# Is a Schwarzschild solution possible in 3 space-time dimensions?

Reading the book "Supergravity" from Freedman & van Proeyen I stumbled over the assertion that in 3D GR the vacuum solution $$R_{\mu\nu} =0$$ leads to a vanishing 4-rank curvature tensor $$R_{\mu\nu\rho\sigma}=0$$, therefore there are no gauge invariant degrees of freedom in the 3D GR vacuum case (I indeed found in Landau & Lifshitz's volume II chapter 93 a relation between the 3D Ricci-tensor and the 3D full 4-rank curvature tensor that confirms that).

But would that implicate that an axial-symmetric solution of the vacuum EFEs $$R_{\mu\nu} =0$$ in 3D=(1 time + 2space) would be trivial, i.e. not Schwarzschild ( replace if appropiate $$r^2 d\Omega^2 \rightarrow r^2 d\phi^2$$ instead of $$r^2 d\Omega^2= r^2 (d\theta^2 +\sin^2 \theta d\phi^2$$)) ?

No it is not. The Weyl tensor vanishes by definition in three dimensions, Einstein's equations (in the absence of matter) impose:

$$R_{\mu\nu} = 0 \rightarrow R=0$$

and since $$Riemann = Weyl + Ricci$$ no geometry can be formed.

The solution in three dimensional spacetime is the BTZ black hole (https://arxiv.org/abs/hep-th/9204099v3) which includes a cosmological constant thus the Ricci tensor is no longer equal to zero.

A derivation: Consider $$2+1$$ Gravity and a cosmological constant term: $$$$S = \int d^3 x \sqrt{-g} \big(R -2Λ\big)$$$$ Einstein's equation read: $$$$G_{\mu\nu} + \Lambda g_{\mu\nu} = 0$$$$ and in the form of differential equations, imposing a two degree of freedom metric: $$$$ds^2 = -b(r)dt^2 + f(r)dr^2 + r^2 d\theta^2$$$$ we get: $$$$\frac{2 \Lambda -\frac{f'(r)}{r f(r)^2}}{2 b(r)} =0$$$$ $$$$-\frac{\frac{b'(r)}{r b(r)}+2 \Lambda f(r)}{2 f(r)^2} =0$$$$ $$$$\frac{b(r) \left(b'(r) f'(r)-2 f(r) b''(r)\right)+f(r) b'(r)^2-4 \Lambda b(r)^2 f(r)^2}{4 r^2 b(r)^2 f(r)^2} =0$$$$ The first one is a differential equation for $$f(r)$$: $$2 \Lambda -\frac{f'(r)}{r f(r)^2} =0 \Rightarrow \Big(\Lambda r^2 + \cfrac{1}{f(r)}\Big)' =0 \Rightarrow$$ $$$$f(r) = \cfrac{1}{C - Λr^2}$$$$ where $$C$$ is a constant of integration. Now we can obtain $$b(r)$$ from the second equation: $$\frac{b'(r)}{r b(r)}+2 \Lambda f(r)=0 \Rightarrow (\ln(C-\Lambda r^2))' - (\ln b(r))'=0 \Rightarrow$$ $$$$b(r) = C - \Lambda r^2$$$$ Now, if we set $$C=-M$$ and $$\Lambda = -1/l^2$$, where $$l$$ the AdS radius we obtain the BTZ Black hole: $$$$b(r) = \cfrac{r^2}{l^2} -M = \cfrac{1}{f(r)}$$$$ We can see that this solution satisfies the gauge $$g_{tt}g_{rr} = -1$$. The obtained configurations satisfy the last Einstein equation.

• Thank you for the quick answer. What does BTZ mean ? Oct 12, 2020 at 14:35
• Bañados, Teitelboim & Zanelli Oct 12, 2020 at 14:40
• If i find time i'll sketch a derivation. Oct 12, 2020 at 14:41

As mentioned in ApolloRa's answer, in 2+1 dimensions there exist no asympototically flat black hole solutions. However, you can still solve the Einstein Field Equations to find the metric of a non-spinning point mass $$M$$. The answer is given by

$$ds^2 = -dt^2 +\frac{1}{(1-4GM)^2}dr^2 + r^2 d\phi^2$$

As you can easily check this metric is flat for all $$r>0$$. However, it has a singular curvature at $$r=0$$. This can be confirmed by calculation the holonomy along a curve around the origin. If you parallel transport a vector around the origin, you will find that it has been rotated by $$8\pi GM$$ radians when it returns to its original position.

The spatial part of this metric, is that of a cone, which gives this type of singularity it name, a conical singularity.

• Thank you: Just a stupid question: in the denominator $1-4GM$ there is no $r$, is that meant like that ? And if there is a singular curvature in this metric, in a polar coordinates the curvature would be also singular ? Oct 12, 2020 at 15:19
• @FredericThomas Yes, the coefficient of $dr^2$ is simply a constant. Oct 12, 2020 at 15:22
• The curvature in this solution can be thought of (sorta) as a delta function at the origin. Oct 12, 2020 at 15:24
• @FredericThomas With one fewer space dimension than usual, rhe Newtonian acceleration is $-GM/r$ instead of $-GM/r^2$, so $GM/c^2$ is dimensionless.
– J.G.
Oct 13, 2020 at 11:32