# Divergence Free Vector Fields that are undefined at the origin

I am aware of vector fields which are undefined at the origin but whose divergence everywhere else is 0. In particular, my students have already seen the inverse square vector field, i.e. $\vec{F}=\frac{\vec{r}}{||\vec{r}||^3}$ where $\vec{r}=\langle x,y,z \rangle$ and the vector field for an ideal electric dipole $\vec{E}=\nabla \left(\frac{z}{||r||^3} \right)$.

Are there still other examples of divergence free vector fields that blow up at the origin? I want to hammer the concept of computing the flux of these vector fields across solids which enclose the origin by constructing a smaller solid inside (whose flux we can easily compute) and applying the divergence theorem to the solid sandwiched in between.

Thanks!

• Hmm, you could look into the magnetic vector potential $\vec{A}$. – WalyKu Apr 30 '14 at 17:40

Since divergence $\nabla$ is linear operator, the simplest way to construct another examples of such an field from the known example would be simply adding adding well-behaved divergence-free field to this one.
Another way would be directed derivative of this field. This works, because if $\nabla \cdot \vec F = 0$, then
$$\nabla \cdot [(\vec a \cdot \nabla) \vec F] = (\vec a \cdot \nabla) \nabla \cdot \vec F = 0$$
so, if $\vec F$ is divergence-free, then $(\vec a \cdot \nabla) \vec F$ is also divergence-free for some constant vector $\vec a$, but it still blows up at origin if $\vec F$ blows up at origin. Needless to say, one can repeat this arbitrarily many times, so
$$(\vec a_1 \cdot \nabla)(\vec a_2 \cdot \nabla)...(\vec a_n \cdot \nabla) \vec F$$