Download Chapter 3 Electromagnetic Theory, Photons, and Light

Lecture 5
Chapter 3
Electromagnetic Theory, Photons,
and Light
Maxwell equations
Gauss’s
 
 B  dS  0
Gauss’s
  1
 E  dS   q
S
S


In matter

 
d  
Faraday’s CE  dl   AB  dS
dt

 

E  
Ampère  dS
B  dl     J  

A
Maxwell’s C
t 

+ 

  fields are defined through interaction with charges
Lorentz force: F  qE  qv  B
Maxwell equations: free space, no charges
Current J and charge  are zero
Integral form of Maxwell equations in free space:
 
no magnetic ‘charges’
SB  dS  0
 
no electric charges
SE  dS  0

 
 dB  
changing magnetic field
E  dl    
 dS 

C
A
creates changing electric

 dt
field

 
E 
changing electric field
CB  dl  0 0 A t  dS
creates changing magnetic
field
There is remarkable symmetry between electric and magnetic fields!
Maxwell equations: differential form
(free space)
 
E  0
 
B  0

 
B
E  
t

 
E
  B  0 0
t
 
 ˆ  ˆ
ˆ


i

j k
Notation:
x
y
z
  2
2
2
2
Laplacian:       2  2  2
x
y
z
 

E x E y E z
  E  div ( E ) 


0
y
x
z
   E z E y   E x E z   E y E x 
  E  



iˆ  
kˆ
 ˆj  
z   z
x   x
y 
 y
 

  E  curl (E )
E E
B
z

y

x
y
z
t
B y
E x E z


t
x
z
E y E x
B
 z

t
y
x
Electromagnetic waves
(free space)

 
B
E  
t

 
E
  B  0 0
t
Changing E field creates B field
Changing B field creates E field
Is it possible to create self-sustaining
EM field?
Can manipulate mathematically into:


2
2


 E
 B
2
2
 E  0 0 2
 B  0 0 2
t
t
Electromagnetic waves


 E
2
 E  0 0 2
t
 
 ˆ  ˆ
j k
  iˆ 
x
x
x
  2
2
2
2
    2  2  2
z
x
y


 B
2
 B  0 0 2
t
2
2
2
2





1



2
Resembles wave equation:    2  2  2  2 2
z
x
y
v t
2
2
2 Ex 2 Ex 2 Ex
2 Ex


 0 0 2
2
2
2
x
y
z
t
 2 Bx  2 Bx  2 Bx
 2 Bx


 0 0 2
2
2
2
x
y
z
t
2Ey
2Ey
 2 By
t
x
x
2

2Ey
y
2

2Ey
z
2
 0 0
2
 2 Ez
 2 Ez  2 Ez  2 Ez


 0 0 2
2
2
2
x
y
z
t
2

 2 By
y
2

 2 By
z
2
 0 0
 2 By
t 2
 2 Bz
 2 B z  2 Bz  2 B z


 0 0 2
2
2
2
x
y
z
t
Each component of the EM field obeys the scalar wave equation,
1
provided that
v
 0 0
Light - electromagnetic wave?
Maxwell in ~1865 found that EM wave must move at speed v 
1
 0 0
At that time permittivity 0 and permeability 0 were known from
electric/magnetic force measurements and Maxwell calculated
v
1
 0 0
 310,740 km/s
Speed of light was also measured by Fizeau in 1849: 315,300 km/s
Maxwell wrote:
This velocity is so nearly that of light, that it seems we have strong reason to
conclude that light itself (including radiant heat, and other radiations if
any) is an electromagnetic disturbance in the form of waves propagated
through the electromagnetic field according to electromagnetic laws.
celer (lat. - fast)
Exact value of speed of light: c = 2.997 924 58 × 108 m/s
Electromagnetic wave
Assume: reference frame is chosen so that E=(Ex,0,0)
longitudinal wave, propagates along x
 
E  0
 
B  0
E x E y E z
0


z
x
y

 
B
E  
t

 
E
  B  0 0
t
E x
0
x
Ex does not vary with x
This cannot be a wave!
Conclusion: it must be transverse wave,
i.e. Ex=0. Similarly Bx=0.
Since E is perpendicular to x, we must
specify its direction as a function of time
Direction of vector E in EM wave is called polarization
Simple case: polarization is fixed, i.e. direction of E does not change
Polarized electromagnetic wave
We are free to chose y-axis so that E field propagating along x is
polarized along y: (0, Ey ,0).
 
E y
Bz
E  0
Bx
Ez E y



 
t
y
z
x
t
B  0
By
Ex Ez



Also: Bx=By=const (=0)
 
B
t
z
x
E  
E y Ex
Bz
t



x
y
t
 
E
  B  0 0
t
E-field of wave has only y component
B-field of wave has only z component
(for polarized wave propagating along x)
In free space,
the plane EM wave is transverse
Harmonic polarized electromagnetic wave
Harmonic functions are solution for wave equation:
E y  x, t   E0 y cos t  x / c    
polarized along y axis
propagates along x axis
Find B:
E y
Bz

x
t
Bz   
E y
x
dt
1
Bz  x, t   E0 y cos t  x / c    
c
E y  cBz
This is true for any wave:
- amplitude ratio is c
- E and B are in-phase
Harmonic polarized electromagnetic wave
Electromagnetic waves
* direction of propagation is in
the direction of cross-product:
 
EB
* EM field does not ‘move’ in
space, only disturbance does.
Changing E field creates
changing B field and vice versa
Energy of EM wave
It was shown (in Phys 272) that field
energy densities are:
0 2 u  1 B2
uE  E
B
2 0
2
Since E=cB and c=(00)-1/2:
uE  uB
- the energy in EM wave is shared equally
between electric and magnetic fields
Total energy: u  uE  uB   0 E 
2
1
0
B 2 (W/m2)
The Poynting vector
EM field contains energy that propagates
through space at speed c
Energy transported through area A in time
t: uAct
Energy S transported by a wave through
unit area in unit time:
E
c2
1
1
uAct
2
 0 EB  EB
S
 uc  c 0 E  c 0 E cB  
 0 0
0
At
John Henry Poynting
(1852-1914)
The Poynting vector:
 1  
S
EB
0
power flow per unit area for a
wave, direction of propagation
is direction of S.
(units: W/m2)
The Poynting vector: polarized harmonic wave
 1  
S
EB
Polarized EM wave:
 
 
E  E0 cos k  r  t
 
 
B  B0 cos k  r  t




Poynting vector:
 
 1  
2
S
E0  B0 cos k  r  t
0

0



This is instantaneous value: S is oscillating
Light field oscillates at ~10 15 Hz most detectors will see average value of S.
Irradiance (used to be called Intensity)
Average value for periodic function:
need to average one period only.
It can be shown that average of cos2 is:
And average power flow per unit time:
Irradiance:
c 0 2
I S T 
E0
2

 
 1  
2
S
E0  B0 cos k  r  t
0



cos 2 t   1 2
T
1
c 0 2
S T 
E0 B0 
E0
2 0
2
Alternative eq-ns:
c
2
I  c 0 E

B2
T
0
For linear isotropic
dielectric:
T
I  v E 2
T
Irradiance is proportional to the square of the amplitude of the E field
Usually mostly E-field component interacts with matter, and we
will refer to E as optical field and use energy eq-ns with E
Optical power  radiant flux  total power falling on some area (Watts)
Spherical wave: inverse square law
Spherical waves are produced by point sources.
As you move away from the source light intensity
drops
Spherical wave eq-n:
A
 r , t   cosk r  vt 
r


 
 
 E0
 B0
E
cos k  r  t B 
cos k  r  t
r
r




 1  E0 B0  2  
   cos k  r  t
S
0  r
r 

I S
T


c 0 1 2

E0
2
2 r
Inverse square law: the irradiance from a point source drops as 1/r2

Radiation pressure
Using classical EM theory Maxwell showed that radiation pressure
equals the energy density of the EM waves:
P u
0
2
E 
2
1
2 0
B
2
S t 
P t  
c
S  uc
This is the instantaneous pressure that would be exerted on a
perfectly absorbing surface by a normally incident beam
Average pressure:
P t  T 
S t  T
c
I

c
(N/m2)
* for reflecting surface pressure doubles
Radiation pressure application
Star wars episode 2
NASA to Launch World's Largest
Solar Sail in November 2014:
Sunjammer
http://www.sunjammermission.com
http://www.livescience.com/32593how-do-solar-sails-work-.html
Example problem
A laser pointer emits light at 630 nm in xy plane at =450 to axis x
(counter clock-wise). The light is polarized along axis z , beam
cross-section is A=1 mm2 and its power is P=1 mW.
1. Write an equation of E and B components of this EM wave for the
region of the beam.
z E
 
 
E  E0 cos k  r  t
y
B
Find :   2  2c 
Find k: k  2   iˆ cos   ˆj sin 
Find E0:
P
c 0 2
x
I
E0
Irradiance:
A
2

E0  2 P Ac 0
E0  2 P Ac 0 k̂




Electric field: E 



 2c 
2P
 2 ˆ
ˆ
i cos   j sin   r 
t
k̂ cos 
Ac 0
 
