# Differentiability of electric field due to bounded volume charge distribution

In books on electromagnetism, one often sees expressions of Maxwell's equations like $$\nabla \cdot \mathbf{E}$$ and $$\nabla \times \mathbf{E}$$. These expressions make sense if $$\mathbf{E}$$ (which is due to bounded volume charge distribution) is differentiable. I ask this question because in all the textbooks on electromagnetism which I have seen, expressions like $$\nabla \cdot \mathbf{E}$$ and $$\nabla \times \mathbf{E}$$ are used and nowhere do they prove the differentiability of $$\mathbf{E}$$. How can it be justified?

Is the differentiability of $$\mathbf{E}$$ such a trivial case? If yes, why is it so? If no, why do the books ignore discussing the differentiability of $$\mathbf{E}$$?

Maxwell's equations continue to hold even when the fields are not differentiable in the usual sense as they can be interpreted in terms of "weak" or distributional derivatives. For example, the electric field jumps discontinuously across a surface charge distribution, but $$\nabla \cdot {\bf D}= \rho$$ remains true with $$\rho(x,y,z)=\sigma(x,y) \delta(z)$$. This is the case in most of physics, which is why you seldom see differentiably conditions in discussions of vector calculus in physics texts. There are exceptions of course, so caution is always required.

• Discontinuities (where fields are not differentiable) can be figured out (eg using Gauss law) and the technique you mention can be used. But how can we ensure that in the rest of the points (where it is usually considered differentiable), the field is differentiable so that $\nabla \cdot \mathbf{E}$ and $\nabla \times \mathbf{E}$ are meaningful. – N.G.Tyson Apr 29 at 12:34
• Away from sources the fields are usually solutions of some kind of laplace equation (may be a vector Laplacian) or derivatives of such solutions. But solutions of laplace equation are always smooth ($C^\infty$) functions. This is just as in complex analysis where $u$ being a solution of Laplace means that it is the real part of an analytic function and for analytic functions once-differentiable $\Rightarrow$ infinitely differentiable. – mike stone Apr 29 at 12:41
• ok and what about within the source where $\vec{E}$ is considered differentiable – N.G.Tyson Apr 29 at 12:46
• Compute the electric field in a ball of uniform charge, for example. Inside the ball the potential is a nice smooth quadratic function of $r$. Its second derivative is discontinuous on the surface where $\rho$ is discontinuous, but otherwise the potential is smooth. The issue of differentiability is really rather uninteresting which is why is not discussed in physics books. If you are a mathematician, though, you need the full machinary of Sobolev spaces to do a good job of seeing when the distributional solution of any PDE is bona-fide conventional non-distribitional function. – mike stone Apr 29 at 15:29

Differentiability of the EM fields, like for many other quantities introduced in Physics, is not a property of the world but it is part of the mathematical model we find useful to describe the world.

As such, it is a property whose validity may be judged on an experimental basis. Until the model is in agreement with experiments, the property is valid. As soon we find a significant departure from the experiments we should be ready to change our model.

Actually, for EM fields we know that continuity properties may hold only on a coarse grained scale where the spatial average property of any real measurement allows to deal with smoothly varying quantities. However at the level of a microscopic QFT description a classical EM field emerges only when we can neglect the effect on the local values of the fluctuations of the underlying quantum fields.

Notice, that even much before QM or QFT entered into the toolbox of physicists, the smoothness of macroscopic fields was considered as a consequence of averaging over spatial regions large with respect to typical atomic length-scales but small with respect to the typical length of variation of the fields.

This is exactly the analog of a classical trajectory. We know that it does not exist at the QM level, still it is a very good approximation, with all possible smoothness, if we have to deal with planetary orbits.