[Physics] Do Maxwell’s equations imply that still charges produce electrostatic fields and no magnetic fields


Suppose we have a charge distribution $\rho$, whose current density is J$=\vec{0}$ everywhere; the continuity equation implies $\frac{\partial \rho}{\partial t}=0$, i.e., the charges don't move and the density is always the same. We'd expect such a distribution to produce a static electric field and no magnetic field. If we plug our variables into Maxwell's equations we get $$\nabla\cdot \textbf{E}=4\pi \rho $$ $$ \nabla \cdot \textbf{B}=0$$ $$\nabla \times \textbf E= -\frac{1}{c}\frac{\partial \textbf{B}}{\partial t}$$ $$\nabla \times \textbf B= \frac{1}{c}\frac{\partial \textbf{E}}{\partial t}$$

But how does one go from this to $\bf B=0$ and $\frac{\partial \bf E}{\partial t}=0$?

Best Answer

A system with no moving charges is consistent with there being only a static electric field and no magnetic field. However, it does not require there 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.