Slide 1

Invariants of the field Section 25

Slide 2

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

Slide 3

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

Slide 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

Slide 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

Slide 6

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

Slide 7

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.

Slide 8

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

Slide 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: 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 Doesnt work when both invariants are zero, e.g. EM wave: E = H and E perpendicular to H conditions are invariant.

Slide 10

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

Slide 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

Slide 12

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.

Slide 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.

Slide 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

Slide 15

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.

Slide 16

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.