Birkhoff's Theorem

Birkhoff's Theorem states that the Schwarzschild geometry is the only spherically symmetric solution of Einstein's equation. This result is remarkable, in that the Schwarzschild geometry has a timelike symmetry (Killing vector), even though this was not assumed; spherically symmetric vacuum solutions of Einstein's equation are automatically time independent! We outline the proof of Birkhoff's Theorem below.

Step 1: Spherical symmetry implies that our spacetime can be foliated by 2-spheres, each with line element \begin{equation} ds^2 = r^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \end{equation} We will use $r$ as our third coordinate, and call the remaining coordinate $t$. Spherical symmetry further implies that we can choose the $r$ and $t$ directions to be orthogonal to our spheres. The full line element therefore takes the form 1) \begin{equation} ds^2 = -A^2\,dt^2 + B^2 dr^2 + 2 C\,dt\,dr + r^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \end{equation} where $A$, $B$, and $C$ are functions of $t$ and $r$ only.

Step 2: We can write the metric in the $(t,r)$-plane as the difference of squares. With $\theta$ and $\phi$ constant we have \begin{align} ds^2 &= -A^2\,dt^2 + B^2 dr^2 + 2 C\,dt\,dr \nonumber\\ &= - \left(A\,dt - \frac{C}{A}\,dr\right)^2 + \left(B^2 - \frac{C^2}{A^2}\right)\,dr^2 \end{align} But any differential in two dimensions is a multiple of an exact differential, so we must have 2) \begin{equation} A\,dt - \frac{C}{A}\,dr = P\,dt' \end{equation} for some $P$ and $t'$. We can therefore write the line element as \begin{equation} ds^2 = - P^2 \,dt'^2 + Q^2 \,dr^2 \end{equation} where $Q^2 = B^2-\frac{C^2}{A^2}$ (which is positive since the signature is invariant), and where $P$ and $Q$ are now to be thought of as functions of $t'$ and $r$. We henceforth drop the primes; this argument shows that we could have simply assumed that $C=0$ (in which case $P=A$ and $Q=B$).

Step 3: Compute the Einstein tensor for the line element \begin{equation} ds^2 = - A^2 \,dt^2 + B^2 \,dr^2 + r^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \end{equation} where $A$ and $B$ are functions of $t$ and $r$. The nonzero components are: \begin{align} G^t{}_t &= - \frac{1}{r^2B^2} \left(B^2-1+\frac{2r}{B}\Partial{B}{r}\right) \\ G^t{}_r &= - \frac{2}{r\,A^2B}\Partial{B}{t} \\ G^r{}_r &= - \frac{1}{r^2B^2} \left(B^2-1-\frac{2r}{A}\Partial{A}{r}\right) \\ G^\theta{}_\theta = G^\phi{}_\phi &= \frac{1}{r^2B^2} \left( -\frac{r}{B}\Partial{B}{r} + \frac{r}{A}\Partial{A}{r} +\frac{r^2}{A}\frac{\partial^2 A}{\partial r^2} -\frac{r^2}{B}\frac{\partial^2 B}{\partial t^2} \right. \nonumber\\ & \qquad\qquad \left. -\frac{r^2}{AB}\Partial{A}{r}\Partial{B}{r} +\frac{r^2}{AB}\Partial{A}{t}\Partial{B}{t} \right) \end{align}

Step 4: Setting $G^t{}_r=0$, we see that $B$ must be a function of $r$ alone.

Step 5: Setting $G^t{}_t-G^r{}_r=0$, we obtain \begin{equation} \frac{1}{B}\Partial{B}{r} = -\frac{1}{A}\Partial{A}{r} \label{ABid} \end{equation} from which it follows that \begin{equation} B=\frac{\lambda}{A} \end{equation} where $\lambda$ is an integration “constant”, which however might be a function of $t$. However, by setting $dt'=\frac{dt}{\lambda}$, we can absorb this integration constant into the definition of $t$. We can therefore safely assume that $\lambda=1$, so that \begin{equation} B=\frac{1}{A} \label{ABrel} \end{equation} and the line element takes the form \begin{equation} ds^2 = - A^2 \,dt^2 + \frac{1}{A^2} \,dr^2 + r^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \label{SchwAB} \end{equation} where $A$ is a function of $r$ alone.

Step 6: Setting $G^t{}_t=0$ and using (\ref{ABid}) and (\ref{ABrel}) we obtain \begin{equation} \frac{1}{A}\frac{dA}{dr} = -\frac{1}{B}\frac{dB}{dr} = \frac{B^2-1}{2r} = \frac{1-A^2}{2r\,A^2} \end{equation} which can be separated to yield \begin{equation} \frac{A\,dA}{1-A^2} = \frac{dr}{2r} \end{equation} Integrating both sides now leads to \begin{equation} -\ln(1-A^2) = \ln(ar) \end{equation} where $a$ is an integration constant, or equivalently \begin{equation} A^2 = 1 - \frac{1}{ar} \label{AAeq} \end{equation} The integration constant can be related to mass by comparing the Newtonian limit of this geometry to the (Newtonian) gravitational field of a point particle, leading to \begin{equation} \frac{1}{a} = 2m \label{amlim} \end{equation} Inserting (\ref{ABrel}), (\ref{AAeq}), and (\ref{amlim}) into (\ref{SchwAB}), we recover the Schwarzschild line element in its usual form.

Step 7: Finally, inserting $G^t{}_t=0$ into $G^\theta{}_\theta=G^\phi{}_\phi$ and using (\ref{ABid}) along with the relationship \begin{equation} \Partial{}{r} \left( \frac{1}{A}\Partial{A}{r} \right) = \frac{1}{A}\frac{\partial^2 A}{\partial r^2} - \frac{1}{A^2} \left(\Partial{A}{r}\right)^2 \end{equation} shows that these components vanish identically, thus completing the proof.

1) By choosing $A^2$ and $B^2$ as coefficients, we are making certain assumptions about signature, which turn out to imply that $r>2m$. These assumptions do not affect the argument in any way; a more general derivation would simply rename these coefficients as $A$ and $B$.
2) One way to see this is to use existence and uniqueness properties of differential equations. Assuming that the differential equation $\frac{dt}{dr}=\frac{C}{A^2}$ has a solution $t=T(r)$, then \begin{equation} d( t-T(r) ) = dt - \frac{C}{A^2}\,dr = \frac{A\,dt-\frac{C}{A}\,dr}{A} \end{equation} and we can set $t'=t-T(r)$ and $P=A$.

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