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In Kerr spacetime, given the energy-momentum tensor $T^{ab}$ of a field, what is the energy flux (as measured at infinity) $$ \frac{d^2E}{dt d\Omega} $$ i.e., the amount of energy passing through the horizon per unit time and solid angle (in Boyer-Lindquist coordinate)? Or the total flux?

I know $j^\mu=T^\mu_0$ is a conserved current ($\nabla_\mu j^\mu=0$), but how exactly can this be related to the energy flux?

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  • $\begingroup$ In Kerr spacetime, given the energy-momentum tensor $T^{ab}$ of a field … You already lost me. Kerr spacetime is a vacuum solution, it has $T^{ab}\equiv 0$. Do you mean that there is a “test” field on Kerr background that does not modify geometry, or something else? $\endgroup$
    – A.V.S.
    Commented Jan 12, 2023 at 7:26
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    $\begingroup$ @A.V.S. yeah, assuming it's a test field, in a fixed Kerr background. Then I suppose the flux is well-defined? $\endgroup$
    – hao123
    Commented Jan 12, 2023 at 10:28
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    $\begingroup$ Where can I find discussions of this? $\endgroup$
    – hao123
    Commented Jan 12, 2023 at 10:35
  • $\begingroup$ Have you looked at discussion in Wald's GR book at p. 328? $\endgroup$
    – A.V.S.
    Commented Jan 12, 2023 at 13:13
  • $\begingroup$ @A.V.S. thanks! this is what I need. So the flux is $j^\mu n_\mu$ where $n^\mu$ is the normal vector $n^\mu=-(1,0,0,\Omega_H)$. still I don't quite understand why $n^\mu$ has this specific normalization, since it's null. Is it fixed by requiring the definition of energy is consistent with observer at inifinity? $\endgroup$
    – hao123
    Commented Jan 13, 2023 at 16:15

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