Feb. 7, 2011
Plane EM Waves
The Radiation Spectrum:
Fourier Transforms
Vector Wave Equations for E and B:

2

1E
2

E
2 2 
0
c
t

2

1B
2

B
2 2 
0
c
t
For solutions to the 3-Dimensional wave equation, use
complex notation



i
(
k

r


t
)
ˆ
E

a
E
e
1
0



i
(
k

r


t
)
ˆ
B

a
B
e
2
0
where
ˆ1,a
ˆ2 are
a
unit vecto
rs

r(x,y,z) Cartesian
position
ector
v
E
complex
constants
0,B
0 are
ˆkn
ˆ wave
k
vector
angular
frequency
Before we go further, let’s review complex numbers
imaginary
Argand Diagram
x+iy
y
r z
x
Complex number
Complex conjugate
z

x

iy

z
x

iy
i 
1
x = real part of z
y = imaginary part of z
real
In polar coordinates
 

x

r
cos
y

r
si
z

x

iy

r
(co

i
si
) n
so
The Euler Formula


e

cos

i
sin
i
implies


z

re

r
cos

ir
sin
i
r = magnitude of z
θ = phase angle of z
Re(z) = real part of z = rcosθ
Im(z) = imaginary part of z = r sinθ
Since exponentials are so easy to integrate and differentiate,
it is convenient to describe waves as


(
z
,
t
)

Ae
i
(
kz

t
)
Where A is a real constant
To get the physically meaningful quantity, which must be a real number,
one solves the wave equation and then takes the REAL part of the
solution.
This is OK, since the wave equation is linear, so that the real part of
Ψ and its imaginary part are each separately solutions.
So for example you can write
 
i
(
k


t
)
z
E

Re
E
e
0


Re
alimaginar
where
E

E
i
E
i
a
s
com
cto
0
0
0


then


r
r real r imaginary
E  Re E 0  iE 0
cos(kz  t)  isin(kz  t)
r real
r imaginary
 E 0 cos(kz  t)  E 0
sin(kz  t)
vector

vector
Solution to the wave equations:



i
(
k

r


t
)
ˆ
E

a
E
e
1
0



i
(
k

r


t
)
ˆ
B

a
B
e
2
0
E
,
B
are
comp
scal
0
0


k

wave
vecto

k
n


The waves travel in direction
n or k

k
surfaces of constant phase travel with time in direction






i
(
k

r


t
)
i
(
k

r


t
)
ˆ
ˆ
B

a
B
e
E

a
E
e
2
0
1
0
These must also satisfy Maxwell’s Equations
r r

i( k  r t )
ˆ

E

0   a1 E 0e
0

Recall the definition of the divergence:



E
E
E
y 
x


E
  z

x 
y
z

ˆ
E

(
a
)
E
e
x


E

x
ˆ

(
a
)E
k
ie

x


i
(
k

r

t
)
1
x0


i
(
k

r

t
)
1
x 0x
So


E

0

ˆ
i
k
a
E
0
1
0



i
(
k

r


t
)
ˆ
E

a
E
e
1
0



i
(
k

r


t
)
ˆ
B

a
B
e
2
0
“Can show” the other equations are


ˆ
ˆ
k
a
B
0
i
k
a
E
0 i
2
0
1
0


i

i

ˆ
ˆ
i
k

a
B

a
E
ˆ
ˆ
i
k

a
E

a
B
2
0
1
0
10
20
c
c


ˆ
ˆ
Since
k

a

0
and
k

a

0
1
2


ˆ
ˆ
both
a
(that
is
E
)
and
a
(that
is
B
)
1
2

are
transverse
to
the
direction
of
propag
n,
k



i
(
k

r


t
)
ˆ
E

a
E
e
1
0



i
(
k

r


t
)
ˆ
B

a
B
e
2
0


ˆ
ˆ
k
a
B
0
i
k
a
E
0 i
2
0
1
0


i

i

ˆ
ˆ
k

a
B
E
ˆE
ˆB i
i
k

a
 a
2
0 a
1
0
10
Also
c
20

E0  B0
kc

E
E
0 
0
k
c
 
2
c

B0  E0
kc
2
22


c
k
Require k>0 and ω>0  ω = c k
Hence,
E0=B0



i
(
k

r


t
)
ˆ
E

a
E
e
1
0



i
(
k

r


t
)
ˆ
B

a
B
e
2
0


ˆ
ˆ
k
a
B
0
i
k
a
E
0 i
2
0
1
0
r

i
i

ˆ
ˆE
ˆB ik  a2B0   aˆ1E 0
i
k

a
 a
10
c
20
c

ˆ10 is
ˆ2
Since
k
a
in
direction
a


ˆ20 is
ˆ1
and
k
a
opposite
a


ˆ1(that
ˆ2(that
a
is
E
)and
a
is
B
)
must
be
perpendicu
lareach
toother
SO


Eand
Bare
always
perpendicu
lar
tothe
direction
of
propagatio
n,
and
always
perpendicu
lareach
to
other
Qualitative Picture: For “one” wave with one λ
In real situations, one wants to consider the superposition of many waves
like this – and the more general case where the direction of E
(and hence B) is random as the wave propagates.
Phase Velocity v. Group Velocity
The speed at which the sine moves is the phase velocity

vphase
 c
k
The group velocity is

vg 
k
This is usually discussed when you have several waves superimposed,
which make a modulated wave:
the modulation envelope travels with the group velocity
In a dispersive medium ω=ω(k) so
However, in a vacuum, vgroup= c

v necessa
y
g
k
Group and Phase Velocities
The Radiation Spectrum
Joseph Fourier
The Radiation Spectrum
The spectrum depends on the time variation of the electric field
(or, equivalently, the magnetic field)
It is impossible to know what the spectrum is, if the electric field is
only specified at a single instant of time. One needs to record the
electric field for some sufficiently long time.
The spectrum (energy as a function of frequency) is related to the
E-field (as a function of frequency) through the Poynting Vector.
The E-field (as a function of frequency) is related to the E-field (as
a function of time) through the Fourier Transform


1  i
E
(

) 
E
(
t
)
etdt
2



Likewise,


1 
i

t
E
(
t
)
 
E
(

)
ed

2

ω = angular frequency


see Bracewell’s book:
FT and Its Applications
Fourier Transforms
A function’s Fourier Transform is a specification of the
amplitudes and phases of sinusoidals, which, when
added together, reproduce the function
Given a function F(x)
The Fourier Transform of F(x) is f(σ)

f(

)
F
(
x
)
e
dx

2

ix



The inverse transform is

F
(
x
)


)
e
f(



2

i
x
dx
note change in sign
Not all functions have Fourier Transforms.
F.T. sometimes called the “power spectrum”
e.g. search for periods in a variable star
Visualizing the Fourier Transform:
Visualizing the F.T
.
Suppose you have a complex function:
e

cos
x

i
sin
x
ix
Recall Euler’s formula:
FT(F(x)) =
F
(
x
)

F
(
x
)

iF
(
x
)
R
I






f(

)

F
(
x
)
cos(
2

x

)
dx

i
F
x
)
cos(
2

x

)
dx
R
I(









i
F
(
x
)
sin(
2

x

)
dx

F
x
)
sin(
2

x

)
dx
R
I(

Notes:
When F(x) is real (FI=0) the fourier transform f(σ) can still be complex.
For fixed σ, these integrals involve multiplying F by a sine (or cosine)
with period 1/σ and summing the area underneath the result.
Changing the frequency of the sines and cosines and repeating the
process gives f(σ) at a second value of σ, and so on.
Some examples
b( ) 
(1) F.T. of box function
 xand

B
(x
)
0for
x

2



1 for

x
2
2
2

2 ix
B(x)e
dx




2
2 ix
e
dx

 2

e
 2
 

2ix 
2 ix
2

1
2 i( 2 )
2 i( 2 )

e
e
2ix
sin( )


  sinc( )

“Ringing” -- sharp discontinuity  ripples in spectrum
When ω is large, the F.T. is narrow:
first zero at

other zeros at
1

 
1

(2)
Gaussian
F.T. of gaussian is a gaussian with narrower width
G
(
x
)
1
2 2

x
/
e





FT

g
(
)

e
2
2
2

Dispersion of G(x)  β
Dispersion of g(σ) 
1

(3)
(x
x
)
1
delta- function

(
x

x
)

0
f
x
or

x
1
1
x
x1

Note:
(x)dx1




(
x

x
)
F
(
x
)
dx

F
(
x
)

(
x

x
)
dx


1



F
(
x
)
1
1


1
FT of thedeltafunction:

f ( )    (x  x1)e
2ix
dx

2ix1
e

 (x  x )dx
1

2ix1
e
Amplitude of F.T. of delta function = 1 (constant with sigma)
Phase = 2πxiσ  linear function of sigma
(4)

F
(
x
)

(
x

x
)

(
x

x
)
1
1
x
-x1
+x1
0


2
cos(
2

x

)



2

ix

1
FT

f
()

e 
e
2
ix
1
1
So, cosine with wavelength
1
x1
transforms to delta functions at +/ x1
(5)

F
(
x
)

(
x

x
)

(
x

x
)
1
1
-x1
x
0
x1


FT

f
(
)

2
i
sin
2
x
)
1
Summary of Fourier
Transforms