Re: Black hole equations.
Post by: Kryptid on 15/03/2018 20:52:58
I plugged this equation into a spreadsheet, and got a different answer from what you got.
It looks like the "√" is missing when you do the calculation?

Hmm, let me try to do this bit by bit and see what I get.

r = √((2GM)/(ve2))
r = √(((2 x (6.67 x 10-11 m3kg-1s-2) x (5.972 x 1024 kg))/(11,182 meters per second)2)
r = √(((2 x (6.67 x 10-11) x (5.972 x 1024))/(125,037,124))
r = √(((1.334 x 10-10) x (5.972 x 1024)/(125,037,124))
r = √((796,024,480,000,000)/(125,037,124))
r = √(6,366,305)
r = 2,523.15 meters

It looks like you're right. For some reason, I always seem to have trouble when I rearrange equations. Does anyone recognize where I may have made my error?

EDIT: I think the error is that there should have been no square-root function in the equation to begin with. So the correct form is:

r = (2GM)/(ve2)
Title: Re: Black hole equations.
Post by: guest39538 on 15/03/2018 22:41:23

* ve.jpg (32.62 kB . 740x464 - viewed 1992 times)

ve = > 0 mph


It is a shame the air was not denser up top, you would need less  speed to escape.  I have to say I do not think your calculation has anything to do with an event horizon, it seems to me it is the thrust needed to push against the less dense air up top.   More speed is more thrust. 
Title: Re: Black hole equations.
Post by: Kryptid on 15/03/2018 23:08:52
It is a shame the air was not denser up top, you would need less  speed to escape.  I have to say I do not think your calculation has anything to do with an event horizon, it seems to me it is the thrust needed to push against the less dense air up top.   More speed is more thrust. 

Yes, you could theoretically climb a sufficiently high set of stairs to leave the Earth's atmosphere without reaching escape velocity. That's what a hypothetical "space elevator" would do. However, that's not what escape velocity is about. The escape velocity is the minimum speed that an object has to travel at for its momentum alone to allow it to escape an object's gravitational field indefinitely. In other words, it's the minimum speed that you'd have to push something to so that it never comes to a stop and then falls back to the Earth.

Since this concept does apply to light (light relies only on its momentum to move, as photons are not like tiny rockets with thrusters), then light inevitably gets stuck if the escape velocity is too high. That's one reason light can't escape an event horizon. So if we can see something, then we know it isn't behind any such horizon. Hence, the fact that we can see Cygnus X-1 means that the star must be orbiting outside the event horizon of its black hole companion.

The idea of building a "staircase" to climb out of a black hole doesn't work either, but for reasons that are more difficult to understand. Apparently,space is so twisted inside of an event horizon that you are unable to travel in a direction that leads out of it. You must move inevitably towards the singularity. It's like being trapped inside of an ever-shrinking box. Perhaps someone more well-versed in that area of physics can chime in to give a better explanation.
Title: Re: Black hole equations.
Post by: guest39538 on 16/03/2018 02:08:12
Yes, you could theoretically climb a sufficiently high set of stairs to leave the Earth's atmosphere without reaching escape velocity. That's what a hypothetical "space elevator" would do. However, that's not what escape velocity is about. The escape velocity is the minimum speed that an object has to travel at for its momentum alone to allow it to escape an object's gravitational field indefinitely. In other words, it's the minimum speed that you'd have to push something to so that it never comes to a stop and then falls back to the Earth.
I would of though it would have to be a continual force rather than a speed to escape a gravitational field.   

All BH's are finitely dense? 

There is finite mass in the observable Universe?

How can you detect a BH if they do not emit  light?   

How do you know BH's are just not a body that is black in colour?

Title: Re: Black hole equations.
Post by: Kryptid on 16/03/2018 04:54:11
I would of though it would have to be a continual force rather than a speed to escape a gravitational field.

Fortunately, such a thing isn't necessary (otherwise space travel would be much harder). Since gravity is a force that becomes weaker as you move further from its source, less and less energy is required to keep moving as you more further away. The total energy requirements converge on a finite value, which is equal to the maximum possible gravitational potential energy for the object that is trying to escape the gravitational field. If your total kinetic energy is equal to or above this value, then the gravitational field can slow you down over time as you move away from it, but it can never bring your speed all the way to zero.

Quote
All BH's are finitely dense?

This is an as-yet unresolved question. Relativity predicts an infinitely-dense singularity at the center of a black hole. However, most physicists seem to agree that relativity fails to make the correct prediction in such extreme environments. That's because quantum physics becomes important for predicting the behavior of very small objects and we all know about the infamous "feud" between relativity and quantum physics. My personal bet is that black holes collapse to a small, but finite, size.

Quote
There is finite mass in the observable Universe?

Yes, on the order of 1060 kilograms if memory serves correctly.

Quote
How can you detect a BH if they do not emit  light?
 

Black holes are usually detected by their gravitational effects. If a star is seen to orbit an invisible companion, then the mass of that companion can be estimated by measuring the speed and orbital period of the star. If the observed mass is above about 3 solar masses, then we know it's too massive to be a neutron star (neutron degeneracy pressure fails at such extremes). Although there are some theoretical stars, such as electroweak stars, quark stars and preon stars that might be heavier and more compact than even neutrons stars, they will have their own upper limits before they collapse into black holes.

Another, more definitive, way to detect black holes is via gravitational waves. When two black holes merge, they emit gravitational waves in a very particular pattern. Relativity told us exactly what such a signature should look like to a detector. In 2015, the LIGO observatory detected exactly such a gravitational wave pattern. The calculated masses of the two black holes involved in the merger was about 14.2 and 7.5 solar masses: https://en.wikipedia.org/wiki/GW151226 (https://en.wikipedia.org/wiki/GW151226)

Quote
How do you know BH's are just not a body that is black in colour?

That's because such bodies are not stable. They are too massive to support themselves against collapse. We know that the invisible companion to the star in Cygnus X-1 cannot be, say, a 14.8-solar mass ball of black carbon because carbon (or any such form of atomic matter) would not have sufficient pressure to counteract the enormous gravitational forces acting on it. White dwarfs and neutron stars are ruled out because (1) they too would collapse under such extreme masses) and (2) the Universe is not old enough for them to have cooled off sufficiently to be black.

That's not to say that the exact nature of black holes is settled. There are still alternative models to the classical relativistic black hole that are seriously considered by some physicists. Some examples of these are:

- Gravastar: https://en.wikipedia.org/wiki/Gravastar (https://en.wikipedia.org/wiki/Gravastar)
- Dark-energy star: https://en.wikipedia.org/wiki/Dark-energy_star (https://en.wikipedia.org/wiki/Dark-energy_star)
- Fuzzball: https://en.wikipedia.org/wiki/Fuzzball_(string_theory) (https://en.wikipedia.org/wiki/Fuzzball_(string_theory))
- MECOs: https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object (https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object)

Another recently-predicted object that wouldn't be quite a black hole is a star supported by vacuum polarization: https://gizmodo.com/could-the-weirdness-of-quantum-physics-produce-a-new-ki-1823657773 (https://gizmodo.com/could-the-weirdness-of-quantum-physics-produce-a-new-ki-1823657773)
Title: Re: Black hole equations.
Post by: guest39538 on 16/03/2018 10:12:47
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.

Title: Re: Black hole equations.
Post by: guest39538 on 16/03/2018 10:41:08
Could a BH be transparent? 
Title: Re: Black hole equations.
Post by: Kryptid on 16/03/2018 16:25:42
How can something be infinitely dense without being infinitely large?

An infinitely-dense object need not take up any space in order to have a finite mass, so there's no reason to say that it must have a size at all.

Quote
I would of thought absolute dense existed but not infinitely dense which sounds rather strange.

I'm not sure what you mean when you say "absolute dense".

Quote
Could a BH be transparent?

Actually, there is a way in theoretical physics that at least some star-black hole binaries might be explained in such a way. There is a concept called "mirror matter" which behaves in a manner identically to normal matter, except its particle interactions are right-handed instead of left-handed (honestly, I'm not exactly certain what that means. The idea arose from the observation that the weak nuclear force violates parity). As a consequence of having different particle interactions than normal matter, mirror matter could only interact with normal matter through gravitation. As such, mirror matter would be invisible and intangible.

As a hypothetical black hole replacement, imagine a blue main-sequence star composed of mirror matter with 7 times the mass of the Sun and a 2.8 million kilometer radius. Around this mirror matter star orbits a regular star similar to our Sun that orbits at a distance of 5 million kilometers. Since the sun-like star orbits outside of the radius of the mirror matter star, shell theorem tells us that it should orbit it in exactly the same way that it would orbit a black hole of equal mass. With our telescopes, we could deduce that the visible star is orbiting an invisible companion that was much too heavy to be a neutron star. It would be easy to then designate the invisible star as a black hole even though it was not one.

However, there are ways to tell the difference between mirror matter stars and black holes (at least in some cases). One is by examining accretion disks. For Cygnus X-1, the accretion disk around the invisible companion is about 15,000 kilometers in radius. This would be well inside of a hypothetical mirror star of equal mass. Because of this, the accretion disk would be significantly larger and more diffuse than we would expect for a black hole of 14.8 solar masses. It's rather like how you become lighter when you go deep inside of the Earth. Gravity pulls you down less both because there is less mass below you and because there is now some mass above you. So what you would expect of a mirror matter star is that the calculated mass of the "black hole" would be less for the accretion disk than it would be for the orbiting, visible companion.

Another way to tell the difference would be through gravitational lensing. If you were fortunate enough to be in a position where the binary was passing in front of a source of light, you could see how the invisible partner bends that light (at least in theory). A black hole would bend light much more strongly than a mirror matter star of equal mass.
Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 11:29:35
An infinitely-dense object need not take up any space in order to have a finite mass, so there's no reason to say that it must have a size at all.
Ok, I already have an objective argument, saying something is infinitely  dense is illogical.

At some point there is a point where an objects density is at its absolute density.  There can be no such thing as an infinite density.   This is really poor use of the English language and contradicts the very meaning of infinite.

infinite
ˈɪnfɪnɪt/Submit
adjective
1.
limitless or endless in space, extent, or size; impossible to measure or calculate.

noun
1.
a space or quantity that is infinite.


There is a limit to all density, once there is no space left  between any point of the object is the Dmax.  Where D is density.


A volume of space is finite  , if all points of the volume are occupied, you can't get any denser than that.

Dmax =   V(k) - k where k is space and V is volume

Take note, we don't actually take away space, we occupy it with something.


Maybe in your terms


Dmax =  4/3πr (k) -  4/3πr (k)
Title: Re: Black hole equations.
Post by: The Spoon on 17/03/2018 11:40:50
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.


You dont seem to understand the difference between density and mass.
Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 11:42:33
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.


You dont seem to understand the difference between density and mass.
Yes I do , do you?


added- That was a terrible counter argument against what I wrote
Title: Re: Black hole equations.
Post by: The Spoon on 17/03/2018 11:46:21
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.


You dont seem to understand the difference between density and mass.
Yes I do , do you?


Yep, that is why I commented on your post.

Also, your made up equation 'Dmax =   V(k) - k where k is space and V is volume' what on earth is that supposed to mean?
Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 11:48:33
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.


You dont seem to understand the difference between density and mass.
Yes I do , do you?


Yep, that is why I commented on your post.

Also, your made up equation 'Dmax =   V(k) - k where k is space and V is volume' what on earth is that supposed to mean?
Yes I made up , but it is not made up,

D = density

max = maximum

V = volume

k = space


It is not difficult to read or understand.


Try it this way

R = 1m

R - R = 0k =  D max

Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 11:57:00
In simple terms, two points adjoined to a third point make up the smallest volume, maximum density , three dimensional measure.
Quite clearly 0 is finite and absolute.

Quote
not dependent on, conditioned by, or relative to anything else; independent: an absolute term in logic; the absolute value of a quantity in physics


The above quote is what all my work aims for.   This is why to me it is important science understands, I am trying to give absolute answers.
Title: Re: Black hole equations.
Post by: The Spoon on 17/03/2018 12:04:24
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.


You dont seem to understand the difference between density and mass.
Yes I do , do you?


Yep, that is why I commented on your post.

Also, your made up equation 'Dmax =   V(k) - k where k is space and V is volume' what on earth is that supposed to mean?
Yes I made up , but it is not made up,

D = density

max = maximum

V = volume

k = space


It is not difficult to read or understand.


Try it this way

R = 1m

R - R = 0k =  D max


It is not that it is hard to understand. When you say k=space, what exactly do you mean by that. Also, the next equation you made up, what does R stand for, what does m stand for, how do you know that R - R = 0k (by which you mean from the previous 0 space I think. How does this actually relate to density?
Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 12:12:18
Relativity predicts an infinitely-dense singularity at the center of a black hole.

How can something be infinitely dense without being infinitely large?   

I would of thought absolute dense existed but not infinitely dense which sounds rather strange.


You dont seem to understand the difference between density and mass.
Yes I do , do you?


Yep, that is why I commented on your post.

Also, your made up equation 'Dmax =   V(k) - k where k is space and V is volume' what on earth is that supposed to mean?
Yes I made up , but it is not made up,

D = density

max = maximum

V = volume

k = space


It is not difficult to read or understand.


Try it this way

R = 1m

R - R = 0k =  D max


It is not that it is hard to understand. When you say k=space, what exactly do you mean by that. Also, the next equation you made up, what does R stand for, what does m stand for, how do you know that R - R = 0k (by which you mean from the previous 0 space I think. How does this actually relate to density?
By k= space I mean k equals space, k is the space, I am using k instead of the word space. 

R is a real coordinate space k .

How do I know R - R = 0K ?

Because I have took away the space with my equation. 

However, this particular discussion is on about density, it refers to density, what you have to consider though is the volume of a space is not actual took away, it is just occupied and took away that way.  I am filling all points in a volume so there is no free points , logically there is no free space points  inside a full volume.   You can't logically get any denser than a full space.   

Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 12:15:50
p.s I am using the word density correctly

density
ˈdɛnsɪti/Submit
noun
1.
the degree of compactness of a substance.


If you want to discuss density in some weird way , please define your terms of use.
Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 12:24:57
Let us look at the science ''legalese'' version of the definition of density

Quote
Density is the mass per unit volume. This means that the density of any solid, liquid or gas can be found by dividing its mass in kilograms by its volume in cubic metres. Density can be found using the equation: The unit for density is kg m.
Meaningless , let me translate into realism terms

Density is the mass( a word we use that means nothing)  per unit volume.


Kilogram per unit volume, I think not. Mass is force related and nothing to do with density. Kilogram is the result of acting forces.

added- Proof, an electron has no mass to other electrons but it is still has density.  A Proton has no mass to other Protons but it still has density. 

What you are talking about also has a max density and it is a+b which is also absolute logic.   


a+b =V1

(a+b) - (a+b) = V1

The same see....





Title: Re: Black hole equations.
Post by: The Spoon on 17/03/2018 12:54:47
Let us look at the science ''legalese'' version of the definition of density

Quote
Density is the mass per unit volume. This means that the density of any solid, liquid or gas can be found by dividing its mass in kilograms by its volume in cubic metres. Density can be found using the equation: The unit for density is kg m.
Meaningless , let me translate into realism terms

Density is the mass( a word we use that means nothing)  per unit volume.


Kilogram per unit volume, I think not. Mass is force related and nothing to do with density. Kilogram is the result of acting forces.

added- Proof, an electron has no mass to other electrons but it is still has density.  A Proton has no mass to other Protons but it still has density. 

What you are talking about also has a max density and it is a+b which is also absolute logic.   


a+b =V1

(a+b) - (a+b) = V1

The same see....






Ah, I see. You accuse me of wanting to discuss density 'in a weird way' then you decide you don't like how density is defined because it doesnt fit in with your daft ideas so decide to redefine it so that it does. If you had ever actually measured the density of something you would realise how ridiculous your above post is. 
Title: Re: Black hole equations.
Post by: Kryptid on 17/03/2018 16:03:34
Ok, I already have an objective argument, saying something is infinitely  dense is illogical.

At some point there is a point where an objects density is at its absolute density.  There can be no such thing as an infinite density.   This is really poor use of the English language and contradicts the very meaning of infinite.

infinite
ˈɪnfɪnɪt/Submit
adjective
1.
limitless or endless in space, extent, or size; impossible to measure or calculate.

noun
1.
a space or quantity that is infinite.


There is a limit to all density, once there is no space left  between any point of the object is the Dmax.  Where D is density.


A volume of space is finite  , if all points of the volume are occupied, you can't get any denser than that.

Dmax =   V(k) - k where k is space and V is volume

Take note, we don't actually take away space, we occupy it with something.


Maybe in your terms


Dmax =  4/3πr (k) -  4/3πr (k)

Points, by definition, are dimensionless and therefore have no volume. Therefore, you can fit an unlimited number of points inside of any finite volume. At least mathematically. In the real world, you may indeed be correct in asserting that an upper limit to density exists. If so, it is probably around the Planck density, which is 5.155 x 1096 kilograms per cubic meter.

After reexamining the phrase "infinite density", I'm not sure it would even be appropriate to call it that. Density is a measure of mass per unit volume. If you put volume down as zero, then you end up trying to divide by zero. You can't do that in math. The answer is undefined.
Title: Re: Black hole equations.
Post by: guest39538 on 17/03/2018 18:58:09
Another way to tell the difference would be through gravitational lensing. If you were fortunate enough to be in a position where the binary was passing in front of a source of light, you could see how the invisible partner bends that light (at least in theory). A black hole would bend light much more strongly than a mirror matter star of equal mass.
It was gravitational lensiing that made me think about transparency , the electromagnetic radiation surrounding the BH , ''refracting'' off the BH almost turning but not quite into a visible wavelength.
Title: Re: Black hole equations.
Post by: Kryptid on 17/03/2018 21:17:37
It was gravitational lensiing that made me think about transparency , the electromagnetic radiation surrounding the BH , ''refracting'' off the BH almost turning but not quite into a visible wavelength.

Gravitational lensing only changes the trajectory of the radiation. It doesn't change its wavelength.
Title: Re: Black hole equations.
Post by: guest39538 on 18/03/2018 11:25:37
It was gravitational lensiing that made me think about transparency , the electromagnetic radiation surrounding the BH , ''refracting'' off the BH almost turning but not quite into a visible wavelength.

Gravitational lensing only changes the trajectory of the radiation. It doesn't change its wavelength.
How do you see it / detect it , if it does not change wave-length?
Title: Re: Black hole equations.
Post by: Kryptid on 18/03/2018 14:04:24
How do you see it / detect it , if it does not change wave-length?

It creates visual distortions. You can see some stars that are behind the Sun, for example, because the Sun bends their light around it.
Title: Re: Black hole equations.
Post by: guest39538 on 18/03/2018 14:14:06
How do you see it / detect it , if it does not change wave-length?

It creates visual distortions. You can see some stars that are behind the Sun, for example, because the Sun bends their light around it.
Yes visual distortion, a change in the electromagnetic radiation, the invert creating an almost visible wave-length.  It is similar to but not exact like heat haze off a hot road interacting with air to form an almost visible light and visible air. 

If you can see it distorting, the clarity of transparency is becoming translucent , somewhere close to the infra red range but hardly a wave , more of a ''drip''. (maybe)

added- Like water . The water is visible light.


added- The wave-length of water is your answer to see the BH's better

Title: Re: Black hole equations.
Post by: Kryptid on 18/03/2018 14:20:18
creating an almost visible wave-length.

Again, this does not change the wavelength of the light that is being distorted.
Title: Re: Black hole equations.
Post by: guest39538 on 18/03/2018 14:33:46
creating an almost visible wave-length.

Again, this does not change the wavelength of the light that is being distorted.
You think it doesn't ,  that does not mean it doesn't.   Objectively it does, it is like saying water does not affect the wave-length of light.  You can see water because you can see the light reflected by the molecules of the water.   

It is no difference, if a permeating photon encounters a spacial field with a greater permeability, then E=mc and the frequency you will observe is dependent to the permeability of the field.
Title: Re: Black hole equations.
Post by: Kryptid on 18/03/2018 16:42:07
You think it doesn't ,  that does not mean it doesn't.   Objectively it does, it is like saying water does not affect the wave-length of light.

Water doesn't affect light's wavelength.

Quote
You can see water because you can see the light reflected by the molecules of the water.

Yes, but that doesn't mean that the wavelength has changed at all.

Quote
It is no difference, if a permeating photon encounters a spacial field with a greater permeability, then E=mc and the frequency you will observe is dependent to the permeability of the field.

Gravitational lensing can't affect the wavelength of light. That would violate conservation of energy. A change in wavelength indicates a change of energy. There's nowhere for the energy that the light once had to go.
Title: Re: Black hole equations.
Post by: guest39538 on 18/03/2018 17:32:15
Water doesn't affect light's wavelength.
Based on that logic , we should not be able to see water.   Evidently though we can, so with all due respect, science needs to consider how we can see water if there is no visible wave-length to see. 


Tip:  The rain is translucent compared to air.
Title: Re: Black hole equations.
Post by: Colin2B on 18/03/2018 17:36:57
Water doesn't affect light's wavelength.
Based on that logic , we should not be able to see water.   Evidently though we can, so with all due respect, science needs to consider how we can see water if there is no visible wave-length to see. 
@Kryptid didnt say there was no visible wavelength, he said water doesnt affect (change) it.
Title: Re: Black hole equations.
Post by: guest39538 on 18/03/2018 17:40:25
Water doesn't affect light's wavelength.
Based on that logic , we should not be able to see water.   Evidently though we can, so with all due respect, science needs to consider how we can see water if there is no visible wave-length to see. 
@Kryptid didnt say there was no visible wavelength, he said water doesnt affect (change) it.

I see, sorry.   The water affects light compared to air, I can not see air or any sort of light in/off air, the difference being we can see water, so the water must be affected the light to create a visible wave-length, translucent being a sort of visible wave-length?

Title: Re: Black hole equations.
Post by: Colin2B on 18/03/2018 17:58:23
The water must be affected the light to create a visible wave-length, translucent being a sort of visible wave-length?
No, translucent is not a sort of visible wavelength.
Come on, you are a fisherman you must have spent a long time looking at water.
Think about the reflections from the surface, the colour of the water, the mud particles suspended in it, and you have why you can see that it is there.
Title: Re: Black hole equations.
Post by: PmbPhy on 18/03/2018 18:29:37
Quote from: Thebox
Like I said, you can't challenge me on time because I am the ''master'' on time and space and if Einstein was here today he would concede to me.

Dear God,

I thought I heard it all until today when I read the above statement. I have now heard it all. As a result I am more highly disappointed with people than ever before, in general. I am therefore ready to assume room temperature. Please take me when you find the time.

Your friend, the atheist,
Pete
Title: Re: Black hole equations.
Post by: guest39538 on 18/03/2018 18:48:42
The water must be affected the light to create a visible wave-length, translucent being a sort of visible wave-length?
No, translucent is not a sort of visible wavelength.
Come on, you are a fisherman you must have spent a long time looking at water.
Think about the reflections from the surface, the colour of the water, the mud particles suspended in it, and you have why you can see that it is there.
I have thought about muddy muddy water etc, but I have also thought about clear water.
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 00:24:23
I have thought about muddy muddy water etc, but I have also thought about clear water.

You can see clear water because it does not allow all light to pass through it. Some of it is reflected back towards your eyes. Surely you've seen the Sun glint off of a lake before? You can also see it because water deflects the light's path. Look at the picture of the straw in the following link: https://www.popsci.com/why-does-this-straw-look-like-its-broken (https://www.popsci.com/why-does-this-straw-look-like-its-broken). The straw is just as red in the water as it is out of the water. The wavelength (color) of the light has not changed.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 13:54:56
I have thought about muddy muddy water etc, but I have also thought about clear water.

You can see clear water because it does not allow all light to pass through it. Some of it is reflected back towards your eyes. Surely you've seen the Sun glint off of a lake before? You can also see it because water deflects the light's path. Look at the picture of the straw in the following link: https://www.popsci.com/why-does-this-straw-look-like-its-broken (https://www.popsci.com/why-does-this-straw-look-like-its-broken). The straw is just as red in the water as it is out of the water. The wavelength (color) of the light has not changed.
You are looking at water with a vagueness, try being underwater in a swimming pool.   You can see the entire length of water between you and the side of the pool. 


You say the surface is reflecting light so you can see the water, what is the reflected  visible wave-length?
Title: Re: Black hole equations.
Post by: The Spoon on 19/03/2018 14:38:22
I have thought about muddy muddy water etc, but I have also thought about clear water.

You can see clear water because it does not allow all light to pass through it. Some of it is reflected back towards your eyes. Surely you've seen the Sun glint off of a lake before? You can also see it because water deflects the light's path. Look at the picture of the straw in the following link: https://www.popsci.com/why-does-this-straw-look-like-its-broken (https://www.popsci.com/why-does-this-straw-look-like-its-broken). The straw is just as red in the water as it is out of the water. The wavelength (color) of the light has not changed.
You are looking at water with a vagueness, try being underwater in a swimming pool.   You can see the entire length of water between you and the side of the pool. 


You say the surface is reflecting light so you can see the water, what is the reflected  visible wave-length?
Try reading all of his answer.

'You say the surface is reflecting light so you can see the water, what is the reflected  visible wave-length?'
Given that he has stated light is reflected so that you can see the surface, why do you not think it would be visible wavelngth light?
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 18:26:41
You are looking at water with a vagueness, try being underwater in a swimming pool.   You can see the entire length of water between you and the side of the pool.

So what?

Quote
You say the surface is reflecting light so you can see the water, what is the reflected  visible wave-length?

It's whatever wavelength happened to strike the surface of the water. That should be obvious.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 18:43:23
You are looking at water with a vagueness, try being underwater in a swimming pool.   You can see the entire length of water between you and the side of the pool.

So what?

Quote
You say the surface is reflecting light so you can see the water, what is the reflected  visible wave-length?

It's whatever wavelength happened to strike the surface of the water. That should be obvious.
It is not a wave-length between 400-700nm that is for sure, that should be obvious .
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 18:45:14
It is not a wave-length between 400-700nm that is for sure, that should be obvious .

What makes you say that? Visible light from the Sun reflecting off of objects is how we see them.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 18:45:23

Given that he has stated light is reflected so that you can see the surface, why do you not think it would be visible wavelngth light?

 Quite clearly because we can see the water.   
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 18:48:11
It is not a wave-length between 400-700nm that is for sure, that should be obvious .

What makes you say that? Visible light from the Sun reflecting off of objects is how we see them.
We only see with our eyes things that have visible light between 400 - 700 nm,   400 - 700 nm having colour, the water has no colour but we can see it,   so if we can see it , it must be reflecting visible light in a wave-length other than 400-700nm.

Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 18:50:09
It is not a wave-length between 400-700nm that is for sure, that should be obvious .

What makes you say that? Visible light from the Sun reflecting off of objects is how we see them.
We only see with our eyes things that have visible light between 400 - 700 nm,   400 - 700 nm having colour, the water has no colour but we can see it,   so if we can see it , it must be reflecting visible light in a wave-length other than 400-700nm.

That's called white light, which is a mixture of all the colors of light.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 18:59:28
It is not a wave-length between 400-700nm that is for sure, that should be obvious .

What makes you say that? Visible light from the Sun reflecting off of objects is how we see them.
We only see with our eyes things that have visible light between 400 - 700 nm,   400 - 700 nm having colour, the water has no colour but we can see it,   so if we can see it , it must be reflecting visible light in a wave-length other than 400-700nm.

That's called white light, which is a mixture of all the colors of light.


No it is not white light, it is clear , I can see white and it is different than clear.   White light is a mixture of frequencies I would not disagree, but the light  permeating between objects is not white .
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 19:04:46
No it is not white light, it is clear , I can see white and it is different than clear.   White light is a mixture of frequencies I would not disagree, but the light  permeating between objects is not white .

Very well, if you want to be technical about it. Still, when light of all different visible frequencies reaches your eyes, your brain interprets that as white. That's what's happening when sunlight reflects off of water. Water reflects light of all visible frequencies.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 19:09:07
No it is not white light, it is clear , I can see white and it is different than clear.   White light is a mixture of frequencies I would not disagree, but the light  permeating between objects is not white .

Very well, if you want to be technical about it. Still, when light of all different visible frequencies reaches your eyes, your brain interprets that as white. That's what's happening when sunlight reflects off of water. Water reflects light of all visible frequencies.
LOl, no it is does not,   water is not white either.   Let me explain why you can see the raindrop , the surroundings allow you to see the raindrop, the raindrop distorts your line of sight,  you see the sky and clouds through the raindrop. 

At night you can't see the raindrop, because the sky is dark . 

Try to understand that lensing is the raindrop distorting the light.
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 19:14:45
LOl, no it is does not,   water is not white either.   Let me explain why you can see the raindrop , the surroundings allow you to see the raindrop, the raindrop distorts your line of sight,  you see the sky and clouds through the raindrop. 

At night you can't see the raindrop, because the sky is dark . 

Try to understand that lensing is the raindrop distorting the light.

Yes, the light distortion is part of the reason you can see water. Although water is not opaque like white paint, it still does have a finite reflectivity and does reflect at least some portion of the light that strikes its surface. Otherwise, you wouldn't be able to see your reflection in the water.

I guess this thread has stopped being about black holes now?
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 19:19:03
LOl, no it is does not,   water is not white either.   Let me explain why you can see the raindrop , the surroundings allow you to see the raindrop, the raindrop distorts your line of sight,  you see the sky and clouds through the raindrop. 

At night you can't see the raindrop, because the sky is dark . 

Try to understand that lensing is the raindrop distorting the light.

Yes, the light distortion is part of the reason you can see water. Although water is not opaque like white paint, it still does have a finite reflectivity and does reflect at least some portion of the light that strikes its surface. Otherwise, you wouldn't be able to see your reflection in the water.

I guess this thread has stopped being about black holes now?
Not at all, lensing is light related, we have discussed the distortion of light.   

A BH is either

1) opaque and dark in colour

2) transparent and clear in colour ( such as a field).
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 19:21:24
A BH is either

1) opaque and dark in colour

2) transparent and clear in colour ( such as a field).

Light can't pass through a black hole, so option number one is correct.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 19:24:06
A BH is either

1) opaque and dark in colour

2) transparent and clear in colour ( such as a field).

Light can't pass through a black hole, so option number one is correct.
Why can't light pass through a BH?   Surely both options give the same appearance, 

a dark opaque object would absorb light surely? 

Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 19:25:47
Why can't light pass through a BH?   Surely both options give the same appearance, 

a dark opaque object would absorb light surely? 

Its gravity is too strong to allow any light to escape that crosses its event horizon.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 19:29:31
Why can't light pass through a BH?   Surely both options give the same appearance, 

a dark opaque object would absorb light surely? 

Its gravity is too strong to allow any light to escape that crosses its event horizon.
Interesting, but surely if that was the case, BH's would be showing up as hot spots all over the Universe?   

+hf/S=>T   


It can't gain without loss and not gain temperature.
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 19:30:40
Interesting, but surely if that was the case, BH's would be showing up as hot spots all over the Universe?   

+hf/S=>T   

Heat cannot get out of their event horizon either.
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 19:33:39
Interesting, but surely if that was the case, BH's would be showing up as hot spots all over the Universe?   

+hf/S=>T   

Heat cannot get out of their event horizon either.
Then in time they might turn super nova?

Can we look at Bh's in being a dark star?
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 19:41:05
Then in time they might turn super nova?

Explosions can't escape the horizon. The late Stephen Hawking showed that quantum processes at the event horizon, however, can allow a black hole to "evaporate" over time: https://en.wikipedia.org/wiki/Hawking_radiation (https://en.wikipedia.org/wiki/Hawking_radiation)

Quote
Can we look at Bh's in being a dark star?

Interestingly enough, "dark star" is actually one of the earliest phrases used to describe what would eventually become known as a black holes. The assumptions behind a dark star were a little different, however, because only Newtonian mechanics were known at the time. They were assumed to be solid objects and the concept of an event horizon had not yet been discovered. Light would travel up a certain distance from their surface before inevitably being pulled back down again. This would allow you to see a dark star's emitted light if you were close enough to it (in contrast to relativity's black holes, which do not let light out of their horizons at all).
Title: Re: Black hole equations.
Post by: guest39538 on 19/03/2018 20:13:00
Then in time they might turn super nova?

Explosions can't escape the horizon. The late Stephen Hawking showed that quantum processes at the event horizon, however, can allow a black hole to "evaporate" over time: https://en.wikipedia.org/wiki/Hawking_radiation (https://en.wikipedia.org/wiki/Hawking_radiation)

Quote
Can we look at Bh's in being a dark star?

Interestingly enough, "dark star" is actually one of the earliest phrases used to describe what would eventually become known as a black holes. The assumptions behind a dark star were a little different, however, because only Newtonian mechanics were known at the time. They were assumed to be solid objects and the concept of an event horizon had not yet been discovered. Light would travel up a certain distance from their surface before inevitably being pulled back down again. This would allow you to see a dark star's emitted light if you were close enough to it (in contrast to relativity's black holes, which do not let light out of their horizons at all).
Could it be possible that a BH does emit light, but at a frequency we cannot yet detect?
Title: Re: Black hole equations.
Post by: Kryptid on 19/03/2018 22:17:28
Could it be possible that a BH does emit light, but at a frequency we cannot yet detect?

Technically, Hawking radiation would be just that. For most black holes, the radiation emitted is so weak that we wouldn't detect it. However, that's coming from outside of the horizon, not inside of it.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 14:29:15
Could it be possible that a BH does emit light, but at a frequency we cannot yet detect?

Technically, Hawking radiation would be just that. For most black holes, the radiation emitted is so weak that we wouldn't detect it. However, that's coming from outside of the horizon, not inside of it.
Thank you for the education on BH's , I can see the objective existence.


Do BH's have a polarity?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 20:28:43
Thank you for the education on BH's , I can see the objective existence.

You're welcome. I've actually enjoyed this conversation.

Quote
Do BH's have a polarity?

They can indeed have electric charge. It's one of only three properties they are predicted to have: mass, spin and charge. All black holes have mass (for obvious reasons) and almost all black holes should have spin (since they form from the collapse of spinning stars), but most would have little or no electric charge. This is because any charged black hole would attracted oppositely-charged particles which would neutralize the hole.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 20:32:07
Thank you for the education on BH's , I can see the objective existence.

You're welcome. I've actually enjoyed this conversation.

Quote
Do BH's have a polarity?

They can indeed have electric charge. It's one of only three properties they are predicted to have: mass, spin and charge. All black holes have mass (for obvious reasons) and almost all black holes should have spin (since they form from the collapse of spinning stars), but most would have little or no electric charge. This is because any charged black hole would attracted oppositely-charged particles which would neutralize the hole.
I have also enjoyed this conversation thanks.   

In consideration of charge,  if a BH had a greater charge than an orbiting body, would the charge of the orbiting body be attracted to the charge of the BH?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 20:40:56
In consideration of charge,  if a BH had a greater charge than an orbiting body, would the charge of the orbiting body be attracted to the charge of the BH?

That would depend on the strength of the fields in question, the polarizability of the orbiting body, the distances involved, and the net charge on the orbiting body. If the body was made of something highly polarizable, like metal, it's possible that a very strong electric field around a black hole could induce a dipole in the metal body that would increase the net attraction between the two objects. The electric field around the black hole would probably need to be absurdly strong in order have any such significant effect like that.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 20:49:01
In consideration of charge,  if a BH had a greater charge than an orbiting body, would the charge of the orbiting body be attracted to the charge of the BH?

That would depend on the strength of the fields in question, the polarizability of the orbiting body, the distances involved, and the net charge on the orbiting body. If the body was made of something highly polarizable, like metal, it's possible that a very strong electric field around a black hole could induce a dipole in the metal body that would increase the net attraction between the two objects. The electric field around the black hole would probably need to be absurdly strong in order have any such significant effect like that.

Space itself is not known to have  any gravitational affect on an object, the object having no inertia in regards to space.  An electrical field has an infinite radius?

Little force would be needed to move a body that had no inertia?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 20:50:11
Space itself is not known to have  any gravitational affect on an object, the object having no inertia in regards to space.  An electrical field has an infinite radius?

Little force would be needed to move a body that had no inertia?

I'm not sure what you're getting at. Both the black hole and the orbiting body have inertia.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 20:53:24
Space itself is not known to have  any gravitational affect on an object, the object having no inertia in regards to space.  An electrical field has an infinite radius?

Little force would be needed to move a body that had no inertia?

I'm not sure what you're getting at. Both the black hole and the orbiting body have inertia.
relative to what?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 20:56:31
relative to what?

Relative to anything that applies a force to them (which in this case, would be each other via gravitational and electric fields).
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:00:59
relative to what?

Relative to anything that applies a force to them (which in this case, would be each other via gravitational and electric fields).
Actually, do you not think the inertia is relative to velocity?    The orbiting bodies velocity being resistant to change of speed or direction? 

The BH and the orbiting body as a set,  having no inertia when compared to external forces ? 

In other words an external force could move the BH and orbiting body around in space?

If the orbiting body stopped orbiting the BH, the electrical attraction would pull the body into the BH?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 21:06:06
Actually, do you not think the inertia is relative to velocity? The orbiting bodies velocity being resistant to change of speed or direction?

Only in as much as an increase in velocity causes relativistic mass gain.

Quote
The BH and the orbiting body as a set,  having no inertia when compared to external forces ?

Of course they would. Anything that tries to act on them would experience resistance.

Quote
In other words an external force could move the BH and orbiting body around in space?

Yes.

Quote
If the orbiting body stopped orbiting the BH, the electrical attraction would pull the body into the BH?

The gravity alone would be enough to pull it in, but the electric attraction would help.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:17:26

The gravity alone would be enough to pull it in, but the electric attraction would help.


Would the electrical attraction be enough to pull it in , if the gravity did not exist?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 21:18:38
Would the electrical attraction be enough to pull it in , if the gravity did not exist?

Without gravity, the black hole couldn't even exist.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:21:43
Would the electrical attraction be enough to pull it in , if the gravity did not exist?

Without gravity, the black hole couldn't even exist.

Why couldn't it ?   

An electron is attracted to a Proton by polarity, there does not need to be gravity at all?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 21:25:39
An electron is attracted to a Proton by polarity, there does not need to be gravity at all?

The attraction between a proton and electron is far from sufficient to form a black hole. The whole reason that black holes form from massive, dying stars is because gravity can only be attractive and thus builds up to higher and higher levels as you add more mass. The force eventually overwhelms any mechanism that could oppose the collapse. Electric charges don't work that way.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:29:47
An electron is attracted to a Proton by polarity, there does not need to be gravity at all?

The attraction between a proton and electron is far from sufficient to form a black hole. The whole reason that black holes form from massive, dying stars is because gravity can only be attractive and thus builds up to higher and higher levels as you add more mass. The force eventually overwhelms any mechanism that could oppose the collapse. Electric charges don't work that way.
I must be missing some point of science because in my schematic gravity is not needed or exists?

 [ Invalid Attachment ]





Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 21:32:44
I must be missing some point of science because in my schematic gravity is not needed or exists?


* schematic.jpg (14.83 kB . 740x464 - viewed 1455 times)

Yeah. One thing you're missing is that objects don't have to contain any electric charge in order to have mass.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:34:47
I must be missing some point of science because in my schematic gravity is not needed or exists?


* schematic.jpg (14.83 kB . 740x464 - viewed 1455 times)

Yeah. One thing you're missing is that objects don't have to contain any electric charge in order to have mass.
The point is though , objects do contain the properties of electric charge but it is neutralised by the combining of the two individual mono-poles.   This does not mean the forces of charge stop acting, why would it?
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:36:17
Why would a negative or positive charge be attracted to a neutral object if the neutral object had no acting charges?
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 21:39:42
The point is though , objects do contain the properties of electric charge but it is neutralised by the combining of the two individual mono-poles.   This does not mean the forces of charge stop acting, why would it?

Not all objects contain smaller electric charges. Electrically-neutral black holes don't. Z-bosons don't. Photons don't. Neutrinos don't. Yet all are attracted to gravitational fields.

Quote
Why would a negative or positive charge be attracted to a neutral object if the neutral object had no acting charges?

It wouldn't be unless the neutral object could be polarized (and not all objects contain electric charge, so not all objects can be polarized).
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:45:19
The point is though , objects do contain the properties of electric charge but it is neutralised by the combining of the two individual mono-poles.   This does not mean the forces of charge stop acting, why would it?

Not all objects contain smaller electric charges. Electrically-neutral black holes don't. Z-bosons don't. Photons don't. Neutrinos don't. Yet all are attracted to gravitational fields.

Quote
Why would a negative or positive charge be attracted to a neutral object if the neutral object had no acting charges?

It wouldn't be unless the neutral object could be polarized (and not all objects contain electric charge, so not all objects can be polarized).
Anything that is neutral could be plus and neg ,  that means nothing. Consider lightning, it suck up the neg or pos energy of the ground.  it is attracted to the neg or pos of the ground.

 [ Invalid Attachment ]



Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 21:52:44
Anything that is neutral could be plus and neg

I'm afraid not. Objects which contain both positive and negative charges have magnetic moments and are capable of participating in electromagnetic interactions. The neutron, for example, is neutral but contains internal electric charges which give it a magnetic moment. Such is not true for z-bosons, neutrinos or photons. All of those particles are either without any magnetic moment at all or have one that is too small to measure. Neutral black holes should not have magnetic moments either, as they are not composed of anything more fundamental than themselves. Neutrinos are also known not to interact with the electromagnetic force at all, further showing that they do not contain any electric fields.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 21:55:55
Anything that is neutral could be plus and neg

I'm afraid not. Objects which contain both positive and negative charges have magnetic moments and are capable of participating in electromagnetic interactions. The neutron, for example, is neutral but contains internal electric charges which give it a magnetic moment. Such is not true for z-bosons, neutrinos or photons. All of those particles are either without any magnetic moment at all or have one that is too small to measure. Neutral black holes should not have magnetic moments either, as they are not composed of anything more fundamental than themselves. Neutrinos are also known not to interact with the electromagnetic force at all, further showing that they do not contain any electric fields.
What if our equipment is just simply not advanced enough to measure the electrical properties of any sub atomic particle?

The affects so minute . they are almost negligible.
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 22:01:41
What if our equipment is just simply not advanced enough to measure the electrical properties of any sub atomic particle?

The affects so minute . they are almost negligible.

That has its own consequences. If you propose that gravity is actually the result of electromagnetic attraction, then more total electric charges result in stronger gravitational fields. This, in turn, means that objects that contain more total charge should be more massive than those with less total charge (since mass is seen to correlate with gravity). If this was true, then particles that are more massive than the proton must necessarily contain more electric charges than it does and in turn must have a higher magnetic moment than the proton (more electric charge results in stronger magnetic fields).

The z-boson violates this as it is almost 100 times heavier than the proton. If its very large mass was caused by a large amount of internal charge (as your model posits), then it must also have a magnetic moment much higher than that of the proton. Yet the opposite is true. The proton has a measurable magnetic moment while the z-boson does not.
Title: Re: Black hole equations.
Post by: guest39538 on 20/03/2018 22:24:06
What if our equipment is just simply not advanced enough to measure the electrical properties of any sub atomic particle?

The affects so minute . they are almost negligible.

That has its own consequences. If you propose that gravity is actually the result of electromagnetic attraction, then more total electric charges result in stronger gravitational fields. This, in turn, means that objects that contain more total charge should be more massive than those with less total charge (since mass is seen to correlate with gravity). If this was true, then particles that are more massive than the proton must necessarily contain more electric charges than it does and in turn must have a higher magnetic moment than the proton (more electric charge results in stronger magnetic fields).

The z-boson violates this as it is almost 100 times heavier than the proton. If its very large mass was caused by a large amount of internal charge (as your model posits), then it must also have a magnetic moment much higher than that of the proton. Yet the opposite is true. The proton has a measurable magnetic moment while the z-boson does not.
The Z bosons are made up of W bosons?

F
Title: Re: Black hole equations.
Post by: Kryptid on 20/03/2018 22:50:41
The Z bosons are made up of W bosons?

F

That doesn't avoid the problem, since W bosons have electric charge equal in magnitude to that of the proton. Therefore, a pair of W bosons should still have a high magnetic moment. Secondly, a pair of W bosons would weigh more than a Z boson (W bosons weigh 80.4 GeV while the Z boson weighs 91.2 GeV).

EDIT: I've actually made a mistake. After doing some further research, it turns out that particles with internal electric charges can actually have a magnetic moment of zero. Alpha particles are an example of such a particle.

As I've said before, I urge you to calculate how strong the attractive force between two objects would be using your proposed model of gravity and see if it matches the actual force of gravity. You can start with something very simple like a pair of hydrogen atoms that are separated by some given distance (say, one nanometer). Use Coulomb's law to calculate the force acting between each electric charge and see what it all adds up to. Here is a link that can help you with your calculations: https://socratic.org/questions/what-is-the-coulomb-force-between-two-electrons-that-are-on-opposite-sides-of-th (https://socratic.org/questions/what-is-the-coulomb-force-between-two-electrons-that-are-on-opposite-sides-of-th)
Title: Re: Black hole equations.
Post by: guest39538 on 22/03/2018 08:49:07
The Z bosons are made up of W bosons?

F

That doesn't avoid the problem, since W bosons have electric charge equal in magnitude to that of the proton. Therefore, a pair of W bosons should still have a high magnetic moment. Secondly, a pair of W bosons would weigh more than a Z boson (W bosons weigh 80.4 GeV while the Z boson weighs 91.2 GeV).

EDIT: I've actually made a mistake. After doing some further research, it turns out that particles with internal electric charges can actually have a magnetic moment of zero. Alpha particles are an example of such a particle.

As I've said before, I urge you to calculate how strong the attractive force between two objects would be using your proposed model of gravity and see if it matches the actual force of gravity. You can start with something very simple like a pair of hydrogen atoms that are separated by some given distance (say, one nanometer). Use Coulomb's law to calculate the force acting between each electric charge and see what it all adds up to. Here is a link that can help you with your calculations: https://socratic.org/questions/what-is-the-coulomb-force-between-two-electrons-that-are-on-opposite-sides-of-th (https://socratic.org/questions/what-is-the-coulomb-force-between-two-electrons-that-are-on-opposite-sides-of-th)
I will try to work that out, in the meantime


q1 + q2 = m

How do we measure the force of gravity again?
Title: Re: Black hole equations.
Post by: Kryptid on 22/03/2018 16:00:16
q1 + q2 = m

How do we measure the force of gravity again?

F = G((M1M2)/r2)

Where

F is the force of gravitational attraction (in newtons)
G is the gravitational constant (6.67410−11 m3⋅kg−1⋅s−2)
M1 is the mass of the first object (in kilograms)
M2 is the mass of the second object (in kilograms)
r is the distance between them (in meters)
Title: Re: Black hole equations.
Post by: guest39538 on 22/03/2018 16:17:17
G is the gravitational constant (6.67410−11 m3⋅kg−1⋅s−2)
What does a gravitational constant mean? 

I don't understand  this part.

F=(6.67410−11 m3⋅kg−1⋅s−2)(m1m2)/r


Am I reading that correctly?

Or are you saying G = m1m2/r
Title: Re: Black hole equations.
Post by: guest39538 on 22/03/2018 16:24:09
My brain is overloading this is not computing,


kilogram * kilogram divided by radius squared is just not computing in my brain as being associative to gravity.  Kilogram is the consequence of G, it sounds messy to me. Scratches head.
Title: Re: Black hole equations.
Post by: guest39538 on 22/03/2018 16:31:27
I get


GF = (F1 + F2)  + (F1 + F2)


((F1 + F2)  + (F1 + F2)) > (F1 + F2)

My calculation is calculating from a point perspective, your calculation is the force over a radius?





Title: Re: Black hole equations.
Post by: Kryptid on 22/03/2018 21:18:23
What does a gravitational constant mean?

It's a measure of how much gravity is produced by a given amount of mass.

Quote
I don't understand  this part.

F=(6.67410−11 m3⋅kg−1⋅s−2)(m1m2)/r

Am I reading that correctly?

You don't need to completely understand the units being used here just so long as you can do the calculation. The result will be in newtons anyway, which is easy to understand.

Quote
Or are you saying G = m1m2/r

Nope, the first version is the correct one. G is a constant.

My brain is overloading this is not computing,

kilogram * kilogram divided by radius squared is just not computing in my brain as being associative to gravity.  Kilogram is the consequence of G, it sounds messy to me. Scratches head.

The strength of gravity felt at a given location is determined by both mass and distance, hence the kilogram and meters.

At any rate, I already did the calculation for the amount of gravitational force between two hydrogen atoms separated by 1 nanometer. The result was 1.869 x 10-46 newtons.

I get


GF = (F1 + F2)  + (F1 + F2)


((F1 + F2)  + (F1 + F2)) > (F1 + F2)

My calculation is calculating from a point perspective, your calculation is the force over a radius?

Yes, the force over a given distance is what I'm looking for. How much attractive force does your N-field produce between two hydrogen atoms separated by 1 nanometer? That's the calculation you need to do.
Title: Re: Black hole equations.
Post by: guest39538 on 22/03/2018 21:45:48
What does a gravitational constant mean?

It's a measure of how much gravity is produced by a given amount of mass.

Quote
I don't understand  this part.

F=(6.67410−11 m3⋅kg−1⋅s−2)(m1m2)/r

Am I reading that correctly?

You don't need to completely understand the units being used here just so long as you can do the calculation. The result will be in newtons anyway, which is easy to understand.

Quote
Or are you saying G = m1m2/r

Nope, the first version is the correct one. G is a constant.

My brain is overloading this is not computing,

kilogram * kilogram divided by radius squared is just not computing in my brain as being associative to gravity.  Kilogram is the consequence of G, it sounds messy to me. Scratches head.

The strength of gravity felt at a given location is determined by both mass and distance, hence the kilogram and meters.

At any rate, I already did the calculation for the amount of gravitational force between two hydrogen atoms separated by 1 nanometer. The result was 1.869 x 10-46 newtons.

I get


GF = (F1 + F2)  + (F1 + F2)


((F1 + F2)  + (F1 + F2)) > (F1 + F2)

My calculation is calculating from a point perspective, your calculation is the force over a radius?

Yes, the force over a given distance is what I'm looking for. How much attractive force does your N-field produce between two hydrogen atoms separated by 1 nanometer? That's the calculation you need to do.
Shrugs shoulders and looks lost :D

(8.9875109 N m C−2)^2 + (8.9875109 N m C−2)^2

I have no idea what I am doing with this math , but the above is what I need i think, cant make the  -  ,  go high either or the 9
Title: Re: Black hole equations.
Post by: Kryptid on 22/03/2018 22:15:51
Shrugs shoulders and looks lost :D

(8.9875109 N m C−2)^2 + (8.9875109 N m C−2)^2

I have no idea what I am doing with this math , but the above is what I need i think, cant make the  -  ,  go high either or the 9

You do superscripts with [ sup][ /sup] and subscripts with [ sub][ /sub]. Just don't put the spaces in there like I did in this particular post.

When I get back, I'll post a step-by-step of how I would do the calculations.
Title: Re: Black hole equations.
Post by: guest39538 on 22/03/2018 22:22:13
Shrugs shoulders and looks lost :D

(8.9875109 N m C−2)^2 + (8.9875109 N m C−2)^2

I have no idea what I am doing with this math , but the above is what I need i think, cant make the  -  ,  go high either or the 9

You do superscripts with [ sup][ /sup] and subscripts with [ sub][ /sub]. Just don't put the spaces in there like I did in this particular post.

When I get back, I'll post a step-by-step of how I would do the calculations.
Ok thank you and here is the way I do step by step calculations.


* m1.jpg (19.27 kB . 740x464 - viewed 1390 times)

I am not sure what this means yet, but I get 0.25 as a result for m1 by doing 0.5

1/4 = g ? 




Title: Re: Black hole equations.
Post by: Kryptid on 23/03/2018 02:37:35
Ok thank you and here is the way I do step by step calculations.


* m1.jpg (19.27 kB . 740x464 - viewed 1390 times)

I am not sure what this means yet, but I get 0.25 as a result for m1 by doing 0.5

1/4 = g ? 

So what does that translate to in actual units?

Here are a couple of calculations. The first one is for the force of gravitational attraction between two hydrogen atoms 1 nanometer apart:

F = G((M1M2)/r2)
F = (6.674 x 10−11 m3⋅kg−1⋅s−2)((1.674 x 10-27 kg)(1.674 x 10-27 kg)/(10-9)2)
F = (6.674 x 10−11)((2.802 x 10-54)/(10-9)2)
F = (6.674 x 10−11)((2.802 x 10-54)/(10-18))
F = (6.674 x 10-11)(2.802 x 10-36)
F = 1.87 x 10-46 newtons

This second one is the force of electrical attraction between a proton and an electron separated by a distance of 1 Bohr radius (approximately the radius that an electron is most likely to be found at in a hydrogen atom):

F = k((Q1Q2)/r2)
F = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(1.602 x 10-19 C)/(5.291 x 10-11 m)2)
F = (8.9875 x 109)((2.566 x 10-38)/(5.291 x 10-11)2)
F = (8.9875 x 109)((2.566 x 10-38)/(2.799 x 10-21))
F = (8.9875 x 109)(9.1676 x 10-18)
F = 8.239 x 10-8 newtons
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 12:00:33
So what does that translate to in actual units?
Honestly I have no idea at this time, I think I need to work out what ''your'' numbers mean firstly.   Before I have took Pi , recalculated Pi to use / work out m/s  . I can't remember now I did that but it is on the internet somewhere in a thread.

I am not sure , perhaps I will have to invent my own values.   

Sorry I could  not be more helpful, my formulas work in my mind but I do struggle to give the formulas an end value.


Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.


Added- Eureka,

G= (F=ma)

An object at relative rest is still accelerating ''downwards'', 


G = (q1+q2)a

That might be squared or times 2.

G= (q1+q2)a

or

G= (q1+q2)a


r= 0 so I need no radius in my calculation.

added-

F=g(ma)


added- M1 is only half the mass of the measure.

1kg = 0.5kg


added- invention

Double sided/faced set of scales .

I claim intellectual rights and copy rights to my double sided/faced scales.  Under common law I said it first, so anyone nicking my idea will be sued under this formal disclaimer and oral patent.
My intellectual rights are protected by this thread and forum, anyone using my idea will be using my intellectual rights so therefore cannot copy my idea.

S.P.Leese

23/03/2018





 
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 12:29:36
 [ Invalid Attachment ]
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 12:38:44
Typo error , re-done

 [ Invalid Attachment ]



Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 12:43:44
 [ Invalid Attachment ]
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 12:47:21
A 1 kg object in free fall


F=1/2(m)a2

So at relative rest a 1 kg object is applying 4.905N

From a 1 m fall , F = 9.81N
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 12:58:30
 [ Invalid Attachment ]
Title: Re: Black hole equations.
Post by: Kryptid on 23/03/2018 16:26:22
Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 16:53:20
Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.
I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. 


Coulombs constant , might as well be Mandarin , I can't read it.
Title: Re: Black hole equations.
Post by: Kryptid on 23/03/2018 19:28:20
Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.
I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. 


Coulombs constant , might as well be Mandarin , I can't read it.


Then I'll try to do it for you. I'll just need some information first. How far apart are the proton and electron in your model of the hydrogen atom? Is it in agreement with the current model (i.e. approximately the Bohr radius) or is it some different value?
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 19:51:44
Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.
I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. 


Coulombs constant , might as well be Mandarin , I can't read it.


Then I'll try to do it for you. I'll just need some information first. How far apart are the proton and electron in your model of the hydrogen atom? Is it in agreement with the current model (i.e. approximately the Bohr radius) or is it some different value?
r=0

In my model the Proton and Electron are merged and have a void in the center which is a dense field.   Sort of like a maltesers.
Title: Re: Black hole equations.
Post by: Kryptid on 23/03/2018 22:20:17
In my model the Proton and Electron are merged and have a void in the center which is a dense field.   Sort of like a maltesers.

...okay then.

I can already tell you that the attractive force will be exactly zero in that case. If the proton in Atom 1 is the exact same distance from the proton in Atom 2 as it is from the electron in Atom 2, then the attractive and repulsive forces will be exactly equal to each other and cancel out. If you insist on it, I can do the calculation anyway if you don't believe me. It really shouldn't be necessary though...
Title: Re: Black hole equations.
Post by: guest39538 on 23/03/2018 23:16:07
In my model the Proton and Electron are merged and have a void in the center which is a dense field.   Sort of like a maltesers.

...okay then.

I can already tell you that the attractive force will be exactly zero in that case. If the proton in Atom 1 is the exact same distance from the proton in Atom 2 as it is from the electron in Atom 2, then the attractive and repulsive forces will be exactly equal to each other and cancel out. If you insist on it, I can do the calculation anyway if you don't believe me. It really shouldn't be necessary though...
No, the attractive force is not 0, the electron and the proton while merged retain their attractive force to retain form, if there was 0 attractive force they would not sustain being merged. 

When r=0 between two atoms , the attractive force remains between electrons and protons of each atom, the fields form bonds.

Quote
then the attractive and repulsive forces will be exactly equal to each other and cancel out

They do , an atom cannot pass through an atom because it is equally repulsive as attractive.  They can only be attracted to r=0 .  They may be able to ''squash'' other atoms out of form.

Title: Re: Black hole equations.
Post by: Kryptid on 23/03/2018 23:40:17
No, the attractive force is not 0, the electron and the proton while merged retain their attractive force to retain form, if there was 0 attractive force they would not sustain being merged.

I'm talking about zero attractive force between the two different atoms, not between the proton and the electron in the same atom. I'm also talking about net attraction, not gross attractions. Sure, the proton in Atom 1 will be attracted to the electron in Atom 2, but it is also being repelled by the proton in Atom 2. Since protons have electric charge equally strong to that of electrons, and since you have defined that protons and electrons are at the same place in the same atom, then the attraction and repulsion between the atoms have to be equal to each other. They are pulling on each other just as strongly as they are pushing away from each other. The net force has to be zero.

Quote
When r=0 between two atoms , the attractive force remains between electrons and protons of each atom, the fields form bonds.

The repulsive forces remain between the protons in different atoms and between the electrons in different atoms as well.

Quote
They do , an atom cannot pass through an atom because it is equally repulsive as attractive.

Then you agree that two stationary hydrogen atoms separated by a distance of 1 nanometer would not move each other. The only way they could move towards each other would be if the attraction between the two atoms is greater than the repulsion between them. That can't happen if the forces of attraction and repulsion are equal.
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 01:12:25
I'm talking about zero attractive force between the two different atoms, not between the proton and the electron in the same atom. I'm also talking about net attraction, not gross attractions. Sure, the proton in Atom 1 will be attracted to the electron in Atom 2, but it is also being repelled by the proton in Atom 2. Since protons have electric charge equally strong to that of electrons, and since you have defined that protons and electrons are at the same place in the same atom, then the attraction and repulsion between the atoms have to be equal to each other. They are pulling on each other just as strongly as they are pushing away from each other. The net force has to be zero.
I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.
Title: Re: Black hole equations.
Post by: Kryptid on 24/03/2018 02:49:15
I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.

So how do the attractive forces between the two atoms become stronger than the repulsive forces between the same two atoms? The repulsive force between two protons should be exactly equal to the attractive force between a proton and an electron (at the same distance apart).

Let me do the equations. I'll consider a pair of hydrogen atoms one nanometer apart. I'll label the proton in the first atom P1, the proton in the second atom P2, the electron in the first atom as E1 and the electron in the second atom as E2. Between these two atoms, the possible interactions are P1P2 (proton-proton repulsion), P1E2 (proton-electron attraction), P2E1 (proton-electron attraction) and E1E2 (electron-electron repulsion). Take note that P1E1 and P2E2 are not considered because they are interactions within the atoms, not between the atoms. Now let's calculate the force involved in each interaction. Positive numbers indicate repulsion, negative numbers indicate attraction:

FP1P2 = k((Q1Q2)/r2)
FP1P2 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(1.602 x 10-19 C)/(10-9 m)2)
FP1P2 = (8.9875 x 109)((2.566 x 10-38)/(10-9)2)
FP1P2 = (8.9875 x 109)((2.566 x 10-38)/10-18)
FP1P2 = (8.9875 x 109)(2.566 x 10-20)
FP1P2 = 2.306 x 10-10 newtons

FP1E2 = k((Q1Q2)/r2)
FP1E2 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)
FP1E2 = (8.9875 x 109)((-2.566 x 10-38)/(10-9)2)
FP1E2 = (8.9875 x 109)((-2.566 x 10-38)/10-18)
FP1E2 = (8.9875 x 109)(-2.566 x 10-20)
FP1E2 = -2.306 x 10-10 newtons

FP2E1 = k((Q1Q2)/r2)
FP2E1 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)
FP2E1 = (8.9875 x 109)((-2.566 x 10-38)/(10-9)2)
FP2E1 = (8.9875 x 109)((-2.566 x 10-38)/10-18)
FP2E1 = (8.9875 x 109)(-2.566 x 10-20)
FP2E1 = -2.306 x 10-10 newtons

FE1E2 = k((Q1Q2)/r2)
FE1E2 = (8.9875 x 109 N m2 C−2)((-1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)
FE1E2 = (8.9875 x 109)((2.566 x 10-38)/(10-9)2)
FE1E2 = (8.9875 x 109)((2.566 x 10-38)/10-18)
FE1E2 = (8.9875 x 109)(2.566 x 10-20)
FE1E2 = 2.306 x 10-10 newtons

Now let's add up all of the forces and see what the net attraction or repulsion is:

Fnet = FP1P2 + FP1E2 + FP2E1 + FE1E2
Fnet = (2.306 x 10-10 newtons) + (-2.306 x 10-10 newtons) + (-2.306 x 10-10 newtons) + (2.306 x 10-10 newtons)
Fnet = 0 newtons

There's the math for you. There is no net electrostatic force between two hydrogen atoms when they have the structure that you propose that they have.
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 14:37:06
I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.

So how do the attractive forces between the two atoms become stronger than the repulsive forces between the same two atoms? The repulsive force between two protons should be exactly equal to the attractive force between a proton and an electron (at the same distance apart).

Let me do the equations. I'll consider a pair of hydrogen atoms one nanometer apart. I'll label the proton in the first atom P1, the proton in the second atom P2, the electron in the first atom as E1 and the electron in the second atom as E2. Between these two atoms, the possible interactions are P1P2 (proton-proton repulsion), P1E2 (proton-electron attraction), P2E1 (proton-electron attraction) and E1E2 (electron-electron repulsion). Take note that P1E1 and P2E2 are not considered because they are interactions within the atoms, not between the atoms. Now let's calculate the force involved in each interaction. Positive numbers indicate repulsion, negative numbers indicate attraction:

FP1P2 = k((Q1Q2)/r2)
FP1P2 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(1.602 x 10-19 C)/(10-9 m)2)
FP1P2 = (8.9875 x 109)((2.566 x 10-38)/(10-9)2)
FP1P2 = (8.9875 x 109)((2.566 x 10-38)/10-18)
FP1P2 = (8.9875 x 109)(2.566 x 10-20)
FP1P2 = 2.306 x 10-10 newtons

FP1E2 = k((Q1Q2)/r2)
FP1E2 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)
FP1E2 = (8.9875 x 109)((-2.566 x 10-38)/(10-9)2)
FP1E2 = (8.9875 x 109)((-2.566 x 10-38)/10-18)
FP1E2 = (8.9875 x 109)(-2.566 x 10-20)
FP1E2 = -2.306 x 10-10 newtons

FP2E1 = k((Q1Q2)/r2)
FP2E1 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)
FP2E1 = (8.9875 x 109)((-2.566 x 10-38)/(10-9)2)
FP2E1 = (8.9875 x 109)((-2.566 x 10-38)/10-18)
FP2E1 = (8.9875 x 109)(-2.566 x 10-20)
FP2E1 = -2.306 x 10-10 newtons

FE1E2 = k((Q1Q2)/r2)
FE1E2 = (8.9875 x 109 N m2 C−2)((-1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)
FE1E2 = (8.9875 x 109)((2.566 x 10-38)/(10-9)2)
FE1E2 = (8.9875 x 109)((2.566 x 10-38)/10-18)
FE1E2 = (8.9875 x 109)(2.566 x 10-20)
FE1E2 = 2.306 x 10-10 newtons

Now let's add up all of the forces and see what the net attraction or repulsion is:

Fnet = FP1P2 + FP1E2 + FP2E1 + FE1E2
Fnet = (2.306 x 10-10 newtons) + (-2.306 x 10-10 newtons) + (-2.306 x 10-10 newtons) + (2.306 x 10-10 newtons)
Fnet = 0 newtons

There's the math for you. There is no net electrostatic force between two hydrogen atoms when they have the structure that you propose that they have.
That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.

Imagine weighing two identical mass on pan scales to measure 0. 


* force scales.1.jpg (15.89 kB . 731x461 - viewed 1288 times)


The Universe works on a critical balanced system.
Title: Re: Black hole equations.
Post by: Kryptid on 24/03/2018 14:43:00
That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.

The net force absolutely is zero. If there is more force on the attractive side than the repulsive side, please explain where it comes from. The numbers certainly don't show it.

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Imagine weighing two identical mass on pan scales to measure 0.

That's a great analogy, actually. Even though both masses are experiencing a force pulling them down towards the Earth, they are keeping each other from tipping the scales in either direction. Their forces are equal so that the scale doesn't move. Same thing with two hydrogen atoms. The attractions and repulsions are equal, so there is no movement.
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 14:54:36
That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.

The net force absolutely is zero. If there is more force on the attractive side than the repulsive side, please explain where it comes from. The numbers certainly don't show it.

Quote
Imagine weighing two identical mass on pan scales to measure 0.

That's a great analogy, actually. Even though both masses are experiencing a force pulling them down towards the Earth, they are keeping each other from tipping the scales in either direction. Their forces are equal so that the scale doesn't move. Same thing with two hydrogen atoms. The attractions and repulsions are equal, so there is no movement.
The attractive force and repulsive force is equal of the atoms, if the field density of the atoms were greater the atom would not displace through the field to adjoin the other atom. The fields repulsion is not as great as the atoms attraction, the atoms repulsion solidity means 0 radius but solidity.

Yes you understood the analogy , hopefully now can see why I think gravity be related to this. 

I think you measure the net charge as zero, but you measure the net force of the charge as gravity.  I did say ages ago I thought science already had the gravity mechanism answer.

Added- Hence N-field particle and n-field.
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 15:20:28
A simple experiment to show something

Equipment

Two people

One pen


Method

Person one - On their left hand draw a plus sign

On their right hand draw a negative sign

Lock their thumbs together so hands are joined


Person Two - On their left hand draw a negative sign

On their right hand draw a positive sign

Lock their thumbs together joining hands


Person one and two

Lock your hands with your fingers to the other persons hands,


You both are now observing bonding.


I give permission for schools to use this to show children how bonding works , an easy way to explain molecular construction.


You left hand is attracted to your right hand and vice versus.  Your left hand is attracted to their right hand and your right hand is attracted to their left hand.  So on......
Title: Re: Black hole equations.
Post by: Kryptid on 24/03/2018 19:53:24
The fields repulsion is not as great as the atoms attraction

This is the part that you have not demonstrated. The equations show quite clearly that the attractions and repulsions are equal. Where are you getting this extra attraction from? It certainly isn't from electrostatic attraction.
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 20:38:04
The fields repulsion is not as great as the atoms attraction

This is the part that you have not demonstrated. The equations show quite clearly that the attractions and repulsions are equal. Where are you getting this extra attraction from? It certainly isn't from electrostatic attraction.
You are not considering this correctly , consider two rubber balls and squeeze them together, that is the repulsion because of field density N-field particle.   


Move them a length apart , the spacial field between them is not dense enough to stop them being drawn together by force to a 0 radius.

Now consider magnetic suspension, the field is dense enough to stop the object falling.   

added - i.e Quantum field physicality.  ( Q.F.P)

Title: Re: Black hole equations.
Post by: Kryptid on 24/03/2018 21:44:04
You are not considering this correctly

So you're saying that Coulomb's law is wrong? That's quite a bold assertion.

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consider two rubber balls and squeeze them together, that is the repulsion because of field density N-field particle.

Atoms aren't rubber balls, but I don't really care about repulsion at this time. I'm trying to understand why you think two forces acting in equal and opposite directions somehow adds up to make net attraction...

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Move them a length apart , the spacial field between them is not dense enough to stop them being drawn together by force to a 0 radius.

What force? It isn't electrostatic. I've already shown mathematically via Coulomb's law that it can't be.
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 22:13:56
So you're saying that Coulomb's law is wrong? That's quite a bold assertion.
No, you read that wrong

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What force? It isn't electrostatic. I've already shown mathematically via Coulomb's law that it can't be.

It is , no net charge does not mean no net force.  You must be reading wrongly.

Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 22:20:32
 [ Invalid Attachment ]
Title: Re: Black hole equations.
Post by: guest39538 on 24/03/2018 22:36:37
0.5 = 0.25 = 1/4
*2+*2 =*4
0.5/4=0.0681

No idea at this time what this means.
Title: Re: Black hole equations.
Post by: Kryptid on 25/03/2018 02:57:32
We've been through this song and dance before. I'm not doing it again. I'm out.
Title: Re: Black hole equations.
Post by: guest39538 on 25/03/2018 05:48:45
We've been through this song and dance before. I'm not doing it again. I'm out.
Thank you for trying to understand , I thought you had at one point.  I guess I will have to wait until somebody else comes along .

Title: Re: Black hole equations.
Post by: guest39538 on 25/03/2018 12:34:47
0.5 = 0.25 = 1/4
*2+*2 =*4
0.5/4=0.0681

No idea at this time what this means.

Quote
The objective of a heuristic is to produce a solution in a reasonable time frame that is good enough for solving the problem at hand. This solution may not be the best of all the solutions to this problem, or it may simply approximate the exact solution. But it is still valuable because finding it does not require a prohibitively long time.

1010       0.000     0.0681

https://books.google.co.uk/books?id=gwUwIEPqk30C&pg=PA332&lpg=PA332&dq=0.0681&source=bl&ots=GLlZIrS9i6&sig=y0EKDDgS7DdHanDDCjNQy-ZVAYg&hl=en&sa=X&ved=0ahUKEwjHjvyyrofaAhXHbVAKHdHHBfMQ6AEIPTAF#v=onepage&q=0.0681&f=false