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I was doing a problem regarding field theory. I am given the following lagrangian density: $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi_i\partial^\mu\phi_i-\frac{m^2}{2}\phi_i\phi_i$$ for three scalar fields. I want to determine the equations of motion for the field $\phi_i$. I used the Euler Lagrange equation:$$\frac{\partial\mathcal{L}}{\partial\phi_j}-\partial_\mu[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_j)}]=0.$$ My question is, is there a difference between using the Euler Lagrange equation the way I wrote above and the Euler Lagrange equation with an upper $\mu$ index: $$\frac{\partial\mathcal{L}}{\partial\phi_j}-\partial^\mu[\frac{\partial\mathcal{L}}{\partial(\partial^\mu\phi_j)}]=0~?$$ I ask this because by the end of my calculations, if I use the former equations , I get $$\partial_\nu\partial^\nu\phi_j+m^2\phi_j=0$$ while the latter equations give: $$\partial^\nu\partial_\nu\phi_j+m^2\phi_j=0.$$ Are both the Klein-Gordon Equation or just the last one?

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As mentioned in the comments by WarreG and user2723984, since the full expression of the two in terms of the metric tensor is $$ \partial_\nu \partial^\nu=\partial_\nu g^{\mu\nu}\partial_\mu=\partial^\mu\partial_\mu, $$ the two equations are identical.

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    $\begingroup$ Isn't it a bit dangerous to write $\partial_\nu g^{\mu\nu}\partial_\mu$ which easily can be interpreted as $(\partial_\nu g^{\mu\nu}) \partial_\mu$? $\endgroup$ – md2perpe Jan 11 at 20:51
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    $\begingroup$ $g_{\mu\nu}$ is constant, if it wasn't the equation wouldn't be correct either way. $\endgroup$ – MannyC Jan 11 at 21:07
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I would say the answer to your question is: it depends on the metric. Yes, because if your metric is Minkowski and Cartesian then $$\partial_{\mu}\partial^{\mu} = \partial_{\mu}\left(g^{\mu\nu}\partial_{\nu}\right)=\underbrace{\left(\partial_{\mu}g^{\mu\nu}\right)}_{0}\partial_{\nu}+g^{\mu\nu}\left(\partial_{\mu}\partial_{\nu}\right)=g^{\mu\nu}\partial_{\mu}\partial_{\nu}=\partial^{\nu}\partial_{\nu}.$$ But for example, if your metric is Minkowski and spherical? In this case $\partial_{\mu}g^{\mu\nu}\neq0$ and $\partial^{\mu}\partial_{\mu}\neq\partial_{\mu}\partial^{\mu}$.

As you can see, $\partial_{\mu}\partial^{\mu}$ and $\partial^{\mu}\partial_{\mu}$ are equivalent as long as you understand that you are dealing with Cartesian coordinates.

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