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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!

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  • $\begingroup$ Have a look at physics.stackexchange.com/q/134948 $\endgroup$ – anna v Jan 22 '15 at 5:03
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    $\begingroup$ I think that Hawking said in his original paper that this was a dangerous analogy. $\endgroup$ – jinawee Jan 22 '15 at 12:22
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    $\begingroup$ 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. $\endgroup$ – Edward Hughes Mar 3 '16 at 10:42
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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

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  • $\begingroup$ 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! $\endgroup$ – blue sky Jan 22 '15 at 20:10
  • $\begingroup$ 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. $\endgroup$ – anna v May 30 '15 at 14:59

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