Five dimensional empty AdS$_5$ space has mass $$ E = \frac{3 \pi \ell^2}{32 G}. $$

Is the above equation correct?

Let's do some dimensional analysis to confirm. In natural units, in 5 dimensions $[G] = -3$ where $[...]$ is the mass dimension. Also $[\ell]=-1$. Therefore $\left[ \frac{\ell^2}{G} \right] = 1$. So the dimensions seem to work out fine.

Here's my second question:

The limit of $\ell \to \infty$ of $AdS_5$ space is flat space. Isn't it weird that the mass diverges in this limit? I would have presumed that mass should vanish in this limit, since flat space has vanishing mass? Are we using two different definitions of mass?

EDIT: Due to some requests in the comments, I will include the derivation of the formula above.

I use the boundary stress tensor given by Brown-York (derived from the Einstein action alongwith the Gibbons-Hawking boundary term. $$ t_{ij} = \frac{1}{8\pi G} \left[ K_{ij} - \gamma_{ij} K + \frac{2}{\sqrt{-\gamma}} \frac{\delta S_{ct}}{\delta \gamma^{ij}} \right] $$ Here $K_{ij} = \nabla_{(i} n_{j)}$ is the extrinsic curvature. $n^\mu$ is the unit normal vector to the boundary. $\gamma$ is the boundary metric and $K = \gamma^{ij} K_{ij} $. $S_{ct}$ is the counterterm action included to make all the B-Y charges finite, which are defined as $$ Q_\xi = \int_{\cal B} d^d x \sqrt{\sigma} u^i \xi^j t_{ij} $$ Here $\sigma_{ab}$ is the metric on a spatial hypersurface and $u^i$ is a time-like unit vector normal to the hypersurface. $\xi^j$ is a Killing vector of the boundary metric.

Now, for $AdS_5$, the counterterm action is given by $$ S_{ct} = -\frac{3}{\ell} \int d^4 x \sqrt{-\gamma} \left( 1 + \frac{\ell^2}{12} R(\gamma) \right) $$ The B-Y tensor is then $$ t_{ij} = \frac{1}{8\pi G} \left[ K_{ij} - \gamma_{ij} K - \frac{3}{\ell} \gamma_{ij} + \frac{\ell}{2} \left( R_{ij} - \frac{1}{2} \gamma_{ij} R \right) \right] $$ We can now work in the Fefferman-Graham coordinates for $AdS_5$ space where the metric is $$ ds^2 = \frac{\ell^2 d\rho^2}{4\rho^2} - \frac{(1+\rho)^2}{4\rho} dt^2 + \frac{\ell^2 ( 1 - \rho)^2}{4\rho} d\Omega_3^2 $$ Thus $$ t_{ij} = - \frac{ \rho }{4\ell \pi G} \left( \gamma_{ij}^{(0)} + \gamma_{ij}^{(2)} \right) + {\cal O}(\rho^2) $$ where $$ \gamma_{ij}^{(0)} dx^i dx^j = -\frac{1}{4} dt^2 + \frac{\ell^2}{4} d \Omega_3^2 $$ $$ \gamma_{ij}^{(2)} dx^i dx^j = - \frac{1}{2} dt^2 - \frac{\ell^2}{2} d \Omega_3^2 \\ $$ We also have $$ u = \frac{2 \sqrt{\rho}}{1+\rho} \partial_t,~~ \xi = \partial_t,~~\sqrt{\sigma} = \frac{\ell^3 (1 - \rho)^3 }{8 \rho^{3/2} } \sin^2\theta \sin \phi $$ Plugging all this in, we find that the B-Y charge corresponding to the Killing vector $\partial_t$ is $$ Q_t = \frac{3 \pi \ell^2 }{32 G} $$ This is where I got the formula from. I interpret this as the mass of $AdS_5$ space.

Disclaimer - I have intentionally left out several computations to reduce the length of the problem. I have not referred any paper and all computations have been done by me.


As has been shown in this paper, pointed out by Matthew in the comments, the expression found is indeed correct and can be understood from a holographic point of view. I now reproduce the argument from relevant section (number 5) of the paper:

It seems unusual from the gravitational point of view to have a mass for a solution that is a natural vacuum, but we will show that this is precisely correct from the perspective of the AdS/CFT correspondence.

We use the conversion formula to gauge variables: $$\frac{\ell^3}{G}=\frac{2N^2}{\pi}$$

Then the mass of global AdS$_5$ is $$M=\frac{3N^2}{16\ell} $$ The Yang-Mills dual of AdS$_5$ is defined on the global AdS$_5$ boundary with topology $S^3\times R$. A quantum field theory on such a manifold can have a non-vanishing vacuum energy - the Casimir effect. In the free field limit, the Casimir energy on $S^3\times R$ is: $$E_\text{cas}=\frac{1}{960r} (4n_0+17n_{1/2}+88n_1)$$ where $n_0$ is the number of real scalars, $n_{1/2}$ the number of Weyl fermions and $n_1$ the number of gauge bosons, and $r$ is the radius of $S^3$. For $SU(N)$, $\mathcal{N}=4$ super Yang-Mills $n_0=6(N^2-1)$, $n_{1/2}=4(N^2-1)$ and $n_1=N^2-1$ giving: $$E_\text{cas}=\frac{3(N^2-1)}{16r} $$ To compare with [the expression for the mass], remember that $M$ is measured with respect to coordinate time while the Casimir energy is defined with respect to proper boundary time. Converting to coordinate time by multiplying by $\sqrt{-g_{tt}}=\frac{r}{\ell}$ gives the Casimir “mass”: $$M_\text{cas}=\frac{3(N^2-1)}{16\ell} $$ In the large $N$ limit we recover the earlier expression for the mass of AdS$_5$. $$M=\frac{3\pi\ell^2}{32G}$$

This is an CW answer based on comments by other users, supplemented with the relevant results from a paper on the topic. I have written this answer to get it out of the 'unanswered' tab.


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