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There is well known gravitational redshift of real photons. What about gravitational redshift of virtual photons of charged neutron star? Is electrostatic force become weaker while mass of charged neutron star is growing?

How can electrostatic field of a charged particle swallowed by a black hole escape from the event horizon? How can charged black holes exist at all?

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Dear voix, I doubt one can give a meaningful answer to your question because it's not a question: it's a sequence of misunderstandings. The term "red shift" means a change of the frequency of the photons or radiation, not a change of their intensity or number. The electrostatic force - because it's static - boils down to zero-frequency photons, so the red shift has clearly no impact on them. The force produced by virtual photons is not proportional to the "number" of virtual photons - which is really an ill-defined notion - or the "intensity" of individual photons - which is nonsense, too. –  Luboš Motl Mar 6 '11 at 18:11
    
    
@Lubos: while you are right about the misunderstandings, there is a simple way to understand the effect of redshift on virtual photons--- quantize the field using modes for a background gravitational field. Then the "virtual" photons (meaning temporary excitations of EM modes) feel the gravitational redshift. –  Ron Maimon Sep 3 '11 at 23:19

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There is a difference between the value of a field, and the excitations of that field. If you have a speherically symmetric charge distribution collapse to a black hole (while maintaining the spherical symmetry), there is no electric nor gravitational radiation--Gauss's Law and Birkhoff's Theorem work in tandem to keep the gravitational and electric forces constant far from the object, and since there is nothing to propogate, nothing needs to propogate. The field already lives out in the place where information about where the charge distribution needs to go.

What is a photon, then? A photon is a propogating disturbance in the electromagnetic field. You need to have an accelerating charge in order to produce such a disturbance, and there's no need to violate causality at the black hole horizon in order for a particle to feel force from a field that already exists locally--if, say, an electron were coming close to a charged black hole, it would see the field that existed there, scatter off of the field, and emit a photon. (and of course, it's new field would be set up and propogate to the black hole, and scatter the black hole--but there's no problem with information falling IN to a black hole, which is allowed to externally display it's mass, charge, angular momentum and linear momentum, all of which get inprinted on the horizon as matter falls in).

So, in summary, it's not necessary for a photon to be emitted in order for a second particle to feel a force--electrostatic forces in fact don't involve photons at all (at least, until the scattering event begins).

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In quantum field theory, we certainly do describe the force in electrostatic situations as being due to virtual photons. –  Ted Bunn Mar 6 '11 at 17:41
    
@Ted Bunn--but in the lab frame, you would interpret the virtual photon exchange as the incoming particle scattering off of the electrostatic field, emitting a photon, which would then scatter the particle at rest in the lab frame. There's no need for the lab frame photon to leave the lab frame particle. The picture I talked about involved photon exchange. –  Jerry Schirmer Mar 6 '11 at 17:58
    
And also, that's not how you treat things like the Hydrogen atom in QED, where you set up a background E-field and then quantize around that. If we're considering a particle scattering off of a black hole, the situation is much closer that that sort of thing than something like Babha scattering. –  Jerry Schirmer Mar 6 '11 at 18:02
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@Jerry is right. Here is an excellent descrption of virtual particles -in particular, why we should NOT think of them as particles at all. –  FrankH Oct 13 '11 at 13:15
    
Thanks for the link! Pretty interesting! –  Manishearth Feb 10 '12 at 16:36

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