Supposedly, "Any divergence-free vector field can be expressed as the curl of some other divergence-free vector field" over a simply-connected domain.

So, what is one such vector potential which works for half of the Coulomb field?

To be clear, I want a vector potential whose curl equals the vector field $\mathbf R/|\mathbf R|^3$ for $z>0$ (for any $x$ and any $y$). $\mathbf R$ is the position vector $(x,y,z)$.

I know the scalar potential method is usually used instead of this, but am curious about how ugly a vector potential would look. If this gets answered, it should then be easy to answer this.

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    $\begingroup$ This is mathematically the same as the question of how magnetic monopoles are constructed. See, for example, this question and references therein. $\endgroup$ Jan 31, 2017 at 23:47
  • $\begingroup$ I think I'm asking for something different since I only want the monopole effects over half of space...I could not find the math which answers my question in the references, so please point me directly if I missed it. $\endgroup$
    – bobuhito
    Feb 1, 2017 at 0:09
  • $\begingroup$ The initial title is somewhat misleading - it leads one to expect a standard vector potential. I've edited for clarity. $\endgroup$ Feb 2, 2017 at 1:28

1 Answer 1


In cylindrical coordinates, this potential (pointed angularly about the z-axis) works: $$\vec{C} = \frac{1-\frac{z}{\sqrt{r^2+z^2}}}{r} \hat{\phi} $$

By choice, the curl also contains an impulse along the singular line (x=0, y=0, z<0) to zero the divergence. So, a solenoid sleeve from infinity can thereby generate the Coulomb field. In hindsight, I guess this is obvious, but it's interesting to do away with electric monopoles and consider an electron to be tied to infinity (or maybe a nearby positron) this way.

  • $\begingroup$ Your expression is not clear, could you please edit it using MathJax for clarity (the SE standard for maths). Thanks. $\endgroup$ Feb 2, 2017 at 18:43

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