# 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?

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).

• Sorry, but does this phrase "...This is the massive electrodynamics model, and this limit shows why it is renormalizable..." say that the pure massive fermion QED (if we turn on the fermions and turn off the scalar sector) is renormalizable? – Name YYY Dec 29 '16 at 10:36

Starting with the Lagrangian for a massive $U(1)$ vector boson $A_\mu$ which like you said has 3 DOF:

$\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 A^\mu A_\mu$

now if we change variables to $A_\mu\rightarrow A_\mu - \partial_\mu \theta$ and we have (Note that $F^{\mu \nu}$ and hence $F_{\mu \nu}F^{\mu \nu}$ is invariant under this transformation.):

$\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 (A_\mu - \partial_\mu \theta ) (A_\mu - \partial_\mu \theta )$

and rescale $\theta \rightarrow \frac{1}{m}\theta$

$\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - (m A_\mu - \partial_\mu \theta ) (m A_\mu - \partial_\mu \theta )$

$= - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 A^\mu A_\mu +2m \partial_\mu \theta A^\mu -\partial_\mu \theta \partial^\mu \theta$

The crucial point now is that the Lagrangian has a gauge redundancy ($A_\mu \rightarrow A_\mu +\frac{1}{m} \partial_\mu \psi$, $\theta \rightarrow \theta + \psi$) which decreases the DOF of $A_\mu$ by 1, down to 2 DOF. You might object that the mass term is still there - but there is a mixing term between $A_\mu$ and $\theta$ so its not correct to think of these as independent propagating DOF.

Finally, at very high energies any operator with a positive mass dimension is irrelevant so we would have

$\mathcal{L} \approx - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - \partial_\mu \theta \partial^\mu \theta$

and $A_\mu$ has only 2 DOF and there is another (free) DOF that is decoupled. The high energy limit is the massless limit since we need a scale to compare the mass to.

As far as classical verses quantum, my analysis was essentially classical, but it goes through for the quantum case with the caveat that we will have to add gauge-fixing terms for the gauge redundancy that I have introduced.

You might ask, well what happens at low energy - well then, we should have really stuck with the original Lagrangian we started with, since that works well when the mass term is relevant and we would just have a vector field with 3 DOF.

As for your question regarding what happens to the electron in all this, the coupling to the original gauge field will be of the form

$A_\mu J^\mu$

which under the change of variables $A_\mu\rightarrow A_\mu - \partial_\mu \theta$ goes to

$A_\mu J^\mu - \partial_\mu \theta J^\mu$

under integration by parts the last term is

$\theta \partial_\mu J^\mu$

and $\partial_\mu J^\mu=0$ for a conserved current, which is logically necessary if we want to couple a massless gauge field to it.

Note that this is in stark contrast to a nonabelian gauge theory with mass terms since the would-be goldstone bosons form a non-linear sigma model which is non-renormalizable. See What evidence is there for the electroweak higgs mechanism?

As a final word, I think Ron's response is more concise and answers your original question much better. I would delete my response, but it seems as though some people have found it useful so Ill leave it.

• But in electromagnetism, the electron already defines a mass scale, so you cannot rescale without losing the content of QED. – Arnold Neumaier Oct 24 '12 at 16:34
• @ArnoldNeumaier - I'm not sure what you mean. I feel the heart of the question is about DOF in a U(1) gauge field, which my toy model addresses. If I couple my gauge field above to a fermion, isn't it just a spectator to my discussion above? – DJBunk Oct 24 '12 at 17:02
• As I understood, the question is about whether massive QED has ordinary QED as massless limit. As ordinary QED is used in the low energy region, one cannot argue with irrelevant operators. As you write, there are 3 DOF in thiscase. The question is why at low energies but $m\to 0$ (while keeping the electron mass fixed) the transversal degree of freedom fades out. I gave a quick answer without detials, but don't have time right now for a thorough answer. – Arnold Neumaier Oct 24 '12 at 17:15
• Well, I am happy with the not-interacting version, this is, without electron. To put the electron in, is a plus, but it amounts to study its interaction with the extra scalar field, doesn't it? – arivero Oct 26 '12 at 0:25

QED is the massless limit of massive QED. The transversal modes are more and more suppressed as the mass gets tiny. This can be seen already in the noninteracting version, where $e=0$, so that the photon field decouples.