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 ...

  Finding superpotentials and central charges in $AdS_3$

+ 2 like - 0 dislike
1725 views

In text "Covariant theory of asymptotic symmetries, conservation laws and central charges" is given an example of finding central charges and superpotential (among other things).

I am interested in $AdS_3$ case, since there is a lot of literature on this space-time, and I'm going to need to preform the same analysis for 4 dimensional Near-Horizon Extremal Kerr (NHEK) metric, so I would like to know how to do that in a simpler case.

In a given example, on page 48, after specifying boundary conditions, they go on to find the linear part of $\delta[(1/16\pi)\sqrt{-g}(R-2\Lambda)/\delta g_{\mu\nu}]$, which is, as far as I understand it, a kind of variation of Lagrangian.

The formula is given:

$\mathcal{H}^{\mu\nu}[h;\bar{g}]:=\frac{\sqrt{-g}}{32\pi}\left[-h\bar{R}^{\mu\nu}+\frac{1}{2}h\bar{R}\bar{g}^{\mu\nu}+2h^{\mu\alpha}\bar{R}_\alpha^\nu+2h^{\nu\beta}\bar{R}_\beta^\mu-h^{\mu\nu}\bar{R}-h^{\alpha\beta}\bar{R}_{\alpha\beta}\bar{g}^{\mu\nu}+\bar{D}^\mu\bar{D}^\nu h+\bar{D}^\lambda\bar{D}_\lambda h^{\mu\nu}-2\bar{D}_\lambda\bar{D}^{(\mu}h^{\nu)\lambda}-\bar{g}^{\mu\nu}(\bar{D}^\lambda\bar{D}_\lambda h-\bar{D}_\lambda\bar{D}_\rho h^{\rho\lambda})+2\Lambda h^{\mu\nu}-\Lambda \bar{g}^{\mu\nu}h\right]$

I also have the background metric ($\bar{g}_{\mu\nu}$) which is that of $AdS_3$, $\bar{D}$ is covariant derivative, $h$ is trace, given by $h=\bar{g}^{\mu\nu}h_{\mu\nu}$. Ricci tensor and scalar are known.

Now, what confuses me is: how did they get the results (for example $\mathcal{H}^{tt}\to\mathcal{O}(r^-{3})$)? I don't understand this, since they only give the boundary conditions as leading orders in $r$ ($h_{\mu\nu}\to \mathcal{O}(r^{m})$, $m\in\mathbb{Z}$). I can raise and lower indices, I know about summation, and I can find out what terms need to be present in this expression.

But how do I preform calculation with $\mathcal{O}$ notation?

I've been baffled by this every time I read similar articles. They all do these calculations, but I cannot find a single example where everything is explained in detail :\

So any help on clarifying this is welcomed. Any mathematics books that explain this or something...

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user dingo_d
asked Sep 12, 2013 in Theoretical Physics by dingo_d (110 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

You have : $(6.37, 6.38)$:

$\bar R_{tt} \sim \bar R_{\theta\theta} \sim \bar g_{tt} \sim \bar g_{\theta\theta} \sim r^{2},\bar R_{rr}\sim \bar g_{rr}\sim r^{-2} \tag{1}$

$\bar R^{tt} \sim \bar R^{\theta\theta} \sim \bar g^{tt} \sim \bar g^{\theta\theta} \sim r^{-2},\bar R^{rr}\sim \bar g^{rr}\sim r^{2} \tag{2}$

$h \sim r^{-2}, \bar g \sim r^2, \bar R \sim \bar R^\alpha_\beta \sim r^0, \tag{3}$

$h^{tt} \sim h^{\theta\theta} \sim r^{-4}, h^{rr}\sim r^{0}\tag{4}$

$\bar D_r \sim r^{-1}, \bar D_r \bar D_r \sim r^{-2}, \bar g^{rr}\bar D_r \bar D_r \sim r^{0}\tag{5}$

By "$\sim$" applyed to $h$ terms, I indicated the worst possible $r$ dimension.

Looking for instance at $\mathcal{H}^{tt}$, we have typical terms :

$\mathcal{H}^{tt} \sim \sqrt{\bar g}(h \bar R^{tt}+...)$

So, you have : ${H}^{tt} \sim (r^1)(r^{-2})(r^{-2})\sim (r^{-3})\tag{6}$ (You could check that the terms $...$ have the same worst dimension ($r^{-4}$))

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Trimok
answered Sep 12, 2013 by Trimok (955 points) [ no revision ]
Oh, so there's no real 'calculation' with that. Just seeing the general dependence on $r$?

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user dingo_d
I am interested in this, because they do the same calculation for superpotentials, which are used to calculate the central charge. For instance the derivatives $\bar{D}_\sigma(h^{\mu\sigma}\xi^\nu)$ Do they leave that $h_{\mu\nu}$ until the very end?

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user dingo_d
Oh, and how did you get $\bar{D}_r\sim r^{-1}$? To what did you apply it? I tried with $h$ and got $\bar{D}_r h\sim r^{-3}$, and $\bar{D}_r h_{rr}\sim r^{-5}$? Is it because of $\bar{g}^{rr}\bar{D}_r\bar{D}_r\sim r^0$, and $\bar{g}^{rr}\sim r^2$? So that means that $\bar{g}^{tt}\bar{D}_t\bar{D}_t\sim r^{0}\Rightarrow \bar{D}_t\bar{D}_t\sim r^2$?

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user dingo_d
@dingo_d : Note that $\tilde D_r$ involves standard derivative $\partial_r (\sim r^{-1})$ and $\Gamma_{rx}^y$ also of order $\sim r^{-1}$ $(6.36)$. So $\tilde D_r \sim r^{-1}$. With $\tilde D_t$ or $\tilde D_\theta$, this is more subtle because the $\Gamma_{tx}^y$ and $\Gamma_{\theta x}^y$ have not homogeneous dimension in $r$ $(6.36)$. However we can compute, for instance $\bar{g}^{tt}\bar{D}_t\bar{D}_t h^{tt}$, we find terms as $\bar{g}^{tt}\Gamma_{tt}^r \partial_rh^{tt}, \bar{g}^{tt}\Gamma_{tt}^r\Gamma_{tr}^th^{tt},\bar{g}^{tt}\Gamma_{tr}^t\Gamma_{tr‌​}^t h^{rr}$, all have order $r^{-4}$

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Trimok
@dingo-d : You have a term $\bar{g}^{tt}\Gamma_{t\theta}^t\Gamma_{t\theta​}^t h^{\theta \theta}$ also, in the above list.

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Trimok
I'll try to do the whole calculation by hand and see if I get the same result :) Thanks for guidance :)

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user dingo_d
@dingo_d : BTW, see the Wikipedia about the big O notation

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Trimok

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$\hbar$ysi$\varnothing$sOverflow
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
...