# Birkhoff's theorem in general setup ($r^2 \rightarrow e^{2\gamma(r)}$)

I am trying to prove Birkhoff's theorem in general setup.

Birkhoff's theorem states that every spherically symmetric vacuum solution to Einstein's field equation is the Schwarzschild. (i.e., it says spherically symmetric vacuum solutions admits static, asymptotically flat)

First I know from the classical textbook Carroll \begin{align} ds^2 = - e^{2\alpha(t,r)} dt^2 + e^{2\beta(t,r)} dr^2 + r^2 d\Omega^2 \end{align} and computing Ricci tensors i.e., $$R_{01}=0$$ $$\beta(t,r) = \beta(r)$$ and the other combinations gives $$\alpha = -\beta$$.

I think $$r^2$$ terms can also be generalized to some functions, so I tried the following metric form \begin{align} ds^2 = - e^{2\alpha(t,r)} dt^2 + e^{2\beta(t,r)} dr^2 + e^{2\gamma(r,t)} d\Omega^2 \end{align} but the situation is not the same. For this case \begin{align} R_{tr} = 2 \gamma' (\dot{\beta} - \dot{\gamma}) + 2 \alpha' \dot{\gamma} - 2 \dot{\gamma}' \end{align} where prime means r derivative and dot means time derivatives. Because of time dependency of $$\gamma$$, from this we can not factor out $$\beta(r,t) = \beta(r)$$ anymore.

I guess maybe the "t" dependence on the angular part is problematic, so my next metric form is

\begin{align} ds^2 = - e^{2\alpha(t,r)} dt^2 + e^{2\beta(t,r)} dr^2 + e^{2\gamma(r)} d\Omega^2 \end{align} and compute corresponding Ricci tensors. [This metric seems fine because it is nothing but isotropic form of Carroll's textbook]

\begin{align} &R_{tt} = e^{2(\alpha-\beta)} \left( 2\gamma' \alpha' + (\alpha')^2 - \alpha' \beta' + \alpha'' \right) + \dot{\alpha} \dot{\beta} - (\dot{\beta})^2 -\ddot{\beta} \\ &R_{tr}= 2 \gamma' \dot{\beta} \\ & R_{rr} = - 2 (\gamma')^2 - 2 \gamma'' + 2 \gamma' \beta' + (- (\alpha')^2 + \alpha' \beta' - \alpha'') + e^{2(\beta - \alpha)} ( -\dot{\alpha} \dot{\beta} + (\dot{\beta})^2 + \ddot{\beta}) \\ & R_{\theta \theta} = 1 - e^{2(\gamma-\beta)} \left( - 2 (\gamma')^2 - \gamma'' - \gamma'(\alpha' - \beta') \right) \\ & R_{\varphi \varphi} = \sin^2(\theta) R_{\theta \theta} \end{align}

One trivial results from $$R_{tr}=0$$ is $$\beta(t,r) = \beta(r)$$. So I was happy

From $$R_{tt}=0$$ we have \begin{align} 2\gamma' \alpha' + (\alpha')^2 - \alpha'\beta' + \alpha'' =0 \end{align} And from \begin{align} 0= e^{2(\beta-\alpha)} R_{tt} + R_{rr} = - 2(\gamma')^2 - 2 \gamma'' + 2 \gamma' (\alpha'+\beta') \end{align} and from $$R_{\theta \theta}=0$$, \begin{align} e^{2(\beta-\gamma)} = - 2 (\gamma')^2 - \gamma'' - \gamma'(\alpha' - \beta') \end{align}

Of course, I know $$e^{2\gamma} =r^2$$ is one solution for $$\gamma$$ and in that case, it reduces the Schwarzschild metric, but I am worried whether that $$e^{2\gamma} = r^2$$ is a unique solution for $$\gamma$$.

At this level, It seems there are no ways to determine $$e^{2\gamma} =r^2$$ from the equations. Can anyone comments on this $$e^{2\gamma}$$ setup?

• I am only commenting to say that Birkhoff is more complicated than this, a more prudent thing to say would be that a spherically symmetric vacuum solution is always locally isometric to an open subset of the maximally extended Schwarzschild solution, and there is a rather complete, general and rigorous proof of this in the appendix of Hawking-Ellis. – Bence Racskó Nov 11 '20 at 20:42

If we follow Carroll's proof of Birkhoff's theorem the radial coordinate $$r$$ has no prior definition. At this stage of the proof $$r^2$$ is identified as the coefficient in front of $$d\Omega^2$$ in the metric \begin{align} ds^2~=~&g_{aa}(a,b)~da^2 +2g_{ab}(a,b)~ da~db \cr &+g_{bb}(a,b)~ db^2 +r^2(a,b)d\Omega^2.\end{align} \tag{5.29/7.4} We next perform a coordinate transformation $$(a,b,\theta,\phi)\to(a,r,\theta,\phi)$$ so that the metric becomes \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} Later on the time coordinate $$t$$ is identified as some function $$t(a,r)$$, cf. e.g. this related Phys.SE post.