I Kerr Black Hole: Superradiance Flux - Show Negative when 0<ω<mΩH

[ergospherical]

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 $\chi = \dfrac{\partial}{\partial \phi}$

 

[robphy]

An old post asking about the same problem
https://www.physicsforums.com/threads/super-radiance-of-electromagnetic-waves.696317/

 

[ergospherical]

Haha, well I'm glad I'm not the only one who found the hint to be cryptic.
Can you see how to do it? I might try again tomorrow but I've spent slightly too long fiddling around, lol.