Download 3. Maxwell`s Equations, Light Waves, Power, and Photons

Maxwell's Equations and Light Waves
Longitudinal vs. transverse waves
Derivation of wave equation from Maxwell's Equations
Why light waves are transverse waves
Why we neglect the magnetic field
Photons and photon statistics
The equations of optics are
Maxwell’s equations.
 E  0
 B  0
B
 E  
t
E
  B  me
t
where E is the electric field, B is the magnetic field,
e is the permittivity, and m is the permeability of the
medium.
As written, they assume no charges (or free space).
Derivation of the Wave Equation
from Maxwell’s Equations
Take   of:
B
 E  
t
B
  [  E]    [ ]
t
Change the order of differentiation on the RHS:

  [  E]   [  B]
t
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
But:
E
  B  me
t
Substituting for  B , we have:

E

[ E] [ B]    [  E]   [ me ]
t
t
t
Or:
 E
  [  E]   me 2
t
2
assuming that m
and e are constant
in time.
Derivation of the Wave Equation
from Maxwell’s Equations (cont’d)
Identity:
r
r r
r r r
r
2
  [  f ]  (  f )   f
Using the identity,
2E

[ E]  me 2
t
becomes:
2E

2
(E) E   me 2
t
If we now assume zero charge density: r = 0, then
E  0
and we’re left with the Wave Equation!
2 E  me  E
t 2
2
where µe  1/c2
Why light waves are transverse
Suppose a wave propagates in the x-direction. Then it’s a function
of x and t (and not y or z), so all y- and z-derivatives are zero:
Ey Ez By Bz



0
y z y z
Now, in a charge-free medium,
that is,
E 0 andΚΚB0
Ex  Ey  Ez 0 Bx  By  Bz 0
x y z
x y z
Substituting the zero
values, we have:
Ex 0 and Bx 0
x
x
So the longitudinal fields are at most constant, and not waves.
The magnetic-field direction in a light wave
Suppose a wave propagates in the x-direction and has its electric field
along the y-direction [so Ex = Ez= 0, and Ey = Ey(x,t)].
What is the direction of the magnetic field?
Use:








x



E y Ex Ez E y E
E
B
z
  E 

,

,

t
y z z x x y
So:
In other words:







E y 
B
  0,0,
t
x 

E y
B
z


t x
And the magnetic field points in the z-direction.
The magnetic-field strength in a light wave
Suppose a wave propagates in the x-direction and has its electric
field in the y-direction. What is the strength of the magnetic field?
Start with:
E y
B
z


t x





and E r,t  E0 exp i kx w t

y 









 

t
We can integrate:
Ey
Bz (x,t)  Bz (x,0) 
dt
x
0
Take Bz(x,0) = 0
So:


Bz (x,t)  ik E0 expi(kxwt)
iw


But w / k = c:
Bz (x,t) 1c Ey(x,t)
Differentiating Ey with
respect to x yields an ik,
and integrating with
respect to t yields a 1/-iw.
An Electromagnetic Wave
The electric and magnetic fields are in phase.
snapshot of the
wave at one time
The electric field, the magnetic field, and the k-vector are all
perpendicular:
E  B k
The Energy Density of a Light Wave
The energy density of an electric field is:
The energy density of a magnetic field is:
1
Using B = E/c, and c 
we have:
em

1 2
UE  eE
2
11 2
UB 
B
2m
, which together imply that
B  E em

11 2
1 2
UB 
E em  e E  U E
2m
2
Total energy density:
U  U E  U B  e E2
So the electrical and magnetic energy densities in light are equal.
Why we neglect the magnetic field
The force on a charge, q, is:
Felectrical Fmagnetic
F  qE  qv  B
Taking the ratio of
the magnitudes
of the two forces:
Since B = E/c:
Fmagnetic
Felectrical
Fmagnetic
Felectrical
qvB

qE
where v is the
charge velocity
v  B  vBsin 
 vB
v

c
So as long as a charge’s velocity is much less than the speed of light,
we can neglect the light’s magnetic force compared to its electric force.
The Poynting Vector:
S = c2 e E x B
The power per unit area in a beam.
U = Energy density
A
Justification (but not a proof):
Energy passing through area A in time Dt:
=
U V
=
U A c Dt
So the energy per unit time per unit area:
c Dt
=
U V / ( A Dt ) = U A c Dt / ( A Dt ) = U c = c e E2
=
c2 e E B
And the direction E  B  k is reasonable.
The Irradiance (often called the Intensity)
A light wave’s average power
per unit area is the irradiance.
1
S(r ,t) 
T
tT /2

S(r ,t ') dt '
tT /2
Substituting a light wave into the expression for the Poynting vector,
S  c 2 e E  B, yields:
real amplitudes
S(r ,t)  c 2 e E0  B0 cos 2 (k  r  w t   )
The average of cos2 is 1/2:
 I(r ,t) 
S(r ,t) 
 c 2 e E0  B0 (1/ 2)
The Irradiance (continued)
Since the electric and magnetic fields are perpendicular and B0 = E0 / c,
I

1
2
c e E0  B0
2
I 
or:
becomes: I
1 ce E
0
2
~
2

1
2
c e E0
2
because the real amplitude
squared is the same as the
mag-squared complex one.
where:
2
E0 
E0 x E0*x  E0 y E0*y  E0 z E0*z
Remember: this formula only works when the wave is of the form:
 


E r ,t  Re E0 exp  i k  r  w t 
that is, when all the fields involved have the same
k  r  wt
Sums of fields: Electromagnetism is linear,
so the principle of Superposition holds.
If E1(x,t) and E2(x,t) are solutions to the wave equation,
then E1(x,t) + E2(x,t) is also a solution.
Proof:
 2 (E1  E2 )  2 E1  2 E2


2
2
x
x
 x2
and
 2 (E1  E2 )  2 E1  2 E2
 2 
2
t
t
t 2
 2 (E1  E2 ) 1  2 (E1  E2 )   2 E1 1  2 E1    2 E2 1  2 E2 
 2
 2  2
 2  2
0
2
2
2 
2 
x
c
t
c  t    x
c  t 
  x
This means that light beams can pass through each other.
It also means that waves can constructively or destructively interfere.
The irradiance of the sum of two waves
If they’re both proportional to
exp  i(k  r  w t)  , then the irradiance is:
I  12 ce E0  E0*  12 ce  E0 x E0 x *  E0 y E0 y *  E0 z E0 z * 
Different polarizations (say x and y):
I  12 ce  E0 x  E0 x *  E0 y  E0 y *   I x  I y
Same polarizations (say E0x
Intensities
add.
 E1  E2):
 
I  I  ce Re E  E  I
I  12 ce  E1  E1*  2 Re E1  E2*  E2  E2* 
Therefore:
*
1
1
2
2
Note the
cross term!
The cross term is the origin of interference!
Interference only occurs for beams with the same polarization.
The irradiance of the sum of two waves
of different color
We can’t use the formula because the k’s and w’s are different.
So we need to go back to the Poynting vector, S (r ,t)  c
2
e EB
S (r ,t)  c 2e  E1  E2    B1  B2 
 c 2e
E  B
1
1
 E1  B2  E2  B1  E2  B2
E10 cos(k1  r  w1 t  1 )  B20 cos(k2  r  w 2 t  2 )
This product averages to zero, as does
Different colors:
I  I1  I 2
E2  B1
Intensities add.
Waves of different color (frequency) do not interfere!

Irradiance of a sum of two waves
Same polarizations
Same
colors
Different
colors
I  I1  I 2 

ce Re E1  E2
I  I1  I 2
*

Different polarizations
I  I1  I 2
I  I1  I 2
Interference only occurs when the waves have the same color and
polarization.
Light is not only a wave, but also a particle.
Photographs taken in dimmer light look grainier.
Very very dim
Bright
Very dim
Very bright
Dim
Very very bright
When we detect very weak light, we find that it’s made up of particles.
We call them photons.
Photons
The energy of a single photon is: hn or
w
= (h/2p)w
where h is Planck's constant, 6.626 x 10-34 Joule-seconds
One photon of visible light contains about 10-19 Joules, not much!
F is the photon flux, or
the number of photons/sec
in a beam.
F = P / hn
where P is the beam power.
Counting photons tells us a lot about
the light source.
Random (incoherent) light sources,
such as stars and light bulbs, emit
photons with random arrival times
and a Bose-Einstein distribution.
Laser (coherent) light sources, on
the other hand, have a more
uniform (but still random)
distribution: Poisson.
Bose-Einstein
Poisson
Photons have momentum
If an atom emits a photon, it recoils in the opposite direction.
If the atoms are excited and then emit light, the atomic beam spreads
much more than if the atoms are not excited and do not emit.
Photons—Radiation Pressure
Photons have no mass and always travel at the speed of light.
The momentum of a single photon is: h/l, or
Radiation pressure = Energy Density
(Force/Area = Energy/Volume)
When radiation pressure cannot be neglected:
Comet tails (other forces are small)
Viking space craft (would've missed
Mars by 15,000 km)
Stellar interiors (resists gravity)
PetaWatt laser (1015 Watts!)
k
Photons
"What is known of [photons] comes from observing the
results of their being created or annihilated."
Eugene Hecht
What is known of nearly everything comes from observing the
results of photons being created or annihilated.