Invariants of the field Section 25

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

There are only two invariants (HW) H 2 – E 2 = 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

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

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

Invariance of H 2 – E 2 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

Invariance of H 2 – E 2 gives a fourth theorem If E>H (or H>E) in any inertial reference system, the same holds in all inertial systems.

Lorentz transforms can be found to give E and H arbitrary values subject to two conditions: H 2 – E 2 = invariant scalar E.H = invariant pseudo scalar

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:  H 2 – E 2 = H 0 2 – E 0 2  ± E H = E 0.H 0  + sign if E 0 & H 0 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.

If E and H are perpendicular, we can usually find a frame in which E = 0 (when E 2 < H 2 ), i.e. pure magnetic. Or, H = 0 (when E 2 > H 2 ), i.e. pure electric. In other words, we can always make the smaller field vanish by suitable transform. Except when E 2 = H 2, e.g. electromagnetic wave

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

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

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.

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

The square of F is given by just two invariants F 2 = (E + iH).(E + iH) = (E 2 – H 2 ) + 2 i E.H The real and imaginary parts are the only two independent invariants of the tensor F ik.

If F 2 is non-zero, then F = a n a is a complex number n is a complex unit vector, n 2 =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.