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The Wave Equation:
The displacement of a wave propagating in space is a function both of position and of
time:    ( x, t ) . Let the wave have propagation speed c. Then we can make a Galilean
transformation x  x  ct , to observe the disturbance of the medium as the wave
progresses through space. This removes the dependence of  on time:
 ( x, t )   ( x )  f ( x  ct ) , say.
Here we assume that the wave is travelling in the positive x sense (w.r.t. some inertial
frame), so we have that c  0 .
 ( x, t )  f ( x  ct ) .
(1)
A wave propagating with constant velocity c has a profile that is uniform in space and
constant in time. Thus,  x  x, t  t    ( x, t ) .
We can derive a wave equation from these few considerations which relates the change of
 with the change of x and of t.
x   x  ct
  ( x, t )   ( x )
  x  


x x  x x 
  x 


 c
t
x  t
x 
2
2


 
 2
 2
 
2  


c


c


c


c

c


t 
x  
tx 
x t
x  t
t 2
x  2
 2
1  2

x  2 c 2 t 2
By Galilean invariance, this is also the wave equation in the first inertial frame:
 2
1  2

x 2 c 2 t 2
(2)
A solution to this equation is of the form
 x, t   A expik x  ct    
where A, k, and  are constants.
(3)
We want a real solution, so take
 ( x, t )  A sin k x  ct )   .
Note that initial conditions determine :  x  0, t  0  A sin  .
Let the properties of the wave be given: let  be the wavelength and f. An

elementary argument gives us c = f. Thus, kc = kf = k
.
2
Consider equation (3): We must have that  x  , t    x, t  . Thus,
exp i kx  k  kct     exp i kx  kct   
exp i k   1  exp 2i 
k
2

Definition: The quantity k is called the propagation number of the given wave.
Thus we have completely characterized the one-dimensional wave from some simple
considerations and obtained the formula
 ( x, t )  A sin k x  ct )    

c  f


  2f

2

k


(4)
Definition: We define the quantity , called the phase:
  kx  kct  
(5)
We have,
  
  k
 x t
  
  
 t  x
    t   x 
       1
 t  x  x    t
  t 
   1
k  x 

 x 
     c
k
 t 
(6)
 x 
Definition:   is the speed of propagation of the condition of constant phase.
 t  
Plane Waves:
Recall the equation of a plane in Cartesian coordinates:
 : r  r0   k  0
In Cartesian coordinates,
xkx  yk y  zk z  a
a  x0 k x  y 0 k y  z 0 k z
Thus, k  r  a is the equation of the plane  .
We can now construct a function defined on a set of planes each with normal vector k,
which varies sinusoidally in space. The function will be scalar-valued, but will have a
vector as an argument. This is the function  (r )  A sin( k  r ) , or,
 r   A exp( ik  r)
(7)
We insist that this function be spatially repetitive:




 r    r  kˆ    r 
k 

k 
(8)
From (7), this requirement is the same as
A exp( ik  r )  A exp i k  r  k 
 exp( ik )  1  exp( 2i )
k
2

Definition: The vector k, whose magnitude k is the propagation number, is called the
propagation vector.
We introduce a time-dependence into equation (7) and get
 r, t   A expik  r  t 
(9)
As before, the phase is that quantity  k  r  t . A wave front is a surface joining all
points of equal phase.
The phase is constant in time and uniform in space. Thus,
 rk  rk , t  t    rk , t 
 A exp i rk k  rk k  t  t   A exp i rk k  t 
exp i rk k  t   1

rk k  t  0
r

   c
t
k

rk is the component of the position vector r in the k direction and so rk = rkk. The result
dr
is that the magnitude of the wave velocity, k , is equal simply to c.
dt
To derive a three-dimensional analogue of equation (2), we recall the direction cosines:
Thus, for
 r, t   A exp i k x x  k y y  k z z  t   A exp ik x  y  z  t  , we have
 2
  2 k 2
2
x
 2
   2 k 2
2
y
 2
  2 k 2
2
z
 2
  2
t 2
 2  2  2
1  2
 2  2  2  2 2
x
y
z
c t
Note the symmetry between the variables x, y and z in equation (10).
(10)
Light in Matter:
From Maxwell’s equations, we have
c2 
1
(11)
 0 0
If we consider a homogeneous, isotropic dielectric in a region of space, equation (11) is
modified and we get that
v2 
1
(12)

Note the convention: c denotes the speed of propagation of electromagnetic (em)
radiation in vacuo; v denotes the speed of propagation of em radiation in any other
medium.
and  are related linearly to  and to respectively by dimensionless constants:
  K E 0 

  K M 0 
Thus, equation (12) becomes v 
(13)
1
K E K M  0 0

c
KE KM
.
c
 KE KM
v
n is called the absolute index of refraction of the dielectric.
We define the quantity n 
(14)
For materials that are transparent to visible em radiation (i.e., light), KM is almost unity,
since these materials – in particular, glass – are not magnetic. Thus,
n  KE
(15)
This equation is known as Maxwell’s relation.
Now KE is constant, so equation (15) suggests that n is constant, once the material
properties of the dielectric are fixed. However, it is an experimental fact that n depends
on the frequency of the incident em radiation. This dependency is called dispersion.
Maxwell’s equations ignore dispersion. Clearly, it must be considered.
Firstly, we consider the different ways in which em radiation, or, equivalently, photons,
interacts with a given dielectric. This provides the key to the physical basis for the
frequency-dependence of n. We consider the interaction of an incident em wave with the
array of atoms which constitutes the dielectric. An atom reacts to the incoming radiation
in two ways. Depending on the frequency, the incident photon may simply be scattered –
redirected without being altered. If, on the other hand, the energy of the incoming photon
matches that of one of the excited states, the atom will absorb the photon, and is raised to
a higher energy level. In gases under pressure and in dense materials, it is probable that
this energy will be dissipated by random atomic motion, before a photon can be emitted.
(This dissipation is analogous to a damping force in an oscillating system.) This
absorption is thus known as dissipative absorption.
In contrast to this atomic excitation, a process called non-resonant scattering occurs
when incoming em radiation of frequencies lower than the frequencies necessary for
absorption interacts with the atoms. The energy of the photon is too small to cause a
transition of the atom to any excited state. However, the em field of the incident light can
drive the electron cloud of the atom into oscillation (more on this later). There is no
atomic transition: the electron remains in its ground state but the electron cloud vibrates
slightly at the frequency of the incident light (analogous to the driving frequency in an
oscillator). When the electron cloud starts to vibrate w.r.t. the positive nucleus, the
system constitutes an oscillating electric dipole and as such will immediately begin to
radiate at that same frequency. The resulting emitted photon has the same frequency –
and thus energy – as the incident one. Therefore, this process of scattering is completely
elastic.
We also have to consider polarization. When a dielectric is subject to an external E field,
the internal charge distribution is disturbed. This corresponds to the generation of many
electric dipole moments, which in turn contribute to the electric field. This is
polarization. Polarization is characterized by the dipole moment per unit volume due to
the E field, called the electric polarization, denoted by the vector P. It is found that
   0 E  P
(16)
There are in fact several kinds of polarization:


Orientational Polarization: A molecule that itself has a dipole moment is subject to
orientational polarization. These molecules are called polar molecules. Usually, a
collection of polar molecules will be such that the orientation of the polarization is
random – the randomness being due to thermal effects. On application of an E field,
these dipoles align. An example is the water molecule.
Electronic Polarization: In non-polar molecules and atoms, no such “internal”
dipoles exist. But on application of an E field, the electron cloud of each atom /
molecule shifts relative to the nucleus – thereby producing a dipole moment. This is
the most significant sort of polarization here.

Ionic Polarization: If a collection of ionic molecules (e.g., NaCl) is subjected to an
external E field, the positive and negative ions undergo a shift relative to each other,
thereby producing dipole moments, which will be aligned in the external field.
If the dielectric is subjected to an incident harmonic em wave, its internal structure will
experience time-dependent (external) forces and / or torques. These forces are due to the
E field of the incident wave only, since FM  qv  B , and so is negligible for v  c
Consider now the electron cloud of a nucleus in the external E field. Since the electrons
have small inertia, they will be sensitive to the applied E field. We can assume that the
electron cloud has some stable equilibrium position w.r.t. the nucleus (due to a minimum
of the potential function, where the potential function arises because of the Coulomb
interaction between the positive nucleus and the negative electron cloud). Any small
disturbance from equilibrium will result in a restoring force proportional to the
displacement from equilibrium. So we assume that a restoring force of the F = -kx acts
on the system. Once disturbed, any electron in the cloud will oscillate with natural
k
frequency  0 
. Further, if the electron oscillates at this frequency, we assume
me
that it will emit a photon at precisely this frequency.
We therefore think of the electron cloud as an oscillating system; as though it were a
collection of particles attached to the nucleus by springs of spring constant k – i.e., a
collection of coupled oscillators. Applying the external E field will result in the system’s
being driven at frequency , where this arises E  E0 cos t .
Thus, the force on each electron due to E is FE  qe E0 cos t . Therefore, the equation of
motion for each particle in the system is
me
d 2x
  02  qe E0 cos t
dt 2
(17)
We assume the (steady-state) solution x  x0 cos t . Notice that after transience has
ended, the system oscillates at the frequency of the external field.
qe / me
E0
 02   2
(18)
q e / me
E 0 cos t
 02   2
(18’)
We find x0 to be x0 
Thus,
x(t ) 
Recall some facts about dipoles:

 1  d  qE
 1  qEd sin 
 tot   1   2  2qEd sin   qeD sin 
D = 2d
p  qD

  pE
If there are N molecules per unit volume, then the dipole moment per unit volume, or
electric polarization, P, is
P  qDN
(19)
In our case, we have that P = qexN.
Thus,
P  qe xN  qe
q e / me
E0 costN , from (18’)
 02   2
Now P = (E.
qe2 / me
Thus, E    0   2
E
0   2
q e2 N
  0 
me  02   2


  K E 0
q e2 N
KE  1
 0 me  02   2


K E  n2
q e2 N
 n   1 
 0 me  02   2


We shall write
n 2    1 
qe2 N
 0 me  02   2


(20)
Remarks:




Definition: We say that  is a resonant frequency.
For    0 , we have that  02   2  0 , so n 2  0 . This implies a complex value for
n, on which more later.
For    0 , we have that  02   2  0 , so n 2  0 .
We can rearrange equation (20) to express n as a function of .
 0 me 2
4 2 0 me 2  1 1 
1
2
2  0 me
2
2
 2  0    4 2
f0  f 
c  2  2   0 2  2 ,
2
2
n  1 qe N
qe N
qe N
 0  
1

  2  0 2 
2

n 1
(21)
4 2  0 me 2 


c

qe2 N




Various graphs are plotted for the values of wavelength and of n listed below:
Nanometres
n
728.135
706.519
667.815
587.562
504.774
501.567
492.193
471.314
447.148
438.793
414.376
412.086
402.619
388.865
1.53460
1.53520
1.53629
1.53954
1.54417
1.54473
1.54528
1.54624
1.54943
1.55026
1.55374
1.55402
1.55530
1.55767
Equation (20) is not useful practically as it is. For, we must firstly generalize: a system
consisting of many oscillators will have many natural frequencies. Thus, for N molecules
per unit volume, with each molecule having a certain number of natural frequencies, we
should have
n 2    1 
Nqe2
 0 me

j
2
0j
1
 2
(21)
This is not entirely correct: we should have
n 2    1 
where

j
Nqe2
 0 me

j
fj
2
0j
 2
(22)
fj 1
This fact is simply the statement that while the system has many natural frequencies,
some are more dominant than others.
The fj’s have another, quantum-mechanical interpretation. The frequencies  0 j are called
the characteristic frequencies at which an atom may absorb or emit radiant energy. The
fj terms reflect the fact that it is more probable that an atom will absorb / emit radiant
energy at some frequencies (called modes), than at others. As such the terms fj terms are
known as transmission probabilities.
Note that when    0 j , the function n 2   is discontinuous, which contradicts
experience. This is because we have failed to consider a damping force acting on each
oscillator (electron). Atoms and molecules in a gas under any signicicant pressure, and in
liquids and solids – and hence in close proximity – experience a “damping” force. The
effect of the “damping” is energy dissipation due to “heat” loss (i.e. due to random
molecular motion).
Thus, if we include a velocity-dependent damping force, m e 
d 2x
dx

  02 x  qe E0 cos t
2
dt
dt
dx
, equation (17) becomes
dt
(23)
Solving in the complex plane gives
d 2z
dz

  02 z  qe E0 exp it 
2
dt
dt
(23’)
We choose the steady-state solution z  x0 exp it    , where x0 is real.


This gives x0  02   2   i exp it   qe E0 exp it  exp  i  . Solving algebraically
for x0 in the complex plane gives the following Argand diagram:
Hence, x0 
2

qe E0 / me
2
0
x(t )  x0 cos t 

P  q e xN  q e N


2
e

 2
2
0
 2
n 2    1 

   
2
q e / me

2
0
   
2
2
q e / me
Nq / me
2
0
(24)
   

2 2
2
   
2 2
2
E 0 cos t
2
E (t )     0 E (t )
    0 
Nqe2 /  0 me
2
0

Since n  K E
   
2 2
2
Thus, the general formula for the coupled oscillators with the weighting terms fj will be
n 2    1 
Nqe2
 0 me

j

fj
2
0j

    
2 2
(25)
2
j
Note that equation (25) is similar to our earlier derivation, in equation (22).
convenience, we shall write (25) in the form
n 2    1 
Nqe2
 0 me

j
where here i denotes

fj
2
0j
 2
    
2
2
j
 1 , and is not a summation index.
For
(25’)
In the limit of small damping, with  being small, equation (25) is is approximated by
n 2    1 
Nqe2
 0 me

j
fj
2
0j
 2
(25’)
Setting  equal to zero also gives this result. Thus (25) reduces to the case of no
damping, as in equation (22).
In this way, we have resolved the problem of the discontinuities at the characteristic
frequencies  0 j .
Unfortunately, this is not the end of the story, since equation (25’) is not fully right, since
it fails to consider the potential functions between the molecules themselves. This is all
very well in gases, where we can model molecules as points that do not interact with each
other, but in solids and in liquids, these potentials are important. In fact, equation (25’)
should be
Nqe2
n2 1

n 2  2 3 0 me

j
fj
2
0j
  2  i j
(26)
(Note the 3 factor!)
All of the above analysis works equally well for bound ions. But ionized molecules have
a greater inertia than electrons and, consequently, the effects of ions on the index of
refraction as a function of frequency is comparatively unimportant (since from any of the
1
above equations, we see that n 2  .
m
We restrict our attention now to the case of equation (26) where  02j   2   , so that
the equation
Nqe2
n2 1

n 2  2 3 0 me

j
fj
2
0j
 2
(26’)
describes the frequency-dependence of n satisfactorily.
(26’) tells us a lot:

Substances such as glass have their characteristic frequencies  0 j outside of the
visible range of the spectrum. In particular, the characteristic frequencies of glass are
in the UV range, so glass tends to absorb radiant energy in this range and is thus
opaque to UV radiation.

If  0 j
Nqe2
n2 1

  , we have that 2
n  2 3 0 me
fj

j
2
0j
, and so for a given frequency 
much less than the characteristic frequencies, n is nearly constant. This explains why
the values for n do not change much for glass over the range of visible light, since, as
we have noted, the characteristic frequencies of glass are in the UV range.
Cf, nred = 1.5346; nviolet = 1.5577. Thus, n = 0.0231.
Thus, n versus  is nearly a straight line in regions such as    ,     , for
suitable .

As    0 j , n increases greatly toward a value corresponding to resonance. In the
limit of resonance, the amplitude – and n – are very large. However, as    0 j , the
fj
Nqe2
term    in equation (25’), viz. n    1 
decreases,

 0 me j  02   2  i
and so the damping term , dominates, and so n(will tend to decrease again.
Thus, we have the following:
2
2
0j
Definition: (refer to figure)
The regions where n increases gradually with as  increases towards a resonant
frequency are called regions of normal dispersion. From the basic equation for the
E 0 q e / me
amplitude A as a function of , viz,. A  
, we see that A, and
2
2
2 2
 0 j     j 

hence n has a maximum at  2   02j 
2
2

, just before the resonant frequency j.
The regions surrounding the  0 j ’s are called absorption bands.
dn
 0 , anomolous absorption takes place. Anomolous absorption takes place
d
in the region of the absorption bands.
When
Application: The prism

All of this explains why the red is deviated less than the blue: We have, more or less,
1
that n  . Thus, for a longer wavelength, such as that of red light, the index of

refraction is smaller. By Snell’s Law, this means that the deviation is less.
Therefore, red light, on going through a prism, is deviated less than violet light.
IN sum, over the visible region of the spectrum, electronic polarization is what
determines n(). Classically, one imagines electron-oscillators vibtrating at the
frequency of the incident wave. When the wave’s frequency is appreciably differentf rom
a characteristic frequency, the oscillations are small, and there is little dissipative
absorption. At resonance, however, the oscillator amplitudes are increased, and the
external field imparts an increased amount of energy to the electrons. This energy is
dissipated through the thermal agitation of the eletrons, and so the energy from the
external field is effectively lost. This is dissipative absorption, and occurs at an
absoroption band. The material, while being essentially transparent at other frequencies,
is opaque to incident light at the characteristic frequencies.
The end.
Recall our basic equation for absorption in a rare solid:
qe2 N
n    1 
 0 me  02   2
2


(27)
Recall also the modified equation
n 2    1 
Nqe2
 0 me

j

fj
2
0j

    
2 2
(28)
2
j
We have already noted that for    0 , this relation leads to negative n2 and so to a
complex value for n, the index of refraction. This is the case for metals: for metals, we
usually encounter a complex quantity n. However, equations (27) and (28) do not apply
directly to metals. Firstly metals are dense and these equations are for rare materials.
Secondly, our model of the electron cloud and the nucleus as a classical oscillator does
not hold true here: in a metal, many electrons are unbounded (“free”), and this gives rise
to conduction. Thus, some electrons experience no restoring force since they are not
bounded in a potential well. This point is the key to explaining why metals are so good at
absorbing incident light.
Firstly, it will be convenient to consider the complex index of refraction as the sum of a
real and an imaginary part, which we shall write as
n~  nR  in I
(29)
This is the case for metals.
Consider now a wave that propagates in the y direction of a given inertial frame and is
described by the wavefunction
E  E0 cost  ky
n~ 

E  E0 cos t 
y
c 

c
The last equation follows from the identity n~  .
v
  n~ 
Thus, E  E 0 cos   t  y  .
c 
 
(30)
~  n  in , we get
Writing this in exponential form and using the fact that n
R
I

y 
E  E 0 exp i t   n R  in I   
c 


n 
E  E 0 i t  R
c

 n
y  I
c


y


n  
 n 
E  E 0 exp   I y  exp i t  R y 
c 
c



Thus, on taking the real part, we have

n  
 n   
E  E 0 exp   I y  cost  R y 
c 
 c  

(31)
thus, the wave advances along the (30)
nR
. However, as time
c
evolves, the amplitude of the E field is decreasing exponentially. We say that the
amplitude of the E field is being attenuated. This attenuation is given by the exponential
Thus, the wave advances along the y-direction with speed
 n 
exp   I y 
c 

(32)
Since irradiance (time-averaged intensity, given by I) is proportional to the square of
amplitude, we must have that
I  I 0 exp  y 
(33)
2n I 
c
(34)
Where  
 is called the absorption coefficient or the attenuation coefficient.
I0 is the irradiance at y = 0, i.e. at the interface of the media.
We see that  determines the degree to which a material will be transparent (or opaque)
to incident radiation. We note that  depends on the frequency of the inciden tradiation.
The penetration depth, p, is that distance travelled through the medium by the radiation
such that the value of I is reduced by a factor of 1 / e. I.e.,
I  y  p 
I0
e
***
Let  denote the wavelength in vacuo. Recall that frequencies do not change in going
from one medium to another. For, consider a light ray going through a block of glass:
The requirement that f = f0 is the statement that the number of wavefronts entering the
block (at A) per unit time is equal to the number of wavefronts leaving the block (at B)
per unit time. A wave cannot be destroyed on undergoing refraction.
For incident radiation in the ultraviolet range ( ~ 100nm), the penetration depth of
copper is 0.6nm. For radiation in the infrared range ( ~ 10,000 nm), the penetration
depth of copper is ~ 6nm. Thus, for copper, the penetration depth increases with
inceasing wavelength.
These penetration depths are very small. In general, this holds true for metals. Thus,
metals are opaque to incident radiation. This in fact corresponds to high reflectance.
Metals that are silvery grey are precisely that way because they reflect most incident
(visible) radiation. In fact, they reflect up to 95% of the incident light and are thus
colourless.
***
Extending the Absorption Equation to Metals:

Consider the metal to be a collection of driven, damped oscillators. Some free
electrons are present (i.e. are not bound to any nucleus) and thus have no restoring
force associated with them. Others are bound to atoms, and behave in a way
consistent with equation (27), or the modified versions thereof. However, the
conduction of electrons (the movement of the free electrons through the metal) is the
dominant behaviour and determines the optical properties of the metal.

Recall that the displacement of an electron from its equilibrium position in the
nuclues, due the application of an external E field, is given by
q
1
(35)
x(t )  e 2
E (t )
me  0   2
If the restoring force is absent, we have that  = 0, and so we have that
q 1
(36)
x(t )   e 2 E (t )
me 
Thus, the displacement is out of phase with the applied field E by an amount . This
is different from the case of transparent media, where the the frequencies in the
visible range, we had    0 and consequently, x and E were in phase.
Free electrons thus vibrate out of phase with the applied E field, and so emit waves
(radiation) that cancels with the incident radiation. This results in the decay of an
incident wave in a dense material. This explains why a dense material, where
conduction takes place, is opaque to incident radiation.
Consequently, equation (28) is modified readily:



Nqe2
n    1 
 0 me
2


fj
fe




2
2
2
j  0 j    i j  
    i e

where the first bracketed term, viz.,
(37)
fe
, is the contribution from the unbounded
   i e 
2
electron, for which  0e = 0.
Equation (37) tells us that both the free electrons and the bound electrons contribute to
absorption. Thus, if a metal has a particular colour, this is because selective absorption is
taking place, as a result of absorption both by the conducting electrons and the oscillating
molecules.
We have noted that for copper, the penetration depth is increased with increasing
wavelength. This is because , the attenuation coefficient, increases with increasing
wavelength, which is due in turn to the fact that nI increases with increasing wavelength.
This means that copper and gold are opaque to light of longer wavelength, and rather than
actually absorbing this light, it is reflected..
If we look at the behaviour of n as a function of  for absorption dominated by the
conducting electrons and for larve values of , equation (37) tells us that
Nqe2
n    1 
 0 me  2
2
(38)
Further, we can think of a collection of free electrons and positive ions in a metal as a
plasma, which oscillates – due to the incident radiation – at a freqeucny  p 
 p
Thus, n    1  

2
Definition:
p 
Nqe2
.
 0 me
2

 .

Nqe2
is called the plasma frequency.
 0 me
If    p , we have that n 2  0 , so n is complex.
If    p , we have that n 2  0 , so n is real.
p is thus a critical frequency that determines whether or not n will be real or complex.
If    p , n is real, so nI = 0. Thus, the behaviour of the radiation in the medium in this
case is not subject to attentuation and so the medium is transparent. On the other hand, if
   p , n is complex. This corresponds to opacity.
Polarization:
Definition:
EM radiation is said to be linearly polarized if it oscillates in one plane
only. The radiation is then said to reside in a plane of oscillation.
Consider now two waves which are polarized and which reside in two orthogonal planes:
E1  E2  0 . We may rotate our coordinate frame so that E1 is along the x-axis and E2 is
along the is along the y-axis. The direction of propagation of the two waves is along the
z axis. Since these are transverse waves, we have that E i  k i is zero for both waves,
where ki is the propagation vector.
Thus, we have the following:
E1  E x  ˆiE 0 x coskz  t 
E  E  ˆjE coskz  t   
y
2
0y
E y  E 0 y coskz  t     E 0 y coskz  t  cos   sin kz  t sin  
Ey
 coskz  t  cos   sin kz  t sin 
E0 y
Ey
E0 y
Ey
E0 y

Ex
cos   coskz  t  cos   sin kz  t sin    coskz  t  cos    sin kz  t sin 
E0 x

Ex
cos    sin kz  t sin 
E0 x
sin   1  cos 2 


1
sin kz  t   1  cos 2 kz  t  2
 E

 Ey
Ex
2
2
1   x




sin
t


kz
cos

1


cos



  E 0 x

 E 0 y E 0 x
2
2


 Ey

E
 0y
 Ey
  Ex 
2

 


cos
2


cos
E
 E 
 0y
  0x 
 Ey

E
 0y
  Ex
 
 E
  0x
2
2
E

  2 cos   y
E

 0y
2
 E x

 E
 0 x
 E x

 E
 0 x



2

 sin 2 

E

  sin 2   sin 2   x
 E0 x




2

  sin 2 

(39)
This is the equation of an ellipse, oriented by some angle  about the x-axis.
Taking a snapshot along the z-axis at any time t gives us the magnitude of the vector
E1+ E2 at that instant. This is the radius vector extending from the origin to the perimeter
of the ellipse.

If we set   2n  1 , where n is any integer, equation (39) takes on a more
2
recognizable form:
 Ey

E
 0y
2
  Ex 
 
 1
 E 
  0x 
2
(40)
Now the angle  is equal to zero:
If E0 x  E0 y  R , equation (40) becomes that of a circle:
E x2  E y2  R 2

   2n

Further, 
  2n  1

(41)

Ex 
E0 x 
E0 y 
 Ey 
Ex 
E0 x 
 Ey 
E0 y
This is the case of linear polarization.
(42)