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b) Show that the time averaged flux of ##J^a = -{T^a}_b \xi^b## across the horizon of a Kerr black hole is negative when ##0 \leq \omega \leq m\Omega_H ##. Given that ##dF = 0## i.e. ##\nabla_{[a} F_{bc]} = 0##,\begin{align*}
-2\nabla_{[a} (F_{b]c} w^c) &= F_{ac} \nabla_b w^c + F_{cb} \nabla_a w^c - w^c (\nabla_b F_{ca} + \nabla_a F_{bc}) \\
&= F_{ac} \nabla_b w^c + F_{cb} \nabla_a w^c + w^c \nabla_c F_{ab} \\
&= L_w F_{ab}
\end{align*}It is hinted to use this equation to relate ##F_{ab} \xi^b## to ##F_{ab} \chi^b##, but how? The tensor ##T## is ##T_{ab} = \nabla_a \phi \nabla_b \phi - \dfrac{1}{2} g_{ab} (\nabla_c \phi \nabla^c \phi + m^2 \phi^2)## so the time-averaged flux is ##\langle J_{a} (-\chi^a) \rangle = \langle (\chi^a \nabla_a \phi)(\xi^b \nabla_b \phi) \rangle##
edit: ##\xi = \dfrac{\partial}{\partial t}## and ##\chi = \dfrac{\partial}{\partial \phi}##
-2\nabla_{[a} (F_{b]c} w^c) &= F_{ac} \nabla_b w^c + F_{cb} \nabla_a w^c - w^c (\nabla_b F_{ca} + \nabla_a F_{bc}) \\
&= F_{ac} \nabla_b w^c + F_{cb} \nabla_a w^c + w^c \nabla_c F_{ab} \\
&= L_w F_{ab}
\end{align*}It is hinted to use this equation to relate ##F_{ab} \xi^b## to ##F_{ab} \chi^b##, but how? The tensor ##T## is ##T_{ab} = \nabla_a \phi \nabla_b \phi - \dfrac{1}{2} g_{ab} (\nabla_c \phi \nabla^c \phi + m^2 \phi^2)## so the time-averaged flux is ##\langle J_{a} (-\chi^a) \rangle = \langle (\chi^a \nabla_a \phi)(\xi^b \nabla_b \phi) \rangle##
edit: ##\xi = \dfrac{\partial}{\partial t}## and ##\chi = \dfrac{\partial}{\partial \phi}##
