I 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##?
 
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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
 
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