May 4, 2010
1
1
J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Section 7
Electromagnetic Waves

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A basic feature of Maxwell equations for the EM field is the existence of travelling wave solutions which represent the transport of energy from one point to another.
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The simplest and most fundamental EM waves are transverse , plane waves.
In a region of space where there are no free sources ( ρ = 0,
~
= 0 ),
Maxwell’s equations reduce to a simple form given
~
~
= 0
= 0 ,
,
~
+
1 c ∂ t
~
−
µ c ∂ t
= 0
= 0 (1) where
~ and
~ are given by relations
~
= E and
~
=
1
µ
(2) where is the electric permittivity and µ the magnetic permeability
which assumed to be independent of the frequency.
Electromagnetic Waves

Maxwell’s equations can be written as
∇ 2 ~
−
µ c 2
∂
2 ~
∂ t 2
= 0 and ∇ 2 ~
−
µ c 2
∂
2 ~
∂ t 2
= 0 (3)
In other words each component of the form:
∇ 2 u −
1 v 2
∂ 2 u
∂ t 2
~
= 0 and
~ where obeys a v = √ wave equation c
µ of
(4) is a constant with dimensions of velocity characteristic of the medium.
The wave equation admits admits plane-wave solutions : u = e i
~
· ~ − i ω t
~
( x , t ) =
~ e ik ~ · x − i ω t and
~
( x , t ) =
~ e ik ~ · x − i ω t
(5)
(6) where the relation between the frequency ω and the wave vector
~ is k =
ω v
=
√
µ
ω c or
~
·
~
=
ω v
2 also the vectors ~ ,
~ and
~
are constant in time and space.
Electromagnetic Waves
(7)

If we consider waves propagating in one direction, say x -direction then the fundamental solution is: u ( x , t ) = Ae ik ( x − vt )
+ Be
− ik ( x + vt )
(8) which represents waves traveling to the right and to the left with propagation velocities v which is called phase velocity of the wave.
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From the divergence relations of (1) by applying (6) we get
~ ·
~
= 0 and n ·
~
= 0 (9)
This means that
~
(or
~
) and
~
(or
~
) are both perpendicular to the direction of propagation ~ . Such a wave is called transverse wave .
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The curl equations provide a further restriction
~
=
√
µ n ×
~ and
~
= − √
1
µ
~ × (10)
The combination of equations (9) and (10) suggests that the vectors
~ , and
~ form an orthonormal set .
Also, if ~
is real, then (10) implies that that
~ and
~ have the same phase.
Electromagnetic Waves

It is then useful to introduce a set of real mutually orthogonal unit vectors ( ~
1
, ~
2
, ~ ) .
In terms of these unit vectors the field strengths
~ and
~ are
~
= ~
1
E
0
,
~
= ~
2
√
µ E
0
(11) or
~
= ~
2
E
0
0
,
~
= − ~
1
√
µ E
0
0
(12)
E
0 and E
0
0 are constants, possibly complex.
In other words the most general way to write the electric/magnetic field vector is:
~
=
~
=
( E
0
√
µ ( E
0
~
~
2
1
+ E
0
0
~
− E
0
0
~
1
2
) e ik ~ · x − i ω t
) e ik ~ · x − i ω t
(13)
(14)
Electromagnetic Waves

Thus the wave described by (6) and
(11) or (12) is a transverse wave
propagating in the direction ~ .
Or that E and B are oscillating in a plane perpendicular to the wave vector k , determining the direction of propagation of the wave.
The energy flux of EM waves is described by the real part of the complex Poynting vector
~
=
1
2 c
4 π
~
× H
∗
=
1 c
2 4 π h
~
R
× H
R
+ E
I
× H
I
+ i E
I
× H
R
− E
R
× H
I i where
~ and evaluated.
2
~ are the measured fields at the point where
~ is
2
Note : we use the magnetic induction
~ because although
~ is the applied
induction, the actual field that carries the energy and momentum in media is
~
.
Electromagnetic Waves

The time averaged flux of energy is:
~
= c
8 π r
µ
| E
0
|
2
(15)
The total time averaged density (and not just the energy density associated with the electric field component) is:
1 u =
16 π
~ ·
∗
+
1
µ
~
·
∗
=
8 π
| E
0
| 2
(16) v = c / µ .
3 ( Prove the above relations )
Project:
What will happen if ~ is not real?
What type of waves you will get?
What will be the form of E ?
E
R
3
Note: To prove the above relations use h cos
2
= (
~
+
∗
) / 2 we get h E
2
R i =
~
· ∗
/ 2 .
x i = 1 / 2 and since
Electromagnetic Waves

The plane wave (6) and (11) is a wave with its electric field vector always
in the direction ~
1
. Such a wave is said to be linearly polarized with polarization vector ~
1
. The wave described by (12) is linearly polarized
with polarization vector
The two waves :
~
2 and is linearly independent of the first.
E
1
= ~
1
E
1 e i
~
· ~ − i ω t
, E
2
= ~
2
E
2 e i
~
· ~ − i ω t with (17)
B i
=
√
µ
~
× E i
, i = 1 , 2 k
Can be combined to give the most general homogeneous plane waves propagating in the direction
~
= k n ,
~
( ~ , t ) = ( ~
1
E
1
+ ~
2
E
2
) e i
~
· ~ − i ω t
~
( x , t ) = h
~
1
| E
1
| + ~
2
| E
2
| e i ( φ
2
− φ
1
) i e i
~
· ~ − i ω t + i φ
1
(18)
(19)
The amplitudes E
1
= | E
1
| e i φ
1 and E
2
= | E
2
| e i φ
2 are complex numbers in order to allow the possibility of a phase difference between waves of different polarization.
Electromagnetic Waves

If the amplitudes E
1
= | E
1
| e i φ
1 and E
2
= | E
2
| e i φ
2 have the same phase
(18) represents
a linearly polarized wave with the polarization vector making an angle θ = tan
− 1 ( < ( E
2
) / = ( E
1
)) (which remains constant as the field evolves in space and time) with ~
1
E = p
E 2
1
+ E 2
2
.
and magnitude
If E
1 and polarized
E
2 have the different phase
the wave (18) is
elliptically and the electric vector rotates around
~
.
Electromagnetic Waves

• E
1
• φ
1
= E
2
− φ
2
= E
0
= ± π/ 2 and the wave becomes
~
( ~ , t ) = E
0
( ~
1
± i ~
2
) e i
~
· ~ − i ω t
(20)
At a fixed point in space, the fields are such that the electric vector is constant in magnitude, but sweeps around in a circle at a frequency ω .
The components of the electric field, obtained by taking the real part of
(20)
E x
( ~ , t ) = E
0 cos( kz − ω t ) , E y
( ~ , t ) = ∓ E
0 cos( kz − ω t ) (21)
For the upper sign ( ~
1
+ i ~
2
) the rotation is counter-clockwise when the observer is facing into the oncoming wave. The wave is called left circularly polarized in optics while in modern physics such a wave is said to have positive helicity .
For the lower sign ( ~
1
− i ~
2
) the wave is right circularly polarized or it has negative helicity .
Electromagnetic Waves

An alternative general expression for complex orthogonal vectors
~ can be given in terms of the
~
±
=
1
√
2
( ~
1
± i ~
2
) (22) with properties
~
∗
±
· ~
∓
= 0 , ~
∗
±
· ~
3
= 0 , ~
∗
±
· ~
±
= 1 .
Then the general representation of the electric vector
(23)
~
( ~ , t ) = ( E
+
~
+
+ E
−
~
−
) e i
~
· ~ − i ω t
(24) where E
− and E
+ are complex amplitudes
If E
− and E
+ have different amplitudes but the same phase
eqn (24)
represents an elliptically polarized wave with principle axes of the ellipse in the directions of ~
1 and ~
2
.
The ratio of the semimajor to semiminor axis is | (1 + r ) / (1 − r ) | , where
E
−
/ E
+
= r .
Electromagnetic Waves

The ratio of the semimajor to semiminor axis is | (1 + r ) / (1 − r ) | , where
E
−
/ E
+
= r .
If the amplitudes have a phase difference between them E
−
/ E
+
= re i α then the ellipse traced out by the
~ vector has its axes rotated by an
, angle α/ 2 .
Figure: The figure shows the general case of elliptical polarization and the ellipses traced out by both
~ and
~ at a given point in space.
Note : For r = ± 1 we get back to a linearly polarized wave.
Electromagnetic Waves

Figure: The figure shows the linear, circular and elliptical polarization
Electromagnetic Waves

The polarization content of an EM wave is known if it can be written in
the form of either (18) or (24) with known coefficients
( E
1
, E
2
) or
( E
−
, E
+
) .
In practice, the converse problem arises i.e.
given a wave of the form (6),
how can we determine from observations on the beam the state of polarization?
A useful tool for this are the four Stokes parameters . These are quadratic in the field strength and can be determined through intensity measurements only.
Their measurements determines completely the state of polarization of the wave.
For a wave propagating in the z -direction the scalar products
~
1
·
~
, ~
2
·
~
, ~
∗
+
·
~
, ~
∗
−
· (25) are the amplitudes of radiation respectively, with linear polarization in the x -direction, linear polarization in the y -direction, positive helicity and negative helicity .
The squares of these amplitudes give a measure of the intensity of each type of polarization .
The phase
information can be taken by using cross products
Electromagnetic Waves

In terms of the linear polarization bases ( ~
1
, ~
2
) , the Stokes parameters are: s s s
0
1
2 s
3
= | ~
1
·
~
| 2
+ | ~
2
·
~
| 2
= | ~
1
·
~
|
2
− | ~
2
·
~
|
2
= a
2
1
+ a
2
2
= a
2
1
− a
2
2
= 2 < h
( ~
1
·
~
)
∗
( ~
1
·
~
) i
= 2 a
1 a
2 cos( δ
1
− δ
2
)
= 2 = h
( ~
1
·
~
)
∗
( ~
1
·
~
) i
= 2 a
1 a
2 sin( δ
1
− δ
2
)
(26)
where we defined the coefficients of (18) or (24) as magnitude times a
phase factor:
E
1
= a
1 e i δ
1 , E
2
= a
2 e i δ
2 , E
+
= a
+ e i δ
+ , E
−
= a
− e i δ
− (27)
Here s
0 and s
1 contain information regarding the amplitudes of linear polarization , whereas s
2 and s
3 say something about the phases .
Knowing these parameters (e.g by passing a wave through perpendicular polarization filters) is sufficient for us to determine the amplitudes and relative phases of the field components.
Electromagnetic Waves

In terms of the linear polarization bases ( ~
+
, ~
−
) , the Stokes parameters are: s
0 s
1 s
2 s
3
= | ~
∗
+
·
~
| 2
+ | ~
∗
−
·
~
| 2
= a
2
+
+ a
2
−
= 2 < h
( ~
∗
+
·
~
)
∗
( ~
∗
−
·
~
) i
= 2 a
+ a
− cos( δ
−
− δ
+
)
= 2 = h
( ~
∗
+
·
~
)
∗
( ~
∗
−
·
~
) i
= 2 a
+ a
− sin( δ
−
− δ
+
)
= | ~
∗
+
·
~
|
2
− | ~
∗
−
·
~
|
2
= a
2
+
− a
2
−
(28)
Notice an interesting rearrangement of roles of the Stokes parameters with respect to the two bases.
The four Stokes parameters are not independent since they depend on only 3 quantities a
1
, a
2 and δ
1
− δ
2
. They satisfy the relation s
2
0
= s
2
1
+ s
2
2
+ s
2
3
.
(29)
Electromagnetic Waves

The reflection and refraction of light at a plane surface between two media of different dielectric properties are familiar phenomena.
The various aspects of the phenomena divide themselves into two classes
I Kinematic properties :
I
I
Angle of reflection = angle of incidence sin i n
0
Snell’s law : = where i , r are the angles of incidence sin r n and refraction, while n , n
0 are the corresponding indices of refraction.
I Dynamic properties :
I
I
Intensities of reflected and refracted radiation
Phase changes and polarization
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The kinematic properties follow from the wave nature of the phenomena and the need to satisfy certain boundary conditions (BC).
But not on the detailed nature of the waves or the boundary conditions.
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The dynamic properties depend entirely on the specific nature of the EM fields and their boundary conditions.
Electromagnetic Waves

Figure: Incident wave
~ strikes plane interface between different media, giving rise to a reflected wave
00 and a refracted wave
0
. The media below and above the plane z = 0 have permeabilities and dielectric constants n = µ
µ , and n and
0
=
µ ’ ,
µ
’
0 0 .
respectively. The indices of refraction are
Electromagnetic Waves

According to eqn (18) the 3 waves are:
INCIDENT
~
= E
0 e i
~
· ~ − i ω t
,
~
=
√
µ
~
× k
REFRACTED
0
= E
0
0 e i k
0
· ~ − i ω t
,
0
= p
µ 0 0
0 × k 0
0
REFLECTED
00
=
00
0 e i k
00
· ~ − i ω t
,
0
= p
µ 0 0
00 × k 00
00
The wave numbers have magnitudes:
|
~
| = | k
00 | = k =
ω √
µ , | k
0 | = k
0 c
=
ω p
µ 0 0 c
(30)
(31)
(32)
(33)
Electromagnetic Waves

AT the boundary z = 0 the BC must be satisfied at all points on the plane at all times , i.e. the spatial & time variation of all fields must be the same at z = 0 .
Thus the phase factors must be equal at z = 0
~
· ~ z =0
=
00 · ~ z =0
=
0 · ~ z =0
(34) independent of the nature of the boundary conditions.
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Eqn (34) contains the
kinematic aspects of reflection and refraction .
Note that all 3 wave vectors must lie in a plane. From the previous figure we get k sin i = k
00 sin r
0
= k
0 sin r (35)
Since k = k
00
, we find that i = r
0
; the angle of incidence equals the angle of reflection.
Snell’s law is: sin i sin r
= k
0 k
= s
µ 0 0
µ
= n
0 n
(36)
Electromagnetic Waves

The dynamic properties are contained in the boundary conditions :
• normal components of
~
= E and
~ are continuous
• tangential components of
~ and
~
= [ c / ( ωµ )]
~
×
~ are continuous
In terms of fields (30)-(32) these boundary conditions at
z = 0 are: h h k × E
0
+
E
0
+ E
0
00
00
× E
00
0
−
−
0
E
0
0 i
· ~ = 0
0
× E
0
0 i
· ~ = 0
1
µ
~
× E
0
+
00
× E
0
00
−
E
0
+
00
0
−
1
µ 0
0
× E
0
0
0
0
× ~ = 0
× ~ = 0
(37)
Two separate situations, the incident plane wave is linearly polarized :
• The polarization vector is perpendicular to the plane of incidence (the plane defined by
~ and n ).
• The polarization vector is parallel to the plane of incidence.
• The case of arbitrary elliptic polarization can be obtained by
appropriate linear combinations of the two results.
Electromagnetic Waves

~
• Since the
~
-fields are parallel to the
surface the 1st BC of (38) yields nothing
•
The 3rd and 4th of of (38) give
( how?
):
E
0
+ E
00
0
− E
0
0
= 0 r
µ
( E
0
− E
0
00
) cos i − s
0
µ 0
E
0
0 cos r = 0 (38)
• The 2nd, using Snell’s law, duplicates the 3rd.
( prove all the above statements )
Figure: Reflection and refraction with polarization perpendicular to the plane of incidence. All the
~
-fields shown directed away from the viewer.
Electromagnetic Waves

~
The relative amplitudes of the refracted and reflected waves can be found
from (38)
E
0
0
E
0
E
E
00
0
0
=
=
2 n cos i n cos i +
µ
µ 0 p n 0 2 − n 2 sin
2 i
=
2
1 +
µ tan i
µ
0 tan r n cos i −
µ
µ
0 p n 0 2 − n 2 sin
2 i n cos i +
µ
µ 0 p n 0 2 − n 2 sin
2 i
=
1 −
1 +
µ tan i
µ
0 tan r
µ tan i
µ 0 tan r
=
=
2 sin sin( i r cos
+ r ) i
µ = µ 0 sin( r − i ) sin( i + r )
(39)
µ = µ 0
Note that p n 0 2 − n 2 sin
2 i = n
0 cos r but Snell’s law has been used to express it in terms of the angle of incidence.
For optical frequencies it is usually permitted to put µ = µ
0
.
Electromagnetic Waves

~
Boundary conditions involved:
• normal
~
: 1st eqn in (38)
• tangential
~
: 3rd eqn in (38)
• tangential
~
: 4th eqn in (38)
The last two demand that
( E
0
− E
0
00
) cos i − E
0
0 cos r = 0 r
µ
( E
0
+ E
00
0
) − s
0
µ 0
E
0
0
= 0 (40)
Figure: Reflection and refraction with polarization parallel to the plane of incidence.
Electromagnetic Waves

~
The condition that normal
~ is continuous, plus Snell’s law, merely dublicates the 2nd of the previous equations.
The relative amplitudes of refracted and reflected fields are therefore
( how?
)
E
0
0
E
0
E
E
00
0
0
=
=
=
=
2 nn
0 cos i
µ
µ 0 n 0 2 cos i + n p n 0 2 − n 2 sin
2 i
1 +
2 n
0 n
0
0 tan i tan r
= sin( i
2 sin
+ r r cos
) cos( i i
− r )
µ
µ 0 n
0 2 cos i − n p n 0 2 − n 2 sin
2 i
µ
µ
0 n 0 2 cos i + n p n 0 2 − n 2 sin
2 i
1 −
1 +
0
0 tan i tan r tan i tan r
= tan( tan( i i −
+ r r
)
)
µ = µ 0
µ = µ 0
(41)
Electromagnetic Waves

i
For normal incidence i = 0
both (39) and (41) reduce to (
how?
)
E
0
0
E
0
E
E
00
0
0
=
=
1 +
2 q
µ 0
µ
0
→ n 0
2 n
+ n
− 1 + q
µ 0
µ 0
1 + q
µ 0
µ 0 n
0 − n
→ n 0 + n
EXERCISES:
What are the conditions for:
I Total reflection
I Total transmision
(42)
Electromagnetic Waves