Chapter 3
Classical Statistics of Maxwell-Boltzmann
Maxwell-Boltzmann statistics describes the statistically distribution of
particles over various energy levels “states” in thermal equilibrium.
Three hypotheses:
1- the system contains a large number of particles,
2- the particles are identical and distinguishable,
3- Pauli principle not applicable.
1- Computation of Most Probable Macrostate Distribution:
For an isolated system, the volume V, energy U and total number of particles
N are fixed
N   ni
i 1
and U   ni i
i 1
The number of ways to distribute n1 particles to level 1, n2 particles to level
2 and so on is
 (n1 , n2 ,...) 
N!
 ni!
i 1
Level “i” has degeneracy “gi” so any of the ni particles can enter into any of
the gi levels
Number of ways  gini
Note: in physics, two or more different quantum states are said to be
degenerate if they are all at the same energy level. Statistically this means
that they are all equally probable of being filled. Conversely, an energy level
is said to be degenerate if it contains two or more different states.
1
Example 1: ni = 3 and gi = 4
k
Macrostates
(n1,n2,n3,n4)
1
2
3
4
5
6
7
8
9
10
(3,0,0,0)
(2,1,0,0)
(2,0,1,0)
(2,0,0,1)
(1,2,0,0)
(1,1,1,0)
(1,1,0,1)
(1,0,2,0)
(1,0,1,1)
(1,0,0,2)
k 
3!
n1!n2!n3!n4!
1
3
3
3
3
6
6
3
6
3
k
Macrostates
(n1,n2,n3,n4)
11
12
13
14
15
16
17
18
19
20
(0,3,0,0)
(0,2,1,0)
(0,2,0,1)
(0,1,2,0)
(0,1,1,1)
(0,1,0,2)
(0,0,3,0)
(0,0,2,1)
(0,0,1,2)
(0,0,0,3)
k 
3!
n1!n2!n3!n4!
1
3
3
3
6
3
1
3
3
1
20
number of ways :  k  1  4  3  12  6  4  64  43  gini
k 1
We have 20 macrostates in level “i” directly linked to 64 microstates (=
number of ways).
For all levels,
Number of ways   gini
i 1
It follows that
gini
 (n1 , n2 ,...)  N!
i 1 ni !
Maximize  (i.e.   ~  ) by keeping fixed U and N. These constraints are
introduced through Lagrange multipliers. For a fixed V, it means that the
energy levels,  i , are unchanged.
At maximum, d   0  d (ln  ) 
d

 0 . Hence we can choose to
maximize ln(  ) instead of  itself, this turns the products into sums
2
(equation above). Since the logarithm is a monotonic function of its
argument, the maxima of  and ln(  ) occur at the same point.
ln(  )  ln( N!)   ni ln( gi )  ln( ni !)
i 1
Applying Stirling’s law to ln( n!) : ln( ni !)  ni ln( ni )  ni
ln(  )  ln( N!)   ni ln( gi )  ni ln( ni )  ni 
i 1
 ln(  )
1
 ln( gi )  ln( ni )  ni  1  ln( gi )  ln( ni )
ni
ni
Now, we introduce the constraints
f1   N   ni  0 
f1
1
ni
f 2  U    i ni  0 
f 2
  i
ni
i 1
i 1
Introducing Lagrange multipliers (see chapter 2 of classical mechanics (2))
 ln(  )
f
f
 1   2  0
ni
ni
ni
ln( gi )  ln( ni )    i  0
g 
ln  i     i
 ni 
ni
 e   i
gi
n 
 ln  i     i
 gi 
 ni  gi e e   i
e  ??
N   ni   gi e e   i
i 1
i 1
3
N  e  gi e   i
 e 
i 1
N
 gi e  i
i 1
ni 
Ngi
e   i
  i
 gi e
i 1
e   i
ni  Ng i
,
Z sp
Z sp   g i e   i
i 1
where:
ni is the number of particles in level i,
 i is the energy of the i-th level,
gi is the degeneracy factor, or number of degenerate microstates which have
the same energy level  i ,
N is the total number of particles,
e(...) is the exponential function
Z sp is the partition function for a single particle
We will prove latter that  is equal to 1 kBT where kB is the Boltzmann
constant ( kB  1.38  1023 JK 1mol 1 ) and T is the absolute temperature.
2- Partition Function
Partition function describes the statistical properties of a system in
thermodynamic equilibrium. It is a function of temperature and other
parameters, such as the volume enclosing a gas. Most of the aggregate
thermodynamic variables of the system, such as the total energy, free
4
energy, entropy, and pressure, can be expressed in terms of the partition
function or its derivatives.
2-1- Definition
As a beginning assumption, assume that a thermodynamically large system
is in constant thermal contact with the environment, with a temperature T,
and both the volume of the system and the number of constituent particles
fixed. This kind of system is called a canonical ensemble. Let us label with s
(s = 1, 2, 3, ...) the exact states (microstates) that the system can occupy, and
denote the total energy of the system when it is in microstate s as  s .
The canonical partition function is
Z sp   e   s
s
In systems with multiple quantum states s sharing the same  s , it is said that
the energy levels of the system are degenerate. In the case of degenerate
energy levels, we can write the partition function in terms of the contribution
from energy levels (indexed by i) as follows:
Z sp   g i e   i
i
where gj is the degeneracy factor, or number of quantum states s which have
the same energy level defined by  i   s .
2-2- Meaning and Significance
It may not be obvious why the partition function, as we have defined it
above, is an important quantity. First, let us consider what goes into it. The
partition function is a function of the temperature T and the microstate
5
energies 1, 2, 3, etc. The microstate energies are determined by other
thermodynamic variables, such as the number of particles and the volume, as
well as microscopic quantities like the mass of the constituent particles. This
dependence on microscopic variables is the central point of statistical
physics. With a model of the microscopic constituents of a system, one can
calculate the microstate energies, and thus the partition function, which will
then allow us to calculate all the other thermodynamic properties of the
system.
The partition function can be related to thermodynamic properties
because it has a very important statistical meaning. The probability, Pi, that
ni particles occupy level i is
ni
1

gi e  i
N Z sp
Pi 
The partition function thus plays the role of a normalizing constant, ensuring
that the probabilities sum up to one:
1
 P  Z  g e 

i
i
i
i
sp
i

1
Z sp  1
Z sp
The probability Ps that the system occupies microstate s is
PS 
1   s
e
, and
Z sp
1
 P  Z  e 

s
s
sp
s
s

1
Z sp  1
Z sp
This is the reason for calling Zsp the "partition function": it encodes how the
probabilities are partitioned among the different microstates, based on their
6
individual energies. This notation also implies another important meaning of
the partition function of a system: it counts the (weighted) number of states a
system can occupy. Hence if all states are equally probable (equal energies)
the partition function is the total number of possible states. Often this is the
practical importance of Zsp.
Example 2: A system possesses three energy levels (1=0, 2=100kB, and
3=200kB) and degeneracy in each level as g1=1, g2=3, and g3=5. Calculate,
at T=100K, the partition function (Zsp), the relative population in each level
(Pi), and the average energy ( U ).
3
Z sp   gi e   i   gi e  i
 1  e  0 / 100k B  3  e 100k B / 100k B  5  e  200k B / 100k B
100k B
i 1
i
Zsp  1  3e1  5e2  2.78
Pi 
ni
1

gi e   i
N Z sp
 P1 
1  e 0 / 100k B
3  e 100k B / 100k B
 0.360, P2 
 0.397, and
2.78
2.78
P3 
U 
5  e  200k B / 100k B
 0.243
2.78
3
3
1 3
ni
n




Pi i  0.360  0  0.397  100k B  0.243  200k B  88.3k B
ii 

i
N i 1
i 1 N
i 1
We conclude that in general
U   Pi i
i 1
2-3- Partition functions of subsystems
7
Suppose a system is subdivided into N sub-systems with negligible
interaction energy. If the partition functions of the sub-systems are
Z sp1 , Z sp 2 ,… Z spN then the partition function of the entire system is the product
of the individual partition functions:
N
Z N   Z spj
j 1
If the sub-systems have the same physical properties, then their partition
functions are equal, Z sp1  Z sp2  ...  Z spN  Z sp , in which case
Z N  (Z sp ) N
However, there is a well-known exception to this rule. If the sub-systems are
actually identical particles, in the quantum mechanical sense that they are
impossible to distinguish even in principle, the total partition function must
be divided by a N! (N factorial):
ZN 
( Z sp ) N
N!
This is to ensure that we do not "over-count" the number of microstates.
Example 3: Consider three sub-systems, each one possesses one particle and
two energy levels (E1 = 0, and E2 =  ) with g1 = g2 = 1.
- Calculate the partition function, Z sp , of one sub-system.
N=1
(n1,n2)

Z sp  1  e  0  1  e    gi e  i
i
i
8
(1,0)
1
0
(0,1)
1

Z sp  1  e 
- Show that the partition function, Z 3 , of the system (system constituted by
the three sub-systems and have the same physical properties) is equal to
(Z sp )3 .
N=3
(n1,n2)

i
Z 3  1  e   0  3  e    3  e 2   1  e 3
(3,0)
1
0
Z 3  1  3e    3e 2   e 3
(2,1)
3

Z3  (1  e  )3  (Zsp )3
(1,2)
3
2
(0,3)
1
3
Example 4:
A system possesses two identical and distinguishable particles, and three
energy levels (E1=0, E2=  , and E3= 2 ) with g1=2, g2=1, and g3=1.
- Show that Z sp   e    gi e 
s
s
i
i
( gi is the number of degenerate microstates which have the same energy
level  i =  s )
 g e 

i
i
 2e  0  e   e  ( 2 )  2  e   e2 
i
Macrostate
(ni)(n1,n2,n3)
Nb. Microstates
 (ni )  N!
i 1
g ini
ni !
i   s
(1,0,0)
2
0
(0,1,0)
1

(0,0,1)
1
2
9
 e 

s
 2  e  0  1  e   1  e  ( 2 )  2  e   e2    gi e  i
s
i
2


- Show that Z 2    gi e  i    e  s
s
 i


Z 2  2  e   e 2 
 e 

s

2
 4  4e   5e 2   2e3  e 4 
 ??
s
Nb. Microstates
(2,0,0)
1
0
1
0
1
0
1
0
1

1

1

1

1
2
1
2
1
2
1
2
(0,2,0)
1
2
(0,1,1)
1
3
1
3
1
4
(1,1,0)
(1,0,1)
(0,0,2)
 e 

s
i   s
Macrostate
 e   ( 0 )  e   ( 0 )  e   ( 0 )  e   ( 0 )  e   (  )  e   (  )  e   (  )  e   (  )  e   ( 2 )  e   ( 2 )
s
 e   ( 2 )  e   ( 2 )  e   ( 2 )  e   (3 )  e   ( 3 )  e   ( 4 )
 e 

s
 4  4e   5e2   2e3  e4  then
s
 e    g e 

i
s
10

s
i
i
In general,
Z sp   gi e  i   e  s
i
s
Z N  ( Z sp ) N   e  s
s
2-4- Relation to thermodynamic variables
In this section, we will state the relationships between the partition function
and the various thermodynamic parameters of the system.
Z sp

Z sp


 
  i 
g
e

i

  i 1

   i g i e   i
i 1
  
 Z sp 
  i   i 
  i  


  
g
e



g
e




i
i


 V T , N
i 1
 T , N
 V T , N  V  i 1
 Z sp 
    

     g i e i  i 
 V T , N
i 1
 V T , N
- Calculation of Average energy:
1
U 
N
U 
U 
e   i
1
ni  i , and ni  Ng i
U 

Z sp
Z sp
i 1
1 
Z sp 

g
i 1

ln( Z sp )

i
e   i  
1 Z sp
Z sp 

11

i 1
i
g i e   i
- Calculation of Internal Energy:
U  NU
 U  N


ln( Z sp )


- Calculation of Entropy:
e   i
We know that for all levels: ni  Ng i
Z sp

g i Z sp  i

e
ni
N
ni
~ (n , n ,...)  N! gi
 

1 2
i 1 ni !
ln( )  ln( N!)   ni ln( gi )  ln( ni !)
i 1
ln( )  N ln( N )  N   ni ln( gi )  ni ln( ni )  ni 
i 1
g 
ln( )  N ln( N )  N   ni ln  i    ni
i 1
 ni  i 1
 Z sp  i 
ln( )  N ln( N )  N   ni ln 
e   N
N
i 1




 Z sp 
  ni (  i )
ln( )  N ln( N )   ni ln 
i 1 
 N 

 Z sp 
 ni    ni  i
ln( )  N ln( N )  ln 
N
i 1
i 1




N
U
ln( )  N ln( N )  N ln( Z sp )  N ln( N )  U
ln( )  N ln( Z sp )  U
S  kB ln( )
12
S  k B N ln( Z sp )  k B U
- Calculation of 
TdS  dU  pdV

S  k B N ln( Z sp )  k B U
1  S 


T  U V
  ln( Z sp ) 
 S 
  
  k B   k BU 
 
  k B N 


U

U
 U V

V

V
 ln( Z sp )   
1
 S 
  

  kB N

  k B   k BU 
  kB  

T
 U V
 U V


 U V
 U / N

1
k BT
- Calculation of the Helmholtz free energy:
A  U  TS

A  U  T k B N ln( Z sp )  k B U 
A  U  NkBT ln( Z sp )   k BT U

1 / 
A   NkBT ln( Z sp )
- Calculation of the pressure:
 A 
 A 
dA  SdT  pdV    dT  
 dV

T

V
 V

T
 A 
p  

 V T
A   NkBT ln( Z sp )
13
  ln( Z sp ) 

p  NkBT 

V

T
3- Applications of Maxwell-Boltzmann Statistics to classical perfect gas
To apply the Maxwell-Boltzmann statistics to the classical perfect gas, we
require a few conditions:
- The gas is monoatomic; hence we consider only the translational kinetic
energy (KE),
- There is no interaction with external fields, e.g. gravity, magnetic fields,
- The energy is quantized as in the case of particle in a box in quantum
mechanics,
- The discrete levels can be replaced by a continuous one if the spacing is
small as compared to T, i.e.,   which is the classical limit.
- The atoms are pointlike (occupying no volume).
- There is no internal structure for the particles, i.e., no excited electronic
states and particles are treated as single particles.
From quantum mechanics,
 2 2 2
i 
(nx  ny2  nz2 )
2
2mL
where volume V = L3 and nx, ny, and nz are integers.
Now we compute the partition function Zsp and in this case gi = 1, because
all the states are summed over.
  s
  e  s
So, for all states: Z sp   e
s 1
s 1
14
k BT


 2 2
Z sp   exp  
(nx2  n y2  nz2 ) 
2
nx n y nz
 2mL k BT




 2 2
 2 2
 2 2
2
2
2




Z sp   exp  
n
exp

n
exp

n
x 
2
 2mL2 k T y 
 2mL2 k T z 
2
mL
k
t
nx
n
n
B
B
B

 y

 z


Z sp  

0

 2 2
2

 dnx
exp  
n
x
2
2
mL
k
T
B




0

 2 2
2

 dn y
exp  
n
y
2
2
mL
k
T
B




0

 2 2
2

 dnz
exp  
n
z
2
2
mL
k
T
B


12
  2 2 
 nx
To evaluate the integral, let x  
2
 2mL k BT 
12
  2 2 
 dnx
dx  
2
2
mL
k
T
B 







0
0
0


0
12
 2mL2 k BT 
 dx
 dnx  
2 2







exp  
nx2  dnx  
2
0
 2mL k BT 
2
2

2
12
12
 

2 2


  
  2mL2 k BT 
 nx   
 dx
 
2
2 2
2
mL
k
T





B 
 
  
12

 2mL2 k BT 
 2 2
2

exp  
nx  dnx  
2
2 2
2
mL
k
T




B


12

 2mL2 k BT 
 2 2
2

exp  
nx  dnx  
2
2 2
2
mL
k
T




B




0
exp(  x 2 ) dx
1 
exp(  x 2 )dx



2 

 
12
 1 2mL2kBT 

 2 2
2
exp  
nx  dnx  
 
2 2
2
2
mL
k
T
4


h
2


B




 2m  k BT 
Z sp  

h2


32
 2m  k BT 
L V 

h2


32
3
3
3  2m k B 
ln( Z sp )  ln( V )  ln( T )  ln 

2
2  h2 
- Pressure
15
 2m kB T 

 L
h2


12
  ln( Z sp ) 

p  Nk BT 

V

T

p
Nk BT
V
pV  NkBT
Since pV  nRT
 NkB  nR  k B 
n
1
R
R
N
NA
where NA is the Avogadro’s number and n the number of moles.
kB 
R
NA
- Internal energy
U  N

1
k BT


ln( Z sp )


 U  N

1

T
k BT 2


 ln( Z sp ) T
T

T
 k BT 2

 ln( Z sp ) 3  1 
3
3  2m k B 
ln( Z sp )  ln( V )  ln( T )  ln 

  

2
2  h2 
T
2T 
 3 
U  NkBT 2 

 2T 
U
3
3
Nk BT  nRT
2
2
For each atom, U 
U 3
 k BT
N 2
- Specific heat
 U 
CV  

 T V , N
and U 
3
NkBT
2
 CV 
3
NkB
2
CV is constant and independent of temperature in an ideal gas.
16
CV 
3
nN Ak B
2
CV 
3
nR
2
- Entropy
Also the entropy is given by: S  k B N ln( Z sp )  k B U
S  Nk B ln( Z sp ) 
S  NkB ln( V ) 
U
T
3
3
 2m  k B  3
NkB ln( T )  NkB ln 
  NkB
2
2
2
 h
 2
17