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The power emitted by a Schwarzschild black hole via Bekenstein-Hawking radiation is usually given for an observer at spatial infinity.

What is the emitted power for an observer hovering just above its horizon at a radial distance $r$? (Here, "emitted" is meant to describe only the flow of energy away from the black hole.)

An observer nearer to the horizon will see a higher temperature $T$, due to red shift. In all cases, the emitted power $P(r)$ at radius $r$ is expected to be given by the black hole horizon surface times $T^4$ times the Stefan-Boltzmann constant.

The emitted power should thus increase when getting closer to the horizon. What is the value of $P(r)$?

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    $\begingroup$ you realize I hope that no observer can stand on the horizon. $\endgroup$
    – anna v
    Commented Nov 6, 2022 at 11:24
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    $\begingroup$ seems complicated techexplorist.com/… $\endgroup$
    – anna v
    Commented Nov 6, 2022 at 11:29
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    $\begingroup$ I'm sure you will have to define the state of motion of the observer (they cannot be stationary at the horizon). $\endgroup$
    – ProfRob
    Commented Nov 6, 2022 at 11:31
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    $\begingroup$ Note that hawking radiation does not come from the horizon. It is seen at infinity but tracing back the trajectory is meaningless. $\endgroup$
    – Jeanbaptiste Roux
    Commented Nov 6, 2022 at 11:49
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    $\begingroup$ @PeterShor "I'm not sure I entirely believe it" the problem is then you, not the fact... $\endgroup$
    – Jeanbaptiste Roux
    Commented Nov 8, 2022 at 14:48

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For the temperature, there is just a redshift factor. The temperature for an observer hovering at radial coordinate $r$ is $$T_r = \frac{T_H}{\chi},$$ where $T_r$ is the temperature at $r$, $T_H$ is the temperature at infinity, and $\chi = \sqrt{- \chi^a \chi_a}$, where $\chi^a$ is the Killing field. For Schwarzschild, $$\chi = \sqrt{1 - \frac{2M}{r}},$$ in units with $G = c = 1$.

This is discussed in books about QFT in curved spacetime and black hole thermodynamics. For example, see Wald's book, Eqs. (5.3.3) and (7.2.10). This ends up matching what one would expect by considering the proximities of the black hole to look like Minkowski spacetime from the point of view of an accelerated observer and then computing the temperature according to the Unruh effect.

As mentioned in the comments to the question, it doesn't make sense to speak of particles in the vicinity of the black hole. They are only defined at infinity. Still, a thermal state is a thermal state and one can still compute the stress tensor associated to the quantum field and see a flux of negative energy into the black hole. These notions are also mentioned in QFTCS books.

Knowing the temperature measured locally, I don't see any immediate problems with using the Stefan–Boltzmann formula to obtain the power from the temperature.

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  • $\begingroup$ Energy conservation does not imply power invariance! I changed the question to make this clearer. $\endgroup$
    – KlausK
    Commented Nov 8, 2022 at 5:19
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    $\begingroup$ If the temperature changes how is the luminosity fixed? $\endgroup$
    – ProfRob
    Commented Nov 8, 2022 at 7:36
  • $\begingroup$ @KlausK Oops, you're correct. At each distance $r$ you'll have a different notion of power. However, do you see any problem with using the Stefan–Boltzmann formula with the updated temperature I gave? $\endgroup$
    – Níckolas Alves
    Commented Nov 8, 2022 at 19:54
  • $\begingroup$ @ProfRob My mistake. I'll correct it in the text. $\endgroup$
    – Níckolas Alves
    Commented Nov 8, 2022 at 19:55
  • $\begingroup$ I thought the question was about the net flow of radiation away from the black hole at a specific radius. Hawking radiation is directional, away from the black hole. Unruh radiation is isotropic. Presumably the radiation observed by an observer hovering at radius $r$ is only partially directional, because it's a mix of Unruh and Hawking radiation. $\endgroup$
    – Peter Shor
    Commented Nov 10, 2022 at 20:43

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