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I'm new to classical field theory, so I have a few basic questions:

From the derivation of the Euler-Lagrange equations, we have the following: \begin{align} \delta S[\phi]&=\int d^4x\delta L(\phi, \partial_{\mu}\phi)\\&= \int d^4x\Big[\frac{\partial L}{\partial\phi}\delta\phi+\frac{\partial L}{\partial(\partial{\mu}\phi)}\delta(\partial_{\mu}\phi)\Big]\end{align}

  1. $S$ is a functional of $\phi$, but it doesn't look like $L$ is, but isn't it also a function of the field $\phi$?

  2. I understand the chain rule, but varying $L,$ which is not a functional but an ordinary function, yields a partial derivative. Why is that? Is there some identity that for an ordinary function $f(x), \delta f(x)=\frac{\partial f}{\partial x}\delta x$ (assuming $x$ is parametrized by, say $t$)?

  3. Is there any mathematical justification for the apparent commutation $\delta \partial_{\mu}\phi=\partial_{\mu}\delta\phi$.

Thank you in advance!

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  • $\begingroup$ $S$ is a functional and $L$ is a function of $\phi$ and its derivatives. So in four spacetime dimensions $L$ is a function of 5 arguments, which is how you arrive at the second line. You never need to use $\delta (\partial_\mu\phi)$ in order to derive the Euler-Lagrange equations. All you need is that $\delta\phi$ and its derivatives $\partial_\mu\delta\phi$ vanish at infinity. $\endgroup$ – user178876 Mar 22 at 16:16
  • $\begingroup$ The answer to your 3rd. question is the following: $\delta A(\phi)$ means changing of $A$ when changing fields $\phi$. Therefore it is not affecting space-time points, so it commutes with the derivative. Related to the other 2, @marmot has already answered you, but I can complete it a little bit more: $L$ is a functional of $\phi$ since it depends on it (plus derivatives) and these fields are functions. Then when you take $\delta$ you vary respect the fields and its derivatives and then you can use what I told you at the beginning $\endgroup$ – Vicky Mar 22 at 16:20

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