If magnetic monopoles exist, they are predicted to have large charges--or equivalently, a large coupling constant, which means that perturbative models don't converge.

While I get that that makes it very difficult to calculate things like the binding energy of monopolium, is it at least possible to make qualitative statements about what magnetostatics, the strongly-coupled analog of electrostatics, looks like? E.g., can you still reasonably approximate the strength of a monopole field as $\frac{1}{r^2}$ down to arbitrarily small lengths?

  • $\begingroup$ Yes; what of it? $\endgroup$ Sep 18, 2021 at 19:55

1 Answer 1


The magnetic monopole acts exactly like an electric monopole. In magnetostatics, the equation of interest


$$\nabla \cdot \vec{B} = \rho$$

where $\rho$ is the magnetic charge density, has a solution:

$$\vec{B} = \frac{1}{4\pi}\frac{q}{r^2}$$

for point particles, i.e: $\rho = \delta(r)$, and $q$ is the magnetic charge. In fact, this is exactly what we want. We designed the magnetic monopole to act like a electric monopole because we saw it would be nice to have symmetry between them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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