I've looked in several books and they all show how to obtain electrical interactions by forcing local gauge invariance of any complex scalar field Lagrangian (like Klein-Gordon or Dirac). I manage to separate the new Lagrangian into the original one (the free Lagrangian) and the interaction part of it. But how I get Maxwell's Lagrangian from this? As the fields that are inserted to keep the Lagrangian gauge invariant should have it's dynamics described by Maxwell's. So, how to does the term $F^{\mu\nu}F_{\mu\nu}$ enter the new Lagrangian naturally?



1 Answer 1


The lagrangian of the gauge field is independent of that of the scalar field. You have to "guess" it. The reason we pick this one is twofold: 1-it is the one which gives the Maxwell equations, so when you try to describe E&M, that looks like a good guess; 2- if you think of all the terms that are both Lorentz invariant, parity invariant and gauge invariant, that's the term with the less number of fields and derivatives in it. That will thus be the dominant one to describe low energy processes.

  • $\begingroup$ So if I start with complex Klein-Gordon's Lagrangian I mean to describe particles with spin-0. It's naturally invariant by global gauge transformation and it's still a free particle Lagrangian. If I force it to be local gauge invariante a new field appear, $A^{\mu}$. After all manipulations I find that the interaction Lagrangian is $\mathcal{L}_{int}=iA^{\mu}(\phi^{*}\partial_{\mu}\phi-\phi\partial_{\mu}\phi^{*})+\phi^{*}\phi A^{\mu}A_{\mu}$. I understand that this is how the matter interacts with the new field. Can I identify it as the EM field? Or it needs the $F^{\mu\nu}F_{\mu\nu}$? $\endgroup$
    – Pierre
    Mar 1, 2014 at 1:05
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    $\begingroup$ @Pierre: Short answer, you need the $F^2$ term. If you don't put it, the gauge field has no dynamics (it does not propagate). So you have to put it by hand if you want some interesting physics. You know it's E&M because that's why you put the gauge invariance at the beginning (the reasoning goes backward. It's because we know the answer that we do it that way.) $\endgroup$
    – Adam
    Mar 1, 2014 at 2:58
  • $\begingroup$ Thanks for your answer! I is very illuminating. May I ask you what would be the formal way to derive the Lagrangian $\mathcal{L}=\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$ in the view of field theory? I mean that I don't want the "covariant notation to Lagrangian" procedure but defining the characteristics that the EM Lagrangian should have and by that arriving to it. May you suggest me a text book on the matter? Thanks again! $\endgroup$
    – Pierre
    Mar 2, 2014 at 0:46
  • $\begingroup$ @Pierre: I don't have good references in mind, but you may look for "effective field theory" on google. The way to construct it is to recognize that the only object that you can construct with $A_\mu$ which is linear and gauge invariant is $F_{\mu \nu}$. From that, one can show that all terms in the Lagrangian can be constructed out of $F_{\mu \nu}$ and $\epsilon_{\mu\nu\sigma\lambda}F^{\sigma\lambda}$ (which is not parity invariant). Therefore, the dominant one at low energy is $F^2$. $\endgroup$
    – Adam
    Mar 2, 2014 at 4:00

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