What vector field $\vec{F}$ defined on the unit sphere has $\vec{\nabla} \times \vec{F} = \vec{r}$?

Question

I want to apply Stokes' theorem in a problem whose geometry dictates that the "curl $\vec{F} \cdot \vec{n}$" integrand of the surface integral must equal one. Thus curl $\vec{F}$ must equal $\vec{n}$, and since my surface is the sphere, $\vec{n}$ is the radial vector on the surface of the sphere. the last step before I convert this surface integral to a path integral (using Stokes' theorem) is to define the vector field $\vec{F}$ whose curl is the radial vector $\vec{r}$. I'm rusty at div, grad, curl and all that, and probably overlooking something obvious. I know that the radial vector is the polar angle unit vector $\hat{\theta}$ cross the azimuthal unit vector $\hat{\phi}$, but I need the vector field $\vec{F}$ whose curl is $\vec{r}$, not the two vectors whose cross product is $\vec{r}$. I tried directly integration the expression for curl $\vec{F} = \vec{r}$ to guess what $\vec{F}$ must be and I got close but no cigar. Any help appreciated...

Answer 1

There is no vector field $\bf F$ that satisfies $$\nabla \times {\bf F} = {\bf \hat{r}} ,$$ where $\bf \hat r$ denotes the unit radial vector field on $\Bbb R^3 - \{ 0 \}$:

If there were, taking the divergence would give $$\nabla \cdot {\bf r} = \nabla \cdot (\nabla \times {\bf F}) = 0 .$$

On the other hand, the usual formula for divergence in spherical coordinates gives $$\nabla \cdot {\bf r} = \frac{2}{r} \neq 0 ,$$ where $r$ is the radial spherical coordinate.