Criterion for a black hole in Anti-de Sitter background

Question

Consider a Schwarzschild-Anti de Sitter (SAdS) metric

$$ds^2=-(1-\frac{2M}{r}+ k\, r^2 )\, dt^2+\frac{dr^2}{1-\frac{2M}{r}+k \,r^2}+r^2 d\Omega_2^2, $$

with $M,k>0$. This solution has only one horizon, say $r_+>0$. Notice that $r_+\neq 2M$. Is $r_+$ the horizon of a black hole?

EDIT: I have tried to clarify / improve my question to the current version because the previous version seemed to be somewhat misleading.

EDIT II: I guess the answer is given by staring at the associated Penrose diagram, from which it can be easily obtained that there is no escape from inside the horizon $r_+$...

Answer 1

Answer to OP v(2):

Yes, "$r^+$" in your language is the horizon of the black hole. I am writing the black hole in $AdS_4$'s metric as

$$ds^2 = f(r)dt^2 - \frac{dr^2}{f(r)}-r^2 d\Omega^2,$$

where $f(r) = 1 - \frac{2M}{r} + \frac{r^2}{a^2}.$

The black hole's event horizon is given by the real roots of $g^{rr} = f(r) = 0$ (note the upper indices), which in this case is given by

$$r^3 +a^2r - 2Ma^2 = 0.$$

Note that $f(r)$ has the following behaviour:

1. $f(r) \to -\infty$, as $r \to 0^+$.
2. It has not extrema, $f'(r)=0 \implies r^3 = -Ma^2 <0.$
3. It has one inflection point: $f''(r) = 0 \implies r = (2Ma^2)^{1/3}.$

Therefore the black hole has only one event horizon given by:

$$r_h = (Ma^2)^{1/3} \left[ \left( 1 + \sqrt{1+\frac{a^2}{27M^2}}\right)^{1/3} + \left( 1 - \sqrt{1+\frac{a^2}{27M^2}}\right)^{1/3} \right].$$

One can invert this equation to obtain the expression for the mass:

$$M = \frac{r_h}{2} \left( 1 + \frac{r_h^2}{a^2}\right),$$

which is basically an inverse of $r_h$.

Relationship of $r_h$ with $r_{Schwarz} = 2M$

Let us compute the deviation of $r_h$ from $r_{Schwarz} = 2M$, if $\frac{M}{a} \ll 1$. Using the relation between $M$ and $r_h$, one can write

$$\frac{M}{a} = \frac{r_h}{2a}\left( 1 + \left(\frac{r_h}{a}\right)^2\right) \ll 1 \implies \frac{r_h}{a} \approx \frac{M}{a} \ll 1.$$

As a result we can write:

$$r_h = \frac{2M}{1+\frac{r_h^2}{a^2}} \approx 2M \left( 1 - \frac{r_h^2}{a^2}\right) = r_{Schwarz} \left( 1 - \frac{4M^2}{a^2}\right)$$

Clearly

1. $r_h \to r_{Schwarz}$ as $a \to \infty$.

2. $r_h \to 0$ as $M \to 0$.

3. $r_h \to (2Ma^2)^{1/3}$ as $M \to \infty$.

Answer to OP v(1):

1. A Schwarzchild black hole in AdS does not have stress energy, i.e. $T_{\mu\nu} = 0$. It satisfies the sourceless Einstein equations with a negative cosmological constant. So there is no "mass distribution" inside the horizon as you mentioned as the stress energy components are all zero everywhere (apart from the singularity). See the Wikipedia page for minor details. For a deeper discussion, see section 4 of these notes where they start with the AdS metric (they start with a spherically symmetric asymptotic AdS solution) and compute all curvature components. Equations (4.7) and (4.8) demonstrate that the stress energy components are all zero. Consequently there are no stress energy components and as a result no "matter distribution" in the interior.

2. Birkhoff theorem states that there is an unique form solving the vacuum Einstein equations given spherical symmetry, which is the Schwarzchild solution. See this excellent answer for more details.

3. Geometry wise the mass $M$ of any black hole is understood in the ADM formalism. The correct way to understand the mass of a black hole is not to think that there is a "mass distribution" sitting in the interior, but there is an energy associated with the spacetime itself, as beautifully explained in the ADM formalism. In this formalism, the ADM mass is a way to define the energy of $d$ dimensional spacelike slices in $d+1$ dimensional spacetime. This definition naturally comes out of the $d+1$ ADM decomposition of the Einstein-Hilbert Lagrangian. See these notes for more details on the ADM decomposition.