# How do divergences of vector fields generate distributions?

Just to clarify the title before I start, there are some "fuzzy" words that I want to get out of the way:

1. Divergence here is in the sense of the divergence theorem, the operator sometimes written $\operatorname{div} \vec F = \nabla \cdot \vec F = \partial_i F^i.$
2. Distributions here are in the sense of generalized functions used to model, e.g., the Dirac $\delta$-function: linear functionals $(\mathbb R^n \to \mathbb R) \to \mathbb R$.

In a post on Physics.SE a user was asking about the first Maxwell equation, $$\operatorname{div} \vec E = 4\pi\rho.$$The central point of confusion is this: we are using this to generalize Coulomb's law, which states that in 3D, for a point source of charge $q$ located at a point $\vec r'$, the resulting electric field is $$\vec E_q(\vec r) = q~\frac {\vec r - \vec r'}{|\vec r - \vec r'|^3},$$ but unfortunately here $\vec E_q$ is an ordinary function (albeit from $\mathbb R^3 \to \mathbb R^3$) and its divergence, defined traditionally, is also an ordinary function, $\operatorname{div} \vec E_q : (\mathbb R^3 - \{\vec 0\})\to \mathbb R.$ However the right hand side above is not an ordinary function.

The right hand side above can, however, be interpreted as a distribution where the distribution-product for "ordinary functions" $\langle f, g\rangle = \int_{\mathbb R^3} dx~dy~dz~ f(x, y, z) ~g(x, y, z).$ We can apparently more or less extend this to vector spaces by just allowing some unit vectors $\hat x, \hat y, \hat z$ and assuming that they and their products commute on both sides with distribution products $\langle,\rangle.$

The problem here is essentially a type error in the physics: in this scenario $\rho$ can only be interpreted as a distribution, which seems to imply that either $\operatorname{div}$ maps vector fields to distributions, or else $\vec E$ is a distribution and $\operatorname{div}$ maps distributions to distributions.

Either way, we need a slightly different definition for divergence from the physicist's pragmatic version. How can we prove the divergence theorem under a type-corrected definition for $\operatorname{div}$ that can properly return a distribution for the Coulomb field $\vec E_q$? Are there any subtleties that appear as a result of the new definition?

• This is a great question. One could argue that the electric field should really be a vector distribution, looking at the definition (force on a test charge, which basically is the action on a test functio ). The problem then is that the force on a point charge has the same type error you mentioned in this question. Although this is not sooo bad, since we know that electrodynamics with point charges in inconsistent anyway. – lalala Dec 2 at 10:39

In THIS ANSWER, I discuss regularizing the Electric Field of a point charge to assign meaning to the Dirac Delta for use in the Divergence Theorem. This provides a rigorous way forward where Dirac Delta is interpreted in terms of the limit of the regularized function $\vec \psi$ given by

$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 1$$

Taking the divergence of $(1)$ reveals that

$$\nabla \cdot \vec \psi(\vec r; a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$$

Now, in the Aforementioned Answer, I showed that for any sufficiently smooth test function $\phi$, we have that

$$\lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV= \begin{cases} 0&, \text{V does not include the origin}\\\\ 4\pi \phi(0)&,\text{V includes the origin} \end{cases}$$

and it is in this sense that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$