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Euclidean Methods for BTZ black Hole

  1. Aug 16, 2017 #1
    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.

     
  2. jcsd
  3. Aug 21, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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