In AdS/CFT, when the spacetime is a planar AdS black hole with dimension ($d+1$), the corresponding energy of boundary field theory is proportional to the black hole mass parameter. For example when $d=4$ the metric is $$ ds^2=-f(r)dt^2+dr^2/f(r)+r^2(dx^2+dy^2) $$ with $$ f(r)=\frac{r^2}{l^2}-\frac{M}{r} $$ with AdS radius $l$. Then the energy of boundary field theory is proportional to the parameter $M$. My question is that if the boundary is not planar, but of the sphere topology, for example the metric takes the form $$ ds^2=-g(r)dt^2+dr^2/g(r)+r^2(d\theta^2+sin^2\theta d\phi^2) $$ with $$ g(r)=1+\frac{r^2}{l^2}-\frac{M}{r} $$ Is the energy of boundary field still proportional to $M$ in $g(r)$? Can anyone help to answer this question or give some references.
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1$\begingroup$ I don't know the answer of the top of my head, but I can tell you how to compute it. You'll have to compute the Brown-York stress tensor, $T_{ij}$. Energy is defined then as $E \sim \int n^i \xi^j T_{ij}$ where the integral is over the boundary, $n^i$ is the unit normal, and $\xi^i$ is a global time-like Killing vector w.r.t. which energy is being defined. In the case here, $\xi = \partial_t$. $\endgroup$– PraharCommented Mar 27, 2016 at 13:32
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So whether or not the parameter you're calling M is the mass depends on dimension. In 4D it is proportional to the mass, but in 5D for example it is not. And in general, the question of what is the mass of a given asymptotically AdS spacetime cannot be just extract from the subleading term in the $g_{tt}$ component.
If you were to compute the Brown-York stress tensor directly, as Prahar suggested, it would be divergent. So one must have some way of regulating the divergences and extracting a physical result. There are many methods for doing this, I think that this one is the most straightforward (http://arxiv.org/abs/hep-th/9902121), although be ware, they use some non-standard conventions for the curvatures that result in some sign flips.
answered Mar 27, 2016 at 14:01
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$\begingroup$ of course the BY stress tensor only makes sense if the appropriate counter-term action is added. $\endgroup$– PraharCommented Mar 27, 2016 at 14:17
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$\begingroup$ Well in the case of AdS boundary conditions, that' s where all the meat of the problem is and they work it out in the paper I cited. $\endgroup$ Commented Mar 27, 2016 at 14:28