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I am trying to construct the Penrose diagram of a black hole in AdS space. Now, I thought I was on a good track, my diagram looked like this:

enter image description here

The grey lines are the surfaces of constant $t$ and constant $r$ of the coordinate chart in which the metric reads

$$g = - f(r) dt^2 + f(r)^{-1} dr^2 \quad , \qquad f(r) = 1 - 2m/r + r^2.$$

Now this looks good so far. However, according to my calculations, a surface of constant r within the event horizon looks like the blue line.

enter image description here

This seems awfully wrong. I cannot pinpoint the error in my calculations, so please have a look at them. I proceed as follows:


1. Tortoise coordinate: Calculate the tortoise coordinate, which is defined via the equation $dr^{\ast} = f(r)^{-1} dr$.

$$r^{\ast}(r) = \frac{-1}{f'(r_+)} \bigg( \ln \sqrt{1 + \frac{3rr_+ + 1}{(r - r_+)^2}} - \frac{A^2-2}{A} \arctan \frac{2r + r_+}{A} + \mathrm{const} \bigg)$$

with $r_+$ being the event horizon and with $A = \sqrt{r_+ f'(r_+) + 3}$.


2. Eddington-Finkelstein coordinates: Define $u = t - r^{\ast}$ and $v = t + r^{\ast}$, such that

$$g = - f dudv$$

This metric is degenerate at the event horizon, so another coordinate change is in order:


3. Another coordinate change: Define $\hat{u} = - \operatorname{sgn}(f) e^{- \eta u}$ and $\hat{v} = e^{\eta v}$, where $\eta$ is chosen such that the degeneracy vanishes. Choose $\eta = f'(r_+)/2$, the metric then is

$$ g = - \operatorname{sgn}(f) \frac{f}{\eta^2} e^{-2\eta r^{\ast}} d\hat{u} d\hat{v} $$


4. Limits: The value of the constant in $r^{\ast}$ is still unspecified. At the event horizon, the $r^{\ast}$ diverges and so the constant may be chosen for both sides of the event horizon independently (denote $C_{1,2}$):

$$ \lim_{r \rightarrow \infty} \hat{u} \hat{v} \overset{!}{=} -1 \quad \Longrightarrow \quad C_1 = \frac{A^2 - 2}{A} \frac{\pi}{2}$$

$$\lim_{r \rightarrow 0} \hat{u} \hat{v} \overset{!}{=} 1 \quad \Longrightarrow \quad C_2 = -\ln \sqrt{ 1 + \frac{1}{r_+^2}} + \frac{A^2-2}{A} \arctan \frac{r_+}{A} $$


5. Compactify: Define $\hat{\hat{u}} = \arctan \hat{u}$ and $\hat{\hat{v}} = \arctan \hat{v}$, then

$$g = - \operatorname{sgn}(f) \frac{f}{\eta^2} \frac{e^{-2\eta r^{\ast}}}{\cos^2 \hat{\hat{u}} \cos^2 \hat{\hat{v}}} d\hat{\hat{u}} d\hat{\hat{v}}$$


6. Rotate: Define $\hat{t} = (\hat{\hat{u}} + \hat{\hat{v}})/2$ and $\hat{x} = -(\hat{\hat{u}} - \hat{\hat{v}})/2$. The metric is

$$g = \operatorname{sgn}(f) \frac{f}{\eta^2} \frac{e^{-2\eta r^{\ast}}}{\cos^2 (\hat{t}-\hat{x}) \cos^2 (\hat{t}+\hat{x})} \big( - d\hat{t} + d\hat{x} \big)$$


7. Finally: The surfaces of constant $r$ (or $t$, respectively), are the parametrised curves such that

$$\boxed{\hat{t} = \frac{1}{2} \Big( \arctan \big[ - \operatorname{sgn}(f) e^{-\eta (t - r^{\ast})} \big] + \arctan \big[ e^{\eta(t + r^{\ast})} \big] \Big) \Big|_{r=const.}}$$

$$\boxed{\hat{x} = \frac{-1}{2} \Big( \arctan \big[ - \operatorname{sgn}(f) e^{-\eta (t - r^{\ast})} \big] - \arctan \big[ e^{\eta(t + r^{\ast})} \big]\Big) \Big|_{r=const.}}$$

However, these equations produce the blue curves that seem so wrong. Any ideas where I have gone wrong?

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  • $\begingroup$ Set Λ=0 and see if the Schwarzschild limit produces the known results, if not we'll take a further look at it. The asymmetry alone is not concerning, since in negative space r<0 the sign of M and Λ is effectively flipped. $\endgroup$
    – Yukterez
    Commented Nov 21, 2023 at 17:48
  • $\begingroup$ Thank you very much for your comment/hint. I reformulated my problem, such that the limits you mentioned are easily calculated, and by this reformulation I was able to solve the initial problem myself. I posted the (corrected) approach as an answer, credit, however, goes to you. $\endgroup$
    – Octavius
    Commented Nov 26, 2023 at 19:26

1 Answer 1

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Yukterez' comment brought me on the right track to figuring out the problem myself. Below, the correction to the construction of my Penrose diagram is explained:


In Step 1., where the tortoise coordinate is calculated, use instead a different, but - if I am not mistaken - equivalent form of $r^{\ast}$, namely the one also used in [1] (see footnote). It is

$$ r^{\ast}(r) = c_0 + c_1 \bigg[ \ln \Big| 1 - \frac{r}{r_+} \Big| - \ln \sqrt{1 + \frac{r(r+r_+)}{r_+^2 + a^2}} + c_2 \arctan \Big( \frac{r \sqrt{3r_+^2 +4a^2}}{2r_+^2 + 2a^2 + rr_+} \Big) \bigg]$$

with $a$ being the AdS radius and where

$$c_1 = \frac{r_+a^2}{3r_+^2 + a^2} \quad , \qquad c_2 = \frac{3r_+^2 + 2a^2}{r_+ \sqrt{3r_+^2 + 4a^2}} .$$

The constant $c_0$ is yet not specified. Looking at the Schwarzschild limit $a \rightarrow \infty$, as Yukterez suggested in his comment, produces the correct and expected Schwarzschild result up to the constant $c_0$. Then, with the exception of Step 4, all steps explained in the initial question can be carried out. Step 4 itself produces the limits according to what constant $c_0$ has been chosen. So, what value of $c_0$ is suitable? Consider the following possibilities:

$\bullet$ Choosing $c_0 = 0$ produces a Penrose diagram as can be seen in Figure A.

$\bullet$ Choosing $c_0$ such that $r^{\ast}(\infty) = 0$ produces the Penrose diagram depicted in Figure B.

$\bullet$ Choosing $c_0$ such that $r^{\ast}$ coincides with the expression in the initial question, probably produces the result shown in the initial question - although I did not recalculate it to check.

enter image description here

The surfaces of constant $t$ and $r$ are symmetric and as expected; everything looks in order. Special thanks to Yukterez for his hint.


[1] M. Socolovsky, Schwarzschild Black Hole in Anti-De Sitter Space, Advances in Applied Clifford Algebras, 28 (2018).

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