# Gauge-invariance of the Maxwell Lagrangian in Srednicki's book

When it comes to the demonstration of the gauge-invariance of the Lagrangian of the Maxwell-theory Srednicki's book proceeds as follows:

$${\cal L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + J^\mu A_{\mu} \tag{54.21}$$

I am interested in the second term $$J^\mu A_\mu$$. Under a gauge transformation the 4-vector potential transforms as $$A'_\mu =A_{\mu} - \partial_\mu \Gamma$$ In order to show gauge invariance one has to show that

$$J^\mu ( A'_{\mu} - A_{\mu}) = -J^\mu \partial_\mu \Gamma = (\partial_\mu J^\mu)\Gamma -\partial_\mu(J^\mu \Gamma)$$

can be "neglected". The argumentation of Srednicki is that the second term can be neglected since upon considering the action the corresponding integral over $$d^4x$$ vanishes and the first term disappears because of current conversation. The latter, however, would mean that the Lagrangian is only gauge-invariant for the $$A_\mu$$-field configuration that fulfills the Maxwell-equations, otherwise the current would be not conserved. Is the Maxwell-Lagrangian (54.21), in particular the coupling term, gauge-invariant for all field configuration or only for the special one that fulfills the field-equations?

In particular in view of the comments given to the post Is the Dirac Lagrangian locally gauge invariant without gauge field $A$? I am confused.

EDIT Actually, I don't know if a background current would automatically fulfill the equation $$\partial_\mu J^\mu=0$$, but at least if the Maxwell-equations are fulfilled, i.e. $$\partial_\nu F^{\mu\nu} = J^\mu$$ by taking a second derivative we get: $$0=\partial_\mu\partial_\nu F^{\mu\nu}= \partial_\mu J^{\mu}$$. So from this perspective the current conservation seems to a result of the Maxwell-equations. And this is what Srednicki mentions the page that just precedes that of expression (54.21). That suggests strongly that the current conservation only is valid if the Maxwell-equations are fulfilled.

• In this case the current is not regarded as being an additional degree of freedom. It is "given" in the Lagrangian, and it being conserved does not depend on the Maxwell equations being fulfilled. On the other hand, J is given in terms of the Dirac field in the Dirac theory, and this current being conserved does depend on the equations of motion. – Krup'a Jan 16 at 14:36
• @Chiral Anomaly I edited my post correspondingly. – Frederic Thomas Jan 16 at 16:03

the current conservation only is valid if the Maxwell-equations are fulfilled,

the logic sould be the other way around, like this:

the Maxwell-equations are fulfilled only if the current conservation is valid.

Gauge-invariance does not depend on the behavior of $$A$$. It only depends on the behavior of $$J$$. The behavior of $$J$$ is constrained by consistency with $$A$$'s equation of motion, and as long as $$J$$ satisfies that consistency condition, the Lagrangian is gauge-invariant whether or not $$A$$ satisfies the equation of motion.

• By "consistency condition" you mean current conservation $\partial_\mu J^\mu=0$? And in classical electrodynamics $\partial_\mu J^\mu=0$ is kind of ad hoc assumed, since it cannot be obtained from the Noether theorem of another field, as the "other field" is not there. – Frederic Thomas Jan 17 at 13:31
• @FredericThomas Yes, by consistency condition I meant $\partial_\mu J^\mu=0$. It's not assumed, it's required (for consistency with Maxwell's equations). In a more complete model that includes the matter fields, it could be obtained from Noether's theorem, but even without those other fields, consistency with Maxwell's equations require it. But the important point is that it's a condition on $J$, not on $A$: the $A$ in the Lagrangian is arbitrary, and the Lagrangian is gauge invariant for any $A$ as long as $J$ satisfies $\partial_\mu J^\mu=0$. – Chiral Anomaly Jan 17 at 15:06
• @FredericThomas By the way, a similar thing occurs in general relativity. Even if we don't include the equations of motion for the "other fields," consistency with the gravitational field equation still requires the stress-energy tensor to satisfy $\nabla_a T^{ab}=0$. This is pretty typical: background data (things that don't come with their own equations of motion) must often satisfy certain conditions in order to be consistent with the equations of motion for the dynamic fields. – Chiral Anomaly Jan 17 at 15:06
• Thank you for the 2 precedent comments! That was very instructive, in particular the hint to general relativity. – Frederic Thomas Jan 17 at 15:48

For what it's worth, the off-shell Maxwell action (54.21) is gauge invariant if the background sources $$J^{\mu}$$ satisfy the continuity equation.

OP's question seems to be spurred by the ambiguity coming from the fact that one should specify which fields (among the gauge fields and matter fields) are treated quantum mechanically (off-shell) and which fields are treated as a classical (on-shell) background.