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I Why magnetic charge is pseudoscalar

  1. Jun 6, 2016 #1
    "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)
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  3. Jun 6, 2016 #2

    Simon Bridge

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    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.
  4. Jun 7, 2016 #3


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    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.
  5. Jun 7, 2016 #4
    .. 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)?
  6. Jun 7, 2016 #5


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    Well, in classical physics parity is not a well-defined observable. Here it's just meant that the Hamiltonian of the theory is invariant under space reflections (more precisely it's sufficient that the variation of the action is invariant under space reflections). In quantum theory you can have states of definite parity in models that are space-reflection symmetric. Then the Hamiltonian commutes with the parity operator (space-reflection operator) and thus a parity eigenstate stays a parity eigenstate under time evolution (in the Schrödinger picture).
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