Blackbody Radiation Theory
I.
History of Blackbody Radiation
A.
What is a blackbody?
A blackbody is an object that absorbs all radiation that is incident upon it.
When radiation falls upon an object, some of the radiation may be absorbed,
reflected, or transmitted. Most objects that appear black are poor reflectors of
optical radiation but they are not blackbodies since they still reflect some
radiation.
A good approximation of a blackbody is a cavity. In the cube below, the radiation
entering the cavity is continually reflected around the cavity. Thus, the cavity acts
like a minnow trap. It is easy for the light to enter but not exit. This is why a
cavity always looks black at room temperature regardless of the color of the
inside of the cavity.
Front View
B.
Side View of Cavity
Emission Spectra
When a body is at thermal equilibrium, it must re-emit all the radiation that it
absorbs. Blackbodies are interesting because the color of their emission depends
only on their temperature and not upon their shape or composition. Thus, their
emission spectrum can be used as a pyrometer (temperature gauge). The
processing of steel became of great importance during the industrial revolution of
the late 1800's. Thus, blackbody pyrometers were extremely useful tools for
determining the temperature of steel ovens and other processing equipment.
Although light bulb filaments and many other regular objects are not true black
bodies, their temperature can be approximated from their emission spectra by
assuming that the object is a blackbody. Thus, blacksmith's used the color of the
heated steel to determine the processing of steel when making buggy springs,
horse shoes etc.
C.
Experimental Emission Spectra
Blackbody Emission
Energy Density
Higher T
0
0
Frequency
We can see from the graph that the maximum emission frequency shifts to higher
frequency as the temperature is raised.
We can also see that the total energy emitted (area under curve) increases at
increasing temperature.
D.
Stephan-Boltzman Equation
The power density emitted by a blackbody is proportional to the 4th power
of its temperature.
P σ T4
A
σ  5.670x108 W K/m 2
E.
Wien's Displacement Law
The peak of the blackbody emission spectra is inversely proportional to the
temperature.
λ max  2.898 mm  K
T
F.
Ultra-violet Catastrophe
The theory of blackbody radiation would appear to be a straight forward
application of classical electromagnitism and classical physics. Although classical
physics produced a workable theory at low frequency (long wavelengths), it failed
at high frequencies (short wavelengths). Since the theory predicted an infinite
amount of energy radiated at short wavelengths, it was called the "ultra-violet
catastrophe."
II.
Density of States For An E&M Wave In A Blackbody Radiator
Since the emission spectrum of a blackbody radiator is independent of the shape
of the blackbody, we are free to choose the simplest shape possible in deriving the
density of state function. Thus, we choose a cube with sides of length L.
L
Blackbody (3-D Cube )
L
1-D E&M Wave Boundary
(analogous to waves on string)
A.
Wave Equation
We start by writing the wave equation for a wave in free space. We developed this
equation using Maxwell's Equations in PHYS2424.

 1 2 E
2 E  2
0
c  t2
where
2
2
2
 2   î   ĵ   k̂
x 2 y 2 z 2
This equation is a separable differential equation and can be solved by separation
of variables. This is a standard math technique that you will learn later in your
course work.. For now, we are only concerned with the fact that the electric field
can be written as the product of three spatial functions (X,Y, Z) and a time
function as shown below:
E  E o Xx  Yy Zz  e iω t
B.
Using Boundary Conditions
From PHYS2424, we know that the electric field inside a perfect conductor in
electrostatic equilibrium is ZERO!
Thus, we have six boundary conditions that must be imposed upon our standing
wave solution. These six conditions are due to the six metal conductors that form
the faces of our cube.
1)
X(0) = 0
2)
X(L) = 0
3)
Y(0) = 0
4)
Y(L) = 0
5)
Z(0) = 0
6)
Z(L) = 0
By looking at the figure of the standing wave on a string, you will notice that
conditions 1,3, and 5 are satisfied by a sine function. Thus, we have that
X(x) = Sin ( kx x)
Y(y) = Sin ( ky y)
Z(z) = Sin (kz z)
We also see that only certain wave numbers (k's) will satisfy the boundary
conditions 2,4, and 6. The argument of the sine function must change by an
integer number of  radian as the spatial variable changes by L. This corresponds
to the conditions that
k x L  jx π where jx is either 0,1, 2, 3,....
k y L  jy π where jy is either 0,1, 2, 3,....
k z L  jz π where jz is either 0,1, 2, 3,....
Thus, we have that the electric field in one direction can be written as
 jx
E  E o sin

D.
π x   jy π y   jz π z  ω t
sin
sin
e
L   L   L 

Wave Number Vector - k
The individual wave numbers (kx ky, kz) found in the previous section are
components of the three dimensional wave vector as shown below:
kz
kx
ky
The magnitude of this vector, k, is called the wave number and is connected to
the wave length, , by the equation that we developed in PHYS2424:
k  k x 2  k y 2  k z 2  2π
λ
We can now convert this to a connection between j and  by
k kx  ky  kz 
2
2
2
2
π
 
 
L
π
 
 
L
2
j
2
2
 2
2  j 2 

j
 j
y
z 
 x
 2π 
  
 λ 
 2π 
  
 λ 
2
2
j  2L .
λ
Thus, only certain wavelengths exist in the cavity as determined by the magnitude
of the j-vector. We will study this material again when we deal with wave guides
in the junior E&M class.
E.
Density of States Function
We now wish to determine the number of available states with wavelengths such
that j is between some value j to j+dj. A spherical shell of radius j and thickness dj
has a volume of
V  4 π j2 dj
j
However, all components of j must be positive so we are restricted to only 1/8th
of the sphere's volume.
1
π j2 dj
2


V   4 π j dj 
2
8
We must also account for the fact that an E&M wave can have two different
polarization states. Thus, the total number of states with wave numbers between j
and j + dj is
Gjdj  π j2 dj
We now convert this into the number of states in the cavity between the
frequency  and  + d using the relationship c  λν . Substituting the
relationship into our previous results, we have that
j 2L ν
c
dj 2L dν
c
Gν  dν 
2

 4πL

 c2

Gν  dν 
 
 2  2 L 
dν
ν 
  c 

2

8π ν

 c3



 dν


L3
We now divide by the volume of the cube to obtain the density of state function:
gν  dν 
2

8π ν

 c3






dν
DENSITY OF STATES
This is the number of allowed energy states per unit volume of the cavity that
emit radiation between the frequency  and  + d. Everyone used this result in
their work even Max Planck.
III.
Calculating Blackbody Energy Spectra
To calculate the energy per volume emitted by the blackbody, you multiply the
average energy emitted per state by the number of states per volume. Thus, we
have that
u νdν   ε  gvdv
Thus, the problem with the classical result had to either reside in the calculation
of the density of states using Maxwell's electromagnetic theory or the average
energy calculation using classical mechanics (Maxwell-Boltzman distribution /
equipartition theorem).
IV.
Classical Physics (Rayleigh - Jeans Theory)
A.
The atoms of the blackbody radiator are considered to be classical harmonic
oscillators whose energy is given by
ε  1 k x2
2
Noting that a classical harmonic oscillator has two degrees of freedom and using
the equipartition theorem (from Maxwell-Boltzman Statistics), we have the
average energy of the oscillator states as


 ε   2  1 k T   k T
2

Since the oscillators are in thermal equilibrium with their environment, this is also
the average energy of the radiation they emit.
B.
Rayleigh - Jeans Radiation Formula
u ν  dν   ε  gν  dν 
2


8π ν k T 

 dν
3


c


The formula can also be written in terms of wavelength by noting that
ν c
λ
dν  c dλ .
λ2
Substituting this into our Rayleigh - Jeans formula, we obtain
u λ dλ 
2


 8 π c k T   c


 λ 2 c 3   λ 2



dλ  8 π k T dλ

λ4

Thus, the classical theory suggested that energy density was proportional to the
square of the frequency or inversely to the fourth power of the wavelength.
Theory fits the experimental data at low energy but predicts infinite energy as
   or equivalently   0.
C.
Max Planck's Solution
Planck was a former student of Kirchhoff 's and had developed a research
program on applying the Second Law of Thermodynamics to problems at a time
when the law was not widely applied. Planck's application of thermodynamics
convinced him that the average energy calculation was incorrect.
Planck decided that the energy of the harmonic oscillator states shoul be
represented by
ε  n h ν where n is an integer 0, 1, 2, ....
Planck's arguments for this statement are beyond the scope of this course. The
interested student can find a discussion in The Quantum Physicists and an
Introduction to Their Physics by William H. Cooper.
We now calculate the average energy for an energy state using this relationship
between energy and frequency.

 n h ν  e
h ν 
kT 
n


n0
ε 
 n h ν 
 kT 


 e

n 0
The bottom summation can be found in a math handbook and is given by

h ν 
k T 
n
 e


n0
1

h ν 

k T

1 e

We can use the following Calculus trick to help evaluate the top summation
d e n x
  n e n x
dx
Where we will define x to be
x

 n h ν  e
h ν 
kT 
n


n 0

 n h ν  e
n0
h ν 
kT 
n


hν
kT

h ν  n e
h ν 
kT 
n


n 0

 n x 
   n x 
  h ν  de
 h ν d  e
dx
dx 
n0

n  0






 n h ν  e
h ν 
kT 
n


n0

 n h ν  e
h ν 
kT 
d  1 
h ν
dx  1 e  x 
n


n0

 n h ν  e
h ν
h ν 
kT 
n


n0




2

  e - x 
 x   

1 e

1
hν



kT
-
 h νe



 hν  

 kT 
 1 e





2
We now substitute our results back into average energy equation and obtain

ε 
 n h ν  e
h ν 
kT 
n


n0

 e
n 0


 n h ν 
 kT 
hν

 
 kT 


hν e
hν



kT
hν


1 e
e
hν




kT
1
Planck now multiplied the average energy by the classical density of states to
obtain his radiation formula of






2 


hν
8π ν 

 dν
u ν  dν   ε  gν  dν  

3


 c

hν



 


 e  k T  1 


uν  dν 
8 π h ν3
  h ν 
  
c3  e k T 




1


dν
We see that the exponential in the denominator prevents the ultra-violet
catastrophe by causing the intensity to decrease exponentially at high
frequencies.
We also can show that Planck's result give the Rayleigh - Jean's result for low
frequency by expanding the exponential function as follows
e
uν  dν 
h ν


k T


1
hν
kT
for ν 
kT
h
8 π h ν3
8π k T ν2
dν 
dν
3


c
h
ν
c 3 1
1
kT 

Planck was aware that a classical harmonic oscillator should have a continuous
energy distribution. However, he thought that the ultra-violet catastrophe might be
due to a convergence problem with the integration. Thus, he made the energy
levels discrete so that the average energy calculation involved a sum. He
intended to obtain the continuous energy distribution of the classical harmonic
oscillator by letting the parameter h approach zero in his final result. However,
letting h go to zero resulted in the ultra-violet catastrophe. Thus, h had to be left
finite. An excellent fit of experimental blackbody spectra could be obtained using
the value for h of
h 1240 eV  nm
c
This constant is called Planck's constant!!
FACT: Planck was a classical physicist and didn’t believe that this constant had
any great physical significance. In particular, he didn't believe that the oscillator's
energy levels were actually quantized! He had used the wrong energy statistics
(Classical Statistics) with the wrong energy to get the right answer! A young
patent clerk named Albert Einstein had a much different view of the constant and
would make it central to his solution of a problem that would win him the Nobel
prize!