Confusion with Index Notation for Fully Contracted Electromagnetic Field Tensor I am going through the derivation of Coulomb's law from classical field theory and the very first step that is performed is re-expressing the Lagrangian in terms of derivatives of the vector potential, such that
$$L=-\frac{1}{4}F_{\mu \nu}^2-A_\mu J^\mu$$
goes to
$$L=-\frac{1}{2}(\partial_\mu A_\nu)^2+\frac{1}{2}(\partial_\mu A_\mu)^2-A_\mu J^\mu$$
via the substitution 
$$F_{\mu \nu}=\partial _\mu A_\nu -\partial_\nu A_\mu$$
The first term in the re-expression of the Lagrangian is straight-forward; however, I do not understand how the second term comes about, as it implies that
$$\partial_\mu A_\nu \partial^\nu A^\mu=(\partial_\mu A_\mu)^2 $$
How is this so? 
 A: You need to be super careful about your indices.  As written, the second expression in your question is a bit sloppy (which is fine if you know what you're doing, but can cause some confusion).  Just to retrace the work you've already done,
$$L = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} - A_\mu J^\mu$$
letting $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, 
$$ L = - \frac{1}{4}(\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu) - A_\mu J^\mu $$
$$ = -\frac{1}{4}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) -(\partial_\mu A_\nu)(\partial^\nu A^\mu) - (\partial_\nu A_\mu)(\partial^\mu A^\nu) + (\partial_{\nu}A_\mu)(\partial^\nu A^\mu)] - A_\mu J^\mu$$
The first and last term in the square brackets are equal, and the middle two are equal as well, so this is just
$$ L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial _\mu A_\nu)(\partial^\nu A^\mu)] - A_\nu J^\mu$$

To answer your question, we can examine the second term in square brackets.  Notice that
$$(\partial_\mu A_\nu)(\partial^\nu A^\mu) = \partial_\mu \left(A_\nu \partial^\nu A^\mu\right)- A_\nu \partial^\nu \partial_\mu A^\mu$$
and
$$A_\nu \partial^\nu \partial_\mu A^\mu = \partial^\nu(A_\nu \partial_\mu A^\mu) - (\partial_\nu A^\nu)(\partial_\mu A^\mu) $$
So rearranging, we find that
$$(\partial_\mu A_\nu)(\partial^\nu A^\mu) = \partial_\mu(A_\nu \partial^\nu A_\mu) - \partial^\nu(A_\nu \partial_\mu A^\mu) + (\partial_\mu A^\mu)^2$$
Remember that when we find the equations of motion for our field, the overall derivative terms (i.e. "surface terms") do not contribute, so we can ignore them.  This leaves us with the equivalent (though not strictly equal) Lagrangian
$$ L' = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A^\mu)^2] - A_\mu J^\mu$$
