PHYS 342: Electromagnetic Waves
Guillaume Gervais
2007
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Contents
1 Review of Electrostatics and Magnetostatics 6
1.1 Coulomb’s Law and Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.1 Coulomb Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.2 The Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.3 Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.1.4 Energy of the Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.1.5 Electrostatics and Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.6 Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.2 Electric Field in Matter and Electric Dipole . . . . . . . . . . . . . . . . . . . . . . . 12
1.2.1 Multipole expansion (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.2.2 Energy of the electric dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.2.3 Dialectrics and Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.2.4 Electric Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.3 Magnetostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.3.1 Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.3.2 Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.3.3 Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.3.4 Magnetic Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.3.5 Magnetic Fields in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.3.6 Historical Aside: Etymology and William Gilbert (1544-1603) . . . . . . . . . 23
1.4 Maxwell’s Equations for Static Electromagnetism . . . . . . . . . . . . . . . . . . . . 24
2 Electrodynamics 25
2.1 Faraday Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.1.1 Electromotive Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.1.2 Faraday Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
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2.1.3 Historical Aside: Michael Faraday (1791-1867) . . . . . . . . . . . . . . . . . 27
2.2 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.2.1 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.3 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.3.1 Displacement Current (In Vacuum) . . . . . . . . . . . . . . . . . . . . . . . . 30
2.3.2 Maxwell’s Equations (with dynamics) . . . . . . . . . . . . . . . . . . . . . . 32
2.3.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.4 Electromagnetic Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.4.1 Potentials for dynamic case (∂ρ∂t 6= 0) . . . . . . . . . . . . . . . . . . . . . . . 36
2.4.2 Gauge Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.4.3 Equations of motion for electromagnetic potentials . . . . . . . . . . . . . . . 39
2.4.4 Energy and Momentum in EM theory: Poynting’s Theorem . . . . . . . . . . 41
3 Electromagnetic Waves 44
3.1 Waves in Vacuum (General Case) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.1.1 Review of the Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.1.2 Sinusoidal Waves in 1D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.1.3 Boundary Conditions and Interfaces . . . . . . . . . . . . . . . . . . . . . . . 48
3.1.4 Polarization of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.1.5 Linear Combination and Power Transform . . . . . . . . . . . . . . . . . . . . 52
3.2 Electromagnetic Waves (Free Space) . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.2.1 Wave Equation for EM field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.2.2 Plane wave solution in free space . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.2.3 Polarization of EM waves (for a plane wave) . . . . . . . . . . . . . . . . . . . 56
3.2.4 Energy and Momentum of EM waves . . . . . . . . . . . . . . . . . . . . . . . 58
3.2.5 Electromagnetic Waves in Matter . . . . . . . . . . . . . . . . . . . . . . . . . 59
3.3 Reflection and Refraction at Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 61
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3.3.1 Normal Incidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.3.2 Oblique Incidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
3.3.3 Brewster angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.3.4 Total internal reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
3.4 Absorption and Dispersion in Conductors . . . . . . . . . . . . . . . . . . . . . . . . 71
3.4.1 Dispersion of a wave packet in general . . . . . . . . . . . . . . . . . . . . . . 71
3.4.2 Drude Model of the Electric Susceptibility in Matter . . . . . . . . . . . . . . 73
3.4.3 Complex Refraction Index and Anamolous Dispersion . . . . . . . . . . . . . 76
3.4.4 Dispersion and conductivity in a conductor . . . . . . . . . . . . . . . . . . . 77
3.4.5 Propagation in a Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
3.4.6 Historical Aside: James Maxwell (1821-1879) . . . . . . . . . . . . . . . . . . 81
4 Guided Waves 82
4.1 Transverse Electric and Magnetic Modes . . . . . . . . . . . . . . . . . . . . . . . . . 82
4.1.1 Helmholtz equation for the transverses components . . . . . . . . . . . . . . . 82
4.2 Electromagnetic Waves Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
4.2.1 TEM modes propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
4.3 Rectangular Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
4.3.1 TE Modes (Ez = 0, Bz 6= 0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
4.3.2 TM Modes (Ez 6= 0, Bz = 0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.3.3 Dominant modes and velocities . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.4 Circular wave guides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.4.1 Hollow circular waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.5 Electromagnetic Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.5.1 What is an EM cavity? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.5.2 Quality Factor of a Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5 Electromagnetic Radiation 97
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5.1 Fields of Moving Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
5.1.1 Review of Inhomogeneous Wave Equation . . . . . . . . . . . . . . . . . . . . 97
5.1.2 Retarded Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
5.1.3 Lienart-Wiechert Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
5.1.4 Fields of a Moving Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . 100
5.2 Radiation of moving point charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.2.1 Radiation in general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.2.2 Radiation and Larmor Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.2.3 Breakdown of Rutherford’s picture of the atom . . . . . . . . . . . . . . . . . 105
5.3 Dipolar Radiation (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.3.1 Electric Dipole Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.3.2 Scattering by electrons: why is that sky blue? (optional) . . . . . . . . . . . . 108
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1 Review of Electrostatics and Magnetostatics
1.1 Coulomb’s Law and Gauss’ Law
1.1.1 Coulomb Law
q1
F12
r12
q2
F21
Coulomb measured the electric force using a torsional pendulum. He found
~F12 =q1q2
4πεor212
r12
with
r12 =∣∣∣∣~r1 − ~r2∣∣∣∣
ˆr12 =
(~r1 − ~r2
)r12
where εo ≡ permeativity of free space (= 8.85× 10−12 C2/Nm2)
Notes about Coulomb Law:
1. Central force potential
2. F ∝ q1q2
3. F ∝ r−(2+ε) where ε < 5.8×10−16
Superposition principle for ensemble of charges: the total force exerted on a charge q1 is given bythe sum of the force exerted by all the other charges qi.
~F1,T ot =q1
4πε0
N∑i=2
qir21i
r1i
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1.1.2 The Electric Field
Electric field, ~E(~x) is the force per unit charge exerted at position ~x.
~E ≡ limq→0
~F
qso,
~F = q ~E
qi
r
rir-ri
!(r )i
r
rir-ri
Discrete charge distribution Continuous charge distribution
We call the variable ~x′ the field point variable, whereas ~x is the source point variable.
Discrete distribution of charges:
~E(~x) =1
4πεo
∑i
qi∣∣∣∣~x− ~x′i
∣∣∣∣3(~x− ~x′i)
Continuous distribution of charges:
~E(~x) =1
4πεo
∫ρ(~x′)∣∣∣∣~x− ~x′∣∣∣∣3
(~x− ~x′) ~dx′
.
Here ~dx ≡ d3x ≡ dV ≡ dxdydz is the differential volume element.
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1.1.3 Gauss’ Law
q
S
Gaussian Surface
We define the electric flux ΦE going through a surface, S, as the integral
ΦE
[S
]=
∮s
~E · ~ds
Gauss’ Law states that:
∮s
~E · ~ds =∑
i
Qi
εo
where∑
iQi ≡ QT is the total charge enclosed inside the closed surface, S.
For a continuous charge distribution, ρ(~x), we have that since mathematically,
∮s
~E · ~ds =∫
V
~∇ · ~E · ~dx
then
∮s
~E · ~ds =∫
V
~∇ · ~E · ~dx =1εo
∫Vρ ~dx
which therefore gives the differential form of the Gauss’ Law as
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~∇ · ~E =ρ(~x)
εo
Notes about Gauss Law:
1. Since the electric field is an inverse square law, we can define a scalar function φ, or potentialsuch that ~E = −~∇φ, where the electric potential φ is φ = q
4πε0r for a simple charge. Gauss’differential form can therefore be rewritten as
∇2φ =−ρ
εo
which is known as Poisson’s Equation. In the case of continuous distribution ρ, theelectrostatic potential is given by
φ =∫
ρ(~x′)∣∣∣∣~x− ~x′∣∣∣∣ ·d~x′
2. Electric Force is Conservative: Since the electric field derives from a gradient of a scalarfunction, we must have that
~∇× ~E = ~0
i.e. the curl of any gradient of a scalar function must be identically zero, or∮~E·d~l =
∫S(~∇× ~E)·d~s = 0
3. Important Identity
∇2
(1∣∣∣∣~x− ~x′∣∣∣∣
)=
1εoδ(~x− ~x′)
1.1.4 Energy of the Electric Field
The potential energy for a discrete charge distribution is given by:
U =12
∑i
qiφ(~xi) =12
14πε0
∑i,j
qiqj∣∣∣∣~xi − ~xj
∣∣∣∣9
or, for a continuous distribution:
U =12
∫d~xρ(~x)φ(~x) =
−εo2
∫d~xφ∇2φ
since ∇2φ = −ρε0
. We can rewrite using the identity φ∇2φ = ~∇ · (φ~∇φ)− ~∇φ · ~∇φ as
U =εo2
∫d~x · ( ~E)2 +
εo2
∫∞d~s · ~Eφ
Here the second term, the surface integral, as obtained using Gauss divergence theorem and doesnot contribute since at infinity, the electric field is falling off as 1
r2 , the potential as 1r and the
surface grows only as r2. Far from the source, the resulting 1r behaviour dominates and the integral
does not contribute.
Finally, if we define an Energy Density as E = εo/2∣∣∣∣ ~E∣∣∣∣2
then we can write the potential energy as
U =
∫V
Ed~x =
∫V
εo
2
∣∣∣∣ ~E∣∣∣∣2d~x
1.1.5 Electrostatics and Conductors
Imagine a closed surface S in which charges are flowing in and out through it:
S
qiqjqk
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We define the current as
I =∫
s
~J ·d~s
where ~J is the current density, ie the total charge, Q, passing through surface ~S per unit area,per unit time. We can rewrite the above integral as
∫S
~J ·d~s =∫
V(~∇ · ~J)d~x = − ∂
∂t
∫Vρ(~x) ~dx
which therefore gives in the differential form
~∇ · ~J +∂ρ
∂t= 0
and is known as a continuity equation. The current density is given by ~J = ρ~v where ~v isthe velocity of the charges, and for a distribution of discrete charges can be expressed as ~J(~x) =∑
i qi~viδ(~x− ~xi)
A material is said to have free charges if these charges move freely when an electric field is applied.In a conductor, there exists a linear relation between the applied field and the resulting currentflow, and so we define the physical quantity conductivity, σ, as the ratio between the appliedelectric field and the resulting current density, ie.
~J = σ ~E
which is simply Ohm’s Law.
1.1.6 Capacity
Because of the principle of superposition, the potential of a conductor is proportional to the chargesin the system. If we have N charges, Qi, located on N conductors, we can define the capacity as
Qi =∑
j
Cijφj
where Cij is termed the “coefficient of capacity” for i = j, and the “coefficient of electrostaticinfluence” for i 6= j.
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Note that for a simple conductor we have
Q = Cφ
Using the definition of capacity, we can write the potential energy as
U =12
∑i
Qiφi =12
∑i,j
Cijφiφj
or, for a simple conductor
U =1
2Cφ2
1.2 Electric Field in Matter and Electric Dipole
In an insulator, the electrons are not free to move as in a metal, but instead are bound to atoms(or molecules). Therefore, If we apply an electric field there will be no current, but instead only adistortion in the local electronic density. This gives rise to a macroscopic polarization which willbe an average over all polarized molecules. We define the polarization ~P per unit volume as
~P = ~p(~x)N
where ~p(~x) is the dipole moment of the the molecule, and N is the number of molecules in thesolid per unit volume. For two simple charges +q and −q located on an axis and separated by adistance a, the dipole moment is simply given by ~p = q~a. For an arbitrary distribution of chargesρ, the dipole moment is given by
~p(~x) =
∫V
~dx′ρ(~x′)~x′
.
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1.2.1 Multipole expansion (optional)
We can expand the electrostatic potential in a series of Legendre Polynomials
φ(~x) =∞∑l=0
φ(2l)(~x)
where φ(2l)(~x) is the potential generated by a multipole, and is of the form
φ(2l)(~x) =1
4πεorl+1
∫V
~dx′ρ(~x′)r′lPl(cosγ)
where Pl(t) are the Legendre Polynomial of order l, and where we have defined r ≡ |~x| and r′ ≡ |~x′|.
Note that the above relation results from the following mathematical expansion
1∣∣∣∣~x− ~x′∣∣∣∣ =
∞∑l=0
r′l
rl+1Pl(t)
Monopole case, for l = 0:
φ(1)(~x) =1
4πεor
∫Vd~xρ(~x′) =
q
4πεor
Dipolar case, for l = 1:
φ(2)(~x) =1
4πεor3
∫V
~dx′ρ(~x′)r′cos(θ) =1
4πεor3~p · ~x
i.e. this gives
φ(2)(~x) =1
4πεor3~p · ~x
where the dipole moment, ~p, has been defined as:
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~p(~x) =
∫V
~dx′ρ(~x′)~x′
For simple charges
a
+q -q
1.2.2 Energy of the electric dipole
In the presence of an externally applied electric field, ~E, the dipole energy is given as
U = − ~E · ~p
The torque, τ , on the dipole is
τ = ~p× ~E
and the electric field generated by the dipole is
~Edip ≡ −~∇φ(2)(~x)
~Edip =1
4πεor3
[3(~p · ~x)~x− ~p
]
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1.2.3 Dialectrics and Polarization
We call the average polarization ~P (~x) = ~p(~x)N where ~p(~x) is the polarization of each moleculeinside the dialectric material. The applied electric field will create an overall charge in the materialthat is bound to it. We define this bound charge as
ρ′ = −~∇ · ~P
where the we use ρ to denote the bound charge, as oppose to ρ which describes the free chargeinside the material.
When the polarization, ~P (~x) is discontinuous, such as at an interface between a dialectric andvacuum, a surface charge distribution is created
∫V−~∇ · ~P ~dx =
∮Sd~s · ~P = ~P · ~nA
where A is the surface area at the interface.
Therefore, at the interface, there will be a superficial density of bound charge, ρs, given by
ρ′s = ~P · n
1.2.4 Electric Displacement
We shall define the total charge ρtot inside the material as the one that includes both the contri-bution from free electrons, and the bound charge:
ρtot = ρ+ ρ′
where here ρ is the free charge and ρ′ is the “bounded” charge.
Note that from Gauss law it is the total charge that must be taken into account, therefore:
~∇ · ~E =1εo
(~ρ+ ~ρ′) =1εo
(ρ− ~∇ · ~P )
15
so that
~∇ ·(~E +
~P
εo
)=
ρ
εo.
We define the electric displacement, ~D as
~D ≡ εo~E + ~P
so that we can rewrite Gauss’ law in a matter as:
~∇ · ~D = ρ
where ρ is the “free charge” that is not bound to the dialectric.
Note here that ~E is a fundamental field, whereas, ~D is a derived field that is dependent on thepresence of the dialectric.
1.3 Magnetostatics
In electrostatics, we were concerned by the study of stationnary charges, i.e. with constantelectric fields, and in analogy magnetostatics refers to the study of steady currents, therebygenerating constant magnetic fields.
1.3.1 Biot-Savart Law
In analogy to the Coulomb force, two constant currents flowing in wires will generate a force knownas the Biot-Savart law. For two parallel conductors, each carrying a current, I, we have
d~F12 =µoI1I2
4πd~l1 × (d~l2 × r12)
r212
16
I2
I1
such that
~B( ~x1) =µoI2
4π
∮c
d~l2 × r12
r212
~F12 = I1
∮c
d~l1 × ~B(~x1)
We note that for a point charge in motion with a constant velocity ~v, the Biot-Savart law recoversthe form
~Fmag = q~v × ~B
known as the Lorentz magnetic force.
1.3.2 Vector Potential
We can generalize the Biot-Savart law for an arbitrary current configuration, ~J(~x) to
~B(~x) =µo
4π
∫V
~dx′~J(~x′)× (~x− ~x′)
|~x− ~x′|3
In magnetostatics, we impose that the charge are constant in time, so that ∂ρ/∂t = 0, and hence~∇ · ~J = 0, so we can write
~B = ~∇× ~A
where ~A is the Vector Potential, given by
17
~A(~x) =µo
4π
∫V
~dx′~J(~x′)
|~x− ~x′|
Note that for any vector, ~F , ∇·(~∇× ~F ) = 0 is always mathematically satisfied, so that the definitionfor ~A will satisfy
~∇ · ~B = 0
;
i.e. unlike in electrostatics, there is no single magnetic monopole.
1.3.3 Ampere’s Law
The analogue of Gauss’ law in magnetostatics is given by Ampere’s law
∮c
~B · d~l = µo
∫S
~J · d~s =∑
i
µoIi
or, using Stoke’s theorem for a vector ~f(~r)
∮c
~f(~r) · d~l =∫
S(~∇× ~f(~r)) · d~s
we can rewrite Ampere’s Law in a differential form as
(~∇× ~B) = µo~J
.
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1.3.4 Magnetic Dipole
Similar to the electrostatic scalar potential, we can expand the magnetic vector potential, ~A, in amultipole series, A(2l)(~x)
~A(~x) =µo
4π
∞∑l=0
1rl+1
∫V
~dx′ ~J(~x′)Pl(cosγ)
We can show that the first term in the series, ~A(1)(~r) = 0 since ~∇ · ~J = 0.
The second term in the series is the dipolar term, given by
~A(2)(~x) =µo
4π
~m× ~x
r3
where we have defined r ≡ |~x| and the magnetic dipole moment ~m, as
~m =
∫V
~M(~x) ~dx =1
2
∫V
~x× ~J(~x) ~dx
.
We can show that the component of the magnetic field due to the dipolar term in the vectorpotential is
B(2)(~x) =µo
4π
1
r3
3(~m · ~x)− ~m
Sm
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For a simple current loop we have that
~m =12
∫V
~dx(~x× ~J) =I
2
∫c~x× d~l = I ×Area
withI =
∫s
~J · d~s
and so
|~m| = I × Area
~m = I
∫s
d~s
For a discrete distribution ~J(~x) =∑
i qi~viδ(~x− ~xi), we have that
~m =12
∑i
qi(~xi × ~vi) =∑
i
qi2m
Li
where Li is the angular momentum of charge i, and m is the mass of charge i, and thefore
~m =q
2m~L
.
It is easy to show that the force and potential energy of a magnetic dipole in a constant magneticfield ~B is given by
~F = ~∇(~m · ~B)
and
U = −~m · ~B
20
and finally the torque on the dipole given by
τ = ~m× ~B
.
1.3.5 Magnetic Fields in Matter
Electric fields applied to a material will give rise to a not overall polarization, and likewise magneticfields applied in matter will create magnetization, ~M , i.e. local magnetic dipolar moments will beinduced by the external field. The vector potential ~A in matter will thus contain two contributions:
A(~x) =µo
4π
∫V
~dx′ ~J(~x′)
|~x− ~x′|︸ ︷︷ ︸Free
+~M(~x′)× (~x− ~x′)
|~x− ~x′|3
︸ ︷︷ ︸
Bound
The bound term can be rewritten as
∫Vd~x′ ~M(~x′)× ~∇′
(1
|~x− ~x′|
)=
∫V
~dx~∇× ~M(~x)
|~x− ~x′|
so that we can define a density of bound current, ~J ′
~J ′ = ~∇× ~M
and, similarly as in electrotstatics, at interfaces there will be a superficial density of bound current
~K′ = ~M × n
Ampere’s law in matter should therefore be rewritten to include the sum of all currents as
~∇× ~B = µo~J + ~∇× ~M
21
and similarly as for the electric displacement, we define the magnetic field in a medium, ~H, by theequation
µo~H = ~B − µo
~M
i.e. ~H is the sum of the applied field, ~B, plus the effect of the magnetization induced in the medium,~M . Ampere’s law in matter is therefore rewritten as
~∇× ~H = ~J
~H =~B
µo
− ~M
22
1.3.6 Historical Aside: Etymology and William Gilbert (1544-1603)
While the etymology of the word magnet is still disputed today (it could be from a region eastof Thessaly, or perhaps the name of a shepherd, Magnes, who noticed the iron in his shoes beingattracted to the ground on Mt. Ida), its contextual counterpart electricity has much more definiteroots. ηλεκρτoν was the greek word for amber, a material many noticed would induce an attractiveforce between other objects after some rubbing. For many centuries, these two effects were thoughtto be different manifestations of the same thing (what that thing was, wasn’t really clear). It wasn’tuntil 1600 that William Gilbert published his only book: De Magnete, Magneticisque Corporibus,et de Magno Magnete Tellure (On the Magnet and Magnetic Bodies, and on That Great Magnet theEarth), that the mysteries shrouding these strange effects began to be cleared. Gilbert insisted thatthe two phenomena should be considered separately. Though not taken any chances on claimingto know the cause, he, in perhaps what could be considered the first true example of experimentalphysics, demonstrated many properties of the magnet, or lodestone. These included the use of amagnetic needle in determining one’s location on the earth, as well as firmly putting to rest thepopular notion that goat’s blood could replenish a magnet’s power. Many of Gilbert’s immediatesuccessors, like Galileo and Kepler, recognized the import of his work. However, he is often neglectedin modern treatments of the history of science.
Figure 1: Gilberts orb of the earth showing declinations of an iron rod
[See also: Cambridge Scientific Minds, eds. Harman and Mitton, ch 1 by Stephen Pumfrey, 2002]
23
1.4 Maxwell’s Equations for Static Electromagnetism
Finally, we have Maxwell’s Equations for static electromagnetism:
1.) Maxwell’s equations in matter:
~∇ · ~D = ρ~∇× ~E = 0~∇ · ~B = 0
~∇× ~H = ~J
2.) Maxwell’s equations in vacuum:
~∇ · ~E =ρ
εo
~∇× ~E = 0~∇ · ~B = 0
~∇× ~B = µo~J
24
2 Electrodynamics
2.1 Faraday Law
2.1.1 Electromotive Force
We define the electromotive force of a closed circuit as
E =
∮C
~E · d~l
which has the units of a potential, or volts. When charges are stationary ∂ρ/∂t = 0, as in electro-statics and magnetostatics, the circulation of the electric field over a closed loop is
E =∮
C
~E · d~l =∫
S(~∇× ~E) · d~s = 0
since ~∇ × ~E = 0 when ~E = −~∇φ (Maxwell’s equations for static electromagnetic fields). Inelectrodynamics, charges and current are not steady nor stationary and consequently ∂ρ/∂t 6= 0,i.e. the charge and current are now time dependent and so will be the electric and magnetic fields.Hence, if the electric field is allowed to vary in time, the circulation over a closed path will nowgives
~∇× ~E 6= 0
i.e. a net potential difference will appear, and the electric field is not irrotational anymore.
2.1.2 Faraday Law
We define the magnetic flux, φB, though a surface, S, as:
φB =∫
S
~B · d~s =∮
C
~A · d~l
since ~B = ~∇ × ~A and we have used Stoke’s theorem. Faraday Law, which is empirical in natureand dates from ∼1830, states that:
25
∮~E · d~l ≡ E = −∂φB
∂t
“A changing magnetic field in time induces an electric field”
This is very important, as the seemingly disconnected electricity and magnetism are now linkedto one another through the time dependence of the magnetic flux. The induced electric field isopposite to the variation in the magnetic field, which is also known as Lens’ Law.
We can rewrite
∮C
~E · d~l = − ∂
∂t
∫S
~B · ~ds =∫
C(~∇× ~E) · d~s
and so Faraday law in differential form is now:
~∇× ~E = −∂ ~B
∂t
26
2.1.3 Historical Aside: Michael Faraday (1791-1867)
Upon a first glance, the casual observer might mistake any modern Electricity and Magnetismtextbook for a solely mathematical explication. Open any page of J.D. Jackson’s account of thesubject, for example, and you will most likely be confronted with at least 1 equation, if not twenty.Given this, the achievements of Michael Faraday should be considered all the more noteworthy.Having almost no formal training in an academic sense, and certainly no facility with algebra orcalculus, Faraday, through rigorous and brilliant experimental work, was able to lay the groundworkof a discipline that would eventually become that hallmark of the modern age: electricity. He wasemployed at age 13 by a bookbinder as an errand boy. The books that passed his way whileemployed there would serve as the basis for his self-administered educational career. Through aseries of almost chance events, he secured his next position as an assistant in the lab of the chemistHumphry Davy at the Royal Institution. Thus began a long and most fruitful career for Faradayas a natural philosopher. (Disciplines were just forming at this point in the history of academicscience, and Faraday preferred not to be called a physicists or chemist. His work shouldn’t beconfined to either department.) Just eight years after starting his scientific career, working onrecent phenomena discussed by Ørsted and Ampere, he recorded the first sketches for what wouldbecome the electric motor, shown here from his journal of December 1821. 15 years later Faradaymade known his intuitions that the phenomenon of radiating light was intricately related to theelectric and magnetic fields he was so familiar with. It would however, take James Maxwell, ayounger and much more mathematically minded physicist, to fully uncover this relationship, whichhe did in 1864.
Faraday (right) and Chemistry Prof, John. Daniell, ca 1843
Faraday's original sketch for the `electromagnetic rotation apparatus'
[For more, see: Michael Faraday and the Royal Institution, John Meurig Thomas, 1991]
27
2.2 Inductance
The Faraday law states that an electric field will be induced by a time-dependent magnetic field,or flux, with a loop C. Such a loop, or “circuit” will be characterized by an “inductance” L whichwill depend on the geometry of the loop.
2.2.1 General Case
Consider a loop C1 in which a current I1 is flowing. In such a loop, the magnetic field generated isproportional to the current
~B1 =µo
4πI1
∮d~l × ~rr2
⇒ ~B1 ∝ I1
This magnetic field will extend beyond the loop, and could influence other loops located in thevicinity of loop 1. Consider a second loop C2 adjacent to C1. On that loop, the magnetic flux goingthrough the loop (generated by C1) is
φ2 =∫
S
~B · d~s2 ∝ I1
so it will be directly proportional to the current I1. In general we can write the following linearrelation between fluxes on a loop i created by currents j as
φ(i)B =
∑j
MijIj
where the Mij are mutual inductance coefficients.
C2C1
I1
2B
1B
28
Lets consider the vector ~xi which is a coordinates of a point on a loop Ci. The vector potential atposition ~xi , ~A(~xi), due to a current in loop Cj is given by
~A(~xi) =Ijµo
4π
∮Cj
dlj|~xi − ~xj |
where d~lj is the vector element on Cj at a coordinate ~xj .Since
φB =∫
S
~B · d~s =∫
C
~A · d~l
we have that the mutual inductance is given by
Mij =µo
4π
∮ ∮d~li · d~lj|xi − xj|
which is known as the Neumann formula.
We can note the following properties of the mutual inductance coefficients:
1. Mij is purely geometric; Mij = Mji
2. Mii ≡ L ≡ coefficient of self inductance
In case 2, i.e. the inductance of a current loop on itself, we have the following relation between themagnetic flux and the current:
φB = LI
E = −LdI
dt
which can be summarized by the following statement:
“A changing current, dI/dt, induces an EMF in the loop itself.”
This circuit, or current loop, will have energy stored, and so if we consider a circuit with aninductance L, Ohms Law tells us that
29
Power = E I ≡ V I = L
(dI
dt
)I =
L
2d(I2)dt
so the total work stored into the circuit, or inductor, will be given by
W =∫ ∞
0
L
2d(I2)dt
dt =LI2
2
W =LI2
2
which is the energy stored inside the inductor.
2.3 Maxwell’s Equations
2.3.1 Displacement Current (In Vacuum)
In a static system ∂ρ/∂t = 0, and Faraday law showed us that a time-dependent magnetic field willinduce an electric field inside a circuit, or loop. In electrodynamics
∂ρ
∂t6= 0
and so the continuity equation, which is always valid
~∇ · ~J +∂ρ
∂t= 0
describes that the changes in charges must be compensated by changes in current flow. Let’sconsider Gauss’ Law
~∇ · ~E = ρ/εo
and so we can rewrite the continuity equation as
~∇ · ~J + εo∂
∂t(~∇ · ~E) = 0.
30
We also know from Ampere’s Law that the magnetic field is related to the current density J by
~∇× ~B = µo~J
and by taking the divergence on both side, we now find the situation where
~∇ · (~∇× ~B) = µo(~∇ · ~J) 6= 0.
But the divergence of a curl must always be zero. So we are now force to modify Ampere law andmake the following addition:
µo~J → µo
~J + µoεo∂ ~E
∂t
and
~∇× ~B = µo~J + µoεo
∂ ~E
∂t.
Now we can easily verify that the divergence of the curl will indeed be zero,
∇(~∇× ~B) = µo~∇ · ~J + µoεo
∂
∂t(~∇ · ~E
= µo
~∇ · ~J + εo
∂
∂t(~∇ · ~E)
︸ ︷︷ ︸
=0
.
This modification to Ampere’s law lead to an important consequence on the magnetic field, namely:
“A changing electric field induces a magnetic field”.
The term
31
~JD ≡ εo∂ ~E
∂t
is sometimes called the displacement current.
2.3.2 Maxwell’s Equations (with dynamics)
In can now write the four Maxwell’s equations of electrodynamics:
In Free Space:
~∇ · ~E =ρ
εo
~∇× ~E = −∂~B
∂t~∇ ·B = 0
~∇× ~B = µo~J + µoεo
∂ ~E
∂t
and similarly, we can express them in matter, recalling the definity for the electric field displacement~D and magnetic field in matter ~H:
~D = εo ~E + ~P
−~∇ · ~P = ρ′
~H =~B
µo− ~M
~∇× ~M = ~J ′
where ρ′ is the bound charge density, ρ is the free charge density, such that the total charge is
ρtot = ρ+ ρ′
, and similarly is ~J ′ is the bound current density ~J the free current density such that the totalcurrent is
~Jtot = ~J + ~J ′
.
Maxwell’s equations in a medium becomes
32
~∇ · ~D = ρ
~∇× ~E = −∂ ~B
∂t~∇ · ~B = 0
~∇× ~H = ~J +∂ ~D
∂t
For the special case of a medium that is linear and isotropic, i.e that will respond linearly to fields,and which response will not depend on orientation of the fields, we can write the polarization andmagnetization vector as:
~P = εoχe~E
~D = ε ~E~M = χm
~H
~H =~B
µ
where χe is the electric susceptibility and χm is the magnetic susceptibility. For suchmedium, we can then express the the permeativity of the medium by
ε = εo(1 + χe)
and its magnetic permeability as
µ = µo(1 + χm)
2.3.3 Boundary conditions
Let an interface be described by a surface, S, where n is the unit vector perpendicular to thesurface.
In general the fields, ~E, ~D, ~B, ~H, will be discontinuous at a boundary of two different media,described by ε1, µ1 and ε2, µ2.
For the electric field we have the following boundary condition at the interface of two media:
33
n
S1
1!
1µ
2 2!
2µ
( ~D1 − ~D2) · n = ρs (1)
( ~E1 − ~E2)× n = ρs (2)
i.e., the parallel component of ~E (w.r.t. interface plane) is continuous and the perpendicularcomponent of ~D is discontinuous (here, n is a surface vector of the interface, perpendicular to itsplane, and ρs is the superficial density of free charges).
We can demonstrate the first boundary condition by considering Gauss’ theorem:
From Gauss’ Law ∫V
~∇ · ~D ~dr = ( ~D1 − ~D2) · nA = ρsA
which gives directly
( ~D1 − ~D2) · n = ρs
.
Similarly, we can show the second boundary condition by considering a small amperian currentloop taken across the interface
∮C
~E~dl = − ∂
∂t
∫s
~B · d~s
which, integrating the left side, gives
( ~E1 − ~E2) ·~l = − ∂
∂t
∫~B · d~s
34
and in the limit of an infinitely narrow loop
limδ→0
− ∂
∂t
∫~B · d~s→ 0
we obtain the boundary condition for the parallel component of the electric field
( ~E1 − ~E2)× n = 0
Similarly, we can show from ~∇· ~B = 0 that at the interface between two media of different magneticproperties, we have that
( ~B1 − ~B2) · n = 0 (3)
since ~∇ · ~B = 0 (no magnetic charge!)
But, the parallel component of ~H is discontinuous by ~K, the superficial density of free current
( ~H1 − ~H2)× n = ~K (4)
For linear media, ~D = ε ~E, ~B = µ ~H, so the boundary conditions yield:
(ε1~E1 − ε2
~E2) · n = ρs
( ~E1 − ~E2)× n = 0
( ~B1 − ~B2) · n = 0( ~B1
µ1
−~B2
µ2
)× n = ~K
35
2.4 Electromagnetic Potentials
2.4.1 Potentials for dynamic case (∂ρ∂t 6= 0)
We recall that both the electric and magnetic fields derive from potentials, in one case a scalarpotential φ and in the other a vector potential ~A. In electro-magnetostatics, when ∂ρ
∂t = 0 they arerelated to the fields by the following relations:
~∇× ~E = ~0~E = −~∇φ~∇ · ~B = 0
~B = ~∇× ~A
and as we have seen previously, the curl of a gradient and the divergence of a curl must always be(mathematicallly) zero.
But in electrodynamics, once we introduce time-dependent sources, we are are now confronted withthe following situation:
~∇× ~E = −∂~B
∂t= −∂(~∇× ~A)
∂tand thus the Faraday law reads
~∇× ( ~E +∂ ~A
∂t) = ~0
which is NOT zero for ~E = −~∇φ. To remedy to this, we must re-write the scalar potential, φ, andvector potential, ~A and relate them to the fields in the following way:
~E = −~∇φ− ∂ ~A
∂t~B = ~∇× ~A
so that we correctly have
~∇× ~E = −∂(~∇× ~A)∂t
⇒ ~∇×(~E + ∂ ~A
∂t
)= ~0
⇒ ~∇×(−~∇φ− ∂ ~A
∂t + ∂ ~A∂t
)= ~0
36
since the curl of gradient scalar function is always zero. Importantly, we note that if the potentialvector A is now time-dependent, which is true for time-dependent magnetic field, itwill act as a potential source for the electric field.
2.4.2 Gauge Transformations
The correspondence between fields and potentials is not unique. In fact, we can always choose apotential arbitrarily for as long as Maxwell’s equations are satisfied. We can therefore freely modifythe potentials by the addition of a function, ξ, without ever modifying the fields ~E, or ~B. Thistransformation to the potentials is called a Gauge Transformation.
φ′ → φ− ∂ξ
∂t
~A′ → ~A + ~∇ξ
and we show that this transformation does not modify the fields in any way:
i) Scalar potential:
~E′ = −~∇φ′ − ∂ ~A′
∂t
= −~∇φ+ ~∇(∂ξ
∂t
)− ∂ ~A
∂t+ ~∇
(∂ξ
∂t
)= −~∇φ− ∂ ~A
∂t
= ~E
ii) Vector potential:
~B′ = ~∇× ~A′
= ~∇× ( ~A+ ~∇ξ)
= ~∇× ~A+:0~∇× ~∇ξ
= ~B
Gauge transformations are very important for other fields of physics such as for example in quantumelectrodynamics. This non-uniqueness of potentials imply that we can impose certain conditions
37
on potentials without changing ~E and ~B. These are called gauges. There are two gauges thatare particularly useful in classical electrodynamics, the so-called Lorentz and Coulomb (transverse)gauge:
1.) Lorentz Gauge:
In the Lorentz gauge, we impose the following condition to the potentials:
~∇ · ~A + µoεo∂φ
∂t= 0
If the potentials do not satisfy this condition, we can always find a function, ξ, so that φ′, or ~A′,does satisfy it. Let’s transform the potentials for this condition according to a gauge transformationby the function ξ:
~∇ · ~A′ + µoεo∂φ′
∂t= ~∇ · ~A+ µoεo
∂φ
∂t+∇2ξ − εoµo
∂2ξ
∂t2
and so we see that we can always impose the Lorentz condition on the potential φ′ and ~A′ providedthat the function ξ satisfied the equation
∇2ξ − µoεo∂2ξ
∂t2= −~∇ · ~A + µoεo
∂φ
∂t
i.e. all ξ satisfying the wave equation (left hand side) with a source (right hand side) willbe a “Lorentz gauge”.
2.) Coulomb Gauge:
In the Coulomb, also called transverse gauge, the following condition is imposed on the potentialvector ~A:
~∇ · ~A = 0
and we can readily show that in that case, we can always make a transformation so that thepotential vector is in the transverse gauge provided that we solve the Poisson euquation for ξ :
38
~∇ · ~A′ = ~∇ · ~A+∇2ξ
⇒ ξ(~x) = 14π
∫d~x′ −
~∇· ~A(~x)
|~x−~x′|
Remember! In electrostatic, the potential satisfies to a Poisson equation of the form
∇2φ = − ρ
εo
which has for solution
φ(~x) =1
4πεo
∫vd~x′
ρ(~x′)
|~x− ~x′|. so by analogy, we know the solution for the gauge transformation function ξ.
2.4.3 Equations of motion for electromagnetic potentials
In electrodynamics, the introduction of the time-dependence for the sources, and hence for thefields, has linked the electric and magnetic fields in a dynamical way, and through their depen-dence. We have seen that we can now induce an electric field by varying a magnetic field (Faraday)and conversely, we can generate a magnetic field by varying an electric field. We shall now see thatboth potential, scalar φ and vector ~A, can be treated very similarly within the Lorentz gauge andthat in case, they will obey to the same dynamical wave equation.
Let’s consider Gauss’ law
~∇ · ~E = ρεo
⇒ ~∇ ·(−~∇φ− ∂ ~A
∂t
)= ρ
εo
and so the potentials must obey to:
∇2φ +∂(~∇ · ~A)
∂t= − ρ
εo
(*)
39
If are now considering Ampere law:
~∇× (~∇× ~A) = µo~J + µoεo
∂
∂t
(−~∇φ− ∂ ~A
∂t
)and making use of the mathematical identity
~∇× (~∇× ~A) = ~∇(~∇ · ~A)− ~∇2 ~A
and so we can rewrite it in the following form:
~∇(~∇ · ~A)− ~∇2 ~A+ µoεo∂
∂t~∇φ+ µoεo
∂2 ~A
∂t2= µo
~J
.
Let’s impose the Lorentz gauge
~∇(~∇ · ~A) + µoεo∂
∂t~∇φ→ 0
an so, in that the case, the potential vector ~A obeys to the following equation
∇2 ~A− µoεo∂2 ~A
∂t2= −µo
~J
which is simply a wave equation with a source. If we go back to the equation (*), andreplacing the Lorentz condition ~∇(~∇ · ~A) = −µoεo
∂∂t~∇φ we obtain the following equation for the
scalar potential φ
∇2φ− µoεo∂2φ
∂t2= − ρ
εo
In the Lorentz gauge, both the potential ~A and scalar potential φ are treated on an equal footingand they both satisfy to a wave equation with a source. It is typical to define the wave operator,or d’Alembertian, as
2 = ∇2 − 1v2φ
∂2
∂t2= ∇2 − µoεo
∂2
∂t2
40
where vφ is the phase velocity of the wave, and which will be discussed in great details in chapter3. We can therefore re-write the following equations for ~A, and φ:
2φ = − ρ
εo
2 ~A = −µo~J
i.e. the same wave equation but with a distinct source in each case.
2.4.4 Energy and Momentum in EM theory: Poynting’s Theorem
We now consider moving charges in free space that would be subjected to both an electric and amagnetic field. We are interested in the sum of the resulting ~E field with Ampere’s Law, with thatof the ~B field with the Faraday law. We therefore consider the quantity:
~B · (~∇× ~E) = − ~B · ∂~B
∂t(1)
and
~E · (~∇× ~B) = µo~E · ~J + µoεo ~E ·
∂ ~E
∂t(2)
and if we subtract (1)− (2) then we obtain
~B(~∇× ~E)− ~E(~∇× ~B) = −µo~E · ~J − 1
2∂ ~B2
∂t − µoεo
2∂ ~E2
∂t
⇒ 1µo
~B(~∇× ~E)− ~E(~∇× ~B)
= −1
2∂∂t
~B2
µo+ εo ~E
2
− ~E · ~J
and if we make use of the following mathematical identity
~∇ · ( ~E × ~B) = ~B · (~∇× ~E)− ~E(~∇× ~B)
it gives
41
1µo
~∇ · ( ~E × ~B) = −12∂
∂t
~B2
µo+ εo ~E
2
− ~E · ~J
and finally
~E · ~J = −1
2
∂
∂t
~B2
µo
+ εo~E2
− 1
µo
~∇ · ( ~E × ~B).
We already recognize on the right hand side the magnetic and electric energy density given by 12
B2
µo
and εo2 E
2, respectively. We are interested in the total energy stored over the whole volume, so wecan integrate both side of the equation over the whole volume, or sphere, taken at infinity
∫V
( ~E · ~J)d~x = −12∂
∂t
∫V
~B2
µo+ εo ~E
2
d~x− 1
µo
∫V
~∇ · ( ~E × ~B)d~x.
The term given by ~E · ~J is particularly interesting, for we can express for discrete charges as
∫V
( ~E · ~J)d~x =∑
i
qi~vi~E
=∑
i
vi~Fi
=∂Ki
∂t
i.e. it is given by the time-derivative of the kinetic energy of the charged particle, or the power.Making use of that, and of Gauss divergence theorem, we can rewrite the volume integrals as
∂K
∂t= − ∂
∂t
1
2
∫ ( ~B2
µo
+ εo~E2
)d~x
− 1
µo
∮S
( ~E × ~B) · d~s
which is known as the Poynting’s Work-Energy Theorem. The first term of the equationdescribes the total energy stored in the field while the second term is the energy flux that is carriedout. The Poynting theorem states that the
“Work done on the charges by the electromagnetic force is equal to the decrease in energy stored inthe field less the energy that flowed out through the surface”
42
We define the Poynting Vector (or “Energy Flux Density”) as
~S ≡ 1
µo
~E × ~B
and the total electromagnetic field energy density is
E =1
2
( ~B2
µo
+ εo~E2
).
Making use of these notations, the Poynting theorem can be re-written in a more compact differ-ential form
− ~E · ~J =∂E∂t
+ ~∇ · ~S
The Poynting theorem therefore defines a rate equation for the energy stored inside the fields mustbe compensated by the energy flux carried away at a sphere taken at infinity.
It is also usual to define the following quantity for the electromagnetic field
1. Density of momentum for the E-M field:
~P = µoεo~S
i.e. the Poynting vector represents the momentum carried away by the electro-magnetic fieldsand similarly, we can define
2. Angular momentum density:
~λ = ~r × ~P = εo(~r × ( ~E × ~B))
43
3 Electromagnetic Waves
3.1 Waves in Vacuum (General Case)
3.1.1 Review of the Wave Equation
A wave is simply a “disturbance of a continuous medium that propagates with a fixed shape anda constant velocity”.
Let Ψ be a function, Ψ = Ψ(x, t) which depends on position, x, and time, t. In general, one canshow that Ψ will be a function which describes a wave propagation if it is a solution of the classicalwave equation:
∂2Ψ
∂x2=
1
v2
∂2Ψ
∂t2
where v is the velocity of propagation of the wave.
Lets introduce the variables, u1 = x− vt and u2 = x+ vt, then we have
∂
∂u1=
(∂
∂x− 1v
∂
∂t
)∂
∂u2=
(∂
∂x+
1v
∂
∂t
)
and so we can write the wave equation as
∂2Ψ
∂u1∂u2
= 0
which has solutions of the form
Ψ = Ψ1(u1) + Ψ2(u2)
or, writing in x, t coordinates
44
Ψgeneral = Ψ1(x− vt) + Ψ1(x + vt)
This solution, known as the “D’Alambert Solution” is the sum of right traveling and left travelingwaves, which is the most general solution of the wave equation.
3.1.2 Sinusoidal Waves in 1D
Assume a solution to the wave equation of the form
Ψ = Acos[k(xvt) + δ]
where:
k → wave vector = 2πλv → propagating speedA → amplitudeδ → phase
!
v
Travelling Wave
A
"k
x
y
First, verify this solution indeed satisfies the wave equation:
∂2Ψ∂x2
= −Ak2cos[k(x− vt) + δ] = −k2Ψ
45
and
∂2Ψ∂t2
= −A(vk)2cos[k(x− vt) + δ] = −(vk)2Ψ
Therefore we have
∂2Ψ∂x2
= −k2Ψ
and
1v2
∂2Ψ∂t2
= −k2Ψ
and so
∂2Ψ∂x2
=1v2
∂2Ψ∂t2
which is the wave equation.
We call k the wave number (in 1D), or ~k the wave vector (in 2D), and we have
k =2πλ
since, when x→ x+ 2π/k, the cosine has executed a complete cycle, ie.
Ψ(x+
2πk, vt
)= Acos
[k
(x+
2πk− vt
)+ δ
]= Ψ(x, vt)
Similarly, for a full cycle, we define the period as
T =2πkv
So that
Ψ(x, v(t+ T )) = Ψ(x, vt)
The frequency of the wave is simply the inverse of the period, ν = 1/T , and so we have the relations
46
ν =1
T=
kv
2π=
v
λ
Recall that for circular motion, we define the angular frequency, ω, as
ω = 2πν = kv
We can therefore rewrite the wave equation in terms of wave vector, k, and angular frequency, ω,as
Ψ(x, t) = Acos(kx− ωt + δ)
Note that for a right travelling wave:
ΨR(x, t) = Acos[k(x− vt) + δ]= Acos[kx− ωt+ δ]
for a left travelling wave:
ΨL(x, t) = Acos[k(x+ vt)− δ]= Acos[kx+ ωt− δ]
where the sign convention for the δ term is for a delay. But, the cosine function is an even function,ie. cos(θ) = cos(−θ), and so we can rewrite ΨL as
ΨL(x, t) = Acos[kx+ ωt− δ] = Acos[−kx− ωt+ δ]
or, in other words
ΨL(x, t) = [ΨR(x, t)]k→−k
47
The inversion of k → −k transforms the wave from a right travelling to a left travelling wave andvice-versa.
Complex Notation:
Euler’s Formula states
eiθ = cosθ + isinθ
and so we can write the sinusoidal solution as
Ψ(x, t) = Acos(kx− ωt+ δ) = ReAei(kx−ωt+δ) ≡ ReΨ
Where ReΨ is the real part of the complex solution
Ψ ≡ Aei(kx−ωt)
with complex amplitude defined by
A = Aeiθ
where the phase, δ, is now absorbed in the A. Note that we use complex notation because the eiθ
are easier to manipulate mathematically than sin and cos.
3.1.3 Boundary Conditions and Interfaces
Wave Boundaries/Interfaces
x
v 1
k1 k2v 3
v 2
48
ΨI(x, t) = AIei(k1x−ωt), x < 0 → incident wave
ΨR(x, t) = ARe−i(k1x−ωt), x < 0 → reflected wave
ΨT (x, t) = AT ei(k2x−ωt), x < 0 → transmitted wave
Note that the frequency, ω, will be the same for all the waves. Therefore, since ω = kv, kv mustbe constant and so
k1v1 = k2v2
giving the result
k2
k1
=v1
v2
Boundary Conditions (at x=0):
Ψ(0−, t) = Ψ(0+, t) (1)
ie., wave must be continuous across interface.
In general, for a string for example, the 1st derivative of Ψ must also be continuous across aninterface for, if the slope wasn’t equal, it would give rise to a net force on the string. Therefore wealso have
∂Ψ
∂x
∣∣∣∣o+
=∂Ψ
∂x
∣∣∣∣o−
(2)
Applying boundary condition (1) to the wavefunctions yield (at x=0)
AI + AR = AT (i)
and boundary condition (2) yields
49
k1(AI − AR) = k2AT
or
(AI − AR) =k2
k1Ar (ii)
Subtracting (i)-(ii) gives
2AR = AT
(1− k2
k1
)
and adding (i)+(ii) gives
2AI = AT
(1 +
k2
k1
)giving
AT =2k1
k1 + k2
AI
and
2AR =2k1
(k1 + k2)
(k1 − k2
k1
)AI
AR =
(k1 − k2
k1 + k2
)AI
or, in terms of v1,2 we have
50
AR =
(v2 − v1
v2 + v1
)AI
AR =
(2v2
v2 + v1
)AI
Which gives the amplitudes of the reflected and transmitted waves as a function of the time incidentwave front.
3.1.4 Polarization of Waves
Wave Propagation
displacement
propagation
In general there are two types of waves:
1. If displacement is perpendicular with the direction of propagation, the wave is said to betransverse (e.g. string)
2. If the displacement is along direction of propagation, Longitudinal (e.g. sound).
EM waves are TRANSVERSE!
Polarization:
For transverse waves, the displacement can take a different position in space with respect to direc-tion of polarization.
We define the polarization vector, ~n, which for a wave propagating along the ~z, would be lying inthe x-y plane, i.e.
n · z = 0
and
n = cosθx + sinθy
51
So that, in general, a transverse wave propagating along ~z-axis can be written as
Ψ(z, t) = Acosθei(kz−ωt)x + Asinθei(kz−ωt)y
3.1.5 Linear Combination and Power Transform
The sinusoidal solution of the wave equation can be written using complex notation as
Ψ = Aei(kx−ωt)
In fact, any wave can be decomposed into a linear combination of sinusoidal waves:
Ψ(x, t) =∫∞−∞ A(k)ei(kz−ωt)dk
with ω = ω(k)
The coefficient A(k) can be obtained from the fourier transform.
3.2 Electromagnetic Waves (Free Space)
3.2.1 Wave Equation for EM field
We consider the Maxwell’s Equations with no source an in free space (vacuum). Although nocharge sources are present, fields may exist in the form of EM waves, i.e. charge and ∂ρ/∂t can befar away, and yet can propagate through the vacuum with constant energy into places where nocharges are present.
Maxwell’s equations tell us
~∇ · ~E = 0 (i)
~∇ · ~B = 0 (ii)
~∇× ~E = −∂ ~B∂t
(iii)
~∇× ~B = µoεo∂ ~E∂t
(iv)
52
We now take the curl of (ii)
~∇× (~∇× ~E) = ~∇:0(~∇ · ~E)− ~∇2 ~E
= ~∇×(−∂
~B
∂t
)= −∂
~∇× ~B
∂t
= −µoεo∂2 ~E
∂t2
which gives the result
~∇2 ~E = µoεo∂2 ~E
∂t2
Similarly, if we take the curl of (iv)
~∇× (~∇× ~B) = ~∇:0(~∇ · ~B)
= ~∇2 ~B
= µoεo∂(~∇× ~E)
∂t
= −µoεo∂2 ~B
∂t2
giving
∇2 ~B = µoεo∂2 ~B
∂t2
So, we have shown that Maxwell’s Equations in free space generate ~E and ~B fields that obey thewave equation
≡ ~∇2 − 1v2
∂2
∂t2
53
such that
~E = 0
~B = 01
v2= µoεo
where v is the speed of propagation. Note that in free space
v ≡ 1√εoµo
≡ c
ie., the speed of the wave is the speed of light.
Maxwell hence discovered that light is an electromagnetic phenomenon, and a travelling wave!
In free space and EM wave travels at the speed of light. In a medium however the speed is modified.If the medium is linear such that, for example,
~D = ε ~E
~H =1µ~B
Then the wave has propagation speed defined by
v = 1√εµ≡ c
n
n ≡√
εµεoµo
where n is the index of refraction of the material.
For most materials µ ∼ µo, and in all materials ε > εo and so
n ∼ √εR
εr = εεo
v < c
54
3.2.2 Plane wave solution in free space
In last section, we have seen that Maxwell’s equations in free space provides us with a classicalwave equation for both the electric and magnetic field. The solution of the wave equation will giveus the electromagnetic field as a three-dimensional wave, propagating along a specific direction ~k,the wave-vector for the wave, which in magnitude describe the spatial extent of the wave and itsdirection will provide the direction of propagation for the wave.
An electromagnetic wave is said to be monochromatic if it has a well-defined frequency ω. Forall intent and purposes, in this course, the electromagnetic waves will always be monochromatic.However, the wave vector ~k needs not be unique, since for a given ω, it must only obey in magnitude|~k| = ω
c , but obviously can point anywhere in space and still obey to that relation. We shall call aplane wave an electromagnetic wave that has a well-defined wave vector ~k that is unique.
The plane wave solution of the wave equation of the electromagnetic field is therefore
~E(~x, t) = Re ~E0ei(~k·~x−ωt)
~B(~x, t) = Re ~B0ei(~k·~x−ωt)
where the amplitudes ~E0 and ~B0 are complex (they include the phase shift previously discussed),but as usual, the solution for the fields must be real.
These amplitudes, if we write ~k = |~k|k must obey to the following relations
~E0 · k = 0
~B0 =|~k|ω
(k × ~E0)
where the first relation can be readily derived from ~∇ · ~E = 0 and the second from ~∇× ~E = −∂ ~B∂t .
These relations are important for if we fix the wave vector along the z-axis, i.e. ~k = kz, then wemust have that ~E0 · z = 0 = ~B0 · z, or in another words, the z-component of the electro-magneticfield must be zero. The second condition implies that the electric field and magnetic field mustbe orthogonal between them, and also mutually with the wave vector ~k. Therefore, ~E, ~B and ~kforms an oriented tryad and the electromagntic field is transverse, the fields oscillate in direction
55
perpendicular to the direction of motion of the wave.
The second relation discussed above also implies that, since in free spave ωk = c, the magnitude of
the fields must related by
~|B0|~|E0|
=1c
which means that the magnetic part of the electromagnetic field will reduced in magnitude by anamount of “c” compared to the electric field.
3.2.3 Polarization of EM waves (for a plane wave)
Polarization is defined as the “direction of the vector ~E in space as a function of time at a givenpoint”.
For a general plane wave we can write the fields, ~E and ~B as
~E(~x, t) = Re ~Eoei(~k·~x−ωt)n
~B(~x, t) = Re1
c~Eoe
i(~k·~x−ωt)k × n
We can decompose ~Eo, which is complex, into a basis of 2 unit vectors, ε1 and ε2 which forms,along with k, an orthonormal basis. We then get
~Eo = E1ε1 + E2ε2
where ε1, ε2 ∈ C are complex.
Consider the following cases:
i)Relative phases between E1 and E2 differ by π (or integer multiple of)
E1 = eiα|E1|E2 = ±eiα|E2|
choosing ε1 ≡ x, ε2 ≡ y and k ≡ z then
56
Ex = |E1|cos(kz − ωt + α)
Ey = ±|E2|cos(kz − ωt + α)
The direction of the electric field is constant in time and
|Ex||Ey|
= constant
The wave is said to be linearly polarized.
k, z
!1
"1"
"2
!2
ii) Relative phases between E1 and E2 differ by odd multiple of π/2
• If |E1| = |E2|, then ~E will, as a function of time, trace a circle in the plane perpendicular to k,and the wave is said to have circular polarization.
If ~E turns clockwise when looking from ~k it is labelled right circular. If ~E turns counter-clockwisewhen looking from ~k it is labelled left circular.
• If |E1| 6= |E2| then the wave will have elliptical polarization.
k, z
!1
!2
"
iii) Other phases will be examined as a homework exercise.
57
3.2.4 Energy and Momentum of EM waves
According to Poynting’s Theorem, the energy density and energy flux density for the electromag-netic field in general is given by
E =1
2
(εo| ~E|2 +
1
µo
| ~B|2)
~S =1
µo
( ~E × ~B)
We are now interested in calculating the energy density and Poynting vector but in the case ofmonochromatic plane waves propagating in free space. Defining the directing of the electric fieldalong an arbitrary axis labeled n, in this case:
~E(~x, t) = Re ~Eoei(~k·~x−ωt)n
~B(~x, t) = Re
1c~Eoe
i(~k·~x−ωt)k × n
| ~Bo|2 =
| ~Eo|2
c2
which will yield the following
E =12
(εo| ~E0|2 +
| ~B0|2
µo
)cos2(kx− ωt+ α)
=12(εo| ~E0|2 + εo| ~E0|2)cos2(kx− ωt+ α)
= εo| ~E0|2cos2(kx− ωt+ α).
This energy density is therefore oscillating periodically as the wave travel in space. We are interestedin the time-averaged energy density that will be stored in the fields, and for this we need tocalculate the time average over one period, so
< E > = < εo| ~Eo|2cos2(kx− ωt+ α) >
=12εo| ~Eo|2
58
since the time averaging of cos2(x) = 12 . Similarly, the Poynting vector becomes
~S =1µoc
| ~Eo|2kcos2(kx− ωt+ α)
= cεo| ~Eo|2cos2(kx− ωt+ α)k
and its time-average is
< ~S >=12cεo| ~Eo|2
which can be rewritten in term of the energy density of the electromagnetic wave
~S = cE k
.
The momentum density carried by the EM field is given by
~P =~S
c2=E k
c
3.2.5 Electromagnetic Waves in Matter
Let’s consider a region inside matter, i.e. in a medium characterized by εµ, but free of chargesand currents. In this case, the Maxwell’s equation are:
~∇ · ~D = 0
~∇× ~E = −∂ ~B
∂t
~∇ · ~B = 0
~∇× ~H =∂ ~D
∂t
59
If in addition the medium is linear and isotropic, we can rewrite the Maxwell’s equations using
~D = ε ~E
~B = µ ~H
so that for a homogeneous medium, ie. ε, µ 6= f(~x), Maxwell’s equations reduce to
~∇ · ~E = 0
~∇× ~E = −∂ ~B
∂t
~∇ · ~B = 0
~∇× ~B = µε∂ ~E
∂t.
These are the exact same equations than in free space but with µo, εo → µ, ε in the last equations.Therefore, similarly as in free space, the Maxwell’s equation will obey a wave equation but withthe the speed of propagation for light in a linear and homogeneous medium given by
v =1√
εµ≡ c
n
n ≡√
εµ
εoµo
where n is the index of refraction for the medium and ω
|~k|= c
n .For most magnetic materials, we
have usually that µ ∼ µo → n ∼ √εr, and in that case
εr = 1 + χE ≡ε
εoε ≡ εo(1 + χE)
where χE is the electric susceptibility for the medium. Similarly, the energy density, E , andPoynting’s Vector, ~S will be given in the medium by
60
E =12
(ε| ~E|2 +
| ~B|2
µ
)~S =
1µ
( ~E × ~B)
3.3 Reflection and Refraction at Interfaces
We now have developed a framework, using plane wave, for the propagation of electromagneticwaves in free space, but also in a medium free of charges and current. In this section, we willexplore the propagation of electromagnetic waves at the interfacial region between two media withindex of refraction n1 and n2.
3.3.1 Normal Incidence
We consider first the case where the wave vector of the incident propagating wave ~ki is normal toan interface defined by z = 0. In that case, there will be a reflected and transmitted waves all witha propagating wave vector perpendicular to the interface.
kr
ki kt
n1 n2
Let’s consider a plane wave going from medium 1 to medium 2 with corresponding indicies ofrefraction n1 and n2. Since there is no free charges or current at the interface, the boundaryconditions will impose that all components of electric field and displacement and magnetic field arecontinuous.
Boundary Conditions
~E||, ~H|| are continuous~B⊥, ~D⊥ are continuous
So that we can write by considering the z = 0 interface
61
ε1Ez(0−) = ε2Ez(0+)
~H||(0−) = ~H||(0+)
~E||, Bz continuous at z = 0.
We consider that the incident plane wave is normal to the interface, and similarly to the transmittedand reflected wave, i.e.
~ki ≡ ~k1 = |~k1|z~kr = −|~k1|z~kt = |~k2|z
We can therefore write the following fields:
~Ei = E0i e
i(k1z−ωt)x
~Er = E0re
i(−k1z−ωt)x
~Et = E0t e
i(k2z−ωt)x
E0i , E
0r , E
0t are the amplitudes of incident, reflected and transmitted waves that are to be deter-
mined.
Note:
1. we have supposed an electric field with a polarization following x
2. the angular frequency ω must be the same for all waves, but, ~k is different since it depends onthe medium properties (n). The frequency is the same for all waves because the discontinuityis in space, not in time.
Since (in magnitude) the magnetic fields is related to the electric field through B = |~k|ω E, the
corresponding incident, reflected and transmitted magnetic fields are:
62
~Bi =E0
i
v1
ei(k1z−ωt)y
~Br = −E0r
v1
ei(−k1z−ωt)y
~Bt =E0
t
v2
ei(k2z−ωt)y
where we have used defined the speed of propagation in the medium vi and the fact that ~B = |~k|ω k× ~E
vi =1
√εiµi
=c
ni=
ω
|~ki|
The following boundary conditions give
~E|| continuous ⇒ E0i + E0
r = E0t
~H|| continuous ⇒ 1v1µ1
(E0i − E0
r ) =1
v2µ2E0
t
which gives:
E0r =
(1− β
1 + β
)E0
i
E0t =
(2
1 + β
)E0
i
where we have defined
β ≡ µ1v1µ2v2
=µ1n2
µ2n1
.
If we further assume µ1 ∼ µ2 ∼ µo then they reduce to
63
E0r =
(n1 − n2
n2 + n1
)E0
i
E0t =
(2n1
n2 + n1
)E0
i
Note:
1. v2 > v1 (n2 < n1) → Reflected waves is in-phase with incident
2. v2 < v1 (n2 > n1) → Reflected waves is out of phase with incident (Er is of opposite signwith respect to Ei and so out of phase by π).
3. v2 = v1 (n2 = n1) → No reflected waves, but Et = Ei.
We define the intensity of energy as
Iα =12εvE2
α
and we define the reflection and transmission coefficients as the ration of the intensities
R ≡ IrIi
R ≡(E0
r
E0i
)2
=(n1 − n2
n1 + n2
)2
T ≡ ItIi
=ε2v2ε1v1
(E0
t
E0i
)2
=4n1n2
(n1 + n2)2.
These coefficients, R and T, measure the fraction of the incident energy that is reflected andtransmitted at the interface, and we can verify that, owing to energy conversation, they obey therelation
R + T = 1
.
64
3.3.2 Oblique Incidence
In this section, we consider the case where the the wave vector for the incident wave is not normalto the interface, i.e. there will be an angle θi 6= 0 with respect to the normal of the interface.
z
kr
ki
!i
!r !t
kt
n1 n2
Again, we suppose a monochromatic plane wave of the form
~Ei(~x, t) = ~E0i e
i(~ki·~x−ωt)
~Er(~x, t) = ~E0r e
i(~kr·~x)−ωt)
~Et(~x, t) = ~E0t e
i(~kt·~x)−ωt)
and Maxwell’s equations impose that the magnetic field be of the form
~Bi(~x, t) =1v1ki × ~Ei
~Br(~x, t) =1v1kr × ~Er
~Bt(~x, t) =1v2kt × ~Et.
Important Notes:
1. We have now that the wave vector ~kα makes an angle θα with the normal vector, n, of theinterface.
2. The plane defined by ~ki,~kr, and~kt is called the plane of incidence.
The boundary conditions must be satisfied everywhere at the interface (ie. at z = 0) and so thisimply that the phase factor ei(~kα·~x−ωt) must be the same for ~Ei, ~Er, ~Et at z = 0. Since all wavevectors are lying in a same plane, and choosing the direction y perpendicular to the plane, and xin the plane of incidence, the requirement that at z = 0 all phase must be equal is equivalent to
65
~ki · x = ~kr · x = ~kt · x
which means that the x-component of the wave vector must be equal at the interface. But (~ki)x =|~ki|sinθi, and similarly for the reflected and transmitted waves, so the condition above becomes:
kisinθi = krsinθr = ktsinθt
.
We recall the dispersion relation for an electromagnetic wave in a medium free of charges andcurrent
ω(~k)
|~k|=c
n
⇒ ω
c=|~k|n
and therefore in magnitude the incident and reflected wave vector must be equal
⇒ ki = kr
whereas the transmitted wave vector in magnitude is given by
⇒ kt
n2=ki
n1.
As a consequence of the boundary condition imposed at z = 0 which requires that kisinθi = krsinθr = ktsinθt,we recover the Law of Snell-Descartes:
θi = θr
and
n1 sin θi = n2 sin θt
A wave, in general, can be considered as the superposition of two polarizations: one linear polar-ization with ~Ei that is perpendicular to the plane of incidence, and one that is linear with ~Ei insideor parallel to the incident plane. So we can therefore write in general:
~Etot = ~E⊥ + ~E||
66
.
Our goal is now to solve for ~Ei, ~Er, ~Et for both cases, i.e with the incident electric field having apolarization that is perpendicular and parallel to the plane of incidence.
i) Perpendicular Polarization
For the perpendiculat polarization, we consider the electric field to be polarized perpendicularlywith respect to the plane of incidence.
z
kr
ki
Ei
Hi
Et
Ht!i
!r !t
kt
n1 n2
Er
Hr
Boundary conditions imply:
Continuity of ~E||:
E0i + E0
r = E0t (1)
Continuity of ~H||:
1µ1v1
(E0i − E0
r ) cos θi =1
µ2v2E0
t cos θr (2)
and defining
β =µ1v1µ2v2
=µ1n2
µ2n1
then from equations (1) and (2) we get the Fresnel Equations for perpendicular polarization
E0r =
cos θi − β cos θt
cos θi + β cos θt
E0i
E0t =
2 cos θi
cos θi + β cos θt
E0i
67
or, for the special case where µ1 = µ2
E0r ' sin(θt − θi)
sin(θt + θi)E0
i
E0t ' 2 cos θi sin θt
sin(θt + θi)E0
i
.
(Note that the explicit algebra for E0r , E
0t is left as an exercise)
ii) Parallel Polarization
In the case of parallel polarization, the electric field of the electromagnetic wave is lying in theplane of incidence.
z
kr
ki Ei
Hi
Et
Ht
!i
!r !t
kt
n1 n2
Er
Hr
Boundary Conditions are:
Continuity of ~E||
(E0i − E0
r )cosθi = E0t cosθt
Continuity of ~H||:
1µ1v1
(E0i + E0
r ) =1
µ2v2E0
t (2)
and again, using the same definition for β as above we can combine these two equations to get theFresnel equations for parallel polarization:
68
E0r =
cos θi − β−1 cos θt
cos θi + β−1 cos θt
E0i
E0t =
2 cos θi
cos θt + β−1 cos θi
E0i
or, for the special case where µ1 = µ2
E0r ' tan(θi − θt)
tan(θi + θt)E0
i
E0t ' 2 cos θi sin θt
sin(θi + θt) cos(θt − θi)E0
i
3.3.3 Brewster angles
Let’s consider the Fresnel equations for parallel polarization and µi = µ2
E0r '
tan(θi − θt)tan(θi + θt)
E0i
In that case, we note that if θi + θt = π/2 then the reflected wave amplitude goes to zero. Theincident angle which satisfies this condition is termed the Brewster Angle, or angle of totalpolarization, θp
tan θp =sin θp
cos θp
=sin θp
sin(π/2− θp)
=n2sinθt
n1 sin(π/2− θp)
=n2
n1
tan θp =n2
n1
69
For glass, n2 ∼ 1.5 and for air n2 ∼ 1 giving θp ∼ 56o.
For θp ∼ 56o, only the polarization of light where ~E is normal (perpendicular) to theincidence plane will be reflected. The Brewster angle therefore provides a simple way topolarize light.
3.3.4 Total internal reflection
Now consider a situation were n1 > n2, i.e. as in the case where light passes from glass to air.
n1 sin θi = n2 sin θt
There is a critical angle, θi ≡ θc where the refracted wave does not seem to exist, i.e. for sin θt =sinπ/2 = 1:
θc = sin−1
(n2
n1
)
which is the critical angle for total internal reflection.
What is happening at θi = θc? Let’s investigate the transmitted component of the electric field:
~Et = ~E0t e
i(ktx sin θt+ktz cos θt−ωt)
but owing to the trigonometric relation
cos2 θt + sin2 θt = 1
and the law of Snell-Descartes
n1 sin θi = n2 sin θt
from these two equations we get
cos2θt = 1− sin2θi
(n2/n1)2= 1− sin2θi
sin2θc
70
cosθt =
√1− sin2θi
sin2θc
which is ∈ Re for θi < θc, and ∈ C for θi > θc.
For θi > θc then we can define cos θt ≡ iξ, ξ =√
sin2θisin2θc
− 1, and the transmitted wave becomes
~Et = ~E0t e−ktξzei(ktx sin θt−ωt)
.
An incident wave crossing the interface at an angle beyond the critical angle θi > θc will have itsamplitude decay along the interface over a characteristic length δ defined by
δ ≡(
1ktξ
).
3.4 Absorption and Dispersion in Conductors
3.4.1 Dispersion of a wave packet in general
We say that there is “dispersion” when the index of refraction (or ε) of the medium is a functionof frequency. In general, n is always a function of ω (i.e. n = n(ω)), but this dependency can bemore, or less, pronounced.
In a region of the EM spectrum where dispersion is pronounced, a wave packet formed by asuperposition of monochromatic plane waves of near-equal frequencies will have its shape distortedin time. Because phase velocity is a function of ω, v = v(ω), the packet will be dispersed. Similarly,a wave is more or less refracted following n, thus depending on ω.
Consider a wave packet, Ψ(x, t), which is a continuous superposition of plane waves with amplitudesA(k). We assume plan wave, and Ψ is noted here to denote either ~E or ~B.
Ψ(x, t) =∫dk
2πA(k)ei(~k·~x−ωt)
If we have no dispersion, then ω = kv, v is the phase velocity. Lets define
ξ = x− vt
as D’Alembert’s solution. Then
71
Wave Dispersion
Ψ(x, t) =∫dk
2πA(k)eikξ
= A(ξ)= Ψ(x− vt, 0)
where A(ξ) is the Fourier transform of A.
But what if ω is a linear function of k? We still have v = ω/k for phase velocity, but the ratio ω/know depends on frequency. Lets write
ω(k) ∼= ωo + vg(k − ko) +12a(k − ko)2
where vg is the group velocity, given by
vg =(dω
dk
)k0
Then we have, if we ignore for now the quadratic term in ω(k)
Ψ(x, t) =∫dk
2πA(k)eikξ · e−i(ωo−vgko)t
= A(ξ)e−i(ωo−vgko)t
= Ψ(x− vgt, 0)e−i(ωo−vgko)t
72
i.e., Ψ(x, t) is the product of a wave propagating with speed vg (1st term) and an oscillating functionin time (2nd term). Note that vg, the “group velocity”, i.e. the velocity of the wave packet as awhole.
Wave Packet
vg
v
In general, a medium will be dispersed if the index of refraction depends on thefrequency: n = n(ω), i.e. when:
c
n=ω
k
and we define the phase velocity vφ and group velocity vg for a dispersive medium as:
vφ = ωk
vg = dωdk
3.4.2 Drude Model of the Electric Susceptibility in Matter
We are now interested in the deriving a microscopic model for the dialectric constant in materialswhich shall provide us insights as to whether or not the medium is dispersive and attenuating.The simplest model is know as the ‘Drude model’ and it gives us a general form for the frequencydependence of the dialectric constant in a medium. The model considers the atom as a harmonicoscillator or as an ensemble of oscillators, i.e. the electron cloud is harmonically bound tothe nucleus with a characteristic frequency ω.
We want to study the response of such an oscillator upon the application of an oscillating electricfield with frequency ω.
~E = ~Eoei(~k·~x−ωt)
73
Drude Model
E
electroncloudNucleus
The electric field gives a force, ~F = −e ~E on the electric cloud, which in turn has a restoring force−mω2
o~x. We also add a dissipative friction force proportional to the speed of the electron cloud.∑ ~Fi = m~a gives
m~x+mγ~x+mω2o~x = −e ~Eoe
−iωt
which is the equation of a forced harmonic oscillator where m is the electron mass, γ is the frictioncoefficient/unit mass, and ωo is the characteristic frequency of the oscillator.
The general solution to this equation is given by
~x(t) = Transient︸ ︷︷ ︸neglected
×gstationary
The stationary solution is of the form
~x(t) = ~xoe−iωt
Substituting into the equation of motion one gets
−ω2m~xo − imωγ~xo +mω2o~xo = −e ~Eo
which gives
~xo =−e
m
~Eo
ω2o − ω2 − iγω
74
This equation is equivalent to an oscillating dipole ~p = ~p0e−iωt,
~p = q~x = −e~x(t) ≡ ~poe−iωt
so that the dipole magnitude is given by
~po =e2
m
~Eo
ω2o − ω2 − iγω
.
If we now suppose a medium containing several elements with relative importance or fraction fα,each of which with oscillators frequency ωα and damping γα per unit volume, the polarization,~P = ~p ·N(~r) is simply given by
~P =∑α
Nfαe2/m
ω2α − ω2 − iωγα
~E
and since for a linear and isotropic medium the electric susceptibility is given by ~P = ε0χe~E, we
obtain the frequency-dependent susceptibility
χe(ω) =∑α
Nfαe2/mε0
ω2α − ω2 − iωγα
.
We note that the electric susceptibility is now complex, and therefore so will be in that model forthe dialectric constant εr = ε
ε0= (1 + χe)
εr = 1 +Ne2
mε0
∑α
fα
ω2α − ω2 − iωγα
.
Note:
1. We call ε the permittivity of the material. In a linear medium, we must have that ~D = ε ~E,whereas εr is called the relative permittivity or dialectric constant of the material.
2. The tilde, i.e εr indicates that the dialectric constant is now complex, i.e.
ε ≡ Re(ε) + iIm(ε)
. Usually, the complex part can be neglected, but if the frequency of the electromagneticwaves is such that ω ' ωα, it will play an important role.
75
3. There could now be an out of phase component between ~D and ~E since
~D = ε ~E
3.4.3 Complex Refraction Index and Anamolous Dispersion
If the dialectric constant is complex, as it is the Drude model, the electromagnetic wave in themedium will be attenuated since the wave vector, ~k will now has an imaginary component as well.
We recall that the wave equation for the electric field is
∇2 ~E = εµo∂2 ~E
∂t2
and so the plane wave solution will admit the following dispersion relation
k =√εµoω
and so the wave vector will now be complex, which we can define as k = k + iκ and the electro-magnetic wave will be attenuated according to
~E(z, t) = ~E0(z, t)e−κzei(kz−ωt)
.
It is usual to define the absorption coefficient for the intensity of the electromagnetic wave whichvaries as the square of the electric field amplitudes, E2, and hence as e−2κz. We shall denote byα ≡ 2κ the attenuation coefficient. Conversely, we can define an attenuation length δ, or in aconductor a penetration depth for the electromagnetic wave as δ ≡ κ which is the characteristiclength over which the wave will be attenuated by 1/e. The electromagnetic wave will propagatewill propagate with the dispersion relation ω
k = cn , where k is the real part of k. Similarly, we can
define the complex index of refraction n ≡√εr so that k = ω
c n.
We are now interested in calculating the frequency dependence of k and κ, and to achieve thiswe need to calculate
√εr. For a dilute gases of oscillators, the frequency-dependent term in the
dialectric constant of the Drude model is usually much smaller than 1, so we can approximate thesquare root as
√1 + x ' 1 + x
2 , for x 1. And so we obtain for the dispersion in general:
76
k =ω
c
√εr '
ω
c(1 +
Ne2
2mε0
∑α
fα
ω2α − ω2 − iωγα
).
and it can be easily shown that, separating the real and imaginary part of k that we obtain thefollowing relation for n, the real part of n, and the attenuation coefficient α:
n =ck
ω' 1 +
Ne2
2mε0
∑α
fα(ω2α − ω2)
(ω2α − ω2)2 + ω2γ2
α
α = 2κ ' Ne2ω2
mε0c
∑α
fαγα
(ω2α − ω2)2 + ω2γ2
α
.
Let’s examine the above relations for a single type of elements, near a mode frequency ωα andattenuation γα. We can see that the index of refraction n goes to a finite value with ω → 0. Itwill then increase up to ω = ω1 < ωα, frequencies at which dn
dω = 0. In that region, the dispersionis said to be normal, since dn
dω > 0. In that regime, blue light is more refracted than red light andthe attenuation of light is relatively small. The attenuation is always positive and will increasefrom zero to a maximum at ω = ωα, and will drop for ω > ωα. The index of refraction will beexactly n = 1 at ω = ωα and will then drop to values less than 1. In that regime, where n < 1,the phase velocity of the electromagnetic wave will be greater than the speed of lightc. It will then be at a minimum at ω = ω2 > ωα and then goes towards n = 1. In the range offrequencies ω1 < ω < ω2, the dispersion is said to be anomalous, for dn
dω < 0. The attenuation inthis regime is very large, and the medium is relatively opaque.
3.4.4 Dispersion and conductivity in a conductor
In a conductor, there is a density of free electrons that are not bound to the atoms (or lattice)and so we can model it using the Drude model and setting the resonant frequency ωα = 0. Now,N is the density of free electrons (#/unit volume). The complex conductivity in a conductor istherefore
εr = 1− Ne2/Mε0
ω2 + iγω
77
where now the sum over all elements is gone since we are now considering only the conductionelectrons. Starting from the Drude model, we can write the equation of motion for the freeelectrons as
m~x+mγ~x = −e ~Eoe−iωt
which has for solution
~x = +(e/M) ~Eω2+iωγ
~v ≡ ~x = −(e/M) ~Eγ−iω.
From the definition of the current density ~J = −Ne~v, we can therefore write
~J = −Ne~v =N(e2/M) ~Eγ − iω
and we define the complex conductivity σ as
σ =N(e2/M)
γ − iω
such that we have
~J = σ ~E.
We shall refer to σ as the AC-conductivity of a conductor.
Note:
1. In the DC limit, we put ω = 0, we therefore we obtain the DC conductivity
σdc =Ne2
mγ
2. ~J is not necessarily in phase with ~E since σ ∈ C and ~J = σ ~E. That implies that current andthe electric field are not necessarily in phase.
78
3.4.5 Propagation in a Conductor
In free space, we have derived the propagation of electromagnetic waves assuming that the densityof electrons ρ and current density ~J was zero. In a conductor, however, there is always a finitedensity of free electrons which, if an electric field is applied, will generates a current density~J = ρ~v where ρ is the density free electrons. This latter is dictated in most cases by Ohm’s lawwhich states that current density of free electrons must be proportional to the electric field, ~J = σ ~E.The Maxwell’s equation in that case must be modified to account for these free electrons in thefollowing way:
~∇ · ~E = ρ/ε
~∇× ~E = −∂ ~B
∂t
~∇ · ~B = 0
~∇× ~B = µε∂ ~E
∂t+ µσ ~E.
and the continuity equation must satisfy ~∇ · ~J = −∂ρ∂t , which upon replacing Ohm’s law gives
∂ρ∂t = −σ~∇ · ~E = −σ
ε ρ. Solving this 1st order differential equation for ρ gives
ρ(t) = e−(σε)tρ(0)
or, in other words, the free density of charges will dissipate in a characteristic time τ ≡ σε and
thus is very short for a good conductor τ 1/ω and very long for a poor conductorτ 1/ω. This defines a transient time for the charges to flow to the edge, and so we are interestedin the solution of Maxwell’s equations at a time t τ which will be given by
~∇ · ~E = 0
~∇× ~E = −∂ ~B
∂t
~∇ · ~B = 0
~∇× ~B = µε∂ ~E
∂t+ µσ ~E.
and it is easy to show that they lead to the following wave equation for the electromagnetic fields:
79
~∇2 ~E = µε∂2 ~E
∂t2+ µσ
∂ ~E
∂t
~∇2 ~B = µε∂2 ~B
∂t2+ µσ
∂ ~B
∂t
and assuming a plane wave solution as usual of the form ~E(z, t) = ~E0ei(kz−ωt), with k complex, weobtain the following dispersion relation:
k2 = ω2
(µε+
iσµ
ω
)where as perviously we have defined k ≡ k+ iκ. The electromagnetic field will then be attenuatedas
~E(z, t) = ~E0e−κzei(kz−ωt)
and so the electromagnetic field will penetrate the conductor over a distance, called skin depth,corresponding to a decrease in amplitude by 1/e, or
δ ≡ 1κ.
The explicit form of the real and imaginary part of k, with respect to the dispersion relation givenabove, will be left as an exercise.
80
3.4.6 Historical Aside: James Maxwell (1821-1879)
James Clerk Maxwell needs little introduction. His contributions to physics are amongst the mostnotable in the entire discipline. What perhaps is less known about the man, was his activities andinterests outside the gridiron of analytic derivations and physical formulae. Shown here is the firstpage from one of his many poems composed throughout his life. This one from childhood, in 1845.
This poem recounts a knight’s seduction by a vampire. Other titles include: To the Chief MusicianUpon Nabla, an ode to one Professor Tait, who first discussed Hamilton’s del operator, ∇ (thenknown as nabla), and also A Problem in Dynamics, which begins:
An inextensible heavy chainLies on a smooth horizontal plane,An impulsive force is applied at A,Required the initial motion of K...
[see The Life of James Clerk Maxwell, by Lewis Campbell and William Garnett, 1882, for more]
81
4 Guided Waves
4.1 Transverse Electric and Magnetic Modes
In free space we have shown that Maxwell’s equations obey a wave equation where the electric,magnetic and k-vector are forming an orthogonal tryad. We will now consider the guided trans-mission of EM waves in the interior of a hollow pipe. Here, we are interested in a hollow conductorwith walls that are perfectly conducting so that both the electric and magnetic fields must van-ish at the material surface. The waves that will propagate inside the hollow pipe will be confined,and in general such waves will not be transverse, so a component of the electric or magnetic fieldalong the propagation axis may exist.
4.1.1 Helmholtz equation for the transverses components
We assume a structure with a translational symmetry following the z axis and an arbritary cross-section
z
Hollow Conductor
and we are assuming that the wave is a monochromatic plane wave which propagates along thez-direction, i.e. ~k = kz.
~E(~x, t) = ~Eo(x, y)ei(kz−ωt)
~B(~x, t) = ~Bo(x, y)ei(kz−ωt)
i.e. ~Eo, ~Bo are not a function of z per symmetry argument. Their explicit form is thus ~Eo =Ex(x, y)x+ Ey(x, y)y + Ez(x, y)z and ~Bo = Bx(x, y)x+By(x, y)y +Bz(x, y)z.
As always, ~E and ~B must satisfy the following Maxwell’s equations
82
~∇ · ~E = 0
~∇× ~E = −∂ ~B
∂t~∇ · ~B = 0
~∇× ~B =1
c2
∂ ~E
∂t
and the problem is now to solve Maxwell’s equations for ~Eo, ~Bo which only depends on transversevariables (x,y).
We also note that ~Eo, ~Bo must satisfy the boundary conditions at the inner wall
~E|| = 0
~B⊥ = 0
where we have assumed the material to be a perfect conductor.
In general, a guided wave will not be transverse, i.e. we must include the Ez and Bz componentsso we write
~Eo = Exx+ Eyy + Ez z
~Bo = Bxx+Byy +Bz z
and so the goal is to show that if Ez and Bz are known, that the transverse components aredetermined too. If that is the case, the problem is reduced to solve for the ttransverse componentonly.
From Maxwell’s equations we have
~∇× ~E +∂ ~B
∂t= 0 (i)
~∇× ~B − 1c2∂ ~E
∂t= 0 (ii)
Examining the vector components of equation (i) component by component yields
83
(~∇× ~E)z =∂Ey
∂x− ∂Ex
∂y= iωBz
(~∇× ~E)x =∂Ez
∂y− ∂Ey
∂z= iωBx
(~∇× ~E)y =∂Ex
∂z− ∂Ez
∂x= iωBy
and since the partial derivation ∂Ey
∂z = ikEy and ∂Ex∂z = ikEx
z :∂Ey
∂x− ∂Ex
∂y= iωBz
x :∂EZ
∂y− ikEy = iωBx
y : −∂EZ
∂x+ ikEx = iωBy
and, similarly, from (ii) we have
z : ∂By
∂x− ∂Bx
∂y= −iω
c2Ez
x : ∂BZ
∂y− ikBy = −iω
c2Ex
y : −∂BZ
∂x+ ikBx = −iω
c2By
Our goal is now to solve for the transverse components (Ex, Ey) and (Bx, By) as a function oflongitudinal components Ez, Bz. From the relations above, we can easily show that they result into
Ex =i
(ω/c)2 − k2
(k∂Ez
∂x+ ω
∂Bz
∂y
)Ey =
i
(ω/c)2 − k2
(k∂Ez
∂y− ω
∂Bz
∂x
)
Bx =i
(ω/c)2 − k2
(k∂Bz
∂x− ω
c2∂Ez
∂y
)By =
i
(ω/c)2 − k2
(k∂Bz
∂y+ω
c2∂Ez
∂x
)84
But every components of the fields must satisfy to ~∇ · ~E = ~∇ · ~B = 0, so
~∇ · ~E =∂Ex
∂x+∂Ey
∂y+∂Ez
∂z= 0
Which gives, using the equations for Ex and Ey, and with ∂Ez∂z = ikEz
i
(ω/c2)− k2
k∂2Ez
∂x2+ k
∂2Ez
∂y2+
ω∂2Ez
∂x∂y−
ω∂2Ez
∂x∂y
+ ikEz = 0
and therefore we obtain the following equation for Ez
[∂2
∂x2+
∂2
∂y2+ (ω/c2)2 − k2
]Ez = 0
or, as it more usually written
[∂2
∂x2 + ∂2
∂y2 + γ2
]Ez = 0
γ2 ≡ (ω/c)2 − k2
known as the Helmholtz equation. Similarly, its can easily be shown that
[∂2
∂x2+
∂2
∂y2+ γ2
]Bz = 0
The wave guide problem is now reduced to solving the Helmholtz equation for Ez and Bz withthe appropriate boundary condition coming from the in x-y plane, and with the conducting (ordialectric) surface separating the two media. Once the Ez or Bz components are known, all othersare easily determined with the equations above.
Helmholtz euqation + Boundary Conditions −→ Ez, Bz −→ (Ex, Ey), (Bx, By)
85
4.2 Electromagnetic Waves Modes
In free space, as we have seen before, electromagnetic waves are always transverse i.e. the ~E and~B fields are mutually perpendicular with ~k. For a guided wave, this no longer true and we mustconsider different propagating modes (for a given frequency ω). We shall consider several modes ofpropagation, some where Ez = 0 that we call transverse electric (TE), some with Bz = 0 known astransverse magnetic (TM) and the case where Ez, Bz = 0 known as TEM.
1.) TEM mode (Transverse electric and magnetic) Here, ~E and ~B are perpendicular to ~k, as infree space. We shall see that this is impossible in a closed waveguide.
2.) TM mode (Transverse magnetic) In this case, ~B is perpendicular to the direction of propa-gation, but Ez 6= 0. We must solve [∂2
x + ∂2y + γ2]Ez = 0
3.) TE mode (Transverse Electric) In this case, ~E is perpendicular to the direction of propagation,but Bz 6= 0. We must solve [∂2
x + ∂2y + γ2]Bz = 0
4.2.1 TEM modes propagation
In the case of TEM modes, we have Ez = Bz = 0. The Maxwell’s equation will impose that
~∇ · ~E = 0 → ∂Ex
∂x+∂Ey
∂y= 0
~∇× ~E = −∂Bz
∂t= 0 → ∂Ex
∂y− ∂Ey
∂x= 0
~Eo = Exx+ Eyy has zero divergence and zero curl. It can therefore
~Eo = −~∇ψ
where ψ is a scalar potential that satisfies the Laplace equation; ∇2ψ = 0. However, the boundaryconditions at surface requires that it is an equipotential and since the Laplace equation ∇2ψadmits no local minima or maxima, ψ must be defined constant throughout the hollow conductorand therefore no wave propagation occurs (the electric field inside! is zero everwhere!).
However, note that TEM modes can exist outside of a hollow conductor. For example, propagatingmodes surrounding a conducting wire which would propagate at speed c = ω/|~k|
4.3 Rectangular Waveguide
We are now considering the case where the boundary is given by a hollow pipe, with walls perfectlyconducting, with a square cross section. The dimension of cross section are denotes by a and bin the x and y direction, respectively.
86
k
TEM modes can’t exist inside a hollow conductgor
Outside: TEM mode is possible
Inside: E=0, no TEM mode
z
x
y
Rectangular Wave Guide
a
b
Boundary conditions at the surface are:
~E|| = 0 at surface~B⊥ = 0 at surface
4.3.1 TE Modes (Ez = 0, Bz 6= 0)
We first consider the TE modes (Ez = 0, Bz 6= 0) where we must solve the Helmholtz equation forthe magnetic field along z:
[∂2
∂x2 + ∂2
∂y2 + γ2
]Bo
z = 0
γ2 = ω2
c2− k2
Since the boundary condition can easily be expressed in cartesian coordinates, we can solce theHelmholtz equation in these coordinates by separating the variables. We write
87
Bz(x, y) = X(x)Y (y)
so that, substituting back into the Helmholtz equation gives
Yd2X
dx2+X
d2Y
dy2+ γ2XY = 0
1X
d2X
dx2︸ ︷︷ ︸=−k2
x
+1Y
d2Y
dy2︸ ︷︷ ︸=−k2
y
+γ2 = 0
−k2x +−k2
y + γ2 = 0
where kx and ky are constant. So the general solution is of the form
X(x) = Asin(kxx) +Bcos(kxx)
for which we must apply the boundary conditions : first, Bx must vanish at the surface i.e. atx = 0, a. Recalling that
Bx =1
(ω/c)2 − k2
(k∂Bz
∂x− ω
c
70 for TE mode
∂Ez
∂y
)it implies that at the boundary
∂Bz
∂x= 0 → ∂X
∂x= 0 → X ′ = Akxcoskxx−Bkxsinkxx = 0
The boundary condition at x = 0, since cos(0) 6= 0, implies that
A ≡ 0
and the the boundary condition at x = a imposes that
kxa = mπ
kx = mπa
where m = 0, 1, 2, 3 . . ..
And similarly for Y (y), and the boundary conditions at y = 0, b where ∂By
∂y = 0
88
ky =nπ
bn = 0, 1, 2 . . .
We now notice that the k − vectors along x and y are now discrete, and are labeled according tothe m,n indices. The complete solution for Bz(x, y) is therefore
Bz = B0cos
(mπx
a
)cos
(nπy
b
)m, n ∈ N
where we define B0 the amplitude of the mode. Finally, to obey the Helmholtz equation, the kx
and ky must obey
γ2 = π2
(m2
a2+
n2
b2
)= ω2/c2 − k2
and so the wavevector will be discretized according to
k =
√(ω
c
)2
− π2
(m2
a2+
n2
b2
).
We will define the mode frequencies ωmn as
ωmn ≡ πc
√m2
a2+n2
b2
so that
k =1c
√ω2 − ω2
mn.
We immediately see that there is a frequency cutoff ≡ ωcmn for which if ω < ωc
mn ⇒ k ∈ C ⇒ k =Rek+ iκ. That means that the wave will not propagate below that frequency cutoff, and insteadbe exponentially damped i.e if ωc
mn the cutoff, then when ω < ωcmn, the EM wave and does not
propagate.
89
If a > b, the cutoff is for the TE modes is the TE10 mode associated wth the ω10 frequency
ω10 =πc
a
NOTE: Given Bz, we can easily obtain Ex, Ey, Bx, By using the equations given above. We callthis solution the TEmn solution.
4.3.2 TM Modes (Ez 6= 0, Bz = 0)
The approach to finding the TM solutions is the same as for TE modes, but here we set Bz = 0and we are solving for Ez with the conditions that Ez must vanish at the surfaces. By solving theHelmholtz equation in a similar way by separation of variables, the solution is
Ez = E0sin
(mπx
a
)sin
(nπy
b
)ωmn = πc
√m2
a2 + n2
b2
but we must note that here both indices m AND n must be non-zero for waves to propagate.The lowest frequency of the TM mode will therefore be the TM11 mode.
4.3.3 Dominant modes and velocities
ωmn = πc
√m2
a2+n2
b2
Assuming that a > b, then the lowest cutoff frequency is ω10 = πc/b. The TE10 is therefore thedominant mode. We typically design waveguides such that only the dominant mode propagates,and we set its frequencies at the propagating frequencies of the waves that we want to propagate.
A hollow waveguide cane be considered as propagating electromagnetic waves dispersively. Let’scalculate the phase and group velocity for the rectangular wave guide:
Phase velocity:
vφ ≡ω
k=
c√1− (ωmn
ω )2> c
90
assuming that ω > ωcmn. The phase velocity inside a wave guide is there fore greater than the speed
of light.
Group velocity:
vg =1
dk/dω= c
√1− (
ωmn
ω)2
The group velocity can be expressed as a simple ratio of the speed of light and the phase velocity,and we for the rectangular wave guide with perfectly conducting walls, the following relation holdsvgv
2φ = c2.
Group Velocity
!
v"
vg
velo
city
c
!mn
We note that vφ always > c, whereas vg always < c.
4.4 Circular wave guides
4.4.1 Hollow circular waveguide
Consider the TM modes and set ε = µ = 1. We must solve the Helmholtz equation
[~∇2 + γ2]Eoz = 0
in cylindrical coordinates. We must therefore solve
1r
∂
∂r
(r∂Ez
∂r
)+
1r2∂2Ez
∂φ2+ γ2Ez = 0 (*)
91
with the boundary condition Ez(r = a,∀φ) = 0.
The solution of (*) is the bessel function Jm(r).
Ez(r, φ) ∝ Jm(γr)cos(mφ+ γ) m = 0, 1, 2 . . .
where γ is a meaningless phase. The boundary conditions imply
Jm(γa) = 0 ⇒ γ =xmn
a
where J(xmn) = 0. ie., xmn are the Bessel function zeros, or the nth roots of the order m Besselfunction. The dispersion relation becomes
ω2 = ω2mn + c2k2
ωmn = cxmn
a
where m ≥ 0, n ≥ 1.
The general solution for the TM modes can therefore be written as
Ez(r, φ) = EomnJm
(xmnr
a
)cos(mφ + αmn)
For a mn mode, m indicates the order of the Bessel function, n are the roots of mth order Besselfunction.
TE modes: Similarly, the TE modes are given with boundary conditions that ∂Bz∂x,y = 0 at r = a,
and so
J ′m(γa) = 0 ⇒ γ =ymn
a
where J ′(ymn) = 0 and ymn are the nth roots of the derivative of Jm.
92
Bz(r, φ) = BomnJm
(ymnr
a
)cos(mφ + βmn)
ω2 = N2mn + c2k2
ωmn = cymn
a
Once the Ez (or Bz) components are known, we can calculate the transverse components of thefield by using (ε = µ = 1)
Er =1γ2
iω
c
1r
∂Bz
∂φ+ ik
∂Ez
∂r
Eφ =
1γ2
− iωc
∂Bz
∂φ− ik
1r
∂Ez
∂φ
Br =1γ2
− iωc
1r
∂Ez
∂φ+ ik
∂Bz
∂r
Bφ =
1γ2
iω
c
∂Ez
∂φ+ ik
1r
∂Bz
∂φ
4.5 Electromagnetic Cavity
4.5.1 What is an EM cavity?
An electromagnetic cavity is simply a conductor closed on all sides, and for which the EM field canoscillate at some frequencies. No monochromatic wave can travel in it, but, we can have standingwaves at discrete frequencies ωmnr, forming inside it.
The problem is to find the natural frequencies of the cavity, as well as the configuration of the ~E, ~Bfields for each oscillating mode.
We consider a rectangular waveguide to which we add two surfaces, one at each end at distancez = 0, d. Since, we know the solution for an unbounded conductor waveguide, we can consider thesolution as the sum (along z) of an incident plus reflected wave in z direction which interfere witheach other and that of the wave-guide along the x and y directions. The interference along z willcreate standing waves.
Consider TE modes, ie Ez = 0. In the z-direction, at the new interfaces, the Bz component mustobey to the new boundary conditions along z
93
z
x
y
Finite Rectangular Wave Guide
a
b
Bz = 0 at z = 0, d
⇒ 12i
(Bzeikz −Bze
−ikz) = Bzsinkz
Applying the boundary conditions: z = 0 gives simply Bzsinkz = 0 (i.e. this boundary conditionsgives no new information) but at z = d, it gives Bzsinkd = 0, which will be true only whenkd = rπ, r ∈ N
The dispersion relation is, for the wave guide
ω2 = ω2mn + c2k2
where we now have c2k2 to be discrete, i.e. according to
k2 =(rπ
d
)2
which gives finally
ωmnr =
√ω2
mn +
(rπc
d
)2
And the solution for Bz is
94
Bz = Bmnrcos
(mπx
a
)cos
(nπy
b
)sin
(rπz
d
)
Similarly, it is easy to show that for TM modes, the solution is:
Ez = Emnrsin
(mπx
a
)sin
(nπy
b
)cos
(rπz
d
)
4.5.2 Quality Factor of a Cavity
ω2mn = πc
√m2
a2+n2
b2+r2
d2
A priori it seems impossible to excite an EM cavity unless the excitation frequency is exactly thatof the cavity resonance, given by the frequencies ωmnr. In practice, however, there will be someenergy losses which will make the spectrum of the cavity going from a theoretical delta functionδ(ω − ωmnr) to that of a spectrum of peaks with finite widths owing to the finite conductivity ofthe walls.
We define the quality factor, Q of a cavity as
Q ≡ ωo × EnergyPower Loss
where ωo is the frequency of the mode considered.
Let < U > be the time-averaged energy of the cavity, then its time derivative is simply the powerloss at the walls, and from the definition of the Q − factor is given by (where the minus sign isreminiscent to that it is an energy loss)
d<U>dt = −ωo<U>
Q
⇒ < U(t) >= Uoe−ωot/Q
or in other words, the energy decreases exponentially in time due to losses. We recall that for anEM field, the energy stored is
ε =12
| ~B|2
µo+ εo| ~E|2
95
so that
< U >∼ | ~E, ~B|2
⇒ E(t) = Eoe−ωot/2Q
i.e. electric and magnetic fields damped exponentially in time as well. We are interested in what’shappening in the spectrum of frequencies ωmnr, and to gain that information, we shall Fouriertransform the time-dependent field
E(ω) =∫ ∞
odtEoe
−ωot/2Qe−i(ω−ωo)t
Which has for solution the following spectrum in frequencies:
|E(ω)|2 ∝ 1
(ω − ωo)2 + (ω/2Q)2
where w0 is the frequency of a given mode under cosideration. This functional form is known as a“Lorentzian curve” which is characterized by its width, Γ, at half height, given here by
Γ =ωo
Q
and therefore the larger the quality factor Q is, the more closely will the frequency spectrum aroundωo resemble a δ-function. To calculate the Q-factor of the cavity, one typically consider the energyloss for a given mode by the ohmic loss from
U =1
2
∫d~x
εo
~E|2 +1
µo
| ~B|2
where the ~E, ~B fields are being damped by the conductors having a finite conductivity σ at thewalls.
96
5 Electromagnetic Radiation
5.1 Fields of Moving Charges
5.1.1 Review of Inhomogeneous Wave Equation
We recall that scalar and vector potentials for electrodynamics are given by
~E = −~∇φ− ∂ ~A
∂t~B = ~∇× ~A
And we have shown previously that they obey a wave equation with a source (In the Lorentz gauge):
2φ = − ρ
εo
2 ~A = −µo~J
with
2 ≡ ~∇2 − µoεo∂2
∂t2
Our goal is now to solve these equations when ∂∂t 6= 0.
5.1.2 Retarded Potentials
When charges are static, such as when ∂ρ∂t = 0, we have seen that the equation of motion for
electromagnetic potentials reduces to the Poisson equation i.e.
∇2φ = − ρεo
and∇2 ~A = −µo
~J
which have the following corresponding solutions
97
Φ(~x) = 14πεo
∫d~x′ ρ(~x′)
|~x−~x′|
~A(~x) = µo
4π
∫d~x′
~J~x′
|~x−~x′|
Imagine now that the charges aren’t static anymore, i.e. so there will be a ‘time lag’ betweenpropagation of information which travels at speed of light to the observe. The delay for theinformation to travel from the source location (variable x′) to the observer (variable x) locationwhere we wish to calculate the fields ~E and ~B is known as the retarded time, and is defined as
tr ≡ t− r
c
where
r ≡ |~x− ~x′|
So for moving charges, the potentials should be generalized as
Φ(~x, t) =1
4πεo
∫ρ(~x′, tr)
rd~x′
~A(~x, t) =µo
4π
∫ ~J(~x′, tr)
rd~x′
where ρ(~x, tr) is the charge density that prevailed at ~x′ and retarded time, tr. These are known asthe retarted potentials, and are reminiscent from the fact that fields are not instantaneous.
Similarly, we can define an advanced time, ta as
ta ≡ t +r
c
and it follows that we can define advanced potentials
98
Φ(~x, t) =1
4πεo
∫ρ(~x′, tr)
|~x− ~x′|d~x′
~A(~x, t) =µo
4π
∫ ~J(~x′, tr)
|~x− ~x′|d~x′
5.1.3 Lienart-Wiechert Potentials
We shall now consider the potentials of a point charge moving on a specific trajectory. By pointcharge we mean that the charge has no distribution, no size. We define its path or trajectory at atime, t, by the vector
~ω(t) = position of q at time t
so that the retarded time is determined from the equation
|~x− ~ω(tr)| = c(t− tr)
where |~x− ~ω(tr)| is the distance over which the “field effect” propagates, and c(t− tr) is the timefor the “field effect” to travel.
So, we shall define (~x− ~ω(tr)) = ~R the vector from retarded position to the field observer at ~x.
It can be shown that this “retarded position” will induces a charge
∫V
ρ(~x′, tr)d~x′ =q
1− R · ~v/c
where, ~v is the velocity of the point charge at retarded time tr, and ~R is the vector from theretarded position to field point ~x.This in turn will modify the potentials, φ and ~A which are knownas the Lienart-Wiechert potentials for a moving point charge:
φ(~x, t) = 14πεo
qc
(Rc−~R·~v)
~A(~x, t) = µo
4πqc~v
(Rc−~R·~v)= ~v
c2φ(~x, t).
99
5.1.4 Fields of a Moving Point Charge
From the retarded potentials φ(~x, t) and ~A(~x, t) for a moving point chargem, the electromagneticfields created by a moving charge can be calculated from the field equations
~E = −~∇φ− ∂ ~A
∂t, ~B = ~∇× ~A
.
We recall that we had defined the vector ~R = ~x − ~ω(tr) and so we can defined the velocity as~v =~ω(tr) where as usual ~x is the posiiton of the field observer, ω(tr) is the position at time tr. The~R and~ω(tr) must be taken at retarded time t−r determined by |~x−~ω(tr)| = c(t− tr) = f(~x, t) andthe calculations of the gradient operator is rather cumbersome, so we shall only give the resultshere for the fields, and not the derivation. Making use of the notation
~u ≡ cR− ~v
the fields due to a moving charge are given by
~Empc(~x, t) =q
4πεo
R
(~R · ~u)3[(c2 − v2)~u + (~R× (~u× ~a))]
~Bmpc(~x, t) =1
c~R× ~Empc(~x, t)
where ~a ≡ ~dv/dt is the acceleration of the point charge at retarded time, tr. We note that~B is perpendicular to ~E and to the ~R (vector to retarded position).
Note that in the special case where if ~v = 0 = ~a, i.e. when charges are not moving, then ~u → cRand so the electric field is
~Empc(~x, t) → q
4πεoRc3R
(~R · ~u)3
=q
4πεo
~Rc3
R3c3
=q
4πεoR
R2
and so
lim~u→0,~a→0
~Empc(~x, t) =qR
4πεoR2.
100
The electric field hat is recovered in the limit of no velocity and no acceleration is that of staticcharges, as it should.
The electric field of moving charge ~Empc, given by
~Empc(~x, t) =q
4πεoR
(~R · ~u)3[(c2 − v2)~u+ (~R× (~u× ~a))]
effectively contains two terms, one that depends on the velocity of the charges ~u ≡ cR− ~v, where~v ≡ d~ω(tr)/dt, and which gives the field in (c2 − v2)~u. This term is the generalized Coulombfield, and is also termed the ‘velocity field’ Ev ∝ 1/R2. The second term depending on theacceleration of the moving charges ~a, and given by ~R× (~u×~a) is termed the ‘acceleration field’, Ea ∝ 1/R. The fact that the velocity field falls much faster than the acceleration field will haveimportant consequences on the radiation of moving charges, as we shall see in the next section.Finally, we note that similar results are obtained for the magnetic field ~Bmpc since it is given by~Bmpc = R
c × ~E(~x, t).
5.2 Radiation of moving point charges
5.2.1 Radiation in general
Radiation of moving charge is a consequence of electrodynamics and is a classical phenomenon.We shall see that only charges that are undergoing acceleration can radiate, or lose energy in byemitting electromagnetic waves. To see this, we revisit the Poynting theorem that stated that
∂κi
∂t= − ∂
∂t
12
∫d~x
(| ~B|2
µo+ εo| ~E|2
)− 1µo
∮~E × ~B · d~s
so electromagnetic waves in vacuum propagates out to infinity carrying energy with them.
INSERT FIGURE “RadiationWaves”
The radiation of charges will be described in term of power radiated, or energy loss per unit time,and is given by the Poynting vector P =
∮~S ·d~s, but the charges will carry some energy with them.
To obtain the total power radiated, one must integrate over a sphere at infinity, so that ~R → ∞,and so
Prad = limR→∞
∮~S(~R) · d~s
We have seen that the electric field for moving point charges can be separated into a velocityfield ~Ev and an acceleration field ~Ea. Let’s consider first the dependence with respect to the fieldobserver distance. We note that
101
~E = ~Ev + ~Ea
∼(
1r2
)v
+(
1r
)~a
so that the power radiated at infinity which is the Poynting vector times the surface of a sphere atinfinity with surface S ∝ r2), which is given by the square of the field will go as
~S = ~Sv + ~S~a
Prad ∼
*0(1r4
)~v
· r2 +(
1r2
)~v
· r2
So only the acceleration field will radiate at infinity and hence we need only to consider the term
~Ea =q
4πεoR
(~R · ~u)3[~R× (~u× ~a)]
which is perpendicular to the vector ~R, and the corresponding Poynting vector is
⇒ ~Sa =1µoc
[| ~Ea|2R− (*
0R · ~Ea) ~Ea] =
| ~Ea|2Rµoc.
Recall that ~u = cR − ~v = c(R − ~v/c). For a non-relativistic moving charge with |~v/c| << 1,we can make the approximation
~u ∼ cR
or, in other words at a time tr, the charge is almost ‘at rest’ so that
(~R · ~u)3 ∼ c3R3
which gives for the acceleration field
~Ea =q
4πεoR
c3R3[~R× (cR× ~a)]
=q
4πεoc2R[R× (R× ~a)]
102
but c2 = 1/εoµo and so we get
~Ea =µoq
4πR[(R · ~a)R− ~a].
So, in the far field limi, at infinity, the power radiated is
Prad → [S(r)][4πr2]
and for a moving point charge, Sv ∼ 1/R4, Sa ∼ 1/R2, so there will be no radiation if a charge ismoving with a uniform velocity, however, radiation will occur if ~a 6= 0.
5.2.2 Radiation and Larmor Formula
To calculate the total power radiated by a particale at time tr, we need to consider a sphere ofradius R centered at position of the particle at time tr, keeping in mind that the radiation will takea time (t− tr) = R/c to reach the sphere.
So,
~Sa =|Ea|2Rµoc
|Ea|2 =µ2
oq2
(4πR)2((R · ~a)R− ~a)((R · ~a)R− ~a)
=µ2
oq2
(4πR)2(|~a|2 − 2(~a · R)2 + (R · a)2)
=µ2
oq2
(4πR)2(|~a|2 − (R · a)2)
=µ2
oq2
(4πR)2a2(1− cos2θ)
=µ2
oq2
(4πR)2a2sin2θ
And therefore
~Sa =µoq
2
c(4πR)2a2sin2Rθ ∝
(sinθ
R
)2
103
The total Power is given by
P =∮~sa · d~s
=µoq
2a2
(4π)2c
∫sin2θ
R
2
R2sinθdθdφ
=µoq
2a2
(4π)2c
∫ π
0sin3θdθ
∫ 2π
0dφ
=µoq
2a2
(4π)2c432π
=µoq
2a2
6πc
P =µoq
2a2
6πc
which is the Larmor Formula for a moving point charge (v/c << 1)
The Larmour formula is only valid if v/c << 1. As the velocity of the moving charge increases,one needs to take into account the v/c that we neglected in ~u = cR− ~v = c(R− ~v/c).
We can show that in this case, the Larmour formula will be generalized by
P =µoq
2γ6
6πc
[a2 −
∣∣∣∣~v × ~a
c
∣∣∣∣2]γ =
1√1− v2/c2
known as Lieard’s generalization of Larmour Formula Note that as γ → 1, P → Plarmor.
Special case 1: Brehmsstrahlung: If ~v is colinear to ~a, |~v × ~a| = va sin θ → 0 then:
P =µ0q
2γ6
6πca2
INSERT FIGURE ”bremsstrahlung”
104
Emitted radiation is known as bremsstrahlung (braking radiation).
Special case 2: synchrotron radiation: When ~v⊥~a, |~v × ~a| = va sin θ = va. This gives rise toa circular orbit.
INSERT FIGURE ”Sychrotron”
P =µ0q
2γ6a2
6πc[1− v2/c2] =
µ0q2γ6a2
6πc1γ2
P =µ0q
2γ4a2
6πc
Which goes as γ4. This is called sychrotron radiation.
5.2.3 Breakdown of Rutherford’s picture of the atom
In Ritherford’s picture of the atom, the electron orbit is in a classical orbit around the nucelas.Let’s consider the simplest possible case, the hydrodgen atom:
INSERT FIGURE: ”hydrogen”
Assuming that the e− is orbiting classically, i.e. according to Newtonian mechanics and classicalelectromagnetic theory, how long would the e− remain on a stable orbit? We have shown that since|~a| 6== 0 there will be power radiated equal to:
P =µ0q
2γ4a2
6πcfor ~v⊥~a
if non-relativistic, i.e. v/c 1 and γ = 1/√
1− 1v2/c2
' 1, we get:
P =µ0e
2a2
6πc= −dE
dt
Remembering that radiated power is energy lost by the system. Further, we assume that theenergy loss per revolutio nis small compared with the total energy of the atom. In the field, whichis generated by the Coulomb potential of the hydrogen nucleus, the total energy and accelerationof the electron are:
E =mv2
2− e2
4πε0ra =
v2
r
105
Recall the virial theorem: 2〈T 〉 = n〈V 〉 where n = −1 is the power of the potential’s r-dependence.Therefore:
〈T 〉 = −〈V 〉2
〈E〉 =〈V 〉2
〈V 〉2
=e2
8πε0r⇒ v2 =
e2
4πε0rm
〈E〉 =mv2
2− e2
4πε0r= − e2
8πε0r
and
a =e2
4πε0r2m
So thatd〈E〉dt
dr
dt=
e2
8πε0r2dr
dt= −µ0e
2a2
6πc= −µ0e
2
6πc
(e2
4πε0r2m
)2
dr
dt= −µoe
2
6πc
(e2
4πεor2m
)2 4πεor2
e2
= −µoe2
3πc
(e2
4πεor2m
)= − µoe
4
3πc(4πεo)r2m2
or
dt =3πc(4πεo)m2
µoe4r2dr
So that the total time the electron will spend in orbit will be:
τ =∫ τ
0dt = −3πc(4πεo)m2
µoe4
∫ 0
a0
r2dr
Where a0 is the Bohr radius. We’re integrating over the radial distance as the electron falls fromits initial radius into the nucleus (r = 0). This gives:
τ =πc(4πεo)m
2
µoe4a3
0
106
Numerically, this gives:
τ ' 4π2 · 10−11 · 10−60 · 108a30
4π · 10−7 · 5 · 10−76∼ a3
0
10−20' 10−10s
,
a very short lifespan indeed!
5.3 Dipolar Radiation (optional)
5.3.1 Electric Dipole Radiation
INSERT FIGURE ”electric dipole”
Imagine an (oscillating) dipole with a time-varying charge according to q(t) = q0eiωt (the physical
charge being given by the real part) with a dipolar moment ~p(t) = ~p0eiωt.
We make the following approximations:
i) d r : the observer sees a dipole, not the individual charges;
ii) d c/ω (d λ) : distance much smaller than the wavelength of of the oscillating dipole;
iii) r c/ω (r λ) : we observe the fields far from the source. We call this the ”radiation zone”.
Approximations i) and ii) are the perfect dipole approximation. When apply all of these approxi-mations to retadrded potientials, the retarded potentials for oscillating dipole:
Φ(r, θ, t) = − p0ω
4πεoc
(cos θr
)sin[ω(t− r/c)]
~A(r, θ, t) = −µop0ω
4πrsin[ω(t− r/c)]z
Using the field equations ~E = −~∇Φ− ∂ ~A∂t and ~B = ~∇× ~A, the electric and magnetic fields for
an oscillating electric dipole become:
~E = −µop0ω2
4πcos[ω(t− r/c)]θ
~B = −µop0ω2
4πccos[ω(t− r/c)]φ
As usual, the power radiated is obtained by integrating the Poynting vector ~S = 1µo
~E× ~B at infinity.Hence:
107
~S =µo
c
[p0ω
2
4π
(sin θr
)cos[ω(t− r/c)]
]2
r
and if we average it in time (remember 〈cos2 ωt〉 = 1/2〉),
〈~S〉 =µop
20ω
4
32π2c
sin2 θ
r2r
Such that the differential radiated power for an electric dipole is:
(dP
dΩ
)e.dip.
=µop
20ω
4
32π2csin2 θ
Note:
i)(
dPdΩ
)e.dip.
∼ sin2 θ i.e. no radiation for θ = 0.
INSERT FIGURE ”radiation, theta, zero, blah”
ii)(
dPdΩ
)e.dip.
∼ ω4 ...the fourth power of omega!
5.3.2 Scattering by electrons: why is that sky blue? (optional)
We want to study the scattering of EM waves (light!) by free (Thomson) or bound (Rayleigh)electrons. Picture this process:
INSERT FIGURE ”scattering”
Let’s assume the incident wave is linearly polarized with an electric field of the form:
~E = ~E0ei(~k·~x−ωt)
Recall the Drude model in section 3.3.2: we had developed a theory of the electron gas and foundthat the electrons form an oscillating dipole in response to the driving electric field:
~p0 =θ2
m
~E0
ω20 − ω2 − iωγ
~p = ~p0eiωt
The oscillating dipole will radiate EM waves according to:
108
(dP
dΩ
)e.dip.
=µop
20ω
4
32π2csin2 θ =
(dP
dΩ
)e.dip.
=µoω
4
32π2c
| ~E20
(ω20 − ω2)2 + ω2γ2
sin2 θ
A fraction of the incident EM wave energy is absorbed by the motion of the electrons and givenback in the form of radiation. We define the differential cross section as the power emitted in agiven direction per unit solid angle per unit of incident energy flux:
(dσ
dΩ
)e.dip.
=(dP/dΩ)
〈~S〉i
with:
〈~S〉 =1µo〈 ~E × ~B〉 =
1µoc
| ~E0|2 = cεo| ~E0|2
so,
(dσ
dΩ
)e.dip.
=µoe
4ω4
32π2m2c2εo| ~E|2| ~E|2 sin2 θ
(ω20 − ω2)2 + ω2γ2
=µoc
2r20ω4 sin2 θ
εo32π2[(ω20 − ω2)2 + ω2γ2]
Where we have used e2/mc2 ≡ r0, the classical radius of the electron ' 3× 10−13cm.
(dσ
dΩ
)e.dip.
=µoc
2
εo32π2r20 sin2 θ
ω4
(ω20 − ω2)2 + ω2γ2
The total cross-section is given by:
σtot =∫
Ω
(dσ
dΩ
)dΩ
=µoc
2
εo12πr20
ω4
(ω20 − ω2)2 + ω2γ2
Case i) Free electrons ⇒ ω = 0; γ = 0:
109
⇒ σtot =µoc
2
εo12πr20 6= f(ω)
This is Thomson scattering.
Case ii) Bound electrons, but with ω ω0:
⇒ σtot =µoc
2
εo12πr20
(ω
ω0
)4
This is Rayleigh scattering. Note the ω4 dependence!
So why is that sky blue?
Well, scattering of light in the atmosphere is well approximated by scattering off bound electronsin atmospheric molecules.
σ ∼ ω4 ∼ (1/λ)4
⇒ short wavelength (blue) light is therefore scattered to a much higher degree than the longer(red) wavelengths. Remember, the light coming from the sun contains a very broad spectrum, buton earth we see the light that is scattered by the atmosphere in addition to the direct rays of thesun (otherwise the rest of the sky would be pitch black!).
INSERT FIGURE sun
At sunrise and sunset, as seen from sea level, the sunlight will travel a longer distance through theatmosphere to reach your eyes. The blue components have mostly been scattered away, leaving thecharacteristic sunset reds, oranges and pinks.
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