In the textbook Condensed Matter Field Theory by Altland and Simons, it is said that the Maxwell Lagrangian $\mathcal{L}$ coupled to a four-current $j^\mu$ satisfying $\partial_\mu j^\mu = 0$ is the unique Lagrangian that is gauge invariant and Lorentz invariant up to quadratic order in $A_\mu$, i.e., the Lagrangian must have the form $$ \mathcal{L} = c_1 F_{\mu \nu}F^{\mu \nu} + c_2 A_\mu j^\mu $$ for some coefficients $c_1$ and $c_2$ (which can later be fixed by requiring that the Lagrangian reproduces Maxwell's equations). My question is: why is the term $$c_3 (\partial_\mu j_\nu)F^{\mu \nu}$$ not included, where $c_3$ is an arbitrary constant? If one adds this term to the Lagrangian, it contributes the term $c_3 \partial^\mu \partial_\mu j_\nu$ to the equations of motion, which doesn't seem problematic to me.

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    $\begingroup$ I guess that since you're not describing the dynamics of $j^\mu$ (it is an external field, not a dynamical variable of the theory) you can just absorb the new term in the $c_2$ term by redefining $j^\mu$ (but I'm just guessing) $\endgroup$ Commented Jun 18 at 18:14

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Lagrangian term $$ (\partial_\mu j_\nu)F^{\mu \nu} $$ is of mass-dimension 6.

This is not a problem per se according to the effective field theory paradigm. But such terms are usually suppressed by the factor of $$ \frac {p^2}{\Lambda^2} $$ where $\Lambda$ is some large cutoff energy scale. In the context of QFT, $\Lambda$ is in the ball park of GUT scale or Planck scale.

  • $\begingroup$ Ok, so I guess the Lagrangian is "fixed" if you only consider renormalizable operators for $d = 4$, but it's not fixed if you're only considering classical effects. The authors also mention the criterion of the Lagrangian being "simple," so I guess this explanation makes sense. Thanks. $\endgroup$ Commented Jun 18 at 18:34
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    $\begingroup$ Maxwell's action is clearly not the only one you can write down that is consistent with gauge symmetries. For instance, any term of the form $(F^{\mu\nu} F_{\mu\nu} )^n$ for any $n \geq 0$ is allowed. However, the argument by @MadMax gets rid of all these terms, except for $n=0$ and $n=1$. $\endgroup$
    – Prahar
    Commented Jun 19 at 4:58

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