5
$\begingroup$

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.

Ps: Someone knows a good lecture about AdS black hole?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Andrea89 Jun 15 '16 at 15:50
  • $\begingroup$ Do you mean by ADM the AdS spacetime? ADM is a method in general relativity. $\endgroup$ – Lawrence B. Crowell Jun 15 '16 at 21:26
  • $\begingroup$ 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? $\endgroup$ – Andrea89 Jun 15 '16 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.