6
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.