Maxwell's Equations & Light Waves Longitudinal vs. transverse waves Derivation of wave equation from Maxwell's Equations Why light waves are transverse.
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Maxwell's Equations & 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 areMaxwell’s equations.
E is the electric field, B is the magnetic field, is the permittivity, and is the permeability of the medium. As written, they assume no charges (or free space).
E 0
E
B
t
B 0
B
E
t
Derivation of the Wave Equation from Maxwell’s Equations
Take of:
Change the order of differentiation on the RHS:
E
B
t
[
E]
[
B
t]
[
E]
t
[
B]
Derivation of the Wave Equation from Maxwell’s Equations (cont’d)
But:
Substituting for , we have:
Or:
[
E]
t
[ E
t]
[
E] 2
E
t2
[
E]
t[
B]
assuming that and are constant in time.
B
E
t
B
Derivation of the Wave Equation from Maxwell’s Equations (cont’d)
Using the identity,
becomes:
If we now assume zero charge density: = 0, then
and we’re left with the Wave Equation!
[
E] 2
E
t2
(
E) 2
E 2
E
t2
E 0
2E 2
E
t2
Identity: r[
r
rf ]
r(
r
rf ) 2
rf
where µ 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:
Now, in a charge-free medium,
that is,
Eyy
Ezz
Byy
Bzz
0
E0 andÊÊ
B0
Exx
Eyy
Ezz
0 Bxx
Byy
Bzz
0
Exx
0 and Bxx
0Substituting the zero values, we have:
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:
So:
In other words:
And the magnetic field points in the z-direction.
Bt
0,0,Eyx
Bzt
Eyx
Bt
E Ez
yEyz
,Exz
Ezx
,Eyx
Exy
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?
The magnetic-field strength in a light wave
We can integrate: Bz(x,t) Bz(x,0) Eyx
0
t
dt
Take Bz(x,0) = 0
Differentiating Ey with respect to x yields an ik, and integrating with respect to t yields a 1/-i.
Eyr ,t
E0exp i kx t
Start with: Bzt
Eyx
and
Bz(x,t) ik i
E0exp i(kx t)
Bz(x,t)1
c Ey(x,t)
So:
But / k = c:
An Electromagnetic Wave
The electric field, the magnetic field, and the k-vector are all perpendicular:
The electric and magnetic fields are in phase.
E
B
k
snapshot of the wave at one time
The Energy Density of a Light Wave
The energy density of an electric field is:
The energy density of a magnetic field is:
Using B = E/c, and , which together imply that
we have:
Total energy density:
So the electrical and magnetic energy densities in light are equal.
UE
1
2E2
UB
1
2
1
B2
U
B
1
2
1
E2 1
2E2 U
E
U UEU
BE2
c
1
B E
Why we neglect the magnetic field
The force on a charge, q, is:
Taking the ratio of the magnitudesof the two forces:
Since B = E/c:
F q
E q
v
B
Fmagnetic
Felectrical
qvB
qE
Fmagnetic
Felectrical
vc
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.
where is the charge velocity
v
F
electrical F
magnetic
v
B vBsin
vB
The Poynting Vector: S = c2 E x B
The power per unit area in a beam.
Justification (but not a proof):
Energy passing through area A in time t:
= U V = U A c t
So the energy per unit time per unit area:
= U V / ( A t ) = U A c t / ( A t ) = U c = c E2
= c2 E B
And the direction is reasonable. E
B
k
A
c t
U = Energy density
The Irradiance (often called the Intensity)
A light wave’s average power per unit area is the irradiance.
Substituting a light wave into the expression for the Poynting vector,
, yields:
The average of cos2 is 1/2:
S c2
E
B
S (
r ,t) c2
E
0
B
0cos2(
k
r t )
S (
r ,t)
1
T
S (
r ,t ') dt '
t T /2
tT /2
real amplitudes
I (r ,t)
S (
r ,t)
c2 E
0
B
0(1 / 2)
The Irradiance (continued)
Since the electric and magnetic fields are perpendicular and B0 = E0 / c,
I 12
c2 E
0
B
0
I 1
2 cE
0~
2
E
0
2
E
0x E
0x*
E
0 y E
0 y*
E
0z E
0z*
Remember: this formula only works when the wave is of the form:
where:
becomes:
E
r ,t Re
E
0exp i
k
r t
that is, when all the fields involved have the same k
r t
because the real amplitude squared is the same as the mag-squared complex one.
I 12
c E
0
2
or:
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: and
This means that light beams can pass through each other. It also means that waves can constructively or destructively interfere.
2(E1 E
2)
x2
2 E1
x2
2 E2
x2
2(E1 E
2)
t2
2 E1
t2
2 E2
t2
2(E1 E
2)
x2
1
c2
2(E1 E
2)
t2
2 E1
x2
1
c2
2 E1
t2
2 E2
x2
1
c2
2 E2
t2
0
Same polarizations (say ):
The irradiance of the sum of two waves
If they’re both proportional to , then the irradiance is:
I 1
2c
E
0
E
0* 1
2c
E
0x E
0x*
E
0 y E
0 y*
E
0z E
0z*
I 1
2c
E
0xE
0x*
E
0 yE
0 y* I
x I
y
I 1
2c
E
1E
1* 2Re
E
1E
2*
E2E
2*
I I1 c Re
E
1E
2* I
2
Different polarizations (say x and y):
The cross term is the origin of interference! Interference only occurs for beams with the same polarization.
Therefore: Note the cross term!
Intensities add.
exp i(k
r t)
E
0x
E
1
E
2
The irradiance of the sum of two waves of different color
Waves of different color (frequency) do not interfere!
I I1 I
2Different colors: Intensities add.
E
10cos(
k
1r
1t
1)
B
20cos(
k
2r
2t
2)
We can’t use the formula because the k’s and ’s are different.
So we need to go back to the Poynting vector, S (
r ,t) c2
E
B
S (
r ,t) c2
E
1
E
2
B
1
B
2
c2
E
1
B
1
E
1
B
2
E
2
B
1
E
2
B
2
This product averages to zero, as does E
2
B
1
Irradiance of a sum of two waves
I I1 I
2
c ReE
1E
2*
Different colors
Different polarizations
Same colors
Same polarizations
I I1 I
2
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 Very dim Dim
Bright Very bright 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: h or = (h/2)
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!
is the photon flux, or the number of photons/secin a beam.
= P / h
where P is the beam power.
Counting photons tells us a lot aboutthe 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, onthe 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/, 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.
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