Interpreting the inhomogeneous wave equations for $\vec{E}$ and $\vec{B}$ Starting from Maxwell's equations,
\begin{align}
\nabla \cdot \vec{E} & = \frac{\rho}{\epsilon_0} & \nabla \cdot \vec{B} & = 0 \\
\nabla \times \vec{E} & = - \frac{\partial \vec{B}}{\partial t} & \nabla \times \vec{B} & = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}
\end{align}
it is standard practice in any electromagnetism course to derive a wave equation to show that waves propagate through these fields, and that light is that wave. Simplifications are frequently made, like using the vector potential or looking at the homogeneous case ($\rho = 0$ and $\vec{J} = 0$).
The inhomogeneous wave equation for $\vec{B}$ is fairly straightforward to produce and interpret (take the time derivative of the $\nabla \times \vec{E}$ equation, then substitute for $\frac{\partial \vec{E}}{\partial t}$ by solving for it in the $\nabla \times \vec{B}$ equation). After using the fact that $\nabla\cdot \vec{B} = 0$ you get:
$$\frac{1}{c^2}\frac{\partial^2 \vec{B}}{\partial t^2} - \nabla^2 \vec{B}=  \mu_0 \nabla \times \vec{J},$$
a nice, clean, wave equation that has $\nabla \times \vec{J}$ as the source term.
With the wave equation for $\vec{E}$, though, things don't look as clean. Just reverse the order of how the two curl equations were manipulated to get the $\vec{B}$ wave equation to get
$$\frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2} - \nabla^2 \vec{E} = -\frac{1}{\epsilon_0}\nabla \rho - \mu_0 \frac{\partial \vec{J}}{\partial t}.$$
Again, a nice, clean wave equation, but this time I'm finding the source term harder to interpret from a physical point of view. In terms of special relativistic tensors, we can render both equations as a single equation to be
$$\partial_\alpha \partial^\alpha F_{\mu\nu} = \mu_0 \eta_{\nu\alpha} \partial_\mu J^\alpha - \mu_0 \eta_{\mu\alpha}\partial_\nu J^\alpha ,$$
where $J^0 \equiv c\rho$, and $\operatorname{sig}(\eta_{\mu\nu}) = (+,-,-,-)$. This suggests that the right hand side of the $\vec{E}$ wave equation is part of the $4$-dimensional generalization of the curl - specifically the mixed (time-like, space-like) components of an anti-symmetric tensor formed with $J^\mu$ and $\frac{\partial}{\partial x^\nu}$. The pure space-like parts of that tensor are easy to visualize because studying E&M gives lots of practice working with ordinary curls (a tendency of a vector field to circulate around a point). Is there a similarly intuitive and visualizable description for $$\frac{\partial J^0}{\partial x^i} + \frac{\partial J^i}{\partial x^0}?$$
 A: One partial answer is to reformulate the expression in terms of differential forms (for one thing, it's a lot easier to get the signs right!). The inhomogenous Maxwell's equations are $d(\ast F) = \ast J$ or equivalently $\ast d(\ast F) = \pm J$, where the $\pm = (-1)^D$ depends on the number of spacetime dimensions $D$. The inhomogeneous wave equation is $d(\ast d (\ast F)) = \pm dJ$, giving us a geometrical picture if one is comfortable picturing higher differential forms - your mileage may vary.
Note that the confusing $\nabla \rho + \dot{\vec{J}}$ term sourcing the $\vec{E}$ field wave equation is exactly analogous to the expression $-\vec{E} = \nabla \phi + \dot{\vec{A}}$ in terms of the gauge potentials. The reason for that particular linear combination of partial derivatives is so the $\vec{E}$ (and $\vec{B}$) will be left invariant under gauge transformations of the form $A \to A + d\lambda$ for any scalar $\lambda$. Similarly, the combination of partials in the inhomogeneous wave equation conspire to leave the equation invariant under the addition of any total divergence to $J$. But unlike with the true gauge symmetry of $A_\mu$, adding a four-divergence to $J_\mu$ changes the real physics. So the wave equation for $F_{\mu \nu}$ has extraneous solutions that don't actually satisfy Maxwell's equations. Any physical intuition from that equation should probably be taken with a grain of salt, because it has both physical and unphysical solutions. Put another way, the wave equation for $F_{\mu\nu}$ only tells you that the one-form $\ast d(\ast F) - J$ is closed, while Maxwell's equations tell you that it is in fact identically zero.
