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Correct me if I am wrong: in the Hawking Radiation mechanism, when a virtual particle-antiparticle pair gets created at the edge of the black hole, a black hole could sometimes eat up one of the particles before it could annihilate with the other particle, and this causes the other particle to radiate away, and this process causes a loss in the mass of the black hole.

What I don't understand is: if a black-hole also absorbs a particle from the pair, shouldn't the black hole actually be gaining mass? Since the black-hole hasn't lost anything but instead gains a particle?

Danu
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Hritik Narayan
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    The black hole initially lost the gravitational energy that was needed to create the pair. The pair-creation model is a bad description of Hawking radiation, which for macroscopic black holes is really photons. The second particle that gets created above the event horizon doesn't have nearly enough energy to escape. It does, however, produce photons above the event horizon, some of which can escape after being red-shifted very strongly. What we would see is therefor black body radiation escaping, but as long as black holes are much colder than the universe not even that can happen. – CuriousOne Dec 23 '14 at 13:20
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    @CuriousOne: That probably could be it's own answer ;) – Kyle Kanos Dec 23 '14 at 13:28
  • Many possible duplicates: e.g. http://physics.stackexchange.com/q/30597/ http://physics.stackexchange.com/q/22498/ – ProfRob Dec 23 '14 at 13:44
  • case of similar titles and unsimilar content xD @RobJeffries – Hritik Narayan Dec 23 '14 at 13:49
  • and http://physics.stackexchange.com/q/26605/ – ProfRob Dec 23 '14 at 15:38
  • This question does not have an answer at the alleged duplicate. – Warren Dew Mar 16 '18 at 02:50

2 Answers2

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The black hole initially lost the gravitational energy that was needed to create the pair. The pair-creation model is a bad description of Hawking radiation, which for macroscopic black holes is really photons. The second particle that gets created above the event horizon doesn't have nearly enough energy to escape. It does, however, produce photons above the event horizon, some of which can escape after being red-shifted very strongly. What we would see is therefor black body radiation escaping, but as long as black holes are much colder than the universe not even that can happen. See http://en.wikipedia.org/wiki/Hawking_radiation for the details.

CuriousOne
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  • I don't understand why you say "The Black Hole initially lost the gravitational energy that was needed to create the pair." I thought the virtual particles were created near the Black Hole in empty space. – John Fistere Oct 09 '15 at 04:24
  • @JohnFistere Late response, but the energy really comes from the gravitational field, which exists outside the event horiz0n and in space. This field is what lets us measure the "mass" of the black hole. Loosely speaking, the field, not the mass, "is" the gravity of the black hole. Since there's energy in the field permeating a region near the horizon, pair-creation is possible by "borrowing" that energy to form particles. But if some of that energy escapes (the photons of the Hawking radiation), that energy cannot be returned to the field, so the field loses energy.. – zibadawa timmy Mar 18 '17 at 06:32
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According to p. 303-304 of the book Gravity from the Ground up by physicist Bernard Schutz, viewable on google books here, it's because in terms of the pair-production explanation for Hawking radiation, one member of the pair actually has negative energy and thus causes the black hole to lose mass (negative mass/energy falling into a black hole can also cause it to lose mass and decrease in radius in classical general relativity, see the second paragraph of my answer here). From those pages:

Quantum theory allows uncertainties and fluctuations that are not allowed in non-quantum physics. Temporary fluctuations can produce photons of negative energy. In order to preserve the total energy, negative-energy photons form in pairs with positive-energy partners. These pairs almost immediately re-combine and disappear, since the quantum theory has to get rid of the negative-energy photons quickly in order to produce macroscopic physics of positive energy. But negative energy does exist for short times, in these quantum fluctuations.

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How can black holes emit radiation? It should be no surprise that the answer lies in quantum uncertainty. All over spacetime the quantum electromagnetic field is undergoing the little negative-energy fluctuations that we considered above. Normally they are harmless and invisible, because the negative-energy photons disappear as quickly as they form. But near the horizon of a black hole, it is possible for such a photon to form outside the hole and cross into it.

Once inside, it is actually viable: as we remarked earlier, it is possible to find trajectories for photons inside the horizon that have negative total energy. So such a photon can just stay inside, and that leaves its positive-energy partner outside on its own. It has no choice but to continue moving outwards. It becomes one of the photons of the Hawking radiation.

In this answer John Rennie gives some more explanation of the mathematical derivation of Hawking radiation that this verbal description is meant to serve as shorthand for; I'm sure you need a good technical understanding of the mathematics of quantum field theory to really understand it though, verbal descriptions can only give you a flavor.

Community
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Hypnosifl
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  • So the particle-antiparticle pair which I considered is actually considered to be an Energy-Negative Energy pair? i.e. does this rely on the claim that antiparticles have negative mass-energy? – Hritik Narayan Dec 23 '14 at 14:50
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    No, the explanation refers specifically to photon pairs, and photons lack charge so there is no "antiparticle" version of a photon (or you could say that photons are their own antiparticles). Antimatter is not currently believed to have negative mass-energy. – Hypnosifl Dec 23 '14 at 14:58
  • I don't think the excerpt from the book is a very good explanation. Sounds a little bit like a negative energy sea analog to the Dirac equation? I thought we could do better than that these days. – CuriousOne Dec 23 '14 at 17:14
  • @CuriousOne - I found a link to an answer by John Rennie which goes into more detail and added a link to the end of my answer. It sounds like the book's description is a shorthand for some detailed mathematical derivation in modern quantum field theory, so presumably it wouldn't involve the "negative energy sea" concept of Dirac since I think that is seen as outmoded in modern QFT. – Hypnosifl Dec 23 '14 at 17:26
  • @Hypnosifl: I hope not and I would be surprised if that's how the actual calculation is really done. As always the handwaving explanations are fraught with problems... – CuriousOne Dec 23 '14 at 21:42