Deriving Birkhoff's Theorem

Question

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 1

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$^2$

$$x^0~\equiv~ t, \quad x^1~\equiv~ r~>~0, \quad x^2~\equiv~\theta, \quad\text{and} \quad x^3~\equiv~\phi.$$

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 $$\begin{align}0~=~&\Lambda\left(-\delta^t_t+\delta^r_r \right)\stackrel{(E)}{=}~-R^t{}_t+R^r{}_r\cr ~\stackrel{(A)}{=}~& e^{-2\alpha} R_{tt}+e^{-2\beta}R_{rr} ~=~\frac{2}{r}e^{-2\beta}\partial_r(\alpha+\beta) \end{align}$$ 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 $$ re^{-2\beta}~=~r-R_S-\frac{\Lambda}{3}r^3 $$ for some real integration constant $R_S$. In other words, we have derived the Schwarzschild-(anti)de Sitter solution, $$e^{2\alpha}~=~e^{-2\beta}~=~1-\frac{R_S}{r}-\frac{\Lambda}{3}r^2.$$

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

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

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.

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$^1$ 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 $t(a,r)$.

$^2$ Notation and conventions. The metric signature is $(-,+,\ldots,+)$. We work in units where the speed of light $c=1$ is one.

Answer 2

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 $g_{11}$.

Answer 3

In this answer we generalize Birkhoff's theorem to arbitrary spacetime dimension $d>2$ with a cosmological constant $\Lambda$ and an electric charge $Q$.

The Einstein's equations read $$\begin{align} R_{\mu\nu}-\left(\frac{R}{2}-\Lambda \right)g_{\mu\nu}~=~&\kappa T_{\mu\nu},\qquad \kappa~\equiv~\frac{8\pi G}{c^4}, \cr ~\Downarrow~&\cr \frac{2-d}{2}R+d\Lambda ~=~&\kappa T, \cr R_{\mu\nu}-\frac{2\Lambda}{d-2} g_{\mu\nu} ~=~& \kappa\left(T_{\mu\nu}-\frac{T}{d-2}g_{\mu\nu} \right). \end{align}\tag{E} $$

In a spherically symmetric coordinate system there are first and foremost the time coordinate $t$ and the radial coordinate $r>0$. In order to treat all the angular coordinates on $\mathbb{S}^{d-2}$ on equal footing, let us use stereographic coordinates $Y^a$, $a\in\{1,\ldots, d\!-\!2\}$. The spherically symmetric ansatz for the metric tensor and the gauge 1-form becomes$^1$ $$\begin{align} ds^2~=~&-e^{2\alpha(r,t)}dt^2 + e^{2\beta(r,t)}dr^2 +4r^2\frac{dY^adY^a}{(1+Y^bY^b)^2},\cr A~=~&\underbrace{A_t(r,t)}_{=-\phi(r,t)}dt+A_r(r,t)dr,\cr &\text{Radial gauge}:~~ A_r(r,t)~=~0. \end{align} \tag{A}$$ The corresponding non-zero components of the Levi-Civita Christoffel symbols are $$\begin{align} \Gamma^t_{tt}~=~&\dot{\alpha}, \qquad \Gamma^t_{tr}~=~\Gamma^t_{rt} ~=~\alpha^{\prime}, \qquad \Gamma^t_{rr}~=~e^{2(\beta-\alpha)}\dot{\beta}, \cr \Gamma^r_{tt}~=~&e^{2(\alpha-\beta)}\alpha^{\prime},\qquad \Gamma^r_{tr}~=~\Gamma^r_{rt} ~=~\dot{\beta}, \qquad \Gamma^r_{rr}~=~\beta^{\prime}, \cr \Gamma^r_{ab}~=~&-\frac{4r e^{-2\beta}\delta_{ab}}{(1+Y^cY^c)^2}, \qquad \Gamma^a_{br}~=~\Gamma^a_{rb} ~=~\frac{\delta^a_b}{r}, \cr \Gamma^c_{ab}~=~&2\frac{\delta_{ab}Y^c-\delta^c_bY^a-\delta^c_aY^b}{1+Y^dY^d} , \end{align}$$ where dot (prime) denotes differentiation wrt. $t$ ($r$), respectively. The non-zero components of the Ricci tensor are $$\begin{align} R_{tt}~=~&-\left[\ddot{\beta}+\dot{\beta}^2-\dot{\alpha}\dot{\beta}\right]\cr &+e^{2(\alpha-\beta)}\left[\alpha^{\prime\prime}+(\alpha^{\prime})^2 -\alpha^{\prime}\beta^{\prime} +\frac{d-2}{r}\alpha^{\prime}\right],\cr R_{rt}~=~&\frac{d-2}{r}\dot{\beta},\cr R_{rr}~=~&-\left[\alpha^{\prime\prime}+(\alpha^{\prime})^2 -\alpha^{\prime}\beta^{\prime} -\frac{d-2}{r}\beta^{\prime}\right] \cr &+e^{2(\beta-\alpha)}\left[\ddot{\beta}+\dot{\beta}^2-\dot{\alpha}\dot{\beta}\right],\cr R_{aa}~=~&4\frac{e^{-2\beta}\left[r(\beta-\alpha)^{\prime} -(d-3)\right] +(d-3)}{(1+Y^bY^b)^2} \qquad(\text{no sum over }a). \end{align}$$ The scalar curvature is $$\begin{align} R~=~&2e^{-2\alpha}\left[\ddot{\beta}+\dot{\beta}^2-\dot{\alpha}\dot{\beta}\right]\cr & -2e^{-2\beta}\left[\alpha^{\prime\prime}+(\alpha^{\prime})^2 -\alpha^{\prime}\beta^{\prime} +\frac{d-2}{r}(\alpha-\beta)^{\prime} +\frac{(d-2)(d-3)}{2r^2}\right]\cr & +\frac{(d-2)(d-3)}{r^2}. \end{align}$$

The Lagrangian E&M density is $$ {\cal L}~=~-\frac{\sqrt{|g|}}{4\mu_0}F_{\mu\nu}F^{\mu\nu} +J^{\mu}A_{\mu}, \qquad F^{\mu\nu}~:=~g^{\mu\lambda}F_{\lambda\kappa}g^{\kappa\nu}, $$ so the metric/Hilbert SEM tensor becomes $$\begin{align} T_{\mu\nu} ~=~&-\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}}\cr ~=~&\frac{1}{\mu_0}F_{\mu\lambda}F_{\nu}{}^{\lambda} -\frac{1}{4\mu_0}g_{\mu\nu}F_{\lambda\kappa}F^{\lambda\kappa}.\end{align}$$ Because of the ansatz $(A)$ the SEM tensor takes a diagonal form $$\begin{align} T^{\mu}{}_{\nu}~=~& \left( \delta_t^{\mu}\delta^t_{\nu} +\delta_r^{\mu}\delta^r_{\nu} -\frac{1}{2}\delta^{\mu}_{\nu} \right) \frac{1}{\mu_0}F_{tr}F^{tr}, \qquad F_{tr}~=~\phi^{\prime}~=~-E_r,\cr ~\Downarrow~&\cr T^t{}_t~=~& T^r{}_r, \cr T~=~& \left(2 -\frac{d}{2} \right)\frac{1}{\mu_0}F_{tr}F^{tr}.\end{align} $$

The argument is now as follows.

1. From the diagonal form of $g_{\mu\nu}$ and $T_{\mu\nu}$, and hence of the Ricci tensor, $$0~\stackrel{(E)}{=}~R_{tr}~=~\frac{d-2}{r}\dot{\beta}$$ follows that $\beta$ is independent of $t$.

2. From $$\begin{align}0~=~&\kappa\left(-T^t_t+T^r_r \right)+ \frac{2\Lambda-\kappa T}{d-2}\left(-\delta^t_t+\delta^r_r \right)\cr ~\stackrel{(E)}{=}~&-R^t{}_t+R^r{}_r\cr ~\stackrel{(A)}{=}~& e^{-2\alpha} R_{tt}+e^{-2\beta}R_{rr}\cr ~=~&\frac{d-2}{r}e^{-2\beta}(\alpha+\beta)^{\prime} \end{align}$$ follows that $(\alpha+\beta)^{\prime}=0$. 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. Maxwell equations in curved spacetime reads $$\partial_{\mu} (\sqrt{|g|}F^{\mu\nu}) ~=~ -\mu_0 J^{\nu}.$$ No current density $J^r=0$ implies that the electric field $$F^{tr}~\stackrel{(B)}{=}~-F_{tr}~=~E_r$$ does not depend on time $t$. Gauss' law reduces to flat spacetime$^2$ $$\begin{align} \int_{[t_i,t_f]\times r\mathbb{S}^{d-2}}d^{d-1}x~\sqrt{|g|}F^{rt} ~=~&\int_{[t_i,t_f]\times r\mathbb{B}^{d-1}} d^dx~\partial_{\mu} (\sqrt{|g|}F^{\mu t})\cr ~=~&-\mu_0 Q \underbrace{\sqrt{|g_{tt}(r)|}(t_f-t_i)}_{=\tau_f-\tau_i} \cr ~\Downarrow~&\cr r^{d-2} E_r ~=~&\frac{\mu_0 Q}{\Omega_{d-2}}\cr ~\Downarrow~&\cr \phi~=~&\left\{\begin{array}{lcl} \frac{\mu_0Q}{\Omega_{d-2}}\frac{r^{3-d}}{d-3} &{\rm for}& d>3,\cr -\frac{\mu_0Q}{2}\ln(r) &{\rm for}& d=3, \end{array}\right. \end{align}$$ where $$\Omega_{d-1}~:=~{\rm Vol}(\mathbb{S}^{d-1}) ~=~2\frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}$$ is the volume of $\mathbb{S}^{d-1}$.

6. From $$\begin{align} \frac{\kappa\mu_0Q^2}{\Omega_{d-2}^2}\frac{r^{2(2-d)}}{d-2} g_{aa} ~=~&\left(\frac{1}{2} +\frac{2-d/2}{d-2}\right)\frac{\kappa E_r^2}{\mu_0} g_{aa}\cr ~=~&\kappa\left(T_{aa}-\frac{T}{d-2}g_{aa}\right)\cr ~\stackrel{(E)}{=}~&-\frac{2\Lambda}{d-2} g_{aa} +R_{aa}, \cr ~\Downarrow~&\cr \left(\frac{\kappa\mu_0Q^2}{\Omega_{d-2}^2}r^{4-2d}+2\Lambda\right)\frac{r^2}{d-2}~\stackrel{(B)}{=}~&\left(\frac{\kappa\mu_0Q^2}{\Omega_{d-2}^2}r^{4-2d}+2\Lambda\right)\frac{g_{aa}}{d-2}\frac{(1+Y^bY^b)^2}{4}\cr ~=~& R_{aa}\frac{(1+Y^bY^b)^2}{4}\cr ~=~& e^{-2\beta}\left[r(\beta-\alpha)^{\prime} -(d-3)\right] +(d-3) \cr ~=~& -r(e^{-2\beta})^{\prime} -(d-3)e^{-2\beta} +(d-3)\cr ~=~& -r^{4-d}(r^{d-3}e^{-2\beta})^{\prime} +(d-3), \end{align}$$ (no sum over $a$), it follows that $$ r^{d-3}e^{-2\beta}~=~ \left\{\begin{array}{lcl} r^{d-3}-R_S^{d-3}+ \frac{\kappa\mu_0Q^2}{\Omega_{d-2}^2} \frac{r^{3-d}}{(d-3)(d-2)} -\frac{2\Lambda r^{d-1}}{(d-1)(d-2)}&{\rm for}& d>3,\cr C-\frac{\kappa\mu_0Q^2}{4}\ln(r)-\Lambda r^2&{\rm for}& d=3, \end{array}\right. $$ for some real integration constants$^3$ $R_S^{d-3}$ and $C$. In other words, we have derived the higher-dimensional Reissner-Nordström-(anti)de Sitter solution aka. the Tangherlini solution $$e^{2\alpha}~=~e^{-2\beta}~=~ \left\{\begin{array}{lcl} 1-\frac{R_S^{d-3}}{r^{d-3}} +\frac{\kappa\mu_0Q^2}{(d-3)(d-2)\Omega_{d-2}^2} \frac{1}{r^{2d-6}} -\frac{2\Lambda r^2}{(d-1)(d-2)} &{\rm for}& d>3,\cr C-\frac{\kappa\mu_0Q^2}{4}\ln(r)-\Lambda r^2&{\rm for}& d=3, \end{array}\right. $$ cf. Ref. 2. Remarkably the corresponding $d=3$ Newtonian $\ln(r)$ potential is nowhere to be found, cf. e.g. this Phys.SE post.

References:

1. M. Blau, Lecture Notes on GR; Chapter 31.

2. F.R. Tangherlini, Il Nuovo Cimento 27 (1963) 636–651.

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$^1$ Notation and conventions. The metric signature is $(-,+,\ldots,+)$. We work in units where the speed of light $c=1$ is one.

$^2$ In $d=3$ it turns out that the spacetime is a wedge, so that the true charge (mass) is only a corresponding fraction of $Q$ ($M$), respectively.

$^3$ For $d>3$ by comparing with the asymptotic Newtonian limit $r\to\infty$, we may identify the integration constant $$ R_S^{d-3} ~=~\frac{2\kappa M}{(d-2)\Omega_{d-2}},\qquad C~=~\kappa M,$$ where $M$ is the mass/total energy of the black hole.