# Chapter 16 ELECTROMAGNETIC WAVES IN MATTER ```Chapter 16 ELECTROMAGNETIC WAVES
IN MATTER
• Introduction
• Energy and momentum in electromagnetic waves
• Angular distribution and polarization
• Scattering of electromagnetic waves
• Electromagnetic waves in matter
• EM waves at boundaries
• Summary
 2 
 2 
=  
2

2
INTRODUCTION
The complete description of electromagnetism requires
the four Maxwell equations plus the Lorentz force relation that defines the electric and magnetic fields. That
is:
Φ =
I
→
− −
→
1
E &middot; S =

0
 
Φ =
I
I

 
→
− −
→
 E &middot; l = −
I


→
− −
→
 B &middot; dl = 0




=  
(2)


Equations 1 and 2 describe a system of coupled
electric and magnetic fields where an oscillating 
field generates a travelling  wave and an oscillating
 field generates a travelling  wave. Also it was
shown that the non-zero derivatives of  and  with
respect to the z direction do correspond to a travelling
wave in the z direction by derivation of the following
wave equations
−
Z
Z


 



Inserting the known values  = 8854&times;10−12  2  2
and  = 410−7 2 gives that the velocity of an
electromagnetic wave in vacuum is  = 299792458 &times;
108  These waves are transverse waves comprising
perpendicular electric and magnetic oscillatory fields
that are in phase and perpendicular to the direction of
propagation. The changing E field induces the perpendicular B field, while this changing B field induces the
corresponding E field. The amplitudes of the E and B
fields are related in that 0 = 0 and the wave travels
→ −
−
→
in the E &times; B direction. The prediction of transverse
electromagnetic waves is a general feature of Maxwell’s
equations and was demonstrated.
−→
→
B −
&middot; S

→
−
−
→
→
E −
) &middot; S
 ( j + 0



−
→
→ → −
−
→
F = ( E + −
v &times; B)
It was shown in chapter 15 that Maxwell’s equations for vacuum that is, with zero current and charge
→
−
densities, j =  = 0 led to the fact that the oscillating electric and magnetic fields of a plane wave must be
transverse to the direction of propagation of the wave.
Moreover, Maxwell’s equations lead to the following
relations for a plane electromagnetic wave travelling
in the z direction.


=−


 2 
 2 
=


(4)


 2
2
These are examples of the general wave equation in
one dimension
1 2Ψ
2Ψ
=
 2
 2 2
where both the electric and magnetic waves travel at
a common velocity in vacuum, called c:
1
≡= √
 

→
→ −
−
B &middot; S = 0
Z
(3)
(1)
ENERGY AND MOMENTUM IN
ELECTROMAGNETIC WAVES
A remarkable feature of electromagnetic waves is that
energy is transmitted through vacuum, that is the travelling electric and magnetic fields carry energy. The
existence of life on earth is a direct result of the transmission of energy from the Sun to the Earth via electromagnetic waves in vacuum. The energy transmitted
via an electromagnetic wave can be related to the 
and  fields in the wave.
119
Figure 1 An electromagnetic wave incident upon a
point charge that initially is at rest.
The instantaneous power transmitted by the wave
usually is written as the Poynting vector S where:
−
→
→ −
→
1−
E&times;B
S ≡
0
→
−
The Poynting vector S gives the power, in watt/m2 
→ −
−
→
radiated in the direction E &times; B
Consider a harmonic EM wave of the form:
→
−
→
−
E = 0bi sin( − ) and B = 0bj sin( − )
Figure 2 The radiation pattern of the Poynting vector
Inserting these into the Poynting vector and using the
fact that the average of a sin2 function equals 1/2,
gives:
|S| =
1 1
1
0 0 =
 
2 0
0
0
0
where  = √
and  = √

2
2
An electromagnetic wave carries momentum as well
as energy. The momentum carried by an electromagnetic wave can be seen by considering the force on a
charge in the EM wave and remembering that force
is rate of change of momentum. A charge q, mass
−→
m, will experience a force F = bi in the electric
−
→
→
field leading to a velocity −
v = E  at a time t. The
moving charge then will experience a magnetic force
2 −
→ −
→
→
−
−−→
→
v &times; B =  E &times; B That is, the net force
F =  −
is in the direction of the electromagnetic field. This
force per unit area is called radiation pressure.
ANGULAR DISTRIBUTION AND
POLARIZATION
For an electric dipole radiator it is obvious that the
radiated electric field vector Ē will point along the
 axis in figure 2 while the magnetic field vector B̄
vector will be in the  −  plane circulating around the
→
−
→
−
 axis. Moreover, both the transverse E and B fields
will be zero for a wave propagating along the  axis
with  =  = 0 As a consequence the Poynting vector
of the radiated field has an angular distribution of the
form
→ sin2 
−
S ∝
2
120
Figure 3 Polarization analyzers.
where  is the angle with respect to the  axis. Thus
→
−
the electromagnetic wave is polarized with the E vector oriented parallel to the  axis. This behaviour was
demonstrated by the microwave demonstration which
emitted linearly-polarized waves, that is, waves with
the electric field vector parallel to the axis of the electric dipole radiator.
If a grid comprising parallel wires of a good conductor, is inserted between the source and detector with
→
−
the wires parallel to the E field vector then the wave
is reflected by the current induced in the wires.
Whereas, if the orientation of the wires is perpendicular to the electric field vector then the wave is
transmitted as demonstrated and illustrated in figure
3. The grid serves as a polarization analyzer and is
analogous to use of polaroid material to analyze the
polarization of light.
SCATTERING OF EM WAVES
Electromagnetic waves are scattered by free or bound
charges. These charges oscillate along the direction of
the electric field vector of the incident electromagnetic
wave. These oscillating charges reradiate like small
electric dipoles. The coherent oscillations of free elec-
ELECTROMAGNETIC WAVES IN
MATTER
In a dielectric the total real charge density  =   +
 where   is the applied free charge density and
 is the polarization charge in the dielectric. In the
discussion of dielectrics it was shown that the polarization charges can be absorbed into the dielectric constant allowing an alternate expression of Gauss’s Law
as:
I
Z
→
→ −
−
1
 E &middot; S =
  


 
Figure 4 Scattering of sunlight in the upper atmosphere.
This can be written as:
I
Z
→
→ −
−
  E &middot; S =   

 
trons in a metal or plasma lead to the reflection of the
incident wave at frequencies below the plasma oscillation frequency. Above the plasma oscillation frequency
the electromagnetic radiation is transmitted and scattered by reradiation from charges oscillating in the incident electromagnetic field.
A nice example of scattering of electromagnetic
waves is scattering of sunlight by the atmosphere. The
bound electrons in the atoms have resonances in the
ultraviolet region. Because the eye is sensitive only to
light, that spans frequencies just below the resonance
frequency for these bound electrons, it can be shown
bound electrons, charge q and mass m, is:
−
→
4
4
sin2 
b
S ∝ 2 02 2
r

( 0 −  2 )2 2
where  0 is the resonance frequency of the bound electrons,  the incident frequency, and  is the angle with
respect to the electric field vector of the incident wave.
Note that the radiation peaks perpendicular to the oscillating electric dipoles and that the reradiated power
increases as  4  As a consequence, blue light is scattered much more than red light. This is the reason
that the sky looks blue and the setting sun looks red
because the blue component has been scattered away
due to the long path length travelled by the sunlight
in the atmosphere. Sunlight transmitted through the
upper atmosphere is reradiated downwards preferentially for the horizontal component of the incident light
wave, thus the scattered light is plane polarized as illustrated in figure 4. The demonstration illustrates
this phenomena. The atmosphere becomes opaque
near resonances such as the UV region which is helpful
in reducing the risk of skin cancer. At high frequencies,
well above the UV, the scattering becomes small.
This formula can be simplified by defining the permittivity 
 ≡  
Note that only the applied charge density need be
known when applying this alternate form of Gauss’s
Law.
Similarly the total current density in magnetic ma→
−
−−→ −−→
→
−
terials can be split as j = j  + j where j  
→
−
is the applied current density and j  is the atomic
bound current density. It was shown that the eﬀects of
magnetisation can be absorbed into the relative permeability  allowing Amp&egrave;re’s Law to be written as:
−
→
Z
−−→ −
−
→
B →
&middot;
dl
=

0  j  &middot; S





I


This can be rewritten as:
Z
→
−
I
−−→ −
→
−
→
B
&middot;
dl
=

 j  &middot; S
 0 



This can be simplified by defining permeability :
 =  
Two other terms must be added to the current density,
−
→
namely Maxwell’s displacement current j  and the
−−→
polarization current density j that occurs in a dielectric when the bound polarization charges are flowing as the polarization is changing. It can be shown
that the polarization current can be absorbed into the
dielectric constant giving an alternate expression of the
Amp&egrave;re-Maxwell equation:
I
→
−
−
→
Z
−−→  E −
−
→
B →
(
j
+
&middot;
dl
=
) &middot; S

  






Thus Maxwell’s equations, can be expressed in a
form where both the bound charges and bound currents were absorbed implicity into the permittivity 
121
and permeability  giving an alternate expression of
these laws.
Maxwell’s laws-Alternate form
I
Z
→
→ −
−
Φ =
 E &middot; S =
  

 
Φ =
I


I

 
→
− −
→
 E &middot; l = −
I


→
− −
→
B &middot; S = 0
Z
 



Figure 5 Schematic illustration of the frequency response of the dielectric constant  of a typical dielectric.
The dielectric constant equals the square of the refractive
index, n.
−→
→
B −
&middot; S

−
→
→
−
Z
−−→  E −
−
→
B →
(
j
+
&middot;
dl
=
) &middot; S

  

 



These equations are similar to those for waves in
vacuum except that 0 and 0 are replaced by  and 
As shown in chapter 15, the two flux equations require
that electromagnetic waves be transverse. Following
the same procedure as used for waves in vacuum it is
easy to see that for a plane wave propagating in the
 direction the two circulation equations lead to the
relations


=−
(1)




= 
(2)
−


and from these one obtains the wave equations:
 2 
 2 
=

 2
2
 2 
 2 
=

 2
2
Therefore the velocity v of the EM wave is:
(3)
(4)

1
1
=√
=√ =√
√

  0 0
 
Note that the dielectric constant  always is greater
than unity while the relative permeability  is essentially unity except for ferromagnetic materials where
it is a large number. Thus the velocity  ≤ 
Refractive index
In the discussion of optics the refractive index n is
defined as:
 √
 ≡ =  

For most transparent materials  = 1 and thus the
√
refractive index  =  
The refractive index of a material depends on the
frequency of the radiation as illustrated schematically
122
in figure 5. As an example, consider water. For a
constant electric field, the dielectric constant of water is 81, implying a refractive index of 9. However,
this large dielectric constant is due to alignment of the
large polar water molecule that has a large moment of
inertia. Above a moderately low frequency, the slowly
moving polar molecules cannot move fast enough to
follow the oscillating electric field and thus the alignment of the polar molecule does not contribute to the
oscillating polarization electric field in the dielectric.
As a consequence, the dielectric constant decreases
considerably. At a higher frequency the ionic polarization of the molecules ceases to contribute to the
polarization, reducing the dielectric constant further,
when the molecular polarization cannot follow the oscillating electric field. As a consequence, in the visible region of the electromagnetic spectrum, the refractive index of water equals 133, that is the dielectric
constant is reduced to 177. At still higher frequencies the atomic polarization of the more weakly-bound
electrons cease to contribute again because they cannot follow the rapid oscillations. Eventually, for hard
X rays and gamma rays, even the atomic polarization
eﬀects become negligible and the refractive index approaches unity.
The frequency dependence of the refractive index
is further complicated in that the eﬀects are enhanced
when the electromagnetic frequency approaches resonant frequencies of bound electrons resulting in localized frequency regions where the refractive index increases as illustrated in figure 4. For example, there
are many resonances of bound atomic electrons in the
ultra-violet region of the EM spectrum. These cause
the dielectric constant, and thus, the refractive index,
to increase with frequency from the red to the blue
regions of the spectrum. As a consequence, blue light
travels slower than red light in glass or water. This
frequency dependence of the refractive index is called
dispersion since it leads to dispersion of the diﬀerent
frequencies in a glass prism or rainbow as will be discussed later. This discussion is overly simple in that
Figure 6 Boundary conditions for electric and magnetic
fields at a plane surface.
it ignores phase eﬀects and damping.
EM WAVE AT A BOUNDARY
Wavelength
Electromagnetic waves in material travel at a velocity
 =  and the wavelength  and frequency  are related by  =  =   At a boundary, the frequencies of
the waves in the two materials must be the same. Thus
if the refractive indices of the materials are diﬀerent,
so are the wavelengths in these materials. That is,
=


=
1 1
2 2
thus:
1 1 = 2 2
Figure 7 Electromagnetic wave incident normally upon
a plane interface between two materials of refractive
index n1 and n2 
travelling to the right. Assume that the incident  ,
reflected  , and transmitted  electric vectors point
upward, then using the boundary condition that k is
continuous implies that:
 +  = 
(1)
Assuming that the electric field vector points upwards
implies that the magnetic field points out of the page
for the incident and transmitted waves which are travelling to the right, and into the page for the left-going

reflected wave. Requiring that k be continuous across
the boundary implies:
1
1
( −  ) =

1
2
The relation between  and  for a wave is  = 
Thus:



=
=



The wavelength is shorter in materials with a large
refractive index. That is the wave does not travel as
far in a given time because the velocity is lower.
Using this to rewrite the boundary condition for 
gives:
2
1
( −  ) =

(2)
1
2
Boundary conditions
Solving equations 1 and 2 gives the electric field vector
for the reflected wave:
Remember that polarization of a dielectric causes surface charge density that partially cancels the normal
component of E in the dielectric but leaves the parallel component unchanged, as shown in figure 6. By
contrast, the surface currents due to magnetization of
matter enhances the parallel component of the B field
inside the material but leaves the normal component
of B unchanged.
The use of boundary conditions allows detailed calculation of the behaviour of EM waves at boundaries.
As an example, consider the simplest possible case of
an EM wave incident normally to a boundary as shown
in figure 7. For normal incidence, only the parallel
components of  and  are involved. In the left-hand
medium, refractive index 1  one has two waves, the
incident wave moving to the right and the reflected
wave moving to the left. In the right-hand medium,
refractive index 2  there is only the transmitted wave
( 1 −

= 11

( +
1
2
2 )
2
2 )
≈
1 − 2
1 + 2
and the electric field vector for the transmitted wave:
2 1

21
= 1 1 2 ≈

( +  )
1 + 2
1
2
where the approximation applies since typically 1 =
1
2

2 = 0  Note that 
 is positive if 1  2 while it
1
2
is negative if     This is analogous to reflections
1
2
of waves on a rope, as illustrated in figure 8. When
the right-hand rope is heavier that the left-hand rope,
it does not move as easily as the left-hand rope so the
amplitude of the reflected wave is opposite in sign to
the amplitude of the incident wave leading to partially
cancellation of the motion at the joint between the
123
Conductive materials
Figure 8 Reflection and transmission at the interface
between ropes of diﬀerent mass per unit length.
two ropes and a smaller amplitude for the transmitted
wave. On the other hand, when the right-hand rope
is lighter than the left-hand rope, the reflected wave
and incident wave have the same phase at the junction
leading to an amplitude of the transmitted wave that
is the sum of the reflected and incident amplitudes.
The intensity of the reflected and transmitted waves
→
−
is obtained by evaluating the Poynting vectors S =
→
−
→
−
1
 E &times; B for the waves, where  is the permeability
of the material. Substituting  = 
 in the Poynting
vector gives for the magnitude:
=
 2


This gives that the reflectivity  of the surface, defined
as the ratio of reflected and incident Poynting vectors,
is:
&macr;
&macr;2 &macr;&macr; 1
&macr;
&macr;
&macr;
2 &macr;2
&macr; 1 − 2 &macr;2
 &macr;&macr;  &macr;&macr;
&macr; ( 1 − 2 ) &macr;
&macr;
&macr;
=
=&macr;
= &macr; 1
&macr; ≈&macr;
&macr; ( + 2 ) &macr;

 &macr;
1 + 2 &macr;
1
2
Similarly, the transmission  , is given by:
 =
2 1

=

2 1
Note that:
&macr;
&macr;
4 1 2
&macr;  &macr;2
41 2
1
2
&macr;
&macr; =
≈
2
&macr;  &macr;
( 1 + 2 )2
(
1 + 2 )
1
2
 +=1
As an example, consider 1 = 1 and 2 = 15 typical
of glass. Then this gives  = 4% and  = 96%. For
a typical 4−element camera lens there are 8 surfaces
implying 32% of the light can be reflected which can seriously degrade the contrast if anti-reflection coatings
are not used. When the two media have the same value
of  the reflectivity is zero and transmission 100%.
The reason for discussing the derivation of the transmission and reflection coeﬃcients for normal incidence,
is to show that electromagnetism explains the details
of reflection and refraction. One can derive more complicated relations relating the incident, reflected and
transmitted waves for non-normal incidence using the
same boundary conditions.
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A good conductor is an equipotential for constant electric fields implying that the electric field is zero inside
the good conductor. This implies that a low frequency
electromagnetic wave is reflected and not transmitted
by a good conductor. The electric field in conductors
continues to be zero below the plasma frequency. For a
good metallic conductor the plasma frequency, which
depends on the density of charge carriers, is in the high
ultraviolet region while it is between the AM and FM
radio bands for the earth’s ionosphere. This explains
why AM radio waves are reflected by the ionosphere
and bounce around the earth because they are below
the ionosphere plasma oscillation frequency. However,
FM, TV, light waves, and higher frequencies are above
the plasma frequency and thus such waves are transmitted by the ionosphere. The much higher plasma
frequency in metals explains why radio waves and even
light waves are reflected well by a good conductor. Radio and microwaves penetrate only a very small distance into a good conductor, for example, microwaves
penetrate only the order of a micrometer in aluminum,
thus even a thin layer of kitchen foil is suﬃcient to reflect microwaves or light as demonstrated.
SUMMARY
This is the completion of the discussion of electromagnetism. We see that the laws of electricity and
magnetism lead to Maxwell’s equations, which predict