The mass of magnetic monopoles

Question

If there would be magnetic monopoles somewhere, such a way to bring Maxwell equations in a symmetric way, there should be a law of force describing the movement of such particles. If we imagine something similar to Lorentz force \begin{equation} \vec F = q_m \vec B \end{equation} then, the force over a magnetic monopole would be an axial vector. Now, I don't know any example of axial force in physics (is there some?), but if we apply Newton's second law

\begin{equation} m_m \vec {d^2\vec r\over dt^2} = \vec F \end{equation}

We get an axial acceleration, which is not possible.

Is it wrong to assume a symmetric force law for magnetic monopoles?

I tried to think if magnetic monopole mass is a pseudoscalar, but then It would not be a mass in the sense we know.

I thought that maybe to think about it using Newtonian mechanics is the problem, but trying to think about it using energy concepts.

We know the energy of a dipole is given by

$$ U = -\vec m\cdot \vec B $$

where $\vec m$ is the magnetic dipole. Usually, both are axial vectors, so again no problem. But if there is magnetic monopoles, we could use two of then to build a magnetic dipole, an then

$$ \vec m = q_m \vec d $$

where $\vec d$ is the distance between the monopoles. Again, we found a situation where:

1. We get a polar magnetic dipole, which implies a pseudoescalar potential energy

2. We define the magnetic charge as being a pseudoescalar.

Where is the wrong assumption here? Is it true that we should assume pseudoscalars for magnetic monopole mass/charge, or the force and energy laws aforementioned should change?

Answer 1

The force is a vector because magnetic charge is a pseudo scalar and the magnetic field is a pseudovector. You can derive this from Maxwell's equations: $$ \begin{align} \nabla\cdot E &= \rho_e & \nabla\cdot B &= \rho_m \\ \nabla\times E &= -j_m-\partial_tB & \nabla\times B &= j_e+\partial_tE \end{align} $$ Energy and momentum balance are: $$ \partial_t\left(\frac12E^2+\frac12B^2\right)+\nabla\cdot\left(E\times B\right)+E\cdot j_e+B\cdot j_m = 0 \\ \partial_t\left(E\times B\right)+\nabla\cdot\left(\frac12E^2-E\otimes E+\frac12B^2-B\otimes B\right)+\rho_eE+j_e\times B+\rho_mB+E\times j_m = 0 $$ As usual, the absolute parity is conventional, but the relative parity is fixed. Fixing electric charge to be a scalar, $E,j_e$ are vectors while $\rho_m$ is a pseudo scalar and $B,j_m$ are pseudo vectors.

You can identify the Lorentz force, which is a vector (so mass is still a scalar in Newton's second law): $$ F_m = q_m(B+E\times v) $$ which is to contrast with the usual electric counterpart: $$ F_e = q_e(E+v\times B) $$

You should be careful when making analogies with magnetic current loops. There is a subtle difference between Gilbert dipoles (two opposing magnetic charges) and Ampère dipoles (current loop), even if they produce similar magnetic fields:

• Gilbert: $\rho_m = -m\cdot\nabla\delta$
• Ampère: $j_e = \nabla \delta\times m$

Yes, when subjected to an external field $B$ they both have the same energy: $$ U = -m\cdot B $$ but the forces are slightly different:

• Gilbert: $F = (m\cdot \nabla) B$
• Ampère: $F = \nabla(m\cdot B)$

For static fields, the difference is not noticeable since both forces agree in regions where there are no electric currents generating $B$. You can trace this back to the differing magnetic field of each dipole. There is an extra Dirac delta contribution for the Gilbert dipole, which you do not have for the Ampère dipole.