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This question already has an answer here:

I am trying to formally learn electrodynamics on my own (I only took an introductory course). I have come across the differential form of Gauss's Law.

$$ \nabla \cdot \mathbf E = \frac {\rho}{\epsilon_0}.$$

That's fine and all, but I run into what I believe to be a conceptual misunderstanding when evaluating this for a point charge.

I know the math looks better in spherical coordinates, but I will be using Cartesian.

So when I calculate the divergence I obtain:

$$ \nabla \cdot \mathbf E = \nabla \cdot kQ\langle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\rangle = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

This can further be simplified:

$$\frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3-3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

Now instinctively I would say that 3-3 is zero and then the while thing is zero everywhere. I am confused as to why (purely mathematically) this expression is not equal to zero at the origin. I completely understand why it physically has to be that way. And I also understand that it is modeled with the delta dirac function. But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?

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marked as duplicate by Qmechanic 3 hours ago

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  • $\begingroup$ Hint: You are applying the rules of differentiation where the field is not defined/singular/not differentiable. $\endgroup$ – Qmechanic Jan 9 '18 at 10:21
  • $\begingroup$ Yes! Very true. Thank you. I should have thought of that. $\endgroup$ – Ben Jan 9 '18 at 11:24
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    $\begingroup$ It is not customary to use angle brackets to denote components of a vector. Better use () or []. $\endgroup$ – Ján Lalinský Jan 10 at 14:14
  • $\begingroup$ I picked up the angle bracket habit from mathematicians. $\endgroup$ – Ben Feb 19 at 23:11
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What you want to compute is essentially $$\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3}$$ at the origin. Of course, that doesn't exist as a function since the field is singular. On the other hand, you have already shown that it vanishes everywhere else.

So you need to interpret the expression in a weak sense, i.e. as a distribution, and consider the integral $$\int_{B_\epsilon}\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3} \,\text{d}^3x$$ over some volume containing the origin, conveniently chosen as a ball of radius $\epsilon$, convert it to a surface integral which does not include the singularity and see that the result is finite.

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  • $\begingroup$ You used a phrase here that I would like to ask you about. You said "that doesn't exist as a function since the field is singular." I have not studied (much to my dismay) abstract algebra. When mathematicians use the phrase vector FIELD, are they actually talking about a type of field in the AA sense? $\endgroup$ – Ben Jan 11 '18 at 10:15
  • $\begingroup$ @Ben: No, I'm using it in the simple sense of "function defined at every point in space": The electric field assigns a vector to every point $x$, i.e. it is a function of a variable $x\in \mathbb{R}^3$, and in this case, the function is sigular and not differentiable at $x=0$. $\endgroup$ – Toffomat Jan 11 '18 at 12:04
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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}

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  • $\begingroup$ Dear Frobenius. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) $\endgroup$ – Qmechanic 2 hours ago
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But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?

Divergence includes partial derivatives with respect to x, y & z of the respective components of the electric field. Now, we look at the definition of partial derivatives which is often called as the first principle: $$\frac{\partial f(x,y,z)}{\partial x}_{x,y,z = 0} = \lim_{ x\to0} \frac{f(x,0,0)-f(0,0,0)}{x-0}$$ In our case, if we take the partial derivative of, say x component of the electric field, with respect to x then we would use the definition, but $f(0)$ or the x component of electric field at the origin is undefined. It is called as a singularity. And still, we want to respect gauss's theorem of divergence which is kind of very elegant & intuitive. So, to find a way out of it, we physicists (actually, Paul Dirac) came up with Delta functions (to which mathematicians are unhappy).

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