I'm looking at the Lagrangian density for the free scalar field:
$ \mathcal{L} = \frac{1}{2} ( \partial_{\mu} \phi \partial^{\mu} \phi – m^2 \phi^2)$
and I'm trying to figure out how to write down the classical equations of motion using the Euler-Lagrange equations.
My question is:
How should I think about
$ \partial_{\mu} \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)} = \partial^2\phi$?
What I mean is: in the Lagrangian density I have a derivative once with a raised and once with a lowered index. When taking the derivative of $\mathcal{L}$ by $\partial_{\mu}\phi$ this doesn't seem to matter and I get $ \partial^{\mu} \phi $ twice. Since $ g^{\mu\nu} \partial_{\nu} = \partial^{\mu}$ there seems to be something missing from my perspective.
Thank you!
Edit: $g_{\mu\nu}$ is supposed to denote the Minkowski tensor.
Edit: typo in description of Lagrangian density, irrelevant for question though.
Best Answer
It's probably a bit late, but I just came upon this question. Let's solve it explicitly:
$$\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} =\frac{1}{2}\partial_\mu\frac{\partial}{\partial(\partial_\mu\phi)}\left(\partial_\nu\phi\partial^\nu\phi-m^2\phi\right)$$
Note the different indices in $\mathcal{L}$. It's important to be thoroughly explicit when calculating derivatives in tensor notation!
Obviously, the derivative of the $m^2\phi$ term just vanishes since it has nothing to do with $\partial_\mu\phi$. Let's single out the first term and work on it:
$$ \frac{\partial}{\partial(\partial_\mu\phi)}\partial_\nu\phi\partial^\nu\phi=\frac{\partial(\partial_\nu\phi)}{\partial(\partial_\mu\phi)}\partial^\nu\phi+\partial_\nu\phi\frac{\partial(\partial^\nu\phi)}{\partial(\partial_\mu\phi)} $$
By tensor derivative convention, we have $$\frac{\partial(\partial_\nu\phi)}{\partial(\partial_\mu\phi)}\equiv\delta_\nu^\mu$$ so the first term above becomes $$ \frac{\partial(\partial_\nu\phi)}{\partial(\partial_\mu\phi)}\partial^\nu\phi =\delta_\nu^\mu\partial^\nu\phi=\partial^\mu\phi$$ As for the second term, we have to lower the index on the numerator to be able to work with it: $$ \frac{\partial(\partial^\nu\phi)}{\partial(\partial_\mu\phi)} =g^{\omega\nu}\frac{\partial(\partial_\omega\phi)}{\partial(\partial_\mu\phi)}=g^{\omega\nu}\delta^\mu_\omega=g^{\mu\nu}$$ And thus $$ \partial_\nu\phi\frac{\partial(\partial^\nu\phi)}{\partial(\partial_\mu\phi)} =\partial_\nu\phi \ g^{\mu\nu}=\partial^\mu\phi$$
Putting everything together we have $$ \partial_\mu\frac{\partial\mathcal{\mathcal{L}}}{\partial(\partial_\mu\phi)}=\frac{1}{2}\partial_\mu\left(\partial^\mu\phi+\partial^\mu\phi\right) =2\cdot\frac{1}{2}\partial_\mu\partial^\mu\phi=\partial^2\phi$$
Hope the step-by-step helps, hit me up if you have any other questions!