Download Section 25

Invariants of the field
Section 25
Certain functions of E and H are invariant under Lorentz
transform
• The 4D representation of the field is Fik
• Fik Fik = an invariant scalar
• (1/2)eiklm Fik Flm = 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
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
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 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
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.
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
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:
The conditions E = H and E perpendicular to H are invariant.
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
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 Fik given (or of any antisymmetric 4tensor), are the only ones.
Consider a Lorentz transform of F = E + iH
along the X axis. (Homework)
Where
Rotation matrix
A 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
• F2 = (E + iH).(E + iH) = (E2 – H2) + 2 i E.H
• The real and imaginary parts are the only two independent invariants of
the tensor Fik.
If F2 is non-zero, then F = a n
• a is 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.