Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Which principles do you want to start from? – user1504 Aug 15 '12 at 18:26
Related: – Qmechanic Aug 15 '12 at 18:32
up vote 2 down vote accepted

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.

share|cite|improve this answer
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 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

I'm almost 100% sure the Lagrangian is an assumption of the theory. It cannot be derived. I don't have any references for this claim. I just know that from every course I've been taught and every book I've read, the Lagrangian (assuming it is being used at all) is where you start. It is the "first principle" in this case.

share|cite|improve this answer
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
I don't see how that Lagrangian is "derivable" either. Of course, we write it as such so that the Euler-Lagrange equations give us the classical equations of motion. But, I wouldn't consider that a derivation. Really it's just substituting one assumption for another. – mcFreid Aug 15 '12 at 19:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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