= Solution
The normal to a surface of constant $r$ is $n_a=(dr)_a$, and the inverse ingoing metric gives
$$
n^an_a=g^{rr}=f(r).
$$
Hence $r=r_+$ is a <null hypersurface>. Let
$$
\xi=k+\Omega_+m=\partial_v+\Omega_+\partial_\chi,
\qquad \Omega_+=\Omega(r_+)=\frac{r_-}{Lr_+}.
$$
This constant linear combination of <Killing vector fields> is Killing. Its squared norm is
$$
\xi^2=-f+r^2(\Omega-\Omega_+)^2,
$$
which vanishes at $r_+$. More strongly, lowering its index in the ingoing metric gives $\xi_a=(dr)_a$ on the horizon. It is therefore the null normal and generator, so the surface is a <Killing horizon>.
Using $\nabla_a(\xi^2)=-2\kappa\xi_a$ on a Killing horizon, the squared term contributes no first derivative at $r_+$ and
$$
\kappa=\frac12f'(r_+).
$$
Differentiating $f$ at its simple outer root gives
$$
f'(r_+)=\frac{2(r_+^2-r_-^2)}{L^2r_+},
$$
so the <surface gravity> is
$$
\boxed{\kappa=\frac{r_+^2-r_-^2}{L^2r_+}}.
$$
It vanishes in the extremal case $r_+=r_-$.
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