# 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? Apr 10 '15 at 12:26

For the last part, he did this for the general retarded potentials (if I remember well) and he says to us that it is easier to calculate the potentials of a specific problem and then differentiate them for that specific problem.

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.

One way to look at it is that the electric field is the time-space components of the field strength tensor, while the magnetic field is the space-space components of the field strength tensor. Since the charge density is the time component of the four-vector current density, it (hopefully) seems plausible that the space-space components of the field strength would not get any contribution from such terms.

To see why this is the case, note that we can write out the four-vector potential as \begin{align*} A_\mu &= \frac{\mu_0}{4 \pi} \int \frac{ J_\mu (\vec{r}', t_r)}{R} \, d\tau'. \end{align*} The field strength tensor is then \begin{align*} F_{\mu \nu} &= \partial_\mu A_\nu - \partial_\nu A_\mu \\ &= \frac{\mu_0}{4 \pi} \int \left[\frac{ \dot{J}_\nu (\vec{r}', t_r)}{R} (\partial_\mu t_r) - \frac{ \dot{J}_\mu (\vec{r}', t_r)}{R} (\partial_\nu t_r)- \frac{ J_\nu (\vec{r}', t_r)}{R^2} (\partial_\mu R) + \frac{ J_\mu (\vec{r}', t_r)}{R^2} (\partial_\nu R) \right] \, d \tau' \end{align*} Now, $$(\partial_\mu t_r) = (1, - \hat{R}/R^2)/c$$ and $$\partial_\mu R = (0, \hat{R})$$. The "electric part" of $$F_{\mu \nu}$$ will be given by the components $$F_{0i}$$. In this case, the third term in our expression above will vanish (since $$\partial_0 R = 0$$), and the first, second, and fourth terms correspond to the second, third, and first terms of your electric field expression for Jefimenko's equations. The terms involving $$\rho$$ are those that depend on $$J_0 = c \rho$$, which will in this case be the second and fourth terms above.

For the "magnetic part", we look at the components of $$F_{\mu \nu}$$ with $$\mu = i$$, $$\nu = j$$. In this case, all of the terms contribute. But because each component of a cross product is secretly two terms (if $$\vec{A} = \vec{B} \times \vec{C}$$, then $$A_x = B_y C_z - B_z C_y$$), we can pair off each of these two terms into two cross products. If we do this, we end up with Jefimenko's expression for the magnetic field. We also see (to answer your original question) that none of the terms that contribute to the magnetic field involve $$J_0 = c \rho$$, and so $$\rho$$ does not appear on our expression for $$\vec{B}$$.

(As an aside, it pains me to write out a nice relativistic equation where the arguments of the functions are decomposed into $$\vec{r'}$$ and $$t_r$$. If I have time at some later date, I might try rewriting this whole thing in terms of the retarded Green's function instead and see if it looks any nicer.)