electromagnetic waves - institutkokkotas/teaching/field_theory_files/... · plane electromagnetic...

Electromagnetic Waves

May 4, 20101

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

Maxwell Equations

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

~∇ · ~E = 0 , ~∇× ~E +1

c

∂~B

∂t= 0

~∇ · ~B = 0 , ~∇× ~B − µε

c

∂~E

∂t= 0 (1)

where ~D and ~H are given by relations

~D = ε~E and ~H =1

µ~B (2)

where ε is the electric permittivity and µ the magnetic permeabilitywhich assumed to be independent of the frequency.


Plane Electromagnetic Waves

Maxwell’s equations can be written as

∇2~B − µε

c2

∂2~B

∂t2= 0 and ∇2~E − µε

c2

∂2~E

∂t2= 0 (3)

In other words each component of ~B and ~E obeys a wave equation ofthe form:

∇2u − 1

v2

∂2u

∂t2= 0 where v =

c√µε

(4)

is a constant with dimensions of velocity characteristic of the medium.The wave equation admits admits plane-wave solutions:

u = e i~k·~x−iωt (5)

~E (~x , t) = ~Ee ik~n·~x−iωt and ~B(~x , t) = ~Be ik~n·~x−iωt (6)

where the relation between the frequency ω and the wave vector ~k is

k =ω

v=√µεω

cor ~k · ~k =

(ωv

)2

(7)

also the vectors ~n, ~E and ~B are constant in time and space.


If we consider waves propagating in one direction, say x-direction thenthe fundamental solution is:

u(x , t) = Ae ik(x−vt) + Be−ik(x+vt) (8)

which represents waves traveling to the right and to the left withpropagation 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 thedirection of propagation ~n. Such a wave is called transverse wave.? The curl equations provide a further restriction

~B =√µε~n × ~E and ~E = − 1

√µε~n × ~B (10)

The combination of equations (9) and (10) suggests that the vectors ~n, ~Eand ~B form an orthonormal set.

Also, if ~n is real, then (10) implies that that ~E and ~B have the same

phase.


It is then useful to introduce a set of real mutually orthogonal unitvectors (~ε1,~ε2,~n).

In terms of these unit vectors the fieldstrengths ~E and ~B are

~E = ~ε1E0 , ~B = ~ε2√µεE0 (11)

or

~E = ~ε2E′0 , ~B = −~ε1

√µεE ′0 (12)

E0 and E ′0 are constants, possibly complex.

In other words the most general way to write the electric/magnetic fieldvector is:

~E = (E0~ε1 + E ′0~ε2)e ik~n·~x−iωt (13)

~B =√µε(E0~ε2 − E ′0~ε1)e ik~n·~x−iωt (14)


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

Or that E and B are oscillating in aplane perpendicular to the wavevector k, determining the directionof propagation of the wave.

The energy flux of EM waves is described by the real part of thecomplex Poynting vector

~S =1

2

c

4π~E × ~H∗ =

1

2

c

4π

[~ER × ~HR + ~EI × ~HI + i

(~EI × ~HR − ~ER × ~HI

)]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 appliedinduction, the actual field that carries the energy and momentum in media is ~H.


The time averaged flux of energy is:

~S =c

8π

√ε

µ|E0|2~n (15)

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

u =1

16π

(ε~E · ~E∗ +

1

µ~B · ~B∗

)=

ε

8π|E0|2 (16)

The ratio of the magnitude of (15) to (16) is the speed of energy flow i.e.v = c/

√µε. 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 〈cos2 x〉 = 1/2 and since~ER = (~E + ~E∗)/2 we get 〈~E 2

R〉 = ~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 alwaysin the direction ~ε1. Such a wave is said to be linearly polarized withpolarization vector ~ε1. The wave described by (12) is linearly polarizedwith polarization vector ~ε2 and is linearly independent of the first.

The two waves :

~E1 = ~ε1E1ei~k·~x−iωt , ~E2 = ~ε2E2e

i~k·~x−iωt

with (17)

~Bi =√µε~k × ~Ei

k, i = 1, 2

Can be combined to give the most general homogeneous plane wavespropagating in the direction ~k = k~n,

~E (~x , t) = (~ε1E1 + ~ε2E2) e i~k·~x−iωt (18)

~E (~x , t) =[~ε1|E1|+ ~ε2|E2|e i(φ2−φ1)

]e i~k·~x−iωt+iφ1 (19)

The amplitudes E1 = |E1|e iφ1 and E2 = |E2|e iφ2 are complex numbers in

order to allow the possibility of a phase difference between waves of

different polarization.Electromagnetic Waves

LINEARLY POLARIZEDIf the amplitudes E1 = |E1|e iφ1 and E2 = |E2|e iφ2 have the same phase(18) represents a linearly polarized wave with the polarization vectormaking an angle θ = tan−1 (<(E2)/=(E1)) (which remains constant asthe field evolves in space and time) with ~ε1 and magnitudeE =

√E 2

1 + E 22 .

ELLIPTICALLY POLARIZEDIf E1 and E2 have the different phase the wave (18) is elliptically

polarized and the electric vector rotates around ~k .


Circular Polarization

• E1 = E2 = E0

• φ1 − φ2 = ±π/2 and the wave becomes

~E (~x , t) = E0 (~ε1 ± i~ε2) e i~k·~x−iωt (20)

At a fixed point in space, the fields are suchthat the electric vector is constant inmagnitude, but sweeps around in a circle ata 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 (~ε1 + i~ε2) the rotation is counter-clockwise when theobserver is facing into the oncoming wave. The wave is called leftcircularly polarized in optics while in modern physics such a wave is saidto have positive helicity.

For the lower sign (~ε1 − i~ε2) the wave is right circularly polarized or it

has negative helicity.


Elliptically Polarized EM Waves

An alternative general expression for ~E can be given in terms of thecomplex orthogonal vectors

~ε± =1√2

(~ε1 ± i~ε2) (22)

with properties

~ε∗± · ~ε∓ = 0 , ~ε∗± · ~ε3 = 0 , ~ε∗± · ~ε± = 1 . (23)

Then the general representation of the electric vector

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

where E− and E+ are complex amplitudesIf E− and E+ have different amplitudes but the same phase eqn (24)represents an elliptically polarized wave with principle axes of theellipse in the directions of ~ε1 and ~ε2.

The ratio of the semimajor to semiminor axis is |(1 + r)/(1− r)|, where

E−/E+ = r .


The ratio of the semimajor to semiminor axis is |(1 + r)/(1− r)|, whereE−/E+ = r .If the amplitudes have a phase difference between them E−/E+ = re iα,

then the ellipse traced out by the ~E vector has its axes rotated by anangle α/2.

Figure: The figure shows the general case of elliptical polarization andthe 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.


Polarization

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


Stokes Parameters

The polarization content of an EM wave is known if it can be written inthe 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 ofpolarization?A useful tool for this are the four Stokes parameters. These arequadratic in the field strength and can be determined through intensitymeasurements only. Their measurements determines completely the stateof polarization of the wave.For a wave propagating in the z-direction the scalar products

~ε1 · ~E , ~ε2 · ~E , ~ε∗+ · ~E , ~ε∗− · ~E (25)

are the amplitudes of radiation respectively, with linear polarization inthe x-direction, linear polarization in the y -direction, positive helicityand negative helicity.The squares of these amplitudes give a measure of the intensity of eachtype of polarization.

The phase information can be taken by using cross productsElectromagnetic Waves

In terms of the linear polarization bases (~ε1, ~ε2), the Stokes parametersare:

s0 = |~ε1 · ~E |2 + |~ε2 · ~E |2 = a21 + a2

2

s1 = |~ε1 · ~E |2 − |~ε2 · ~E |2 = a21 − a2

2

s2 = 2<[(~ε1 · ~E )∗(~ε1 · ~E )

]= 2a1a2 cos(δ1 − δ2) (26)

s3 = 2=[(~ε1 · ~E )∗(~ε1 · ~E )

]= 2a1a2 sin(δ1 − δ2)

where we defined the coefficients of (18) or (24) as magnitude times aphase factor:

E1 = a1eiδ1 , E2 = a2e

iδ2 , E+ = a+e iδ+ , E− = a−e iδ− (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.


Stokes Parameters

In terms of the linear polarization bases (~ε+, ~ε−), the Stokes parametersare:

s0 = |~ε∗+ · ~E |2 + |~ε∗− · ~E |2 = a2+ + a2

−

s1 = 2<[(~ε∗+ · ~E )∗(~ε∗− · ~E )

]= 2a+a− cos(δ− − δ+) (28)

s2 = 2=[(~ε∗+ · ~E )∗(~ε∗− · ~E )

]= 2a+a− sin(δ− − δ+)

s3 = |~ε∗+ · ~E |2 − |~ε∗− · ~E |2 = a2+ − a2

−

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

s20 = s2

1 + s22 + s2

3 . (29)


Reflection & Refraction of EM Waves

The reflection and refraction of light at a plane surface between twomedia 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 incidenceI Snell’s law: sin i

sin r = n′

n where i , r are the angles of incidenceand refraction, while n, n′ are the corresponding indices ofrefraction.

I Dynamic properties:

I Intensities of reflected and refracted radiationI Phase changes and polarization

? The kinematic properties follow from the wave nature of thephenomena 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.


Figure: Incident wave ~k strikes plane interface between different media,giving rise to a reflected wave ~k ′′ and a refracted wave ~k ′. The mediabelow and above the plane z = 0 have permeabilities and dielectricconstants µ, ε and µ’, ε’ respectively. The indices of refraction aren =√µε and n′ =

√µ′ε′.


According to eqn (18) the 3 waves are:

INCIDENT

~E = ~E0ei~k·~x−iωt , ~B =

√µε~k × ~E

k(30)

REFRACTED

~E ′ = ~E ′0ei~k′·~x−iωt , ~B ′ =

√µ′ε′

~k ′ × ~E ′

k ′(31)

REFLECTED

~E ′′ = ~E ′′0 e i~k′′·~x−iωt , ~B ′ =√µ′ε′

~k ′′ × ~E ′′

k ′′(32)

The wave numbers have magnitudes:

|~k| = |~k ′′| = k =ω

c

√µε , |~k ′| = k ′ =

ω

c

√µ′ε′ (33)


AT the boundary z = 0 the BC must be satisfied at all points on theplane at all times, i.e. the spatial & time variation of all fields must bethe same at z = 0.Thus the phase factors must be equal at z = 0(

~k · ~x)

z=0=(~k ′′ · ~x

)z=0

=(~k ′ · ~x

)z=0

(34)

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 figurewe get

k sin i = k ′′ sin r ′ = k ′ sin r (35)

Since k = k ′′, we find that i = r ′; the angle of incidence equals the angleof reflection.

Snell’s law is:

sin i

sin r=

k ′

k=

√µ′ε′

µε=

n′

n(36)


Reflection & Refraction of EM Waves

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

ε(~E0 + ~E ′′0

)− ε′~E ′0

]· ~n = 0[

~k × ~E0 + ~k ′′ × ~E ′′0 − ~k ′ × ~E ′0]· ~n = 0(

~E0 + ~E ′′0 − ~E ′0)× ~n = 0 (37)[

1

µ

(~k × ~E0 + ~k ′′ × ~E ′′0

)− 1

µ′

(~k ′ × ~E ′0

)]× ~n = 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.


~E : Perpendicular to the plane of incidence

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

E0 + E ′′0 − E ′0 = 0√ε

µ(E0 − E ′′0 ) cos i −

√ε′

µ′E ′0 cos r = 0 (38)

• The 2nd, using Snell’s law, duplicatesthe 3rd.(prove all the above statements)

Figure: Reflection and refractionwith polarization perpendicular tothe plane of incidence. All the~E -fields shown directed away fromthe viewer.


~E : Perpendicular to the plane of incidence

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

E ′0E0

=2n cos i

n cos i + µµ′

√n′2 − n2 sin2 i

=2

1 + µ tan iµ′ tan r

=

[2 sin r cos i

sin(i + r)

]µ=µ′

E ′′0E0

=n cos i − µ

µ′

√n′2 − n2 sin2 i

n cos i + µµ′

√n′2 − n2 sin2 i

=1− µ tan i

µ′ tan r

1 + µ tan iµ′ tan r

=

[sin(r − i)

sin(i + r)

]µ=µ′

(39)

Note that√

n′2 − n2 sin2 i = n′ cos r but Snell’s law has been used toexpress it in terms of the angle of incidence.

For optical frequencies it is usually permitted to put µ = µ′.


~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

(E0 − E ′′0 ) cos i − E ′0 cos r = 0√ε

µ(E0 + E ′′0 )−

√ε′

µ′E ′0 = 0 (40)

Figure: Reflection and refraction withpolarization parallel to the plane ofincidence.


~E : Parallel to the plane of incidence

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

E ′0E0

=2nn′ cos i

µµ′ n′2 cos i + n

√n′2 − n2 sin2 i

=2 n′ε

nε′

1 + ε tan iε′ tan r

=

[2 sin r cos i

sin(i + r) cos(i − r)

]µ=µ′

E ′′0E0

=

µµ′ n′2 cos i − n

√n′2 − n2 sin2 i

µµ′ n′2 cos i + n

√n′2 − n2 sin2 i

=1− ε tan i

ε′ tan r

1 + ε tan iε′ tan r

=

[tan(i − r)

tan(i + r)

]µ=µ′

(41)


Normal incidence i = 0

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

E ′0E0

=2

1 +√

µε′

µ′ε

→ 2n

n′ + n

E ′′0E0

=−1 +

√µε′

µ′ε

1 +√

µε′

µ′ε

→ n′ − n

n′ + n(42)

EXERCISES:What are the conditions for:

I Total reflection

I Total transmision


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