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I have read the other posts such as divergence of $\,\hat{r}$ divided by $r^{2}$:

  • But my question is very basic - where does the extra $r^{2}$ come from in the below proof $?$.
  • I realize why $\frac{1}{r^{2}}$ is taken out of the derivative, but somehow a new $r^{2}$ suddenly appears within the parenthesis below to cancel out the denominator.
  • I have seen this proof in multiple textbooks, but they don't explain where the new $r^{2}$ (which I bolded below) comes from.
  • Can someone explain in simple terms $?$. After all, isn't $\hat{r}$ just the unit vector in (usually) three dimensions $?$.
  • Even if you rewrite $\hat{r}$ as $\frac{\vec{r}}{r}$ you don't suddenly get $r^{2}$.

$$ \nabla\cdot f = \frac{1}{r^2} \frac{\partial} {\partial r}\left(r^{2}f_{r}\right) = \frac{1}{r^{2}}\frac{\partial}{r} \left({\bf r^{2}}\, \frac{1}{r^{2}}\right) = 0 $$

I think it may be helpful if I just show the portion of the textbook. Here is the excerpt from Griffiths, Introduction to Electrodynamics (4th ed.): Snapshot of Textbook

So the function (f) in the above equation is simply $\frac{\hat r}{r^2}$ That's it. The extra ${r^2}$ that I bolded above just 'appears' and I don't understand why.

Rasputin
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  • You will get both $r^2$ terms (one inside, one outside of $\partial_r$) when you apply the standard method of coordinate transformation from Cartesian to spherical. In the link I showed Cartesian to cylindrical. – Kurt G. Apr 27 '24 at 20:49
  • But why convert? In other words, in the book I am reading (Introduction to Electrodynamics), the initial function is $\frac{\hat r}{r^2}$. And then the author "proves" that the Divergence of this function is 0 at the origin using the above manipulations, but at no point says anything about converting anything to anything else. I don't see what it is about $\hat r$ that automatically assumes anything other than Cartesian coordinates. – Rasputin Apr 27 '24 at 22:59
  • Unless you misread that book and the author proves that the divergence of the Coulomb-Newton potential is a Dirac delta at the origin I'd be interested to know who its author is. $r^2$ is a spherical coordinate and $\hat r$ is a spherical coordinate unit basis vector. Divergence has to do with Gauss' theorem and we want that to work when we change variables. That's why we convert. Thinking the divergence of an $f(r)$ should be $\partial_r f(r)$ just because that looks less complicated has little to do with maths. – Kurt G. Apr 28 '24 at 04:19

2 Answers2

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If you have a radial function times $x$, i.e. $g(x) = xf(|x|)$, then

$$ \nabla \cdot g(x) = \sum_{i=1}^{n} \frac{\partial g}{\partial x_i}(x) = \sum_{i=1}^{n}f(|x|)+ x_i f'(|x|)\frac{\partial (|x|)}{\partial x_i} = nf(|x|)+ f'(|x|)\sum_{i=1}^{n}\frac{x_{i}^2}{|x|} = nf(|x|) + |x|f'(|x|) $$

Writing $|x| = r$

\begin{equation}\tag{1} \nabla \cdot g = \frac{1}{r^{n-1}}(nr^{n-1}f + r^nf') = \frac{1}{r^{n-1}}\partial_r (r^n f) \end{equation}

In your case $n=3$ and $f(r) = r^{-3}$ (because $\widehat{r} = r/|r|$).

If we want the formula to be like you mentioned, we should write $$ g(x) = \frac{xf(|x|)}{|x|} =x \frac{f(|x|)}{|x|} $$ applying this in $(1)$ we have

\begin{align*} \nabla \cdot g (r) = \frac{1}{r^{n-1}}\partial_r \left(r^n \frac{f(r)}{r}\right) = \frac{1}{r^{n-1}}\partial_r \left(r^{n-1}f\right) \end{align*}

Raul Fernandes Horta
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  • @TedShifrin I edited it – Raul Fernandes Horta Apr 27 '24 at 21:26
  • I am reading a book about Electrodynamics and the answer the author gives is that the Divergence is 0 at the origin and the equations in my question are the proof of that proposition. The reason it is zero is because the bolded $r^2$ cancels out the other $r^2$ in the denominator, and the Divergence of 1/1 is 0. But your equations above doesn't result in 0. I'm sure I am missing something obvious but that is what I am trying to figure out. Why does this random $r^2$ show up? The author must have believed it was so obvious it needed no explanation. But I have no clue what it is. – Rasputin Apr 27 '24 at 23:05
  • @Rasputin like I mentioned, in your case $f(r) = r^{-3}$ so inside the derivative becomes $r^{3} r^{-3} = 1$, so it becomes $0$ when you differentiate – Raul Fernandes Horta Apr 28 '24 at 00:17
  • @RaultFernandesHorta There is no $r^3$ in my question or the textbook excerpt I included above. I agree that $\frac{1}{r^2}$ multiplied by $r^2$ and $\frac{1}{r^3}$ multiplied by $r^3$ are both '1' and the derivative '1' is '0', but that is not my question. My question is how $\hat{r}$ somehow becomes $\frac{1}{r^2}$ multiplied by $r^2$. – Rasputin May 01 '24 at 01:09
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Lest see some $\ds{\underline{use\!f\!ul\ identities}}$: $$ \begin{array}{rrcl} \ds{\LARGE\bullet} & \ds{\nabla^{2}} & \ds{\equiv} & \ds{\nabla\cdot\nabla} \\[1mm] \ds{\LARGE\bullet} & \ds{r} & \ds{\equiv} & \ds{\verts{\vec{r}}} \\[1mm] \ds{\LARGE\bullet} & \ds{\hat{r}} & \ds{\equiv} & \ds{\vec{r} \over r} \\[1mm] \ds{\LARGE\bullet} & \ds{\nabla\on{f}\pars{r}} & \ds{=} & \ds{\on{f}'\pars{r}\,\hat{r}} \\[1mm] \ds{\LARGE\bullet} & \ds{\nabla\cdot\bracks{\on{f}\pars{\vec{r}} \vec{\bf A}\pars{\vec{r}}}} & \ds{=} & \ds{\bracks{\nabla\on{f}\pars{\vec{r}}}\cdot\vec{\bf A}\pars{\vec{r}} + \on{f}\pars{\vec{r}}\nabla\cdot\vec{\bf A}\pars{\vec{r}}} \end{array} $$ \begin{align} & ----------------------------- \\ & \mbox{Therefore,} \\[2mm] & \color{#44f}{\left.\nabla^{2}\pars{1 \over r}\right\vert_{r\ \not=\ 0}} \sr{\rm by\ def.}{=} \nabla\cdot\nabla\pars{1 \over r} = \nabla\cdot\bracks{\totald{\pars{1/r}}{r}\,\hat{r}} \\[5mm] = & \ \nabla\cdot\pars{-\,{1 \over r^{2}}\,{\vec{r} \over r}} = -\nabla\cdot\pars{{1 \over r^{3}}\,\vec{r}} = -\bracks{\vec{r}\cdot\nabla\pars{1 \over r^{3}} + {\nabla\cdot\vec{r} \over r^{3}}} \\[5mm] = & \ -\bracks{\pars{-\,{3\hat{r} \over r^{4}}\cdot\vec{r}} + {3 \over r^{3}}} = -\bracks{-\,{3 \over r^{3}} + {3 \over r^{3}}} = \bbx{\color{#44f}{\LARGE 0}} \end{align} In addition, \begin{align} \color{#44f}{\int_{V}\nabla^{2}\pars{1 \over r}\dd^{3}\vec{r}}\ & =\ \overbrace{\int_{V}\nabla\cdot\nabla\pars{1 \over r}\dd^{3}\vec{r} = \int_{S}\nabla\pars{1 \over r}\cdot\dd\vec{S}} ^{\ds{Gauss's\ Divergence\ Theorem}} \\[5mm] & = -\int_{S}\overbrace{{\hat{r}\cdot\dd\vec{S} \over r^{2}}}^{\ds{\dd\Omega_{\vec{r}}}} = \bbx{\color{#44f}{-\,4\pi}} \\ & \end{align}

Felix Marin
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