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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 $P(r)$?

KlausK
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    you realize I hope that no observer can stand on the horizon. – anna v Nov 06 '22 at 11:24
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    seems complicated https://www.techexplorist.com/formula-calculating-hawking-radiation-event-horizon-black-hole/29655/ – anna v Nov 06 '22 at 11:29
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    I'm sure you will have to define the state of motion of the observer (they cannot be stationary at the horizon). – ProfRob Nov 06 '22 at 11:31
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    Note that hawking radiation does not come from the horizon. It is seen at infinity but tracing back the trajectory is meaningless. – Jeanbaptiste Roux Nov 06 '22 at 11:49
  • @Jeanbaptiste: Until what distance to the horizon can the radiation be traced back? – KlausK Nov 06 '22 at 17:01
  • @ProfRob: I edited the question. – KlausK Nov 06 '22 at 17:06
  • @anna v: Indeed, but that paper is for Einstein-dilaton-Gauss-Bonnet black holes. Schwarzschild black holes would be sufficient... – KlausK Nov 06 '22 at 17:07
  • Wow, there are people editing comments... incredible. – KlausK Nov 08 '22 at 05:38
  • @JeanbaptisteRoux: This "Hawking radiation does not come form the horizon" is a statement that has been repeated incessantly, but I'm not sure I entirely believe it. If you cover up the horizon (by having a neutron star whose radius is slightly greater than the Schwarzschild radius, for example), no Hawking radiation is emitted. So where could it be coming from if not from the horizon? – Peter Shor Nov 08 '22 at 13:25
  • Related: https://physics.stackexchange.com/q/630787/123208 & links therein. – PM 2Ring Nov 08 '22 at 13:56
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    @PeterShor "I'm not sure I entirely believe it" the problem is then you, not the fact... – Jeanbaptiste Roux Nov 08 '22 at 14:48
  • @JeanbaptisteRoux: If the Hawking radiation doesn't come from the horizon, please explain to me why a neutron star slightly larger than the Schwarzschild radius doesn't give off Hawking radiation. – Peter Shor Nov 08 '22 at 16:35
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    @PeterShor I will not give a lecture on QFT in curved space-time. Did you even read Hawking's original paper? Also, what does a neutron star have to do with the Schwarzschild radius of a black hole? Do you understand that you don't understand? – Jeanbaptiste Roux Nov 08 '22 at 17:14
  • @JeanbaptisteRoux: My point is that the horizon must have *something* to do with generating the Hawking radiation. The space around a neutron star is curved in exactly the same way as a black hole, except that there is no horizon. And there is no Hawking radiation. – Peter Shor Nov 08 '22 at 17:30
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    @JeanbaptisteRoux: No horizon means no black hole radiation. True, the radiation does not originate at the horizon, but slightly above it. In practice, the difference is negligible for any macroscopic black hole. (P.S. Do you know who Peter Shor is?) – KlausK Nov 09 '22 at 11:48
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    @PeterShor Please quit at-ing me out, I don't want to argue with you since you don't want to understand that you actually don't understand what hawking radiation is. – Jeanbaptiste Roux Nov 09 '22 at 12:12

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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.

Níckolas Alves
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  • Energy conservation does not imply power invariance! I changed the question to make this clearer. – KlausK Nov 08 '22 at 05:19
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    If the temperature changes how is the luminosity fixed? – ProfRob Nov 08 '22 at 07:36
  • @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? – Níckolas Alves Nov 08 '22 at 19:54
  • @ProfRob My mistake. I'll correct it in the text. – Níckolas Alves Nov 08 '22 at 19:55
  • 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. – Peter Shor Nov 10 '22 at 20:43
  • @PeterShor that depends on the quantum state of the field. In the so-called Unruh state, there is only thermal radiation coming from the black hole, not from infinity (however, this state is ill-behaved near the white hole horizon in an eternal Schwarzschild black hole). In the so-called Hartle-Hawking state, radiation is indeed isotropic, but it includes a flow of radiation from the white hole horizon, and hence it doesn't make sense in a black hole formed out of stellar collapse. In either case, the redshift argument applies, because it is the correction between the "notion of time" at + – Níckolas Alves Nov 10 '22 at 21:27
  • infinity (where the temperatures are computed) and at the point where the observer is actually making the measurements. These notions are discussed in Wald's book on Chaps. 5 and 7. It might also be interesting to point out that the Unruh state is associated to what's often called the "Hawking effect" (present in Hawking's original paper), while the Hartle-Hawking state is associated to the "Unruh effect in curved spacetime" (a generalization of the flat spacetime case in which acceleration induces thermality). – Níckolas Alves Nov 10 '22 at 21:28