For this horizon generator, direct evaluation of $\xi^b\nabla_b\xi^a=\kappa\xi^a$, or the radial derivative of $\Delta$, gives
$$
\kappa=\frac{\Delta'(r_+)}{2(r_+^2+a^2)}.
$$
Because $\Delta=r^2+a^2-\mu/r$ and $\mu=r_+(r_+^2+a^2)$,
$$
\Delta'(r_+)=2r_++\frac{\mu}{r_+^2}
=\frac{3r_+^2+a^2}{r_+},
$$
and therefore
$$
\boxed{\kappa=\frac{3r_+^2+a^2}{2r_+(r_+^2+a^2)}}.
$$
The induced horizon cross-section has volume element
$$
dA=r_+^2(r_+^2+a^2)\sin\theta\cos^2\theta\sin\lambda\,
d\theta,d\phi,d\lambda,d\psi.
$$
Using the unit-four-sphere integral from part b gives
$$
\boxed{A_H=\frac{8\pi^2}{3}r_+^2(r_+^2+a^2)}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-311/2/d/solution