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.


1 Answer 1


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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