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.


  • $\begingroup$ Hmm, you could look into the magnetic vector potential $\vec{A}$. $\endgroup$ – 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

\begin{equation} \nabla \cdot [(\vec a \cdot \nabla) \vec F] = (\vec a \cdot \nabla) \nabla \cdot \vec F = 0 \end{equation}

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

\begin{equation} (\vec a_1 \cdot \nabla)(\vec a_2 \cdot \nabla)...(\vec a_n \cdot \nabla) \vec F \end{equation}

also satisfies the desired property. The same is true for almost all linear combinations (for some linear combinations it might not blow up at the origin, it will only be divergence-free).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.