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I just started in Griffith's Introduction to electrodynamics and I stumbled upon the divergence of $\frac{ \hat r}{r^2}$ , now from the book, Griffiths says:

enter image description here

Now what is the paradox, exactly? Ignoring any physical intuition behind this (point charge at the origin) how are we supposed to believe that the source of $\vec v$ is concentrated at the origin mathematically? Or are we forced to believe that because there was a contradiction with the divergence theorem?

Also how would the situation differ if $\vec v$ was the same vector function but not for a point charge? Or is it impossible?

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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.

enter image description here

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 $\nabla.\vec v = 0$ 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, $\nabla.\vec v \rightarrow \infty$ at that point.

However, if you apply the divergence theorem, you will find $$\int \nabla.\vec v \ \text{d}V = \oint \vec v.\text{d}\vec a = 4 \pi$$ 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.

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

The point charge $\:q\:$ being at rest at $\:\mathbf{r}_{0}\:$ we have \begin{equation} \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} \end{equation} Now, \begin{equation} \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} \end{equation} and 1,2 \begin{equation} \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} \end{equation} so \begin{equation} \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} \end{equation}

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

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

\begin{equation} \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} \end{equation}

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
\begin{equation} f(x)\boldsymbol{\equiv}\delta\left(x\boldsymbol{-}x_{0}\right) \tag{p-03}\label{p-03} \end{equation} 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
\begin{equation} F(\mathbf{r})\boldsymbol{\equiv}\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right) \tag{p-05}\label{p-05} \end{equation} 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} \begin{equation} 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} \end{equation} Based on the identity \begin{equation} \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} \end{equation} for the rhs of \eqref{p-06} we have for $\;\mathbf{r}\boldsymbol{\ne} \mathbf{r}_{0}$ \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\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} \end{equation} Note first that \begin{equation} \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} \end{equation} since \begin{equation} \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} \end{equation} and second \begin{equation} \boldsymbol{\nabla}\boldsymbol{\cdot}\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\vphantom{\tfrac12}\right)\boldsymbol{=}3 \tag{p-11}\label{p-11} \end{equation} So \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} From Gauss's Theorem \begin{equation} \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} \end{equation} 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 \begin{equation} \mathbf{n}\boldsymbol{=}\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert} \tag{p-15}\label{p-15} \end{equation} is normal outwards to the surface $\;\mathcal S\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$ so \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} So \begin{equation} \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} \end{equation} 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, \begin{equation} \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} \end{equation}


Related : Divergence of Electric Field Due to a Point Charge.

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  • $\begingroup$ Spectacular and wonderful demonstration. Why don't I get something like that in my questions? :-( My compliments to you. $\endgroup$ – Sebastiano Jun 26 at 12:01
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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!

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The word "paradox" is not really justified, this is just caused by sloppy math, ignoring that our field isn't defined at the origin. The divergence truly is equal to zero everywhere the function is defined and thus the calculation is valid. Our calculation doesn't say anything about the origin. There, special care is needed. In terms of "regular functions", you can't say anything, discrete contributions have to be counted as special cases. But with generalization to distributions, you can say it's a delta function, as this gives you a way to describe discrete contributions, too.

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