Invariants of the field Section 25 - PowerPoint Presentation

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Invariants of the field

Section 25

Certain functions of

E

and H are invariant under Lorentz transform

The 4D representation of the field is FikFik Fik = an invariant scalar(1/2)eiklm Fik Flm = an invariant pseudo scalarDual of antisymmetric tensor Fik is an antisymmetric pseudo tensorInvariant 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

– E2 = invariant scalarE.H = invariant pseudo scalarE is a polar vector: components change sign under inversion or reflectionH 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 positiveFor obtuse angles E.H is negative

Invariance of H

2

– 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

Invariance of H

2

– E2

gives a fourth theoremIf 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

– E2 = invariant scalarE.H = invariant pseudo scalar

We can

usually

find a reference frame where E and

H are parallel at a given pointIn this systemE.H = E H Cos[0] =E H, orE.H =E H Cos[180] = - E HValues 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 frameDoesn’t work when both invariants are zero, e.g. EM wave: The conditions E = H and E perpendicular to H are invariant.

If

E

and H are perpendicular, we can

usually find a frame in whichE = 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

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, which here is zero.

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. (Homework)

Where

Rotation matrixA 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 space6 angles of rotation in 4D 3 complex angles in 3DThe 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) = (E2 – H2) + 2 i E.HThe real and imaginary parts are the only two independent invariants of the tensor Fik.

If F

2

is non-zero, then F

= a na is a complex numbern is a complex unit vector, n2 = 1A suitable complex rotation in 3D will point n along one coordinate axisThen n becomes realAnd 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.