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Let's says there is a fundamental particle:

  • That is so massive that it is a black hole by itself (Compton wavelength < Schwarzschild radius)
  • That carries a conserved quantum number (e.g. charge of an exotic interaction) which no lighter particle carries

Would it be able to emit Hawking radiation? If not, does it contradict with the classical arguments (entropy analogy, pair creation at the horizon etc.) regarding the origin of Hawking radiation?

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2 Answers 2

As John Rennie already put it, Hawking radiation is an semi-classical effect derived by treating the spacetime classically, including the source terms if any, and quantizing fields in the curved background. In addition general relativity has problems with distributional sources, so it is not clear how to treat elementary particles in this case. So one cannot proceed with Hawking's arguments in the case you describe.

Nevertheless, one can try to think of it in very qualitative terms. You relate the Compton lenght with Schwarzschild radius, but remember that in general the particle should have spin, and as you already put it, some charge. Considering only electromagnetical interactions we know that a black hole should be described by the Kerr-Newman metric, that is defined by three parameters: mass $M$, angular momentum $J$ and charge $Q$. For the metric to describe a black hole the parameters must satisfy the inequality $Q^2 +(J/M)^2\leq M^2$, otherwise we have a naked singularity. Curiously all known elementary particles in the standard model (with the exception of the Higgs boson) violate the aforementioned inequality, and therefore naively GR ascribe to then the metric of a naked singularity, assuming of course that the neglected weak and strong charges do not alter this radically.

As for the Higgs boson it is (un)fortunate that charged scalar fields violate uniqueness results for black holes, so we really can't say much about it. In any case what I wished to bring forth with this discussion is that you cannot look only at the relation between Compton lenght and Schwarzschild radius, but must consider the part that spin and the charges have to play in this. In particular if you think this argument has some sense you recognize that in the presence of charge and spin you need an even larger mass to be in the black hole side of the inequality, and therefore you go even further (that is to larger curvatures) from the case discussed in Hawking effect. This goes to show how far we are from being able to tackle this problem with any confidence.

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This is really a comment, but it got too long for the comment box.

The problem is that the Hawking calculation is semi-classical. That is, it assumes the spacetime curvature is given by the (classical) Einstein equation. Once the radius of the event horizon decreases into the quantum regime the approximations Hawking used are no longer valid. You would need a proper quantum gravity calculation to make any progress.

In fact I'm not sure that your concept of a particle would be a valid description either. The only discussion I've seen of this type of physics was a string theory talk describing the final stages of black hole evaporation $^1$. This was some time ago, and to be honest I understood very little of the talk anyway, but my recollection is that the final stage of the black hole evaporation left behind a string in a highly excited state. The point is that at these sorts of energies it may well be that quantum field theory is not an adequate description of matter, so you can't postulate a particle with well defined properties. That would make your question as it stands somewhat devoid of physical meaning.

$^1$ Whether string theory is the correct description or not no-one knows - as I recall the presenter of the talk admitted it was highly speculative.

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