Chapter 4
Time Varying Fields, Simple Waves
4.1
Introduction
In this Chapter, the charge density
The charge conservation law
and current density J are generalized to be varying with time.
@
+ r J = 0;
@t
(4.1)
imposes a constraint between the two quantities (charge density in C m 3 ) and J (the current
density in A m 2 ) and various electrodynamic laws must be consistent with this basic law. A time
varying magnetic ‡ux induces an electric …eld through Faraday’s law,
I
E dl =
C
d
dt
Z
B dS:
(4.2)
S
Likewise, a time varying electric …eld induces a magnetic …eld through the displacement current,
I
d
B dl = 0 "0
dt
C
Z
E dS;
(4.3)
S
even in the absence of the current J = 0. As is well known, the displacement current played a
crucial role in Maxwell’s prediction that electromagnetic waves can propagate through vacuum.
The generalized set of Maxwell’s equations,
r E=
r
E=
;
(4.4)
@B
;
@t
(4.5)
"0
r B = 0;
r
B=
J + "0
0
1
(4.6)
@E
@t
;
(4.7)
is consistent with the charge conservation law, since the divergence of the LHS of the last equation
indeed vanishes (recall that r (r B) =0 identically) which requires that
r J + "0
4.2
@
@
r E=r J+
= 0:
@t
@t
(4.8)
Faraday’s Law
In 1831, Faraday discovered that an electric current was induced along a conductor loop when a
magnetic ‡ux enclosed by the loop changed with time. This important discovery gave an answer
to the old question prior to Faraday’s time whether a magnetic …eld could induce an electric …eld
because it had been known that an electric …eld, via an electric current, could induce magnetic
…eld. What Faraday found was that an electric …eld (or electromotive force, emf) was induced by
a time varying magnetic ‡ux. The integral form of the Faraday’s law,
emf =
I
E dl =
d
dt
Z
B dS;
(4.9)
S
was later put into a di¤erential form by Maxwell,
r
E=
@B
:
@t
(4.10)
The negative sign is due to Lenz and indicates that the emf is so induced as to oppose the change in
the magnetic ‡ux. It should be noted that many of the experiments originally done by Faraday were
actually due to motional emf in which motion of conductors across a magnetic …eld was responsible
for generation of emf without apparent time variation of the magnetic …eld itself. An object moving
in a magnetic …eld experiences an e¤ective electric …eld given by
Eef f = v
B:
(4.11)
This may be seen by noting that change in the magnetic ‡ux enclosed by a loop consists of two
parts, one due to time variation of the magnetic …eld,
d
1
=
Z
@B
dSdt;
@t
and the other due to the change in the shape of the loop,
d
2
=
=
=
Z
I
B
@S
dt
@t
B (v dl)dt
I
(v B) dldt:
2
Figure 4-1: Induction of emf in time varying magnetic …eld (upper …gures) and motional emf in
static magnetic …eld (lower …gure).
Then the total ‡ux change is
d
=
dt
Z
@B
dS
@t
I
(v
B) dl;
and the emf induced along the loop is given by
emf =
I
E dl =
Z
@B
dS +
@t
I
(v
B) dl:
(4.12)
An emf can be generated by letting a conductor move across a stationary magnetic …eld as done in
most electric generators.
4.3
Displacement Current and Wave Equation
As shown in Introduction, the displacement current density
"0
@E
;
@t
plays a crucial role for the Maxwell’s equations to be consistent with the charge conservation law,
@
+ r J = 0:
@t
3
The magnetic …eld induced by the displacement current may be best visualized in a capacitor being
slowly charged as shown in Fig.4-2. The current ‡ows on the surface of electrodes. The radially
Figure 4-2: The surface currents on the capacitor electrode are consistent with the magnetic …elds
both outside and inside.
outward surface current on the outer surface of the upper plate,
I
Js1 =
2
;
(4.13)
is consistent with the boundary condition for the magnetic …eld,
n1 H = Js1 ;
where
H=
I
2
e ;
(4.14)
(4.15)
is the magnetic …eld expected from the Ampere’s law. In the space between the electrodes, no
conduction current exists but there exists an azimuthal magnetic …eld,
B ( )=
1
2
0 "0
@Ez
;
@t
(4.16)
as if there were a uniform conduction current equal to
Jz = "0
@Ez
:
@t
(4.17)
The magnetic …eld is required to exist to satisfy the boundary condition at the inner electrode
surface,
n2 H = Js2 ;
(4.18)
4
where
I
;
2 a2
Js2 =
(4.19)
is the radially inward surface conduction current on the inner surface of the upper plate. Since
I=
and
Ez =
dq
;
dt
(4.20)
q
;
"0 a2
(4.21)
@Ez 1
:
@t 2
(4.22)
we …nd
H =
"0
This is consistent with the Maxwell’s equation,
1 @
@Ez
( H ) = "0
;
@
@t
(4.23)
@Ez
:
@t
(4.24)
or its integral form,
2
2
H =
"0
As is well known, the displacement current was instrumental for Maxwell to predict that electromagnetic …elds obey a wave equation. To see what wave equations the electromagnetic …elds
should satisfy, let us take a curl of the Faraday’s law,
r
r
@B
;
@t
@
r
@t
E =
(r
E) =
(4.25)
B:
The LHS can be expanded as
r
(r
E) = r(r E)
r2 E:
Since
r E=
"0
; r
B=
0
J + "0
@E
@t
;
Eq. (4.25) reduces to the following inhomogeneous wave equation,
r2
1 @2
c2 @t2
E=
1
r +
"0
0
@J
:
@t
(4.26)
where c is the speed of light in vacuum,
c2 =
1
"0
5
:
0
(4.27)
Likewise, the magnetic …eld obeys
1 @2
c2 @t2
r2
B=
0r
(4.28)
J:
If the displacement current were absent (which, incidentally, is equivalent to the assumption that
c ! 1), both …elds would merely satisfy vector Poisson’s equation,
r2 E =
1
r +
"0
r2 B =
0
0r
@J
;
@t
J;
which do not exhibit any propagation nature with a …nite speed.
Equation (4.26) can be solved symbolically as
E=
r
2
1
r +
"0
1
1 @2
c2 @t2
0
@J
@t
;
(4.29)
where
1
;
1 @2
2
r
c2 @t2
is the propagator integral operator which yields a retarded solution for the electric …eld,
E(r; t) =
=
Z
1
r0
4 "0
Z
1
r
4 "0
(r0 ; t
) 0
dV
0
jr r j
(r0 ; t
) 0
dV
0
jr r j
where
=
jr
0
4
Z
@
4 @t
0
c
r0 j
1
jr
Z
@
r0 j @(t
)
J(r0 ; t
)dV 0
J(r0 ; t
) 0
dV ;
0
jr r j
;
In Eq. (4.31),
1
4 "0
Z
(r0 ; t
) 0
dV ; A (r; t) = 0
0
jr r j
4
6
(4.31)
(4.32)
is the time required for electromagnetic disturbances to propagate over a distance jr
the source at r0 and the observer at r: Note that
Z
r0 (r0 ; t
) 0
dV
jr r0 j
Z
(r0 ; t
)
r r0
=
r0
(r0 ; t
) dV 0
jr r0 j
jr r0 j3
Z
(r0 ; t
) 0
= r
dV :
0
jr r j
(r; t) =
(4.30)
Z
J(r0 ; t
) 0
dV ;
0
jr r j
r0 j between
are the retarded scalar and vector potential, respectively. For a single charged particle moving at
a velocity vp (t) ; the scalar potential becomes
(r; t) =
=
Z
e
4 "0
e
4 "0 (1
rp t 1c jr r0 j
dV 0
jr r0 j
1
;
n ) jr rp j ret
r0
(4.33)
where
jret means every time dependent quantity is to be evaluated at the retarded time t0 to be
determined implicitly from
1
t t0
r rp t0 = 0:
c
Likewise, the vector potential due to a moving charge is
A (r; t) =
e
4
0
(1
n
v
) jr
rp j
:
(4.34)
ret
These potentials were …rst formulated by Lienard and Wiechert. We will use the potentials in
Chapter 8 in formulating the radiation electromagnetic …elds due to moving charges.
Equation (4.26) can also be written as
E =
r2
=
r2
1
1 @2
c2 @t2
1
1 @2
c2 @t2
1
r +
"0
0
@J
@t
1
r +
"0
0
@Jl
+
@t
0
@Jt
@t
;
(4.35)
where Jl is the longitudinal component of the current density satisfying
@
+ r Jl = 0;
@t
(4.36)
and Jt = J Jl is the transverse component which does not a¤ect the charge density because of
the identity (or de…nition)
r Jt = 0:
(4.37)
Since r
Jl = 0; it follows
r(r Jl ) = r2 Jl ;
and a symbolic solution for the longitudinal current density is
Jl =
1 @
r :
r2 @t
7
(4.38)
Substituting into Eq. (4.35), we obtain
E=
1
r +
"0 r2
r
2
1
1 @2
c2 @t2
0
@Jt
:
@t
(4.39)
This formulation is consistent with the choice of Coulomb gauge for the scalar and vector potentials
as we will see in more detail in Chapter 6. Solution for the electric …eld is
E(r; t) =
1
r
4 "0
Z
(r0 ; t)
dV 0
jr r0 j
@
4 @t
0
Z
Jt (r0 ; t
) 0
dV :
0
jr r j
(4.40)
Note that in this formulation, the Coulomb electric …eld (the …rst term in the RHS) is non-retarded
and instantaneous. This unphysical result is a consequence of the Coulomb gauge which singles
out the transverse current Jt . In fact, the non-retarded longitudinal electric …led is cancelled by
a term contained in the last term and all physically observable electromagnetic …elds are retarded
because of the …nite propagation speed c:
4.4
Fields and Potentials
By now, it is clear that there exists two kinds of electric …eld, one de…ned in terms of the scalar
potential,
E1 = r ;
(4.41)
and another originating from time varying magnetic …eld,
r
@B
=
@t
E2 =
@
r
@t
A:
(4.42)
The …eld E1 is longitudinal because r E1 = 0: In Eq. (4.42), the longitudinal components of
E and A vanish. In fact Maxwell’s equations do not impose any conditions on the longitudinal
component of A; and choice of r A is arbitrary. Therefore, we are allowed to assume
E =
r
=
r
@
A
@t
@
Al
@t
@
At ;
@t
where Al and At are the longitudinal and transverse component of the vector potential. Substitution in r E = ="0 yields
@
r E = r2
r Al = :
(4.43)
@t
"0
In Coulomb gauge, Al is so chosen as to satisfy
Coulomb gauge: r Al = 0:
8
(4.44)
(Note that Al may not be zero identically.) Eq. (4.43) becomes
r2
C
(r; t)
;
"0
(r; t) =
(4.45)
where C is the scalar potential in the Coulomb gauge. It should be emphasized that both C
and are time dependent and in Coulomb gauge the scalar potential responds to a change in the
charge density instantaneously. In Lorenz gauge, r Al is assigned as
Lorenz gauge: r Al +
1 @ L
= 0:
c2 @t
(4.46)
In this case, the scalar potential obeys a wave equation,
1 @2
c2 @t2
r2
L
=
"0
:
(4.47)
This yields a retarded solution,
1
L (r; t) =
4 "0
Z
1
1
r
c
r0 ; t
r0 j
jr
r0
dV 0 :
(4.48)
Proof is straightforward if the following are noted:
r2
r2
r0 ; t
1
jr
1
r
c
r0 j
r0
=
4
=
r0
r
1 @2
c2 @t2
;
1
r
c
r0 ; t
r0
:
For the vector potential, we rewrite the Ampere-Maxwell’s law
r
B=
J + "0
0
@E
@t
;
in terms of the potentials as
r2 A
1 @2A
c2 @t2
r (r A) +
1 @
c2 @t
=
0 J:
(4.49)
In Coulomb gauge with r A =0; this reduces to
r2 A
r2 A
1 @2A
c2 @t2
1 @2A
c2 @t2
=
0J
=
0 Jt ;
+
1 @
r
c2 @t
since the longitudinal component of the current Jl vanishes through the continuity equation
@
+ r Jl = 0:
@t
9
(4.50)
Note that
@
@
= "0 r E =
@t
@t
and thus
"0
Jl
@
r
@t
"0
@ 2
r
@t
C;
= 0:
C
In Coulomb gauge, the vector potential is transverse.
In Lorenz gauge, Eq. (4.49) becomes a wave equation for A ;
1 @2A
=
c2 @t2
r2 A
(4.51)
0 J;
where J = Jl + Jt ; A = Al + At : In Lorenz gauge, the potentials are symmetric in the sense that
both and A satisfy the same wave equation,
r2
1 @2
c2 @t2
=c
A
!
=
c
J
0
!
;
(4.52)
and appropriately form a covariant four vector ( =c; A) : In contrast, the formulation of electromagnetic …elds in terms of Coulomb gauge is not invariant under the Lorentz transformation. All
potentials and …elds are retarded in Lorenz gauge while in the Coulomb gauge, the scalar potential
is non-retarded. The appearance of non-retarded scalar potential is due to the choice r A = 0; or
the assumption that the vector potential is purely transverse.
4.5
Poynting Vector: Energy and Momentum Conservation
The complete set of Maxwell’s equations is
r E=
r
E=
"0
;
@B
;
@t
r B = 0;
r
B=
0
J + "0
@E
@t
:
It is noted that the charge density contains all kinds of charges, free charges, bound charges, etc.
Likewise, the current density J contains all kinds of current, conduction currents, magnetization
currents, etc. If free charges are singled out, the …rst equation can be written in terms of the
displacement vector D;
r D = free ;
(4.53)
where
D = "0 E + P = "E;
10
(4.54)
with P being the polarization vector, or the electric dipole moment density. The current density
is likewise decomposed into the part due to the motion of free charges and the other due to time
variation of the polarization vector,
@
r P + r Jp = 0;
@t
from which
Jp =
@P
:
@t
(4.55)
Therefore, the magnetic induction equation can be rewritten as
r
B=
0
J + "0
@E @P
+
@t
@t
=
0
J+
@D
@t
;
(4.56)
where the current density J consists of conduction and magnetization currents,
J = Jc +Jm = Jc +r
M:
If the conduction current is singled out, Eq. (4.56) can be rewritten in terms of the vector H;
r
H = Jc +
where
H=
B
@D
;
@t
(4.57)
M:
0
In macroscopic applications, free charges and conduction currents are the quantities that can be
controlled by external means and Eqs. (4.53) and (4.57) are often more convenient than the original
forms.
The Maxwell’s equations are also consistent with energy conservation law. To show this, we
de…ne the Poynting vector by
E H; (W m 2 ):
(4.58)
As its dimensions imply, the Poynting vector indicates the ‡ow of electromagnetic energy per unit
area per unit time, that is, power density. The divergence of the Poynting vector is
r (E
H) = (r E) H (r H) E
@D
@B
=
H
Jc +
E
@t
@t
1
@ 1
=
H 2 + "E 2
E J;
@t 2
2
provided that the permittivity " and permeability
11
are independent of the frequency !: If not, the
energy densities should be modi…ed as
1 @ [!" (!)] 2 1 @ [! (!)] 2
E ;
H ;
2
@!
2
@!
(4.59)
respectively. We will return to this problem shortly in the following Section. Integrating over a
volume, we …nd
Z
V
r (E
H) dV =
I
(E
d
(Um + Ue )
dt
H) dS =
S
where
Ue =
1
2
Z
"E 2 dV;
Um =
1
2
Z
Z
E JdV;
H 2 dV;
(4.60)
(4.61)
are the total electric and magnetic energies stored in the volume and
Z
E JdV; (J s 1 = W)
(4.62)
is the rate of electromagnetic energy conversion into other forms of energy, e.g., creation of heat
through Joule dissipation and acceleration of charged particles. Therefore,
I
(E H) dS;
(4.63)
S
can be interpreted as the electromagnetic power ‡ow into the volume.
For a system consisting of charged particles, the momentum conservation can be shown in a
similar way. The mechanical momentum Pm follows the equation of motion,
dPm
=
dt
Z
( E+J
B) dV; (N).
(4.64)
Substituting
= "0 r E
and
J=
1
0
r
"0
B
@E
;
@t
we …nd
dPm
dt
=
=
where
Z
Z
1
"0 (r E)E +
0
!
r T
!
T = Tij = "0 Ei Ej
r
1 @
E
c2 @t
1
"0 E 2
2
ij
+
"0
@E
@t
B dV
H dV;
1
0
12
B
Bi Bj
(4.65)
1
2
B2
0
ij ;
(N m
2
)
(4.66)
is the Maxwell’s stress tensor and use is made of the following vector identities,
rE 2 = r(E E) = 2E (r
E) + 2 (E r) E;
r (EE) = (r E)E + (E r) E
rB 2 = r(B B) = 2B
(r
B) + 2 (B r) B
r (BB) = (r B)B + (B r) B = (B r) B:
Eq. (4.65) suggests that the vector
1
E
c2
3
H; (N s m
)
(4.67)
can be regarded as the electromagnetic momentum density and
1
E
c
H; (N m
2
)
(4.68)
as the electromagnetic momentum ‡ux density. For a system of charged particles, momentum
conservation thus requires inclusion of the electromagnetic momentum as well as mechanical momentum.
From the momentum density in Eq. (4.67), the angular momentum density associated with
electromagnetic …elds is naturally de…ned by
1
r
c2
(E
H); (J s m
3
)
(4.69)
and the total angular momentum associated with electromagnetic …elds by
Z
1
L = 2 r (E H)dV ; (J s).
c
(4.70)
Its ‡ux density is
1
R= r
c
(E
H); (J m
2
):
(4.71)
Both the momentum and angular momentum densities are proportional to the Poynting vector,
namely, the energy ‡ow. If for example a system is losing energy through radiation of electromagnetic energy, the system is necessarily losing momentum and angular momentum as well. Consider
a charged particle undergoing circular motion, e.g., electron in a magnetic …eld. Since the electron is continuously accelerated by the centripetal force, it radiates electromagnetic energy. At
the same time it loses its angular momentum to radiation. Therefore, it is natural to expect that
electromagnetic …elds (with proper polarization) carry an angular momentum with them.
13
4.6
Plane Electromagnetic Waves
One special mode of electromagnetic waves in free space is a plane wave in which the amplitude
of electric and magnetic …eld remains constant. Without loss of generality, we may assume wave
propagation in the z direction and an electric …eld in the x direction,
E(z; t) = E0 ei(kz
!t)
ex :
In free space, the electric …eld satis…es the wave equation,
@2
@z 2
1 @2
c2 @t2
provided
E(z; t) = 0;
(4.72)
!
= c:
k
From
r
E=
(4.73)
@B
;
@t
we …nd
E = !B;
k
(4.74)
where k = kez : The magnetic …eld is thus in the y direction,
B=
k
E0 ei(kz
!
!t)
ey ;
(4.75)
and its amplitude is
B0 =
E0
E0
E0
; or H0 =
;
=
c
c 0
Z
where
Z=
r
0
"0
(4.76)
= 376:8; ( )
(4.77)
is the impedance of free space.
The electric and magnetic energy densities associated with a plane wave are the same, for
1
1
1
"0 E 2 = "0 Z 2 H 2 =
2
2
2
0H
2
:
(4.78)
This equipartition of wave energy is similar to that in mechanical waves in which kinetic and
potential energies are equal. The total wave energy density is therefore given by
u=2
1
"0 E 2 = "0 E 2 ; (J m
2
14
3
)
(4.79)
and the Poynting ‡ux may be written in terms of either the electric or magnetic …eld as
Sz = Ex Hy = c"0 Ex2 =
Ex2
; (W m
Z
2
)
(4.80)
or
Sz = c
2
0 Hy
= ZHy2 :
(4.81)
For a harmonic wave with an amplitude E0 , the average (rms) wave energy density is given by
1
uave = "0 E02 ; (W m
2
2
)
(4.82)
and corresponding rms Poynting ‡ux is
1
1 E02
Szave = c"0 E02 =
; (W m
2
2 Z
2
):
(4.83)
Electromagnetic waves radiated by a localized source approach plane waves at a su¢ ciently
large distance but they can never be pure plane waves. Plane waves can be constructed from two
circularly polarized waves with opposite helicities, one rotating with positive helicity and another
with negative helicity. Helicity of electromagnetic waves is related to the angular momentum
associated with the waves. Evidently, a plane polarized wave carry zero angular momentum. A
more general theory of electromagnetic radiation will be developed in Chapter 5.
In a dielectric medium, the Maxwell’s equations are modi…ed as
r
r
@B
;
@t
E=
B=
0
@("E)
;
@t
(4.84)
where " is the permittivity which in general depends on the wave frequency and spatial position
and also the electric …eld. The origin of the permittivity is in the current induced by the electric
…eld in a material medium. In the magnetic induction equation
r
B=
0
J + "0
@E
@t
;
if the current density is proportional to the electric …eld through a conductivity ; J = E; we have
r
B=
0
E + "0
@E
@t
=
0"
@E
;
@t
(4.85)
where the permittivity is given by
" = "0 1 + i
!"0
:
(4.86)
It should be noted that the current density is due to deviation of electron orbit from bound harmonic
15
motion in molecules. The equation of motion for an electron placed in an oscillating electric …eld is
m
d2
+ ! 20 x =
dt2
eE0 e
i!t
;
(4.87)
where m is the electron mass and ! 0 is the frequency of bound harmonic motion. The current
density is
dx
i! eE
J = ne
= 2
;
dt
!
! 20 m
and the conductivity becomes
=
i!
!2
ne2
;
! 20 m
(4.88)
where n is the number density of electrons. Then the permittivity is given by
"(!) = "0
! 2p
1
! 20
!2
!
;
(4.89)
where ! p is the “plasma” frequency de…ned by
! 2p =
ne2
:
"0 m
(4.90)
The phase velocity of electromagnetic waves in a dielectric is
and the group velocity is
!
1
=p
k
"(!)
d!
=
dk
;
(4.91)
0
!
1
:
1 ! d" k
1+
2 "(!) d!
(4.92)
(At frequencies remotely separated from the resonance frequency ! 0 ; the group velocity coincides
with the energy propagation velocity. Near the resonance, however, the group velocity exceeds c and
it loses the meaning of energy propagation velocity. The concept of signal velocity was introduced
by Brillouin.) The impedance is accordingly modi…ed as
E
Z(!) =
=
H
r
0
"(!)
:
By de…nition, the group velocity is equal to
d!
Poynting ‡ux
=
;
dk
Energy density
16
(4.93)
where the Poynting ‡ux is
E2
S=
=
Z
s
"(!)
E 2 ; (W m
2
):
0
Therefore the wave energy density in a dielectric medium is
u =
=
=
1+
1 ! d"
2 "(!) d!
"(!)E 2
1 d
1
[!"(!)]E 2 +
H2
2 d!
2 0
1
1
1 ! 2p (! 2 + ! 20 )
"0 E 2 +
"0 E 2 +
2
2
2
2 (!
!0)
2
0H
2
; (J m
3
)
where the relationship
1
1
"(!)E 2 =
2
2
0H
2
;
(4.94)
is used. This result is valid only if the group velocity can be regarded as energy propagation velocity
which may not hold near the resonance ! ' ! 0 if the dielectric is dissipative.
The origin of the additional factor
1 d" 2
! E ;
2 d!
in the electric energy density is due to electron kinetic and potential energies in an oscillating
electric …eld. From the equation of motion of a bound electron,
@2
+ ! 20 x =
@t2
e
E;
m
we readily …nd the electron kinetic energy density,
1
1 ne2
!2
1 ! 2 ! 2p
2
nmv 2 =
E
=
"0 E 2 ;
2
2 m (! 2 ! 0 )2
2 (! 2 ! 0 )2
(4.95)
and potential energy density
1 ! 20 ! 2p
1
nm! 20 x2 =
"0 E 2 :
2
2 (! 2 ! 0 )2
(4.96)
Therefore, the total energy density associated with the electric …eld is
1
1 ! 2 ! 2p
1 ! 20 ! 2p
2
"0 E 2 +
"
E
+
"0 E 2 ; (J m
0
2
2 (! 2 ! 0 )2
2 (! 2 ! 0 )2
3
)
(4.97)
which is consistent with that conveniently calculated from
1 d
[!"(!)]E 2 :
2 d!
(4.98)
Since the wave under consideration is strongly dispersive, there is no simple energy equipartition
as in the case of nondispersive waves. Note that the electron kinetic and potential energies are the
17
result of forced oscillations.
The fact that
E
@D
@ ("E)
=E
;
@t
@t
(4.99)
is not always equal to
1 @E 2
"
;
2 @t
(4.100)
can be seen if we realize that the permittivity " is in general frequency dependent. Waves become
dispersive and wave amplitude is bound to decrease although slowly. In this case,
@ ("E)
@t
=
i!" (!) E0 e
i!t
' ( i! 0 + ) " (! 0 ) + i
=
where
i! 0 " (! 0 ) E0 e
i!t
@"
E0 e
@! 0
+ " (! 0 ) + ! 0
i!t
@" dE0
e
@! 0 dt
i!t
;
(4.101)
(< 0) is the damping rate of the amplitude,
dE0
= E0 :
dt
(4.102)
Therefore, time variation of electric energy density should be calculated from
E
@ ("E)
1
@" dE 2
=
" (! 0 ) + ! 0
;
@t
2
@! 0 dt
(4.103)
and the electric energy density becomes
ue =
1
@"
" (! 0 ) + ! 0
E2:
2
@! 0
(4.104)
Likewise, the magnetic energy density should be generalized as
um =
1
2
(! 0 ) + ! 0
@
H 2:
@! 0
(4.105)
These expressions were originally formulated by von Laue.
4.7
Wave Re‡ection and Transmission - Normal Incidence
Re‡ection and transmission of plane electromagnetic waves at a boundary of two dielectrics can be
conveniently analyzed in terms of impedance mismatch. Let a plane wave of amplitude Ei in air be
incident normally on a ‡at dielectric surface of impedance Z2 as shown in Fig.4-3: The impedance
of air is very close to the vacuum impedance, Z1 = 377 : The incident Poynting ‡ux is split into
18
those of re‡ected (Er ) and transmitted (Et ) waves,
E2 E2
Ei2
= r + t ; (W m
Z1
Z1
Z2
2
):
(4.106)
At the boundary, continuity of the electric …elds, which are all tangential to the surface, yields
Figure 4-3: A plane wave incident normal to a dielectric boundary.
Ei + Er = Et :
(4.107)
Solving these equations for Er and Et ; we …nd
Er =
Z2 Z1
Ei;
Z2 + Z1
(4.108)
Et =
2Z2
Ei :
Z2 + Z1
(4.109)
In terms of the magnetic …elds,
Z1 Hi2 = Z1 Hr2 + Z2 Ht2 ;
(4.110)
Hi + Hr = Ht ;
(4.111)
yield
Hr =
Z1 Z2
Hi;
Z1 + Z2
Ht =
2Z1
Hi :
Z1 + Z2
(4.112)
Note that the polarity of either electric or magnetic …eld of the re‡ected wave is reversed. This
is understandable because the incident and re‡ected …elds must satisfy the following vectorial
relationships,
k Ei = ! 0 Hi ;
19
k
Er = !
0 Hr :
Either Er or Hr must change sign on re‡ection for the re‡ected Poynting ‡ux to be in opposite
direction relative to the incident ‡ux.
It should be pointed out that the analysis presented above is solely based on energy conservation and no consideration was given to momentum conservation. In fact, if only the momenta
associated with the three waves (incident, re‡ected, and transmitted) are considered, momentum
is not conserved. Since the momentum ‡ux densities associated with each wave are
1 Ei2
= "1 Ei2 ;
c1 Z1
Re‡ected wave : "1 Er2 ;
Incident wave :
(4.113)
Transmitted : "2 Et2 ;
the following momentum unbalance emerges,
"1 (Ei2 + Er2 )
"2 Et2 =
2(Z12 Z22 )
2(Z1 Z2 )
"1 Ei2 =
"1 Ei2 :
2
(Z1 + Z2 )
Z1 + Z2
(4.114)
This is actually taken up by the dielectric body as mechanical momentum. Recall that an in…nitely
massive body can absorb momentum without absorbing energy. If Z2 > Z1 ; the dielectric is pushed
to the left. In fact, at the boundary, the electric energy densities are discontinuous, and the
di¤erence is
1
"2 Et2
2
1
2Z22
"1 Et2 = ("2 "1 )
E2
2
(Z1 + Z2 )2 i
2(Z12 Z22 )
=
"1 Ei2
(Z1 + Z2 )2
2(Z1 Z2 )
=
"1 Ei2 :
Z1 + Z2
This appears as a force per unit area on the boundary surface acting from the higher energy density
side to the lower because electric force in the direction perpendicular to the …eld appears as pressure.
Re‡ection at a conductor surface can be analyzed in a similar manner by modifying the impedance appropriately. For a medium having a conductivity ; the impedance is given by
Z=
r
i! 0
:
i!"0
This can be seen from the Maxwell’s equation in the presence of conduction current,
r
@E
@t
i!"0 E;
H = J + "0
=
E
20
(4.115)
and the e¤ective permittivity in a conductor is de…ned by
"e¤ = "0
i!
:
Therefore, the impedance in a conductor is
Z=
r
0
"e¤
=
r
i! 0
:
i!"0
For ordinary conductors, the conduction current far dominates over the displacement current even
in microwave frequency range. Then,
Z'
r
i!
0
= (1
r
!
i)
2
0
:
(4.116)
A complex impedance indicates strong dissipation of electromagnetic energy. The magnitude of
the impedance
r
! 0
jZj =
;
(4.117)
p
and electromagnetic waves
is much smaller than the free space impedance Z0 =
0 ="0 ' 377
incident on a conductor surface su¤ers strong re‡ection. However, re‡ection can never be complete.
Bath room mirrors coated with aluminum has power re‡ection coe¢ cient of about 90% at optical
frequency ! ' 1014 rad/sec.
Example 1 Impedance Matching
A dielectric …lm a quarter wavelength thick coated on a surface of another dielectric (say, optical
glass) can eliminate re‡ection of electromagnetic waves normally incident if the impedance of the
…lm is chosen to be a geometric mean,
Zf =
p
Z0 Zg :
(4.118)
This condition follows from cancellation between two re‡ected waves, one at the air-…lm boundary
and another at the …lm-glass boundary,
Zf Z0
Zg Zf
+ ei
= 0;
Zf + Z0
Zg + Zf
f
f
where the phase factor ei = 1 is due to additional propagation distance
2=
of the wave
4
2
re‡ected at the …lm-glass boundary. Here, f is the wavelength in the dielectric …lm. Solving
Zf Z0
Zg Zf
=
;
Zf + Z0
Zg + Zf
21
for Zf ; we …nd
p
Z0 Zg :
Zf =
Likewise, re‡ection from a conductor plate can be avoided by placing a thin conducting plate
of thickness d at a quarter wavelength in front of the conductor surface if the conductance of the
plate is chosen to be
1
=
:
(4.119)
Z0 d
4.8
Re‡ection and Transmission at Arbitrary Incident Angle
For an arbitrary incident angle 1 ; re‡ection and transmission at a dielectric boundary can be
analyzed by exploiting the boundary conditions for the electric and magnetic …elds as follows. An
incident wave continues to be assumed plane polarized. The refracted angle 2 is related to the
incident angle 1 through the well known Snell’s law,
sin
sin
1
=
2
n2
;
n1
(4.120)
p
p
where n1 = "1 ="0 and n2 = "2 ="0 are the indices of refraction of respective media. The Snell’s
law follows from the conservation of the wavenumber parallel to the boundary,
k1 sin
1
= k2 sin
2;
(4.121)
and the change in the wave propagation velocity,
!
1
=p
k1
"1
or
!
1
=p
k2
"2
;
0
;
(4.122)
0
k2
k1
=
:
n1
n2
(4.123)
The normal wavenumber in the medium 2 is
kz = k2 cos
2
= k2
s
n1
n2
1
2
sin2
1:
This becomes pure imaginary when
sin
1
>
n2
; n2 < n1 ;
n1
This is the condition for total re‡ection which occurs when a wave is incident on a medium with a
smaller index of refraction (e.g., from glass to air), n2 < n1 : An imaginary wavenumber indicates
exponential damping in the region n2 from the surface,
E0 e
kz z
22
;
where kz is the damping factor in the axial (z) direction,
kz = k2
s
2
n1
n2
sin2
1
1:
The wavenumber component along the surface is
kk = k2
n1
sin :
n2
If the second medium is vacuum (or air) n2 = 1; k2 = !=c = k0 ; and
kk = k0 n1 sin :
In total re‡ection, electromagnetic …elds in the region of smaller index of refraction exponentially
decay from the surface. Such waves are called “evanescent.” Important applications of evanescent
waves are being found in high resolution microscopy even with visible light.
In analysis of wave re‡ection and transmission at a boundary between di¤erent media, use of
the continuity of tangential electric and magnetic …elds, Et ; Ht ; is su¢ cient because continuity of
normal components Dn ; Bn is redundant. From Maxwell’s equation,
r
H=
@D
; or k
@t
H=
!D;
it is evident that continuity of Dn demands continuity of n (k H) = k (n H). However, the
tangential component of H; n H; is continuous and the wavenumber parallel to the boundary is
also continuous. (Note that in k (n H) = (k n) H; only the parallel component of k appears.)
The latter (continuity of k n) follows from the conservation of wave momentum parallel to the
boundary which manifests itself in the form of well known Snell’s law. Likewise, from
r
E=
@B
; or k
@t
E = !B;
we see that continuity of Bn automatically follows because of continuity of tangential component
of the electric …eld.
4.8.1
H in the Incident Plane (E tangential to the boundary)
We …rst consider the case in which the magnetic …eld of the incident wave is in the incident plane,
or the electric …eld is parallel to the boundary surface. The re‡ected and refracted electric …elds
are also parallel to the surface and the continuity of the tangential component of the electric …eld
gives
Ei + Er = Et :
(4.124)
23
Figure 4-4: A plane wave with incident angle 1 when the magnetic …eld is in the incidence plane
and the electric …eld is parallel to the boundary surface.
Since for each wave, the following vectorial relationship holds,
E=
k
0 !H;
(4.125)
the continuity of tangential component of the H …eld yields
k1 cos
1 (Ei
Er ) = k2 cos
2 Et :
(4.126)
Recalling the Snell’s law, we thus …nd
Er =
sin(
sin(
Et =
2 cos
sin(
In the limit of normal incidence (small angles
Er =
sin(
sin(
1)
2
1
+
2)
Ei '
sin
sin
2
1)
1+
2)
1 sin 2
1
+
2)
1 ; 2 ),
sin
1 + sin
2
where
Zi =
Ei ;
r
1
Ei :
2
0
(4.128)
we recover
Ei =
"i
(4.127)
;
n 1 n2
Z2 Z1
Ei =
Ei ;
n2 + n1
Z2 + Z1
(4.129)
(4.130)
is the impedance of respective media.
When the conditions for total re‡ection are met, the impedance de…ned by the ratio between
24
the tangential component of the electric and magnetic …eld,
Z2TE =
E2
H2 cos
r
i
=
2
0
"2
1
s
n1
n2
;
2
sin2
1
(4.131)
1
is also pure imaginary which indeed ensures total re‡ection of electromagnetic waves,
Z1
i Z2TE
= 1:
TE
Z1 + i Z2
In total re‡ection, the phase di¤erence between the incident and re‡ected waves is
TE
4.8.2
= 2 tan
1
Z1
Z2TE
!
= 2 tan
1
p
sin2
1
(n2 =n1 )2
cos
1
!
:
(4.132)
E in the Incident Plane (H tangential to the boundary)
Figure 4-5: The electric …eld is in the incidence plane and the magnetioc …eld is paralllel to the
boundary.
In this case, the magnetic …eld is parallel to the boundary plane and the continuity of tangential
component of the magnetic …eld is simply
Hi + Hr = Ht :
(4.133)
For each wave, the electric …eld is related to the magnetic …eld through
E=
1
k
!"
25
H:
(4.134)
Therefore, continuity of the tangential component of the electric …eld yields
1
cos
n1
1 (Hi
1
cos
n2
Hr ) =
2 Ht
=
1
cos
n2
2 (Hi
+ Hr ):
(4.135)
Solving for Hr ; we …nd
Hr =
=
sin 1 cos
sin 1 cos
tan( 1
tan ( 1 +
sin
+
sin
1
2)
Hi :
2)
1
2 cos 2
2 cos 2
Hi
(4.136)
(4.137)
The transmitted magnetic …eld is
Ht =
1+
tan(
tan (
1
2)
1+
2)
The re‡ected electric …eld is
Er =
Hi =
tan(
tan (
1)
2
1
2 sin 1 cos
1 + 2 ) cos(
sin(
+
2)
1
1
2)
Hi :
Ei ;
(4.138)
and transmitted electric …eld is
Et =
2 sin 2 cos
sin( 1 + 2 ) cos(
1
2)
1
Ei :
(4.139)
If
1
+
=
2
2
;
(4.140)
the re‡ected wave vanishes completely. Under this condition, Snell’s law becomes
n2
sin
=
n1
sin
1
=
2
sin
cos
1
= tan
1;
(4.141)
1
and this particular angle
B
= tan
1
n2
n1
;
(4.142)
is called the Brewster’s angle. If incident wave is plane polarized with the magnetic …eld oriented
parallel to, say, a surface of glass, re‡ection can be avoided at the Brewster’s angle. This principle
is often exploited in designing re‡ecting mirrors in lasers so that output laser beam has a high
degree of planar polarization.
Because the re‡ection coe¢ cient depends on wave polarization, unpolarized wave with random
polarization becomes partially polarized on re‡ection and transmission at a dielectric boundary.
The magnetic re‡ection coe¢ cient derived in Eq. (4.136) will play an important role in analyzing
the transition radiation discussed in Chapter 8.
When the conditions for total re‡ection are met, the impedance de…ned by the ratio between
26
the tangential component of the electric and magnetic …eld is
Z2TM
E2 cos
=
H2
2
=
r
i
0
"2
s
n1
n2
2
sin2
1
1:
(4.143)
The phase di¤erence between the incident and re‡ected waves is
TM
4.9
= 2 tan
1
Z1
Z2TM
!
= 2 tan
1
!
p
sin2 1 (n2 =n1 )2
:
(n2 =n1 )2 cos 1
(4.144)
Circularly and Elliptically Polarized Plane Waves
Planar polarization discussed in the preceding section is a highly idealized mode of propagation of
electromagnetic waves. A plane polarized wave can be decomposed into two circularly polarized
plane waves of opposite helicity, one with positive helicity and another with negative helicity.
Helicity of an electromagnetic wave is closely related with the angular momentum carried by the
wave.
Circularly polarized waves propagating in the z-direction can be described by the electric …eld
vectors,
E (z; t) = E0 (ex iey )ei(kz !t) ;
(4.145)
where the positive sign is for positive helicity and minus sign is for negative helicity. The sum of
these two waves of opposite helicity trivially yields a plane polarized wave with the electric …eld in
the x-direction. Corresponding magnetic …elds are
B
=
=
1
k E
!
k
E0 (ey
!
iex )ei(kz
!t)
;
(4.146)
and Poynting ‡ux is
S=E
H =2
E02
ez ;
Z
(4.147)
where the factor 2 accounts for the two independent modes of equal amplitude.
A general form of mixed helicity may be written as
E(z; t) = (E1 ex + E2 iey )ei(kz
!t)
;
(4.148)
where E1 and E2 are complex amplitude. Corresponding magnetic …eld,
B(z; t) =
k
(E1 ey
!
E2 iex )ei(kz
!t)
;
(4.149)
is of course normal to the electric …eld,
E B = 0:
27
(4.150)
However, the scalar product E B in general does not vanish.
Figure 4-6: Trace of the head of the rotating electric …eld vector associated with a circularly
polarized wave with positive helicity. In the case of negative helicity, the direction of rotation
relative to the wavevector is reversed.
Example 2 Re‡ection and Transmission of Circularly Polarized Wave
Consider a circularly polarized wave incident at an angle i to a ‡at surface of a dielectric. The
electric …eld of the incident wave may be decomposed into two components, one in the incident
plane and another perpendicular to the incident plane,
Ei = Ek0 + iE?0 :
The re‡ected wave of the parallel component is
Ekr =
tan(
tan(
1)
2
1
+
Ek0 ;
2)
while the re‡ected perpendicular component is
sin(
sin(
E?r =
1)
2
2
+
1)
E?0 :
The transmitted (refracted) components are
Ekt =
2 sin 1 cos
sin( 1 + 2 ) cos(
E?t =
2 cos
sin(
2
1
1 sin 2
1
+
2)
2)
Ek0 ;
E?0 :
Both re‡ected and transmitted waves are elliptically polarized. In particular, if the incident angle is
at the Brewster’s angle, 1 = =2 2 ; Ekr vanishes and the re‡ected wave becomes plane polarized.
28
On total re‡ection of a circularly polarized electromagnetic wave, the re‡ected wave becomes
elliptically polarized because the phases of TE and TM components di¤er as evident from Eqs.
(4.132) and (4.144).
4.10
Stokes’Parameters
Consider an electromagnetic wave propagating in the z direction with electric …eld components
E(r; t) = (Ex ex + Ey ey: )ei(kz
!t)
:
(4.151)
The amplitudes Ex and Ey may be complex allowing for …nite phase di¤erence,
Ex = jEx j ei x ; Ey = jEy j ei y :
(4.152)
is …nite,
If = x
y = 0; the …led is a simple superposition of two linearly polarized waves. If
the wave is in general elliptically polarized. In optics, direct measurement of the phase di¤erence
is not easy. What is normally measured is the intensity or the quadratic quantities of the electric
…eld,
jEx j2 ; jEy j2 ; jEx j jEy j cos( x
(4.153)
y ); and jEx j jEy j sin( x
y ):
Stokes’parameters are de…ned, with I = jEx j2 + jEy j2 the total intensity, by
s0 =
s1 =
s2 =
s3 =
1
I
1
I
2
I
2
I
jEx j2 + jEy j2 = 1;
jEx j2
jEy j2 ;
jEx j jEy j cos(
jEx j jEy j sin(
x
x
(4.154)
y );
y );
and satisfy
s21 + s22 + s23 = 1;
(4.155)
if the waves are purely coherent. If not, s2 and s3 are to be modi…ed as
s2 =
2
2
Re Ex Ey ; s3 = Im Ex Ey ;
I
I
(4.156)
where the bar indicates time average. For incoherent waves,
s21 + s22 + s23 < 1:
(4.157)
”Natural light” is characterized by a collection of many waves with random phases and complete
depolarization, s1 = s2 = s3 = 0 even though each wave may be highly monochromatic. For a
plane wave polarized in the x direction,
s1 = 1; s2 = s3 = 0:
29
For a plane wave polarized in the direction
= =4 (along the plane x = y),
s1 = s3 = 0; s2 = 1:
For a circularly polarized wave with positive (negative) helicity,
s1 = s2 = 0; s3 =
1:
In experiments, Stokes’ parameters can be determined by rotating a polarizer plate. jEx j2
and jEy j2 can be readily found by aligning the polarization direction along x and y direction,
respectively. If the polarizer is at angle from the x axis, the intensity measured at that angle is
I( ) = (jEx jei
x
cos + jEy jei
y
sin ) jEx je
i
x
= jEx j2 cos2 + jEy j2 sin2 + 2 jEx j jEy j cos sin cos(
By choosing
i
cos + jEy je
y
sin
y ):
x
= =4; for example, we have
I
=
=
4
1
jEx j2 + jEy j2 + 2jEx jjEy j cos(
2
y)
x
:
(4.158)
Finally, exploiting a uniaxial (or biaxial) crystal, a so-called quarter wavelength plate can be fabricated to induce =2 relative phase delay between Ex and Ey due to the di¤erent propagation
velocities of ordinary and extraordinary modes. The intensity in this case is
I 0 ( ) = (jEx jei
x
cos + ijEy jei
y
sin ) jEx je
= jEx j2 cos2 + jEy j2 sin2
If
i
x
cos
ijEy je
2 jEx j jEy j cos sin sin(
i
y
sin
y)
x
= =4; this reduces to
I0
=
=
4
Therefore, jEx jei x and jEy jei
suring the four intensities,
y
1
jEx j2 + jEy j2
2
2jEx jjEy j sin(
y)
x
:
(4.159)
and corresponding Stokes’parameters can be determined by mea-
I ( = 0) ; I
=
2
; I
=
4
and I 0
=
4
:
The tensor de…ned by
Iij = Ei Ej ; (i; j = x; y)
(4.160)
is called polarization tensor. Unless the …eld is purely coherent, the time averaged intensity Ei Ej
may still vary slowly with time. Iik is Hermitian and thus can be diagonalized through the eigen-
30
values
1
and
2
which are the roots of
det (Iij
ij )
= 0:
(4.161)
Corresponding two polarization eigenvectors n(1) and n(2) can be determined from
(1)
Iij nj
(1)
1 ni ;
=
(2)
Iij nj
=
(2)
1 ni :
(4.162)
In terms of the eigenvectors, the polarization tensor can be written in the form
!
I =
(1) (1)
1n n
+
(2) (2)
2n n :
(4.163)
As an example, let us consider superposition of two plane polarized waves, one in the x direction
and another in the direction (cos ; sin ) in the x y plane. The relative phase between the two
waves is assumed to be random and the intensities of the waves are I1 and I2 : The electric …eld is
E = (E1 + E2 ei cos ; E2 ei sin );
where I1 = E12 ; I2 = E22 ; and
is random phase. Then
I1 + I2 cos2
I2 cos sin
Iij =
Eigenvalues are
1;2
1
=
2
min =
I1 + I2
max
=
I2 cos sin
I2 sin2
q
(I1 + I2 )2
The ratio
=
(4.164)
I1 + I2
I1 + I2 +
p
p
!
:
(4.165)
4I1 I2 sin2
:
(I1 + I2 )2
4I1 I2 sin2
(I1 + I2 )2
4I1 I2 sin2
(4.166)
;
(4.167)
may be called the degree of depolarization.
= 1 corresponds to completely unpolarized state,
while = 0 corresponds to a plane polarized wave. When I1 = I2 = I; the polarization tensor
reduces to
!
1 + cos2
cos sin
Iij = I
:
(4.168)
cos sin
sin2
Eigenvalues are
1;2
=1
cos ;
(4.169)
and the degree of depolarization is
=
1 cos
:
1 + cos
(4.170)
The eigenvectors are
n(1) =
cos ; sin
2
2
; n(2) =
31
sin ; cos
2
2
:
(4.171)
Derivation of the eigenvectors is left for exercise.
The Stokes’parameters si and the polarization tensor are related through
1
Iij = I
2
ij
+
3
X
m=1
where
(1)
ij
=
(m)
sm ij
1
0
0
1
!
;
(2)
ij
!
=
1 + s1 s2
s2 + is3 1
1
= I
2
0 1
1 0
!
(3)
ij
;
=
0
i
!
is3
s1
i
0
!
;
;
(4.172)
(4.173)
are Pauli’s spin matrices. When s1 = 1; s2 = s3 = 0; Iij reduces to
Iij = I
1 0
0 0
!
;
which describes a plane polarized wave in the x direction. When s2 = 1; s1 = s3 = 0;
1 1
1 1
1
Iij = I
2
!
:
This describes the case of plane polarization in the direction
1
Iij = I
2
1
i
i
1
!
= =4: When s3 = 1; s1 = s2 = 0;
;
which describes circular polarization with positive helicity. Finally, s3 =
corresponding
!
1 i
1
Iij = I
;
2
i 1
1; s1 = s2 = 0 and
describes circular polarization with negative helicity.
4.11
Propagation along a Conductor Rod
In this section, we analyze propagation of electromagnetic waves along a conductor rod of radius
a and conductivity : A simple transverse Magnetic (TM) mode is considered with the following
…eld components, E ; Ez ; and H : The axial electric …eld Ez outside the conductor rod satis…es the
scalar wave equation,
@2
1 @
@2
+
+
@ 2
@
@z 2
Ez ( ; z; t) =
32
1 @2
Ez ( ; z; t):
c2 @t2
(4.174)
Figure 4-7: Field pro…les of electromagentic wave propagating along a conductor rod.
For a harmonic wave with time dependence e
i!t ;
this reduces to
@2
1 @
@2
!2
+
+
+
@ 2
@
@z 2
c2
Ez ( ; z) = 0:
(4.175)
Furthermore, since the wave is propagating along the rod, we may single out the z dependence in
the form eikz and reduce the wave equation to
d2
1 d
+
+
2
d
d
where
2
=
!
c
2
2
Ez ( ) = 0;
k2 :
(4.177)
Elementary solutions to Eq. (4.176) are the Bessel functions J0 (
binations,
(1)
H0 ( ) = J0 ( ) + iN0 ( );
(2)
H0 (
) = J0 (
(4.176)
)
iN0 (
);
); N0 (
) and their linear com(4.178)
(4.179)
known as Hankel functions of the …rst and second kind, respectively. Their asymptotic behavior at
large argument is
r
h
i
2
(1)
H0 (x) !
exp i x
;
x
4
r
h
i
2
(2)
H0 (x) !
exp i x
;
x
4
indicating radially outward and inward propagation, respectively. In the present case, both k
and are complex because of dissipation in the rod. It is reasonable to assume radially outward
33
propagation (radiation from the rod) and we choose
(1)
Ez ( ) = E0 H0 (
):
(4.180)
From r E = 0; we …nd the radial component of the electric …eld,
ik
E ( )=
and from r
E = i!
0 H;
);
(4.181)
the azimuthal component of the magnetic …eld,
H ( )=
where use is made of
(1)
E0 H1 (
i!"0
E0
(1)
H1 (
);
(4.182)
d
(1)
(1)
xH1 (x) = xH0 (x);
dx
d (1)
(1)
H (x) = H1 (x):
dx 0
The boundary condition at the rod surface is
Ez ( = a)
=
H ( = a)
or
r
Z=
i!
0
(surface impedance);
(4.183)
(1)
H0 ( a)
(1)
H1 ( a)
= i!"0 Z:
(4.184)
For a given wave frequency !; this equation determines and thus k; the axial wavenumber and
in this respect, it is a dispersion relation.
For a small impedance Z; approaches 0, and the axial wavenumber k becomes
k=
!
;
c
as expected. Noting
2
(1)
lim H0 (x) ! 1 + i (ln(x=2) +
x!0
E) ;
(1)
lim H1 (x) !
x!0
21
;
x
(4.185)
we see that the axial electric …eld becomes negligible and the transverse …elds approach those of
TEM mode,
E ( )
a
E ( ) = E0 ; H ( ) =
;
(4.186)
Z0
p
where Z0 =
0 ="0 is the impedance of free space. The …elds in coaxial cables used for transmission
of electromagnetic waves can be approximated by those given above,
E =
V
1
;
ln(b=a)
H =
34
E
I
=
;
Z
2
(4.187)
where V is the potential di¤erence between the inner and outer conductor with radii a and
p
b;respectively, I is the current and Z =
0 =" with " the permittivity of the insulating material …lling the cable. The characteristic impedance of the cable is
Zcable
p
V
=
=
I
0 ="
2
b
a
ln
:
(4.188)
Similarly, the impedance of a parallel wire transmission line with conductor radius a and separation
distance D is approximately given by
Z'
4.12
p
0 ="0
D
a
ln
;
D
a:
(4.189)
Skin E¤ects in Conductors
In a conductor, the conduction current dominates over the displacement current and a simple Ohm’s
law,
J = E;
(4.190)
may be assumed where
Maxwell’s equations
(S/m) is the conductivity. Eliminating the electric …eld between the
r
@B
; r
@t
E=
B'
0
E;
yields the following di¤usion equation for the magnetic …eld,
r2 B =
0
@B
:
@t
If the …eld is oscillating at a frequency !; we obtain the following ordinary di¤erential equation,
r2 B + i!
0
B = 0:
Penetration of the magnetic …eld into a conductor slab can be described by
d2 By
+ i!
dz 2
0
By = 0:
(4.191)
This has a bounded solution
B(z) = B0 eikz ;
where
k=
p
i!
0
1 + ip
= p
!
2
0
:
(4.192)
The magnetic …eld decays exponentially from the conductor surface and the quantity
=
s
2
!
0
35
; (m)
(4.193)
is called the skin depth. Damping of the electromagnetic …elds is evidently due to Ohmic dissipation
in the conductor.
For a cylindrical conductor rod with radius a( c=!) in which an axial current Jz ( )e i!t is
excited, Jz obeys
1 d
d2
+
+ i! 0
Jz ( ) = 0;
(4.194)
2
d
d
This has a bounded solution
Jz ( ) = E0
J0 (k )
;
J0 (ka)
(4.195)
where E0 is the electric …eld at the rod surface. The impedance per unit length of the rod can thus
be de…ned by
Z
E0
k J0 (ka)
=
=
; (Ohms/m):
(4.196)
l
I
2 a J1 (ka)
where use is made of
I=2
Z
a
a
J1 (ka)
E0
:
k
J0 (ka)
Jz ( ) d = 2
0
In the low frequency limit jkj a
(4.197)
1; series expansion of the Bessel functions can be exploited,
J0 (x) ' 1
x2
x
; J1 (x) '
4
2
1
x2
8
:
The impedance reduces to
Z
1
=
l
a2
where
i!
0
8
;
(4.198)
Li
= 0 (H/m)
l
8
(4.199)
is the well known internal inductance of a cylindrical rod carrying a uniform current (no skin e¤ect).
In the high frequency limit (strong skin e¤ect), the ratio J0 (ka)=J1 (ka) approaches i; and the
impedance is
Z
1
=
(1 i):
(4.200)
l
2 a
This is also reasonable since with strong skin e¤ect, the current ‡ow is limited in a thin layer with
an area 2 a : Note that the inductive reactance is identical to the resistance in this limit.
If a conductor cylinder is placed in an oscillating axial magnetic …eld Hz0 e i!t as in inductive
rf heating, the current ‡ows in the azimuthal direction. J ( ) satis…es
1 d
d2
+
2
d
d
1
2
+ i!
0
J ( ) = 0;
< a:
An appropriate solution is
J ( )=k
J1 (k )
Hz0 ;
J0 (ka)
36
< a;
(4.201)
where k is still given by
k=
4.13
p
i!
0
:
Skin E¤ect in a Plasma
A charge neutral plasma contains equal amount of positive and negative charge density. When
placed in an electric …eld oscillating at a high frequency, a plasma current due to electron motion
is induced. The conductivity can be found by letting ! 0 = 0 in Eq. (4.88) (because in a plasma,
electrons are free),
i ne2
=
;
(4.202)
! m
and corresponding permittivity
" = "0
! 2p
!2
1
!
In a collisional plasma with electron collision frequency
" = "0
1
:
c;
! 2p
! (! + i c )
(4.203)
this is modi…ed as
!
;
(4.204)
as can be easily worked out by introducing a …nite collision frequency in the equation of motion.
For a frequency much smaller than the plasma frequency, the permittivity becomes negative,
"'
"0
! 2p
;
!2
(4.205)
and wave propagation is forbidden. The wavenumber is complex in this case
k=i
!p
;
c
(4.206)
and the wave amplitude decays in a manner
E0 ei(kz
!t)
= E0 exp
!p
z e
c
i!t
;
(4.207)
where z is the distance in the plasma from the vacuum-plasma boundary. This means that an
electromagnetic wave incident on a plasma cannot penetrate into plasma except for a distance of
the order of the skin depth de…ned by
c
=
:
(4.208)
!p
The wave is completely re‡ected if the plasma is collisionless. Re‡ection of low frequency radio
waves by the ionoshperic plasma and is a well known example.
37
In a collisional plasma with !
!p;
c;
the skin depth is modi…ed as
c
'
!p
r
!
:
2 c
(4.209)
Derivation of this formula is left for exercise.
Strictly speaking, the collisionless skin depth = c=! p is valid for a cold plasma with negligible
electron temperature. To implement the e¤ects of …nite electron temperature, it is necessary to
employ a kinetic theory to …nd a conductivity : The electron velocity distribution function f
(r; v; t) obeys the kinetic equation,
@f
+ v rf
@t
e
(E + v
m
B)
@f
= 0:
@v
(4.210)
As it is, it is a nonlinear equation because the electromagnetic …elds E and B associated with a
wave a¤ect the distribution function f: Let us assume a wave propagating along the z direction
with electric …eld polarized in the x direction Ex (z; t) = E0 ei(kz !t) : The distribution function may
be linearized as f = f1 + FM ; where f1 is the perturbation and FM is unperturbed Maxwellian
distribution. Noting
mv
@FM (v 2 )
=
FM (v 2 );
(4.211)
@v
Te
and thus
(v
B)
we …nd
f1 =
@FM (v 2 )
= 0;
@v
e
Ex
Te
vx FM :
i (kvz !)
(4.212)
The current density can be found from the …rst order moment,
Z
Je = ne vf1 dv:
(4.213)
Only Jx is nonvanishing, and given by
Jx =
=
=
where
= !=kvT e with vT e =
p
ne2
Ex
i
Te
Z
vx2
kvz
!
FM dv
Z 1
ne2
1
e
i
Ex p
mkvT e
1 t
2
ne
i
Ex Z ( ) ;
mkvT e
t2
dt
(4.214)
2Te =m being the thermal velocity of electrons and Z ( ) is known
38
as the plasma dispersion function. The permittivity is therefore given by
" = "0
!
! 2p
Z( ) :
1+
!kvT e
If the electron temperature is negligible
(4.215)
1; the function Z ( ) approaches
Z( )'
1
;
and we recover the case of cold plasma,
" = "0
In the opposite limit
1;
1
! 2p
!2
!
:
p
Z( )'i ;
and
! 2p
p
" ' i "0
:
!kvT e
The damping factor (inverse skin depth) in this limit is given by
Im k =
1
1=6
'
2
!p
c
2=3
!
vT e
1=3
:
(4.216)
This is often called anomalous skin e¤ect. (This is probably misnomer because there is nothing
anomalous in the derivation.)
4.14
Waves in Anisotropic Dielectrics
Some crystals exhibit anisotropy in polarizability. The permittivity in such media becomes a tensor,
and the displacement vector D and electric …eld E are related through a dielectric tensor ";
D = " E; or Di ="ij Ej ;
(4.217)
where " is a diagonal tensor consisting of three permittivities in each axial direction, x; y; and z;
1
"x 0 0
B
C
" = @ 0 "y 0 A :
0 0 "z
0
(4.218)
In uniaxial crystals, polarization occurs preferentially along one axis, say, "x = "y 6= "z : In
anisotropic media, the phase and group velocities are in general oriented in di¤erent directions
and the well known double refraction phenomenon (already known in the 17th century) can be
39
explained in terms of the dielectric anisotropy.
The Maxwell’s equations to describe wave propagation in an anisotropic medium in which there
are no sources ( free = 0; J = 0) are
r
@B
; r
@t
E=
B=
0
@D
=
@t
Then
r
r
E=
@2E
;
@t2
0"
r (r E) r2 E+
0"
0"
@2E
= 0:
@t2
@E
:
@t
(4.219)
(4.220)
(4.221)
Note that in an anisotropic medium, the divergence of the electric …eld, r E; does not necessarily
vanish, although
r D = free = 0;
(4.222)
must hold. After Fourier decomposition, Eq. (4.221) reduces to
k(k E)
k2 E + !2
0"
E = 0;
(4.223)
or
(k 2
!2
ki kj
ij
0 "ij )Ej
= 0:
(4.224)
The dispersion relation of electromagnetic waves in a uniaxial crystal is therefore given by the
determinant,
det(k 2 ij ki kj ! 2 0 "ij ) = 0:
(4.225)
Introducing ^"ij = "ij ="0 and the index of refraction,
n=
c
k;
!
(4.226)
we rewrite the dispersion relation as
det(n2
ij
ni nj
^"ij ) = 0:
(4.227)
In uniaxial crystals, polarizability along one axis (say, z-axis) di¤ers from those along other
axes. We assume
"x = "y "? ; " z "k :
(4.228)
Since the crystal is symmetric about the z-axis, the wave vector k can be assumed to be in the
x z plane without loss of generality,
kx = k? ;
40
kz = kk ;
(4.229)
or
n? =
ck?
;
!
nk =
ckk
:
!
(4.230)
The dispersion relation then becomes
n2k
^"? 0
0
n? nk
n? nk
n2
0
^"? 0
n2?
= 0:
(4.231)
^"k
Expanding the determinant, we …nd
n2 = ^"? ;
(4.232)
and
^"k n2k + ^"? n2?
^"k^"? = 0:
(4.233)
The …rst mode of propagation is independent of the propagation angle as if in an isotropic medium.
It is called the ordinary mode because it maintains the properties of electromagnetic waves in
isotropic media. The electric …eld is in the y-direction and the dispersion relation in Eq. (4.232) is
equivalent to the following di¤erential equation,
@2
@2
+
@x2 @z 2
0 "? !
2
Ey (x; y) = 0:
(4.234)
In this mode, the phase and group velocities are in the same direction (in the direction of k)
although in magnitude they di¤er because of the frequency dependence of the permittivity "? (!):
The second mode has a peculiar property and for this reason is called extraordinary mode. Let
the angle between k and the z-axis be : The dispersion relation in Eq. (4.233) becomes
ck
!
^"k cos2 + ^"? sin2
The phase velocity given by
s
!
cos2
=c
k
^"?
+
2
^"k^"? = 0:
sin2
ek ;
^"k
(4.235)
(4.236)
is directed along k: The group velocity is
d!
dk
=
=
@!
1 @!
ek +
e
@k
k@
!
c2
1
ek +
k
!=k ^"k
1
^"?
sin cos e :
(4.237)
This is evidently not parallel to the phase velocity unless ^"k = ^"? (isotropic dielectric). The
magnitude of the group velocity is always larger than that of the phase velocity.
Electromagnetic waves in a plasma con…ned by a magnetic …eld is another example of anisotropic
medium which accommodates variety of waves. A simple case of cold plasma will be considered in
41
Problem 4.11.
Figure 4-8: Double refraction of randomly polarized incident wave by a uniaxial crystal. Propagation of the O-mode (ordinary mode) is una¤ected but that of the X-mode (extraordinary mode) is.
The group velocity of the X-mode in the crystal deviates from the incident direction.
An important consequence of the presence of the ordinary and extraordinary modes is the well
known double refraction caused by some crystals. Consider a light beam with random polarizations
incident normal to a surface of a uniaxial crystal as shown in Fig. 4-8. Unless the optical axis (zaxis in the geometry assumed) coincides with the direction of the incident beam, the beam is split
into two beams at the surface. In the crystal, one beam propagates along the direction of the
incident beam and another at an angle. The phase velocities of the ordinary and extraordinary
modes are in the same direction as the incident beam but the group velocity of the extraordinary
mode deviates from the incident direction. Wave energy propagates at the group velocity and thus
beam splitting occurs.
Some isotropic dielectrics can become uniaxial media if placed in an electric …eld. This is
because the permittivity is in general nonlinear and the component in the direction of the …eld
becomes …eld-dependent,
" = "(0) + E n ;
(4.238)
where is a constant, n = 1 is for Pockel’s e¤ect and n = 2 is for Kerr e¤ect. Pockel’s e¤ect in
some liquids is widely used for laser switching and optical modulation.
42
Problems
4.1 A sphere of radius a carries a charge q which is decreasing due to emission of charge in
every radial direction. Show that there is no magnetic …eld even though there exists a radial
conduction current density.
4.2 A cylindrical permanent magnet with radius a and a uniform axial magnetization Mz is
rotating at an angular frequency ! about its axis. What is the emf induced in the closed
circuit shown?
Problem 4.2. Unipolar generator.
4.3 The permittivity of an unmagnetized plasma is given by
"(!) = "0
1
! 2pe
!2
!
;
where ! pe is the electron plasma frequency,
! pe =
s
ne2
:
" 0 me
Show that the energy density of a plane electromagnetic wave in a plasma is
1
1 ! 2pe
1
u = "0 E 2 + "0 2 E 2 +
2
2 !
2
and interpret the term
1 ! 2pe 2
"0
E :
2 !2
43
0H
2
= "0 E 2 ;
4.4 An isotropic dielectric has a permittivity in the form
"(!) = "0
1
! 2p
!2
! 20
!
:
Show that the group velocity does not exceed c: What if a …nite dissipation is allowed,
! 2p
! 2 + 2i !
1
"(!) = "0
! 20
!
;
where is a damping constant? For simplicity, assume ! p = ! 0 and plot Re (d!=dk) as a
function of ! for various =! 0 :
4.5 Show that if a dielectric medium is loss free, its dielectric tensor should be Hermitian,
"ij = "ji :
Note: In the absence of external magnetic …eld, the tensor is symmetric "ij = "ji : In this
case, the loss-less condition is Im ("ij ) = 0:
4.6 Find the permittivity and thickness of a dielectric …lm to be coated on a glass surface to
eliminate re‡ection of light wave of wavelength = 550 nm. Assume that the glass has an
index of refraction of n = 1:5:
4.7 A light beam linearly polarized is incident on a glass (nglass = 1:5) surface at an angle i = 45 :
Find the amplitude and polarization of the re‡ected and refracted waves. Consider both cases
of polarization, electric …eld in the incident plane and magnetic …eld in the incident plane.
4.8 A light beam of circular polarization in glass is incident on a ‡at glass-air boundary at an
angle i = 60 : Find the polarization of totally re‡ected beam.
4.9 A uniaxial crystal has "xx = "yy = 1:3"0 and "zz = 1:5"0 : The optical axis is at angle 70 to
a ‡at surface. A light beam of random polarization is incident normal to the surface. Find
the propagation direction of the extraordinary mode in the crystal.
4.10 A uniaxial crystal has the following permittivity tensor,
0
1
3"0 0
0
B
C
!
" = @ 0 3"0 0 A :
0
0 2"0
A light beam in air is incident on a ‡at surface(x y plane)of the crystal at an angle i
from the normal which coincides with the optical axis z: Show that the refraction law for the
44
extraordinary mode is
tan
r
p
^"? sin
=q
^"k ^"k
i
sin2
;
i
where ^"? = 3; ^"k = 2: What is the refraction law for the ordinary mode?
4.11 Find the impedance per unit length of a copper wire of radius 3 mm at f = 60, 103 and 106
Hz. Copper conductivity is = 5:8 107 S/m at room temperature.
4.12 Using the equation of motion for electrons in a cold magnetized plasma,
me
@v
=
@t
e (E + v
B0 ) ;
where B0 (external magnetic …eld) is in z direction, show that the dielectric tensor is given
by
3
2
! 2p
! 2p
i
0
7
6 1 !2
2
2)
!(! 2
7
6
2
2
7
6
!
!
!
p
ps
7
" = "0 6
1
0
7
6 i
2)
2
!2
7
6 !(! 2
4
! 2p 5
0
0
1
!2
p
where ! p = ne2 =me "0 is the plasma frequency and e = eB0 =me is the cyclotron frequency.
The wavevector k may be assumed to be k = k? ex +kk ez without loss of generality because of
axial symmetry. Note that the tensor is Hermitian. (In this analysis, the electron temperature
is ignored.)
4.13 A laser beam passes through a glass window of refractive index 1.50 into water with refractive
index 1.33. The beam is E-polarized (electric …eld in the incident plane). Design the glass
window to avoid re‡ection at both surfaces.
45