homework and exercises - Deriving Birkhoff's Theorem

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?

Answer

The Birkhoff's Theorem in 3+1D is e.g. proven (at a physics level of rigor) in Ref. 1 and Ref. 2. (An elegant equivalent 1-page proof of Birkhoff's theorem is given in Refs. 3-4.) Imagine that we have managed to argue$^1$ that the metric is of the form of eq. (5.38) in Ref. 1 or eq. (7.13) in Ref. 2:

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

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 $$x^0\equiv t, \quad x^1\equiv r, \quad x^2\equiv\theta, \quad\text{and} \quad x^3\equiv\phi.$$ The Einstein's equations in vacuum read

$$R_{\mu\nu}~=~\Lambda g_{\mu\nu}~.\tag{E} $$

The argument is now as follows.

1. 
   

   From $$0~\stackrel{(E)}{=}~R_{tr}~=~\frac{2}{r}\partial_t\beta$$ follows that $\beta$ is independent of $t$.

   
   


2. 
   

   From $$0~\stackrel{(A)}{=}~\Lambda\left(e^{2(\beta-\alpha)} g_{tt}+g_{rr} \right) ~\stackrel{(E)}{=}~ e^{2(\beta-\alpha)} R_{tt}+R_{rr}~=~\frac{2}{r}\partial_r(\alpha+\beta) $$ 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 $$ds^2~=~-e^{-2\beta}dT^2 + e^{2\beta}dr^2 +r^2 d\Omega^2.\tag{B}$$

   
   


4. 
   
   

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

   
   


5. 
   

   From $$\Lambda r^2~\stackrel{(B)}{=}~\Lambda g_{\theta\theta} ~\stackrel{(E)}{=}~ R_{\theta\theta} ~=~1+e^{-2\beta}\left(r\partial_r(\beta-\alpha)-1\right) ~=~1-\partial_r(re^{-2\beta}), $$ it follows that $$ re^{-2\beta}~=~r-R-\frac{\Lambda}{3}r^3 $$ for some real integration constant $R$. In other words, we have derived the Schwarzschild-(anti)de Sitter solution, $$e^{2\alpha}~=~e^{-2\beta}~=~1-\frac{R}{r}-\frac{\Lambda}{3}r^2.$$

   
   

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

$$ds^2~=~-\left(1-\frac{R}{r}-\frac{\Lambda}{3}r^2\right)e^{2f(t)}dt^2 + \left(1-\frac{R}{r}-\frac{\Lambda}{3}r^2\right)^{-1}dr^2 +r^2 d\Omega^2.\qquad\tag{C}$$

It is interesting that the metric $(C)$ is the most general metric of the form $(A)$ that satisfies Einstein's vacuum equations. 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.

   
   


3. 
   

   Eric Poisson, A Relativist's Toolkit, 2004; Section 5.1.1.

   
   



4. 
   

   Eric Poisson, An Advanced course in GR; Section 5.1.1.

   
   

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$^1$ Here we for convenience show how Ref. 1 and Ref. 2 reduce from
$$ ds^2~=~g_{aa}(a,r)~da^2 +2g_{ar}(a,r)~ da~dr +g_{rr}(a,r)~ dr^2 +r^2d\Omega^2\qquad \tag{5.30/7.5}$$ to $$ ds^2~=~m(r,t)~dt^2 +n(r,t)~dr^2 +r^2d\Omega^2. \tag{5.37/7.12}$$ Proof: Define a function $$n~:=~g_{rr}-\frac{g_{ar}^2}{g_{aa}}$$ and an inexact differential $$ \omega~:=~da+\frac{g_{ar}}{g_{aa}}dr.$$ Then eq. (5.30/7.5) reads $$ ds^2~=~g_{aa}\omega^2 +n~dr^2 +r^2d\Omega^2.$$ The function $\sqrt{m}$ in eq. (5.37/7.12) can be viewed as an integrating factor to make the differential $\sqrt{\frac{g_{aa}}{m}}\omega$ exact, i.e. of the form $dt$ for some function $t(a,r)$.