Euclidean Methods for BTZ black Hole

[craigthone]

This is an exercise from Hartman's lecture 6th. Using the Euclidean method to calculate the BTZ black hole mass entropy. The BTZ metric is given by
$$ ds^2=(r^2-8M)d\tau^2 +\frac{dr^2}{r^2-8M}+r^2d\phi^2$$
and $\tau \sim \tau+\beta, \beta=\frac{\pi}{\sqrt{2M}}$.
Then we calculate the Euclidean action
$$ S_E=-\frac{1}{16π}\int_M \sqrt{g}(R+2)-\frac {1} {8π} \int_{\partial M} \sqrt { h } K+\frac {a} {8π} \int_{\partial M} \sqrt { h } $$

For the BTZ black hole solution, we can calculate
$$\sqrt{g}=r, \sqrt{h}=r \sqrt{r^2-8M}$$
$$ n_{\alpha}=(r^2-8M)^{-1/2}\partial _{\alpha}r$$
$$R=-6, K=\frac{\sqrt{r^2-8M}}{r}+\frac{r}{\sqrt{r^2-8M}}$$
And then we have
$$ -\frac{1}{16π}\int_M \sqrt{g}(R+2)=\frac{\beta}{4}r^2_0$$
$$-\frac {1} {8π} \int_{\partial M} \sqrt { h } K=-\frac{\beta}{4}[2r^2_0 -8M ] $$
$$\frac {a} {8π} \int_{\partial M} \sqrt { h }=a \frac{\beta}{4}[2r^2_0 -4M ] $$
where the boundary is at $r=r_0$
In order to cancel the divergent part of the action, we take $a=1$.
Then the Euclidean action is
$$S_E=\beta M=\frac{\pi^2}{2\beta}$$
The black hole energy is $$E=\frac{\partial}{\partial \beta}S_E=-M $$
This is awkard since we know that $E=M$ for the black hole.

Who can help me out. Thanks in advanced.

 

[mark726]

Hello,

I would be happy to help you with this exercise. First, let me explain the steps you have taken so far.

You have correctly identified the BTZ black hole metric and the periodicity of the Euclidean time coordinate, which is necessary for calculating the black hole mass entropy. Then, you have used the Euclidean action formula for gravity, which includes the Einstein-Hilbert term, the extrinsic curvature term, and the Gibbons-Hawking term. These terms are necessary for a well-defined variational principle and to avoid boundary divergences. You have also correctly calculated the values of the metric, the extrinsic curvature, and the Ricci scalar for the BTZ black hole solution, and substituted them into the action formula.

However, there seems to be a mistake in your calculation of the Euclidean action. The correct expression for the action should be:
$$S_E = \frac{\beta}{4}r_0^2 - \frac{\beta}{4}(2r_0^2 - 8M) + \frac{\beta}{4}(2r_0^2 - 4M) = \frac{\pi^2}{2\beta}$$
This is because the extrinsic curvature term should have a minus sign, and the Gibbons-Hawking term should have a plus sign. This mistake leads to the incorrect result for the black hole energy, which should be:
$$E = \frac{\partial}{\partial \beta}S_E = M$$
This is consistent with the fact that the black hole energy is equal to its mass. So, you were on the right track, but just made a small error in your calculation.

I hope this helps! Let me know if you have any further questions or if you need any clarification.