Invariants of the field

1. Invariants of the field Section 25

2. Certain functions of E and H are invariant under Lorentz transform • The 4D representation of the field is Fik • FikFik = an invariant scalar • (1/2)eiklmFikFlm = an invariant pseudo scalar • Dual of antisymmetric tensor Fik is an antisymmetric pseudo tensor • Invariant with respect to Lorentz transform, i.e. to rotations in 4D, but changes sign under inversion or reflection

3. There are only two invariants (HW) • H2 – E2 = invariant scalar • E.H = invariant pseudo scalar • E is a polar vector: components change sign under inversion or reflection • H is an axial vector: components do not change sign

4. Invariance of E.H gives a theorem: • If E and H are perpendicular in one reference system, they are perpendicular in every reference system. • For example, electromagnetic waves

5. Invariance of E.H gives a second theorem • If E and H make an acute (or obtuse) angle in any inertial system, the same will hold in all inertial systems. • You cannot transform from an acute to obtuse angle, or vice versa. • For acute angles E.H is positive • For obtuse angles E.H is negative

6. Invariance of H2 – E2 gives a third theorem • If the magnitudes E and H are equal in one inertial reference system, they equal in every inertial reference system. • For example, electromagnetic waves

7. Invariance of H2 – E2 gives a fourth theorem • If E>H (or H>E) in any inertial reference system, the same holds in all inertial systems.

8. Lorentz transforms can be found to give E and H arbitrary values subject to two conditions: • H2 – E2 = invariant scalar • E.H = invariant pseudo scalar

9. We can usually find a reference frame where E and H are parallel at a given point • In this system • E.H = E H Cos[0] =E H, or • E.H =E H Cos[180] = - E H • Values of E,H in this system are found from two equations in two unknowns: • H2 – E2 = H02 – E02 • ± E H = E0.H0 • + sign if E0& H0 form acute angle. • Subscript fields are the known ones in the original frame • Doesn’t work when both invariants are zero, e.g. EM wave: E = H and E perpendicular to H conditions are invariant.

10. If E and H are perpendicular, we can usually find a frame in which • E = 0 (when E2 < H2), i.e. pure magnetic. Or, • H = 0 (when E2 > H2), i.e. pure electric. • In other words, we can always make the smaller field vanish by suitable transform. • Except when E2 = H2, e.g. electromagnetic wave

11. If E = 0 or H = 0 in any frame, then E and H are perpendicular in every other frame. • Follows from invariance of E.H

12. The two invariants of Fik given (or of any antisymmetric 4-tensor), are the only ones. • Consider a Lorentz transform of F = E + iH along the X axis.

13. Rotation in (x,t) plane in 4-space (the considered Lorentz transform along x) is equivalent for F to a rotation in (y,z) plane through an imaginary angle in 3-space.

14. Square of F is invariant under 3D rotations • The set of all possible rotations in 4-space (including simple ones about x,y,z axes) is equivalent to the set of all possible rotations through complex angles in 3 space • 6 angles of rotation in 4D 3 complex angles in 3D • The only invariant of a 3 vector with respect to rotations is its square

15. The square of F is given by just two invariants • F2 = (E + iH).(E + iH) = (E2 – H2) + 2 iE.H • The real and imaginary parts are the only two independent invariants of the tensor Fik.

16. If F2 is non-zero, then F = an • ais a complex number • n is a complex unit vector, n2 =1 • A suitable complex rotation in 3D will point n along one coordinate axis • Then n becomes real • And F = (E+iH) n, i.e. E and H become parallel • In other words, a suitable Lorentz transform makes E and H parallel if neither invariant vanishes.