Is there a more straightforward way to understand $\frac{\partial L}{\partial ( \partial_\mu \phi )} = \partial^\mu \phi$?

Question

I am wondering if there is a more straightforward way to understand $\frac{\partial\mathcal L}{\partial(\partial_\mu \phi)} = \partial^\mu \phi$ for a real scalar field $\mathcal{L} = \frac{1}{2} (\partial^\mu\phi)(\partial_\mu\phi) - V(\phi) $.

Here's what I did by expanding the indices:

\begin{align} \frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)} &= \frac{\partial}{\partial (\partial_\mu\phi)} \left[ \frac{1}{2} (\partial^\nu\phi)(\partial_\nu\phi) \right] \\ &= \frac{1}{2} \left[ \frac{\partial (\partial^\nu\phi)}{\partial (\partial_\mu\phi)} (\partial_\nu\phi) + (\partial^\nu\phi) \frac{\partial (\partial_\nu\phi)}{\partial (\partial_\mu\phi)} \right]\\ &= \frac{1}{2} \left[ \delta^\nu_\mu (\partial_\nu\phi) + (\partial^\nu\phi) \delta^\nu_\mu \right] \\ &= \partial^\mu \phi. \end{align}

Does this seem correct? Is there a more straightforward way to see why there is a raise in index from $\partial_\mu$ to $\partial^\mu$ when we take the derivative?

Answer 1

I've always found things like $\frac{\partial\mathcal L}{\partial(\partial_\mu \phi)}$ to cause confusion are errors when calculating functional derivatives--- especially with up and down indices. It's much easier to always go back to the basic definition of the functional derivative $$ \delta S= \int d^dx \left(\frac{\delta S}{\delta \phi} \right)\delta \phi. $$ For your case $$ \delta S = \delta \int d^dx \frac 12 g^{\mu\nu}\partial_\mu \phi \partial_\nu \phi\\ = \int d^dx \frac 12 g^{\mu\nu}(\partial_\mu \delta \phi) \partial_\nu \phi+ \int d^dx \frac 12 g^{\mu\nu}\partial_\mu \phi (\partial_\nu \delta \phi)\\= -\int d^dx \frac 12 g^{\mu\nu}(\delta \phi) \partial^2_{\nu\mu} \phi- \int d^dx \frac 12 g^{\mu\nu}\partial^2_{\nu\mu} \phi (\delta \phi)\\= \int d^dx (- g^{\mu\nu}\partial^2_{\mu\nu}\phi) (\delta \phi)\\ = \int d^dx(-\partial^\mu\partial_\mu \phi) \delta \phi. $$