Derivative with respect a partial derivative in lagrangian density

Question

I am reading about field theory and lagrangian densities, and I found the following lagrangian density in my book: $$ \mathcal{L} = \dfrac{1}{2} (\partial_\mu \phi)^2 - \dfrac{1}{2} m^2 \phi^2\ , $$ where it states that $(\partial_\mu \phi)^2$ means $\partial_\mu \phi\, \partial^\mu \phi$. Then, the solution to one problem contains the expression $$ \dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} = \partial^\mu \phi\ , $$ but, apparently, I'm not able to see why that derivative leads to that. I tried to compute it step by step using the expansion $$ (\partial_\mu \phi)^2 = \eta^{\mu\nu} \partial_\mu \phi\, \partial_\nu \phi\ , $$ but I'm not sure if this is the right way. What would be the correct way of calculating that derivative?

Answer 1

You are on the right track, friend. Just derive the expression you just wrote

$$ \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)} = \dfrac{1}{2}\dfrac{\partial}{\partial(\partial_{\mu}\phi)}(η^{\rho\lambda}∂_{\rho}ϕ∂_{\lambda}ϕ ) = \dfrac{1}{2}(\eta^{\rho\lambda}\delta_{\rho}^{\mu}\partial_{\lambda}\phi + \eta^{\rho\lambda}\partial_{\rho}\phi\delta_{\lambda}^{\mu}) = \dfrac{1}{2}(\eta^{\mu\lambda}\partial_{\lambda}\phi + \eta^{\rho\mu}\partial_{\rho}\phi) = \partial^{\mu}\phi $$

with $$ \dfrac{\partial}{\partial(\partial_{\mu}\phi)}(\partial_{\nu}\phi) = \delta_{\nu}^{\mu} $$

Cheers