# How to find that there is a conical singularity in the BTZ black hole?

Considering a non-rotating and non-charged 2+1 dimensional black hole, known as the BTZ black hole which obtained by adding a negative cosmological constant $$\Lambda=-\frac{1}{l^2},l\ne0$$ to the Einstein-Hilbert action, resulting the following metric:

$$ds^2=-\left(\frac{r^2}{l^2}-M\right)dt^2+\left(\frac{r^2}{l^2}-M\right)^{-1}dr^2+r^2d\theta^2,\quad \theta\in[0,2\pi]$$

Where $$M$$ is the black holes mass. There is one horizon to this black hole given as a singularity in the metric in the radius $$r=\sqrt{M}l$$ But this is not a real curvature singularity of the system as one can make a coordinate transformation in which the horizon is no longer constitute singularity as in Schwarzschild black hole. In order to find intrinsic curvature singularity one needs to appeal to the Kretschmann scalar of the metric given by: $$K = R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$$ where $$R_{\mu\nu\rho\sigma}$$ is the Riemann tensor of the metric and see whenever it have a singularity, its singularity is in particular the real intrinsic curvature singularity of the metric, according to BTZ black hole the Kretschmann scalar is: $$K=-\frac{6}{l^4}$$ which doesn't depends on $$r$$ at all, thus didn't have a curvature singularity.

However, in many sources the BTZ black hole have another kind of singularity known as conical singularity, is there is some parallel test or operator which given a metric and in particular BTZ black hole, can reveal whenever the metric contain a conical singularity?

Conical singularities mean there exist an angular deficit in spacetime. Imagine that you have a 2-d space (as an example take a piece of paper and cut of a wedge, so you obtain a cone if you glue the two sides) and you remove a piece of this 2-d space. As a result, your coordinates that used to descibe your old 2-d space will assume a smaller domain, and now if you try to descibe this new 2-d space with polar coordinates your angular coordinate will also have a reduced domain, due to the cut. So

$$ds^2_{old} = dr^2 + r^2d\theta^2$$

and

$$ds^2_{new} = d\rho^2 + \rho^2 \left(1-\frac{x}{2\pi}\right) d\theta^2$$

where $$x$$ will be the angle deficit.

In the BTZ case, if you perform the coordinate transformation $$t = \tau\sqrt{M}$$ and $$r=\rho/\sqrt{M}$$ you obtain

$$ds^2 = - \left(\frac{\rho^2}{l^2} - M^2\right)d\tau^2 + \left(\frac{\rho^2}{M^2l^2} - 1\right)^{-1} d\rho^2 + \rho^2 \left(1-\frac{\delta \theta}{2\pi}\right) d\theta^2$$

so there exists an angle deficit of $$\delta \theta = (2 (-1 + M) π)/M$$ (measured in radians).

Here are some useful references:

Identifying conical singularities

Cosmic strings and black holes

Angular Deficit of a Conical Singularity

Gravitational field of a global monopole

• Except $ρ$ is the timelike coordinate at $ρ\to 0$ (since this is inside the black hole) so the term “angle deficit” is not applicable here. Commented Mar 24 at 5:04
• @A.V.S. it actually is, but it is not apparent for the coordinate based treatment here. To show that it is an angle deficit, one would need to calculate the holonomy of a closed loop around the singularity. Commented Mar 24 at 13:28
• @TimRias This is a Misner space type singularity. The loop around singularity has no closed loop reference in the (unfactored) AdS to conclude that there is a (finite) deficit. Commented Mar 24 at 16:29
• @TimRias … and anyway that holonomy would produce not angle but rapidity. Commented Mar 24 at 16:34
• @A.V.S. Your absolutely right it is in fact a boost. Commented Mar 25 at 11:25