# Why is there no (time derivative of charge density) in the $B$ field in Jefimenko's equations?

I was going through Griffiths chapter on potentials and fields just to brush up on a few old things. He gets to Jefimenko's equations by this general path:

1. Maxwell's equations.

2. Introduce scalar and vector potentials.

3. Reduce MW's equations to the two equations with only potentials (no fields).

4. Choose the Lorenz gauge, which frames the two equations as 4-D Poisson equations: $$\Box^2V = -\rho/\epsilon, \quad \Box^2 \vec A = -\mu\vec J \,.$$

5. We already know the solutions to the static case ($\dot{\rho},\dot{\vec J} = 0$ means $\Box^2$ becomes $\nabla^2$).

6. He says and then briefly proves that to get the non-static answers, we just use the retarded time $t_r = t - \frac{\left|r - r' \right|}{c}$ in the integrals for the static case.

7. So now we have the potentials for a non-static source, and we just plug them into $\vec E = -\nabla V - \frac{\partial \vec A}{\partial t}$ and $\vec B = \nabla \times \vec A$ which apparently give us Jefimenko's equations (say $R = |r - r'|$, $\hat R$ is the unit vector in the direction of $R$, and $J$ and all $r$'s are vectors):

$$E(r, t) = \frac{1}{4 \pi \epsilon} \int \left[\frac{\rho(r',t_r)}{R^2}\hat{R} + \frac{\dot{\rho}(r',t_r)}{cR}\hat{R} - \frac{\dot{J} (r',t_r)}{c^2R} \right] \mathrm{d} \tau '$$

$$B(r, t) = \frac{\mu}{4 \pi} \int \left[ \frac{J(r',t_r)}{R^2} + \frac{\dot{J} \left( r', t_r \right)}{cR} \right] \times \hat{R} \ \mathrm{d}\tau '$$

So it all makes sense to me, the derivation anyway. But is there a simple intuitive reason for why the $E$ equation has a $\dot J$ term but the $B$ term has no $\dot \rho$ term? I'd suspect it has something to do with the "asymmetry" in Maxwell's equations with respect to $E$ and $B$ (which I also don't really understand), but that doesn't answer much.

Also, what are some concrete examples of the difference between $\dot \rho$ and $\dot J$? What I mean is, it seems like they must usually be very linked, because if you have a changing amount of charge $\dot \rho$ at some point, that implies that there is a changing amount of current $\dot J$ at the point also, because the charge at that spot has to go somewhere (so $\dot J \neq 0$) for $\rho$ to change. I guess I could think of some mythical (to me anyway, because I don't know if this is possible) chemical reaction where charge at a point "disappears" and "reappears" without having to actually move spatially.

I have one more small point of confusion about Jefimenko's equations: Griffiths says

In practice Jefimenko's equations are of limited utility, since it's typically easier to calculate the retarded potentials and differentiate them, rather than going directly to the fields.

Isn't that exactly what he did in the process I described?

• I was pondering that very same question, and found this question here, but unfortunately no answer was posted... Did you find a reasonable answer somewhere else? – Shay Ben Moshe Apr 10 '15 at 12:26

I know it's been quite a long time since this question was posted, but for other curious people, I thought I would add my understanding here as well. The intuitive reason that the $\dot{\rho}$ is not included in $\vec{B}$ is already within the question, that is $\vec{B}$ is only dependent on the vector potential $\vec{A}$, which does not explicitly contain contributions from the charge density $\rho$. Though current may be created from time varying local charge, local charge is not necessary to create current locally (standing electric waves being generated from outside of the field of view of the field location, for example, can create free current $J_f=\sigma E$). Electric field, however, does have an explicit dependence on current density, as seen by the $\frac{\partial A}{\partial t}$ term.
Let's say you do have a current being produced by time varying charge, J=$\frac{d\rho}{dt}$. It is now obvious that B does indeed have a $\dot{\rho}$ term in J, and that $\dot{\rho}$ does not equal nor imply a significant $\dot{J}$. You could have charge varying at a constant rate in time, without any acceleration in its decay/increase.
You will find the answer to this question in my paper about General Classical Electrodynamics, in particular, see §3.2 and §3.6. Either we simplify Maxwell's theory by applying Ockham's razor, such that the $\dot \rho$ term does not occur anymore in Jefimenko's electric field expression, or we generalise classical electrodynamics theory with extra longitudinal far field waves, such that the electric field $\dot \rho$ term in combination with scalar field $\dot \rho$ term can be understood as a longitudinal electric 'far field' wave, which carries energy/information/momentum. Such a term should not occur in the vector magnetic field expression, because the $\dot \rho$ term is in fact a longitudinal electric field wave. Extra scalar 'magnetic' field expressions are required in order to understand the longitudinal electric wave.
The incorrect equation $\vec J = \dot \rho$ should be $~ -\nabla \! \cdot \! \vec J = \dot \rho$ (charge current continuity equation), so you cannot just replace $\vec J$ for $\dot \rho$, in Jefimenko's magnetic field expression.