# Proving Vector Identities

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.

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

• for the first line should the last partial derivative be partial z? – butter chi Oct 28 at 15:37
• Yes, of course it is. – user247327 Oct 28 at 15:41
• @user247327 in the second line, how do you get $3x, 3y$ and $3z$ ? – butter chi Oct 28 at 16:08
• In line 2, should it not read: $$= \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\$$ i.e. The power of a half is inside the bracket. – user827887 Oct 28 at 22:17

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).

• which of your r's are vectors for your answer to B? – butter chi Oct 29 at 14:16
• @butterchi The \mathrm (bold, non-italic) ones are. – J.G. Oct 29 at 14:22
• Is the r meant to have a hat? – butter chi Oct 29 at 14:26
• @butterchi Sometimes, but sometimes not. $\mathrm{r}=r\hat{\mathrm{r}}$. – J.G. Oct 29 at 15:11


$$\ds{\large\pars{B}:}$$ \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, $$\nabla\cdot\pars{\vec{r} \over r^{p}} = 0 \implies \bbx{p = 3} \\$$