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Take a hollow sphere and conduct a process on the inside, which transfers mass into kinetic energy (e.g. we let a big nuclear bomb detonate or something like that). For simplicity, assume that this will happen in a spherically symmetric way. How will the gravitational field look at the outside? In particular, how do the "new" components of the Stress-Energy tensor affect the field?

My (very, very vague) physical intuition says, that in the metric tensor, calculated from the Einstein tensor, the additional contributions from the Stress-Energy tensor will somehow compensate for the loss in mass. However, far out, in the asymptotically flat case, the gravitational field will get weaker due to Newton's Theorem. But how does this relate to Birkhoff's theorem? The weak field limit is obviously not static. The problem is somehow, that the $T^{\mu\nu}$ is in this example not mass-dominated, therefore most of the stuff one generaly does will not work. So, what happens here?

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    $\begingroup$ I'm pretty sure the external asymptotic metric is static; the hypothetical symmetric nuclear explosion does not exhibit a quadrupole moment and therefore has no gravitational field emission. Also it just rearranges the 4D vectors of particles and switches around some of them, keeping the total mass-energy in the volume constant, and I guess the Newtonian limit assumes a re-calculation of the source region in terms of Newtonian mass. As an extreme non-Newtonian case, consider a star's energy worth of pure photons, gravitating. $\endgroup$ – BjornW Jun 11 '15 at 0:00
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Imagine two Schwarzschild solutions one with parameter $M$ the other with parameter $m.$ so you have the metric $$\left(1-\frac{2GM}{c^2r}\right)c^2dt^2-\left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$ and the metric $$\left(1-\frac{2Gm}{c^2r}\right)c^2dt^2-\left(1-\frac{2Gm}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right).$$

For each of them the terms $M$ and $m$ are parameters, parameters that tell you that the curvature far away makes masses go in circles just like Newtonian gravity predicts a mass of mass $M$ or $m$ would. Neither is related to the mass of any things inside. And each solution describes a static metric, which in this case is a way that curvature can vary from place to place in such a way as to cause future curvature to be exactly like it is now. So empty curve can curve itself in this way (amongst many).

OK, now the coordinate $r$ takes a symmetric spherical surface and takes the circumference divided by $2\pi$ (or the surface area divided by $4\pi$ then take the square root, same thing in this case for this metric). If $M>m$ then you could cut out the surface of say $r=2\frac{2GM}{c^2}$ from both solutions and then remove the inner part $r<2\frac{2GM}{c^2}$ from the solution $$\left(1-\frac{2GM}{c^2r}\right)c^2dt^2-\left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$ and remove the outer part $r>2\frac{2GM}{c^2}$ from the solution $$\left(1-\frac{2Gm}{c^2r}\right)c^2dt^2-\left(1-\frac{2Gm}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right).$$

If you do this you can sew them together and you get a new solution to Einstein's Equation but now it is not a vacuum solution, now you need some stress-energy at that surface. And since $M>m$ you don't need exotic matter. An ordinary hard shell of positive mass matter would do. If you shell has some thickness then your two metrics would have to continuously transition over the thickness of the shell.

So it isn't masses at the center that are the sources of spherically symmetric solutions, it is spherical shells of stress energy. When someone says that a spherical solution has a matter source at the center they mean that in a shell region without stress-energy the solution is $$\left(1-\frac{2Gm}{c^2r}\right)c^2dt^2-\left(1-\frac{2Gm}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$ for some $m$ and that solution looks just like the solution of having mass at the center. Its a subtle but important difference. In a shell region of vacuum the solution looks like the solution where all the mass is at the center, but only in that shell. If you move to another shell the solution could look different if there is stress-enrgy there because stress-energy makes spacetime curve differently than it does/can in vacuum or even if there is no stress-energy there at the new shell it might curve like $$\left(1-\frac{2Gm}{c^2r}\right)c^2dt^2-\left(1-\frac{2Gm}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$ for some different value of $m$ if there was stress-energy on any shell region between the two shells.

So really the source is spherical shells of stress-energy, not masses at the center. If there is a hard shell around the center, the center might be empty. If you allowed exotic matter on your shells you might have a wormhole that spherically symmetrically open into a different universe and so there never even was a center, not now and not ever. So just because you have a spherically symmetric shell doesn't even mean there is a center let alone that all the matter has to be there. It's like in Newtonian gravity, the field outside is just like the field of a point mass but really the source is a series of spherical shells of matter. Being just like another field produced by X is different than there being X since different things can produce the same field somewhere else.

OK, so what kinds of stress-energy are allowed? Well, we have to have spherical symmetry. So hard shells are OK, they can have mass and hence energy for that, they can have stress and pressure but only that which respects the spherical symmetry. You can have gases or dust, but again it has to respect the spherical symmetry, so for instance a pressure gradient in the $r$ direction is fine, but not in any other direction.

Now we can answer your question. Notice that outside everything in the region of vacuum, the spacetime is curved in a way so ad to cause its future self. It is curved in a way that fits the vacuum perfectly, it can't and won't change in the entire outside unless and until something out there changes.

Hmm, it won't change until it changes. But it will be forced to change if stress energy gets out there. So if your original sphere expands and gets larger the spacetime outside the new larger sphere won't ever have changed, but the part that is now inside will change. And if there is a part that had matter rush through it but is now a vacuum it will look like $$\left(1-\frac{2Gm}{c^2r}\right)c^2dt^2-\left(1-\frac{2Gm}{c^2r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$ for a smaller value of $m$ than it had before (if ordinary matter rushed through, if exotic matter rushed through it will have a larger value of $m$ now).

That's it. It's what Birkhoff actually predicts, that the metric in a vacuum in a spherical shell is determined by the stress-energy on all the spherical shells inside the current shell.

So the things outside all the matter never changed and never change. Now if your explosion inside produced radiation and any of it escapes and travels radially outwards then you have two problems, firstly you can't do that spherically symmetrically and secondly even having radiation in a region means you aren't a vacuum region so when the radiation passes through a shell the mass parameter value at that shell will go down.

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According to Birkhoff's theorem, any spherically symmetric gravitational field should be produced by a mass at its center. Conversion of mass to kinetic energy inside a spherical shell should not affect the gravitational field outside the shell. Even if the interior of the shell becomes an intense hot group of photons of energy M*c^2, they will gravitate the same as a mass E/c^2.

Since your interior process is conducted "in a spherically symmetric way", the Schwarzschild metric describes the gravitational field as long as the electric charge is zero and the angular momentum is zero. Even if the hollow sphere expanded, the gravitational field would not change - there would be no gravitational waves.

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