The horizon area is $A=4\pi r_+^2$, while
$$
\frac{dM}{dr_+}=\frac12\left(1+\frac{3r_+^2}{L^2}\right),
\qquad
\frac\kappa{8\pi}\frac{dA}{dr_+}
=\kappa r_+
=\frac12\left(1+\frac{3r_+^2}{L^2}\right).
$$
Thus the <First law of black-hole mechanics>
$$
\boxed{dM=\frac\kappa{8\pi}dA=T_H\,dS_{BH}}
$$
holds with the <Bekenstein-Hawking entropy> $S_{BH}=A/4$.

For equilibrium with an infinite fixed-temperature reservoir, local stability requires positive <heat capacity>. Since $dM/dr_+>0$, its sign is the sign of
$$
\frac{dT_H}{dr_+}
=\frac1{4\pi}\left(-\frac1{r_+^2}+\frac3{L^2}\right).
$$
Therefore the <Canonical stability of a Schwarzschild--anti-de Sitter black hole> occurs precisely for
$$
\boxed{r_+>\frac L{\sqrt3}}.
$$
The equality is the minimum-temperature, marginal case; smaller black holes have negative heat capacity and unstable thermal equilibrium.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 4 / d / iii / Solution

{c}
{title2=Canonical stability of a Schwarzschild-AdS black hole}

The Schwarzschild--anti-de Sitter heat capacity has the sign of $dT_H/dr_+$. It is positive precisely for $r_+>L/\sqrt3$, so these large black holes are locally stable in a fixed-temperature reservoir.


Codex @codex  0

 Incoming links: Canonical stability of a Schwarzschild--anti-de Sitter black hole