Divergence of inverse square vector field

Question

After trying to get the divergence of a vector field I got: $${\nabla \cdot \Bigg(\frac{\vec r}{|r|^3}\Bigg)=\frac{\partial}{\partial r}.\frac{1}{r^2}=-\frac{2}{r^3}}$$ But in wikipedia found that this must be $${\nabla \cdot \Bigg(\frac{\vec r}{|r|^3}\Bigg)=4\pi \delta(r)}$$ What is my mistake and where does my intuition get wrong?

Answer 1

First, the divergence in spherical coordinates, expressed in terms of derivatives, would take the form

$$\nabla \cdot \vec A = \frac{1}{r^2} \frac{\partial}{\partial r} [r^2 A^r] + \ldots$$

where $A^r$ is the radial component of the vector field $\vec A$. In this case, that's $1/r^2$, so we naively get 0 for this contribution.

Of course, this formula cannot be valid at the origin (there's a $1/r^2$ in it), so we resort to an alternative definition of divergence, one that is regular at the origin:

$$\nabla \cdot \vec A |_{\vec 0} = \lim_{\delta r \to 0} \frac{1}{4 \pi (\delta r)^3/3} \int_0^{2\pi} \int_0^\pi \vec A \cdot \hat r \, (\delta r)^2 \, \sin \theta \, d\theta \, d\phi$$

This basically follows from the divergence theorem and is a general alternative definition for divergence that eschews derivatives. It also can be done with curl and gradient instead. The limit process, of course, makes this entirely equivalent to a derivative.

Evaluating this definition for $\vec A = \hat r/r^2$ at $r = \delta r$ (as we're integrating over a sphere of this radius, and then taking the limit) gives

$$\left. \nabla \cdot \left( \frac{\hat r}{r^2} \right) \right|_{\vec 0} = \lim_{\delta r \to 0} \frac{3}{\delta r} $$

which obviously diverges (positively, as negative radii have no meaning).

So this is a vector field whose divergence is zero everywhere except the origin, where its divergence...well, diverges. That all certainly sounds like a delta function.

Typically, one uses the divergence theorem directly to verify the stated condition of the delta function: that its integral over any region containing zero is 1. That is, we do

$$4\pi \int_0^r \nabla \cdot \frac{\hat r}{r^2} r^2 \, dr= \int_0^{2\pi} \int_0^\pi \frac{\hat r}{r^2} \cdot \hat r r^2 \sin \theta \, d\theta \, d\phi$$

If $\nabla \cdot \hat r/r^2 = 4 \pi \delta$, then the surface integral on the left should evaluate to $4\pi$ (which it does, obviously). This is not 100% formally sound, but you can always check this by integrating against a test function (a scalar field would do), just as you would in 1d.

Answer 2

A common way to show that $\nabla \cdot \left(\frac{\hat r}{r^2}\right)=4\pi \delta (\vec r)$ is to regularize the function $\left(\frac{\hat r}{r^2}\right)$ in terms of a parameter, say $a$. To that end, let $\vec \psi$ be the regularized function 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, any sufficiently smooth test function $\phi$, we have that

$$\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV&=\lim_{a \to 0}\int_V \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV\\\\ &=0 \end{align}$$

if $V$ does not include the origin.

Now, suppose that $V$ does include the origin. Then, we have

$$\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV&=\lim_{a\to 0}\int_{V-V_{\delta}} \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV+\lim_{a\to 0}\int_{V_{\delta}} \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV\\\\ &=\lim_{a\to 0}\int_{V_{\delta}} \frac{3a^2}{(r^2+a^2)^{5/2}}\phi(\vec r)dV \end{align}$$

where $V_{\delta}$ is a spherical region centered at $\vec r=0$ with radius $\delta$. For any $\epsilon>0$, take $\delta>0$ such that $|\phi(\vec r)-\phi(0)|\le \epsilon/(4\pi)$ whenever $0<|\vec r|< \delta$. Then, we have

$$\begin{align} \lim_{a \to 0}\left|\int_V \nabla \cdot \vec \psi(\vec r; a)(\phi(\vec r)-\phi(0))\,dV\right|&\le \lim_{a\to 0} \int_{V_{\delta}} \left|\phi(\vec r)-\phi(0)\right|\frac{3a^2}{(r^2+a^2)^{5/2}}dV\\\\ &\le \left(\frac{\epsilon}{4\pi}\,4\pi\right)\,\lim_{a \to 0}\int_{0}^{\infty}\frac{3a^2}{(r^2+a^2)^{5/2}}r^2\,dr\\\\ &\le \epsilon \end{align}$$

Thus, we have for any test function $\phi$,

$$\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dV&=4\pi \phi(0) \end{align}$$

and it is in this sense that

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