PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Class Activity - Class 19 March 31, 2006
Name________________SOLUTION____________
Consider a collection of 3N identical harmonic oscillators, all of frequency . The energies that
one these oscillators can take on (measured relative to the ground state) are n = nh.

1
a) Find the partition function (Z) for one of these oscillators. (Hint:  x n 
for x < 1)
n 0
1 x

Z  e
  nh
n0


  e   h

n
n 0

1
1  e   h

If n = (n+½)h, then Z   e
  ( n  12 ) h
n 0
e
  12 h
 e


  h n
n 0
  12 h
e

1  e   h
b) Find the average energy per oscillator for this collection of oscillators.
U


h e   h
h

ln Z  
 ln( 1  e   h ) 
  h
  h
N


1 e
e 1


If n = (n+½)h, then
U




ln Z  
ln( e
N



1
h
2

)  ln( 1  e   h )



  12 h  ln( 1  e   h )

U 1
h e  h h
h
 2 h 

  h
  h

N
1 e
2
e
1
c) What is the average energy of the collection of 3N oscillators?
There are 3N oscillators, so the average energy per oscillator can be written,
U
h
3 Nh
  h
U   h
,
so
3N e  1
e 1
If n = (n+½)h, then
U
h
h

  h
, so
3N
2 e 1
1 
1
U  3 Nh    h

 2 e 1
Turn the paper over. There is more on the back.
d) What form does the answer to part c) take on at low temperatures (i.e. kT << h)?
If kT << h, then  >> h, so h >> 1, and eh >> 1, so eh–eh, so
3 N h
3 N h
U   h
  h  3 N he   h
e
1
e
1 
1 
1
1
 3Nh    h   3N h
If n = (n+½)h, then U  3Nh    h

 2 e  1
2 e 

1
2
 e   h
e) What form does the answer to part c) take on at high temperatures (i.e. kT >> h)?
(Hint: e x  1  x for x << 1)
For kT >> h, h << 1, so eh 1 + h, and eh – 1  (1 + hv) –1 = hv. Then,
3N h
3N h
3N h 3N
U   h



 3NkT
1
e  1 1   h  1  h
kT
1 
1
 h


3
N
 kT 
If n = (n+½)h, then U  3 Nh    h


 2 e  1
 2

