Jefimenko's equations are $$\textbf{E}(\textbf{r}, t_r) = \frac{1}{4\pi\epsilon_0}\int \left[\rho\left(\textbf{r}', t_r\right)\frac{\textbf{r} - \textbf{r}'}{\left|\textbf{r} - \textbf{r}'\right|^3} + \dot{\rho}\left(\textbf{r}', t_r\right)\frac{\textbf{r} - \textbf{r}'}{c\left|\textbf{r} - \textbf{r}'\right|^2} + \dot{\textbf{J}}\left(\textbf{r}', t_r\right)\frac{1}{c^2\left|\textbf{r} - \textbf{r}'\right|}\right] d\tau'$$


$$\textbf{B}(\textbf{r}, t_r) = \frac{\mu_0}{4\pi}\int \left[\textbf{J}\left(\textbf{r}', t_r\right)\times\frac{\textbf{r} - \textbf{r}'}{\left|\textbf{r} - \textbf{r}'\right|^3} + \dot{\textbf{J}}\left(\textbf{r}', t_r\right)\times\frac{\textbf{r} - \textbf{r}'}{c\left|\textbf{r} - \textbf{r}'\right|^2}\right] d\tau',$$

where $t_r = t - \dfrac{\left|\textbf{r} - \textbf{r}'\right|}{c}$ is the retarded time.

Griffiths claims that Coulomb's law holds when "all the source charges are stationary." However, the first of Jefimenko's equations tells us that all that is required for Coulomb's law to hold is $\dot{\rho} = 0$ and $\dot{\textbf{J}} = \mathbf{0}$. I find it noteworthy that neither of the conditions $\dot{\rho} = 0$ and $\dot{\textbf{J}} = \mathbf{0}$ implies the other. While it is true that $$\textrm{No moving charges} \implies \left[\dot{\rho} = 0 \; \textrm{ and } \; \dot{\textbf{J}} = \mathbf{0}\right],$$ the converse doesn't hold, and thus Griffiths provides us with an unnecessarily strong condition for Coulomb's law to be valid.

Griffiths also provides us with an unnecessarily strong condition for the Biot-Savart law to hold, namely the minimal condition for Coulomb's law to hold, $\dot{\rho} = 0$ and $\dot{\textbf{J}} = \mathbf{0}$. From Jefimenko's second equation one sees that the minimal requirement for the Biot-Savart law to hold is just $\dot{\textbf{J}} = \mathbf{0}$. This is quite remarkable considering that one might naively expect any piling-up charges to produce a changing electric field that would produce a magnetic field that would depend on the piling up of charge.

I believe I've made it clear why I don't trust authors to give me the full picture right away. (Perhaps it's justified in Griffiths's case though I'm not content with it.) Jackson claims that Coulomb's law is valid when $\dot{\rho} = 0$, $\dot{\textbf{E}} = 0$, and $\dot{\textbf{B}} = 0$; I'm not sure how he intends to define magnetostatics, which I'm taking to be his condition for the Biot-Savart law to hold, but what is clear is that one thing that characterizes magnetostatics is $\dot{\rho} = 0$. If someone could offer their complete interpretation of how Jackson intends to define magnetostatics, that would be great.

The question: Are Jackson's conditions for Coulomb's law to hold minimal? Are his conditions (whatever they are) for the Biot-Savart law to hold minimal?


1 Answer 1


As you say. the Biot-Savart law holds whenever $\dot{\bf J} \equiv {\bf 0}$, even if ${\bf \nabla} \cdot {\bf J} \neq {\bf 0}$ and so $\rho$ changes linearly over time. The assumption $\dot{\rho} \equiv 0$ is not necessary. Moreover, since ${\bf J}({\bf x}, t) \equiv {\bf J}({\bf x}, t_r)$ in this case, we trivially have that that the instantaneous version of the Biot-Savart law holds exactly.

Your intuition is correct that the charge buildup induces a changing electric field which induces a part of the magnetic field. However, it turns out that even incorporating this induction effect, when $\dot{\bf J} \equiv 0$ the B-S law continues to hold exactly. When you derive the B-S law from Ampere's law, there is a term ${\bf \nabla} \cdot {\bf J}$ that usually gets dropped because we assume the charges aren't building up. If we keep that term, we find that if $\dot{\bf J} \equiv 0$ then it exactly cancels the electromagnetic induction effect from the changing electric field, so that the B-S law still holds exactly.

Much more surprisingly, the assumption $\dot{\rho} \equiv 0$ isn't actually necessary for Coulomb's law to hold either, despite the appearance of $\dot{\rho}$ in the first Jefimenko's equation. As long as $\dot{\bf J} \equiv {\bf 0}$, the second term in the equation turns out to exactly cancel out the retardation effect, and the instantaneous version of Coulomb's law holds exactly, even though the charge density is changing over time! This is one of those situations where EM superficially looks nonlocal but actually remains completely local, and is discussed in detail here.

TLDR: both Coulomb's law and the Biot-Savart law hold exactly if $\dot{\bf J} \equiv 0$, even if $\dot{\rho} \neq 0$.

As for the general definition of "magnetostatics", you already asked that question and I answered it here. Should you describe an electrodynamic situation as "magnetostatic" if the electromagnetic induction from the changing electric field causes the Biot-Savart law to hold exactly? I probably wouldn't, but that's just personal preference. There's no standardized definition of the term "magnetostatic" precise enough to definitively include or exclude this unusual edge case.

  • $\begingroup$ What I meant to say was that you didn't show that $\left[\dot{\rho} = 0, \mathbf{E} = \mathbf{0} \textrm{ and } \mathbf{B} = \textbf{0}\right] \implies \dot{\mathbf{J}} = \mathbf{0}$ and that the converse is false. $\endgroup$
    – user113773
    Feb 26, 2019 at 5:05
  • $\begingroup$ @PiKindOfGuy $\dot{\bf E} = \dot{\bf B} = {\bf 0}$ is enough to show that $\dot{\bf J} = {\bf 0}$: just take the time derivative of Ampere's law. I explained why the converse is false in my previous comment. $\endgroup$
    – tparker
    Feb 26, 2019 at 13:00

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