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Why isn't Hawking radiation emitted by every massive body? (because every massive body causes deformation of space-time, and this causes the vacuum state to be observer dependent, if I'm not wrong)

What makes black holes special in this context? Does it have something to do with the existence of a singularity/event horizon? (Why is that an important ingredient?)

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    $\begingroup$ I think this is covered by my answer to An explanation of Hawking Radiation to the extent that it can be explained in layman's terms. I'm not going to close this as a duplicate, but I think it is effectively a duplicate. $\endgroup$ Commented Jun 14, 2017 at 10:34
  • $\begingroup$ I wanted to know in detail why the horizon is important. Not a layman's explanation but maybe a slightly watered down one. (mainly because I can understand the math behind GR but not QFT). I guess I have to try to get as much as I can from Hawking's paper! $\endgroup$ Commented Jun 14, 2017 at 10:41
  • $\begingroup$ Similar questions: 1) physics.stackexchange.com/questions/9359/… 2) physics.stackexchange.com/questions/2696/… $\endgroup$
    – Avantgarde
    Commented Jun 14, 2017 at 18:12

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It's not unique to black holes. An accelerating observer would also experience a thermal radiation. This is called the Unruh radiation and is essentially equivalent to Hawking radiation. What's important is that when we have different choices of what we can call the vacuum in some spacetime, one particular vacuum state can be a thermal state seen by an observer who has another vacuum choice.

EDIT: What's common between acceleration and black holes is that they have a horizon. The existence of horizon makes you inaccessible of the information on the other side of the horizon. Basically it's this ignorance that leads to a thermal state.

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  • $\begingroup$ Could you clarify the statement in your edit? $\endgroup$ Commented Jun 16, 2017 at 6:35
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    $\begingroup$ @HritikNarayan The accelerating observer is in one wedge of the whole Minkowski spacetime, and he/she has no access of the other wedge. When you consider some quantum fields in Minkowski spacetime, you should really consider the tensor product of two wedges. Since the accelerating observer is ignorant of the other wedge, effectively he/she traces out the other wedge. In thermal field theory, this gives arise to a thermal state. $\endgroup$
    – JamieBondi
    Commented Jun 16, 2017 at 18:26
  • $\begingroup$ That's good to know, but I think the question (at least in my mind) was why a regular massive object which isn't a black hole doesn't evaporate like a black hole does? $\endgroup$ Commented Aug 29, 2021 at 4:29

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