# BTZ Black Hole Central Charge and Conformal Weight

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I have been trying to reproduce a calculation (equation 4.12) in this paper http://arxiv.org/pdf/1107.2678v1.pdf by Carlip reviewing the derivation of the effective central charge of the BTZ Black Hole so that one can use the microcanonical form of the Cardy formula for the entropy. He is evaluating the boundary variation of the Hamiltonian, which for this case is given by:

$\delta H = -\frac{1}{16\pi G} \int d\phi(rn^r\sigma^{\phi \phi}[N\xi^t(D_{\phi}\delta q_{r\phi} - D_r\delta q_{\phi \phi}) + (D_rN\xi^t) \delta q_{\phi\phi} + 2\xi^{\phi} \delta \pi^r_{\phi})$

Here $q_{ij}$ is the spatial metric, $D_i$ is the spatial covariant derivative, $n^r$ is the normal to the boundary ($n^r = N$ where $N$ is the lapse for this case) and $\sigma^{\phi \phi} = r^{-2}$ is the inverse of the induced metric on the boundary. The boundary is at spatial infinity and the boundary conditions are

$g_{tt} = -\frac{r^2}{l^2}$ + order one terms

$g_{rr} = \frac{l^2}{r^2}$ + order $r^{-4}$

$g_{\phi \phi} = r^2$ + order one

$g_{t \phi} =$ order one

The rest of the terms of the metric are required to be on order $r^{-3}$. Carlip finds that the above expression for the variation gives:

$\delta H = -\frac{1}{16\pi G} \int d\phi(\xi^t\delta(\frac{r_+^2 + r_-^2}{l^2}) + \xi^{\phi} \delta (\frac{2r_-r_+}{l}))$

I can easily see where the second term comes from because $\pi^r_{\phi} = \frac{r_-r_+}{l}$ which is also proportional to the angular momentum $J$ but I cannot see how he gets $\frac{r_+^2 + r_-^2}{l^2}$, which is the mass of the black hole. Also, in the non-asymptotic limit the lapse is:

$N = (-8GM + \frac{r^2}{l^2} + \frac{16G^2J^2}{r^2})^{1/2}$

Now I understand why the mass should be in the above variation of the Hamiltonian but I still can't seem to derive it. Any help on seeing where that term comes from would be very helpful!

This post imported from StackExchange Physics at 2014-08-08 06:31 (UCT), posted by SE-user Confused_Grad_Student
asked Aug 8, 2014
Could you reformulate that as a precise question? "I cannot derive a formula" is borderline off-topic as homework and/or unclear what you are asking. Which step is it that troubles you? What are the equations you are talking about (not many people are going to crawl through the paper for that)?

This post imported from StackExchange Physics at 2014-08-08 06:32 (UCT), posted by SE-user ACuriousMind
I feel your pain, I am only in undergraduate and I have encountered many papers that skip a derivation and say "it is quite simple to see."

This post imported from StackExchange Physics at 2014-08-08 06:32 (UCT), posted by SE-user NaturalPhilosopy
@ACuriousMind Sure thing, I've added more information, hopefully that will be enough.

This post imported from StackExchange Physics at 2014-08-08 06:32 (UCT), posted by SE-user Confused_Grad_Student

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