Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  BTZ Black Hole Central Charge and Conformal Weight

+ 5 like - 0 dislike
2448 views

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 7, 2014 in Theoretical Physics by Confused_Grad_Student (25 points) [ no revision ]
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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...