# 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 $\frac{\partial \bf E}{\partial t}=0$?

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.

• That seems weird to me. If, given $\rho$ and J Maxwell's equations have multiple solutions, and thus multiple fields, then which physical parameter dictates which field the charges actually produce? – Nicol Dec 16 '17 at 18:38
• Yes this is important. In many differential equations there are more degrees of freedom than just one solution. – mathreadler Dec 17 '17 at 9:19

The OP is asking whether Maxwell's equations together with the continuity equation and the condition that $\textbf J =0$ is enough to obtain $\textbf B=0$ and $\frac{\partial \textbf E}{\partial t}=0$.

The answer for the first part is negative. It's easy to see that Maxwell's equations alone do not precisely determine the electric and magnetic field. Adding a constant field to a solution will again give a solution: $$\textbf E \to \textbf E + \textbf E_0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\textbf B \to \textbf B + \textbf B_0$$ This corresponds to the physical possibility of a "background field" which cannot be determined solely from the equations themselves.

The answer to the second question is also negative. Take $\textbf E = (xt,-yt+3y,0)$. Then

$$\nabla \cdot \textbf E = 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{but}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\partial \textbf E}{\partial t}=(x,-y,0)\neq 0$$ This solves Maxwell's equations, as a corresponding magnetic field would be $\textbf B = (0,0,xy/c)$. This example may not be particularly significant physically, but it shows that it is not possible to obtain the conditions you're stating purely from the equations. To obtain what you're stating you need more information, mostly in the form of boundary conditions, which is often the most physically relevant.

Further to the existing answers: yes, the solution isn't unique. In fact, even if we strengthen the assumptions to $\rho=0$ so that there are no charges (as occurs in a vacuum), you can still get an electromagnetic wave. In other words, light can travel through a vacuum!

If this feature of differential equations seems odd, it's no odder than the fact that, when there are no forces at all on a body (i.e. $\ddot{x}=0$), it can still be moving. Mathematically, the reason is the same. For the OP's problem we can show that $$c^{-2}\partial_t^2 \mathbf{E}-\nabla^2\mathbf{E}=-4\pi\boldsymbol{\nabla}\rho,\,c^{-2}\partial_t^2 \mathbf{B}-\nabla^2\mathbf{B}=0.$$(If you want to try proving these, use $\boldsymbol{\nabla}\times\boldsymbol{\nabla}\times\mathbf{X}=\boldsymbol{\nabla}(\boldsymbol{\nabla}\cdot\mathbf{X})-\nabla^2\mathbf{X}$.) We can always add to the solution something proportional to the plane wave $\exp i(\mathbf{k}\cdot\mathbf{x}-kct)$ for a constant wavevector $\mathbf{k}$, so you can't rule them out.

When we go back to the original first-order equations, we find such plane-wave solutions can be added but not arbitrarily. For example, if I add $\mathbf{E}_0,\,\mathbf{B}_0$ times the plane wave, we require $\mathbf{E}_0\cdot\mathbf{k}=\mathbf{B}_0\cdot\mathbf{k}=0$.

A short, but I think important, comment to add to toBuzz's Answer and J. G.'s answer, which show that the no magnetic field solution is sound, but not unique, and in general one must consider boundary conditions.

A key word and concept used here to fix solutions that any full answer to the present and similar questions should, IMO, talk about is the Sommerfeld Radiation Condition. The deployment of this condition does make the zero magnetic field solution to the OP's problem unique. In particular, it rules out infinite plane waves that can be added to the solution as in J. G.s answer.

Not only does the Sommerfeld condition remove ambiguity and is a widely assumed "boundary condition at infinity" for electromagnetic problems, it (or something like it) is also needed to deploy Helmholtz's Theorem, which is essentially about the Hodge decomposition of a one form when we rule out the harmonic component.