Electromagnetic 4-potential and basic index contraction I'm trying to learn about relativistic electrodynamics on my own, and I am struggling with derivatives of the 4-potential and index (Einstein) notation.
I think I understand expressions such as $\partial_\mu A^\mu$. The index is repeated and is once up and once down, so I would expand the sum as: $\partial_0 A^0 + \partial_1 A^1 + \partial_2 A^2 + \partial_3 A^3$, which gives me a scalar.


*

*How am I to interpret something like this: $(\partial_\mu
    A_\nu)(\partial^\mu A^\nu)$ ? We are summing over the two indices
this time, which is fine. What confuses me is that we are taking a
covariant derivative of a covariant vector. Does one need to
"convert" $A_\nu$ in the first term to contravariant, like so:
$(\partial_\mu A^\rho \eta_{\nu\rho})(\partial^\mu
    A_\sigma\eta^{\nu\sigma})$?
I guess my doubts arise from the fact that I see a covariant vector
as being an entirely different object from a contravariant one. The
covariant derivative $\partial_\mu = \frac{\partial}{\partial
    x^\mu}$  differentiates with respect to the components of the
contravariant vector $x$. So I don't understand how such an
operation can be applied to a vector that isn't also contravariant.

*How should I interpret terms such as $(\partial_\mu A^\mu)^2$ ? Is it just $\left(\partial_0 A^0 + \partial_1 A^1 + \partial_2 A^2 + \partial_3 A^3\right)^2$ or is there something else going on? 

*According to some textbook, $(\partial_\mu \phi)^2 = \eta^{\mu\nu}\partial_\mu \phi\partial_\nu\phi$, but I don't understand why. For me $\partial_\mu \phi$ is just the derivative of a scalar $\phi$ with respect to some (unspecified) component $\mu$ of a contravariant 4-vector $x$. Instead, judging from the right-hand side, it is to be interpreted as a vector $(\frac{\partial}{\partial x^0},\boldsymbol{\nabla})\phi$ which is then squared. Is it just sloppy notation, or am I being stupid?


Thanks.
EDIT:


*

*Are the following then true?
$$\frac{\partial}{\partial(\partial_\mu A_\nu)} \left(\partial_\mu A_\nu\right) = 1$$
$$\frac{\partial}{\partial(\partial_\mu A_\nu)} \left(\partial^\mu A^\nu\right) = 0$$

*Can I also raise and lower the indices of a partial derivative?

 A: 1) You have the right idea about $\partial_{\mu}A^{\mu}$ and 2) $(\partial_{\mu}A^{\mu})^2$.
For the rest of it, you're goign to have to be careful about covariant and contravariant indices of $\partial_\mu$ and $A^{\mu}$.  Since you are already going with $A$ having vector indices, I think you should stick to that as your "base" version of $A$ for now.  Now, you correctly seem to note that lowering and raising is done with $\eta_{\mu\nu}$ and its inverse $\eta^{\mu\nu}$ (which are the same matrix for Mikowskian coordinates).  Thus, you will find that an expression like $A_{\mu}A^{\mu} = \eta_{\mu\nu}A^{\mu}A^{\nu} = -(A^{0})^{2} + (A^{1})^{2} + (A^{2})^{2} + (A^{3})^{2}$, and you can extend this to most of the other examples you cite.
3) applying $\partial_\mu$ to a scalar $\phi$ gives you a result $\partial_{\mu}\phi$ that is best interpreted as a one form, not as a scalar--that's because you have the free index $\mu$ floating around, and if you change coordinates, you will have to apply the chain rule to get the right answer for your new $\partial_{\mu}\phi$.  this is why you have to apply the inverse metric to "square" $\partial_{\mu}\phi$.  
4) If you are taking variations of vector valued quantities, you should leave a dummy index for your variational term:
$\begin{equation}
\frac{\delta}{\delta A^{\mu}}A^{\nu} = \delta^{\nu}{}_{\mu}
\end{equation}$
This will give you terms like
$\begin{equation}
\frac{\delta}{\delta A^{\alpha}} A^{\mu}A_{\mu} = \frac{\delta}{\delta A^{\alpha}}(\eta_{\mu\nu}A^\mu A^{\nu}) = \eta_{\mu\nu}\delta^{\mu}_{\alpha}A^{\nu} +\eta_{\mu\nu}A^{\mu}\delta^{\nu}_{\alpha} = 2 A_{\alpha}
\end{equation}$
which should seem pretty reasonable from your calculus I based intuition.  Similarly, you should probably think of $\frac{\delta}{\delta \partial_{\alpha}A_{\beta}}\partial_{\mu}A_{\nu} = \delta^{\alpha}_{\mu}\delta^{\beta}_{\nu}$.  If you're taking a variation of the up version of a quantity with respect to the down version of the quantity, just use the metric to raise/lower the target before taking the variation, and fix everything after the fact using raising and lowering conventions and the Kroneker deltas.
