dextercioby said:
I think it's an open matter. It involves the existence of consistent cross-couplings between a massless gauge abelian one-form field and an arbitrary massless field and then interpreting the nonzero coupling constant as the particle's electric charge. Something could however go wrong: the appearance of consistent self couplings of the massless field resulting in the appearence of a mass term...
Daniel.
On the one hand there is the experimental fact; "we see mass without charge, but no electric charge without mass", on the other hand, theoretically, 4 dimensional, massless QED with invariant vacuum is possible. That is, theoretically, nothing wrong with massless fermions.
Weinberg-Witten theorem;
Limits on massless particles,
Phys. Lett. B96 (1980), 59-62.
puts sharp restrictions on the possible massless field, they cannot be completely arbitrarly as you suggest.
Basically, In dimension n\geq4, the spin of a massless particle is classified by a representation of the little group SO(n-2). If a local, conserved, symmetric and gauge-invariant stress tensor exists, the allowed representations for massless fields are the spinor representation(s)[there are two of these if n is even and one if n is odd], and the exterior powers of the fundamental (n-2)-dimensional representation including the trivial(scalar) representation.The theorem shows that Poincare invariant global charges vanish except for massless particles in the trivial or spinor representation;(j=0,\pm1/2).
Without adjusting some parameters to make the particles massless,
Massless particles must be massless for a reason.One possible reason is supersymmetry. But there are othere possible reasons;
Scalar particles are massless when there is a broken symmetry(
Goldstone).
Fermions in
four dimension are massless when they are
chiral, that is, when there is
unbroken chiral symmetry.
If the massless, spin1/2, particles transform in a representationR of some unbroken symmetry group
G, then the massless particles of spin(-1/2) transform, according to the CPT theorem, in the conjugate representation R^*. If R and R^* are distinct, then this spectrum cannot be perturbed in a
G-invariant way to give masses to the fermions. So, nothing will "go wrong" and you won't get a mass term.
This shows that massless QED is possible, that is, we cannot, using the tools of QFT, show that massless charged fermions do not exist.
However, the question, which I don't know the answer to, is; Is classical electrodynamics of massless charge possible?
regareds
sam