For a metric $ds^2=-f(r)dv^2+2\,dv\,dr+r^2d\Omega_2^2$ with a simple Killing horizon, the <surface gravity> of $k=\partial_v$ is $\kappa=f'(r_+)/2$. Here
$$
f'(r)=\frac{2r}{L^2}
+\frac{r_+}{r^2}\left(1+\frac{r_+^2}{L^2}\right),
$$
so
$$
\kappa=\frac1{2r_+}+\frac{3r_+}{2L^2}.
$$
The <Hawking temperature> of the <Schwarzschild--anti-de Sitter black hole> is therefore
$$
\boxed{T_H=\frac\kappa{2\pi}
=\frac{1+3r_+^2/L^2}{4\pi r_+}}.
$$

Solution

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

{c}
{title2=Schwarzschild-AdS black hole}

In four dimensions the Schwarzschild--anti-de Sitter black hole has
$$
f(r)=1+\frac{r^2}{L^2}-\frac{2M}{r}.
$$
Writing the horizon radius as $r_+$ gives $M=r_+(1+r_+^2/L^2)/2$ and
$$
T_H=\frac{1+3r_+^2/L^2}{4\pi r_+}.
$$


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