# How can a black hole have any force charge?

FilipKunc

## Main Question or Discussion Point

how can a black hole have any force charge???

hello! i'm in junior high school, so please explain it to me in a understandable way...
how can a black hole have an electric charge? since the electromagnetic force messengers are photons, and even they can't escape from behind the horizon, than if there was an electron just outside the horizon, it woludn't be attracted to the black hole because its magnetic charge...
and the other thing i can't get is: even if the black hole has a charge, than what happens to it once the black hole evaporates due to hawking's radiation??? it's whole electromagnetically charged interior would be emmited in form of non-charged photons...

one more question about the black holes: when we analyze them from the point of view of the string theory (if i got the theory right...) than there is no singularity. if i remeber it right brian greene wrote in his book that there is a minimun size an object can have in the string theory. if there is a minimum size an object can have, then the black hole has no singularity, since there can't be any point of infinite density...

it would be nice if someone could reply.

Related Beyond the Standard Model News on Phys.org

Originally posted by FilipKunc
one more question about the black holes: when we analyze them from the point of view of the string theory (if i got the theory right...) than there is no singularity. if i remeber it right brian greene wrote in his book that there is a minimun size an object can have in the string theory. if there is a minimum size an object can have, then the black hole has no singularity, since there can't be any point of infinite density...

it would be nice if someone could reply.
Welcome to the PFs, FilipKunc.

I'll leave the first question to someone more knowledgeable than I...I think I know the answer, but I'm not positive.

As to the second question (quoted above), that's really part of the beauty of the string theory approach (I think loop quantum gravity has the same limit, but I'm not sure): there is no singularity anymore. Part of the reason for string theory is to unify General Relativity and Quantum Mechanics, and doing this requires making General Relativity applicable at the smallest possible levels. However, when you get down to a singularity of infinite density, the mathematics of GR (general relativity) breaks down. So, if they (string theorists) can get rid of the singularity (by imposing a minimum size), then they can use GR at all possible levels of size.

Originally posted by FilipKunc
how can a black hole have an electric charge? since the electromagnetic force messengers are photons, and even they can't escape from behind the horizon,
If you want to adopt a particle description of the electromagnetic force, as being mediated by photons, then virtual photons can get out of a black hole because they can travel faster than light. However, we can't observe them, because they're virtual. Real photons, which we can observe, cannot escape.

and the other thing i can't get is: even if the black hole has a charge, than what happens to it once the black hole evaporates due to hawking's radiation??? it's whole electromagnetically charged interior would be emmited in form of non-charged photons...
Hawking radiation can emit particles other than photons, including charged particles. However, this is a still problem, because presumably by the time the hole evaporates, its mass is smaller than the mass of the smallest charged particle; there are no massless charged particles. Nobody understands the final stages of black hole evaporation.

one more question about the black holes: when we analyze them from the point of view of the string theory (if i got the theory right...) than there is no singularity.
It is known that string theory can get rid of some singularities, but there is no proof that it gets rid of all singularities. In fact, some people think it must have some singularities ... see my references in the "Quantum fluctuations and singularities" thread.

FilipKunc

Originally posted by Ambitwistor
If you want to adopt a particle description of the electromagnetic force, as being mediated by photons, then virtual photons can get out of a black hole because they can travel faster than light. However, we can't observe them, because they're virtual. Real photons, which we can observe, cannot escape.

how can they travel faster than light...?
i thought that's not possible. they are called virtual because we can't observe them, but they still have to obey special relativity, don't they ?

Virtual particles don't have to obey all of special relativity. In particular, they don't have to be "on the mass shell" (or "on-shell"), meaning they don't have to satisfy the relativistic relation,

$$(mc^2)^2 = E^2 - (pc)^2$$

which is what constrains a massive particle to travel at less than c, a massless particle to travel exactly at c, etc.

The thing to remember, FilipKunc, is that the virtual particles don't "really exist" in the usual sense of the term. They are just an intermediate stage of a process.

jeff
Ways that charged black holes can form include gravitational collapse of charged bodies, presence of external magnetic fields, differences in the effect of radiative pressure on the electrons and ions of matter accreting onto black holes, and of course charges falling into them.

In realistic situations however, black holes can have little appreciable charge. This is because the charge to mass ratio qe/me and qp/mp of electron and proton is 1021 and 1018 respectively while the ratio of gravitational force to electrostatic force in the interaction of these particles with a BH of charge Q and mass M is of the order qQ/mM. Thus Q/M < (q/m)-1, since otherwise charges of like sign would be repelled from the BH while charges of the opposite sign would fall in and neutralize any electric charge.

Before I remark on physical effects associated directly with the electrical charge on black holes, note that although black hole emissions via the hawking effect will include all particles including photons, this has nothing to do with charges carried by holes. I'll say something more about this below.

From the classical viewpoint, since nothing can emerge from behind an event horizon, any electromagnetic field sourced by the charge on a black hole and living in the region exterior to it must've been emitted early enough before or during gravitational collapse to avoid entrapment (Analogous remarks apply for the gravitational field due to hole mass).

The quantum effect we're interested in here is particle production due to the presence of the electrostatic field produced by the charge on the hole. To understand this, note that the undisturbed quantum physical vacuum is the state without real particles. However, in an external field virtual particles may acquire sufficient energy to become real. Specifically, if the external field contributes enough energy E &ne; 0 to a virtual particle of invariant mass m, and off-shell momentum pvirtual we get p&sup2; virtual &rarr; p&sup2;real = p&sup2; virtual + E&sup2; = m&sup2; resulting in a real particle. Here, this field is produced by the electrical charge of a black hole, and so long as the corresponding potential Vh &equiv; Q/rh on the event horizon satisfies eVh > mc&sup2; with m and q particle mass and charge respectively, pair creation with one particle escaping and the other heading into the singularity will continue. The charge falling into the hole rapidly reduces Q to virtually nill. In fact, for BHs of mass less than about 105 solar masses this happens in a time much shorter than the characteristic time for evaporation so that such BHs can be assumed uncharged during nearly the entire period of their evaporation. Given the smallness of Q in realistic situations, holes must similarly be rather small for these processes to be significant.

The above discussed emissions we're created by the energy of the electrostatic field produced by the hole charge and hence reduce this energy over time. On the other hand, hawking radiation is created by the gravitational field of the hole and thus reduces it's gravitational energy represented by it's "invariant" mass causing an event horizon to shrink.

The question of the ultimate fate of any residual charge is tied up with that of the black hole that carries it, and this is something we can't say much about with certainty.

Originally posted by Ambitwistor
However, this [that "Hawking radiation can emit...charged particles"] is a still problem, because presumably by the time the hole evaporates, its mass is smaller than the mass of the smallest charged particle; there are no massless charged particles.
The characteristic energies of black hole emissions vary inversely to hole size (i.e., smaller holes are hotter) so masses of emitted particles increase as a hole evaporates. Thus it's in the early stages of evaporation when hole temperature is small that low mass particles are emitted: holes lose mass not by the emission of quanta but by the absorption of negative energy. The associated relevant issue is that the hawking effect can lead to gross violations of conservation of lepton and baryon number.

For example, the number of baryons out of which some neutron star was originally composed shouldn't leave imprints at late times on the hole exterior ("black holes have no hair"). This means that the hole should evaporate in a baryon-antibaryon symmetric manner. Thus the expected baryon number after evaporation should vanish. Indeed, one would expect that virtually all of a black hole's energy should be radiated in particles that carry zero baryon number since hole temperature is much less than the mass of any baryon until the very late stages of evarporation. Thus the conservation of baryon number in the process of hole formation and evaporation would require a highly implausible mechanism in that it would have to shut off emission of photons, neutrinos, etc with virtually 100% efficiency. (Keep in mind that in spite of the evaporation of the hole there's still a singularity present at which baryons might "leave" spacetime). Then the fact that quantum black holes enable some laws of particle physics to be transcended means that the presence of holes allow certain otherwise forbidden subatomic processes to occur. If baryon non-conservation was impossible, this would be quite worrisome. However, discussions of such non-conservative processes in connection with a number of ideas, particularly those having to do with grand unification, strongly suggest that such violations should occur.

Originally posted by Ambitwistor
...there are no massless charged particles.
I assume you meant electrically charged particles.

Tom Mattson
Staff Emeritus
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Jeffhunter,

Since you have elected not to receive Private Messages, I have to post this warning here. I have deleted your slanderous post about Jeff. The content of the PM I tried to send is below. Read it.

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Originally posted by Ambitwistor
Hawking radiation can emit particles other than photons, including charged particles. However, this is a still problem, because presumably by the time the hole evaporates, its mass is smaller than the mass of the smallest charged particle; there are no massless charged particles. Nobody understands the final stages of black hole evaporation.
If a black hole evaporates, then presumably it will evaporate to the point where it's mass can no longer sustain the gravitational force necessary to remain a black hole, right? Shouldn't it explode outwards at that point?