3

What are the virtual particle pairs generated during the Hawking radiation? If a photon is emitted by Hawking radiation, what is its negative energy partner which fell into the black hole? Does it have a name? What is a negative energy photon anyway? Antiphoton? When a negative energy photon annihilate with a positive energy photon, can we see anything? Or just nothing since it just goes back to the vacuum? Thanks!

blue sky
  • 39
  • Have a look at http://physics.stackexchange.com/q/134948/ – anna v Jan 22 '15 at 05:03
  • 4
    I think that Hawking said in his original paper that this was a dangerous analogy. – jinawee Jan 22 '15 at 12:22
  • 2
    Seconding jinawee - the picture of virtual particle pairs is categorically not the right way to think about Hawking radiation. Quite obviously it must be wrong, because it is a loop level effect, and loops in QFT have to close, which they don't in this heuristic picture. You're much better off thinking about Hawking radiation as a horizon effect like the Unruh effect. – Edward Hughes Mar 03 '16 at 10:42

1 Answers1

1

To start with, a photon is an antiparticle to a photon. There is no conservation of photons and certainly no negative energy photons.

Any body in space with a temperature larger than the "bath" it is in, i.e. the temperature of cosmic background radiation, will be radiating a black body radiation. The theoretical treatment is a combination of thermodynamics with field theoretical assumptions at the micrpscopic level, and that is the exposition in the wiki article..

The creation and annihilation of particles at the horizon is invoked to explain how particles can escape the gravitational attraction of the black hole at the microscopic level. As photons are antiparticles of photons and can be generated by in falling electrons for example, the only balances needed are directional and energy balances, so that the photon (or another particle) will not be trapped and fall back in.

In the case of the black body radiation of an ordinary body the temperature is such that the escaping radiation is a low energy photon, created by some transitions within the body and escaping from the surface. The energy balance is with the internal energy of the body, which cools incrementally. In the case of the black hole the energy balance is with its gravitational energy .

Here is a Feynman type diagram for the generation of particle/antiparticle by the Hawking radiation

hawking radiation

To calculate the probabilities quantum mechanically would take exact diagrams. These pairs are virtual and they can also have photon vertices , which will have a probability of escaping the horizon and form a real photon spectrum from the black hole. For example:

feynman diagram

anna v
  • 233,453
  • Thank you, Anna, for the quick reply. In your first sentence, you said: "a photon is an antiparticle to a photon. There is no conservation of photons and certainly no negative energy photons." But Hawking said that the virtual particle pair needs to have one with positive energy and one with negative energy. The one with negative energy fall into the black hole so to reduce its mass. The one with positive energy is emitted to become Hawking radiation. How do you explain that? Thanks! – blue sky Jan 22 '15 at 20:10
  • just saw this. The potential energy is negative in doing the energy balance.A feynman diagram that could be calculable needs a gravitational vertex, and the energy is balanced by the gravitational field of the black hole. – anna v May 30 '15 at 14:59
  • @anna v: What attracts the negative-energy particle into the black hole--something positive; but, what is that? – tony Aug 05 '23 at 13:08
  • @tony energy is neither attracted nor repulsed. It obeys conservation laws. example: a particle decays to two particles. From momentum conservation in the center of mass where the decays is at rest , the particles will have equal and opposite momentum, and from energy conservation an amount of energy dependent on their masses. In the HR if it happens that the momentum direction is through the horizon, the energy can be labeled +, whereas the one going backwards gets a -, which means the total energy of the black hole should diminish. – anna v Aug 05 '23 at 15:35
  • @anna v: Thank you. When the black hole evaporates, what happens to the singularity? Is this a tiny (high-density) piece of rock? Could we see it/ reach out and touch it--assuming that we could survive the conditions inside the BH? – tony Aug 07 '23 at 11:52
  • the singularity is a mathematical point in the models used to describe the black hole. It does not have a separate existence afaik – anna v Aug 07 '23 at 13:41
  • this might interest you https://en.wikipedia.org/wiki/Micro_black_hole#Stability – anna v Aug 07 '23 at 13:54
  • @anna v: This topic is frustrating, confusing; but, fascinating. The singularity, then, is a point, in a mathematical model. This point has "zero-volume" and therefore, "infinite-density" which produces the tidal waves of gravitational power. The model works! Some physicists do not believe that "infinite-density" can exist, even in a BH, where the laws of physics do break down. That a collapsed star becomes a high-density [thing] is easy to understand; but, the exact nature of this "unseen", gravity-generating species is what? – tony Aug 11 '23 at 11:52
  • @anna v: Is it that we are at the frontier of current knowledge/ theory? Thanks again. – tony Aug 11 '23 at 11:54
  • @tony If you read about the microholes stability you will see that the whole thing decays fast so the singularity that modeled it disappears. or becomes a new type of particle as some theories state. The same will be true for any black hole , after the mass is radiated away, there is no singularity per se. observations and theory models agree at present. – anna v Aug 11 '23 at 12:01
  • @anna v: When an observer (i) watches person (ii) approaching a BH, (ii)'s motion slows until it is infinitely slow (dilation); (i) effectively, will never see (ii) enter the BH. For (ii), his time remains normal. His clock ticks normally. He falls into the BH in his own normal time. How can both of these things occur? – tony Sep 30 '23 at 15:11
  • @tony general relativiry is not my field, I vaguely understand that it is all in the mathematics and the definitions, sorry Icannot discuss this. – anna v Oct 01 '23 at 07:50
  • @tony the answer by ron maimon here may help https://physics.stackexchange.com/questions/16053/how-does-the-star-that-has-collapsed-to-form-a-schwarschild-black-hole-appear-to . – anna v Oct 01 '23 at 08:36