Birkhoff's Theorem: Exterior Solution to Einstein's Field Equations

[maxverywell]

Birkhoff's theorem says that a spherically symmetric vacuum solution to Einstein's field equations in exterior region is necessarily static.

Now, in wikipedia article it says:

Another interesting consequence of Birkhoff's theorem is that for a spherically symmetric thin shell, the interior solution must be given by the Minkowski metric; in other words, the gravitational field must vanish inside a spherically symmetric shell.

http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity)

How is that a consequence of Birkhoff's theorem which speaks only about the exterior region?

 

[WannabeNewton]

You have to consider a thin spherically symmetric shell configuration wherein the source matter is a dirac delta distribution, or at the least confined to a region $R_1 < r < R_2$. Then the region exterior to $R_2$ will of course have to be Schwarzschild as a consequence of Birkhoff's theorem but note that the region interior to $R_1$ is also spherically symmetric vacuum by hypothesis. Hence Birkhoff's theorem should apply in the interior as well implying that the interior metric is static.

Now the general form of the metric in a static spherically symmetric space-time can be written in appropriate coordinates as $ds^{2} = -A(r)dt^{2} + B(r)dr^2 + r^{2}d\Omega^{2}$. One can then use the vacuum Einstein equations to deduce that $B(r) = A(r)^{-1}, A(r) = 1 + \frac{C}{r}$. Now in the interior case $A(r)$ must be well behaved as $r\rightarrow 0$ because the metric has to be non-degenerate, so it must be the case that $C = 0$ otherwise $A(r)$ will necessarily blow up at $r =0$ and this will make the metric degenerate at $r =0$ but by definition the metric must be non-degenerate everywhere.

 

[TrickyDicky]

You have to consider a thin spherically symmetric shell configuration wherein the source matter is a dirac delta distribution, or at the least confined to a region $R_1 < r < R_2$. Then the region exterior to $R_2$ will of course have to be Schwarzschild as a consequence of Birkhoff's theorem but note that the region interior to $R_1$ is also spherically symmetric vacuum by hypothesis. Hence Birkhoff's theorem should apply in the interior as well implying that the interior metric is static.

Now the general form of the metric in a static spherically symmetric space-time can be written in appropriate coordinates as $ds^{2} = -A(r)dt^{2} + B(r)dr^2 + r^{2}d\Omega^{2}$. One can then use the vacuum Einstein equations to deduce that $B(r) = A(r)^{-1}, A(r) = 1 + \frac{C}{r}$. Now in the interior case $A(r)$ must be well behaved as $r\rightarrow 0$ because the metric has to be non-degenerate, so it must be the case that $C = 0$ otherwise $A(r)$ will necessarily blow up at $r =0$ and this will make the metric degenerate at $r =0$ but by definition the metric must be non-degenerate everywhere.

This is quite elegant, y'know?

 

[WannabeNewton]

I should clarify that the reason the argument "no singularity in the interior metric" works is that the interior is completely empty (devoid of all mass-energy). By "everywhere" I meant everywhere in the interior. See e.g. Ohanian p. 301

Hence the geodesics of the interior metric will be those of flat space-time; in particular a test particle freely falling in the interior that is initially placed at rest will remain at rest so it will "freely float".

 

[maxverywell]

Thanks WannabeNewton!

 

[WannabeNewton]

Np Max. Make note of post #4 above because it's an important detail I forgot to mention in post #2. The argument that there cannot be singularities in the interior metric (hence no degeneracies given the general form of the metric in post #2) only works here because the interior is devoid of mass-energy. The presence of mass-energy would prevent such a simple argument from being made because for example the Kruskal extension into the interior of a Schwarzschild black hole still finds a geometric singularity in the interior.