quantum field theory - What is the massless limit of massive electromagnetism?

Consider electromagnetism, an abelian gauge theory, with a massive photon. Is the massless limit equal to electromagnetism? What does it happen at the quantum level with the extra degree of freedom? And, what does it happen at the classical level? We ca not get an extra massless classical scalar field, do we?

Answer

There is a simple way to understand the massive electrodynamics Lagrangian and limit, which is the Stueckelberg (Affine Higgs) mechanism. This is matematically equivalent to DJBunk's answer, but it is slightly more intuitive physically.

Consider an Abelian Higgs model, with a massless electrodynamic vector potential $A$ and a scalar field with a $\phi^4$ potential

$$ S = \int |F|^2 + |D\phi|^2 + \lambda (\phi^2 - a)^2 $$

Then consider the limit that the charge e on $\phi$ goes to zero while the mass of the Higgs goes to infinity ($a\rightarrow\infty$), in such a way that the product $ea$ stays constant. In this limit, you can write the complex scalar as:

$$ \phi = R e^{i\theta}$$

And the R excitations have a mass that goes as a, and goes to infinity, while the $\theta$ excitations are eaten by the A field and together make a gauge boson of mass ea. This is the massive electrodynamics model, and this limit shows why it is renormalizable--- you can take e to zero in a U(1) gauge theory, because there is no charge quantization.

In this model, it is obvious what the massless limit of massive electromagnetism is: this is the limit that $e=0$ for the Higgs field. In this case, the Higgs is decoupled from the gauge field, and you just have a massless gauge field. The longitudinal degree of freedom just decouples. This is the clearest way to see why it must be so in my opinion.

When the mass is small, the longitudinal degree of freedom, the one that comes from the infinitely heavy Higgs in this model, is almost decoupled, so that the limit is smooth. The theory can be analyzed by starting with the massless electromagnetism and adding the Higgs field as a perturbation (so long as you work in the effective potential formalism).