- #1

- 254

- 27

You should upgrade or use an alternative browser.

- I
- Thread starter tzimie
- Start date

- #1

- 254

- 27

- #2

Simon Bridge

Science Advisor

Homework Helper

- 17,857

- 1,655

The standard model has zero magnetic charges - but any model may be extended in any way you like.

That's not the hard part.

- #3

- 17,482

- 8,480

$$\vec{F}=q_e \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right) + q_m \left (\vec{B}-\frac{\vec{v}}{c} \times \vec{E} \right),$$

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.

- #4

- 254

- 27

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)?

..

- #5

- 17,482

- 8,480

Share: