10
Applications II
Classical Domain
We now address some elaborations for realistic applications in higher dimensions
and having independent dynamic parts. The technical details introduced here are
widely used in statistical mechanics. They are illustrated with classical models of
ideal gases and solids.
Partition Function Revisited
You saw that the partition function is a useful quantity for generating average
energies, standard deviations, and Helmholtz free energy F. In fact, all the
thermodynamic functions of the last chapter can, in principle, be constructed from Z. For
discrete energy levels Ej, we wrote
Z   exp E j .


j
Often, the same energy level can be attained in multiple ways; for example, a hydrogen
atom may have the same energy for 2  1values of angular momentum  . We can group
such degenerate terms together so that Z appears as
Z    j exp  E j  ,
(1)
j
where jrepresents the degeneracy of the jth energy level and the sum is now over
differing energies. Canonical probabilities are then written
wi 
 j exp  Ei 
Z
.
(2)
A bridge to thermodynamics can be made through the simple connection with the
Helmholtz free energy F,
(3)
F = - kT ln Z .
Other thermodynamic potentials can then be constructed from relations like G  F  PV
and F  E  TS .
Independent Systems and Dimensions
When two independent systems have corresponding entropies S1and S2, the
combination of these systems should have a total entropy S given by S = S1 + S2. We say
that entropy and information are “additive.” It is easy to show that an expression for the
combined system is
S  k E1  E2   k ln Z1Z2  .
The additivity of energy is expected, but the partition function for the composite is seen
to be the product of the independent Z’s. The rule can be extended to any number of
independent systems. The composite Z for N independent systems is
Z  Z1 Z2  Z N
(4)
Applications II 2
The same product rule for Z applies when you consider independent motions or
independent dimensions. For example, a molecule that has independent modes of
rotation, vibration and translation may be expressed in obvious notation as
Z  Z rotation Z vibration Z translation
Similarly, if the dynamics can be easily separated into independent motions along axes x,
y, and z, we can write
Z  Z x Z y Zz
1.
Show that for independent systems 1 and 2, the entropy of the composite is given
by S  k E1  E2   k ln Z1Z2  .
2.
Given that the partition function for an ideal gas of N classical particles moving in
the x-direction in a rectangular box of sides Lx, Ly, and Lz is
N
 2m  2 L 
Z x  
 x
   h 
(a) Write the partition function for the gas in three dimensions.
(b) Calculate the pressure of the gas in terms of T and V (V  Lx Ly Lz ) .
1
[ans. P  NkT / V ] Hint: Calculate F and use F V   P .
Independent Particles
We often must treat independent particles. If Z1 is the partition function for a
single distinguishable particle, then the partition function for N such particles is simply
(5)
Z  Z1N
A further adjustment must be made for identical particles.
At the molecular level particles of the same species are indistinguishable from
each other. When there are N particles, they can be rearranged in N! ways each of which
is indistinguishable from the others. (Thus the particles a, b, c can be arranged six ways;
(abc), (bac),...etc.) The remedy for indistinguishable particles is to divide out the excess
factor of N! from Eq(5):
ZN
(6)
Z 1
N!
3.
The partition function for a point particle moving in the x-direction in a
rectangular box of sides Lx, Ly, and Lz is
1
 2 m  2 Lx
Z1, x  

   h
(a) Write the partition function for N of these indistinguishable particles.
(b) Find the partition function for the particle gas in three dimensions. Compare
your result with part (a) of problem 2.
Applications II 3
Density of States
Many systems have continuous energy levels and it is desirable or necessary to
treat the partition function as an integral,
Z   exp E j   exp E d ,


j
where the density of states d is the number of
states in a neighborhood around energy E. A single p
cell in one-dimensional phase space is limited by
the uncertainty relation,
x px  h .
The number of states available to the system is the
h h h
number of accessible cells,
h h
dxdpx
d 
.
h
Similarly, in three dimensions each particle
contributes the factor
dxdydz dpx dp y dpz
(7)
d 
h3
Note that the density of states is the continuum counterpart of the degeneracy factor in
the discrete case of Eq.(1).
Very often it is convenient to convert three-dimensionsonal integrals to spherical
coodinates. In these cases we can write
(8)
dxdydz  r 2 sin drdd
Momentum integrals can be treated in
pz
the same way. In this case x, y, and z
are replaced by px, py, and pz. The
length of the position vector,
p

r  x 2  y 2  z 2 , is then replaced
by the momentum magnitude
p
px2

p2y

pz2
py

and we have
px
dpx dp y dpz  p2 sin dpdd ,
where the angles now relate to the momentum vector p.
4.
Use the diagram to confirm the volume element given in Eq.(9).
(9)
x
Applications II 4
Application: Ideal Gas
An application to ideal gas illustrates a typical calculation in statistical mechanics.
This time we approach it with a more rigorous treatment based on the canonical
distribution and the partition function. First the partition function is evaluated and then
various relations between variables are generated using expressions derived in the last
chapter. A select few such relations are collected here:
F   kT ln Z

E   ln Z

 F 
 F 
P    and S   
 V  T
 T  V
For your convenience, we also include two important definite integrals,

2
 exp  r  dr  2
0
1


(10)
and

1 
2
2
 exp  r  r dr  4 
(11)
3
2
0
The following problem includes derivations of the ideal gas state equations and
uses various thermodynamic properties. Although it is a repetitious example, it requires
techniques necessary for many applications.
5.
(a) Evaluate the single-particle partition function for a point particle of mass m
confined to volume V. Do this two ways; (i) integrate over a three-dimensional
volume element in momentum space as in Eq.(9), and (ii) treat the gas as if it is
confined to a rectangular box and integrate over one dimension at a time. These
calculations are shown on the following page. Be sure you can perform these
(given the integrals above).
(b) Construct the full partition function for a gas of N identical particles described
in part (a).
(c) Derive the average energy expression of the gas and show that its heat
capacity. is given by CV  23 nR where n is the number of moles and R is the gas
constant.
(d) Write the Helmholtz free energy function and derive the ideal gas equation.
(e) Find an entropy expression for the gas. Show that an adiabatic process obeys
2
TV 3  constant.
Applications II 5
Solution to problem 5 (a)
Single-Particle Partition Function by 3-D Integration (take h = 1 for convenience)
Single-Particle Partition Function by 1-D Integration (take h = 1 for convenience)
Notice the difference in integration limits. In one dimension momentum takes negative
and positive values. In three dimensions p represents the magnitude
p
px2  p2y  pz2 and can only take positive values.
Applications II 6
Application: Classical Model of a Solid
A simple classical model treats a solid as a collection of springs all of which have
a common angular frequency . The energy of an individual spring is
H  a 2 p 2  b2 x 2 with a 2  1 2 m and b 2  21m m2
Now the integral over x is not trivial. For a single oscillator, we have

1 
Z   exp  H d  3  exp  a 2 p 2 d 3 p  exp  b2 x 2 d 3 x .
0
h 0
4
(a) Evaluate the single-particle partition function for a harmonic oscillator with
mass m and angular frequency .. Do this by integrating over a three-dimensional
volume element in momentum space and over a three-dimensional volume
element in configuration space (as indicated above).
(b) Construct the full partition function for a collection of N distinguishable
oscillators described in part (a). This is a classical model of a solid.
(c) Derive the average energy expression of the solid and show that its heat
capacity. is given by CV  3nR where n is the number of moles and R is the gas
constant.
This model is adequate at room temperatures and above. Better results will be found by
using quantum mechanical oscillators to model solids.
Application: Equipartition Theorem
The last problem was a special case of the equipartition theorem. Energy
expressions and heat capacities for number of substances can be quickly approximated for
‘high’ temperatures. Each px2 and qx2 term in the Hamiltonian (energy expression)
contributes an average energy of 12 nRT . When the Hamiltonian for an independent
particle has the form
H  a1 p12  a K pK2  b1q12 bM q 2M ,
(12)
where qj is the generalized coordinate associated with momentum pj, then the energy for a
system of N such particles is
NK  M
n K  M 
(13)
E
kT 
RT .
2
2
The degrees of freedom refers to the number of terms on the right hand side of Eq.(12),
K+M. Thus the heat capacity is the product of 12 nR and the number of degrees of
freedom.
Applications II 7
As an example, consider a material consisting of N springs with frequency . The
Hamiltonian for one particle is
H  21m p2  21 m 2r 2
so there are six degrees of freedom and the heat capacity is 3nR. This was the result in
problem 6.
A rigid diatomic molecule may translate and rotate around its center of mass.
Translation contributes three degrees of freedom. Rotation contributes only two degrees
of freedom corresponding to the angular momenta associated with each of two angles.
Consequently, the heat capacity of a rigid diatomic ideal gas is 25 Nk . At high
temperatures a vibration mode may be excited in diatomic molecule. This vibration
simulates a one-dimensional harmonic oscillator contributing two more degrees of
freedom (one from p2 and one from x2 terms) to produce a heat capacity of 72 Nk .
7
The energy of a rigid rotating molecule has the form H  ap2  bp2  cp2 where
the p’s represent momenta and a, b, and c are constants. Use the equipartition
theorem to write the heat capacity of a gas of such molecules. [ans. 25 nR ]