This has been bugging me for some time.

As I understand it, Hawking radiation is the result of the mismatch between the vacuum state of a quantum field as seen by a free falling observer (falling directly toward the black hole) and one that is sitting at a constant radius far away from the black hole.

This is perhaps the result of my naive understanding of the subject, but it doesn't make sense to me to say that a black hole radiates, because it's observer-dependent. How do we know that the thermal bath of particles that we see when we sit at a constant radius actually causes the black hole to lose mass? Is it just a simple energy conservation argument, or is there some subtle process here that I'm missing?

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    $\begingroup$ Related: Primordial Black Hole Formation. $\endgroup$ – dmckee Jun 26 '11 at 1:49
  • $\begingroup$ And fundamentally, the answer is that we don't know for SURE. Experimental observation is the proof of anything, and we haven't seen any direct evidence for evaporating black holes. But general relativity and quantum field theory are both very, very well established, and when combined, the clear prediction is that you get Hawking radiation. $\endgroup$ – Jerry Schirmer Jun 26 '11 at 5:49
  • $\begingroup$ I may be more confused than you are, but I think there is no mismatch of what the falling and the stationary observers will say about the hole evaporation. The stationary has the time to wait and see if it evaporates, while the falling will bi in the hole in finite proper time. So it is meaningless for him to make any prediction about the future of the hole, where he sees any radiation or not on his way down the hole. Also the whole process may be global, and the falling observer is only in a local inertial frame, I am not sure if he can say anything about the posible evaporation at all. $\endgroup$ – MBN Jun 26 '11 at 17:02

Well, here is a simple scenario:

The quantum field theoretic vacuum is seething with pair production of virtual particles everywhere. Take one such virtual pair at the event horizon of the black hole. One part of the pair, if it is going down, is grabbed by the hole and disappears while the other is with its high momentum runs away from the horizon, on shell. Where did it find the energy? Read on

A slightly more precise, but still much simplified, view of the process is that vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). By this process, the black hole loses mass, and, to an outside observer, it would appear that the black hole has just emitted a particle. In another model, the process is a quantum tunneling effect, whereby particle-antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon.

The faraway observer is used to define the "negative energy" of the eaten-up partner.

  • $\begingroup$ In the original paper Hawking describes the mechanism of the pair of particles, one falls in, one escapes and so on. But then he makes the remark that "this mechanism is heuristic only and shouldn't be taken too literary" (quoting from memory). $\endgroup$ – MBN Jun 26 '11 at 17:06
  • $\begingroup$ @MBN right, that is why one talks of "models". In truth one should have a theory of quantized gravity to be working on the horizon of a black hole. $\endgroup$ – anna v Jun 26 '11 at 17:13
  • $\begingroup$ Yes, when there is a theory of quantum gravity it may be clearer. But the radiation and the question are within a model so the question is meaningful and should be possible to answer. The way I understood the question (and would also like to see an explanation) is that: since number of particles is observer dependent, different observers can see an evaporating and a non evaporating hole. But whether there is a hole or not, where spacetime is not flat or is, is not observer dependent. So how come? $\endgroup$ – MBN Jun 26 '11 at 18:14
  • $\begingroup$ @MBN why do you say "the number of particles is observer dependent"? the energy, yes, but the number? I do not think one can count the vacuum pairs in any sense. $\endgroup$ – anna v Jun 26 '11 at 18:53
  • $\begingroup$ I think MBN is talking in the context of the Fock space for an observer. The derivation I've seen for Hawking radiation (Mukhanov and Winitzki) is based on the fact that the vacuum ("zero particle") state of a quantum field is not the same for all observers, and was done for a free field theory. This leads me to another point of my confusion--I've heard the pair production argument, but don't you have to include interactions for pair production (in QED at least)? $\endgroup$ – Desert Coyote Jun 26 '11 at 20:08

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