Virtual particles and Gauss's Law

[PeterPumpkin]

Does Gauss's Law apply to virtual particles?

For example, when computing the field around a real proton, is the net charge in Gauss's Law the proton charge plus the contribution of all the virtual charges within the closed surface?

(I'm thinking about the screening of charges by virtual charges. I realize the contribution of most virtual charges will cancel.)

 

[Bill_K]

For example, when computing the field around a real proton, is the net charge in Gauss's Law the proton charge plus the contribution of all the virtual charges within the closed surface? (I'm thinking about the screening of charges by virtual charges. I realize the contribution of most virtual charges will cancel.)

This is an interesting point, PP. Not just most of the virtual charges cancel, all of them do. They are produced in pairs, and there's exactly as many positive ones as negative ones. They do affect the Coulomb field, but not in the way that's usually visualized. Most people think of the screening as a gradual thing. That is, as you approach the origin you would gradually see more and more charge. Not so. Picture the virtual pairs as little dipoles pointing toward the origin. By symmetry the field they produce adds up to zero. Everywhere except the origin, where it is singular. And in fact if you use QED to calculate the leading radiative correction to a Coulomb field, that's what you get:

V(r) = e²/r + C δ(x)

where C is a constant. The correction to the potential is a three-dimensional delta function. If this is a proton we're talking about, the only electron orbitals affected by the extra term are the S wave electrons, whose wavefunction at the origin does not vanish. The effect produced by this is the well-known Lamb shift.

 

[PeterPumpkin]

Thanks. I was wondering could one calculate the screening from the virtual particle model?

Seconding, I was wondering do all the virtual charges vanish within the gaussian suface? The virtual charges will be polarised and so, if you thinking of the charges that intersect the gaussian surface, there must be more negative inside the surface than outside. That's assuming we're talking about the screening round a proton.

 

[tom.stoer]

Using the temporal gauge A°(x)=0 one can quantize QED using physical states which are in the kernel of the Gauss law, i.e. G(x)|phys> = 0. So the physical Hamiltonian, physical electrons, positrons and photons always respect the Gauss law constraint.