STATISTICAL MECHANICS
PD Dr. Christian Holm
PART 5-6 Some special topics, Thermal Radiation, and
Plank distribution
Equipartition theorem
• Let Xi be Pi or ri
• Generalized equipartition theorem
Special case:
• For every quadratic degree of freedom Xi
in the partition function, with an energy
constitution
• We have
• “one half kBT for every quadratic degree of
freedom”
• Therefore ideal gas:
• 3N-- quadratic degree of freedom
Additive Hamiltonian
• If the Hamiltonian of a system is a sum of
independent terms, the partition function is
a product of independent terms, and thus
the free energy is again a sum of
independent terms.
• We used that for the ideal gas.
• Other example: Particle with translational,
rotational, and vibrational degree of
freedom
Partition function in generalized
coordinates
• The integrals will contain the Jacobian
for the transformation in generalized
coordinates. We are not going to look
into this very much. Let’s just make one
example.
• Dipole in electric field
Dipole in electric field
• The energy is
• The rotational partition function is then
given by
• Explain why are different types of
coordinates are useful.
• Jacobian from spherical coordinates, etc ...
(not really necessary to do
the change of variable)
• Substitute
4
• Via parameter differentiation, we can now
work out the average component of D in
the direction of the field:
Langevin function
• Then we define
• Then we have:
• For small arguments we have
,
hence for small electric fields we have
THERMAL RADIATION AND
PLANK DISTRIBUTION
PLANK DISTRIBUTION FUNCTION
• The Plank distribution function is the first application of thermal
Physics.
• It describes black body radiation and also thermal energy spectrum
of lattice vibrations.
The energy states of quantum harmonic oscillator is given by
 s  s
Where   2f is the frequency of radiation and s is zero or any positive
integer.
We omit the zero point energy 1 .
2
The partition function is then given by
s4
4

z   exp(
s 0
s


s 3
)   exp(  s )
s 0

3
s2
This sum is of the form
with x  exp(    ).
s
x

2
s 1

s0
(A mode of radiation can only be
exited in units of  ).
Which is geometric series with
x smaller than 1, thus
1
x

,

1 x
s
Which implies the partition function
1
Z
.
1  exp(   )

Now let’s calculate s  (average excitation state).
The probability that the system is in the state
s of energy s
by the Boltzmann factor:
exp(  s  )
P( s) 
Z
The thermal average value of
s
is
s exp(  s  )
 s   sP ( s)  
Z
s 0
s 0

Chose y   ,


is given
Then we have,
d
 s exp( sy )   dy  exp( sy )

d 
1
exp(  y )
 
  
dy  1  exp(  y )  1  exp(  y )2
Thus,
exp(  y )
 s 
,
1  exp(  y )
or
 s 
1
exp(   )  1
This is the Planck distribution function for the thermal average
number of photons.
Plank distribution as a
1.0
s ( ) 
function of the reduced
temperature

1
2
.
0.5
s ( )
0
0
0.5
  
1.0
PLANKS LAW AND STEFAN-BOLTZMANN
LAW
•
The thermal average energy in the mode is

    s  
.
exp(   )  1
For high temperature limit (the classical limit):
  
Then we have the approximation
exp(  )  1  
And the classical average energy is
   

Now we want to find out the radiation modes confined with in a
perfectly conducting cavity in the form of a cube of edge L , then there is a set
of modes of the form
Ex  Ex 0 sin t cos( nxx L) sin( n yy L) sin( nzz L) 

E y  E y 0 sin t sin( nxx L) cos( n yy L) sin( nzz L) 

Ez  Ez 0 sin t cos( nxx L) sin( n yy L) cos( nzz L)
(i)
The field must be divergence free:
Ex E y Ez
divE 


0
x
y
z
(ii)
When we insert (i) into (ii)
E x 0 nx  E y 0 n y  E z 0 nz  E 0 .n
Which implies that the electromagnetic field in the cavity is transversely
polarized.
The modes have to satisfy the wave equation
2
2
2
2






Ez
2
c  2  2  2  Ez  2
z 
t
 x y


 c 2 2 nx2  ny2  nz2   2 L2
Let n  nx  ny  nz
2
2
2
2
 n  nc .
L
The frequencies have to satisfy certain conditions.
The total energy in the cavity is
 
U
n

n exp(  n  )  1

n


n
sum over triplet of
positiveintegers
Replace sum by integral over
  (...) 
n
dnx dn y dnz
1 

 4n dn(...) 
2
3
because of only
positivespace of
octant


2
0
n
 U    dn n
exp( n  )  1
0
2
2
for two independent
states of polarization

  c
2
 dn n exp(cn L ) 1

L
1
3
()
0
Let x  cn
L

,



3
x
 U   c L
dx
L c  exp x  1
0


 

2
4
4

15
U
2 4


3 3
V 15h c
This is known as Stefan-Boltzmann law of radiation.
()
Let us write () with the help of
uw :
energy per unit volume per unit frequency range (n  nc L)
  2 c  1 
U
1
L
3
 3  dw w
  dw u w  
. 
V
exp(   )  1  c 
 L L 0


w3
 2 3  dw
 c 0
exp(   )  1


w3
uw  2 3
 c exp(   )  1
This result is known as Planck radiation law.
4
1.6
Maximum
Plot of
x3/(ex
- 1)
with x =ωħ/τ. This
function is involved
1.4
1.2
in the Plank
radiation law for the
1.0
spectral density u.
The temperature of 0.8
a black body may
be found from the
0.6
frequency max at
which the radiant
0.4
energy density is a
maximum, per unit
0.2
frequency range.
0
0
1
2
3
 
4
5
6
7
8
The entropy of the thermal photons at constant volume is
d  dU ,

Then from
4 2V 2
d 
 d
3 3
15 c
And on integration
 4 2V
 ( )  
 45
  
  .
 c 
3