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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-311/4/d/i/solution