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It is well-known that, given a Killing vector $\xi_\mu$ for a metric, this can be dimensionally reduced. This is exactly the case of the Schwarzschild solution due to the rotational symmetry. This procedure is quite general and one can always reformulate a general 2d (or 3d) model, as the Jackiw-Teitelboim model, of general relativity and obtain solutions for the 4d case with the given symmetry properties.A reference about this is this. The effect, as always happens in dimensional reduction, is to introduce another field into the gravity action.

So far, I was not able to see the derivation of the reduced Schwarzschild metric from the dimensionally reduced Einstein equations. This is generally given for granted. So, I would appreciate to see such a derivation or to get some tutorial about this.


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The dimensional reduction in this case is just plugging in Schwarzschild's ansatz for the metric. –  Ron Maimon Oct 6 '12 at 2:09
@RonMaimon: The point here is that I have the dimensional reduced action (or equations if you want). I would like to see, from these equations, how to derive the Schwarzschild metric using Schwarzschild coordinates. It is always taken for granted as in the reference I cite. Even if it would be a homework, I would like to see the solution. The general solution of Jackiw-Teitelboim theory has not such a solution as far as I know. A reference would be enough. –  Jon Oct 6 '12 at 8:17
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1 Answer

up vote 4 down vote accepted

The dimensionally reduced Einstein-Hilbert action of a spherically invariant metric is a special case of two-dimensional dilaton gravity. This type of theories was the subject of a quite extensive research at least in the past decade or so. It has applications in integrable systems, quantization of 1+1 gravity and it (what is relevant to this question) offers an alternative proof of the Birkhoff's theorem (: The Schwarzchild solution is the unique spherically symmetric solution of the vacuum Einstein equations). In many of these works the Schwarzchild solution was obtained as a byproduct as the solution of the 2-d field equations

Technically it is not very easy since it requires the solution of nonlinear partial differential field equations in 2D. In the following reference by Marco Cavaglià the derivation is quite transparent, which can be summarized as follows:

Starting from the general spherically symmetric 3+1 metric:

$ds^2 = \phi^{-\frac{1}{2}}g_{\mu\nu} dx^{\mu}dx^{\nu}+\phi d\Omega^2$

(the indices run over a Lorentzian 2-space). After the substitution in the Einstein-Hilbert action and the integration of the angular coordinates, the following reduced 2d theory is obtained:

$S = \int d^2x \sqrt{-g}[\phi R^{(2)}(g) + V(\phi)]$, with ,


The field equations:

$ (\nabla_{(\mu}\nabla_{\nu)}- g_{\mu\nu}\nabla_{\sigma}\nabla^{\sigma})\phi + \frac{1}{2}g_{\mu\nu}V(\phi)=0$


The nonlinear field equations can be solved by means of a Bäcklund transformation:

$ \nabla_{\mu} \psi = \frac{\nabla_{\mu}\phi }{\nabla_{\sigma}\phi\nabla^{\sigma}\phi}$

Leading to the following linear field equations

$\nabla_{\sigma}\nabla^{\sigma}\psi = 0$


$\nabla_{\sigma} M = 0$


$M = N - \nabla_{\sigma}\phi\nabla^{\sigma}\phi$


$N = \int^{\phi}V(\phi\prime)d\phi\prime$

The linear field equations can be solved in the conformal gauge:

$ds^2 = \rho(u,v)dudv$


$\psi = U(u) + V(v)$, and $M=const.$

The solution for the metric is obtained from the Bäcklund transformation equation:

$\rho^2 = \partial_u\phi \partial_v\phi\partial_u\psi \partial_v\psi$, or

$\rho= (N(\phi)-M)\partial_u\psi \partial_v\psi$,


$\frac{d\psi}{d\phi} = \frac{1}{N-M}$

By performing a coordinate transformation to the independent variables $U$ and $V$ instead of $u$ and $v$ respectively. We obtain:

$\rho= N(\phi(U+V))-M)$

Again changing the coordinates to T=U-V and $\phi(U+V)$, we obtain:

$ds^2 = (N-M)dT^2+(N-M)^{-1}d\phi^2$

which is the Schwarzchild solution for $N=\sqrt\phi$.

Please observe that the metric was not assumed to be stationary, thus this derivation consists of a proof of the Birkhoff's theorem.

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Thank you very much. This is exactly what I was looking for. –  Jon Oct 11 '12 at 12:40
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