# AdS Space Black Holes Penrose Diagram

How do we construct the Penrose diagram for an AdS Space Schwarzschild solution?

We start with the AdS Schwarzschild Metric and then do some transformation to compactify the coordinate ranges? What are those set of coordinate transformation which are done in the case of AdS Black Holes?

How is this solution different from Schwarzschild solution in Flat Space?

The difference lies in spatial infinity: AdS like spacetimes have timelike infinity while Schwarzschild spacetime has a lightlike infinity.

In general the prescription to build Penrose diagrams of spherically symmetric spacetimes goes as:

1) Define the tortoise coordinate except for the constant;

2) Define null coordinates in the 'usual way';

3) Define new null coordinates in order to avoid the coordinate singularity in the event horizon;

4) Reparametrize the null coordinate in order to put it in a finite diagram.

In the case of AdS-Schwarzschild spacetime it is a little bit of hard work:

1) We have a metric of type $g_{\mu \nu }= (-f, f^{-1},r^2,r^2\sin^2 \theta )$ and $f = 1- \frac{2m}{r} + c^2r^2$, $c$ meaning the inverse of the AdS curvature radio. Now, as long as $f$ has always a real positive and two imaginary roots, we have \begin{eqnarray} r_* = \int \frac{1}{f}dr = \frac{(2r_+^2 + 4p)\arctan \left(\frac{r_++2r}{\sqrt{4p-r_+^2}}\right) + r_+ \sqrt{4p-r_+^2} \ln \left( \frac{(r-r_+)^2}{r^2+r_+r+p} \right)}{2c^2(2r_+^2+p)\sqrt{4p-r_+^2}} + constant \end{eqnarray} in which $r_+$ is the event horizon ($f(r_+)=0$) and $p=r_+^2+\frac{1}{c^2}= \frac{2m}{c^2r_+}$. (Assume: $constant \equiv C$).

2) The usual null-coordinates, $u$ and $v$ written as \begin{eqnarray} du=dt-dr_* \\ dv=dt+dr_* \end{eqnarray} lead the metric to \begin{eqnarray} ds^2 = -f du dv + r^2 d \Omega^2 \end{eqnarray} which is still singular in $r=r_+$ as long as $f=0$ there.

3) Defining now \begin{eqnarray} \tilde{U}=-e^{-\eta u}\\ \tilde{V}=e^{\eta v}, \end{eqnarray} we have \begin{eqnarray} ds^2 = \frac{-f}{\eta^2}e^{-2\eta r_*}d \tilde{U} d\tilde{V} +r^2 d\Omega^2 \end{eqnarray} (being $\tilde{U}$ and $\tilde{V}$, also null). If $\eta = \frac{c^2(2r_+^2+p)}{2r_+}$, the coordinate singularity is avoided.

With these new coordinates, and the proper choice of $C$ (in blocks) we have, \begin{eqnarray} \lim_{r\rightarrow \infty} \tilde{U} \tilde{V} = -1 \\ \lim_{r\rightarrow r_+} \tilde{U} \tilde{V} = 0 \\ \lim_{r\rightarrow 0} \tilde{U} \tilde{V} = 1. \end{eqnarray}

4) Finally, define \begin{eqnarray} U = \arctan \tilde{U}\\ V = \arctan \tilde{V} \end{eqnarray}

in which case, the maximal spacetime is placed in a square of size $\pi / \sqrt{2}$. To see this, we have to take the 3 limits above, considering the coordinates $U$ and $V$ as in the file bellow.