Schwarzschild metric in lower-dimensional spaces

I was trying to explain the consequences of the Schwarzschild metric to someone last night and obviously it's pretty difficult to conceptualize in four-dimensional spacetime. Elementary googling has failed me, what are the two-dimensional and / or three-dimensional spacetime equivalents?

• The geometry is completely different in 2+1D spacetimes. Black holes like a Schwarzschild black hole don't exist. There is a type of black hole called a BTZ black-hole but this requires there to be a non-zero cosmological constant. I would avoid straying into lower dimensional spacetimes as it will only confuse matters. – John Rennie Sep 25 '15 at 10:40
• Possible duplicates: physics.stackexchange.com/q/75473/2451 – Qmechanic Sep 25 '15 at 10:59
• I agree with John Rennie, you should avoid lower dimensions. I don't think it's particularly difficult to conceptualize it. IMHO you should ask a direct question about the Schwarzschild metric and avoid wasting your time on a wild goose chase. – John Duffield Sep 25 '15 at 12:38
• You should use 3+1 dimensions, but without talking about the angular part, taking for example a radially ingoing particle. – Rexcirus Sep 25 '15 at 12:42

General relativity in low dimensions gets progressively simpler, making it difficult to really make interesting statements on it. Here is the situation :

2+1 dimensions :

In 2+1 dimensions, the Riemann tensor depends only on the Ricci tensor. So if spacetime is Ricci flat at a point, it is totally flat. This means that when there is a 0 stress energy tensor at that point, there is no gravitational effects.

However, this does not mean that the 2+1 direct equivalent of Schwarzschild is just Minkowski space with a singularity in the middle. It is actually similar to a flat section of a cosmic string metric, with the appearance of a conical singularity. The metric is

$$$$ds^2 = -(dt - \frac{A}{B} d\phi)^2 + dr^2 + B^2 r^2 d\phi^2$$$$

Which is equivalent to Minkowski space via the coordinate transform $$\phi \rightarrow B \phi$$ and $$t \rightarrow t - \frac{A}{B}\phi$$, with the following identification :

$$\begin{eqnarray} (t, \phi) \approx (t - \frac{2\pi A}{B}, \phi + 2 \pi B) \end{eqnarray}$$

B is roughly equivalent to the mass ($$\frac{2 - 2B}{8G} = m$$), and the angular momentum is $$J^{ij} = \frac{A}{4BG}\epsilon^{ij}$$.

Due to the identifications, this spacetime has CTCs going through it. And due to the fact that gravity doesn't propagate in 2D, you can add arbitrarily many pointlike masses of mass $$M_\alpha$$ and spins $$J_\alpha$$, via

$$\begin{eqnarray} ds^2 = (dt + 4G \sum_\alpha J_\alpha \frac{\epsilon_{ij} (x^i - a_\alpha^i) dx^j}{\vert \vec{r} - \vec{a}_\alpha\vert^2}) - \prod_\alpha \vert \vec{r} - \vec{a}_\alpha \vert^{-8GM_\alpha} d\vec{r}.d\vec{r} \end{eqnarray}$$

Since gravity doesn't propagate in a vacuum in 2D, as said in the comment, the usual solution is to add a cosmological constant term $$\Lambda = -\frac{1}{l^2}$$, known as the BTZ black hole (named after Banados, Teitelboim and Zanelli), with the metric

$$\begin{eqnarray} ds^2 = -N^2 dt^2 + r^2 (d\phi + N^\phi dt)^2 + N^{-2} dr^2 \end{eqnarray}$$

With

$$\begin{eqnarray} N^2 &=& -M + \frac{r^2}{l^2} + \frac{J^2}{4r^2}\\ N^\phi &=& -\frac{J}{2r^2} \end{eqnarray}$$

Generically it has two events horizons, but much like in 3D, drops to just one when $$J = 0$$. Those horizons are the zeros of the lapse function N :

$$\begin{eqnarray} r_\pm^2 = \frac{l^2}{2} [M \pm \sqrt{M^2 - \frac{J^2}{l^2}}] \end{eqnarray}$$

If you want more details on the BTZ black holes, "Quantum gravity in 2+1 dimensions" is the reference I use, but check the references on the wiki page Rennie posted.

1+1 dimensions :

Things get a bit more complicated in 1+1 dimensions because of the Gauss-Bonnet theorem, which roughly states that the Hilbert action only depends on the topology of spacetime, hence matter has no influence on spacetime (unless a cosmological term is added, but then it is just $$T_{ab} = \Lambda g_{ab}$$).

A toy model commonly used in this case is

$$\begin{eqnarray} R - \Lambda = T \end{eqnarray}$$

It possess a black hole solution in the form

$$\begin{eqnarray} ds^2 = -(a^2 r^2 - \frac{b}{ar}) dt^2 + \frac{1}{a^2 r^2 - \frac{b}{ar}}dr 2 \end{eqnarray}$$

• Why did this get downvoted? It seems like a valiant attempt to summarise a very complicated subject. – John Rennie Sep 25 '15 at 14:39

As the comments and other answer note, this solution does not exist in the same form in a lower dimensional spacetime. The thing that you probably want to do instead is exploit the symmetry in this spacetime so that you can think about just the parts that have "interesting" behavior. Schwarzschild is spherically symmetric, so that would mean, in this case, focusing on just the characteristics the depend on $r$ and $t$. This is a 2D subspace where you can draw some pictures and build some intuition.