Lecture 3:
Maxwell equations and electromagnetic wave
propagation
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Plane waves
Phase velocity
Polarization
Maxwell equations
Wave equations
Solutions of wave equations
Poynting vector
Fresnel reflection
Total internal reflection
Evanescent wave
Reading: Senior 2.3
Keiser 2.1-2.2
1
Plane linearly polarized waves
•  The electric or magnetic field of a plane linearly
polarized wave traveling in a direction k can be
represented in the general form
A(r, t) = eiA0 exp i(k ● r – ωt)
where r = xex + yey + zez represents a general
position vector and k = kxex + kyey + kzez represents
the wave propagation vector (wavevector).
The magnitude of the wavevector k is k = 2π/λ, which
is known as the wave propagation constant, λ is the
wavelength of the light in a vacuum or free space
ω = 2πυ, where υ is the frequency of the light, υ = c/λ
A0 is the maximum amplitude of the wave
2
Plane waves
y
r
exp i(kr – ωt) @ a particular time t
k
Wave vector k perpendicular to the
plane wavefront
λ
kr = const. + 2π
spatial phase kr = const.
x
3
•  The components of the actual (measurable) electromagnetic field
are obtained by taking the real part of the complex exponential form.
•  For example, if k = kez, and A denotes the electric field E in the
ei = ex direction, then the real measurable electric field is given by
Ex(z, t) = Re (E) = ex E0x cos (kz – ωt)
which represents a plane wave that varies harmonically as it travels
in the z direction.
y
wavefronts (⊥ k)
E-field
polarization (x)
(plane wave
in free space)
z
λ
k = ez k = ez 2π/λ
(wavevector)
4
Phase velocity
•  For a plane optical wave traveling in the z direction, the electric field
has a phase varies with z and t
φ = kz – ωt
For a point of constant phase on the space- and time-varying field,
φ  = constant and thus kdz – ωdt = 0. If we track this point of constant
phase, we find that it is moving with a velocity of
vp = dz/dt = ω/k
phase velocity
•  In free space, the phase velocity vp = c = ω/k = υλ
the propagation constant k = ω/c
(c = 3 × 108 m/s)
5
•  For a wave propagating in a dielectric medium of refractive index n
vp = c/n = ω/k
Propagation constant in a
k = n ω/c
medium of refractive index n
•  The propagation constant in any medium
k = n ko = n 2π/λο
free space
free space
wavenumber wavelength
6
Fields in a linearly polarized plane wave
H
E
H
E
k
k
•  Maxwell equations show that E and H are both perpendicular to
the direction of propagation. Such a wave is called a transverse wave.
•  Furthermore, E and H are mutually perpendicular – E, H, and k form
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a mutually orthogonal triad.
General state of polarization
•
A general state of polarization is described by considering another
linearly polarized wave which is independent of the first wave and
orthogonal to it.
Ey (z, t) = ey E0y cos (kz - ωt + δ)
where δ is the relative phase difference between the waves. The
resultant wave is
E(z, t) = Ex(z, t) + Ey(z, t)
•
If δ is zero or an integer multiple of 2π, the waves are in phase. The
wave is then also a linearly polarized wave with polarization vector
making an angle
E0y
E
θ = tan-1 E0y/E0x
with respect to ex and having a magnitude
E = (E0x2 + E0y2)1/2
•
θ
E0x
Conversely, an arbitrary linearly polarized wave can be resolved into
two independent orthogonal plane waves that are in phase.
8
Elliptical and circular polarization
•  For general values of δ the wave is elliptically polarized. The
resultant field vector E both rotates and changes its magnitude
as a function of the angular frequency ω. For a general value of
δ
(Ex/E0x)2 + (Ey/E0y)2 – 2(Ex/E0x) (Ey/E0y) cosδ = sin2δ
which is the general equation of an ellipse.
•  This ellipse represents the trajectory of the E vector = state of
polarization (SOP)
•  The axis of the ellipse makes an angle φ relative to the x axis
given by
tan 2φ = 2E0xE0y cosδ / (E0x2 – E0y2)
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•  To get a better picture, let us align the principal axis of the ellipse
with the x axis. Then φ = 0, or equivalently, δ = ±π/2, ±3π/2, …, so that
(Ex/E0x)2 + (Ey/E0y)2 = 1
This is the familiar equation of an ellipse.
•  When E0x = E0y = E0 and the relative phase difference δ = ±π/2,
±3π/2, …, then we have circularly polarized light.
Ex2 + Ey2 = E02
which defines a circle
•  Choosing the positive sign of δ,
Ex (z, t) = ex E0 cos(kz - ωt)
Ey (z, t) = -ey E0 sin(kz - ωt)
•  In this case, the endpoint of E traces out a circle at a given point in space.
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•  Consider an observer located at some arbitrary point toward which the wave is
approaching. For convenience, we choose this point at z = π/k at t = 0.
Ex (z, t) = -ex E0, Ey (z, t) = 0
E lies along the –x axis.
At a later time, say t = π/2ω, the electric field vector has rotated through 90o and now lies
along the -y axis. Thus, as the wave moves toward the observer with increasing time, E
rotates counterclockwise at an angular frequency ω. It makes one complete rotation as
the wave advances through one wavelength. Such a light wave is right circularly
polarized.
•  If we choose the negative sign for δ, then the electric field vector is given by
E = E0 [ex cos(kz - ωt) + ey sin(kz – ωt)]
Now E rotates clockwise and the wave is left circularly polarized.
H
k
LHC
H
E
k
RHC
E
11
Electromagnetic fields
•  The electromagnetic field in a medium is generally characterized by
the following four field quantities:
Electric field
Electric displacement
Magnetic intensity
Magnetic field
or magnetic induction
V m-1
E(r, t)
D(r, t)
H(r, t)
C m-2
A m-1
B(r, t)
Wb m-2 or T
(The units are in SI units)
(coulomb C = A•s)
(weber Wb = V•s)
•  E and H are the electric and magnetic field vectors, and D and B are
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the corresponding flux densities.
Maxwell equations
Faraday’s law of induction
curl
equations
∇  × E = -∂B/∂t
divergence
conditions
∇  ● D = ρ
Gauss’ law for the electric field
∇  ● B = 0
Gauss’ law for the magnetic field
(No free poles)
∇  × H = J + ∂D/∂t
Ampere-Maxwell law
where the current density vector J (= σE, σ is the conductivity)
and the charge density ρ represent the sources for the electromagnetic
field.
13
Constitutive relations
•  The flux densities (arise in response to E and H propagating
inside the medium) are related to the field vectors by the
constitutive relations
D(r, t) = εoE(r, t) + P(r, t)
B(r, t) = µoH(r, t) + µ0M(r, t)
where εo = 8.854 × 10-12 F m-1 ≈ (1/36π) × 10-9 F m-1 is the
electric permittivity of free space, µo = 4π × 10-7 H m-1 is the
magnetic permeability of free space,
polarization (electric polarization)
magnetization (magnetic polarization)
P(r, t) C m-2
M(r, t) A m-1
•  For nonmagnetic materials such as silica glass, M = 0, and
thus B = µ0H.
14
Electric susceptibility
•  The relation between E and P is through the electric
susceptibility function χe.
P(r, t) = ε0χe(k, ω) E(r, t)
D = εo (1+ χe(k, ω)) E(r, t) = ε0 εr(k, ω) E(r, t) = ε(k, ω) E(r, t)
where the dielectric constant (relative permittivity) εr is
defined as 1+χe, and the permittivity of the medium ε = εr ε0.
•  For isotropic medium, χe and εr are scalars so that E // P and
D // E. (∇·E = (1/ε) ∇·D = 0 in free space)
•  In general, χe and εr are second-rank tensors (expressed in 3×3
matrices), in which case the medium they describe is
anisotropic. (E not // P, D not // E, in general ∇·E ≠ 0)
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•  The four field vectors are related by the relations:
D = εE
B = µH
where ε is the dielectric permittivity and µ is the magnetic
permeability of the medium
ε = ε0 εr
µ = µ0 µr
where µr and εr are the relative permeability and relative
permittivity for the dielectric medium, and µ0 and ε0 are
the permeability and permittivity of free space
(for non-magnetic materials, µr = 1)
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Maxwell equations in a medium with no sources
•  The basis for the study of electromagnetic wave
propagation is provided by Maxwell equations.
•  For a medium with zero conductivity these vector
relationships may be written in terms of the electric field
E, magnetic intensity H, electric flux density D and
magnetic flux density B as the curl equations and the
divergence conditions:
Curl equations
∇  × E = -∂B/∂t
∇  × H = ∂D/∂t
Divergence equations
∇  ● D = 0
(no free charges)
∇  ● B = 0
(no free poles)
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Wave equations
•  Maxwell equations can be used to obtain the wave
equations that describe light propagation.
∇  × (∇ × E) = -µε ∂2E/∂t2
∇  × (∇ × H) = -µε ∂2H/∂t2
•  Using the vector identity ∇ × (∇ × ) ≡ ∇ (∇ • ) - ∇2( )
•  For homogeneous (source free), isotropic media, ∇·E = (1/ε) ∇·D = 0
We obtain the wave equations:
∇2E = µε ∂2E/∂t2
and
∇2H = µε ∂2H/∂t2
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Scalar wave equations
•  For rectangular Cartesian and cylindrical polar
coordinates the above wave equations hold for each
component of the field vector, every component
satisfying the scalar wave equation:
∇2ψ = µε ∂2ψ/∂t2
where ψ may represent a component of the E or H field
•  If planar waveguides, described by rectangular Cartesian
coordinates (x, y, z), or circular fibers, described by
cylindrical polar coordinates (r, φ, z), are considered,
then the Laplacian operator takes the form:
∇2ψ = ∂2ψ/∂x2 + ∂2ψ/∂y2 + ∂2ψ/∂z2
or
∇2ψ = ∂2ψ/∂r2 + (1/r) ∂ψ/∂r + (1/r2) ∂2ψ/∂φ2 + ∂2ψ/∂z219
Uniform plane waves as a basic solution
•  The basic solution of the wave equation is a harmonic wave.
The most important form of which is a uniform plane wave:
ψ = ψo exp i(k • r – ωt)
where ω is the angular frequency of the field, t is the time,
k is the propagation vector which gives the direction of
propagation and the rate of change of phase with distance, the
components of r specify the coordinate point at which the field is
observed
λ is the optical wavelength in a vacuum. The magnitude of the
propagation vector or the vacuum phase propagation constant k
(where k = |k|) is given by k = 2π/λ
(k is also known as the free space wave number)
20
To see the plane wave as a solution to the wave equation:
Consider a plane wave propagating in free space in the z direction,
E = Eo exp i(kz - ωt)
1-D wave equation
∂2E/∂z2 = µε ∂2E/∂t2 = µ0ε0 ∂2E/∂t2
k2 E = µ0ε0 ω2 E
k2/ω2 = µ0ε0
k2/ω2 = (2π/λ)2 / (2πυ)2 = 1/(λυ)2 = 1/c2 = µ0ε0
21
Phase velocity in free space
•  It is known that εo = 8.854 x 10-12 F m-1 ≈ (1/36π) × 10-9 F m-1
is the electric permittivity of free space, µo = 4π × 10-7 H m-1 is
the magnetic permeability of free space,
(µ0 ε0)-1/2 = (4π × 10-7 × (1/36π) × 10-9)-1/2 = 3 × 108 (F·Hm-2)-1/2
[F = C/V = A· s/V; H = Wb/A = V· s/A; F·H = s2]
•  Maxwell realized that light is an electromagnetic wave from the fact
that the phase velocity derived from the wave equation agrees with the
measured speed of light.
22
Phase velocity in dielectric media
vp = 1/√(µ0ε) = 1 /√(µ0ε0εr)
•  The velocity of light in a dielectric medium is therefore
vp = c /√εr
where we used the relation µ0ε0 = 1/c2 and c is the speed of light.
vp = c / n
n = √εr
⇒ The refractive index n is rooted in the material relative
permittivity.
23
Transverse waves
•  For plane waves in a non-conducting isotropic medium the Maxwell
equations take the following forms
H
k × E = ωµ H
k × H = -ωε E
k●E=0
k●H=0
E
k
•  The three vectors k, E, and H constitute a mutually orthogonal triad.
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Vector Helmholtz equations
•  Using the operator form ∂/∂t = -iω, we find that the wave
equations become
∇2E + k2E = 0
∇2H + k2H = 0
where k = ω√(µ0ε) is the wavenumber, having units of m-1.
•  These two equations are known as the vector Helmholtz
equations.
•  They form the starting point for the analysis of all types of
waveguides that are constructed from linear, homogeneous,
and isotropic materials.
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Energy flow and the Poynting vector
•  Poynting’s theorem states that the time rate of flow of
electromagnetic energy per unit area (or optical power per unit area)
is given by the vector S, called the Poynting vector,
S=E×H
This vector specifies both the direction and the magnitude of the energy
flux. (watts per square meter)
•  Consider the case of plane harmonic waves in which the fields are
given by the real expressions
E = Eo cos (k • r - ωt)
H = Ho cos (k • r - ωt)
26
•  For the instantaneous value of the Poynting vector
S = Eo × Ho cos2 (k • r - ωt)
•  As the average value of the cosine squared is ½, then for the
average value of the Poynting vector
<S> = ½ Eo × Ho
*if the complex exponential form of the wave function for E and H
is used, the average Poynting flux can be expressed as ½ Eo × Ho*
•  As the wavevector k is perpendicular to both E and H, k has the
same direction as the Poynting vector S.
27
Relationships among the directions of E, D, H, B, k,
and S in free space or in simple media
H // B
E // D
k // S
•  In isotropic media the direction of the energy flow is specified
by the direction of S and is the same as the direction of the
wavevector k. (In anisotropic media (e.g. in certain crystals and
semiconductors) S and k are not always in the same direction.) 28
Intrinsic impedance √(µ0/ε)
•  The electric and magnetic field amplitudes are related by the
intrinsic impedance Z = √(µ0/ε). (in units of Ω)
•  In free space, it becomes Zo = √(µ0/ε0) ≈ 120π Ω ≈ 377 Ω. We
can express Z = Z0/n.
•  Consider a uniform plane wave that propagates in the +z
direction. Suppose the polarization is along x, so that its
phasor form is E = axE = ax E0 exp(-ikz), where ax is a unit
vector along x. Apply the curl E equation
∇  x E = (∂E/∂z) ay = -ikE ay = -iωµ0H
⇒ H = H0 exp(-ikz) ay, where E0/H0 = ωµ0/k = √(µ0/ε) ≡ Z.
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•  An alternative expression for the average Poynting flux is
<S> = I k/k
unit vector in the
magnitude of the
average Poynting flux direction of propagation
•  I is called the irradiance or intensity, given by
I = ½ EoHo = (n/2Zo) |Eo|2 ∝ |Eo|2
[W/cm2] = [V2/(Ω·cm2)] = [1/Ω] [V/cm]2
•  Thus the rate of flow of energy is proportional to the square of the
amplitude of the electric field. Z0 is the intrinsic impedance of free
space in units of Ω.
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Reflection and refraction
•  We now review the phenomena of reflection and refraction of light
from the standpoint of electromagnetic theory.
•  Consider a plane harmonic wave incident upon a plane boundary
separating two different optical media.
ki
incident
kr
θi θr reflected
θt transmitted
kt
*The space-time dependence of these
three waves, aside from constant
amplitude factors, is given by
exp i(ki•r – ωt)
exp i(kr•r – ωt)
exp i(kt•r – ωt)
incident
reflected
transmitted
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•  The condition that
ki • r = kr • r = kt • r
is required at the interface for the boundary conditions to be satisfied
at all points along the interface at all times.
•  This implies that the three wavevectors lie in the same plane known
as the plane of incidence.
•  The projections of these three wavevectors on the interface are all
equal so that
ki sin θi = kr sin θr = kt sin θt
where θi is the angle of incidence, θr is the angle of reflection and
θt is the angle of refraction.
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•  Because ki = kr and ki/kt = n1/n2
θi = θr
(Law of reflection)
n1 sin θi = n2 sin θt
(Snell’s law)
Boundary conditions
Medium 1
Medium 2
n × H1
n × H2
B1n
D1n
B2n
D2n
n × E1
n
n × E2
Because B = µoH for optical fields, the tangential component of B and the
normal component of H are also continuous. All of the magnetic field
components in an optical field are continuous across a boundary. Possible
discontinuities in an optical field exist only in the normal component of E or the
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tangential component of D.
TE polarization (s-wave)
•  The electric field is linearly polarized in a direction perpendicular
to the plane of incidence, while the magnetic field is polarized parallel to
the plane of incidence. This is called the transverse electric (TE)
polarization. This wave is also called s-polarized.
n
Hi
ki
n1
n2
x
Ei
Hr
θi
x
Er
θr
θt
kr
x
Et
Ht
kt
34
•  The reflection coefficient, r, and the transmission coefficient, t,
of the TE electric field are given by the following Fresnel equations:
rs ≡ Er/Ei =
ts ≡ Et/Ei =
n1 cos θi – n2 cos θt
n1 cos θi + n2 cos θt
2n1 cos θi
n1 cos θi + n2 cos θt
=
=
n1 cos θi - (n22 – n12 sin2θi)1/2
n1 cos θi + (n22 – n12 sin2θi)1/2
2n1 cos θi
n1 cos θi + (n22 – n12 sin2θi)1/2
•  The intensity reflectance and transmittance, R and T, which are also
known as reflectivity and transmissivity, are given by
Rs ≡ Ir/Ii =
n1 cos θi – n2 cos θt 2
n1 cos θi + n2 cos θt
Ts ≡ It/Ii = 1 - Rs
35
TM polarization (p-wave)
•  The electric field is linearly polarized in a direction parallel to the
plane of incidence while the magnetic field is polarized perpendicular
to the plane of incidence. This is called the transverse magnetic (TM)
polarization. This wave is also called p-polarized.
n
Ei
ki
n1
•
Hi
Er
θi
θr
kr
•
Hr
n2
θt
Ht
Et
•
kt
36
•  The reflection coefficient, r, and the transmission coefficient, t,
of the TM electric field are given by the following Fresnel equations:
rp ≡ Er/Ei =
tp ≡ Et/Ei =
-n2 cos θi + n1 cos θt
n2 cos θi + n1 cos θt
2n1 cos θi
n2 cos θi + n1 cos θt
=
-n22 cos θi + n1(n22 – n12 sin2θi)1/2
=
n22 cos θi + n1(n22 – n12 sin2θi)1/2
2n1n2 cos θi
n22 cos θi + n1(n22 – n12 sin2θi)1/2
•  The intensity reflectance and transmittance for TM polarization are
given by
Rp ≡ Ir/Ii =
-n2 cos θi + n1 cos θt 2
n2 cos θi + n1 cos θt
Tp ≡ It/Ii = 1 - Rp
37
Total Internal Reflection
For θi > θc
sin θi > n2/n1
|rs| =
|rp| =
n1 cos θi - i (n12 sin2θi - n22)1/2
n1 cos θi + i (n12 sin2θi - n22)1/2
-n22 cos θi + i n1(n12 sin2θi – n22)1/2
2
n2 cos θi + i
2
n1(n1 sin2θi
– n2
2)1/2
=1
=1
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total internal reflection
for θ > θc
n1 = 1.5 (internal
n2 = 1.0 reflection)
θB ~ 34o (Brewster angle)
θc ~ 42o
Rp
Rs
θB
θc
39
Phase changes in total internal reflection
•  In the case of total internal reflection the complex values for the
coefficients of reflection, given by the Fresnel coefficients rs and rp,
imply that there is a change of phase which is a function of the angle
of incidence.
•  As the absolute values of rs and rp are both unity, we can write
rs = ae-iα / aeiα = exp –iϕs rp = -be-iβ / beiβ = -exp -iϕp
where ϕs and ϕp are the phase changes for the TE and TM cases, and
the complex numbers ae-iα and –be-iβ represent the numerators in
rs and rp. Their complex conjugates appear in the denominators.
40
2
2
2 1/2
ae-iα = n1 cos θi - i (n1 sin θi - n2 )
be-iβ
2 cos θ - i n (n 2 sin2θ – n 2)1/2
n
i
1 1
i
2
= 2
•  We see that ϕs = 2α and ϕp = 2β. Accordingly, tan α = tan (ϕs/2) and
tan β = tan (ϕp/2).
•  We therefore find the following expressions for the phase changes that
occur in internal reflection:
tan (ϕs/2) = (n12 sin2θi - n22)1/2 / (n1 cos θi)
tan (ϕp/2) = n1(n12 sin2θi – n22)1/2 / (n22 cos θi)
41
Phase change (radian)
Total internal reflection phase shifts
n1 = 1.5, n2 = 1
ϕp
ϕs
Angle of incidence
•  Useful for our discussion on the phase-matching condition of
42
waveguide modes
Evanescent wave
•  In spite of the fact that the incident energy is totally reflected when
the angle of incidence exceeds the critical angle, there is still an
electromagnetic wave field in the region beyond the boundary. This
field is known as the evanescent wave.
•  Its existence can be understood by consideration of the wave
function of the electric field of the transmitted wave:
Et = Et exp i (kt • r – ωt)
Choose the coordinate axis such that the plane of incidence is on the xy
plane and the boundary is at y = 0.
43
exp (-κy) exp i((ki sin θi)x – ωt)
y wavefronts
λ/(n1sin θi)
vp = ω/(ki sinθi)
n2
x
n1
λ/n1
Ei
θi > θc
kt • r = kt x sin θt + kt y cos θt
Er
total
internal
reflection
= kt x (n1/n2) sin θi + kt y (1 – (n1/n2)2 sin2 θi)1/2
= ki x sin θi + i kt y ((n12sin2θi/n22) – 1)1/2
44
•  The wave function for the electric field of the evanescent wave is
Eevan = Et exp (-κy) exp i ((ki sin θi)x - ωt)
where
κ = kt ((n12sin2θi/n22) – 1)1/2
•  The factor exp (-κy) shows that the evanescent wave amplitude drops
off very rapidly in the rarer medium as a function of distance from the
boundary.
•  The oscillatory term exp i ((ki sin θi) x - ωt) indicates that the
evanescent wave can be described in terms of surfaces of constant phase
moving parallel to the boundary with phase velocity ω/(ki sin θi).
•  The evanescent field stores energy and transports it in the direction of
surface propagation, but does not transport energy in the transverse
direction. Therefore, evanescent wave is also known as surface wave.
45
Evanescent wave amplitude normal to the
interface drops exponentially
e-κy
y
n2=1
λ = 600 nm
θi n1=1.5
θi= 42o ≈ θc
1/e
θi= 60o
θi= 44o
Position y (nm)
46