6. Some further Applications of the Matrices F and M
6.1 The Determinants of F and M
Do calculate the determinants of F and M (or let the TI-89 calculate them), and you will see:
det ( F ) = –c2 ·(E·B)2 = det ( M ) and det ( F' ) = –c2 ·(E'·B')2 = det ( M' )
Because of F' = L · F · L-1 we know further det ( F' ) = det ( F ) , and equally we have det ( M' ) = det ( M )
The 4 determinants have the same value, and hence we have (E·B)2 = (E'·B')2
If E and B are orthogonal in any special coordinate frame, then this will hold true in any other frame!
By the way, E·B here denotes the 3d-scalar product of the field vectors, and not the product of two matrices.
6.2 The Product of the Matrices F and M
As we know, not only (E·B)2 is a relativistic invariant, but E·B itself (compare [1-225]).
We get this result directly, if we compute the product of F and M :
F · M = M · F = c · (E·B) · Id , where Id stands for the 4x4-identity-matrix
Looking at the trace of F · M (i.e. the sum of the diagonal elements of F · M ) we find
4· c · (E·B) = Trace ( F · M ) = Trace ( L·F·L-1· L·M·L-1 ) = Trace ( F' · M' ) = 4· c · (E'·B')
and we have proved E·B = E'·B'
6.3 Another Invariant from the Difference F – M
If you calculate det ( F – M ) and det ( F + M ) you will get
det ( F – M ) = det ( F + M ) = – ( E2 – c2 · B2 )2 = det ( M – F ) = det ( M + F ) and hence
det ( F' – M' ) = det ( F' + M' ) = – ( E'2 – c2 · B'2 )2 = det ( M' – F' ) = det ( M' + F')
But we know from matrix algebra, that
det ( F – M ) = det ( L · ( F – M ) · L-1 ) = det ( ( L · F – L · M ) · L-1 ) = det ( ( L · F · L-1 ) – ( L · M · L-1 ) ) = det ( F' – M')
So, the term ( E2 – c2 · B2 )2 = a2 is invariant under Lorentz transformations . As a matter of fact E2 – c2 · B2 and not only the square of it is invariant ! The absolute value of this term being invariant, it can not jump directly from +a to –a, because E' and B' vary continuously as functions of the relative speed v, starting with E = E' and B = B' for v = 0. For an elementary proof consult Freund [1-225f] e.g.
6.4 An elegant Proof for M' = L · M · L-1
F · M = M · F = c · (E·B) · Id shows M beeing very close to the inverse of F ! More exactly, we have
M = c · (E·B) · F-1 and hence M' = c · (E'·B') · (F')-1 = c · (E·B) · (F')-1 according to 6.2
So L · M · L-1 = L · [ c · (E·B) · F-1 ] · L-1 = c · (E·B) · L · F-1 · L-1 = c · (E·B) · (F')-1 = M'
and we have fullfilled our promise of section 3 !
6.5 The Matrices G·F , G·M , F·G and M·G
The matrices G·F, G·M, F·G und M·G are anti-symmetric. Not F is the matrix corresponding to the tensor of the electromagnetic field, but G·F . Sometimes in tensor calculus a second tensor (Mkl oder *Fkl) is introduced, which corresponds to the matrix G·M . Using these anti-symmetric tensors, Maxwells 8 equations can be formulated as nicely as we did:
Most of the textbooks written for high school or undergraduate students do not prove the form-invariance of Maxwell's equations. Examples: , , ,  und  . Some of them restrict themselves to mechanics, as I did in 'Epstein Explains Einstein'. Books  to  are the ones I read after the publication of 'Epstein Explains Einstein'. Thanks to Shadowitz  and Sartori  I became familiar with the diagrams of the Loedel and Brehme type. They seem to be completely unknown in Europe ! But there is an easy way to learn about them: Just download those programs that will run on your type of computer system. Solve your exercises using any type of STR diagrams. It's fun - and it is more than fun !