Maxwell's equations (in differential form) and Jefimenko's equations are in some sense inverses of each other: Maxwell's equations tell you how to use the EM fields $F$ to calculate the current (and charge) density $J$, and Jefimenko's equations tell you how to use the current density $J$ to calculate the EM fields $F$.

Maxwell's equations are only internally consistent if the current is conserved. Put another way: given any EM field $F$, the corresponding current density $J$ that comes out of Maxwell's equations will automatically be conserved.

Q1. What happens if you just blindly plug a non-conserved current $J$ into Jefimenko's equations? Do you still get a seemingly sensible EM field $F$?

Q2. If the answer to Q1 is yes, then what happens if you take that EM field $F$ and plug it into Maxwell's equations to try to get back a current density? The resulting current density $J'$ must be conserved if it comes out of Maxwell's equations, so clearly you can't end up with the same current density $J$ that you started with. This seems to suggest that composing together Maxwell's and Jefimenko's equations gets you some kind of projection map from the space of all possible four-current fields $J$ to the subspace of conserved four-current fields. What is that nature of this projection map? (My guess is that there's some natural decomposition of a general differential form $J$ as the sum of a closed part and a non-closed part, and this projection map simply discards the non-closed part.)

  • $\begingroup$ As a first guess, when a current is not conserved, charge must accumulate somewhere or be drained from somewhere. An increasing/decreasing charge must lead to an increasing/decreasing electric field. $\endgroup$ – mmesser314 Apr 17 at 16:53
  • $\begingroup$ Current isn’t conserved, charge is conserved. You can very well have nonzero current now and zero current later. $\endgroup$ – Dale Apr 17 at 21:28
  • 1
    $\begingroup$ @Dale Yes, but it's standard terminology that the term "conserved current" refers to a four-vector field with zero divergence. It's shorter to say than "current for a conserved charge". $\endgroup$ – tparker Apr 18 at 1:02

If we use arbitrary charge/current densities in Jefimenko's equations, then the resulting fields generally do not satisfy Maxwell's equations. The derivation of Jefimenko's equations assumes that the current is conserved. Here's a quick review of the derivation, which I'll refer back to when answering Q2.

Half of Maxwell's equations (the metric-independent ones) are satisfied by the definition $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$, with no constraints on $A_\mu$. The remaining Maxwell's equations (the metric-dependent ones) are $$ \newcommand{\bfx}{\mathbf{x}} \newcommand{\pl}{\partial} \pl^2 A_\mu - \pl_\mu \pl\cdot A = J_\mu. \tag{1} $$ According to equation (6.66) in ref 1, the distribution $$ G(\bfx'-\bfx,t'-t)\equiv\frac{\delta(t-(t-|\bfx'-\bfx|))}{|\bfx'-\bfx|} $$ satisfies \begin{align} \pl^2 G(\bfx'-\bfx,t'-t) &\equiv (\pl_t^2-\pl_\bfx^2)G(\bfx'-\bfx,t'-t) \\ &=4\pi \delta(t'-t)\delta(\bfx'-\bfx). \tag{2} \end{align} Now consider the gauge $$ \pl\cdot A=0. \tag{3} $$ In this gauge, equation (1) reduces to $\pl^2 A_\mu\propto J_\mu$, which is satisfied by $$ A_\mu(\bfx,t)=\frac{1}{4\pi}\int dt'\,d^3\bfx'\ G(\bfx'-\bfx,t'-t) J_\mu(\bfx',t'). \tag{4} $$ Equation (4) immediately gives Jefimenko's equations as shown on the Wikipedia page (the version written in terms of potentials).

We're not done yet, because we need to check that the condition (3) is consistent with (4). In general, it isn't: they are consistent only if the current is conserved. To see this, apply $\pl^\mu$ to the equation $\pl^2 A_\mu\propto J_\mu$ and then use (3). The result is $0=\pl\cdot J$. This shows that current conservation is a necessary condition for the derivation shown above to be valid. To see that it is also a sufficient condition, start with (4) and use the fact that $G$ depends only on the differences $\bfx'-\bfx$ and $t'-t$ to get $$ \pl^\mu A_\mu=\frac{1}{4\pi}\int dt'\,d^3\bfx'\ G(\bfx'-\bfx,t'-t) \partial^\mu J_\mu(\bfx',t'). \tag{5} $$ This shows that current conservation implies (3). Altogether, the derivation that leads to Jefimenko's equations (4) is valid if and only if $\pl\cdot J=0$.

Now we have the material we need to answer Q2. If we use an arbitrary current in equation (4) and put this back into Maxwell's equation (1), then the identity (2) impiles $$ \pl^2 A_\mu - \pl_\mu \pl\cdot A = \tilde J_\mu $$ with $$ \tilde J_\mu\equiv J_\mu - \frac{1}{4\pi} \pl_\mu \int G\pl\cdot J. \tag{6} $$ The question is whether the modified current (6) is conserved. Apply $\pl^\mu$ to (6) and use (2) to confirm that the answer is yes.

  1. Jackson (1975), Classical Electrodynamics, 2nd edition
  • 1
    $\begingroup$ What about Q2? What one was really my main question. $\endgroup$ – tparker Apr 17 at 18:14
  • $\begingroup$ So this is certainly a reasonable interpretation of my phrasing of Q1, but when I said "seemingly sensible" I really meant something more like "do the integrals even converge at all." The heart of my question is Q2, where I'm asking about the physical interpretation of the EM fields that Jefimenko's equations do produce if you plug in a non-conserved current. $\endgroup$ – tparker Apr 17 at 18:18
  • $\begingroup$ @tparker I updated the answer to address Q2. $\endgroup$ – Chiral Anomaly Apr 17 at 18:31
  • $\begingroup$ @tparker The update shows that the effect is to project the original current onto the space of conserved currents, like you anticipated. I didn't use the word "project," but that should be clear from equation (6): If we apply the same modification to $\tilde J_\mu$ that we applied to $J_\mu$, we get the same $\tilde J_\mu$ back again. So it really is a projection, which I assume is what you meant by discarding the non-closed part (of the hodge dual of $J$). Is that what you were looking for? $\endgroup$ – Chiral Anomaly Apr 17 at 18:37
  • $\begingroup$ Yes, thank you. In retrospect, the final equation (6) is really natural - it effectively subtracts off the EM fields "sourced" by the non-conserved part of $J$ - and I probably should have been able to guess it without doing any math. $\endgroup$ – tparker Apr 17 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.