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Electromagnetic Waves
May 4, 20101
1
J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Section 7
Electromagnetic Waves
Maxwell Equations
? 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.
? The simplest and most fundamental EM waves are transverse, plane
waves.
In a region of space where there are no free sources (ρ = 0, ~J = 0),
Maxwell’s equations reduce to a simple form given
~ =0 ,
~ ·E
∇
~ =0 ,
~ ·B
∇
~
~ + 1 ∂B = 0
~ ×E
∇
c ∂t
~
~ − µ ∂ E = 0
~ ×B
∇
c ∂t
(1)
~ and H
~ are given by relations
where D
~ = E
~
D
~ = 1B
~
and H
µ
(2)
where is the electric permittivity and µ the magnetic permeability
which assumed to be independent of the frequency.
Electromagnetic Waves
Plane Electromagnetic Waves
Maxwell’s equations can be written as
~ −
∇2 B
~
µ ∂ 2 B
=0
2
c ∂t 2
~ −
and ∇2 E
~
µ ∂ 2 E
=0
2
c ∂t 2
(3)
~ and E
~ obeys a wave equation of
In other words each component of B
the form:
c
1 ∂2u
(4)
∇2 u − 2 2 = 0 where v = √
v ∂t
µ
is a constant with dimensions of velocity characteristic of the medium.
The wave equation admits admits plane-wave solutions:
~
= e i k·~x −iωt
~ (~x , t) = Ee
~ ik~n·~x −iωt
E
u
(5)
~ x , t) = Be
~ ik~n·~x −iωt
and B(~
(6)
where the relation between the frequency ω and the wave vector ~k is
ω 2
ω √ ω
k = = µ
or ~k · ~k =
(7)
v
c
v
~ are constant in time and space.
also the vectors ~n, E~ and B
Electromagnetic Waves
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.
? From the divergence relations of (1) by applying (6) we get
~n · E~ = 0
and
~=0
~n · B
(9)
~ ) and B
~ are both perpendicular to the
~ (or B)
This means that E~ (or E
~
direction of propagation n. Such a wave is called transverse wave.
? The curl equations provide a further restriction
~ = √µ ~n × E~
B
1
~
and E~ = − √ ~n × B
µ
(10)
The combination of equations (9) and (10) suggests that the vectors ~n, E~
~ form an orthonormal set.
and B
~ have the same
Also, if ~n is real, then (10) implies that that E~ and B
phase.
Electromagnetic Waves
It is then useful to introduce a set of real mutually orthogonal unit
vectors (~1 ,~2 , ~n).
In terms of these unit vectors the field
~ are
strengths E~ and B
E~ = ~1 E0 ,
~ = ~2 õE0
B
(11)
or
E~ = ~2 E00 ,
~ = −~1 √µE00
B
(12)
E0 and E00 are constants, possibly complex.
In other words the most general way to write the electric/magnetic field
vector is:
~
E
~
B
=
=
√
(E0 ~1 + E00 ~2 )e ik~n·~x −iωt
(13)
E00 ~1 )e ik~n·~x −iωt
(14)
µ(E0 ~2 −
Electromagnetic Waves
Thus the wave described by (6) and
(11) or (12) is a transverse wave
propagating in the direction ~n.
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
h
i
~ ×H
~∗ = 1 c E
~R × H
~R + E
~I × H
~I + i E
~I × H
~R − E
~R × H
~I
~S = 1 c E
2 4π
2 4π
~ and H
~ are the measured fields at the point where ~S is
where E
evaluated.2
~ because although B
~ is the applied
Note : we use the magnetic induction H
~
induction, the actual field that carries the energy and momentum in media is H.
2
Electromagnetic Waves
The time averaged flux of energy is:
r
~S = c
|E0 |2~n
8π µ
(15)
The total time averaged density (and not just the energy density
associated with the electric field component) is:
1
~ ·E
~∗ + 1B
~ ·B
~ ∗ = |E0 |2
E
u=
(16)
16π
µ
8π
The ratio of the magnitude of (15) to (16) is the speed of energy flow i.e.
√
v = c/ µ. 3 (Prove the above relations)
Project:
What will happen if ~n is not real?
What type of waves you will get?
What will be the form of E?
3
Note: To prove the above relations use hcos2 xi = 1/2 and since
~
~ +E
~ ∗ )/2 we get hE
~ R2 i = E
~ ·E
~ ∗ /2.
ER = ( E
Electromagnetic Waves
Linear and Circular Polarization of EM 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 ~2 and is linearly independent of the first.
The two waves :
~1
E
~
= ~1 E1 e i k·~x −iωt ,
with
~ 2 = ~2 E2 e i~k·~x −iωt
E
(17)
~ ~
~ i = √µ k × Ei , i = 1, 2
B
k
Can be combined to give the most general homogeneous plane waves
propagating in the direction ~k = k~n,
~ (~x , t)
E
~
(~1 E1 + ~2 E2 ) e i k·~x −iωt
h
i
~ (~x , t) = ~1 |E1 | + ~2 |E2 |e i(φ2 −φ1 ) e i~k·~x −iωt+iφ1
E
=
(18)
(19)
The amplitudes E1 = |E1 |e iφ1 and E2 = |E2 |e iφ2 are complex numbers in
order to allow the possibility of a phase difference between waves of
different polarization.
Electromagnetic Waves
LINEARLY POLARIZED
If the amplitudes E1 = |E1 |e iφ1 and E2 = |E2 |e iφ2 have the same phase
(18) represents a linearly polarized wave with the polarization vector
making an angle θ = tan−1 (<(E2 )/=(E1 )) (which remains constant as
the field
p evolves in space and time) with ~1 and magnitude
E = E12 + E22 .
ELLIPTICALLY POLARIZED
If E1 and E2 have the different phase the wave (18) is elliptically
polarized and the electric vector rotates around ~k.
Electromagnetic Waves
Circular Polarization
• E1 = E2 = E0
• φ1 − φ2 = ±π/2 and the wave becomes
~ (~x , t) = E0 (~1 ± i~2 ) e i~k·~x −iωt
E
(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)
Ex (~x , t) = E0 cos(kz − ωt) ,
Ey (~x , t) = ∓E0 cos(kz − ωt)
(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
Elliptically Polarized EM Waves
~ can be given in terms of the
An alternative general expression for E
complex orthogonal vectors
1
~± = √ (~1 ± i~2 )
2
(22)
with properties
~∗± · ~∓ = 0 ,
~∗± · ~3 = 0 ,
~∗± · ~± = 1 .
(23)
Then the general representation of the electric vector
~ (~x , t) = (E+~+ + E−~− ) e i~k·~x −iωt
E
(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 iα ,
~ vector has its axes rotated by an
then the ellipse traced out by the E
angle α/2.
Figure: The figure shows the general case of elliptical polarization and
~ and B
~ at a given point in space.
the ellipses traced out by both E
Note : For r = ±1 we get back to a linearly polarized wave.
Electromagnetic Waves
Polarization
Figure: The figure shows the linear, circular and elliptical polarization
Electromagnetic Waves
Stokes Parameters
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 (E1 , E2 ) 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
~ , ~2 · E
~ , ~∗ · E
~ , ~∗ · E
~
~1 · E
(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:
s0
s1
s2
s3
~ |2 + |~2 · E
~ |2 = a 2 + a 2
= |~1 · E
1
2
2
2
2
~
~
= |~1 · E | − |~2 · E | = a1 − a22
h
i
~ )∗ (~1 · E
~ ) = 2a1 a2 cos(δ1 − δ2 )
= 2< (~1 · E
h
i
~ )∗ (~1 · E
~ ) = 2a1 a2 sin(δ1 − δ2 )
= 2= (~1 · E
(26)
where we defined the coefficients of (18) or (24) as magnitude times a
phase factor:
E1 = a1 e iδ1 ,
E2 = a2 e iδ2 ,
E+ = a+ e iδ+ ,
E− = a− e iδ−
(27)
Here s0 and s1 contain information regarding the amplitudes of linear
polarization, whereas s2 and s3 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
Stokes Parameters
In terms of the linear polarization bases (~+ , ~− ), the Stokes parameters
are:
s2
~ |2 + |~∗ · E
~ |2 = a 2 + a 2
= |~∗+ · E
−
+
−
h
i
∗ ~ ∗ ∗ ~
= 2< (~+ · E ) (~− · E ) = 2a+ a− cos(δ− − δ+ )
h
i
~ )∗ (~∗ · E
~ ) = 2a+ a− sin(δ− − δ+ )
= 2= (~∗+ · E
−
s3
~ |2 − |~∗ · E
~ |2 = a 2 − a 2
= |~∗+ · E
−
+
−
s0
s1
(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 a1 , a2 and δ1 − δ2 . They satisfy the relation
s02 = s12 + s22 + s32 .
Electromagnetic Waves
(29)
Reflection & Refraction of EM 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
I
Angle of reflection = angle of incidence
n0
sin i
Snell’s law: sin
r = n where i, r are the angles of incidence
and refraction, while n, n0 are the corresponding indices of
refraction.
Dynamic properties:
I
I
Intensities of reflected and refracted radiation
Phase changes and polarization
? 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.
? The dynamic properties depend entirely on the specific nature of
the EM fields and their boundary conditions.
Electromagnetic Waves
Figure: Incident wave ~k strikes plane interface between different media,
giving rise to a reflected wave ~k 00 and a refracted wave ~k 0 . The media
below and above the plane z = 0 have permeabilities and dielectric
constants µ, and √
µ’, ’ respectively. The indices of refraction are
√
n = µ and n0 = µ0 0 .
Electromagnetic Waves
According to eqn (18) the 3 waves are:
INCIDENT
~ ~
~ = √µ k × E
B
k
~ =E
~ 0 e i~k·~x −iωt ,
E
(30)
REFRACTED
~0 = E
~ 0 e i~k 0 ·~x −iωt ,
E
0
~0 =
B
~k 0 × E
~0
p
µ0 0
k0
~ 00 = E
~ 00 e i~k 00 ·~x −iωt ,
E
0
~0 =
B
p
(31)
REFLECTED
µ0 0
~k 00 × E
~ 00
k 00
(32)
The wave numbers have magnitudes:
ω√
|~k| = |~k 00 | = k =
µ ,
c
ωp 0 0
|~k 0 | = k 0 =
µ
c
Electromagnetic Waves
(33)
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
~k · ~x
= ~k 00 · ~x
= ~k 0 · ~x
(34)
z=0
z=0
z=0
independent of the nature of the boundary conditions.
? 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:
k0
sin i
=
=
sin r
k
s
µ0 0
n0
=
µ
n
Electromagnetic Waves
(36)
Reflection & Refraction of EM Waves
The dynamic properties are contained in the boundary conditions :
~ = E
~ and B
~ are continuous
• normal components of D
~
~
~ are continuous
• tangential components of E and H = [c/(ωµ)]~k × E
In terms of fields (30)-(32) these boundary conditions at z = 0 are:
h i
~0 + E
~ 00 − 0 E
~ 0 · ~n = 0
E
0
0
h
i
~k × E
~ 0 + ~k 00 × E
~ 00 − ~k 0 × E
~ 0 · ~n = 0
0
0
~0 + E
~ 00 − E
~ 0 × ~n = 0
E
(37)
0
0
1 ~ ~
~ 00 − 1 ~k 0 × E
~ 0 × ~n = 0
k × E0 + ~k 00 × E
0
0
µ
µ0
Two separate situations, the incident plane wave is linearly polarized :
• The polarization vector is perpendicular to the plane of incidence (the
plane defined by ~k 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
~ : Perpendicular to the plane of incidence
E
~ -fields are parallel to the
• Since the E
surface the 1st BC of (38) yields nothing
• The 3rd and 4th of of (38) give
(how?):
E0 + E000 − E00 = 0
s
r
0 0
(E0 − E000 ) cos i −
E cos r = 0 (38)
µ
µ0 0
• 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
E
the viewer.
Electromagnetic Waves
~ : Perpendicular to the plane of incidence
E
The relative amplitudes of the refracted and reflected waves can be found
from (38)
E00
2n cos i
2
2 sin r cos i
p
=
=
=
i
E0
sin(i + r ) µ=µ0
1 + µµ0 tan
n cos i + µµ0 n02 − n2 sin2 i
tan r
p
i
n cos i − µµ0 n02 − n2 sin2 i
1 − µµ0 tan
E000
sin(r − i)
tan r
p
=
(39)
=
=
i
E0
sin(i + r ) µ=µ0
1 + µµ0 tan
n cos i + µ0 n02 − n2 sin2 i
tan r
µ
p
Note that n02 − n2 sin2 i = n0 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 = µ0 .
Electromagnetic Waves
~ : Parallel to the plane of incidence
E
Boundary conditions involved:
~ : 1st eqn in (38)
• normal E
~ : 3rd eqn in (38)
• tangential E
~ : 4th eqn in (38)
• tangential B
The last two demand that
(E0 − E000 ) cos i − E00 cos r = 0
s
r
0 0
00
(E0 + E0 ) −
E =0
µ
µ0 0
(40)
Figure: Reflection and refraction with
polarization parallel to the plane of
incidence.
Electromagnetic Waves
~ : Parallel to the plane of incidence
E
~ is continuous, plus Snell’s law, merely
The condition that normal E
dublicates the 2nd of the previous equations.
The relative amplitudes of refracted and reflected fields are therefore
(how?)
E00
E0
=
=
E000
E0
=
=
2nn0 cos i
p
µ 02
n
cos
i
+
n
n02 − n2 sin2 i
0
µ
0
2 nn0
2 sin r cos i
=
i
sin(i + r ) cos(i − r ) µ=µ0
1 + 0 tan
tan r
p
µ 02
2
02
2
µ0 n cos i − n n − n sin i
p
µ 02
2
02
2
µ0 n cos i + n n − n sin i
i
1 − 0 tan
tan(i − r )
tan r
=
tan i
tan(i + r ) µ=µ0
1 + 0 tan r
Electromagnetic Waves
(41)
Normal incidence i = 0
For normal incidence i = 0 both (39) and (41) reduce to (how?)
E00
E0
E000
E0
=
1+
=
2
q
µ0
µ0 →
2n
n0 + n
q 0
−1 + µ
µ0 n0 − n
q 0 → 0
n +n
1 + µ
µ0 EXERCISES:
What are the conditions for:
I
Total reflection
I
Total transmision
Electromagnetic Waves
(42)