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.

  • $\begingroup$ Are we assuming $g^{\mu\nu}$ is constant? $\endgroup$ Mar 30, 2022 at 20:21
  • $\begingroup$ Sorry, by $g_{\mu\nu}$ I meant the Minkowski tensor. $\endgroup$
    – diesmond
    Mar 31, 2022 at 7:12

1 Answer 1


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!

  • $\begingroup$ Thank you so much, this makes it alot clearer! Just a small follow-up/sanity-check: in your second equation, in the first term on the right-hand-side we still assume that the $\partial_{\nu}$ acts on everything to the right of it despite the brackets. Is that correct? $\endgroup$
    – diesmond
    Jun 6, 2022 at 9:21
  • $\begingroup$ @diesmond I'm not sure I follow the question. Consider the second equation just a product rule: $ \partial(a\cdot b)=(\partial a)\cdot b+a\cdot(\partial b) $. It's probably better to look at $\partial_\nu \phi$ as its own entity and not a derivative acting on something. If we write $\partial_\nu\phi=A_\nu$, the RHS of the second equation is $(\partial A_\nu)\cdot A^\nu+A_\nu\cdot(\partial A^\nu)$. $\endgroup$
    – Sotiris
    Jun 7, 2022 at 10:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .