# Notation question for a classical field in QFT

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?

• The Leibniz product rule applies and one of the tenets of tensor theory, relabeling the summation indices (aka dummy ones). – DanielC Sep 23 at 22:29

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 $$\frac{\partial(\partial_\mu \phi)}{\partial (\partial_\nu\phi)} = \delta^\nu_\mu$$.