# The energy of dual boundary field in AdS/CFT

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.

• 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, $\xi = \partial_t$. – Prahar Mar 27 '16 at 13:32

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.