Electromagnetic Waves

Electromagnetic Waves 

May 4, 2010 1 

1 J.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 

∇ · E = 0 , ∇ × E 

1 ∂ 

+ 

c 

B 

∂t 

∇ · B = 0 , ∇ × B − µɛ 

c 

where D and H are given by relations 

∂ E 

∂t 

= 0 

= 0 (1) 

D = ɛ E and H = 1 

µ B (2) 

where ɛ is the electric permittivity and µ the magnetic permeability 

which assumed to be independent of the frequency. 


Plane Electromagnetic Waves 

Maxwell’s equations can be written as 

∇ 2 

µɛ 

B − 

c2 ∂2 B 

∂t 2 = 0 and ∇2 µɛ 

E − 

c2 ∂2 E 

= 0 (3) 

∂t 2 

In other words each component of B and E obeys a wave equation of 

the form: 

∇ 2 u − 1 

v 2 

∂2u = 0 

∂t 2 where 

c 

v = √ 

µɛ 

(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 

E(x, t) = Ee ikn·x−iωt 

and B(x, t) = Be ikn·x−iωt 

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

k = ω 

v = √ µɛ ω 

c 

or k · k = 

ω 

v 

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

2 


(5) 

(6) 

(7)

If we consider waves propagating in one direction, say x-direction then 

the fundamental solution is: 

u(x, t) = Ae ik(x−vt) + Be −ik(x+vt) 

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 

(8) 

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 

B = √ µɛn × E and E = − 1 

√µɛ n × B (10) 

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. 


It is then useful to introduce a set of real mutually orthogonal unit 

vectors (ɛ1,ɛ2,n). 

In terms of these unit vectors the field 

strengths E and B are 

or 

E = ɛ1E0 , B 

√ 

= ɛ2 µɛE0 

E = ɛ2E ′ 0 , B 

√ ′ 

= −ɛ1 µɛE 0 

(11) 

(12) 

E0 and E ′ 0 are constants, possibly complex. 

In other words the most general way to write the electric/magnetic field 

vector is: 

E = (E0 ɛ1 + E ′ 0 ɛ2)e ikn·x−iωt 

B = √ µɛ(E0 ɛ2 − E ′ 0 ɛ1)e ikn·x−iωt 


(13) 

(14)

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 

S = 

2 4π E × H ∗ = 1 c 

2 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 

2 Note : 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. 


The time averaged flux of energy is: 

 

 

c ɛ 

S = 

8π µ |E0| 2n (15) 

The total time averaged density (and not just the energy density 

associated with the electric field component) is: 

u = 1 

 

ɛ 

16π 

E · E ∗ + 1 

µ B · B ∗ 

 

= ɛ 2 

|E0| 

8π 

(16) 

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

v = c/ √ µɛ. 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? 

3 Note: To prove the above relations use 〈cos 2 x〉 = 1/2 and since 

ER = ( E + E ∗ )/2 we get 〈 E 2 R〉 = E · E ∗ /2. 


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 ɛ1. Such a wave is said to be linearly polarized with 

polarization vector ɛ1. The wave described by (12) is linearly polarized 

with polarization vector ɛ2 and is linearly independent of the first. 

The two waves : 

E1 = ɛ1E1e i k·x−iωt , E2 = ɛ2E2e i k·x−iωt 

with (17) 

Bi = √ µɛ k × Ei 

, i = 1, 2 

k 

Can be combined to give the most general homogeneous plane waves 

propagating in the direction k = kn, 

E(x, t) = (ɛ1E1 + ɛ2E2) e i k·x−iωt 

E(x, t) = 

(18) 

 

ɛ1|E1| + ɛ2|E2|e i(φ2−φ1) 

e ik·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. 


LINEARLY POLARIZED 

If the amplitudes E1 = |E1|e iφ1 and E2 = |E2|e iφ2 have the same phase 

(18) represents a linearly polarized wave with the polarization vector 

making an angle θ = tan −1 (ℜ(E2)/ℑ(E1)) (which remains constant as 

the field evolves in space and time) with ɛ1 and magnitude 

E = E 2 1 + E 2 2 . 

ELLIPTICALLY POLARIZED 

If 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 (ɛ1 ± iɛ2) e i k·x−iωt 

(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 (ɛ1 + iɛ2) 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 (ɛ1 − iɛ2) 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 the 

complex orthogonal vectors 

with properties 

ɛ± = 1 

√ 2 (ɛ1 ± iɛ2) (22) 

ɛ ∗ ± · ɛ∓ = 0 , ɛ ∗ ± · ɛ3 = 0 , ɛ ∗ ± · ɛ± = 1 . (23) 

Then the general representation of the electric vector 

E(x, t) = (E+ɛ+ + E−ɛ−) e i k·x−iωt 

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 ɛ1 and ɛ2. 

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

E−/E+ = r. 


(24)

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

E−/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 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. 


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 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 

ɛ1 · E , ɛ2 · E , ɛ ∗ + · E , ɛ ∗ − · 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 


In terms of the linear polarization bases (ɛ1, ɛ2), the Stokes parameters 

are: 

s0 = |ɛ1 · E| 2 + |ɛ2 · E| 2 = a 2 1 + a 2 2 

s1 = |ɛ1 · E| 2 − |ɛ2 · E| 2 = a 2 1 − a 2 2 

 

s2 = 2ℜ (ɛ1 · E) ∗ (ɛ1 · 

E) = 2a1a2 cos(δ1 − δ2) (26) 

 

s3 = 2ℑ (ɛ1 · E) ∗ (ɛ1 · 

E) = 2a1a2 sin(δ1 − δ2) 

where we defined the coefficients of (18) or (24) as magnitude times a 

phase factor: 

E1 = a1e iδ1 , 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 (ɛ+, ɛ−), the Stokes parameters 

are: 

s0 = |ɛ ∗ + · E| 2 + |ɛ ∗ − · E| 2 = a 2 + + a 2 − 

 

s1 = 2ℜ (ɛ ∗ + · E) ∗ (ɛ ∗ − · 

E) = 2a+a− cos(δ− − δ+) (28) 

 

s2 = 2ℑ (ɛ ∗ + · E) ∗ (ɛ ∗ − · 

E) = 2a+a− sin(δ− − δ+) 

s3 = |ɛ ∗ + · E| 2 − |ɛ ∗ − · E| 2 = a 2 + − a 2 − 

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 

s 2 0 = s 2 1 + s 2 2 + s 2 3 . (29) 


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 

◮ Kinematic properties: 

◮ Angle of reflection = angle of incidence 

◮ sin i n′ 

Snell’s law: sin r = n where i, r are the angles of incidence 

and refraction, while n, n ′ are the corresponding indices of 

refraction. 

◮ Dynamic properties: 

◮ Intensities of reflected and refracted radiation 

◮ 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. 


Figure: Incident wave k strikes plane interface between different media, 

giving rise to a reflected wave k ′′ and a refracted wave k ′ . The media 

below and above the plane z = 0 have permeabilities and dielectric 

constants µ, ɛ and µ’, ɛ’ respectively. The indices of refraction are 

n = √ µɛ and n ′ = √ µ ′ ɛ ′ . 


According to eqn (18) the 3 waves are: 

INCIDENT 

REFRACTED 

REFLECTED 

E = E0e i k·x−iωt , B = √ µɛ k × E 

k 

E ′ = E ′ 0e i k ′ ·x−iωt , B ′ = µ ′ ɛ ′ k ′ × E ′ 

E ′′ = E ′′ 

0 e i k ′′ ·x−iωt , B ′ = µ ′ ɛ ′ k ′′ × E ′′ 

k ′′ 

The wave numbers have magnitudes: 

| k| = | k ′′ | = k = ω 

c 

k ′ 

(30) 

(31) 

(32) 

√ 

µɛ , | ′ ′ ω 

k | = k = µ ′ ɛ ′ (33) 

c 


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 = k ′′ 

· x = k ′ 

· x 

(34) 

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 = k ′′ sin r ′ = k ′ sin r (35) 

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

of reflection. 

Snell’s law is: 

sin i 

sin r 

= k′ 

k = 

z=0 

 

µ ′ ɛ ′ 

µɛ 

= n′ 

n 


z=0 

(36)

Reflection & Refraction of EM Waves 

The dynamic properties are contained in the boundary conditions : 

• normal components of D = ɛ 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: 

 

ɛ E0 + E ′′ 

 

0 − ɛ ′ 

 

′ 

E 0 · n = 0 

 

k × E0 

+ k ′′ × E ′′ 

0 − k ′ × E ′ 

0 · n = 0 

 

E0 + E ′′ 

0 − E ′ 

0 × n = 0 (37) 

 

1 

 

k × E0 

+ 

µ 

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 the 

surface the 1st BC of (38) yields nothing 

• The 3rd and 4th of of (38) give 

(how?): 

E0 + E ′′ 

0 − E ′ 0 = 0 

 

ɛ 

µ (E0 − E ′′ 

 

ɛ 

0 ) cos i − 

′ 

µ ′ E ′ 0 cos r = 0 (38) 

• The 2nd, using Snell’s law, duplicates 

the 3rd. 

(prove all the above statements) 

Figure: Reflection and refraction 

with polarization perpendicular to 

the plane of incidence. All the 

E-fields shown directed away from 

the viewer. 


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 

= 

E0 n cos i + µ 

µ ′ 

 

n ′2 − n2 sin 2 i = 

2 

µ tan i 

1 + µ ′ 

2 sin r cos i 

= 

sin(i + r) 

tan r 

µ=µ ′ 

E ′′ 

0 

= 

E0 

n cos i − µ 

µ ′ 

 

n ′2 − n2 sin 2 i 

n cos i + µ 

µ ′ 

 

n ′2 − n2 sin 2 µ tan i 

1 − µ 

= 

i ′ tan r 

µ tan i 

1 + µ ′ 

sin(r − i) 

= 

sin(i + r) 

tan r 

µ=µ ′ 

(39) 

 

Note that n ′2 − n2 sin 2 i = n ′ 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 = 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 

 

ɛ 

µ (E0 + E ′′ 

 

ɛ 

0 ) − 

′ 

µ ′ E ′ 0 = 0 (40) 

Figure: Reflection and refraction with 

polarization parallel to the plane of 

incidence. 


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?) 

E ′ 0 

E0 

E ′′ 

0 

E0 

= 

= 

= 

= 1 − ɛ tan i 

2nn ′ cos i 

µ 

µ ′ n ′2 

cos i + n n ′2 − n2 sin 2 i 

2 n′ ɛ 

nɛ ′ 

ɛ tan i 1 + ɛ ′ 

 

2 sin r cos i 

= 

sin(i + r) cos(i − r) 

tan r 

µ 

µ ′ n ′2 

cos i − n n ′2 − n2 sin 2 i 

µ 

µ ′ n ′2 

cos i + n n ′2 − n2 sin 2 i 

ɛ ′ 

tan r tan(i − r) 

= 

1 + tan(i + r) 

ɛ tan i 

ɛ ′ tan r 

µ=µ ′ 


µ=µ ′ 

(41)

Normal incidence i = 0 

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

E ′ 0 

E0 

E ′′ 

0 

E0 

= 

= 

EXERCISES: 

What are the conditions for: 

◮ Total reflection 

◮ Total transmision 

1 + 

2 

−1 + 

1 + 

 

µɛ ′ 

µ ′ ɛ 

 

µɛ ′ 

µ ′ ɛ 

 

µɛ ′ 

µ ′ ɛ 

→ 2n 

n ′ + n 

→ n′ − n 

n ′ + n 


(42)

Electromagnetic Waves

Create successful ePaper yourself

Delete template?

Save as template?