# Collapsing and expanding shell theorem

## The idea is

1 vote(s)
50.0%

1 vote(s)
50.0%
3. ### Disagrees with important existing theories

0 vote(s)
0.0%
1. Jun 15, 2010

### frodeborli

I've been philosophing about Newtons shell theorem, which tell us that inside a symmetrical sphere, gravity is zero if the sphere is hollow, and linearly declining while moving closer to the center if the sphere is massive and of constant density.

Now, if the sphere was hollow it would naturally collapse unless it was excedingly hard, since the sphere itself is affected by the gravity of every other particle that makes up the hollow sphere.

So expanding the shell theorem, we should see inverted gravity inside the sphere if the sphere is collapsing. Every particle inside the shell will experience negative gravity toward the center of the sphere.

Vice versa; if the sphere is growing, every particle inside it will experience gravity toward the center.

Do anybody here agree with me? If not, please assume anyway, that I am correct in the following reasoning:

Would you agree that if a galaxy were surrounded by a collapsing sphere - the outer stars orbiting the galaxy would rotate slower? They would have "help" in keeping their distance from the central black hole.

If two galaxies were inside the collapsing sphere, would they not be pulled apart from eachother?

Could not any less dense area of space, surrounded by a semi-uniform highly dense collapsing area experience similar effects?

I've been thinking about this for years, and I've not got the mathematical skills to prove this - but I'm certain somebody could help me with some of the math involved.

However, I believe that a configuration of this "sphere" surrounding the observed universe could be found and make Newtonian physics agree more with General Relativity on large distances.

If the surrounding sphere is collapsing, it would mean that at one point we will reach somewhat of a singularity, before the momentum of the collapse fly by and create another expansion period in which all mass left behind will be compressed into the center (because of the expanding shell).

All of this is because we need to incorporate the propagation delay of gravity into the shell theorem.

Now, I am aware that by writing down so many implications of this idea, a lot of people will be very sceptical to it.

Best regards,
Frode Børli, Norway.

2. Jun 15, 2010

### Ich

No. According to Birkhoff's theorem, the gravity inside a shell is zero, no matter whether it's expanding or collapsing. The result holds in GR, too.

3. Jun 15, 2010

### frodeborli

If the shell is 14 billion light years across, and you are 1 billion lightyears away (on the inside) from one side - you will be affected by the increase in gravity from "your side" of the shell before the increase in gravity from the other side reach you. The increase in gravity comes from your side moving towards you.

I do not see how Birkhoff's theorem disagrees with that, unless gravity is instantaneous. Can you please elaborate?

4. Jun 16, 2010

### Chalnoth

It's not quite so simple. This problem is closely related to a problem posed by one Van Flandern, who noted that the speed of gravity in Newtonian Gravity is infinite, and if one naively extrapolates to GR, where the "speed of gravity" is the speed of light, one ends up with predictions that don't match reality.

However, it turns out that you cannot arrive at GR simply by taking Newtonian gravity and tacking on a finite speed of propagation. Here's a rather detailed analysis of why this is wrong, and how GR matches the Newtonian gravity in the appropriate limit, even though Newtonian gravity has infinite speed:
http://arxiv.org/abs/gr-qc/9909087

One way to think about it is that the gravitational field of the object moves along with the object in question, so that you can feel an object's gravitational effects long before it is "within range", so to speak.

5. Jun 16, 2010

### frodeborli

I've read that document now and I'm thinking that it is only correct for an external observer:

An imaginary universe built up by two objects. One object is defined (by us) as being stationary (A). The other object (B) can be freely moved around by us. When we move B closer to A, B will immidiately and without delay detect the gravity from A. A, however will have to wait for the gravity change from B to propagate troughout the universe as it accelerate.

Vice versa changing the defined stationary object to B, A would move toward B and detect A would immediately detect gravity from B. Measurements done from A on B would show that this was correct - also vice versa.

Measurements done from an independant observer would show that both instantaneously affect eachother, but measurements performed on either of the objects would not.

Gravity is a propagating curvature of time-space - propagating at the speed of light, but its effect is instantaneous once propagated. An object moving into an already curved space will instantaneously be affected by it - no matter how far it is away from the source.

If I'm not correct in this logic, then the only other conclusion I can think of is that gravitational waves is not possible unless matter is created or destroyed. Binary stars would not create gravity waves.

My first post in this thread would require that the sphere around the universe was being created or destroyed somehow - for example by antimatter meeting matter where I thought a massive collapsing spherical construct would be.

6. Jun 16, 2010

### Chalnoth

The main point to be made here is that the gravitational anomaly cancellation makes it so that you need a time-dependent force to produce any gravitational radiation. So if you have two test particles far away, for instance, and then start moving one, your deliberate motion of the one object will constitute a time-dependent force, whose effects will travel outward at the speed of light. However, if you just leave the particles alone, for the most part they will simply fall towards one another with very little gravitational radiation, meaning that their respective positions will be felt "instantaneously" everywhere (it's only when the two particles get very very close that the radiation becomes appreciable, because it is only then that the force of attraction between them changes significantly with time).

7. Jun 16, 2010

### frodeborli

If the sum of mass was large enough (surrounding the entire universe), could it not be measurable on a scale of for example 10 million lightyears, the threshold where Newtonian physics start to become inaccurate? Are science certain that it only becomes appreciable when extremely close, no matter how much mass you have? I could, for example say that we have a hundread billion times the combined mass of the entire visible universe surrounding us and collapsing?

But anyway your arguments lead me to another line of thought; assuming a strong flow of increasing gravity must be attributed to mass being created (or to some other time dependant force). We believe the universe was created 14 billion years away from us in time/space. When the universe was created, mass was created. From our point of view, our view of the world should appear to expand due to mass being created 14 billion years away. If matter was created, gravity observed by us should increase in a continous wave until the creation stops - and we would perhaps percieve this as uniform redshift, simply as a result of energy turning into mass.

Further thinking leads to the possibility that ALL observed distance between objects is merely the cause of gravity increasing forever in a continous explosion/creation. Time itself could be just an effect of this creation of matter/increase in gravity from afar.

I'm being very creative (sic) right now, and I need to think more about this. :-)

8. Jun 16, 2010

### Chalnoth

Well, the formulas for the strength of gravitational radiation are relatively straightforward, so you should be able to look them up and use them yourself to see what the effects would be. Of course, they usually use very different geometries than you're talking about here, but if you want to get an extremely rough idea, just plug the numbers in for tremendously huge masses at those large distances and see what comes out.

In the real world, gravitational radiation is usually only significant for objects near neutron stars and black holes.

Well, sort of. Gravitational radiation is an expected product of inflation. I'm not sure if your line of reasoning can be adapted to the one that leads to gravitational radiation from inflation, but in any case we're looking for it now. One of the big thrusts in CMB science today is to detect the signature of these primordial gravitational waves in the cosmic microwave background. There's a small chance that the Planck satellite will be able to detect these waves, but it is more likely we'll have to wait for the results from ground-based and balloon-borne experiments currently being developed.

I'd like to point out, however, that if mass can be said to have been "created" at these early times, then it was created everywhere at about the same time. Furthermore, everything we see today would have been much, much closer than 14 billion light years. If you go all the way back to the earlier part of inflation, for instance, everything we see today would have fit within the size of a proton (perhaps more correctly, all of the matter and other stuff we see today came from a tiny region of inflating but essentially empty space-time smaller than the size of a proton).

This doesn't make any sense to me.

9. Jun 16, 2010

### frodeborli

Which relates to:

A property of us being located at a spot where mass is being created, I'm thinking, is that everything appears to move away from us, because gravity is increasing. The *distance* is not increasing - we are all still at the same spot - if observed by an external observer, but the increasing gravitational effect is making distances appear to increase for us, inside this spot.

I meant that:

1. The big bang is "still happening", relative to us - it is just moving away from us at the speed of light, unless we move in any of the other three dimensions.
2. The expansion of the universe itself does not constitute movement in any dimensions, it is only stretching the dimensions.
3. When we move in any of the three (non-time) dimensions, our outlook on the entire universe becomes more affected by the mass created at the big bang, thus increasing redshift.
4. Any object created at the big bang will, as they move in dimensions other than time, see increasing red shift. If the object only moves in the time dimension, no change in redshift will occur *for it*.
5. The big bang is not a single spot that everything can be backtraced to; it is, as you say, everywhere - but it seems to us that we could backtrack it to a spot because of our time slowing when we move. It would seem so to any object in the universe - but no single object affected by the big bang could ever successfully backtrace the universal expansion.

All of this feels a little far fetched to me, just so that we are clear on that. :-) I need a couple of months of further thinking before I feel confident enough about it to continue arguing for it.

10. Jun 16, 2010

### frodeborli

Back to the topic, regardless of my previous post. Measured from an object within a collapsing sphere, I maintain that you would experience gravity away from the center. Only for an external observer gravity would appear to be instant within the system. This because of the propagation delay of changes in gravity affecting the observer; changes in gravity from the nearest edge of the sphere would reach the observer before gravity from the far edge. An external observer would see the change as instantaneous.

Or does this conflict with observations?

11. Jun 16, 2010

### Chalnoth

Well, again, the anomaly cancellation from a sphere collapsing under its own gravity would ensure that observers inside wouldn't feel anything.

12. Jun 29, 2010

### frodeborli

I've been thinking further on this:

Two objects at a distance have a gravitational field. When one object (A) is considered stationary, and another object (B) moves relative to it, the first object (A) will notice the change in the gravitational field of the other object faster than the speed of light. Only changes in gravity propagates at the speed of light. The curvature of space is already there and it does not propagate.

According to this, all objects inside a collapsing sphere would not notice that the sphere was collapsing - since the gravity of all surrounding particles cancel each out.

But when a sphere collapse, all particles making up the sphere increase in relativistic mass according to special relativity. The increased gravity resulting from the increase in mass propagate at the speed of light.

An object closer to one side of the sphere will notice an increase in gravity from its side of the sphere long before an increase in gravity from the other side of the sphere.

The increase in relativistic mass when the sphere either implode or explode should result in a gravity pull away from the center for all particles inside the sphere.

Please do not say that "anomaly cancellation" also cancels out the effect of relativistic mass increase - unless you explain why, and how it has been observed - or where I am wrong in my logic. :-)

13. Jun 29, 2010

### Chalnoth

I sincerely doubt it. The way I think about it is as follows.

One way to think of it is that gravitational radiation is a way of carrying away information about changes in the gravitational field. If you have a system that moves from an old configuration to a new configuration, there may be some gravitational waves emitted as the system relaxes to the new gravitational field configuration.

Now, if we look at the inside of the sphere, what is the change in the gravitational field? Well, if we consider two hypothetical scenarios where we have static solid shells of matter of equal mass but different radii, the gravitational field inside the spheres is the same: zero. So there is no change in the gravitational field to worry about, and thus if you look at the details of any gravitational waves emitted, they are likely to all cancel.

Outside the sphere, though, the space-time is transitioning from having no gravitational field to obeying a Schwarzschild metric. With this change, there may be some gravitational radiation directed outwards. But I doubt there would be any directed inwards. I'm sure you'd have to delve into the details of GR to be sure, though.

Anyway, with that aside, I'd like to nitpick something. The terminology "relativistic mass" isn't used in modern relativity. It led to too many errors. Today, mass is simply considered to be a property of the matter in question, and independent of velocity. Instead we redefine the energy and momentum to produce a consistent theory. In special relativity, we define:

$$E^2 = p^2c^2 + m^2c^4$$
$$\frac{v}{c} = \frac{pc}{E}$$

Here $E$ is the energy of the particle, $p$ is its momentum, $m$ is its mass, $v$ is its velocity, and $c$ is the speed of light.

In General Relativity you will generally find no mention of any sort of velocity or coordinate-dependent mass term. Instead you have a stress-energy tensor which is in terms of energy density, momentum density, pressure, and stress.

14. Jun 30, 2010

### Ich

You start with Newtonian gravity, add a finite "speed of gravity", take relativisitc mass as the source of gravity, and believe that you have general relativity. This assumption is wrong. Whatever you derive, no matter how stringently, from this assumption is wrong, too. No need to explain why this and that is different in reality.
If you want a correct derivation, you'll have to look up the proof of Birkhof's theorem in a textbook:
"A smooth sphericallly symmetric metric solution of the vacuum Einstein equations is necessarily static." (emphasis mine)
By virtue of said theorem, there is no such (=monopole) radiation.

15. Jun 30, 2010

### Chalnoth

Ah, okay, makes sense. I guess I was thinking incorrectly before: the outside field is the same as well, independent of the radius of the shell. So there wouldn't be any relaxation either inside or outside. Of course, that's just an intuitive argument, but I guess it makes me feel good that this is explicitly what Birkhoff's theorem says.

16. Jul 1, 2010

### frodeborli

Remove every notion of gravity, and only consider curvature of space. This curvature is constant around an object of constant mass. When an object moves into this curvature, it is immediately affected by it and vice versa - there is no notion of "speed" for the curvature of space. It is already there.

When two objects move relative to each other, only their mutual distance change. The effect of changing distance is instant for both objects - the speed of gravity does not apply.

This must be true if Birkhoffs theorem is true, unless I am missing something key here? This has nothing to do with Newtonian gravity, but incidentally this agrees with Newtonian gravity, since Newtonian gravity propagates instantaneously.

But Einstein did introduce relativistic mass, which cause the curvature of space to increase in strength when an object move - as observed by an observer when their relative distance increase or decline. Later the effect was attributed to momentum/kinetic energy because Einstein said it was not good to introduce relativistic mass, but the actual effect is the same?

So, the increased curvature between two objects moving relative to each other propagates at the speed of light, although existing curvature is already there. From that I must conclude that the only way to introduce gravity waves is by objects moving relative to each other, or by mass or energy being created or destroyed.

If you observe/measure a smooth spherical metric solution, then it is no longer symmetric - that is why it must be static when we calculate on it, isn't it? Or that sentence could mean that unless it is static (settled) at the time you calculate on its effects, your calculations will be wrong?

17. Jul 1, 2010

### Chalnoth

Well, if you were to also add in an observer, you're changing the system, and it would be the observer who is producing the change, not the collapsing shell of matter. If you just want to consider a spherically-symmetric collapsing shell of matter, Birkhoff's theorem states that the vacuum solution (i.e. the solution outside of the matter) must be a static (no radiation) solution.

18. Jul 1, 2010

### frodeborli

Yes, so outside the shell I'm speaking of there is nothing. There is no concept of "outside" the universe. But this has nothing to do with what I'm talking about here. The shell might be considered semi symmetric, since I'm talking of a large amount of mass surrounding the visible universe, symmetric or not. However, I believe that the entire universe must be symmetric, since no outside force have ever affected it. A collapsing sphere must thus be symmetric, although not necessarily smooth.

To clarify my point here: say the universe consist of only three particles: a, b and c. We have only one dimension plus time in this imaginary universe. A and c is positioned at position 0 and 100. B is at position 90. Gravity have propagated everywhere. Then we animate the system, allowing gravity to work. Naturally the three particles will collide. Particle b will initially accelerate outward towards c, since c is closer to b.

Imagine a forth particle d at position 30. It will accelerate toward a, a little slower than particle b. The relative distance between b and d will increase, displaying a red shift for each other.

Add another dimension and place particles in a circle around b and d. The effect is still the same, although lessened because there is more mass on the far side of the circle to counter the pull on b from c and on d from a.

Add a third dimension and add particles distributed like a ball, and the effect on b and d is zeroed. But if we animate the system, b and d well initially be unaffected by a and c falling toward each other. As a and c gain energy, space time will curve more. The change in curvature propagate at the speed of light. At one time this increase in curvature around a will reach d, then later it will reach b until finally it reach c.

If c is observing d, at the moment the gravity wave from a reach c, d will immediately begin seeing red shift. The red shift will be lessened due to c coming under effect from the gravity wave.

Since A continue to gain energy, the effect observed by d and c continue. Also the effect is doubled since the same thing is happening when C gain energy. The effect is multiplied since the particles neighboring the particles A and C (the ball) gain energy.

19. Jul 1, 2010

### Chalnoth

I'm sorry, I meant "outside" in the sense of "everywhere except where there is matter". In other words, the interior of the shell as well as the exterior. It doesn't matter.