Computation in QFT

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

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

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 $$\partial_\mu \phi = \partial^\mu \phi$$?

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.
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.
• 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? – user2820579 Apr 22 '19 at 20:27