# Divergence of $\frac{ \hat {\bf r}}{r^2} \equiv \frac{{\bf r}}{r^3}$, what is the 'paradox'?

Now what is the paradox, exactly?

The paradox is that the vector field $$\vec{v}$$ considered obviously points away from the origin and hence seems to have a non-zero divergence, however, when you actually calculate the divergence, it turns out to be zero.

How are we supposed to believe that the source of $$\vec v$$ is concentrated at the origin mathematically?

Most important point to observe is that $$\nabla \cdot \vec v = 0$$ everywhere except at the origin. The diverging lines appearing are from the origin. Our calculations cannot account for that since $$\vec v$$ blows up at $$r = 0$$. Moreover, eq. (1.84) is not even valid for $$r = 0$$. In other words, $$\nabla \cdot \vec v \rightarrow \infty$$ at that point.

However, if you apply the divergence theorem, you will find $$\int \nabla \cdot \vec v \ \text{d}V = \oint \vec v \cdot \text{d}\vec a = 4 \pi$$ Irrespective of the radius of a sphere centred at the origin, we must obtain the surface integral as $$4 \pi$$. The only conclusion is that this must be contributed from the point $$r = 0$$.

This serves as the motivation to define the Dirac delta function: a function which vanishes everywhere except blowing up at a point and has a finite area under the curve.

You must use the Dirac $$\:\delta-$$function and its properties.

The point charge $$\:q\:$$ being at rest at $$\:\mathbf{r}_{0}\:$$ we have $$$$\mathbf{E}\left(\mathbf{r},t\right)\boldsymbol{=}\dfrac{q}{4\pi \epsilon_{0}}\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}} \tag{01}\label{01}$$$$ Now, $$$$\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\boldsymbol{=}\boldsymbol{-}\boldsymbol{\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}-\mathbf{r}_{0}\Vert}\right) \tag{02}\label{02}$$$$ and 1,2 $$$$\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=}\nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=}\boldsymbol{-}4\pi\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right) \tag{03}\label{03}$$$$ so $$$$\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{E}\left(\mathbf{r},t\right)\boldsymbol{=}\dfrac{q\,\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)}{\epsilon_{0}}\boldsymbol{=}\dfrac{\rho\left(\mathbf{r},t\right)}{\epsilon_{0}} \tag{04}\label{04}$$$$

$$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$$

$$\textbf{(1) Proof of the rhs equality of equation \eqref{03} :}$$

$$$$\boxed{\:\: \nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=}\boldsymbol{-}4\pi\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{p-01}\label{p-01}$$$$

Let a real function $$\;f(x)\;$$ of the real variable $$\;x\in\mathbb{R}\;$$ for which \begin{align} f(x)\boldsymbol{=}0 \quad & \text{for any} \quad x\boldsymbol{\ne} x_{0} \quad \textbf{and} \tag{p-02a}\label{p-02a}\\ \int\limits_{\boldsymbol{x_{0}-\varepsilon}}^{\boldsymbol{x_{0}+\varepsilon}}\!\!\!f(x)\mathrm dx\boldsymbol{=}1\quad & \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0 \tag{p-02b}\label{p-02b} \end{align} Under these conditions it seems that this function is not well-defined at $$\;x_{0}$$, may be because of a singularity at this point. But we have good reasons to $$^{\prime\prime}$$believe$$^{\prime\prime}$$ that
$$$$f(x)\boldsymbol{\equiv}\delta\left(x\boldsymbol{-}x_{0}\right) \tag{p-03}\label{p-03}$$$$ since equations \eqref{p-02a},\eqref{p-02b} remind us the defining properties of Dirac delta function on the real axis $$\;\mathbb{R}$$.

For the 3-dimensional case let a real function $$\;F(\mathbf{r})\;$$ of the vector variable $$\;\mathbf{r}\in\mathbb{R}^{\bf 3}\;$$ for which \begin{align} F(\mathbf{r})\boldsymbol{=}0 \quad & \text{for any} \quad \mathbf{r}\boldsymbol{\ne} \mathbf{r}_{0} \quad \textbf{and} \tag{p-04a}\label{p-04a}\\ \iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}F(\mathbf{r})\mathrm d^{\bf 3}\mathbf{r}\boldsymbol{=}1\quad & \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0 \tag{p-04b}\label{p-04b} \end{align} where $$\;\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$$ a ball with center at $$\;\mathbf{r}_{0}\;$$ and radius $$\;\boldsymbol{\varepsilon}$$.

Under these conditions it seems that this function is not well-defined at $$\;\mathbf{r}_{0}$$, may be because of a singularity at this point. But we have good reasons to $$^{\prime\prime}$$believe$$^{\prime\prime}$$ that
$$$$F(\mathbf{r})\boldsymbol{\equiv}\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right) \tag{p-05}\label{p-05}$$$$ since equations \eqref{p-04a},\eqref{p-04b} remind us the defining properties of Dirac delta function in the real space $$\;\mathbb{R}^{\bf 3}$$.

Now, let $$\;F(\mathbf{r})\;$$ be the real function of the lhs of equation \eqref{p-01} $$$$F(\mathbf{r})\stackrel{\textbf{def}}{\boldsymbol{\equiv\!\equiv}}\nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=\nabla\cdot\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\stackrel{\eqref{02}}{\boldsymbol{=\!=}}\boldsymbol{-}\boldsymbol{\nabla\cdot}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right) \tag{p-06}\label{p-06}$$$$ Based on the identity $$$$\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\psi\mathbf{a}\right)\boldsymbol{=}\mathbf{a}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi \boldsymbol{+}\psi\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{a} \tag{p-07}\label{p-07}$$$$ for the rhs of \eqref{p-06} we have for $$\;\mathbf{r}\boldsymbol{\ne} \mathbf{r}_{0}$$ $$$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\boldsymbol{\nabla\cdot}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\boldsymbol{=}\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)\boldsymbol{\cdot}\!\!\!\!\!\!\underbrace{\boldsymbol{\nabla}\left(\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)}_{\eqref{p-09}:\boldsymbol{=-}3\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)\boldsymbol{/}\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 5}}\!\!\!\!\!\!\boldsymbol{+}\left(\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\underbrace{\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)}_{\eqref{p-11}:\boldsymbol{=}3}\boldsymbol{=}0 \tag{p-08}\label{p-08}$$$$ Note first that $$$$\boldsymbol{\nabla}\left(\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\boldsymbol{=-}\dfrac{3}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 4}}\boldsymbol{\nabla}\left(\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert\vphantom{\tfrac12}\right)\stackrel{\eqref{p-10}}{\boldsymbol{=\!=\!=}}\boldsymbol{-}\dfrac{3\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 5}} \tag{p-09}\label{p-09}$$$$ since $$$$\boldsymbol{\nabla}\left(\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert\vphantom{\tfrac12}\right)\boldsymbol{=}\dfrac{\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert} \tag{p-10}\label{p-10}$$$$ and second $$$$\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)\boldsymbol{=}3 \tag{p-11}\label{p-11}$$$$ So $$$$\boxed{\:\: \nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=}0\,,\quad \text{for} \quad\mathbf{r}\boldsymbol{\ne} \mathbf{r}_{0}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{p-12}\label{p-12}$$$$ Now, let a ball $$\;\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$$ with center at $$\;\mathbf{r}_{0}\;$$ and radius $$\;\boldsymbol{\varepsilon}$$. For the volume integral of above function in this ball we have $$$$\iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\mathrm d^{\bf 3}\mathbf{r}\stackrel{\eqref{p-06}}{\boldsymbol{=\!=\!=}}\boldsymbol{-}\iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\boldsymbol{\nabla\cdot}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\mathrm d^{\bf 3}\mathbf{r} \tag{p-13}\label{p-13}$$$$ From Gauss's Theorem $$$$\iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\boldsymbol{\nabla\cdot}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\mathrm d^{\bf 3}\mathbf{r}\boldsymbol{=}\iint\limits_{\mathcal S\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\boldsymbol{\cdot}\mathrm d\mathbf{S} \tag{p-14}\label{p-14}$$$$ where $$\;\mathcal S\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$$ the closed spherical surface with center at $$\;\mathbf{r}_{0}\;$$ and radius $$\;\boldsymbol{\varepsilon}$$, the boundary of the ball $$\;\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)$$.

Now, the unit vector $$$$\mathbf{n}\boldsymbol{=}\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert} \tag{p-15}\label{p-15}$$$$ is normal outwards to the surface $$\;\mathcal S\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$$ so $$$$\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\boldsymbol{\cdot}\mathrm d\mathbf{S}\boldsymbol{=}\mathbf{n}\boldsymbol{\cdot}\mathrm d\mathbf{S}\boldsymbol{=}\mathrm {dS} \tag{p-16}\label{p-16}$$$$ where $$\;\mathrm {dS}\;$$ the infinitesimal area of the infinitesimal spherical patch. Given that $$\;\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert\boldsymbol{=}\boldsymbol{\varepsilon}\;$$ we have $$$$\iint\limits_{\mathcal S\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\:\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\boldsymbol{\cdot}\mathrm d\mathbf{S}\boldsymbol{=}\dfrac{1}{\boldsymbol{\varepsilon}^{\bf 2}}\iint\limits_{\mathcal S\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\mathrm {dS}\boldsymbol{=}\dfrac{1}{\boldsymbol{\varepsilon}^{\bf 2}}\cdot\left(4\pi\boldsymbol{\varepsilon}^{\bf 2}\right)\boldsymbol{=}4\pi \tag{p-17}\label{p-17}$$$$ So $$$$\boxed{\:\: \iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}\nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\mathrm d^{\bf 3}\mathbf{r}\boldsymbol{=}\boldsymbol{-}4\pi\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \quad \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0\:\:} \tag{p-18}\label{p-18}$$$$ The property \eqref{p-12} is identical to \eqref{p-04a} while the property \eqref{p-18} is identical to \eqref{p-04b} except the constant factor $$^{\prime\prime}\boldsymbol{-}4\pi^{\prime\prime}$$. These facts justify the expression via the Dirac delta function, equation \eqref{p-01}.

$$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$$

$$\textbf{(2) Reference :}$$ $$^{\prime\prime}Classical\:\:Electrodynamics^{\prime\prime}$$, J.D.Jackson, 3rd Edition 1999, $$\S$$ 1.7 Poisson and Laplace Equations

The singular nature of the Laplacian of $$\,1/r\,$$ can be exhibited formally in terms of a Dirac delta function. Since $$\,\nabla^{\bf 2}(1/r)\!\boldsymbol{=}\!0\,$$ for $$\,r\!\boldsymbol{\ne}\!0\,$$ and its volume integral is $$\,\boldsymbol{-}4\pi$$, we can write the formal equation, $$\,\nabla^{\bf 2}(1/r)\!\boldsymbol{=}\!\boldsymbol{-}4\pi\delta(\mathbf{x})$$ or, more generally, $$$$\nabla^{\bf 2}\left(\!\dfrac{1}{\vert\mathbf{x}\boldsymbol{-}\mathbf{x'}\vert}\right)\boldsymbol{=}\boldsymbol{-}4\pi\delta\left(\mathbf{x}\boldsymbol{-}\mathbf{x'}\right) \tag{1.31}\label{1.31}$$$$

I am not sure I can answer the question exactly how you meant it, but I can give you some things to think about.

Mathematically, the peculiarity of this situation is caused by the fact that the function is defined on $$\mathbb{R}^3- \{0\}$$, which is homeomorphic to a sphere, whose second (de Rham) cohomology group is $$\mathbb{R}$$. Hence you can have closed 2-forms that are not exact. The flux form associated to your vector field is precisely one of these forms.

Now, you are supposedly in a second year electromagnetism course, I guess? So you probably don’t know the meaning of what I just wrote. Let me put it this way. If you saw complex analysis already, all this is just kind of the residue theorem. If you integrate on a closed loop, you get zero if there’s nothing weird happening inside, or (possibly) non zero if the function diverges somewhere inside the loop, i.e. you have a pole. This is exactly the same thing, but in 3 dimensions, with closed surfaces instead of closed loops, and with flux integrals instead of complex integrals!