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Consider a coupling of a Maxwell field $A$ to some current $4$-vector $j^{\mu}$ through the following term in the action: $A^{\mu} j^{\mu}$. What is needed for this term to leave the equations of motion gauge invariant?

I wrote the Lagrangian as $$ \mathcal{L} = - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} - j^{\mu} A_{\mu} \\ = - \frac{1}{2} (\partial^{\mu} A^{\nu} )(\partial_{\mu} A_{\nu} ) + \frac{1}{2} (\partial^{\nu} A^{\mu})(\partial_{\mu} A_{\nu} - j^{\mu} A_{\mu}. $$ I now let $$A_{\mu} \rightarrow A_{\mu}^{'} = A_{\mu} + \partial_{\mu} \lambda. $$ I plugged this in the Lagrangian and simplified, until I got $$ \mathcal{L}^{'} = \mathcal{L} - \frac{1}{2} (\partial^{\mu} A^{\nu})(\partial_{\mu} A_{\nu} \lambda) + \frac{1}{2} (\partial^{\nu} A^{\mu})(\partial_{\mu} \partial_{\nu} \lambda) - j^{\mu} \partial_{\mu} \lambda $$ where $\mathcal{L}$ is the old Lagrangian. What does this mean? How can I make this extra term such that it leaves the equations of motion gauge invariant ?

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  • $\begingroup$ 1. You talk about equation of motion but you then write only Lagrangians. 2. The Lagrangian only has to be invariant up to a total derivative, and only after use of $\partial_\mu j^\mu = 0$. $\endgroup$ – ACuriousMind Jan 8 '17 at 23:08
  • $\begingroup$ How does then one show gauge invariance? Do I have to write down the action and apply a variation? $\endgroup$ – Kamil Jan 8 '17 at 23:11
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Ignore the kinetic term, as we already know that the Maxwell action is gauge invariant. It follows immediately when written in terms of differential forms, as $F = dA$ and any change $d\alpha$ will have the same $F$ as $d^2 =0$, i.e. the exterior derivative is nilpotent.

Instead, focus solely on the coupling to the conserved current, namely,

$$\mathcal L = j^\mu A_\mu.$$

Under $A_\mu \to A_\mu + \partial_\mu \lambda$, we have a change, $\delta \mathcal L = j^\mu \partial_\mu \lambda$. We'd like to write it in the form of a total derivative, $\delta \mathcal L = \partial_\mu F^\mu$and notice that we can if $j^\mu$ is conserved, that is,

$$F^\mu := j^\mu \lambda, \quad \implies \partial_\mu F^\mu = j^\mu \partial_\mu \lambda + \lambda \partial_\mu j^\mu = j^\mu \partial_\mu \lambda$$

providing that $\partial_\mu j^\mu = 0$, which is true if it arises as a conserved current by Noether's theorem, and is true at least classically, ignoring anomalies.

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  • $\begingroup$ Thank you for the reply. But how do you derive that $\delta \mathcal{L} = j^{\mu} \partial_{\mu} \lambda $ ? $\endgroup$ – Kamil Jan 8 '17 at 23:20
  • $\begingroup$ @Kamil Plug it in... $\mathcal L' = j^\mu (A_\mu + \partial_\mu \lambda) = j^\mu A_\mu + j^\mu \partial_\mu \lambda$ and so $\delta \mathcal L = j^\mu \partial_\mu \lambda$. $\endgroup$ – JamalS Jan 8 '17 at 23:21
  • $\begingroup$ Is it always the case that total derivatives added to an action do not change the equations of motion? $\endgroup$ – Kamil Jan 8 '17 at 23:23
  • $\begingroup$ @Kamil Yes, though there are some subtleties why sometimes, such as in general relativity. If you don't know why total derivatives added to the Lagrangian don't change the action, I would suggest you do some revision on the Lagrangian formalism, before tackling quantum field theory. $\endgroup$ – JamalS Jan 8 '17 at 23:25

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