I'm reading Peskin & Schroeder and they give as a simple example the Lagrangian

$$\mathcal{L} = \frac{1}{2} (\partial_\mu \phi)^2$$

First of all, I'm guessing that $(\partial_\mu \phi)^2$ is abuse of notation for $\partial_\mu\phi\partial^\mu\phi$ . So how do I now calculate the term

$$\partial_\mu\left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)} \right)~?$$

Should I treat the Lagrangian as a product of those two derivative terms, and use the product rule? I need to take the derivative w.r.t $\partial_\mu\phi$, so how should I treat $\partial^\mu \phi$?

  • $\begingroup$ $\partial^\mu \phi = \eta^{\mu\nu} \partial_\nu \phi$ $\endgroup$
    – Slereah
    Dec 24, 2015 at 11:42
  • $\begingroup$ I still don't see how to do it. Does $\frac{\partial}{\partial(\partial_\mu)} (\eta^{\mu\nu}\partial_\nu \phi) = \eta^{\mu\nu}\delta_{\mu\nu}$? $\endgroup$ Dec 24, 2015 at 11:58

1 Answer 1


$\mathscr{L} = \frac{1}{2} (\partial_\mu \phi)^2$ is shorthand notation for $$ \mathscr{L}(\phi(x), \partial_{\mu}\phi(x), x) = g^{\mu\nu}(\partial_{\mu}\phi(x))(\partial_{\nu}\phi(x)) $$ and thus: $$ \frac{\partial\mathscr{L}}{\partial(\partial_{\rho}\phi)}= g^{\mu\nu}\big(\delta_{\mu}^{\rho}\partial_{\nu}\phi(x) + \partial_{\mu}\phi(x)\delta_{\nu}^{\rho}\big). $$ Consequently the equations of motion are: $$ \partial_\rho\left(\frac{\partial\mathscr{L}}{\partial(\partial_{\rho}\phi)}\right)=\partial_\rho\left(g^{\rho\nu}\partial_{\nu}\phi(x) + g^{\mu\rho}\partial_{\mu}\phi(x)\right) = \left(\partial^{\nu}\partial_{\nu} + \partial^{\mu}\partial_{\mu}\right)\phi(x) = 0 $$ which, taking into account the initial $1/2$ become $$ \frac{1}{2}\left(2\ \partial^{\mu}\partial_{\mu}\right)\phi(x) = \Box\,\phi(x) = 0. $$

  • $\begingroup$ Should the deltas not be like this: $\delta^\rho{}_\mu$ ? $\endgroup$ Dec 24, 2015 at 12:41
  • $\begingroup$ Oh yes, sure, I will correct the notation accordingly. $\endgroup$
    – gented
    Dec 24, 2015 at 13:56

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