I'm currently working in 1+1 dimensional spacetime and would like to know if black holes can exist in such a manifold? I think they can because the Schwarzchild metric has the coordinate singularity, not to mention the actual singularity at $r=0$, attached to the $g_{tt}$ and the $g_{rr}$ metric components. However, as the radius is usually oriented in 3 dimensional space, I'm unsure...
$\begingroup$
$\endgroup$
5
-
$\begingroup$ If you take the Schw BH with Schw coordinates with theta and phi equal to constants, then the metric reduces to being in terms of radial and time coordinates, implying a 2D manifold with a BH. Maybe you want to ask a more specific question? $\endgroup$– Daddy KropotkinCommented Dec 30, 2020 at 23:40
-
3$\begingroup$ @Daddy Kropotkin. I do not think that this holds. One has to check. There is no schwarzchild solution in three dimensions for example. The vanishing of Ricci tensor and the by definition vanishing of the weyl tensor mean that no geometry can be formed. Since weyl is again zero in two dimensions and we look for a vaccum solution the Ricci tensor will be again zero meaning that no geometry can be formed. Anyway, just google search the subject. There are black hole solutions in two dimensions in the dilaton gravity theory. $\endgroup$– NooneCommented Dec 30, 2020 at 23:51
-
$\begingroup$ Thanks for the tip. $\endgroup$– PrawwarPCommented Dec 30, 2020 at 23:56
-
1$\begingroup$ @ApolloRa in 2d, the Ricci tensor doesn't need to vanish for the vacuum solution. See my answer for details. $\endgroup$– Prof. LegolasovCommented Dec 31, 2020 at 9:27
-
$\begingroup$ @ PrawwarP, On the existence of 1+1 dimensional black holes: in fact, the 2-dimensional dilaton gravity theories admits black hole (black point) solutions. Please see the following links :[1]sciencedirect.com/science/article/abs/pii/055032139090265F [2]sciencedirect.com/science/article/abs/pii/0550321394902461 A brief review of these (and perhaps a better reference for you) is this one: [3]arxiv.org/abs/1509.05481 $\endgroup$– SG8Commented Mar 4, 2021 at 18:32
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Black holes are solutions of vacuum EFE on a space-time with singularities.
EFE in the vacuum are: $$ G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = 0. $$
These are sometimes written as $$ R_{\mu \nu} = 0, $$ which is completely equivalent in spacetimes of all dimensionalities except for $d = 2$.
This is easy to see. Write down the contraction of the EFE: $$ 0 = G_{\mu \nu} g^{\mu \nu} = R - \frac{1}{2} R d = R \left( 1 - \frac{d}{2} \right) $$ implies that either $R = R_{\mu \nu} = 0$, or $d = 2$. In the latter case there are no restrictions on $R$ whatsoever.
Any 2-dimensional spacetime is a solution of vacuum EFE in 2 dimensions, because that vacuum EFE is an identity, not an equation.
This can be also seen by noticing that by Gauss-Bonnet theorem, the Einstein-Hilbert action in 2 dimensions is a topological invariant proportional to the Euler characterestic $$ S_{EH} = \frac{1}{16 \pi G} \intop_{\mathcal{M}} d^2 x \sqrt{| \det g |} R = \frac{1}{8 G} \chi (\mathcal{M}) $$ that doesn't depend on the metric at all. Its variations w.r.t. the metric are all trivial and no constraining equation on the metric can be obtained via the least action principle.
The conclusion is that in 1+1 dimensions, any Lorentzian metric is a solution to the EFE. You can write down metrics that look like black holes, but those won't be predictions of the theory, rather stuff you put in by hand.
GR looks very different in $d = 2$. Interestingly, $d = 3$ is also a very special case for GR (ask a separate question about this if you want an expanded answer). Only starting with $d = 4$ does GR look like a theory of the gravitational interaction.
answered Dec 31, 2020 at 1:29