Does a time-independent charge distribution necessarily imply an electrostatic situation? If it is specified that a charge distribution $\rho (\textbf{r},t)$ is time-independent, $\rho (\textbf{r},t) = \rho (\textbf{r})$, then it follows from continuity that $\nabla \cdot \textbf{j}(\textbf{r}, t) = 0.$
Now the definition of an electrostatic situation is one in which the $\textbf{E}$ and $\textbf{B}$ fields decouple; that is, Gauss's and Faraday's Laws must reduce to
$$\nabla \cdot \textbf{E} = \rho (\textbf{r})/\epsilon_0 \quad({\rm A})$$
$$\nabla \times \textbf{E} = \textbf{0},\quad({\rm B})$$
which is to say that we need $\frac{\partial \textbf{B}}{\partial t} = \textbf{0}$. This is in turn only true (Ampere-Maxwell Law) if (1) $\frac{\partial \textbf{E}}{\partial t} = \textbf{0}$ and (2) $\textbf{j}(\textbf{r}, t)=\textbf{j}(\textbf{r})$. Now (1) presupposes that (A) and (B) hold, although given the circularity I'm sure there's some uniqueness theorem I can appeal to (Helmholtz or the like). It's (2) that I'm not sure about. In particular, the conditions I give in the first paragraph only let us conclude that $\nabla \cdot \textbf{j}(\textbf{r}, t) = 0.$ Thus my question boils down to, does $\nabla \cdot \textbf{j}(\textbf{r}, t) = 0$ imply that $\textbf{j}(\textbf{r}, t)=\textbf{j}(\textbf{r})$?
Note that I am taking an isolated system in which $\rho (\textbf{r},t) = \rho (\textbf{r})$ is true for all $t$, so there are no free EM fields (if (2) is true).
 A: 2 questions seem to be asked here:
(1)Does a Time independant charge distribution  imply electrostatics?
No, Take a circular ring with osscilating current. This produces EM waves with a time independant charge distribution.
Well what if we also set $\vec{J}$ to be time independant?
Still no, $\vec{J}$ and $\rho$ being time independant does not imply $\frac{\partial \vec{E}}{\partial t},\frac{\partial \vec{B}}{\partial t}$ to be zero, this is because the EM field is not uniquely determined by the location and values of charge and currents, for example the homogenous wave equation
(2)Does the electrostatic condition imply a time independant charge distribution
Yes, setting $\frac{\partial \vec{E}}{\partial t},\frac{\partial \vec{B}}{\partial t}$ to be zero by force;
This imposes certain conditions onto $\rho$ and $\vec{J}$
Gauss law:
$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$
$$\nabla \cdot \frac{\partial \vec{E}}{\partial t} = \frac{1}{\epsilon_0 }\frac{\partial \rho }{\partial t}$$
$$\frac{\partial \rho }{\partial t} = 0$$
Amperes law:
$$\nabla ×  \vec{B} = \mu_0 \vec{J} + \frac{\partial \vec{E}}{\partial t}$$
$$\nabla ×  \vec{B} = \mu_0 \vec{J} $$
$$\nabla ×  \frac{\partial \vec{B}}{\partial t} = \mu_0 \frac{\partial \vec{J}}{\partial t} $$
$$\frac{\partial \vec{J}}{\partial t} = 0$$
This means that when we find the fields in the electrostatic condition, both the charge density and current density are time invariant. Checking our answer, the solution to maxwells equations in this form is coulombs law and biot savart, when the charge density and current density(divergenceless) are time invariant, the fields are also time invariant.
A: The questions hinges on what you mean by an "isolated system."  Because changing $\vec{E}$ and $\vec{B}$ fields are sources for one-another, it is easy to have solutions of Maxwell's Equations with time-independent $\rho$ and $\vec{J}$ but time-dependent fields.  In fact, it is straightforward to set $\rho=0$ and $\vec{J}=0$ and still have dynamical $\vec{E}$ and $\vec{B}$ fields.  That's just the situation in which there are propagating plane electromagnetic waves but no sources.
If you want to rule out the existence of such waves by saying you have an "isolated system," then such solutions do not exist.  However, saying you have an isolated system is not a mathematical statement of precisely what conditions you want to impose.  It will turn out that if you seek to enforce that there are no external propagating electromagnetic waves, then whatever mathematical conditions you impose to enforce that condition will necessarily be equivalent to having only time-independent sources $\rho$ and $\vec{J}$.
If you do not insist that there are no propagating waves, then it is no problem to have a system of currents for which $\rho$ is constant and $\vec{\nabla}\cdot\vec{J}=0$ without the behavior being time-independent.  Any magnetostatic current distribution $\vec{J}(\vec{x},t)=\vec{J}(\vec{x},0)$ can be modified with a time-dependent modulation givng it the overall form, $\vec{J}(\vec{x},t)=\vec{J}(\vec{x},0)f(t)$.  This clearly still satisfies $\vec{\nabla}\cdot\vec{J}=0$, since the spatial derivatives are all simply scaled by $f(t)$. Hence, by the local charge conservation equation, the system still has no charge accumulation, or $\frac{\partial\rho}{\partial t}=0$. In this case, the magnetic field is clearly going to vary with time, simply because $\vec{J}$ does, and thus there will also be an electric field generated by Faraday induction.  These $\vec{E}$ and $\vec{B}$ fields will, in fact, combine to produce outgoing radiation of energy, just as in the cases described above.
