# 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 Lorentz 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 $t_r = t - |r - r'|/c$ 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):

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

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

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?

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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