Transcript Waves EM Maxwell Eqn

Wave Dispersion
EM radiation
Maxwell’s Equations
1
Wave Dispersion
- Simply stated, a dispersion relation is the function ω(k) for an
harmonic wave.
- A dispersion relation connects different properties of the wave
such as its energy, frequency, wavelength and wavenumber.
- From these relations the phase velocity and group velocity of the
wave can be found and thereby refractive index of the medium can
be determine.
2
Non dispersive: All colours moving with same speed
t0
t1
t2
Dispersive: Red moving faster than blue
3
Normal dispersion of visible light
Shorter (blue) wavelengths refracted more than long (red) wavelengths.
Refractive index of blue light > red light.
A medium in which phase velocity is frequency dependent is
known as a dispersive medium, and a dispersion relation
expresses the variation of  as a function of k.
If a group contains number of components of frequencies
which are nearly equal, then:
Group velocity
Phase and Group velocity
vg
P’
6
Non dispersive waves
vp = Constant
Signal is propagated without distortion
(or k)
More generally vp is a function of
  vpk
7
Usually,
dv p
is positive, so that vg < vp
d
dv p
When,
is negative, so that vg > vp
d

When,
is constant, so that vg = vp
k
Normal Dispersion
Anomalous Dispersion
Non-Dispersive medium
(Ex: Free space)
More in electromagnetic waves
Wave packet
(without Dispersion)
Wave packet
(with Dispersion)
Non-dispersive
Dispersive
Wikipedia.org
Wave Packets
Superposition of waves and wave packet formation
y(t) = Sin t
0 < t < 200
12
y(t) = [Sin t + Sin (1.08 t)]/2
0 < t < 200
13
Suppose we have group of many frequency components
lying within the narrow frequency range  …
y(t) = [Sin t + Sin(1.04 t) + Sin (1.08 t)]/3
0 < t < 400
15
y(t) = [Sin t + Sin(1.02 t) + Sin (1.04 t)
+ Sin(1.06 t) + Sin (1.08 t)]/5
0 < t < 400
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y(t) = [Sin t + Sin(1.01 t) + Sin (1.02 t)
+ Sin(1.03 t) + Sin (1.04 t) + Sin (1.05 t)
+ Sin (1.06 t) + Sin (1.07 t)
+ Sin (1.08 t)]/9
0 < t < 800
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y(t) = [Sin t + Sin(1.01 t) + Sin (1.02 t)
+ Sin(1.03 t) + Sin (1.04 t) + Sin (1.05 t)
+ Sin (1.06 t) + Sin (1.07 t)
+ Sin (1.08 t)]/9
0 < t < 400
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http://en.wikipedia.org/wiki/Coherence_%28physics%29
Electromagnetic Radiation
Let‘s first develop the understanding by taking the
example of oscillating charge and/or dipole
oscillator
21
For stationary charges
the electric force field
1
 2
r
Coulomb’s law
© 2005 Pearson Prentice Hall, Inc
© 2005 Pearson Prentice Hall, Inc
Coulomb’s law
What is the electric field produced at a point P by a charge q
located at a distance r?
  q eˆr
E
2
4  0 r
where er is an unit vector from P to the position of the charge
© 2005 Pearson Prentice Hall, Inc
If a charge moves non-uniformly, it radiates
The electric field of a moving point charge
http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge
.html
Electric field
q
E t  
4 0
 er  r  d  er   1 d

 r 2  c dt  r 2   c 2 dt 2 er  
 


2
er  :unit vector directed from q to P at earlier time
q
-
P
The correct formula for the electric field
_
Important features
1. No information can propagate instantaneously
2. The electric field at the time t is determined by the position
of the charge at an earlier time, when the charge was at r’,
the retarded position.
3. First two terms falls off as 1/r’2 and hence are of no interest
at large distances
Correct Expression
(at large distances)
This is electro-magnetic radiation or simply radiation.
It is also to be noted that only accelerating charges produce
radiation.
Electric Dipole Oscillator
© SPK/SB
Radio-wave transmission
TV Antenna
Car Antenna
“Let there be electricity and
magnetism and there is light”
J.C. Maxwell
Vector Analysis
(Refresh)
GRADIENT
For a scalar function T of three variable T(x,y,z), the gradient of T is a
vector quantity given by:
T
T
T
T 
xˆ 
yˆ 
zˆ
x
y
z
- The gradient points in the
direction of the greatest rate of
increase of the function,
- and its magnitude is the slope
(rate of increase) of the graph
in that direction.
DIVERGENCE
For a vector T the divergence of T is given by:
 

 
  T   xˆ 
yˆ  zˆ   Tx xˆ  Ty yˆ  Tz zˆ 
y
z 
 x
 Tx Ty Tz 

 


y
z 
 x
It is a measure of how much the
vector T diverges / spreads out
from the point in question.
CURL
For a vector T the Curl of T is given by:
 


  T   xˆ 
yˆ 
y
z
 x
 xˆ




 x
T
 x
yˆ

y
Ty

zˆ   Tx xˆ  Ty yˆ  Tz zˆ 

zˆ 

z 
Tz 
It is a measure of how much the vector T curls around the point in
question.
DIVERGENCE THEOREM
/ Green’s Theorem / Gauss’s Theorem

 
(


E
)
d


E

d
a


V
S
Integral of a derivative (in this case the divergence) over a volume is
equal to the value of the function at the surface that bounds the
volume.
 (Faucets within th e volume)   (Flow out throug h the surface)
V
S
STOKES’ THEOREM
 
 
(


E
)

d
a

E

d
l


S
P
Integral of a derivative (in this case the curl) over a patch of surface is
equal to the value of the function at the boundary (perimeter of the
patch).
What we know from previous classes?
1) Oscillating magnetic field generates electric field
(Faraday´s law) and vice-versa (modified Ampere´s
Law).
2) Reciprocal production of electric and magnetic
fields leads to the propogation of EM waves with
the speed of light.
Question: WAVES?????? How do we show that a
wave is obtained?
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Our Attempt:
To derive the relevant wave equation
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Consider an oscillating electric field Ey
y
Ey
This will generate a
magnetic field along the zaxis
x
Bz
z
April 6, 2016
42
Faraday’s Law
Y
Ey(x)
Ey(x+x)
The induced electromotive force in any closed
circuit is equal to the negative of the time rate of
change of the magnetic flux through the circuit.
BA
Voltage generated   N
t
C
x
N: Number of turns
B: External magnetic field
A: Area of coil
We know that Faraday´s law in the integral form in given as:
Z

 E.dl    B.ds
t
C
s
where C is the rectangle in the XY plane of length l width x, and S is the open surface
spanning the contour C
April 6, 2016
43
Using the Faraday´s law on the contour C, we get:
 E.dl  
C

 B.ds
t
s


Bz
E y ( x  x)  E y ( x) l  
lx
t
this implies...
E
B

x
t
y
z
Keep this in mind...
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We know that the Ampere´s law with displacement current term can
be written as:

 B.dl     E.ds
t
o
C
o
Y
Ey
s
x
C/
x
Bz(x)
z
Bz(x+x)
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Using the Ampere´s law, for the Contour C/, we get:

 B.dl     E.ds
t
o
o
C
s
E
 B ( x  x)  B ( x)l   
lx
t
y
z
z
0
0
this implies...
E
B

 
x
t
z
0
y
0
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Using the eq. obtained earlier i.e.,
E
B

 
x
t
E
B

x
t
y
z
z
0
 Ey
2
t
y
0
 Ey
2
c
2
2
x
2
where
c 
2
1
 0 0
Form of wave
equation
Note: Similar Equation can be derived for Bz
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Solution of EM Wave equation
Electromagnetic waves
for E field
for B field
In general,
electromagnetic waves
1 
  2
2
c t
Where  represents E or B
2
2
or their components
# A plane wave satisfies wave
equation in Cartesian coordinates
# A spherical wave satisfies wave
equation in spherical polar
coordinates
# A cylindrical wave satisfies wave
equation in cylindrical coordinates
Solution of 3D wave equation
In Cartesian coordinates
  

1
 



2
2
2
2
2
x
y
z
c t
2
2
2
2
2
Separation of variables
 ( x, y, z, t )  X ( x)Y ( y ) Z ( z )T (t )
Substituting for  we obtain
2

1  X 1Y1  Z  1 1T
2
2
2
2 
2 
X x
Y y
Z z
c  T t 
2
2
2
Variables are separated out
Each variable-term independent
And must be a constant
So we may write
1  X  k 2 ; 1  Y  k 2 ;
x
y
2
2
X x
Y y
2
2
2

1  Z  k ; 1  T    2

2
2 
Z z
 T t 
2
2
z
where we use
 c  k k k  k
2
2
2
x
2
y
2
z
2
Solutions are then
X ( x)  e
Z ( z)  e
 ik x x
 ik z z
; Y ( y)  e
; T (t )  e
 ik y y
;
 i t
Total Solution is
 ( x, y, z, t )  X ( x)Y ( y ) Z ( z )T (t )
i[t  ( k x x  k y y  k z z )]
 
i[t  k .r ]
 Ae
 Ae
plane wave
Traveling 3D plane wave
Spherical coordinates (r, θ, φ):
radial distance r,
polar angle θ (theta), and
azimuthal angle φ (phi)
spherical waves
x  r sin  cos 
y  r sin  sin 
z  r cos 
Spherical waves

2 
1

cos   1 
  2
 2 2
 2
 2 2
2
r r r r sin   r sin   r 
2
2
2
2
 f    f
2
Alternatively
2

 2  1   2  
2
  2 
 2 r

r
r r r r  r 
The wave equation becomes
1   r 2 

2
r r  r
 1 
 2 2
 c t
2
1   r 2 

r 2 r  r
2
 1 
 2 2
 c t
u (r )
Put  ( r ) 
r
 1 u u
2 
Then

 2  r
 r u  u
r r r r
r
r

2 

Hence
r
r  r
   u 
  r u

 r  r
2
 u  r  u2  u
r
r r
1   r 2 

r 2 r  r
 (r ) 
Therefore
1   r 2 

2
r r  r
 1  2u

2
 r r
Wave equation transforms to
1 u  1 1 u 
2
2
2
r r
c r t
2
2
 1 
 2 2
 c t
2
u 1 u
2
2
2
r
c t
2
2
u (r )
r
Separation of variables
u (r , t )  R (r )T (t )
Which follows that
1  R  1 1  T  k 2
2
2
2
R r
c T t
2
2
Solutions are
 ikr
R(r )  e ; T (t )  e
Total solution is
u (r )  e
 i t
i ( t  kr )
  kc
Final form of solution
i ( t  kr )
1
 (r )  e
spherical wave
r
General solution
i ( t  kr )
i ( t  kr )
1
1
 (r )  e
 e
r
r
outgoing
waves
incoming
waves
Cylindrical waves
Cylindrical Coordinate Surfaces(ρ, φ, z).
The three orthogonal components, ρ
(green), φ (red), and z (blue), each
increasing at a constant rate. The point is
at the intersection between the three
coloured surfaces.
1 
  2
2
c t
2
2

1  1 

  2
 2 2 2
r r r r  z
2
2
2
2
with angular and azimuthal symmetry, the Laplacian simplifies
and the wave equation
1   
r
r r  r
 1 
 2
2
 v t
2
The solutions are Bessel functions.
For large r, they are approximated as
A
 (r , t ) 
cos( kr  t )
r
Maxwell’s equations
II
Use B in Divergence Theorem

 
 (  E )d   E  da
V
S
No magnetic monopoles
III
Use E in Stokes’ Theorem
 
 
 (  E ) da   E  dl
S
From Faraday’s Law
P
IV
Use B in Stokes’ Theorem
From Ampere’s Law
 
 B  dl   o I
C
Charge conservation is a fundamental law of Physics which is
written as a continuity equation
IV
Maxwell’s equations
Plane EM waves in vacuum
Wave vector k is perpendicular to E
Wave vector k is perpendicular to B
B is perpendicular to E
B, k and E make a right handed
Cartesian co-ordinate system
Plane EM waves in vacuum