Phys 460
1
1.1
Fall 2006
Susan M. Lea
Wavesw basics
The wave equation for a string
Waves occur in many physical systems. A wave is a disturbance that propagates
through the system at a well-determined speed that depends on the physical
properties of the system. All wave disturbances may be decomposed into a sum
of simpler waves- sinusoidal waves. In these waves, the shape of the disturbance,
captured by taking a snapshot of the disturbance in the medium at a ¤xed time,
has the shape of a sine wave. Every point of the system oscillates in time. Thus
we might expect that the system has restoring forces that try to return the
system to its initial equilibrium. Wezll begin by studying a simple mechanical
systemw a string. The restoring force is the tension in the string.
In equilibrium the string lies along the x¡axis: We look at a di¤erential
piece of the displaced string, as shown in the diagram.
The net force on the string has components:
dFx
=
dF y
=
T (x + dx) cos (µ + dµ) ¡ T (x) cos µ
T (x + dx) sin µ ¡ T (x) sin µ
Now let the displacement of the string be small everywhere, fo that µ ¿ 1
everywhere. Then
cos µ
=
sin µ
=
µ2
+ ¢¢¢ ' 1
2
µ3
µ¡
+ ¢¢¢ ' µ
6
1¡
and
dFx
=
dF y
=
T (x + dx) ¡ T (x)
T (x + dx) (µ + dµ) ¡ T (x) µ
1
Each piece of the string moves vertically but not horizontally, so
dFx = 0 ) T (x + dx) = T (x)
The tension is constant along the string. Then the y¡component is
dFy = T dµ = (dm)
@2y
@t2
The time derivatives are partial derivatives: we are looking at a ¤xed value of
x: The mass of the string segment is
dm = ¹dx
The string has to stretch a bit in order to curve, but the mass per unit length
¹ is measured along the undisturbed string. Thus
T dµ = ¹dx
or
@2y
@t2
@2 y
T @µ
=
@t2
¹ @x
The x¡derivatives are also partial because the picture shown is a snapshot
taken at a ¤xed time. Now we work on the angle µ: The slope of the string is
tan µ =
Thus
@y
' µ for µ ¿ 1:
@x
@2 y
T @2y
=
2
@t
¹ @x2
(1)
This is the wave equation for the string. The wave speed is given by
v2 =
T
¹
and v is roughly equal to the square root of the restoring force/inertia. The
general solution to this equation has the form
y = f (x § vt)
as you can easily check by di¤erentiating and stu¢ng in. With the minus
sign, the solution represents a wave propagating in the direction of increasing x;
with the plus sign the pulse propagates in the direction of decreasing x: Since
the wave equation is linear, we can have superpositions of such solutions, for
example
y = f (x + vt) + g (x ¡ vt)
Sinusoidal waves occur when f is a sine (or cosine) wave.
y = A cos [k (x ¡ vt) + Á] = A cos (kx ¡ !t + Á)
2
At ¤xed t; the cosine repeats after a distance ¢x where
k¢x = 2¼
or
2¼
=¸
k
where ¸ is the wavelength of the wave. Similarly, at ¤xed x; the cosine repeats
after a time
2¼
¢t =
=T
!
where T is the wave period. Also
¢x =
!
=v
k
A is the wave amplitude. It is the maximum value of the disturbance anywhere
in the wave. Á is a phase constant. It tells us where the maximum displacement
occurs at t = 0:
The speed v is the phase speed of the wave. The wave phase is the argument
of the cosine. So a ¤xed phase Á0 corresponds to
kx ¡ !t + Á = Á0
or
!
Á ¡Á
Á ¡Á
t+ 0
= vt + 0
k
k
k
so if k is positive, the value of x corresponding to phase Á0 increases at speed v:
Standing waves occur when equal amplitude waves travel in opposite directions:
x=
y stan d
=
=
A cos (kx ¡ !t) + A cos (kx + !t + Á)
µ
¶
Á
2A cos kx +
cos !t
2
Now we have a ¤xed spatial function that oscillates in time.
1.1.1
More math makes it easier
We can use complex numbers to make the math easier. Eulerzs formula tells us
eiµ = cos µ + i sin µ
Thus
A cos (kx ¡ !t + Á)
=
=
=
Re A exp (i [kx ¡ !t + Á])
n¡
´o
¢³
Re Ae iÁ ei(kx¡!t)
³
´
e ikx¡i!t
Re Ae
3
where
e = AeiÁ
A
is the complex amplitude of the wave. With this notation, the general solution
of the wave equation may be written:
Z +1
e (k) eikx¡i!t dk
f (x; t) =
A
¡1
where
! = kv
The "Re " is implied but is not always written. Real physical quantities are
always the real part of complex numbers.
1.2
Boundary conditions: re¢ection and transmission
When a wave reaches a point where the properties of the system change, part of
the wave energy will be re¢ected and part transmitted into the new region. In
the case of 1-D waves on a string, the mass/unit length ¹ is the critical system
property, As ¹ changes, so does v :
s
s
T
T
v1 =
;
v2 =
¹1
¹2
and thus
!1
!2
= v 1;
= v2
k1
k2
The disturbance on the string has three components: the incident wave y i =
A exp (ikix ¡ i!t), the re¢ected wave travelling in the negative x¡direction,
y r = Ar exp (¡ikr x ¡ i! r t) and the transmitted wave yt = At exp (ikt x ¡ i!t t).
The ¤rst two travel at speed v 1 while the transmitted wave travels at speed v2 :
The amplitudes of the re¢ected and transmitted waves are determined by the
boundary conditions at the junction between the strings. To determine the two
amplitudes, we need two boundary conditions. They are:
(1) The string displacement is continuous across the junction. If
this were not true, the string would have an unphysical "break" at the junction.
For simplicity, we place the origin at the junction between the strings.
yi (0; t) + yr (0; t) =
A exp (¡i!t) + Ar exp (¡i!r t) =
y t (0; t)
At exp (¡i! t t)
This boundary condition must hold for all times t: Thus it must be true that
! = !r = !t
The frequency of all the waves is the same. Then it follows that
A + Ar = At
4
(2)
and also
kr =
!
= ki ;
v1
kt =
!
v2
(2) The slope of the string is continuous across the junction. If
this were not true, there would be unbalanced forces on the junction that would
cause rapid acceleration, rapidly restoring the force balance and the continuity
of the slope. (Be careful herex if the strings are tied together so that there is a
massive knot at the junction, then there may be a discontinuity of slope. This
is not the usual case.)
ikiAe¡ i!t ¡ ikr Ate ¡i!t = ikt At e¡i!t
The exponentials cancel, leaving
A
Ar
At
¡
=
v1
v1
v2
(3a)
From equations (2) and (3a) we can solve for the amplitudes:
A ¡ Ar
µ
¶
v1
A 1¡
v2
=
Ar
=
=
v1
(A + Ar )
v2
µ
¶
v1
Ar 1 +
v2
v2 ¡ v1
A
v2 + v1
(4)
and
At
=
=
µ
¶
v 2 ¡ v1
A 1+
v 2 + v1
µ
¶
2v2
A
v 2 + v1
(5)
Note that if v2 = v1 ; there is no junction, and Ar = 0; At = A; as expected.
If the second string is heavier than the ¤rst, ¹2 > ¹1 ; then v 2 < v 1 and Ar is
negative. Recall that these amplitudes are complex numbers, so
Ar = ¡A
v 1 ¡ v2
v1 ¡ v2
= ei¼ A
v 2 + v1
v2 + v1
The re¢ected amplitude has a phase di¤erence of ¼ compared with the incident
amplitude. Taking the real part,
yr = jAj
v1 ¡ v 2
cos (¡ikr x ¡ !t + ¼)
v2 + v 1
and the wave has a phase change of ¼: The re¢ected wave is ¢ipped "upside
down".
Boundary conditions like this (continuity of function, continuity of derivative) apply in most physical systems.
5
1.3
Polarization
The waves on the string are transverse waves: the string itself moves perpendicular (in the y¡direction) to the direction that the wave itself moves (the
x¡direction). The wave is said to be polarized in the vertical (y) direction.
We could also set the string vibrating so that it moves in the z¡direction and
we would say the wave is polarized in the z¡direction. Or we could choose any
direction in the y ¡ z plane. These are linearly polarized waves. Now suppose
we set the string vibrating in the y¡direction, and a quarter period later we also
set it going with equal amplitude in the z¡direction. The total displacement
is
½
·µ
µ
¶¶¸¾
T
~s = A y^ exp (ikx ¡ i!t) + z^ exp
ikx ¡ i! t ¡
4
³
´
= Aeikx¡i!t y^ + z^ei¼=2
=
Aeikx¡i!t (^
y + i z^)
Now we take the real part:
~s = A [y^ cos (kx ¡ !t) ¡ ^z sin (kx ¡ !t)]
Fix attention at one point of the string w say x = 0; then we have
~s (0; t) = A (^
y cos !t + z^ sin !t)
Now the string moves around a circle of radius A in the y ¡ z¡plane. This is
circular polarization.
Contrast the string with waves on a springw these are longitudinal wavesw
the spring coils travel back and forth along the spring- the same direction that
the wave moves. The concept of polarization does not arise for these waves.
1.4
Energy transmission
As the wave travels along the string, it transmits energy. We can see how the
energy is related to the wave properties by looking at the rate at which the
string to the left of a point x = x 0 does work on the string to its right. The
~ on the string to the right, and at the point of
string to the left exerts a force T
contact the string has velocity
~v = y^
@y (x0 ; t)
@t
Thus
~ ¢ ~v = ¡T sin µ @y (x 0 ; t)
P =T
@t
But since the string has small displacement, µ ¿ 1;
sin µ ' tan µ =
6
@y
@x
So, with y = A cos (kx ¡ !t)
P
= ¡T
¯
@y ¯¯ @y (x 0 ; t)
@x ¯x0
@t
= T kA sin (kx0 ¡ !t) !A sin (kx 0 ¡ !t)
= ¹v!A2 sin2 (kx 0 ¡ !t)
Since the sine function is squared, the power is always positive; energy is transmitted to the right continuouslyw this is the direction in which the wave is
propagating. The power depends on the square of the wave ampliudew doubling
the amplitude quadruples the energy transmittedwand also depends on the wave
speed and the wave frequency.
2
Electromagnetic waves in vacuum
Maxwellzs equations in vacuum are (½ = 0; ~j = 0)
~ ¢E
~
r
~ ¢B
~
r
~ £E
~
r
~ £B
~
r
= 0
(6)
= 0
(7)
~
@B
= ¡
@t
~
@E
= "0 ¹0
@t
(8)
(9)
Even though ½ and ~j are zero, the ¤elds are not necessarily zero, because chang~ acts as a source of E;
~ and changing E
~ acts as a source of E:
~ The resulting
ing B
¤elds are wave ¤elds that travel at the speed of light. To demonstrate this,
letzs try to eliminate one of the ¤elds. Start with Amperezs law (9) and take
the curl:
³
´
~ £ r
~ £B
~
r
³
´
~ r
~ ¢B
~ ¡ r2 B
~
r
=
=
~
1 ~
@E
r
£
c2
@t
´
1 @ ³~
~
r
£
E
c2 @t
Now use equation (7) on the left and Faradayzs law (8) on the right:
~ =¡
¡r 2B
~
1 @2B
c2 @ 2t
so we obtain the wave equation with wave speed c :
~ =
r 2B
7
~
1 @2B
2
2
c @ t
(10)
The equation for waves on a string (1) was 1-dimensional in the space variables,
but here we have the r2 operator, so we have derivatives in all 3 space variables.
µ
@2
@2
@2
+
+
@x 2
@y 2
@z 2
¶
2~
~ = 1 @ B
B
c2 @ 2 t
The waves are not constrained to move in one direction, but can move along
any line in space, described by the wave vector ~k: The magnitude of this vector
is related to the wavelength, as with 1¡D waves.
¯ ¯
!
2¼
¯ ¯
k = ¯~k¯ =
=
c
¸
The speed of the waves in vacuum is
c= p
1
= 3 £ 10 8 m/s
"0 ¹0
EM waves occur at all frequencies. The names we give the waves are historical
and re¢ect the methods we use to detect them. They range from gamma rays at
very high frequencies to low frequency radio. See G page 377 and LB page 533.
Visible light occupies a small fraction of the spectrum around º =10 14 ¡ 1015
Hz (or, ¸ = 400-700 nm)
The electric ¤eld satis¤es the same equation, as we can see by starting with
Faradayzs law:
³
´
~ £ r
~ £E
~
r
³
´
~ r
~ ¢E
~ ¡ r2 E
~
r
~
r2 E
2.1
~ £
= ¡r
~
@B
@t
´
@ ³~
~ =¡@
= ¡
r£B
@t
@t
=
Ã
~
@E
"0 ¹0
@t
!
(11)
~
1 @2E
2
2
c @t
Monochromatic plane waves
The term "monochromatic" or "single-color" means that the wave has a single
frequency !: This is an idealization, of course, but it greatly simpli¤es our
math. The wave is plane if a surface of constant phase is a ¢at plane. Letzs
choose to put the z¡axis along the vector ~k; so that the wave is travelling in
the z¡direction. Then we can write the waves as:
~
B
~
E
~ 0 exp (ikz ¡ i!t)
= B
~ 0 exp (ikz ¡ i!t)
= E
Stu¢ng into Gaussz law, we have
~ ¢E
~ = ikE0;z exp (ikz ¡ i!t) = 0
r
8
This can be satis¤ed only if k = 0; which makes no sense, or E 0;z = 0: Thus
~ 0 has no z¡component, and E
~ is perpendicular to ~k: Since r
~ ¢B
~ is also zero,
E
~
we get the same result for B0 : Since both ¤eld vectors are perpendicular to the
direction of wave propagation, EM waves in vacuum are transverse waves. Now
~ 0 : From Faradayzs law, we get
letzs put the x¡axis along the directionof E
~ £E
~
r
~ £ [E 0 x^ exp (ikz ¡ i!t)]
= r
@
= y^
[E 0 exp (ikz ¡ i!t)] = ikE 0 y^ exp (ikz ¡ i!t)
@z
@ ~
~
= ¡ B
0 exp (ikx ¡ i!t) = i! B0 exp (ikx ¡ i!t)
@t
Thus we get
~ 0 = k E 0 y^ = E 0 ^y
B
(12)
!
c
Thus the magnetic ¤eld is perpendicular to the electric ¤eld, has magnitude
~
E 0 =c; and the wave amplitude has the same phase as the E¡wave
amplitude.
~
~
Since E is perpendicular to B; by convention we choose the electric ¤eld vector
to give the direction of polarization. This wave is polarized in the x¡direction:
We can write all of this in a coordinate-free form. If the wave propagates
^ the dot product k^ ¢ ~r gives the distance along the direction of ^k;
in direction k;
so we replace kz with kk^ ¢ ~r = ~k ¢ ~r : Thus
³
´
~ = E
~ 0 exp i~k ¢ ~r ¡ i!t
E
³
´
~ = B
~ 0 exp i~k ¢ ~r ¡ i!t
B
where, from the divergence equations,
~k ¢ E
~ 0 = ~k ¢ B
~0 = 0
and from the curl equation
³
´
~ £E
~ 0 exp i~k ¢ ~r ¡ i!t
r
Thus
¶
µ
¶
µ
¶
@
@
@
@
@
@
Ez ¡
Ey + y^
Ex ¡
E z + z^
Ey ¡
Ex
@y
@z
@z
@x
@x
@y
= [x^ (iky E0;z ¡ ¡kz E0y ) + ^y (ikz E0;x ¡ ikx E 0;z )
³
´
+^
z (ikxE 0;y ¡ iky E 0x )] exp i~k ¢ ~r ¡ i!t
³
´
³
´
~ 0 exp i~k ¢ ~r ¡ i!t = ¡ @ B
~ 0 exp i~k ¢ ~r ¡ i!t
= i~k £ E
@t
³
´
~
~
= i!B 0 exp i k ¢ ~r ¡ i!t
= x^
µ
~k £ E
~ 0 = !B
~0
9
(13)
2.2
Energy and momentum in EM waves
The energy density in the waves ¤elds is
µ
¶
1
B2
2
u =
"0 E +
2
¹0
¢
"0 ¡ 2
=
E + c 2 B2
2
But since we already showed that B = E =c; there is equal energy in the electric
and magnetic components of the wave.
³
´
u = "0 E 2 = " 0 E02 cos2 ~k ¢ ~r ¡ !t + Á
The Poynting vector describes the ¢ux of energy:
Ã
!
~k £ E
~
1 ~
1 ~
~
~
S =
E £B =
E£
¹0
¹0
!
h
³
´i
1 ~ 2 ~ ~ ~
=
kE ¡ E k ¢ E
!¹0
~ = 0; so
But ~k ¢ E
2
2
~ = E k^ = "0 E k^ = cu ^k
S
c¹0
c¹0 "0
This shows explicitly that the energy density in the wave is carried along at the
wave speed c:
If we average over several (or many) wave periods, we get the time averaged
transmitted power per unit area of the wavefront:
<P> =
1
"0 E 02
2c
As with waves on a string, the power goes as the square of the wave amplitude,
depends on the wave speed c; and also on the properties of the medium, here
" 0 : Physicists call this quantity the wave intensity. Astronomers beware w the
term intensity in astronomy means something di¤erent.
The waves carry momentum as well as energy: From Notes 2 equation 6, the
average momentum ¢ux density is
~
~E M > = < S >
<P
c
Now if the wave energy is absorbed at a ¢at screen, momentum is also aborbed,
and thus there is a force exerted on the screen. The force, from Newtonzs
second law, is
d~p
F~ =
dt
10
and the pressure on the screen is the normal force per units area. If the wave
strikes the screen at normal incidence, then
¯
¯
¯ ~ ¯
<
S
>
¯
¯
jF j
E2
1
P =
=
= 2
= " 0 E2
A
c
2c ¹0
2
If the screen re¢ects all the energy and momentum, then the momentum imparted to the screen is twice as big:
Momentum in wave
Momentum in screen
Total
Before re¢ection
~pin
0
~pin
after re¢ection
p~ou t = ¡~p in
p~sc re e n
p~sc re e n ¡ p~in
Setting the total before equal to the total after, we get
~ps cre e n = 2~pin
Things get more interesting when the wave is incident at an angle µ: (See LB
"Digging Deeper" pg 1049.)
Momentum in direction ^k at angle µ to the surface normal n
^ is carried by
the wave. Only the normal component is absorbed or re¢ected:
p n orm al = p cos µ
The momentum in area dA? of the wave impacts area dA of the surface, where
dA? = dA cos µ
11
Thus the momentum absorbed per unit area is
¢p n orm
p cos µ
p
=
=
cos2 µ
¢A
dA? = cos µ
dA?
Thus the radiation pressure is
P =
f
" 0 E 2 cos2 µ
2
where the factor f = 1 for total absorption and = 2 for total re¢ection.
3
Electromagnetic waves in matter
When waves propagate in matter, there are non-zero sources ½ and ~j that must
be included in Maxwellzs equations. These sources a¤ect the speed and other
properties of the waves. Here we will discuss LIH materials that can be de~ and H;
~ thus allowing us to "bury" the bound charges.
scribed by the ¤elds D
Then if the free charge density and free current density are zero, the equations
are:
~ ¢D
~
r
~ ¢B
~
r
~ ¢E
~
= 0 = "r
(14)
= 0
~
@B
@t
~ £E
~
r
= ¡
~
~ £B
r
¹
~
~ £H
~ = " @E
= r
@t
(15)
We ¤nd the wave equation using the same methods as before. Comparing these
equations with equations (6) to (9), we see that the only di¤erence is that " 0
has been replaced by " and ¹0 has been releced by ¹: Thus the wave equation
~ takes the form
for E
2~
2~
~ = "¹ @ E = 1 @ E
r2 E
(16)
2
2
@t
v @t2
where the wave phase speed is
1
v=p
"¹
In most ordinary materials like glass, ¹ ' ¹0 and " > " 0 ; so that
r
r
v
"0 ¹0
"0
=
'
<1
c
"¹
"
and light travels more slowly than in vacuum. The index of refraction of the
material is
r
v
"
n= '
>1
c
"0
12
The Poynting vector is now
~ = 1E
~ £B
~
S
¹
~ and B
~ ¤elds may be found from
The relation between the amplitudes of the E
Faradayzs law:
~
~k £ E
~ 0 = !B
~ 0 = k B0
v
so that
E0
B0 =
(17)
v
The ¤elds are still perpendicular to each other and to ~k; and vary in phase.
4
4.1
Re¢ection and transmission of waves at a boundary
Normal incidence
When a wave reaches a boundary between two di¤erent, LIH media, the wave
is partially transmitted and partially re¢ected. As with waves on a string, each
wave has the same frequency. This ensures that the boundary conditions are
satis¤ed at all times, not just at one time. The boundary conditions on the
¤elds are the same ones that we found in Physics 360:
~ is continuous
normal D
~ is continuous
tangential E
(18)
~ is continuous
normal B
~ is continuous
tangential H
(20)
(19)
(21)
~ and B
~ in all three waves. For
We begin by writing expressions for E
simplicity, let the boundary be the x; y¡plane, with medium 1 occupying the
region z < 0: The wave travels in the +^
z direction (normal incidence). First
~
~
E : We put the x¡axis parallel to E ; so
~i
E
~t
E
=
~r
E
=
=
~ i0 exp (ik1 z ¡ i!t) = E i0 ^x exp (ik1 z ¡ i!t)
E
~ i0 exp (ik2 z ¡ i!t) = E t0 x^ exp (ik2 z ¡ i!t)
E
~ r0 exp (¡ik1 z ¡ i!t) = E r0 x^ exp (¡ik1z ¡ i!t)
E
The minus sign in front of the k1 z term in the re¢ected wave signi¤es that this
wave is going in the ¡z direction The wave numbers are
k1 =
!
p
= ! "1 ¹1
v1
k2 =
!
p
= ! "2 ¹2
v2
and
13
Now in writing the magnetic ¤eld vectors, we must remember that
~k £ E
~ 0 = !B
~0
So
~i
B
~t
B
~r
B
~ i0 exp (ik1 z ¡ i!t) = Bi0 ^z £ ^x exp (ik1z ¡ i!t) = E i0 y^ exp (ik1 z ¡ i!t)
= B
v1
E
~ t0 exp (ik2 z ¡ i!t) = t0 y^ exp (ik2 z ¡ i!t)
= B
v2
E r0
~ r0 exp (¡ik1 z ¡ i!t) = Br0 (¡^
= B
z ) £ x^ exp (ik1 z ¡ i!t) = ¡
y^ exp (¡ik1 z ¡ i!t)
v1
Now we apply the boundary conditions (18) to (21).
~ :
Continuity of normal D
0=0
~ :
Continuity of tangential E
(22)
E i0 + E r0 = E t0
~ :
Continuity of normal B
0=0
~ :
Continuity of tangential H
Bi0 + Br0
¹1
E i0 ¡ E r0
v 1 ¹1
=
=
Bto
¹2
E t0
v 2¹2
(23)
Now we solve. First multiply eqn (23) by v1 ¹1 :
E i0 ¡ E r0 = v1 ¹1
E r0
n2 ¹1
= Et0
v2 ¹2
n1 ¹2
and now add to equation (22):
2E i0 = Et 0
µ
E t0 = 2Ei0
n1 ¹2
n1 ¹2 + n2 ¹1
Thus
n2 ¹1
1+
n1 ¹2
¶
(24)
and then from (22),
E r0
µ
n1 ¹2
= E i0 2
¡1
n1 ¹2 + n2 ¹1
µ
¶
n1 ¹2 ¡ n2 ¹1
= E i0
n1 ¹2 + n2 ¹1
14
¶
(25)
This result is interesting, because it can be negative if n2 ¹1 > n1 ¹2 : As in the
~ vector has a phase change of
case of the waves on a string, it means that the E
¼; or, equivalently, it changes direction. We may simplify these results if, as
is usual, ¹1 ' ¹2 ' ¹ 0: Then
E t0
=
E r0
=
2n1
n1 + n2
µ
¶
n1 ¡ n2
E i0
n1 + n2
E i0
From these results we may calculate the re¢ected and transmitted intensities:
r
1
Ei02
1 "1 2
< Sin c >=
E i0 Bi0 =
=
E
2¹1
2¹1 v 1
2 ¹1 i0
and
<
=
=
=
=
<
Str
<
Sre f
µ
¶2
r
"2 2
1 "2
n1 ¹2
2
E =
4E
¹2 t0
2 ¹2 i0 n1 ¹2 + n2 ¹1
µ
¶2
r
r
1 "1 2
1 "1 2 n1¹2 ¡ n2 ¹1
>=
E r0 =
E i0
2 ¹1
2 ¹1
n1¹2 + n2 ¹1
1
>=
2
r
"r
µ
¶2 µ
¶2 #
r
1 "1 2
"2 ¹1
n1 ¹2
n1 ¹2 ¡ n2 ¹1
St r > + < Sre f >=
E
4
+
2 ¹1 i0
¹2 "1
n1 ¹2 + n2 ¹1
n1 ¹2 + n2 ¹1
" r
µ
¶2 µ
¶2 #
r
1 "1 2 ¹1 "2 ¹2
n1 ¹2
n1 ¹2 ¡ n2 ¹1
E
4
+
2 ¹1 i0 ¹2 "1 ¹1
n1 ¹2 + n2 ¹1
n1 ¹2 + n2 ¹1
"
µ
¶2 #
r
1 "1 2 n2 ¹1
4 (n1 ¹2) 2
n1 ¹2 ¡ n2 ¹1
E
+
2 ¹1 i0 n1 ¹2 (n1 ¹2 + n2 ¹1 )2
n1 ¹2 + n2 ¹1
"
#
r
1 "1 2
4n1 ¹2 n2 ¹1
(n1 ¹2 )2 ¡ 2n1 ¹2 n2¹1 + (n2 ¹1 )2
E
+
2
2 ¹1 i0 (n1 ¹2 + n2 ¹1 )2
(n1 ¹2 + n2 ¹1 )
"
#
r
r
2
2
1 "1 2 (n1 ¹2 ) + 2n1 ¹2 n2 ¹1 + (n2 ¹1 )
1 "1 2
Ei0
=
E = < Sin c >
2
2 ¹1
2 ¹1 i0
(n1 ¹2 + n2 ¹1 )
This result is required by energy conservation.
4.2
Non-normal incidence
When a wave is incident at an angle µ 6= 0; we have to tbe aware that the
polarization of the wave enters into the boundary conditions. The plane that
contains the normal to the surface and the incident ray is called the plane of
incidence. (The ray is normal to the wavefront- see LB p547) If the electric
¤eld vector is perpendicular to the plane of incidence, as in the diagram below,
we say that the wave is polarized perpendicular to the plane of incidence. If
the electric ¤eld vector lies in the plane of incidence, then the wave is polarized
in the plane of incidence.
15
4.2.1
Polarization perpendicular to the plane of incidence
In this case the electric ¤eld vector is entirely tangential to the boundary. We
choose coordinates so that the boundary is the x ¡ y plane. Then
³
´
~ in c = ¡E 0 ^y exp i~k ¢ ~r ¡ i!t
E
³
´
~ re f = ¡E r0 y^ exp i~kr ¢ ~r ¡ i!t
E
and
³
´
~ tra n s = ¡Et0 y^ exp i~kt ¢ ~r ¡ i!t
E
As usual each wave has the same frequency,
¯ ¯ ¯ ¯
!
¯~ ¯ ¯~ ¯
;
¯ kr ¯ = ¯ k¯ =
v1
Further, for points on the boundary,
and
¯ ¯
!
¯~ ¯
¯ k t¯ =
v2
~k ¢ ~r = kr sin µ
~ ) becomes
Thus the boundary condition (19wcontinuity of tangential E
~ in c + E
~ ref
E
´
¡E0 exp i~k ¢ ~r ¡ i!t ¡ E r0 exp i~kr ¢ ~r ¡ i!t
³
´
³
The factors exp (¡i!t) cancel, leaving
~ tra n s
= E
³
´
= ¡Et 0 exp i~kt ¢ ~r ¡ i!t
¡E 0 exp (ikr sin µ) ¡ E r0 exp (ikr r sin µr ) = ¡E t0 exp (ikt r sin µ t )
(26)
This relation must hold at all points on the boundary. Thus we must have
k sin µ = kr sin µ r = kt sin µ t
16
Since k = kr ; we must have sin µ = sin µr ; and further, since the angles of
incidence and re¢ection always lie in the range · µ · ¼=2; we must have
µ = µr
(27)
This is the law of re¢ection. Similarly, since k = !=v 1 and kt = !=v2 ;
sin µ
sin µ t
=
v1
v2
or, equivalently,
n1 sin µ = n2 sin µt
(28)
This is Snellzs Law.
It is important to note that these relations arise from the fact that we have
to satisfy a boundary condition at every point of the boundary. They do not
depend on the details of the boundary condition at all. Thus they hold for
plane waves of any kind, not just EM waves.
With equations (27) and (28) satis¤ed, boundary condition (26) becomes:
E 0 + E r0 = E t0
(29)
~ and H:
~ These cannot give 2 independent
Now we look at the conditions on B
relations, because we only need two equations to ¤nd the two unknowns E r0 and
E t0 : Letzs look at them:
~ in c
B
~ ref
B
and
~ tran s
B
~ :
Normal B
³
´
~ in c = ¡~k £ ~y E0 exp i~k ¢ ~r ¡ i!t
= ~k £ E
³
´
= ¡k (cos µ z^ + sin µ x^ ) £ ~y E 0 exp i~k ¢ ~r ¡ i!t
³
´
= k (cos µ x^ ¡ sin µ ^z) E0 exp i~k ¢ ~r ¡ i!t
³
´
~ re f = ¡~kr £ y^E r0 exp i~kr ¢ ~r ¡ i!t
= ~kr £ E
³
´
= ¡k (¡ cos µ z^ + sin µ ^x) £ ~y Er0 exp i~k ¢ ~r ¡ i!t
³
´
= k (¡ cos µ ^x ¡ sin µ z^) E r0 exp i~k ¢ ~r ¡ i!t
³
´
~ tran s = ¡E t0 ~kt £ y^E t0 exp i~kt ¢ ~r ¡ i!t
= ~kt £ E
³
´
= kt (cos µ t x^ ¡ sin µ t z^) E 0 exp i~k ¢ ~r ¡ i!t
¡k sin µE 0 ¡ k sin µE r0 = ¡kt sin µt E t0
17
Using Snellzs law, we get back equation (29). We do get an independent equa~ :
tion from the boundary condition for tangential H
k
cos µ (E 0 ¡ E r0 ) =
¹1
(E 0 ¡ E r0 ) =
(E 0 ¡ E r0 ) =
kt
cos µ t E t0
¹2
p
v1 ¹1
1 ¡ sin 2 µt
Et 0
v2 ¹2
cos µ
q
1 ¡ n21 sin2 µ=n22
n2 ¹1
E t0
n1 ¹2
cos µ
(30)
We will simplify by taking ¹1 = ¹2 = ¹0 : Then combining equations (29) and
(30), we have
q
0
1
n22 ¡ n21 sin2 µ
A E t0
2E 0 = @ 1 +
n1 cos µ
E t0
=
and then
E r0 = E 0
2E0 n1 cos µ
q
n1 cos µ + n22 ¡ n21 sin 2 µ
n1 cos µ ¡
n1 cos µ +
4.2.2
q
q
n22 ¡ n21 sin2 µ
n22 ¡ n21 sin2 µ
Polarization parallel to the plane of incidence
18
(31)
(32)
~ (again with all ¹s equal) is
Now the boundary condition on tangential H
B0 + Br0
= Bt0
n1 (E 0 + E r0 )
= n2 E t0
(33)
~ we get
while for tangential E
E0 cos µ ¡ Er0 cos µ
E 0 ¡ E r0
= E t0 cos µ t
p
1 ¡ sin2 µ t
= E t0
cos µ
(34)
Again the third non-trivial condition (for normal D) does not give an independent relation. Combining these, we get
0q
1
1 ¡ n21 sin2 µ=n22
n2
A
2E 0 = E t0 @
+
cos µ
n1
E t0
=
=
2E 0 n1 cos µ
q
n2 cos µ + n1 1 ¡ n21 sin2 µ=n22
n22
and then
E r0 = E 0
2E0 n1 n2 cos µ
q
cos µ + n1 n22 ¡ n21 sin 2 µ
n22 cos µ ¡ n1
n22 cos µ + n1
q
q
n22 ¡ n21 sin2 µ
n22 ¡ n21 sin2 µ
(35)
(36)
In both polarizations we ¤nd that the re¢ected and transmitted amplitudes
depend on the angle of incidence, as well as the properties of the two materials.
In fact, the refelected amplitude (36) may actually be zero at an angle given by
q
n22 cos µ = n1 n22 ¡ n21 sin2 µ
square both sides:
¡
¢
n42 cos2 µ = n21 n22 ¡ n21 sin 2 µ
Now use a trick: write 1 = cos2 µ + sin2 µ
n42
cos 2 µ
n21
¢
n22 ¡ 2
2
2
2 n2 ¡ n1 cos µ
n1
Then, provided n2 6= n1 ;
¡
¢
= n22 cos2 µ + sin 2 µ ¡ n21 sin2 µ
=
¡
¢
n22 ¡ n21 sin2 µ
tan µ B =
19
n2
n1
(37)
This angle is called Brewsterzs angle. At this angle of incidence, in this polarization, all of the wave energy is transmitted and none is re¢ected. (Wezll
discuss why this happens later- see also LB §33.3.2).
Note, however, that
nothing special happens at this angle for waves polarized perpendicular to the
plane of incidence. In fact, E r0 in equation (32) is never zero for any µ: Check
this for yourself. What this means is that if we have unpolarized incident light,
equal amplitudes in the two polarizations, at Brewsterzs angle only one of the
two polarizations is re¢ected. The re¢ected light is completely polarized perpendicular to the plane of incidence. This phenomenon is called "polarization
by re¢ection".
Under what conditions do we get a phase change of the re¢ected wave? That
is, when is the sign of E r0 opposite that of E 0 ? Starting from equation (32), we
want to ¤nd when
n1 cos µ
<
n21 cos2 µ
<
q
n22 ¡ n21 sin 2 µ
n22 ¡ n21 sin 2 µ
This is true for
n1 < n2
independent of µ; as we also found for normal incidence.
For the other polarization, Start with equation (36)
q
n22 cos µ ¡ n1 n22 ¡ n21 sin2 µ < 0
or, equivalently,
¡
¢
n42 cos2 µ < n21 n22 ¡ n21 sin 2 µ
Using the same trick as in ¤nding Brewsterzs angle, we ¤nd this reduces to
or
¢
¡ 2
¢ 2
n22 ¡ 2
2
2
2
2 n2 ¡ n1 cos µ < n2 ¡ n1 sin µ
n1
µ < µB
if n2 > n1
and
µ > µ B if n1 < n2
4.2.3
Re¢ection and transmission coe¢cients
The incident power striking unit area of the interface is
³
´
~ in £ H
~ in ¢ z^
S~ in ¢ z^ = E
20
Now we time average to get
Iin
"
Ã
!#
~k
1 2
E e^ £
£ ^e
¢ z^
2 0
!
=
1
E 2 ^k ¢ ^z
2v1 ¹1 0
1
" 1 v1 E 20 cos µ
2
=
=
where
~ 0 = E 0 e^
E
and ^e is the polarization vector. We can compute the re¢ected and transmitted
intensities similarly. Then the re¢ection coe¢cient is de¤ned as
R=
IR
E2
= r20
Iin
E0
and the transmission coe¢cient is
T =
2
IT
v1 Et20 cos µt
n2 E t0
cos µ t
=
=
2
2
Iin
v2 E 0 cos µ
n1 E 0 cos µ
For polarization perpendicular to the plane of incidence, we get
q
0
12
n1 cos µ ¡ n22 ¡ n21 sin 2 µ
A
q
R=@
n1 cos µ + n22 ¡ n21 sin 2 µ
and
T
0
12
2n
cos
µ
1
A n2 cos µ t
q
= @
n1 cos µ
2
n1 cos µ + n22 ¡ n21 sin µ
=
and
R+T
=
=
µ
µ
n1 cos µ ¡
³
4n1 n2 cos µ cos µ t
¶2
q
n1 cos µ + n22 ¡ n21 sin 2 µ
q
n22
¡
n21
¶2
sin µ + 4n1 n2 cos µ cos µt
2
´2
n22 ¡ n21 sin µ
q
q
n21 cos2 µ + n22 ¡ n21 sin2 µ ¡ 2n1 cos µ n22 ¡ n21 sin 2 µ + 4n1 cos µ n22 ¡ n21 sin2 µ
µ
¶2
q
n1 cos µ + n22 ¡ n21 sin2 µ
n1 cos µ +
p
= 1
Verify that this result also holds, as it must, for the other polarization.
21
5
Waves in conductors
An electric ¤eld in a conductor drives a current.
~
~ =E
~j = ¾ E
½
and thus some of the ¤eld energy is converted to kinetic energy of electrons
in the conductor. This leads to absorption of the wave energy. In order to
investigate this, we must include the current in Maxwellzs equations. The only
equation to change is the Ampere-Maxwell law.
~
~
~ £B
~ = ¹~j + ¹" @ E = ¹¾ E
~ + ¹" @E
r
@t
@t
We may still take the free charge density ½f to be zero. For suppose ½f is not
zero. Then from charge conservation,
@½f
@t
=
=
~ ¢ ~j = ¡r
~ ¢ ¾E
~
¡r
³½ ´
f
¡¾
"
This di¤erential equation has the solution
³ ¾ ´
½f = ½0 exp ¡ t
"
so the charge density dies away with a time scale
¿ =
"
¾
which is very small for good conductors. As we discussed before, even if ¾ ! 1;
the timescale does not go to zero, as the dominant factor is then the timescale
for the ¤elds to change, of order (scale of system)/(speed of light).
Now we may rederive the wave equation with the extra term in Amperezs
law. Starting from equation (11), we have
Ã
!
³
´
´
~
@ ³~
@
@E
2~
~
~
~
~
~
r r ¢E ¡r E = ¡
r£B = ¡
¹¾ E + ¹"
@t
@t
@t
which becomes
2~
@
~ + ¹" @ E
¹¾ E
@t
@t2
The wave equation now has an extra term. We may still ¤nd a plane wave
soution of the form:
³
´
~ = E
~ 0 exp i~k ¢ ~r ¡ i!t
E
³
´
~ = B
~ 0 exp i~k ¢ ~r ¡ i!t
B
~ =
r2 E
22
Stu¢ng these expressions into the wave equation and Maxwellzs equations, we
have
¡k2 = ¡i!¹¾ ¡ ¹"! 2
(38)
~k ¢ E
~0
~ ¢B
~0
k
= 0
= 0
~k £ E
~0
~0
= !B
(39)
Thus the wave has the same transverse nature as before. The only thing that
has changed is the relation of k to !: Equation (38) requires that k be complex
if ! is real. So let k = · + i°: Then
k2 = ·2 ¡ °2 + 2i·° = i!¹¾ + ¹"! 2
So we have two equations obtained by setting the real part on the left equal to
the real part on the right, and also by setting the two imaginary parts equal.
· 2 ¡ ° 2 = ¹"! 2
2·° = !¹¾
From the second, we get
°=
!¹¾
2·
(40)
and then from the ¤rst
·2 ¡
³ !¹¾ ´2
2·
¡ ¹"! 2 = 0
This is a quadratic equation for · 2 with solution
q
2
2
2
¹"!
§
(¹"! 2 ) + (!¹¾)
2
·
=
2
q
2
1 § 1 + (¾=!")
= ¹"! 2
2
p
We must get back our previous result k = ! ¹" as ¾ ! 0; so we need the plus
sign:
v
q
u
u
1
+
1 + (¾=!")2
p t
· = ! ¹"
2
and then from (40), we have
¾
°=
2
r v
¹u
u
t
"
1+
23
q
2
1 + (¾=!")
2
The imaginary part of k shows that the wave is damped:
h
i
³
´
³
´
exp i~k ¢ ~r = exp ik^ ¢ ~r (· + i°) = exp i· ^k ¢ ~r exp ¡° ^k ¢ ~r
and taking the real part, we have
³
´
³
´
cos · ^k ¢ ~r exp ¡° ^k ¢ ~r
The wave attenuates exponentially as it propagates. The distance travelled
before the wave amplitude drops to 1=e of its original value is called the skin
depth
1
±=
°
~ and B
~ no longer oscillate in phase.
Faradayzs law (equation (39)) shows that E
~
We may write k in exponential form as
p
°
~k = ^k (· + i°) = k^ · 2 + °2 exp iÁ;
tan Á =
·
where
k
v
q
u
2
u
p
1
+
1 + (¾=!")
¾2 ¹
u
=
·2 + °2 = t¹"! 2
+
2
2 "
1+
v
q
u
2
2
u
p u 2 + (¾=!") + 2 1 + (¾=!") + ¾ 2 =! 2" 2
µ
¶
= ! ¹"u
q
t
2
2 1 + 1 + (¾=!")
and
1
q
´1=4
p ³
2
= ! ¹" 1 + (¾=!")
µ
tan Á =
"! 1 +
Then equation (39) becomes
¡
¢
^k £ e^E 0 keiÁ
B0
¾
q
1 + (¾=!")2
¶
=
~0
B
=
´1=4
p ³
2
! ¹" 1 + (¾=!")
E0 e iÁ
A material is a good conductor as far as wave propagation is concerned if
¾
À1
!"
in which case we ¤nd
B0 '
p
¾!¹E 0 eiÁ
24
2
1 + (¾=!")
so that
E0
v
and the phase shift ± is almost ¼=4: (tan ± ! 1): The ¤elds in the conductor are
primarily magnetic in character. In this limit we also have
r
r
p
1 + ¾=!"
!¾¹
· ' ! ¹"
'
'°
2
2
B0 À
and the wave damps within one wavelength. On the other hand, if ¾=!" ¿ 1;
we have a poor conductor, the phase shift is small, and the amplitudes of the
¤elds are not much di¤erent from those in a non-conducting medium.
5.1
Re¢ection at a conducting surface
There may be a free surface charge density on the surface of a conductor. In
fact, there usually is. But there cannot be a free surface current density as this
~ ; and if ~j = K±
~ (z ¡ zsu rfa ce ) then
would require an in¤nite electric ¤eld (~j = ¾ E
~
~ = K ± (z ¡ zsu rf ac e ) : Such an electric ¤eld cannot exist. Thus we choose to
E
¾
~ normal B
~ and tangential H:
~ Letzs
use the boundary conditions on tangential E;
consider normal incidence for simplicity. Then we have
E0 + Er = Et
and
H0 ¡ Hr
=
!
(E 0 ¡ E r ) =
v1
(E 0 ¡ E r ) =
Ht
r
¾!
E t eiÁ
¹2
r
¾
v1
E t eiÁ
!¹2
Thus we have
2E 0
Et
= Et
=
µ
1 + v1
r
¾ iÁ
e
!¹2
2E0
µ
¶
q
¾
1 + v 1 !¹
eiÁ
¶
2
and then
0
Er = E0 @
1 ¡ v1
1 + v1
As ¾ ! 1; E t ! 0 and E r ! ¡1:
25
q
q
¾
e iÁ
!¹ 2
¾
e iÁ
!¹ 2
1
A
(41)
5.2
Note on computing power using complex numbers
When our amplitudes are complex, as in (41), we must be careful when evaluating physical quantities that involve the square of these numbers. Letzs look
at the power.
~= 1 E
~ £B
~
S
¹0
Since
Re (z1 z2 ) 6= Re (z1 ) Re (z2 )
we must take the real part before multiplying. Thus
³
´
³
´
~ p h ys = 1 E
~ 0 cos ~k ¢ ~r ¡ !t + Á
~ 0 cos ~k ¢ ~r ¡ !t + Á
S
£
B
E
B
¹0
h
³
´
i
1 ~
~ 0 cos 2~k ¢ ~r ¡ 2!t + Á + Á
=
E0 £ B
E
B + cos (ÁE ¡ ÁB )
2¹0
Now we time average, to get
< S~p h ys > =
1 ~
~ 0 cos (Á ¡ Á )
E0 £ B
E
B
2¹0
We can get the same result more easily as follows:
1 ~
~¤ = 1 E
~0 £ B
~ 0 exp (iÁ ¡ iÁ )
E£B
E
B
2¹0
2¹0
and so
~ p h ys > = Re
<S
6
µ
1 ~
~¤
E £B
2¹0
¶
(42)
Dispersive media
In general the properties of a medium, such as " and ¹; are frequency dependent.
This means that waves of di¤erent frequencies travel at di¤erent speeds. A
plane wave carries no information. To transmit a signal we have to vary either
the amplitude of the frequency of the wave- the resulting signal is called AM
or FM ( as on your radio dial). The wave crests travel at the phase speed
v Á = !=k but the envelope that carries the information travels at the group
speed v g = d!=dk: It is v g; not v Á; that must be less than c: In order to get
the signal, many di¤erent frequencies make up the wave packet, and because
they do not all travel at the same speed, over time the signal becomes less well
de¤nedw the pulse is dispersed. Those of you who plan to take Phys 704 in the
spring will learn more about these e¤ects then. For now, letzs see if we can
understand how the frequency dependence arises.
We begin with a simple classical model of an electron in an atom. The
electron behaves as an oscillator with a restoring force
F~ res = ¡m! 20~s
26
where ! 0 is the natural oscillation frequency and ~s is the electron displacement.
There is also damping due to radiation reaction, which we may write as
d~s
F~ d am p = ¡m°
dt
Now when an incident EM wave reaches the electron, it will experience a force
³
´
~ + ~v £ B
~
F~ EM = ¡e E
We have already seen that, for waves in a medium with phase speed v Á ,
B0 =
E0
vÁ
so the magnetic force is smaller than the electric force by a factor v=vÁ which is
typically ¿ 1 since v Á » c: So we will ignore the magnetic force. Now we may
write the equation of motion for the electron:
F~to t
~ 0 cos !t ¡ m!2 ~s ¡ m° d~s
¡eE
0
dt
= m~a
d2 ~s
= m 2
dt
~ 0 : We may put the x¡axis along this
The electron oscillates with ~s parallel to E
direction, to get
³ e
´
¢¢
x + °x_ + ! 20 x = ¡eE0 cos !t = Re ¡ E 0 exp (¡i!t)
m
The solution will be of the form
¡
¢
x = Re x 0 e¡i!t
Stu¤ this in and check:
so
¡
¢
e
¡! 2 ¡ i°! + ! 20 x 0 = ¡ E0
m
x0 =
eE 0 =m
!2 + i°! ¡ ! 20
(43)
and the electron contributes a dipole moment about its original equilibrium
position of
~0
e2
E
~p = ¡e~s = ¡
2
m ! + i°! ¡ ! 20
Summing up over all the particles in the material, where there are n molecules
per unit volume, each of which has f j electrons with frequency ! j and damping
° j ; we get a polarization
2
e
P~ = n~p = ¡n
m
X
j
27
!2
~0
fj E
+ i°j ! ¡ ! 2j
So we now have a model for the susceptibilty of the material
~
P~ = "0 Â eE
Thus
 e = ¡n
e2 X
fj
2
m
! + i° j ! ¡ ! 2j
j
Because  is a complex number, the polarization is out of phase with the driving
~ : The dielectric constant is than
¤eld E
0
1
ne 2 X
fj
A
" = "0 @ 1 +
(44)
"0 m
! 2j ¡ i°j ! ¡ ! 2
j
and consequently the wave equation (16) has plane wave solutions of the form
³
´
~ =E
~ 0 exp i~k ¢ ~r ¡ i!t
E
where
0
¡ 2
¢1
2
2 X f
!
+
i°
!
¡
!
ne
j
j
j
~ 0 = "¹!2 E
~ 0 = "0 ¹! 2 E
~ 0 @1 +
¡k2 E
¡ 2
¢
¡
¢ A
"0 m
2 2 + ° ! 2
!
¡
!
j
j
j
and because " is complex, k must be complex too.
k = kr + iki
Putting the z¡axis along ^k; we have
~ =E
~ 0 exp (¡ki z) exp (ikr z ¡ i!t)
E
The wave is damped. The wave intensity is proportional to E 2 ; and so it decreases as exp (¡2ki z) = exp (¡®z) where ® is the absorption coe¢cient.
Letzs look at the special case where the second term in equation (44) is
small, for example, in gases. Then we can approximate the square root using
the binomal series:
p
x
1+x' 1+
if x ¿ 1
2
So
0
¡ 2
¢1
2
2 X f
!
+
i°
!
¡
!
p
!
ne
j
j
j
k = "¹! = @1 +
¡ 2
¢2 ¡
¢2 A
c
2" 0 m
2
! ¡!
+ ° !
j
so
®=
j
fj° j
! 2 ne2 X
¡ 2
¢
¡
¢
2
c "0 m j ! ¡ ! 2 + ° ! 2
j
j
28
j
(45)
and the refective index is
¡
¢
fj ! 2j ¡ ! 2
ckr
ne 2 X
n=
= 1+
¡
¢
¡
¢
!
2"0 m j !2 ¡ !2 2 + °j ! 2
j
(46)
The denominator of the fraction in these equations becomes very small when
! ' ! j because ° j ¿ ! j ; and thus the fraction becomes large. There are
resonances at the fundamental frequencies of the system. These correspond to
atomic and molecular line transition frequencies. To display the behavior more
clearly, look at a system with only one resonance, and let
ne2
= ! 2p ;
"0 m
!=! j = x;
°j =!j = ´
Then
®=
and
fj ° j
! 2p
! 2 ! 2p
´ fj
=
¡
¢
¡
¢
! 2j c ! 2j 1 ¡ ! 2 =! 2 2 + ° ! 2 =! 4
c!j (1 ¡ x 2 )2 + (´x)2
j
j
j
¡
¢
! 2p
fj 1 ¡ x2
n=1+
2! 2j (1 ¡ x 2 )2 + (´x)2
The plot shows c®=! j (solid line) and n ¡ 1 (dashed line) with ´ = 0:1 (much
larger than actual values, for clarity in the plot) and f j !2p =! 2j = 1
0.5
1
1.5
x
2
2.5
3
The absorption coe¢cient peaks strongly around x = 1; and is very small away
from that region. The width of the peak is determined by ´; which is usually
29
much less than the value of 1/10 used in these plots. The index of refraction n is
greater than 1 for x < 1 but becomes <1 for x > 1: Thus the phase speed of the
waves is actually greater than c for frequencies slightly greater than resonance.
The group speed, however, remains less than c:The refractive index n drops
steeply at the resonancew this e¤ect is called anomalous dispersion. For real
materials, we have to sum the contributions from all the resonances, but graphs
like this describe the behavior near each one.
For frequencies well away from resonance, the e¤ect of damping is negligible
and we have the simpler relation
n = 1+
! 2p
fj
2! 2j (1 ¡ x2 )
Further, if we are well below resonance, x = !=! j ¿ 1 (in transparent materials
like glass !j is usually in the UV), we have
n=1+
¡
¢
! 2p
fj 1 + x2
2
2! j
and we see that n increases with frequency. (See LB §16.5.5).
7
Wave guides
Wezll close this discussion of waves by looking at happens when a wave is con¤ned within a set of boundaries. The wave ¤elds have to satisfy boundary
conditions on the walls as well as the wave equation. Consider a wave guide
made of of a long, straight, perfectly conducting tube with a constant crosssection, whose shape, for the moment, is arbitrary. The electric ¤eld has to
be zero inside the perfectly conducting walls, and since wave ¤elds are time~ must also be zero. Otherwise, the non-zero @ B=@t
~
varying, that means that B
~ through Faradayzs law, and we know there can be no E
~:
would create an E
Then making use of the boundary conditions, we must have
~ ¢ ^n = 0
B
on the guide walls, and
~ =0
n
^£E
~ ¢ ^n because there could be
as well. We do not use the boundary condition on D
a non-zero surface charge density on the wall.
We are interested in waves propagating along the guide, in the z¡direction.
The waves can start o¤ propagating at an angle to the z¡axis, in which case
they will re¢ect at the wall, and propagate down the guide by bouncing back
and forth. The superposition of these waves forms the total disturbance in
the guide. As the waves superpose, they will interfere, and only the waves
that interfere constructively contribute. It is easier to solve a boundary-value
30
problem than to try to sum the appropriate set of re¢ecting waves. (But see LB
Digging Deeper, page 1060 for a simple example of how it goes.) We should
~
note that each of the re¢ecting waves has the form we have come to expect (E
~
~
perpendicular to B perpendicular to k), but when we superpose them to get a
disturbance travelling parallel to the guide (z) axis, the resulting ¤elds can have
a z¡component. (See LB Figure 33.28, for example.) So we have to allow for
~ and B
~ (as
this possibility. We start o¤ with the following assumed form for E
usual, real part is implied).
h
i
~ = E
~ 0 exp (ikz ¡ i!t) = E
~ ? (x; y) + E z (x; y) z^ exp (ikz ¡ i!t)
E
h
i
~ = B
~ 0 exp (ikz ¡ i!t) = B
~ ? (x; y) + Bz (x; y) z^ exp (ikz ¡ i!t)
B
The amplitudes must depend on x; y so as to satisfy the boundary conditions.
Now we stu¤ these into Maxwellzs equations. For simplicity let the guide
interior be vacuum.
³
´
~ ¢E
~ =
~?¢E
~ ? + ikE z exp (ikz ¡ i!t) = 0
r
r
³
´
~ ¢B
~ =
~?¢B
~ ? + ikBz exp (ikz ¡ i!t) = 0
r
r
From these ¤rst two equations we can already see that the z¡components of
the ¤elds act as sources for the perpendicular components:
~?¢E
~?
r
=
~?¢B
~?
r
=
@E x
@Ey
+
= ¡ikE z
@x
@y
@Bx
@By
+
= ¡ikBz
@x
@y
(47)
(48)
From the last two equations, we have
³
~ £E
~
r
³
³
~ £E
~
r
~ £E
~
r
´
´
x
y
´
z
@E z
@E y
@E z
¡
= i!Bx =
¡ ikEy
@y
@z
@y
@Ez
ikEx ¡
= i!By
@x
@E y
@E x
¡
= i!Bz
@x
@y
=
=
=
(49)
(50)
(51)
and similarly
³
³
³
~ £B
~
r
~ £B
~
r
~ £B
~
r
´
´
´
x
y
z
@Bz
!
¡ ikBy = ¡i 2 Ex
@y
c
@Bz
!
= ikBx ¡
= ¡i 2 E y
@x
c
@By
@Bx
!
=
¡
= ¡i 2 E z
@x
@y
c
=
31
(52)
(53)
(54)
We can solve these equations to get the x¡ and y¡components in terms of the
z¡components. From (49)
µ
¶
1 @E z
Ey =
¡ i!Bx
ik
@y
and then from (53)
1
Bx =
ik
Combining these, we have
Ey
=
=
So
Ey
µ
µ
@Bz
i!
¡ 2 Ey
@x
c
¶
·
µ
¶¸
1 @E z
1 @Bz
i!
¡ i!
¡ 2 Ey
ik @y
ik @x
c
2
1 @E z
1 @Bz
!
+ i! 2
+ 2 2 Ey
ik @y
k @x
k c
!2
1¡ 2 2
c k
¶
Ey
µ
¶
! @Bz
1 @E z
= i
¡
k2 @x
k @y
µ
¶
i
@E z
@Bz
=
k
¡!
! 2 =c2 ¡ k 2
@y
@x
(55)
Similarly, we get for the other components:
µ
¶
i
@Ez
@Bz
Ex =
k
+!
! 2=c2 ¡ k2
@x
@y
µ
¶
i
@Bz
! @E z
Bx =
k
¡ 2
! 2=c2 ¡ k2
@x
c @y
µ
¶
i
@Bz
! @E z
By =
k
+ 2
! 2=c2 ¡ k2
@y
c @x
(56)
(57)
(58)
Now if we can just ¤nd the z¡components, we are done! So go back to
equations (47) and (48) and insert the results for the x¡and y¡components.
We simplify the notation by writing
°2 ´
!2
¡ k2
c2
· µ
¶¸
· µ
¶¸
@
i
@E z
@Bz
@
i
@E z
@Bz
k
+
!
+
k
¡
!
@x ° 2
@x
@y
@y ° 2
@y
@x
@ 2 Ez
@ 2 Ez
+
2
@x
@y 2
=
¡ikE z
=
¡° 2E z (59)
and we get the same equation for Bz :
@ 2 Bz
@ 2 Bz
+
= ¡° 2B z
@x 2
@y 2
32
(60)
These are decoupled equations: the solution for E z is independent of Bz and
vice versa. That means we can solve for E z assuming Bz = 0 w these are the
TM or transverse magnetic modes, and then solve for Bz assuming E z = 0 w
these are the transverse electric modes, and then get a general solution as a
linear combination
~ = ®E
~ T E + ¯E
~ TM
E
If both E z and B z are zero we have TEM, or transverse electro-magnetic
modes. For these modes, equation (51) becomes
µ
¶
@Ey
@E x
~ £E
~ )E
~ = ¡rª
~
¡
z^ = 0 = r
@x
@y
and equation (47) becomes
@E x
@E y
~ ¢E
~ = 0 = ¡r2 ª
+
=r
@x
@y
We know that the maxima and minima of the function ª, a solution of Laplacezs
~ £n
equation, must occur on the boundary of the region. But E
^ is zero on the
boundary, which means that ª is constant on the boundary, and thus it is
~ = 0: The only
constant throughout the guide. But if ª is constant then E
exception to this is if the guide has two separate pieces to its boundaryw as in
a coaxial cable.
~ must be
Now it is time to tackle the boundary conditions. Tangential E
zero, and one of the tangential components is E z ; so for the TM mode we must
have
E z = 0 on the walls.
(61)
The boundary condition for Bz is a bit more complicated, so wezll wait for a
speci¤c example.
7.1
Waves in a rectangular guide
Let the guide have a rectangular cross-section measuring a £ b:
7.1.1
TM modes
The E z component satis¤es the di¤erential equation (59) together with the
boundary condition (61). We may solve using separation of variables:
E z = X (x) Y (y)
Then the di¤erential equation is
X 00 Y + XY 00 = ¡° 2 XY
or
X 00
Y 00
+
= ¡° 2
X
Y
33
The two terms on the left of this equation must each be constant if we are to
satisfy the equation for all values of x and y: Further, we want solutions for X
that are zero at two values of x; x = 0 and x = a: Thus we need a negative
separation constant so that the solution is a sine:
X 00 = ¡®2 X ) X = sin ®x and ® =
n¼
a
similarly for Y :
Y 00 = ¡¯ 2 Y ) Y = sin ¯y and ¯ =
Then
³ n¼ ´2
³ m¼ ´2
m¼
b
!2
¡ k2
a
b
c2
At a given frequency !; the corresponding wave number is
r
!2
k=
¡ °2mn
c2
+
= °2mn =
(62)
Clearly ! must be greater than °mn for the wave to propagate (k > 0): Thus
for each mode (value of m and n) there is a minimum frequency at which that
mode will propagate. This is the cut-o¤ frequency for that mode.
r
n2
m2
! cu to ¤,mn = c° mn = c¼
+
a2
b2
There is also a cuto¤ frequency for the guide, corresponding to the lowest cuto¤
frquency for any mode.
! c u to¤, gu id e = !c u to¤ ,lowe st
mo d e
= c°m in
Considering only TM modes, we have a cuto¤ frequency corresponding to m =
n=1
r
1
1
!c u to¤ ,TM = c¼
+ 2
2
a
b
The wave phase speed is
vÁ =
!
!
1
=p
= cp
2
2
2
2
k
! =c ¡ ° mn
1 ¡ c °2mn =!2
and is always greater than c: However the group speed is
vg =
d!
dk
To get this most easily, we start with the expresson for k2 : .
2k
dk
!
=2 2
d!
c
34
So
d!
k
c
= c2 = c
dk
!
vÁ
Thus if v Á > c; vg < c: Information travels down the guide more slowly than in
vacuum. Sometimes the guides are called delay lines, indicating that the guide
slows the propagation of information down the guide.
7.1.2
TE mo des
~ Thus at x = 0
For the TE modes, we use the boundary condition on normal B:
and x = a we must have Bx = 0; while at y = 0 and y = b we must have By = 0:
We need to express these results in terms of Bz ; so we use relations (57) and
(58), remembering that Ez ´ 0 in this mode, so
Bx
=
By
=
¶
@Bz
! 2=c2 ¡ k2
@x
µ
¶
i
@Bz
k
! 2=c2 ¡ k2
@y
i
µ
k
and so are boundary conditons are
@Bz
@x
@Bz
@y
= 0 at x = 0; a
= 0 at y = 0; b
These are Neumann conditions. The solutions are then:
³ n¼x ´
³ m¼y ´
Bz = cos
cos
a
b
with
n2
m2
+
a2
b2
except that now either n or m may be zero (but not both!) giving a lowset
frequency for the TE mode of
° 2mn =
! c u to¤,T E =
c¼
a
for the case a > b: Since this is a lower frquency than for the TM modes, this is
the cuto¤ frequency for the guide.
35