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The Standard Model gives non-zero mass to the electron via the coupling to the Higgs field. Issues of renormalizability aside, this is fundamentally unrelated to the fact that the electron couples to the EM field. However, if the Higgs mechanism did not operate - that is, if there were no vacuum symmetry breaking, the electron field would have no effective mass term. In QFT perturbation theory, this model offers no special difficulty.

My question is, what is the classical limit of this theory, if it has one? Does the electron acquire a purely EM mass? If the low energy renormalized mass is set to zero (is there an obstacle to doing that?) what do the classical field configurations look like? My puzzlement is related to the classical description of the EM field of a relativistic charged particle - namely, that it becomes "flattened" in the direction of motion, just like a Lorentz-contracted sphere. That is, the field is weaker than for a motionless particle in the longitudinal direction, and stronger in the transverse directions. The limiting case is that the field becomes concentrated on a 2-d surface transverse to the particle location, where it has infinite strength - obviously an unphysical situation. So what actually happens?

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This has some bearing on the question in your title

http://arXiv.org/abs/hep-th/9112020

There is an analogous problem in general relativity, the gravitational field generated by a massless particle is called the Aichelburg - Sexl metric.

Couple more comments:

Perturbation theory in the presence of massless charged fields is not unproblematic, there are infrared divergences one has to deal with. These are much more problematic than soft photon divergences. I think Weinberg's field theory book has a discussion of this.

Massless electrons would have an extra chiral symmetry, which should forbid the appearance of mass term.

But, I don't think these quantum field theory issues are directly related to your question, which is essentially one in classical electrodynamics.

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  • $\begingroup$ Thanks. The doc at that link is broken (no pdf or other readable format) and the source is buggy tex. But it's fixable with a little work... $\endgroup$
    – Foster Boondoggle
    Jan 20, 2011 at 3:51
  • $\begingroup$ Wouldn't the chiral symmetry in this case be anomalous? Incidentally, the Jackiw et al paper is essentially classical. The field they derive indeed has the singular form of vanishing field strength everywhere except on the plane orthogonal to the velocity & containing the particle, where it's infinite (but with finite integral over any interval orthogonal to the plane). $\endgroup$
    – Foster Boondoggle
    Jan 20, 2011 at 4:39
  • $\begingroup$ I think you are right, but there are no Abelian instantons, so the axial current is conserved and the electron stays massless. $\endgroup$
    – user566
    Jan 20, 2011 at 4:47
  • $\begingroup$ Yeah, I thought your question was about classical field theory. The Aichelburg - Sexl metric has similar features, it is essentially a shock wave. $\endgroup$
    – user566
    Jan 20, 2011 at 4:49
  • $\begingroup$ So what happens, either classically or in QFT, to avoid the singular behavior? In QFT, maybe total screening, as QGR suggests? $\endgroup$
    – Foster Boondoggle
    Jan 21, 2011 at 4:09
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With a massless charged fermion, the electromagnetic coupling strength will be totally screened, with $1/e^2$ increasing logarithmically with the scale $R$. The electric field will be Coulomb multiplied by this logarithmic factor in $R$.

Actually, the massless charged fermion will be travelling at the speed of light, but this logarithmic screening will still happen.

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