# 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?

• 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. Mar 2, 2012 at 7:07

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

\begin{align}ds^2~=~&-\left(1-\frac{R}{r}-\frac{\Lambda}{3}r^2\right)e^{2f(t)}dt^2 \cr &+ \left(1-\frac{R}{r}-\frac{\Lambda}{3}r^2\right)^{-1}dr^2 +r^2 d\Omega^2.\end{align}\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.

--

$$^1$$ Here we for convenience show how Ref. 1 and Ref. 2 reduce from
\begin{align} ds^2~=~&g_{aa}(a,r)~da^2 +2g_{ar}(a,r)~ da~dr \cr &+g_{rr}(a,r)~ dr^2 +r^2d\Omega^2\end{align} \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 $$t(a,r)$$.

• Notes for later: Ricci tensor: $R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu} = \frac{1}{\sqrt{|g|}}\partial_{\lambda}\left(\sqrt{|g|}\Gamma^{\lambda}_{\mu\nu}\right) -\partial_{\mu}\partial_{\nu}\ln\sqrt{|g|} - \Gamma^{\lambda}_{\mu\kappa}\Gamma^{\kappa}_{\nu\lambda}$. Nov 6, 2014 at 13:46
• Future project: Generalize with arbitrary dimension and electric charge. Sep 9, 2015 at 15:59
• Reissner-Nordström-(anti)de Sitter: $e^{-2\beta}~=~1-\frac{2M}{r}+\frac{Q^2}{r^2} -\frac{\Lambda}{3}r^2$, $A_{\mu}=\frac{Q}{r}\delta^0_{\mu}$ Sep 11, 2015 at 11:16
• Blau's GR notes: blau.itp.unibe.ch/newlecturesGR.pdf Oct 24, 2015 at 23:57
$ds^2 = e^{f(t,r)} dt^2 - e^{g(t,r)} dr^2 - r^2 d\Omega^2$
{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 $g_{11}$.