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 $$\frac{\partial J^0}{\partial x^i} + \frac{\partial J^i}{\partial x^0}?$$

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    $\begingroup$ An observation, which may or may not be useful: The integral of $\nabla \rho$ over any volume completely containing a finite charge density is zero, since $\int_V \nabla f d\tau = \int_{\partial V} f d\vec{a}$ for any smooth function $f$. One can also show (I think) that $\int_V \vec{J} d\tau = - \int_V \vec{r} \dot{\rho} d\tau$ using conservation of charge. We can then conclude that integrating $\epsilon \nabla \rho + \mu \dot{\vec{J}}$ over any volume $V$ containing a finite charge distribution gives $\ddot{\vec{p}}$ for that charge distribution. $\endgroup$ Commented Oct 23, 2017 at 16:59
  • $\begingroup$ Adding the definition of the charge weighted mean velocity ($\vec{J} = \vec{v}_{\mathrm{ave}} \rho$) to the equations gets: $$c^2 \frac{\partial \rho}{\partial x_i} + \frac{\partial J_i}{\partial t} = \left[c^2 \delta_{ij} - (\vec{v}_{\mathrm{ave}})_i (\vec{v}_{\mathrm{ave}})_j \right]\frac{\partial \rho}{\partial x_j} - (\vec{v}_{\mathrm{ave}})_i \rho \frac{\partial (\vec{v}_{\mathrm{ave}})_j}{\partial x_j} + \rho \frac{\partial (\vec{v}_{\mathrm{ave}})_i}{\partial t},$$ after application of the continuity equation $\partial_\mu J^\mu = 0$. The last term is an expected acceleration term… $\endgroup$ Commented Oct 23, 2017 at 23:11

1 Answer 1


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.

  • $\begingroup$ The connection to gauge symmetry pops out faster if you start with the Lorenz gauge vector potential wave equation, and take the derivatives of both sides of the equation necessary to get $\vec{E}$ and $\vec{B}$ wave equations. $\endgroup$ Commented Oct 24, 2017 at 0:09
  • $\begingroup$ @SeanE.Lake Faster than the single line $\ast d(\ast F) = J \implies d(\ast d(\ast F)) = dJ$? $\endgroup$
    – tparker
    Commented Oct 24, 2017 at 0:29
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    $\begingroup$ It's only a single line if the reader understands all of the machinery behind differential forms - a subject I haven't yet mastered, for example. $\endgroup$ Commented Oct 24, 2017 at 0:32
  • $\begingroup$ @SeanE.Lake Gotcha. The point is that while each of your steps in the OP is completely legitimate, they aren't all invertible. So while Maxwell's equations imply the wave equations for $F_{\mu\nu}$ that you gave, the converse is not true - the vast majority of the solutions to those wave equations don't actually solve Maxwell's equations, and so have no physical significance. $\endgroup$
    – tparker
    Commented Oct 24, 2017 at 1:12

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