7. Electromagnetic Waves
7A. Plane Waves
Complex Notation
• Consider Maxwell’s Equations with no sources
• We are going to search for waves of the form
f  r, t   cos  k  r  t  or
f  r, t   sin k  r  t 
• To make things as general as possible, we write
f  r, t   Re  feik r it  
1
2
feik r it  12 f *e ik r it
D  0  B

E   B
t

H  D
t
• To save ourselves work, we will simply keep track of feikr it
– Just remember to take the real part at the end
k D  0  k B
• Space derivatives of expressions like this become ik
k  E  B
• Time derivatives of expressions like this become –i
k  H   D
• Maxwell equations are now
Linear Media
E, D, H, B ~ eikxit
• The reaction of the medium will generally have the same
frequency as the fields only if the material is linear
D  E
• We therefore assume the medium is linear
B  H
• In general,  and  will depend on frequency
k D  0  k B
k  E  B
k  H   D
• It is possible for there to be a phase shift between D and E or B and H
– Similar to a phase shift for a damped, driven harmonic oscillator
• This can be show up as complex  and 
• We will (for now) assume they are both real
k E  0  k B
• Most common situation is  = 0 and  > 0
k  E  B
• With these assumptions, our equations become
k  B    E
Finding the Wave Velocity
• Multiply second equation by 
E, B ~ eikxit
• Substitute third equation
 2B  k  E  k   k  B   k  k  B   k 2B
•
•
•
•
•

1
 2B  k 2B
Use kB = 0

k

We therefore have
We define the phase velocity v and the index of refraction n as
We therefore have

c
v
Recall that c200 = 1, so n   
k
n
•
What does phase velocity mean?
k E  0  k B
k  E  B
k  B    E
1
c
v
n

0  0




E, B ~ Re eik x it  ~ cos  k  x  t   cos kkˆ  x  kvt  cos k kˆ  x  vt 
• It’s the speed at which the peaks, valleys, and nodes move
The Electric Field and Magnetic Flux Density

ik x it
c
E
,
B
~
e
v
k
n
• Electric field will take the form
E  E0 eik x it
– E0 is a constant vector
• From the first equation, we see that E0  k
E0  k
k E  0  k B
k  E  B
k  B    E
• The magnetic field can be
nˆ
ik x it
1
i
k

x

i

t
B

k

E
e
found from second equation
0
B  k  E0 e
c

• Magnetic field is also transverse
• Given k, the index of refraction n, and the constant transverse vector E0, we
have completely described the wave
Time-Averaged Energy Density
E  E0 eik x it
B  B 0 eik x it
• First rewrite B
u   Re  E0e
   E0e
1
8
nˆ
k  E0 eik x it
c
B0   kˆ  E0
• Energy density is
1
2
B

ik x it 2
ik x it
  Re  B0e
1
2

*  ik x it 2
0
E e

ik x it 2
 18  1  B0eikxit  B*0eikxit 
 14  E0  E*0  14  1B 0  B*0  14 Re   E02 e2ik x 2it   1B 02 e2ik x 2it 
• The last terms oscillate at frequency 2 - too fast to measure
• If we do a time average, these terms go away, so
u  14  E0  E*0  14  1B0  B*0  14  E0  E*0  14  1 E0  E*0
u  12  E0  E*0
1
c
v
n

Time-Averaged Poynting Vector
B  B 0 eik x it
E  E0 eik x it
B0   kˆ  E0
• The Poynting vector is
1
ik x it
ik x it
1


Re
E
e

Re
B
e
 0
  0

S  E H   E  B
 
1
4
1
E e
ik x it
0
E e
 14  1 E0  B*0  E*0  B0
• If we time average, we get
S  
1
4
1
E
0
• We note that:
 B  E  B0 
*
0
S 
  B e
 2 Re  E  B e
*  ik x it
0
*
0
1

ik x it
0
0

* ik x it 2
0
B e
2ik x  2it
0

1  
1ˆ 
*
*
ˆ
ˆ


E0  k  E0  E0  k  E0  k
E0  E*0
 2
4 


kˆ u  vkˆ u
• Energy moves in direction of k at phase velocity v



7B. Polarization and Stokes Parameters
The Polarization Vectors
The electric field is transverse E0  k
We define two polarization vectors
ε1 and ε 2
*
k

ε

0

ε

ε
They are chosen to be orthogonal to k and to each other:
i
1
2
If we use real polarization vector 1, typically define the other to be ε  kˆ  ε
2
1
For example, if k is in z-direction, ε1  xˆ , ε 2  yˆ
then we could pick
E0  E1ε1  E2ε2
• An arbitrary wave is then described by two complex numbers
• That means four real parameters
Re  E1  , Im  E1  , Re  E2  , Im  E2 
• The magnetic field is then given by
•
•
•
•
•
B0   kˆ  E0    E1ε 2  E2ε1 
• The intensity (magnitude of
time-averaged Poynting vector) is
I
1
2

  E1  E2
2
2

Linear, Circular, Elliptical Polarization
E0  E1ε1  E2ε2
• If E1 and E2 are proportional with a real proportionality
constant, then we say we have linear polarization
E1 only
E2 only
E1 = E2
• If we let E2 =  iE1 we get circular polarization
• Most general case is called elliptical polarization
Electric field
Magnetic Field
circular
elliptical
Polarization and Stokes Parameters
E0  E1ε1  E2ε2
• Instead of using real polarization vectors, ε   12  ε1  iε 2 
we could use complex ones
– These are also called positive and negative helicity polarizations
E0  Eε   Eε 
• Then we would write
• Any way you look at it, there are four real numbers describing E0
• One of these is the overall phase, corresponding to E  ei E
0
0
– These correspond to tiny time shifts
2
2
2
2
s

E

E

E

E
0
1
2


• The remaining parameters are sometimes
2
2
described in terms of Stokes Parameters
s1  E1  E2  2 Re  E* E 
• Since there are only three independent
s2  2 Re  E1* E2   2 Im  E* E 
parameters, these must be somehow related
s02  s12  s22  s32
s3  2 Im  E1* E2   E  E
2
2
Measuring Polarization and Stokes Parameters
• There are a variety of ways of measuring polarization, but one of the easiest is to
put it through a polarizer
– Blocks all the light of one polarization, lets much of the other polarization
through
• Easiest to only allow through one linear polarization, but you can also make them
to only allow through one circular polarization
Sample Problem 7.1
A pure wave moving in the z-direction is put through a variety of polarizers, and its
intensity measured. The types of polarizers and the resulting intensities measured are
x-polarization: Ix y-polarization: Iy; plus circular polarization: I+
Predict the intensity if you only allowed minus circular polarization I-


2
2
• Recall the intensity is the
I  12   E1  E2
magnitude of the Poynting vector:
• For our three
2
2
1
1
I x  2   E1 , I y  2   E2 , I  
measurements,
we have
2
1
I



E
• We want to know


2
s0  E1  E2  E  E
2
• The Stokes parameter s0 is given by
• From which we can easily see
I x  I y  I  I
• Therefore
I  I x  I y  I
2
2
1
2
2
  E
2
7C. Refraction and Reflection
Boundary Conditions and Waves
•
•
•
•
What happens if our linear medium is not uniform?
We will consider only the case of a planar barrier at z = 0
To simplify, we will assume  = ' = 0
We therefore have
   0n2
n    0 0 
• In each region, we will have waves
E  E0 eik x it , B  k  E  ,   ck n
• We have to match boundary condition at z = 0
E  E , D  D , B   B , H  H
• These must match at all t, x, and y
• Since  = ' = 0, last two conditions simplify to
B  B
    0 n 2
   0n2
Setting Up the Waves
• We will consider a wave coming in from the +z direction in the xz-plane, reflecting
in the xz-plane, and refracting in the xz-plane
• Call the wave number for the incoming, refracted,
and reflected wave k, k', and k", respectively
k
• Call their constant vector E0, E'0, and E"0 respectively
• Then we have
2
ik x x  ik z z it
ik xx ik zz i t
ik x x ik z z i t





n
0
E  E0 e
 E0 e
, E  E0 e
• To make them match on the boundary, we need
   0n2
k
E0 eikx x it  E0 eik xx i t  E0 eik x x i t ,
k 
ik x x it
ik xx i t
ik x x i t
2
2
n E0 z e
 E0 z e
 n E0 z e
,


k  E0 eikx x it   k   E0 eik xx i t    k   E0 eik x x i t  
• These must be valid at all x and all t
     and k x  k x  k x
• The only way to make this work is to have
• Then we have E0  E0  E0 , n2  E0 z  E0 z   n2E0 z , k  E0  k  E0  k  E0
Snell’s Law and Law of Reflection
     and k x  k x  k x
  ck n ,   ck  n ,   ck  n
n
k  k
• Recall we also have
• Combining these,
k
 k  ,
 
we see that
c
n c n
• And therefore
k  k  , nk   nk
• Define the angles as , ', and "
k x k x
• Then we have
   
  sin  
sin  
k 
k
• We also have
k x nk x n
n sin    n sin 
 sin 

sin   
k  nk n
• It is then easy to see that
k z  k  cos    k cos   k z
• We also have
2

n
k
n

kn
2
2

1

sin




k z  k cos  
1  sin  
2
n
n
n
  k
    0 n 2
   0n2
k
  
k 
k z  k z
k z  k n2 n 2  sin 2 
Reflection Amplitudes: Perpendicular Case
• We still have to find the magnitudes of the reflected and refracted waves
• Case I: electric field
E0  yˆ E , E0  yˆ E, E0  yˆ E
perpendicular to the xz-plane:
E0  E0  E0  E  E   E 
• One boundary condition:
• Another boundary condition: k  E0  k   E0  k   E0
E  xˆ kx  zˆ kz   yˆ  E  xˆ kx  zˆ kz   yˆ  E  xˆ kx  zˆ kz   yˆ
 Ekz   E  E  kz  Ekz  2Ekz  E  kz  kz 
• Rewrite using our expressions for k'z
E 
2n cos 
n cos   n2  n2 sin 2 
 Ek z  Ek z  Ek z
k z  k z
2k z
E
E 
E E  
k z  k z
k z  k z
k z  k n2 n 2  sin 2 
E , E  
n cos   n2  n2 sin 2 
n cos   n2  n2 sin 2 
E
Reflection Amplitudes: Parallel Case
E0  E  xˆ cos   zˆ sin   , E0   E   xˆ cos   zˆ sin   ,
• Case II: electric field
parallel to the xz-plane:
E0  E   xˆ cos    zˆ sin   
• One boundary condition:
E0  E0  E0   E  E cos  E cos 
• Another boundary condition:
2
2
2
2




n
E

E
sin



n
E sin    nE  nE   nE 





n  E0 z  E0 z   n E0 z
• First equation times n, plus second times cos:
2nE cos  E  n cos   n cos 
• So we have
2n cos 
n cos   n cos  
n
E 
E
E
E   E   E 
• Solve for E"
n cos   n cos  
n cos   n cos  
n
• Rewrite using our expressions for cos'
E 
2nn cos 
n cos   n n  n sin 
2
2
2
2
n cos   n2  n2 sin 2 
E , E 
n2 cos   n n2  n2 sin 2 
n cos   n n  n sin 
2
2
2
2
E
Brewster’s Angle and Polarization
Perpendicular
Parallel
E
2n cos 

,
2
2
2
E n cos   n  n sin 
E n cos   n2  n2 sin 2 

E n cos   n2  n2 sin 2 
E
2nn cos 

,
2
2
2
2
E n cos   n n  n sin 
E n2 cos   n n2  n2 sin 2 

E n2 cos   n n2  n2 sin 2 
• Are there any cases where nothing is reflected?
• For perpendicular, only if index of refraction matches
• For parallel:
n2 cos  n n2  n2 sin 2 
n4 cos 2   n 2 n2  n 4 sin 2   n 2 n2 cos 2   n 2 n2 sin 2   n 4 sin 2 
n2  n2  n 2  cos 2   n 2  n2  n 2  sin 2 
• Consider light reflected at Brewster’s Angle, defined by
• At this angle, the reflected light is completely polarized
• Evan at other angles, reflected light is partially polarized
 n cos   n sin 
n
tan  P 
n
Total Internal Reflection
•
•
•
•
•
•
Suppose we are going from high index to low index
Snell’s Law
n sin    n sin 
If n sin > n', this would yield sin ' > 1
What do we make of this?
k z  k n2 n 2  sin 2 
We previously found
This implies k'z is
k z  i  ik sin 2   n2 n 2
pure imaginary
• Substituting this in, we find
E  E0 eikx x it e  z
• Wave falls off exponentially in the disallowed region
– The evanescent wave
• The reflection amplitude in each case is
    0 n 2
   0n2
k
E n cos   i n2 sin 2   n2
E  n2 cos   in n 2 sin 2   n2

or

E n cos   i n2 sin 2   n2
E n2 cos   in n 2 sin 2   n2
• These numbers are both complex numbers of magnitude one
k 
Sample Problem 7.2 (1)
Light of frequency  is normally incident from a region of index n to a region of index
n".. In order to avoid reflection, a coating of index n' of thickness d is placed between
them. Show that this works for appropriate choice of n' and d.
• Start by writing down electric field in each region
– Let’s pick polarization in the x-direction
• Fields going both directions in the middle region
E  Exˆ eikz it , E  E1xˆ eik z it  E2 xˆ e ik z it , E  E xˆ eik z it
• We also need magnetic fields from B  nkˆ  E c
zd
z 0
    0 n2
    0 n 2
   0n2
B  nEyˆ eikz it c , B  nE1yˆ eik z it c  nE2 yˆ e ik z it c , B  nE yˆ eik z it c
• Have to match E||, D and B at the boundaries
E  E1  E2 , nE  nE1  nE2 , E1eik d  E2e ik d  E eik d , nE1eik d  nE2e ik d  nE eik d
• Eliminate E"
ik d
 ik d
ik d
 ik d






nE

nE

n
E

n
E
,
n
E
e

n
E
e

n
E
e

n
E
e
1
2
1
2
1
2
1
2
and E
Sample Problem 7.2 (2)
Light of frequency  is normally incident from a region of index n to a region of index
n".. In order to avoid reflection, a coating of index n' of thickness d is placed between
them. Show that this works for appropriate choice of n' and d.
nE1  nE2  nE1  nE2 , nE1eik d  nE2e ik d  nE1eik d  nE2e ik d
• Gather E1 and E2 on either side of the equations
 n  n E2   n  n  E1 ,  n  n E2eikd   n  n E1eikd
• Solve
E2 n  n n  n 2ik d


e
for E2/E1
E1 n  n n  n
• Cross multiply
n2  nn  n  n  n    n2  nn  n  n  n   e 2ik d
• We note that assuming n  n", we can conclude e 2ik d  1
• But it must be real, so
e2ik d  1
k d    2n  1
• We therefore have
n2  nn  n  n  n   n2  nn  n  n  n 
2n2  2nn
n  nn
7D. Wave Packets and Group Velocity
Wave Packets
• No wave is truly monochromatic
– If it were, then the plane wave would go for all time and all space
• To simplify our understanding, let’s work in one dimension
• We’ll combine a number of waves of the form
exp ikx  i  k  t 
– Assume (k) is a known function
• We then make a wave
1 
function by superposing these: u  x, t   2  f  k  exp ikx  i  k  t  dk
1 
ikx
• If you let t = 0, you see that
u  x, 0  
f
k
e
dk



2 
• Or reversing the Fourier transform, we have
1 
 ikx
f k  
u
x
,
0
e
dx



2 
Uncertainty Relation for Arbitrary Waves
1
u  x, t  
2
1
f k  
2
 f  k  exp ikx  i  k  t  dk



u  x, 0  eikx dx
• At any given time, we can define the average position or average wave number


2
2
x u  x  dx
k f  k  dk



x  
,
k


2
2
 u  x  dx
 f  k  dk


• We can similarly define the uncertainty in the position or the wave number

 x  x  u  x

 x  
 u  x  dx
2

2


2
2
k  k  f k 


 k  
 f  k  dk

dx
,
2

2

2

• There is an uncertainty relation between them
 x  k   12
• Same relationship as in quantum mechanics
• Any wave that is finite in extent has some spread in wave number
2
dk
Dispersion and Group Velocity
•
•
•
•
•
1
u  x, t  
f  k  exp ikx  i  k  t  dk

2
 k 
Each mode has a phase velocity given by
vp 
k
– Speed of the peaks and valleys of the modes
If this is bigger than c, can we transmit information faster than light?
Assume we have a nearly monochromatic wave, so f is only non-zero
for a small region of k near k = k0
Assume (k) is well
  k   0   k  k0    k0 
approximated by Taylor series:
Then we have
1 
u  x, t  
f  k  exp ikx  i0t  i  k  k0     k0  t  dk


2
1 ik0  k0 t i0t 

e
f  k  exp ikx  ik   k0  t  dk


2
Dispersion and Group Velocity (2)
1 ik0  k0 t i0t 
u  x, t  
e
f  k  exp ikx  ik   k0  t  dk


2
1 
 ikx
f
k

u
x
,
0
e
 
  dx
• Now substitute


2
1 ik0  k0 t i0t  
 ikx

u  x, t  
e
u
x
,
0
e
dx exp ikx  ik   k0  t  dk




 
2

• Fundamental theorem of Fourier transforms:
• And therefore we have


eikwdw  2  w
ik   k t i t





e
u  x     k0  t , 0 

x

x


k
t
u
x
,0
dx






0

 
• Define the group velocity as v    k   d
g
0
u  x, t   e
ik0  k0 t i0t

0
dk
• Then we have
u  x, t   e


ik0 vg  v p t
u  x  vg t , 0 
0

0
0
More About Group Velocity
• Recall:


c
k n  
• We therefore have
•
•
•
•
d
vg 
dk 0
1
1
c
c
dk


1
d


vg 
vg  

 n     

n     n  
n    n  
 d 
 c d

Under most circumstances, this is the speed at which signals can travel
– Almost always, vg < c
In circumstances where n'() is large and negative, this may be violated
Under such circumstances, Taylor series approximation may be invalid
In situations where n'() is large, usually you get lots of absorption as well
– This leads to additional complications