# Problem with Birkhoffs theorem on the inside of a sphere collapsing by its own weight

## Main Question or Discussion Point

Imagine an empty universe, where nothing exist and time stands still. Then add lots of stars of equal size, distributed in a symmetry around a spot that we call the center of our universe. Since time has not passed, no curvature (gravity) has propagated to affect any of the other stars. No outside force affect this universe, not even you as an observer.

Inside this hollow sphere of evenly distributed stars, we add five stars. One at the Center (C), and for simplicity we call the locations of the other four stars North (N), South (S), East (E) and West (W).

Remember that no time is passing yet, as we are still building our imaginary universe. No curvature of space have propagated from each of the stars, so nothing affects any of our stars yet.

Then we start time, simultaneously, for all stars in the system. What should happen is:

1. Light and gravity should propagate away from each of the stars. After some time, light from each of the stars should reach its closest neighbor. At the same time as light, curvature of space-time have propagated.

Each of the stars are now only affected by their closest neighbors gravity/curvature of space-time. This should trigger a slow collapse of our sphere. As more stars start affecting each other, the collapse should accelerate.

2. At a certain time, gravity from stars making up the sphere in the north region should reach our (N)-star. At the same time, the same happens for the (S), (E) and (W)-stars. The sphere should collapse slightly faster in these directions, but our four stars should begin accelerating toward the sphere-stars.

Now, all our five central stars are moving away from each other, while all stars making up the sphere is moving toward each other.

3. At one time, all stars in our universe will have extended their gravity field so that it affects all other stars in our universe. For the (N)-star, the gravity pull from all stars in the (N)-region equals the gravity pull from all stars in the (E), (S) and (W) region combined. This is Newtons theorem in essence. The gravity cancels out.

For a short time, in step 2 above, all stars inside the universe was moving away from each other. Once curvature of space-time from each star have propagated troughout the universe, all stars are affected by it equally in all directions - making the curvature of space-time flat. There will be no gravitational effect inside the sphere - except gravity from the five stars inside the sphere on each other.

Once the curvature of space-time have propagated, all gravity is instant. When a star moves, the direction of the gravity is immediately reflected for all other observers. Infinitely faster than light. The logic behind this is : once space is curved, everything is affected by the curvature. If one star is defined as stationary, and another star is moving toward it - both stars are affected by the change in curvature - because we can at any time redefine which of the two stars are stationary. This is relativity.

Birkhoffs theorem and Newtons theorem say that inside a hollow symmetric sphere, gravity is zero - and the above logic supports that. But I still have a problem with it.

All moving objects have momentum/energy. Energy contribute to curvature. All stars in the sphere have equal speed - so the effect should be the same - zero gravity. Birkhoffs theorem is still true. BUT:

As all stars in the sphere accelerate, so their momentum/energy increase. Increasing momentum/energy propagate at the speed of light. My intuition tell me that the increase in momentum should affect the closest stars inside the sphere first, and as long as the collapse of the sphere is accelerating - all stars inside should experience an outbound gravitational effect. Visually, this should lead to a redshift effect when watching other stars inside the sphere.

Where am I wrong in this last conclusion, that a sphere that is collapsing by its own gravity will NOT have zero gravity inside - contrary to Birkhoffs theorem.

Newtons theorem does not apply, as gravity propagates instantaneously by Newtons theories.

Chalnoth

The velocity terms in the Einstein Field Equations are extremely non-trivial, so you can't use such intuitive arguments with certainty. Ultimately, you have to go to the math, and the math says that you can't have a spherically-symmetric matter distribution that produces gravitational waves.

The velocity terms in the Einstein Field Equations are extremely non-trivial, so you can't use such intuitive arguments with certainty. Ultimately, you have to go to the math, and the math says that you can't have a spherically-symmetric matter distribution that produces gravitational waves.
This is intuitively correct for an exterior observer of a massive object where you can view all particles in this object as a single particle at its mass center. I am, however, unsure if the velocity of the individual particles is part of the math proving this theorem when a sphere is collapsing. If the sphere is rotating, it does not work - since particles are moving. If the sphere is collapsing, the particles are also moving.

The teorem does not work for a rotating sphere. Who said that it works for inside a collapsing sphere?

Anyway, momentum/energy have direction. Being outside of the sphere is not the same as being inside the sphere. Inside the sphere, all particles move toward you. Outside the sphere, your closest particles move away from you, while the farthest particles move toward you. If the effect is canceled out exterior to the sphere, they can't ALSO be canceled out interior to the sphere WHILE the particles move.

[Edited some incorrect logic :-)]

Last edited:
Chalnoth

The teorem does not work for a rotating sphere; who said that it works for inside a collapsing sphere?
A rotating sphere isn't spherically symmetric. It is axis symmetric.

Anyway, momentum/energy have direction. Being outside of the sphere is not the same as being inside the sphere. Inside the sphere, the closest particle move toward you, while the farthest particles move away from you. Outside the sphere, your closest particles move away from you, while the farthest particles move toward you. If the effect is canceled out exterior to the sphere, they can't ALSO intuitively be canceled out interior to the sphere.
Why not?

A rotating sphere isn't spherically symmetric. It is axis symmetric.
That is a question of definition. If you orbit an object that is "not spinning", you could theoretically notice the same rotational frame dragging - although you will have trouble orbiting it fast enough for any noticeable effect unless the object was very massive.

Frode said:
-snip-
Why not?
Because of both changes in momentum and linear frame dragging having direction and propagation delay. An object moving away from you will not produce the same effect as an object moving toward you.

If objects A, B and C were placed on an imaginary line in space, and A and B were massively heavier than C, objects A and B would fall toward each other. B would move away from C until A and B collide.

If A and C were the massive objects, and B was relatively mass less - no objects would move away from B. Both would move toward it.

In both cases, the massive objects gain momentum as they fall. This increase in momentum causes an increase in curvature of space-time, and this increase in curvature propagate at the speed of light. In the first example, the increase in curvature from A and C will reach B at the same time.

In the second example if the distance between the objects were 1 light-year, the increase in curvature from B will reach C one year before the increase in curvature from A reach C.

Also, the linear frame dragging effect will have different directions inside and outside of the sphere.

I just do not understand how a very distant object can cancel out the effect of a very close object, when no changes on the distant object can reach us faster than light.

A sphere and an observer:

( ) .

As each particle making up the hollow sphere "( )" gain momentum during its collapse, I do not understand how the increase in momentum of ")" can be cancelled out by the increase in momentum of "(" when observed from the ".". Any change in physical properties from ")" must reach the observer "." faster than changes from "(".

Imagine an empty universe, where nothing exist and time stands still. Then add lots of stars of equal size, distributed in a symmetry around a spot that we call the center of our universe. Since time has not passed, no curvature (gravity) has propagated to affect any of the other stars.
Care must be taken when building universes – even imaginary ones. If no time has passed the big bang would not have occurred and there would be no distances between any stars. If there was distance between stars, time would have passed and during that time the stars would have been interacting gravitationally since the onset of gravitation. Starting with a clean slate and instantaneously plopping stars into it that have never interacted with each other is not possible.

Care must be taken when building universes – even imaginary ones. If no time has passed the big bang would not have occurred and there would be no distances between any stars. If there was distance between stars, time would have passed and during that time the stars would have been interacting gravitationally since the onset of gravitation. Starting with a clean slate and instantaneously plopping stars into it that have never interacted with each other is not possible.
Of course not. But we could perhaps use gigantic space ships that were the size of stars instead. So no stars, and instead we have space ships. And we move them stationary relative to each other in the exact same configuration that I was describing. We shut down the engines of these super space ships now, allowing them to act (as stars), and succumb to gravity. Wether time have passed should not matter.

But then the space ships would be interacting during their transition into position. My point is that gravitational intercations can't be stopped and started - neither between stars nor space ships nor whatever. The interactions began at the beginning and continue to the present.

But then the space ships would be interacting during their transition into position. My point is that gravitational intercations can't be stopped and started - neither between stars nor space ships nor whatever. The interactions began at the beginning and continue to the present.
All mass was brought into place from infinity.

Imagine an empty universe, where nothing exist and time stands still. Then add lots of stars of equal size, distributed in a symmetry around a spot that we call the center of our universe. Since time has not passed, no curvature (gravity) has propagated to affect any of the other stars. No outside force affect this universe, not even you as an observer.
And right here is the misunderstanding you seem to have about GR.

The short answer is that your scenario is simply not admissible in GR. When the EM tensor has a certain "value" in a localized area that implies that the curvature "around it" is already in place. One does not exists without the other.

Of course curvature changes in non static spacetimes but we cannot assume a model where spacetime is flat containing massive objects that have not as of yet, so to speak, propagated their curvature in spacetime.