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The simple picture of Hawking radiation is that a pair-antiparticle pair is produced near the event horizon, then one falls into the black hole while the other escapes. Suppose the particles are quarks-antiquarks, which experience quark confinement thanks to QCD. If one of them is swallowed by the black hole, its partner is left alone. Eventually the quark gains enough energy and turns into a hadronic jet.

Is my line of thinking correct? If yes, is it (or QCD in general) taken into account when calculating Hawking radiation?

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3 Answers 3

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Actually working out a rigorous prediction for Hawking radiation, you need to solve the equations for the relevant QFT semi-classically--treating the metric of spacetime as a substitute for the Minkowski metric usually used in QFT. This is similar to deriving the quantum mechanics of the Hydrogen atom without paying attention to the fact that photons exist and the electromagnetic field is quantized.

Anyway, when you work out the Hawking effect for a scalar field, or a fermion or whatever, what you do is start with the Penrose-Carter diagram for the Schwarzschild spacetime. You define a set of linearized plane wave solutions at past infinity, and then use your curved-space equations of motion for the field to evolve forward at time,and then measure what projection onto a set of linearized plane wave solutions that you get at future infinity. Hawking's result in his original paper was that even if you define the "in" state at past null infinity to be the vacuum state (i.e., the state with no particles), then you STILL have a blackbody distribution at future null infinity--thus, we interpret the black hole as radiating. Note how none of this relies upon virtual pair creation, and how it ties pretty directly into ordinary QFT. The bit about pair creation mostly serves to give a nice physical picture as to why there could be radiation at future null infinity.

Now, there are a few reasons why this procedure will not work for quarks. Marek points out the clearest of them--there is no "in" or "out" state that describes a set of free, unbound quarks or gluons. The best we can do is approximate such a state at high energy. So the semi-classical formalism breaks down on some level.

The second reason that this won't work out is more astrophysical than anything. Black holes are known to likely have a very small physical charge, because if they were to acquire a charge of any size, they would have a very large electromagnetic field that would polarize everything around them, suck in oppositely charged ions, and neutralize themselves. If, by some chance, a black hole did gain some sort of color charge by creating an oppositely colored particle near it, the force on that particle would be so great that it would not be able to escape to infinity, anyway, and the black hole would remain uncolored, and we wouldn't see free quarks escaping to infinity, so the effect you describe shouldn't be possible.

If you're asking whether or not we would see bound states of quarks escaping, however, the answer to that question is almost certainly yes. A pion or a proton is a perfectly good state to do semiclassical analysis with, and (I don't know this for sure, but I'm pretty sure that someone has looked at this before, and there isn't a plausible counterargument that I can think of) that we would get perfectly sensible blackbody radiation out of using them as a test particle, so long as the mass of the black hole is low enough (and therefore, its temperature high enough) to not allow the production of such massive particles.

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Thanks for your answer! Can you elaborate your last sentence, esp. the "as the mass of the black hole is low enough to not allow the production of such massive particles" part? –  mtrencseni Dec 12 '10 at 20:11
    
@mtrencseni: Sure--the temperature of the Hawking radiation is inversely proportional to the mass of the black hole--small black holes radiate hotter radiation than large black holes. But, the Boltzmann distribution peaks when the Energy is equal to $kT$. If the temperature is sufficiently small that $kT \ll m_{p}c^{2}$, then there won't be enough energy in the radiation profile to create a particle of mass $m_{p}$. For astrophysical black holes this means, if the Hawking radiation is even detectible, it will mostly be made of photons, which are massless, and perhaps neutrinos. –  Jerry Schirmer Dec 13 '10 at 18:25

I think your reasoning should be correct.

However, as I understand the Hawking radiation (i.e. not much) the effect is derived independently of the microscopic theory. You just assume that particle (which is to escape from around the horizon) needs to accelerate (a lot) in order to overcome black hole's gravity and then notice that accelerating observer will register a black body radiation. This is called Unruh effect.

Now, you can certainly try to derive the effect from the first (i.e. QFT) principles. E.g. this paper claims that it is indeed possible to derive the results from the free QFT. But for realistic QFT (like QCD) this doesn't seem technically feasible and the results you obtain (i.e. Hawking temperature) can depend on the precise theory used.

This paper talks about similar effect but instead from tunneling from under the black hole horizon it discusses tunneling out of the confinement horizon (this is called white hole because confinement means white hadrons; not to be confused with white holes from GR).

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I think the reasoning is correct, but the probability of that happening might be very very low, depending on how much energy is needed to separate the quark/antiquark pair. So it might be unprovable. –  Ebenezer Sklivvze Dec 12 '10 at 14:19

Most calculations of the Hawking effect assume free quantum fields. This assumption breaks down for strongly coupled quantum chromodynamics. As the Hawking temperature is much lower than the QCD deconfinement temperature, there isn't enough energy to hadronize "virtual" quark-antiquark pairs. Instead, the particle just outside the event horizon gets pulled into the black hole by the confining QCD flux tube. It can't escape.

Actually, for astronomical sized black holes, the Hawking temperature is so low only massless gravitons and photons can be radiated. Even neutrinos are too heavy, never mind quarks.

But for sufficiently light mini black holes, we can have the creation of hadron pairs, with one hadron escaping, and the other falling into the black hole.

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