- #1
craigthone
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I know this is some kind of exercise problem, but it isnot widely discussed in general general relativity textbook. Sorry to post it here.
I want to calculate the mass and entropy of non-rotating BTZ black hole using Euclidean method. When I calculate the Euclidean action, I always get an extra minus sign. I think the claculation is standard. Who can help me out? Thanks in advanced.
The BTZ metric $$ds^2=(r^2-8M)d\tau^2+\frac{dr^2}{r^2-8M}+r^2d\phi^2$$
where ##\tau\sim \tau+ \beta , \beta=\frac{\pi}{\sqrt{2M}}##
The Euclidean action for BTZ is $$S_E=-\frac{1}{16\pi} \int_M \sqrt{r}(R+2) -\frac{1}{8\pi} \int_{\partial M} \sqrt{h} K +\frac{a}{8\pi} \int_{\partial M} \sqrt{h}$$
For the BTZ black hole solution, we have
$$\sqrt{g}=r, \sqrt{h}=r\sqrt{r^2-8M}$$
$$ n_{\alpha}=(r^2-8M)^{-1/2}\partial_{\alpha} r $$
$$ R=-6, K=\frac{r}{\sqrt{r^2-8M}}+\frac{\sqrt{r^2-8M}}{r} $$
Then we have
$$ -\frac{1}{16\pi} \int_M \sqrt{r}(R+2) =\frac{\beta}{4}r^2_0$$
$$-\frac{1}{8\pi} \int_{\partial M} \sqrt{h} K=-\frac{\beta}{4}[2r^2_0-8M^2]$$
$$ \frac{a}{8\pi} \int_{\partial M} \sqrt{h}=a\frac{\beta}{4}[2r^2_0-4M^2] $$
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}##, and 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.
I want to calculate the mass and entropy of non-rotating BTZ black hole using Euclidean method. When I calculate the Euclidean action, I always get an extra minus sign. I think the claculation is standard. Who can help me out? Thanks in advanced.
The BTZ metric $$ds^2=(r^2-8M)d\tau^2+\frac{dr^2}{r^2-8M}+r^2d\phi^2$$
where ##\tau\sim \tau+ \beta , \beta=\frac{\pi}{\sqrt{2M}}##
The Euclidean action for BTZ is $$S_E=-\frac{1}{16\pi} \int_M \sqrt{r}(R+2) -\frac{1}{8\pi} \int_{\partial M} \sqrt{h} K +\frac{a}{8\pi} \int_{\partial M} \sqrt{h}$$
For the BTZ black hole solution, we have
$$\sqrt{g}=r, \sqrt{h}=r\sqrt{r^2-8M}$$
$$ n_{\alpha}=(r^2-8M)^{-1/2}\partial_{\alpha} r $$
$$ R=-6, K=\frac{r}{\sqrt{r^2-8M}}+\frac{\sqrt{r^2-8M}}{r} $$
Then we have
$$ -\frac{1}{16\pi} \int_M \sqrt{r}(R+2) =\frac{\beta}{4}r^2_0$$
$$-\frac{1}{8\pi} \int_{\partial M} \sqrt{h} K=-\frac{\beta}{4}[2r^2_0-8M^2]$$
$$ \frac{a}{8\pi} \int_{\partial M} \sqrt{h}=a\frac{\beta}{4}[2r^2_0-4M^2] $$
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}##, and 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.