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Finding superpotentials and central charges in $AdS_3$

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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

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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 (950 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

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