What is the equivalent to $\Box A^\alpha =- \mu_0 J^\alpha$ using differential forms? The set of equations $$\Box A^\alpha = -\mu_0 J^\alpha$$
can be found in section 12.3.5 of Griffiths's book. From what I understand, the real-valued functions on both side of the equations are the coefficients of some $1$-forms with respect to a chart. Thus, I am wondering how the equivalent index-free equation involving differential forms looks like.
 A: If $$\delta \equiv *d*$$
then the Maxwell equations amount to
$$\delta d A = J$$ with $\mu_0 =1$.
Starring the above equation, you get conservation of current. Taking the dual twice will get you to the form you started with, that is modulo some signs. Taking the dual of the above equation you get conservation of current
$$
*\delta d A \sim d*dA = *J
$$
now apply $d$ and get as a consistency on the equations of motion current conservation:
$$
d*J = 0.
$$
It is interesting exercise to consider the action from which those equations of motion come from, and discover how the variational derivative would be expressed in terms of differential forms. My suggestion, without the current, would be
$$
S_{EM} = \int dA \wedge * dA
$$
A: Maxwell's equations can expressed in the language of differential forms.
In terms of tensors, Maxwell's equations can be written as
$$
\nabla_{\nu}F^{\mu\nu} = \mu_0J^{\mu}\,,    \nabla_{[\alpha}F_{\mu\nu]} = 0
$$
where $F_{\mu\nu}$ is the Faraday tensor.
In terms of differential forms, these equations are written as
$$
d{\bf F} = 0,  d{\star}{\bf F} = \mu_0{\bf J}\,,
$$
where ${\bf F} = \frac{1}{2}F_{\mu\nu}dx^{\mu} \wedge dx^{\nu}\,$, $\star$ is the Hodge star operator and ${\bf J} = \frac{1}{3!}{\cal J}_{\alpha\beta\sigma}dx^{\alpha}\wedge dx^{\beta}\wedge dx^{\sigma}$ is the 3-form associated with the current density four-vector. The components of ${\cal J}_{\alpha\beta\sigma}$ are related to those of the current density 4-vector.
The Faraday 2-form, ${\bf F}$, satisfies, ${\bf F} = d{\bf A}$, where ${\bf A} = A_{\mu}dx^{\mu}\,.$
See https://en.wikipedia.org/wiki/Maxwell%27s_equations
and
https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field
Note that there are different sign conventions so be careful when consulting different websites/books on this subject.
A: Recall the inhomogeneous Maxwell equation in natural units:
$\newcommand{\dif}{\mathrm{d}}$
\begin{align}
\mathrm{d}*F={}*J\in\Omega^3
\end{align}
Since $**=-1$ on $\displaystyle{\Omega^1}$, the equation is equivalent to
\begin{align}
{}*\mathrm{d}*F=-J\in\Omega^1.
\end{align}
Now consider some one-form $A$ such that $\dif A=F$, then
\begin{align}\tag{1}
{}*\dif*F={}*\dif*\dif A=-J\in\Omega^1.
\end{align}
In components, this should turn out to be equation $(12.134)$ in Griffiths's book$^1$ and as Griffiths explains in section $12.3.5$, $(12.134)$ can be reduced to $(12.137)$ - the equation in the title of my question - by exploiting the gauge invariance.
To put this into a nutshell, $(1)$ is the equation I was searching for.

$^1$ The PDF is freely available; This answer should be helpful if someone wants to check this
