When is the time derivative of the electric dipole moment 0?

Question

While reading Introduction to Electrodynamics by David J. Griffiths I encountered a problem regarding the time derivative of the electric dipole moment. I wanted to find the conditions when the time derivative of dipole moment is $0$. I concluded that one of the conditions is the magneto-static condition i.e. $\nabla \cdot\mathbf{J} = 0$:

$$\nabla \cdot\mathbf{J}=-\frac{\partial \rho}{\partial t}$$

Then we can express the time derivative of the dipole moment $\mathbf{p}$: $$\frac{\text{d}\mathbf{p}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\int\limits_{V}\rho(\mathbf{r'},t)\mathbf{r'}\, \text{d}V'=\int\limits_{V}\frac{\partial \rho}{\partial t}\mathbf{r'}\, \text{d}V'=\int\limits_{V}-\nabla \cdot\mathbf{J}(\mathbf{r'},t)\:\mathbf{r'}\, \text{d}V'=0$$

Firstly, I was wondering if the reasoning here is sound also if there are any other general conditions for the time derivative of the dipole moment to be 0 or is it rather problem specific in any other case? What happens when the situation is not magnetostatic?

Answer 1

Your reasoning is half correct, the magnetostatic condition $\nabla \cdot\mathbf{J} = 0$ does indeed give rise to constant ${\bf p}$, but the converse is not true.

There are current distributions that do change the charge density $\rho$, but keep ${\bf p}$ unchanged: $$\int\limits_{V}\rho(\mathbf{r'},t)\mathbf{r'}\, \text{d}V'= \text{const.} \tag{1}$$ We call this "non-radiating currents", and a simple example is the purely radial flow, e.g. between two spherical shells:

This current does take charge away from the inner sphere and deposits it on the outer one (and may also do this alternately with some frequency in time) but it would still leave us at any point in time with a spherically symmetric charge distribution, so $(1)$ would always be $0$. More complicated examples can be constructed by taking any current distribution and enclosing it in a perfect Faradey cage. This current distribution combined with the currents in the cage's walls would then again be a non-radiating current.