•More than 99% of what we know about the universe
comes from observing electromagnetic waves
•Other sources
•The Earth, meteorites
•Samples returned from solar wind, Moon, etc.
•Close up observations of nearby planets
•Neutrinos
E  x, y, z , t   E0 sin  kx  t 
•Waves look like:
B  x, y, z , t   B 0 sin  kx  t 
  ck
•Related by:
•Two independent solutions to these equations:
E y 0  cBz 0
E0  cB0
B0
or
E0
E0
B0
c  3.00 108 m/s
Ez 0  cE y 0
•Note that E, B, and
direction of travel are all
mutually perpendicular
•The two solutions are
called polarizations
•We describe polarization by telling which
way E-field points
This page stolen from PHY 114
Wave Equations summarized:
•The quantity k is called the
wave number
•The wave repeats in time
•It also repeats in space f  1 T
  2 f
k  2
E  E0 sin  kx  t 
B  B 0 sin  kx  t 
•EM waves most commonly described
in terms of frequency or wavelength

c
 2 f
k
2

cf

  ck
This page stolen from PHY 114
Wavelength and wave number
•Different types of waves are classified
by their frequency (or wavelength)
 Increasing
f Increasing
cf
Radio Waves
Microwaves
Infrared
Visible
Ultraviolet
X-rays
Gamma Rays
•Boundaries are arbitrary
and overlap
•Visible is 380-740 nm
Red
Vermillion
Orange
Saffron
Yellow
Chartreuse
Green
Turquoise
Blue
Indigo
Violet
Not these
Know these,
in order
These too
This page stolen from PHY 114
The Electromagnetic Spectrum
Dealing with Waves
Before we tackle electromagnetic waves in 3D,
let’s think about some sort of ordinary waves in 1D
•Like, waves on a string
L
•Fixed boundary conditions: (0) =0, (L)=0
•We want to work in infinite length limit, and find the contribution from all waves
•One way to do this is to make the string finite, then take L   later on
•Though this can be done, it is a bit frustrating because
of the boundary conditions
•You only get standing waves
•Can’t really talk about direction of the wave
To get around this problem, a common trick is
to use periodic boundary conditions
•We demand that:
•the two ends match: (0) = (L),
•and their derivatives match: ’(0) = ’(L),
•This allows waves with a direction to them:
Breaking up complicated waves
•Typical wave equation:   x, t   cos  kx  t  ,
•We need periodic boundary conditions:
  0, t     L, t 
cos  kL  t   cos  t 
•This places a restriction on k
•Cosine and sine have period of 2
•kL must be a multiple of 2
L
kL  2 n
n  0, 1, 2,
•Any solution of wave equation can be written as sum
of waves of this type
•Each component of the wave has an energy E(k)
•The total energy is the sum of the contributions

2

E
n

k

k

Etot   E  k   

L
n 
k

•The energy per unit length is then just: E
1
k
tot
  E  nk  
2
L
L n 

 E  nk 
n 
Taking the Infinite Length Limit:
•We want the energy per unit length for an infinite string:


1


Etot
k
lim   kE  nk  
 lim
E  nk  


L  2
2 k 0  n 
L

n 
E(k)

Etot
dk
•In the limit, this is an integral: L   2 E  k 

•We now want to go to 3D:
•Steps are similar
Etot
d 3k
 
E k 
3
3
L
 2 
u
d 3k
 2 
3
E k 
k
k   k x , k y , kZ 
k x L  2 nx ,
L
k y L  2 n y ,
k z L  2 nz .
L
L
P  E   e E kBT
k B  1.3806 10
23
J/K  8.6173 10
5
eV/K
gravity
The application of statistics to the properties of systems
containing a large number of objects
The techniques of statistical mechanics:
•When there are many possibilities, energy will be
distributed among all of them
•The probability of a single “item” being in a given
“state” depends on temperature and energy
Gas molecules in
a tall box:
This page stolen from PHY 215
Statistical Mechanics
Statistical Mechanics: One Mode of EM Field
•We need E(k) for each value of k
•Quantum Mechanics: Electromagnetic waves have energy:
•The probability of each of these possibilities is:
 n  k BT
P

e
•The probabilities must sum to one:
n
1


C   n  k T
1   Pn  C  e  n  kBT
B
e
n 0
n 0

E   n,
n  0,1, 2,
Pn  Ce n  kBT
n 0
•The average amount of energy in the mode k is then given by:
•These sums can be done:*


 n  k BT

ne
E
k

   k BT
E  k     nPn    n0
e
1
 n  k BT
n 0
 n 0 e
I don’t care about
these details
How to do these sums:
 n  k BT
ne
 n 0

E k   

 n  k BT
e
 n 0

1

1 y
y  1 y  y  y 
•Recall: geometric series:

n 0
•Take derivative of this expression:

•Multiply by y
d  1 
1
n 1
2
3
ny  1  2 y  3 y  4 y   


2

dy
1

y
y
n 0
n

 1  y 
ny


2
1

y
n 0
 
n
•Let: y  en  kBT
•Then:  en  kbT 
n 0
E k   
e
 kb T
 kbT
 n  kbT
ne


,

1  e
1  e
e

1
1  e
2
  kbT
n 0
 kb T

2

1  e
e
 kbT
1  e
 kbT

3
 kbT
  kbT

2

e
 kbT
1
Black Body Radiation
u
d 3k
 2 
3
E k  
E k 

e
 k BT
1
•Now put it all together: Energy density for a thermal distribution of EM fields:
•Wait: There are two polarizations for every wave number k:
•Two modes, each with the same energy
u
d 3k
 2 
3
E k  
4 2
k dk 

e
 2 

2
3
k BT
0
•For light, recall:   ck
•Let x  kc k BT
c  k BT 
u 2
  c 

4 
x3dx
0 e x  1
u
4
15
1

2
1


1

 k dk e
2
0
2

k
2
dk
0

e
 k BT
1
ck
ck k BT
1
k BT 

u
15  c 3

2
4
Units:
J/m3
Black Body Radiation: Spectrum
u
1

2

 k dk e
2
0
  kBT 
u
15  c 3
2
ck
ck k BT
1
4
•This is the correct expression for the total energy density
•Sometimes, we will only measure the density over a fixed range of k
•We want energy density per unit wave number:
du 1
ck 3
•Experimentalists rarely do it this way
 2 ck kBT
dk  e
1
•More common ways of measuring it would be
energy per unit frequency or energy per unit wavelength
•Let’s work it out per unit wavelength to see how this works:
2
dk
2
k  2
k
 2

d

du 16 2
c
 5 2 c kBT 
3
d
 e
1
1
ck
2
du du dk
Fudging
the
 2 ck kBT

1  2
minus sign
d  dk d   e
Wien’s Law:
du 16
 5 2
d
 e
2
c
c k BT 
1
2 c
T 
 2.8978 103 m  K
4.96511k B
2900 K
•The spectrum depends on the temperature:
•We can find the peak wavelength:
4500 K
•Color gives you a good idea
of the temperature
•Colors go from dull red (cool)
to electric blue (hot)
5500 K
10,000 K
20,000 K
Sample Problem:
2 c
T 
 2.8978 103 m  K
4.96511k B
The graph at the right shows the
light received from five stars
(a) Which star is the hottest?
(b) Which two stars have the same
surface temperature
(c) What is the temperature of the
green star?
•The overall size of the curve depends on the size and distance of the star
•The peak/color of the star depends on the temperature
•Red star peaks at smallest wavelength / highest temperature
•Blue and black peak at the same wavelength
2.8978 103 m  K
 4110 K
T
9
•Green curve peak is around 705 nm
705 10 m
Stefan-Boltzman Law
•We have formulas for the energy density
•For stars, we want to know rate at which energy escapes
•Watts per square meter
Units: W/m2
How much power comes out of a hole of area A in a
black body at temperature T?
•Energy density is u
•It is light – moving at velocity c
•Half of it is moving the wrong way (½)
•The half that is moving the right way is only moving
4
2
k
T
  B 
1
partly in the right direction (½)
F  4 uc 
c
3
60
 c
•The resulting total power per unit area (flux):
• is called the Stefan-Boltzmann constant
F  T 4
  5.670 108 W/m 2 /K 4
•We can similarly find the flux per unit wavelength:
d F 4 2
c2
 5 2 c k B T 
d
 e
1
Total Luminosity
•A star can be treated very crudely as a sphere with surface temperature T and radius R
•The total power (called luminosity) coming from a star is:
L  AF
L  4 R 2T 4
•Relative luminosity of
two stars:
2
L  R  T 

  
L
 R  T 
•Real stars don’t follow Planck’s Law exactly
•Nonetheless, we can solve luminosity equation
for T, called the effective temperature
4