Taking the divergence of amperes law gives $\nabla \cdot J = 0$

Comparing this to the continuity equation, amperes law requires that for it to be valid $ \partial \rho / \partial t = 0$

Meaning that any sudden breaks in the current cause amperes law to be invalid with describing reality.

However if I have a current that is time varying, such that $ \partial \rho / \partial t $ is still 0 then why is amperes law not enough? why does the $+\partial E / \partial t$ term have to be added. I completely understand why it has to be added in the case of a discontinuity in the wire, as in this case $\nabla \cdot J$ doesn't equal zero.

but for a wire that has a constant charge density everywhere, but has a changing velocity apparently amperes law isn't enough?

My ultimate question is,

What is it about a non zero $\partial j / \partial t$ that means that $ \nabla \cdot J$ doesn't equal 0? as for a non zero dj/dt but zero dp/dt, the divergence of J is still zero and thus amperes law should still hold?

p.s I am not completely clueless on why this has to be added, as in the potential formulation this is crucial as you substitute E in terms of $\phi $ and A into the $\partial E / \partial t$ term, meaning B is dependant on the curl of second time derivative of itself, and if it weren't there the equation would.be different

  • $\begingroup$ You seem to implicitly rely on an idea that "taking the divergence" of an equation produces an equation that is somehow equivalent to the original one, but differentiation is not a bijective operation so this just doesn't work. That the divergence of Ampère's law is fulfilled when $\partial_t \rho = 0$ is not an argument that Ampère's law itself should be fulfilled under that condition. $\endgroup$
    – ACuriousMind
    Commented Nov 8, 2021 at 23:20

1 Answer 1


$\frac{\partial \mathbf{J}}{\partial t} \neq \mathbf{0}$ doesn't imply $\mathbf{\nabla \cdot J} \neq \mathbf{0}$ (as you can see by picking $\mathbf{V} = t \mathbf{v}$ for any field $\mathbf{v}$ with $\mathbf{\nabla \cdot v} = \mathbf{0}$). Instead, $\frac{\partial \mathbf{J}}{\partial t} = \mathbf{0}$ is a separate requirement for Ampère's Law to hold.

A fact that is not that well known is that Maxwell's equations can be fully solved in flat spacetime under the assumption that charge and current densities are localized and the fields vanish at infinity, which is the situations we are usually interested in. In this situation, the electric and magnetic field are given by Jefimenko's equations, which I quote here from Griffiths' book (tags according to 4th edition): $$\begin{align} \mathbf{E}(\mathbf{r},t) &= \frac{1}{4\pi \epsilon_0} \int \left[\frac{\rho(\mathbf{r}', t_r)}{R^2} \hat{\mathbf{R}} + \frac{\dot{\rho}(\mathbf{r}', t_r)}{c R} \hat{\mathbf{R}} - \frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c^2 R}\right] \mathrm{d}\tau', \tag{10.36} \\ \mathbf{B}(\mathbf{r},t) &= \frac{\mu_0}{4\pi} \int \left[\frac{\mathbf{J}(\mathbf{r}',t_r)}{R^2} + \frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c R}\right] \times \hat{\mathbf{R}} \,\mathrm{d}\tau', \tag{10.38} \end{align}$$ where I use SI units and I denote $\mathbf{R} = \mathbf{r} - \mathbf{r}'$ (Griffiths uses a cursive $r$), $t_r = t - \frac{R}{c}$ is the retarded time.

As you might see, the presence of a changing current ($\dot{\mathbf{J}} \neq \mathbf{0}$) means the electric and magnetic field share a term. This fact will lead to the need of a correction to the Ampère's Law in many, but not all, situations — see doi: 10.1119/1.16589, which notices on p.114 that Ampère's Law still holds if the current depends linearly on time and charge density is constant.

These issues are fairly well studied on doi: 10.1119/1.16589, though it focus a bit more on the validity of the Biot–Savart and Couloumb laws. I believe this answer might interest you as well.

  • $\begingroup$ I will most likely accept your answer as you have confirmed what I thought about Div J = 0 doesn't mean that dj/dt must be zero. You say it is a separate condition in of itself on amperes law. Is there any way to derive this condition that you know of, instead of looking at the full jefimenko equations? E.g a cheek divergence of both sides to get the Div J requirement $\endgroup$ Commented Nov 8, 2021 at 23:24
  • $\begingroup$ being abit wishy woshy with the math. it is clear that if (d^2/dt^2) J is non zero then there will be a non zero (d^2/dt^2) B as per amperes law which means that there is a changing electric field as per faradays law ( taking time derivative of faradays law) so it must be included. (doing the same for dj/dt would be similar ish but less intuitive) $\endgroup$ Commented Nov 8, 2021 at 23:31
  • $\begingroup$ Because this is a separate condition imposed by maxwells equations , the original reason I am thinking about this, is the London equation have a non zero dj/dt, yet in the derivation they don't use the maxwell corrected version. so does this mean that they are wrong/approximate solution? $\endgroup$ Commented Nov 8, 2021 at 23:47
  • $\begingroup$ @jensenpaull I tried proving it and failed, and then I recalled that the paper I mentioned actually shows that there are some circumstances in which Ampère's Law holds even though $\mathbf{J}$ changes with time — namely, if the current is linear on time and charge density is constant, Ampère's Law still holds due to some lucky cancellations. This is detailed in the paper, which I find particularly pedagogical. I've corrected my answer $\endgroup$ Commented Nov 8, 2021 at 23:47
  • 1
    $\begingroup$ @jensenpaull Usually university libraries buy access to these papers, so if you are at an Uni. it might be interesting to contact your library. It is also always important to remind to not use Sci-Hub, which is a website providing free access to millions of articles by bypassing the paywalls without any regard to copyright, as described on the site's Wikipedia page $\endgroup$ Commented Nov 9, 2021 at 0:19

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