Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / 1 / ii

Use the right-handed orthonormal basis $({\mathbf{e}}_r,{\mathbf{e}}_{\theta },{\mathbf{e}}_{\varphi })$. The cross product with ${\mathbf{e}}_r$ rotates tangential components by a right angle; taking its cross product again in the opposite order projects a gradient onto the tangent plane. The radial curl of angular gradient fields then depends only on the rotated-gradient part. All coordinate calculations are made away from the polar coordinate singularities and extend wherever the original fields are smooth.

Since $\hat{\mathbf{r}}={\mathbf{e}}_r$, the gradient formula givesThe vector triple product identity gives $({\mathbf{e}}_r\times \nabla G)\times {\mathbf{e}}_r=\nabla G-{\mathbf{e}}_r{G}_r=(G_{\theta }/r){\mathbf{e}}_{\theta }+[G_{\varphi }/(r\sin \theta )]{\mathbf{e}}_{\varphi }$. Adding the radial part yields

The radial component of the curl in spherical coordinates isIt contains no $f_r$, so $R$ cannot contribute. Substituting the tangential components givesThe mixed partial derivatives of smooth $G$ agree, so its terms cancel. Multiplication by $r$, to take the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$, does not change this independence from $R,G$.

The remaining terms giveThis is the Laplace-Beltrami operator on the unit sphere. Taking the dot product with the unit radial vector instead would give $r^{-2}{\Delta }_{\theta ,\varphi }F$; the extra factor $r$ here comes from the position vector in the question.