# Deriving Lagrangian density for electromagnetic field

In considering the (special) relativistic EM field, I understand that assuming a Lagrangian density of the form

$$\mathcal{L} =-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{c}j_\mu A^\mu$$

and following the Euler-Lagrange equations recovers Maxwell's equations.

Does there exist a first-principles derivation of this Lagrangian? A reference or explanation would be greatly appreciated!

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Which principles do you want to start from? –  user1504 Aug 15 '12 at 18:26
–  Qmechanic Aug 15 '12 at 18:32

Ultimately the reasoning must be that (as you stated) it must be constructed so the Euler-Lagrange equations are Maxwell's equations. So in a sense you have to guess the Lagrangian that produces this as is done here for example.

However you can get some guidance from the fact that we need to construct a Lagrangian for a massless non self interacting field. So we need a gauge and lorentz invariant combination of the 4-vector potential which only has a kinetic term (quadratic in derivatives of the fields). You are then not left with many options apart from $F^{\mu\nu}F_{\mu\nu}$. The source term is then trivial to add in if needed.

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What about $\epsilon_{\mu\nu\sigma\tau} F^{\mu\nu} F^{\sigma\tau}$? –  Fabian Aug 15 '12 at 21:18
Well, I said "not left with many options" and indeed the combination you write down would also fit my criteria, as would $\det (F)$, but one may as well try the simplest option first and that turns out to be correct. I note that the combination you give is a pseudoscalar. Can anyone think if there is a reason why that would not be allowed? –  Mistake Ink Aug 16 '12 at 13:22
@MistakeInk - $\epsilon_{\mu \nu \rho \sigma } F^{\mu \nu }F^{\rho \sigma }$is a fine candidate term for the Lagragian, its just that its a total derivative so it doesn't affect the classical EOM and vanishes in perturbation theory. It still does have some consequences though - see en.wikipedia.org/wiki/CP_violation#Strong_CP_problem. As for $Det(F)$ I don't think this term is renormalizable since its equal to $e^{tr \log F}$ which you could expand about some background field value and get arbitrarily high powers of the field strength. –  DJBunk Aug 31 '12 at 21:03
You do however get terms of the form $\log tr (k^2+{F^{\mu \nu}}^2)$ when calculating the effective action in the presence of a background field gauge field. See Chap 16 of Peskin. –  DJBunk Aug 31 '12 at 21:05
Thanks - I guess I'd only ever seen Lagrangians from mechanics, where they are naturally of the form $L=T-V$ and thus what I was calling "derivable". –  mcamac Aug 15 '12 at 18:39