## 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 ) = –c^{2}·(E·B)^{2}= det ( M ) and det ( F' ) = –c^{2}·(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}

IfEandBare orthogonal in any special coordinate frame, then this will hold true in any other frame!

By the way,E·Bhere 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, butE·Bitself (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 provedE·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 ) = – (E^{2}– c^{2}·B^{2})^{2}= det ( M – F ) = det ( M + F ) and hence

det ( F' – M' ) = det ( F' + M' ) = – (E'^{2}– c^{2}·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( E^{2}– c^{2}·B^{2})^{2}= a^{2}is invariant under Lorentz transformations . As a matter of factE^{2}– c^{2}·B^{2 }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, becauseE'andB'varycontinuouslyas functions of the relative speed v, starting withE=E'andB=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 (M^{kl}oder *F^{kl}) 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: [2], [4], [5], [6] und [7] . Some of them restrict themselves to mechanics, as I did in 'Epstein Explains Einstein'. Books [3] to [7] are the ones I read after the publication of 'Epstein Explains Einstein'. Thanks to Shadowitz [3] and Sartori [4] 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 !