How is the mass of the black hole defined in a asymptotcally AdS solution of the black hole? How can I find it? Beacause in asymptotcally flat solution I can read it from the $g_{tt}$ component of the metric.

There are two complementary correspondences to black holes. One is a metric where far from the black hole horizon there is $AdS_3$ spacetime. The other occurs quite oppositely where the $AdS_2\times S^2$ spacetime occurs in the near horizon condition of an extremal black hole.

For the asymptotic condition, usually seen as the AdS black hole one simply has the metric $$ds^2 = -\left(1 - \frac{2m}{r} + a^2r^2\right)dt^2 + \left(1 - \frac{2m}{r} + a^2r^2\right)^{-1}dr^2 + r^2d\Omega^2$$ In the limit the radius $r$ is very large this reduces to $$ds^2 = -r^2(a^2dt^2 - d\Omega^2),$$ that is the metric for $AdS_3$.

It is worth noting what I see as the complementary relationship between black holes and $AdS$ spacetime. The condition for an accelerated observer near the horizon is given by a constant radial distance. We then considerable $$\rho = \int dr \sqrt{g_{rr}} = \int \frac{dr}{\sqrt{1 - 2m/r + Q^2/r^2}}$$ with low and upper limits on integration $r_+$ and $r$. The result is $$\rho = m log[\sqrt{r^2 - 2mr + Q^2} + r - m] + \sqrt{r^2 - 2mr + Q^2}$$ $$= m log[\sqrt{r^2 - 2mr + Q^2} + r - m] + r \sqrt{g_{tt}} - Λ.$$ Here $\Lambda$ is a large number evaluated within an infinitesimal distance from the horizon We write the metric at this position $$ds^2 = \left(1 - \frac{2m}{r(\rho)} + \frac{Q^2}{r(\rho)^2}\right)dt^2 - d\rho^2 - r(\rho)^2dΩ^2.$$ With the near horizon condition we may set $r^2 - 2mr + Q^2 \simeq 0$ in the log so that $$\rho \simeq m log(r - m) + r \sqrt{g_{tt}} – \Lambda.$$ The divergence of the log cancels the arbitrarily large $\Lambda$ $$\rho/r_+ =~ \sqrt{g_{tt}}.$$ We now write the metric as $$ds^2 = \frac{\rho^2}{r_+^2}dt^2 - d\rho^2 - m^2dΩ^2$$ We may now observe that $d\rho^2 = dr^2/g_{tt}^2$ and substitute in $\rho/m$ for $g_{tt}$ for $r_+ = m$ and obtain $$ds^2 = \left(\frac{\rho}{m}\right)^2dt^2 - \left(\frac{m}{\rho}\right)^2dr^2 - m^2dΩ^2.$$ Or $$ds^2 = \left(\frac{\rho}{m}\right)^2dt^2 - \left(\frac{m}{\rho}\right)^2 d\rho^2 - m^2dΩ^2~ for~ \rho \simeq r.$$ This is the metric for $AdS_2$ in the $(t, r)$ variables and a sphere $S^2$ of constant radius $= m$ in the angular variables.

This is considerably easier to see than the approach taken by Carroll and Randall, which admittedly is a more exact derivation. I mention this because is should not escape one's attention that in $10$ dimensions this is $AdS_5\times S^5$ in the $AdS/CFT$ correspondence. This connects with the AdS black hole which reduces by a dimension and the $CFT$ information in the anti-de Sitter spacetime is also on the horizon of the black hole.

• Thanks for the clear explanation. So you are claiming that the ADM is computable as in asymptotically flat black hole, in the the $\sim\frac{2M}{r}$ term? Maybe the answer is just contained in your previous answer but I can't see it. – Andrea89 Jun 15 '16 at 15:50
• Do you mean by ADM the AdS spacetime? ADM is a method in general relativity. – Lawrence B. Crowell Jun 15 '16 at 21:26
• I mean the ADM mass. I'm not an expert about but I know that exist the notion of ADM mass and energy. Probably it makes sense only inside the formalism, I really don't know, but can I compute that mass given an AdS black hole? – Andrea89 Jun 15 '16 at 21:46