Calculating the Area of an RN Event Horizon with Specific Heat Formula

[etotheipi]

By definition $C = T_H \dfrac{\partial S}{\partial T_H} \bigg{)}_Q$ so given $A=4S$ we first need to work out the area of the event horizon. More specifically, let $\Sigma$ be a partial Cauchy surface of constant $v$ in ingoing EF $(v,r,\theta, \phi)$ co-ordinates then $A$ is the area of the 2-sphere $\Sigma \cap \mathcal{H}^+$. This is where I get confused, because since $g = -\dfrac{\Delta}{r^2} dv^2 + 2dv dr + r^2 d\Omega^2$ surely we could just let $(\theta, \phi)$ be co-ordinates on $R$ with $\sqrt{h} = r^2 \sin{\theta}$ and hence $A = \int_{\varphi(R)} d\theta \wedge d\phi \sqrt{h} = 4\pi r^2$, which must be wrong. How do you actually calculate the area of $R$?

 

[etotheipi]

Actually, maybe it's not wrong? We can try to figure out $T_H$ by using the identity $d(\xi^a \xi_a)_{\mathcal{N}} = - 2\kappa \xi$ and for RN we have a timelike Killing vector $k =\partial / \partial v$ so\begin{align*}
d(k^a k_a) = d(g_{ab} k^a k^b) = d\left( \frac{-\Delta}{r^2} \right) = \left(\frac{2\Delta}{r^3} - \frac{1}{r^2} \dfrac{d\Delta}{dr} \right) dr
\end{align*}and since $k = dr$ at $r=r_{\pm}$ we have $\kappa = \dfrac{r_+-r_-}{2r_+^2}$ and $T_H = \dfrac{\kappa}{2\pi}$ so\begin{align*}

T_H = \frac{1}{2\pi} \left( \frac{\sqrt{M^2-e^2}}{2M^2 -e^2 + 2M\sqrt{M^2-e^2}} \right)

\end{align*}I think the rest should be straightforward using $A = 4\pi \left(2M^2 - e^2 + 2M\sqrt{M^2-e^2} \right)$, but I'll try that later because I have a call now