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I'm always a mess with the upstairs and downstairs notation. To be specific, say I want to calculate the Euler-Lagrange equations of

\begin{equation} \mathcal{L} = \frac{1}{2}\partial^\mu\phi \partial_\mu \phi = \frac{1}{2}(\partial_\mu\phi)^2.\tag{1} \end{equation}

So the partial derivative with respect to $\phi$ is zero and

$$ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi. \tag{2}$$

Why the index $\mu$ is upstairs and no downstairs? If I use the metric $(+,-,-,-)$ and $\partial_\mu = (\frac{\partial}{\partial t}, \nabla)$, then $\partial^\mu$ should have a $-\nabla$, right? Why in this case I can make $\partial_\mu \phi = \partial^\mu \phi$?

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You can always write the objects with upper indices as the objects with lower indices contracted with the metric tensor $$\partial^\mu\phi=\eta^{\mu\nu}\partial_\nu\phi$$ Then you can rewrite the Lagrangian as $$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi)$$ Taking the derivative we get $$\frac{\partial\mathcal{L}}{\partial (\partial_\alpha\phi)}=\frac{1}{2}\eta^{\alpha\nu}(\partial_\nu\phi)+\frac{1}{2}\eta^{\mu\alpha}(\partial_\mu\phi)=\partial^\alpha\phi$$ You can expect that the result should have the upper index from the definition of this type of derivative, $$\frac{\partial v^\nu}{\partial v^\mu}=\delta_\mu^\nu,\quad \frac{\partial v_\nu}{\partial v_\mu}=\delta^\mu_\nu$$ The indices should be located in such a way so that $\delta$-symbols were tensors.

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  • $\begingroup$ Thank you, this is what I was thinking but I was not sure if it was correct. $\endgroup$ – user2820579 Apr 22 at 20:14
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At a very basic level, taking a derivative with a down (up) index gives an up (down) index. For instance, let $$ f(v,w) = v^\mu w_{\mu} = v_{\mu} w^\mu. $$ Then $$ \frac{\partial f}{\partial v_\mu} = w^\mu, \quad \frac{\partial f}{\partial v^\mu} = w_\mu. $$ If you understand this example, the QFT computation should not be a problem.

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    $\begingroup$ Is this the reason that for example, the conjugate momentum in the electromagnetic field is "upstairs", i.e. $\pi^i =(\partial\mathcal{L})/\partial(\partial_0A_i)$, where we have taken $\partial_0A_i$ "entirely" downstairs? $\endgroup$ – user2820579 Apr 22 at 20:27
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    $\begingroup$ Yep. It's completely general. $\endgroup$ – Hans Moleman Apr 23 at 2:03

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