Lectures 20,21 (Ch. 32)
Electromagnetic waves
1.
2.
3.
4.
5.
6.
7.
Maxwell’s equations
Wave equation
General properties of the waves
Sinusoidal waves
Travelling and standing waves
Energy characteristics: the Pointing
vector, intensity, power, energy
Generation, transmission and
receiving of electromagnetic waves
Maxwell’s equations
Two Gauss’s laws + Faraday’s law +Amper’s law


q encl
 E  dA  


 B  dA  0


d B
 E  d l   dt
 
d E
 B  d l   ( iencl  dt )
James Clerk Maxwell
(1831 –1879)
Maxwell introduced displacement current, wrote these four equations together,
predicted the electromagnetic waves propagating in vacuum with velocity of light and
shown that light itself is e.m. wave.
1865 Maxwell’s theory of electro-magnetism
1887 Hertz’s experiment
1890 Marconi radio (wireless communication)
Mechanical waves
Transverse waves: oscillation is in the direction
perpendicular to the propagation direction
(waves on the rope, on the surface of water)
Longitudinal waves : oscillation in the direction of
the propagation (sound, spring)
E.M. waves are transverse waves
In mechanical waves there is collective
oscillations of particles.
E and B oscillate in e.m. waves. Matter is not
required. E.M waves may propagate in vacuum.
Wave equation and major characteristics of the wave
 y (t , x )
x
2
1  y (t , x )

v
t
2
2
0
y ( x , t )  A cos(  t  kx )
y ( x  0 , t )  A cos  t
y ( x , t  0 )  A cos kx
2
2
 
,k 
T

   t  kx
  const  d   0
 dt  kdx , v 
dx
dt
v

T


k



k
,   vk
2 
T 2


T
Maxwell’s equations in the absence of charges and currents
take particular symmetric form


 E  d A  0
 B  dA  0


d B
 E  d l   dt
 
d E
 B  d l    dt
Look for solution in the form:
To satisfy Gauss’s laws it is necessary
to have:




E  v, B  v !
If there is a component of E or B
parallel to v Gauss’s laws are not
satisfied . It may be verified
choosing the front of the Gaussian
surface ahead of the wave front.
Faraday’s law:
 Ea   Ba
vdt
 E  vB
dt
Amper’s law:
Ba   Ea
vdt
 B   E
dt
E  v  E  v 
2
2
In vacuum v 
c
v


 0 0

1
 0 0
c
n

,v 
 c  3  10
1

8
m
s
KmK  n
( typically , K m  1  n 
v
1
K , but not always )
( typically n  1  v  c but not always )
Derivation of the wave equation
Look for plane waves: Ey(x,t) and Bz(x,t)
Faraday’s law:
a [ E y ( x   x )  E y ( x )]  
E y
B z
t
 xa
2
B z  E y
 Bz

,

2
x
t
x
xt
2
Amper’s law:
a [  B z ( x   x )  B z ( x )]  
B z
x
  
 Ey
E y
x
E  vB
v
2
1
 Bz

x
xt
1  Ey
  
2

v
2
2
,
2
t
2
1  Bz
0
2

v
2
t
2
t
 xa
 Ey
2
2
t
2
 Bz
E y
0
t
2
E and B in e.m. wave
E y  E 0 cos(  t  kx )
B z  B 0 cos(  t  kx )
or

E 

B 

j E 0 cos(  t  kx )

k B 0 cos(  t  kx )
E y  E 0 cos(  t  kx )
B z   B 0 cos(  t  kx )
or


E  j E 0 cos(  t  kx )


B   k B 0 cos(  t  kx )
This is y-polarized wave. The direction of E oscillations determines polarization of the wave. Do not
confuse polarization of the wave with polarization of dielectric (i.e.separation of charges in E).
The frequency range (spectrum) of e/m. waves
Radio waves, microwaves, IR radiation, light, UV radiation, x-rays and gamma-rays are
e/m waves of different frequencies. All of them propagate in vacuum with v=c=3x108m/s
1

Frequency of e.m.wave does not depend on the medium where it
propagates. It is determined by the frequency of charge oscillations.
T
2
Both the speed of propagation and the wavelength do depend on
1
[ f ]   1 H ( Hertz )

the medium: v=c/n,   vT  v  c ,
s
  0
f 

f
0 
c
f
nf
n
wave length in vacuum
Example. A carbon-dioxide laser emits a sinusoidal e.m. wave that travels in vacuum in the negative x direction. The
wavelength is 10.6μm and the wave is z-polarized. Maximum magnitude of E is 1.5MW/m. Write vector equations for
E and B as functions of time and position. Plot the wave in a figure.


E  k E 0 cos(  t  kx )
 
B  j B 0 cos(  t  kx )
B0 
k 
E0
6

c
2

1 . 5  10 V / m

3  10 m / s
8
2  3 . 17 rad
10 . 6  10
6
 5  10
3
T
 5 . 93  10 rad / m
5
m
  ck  ( 3  10 m / s )  5 . 93  10 rad / m  1 . 78  10 rad / s
8
5
14
)
NB1: Since B=E/c→B (in T) <<E (in V/m
NB2: in general, arbitrary initial phase may be added :


E  k E 0 cos(  t  kx   )


B  j B 0 cos(  t  kx   )
To find initial phase one needs to know either initial conditions E(x,t=0) or
boundary condition E(t,x=0).
•
•
Example. Nd:YAG laser emits IR radiation in vacuum at the wavelength 1.062μm.
The pulse duration is 30ps(picos). How many oscillations of E does the pulse
contain?
T 


1 . 062  10
c
N 
 pulse
T
6
m
3  10 m / s
8

30 ps
3 fs

 3  10
3  10
 12
3  10
 15
s
 15
s  3 fs ( femtos )
 10000
s
The shortest pulses (~100 as (attos),1as=10-18s) obtained today consist of less
then 1 period of E oscillations.They allow to visualize the motion of e in atoms
and molecules.
Ends of string are fixed→nodes on the ends
Max possible wavelength is
determined by the length of string
 max
2
n 
 L   max  2 L  f min 
2L
n
, n  1, 2 ...  f n 
vn
2L
v
 max
 f min n

v
2L
Reflection from a perfect conductor. Standing waves
Total E is the superposition of the incoming and
reflected waves. On the surface of the conductor E
total parallel to the surface should be zero. Perfect
conductor is a perfect reflector with E in ref. wave
oscillating in opposite phase.
E y ( x , t ) in  E 0 cos(  t  kx )
E y ( x , t ) ref   E 0 cos(  t  kx )
cos(    )  cos  cos   sin  sin 
E y ( x , t )  E y ( x , t ) in  E y ( x , t ) ref   2 E sin kx  sin  t
B z ( x , t ) in   B 0 cos(  t  kx )
B z ( x , t ) ref   B 0 cos(  t  kx )
B z ( x , t )  B z ( x , t ) in  B z ( x , t ) ref   2 B 0 cos kx cos  t
E(x)=0 at arbitrary moment of time in the positions
where sinkx=0, that is kx=πn, n=0,1,2,3,..
 x
n
, n  0 ,1, 2 ,..
2
x  0,

2
,,
3
2
, 2  ,...
( nodal planes of E )
If two conductors are placed parallel to each other the nodes of E should be
on the ends just as on the string with fixed ends
 max
2
n 
 L   max  2 L  f min 
2L
n
, n  1, 2 ...  f n 
vn
2L
v
 max

v
2L
 f min n
Example.In a microwave oven a wavelength 12.2cm (strongly absorbed by a water) is used.
What is the minimum size of the oven? What are the other options? Why in the other options
rotation is required?
L min 

 6 . 1cm
2
L  12 . 2 cm one node in the middle
L  18 . 3 cm ...
The Energy Characteristics of e.m. waves
The energy density:
u el 
u 
E
E
2
2
2

2
, u mag 
B
B
2
2
2
B
 E 
2
2
1
, E  vB , v 
2





EB
The Poynting vector is the energy transferred per unite time per unite
cross-section, i.e. power per unite area=the energy flow rate in the
direction of propagation
S 
P

1 dU
, dU  udV  uvdtA
A
A dt
 
 EB

EB
S 
EBv 
, S 



Intensity is the power per unite area averaged over the period of oscillations
For travelling waves:
I 
E 0 B0

E rms 
cos ( t  kx ) 
E
2
2
 E0
2
(1  cos 2 ( t  kx ) ) 
cos ( t  kx ) 
2
 
P   S  dA, U 
A
E 0 B0
E0
2
 Pdt
T
, I 
E 0 B0
E RMS B RMS

2
I 
E 0 B0
2
[S ]  [I ] 
J
2
m s

W
m
2
Standing waves do not transfer the energy:
I 
4 E 0 B0

sin kx cos kx sin  t cos  t 
2 E 0 B0

sin kx cos kx sin 2  t  0
Example. The distance from the sun to the earth is 1.5x1011m.1) What is the power
of radiation of the sun if it’s intensity measured by the earth orbiting satellite is 1.4
kW/m. 2) If the area of the panels of the satellite is 4m and is perpendicular to the
radiation of the sun, what is the power received by satellite?
Psun
 
2
3 W
2
22
2
26
  S  d A  4  R I  1 . 4  10

4

3
.
14

(
1
.
5
)

10
m

4

10
W
2
m
NB: the life on the earth is due to this power of radiation received from the
sun!
Ppanels
 
3 W
2
  S  d A  IA  1 . 4  10

4
m
 5 . 6 kW
2
m
Example
A radio station on the surface of the earth radiates a sinusoidal wave with an average total power 50kW.
Assuming that transmitter radiates equally in all directions, find the amplitudes of E and B detected by a
satellite at a distance 100km.
I 
I 
P
2 R
2
E 0 B0
E0 
B0 
20


5  10
4
6 . 28  10 m
10
E0
c
 7 . 96  10
2
2  0c
2  0 cI  2 . 5  10
E0
2
 8 . 2  10
 11
T
2
V /m
7
W
m
2
E.m. waves are produced by oscillating charge or current

E ~ B ~
sin 
r
,I ~ (
sin 
r
)
2
v
Richard Feynman ( 1918 – 1988)
Optimal size of antenna~λ/2
Optimal position of antenna
(maximizing the induced current in
antenna) corresponds to the wire
parallel to E
Optimal position of antenna
(maximizing the induced current in
antenna) corresponds to the loop
perpendicular to B.
Radiation Pressure
p 
U
EMW carry both energy and
momentum
c
F
Prad 
F 

A
AI
Absorbing plane

S

p

S

S
F 
p

t
Prad 
p
t
I
F 
p
t
Prad 

2I
c
2p
t
1 dU

Ac dt
I
c
P

c
N
m
2
 1 Pa , 1atm  10 Pa
5
Example. Find the force due to a radiation pressure on the
solar panels. I=1.4kW/m2,A=1m2.
c
Reflecting plane

A dt
c
[ Prad ] 
1 dp
Prad 
I
c
1 . 4  10 W / m
3

3  10 m / s
F  Prad A  5  10
8
6
2
 5  10
6
Pa
N
However over long time it influences the satellite orbit!
Comet tails, some stars formation
r
Laser cooling
Nobel Prize,1997
mv
2
~ kT :
2
Steven Chu,Claude Cohen –Tannoudji,Bill Phillips
T ~ 300 K  v ~ 1
km
atom
atom
E2
E2
Photon
E   , p  k
h
2
E1
Photon
 atom
 
K  v ~1
, h  6 . 626  10
 24
kv
 laser
E1
J s
mm
s
Photon
atom
Photons:
7
s
atom
Photon
, T ~ 10
Polarization
Dichroism (dependence of absorption on polarization) is used
for construction of the polarization filters for em waves
A grid of wires is a polarization flter for radio waves
When E in a radio wave is parallel to the wires
the currents are induced in the wires and wave is
absorbed.
Long molecules play a role of wires for light and
used for building of polarization filters (polaroids)
Linear polarized, namely,
y-plz e.m.wave
Axis of the filter. If em wave is polarized along this axis it goes
through without asborption. Linear plz em wave with orthoginal
to this axis in not transmitted (fully absorbed by the filter).
Malus’s law (1809)
Eout
In general case when linear plz wave goes through the filter
only its projection on the axis of the filter goes through.

Ein
E out  E in cos 
I out  I in cos 
2
Unpolarized em wave (random
polarization)
I out  I in cos  
2
I in
2
NB: After the filter em wave
is always linear polarized
along the axis of the filter
Sun, lamp and other
thermal sources
produce unpolarized
light
.
How to check polaroid glasses?
Crossed polaroids do not transmit light
Circular polarization
E y  E 0 cos(  t  kx )
y
Ey
Ez
Ey
E z   E 0 sin(  t  kx )


E t  E 0 [cos(  t  kx ) j  sin(  t  kx ) k ]
x
z
Ez
Left circular
polarization
If E oy  E ox
elliptic polarization
Birefrigent materials: refractive index
depends on polarization: n (  ), n (  )
1
x
x
( k1  k 2 )  x 

c
( n1  n 2 )  x 

2
2