- #1

etotheipi

This is the problem statement:

We can start by writing ##

(\star d \star d \xi)_a = - \nabla^b (d\xi)_{ab} = - \nabla^b \nabla_a \xi_b + \nabla^b \nabla_b \xi_a = 2\nabla^b \nabla_b \xi_a

##. Then with ##\nabla_a \nabla_b \xi_c = R_{cbad} \xi^d = -R_{bcad} \xi^d## we can contract over ##a## and ##b## to find ##\nabla^b \nabla_b \xi_c = -{R^b}_{cbd} \xi^d = -R_{cd} \xi^d## so that \begin{align*}

(\star d \star d \xi)_a = -2R_{ab} \xi^b = -2 \left( 8 \pi \left[ T_{ab} - \frac{1}{2} T g_{ab} \right] \right) \xi^b := 8 \pi J_a'

\end{align*}Because the metric has Lorentzian signature the Hodge star for ##p=1## satisfies ##\star \star d \star d \xi = -(-1)^p d \star d \xi = d \star d \xi##. Then we can use the expression in the question to write down\begin{align*}

\frac{1}{8\pi} \int_{\Sigma} d \star d \xi = \int_{\Sigma} \star J_a' = \frac{1}{8\pi} \int_{S_{\infty}^2} \star d \xi - \frac{1}{8\pi} \int_H \star d \xi

\end{align*}The second term ##\frac{1}{8\pi} \int_{S_{\infty}^2} \star d \xi = \frac{1}{8\pi} \int_{S_{\infty}^2} \star d k + \frac{\Omega_H}{8\pi} \int_{S_{\infty}^2} \star dm= -M + 2\Omega_H J## is negative the Komar mass plus two lots of ##\Omega_H J## so we are left to write\begin{align*}

M = -\int_{\Sigma} \star J_a' + 2\Omega_H J - \frac{1}{8\pi} \int_H \star d\xi

\end{align*}Now I am stuck because I don't know how to integrate ##\star d\xi = \star d k + \Omega_H \star dm## over ##H##. I know that if ##\xi^a \big{|}_{\mathcal{H}^+} = f n^a## for some function ##f## and normal ##n^a = \tilde{f} \partial^a S## to the horizon then the surface gravity satisfies ##\kappa = \xi \cdot \partial \log{|f|}##, but I don't see how it arises from this integral nor how ##A## is defined. Thanks in advance for any hints!

We can start by writing ##

(\star d \star d \xi)_a = - \nabla^b (d\xi)_{ab} = - \nabla^b \nabla_a \xi_b + \nabla^b \nabla_b \xi_a = 2\nabla^b \nabla_b \xi_a

##. Then with ##\nabla_a \nabla_b \xi_c = R_{cbad} \xi^d = -R_{bcad} \xi^d## we can contract over ##a## and ##b## to find ##\nabla^b \nabla_b \xi_c = -{R^b}_{cbd} \xi^d = -R_{cd} \xi^d## so that \begin{align*}

(\star d \star d \xi)_a = -2R_{ab} \xi^b = -2 \left( 8 \pi \left[ T_{ab} - \frac{1}{2} T g_{ab} \right] \right) \xi^b := 8 \pi J_a'

\end{align*}Because the metric has Lorentzian signature the Hodge star for ##p=1## satisfies ##\star \star d \star d \xi = -(-1)^p d \star d \xi = d \star d \xi##. Then we can use the expression in the question to write down\begin{align*}

\frac{1}{8\pi} \int_{\Sigma} d \star d \xi = \int_{\Sigma} \star J_a' = \frac{1}{8\pi} \int_{S_{\infty}^2} \star d \xi - \frac{1}{8\pi} \int_H \star d \xi

\end{align*}The second term ##\frac{1}{8\pi} \int_{S_{\infty}^2} \star d \xi = \frac{1}{8\pi} \int_{S_{\infty}^2} \star d k + \frac{\Omega_H}{8\pi} \int_{S_{\infty}^2} \star dm= -M + 2\Omega_H J## is negative the Komar mass plus two lots of ##\Omega_H J## so we are left to write\begin{align*}

M = -\int_{\Sigma} \star J_a' + 2\Omega_H J - \frac{1}{8\pi} \int_H \star d\xi

\end{align*}Now I am stuck because I don't know how to integrate ##\star d\xi = \star d k + \Omega_H \star dm## over ##H##. I know that if ##\xi^a \big{|}_{\mathcal{H}^+} = f n^a## for some function ##f## and normal ##n^a = \tilde{f} \partial^a S## to the horizon then the surface gravity satisfies ##\kappa = \xi \cdot \partial \log{|f|}##, but I don't see how it arises from this integral nor how ##A## is defined. Thanks in advance for any hints!

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