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If Hawking radiation can escape from black holes, could quarks perhaps become separated from protons despite it being "impossible" for that to happen?

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up vote 6 down vote accepted

Such a process is forbidden by energy conservation: the proton is the lightest baryon (that is the lightest bound state of three quarks).

hawking radiation finds it's energy by reducing the energy of the black hole, but there is not lighter baryon state for the proton to go to.

Baryon number violating proton decay processes are theorized, but have not been observed despite considerable effort by the large, low background, undergraound detector community. Indeed, modern neutrino detectors in effect grew out of the effort to calibrate early proton decay experiments, and now proton decay is a "parisitic" measurement on the neutrino detectors.

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Maybe you should add that quarks can never be free anyway, because of the nature of the strong force. Hawking radiation cannot contain free quarks. – anna v Jan 9 '13 at 5:21
@annav Well, yeah but... The OP seems to have noticed that Hawking radiation gets out of black holes from which "not even light can escape", so I assumed that he had heard "quarks can never be free" and proceeded threw this out as a hail mary pass. So I wanted to provided a different reason why it can't proceed. I freely admit that this is one conjecture on another. You are welcome to write a answer about confinement, of course, and I'd vote for it. – dmckee Jan 9 '13 at 5:28
no need, your answer is good – anna v Jan 9 '13 at 5:42
It should be mentioned that baryon number violation is actually a standard model process (due to instantons), but the rate is so small as to be practically irrelevant. The standard model decay is to leptons, not quarks and it is not a Hawking process at any rate. – Michael Brown Jan 9 '13 at 9:38

dmckee's answer is certainly a reason why quarks can't just tunnel out of protons. However, even if they could tunnel out, the process would be different to that of Hawking radiation. HR arises because the vacuum states of two frames are different. The observer freely falling into the black hole sees no radiation, whereas the observer held stationary outside the event horizon does see it. This is because what free-fall observer perceives as a vacuum (no particles) state is not a vacuum state according to the stationary observer. The two states are related by a Bogoliubov transformation.

In the proton case there is no such relationship between a free fall and non free fall frame in the relevant model, so the emission mechanism (if it existed) would have a different basis.

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protected by Qmechanic Jan 9 '13 at 13:31

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