In 3-space, one *can* interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.

But this is rather illusory. In relativity, the equations look quite different:

$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= 0\end{align*}$$

where $F$ is the electromagnetic field bivector. The vector derivative $\nabla$ can only increase or decrease the grade of an object by 1. Since $F$ is grade 2, the divergence equation describes its relationship with a grade 1 source term (the vector four-current $J$). The curl equation describes its relationship to a grade 3 (trivector) source term (of which there is none).

The reason the 4 Maxwell equations in 3-space come out the way we do is that we ignore the timelike basis vector, which would unify the scalar charge density with the 3-current as the four-current, as well as unify the E field with the B field as a bivector. The relativistic formulation, however, is considerably more sensible, as it correctly presents the EM field as one object of a single grade (a bivector), which can only have two sources (vector or trivector). It just so happens that the EM field has no trivector source.

What if there were trivector sources? Well, as you observe, there would be magnetic charge density (monopoles), but there would also be quite a bit *more*. There would have to be magnetic current as well, which would add an extra term to the $\nabla \times E$ equation to fully symmetrize things.

I guess different authors use different definitions. For me, it is that the E- and B-fields do not have time derivatives, hence curl free, conservative E-fields and B-fields that can depend only on steady currents.

The condition that the divergence of $\partial {\bf E}/\partial t = 0$ is not the same thing. The E-field could be time variable and have this still be true - e.g. in a transverse electromagnetic wave! Clearly that is not a magnetostatic situation either.

The curl of the B-field does not have to be zero in magnetostatics; steady currents are allowed, which obviously means you have to have (uniformly) moving charges. As ${\bf J} = \rho {\bf v}$, then $\partial {\bf J}/\partial t = 0$ implies only that ${\bf v}\partial \rho /\partial t + \rho \partial{\bf v}/\partial t = 0$. So it might be possible to arrange static magnetic fields by having a non-zero rate of change of charge density balanced by accelerating charges to somehow keep the current density constant!
The continuity equation, $\nabla \cdot {\bf J} + \partial \rho/\partial t =0$, tells you that a time-varying charge density would require a current density divergence.

## Best Answer

A system with no moving charges is

consistentwith there being only a static electric field and no magnetic field. However, it does notrequirethere to be no time dependent phenomena. The general solution in this case consists of an electrostatic field, plus freely propagating electromagnetic waves.You can see the consistency of the static fields by setting the time derivatives in your equations to zero. Then there is only a static divergence source for $\vec{E}$, meaning just an electrostatic field. However, it is possible to add the time-dependent fields of one or more propagating waves in top of that.

In general, any system of differential equations is going to have its solutions determined by the equations themselves, along with the boundary conditions. It's the (space and time) boundary conditions in this case that determine whether there are also freely propagating electromagnetic waves present.