This question is apparently quite simple but I can't seem to find an answer to it, so I was hopping anyone could clarify me.

Are the Einstein field equations (EFE) only valid for a 3+1 dimensional space-time?

I've read somewhere, which I can't remember or find, that there were problems with the EFE in a 2+1 dimensions...Why would that be?

What about 1+1?

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    $\begingroup$ Not "problems" just different behaviour. Steve Carlip is your friend. He's written mountains of material on 2+1 dim relativity. $\endgroup$ – twistor59 May 22 '13 at 15:59
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    $\begingroup$ The wikipedia page is short but might still be useful: en.wikipedia.org/wiki/(2%2B1)-dimensional_topological_gravity $\endgroup$ – Olof May 22 '13 at 16:01
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    $\begingroup$ I've heard that there should be no gravitational waves in 2D $\endgroup$ – Dilaton May 22 '13 at 16:05

There is nothing "wrong" with the Einstein field equations in $2+1$ as indicated by the comments, but they do have interesting, significantly restricted behavior in $2+1$ dimensions.

For example, the Wikipedia page referred to by Olof in the comments says that in $2+1$, every vacuum solution is locally either $\mathbb R^{2,1}$, $\mathrm{AdS_3}$, or $\mathrm{dS}_3$. Here's why. In $d+1$ with $d\neq 1$, the vacuum field equations (those with $T_{\mu\nu} =0$) can be manipulated to show that $$ R_{\mu\nu} = \frac{R}{d+1}g_{\mu\nu} $$ On the other hand, one can show (see Weinberg Gravitation and Cosmology eq. 6.7.6) that in $2+1$, the Riemann tensor satisfies $$ R_{\lambda\mu\nu\kappa} = g_{\lambda\nu} R_{\mu\kappa} - g_{\lambda\kappa}R_{\mu\nu} - g_{\mu\nu}R_{\lambda\kappa} + g_{\mu\kappa} R_{\lambda\nu} - \frac{1}{2}(g_{\lambda\nu}g_{\mu\kappa} - g_{\lambda\kappa}g_{\mu\nu})R $$ and combining these results gives $$ R_{\lambda\mu\nu\kappa} = \frac{1}{6}(g_{\lambda\nu}g_{\mu\kappa}-g_{\lambda\kappa}g_{\mu\nu})R $$ which is precisely the Riemann tensor for a maximally symmetric spacetime in $2+1$ which gives the result.

Notice that this behavior is in stark contrast to the vacuum behavior in $3+1$. For example, take the vacuum region outside of a spherically symmetric massive body in $3+1$ (like a black hole). This region is not flat, but in $2+1$ with vanishing cosmological constant any vacuum region outside of a massive body would be. Pretty strange.

  • $\begingroup$ I don't think there are any black hole in 2+1 dimensions if the cosmological constant is zero. For negative cosmological constant there is the BTZ black hole, which is locally AdS. $\endgroup$ – Olof May 22 '13 at 17:52
  • $\begingroup$ @Olof I changed the wording; the intent was not to imply that there are black holes in $2+1$ dimensions. $\endgroup$ – joshphysics May 22 '13 at 18:07
  • $\begingroup$ @PML Sure thing! $\endgroup$ – joshphysics May 22 '13 at 18:41
  • $\begingroup$ Are you shure about the $R^{3,1}$? $\endgroup$ – ungerade May 22 '13 at 21:24
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    $\begingroup$ @ungerade Thanks I changed it to $\mathbb R^{2,1}$. $\endgroup$ – joshphysics May 22 '13 at 22:18

Joshphysics has already given a nice answer showing that in 2+1 dimensional Einstein gravity any metric is locally equivalent to a metric of constant curvature. As dilaton mentioned in a comment this in particular means that there are no local excitations.

The updated question also asks about 1+1 dimensions. In this case the answer is even simpler: the 1+1 dimensional Einstein tensor $R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$ vanishes identically. Hence the Einstein-Hilbert action $\int d^2x \sqrt{-g}R$ is a surface term.


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