[
QUOTE=samalkhaiat]
PROOFS
1) Purely Mathematical;
Assuming that the em field is massless, then the representation theory of Lorentz group (as I said befor) shows that there are only 2 independent components.
2) The covariant 4-potential formalism
A_\mu = (A_{0},A_{1},A_{2},A_{3})
I.E. we have 4 components.
In this formulation, we impose the so-called Lorentz condition;
\partial_{\nu}A^{\nu} = 0
which reduces the number to 3 components.
(mathematically, we impose this condition to exclude the spin zero component of the representation)
You know that the EM theory is invariant under the above-mentioned gauge transformation.The choice of the gauge function imposes the second needed condition on the 4 potential. We are left with only 2.
3) The (
E,
B) 3-vector formulation
Here we have 6 components.
But we also have;
\nabla . E = \rho
\nabla . B = 0
so the number becomes 4.
We can easily show** that;
B^2 - E^2 = inv.
B.E = inv.
these can be used to further reduce the number to only 2.
Are you happy now?
In the em theory, the sources are treated as "given". They impose no restriction on the number of physically significant components of the EM-field.
Seting;
J^{\mu} = (\rho , j) = 0
does not change any thing in any of the above 3 proofs.
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**Consider the complex vector;
V=E + iB
The only invariant of a vector with respect to rotation is its square;
V^2 = E^2 - B^2 + 2iE.B
Thus the real quantities E^2 - B^2 and E.B are the only two independent invariants of the EM-field.
What do these invariants mean?
Lorentz trans. can be used to give
E &
B any arbitrary values, subject only to the condition that
E.E - B.B and
E.B have fixed values.
We can prove;
1) if
E &
B are mutually perpendicular in one frame, that is
E.B = 0, then they are also perpendicular in every other Lorentz frame.
2) if
B>E (or
B<E) in any system, then we will have the same in every other system.
3) if they make an acute angle in any frame, then they will make an acute angle in every other frame.
4)we can always find a frame in which
E &
B are parallel at a given point.
5) if
E.B = 0, then we can always find a frame in which
E = 0 or
B = 0 (depending on [
E.E -
B.B]< or >0) .
6) if, in any frame the field is purely magnetic (
E =
0) or purely electric(
B =
0), then they are mutually perpendicular in every other frame.
7) the case where both invariants are zero is excluded V^2 = 0. In this case,
E &
B are equal and perpendicular in all system.
Good Luck to all PF members.
This is my last post on these forums, as I will need the time to concentrate on the book I am writting.
Best Wishes
Sam
