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The equations of motion for a classical field $\phi$ can be obtained using the Lagrange:

$$ \frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \bigg ( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \bigg )=0 \tag{1}$$

A simple Lagrangian: $$ \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi $$

Has the following equations of motion: $$ \partial_\mu \bigg (\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \bigg ) = 0$$

My confusion is at the moment of calculating: $\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}$, I would think that since we are taking a partial derivative, the result would be: $$ \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} = \frac{\partial }{\partial(\partial_\mu \phi)} \frac{1}{2} \partial_\mu \phi \partial^\mu \phi =\frac{1}{2} \partial^\mu \phi$$

But I know it is $\partial^\mu \phi$. I understand this is a notation confusion, but what is an intuitive way to understand the correct process to take the partial derivative?

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  • $\begingroup$ The Leibniz product rule applies and one of the tenets of tensor theory, relabeling the summation indices (aka dummy ones). $\endgroup$ – DanielC Sep 23 at 22:29
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Note that you have too many $\mu$'s floating around in your expression. The correct way to differentiate is to note that by definition, $\partial^\mu \phi = \eta^{\mu\nu} \partial_\nu \phi$. Therefore,

$$\frac{\partial }{\partial (\partial_\sigma\phi)}\left[ \eta^{\mu\nu}(\partial_\mu \phi)(\partial_\nu\phi)\right] = \eta^{\mu\nu} \delta^\sigma_\mu (\partial_\nu\phi)+\eta^{\mu\nu}(\partial_\mu\phi)\delta^\sigma_\nu $$ $$ = \eta^{\sigma \nu}(\partial_\nu \phi)+\eta^{\mu\sigma}(\partial_\mu\phi) = 2\partial^\sigma\phi$$

where we've used that $\frac{\partial(\partial_\mu \phi)}{\partial (\partial_\nu\phi)} = \delta^\nu_\mu$.

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