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This is a question about Hawking radiation in a fixed Schwarzschild background. The radiation does not back react on the metric and the black hole can not grow or evaporate.

Now I've seen Hawking radiation derived using either a Bogoliubov transformation or by using path integral arguments, so I understand that the vacuum state for a quantum field in the Schwarzchild background will be in a thermal equilibrium state in terms of the Hamiltonian that generates time translations in the Schwarzschild coordinate time.

My question is how is this interpreted as radiation coming from the black hole? This is an equilibrium state that does not change with time and unless I'm missing something there are as many particles moving radially in to the black hole as out from it.

Where are the particles from infinity coming from? Does an observer far from the black hole observe as much radiation as an observer near it?

Basically my confusion is coming from the fact that the state is expressed as a thermal distribution of modes without obvious (to me) spatial dependence. So I'm wondering how people jumped to the interpretation that all of the radiation in this vacuum state is coming from the black hole.

There is a very similar question here: Eternal black holes and Hawking radiation. But I am aware of the fact that this is a fixed background metric, and this still doesn't answer my questions of how the radiation is interpreted.

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    $\begingroup$ You seem to be saying that a black hole thermalizes at the Hawking temperature, but does not radiate. The Hawking radiation is closely related to the Unruh radiation. The following paper states, "We show that this system does not radiate despite the fact that it does in fact thermalize at the Unruh temperature." - arxiv.org/abs/quant-ph/0509151 $\endgroup$
    – safesphere
    Nov 14, 2019 at 17:16
  • $\begingroup$ @safesphere, I'm not confident that I understand the system well enough to claim it doesn't radiate, but yes I'm concerned about the presence of radiation flowing in all directions. Thanks for the paper, it looks interesting. $\endgroup$
    – octonion
    Nov 14, 2019 at 18:32
  • $\begingroup$ @safesphere, I'm not so concerned about the frame dependence. The different frames have different Hamiltonians associated to time translation. So it makes sense to me that a given state would look different expanded in the different energy eigenstates. $\endgroup$
    – octonion
    Nov 14, 2019 at 19:13
  • $\begingroup$ @safesphere, "There is no Hawking radiation in a free falling frame, so it's existence in other frames would violate the postulate of relativity." I don't think the second part of your statement is correct, but that's not what my question is about. $\endgroup$
    – octonion
    Nov 14, 2019 at 21:35
  • $\begingroup$ @safesphere, Yes I agree if the physics are different that would be a problem. But I think if you calculate everything (e.g. the correlation functions) properly in both frames you will the exact same answer. Here's a paper that does it explicitly for Unruh radiation: arxiv.org/abs/1612.03158 $\endgroup$
    – octonion
    Nov 14, 2019 at 22:05

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The local picture usually comes from looking at $\langle T_{\mu\nu} \rangle _{ren}$, the renormalized semiclassical energy-momentum tensor. For a BH formed by stellar collapse it shows a flux of negative energy into the horizon and a thermal Hawking flux out to infinity, as calculated for example by Davies, Fulling, and Unruh 1976.

For a purely Schwarzchild eternal BH with no collapse, this local picture will depend on the boundary conditions (i.e. on the choice of quantum state), and there is no obvious way to say which represents the "real" physics. After all, no actual objects with that geometry are known to exist. States like the one you mention where an equal number of particles fall in as go out are not evaporating, they are in equilibrium. EDIT: The energy-momentum for various states in this background was discussed e.g. by Candelas 1980.

And watch out for treating the Bogoliubov particle calculations as locally meaningful, which can be misleading as shown e.g. by Padmanabhan and Singh 1987.

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  • $\begingroup$ Thank you for the references. Yes I would like to consider a purely Schwarzschild eternal BH in the usual Hartle-Hawking vacuum state. My question is about the interpretation of this particular state. $\endgroup$
    – octonion
    Nov 14, 2019 at 0:27
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    $\begingroup$ In that case the reference you probably want is this one: journals.aps.org/prd/abstract/10.1103/PhysRevD.21.2185. I'll edit to add this. $\endgroup$
    – Joe Schindler
    Nov 14, 2019 at 21:07
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    $\begingroup$ Thanks. To summarize, this last paper seems to suggest at the end of the introduction section that the Unruh vacuum is a better vacuum to consider in the eternal BH rather than the Hartle-Hawking vacuum, since it more closely matches the stellar collapse BH. And it is the Unruh vacuum that looks like a radiating black hole alone, even though the textbooks I've seen calculate the Hartle-Hawking vacuum as a demonstration of Hawking radiation (hence my question). $\endgroup$
    – octonion
    Nov 14, 2019 at 23:13

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