I read about vacuum energy. It explains the Hawking radiation, the black hole necessary radiation:

Physical insight into the process may be gained by imagining that particle-antiparticle radiation is emitted from just beyond the event horizon. Vacuum fluctuations are always created as particle–antiparticle pairs. The creation of these virtual particles near the event horizon of a black hole has been hypothesized by physicist Stephen Hawking to be a mechanism for the eventual "evaporation" of black holes.

Is this "insight" supposed to make us to think that

  1. the virtual pair created is a part of black hole so that when half flies away then the BH mass is reduced? The article on vacuum energy says that the vacuum energy is an underlying background energy that exists in space throughout the entire Universe. So, it is not related to the black holes, though one particle of the couple may create the salute that you may consider as BH evaporation, it actually covers up the fact that the second particle flies into the BH, increasing its mass.

  2. Why do we believe that the receding half of the couple does not fall back to the Black Hole? Yes, it has escaped the horizon but the (pretty strong) BH gravitation is still in action, and it is only few neutons less than $\infty$ because we are still almost at the horizon initially. This means that to actually escape and not to fall back into BH, the speed of the receding particle must be virtually infinite. I doubt that it is likely that you will have many such virtual particles right at the event horizon. It is much much (Almost surely) more likely that "escaped" half of the couple will also eventually fall onto the black hole, increasing its mass.

So, considering the virtual particles of vacuum energy, we find two ways to add mass to the BH. Why do they call it BH (mass) evaporation?

  • $\begingroup$ Even if #2 were true, there exists some radius farther out at which the probably of capture/escape is exactly 50/50, so it's not terribly relevant where exactly that point is (at least in the context of this question). $\endgroup$
    – Dan
    Mar 16 '17 at 18:44

First of all try to understand what vacuum fluctuations are. Virtual pairs exist everywhere appearing and disappearing below the threshold of our detection. Every photon, every charged particle is "dressed" by the vacuum fluctuations as it goes along. The following diagrams are a shorthand, they represent integrals that have to be calculated in order to get measurable predictions.

virtual pairs

The green line is a particle with a mass, the blue squiggly on is a photon.

You cannot evaporate an object by adding mass to it. When you create a couple of real particles out of nothing and send one half to the Black Hole, you provide insight on Hawking evaporation only to very narrow-minded people. They can see that something is emitted from the horizon but this something hides the crucial fact -- another half of this something falling into the BH, increasing its mass.

The analogue you have in mind is wrong. Just over the horizon of the black hole the vacuum fluctuations still exist and are very strong, the energy supplied by the strong gravitational field of the BH. The horizon is the limiting surface from which a real particle cannot escape the gravity of the black hole, but a fluctuation just over the horizon of an electron positron pair, for example , where the momentum of the electron is pointing down whereas the momentum of the positron away from the black hole, the positron has a chance to escape, its mass is very small and there are large tails in the momentum distributions available. The energy for this is supplied by the black hole gravitational potential, by attracting the electron it kicks off the positron. All these have to be calculated with the correct integrals to see if the numbers are such as to lead to black hole evaporation, and they do.

vacuum above horizon

your 2. is confusing. Once the positron is with momentum outside the horizon it is matter of calculations whether it will fall back or not. If it is from the high momentum tail of the distributions of virtual momenta it has the chance to escape, as the mass is very small.

Equivalently one could have just over the horizon virtual electron positron annihilation to two real photons, one absorbed the other kicked out leaving with velocity c. It is mostly the photons that escape, a black body radiation, and reduce the mass of the black hole because of feeding off the gravitational field of the BH.

Anyway the effect for large black holes is very tiny, swamped by infalling radiation. It gets large the smaller the size of the black hole.

  • $\begingroup$ Wow, this puzzled me too (I'm not a scientist) and I was looking for an answer. This is the best one I've seen so far, although I still had to work to wrap my head around it. It sounds like you're saying that the gravitational field around the BH amplifies the vacuum energy around it, so when a particle escapes, it takes that energy away from the black hole. Do I understand correctly? $\endgroup$ Jun 13 '20 at 16:36
  • 1
    $\begingroup$ @LawrenceI.Siden not really. There is no vacuum energy to be amplified. It is the quantum mechanical effect that allows the stealing of gravitational energy from the black hole by allowing interactions with the gravitational field of the black hole to supply the energy so that the particles can become real. $\endgroup$
    – anna v
    Jun 13 '20 at 19:30
  • $\begingroup$ I believe you, but what does it mean "There is no vacuum energy to be amplified."? How could virtual particles even exist if there were no vacuum energy? Please understand, I'm not a scientist. I just like to read about it and listen to lectures. So pretend you are talking to an imbecile if you can bear to humor me. Thanks for understanding. $\endgroup$ Jun 14 '20 at 19:25
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    $\begingroup$ @LawrenceI.Siden virtual particles are a mathematical place holder in the calculations of interactions using feynman diagrams. they are a mathematical mnemonic of quantum numbers , that is their only existence. formulae in a specific calculation. When in socker a player calculates subconsiously how to hit the goal, do all the trajectories he instantaneously considers , exist? Only one does, the one he throws because his subconscious calculations give the best probability of success. In interactions all the probable energy momentum exchanges are mathematically calculated $\endgroup$
    – anna v
    Jun 15 '20 at 3:21
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    $\begingroup$ to give the probability of interaction. $\endgroup$
    – anna v
    Jun 15 '20 at 3:21

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