# Deriving Birkhoff's Theorem

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I am trying to derive Birkhoff's theorem in GR as an exercise: a spherically symmetric gravitational field is static in the vacuum area. I managed to prove that $g_{00}$ is independent of t in the vacuum, and that $g_{00}*g_{11}=f(t)$. But the next question is: Show that you can get back to a Schwarzschild metric by a certain mathematical operation. I am thinking at a coordinate change (or variable change on $r$) to absorb the $t$ dependence of $g_{11}$, but I can't see the right one. Does someone has a tip to share?

Also one further question is to explain if an imploding (spherically) star radiates gravitationally, based on a quantum mechanical argument. I would say that under the isotropy hypothesis we have $[H,J]=0$, so the total angular of the star is conserved, or if gravitons would be emitted, they would carry angular momentum from their spin 2 and violate this conservation, so my answer would be no. Is this wrong?

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user toot
retagged May 1, 2014
You can't get rid of the t dependence in g_{11} by a coordinate transformation--- you need to show that g_{11} is constant. The reason is that a t-dependent r rescaling introduces an off-diagonal t-r term.

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user Ron Maimon

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user Willie Wong

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Here we will only consider the first half of the question(v2). The Birkhoff's Theorem is e.g. proven (at a physics level of rigor) in Ref. 1 and Ref. 2. Imagine that we have managed to argue that the metric is of the form of eq. (5.38) in Ref. 1 or eq. (7.13) in Ref. 2:

$$ds^2~=~-e^{2\alpha(r,t)}dt^2 + e^{2\beta(r,t)}dr^2 +r^2 d\Omega^2. \qquad\qquad(A)$$

It is a straightforward exercise to calculate the corresponding Ricci tensor $R_{\mu\nu}$, see eq. (5.41) in Ref. 1 or eq. (7.16) in Ref. 2. The notation is here $x^0\equiv t$, $x^1\equiv r$, $x^2\equiv\theta$, and $x^3\equiv\phi$. Assuming a vanishing cosmological constant $\Lambda=0$, the Einstein's equations in vacuum read

$$R_{\mu\nu}~=~0~.$$

The argument is now as follows.

1. From $R_{tr}=0$ follows that $\beta$ is independent of $t$.

2. From $e^{2(\beta-\alpha)} R_{tt}+R_{rr}=0$ follows that $\partial_r(\alpha+\beta)=0$. In other words, the function $f(t):=\alpha+\beta$ is independent of $r$.

3. Define a new coordinate variable $T:=\int^t dt'~e^{f(t')}$. Then the metric $(A)$ becomes $$ds^2~=~-e^{-2\beta}dT^2 + e^{2\beta}dr^2 +r^2 d\Omega^2.\qquad\qquad(B)$$

4. Rename the new coordinate variable $T\to t$. Then eq. $(B)$ corresponds to setting $\alpha=-\beta$ in eq. $(A)$.

5. From $R_{\theta\theta}=0$ follows that $$1=e^{-2\beta}(1-2r\partial_r\beta)\equiv\partial_r(re^{-2\beta}),$$ so that $re^{-2\beta}=r-R$ for some real integration constant $R$. In other words, we have derived the Schwarzschild solution, $$e^{2\alpha}~=~e^{-2\beta}~=~1-\frac{R}{r}.$$

Finally, if we switch back to the original $t$ coordinate variable, the metric $(A)$ becomes

$$ds^2~=~-\left(1-\frac{R}{r}\right)e^{2f(t)}dt^2 + \left(1-\frac{R}{r}\right)^{-1}dr^2 +r^2 d\Omega^2.\qquad\qquad(C)$$

It is interesting that the metric $(C)$ is the most general metric of the form $(A)$ that satisfies Einstein's vacuum equations with $\Lambda=0$. The only freedom is the function $f=f(t)$, which reflects the freedom to reparametrize the $t$ coordinate variable.

References:

1. Sean Carroll, Spacetime and Geometry: An Introduction to General Relativity, 2003.

2. Sean Carroll, Lecture Notes on General Relativity, Chapter 7. The pdf file is available here.

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user Qmechanic
answered Mar 19, 2012 by (2,860 points)
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Are you doing this rigorously by using an ansatz for your metric and plugging it into the usual Einstein field equations in vacuum or something "more topological"?

If former, in the most common ansatz,

$ds^2 = e^{f(t,r)} dt^2 - e^{g(t,r)} dr^2 - r^2 d\Omega^2$

you get your metric independence of time t with the

{01}th Ricci tensor component, which sets a time derivative of one of your metric components to 0. Algebraic combinations of the other Ricci tensor componentes give you the relationships between metric component functions $f$ and $g$, somewhere along the way you should get something like $\frac{d}{dt}[f(t,r)-g(t,r)] = 0$. That gives you your time independance of $g_{11}$.

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user Thomas M.
answered Mar 19, 2012 by (10 points)

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