Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 1 down vote favorite
1

What is the simplest physical argument to claim that Proca theory (involving a massive spin-1 boson) is not renormalizable?

share|improve this question
Isn't it "Proca" with one "c"? – DarenW Nov 14 '11 at 7:06
@DarenW: Yes, Alexandru Proca only has one "c". – Simon Nov 14 '11 at 10:38
@Newman: If Ron's answer is acceptable (and it looks ok to me!), then you should accept it. If not, can you comment on it and say what is missing/wrong? – Simon Dec 1 '11 at 2:49

1 Answer

up vote 7 down vote accepted

This argument is going to be challenging, because the theory of a single massive vector interacting like a photon with scalar and spinor matter is renormalizable.

This is an accident of the U(1) gauge theory. It has a Stueckelberg version of the Higgs mechanism, where you take the limit that the Higgs mass goes to infinity while the condensate charge goes to zero. The limit is described on Wikipedia in the page on Higgs mechanism, under Affine Higgs mechanism.

This is an accidental property of massive electrodynamics. The reason the massive gauge field is nonrenormalizablei in general is that for a gauge group other than U(1), the charge cannot be arbitrarily small, so there is no decoupling limit for the Higgs.

The standard argument against a massive vector is that the propagator has a kk/ m2 term in the numerator iwhich means that longitudinal components don't have a falling off propagator, which means that loops involving longitudinal vectors blow up as a power at high k. In gauge theories, gauge inavariance guarantees that longitudinal bosoms aren't produced, but this requires that gauge invariance is broken spontaneously not by the Lagrangian. The Stueckelberg limit means that a mass term doesn't wreck renormalizability for U(1).

share|improve this answer
So far the theory you have described is a free theory since you have decoupled the Higgs. How does one add interaction taking into account that the coupling goes to zero and thus you cannot minimally couple any charged field? I'm probably missing something very trivial. – drake Jul 27 '12 at 22:12
Sorry, what a stupid comment! Different fields can have different charges. What I don't see yet is what happen with charge quantization and magnetic monopoles... – drake Jul 27 '12 at 23:02
1  
@drake: The monopoles go away, and charge cannot be quantized. This is the serious issue with Stueckelberg limit. But any pathology arising from this is only apparent in quantum gravity. In field theory, the existence of this limit is how Stueckelberg both discovered the Abelian Higgs mechanism and proved renormalizability of massive electrodynamics in the 1950s. – Ron Maimon Jul 27 '12 at 23:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.