Electromagnetic Waves

Home , Circular polarization, Elliptical polarization, Linear polarization

May 4, 20101

1J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Section 7 Electromagnetic Waves Maxwell Equations

? A basic feature of Maxwell equations for the EM ﬁeld is the existence of travelling wave solutions which represent the transport of energy from one point to another. ? The simplest and most fundamental EM waves are transverse, plane waves. In a region of space where there are no free sources (ρ = 0, ~J = 0), Maxwell’s equations reduce to a simple form given

1 ∂B~ ∇~ · E~ = 0 , ∇~ × E~ + = 0 c ∂t µ ∂E~ ∇~ · B~ = 0 , ∇~ × B~ − = 0 (1) c ∂t where D~ and H~ are given by relations 1 D~ = E~ and H~ = B~ (2) µ where is the electric permittivity and µ the magnetic permeability which assumed to be independent of the frequency.

Electromagnetic Waves Plane Electromagnetic Waves

Maxwell’s equations can be written as µ ∂2B~ µ ∂2E~ ∇2B~ − = 0 and ∇2E~ − = 0 (3) c2 ∂t2 c2 ∂t2 In other words each component of B~ and E~ obeys a wave equation of the form: 1 ∂2u c ∇2u − = 0 where v = √ (4) v 2 ∂t2 µ is a constant with dimensions of velocity characteristic of the medium. The wave equation admits admits plane-wave solutions: ~ u = eik·~x−iωt (5) E~ (~x, t)= E~eik~n·~x−iωt and B~ (~x, t) = B~eik~n·~x−iωt (6) where the relation between the frequency ω and the wave vector ~k is ω √ ω ω 2 k = = µ or ~k · ~k = (7) v c v also the vectors ~n, E~ and B~ are constant in time and space.

Electromagnetic Waves If we consider waves propagating in one direction, say x-direction then the fundamental solution is:

u(x, t) = Aeik(x−vt) + Be−ik(x+vt) (8) which represents waves traveling to the right and to the left with propagation velocities v which is called phase velocity of the wave. ? From the divergence relations of (1) by applying (6) we get

~n · E~ = 0 and ~n · B~ = 0 (9)

This means that E~ (or E~ ) and B~ (or B~ ) are both perpendicular to the direction of propagation ~n. Such a wave is called transverse wave. ? The curl equations provide a further restriction √ 1 B~ = µ~n × E~ and E~ = −√ ~n × B~ (10) µ

The combination of equations (9) and (10) suggests that the vectors ~n, E~ and B~ form an orthonormal set. Also, if ~n is real, then (10) implies that that E~ and B~ have the same phase.

Electromagnetic Waves It is then useful to introduce a set of real mutually orthogonal unit vectors( ~1,~2,~n). In terms of these unit vectors the ﬁeld strengths E~ and B~ are √ E~ = ~1E0 , B~ = ~2 µE0 (11) or

~ 0 ~ √ 0 E = ~2E0 , B = −~1 µE0 (12)

0 E0 and E0 are constants, possibly complex. In other words the most general way to write the electric/magnetic ﬁeld vector is:

~ 0 ik~n·~x−iωt E =( E0 ~1 + E0 ~2)e (13) ~ √ 0 ik~n·~x−iωt B = µ(E0 ~2 − E0 ~1)e (14)

Electromagnetic Waves Thus the wave described by (6) and (11) or (12) is a transverse wave propagating in the direction ~n.

Or that E and B are oscillating in a plane perpendicular to the wave vector k, determining the direction of propagation of the wave.

The energy ﬂux of EM waves is described by the real part of the complex Poynting vector 1 c 1 c h i ~S = E~ × H~ ∗ = E~ × H~ + E~ × H~ + i E~ × H~ − E~ × H~ 2 4π 2 4π R R I I I R R I

where E~ and H~ are the measured ﬁelds at the point where ~S is evaluated.2

2Note : we use the magnetic induction H~ because although B~ is the applied induction, the actual ﬁeld that carries the energy and momentum in media is H~ . Electromagnetic Waves The time averaged ﬂux of energy is:

c r ~S = |E |2~n (15) 8π µ 0

The total time averaged density (and not just the energy density associated with the electric ﬁeld component) is:

1 1 u = E~ · E~ ∗ + B~ · B~ ∗ = |E |2 (16) 16π µ 8π 0

The ratio of the magnitude of (15) to (16) is the speed of energy ﬂow i.e. √ v = c/ µ. 3 (Prove the above relations)

Project: What will happen if ~n is not real? What type of waves you will get? What will be the form of E?

3Note: To prove the above relations use hcos2 xi = 1/2 and since ~ ~ ~ ∗ ~ 2 ~ ~ ∗ ER = (E + E )/2 we get hER i = E · E /2. Electromagnetic Waves Linear and Circular Polarization of EM Waves

The plane wave (6) and (11) is a wave with its electric ﬁeld vector always in the direction ~1. Such a wave is said to be linearly polarized with polarization vector ~1. The wave described by (12) is linearly polarized with polarization vector ~2 and is linearly independent of the ﬁrst. The two waves :

~ i~k·~x−iωt ~ i~k·~x−iωt E1 = ~1E1e , E2 = ~2E2e with (17) √ ~k × E~ B~ = µ i , i = 1, 2 i k Can be combined to give the most general homogeneous plane waves propagating in the direction ~k = k~n, ~ i~k·~x−iωt E(~x, t) = (~1E1 + ~2E2) e (18) h i ~ ~ i(φ2−φ1) ik·~x−iωt+iφ1 E(~x, t) = ~1|E1| + ~2|E2|e e (19)

iφ1 iφ2 The amplitudes E1 = |E1|e and E2 = |E2|e are complex numbers in order to allow the possibility of a phase difference between waves of different polarization. Electromagnetic Waves LINEARLY POLARIZED iφ1 iφ2 If the amplitudes E1 = |E1|e and E2 = |E2|e have the same phase (18) represents a linearly polarized wave with the polarization vector −1 making an angle θ = tan (<(E2)/=(E1)) (which remains constant as the field evolves in space and time) with ~1 and magnitude p 2 2 E = E1 + E2 . ELLIPTICALLY POLARIZED If E1 and E2 have the different phase the wave (18) is elliptically polarized and the electric vector rotates around ~k.

Electromagnetic Waves Circular Polarization

• E1 = E2 = E0 • φ1 − φ2 = ±π/2 and the wave becomes

~ i~k·~x−iωt E(~x, t) = E0 (~1 ± i~2) e (20)

At a fixed point in space, the fields are such that the electric vector is constant in magnitude, but sweeps around in a circle at a frequency ω. The components of the electric field, obtained by taking the real part of (20)

Ex (~x, t) = E0 cos(kz − ωt) , Ey (~x, t) = ∓E0 cos(kz − ωt) (21)

For the upper sign (~1 + i~2) the rotation is counter-clockwise when the observer is facing into the oncoming wave. The wave is called left circularly polarized in optics while in modern physics such a wave is said to have positive helicity.

For the lower sign (~1 − i~2) the wave is right circularly polarized or it has negative helicity.

Electromagnetic Waves Elliptically Polarized EM Waves

An alternative general expression for E~ can be given in terms of the complex orthogonal vectors 1 ~± = √ (~1 ± i~2) (22) 2 with properties

∗ ∗ ∗ ~± · ~∓ = 0 , ~± · ~3 = 0 , ~± · ~± = 1 . (23) Then the general representation of the electric vector

~ i~k·~x−iωt E(~x, t) = (E+~+ + E−~−) e (24)

where E− and E+ are complex amplitudes If E− and E+ have diﬀerent amplitudes but the same phase eqn (24) represents an elliptically polarized wave with principle axes of the ellipse in the directions of ~1 and ~2. The ratio of the semimajor to semiminor axis is |(1 + r)/(1 − r)|, where

E−/E+ = r. Electromagnetic Waves The ratio of the semimajor to semiminor axis is |(1 + r)/(1 − r)|, where E−/E+ = r. iα If the amplitudes have a phase diﬀerence between them E−/E+ = re , then the ellipse traced out by the E~ vector has its axes rotated by an angle α/2.

Figure: The ﬁgure shows the general case of elliptical polarization and the ellipses traced out by both E~ and B~ at a given point in space.

Note : For r = ±1 we get back to a linearly polarized wave.

Electromagnetic Waves Polarization

Figure: The ﬁgure shows the linear, circular and elliptical polarization

Electromagnetic Waves Stokes Parameters

The polarization content of an EM wave is known if it can be written in the form of either (18) or (24) with known coeﬃcients( E1, E2) or (E−, E+). In practice, the converse problem arises i.e. given a wave of the form (6), how can we determine from observations on the beam the state of polarization? A useful tool for this are the four Stokes parameters. These are quadratic in the ﬁeld strength and can be determined through intensity measurements only. Their measurements determines completely the state of polarization of the wave. For a wave propagating in the z-direction the scalar products ~ ~ ∗ ~ ∗ ~ ~1 · E , ~2 · E , ~+ · E , ~− · E (25) are the amplitudes of radiation respectively, with linear polarization in the x-direction, linear polarization in the y-direction, positive helicity and negative helicity. The squares of these amplitudes give a measure of the intensity of each type of polarization. The phase information can be taken by using cross products Electromagnetic Waves In terms of the linear polarization bases( ~1, ~2), the Stokes parameters are:

~ 2 ~ 2 2 2 s0 = |~1 · E| + |~2 · E| = a1 + a2 ~ 2 ~ 2 2 2 s1 = |~1 · E| − |~2 · E| = a1 − a2 h ~ ∗ ~ i s2 = 2< (~1 · E) (~1 · E) =2 a1a2 cos(δ1 − δ2) (26)

h ~ ∗ ~ i s3 = 2= (~1 · E) (~1 · E) =2 a1a2 sin(δ1 − δ2) where we deﬁned the coeﬃcients of (18) or (24) as magnitude times a phase factor:

iδ1 iδ2 iδ+ iδ− E1 = a1e , E2 = a2e , E+ = a+e , E− = a−e (27)

Here s0 and s1 contain information regarding the amplitudes of linear polarization, whereas s2 and s3 say something about the phases. Knowing these parameters (e.g by passing a wave through perpendicular polarization filters) is sufficient for us to determine the amplitudes and relative phases of the field components.

Electromagnetic Waves Stokes Parameters

In terms of the linear polarization bases( ~+, ~−), the Stokes parameters are:

∗ ~ 2 ∗ ~ 2 2 2 s0 = |~+ · E| + |~− · E| = a+ + a− h ∗ ~ ∗ ∗ ~ i s1 = 2< (~+ · E) (~− · E) = 2a+a− cos(δ− − δ+) (28) h ∗ ~ ∗ ∗ ~ i s2 = 2= (~+ · E) (~− · E) = 2a+a− sin(δ− − δ+) ∗ ~ 2 ∗ ~ 2 2 2 s3 = |~+ · E| − |~− · E| = a+ − a−

Notice an interesting rearrangement of roles of the Stokes parameters with respect to the two bases. The four Stokes parameters are not independent since they depend on only 3 quantities a1, a2 and δ1 − δ2. They satisfy the relation

2 2 2 2 s0 = s1 + s2 + s3 . (29)

Electromagnetic Waves Reﬂection & Refraction of EM Waves

The reﬂection and refraction of light at a plane surface between two media of diﬀerent dielectric properties are familiar phenomena. The various aspects of the phenomena divide themselves into two classes

I Kinematic properties:

I Angle of reﬂection= angle of incidence sin i n0 I Snell’s law: sin r = n where i, r are the angles of incidence and refraction, while n, n0 are the corresponding indices of refraction.

I Dynamic properties:

I Intensities of reflected and refracted radiation I Phase changes and polarization ? The kinematic properties follow from the wave nature of the phenomena and the need to satisfy certain boundary conditions (BC). But not on the detailed nature of the waves or the boundary conditions. ? The dynamic properties depend entirely on the specific nature of the EM fields and their boundary conditions.

Electromagnetic Waves Figure: Incident wave ~k strikes plane interface between diﬀerent media, giving rise to a reﬂected wave ~k00 and a refracted wave ~k0. The media below and above the plane z = 0 have permeabilities and dielectric constants µ, and µ’, ’ respectively. The indices of refraction are √ √ n = µ and n0 = µ00.

Electromagnetic Waves According to eqn (18) the 3 waves are: INCIDENT ~ √ ~k × E~ E~ = E~ eik·~x−iωt , B~ = µ (30) 0 k REFRACTED

0 0 ~0 ~k × E~ E~ 0 = E~ 0eik ·~x−iωt , B~ 0 = pµ00 (31) 0 k0 REFLECTED

00 00 ~00 ~k × E~ E~ 00 = E~ 00eik ·~x−iωt , B~ 0 = pµ00 (32) 0 k00 The wave numbers have magnitudes: ω √ ω |~k| = |~k00| = k = µ, |~k0| = k0 = pµ00 (33) c c

Electromagnetic Waves AT the boundary z = 0 the BC must be satisfied at all points on the plane at all times, i.e. the spatial & time variation of all fields must be the same at z = 0. Thus the phase factors must be equal at z = 0 ~k · ~x = ~k00 · ~x = ~k0 · ~x (34) z=0 z=0 z=0 independent of the nature of the boundary conditions. ? Eqn (34) contains the kinematic aspects of reflection and refraction. Note that all 3 wave vectors must lie in a plane. From the previous figure we get k sin i = k00 sin r 0 = k0 sin r (35) Since k = k00, we find that i = r 0; the angle of incidence equals the angle of reflection. Snell’s law is: s sin i k0 µ00 n0 = = = (36) sin r k µ n

Electromagnetic Waves Reﬂection & Refraction of EM Waves

The dynamic properties are contained in the boundary conditions : • normal components of D~ = E~ and B~ are continuous • tangential components of E~ and H~ = [c/(ωµ)]~k × E~ are continuous In terms of ﬁelds (30)-(32) these boundary conditions at z = 0 are:

h ~ ~ 00 0~ 0i E0 + E0 − E0 · ~n = 0 h~ ~ ~ 00 ~ 00 ~ 0 ~ 0i k × E0 + k × E0 − k × E0 · ~n = 0 ~ ~ 00 ~ 0 E0 + E0 − E0 × ~n = 0 (37) 1 1 ~k × E~ + ~k00 × E~ 00 − ~k0 × E~ 0 × ~n = 0 µ 0 0 µ0 0 Two separate situations, the incident plane wave is linearly polarized : • The polarization vector is perpendicular to the plane of incidence (the plane deﬁned by ~k and ~n ). • The polarization vector is parallel to the plane of incidence. • The case of arbitrary elliptic polarization can be obtained by appropriate linear combinations of the two results.

Electromagnetic Waves E~ : Perpendicular to the plane of incidence

• Since the E~ -ﬁelds are parallel to the surface the 1st BC of (38) yields nothing • The 3rd and 4th of of (38) give (how?):

00 0 E0 + E0 − E0 = 0 s r 0 (E − E 00) cos i − E 0 cos r = 0 (38) µ 0 0 µ0 0

• The 2nd, using Snell’s law, duplicates the 3rd. Figure: Reﬂection and refraction (prove all the above statements) with polarization perpendicular to the plane of incidence. All the E~ -ﬁelds shown directed away from the viewer.

Electromagnetic Waves E~ : Perpendicular to the plane of incidence

The relative amplitudes of the refracted and reﬂected waves can be found from (38)

E 0 2n cos i 2 2 sin r cos i 0 = = = µ p 2 µ tan i E0 02 2 1 + sin(i + r) 0 n cos i + µ0 n − n sin i µ0 tan r µ=µ p 00 µ 02 2 2 µ tan i E n cos i − 0 n − n sin i 1 − 0 sin(r − i) 0 = µ = µ tan r = (39) µ p 2 µ tan i E0 02 2 1 + sin(i + r) 0 n cos i + µ0 n − n sin i µ0 tan r µ=µ p Note that n02 − n2 sin2 i = n0 cos r but Snell’s law has been used to express it in terms of the angle of incidence. For optical frequencies it is usually permitted to put µ = µ0.

Electromagnetic Waves E~ : Parallel to the plane of incidence

Boundary conditions involved: • normal E~ : 1st eqn in (38) • tangential E~ : 3rd eqn in (38) • tangential B~ : 4th eqn in (38)

The last two demand that

00 0 (E0 − E0 ) cos i − E0 cos r = 0 s r 0 (E + E 00) − E 0 = 0 (40) µ 0 0 µ0 0 Figure: Reﬂection and refraction with polarization parallel to the plane of incidence.

Electromagnetic Waves E~ : Parallel to the plane of incidence

The condition that normal E~ is continuous, plus Snell’s law, merely dublicates the 2nd of the previous equations. The relative amplitudes of refracted and reﬂected ﬁelds are therefore (how?)

0 0 E0 2nn cos i = p E0 µ 02 02 2 2 µ0 n cos i + n n − n sin i n0 2 0 2 sin r cos i = n = tan i sin(i + r) cos(i − r) 1 + 0 tan r µ=µ0 p 00 µ n02 cos i − n n02 − n2 sin2 i E0 µ0 = p E0 µ 02 02 2 2 µ0 n cos i + n n − n sin i tan i 1 − 0 tan(i − r) = tan r = (41) tan i tan(i + r) 1 + 0 tan r µ=µ0

Electromagnetic Waves Normal incidence i = 0

For normal incidence i = 0 both (39) and (41) reduce to (how?)

E 0 2 2n 0 = → q 0 0 E0 µ n + n 1 + µ0

q µ0 00 0 E −1 + µ0 n − n 0 = → (42) q 0 0 E0 µ n + n 1 + µ0

EXERCISES: What are the conditions for:

I Total reﬂection

I Total transmision

Electromagnetic Waves