9

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?

Desert Coyote
  • 141
  • 1
    Related: Primordial Black Hole Formation. – dmckee --- ex-moderator kitten Jun 26 '11 at 01:49
  • 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. – Zo the Relativist Jun 26 '11 at 05:49
  • 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. – MBN Jun 26 '11 at 17:02

1 Answers1

5

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.

Peter Mortensen
  • 2,352
anna v
  • 233,453
  • 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). – MBN Jun 26 '11 at 17:06
  • @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. – anna v Jun 26 '11 at 17:13
  • 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? – MBN Jun 26 '11 at 18:14
  • @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. – anna v Jun 26 '11 at 18:53
  • 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)? – Desert Coyote Jun 26 '11 at 20:08
  • @anna_v: I say it because I know it is the case. See for example the Unruh effect. – MBN Jun 26 '11 at 20:38
  • @Desert Coyote the second question is another story. In principle you do have all the diagrams coming in, this would affect the type of particles radiating , and the probabilities would be smaller the higher the masses, but not the phenomenon. – anna v Jun 27 '11 at 04:14
  • @MBN the Unruh effect, as far as I know does not generate particles. In general black body radiation does not. The photon would have to interact with something to pair produce, and a different observer would see this produced pair, though would measure different energies. For the hole,two observers far away would see the same particle coming out from it, though the energy would be different depending on their frame of motion. imo, – anna v Jun 27 '11 at 04:33
  • @anna v: Inertial observer in vacuum sees no particles, accelerated observer sees particles. Clearly the number is different. And in QFT in curved background the number of particles is observer dependent. I don't understand your comment. Are you saying that this is not the case? – MBN Jun 27 '11 at 16:43
  • @MBN GR is not my field, but from reading a bit,it seems that the unruh effect is not all that clear as far as particle creation etc, though the temperature effect seems not to be controversial. The black hole radiation is not controversial as far as I know. – anna v Jun 27 '11 at 19:35
  • The derivation I've seen of Hawking radiation is based on the Unruh effect, and as far as I can tell, they're equivalent effects (the only difference is the geometry, the Unruh effect if for a planar horizon, and the Hawking for a spherical one). In the context of that derivation, whether you see radiation or not depends on the trajectory of the observer. The pair production argument for Hawking radiation bugs me because you need an interacting field theory to make it work, whereas the derivation I've seen only requires free fields. – Desert Coyote Jun 28 '11 at 01:09
  • @Desert Coyote in my opinion one is talking of models, in either case. It seems that the Unruh radiation is not classroom material. There are people with a lot of experience in GR following this forum and as nobody has pitched in with an answer it means to me that there does not exist a decisive one. It must still be a subject of theoretical research. – anna v Jun 28 '11 at 04:31
  • http://www.hep.princeton.edu/~mcdonald/accel/unruhrad.pdf in this link it is claimed that there is some experimental evidence: "The Hawking-Unruh temperature finds application in accelerator physics as the reason that electrons in a storage ring do not reach 100% polarization despite emitting polarized synchrotron radiation " – anna v Jun 28 '11 at 04:35
  • It is not strictly part of GR, it is part of quantum field theory on curved spacetime. So we would need a specialist in that area not general relativity to give us an answer clarifying the confusion. – MBN Jun 28 '11 at 04:55
  • Have a look at this review of a paper, in Lubos' blog http://motls.blogspot.com/2011/06/rubiks-cubes-inside-black-holes.html – anna v Jun 28 '11 at 07:22