View Full Version : maxwell's eqn in invariant form
mathfeel
Dec17-03, 03:43 AM
Maxwell's eqn, in invariant form reads:
F^{\mu \nu}{}_{;\nu} = J^{\mu}
and
F_{\alpha \beta ;\gamma} + F_{\beta \gamma ;\alpha}+F_{\gamma \alpha; \beta} = 0
Can someone give Maxwell's eqn if there is magnetic charge and current? I do not believe the form (matrix element) of F change, however, if it does, please state that as well.
Originally posted by mathfeel
Maxwell's eqn, in invariant form reads:
F^{\mu \nu}{}_{;\nu} = J^{\mu}
and
F_{\alpha \beta ;\gamma} + F_{\beta \gamma ;\alpha}+F_{\gamma \alpha; \beta} = 0
Can someone give Maxwell's eqn if there is magnetic charge and current? I do not believe the form (matrix element) of F change, however, if it does, please state that as well.
The second can be written in terms of the electromagnetic duel tensor D^{\mu \nu} as
D^{\mu \nu}{}_{;\nu} = 0
Instead of setting that equal to zero try setting it proportional to your hypothetical magnetic four current M^\mu like:
D^{\mu \nu}{}_{;\nu} = kM^{\mu}
(Normally I would explicitely put in the constants determied by your system of units for both sets of equations)
I haven't checked into this, but off the top of my head I think this would work. Of course your next job will be to go out and find a magnetic monopole in order to justify having done this.
mathfeel
Dec17-03, 11:46 PM
How do you define the dual tensor?
Originally posted by mathfeel
How do you define the dual tensor?
The electromagnetic duel tensor D_{\mu\nu} is related to the electromagnetic tensor F^{\mu\nu} and the rank 4 Levi-Civita tensor \epsilon_{\alpha\beta\mu\nu} by
D_{\mu\nu} = \frac{1}{2}F^{\alpha\beta}\epsilon_{\alpha\beta\mu \nu}.
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