Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 16

In Cartesian coordinates, the divergence is the sum of the three component derivatives. Therefore

For $r=\sqrt{x^2+y^2+z^2}>0$, the chain rule gives ${\partial }_x(r^{-1})=-x/r^3$, with analogous formulas for the other components. HenceOn the right the numerator denotes the position vector, while its denominator is the cube of its magnitude.

The vector $m$ is constant, so $\nabla (m\cdot r)=m$. Also $\nabla (r^{-3})=-3r/r^5$, by the same chain rule. The product rule therefore yieldsThe three components follow by differentiating $(\alpha x+\beta y+\gamma z)(x^2+y^2+z^2{)}^{-3/2}$.

For constant $m$, the curl of a cross product identity reduces toCommuting mixed second partial derivatives gives $(m\cdot \nabla )\nabla \varphi =\nabla (m\cdot \nabla \varphi )$. Hence the requested sum equals $m\Delta \varphi$, which is zero because $\varphi$ is harmonic. The argument assumes the indicated second derivatives are continuous on the domain.

From the preceding gradient, ${\partial }_x^2(r^{-1})=-r^{-3}+3x^2{r}^{-5}$. Summing the three Cartesian second partial derivatives givesThus the reciprocal radius is harmonic away from the origin. Its negative gradient isThe curl vanishes because $E$ is a gradient field and mixed second derivatives commute. The exclusion of the origin is essential: the potential and field are singular there.