Proving Vector Identities

Question

Let r=$(x,y,z)$ and $r=$||r||.

(A) Prove that $\nabla^2r^3=12r$.

(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?

What I have tried:

For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.

Answer 1

$$ \nabla^2r^3=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(x^2+y^2+z^2)^{3/2}\\ = \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\ =\frac{3(2x^2+y^2+z^2)}{(x^2+y^2+z^2)^{1/2}}+\frac{3(x^2+2y^2+z^2)}{(x^2+y^2+z^2)^{1/2}} +\frac{3( x^2+y^2+2z^2)}{(x^2+y^2+z^2)^{1/2}}\\=3\frac{ (4 x^2+4y^2+4z^2)} {(x^2+y^2+z^2)^{1/2}} =12r $$ $$ \nabla\cdot\big[\frac{(x, y, z)}{(x^2 + y^2 + z^2)^{p/2}}\big] =\frac{3-p}{(x^2+y^2+z^2)^{p/2}}=\frac{3-p}{r^{p }} $$ So $\nabla\cdot\frac{{\bf r}}{r^p}=0$ when $p=3$.

Answer 2

A more concise approach to (A) uses$$\nabla^2r^3=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}r^3\right)=\frac{1}{r^2}\frac{d}{dr}\left(3r^4\right)=12r.$$Similarly, for (B) use$$\nabla(r^{-p})=\frac{d}{dr}(r^{-p})\hat{\mathrm{r}}=-pr^{-p-2}\mathrm{r}$$to deduce$$\nabla\cdot(r^{-p}\mathrm{r})=r^{-p}\nabla\cdot\mathrm{r}+\nabla(r^{-p})\cdot\mathrm{r}=(3-p)r^{-p},$$i.e. the field is solenoidal iff $p=3$ (which would surprise no physicists familiar with the differential form of Gauss's law).

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\large\pars{A}:}$ \begin{align} \nabla^{2}r^{3} & = \nabla\cdot\pars{\nabla r^{3}} = \nabla\cdot\pars{\totald{r^{3}}{r}\,{\vec{r} \over r}} = 3\nabla\cdot\pars{r\,\vec{r}} \\[5mm] & = 3\bracks{\pars{\nabla r}\cdot\vec{r} + r\nabla\cdot\vec{r}} \\[5mm] & = 3\bracks{\pars{\totald{r}{r}\,{\vec{r} \over r}} \cdot\vec{r} + r\pars{3}} \\[5mm] & = 3r + 9r = \bbx{12r} \\ & \end{align}

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$\ds{\large\pars{B}:}$ \begin{align} \nabla\cdot\pars{\vec{r} \over r^{p}} & = \pars{\nabla r^{-p}}\cdot\vec{r} + r^{-p}\,\,\nabla \cdot\vec{r} \\[5mm] & = \pars{\totald{r^{-p}}{r}\,\,{\vec{r} \over r}}\cdot\vec{r} + r^{-p}\pars{3} \\[5mm] & = \underbrace{\pars{-pr^{-p - 1}\,\,\,{\vec{r} \over r}}\cdot\vec{r}}_{\ds{-pr^{-p}}}\ +\ 3r^{-p} \\[5mm] & = \pars{-p + 3}r^{-p} \end{align} Then, $$ \nabla\cdot\pars{\vec{r} \over r^{p}} = 0 \implies \bbx{p = 3} \\ $$