It is demonstrated that the square trace of the electromagnetic tensor is nothing and it is valid: TrF2μν=2c2(E2−c2B2).
=−2c2(B2−E2c2)=2c2(E2−c2B2)
I have seen, also, this explanation of Lorentz invariant E2−c2B2:
After, on the site Why is this invariant in Relativity: E2−c2B2? there are limited informations, mathematical and physical, for the following relationships:
E2−c2B2=0
E2−c2B2>0
E2−c2B2<0
For item 2.) E2−c2B2>0 in Σ. Then there will be a reference system of Σ′ such that ¯B′=0 i.e. the interaction is purely electric. Why?
For item 1.) E2−c2B2=0 in Σ is the case with a plane wave: why? We can also say that if we have a plane wave in an inertial reference Σ we will still find a plane wave in any other inertial reference Σ′.
For item 3.) E2−c2B2<0 in Σ. Both ¯E and ¯B are different from zero in each reference system (otherwise both must be null and therefore there would be no electromagnetic wave). An example is a wire with current? It is correct and why?
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