How to show that the two EM invariants are a complete set?

  • Level: Graduate 
  • Thread starter Thread starter bcrowell
  • Start date Start date
  • Tags Tags
    Complete Em Set
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
6 replies · 2K views
Staff Emeritus
Science Advisor
Insights Author
Messages
6,723
Reaction score
431
In electromagnetism we have these two Lorentz scalars:

##P=B^2-E^2##

##Q=E\cdot B##

WP https://en.wikipedia.org/wiki/Classification_of_electromagnetic_fields claims that these are a complete set of invariants, because "every other invariant can be expressed in terms of these two." How does one prove this? Would the idea be to show that any electromagnetic field tensor can be rendered into one of a set of canonical forms by boosts and rotations? Or maybe you could fiddle with the eigenvalues of the field tensor?

Is the claim only true for invariants that are continuous functions of the field tensor (i.e., continuous functions of its components)?
 
Physics news on Phys.org
I haven't worked this out fully, but the following might do the job. In the general case, where neither P nor Q vanishes, I think you can do a rotation and a boost such that E and B both lie along the x axis. Then clearly there are only two degrees of freedom, corresponding to the x components of the two fields, so there can only be two invariants.
 
bcrowell said:
I haven't worked this out fully, but the following might do the job. In the general case, where neither P nor Q vanishes, I think you can do a rotation and a boost such that E and B both lie along the x axis. Then clearly there are only two degrees of freedom, corresponding to the x components of the two fields, so there can only be two invariants.
I don't think this can be the case. If E.B is non-zero, and it is an invariant, then there is no frame where both E and B lie along the x-axis, because then E.B = 0.
 
phyzguy said:
I don't think this can be the case. If E.B is non-zero, and it is an invariant, then there is no frame where both E and B lie along the x-axis, because then E.B = 0.

Sorry, I don't follow you. If E and B are parallel and both nonzero, then their dot product is nonzero.
 
bcrowell said:
Sorry, I don't follow you. If E and B are parallel and both nonzero, then their dot product is nonzero.

You're right of course. Should teach me to post when I'm tired. Please ignore my comment.