Question

Let r=xi^+yj^+zk^ and r=||r||. Show that:

∇(lnr)=r/r^2.

and

∇×((r^n)r)=0.

Answer 1

We have $\nabla (\ln r)=\nabla (\ln (\sqrt{x^2+y^2+z^2}))=\frac{1}{2}\nabla \ln(x^2+y^2+z^2)$. Now let us calculate the $x-$ coordinate of this gradient (as the other will have the symmetric expression) :

$\nabla(\ln r)_x = \frac{x}{x^2+y^2+z^2}=(\frac{\vec{r}}{|r|^2})_x$

and therefore we have $\nabla (\ln r)= \frac{\vec{r}}{r^2}$ .

Now let us calculate the $x-$coordinate of $\nabla \times (r^n \vec{r})$, the result for other coordinates will follow from the symmetry of the expression of $r(x,y,z)$.

$\nabla\times (r^n \vec{r})_x = \frac{\partial}{\partial y}(r^nz)-\frac{\partial}{\partial z}(r^ny)$, now expanding this expression gives

$\nabla\times (r^n\vec{r})_x = nr^{n-2} yz - nr^{n-2}zy=0$

and therefore $\nabla\times(r^n\vec{r})=0$