# Why magnetic charge is pseudoscalar

• I
"Classic" EM is symmetric against electric and magnetic fields/charges (except magnetic charges are not observed in reality, but this is another story). Still, magnetic field is pseudovector and magnetic charge is pseudoscalar. I suspect where it comes from in classic EM, but I wonder about QFT (I am layman) - is, on the fundamental level, magnetic charge less "fundamental"? Can some extension of the Standard Model (theoretically) have a simple particle with magnetic charge (not a huge and heavy, almost macroscopic, topological defect)

Simon Bridge
Homework Helper
Magnetic charge is a pseudoscalar because it changes sign under parity inversions.
The standard model has zero magnetic charges - but any model may be extended in any way you like.
That's not the hard part.

vanhees71
Gold Member
One assumes that even with magnetic charges electromagnetism should be space-reflection invariant, i.e., parity should be conserved. To see, how the electromagnetic field components behave under parity the most simple argument is to look at the force on a particle with electric and magnetic charge,
$$\vec{F}=q_e \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right) + q_m \left (\vec{B}-\frac{\vec{v}}{c} \times \vec{E} \right),$$
which follows from relativistic covariance by relating the electric charge to the Faraday tensor and the magnetic charge to its dual. Now we assign electric charge to be a scalar. Then from
$$\vec{\nabla} \cdot \vec{E}=\rho_e$$
it's clear that ##\vec{E}## must be a polar vector, which is compatible with the electric part of the Lorentz force law. Since further the velocity of the particle, ##\vec{v}## is a polar vector, ##c## a scalar, ##\vec{B}## must be an axial vector. In order to make also the magnetic part of the force compatible with parity we must assume that ##q_m## (magnetic charge) is pseudo-scalar.

One assumes that even with magnetic charges electromagnetism should be space-reflection invariant, i.e., parity should be conserved...

.. should be conserved for the observables - velocity, trajectory, position. But not for all variables of the theory, correct?
Does it mean that "B" is "just math" (like virtual particles)?
..

vanhees71