In this post, $\hbar=c=1$ units are used throughout.

It is well known that the action of classical electromagnetism is given by $$\mathcal S_{\text{Maxwell}} = \int d^4x\left\{-\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}\right\}$$ which yields a Lagrangian density $$\mathscr{L}_{\text{Maxwell}} = -(4\mu_0)^{-1}F_{\mu\nu}F^{\mu\nu}.$$

In P&S, however, they instead use the density $$\mathscr L = -\frac 1 4F_{\mu\nu}F^{\mu\nu}.$$ This is a bit more natural than the real one, however to me it seems that this is insufficient to get $\mu_0$s and $\varepsilon_0$s into Maxwell's equations. I'm aware that in $c=1$ units $\varepsilon_0 = \mu_0^{-1}$ so we work with $\mu_0$s WLOG, but how does one get $\mu_0$ into the equation using P&S's Lagrangian?


1 Answer 1


He is also using Lorentz-Heaviside units for electromagnetic constants, in which $\varepsilon_0=\mu_0=1$.

  • 3
    $\begingroup$ Ah, I didn't know that was part of the convention. Also, $\varepsilon_0=\mu_0=1$, right? :P $\endgroup$
    – theage
    Commented Jan 18, 2015 at 2:44
  • $\begingroup$ @theage haha obviously, sorry for the mistake. Edited $\endgroup$
    – glS
    Commented Jan 18, 2015 at 10:25

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