Transcript Example of Boltzmann distribution.

Chapter 13
Classical and Quantum Statistics
1
So far, with the exception of the previous chapter, we have
dealt with the 1st and 2nd laws of thermodynamics. In using
these laws to make numerical calculations it is usually
necessary to appeal to experimental measurements. We would
like to calculate all thermodynamic properties of a system from
a microscopic model of that system. We have made a start in
the last chapter and we will make significant progress in the
present chapter.
2
Lagrange Undetermined Multipliers
I would like to give you a trivial example of the use of
Lagrange Undetermined Multipliers. The term is somewhat misleading
because the multipliers can, in fact, be determined. In the following
example it is not necessary to use this sophisticated method and you
should solve the problem in a simple fashion.
Consider the equation ax+by=0……………..(1)
If y=y(x) then y=-(a/b)x However this is not true if both
x and y are independent. The only solution is then a=b=0
Suppose, however, that x and y are not completely independent
but satisfy a constraint condition such as x+2y=0………(2)
What can we say about the coefficients a and b?
The procedure, using Lagrange Multipliers, is to multiply each
constraint condition by a multiplier, which is initially unknown. In the
present case we have one constraint condition and let   multiplier.
giving
( x  2 y)  0 (1)  (2)  0
(a  )x  (b  2) y  0.....(3)
3
Now that the constraints have been implicitly taken into consideration
we treat x and y as two independent variables. We must have
(  a)  0
(b  2 )  0
so
so
  a
b

2
This yields b  2a
Placing this in equation (1) shows that the constraint condition is
satisfied.
4
EXAMPLE: Lagrange Undetermined Multiplier
A cylindrical nuclear reactor has radius R and height H
We wish to minimize the volume of the reactor. V   R2 H
There is a constraint supplied by neutron diffusion theory
2
2
constraint
 2.4048    


dV  2 RHdR   R2 dH
For an extremum dV=0
2 RHdR   R2 dH  0 (1)
2(2.4048 )2
2 2

dR  3 dH  0
3
R
H
(1)  (2)  0
R
     constant
 H
differentiating the constraint
(2.4048 )2
2
dR  3 dH  0 (2)
3
R
H
2
2
(
2
.
4048
)

2 RHdR   R2 dH  
dR   3 dH  0
3
R
H

 2
(2.4048 )2 
2 
2 RH  
 dR   R   3  dH  0
3
R
H 



5
Now we consider R and H to be independent
2
(2.4048 )2

2
2 RH  

0
.....(
3
)

R
  3  0.....(4)
3
R
H
1 2 3 (multiplier is determined)
 R H
From equation (4)

Substituting into equation (3)
2 2
2

R
H2 
(2.4048 )2
(2.4048 )2 R2 H 3
2 RH 

R3
H 
2R
(2.4048 )
6
Boltzmann Statistics.
We consider N distinguishable particles and we can place any
number into a particular state. We wish to determine the
thermodynamic probability for a particular macrostate.
To guide us we consider a simple example: N=3 (A B C)
and take the macrostate in which N1  2 g1  3
We begin with a box labeled N1 and we wish to throw 2 particles into
the box.
1st particle: 3 choices
( N)
2nd particle 2 choices
(N-1)
The total number of choices is (3)(2)
(N)(N-1)
These choices are shown on the next slide
7
Choices
However we can permute the particles in the box
without changing the contents of the box. The number
1st 2nd
of permutations is 2!
( N1 !)
A B
A C
Therefore the number of distinct choices is:
B A
32
N ( N  1)
B
C
2!
N1!
C A
They are AB AC BC
C
B
Now we arrange these particles into the 3 degenerate states.
g1
1st particle has 3 possibilities
2nd particle has 3 possibilities
g1
2
The total number of possibilities is 3
N1
1
g
We will list these possibilities for the case where the box has AB
8
STATES
1
2
AB
AB
3
The total number of possibilities is therefore
AB
A
A
B
B
B
A
A
B
We see that there are 9 possibilities. 32
We also have the AC and BC possibilities, each
with 9 possible distributions among the states
B
B
A
A
32 2
3
2!
N ( N  1) N1
g1
N1!
Now we concentrate on the general case of N
particles and consider placing N1 particles
in energy level 1 which has a degeneracy g1
As above selecting the N1 particles for the
N ( N  1)( N  2)( N  N1  1)
N1!
1
box, we have
possibilities
We can write this as
N ( N  1)( N  2)( N  N1  1)( N  N1 )( N  N1  1)1
N1!
( N  N1 )( N  N1  1)1
9
N!
and this is evidently N !( N  N )!
1
1
Now we consider the possibilities for distributing these particles into
the degenerate states of 1
N
The total number of possibilities is then
N ! g1 1
N1! ( N  N1 )!
Now we go to the  2 level. The procedure is obviously the same
except that we no longer have N particles. The number of available
particles is ( N  N1 )
N
( N  N1 )! g2 2
As above we have, for this level,
N2 ! ( N  N1  N2 )!
For the
3
possibilities
( N  N1  N2 )! g 3 3
N 3! ( N  N1  N2  N 3 )!
N
level
10
Hence the thermodynamic probability is
( N )! g1 1
( N  N1 )! g2 2
( N  N1  N2 )! g3 3
w( N1, N2 , N n ) 

N1!( N  N1 )! N2 ! ( N  N1  N2 )! N 3! ( N  N1  N2  N 3 )!
N
N
N
N
N
N
( N )! g1 1 g2 2 g 3 3
w( N1, N2 , N n ) 

N1! N2 ! N 3!
n
N
gi i
w( N1, N2 , N n )  N ! 
i 1 N i !
Boltzmann statistics
(distinguishable particles)
This is the total number of accessible microstates for a particular
macrostate. The value of w will be different for each particular
macrostate. The greater the value of w, the greater the probability
of occurrence. Remember:
The equilibrium macrostate is the one for which w is a maximum.
11
Example: Before continuing with our discussion of Boltzmann
statistics, let us consider a simple situation.
Consider 6 distinguishable objects (A,B,C,D,E,F) which can
be placed in 6 boxes (1,2,3,4,5,6). We will calculate w for
several macrostates. The degeneracy will be unity for the boxes.
6
6
g iNi
1N i
1
w( N 1 , N 2 ,N n )  N!
 6!
 6!
i 1 N i !
i 1 N i !
i 1 N i !
1
w(1,1,1,1,1,1)  6!
 6!
w(1,1,1,1,1,1)  720
1!1!1!1!1!1!
(most disorderd)
1
w(0,0,0,0,0,6)  6!
1
w(0,0,0,0,0,6)  1
0!0!0!0!0!6!
1
6!
w(1,1,1,1,2,0)  6!

w(1,1,1,1,2,0)  360
1!1!1!1!2!0! 2!
1
6
w(5,1,0,0,0,0, )  6!

w(5,1,0,0,0,0, )  6 12
5!1!0!0!0!0! 1!
n
Now continuing with Boltzmann Statistics:
We vary the N i to find the maximum value for w. This will give the
N i in terms of g i and  i
For convenience we will work with ln(w) instead of w. (The range of
values is much smaller.) This will also permit us to use Stirling’s
fromula, valid for large x: ln(x!)=xln(x)-x
Ni
n
n
n


g
ln( w)  ln  N !  i   ln( N ! )   N i ln( g i )   ln( N i ! )
 i 1 N ! 
i 1
i 1
i 

Apply Stirling’s formula to the last term :
n
n
n
i 1
n
i 1
i 1
ln( w)  ln( N! )   Ni ln( gi )   Ni ln( Ni )   Ni
 Ni 
ln( w)  ln( N! )   Ni ln   N
i 1
 gi 
constants
Now we maximize to obtain the equilibrium distribution.
n
 Ni 

gi  1
 dNi 
d (ln( w))   ln dNi   Ni
Ni  gi
i 1
i 1
 gi 

n
13
n
 Ni 
d (ln( w))   ln dNi  d  Ni
i 1
i 1
 gi 
n
 gi 
d (ln( w))   ln dN i (1)
i 1
 Ni 
n
We have the constraints:
n
N
i 1
Differentiating
i
 N  constant
n
N 
i i
i 1
n
 dN
i 1
i
0
n

i 1
 U  constant
 i dN i  0
We multiply the constraint conditions by the Lagrange multipliers
ln(  ), (  ) (This form of the multipliers is for convenience.)
We also have the condition for a maximum that d(ln(w))=0
n
n
 gi 
ln dNi  ln(  ) dNi     i dNi  0

i 1
i 1
i 1
 Ni 
n
14
The N i are now taken as independent, so the coefficient of each
dN i must vanish. This gives
 gi 
ln   ln(  )  i  0 (2)
 Ni 
Ni  gi e
Solving for N i yields
  i
Hence N i varies with the degeneracy and with the energy of the level.
Now we need to determine the Lagrange multipliers.
n
n
N   Ni    gi e
i 1
N

Z
  i
We write this in the form
i 1
n
with
Z   gi e
i 1
  i
N
  i
N i  gi e (3)
Z
Z comes from zustandsumme, which means a sum over states. It is
usually referred to as a partition function. As we shall see, partition
functions are very useful. It turns out that Z depends only on T and
the parameters that determine the energy levels.
15
Once Z is known it is straightforward to calculate many thermodynamic
properties, such as U, S and P. Of course there are other ways of
calculating these quantities, but the simplest way is to first calculate the
appropriate ln(Z).
We still need to determine β.
16
Now we bring T into statistical mechanics. For a process
taking place between two equilibrium states:
đQ = dU+ PdV and dS = đQ/T so
dS  T1  dU  TP  dV
Considering S as a function of U and V,
dS   US V dU   VS U dV
and comparing we see that
1
T
  US V
(reciprocity relation)
The state variables S and U may be calculated by
statistical methods and so this equation brings the
macroscopic concept of T into statistical thermodynamics.
17
Now we are in a position to discuss the Lagrange multiplier 
We have the following two equations (equations 1 and 2 above)
 gi 
ln   ln(  )  i  0
 Ni 
 gi 
d (ln( w))   ln dN i
i 1
 Ni 
n
Substituting the second equation into the first one
n
n
n
n
i 1
i 1
i 1
i 1
d (ln( w))   i dNi   ln(  )dNi   d   i Ni  ln(  )d  Ni
We are dealing with a closed system with constant V so
n
n
d  Ni  0
and
This gives
d(ln( w))   dU
i 1
 N
i 1
i
i
U
Using the Boltzmann relationship S=kln(w)
1
k
dS   dU
or
 
S
U V
dS=k d(ln(w))
k
18
From the reciprocity relationship:
1
1
  k or  
T
kT
The temperature turns out to be a Lagrange multiplier!
Now that we have determined the Lagrange multipliers,
we write down the following two fundamental equations (see
equation (3)):
i

Ni
N kT
Boltzmann distribution
fi 
 e
(See note on next slide)
g
Z
i
The Boltzmann distribution gives the probability of occupation of a
single state belonging to the ith energy level.
Z 
n
ge
i 1
i

i
kT
The partition function, an explicit function
of T and an implicit function of V (through
the energy levels) contains the statistical
information about the system.
19
NOTE: There is some confusion regarding terminology. The
Ni
N
term Boltzmann distribution for the equation
fi 
gi

Z

e
i
kT
is unfortunate because, among other things, it is then easily
confused with the Maxwell-Boltzmann Distribution. The Boltzmann
distribution applies to systems which have distinguishable particles
and N, V and U are fixed. The Maxwell-Boltzmann Distribution
is applicable only to dilute gases.
20
Consider the case in which the energy levels are very closely
spaced. Then, instead of considering discrete levels with
degeneracies g i , we consider a continuum and replace the
g i by g( )d , the number of states in the energy range
between  and   d
The Boltzmann distribution is then written in the form:

f ( ) 
Ne
 g( )e

kT

Distinguishable particles

kT
d
21
Example of Boltzmann distribution.
Suppose that we have 1000 particles and T=10000K. The available
states and their multiplicities are shown below. The partition
function and the distribution of particles amongst the states are
i

n
calculated using

 i
N
Z 
ge
i 1
(eV)
gi
3
4
2
3
1
2
0
1
kT
i
Ni 
Z
g ie
kT
Z=2.045
Ε(eV) Ni
0
1
2
3
489
307
144
60
22
23
Canonical ensemble
This consists of a set of systems in contact with a large thermal
reservoir. For the Boltzmann distribution V, N and U are fixed.
We now consider a system in which V, N and T are fixed. The
energy of a system is not fixed, but will fluctuate about some
average value due to the continuous interchange of energy
between the system and reservoir. For large N these fluctuations
are small.
The formula giving the probability distribution is one of the
most important in statistical mechanics. It applies to any
system with a fixed number of particles in thermal equilibrium
(V, N, T fixed). We will not provide a derivation of this
probability formula, which is essentially the same as the
Boltzmann formula {See section 13.9 of the textbook.}
Ni gi
 e
N Z

 i
kT
24
e

 i
kT
is called the Boltzmann Factor
We see that at low temperatures only low energy states are
populated. As the temperature increases, the population
shifts to higher energy states.
25
Example: An atmosphere at a temperature of 6000K has, as one
constituent, neutral atoms with the first excited state 1.20eV
above the ground state. The ground state (taken as the reference
energy level of 0eV) is doubly degenerate while the 1st excited
state is 6-fold degenerate. Assuming that there is negligible
population of higher excited states, what fraction of these
molecules are in the 1st excited state? (Use Canonical Distribution.)
1.2eV
0eV
6
2
N
Ni  g ie
Z
Z  2e

0 eV
kT

i
kT
 6e
 (1)
Z 
1.20eV
eV
( 8.6210 5 ( 6103 K ))
K
n
 gie

i
kT
i 1
 2.59


 i
 i
Ni 1
1
Placing in (1)
 g i e kT 
g i e kT
N Z
2.59
N2
1

(6)e
N 2.59
1.20eV
eV
( 8.6210 5 ( 6103 K ))
K
N2
 0.227
N
26
Now we are going to discuss two other important distributions. In
preparation we review some aspects of fundamental particles.
These particles are divided into two classes, depending on their
quantum number called the intrinsic angular momentum or “spin”
Spin quantum number an integer: Bosons (examples are photons,
gravitons, pi mesons,……..)
Spin quantum number an odd number of (1/2):
Fermions (examples are electrons,
quarks, muons,……….)
Fermions obey the Pauli Exclusion Principle: In an isolated system,
no two fermions can occupy the same state.
27
Fermi-Dirac Distribution.
This distribution is for indistinguishable fermions. There can
be no more than one particle in any state. This places the following
restriction on any macrostate: N i  g i
Consider the ith energy level. We wish to place N i particles into
the g i states.
For the 1st particle there are
gi
possibilities.
For the 2nd particle there are ( g i  1) possibilities.

For the ith particle there are ( g i  N i  1)
possibilities.
The total number of possibilities is then ( g i )( g i  1)( g i  N i  1)
Since the particles are indistinguishable, we can permute them in a
particular distribution without obtaining a different distribution.
28
The total number of distinct possibilities is then
gi !
( gi )( gi  1)( gi  N i  1)
or
Ni !( gi  Ni )!
Ni !
The total thermodynamic probability for a particular macrostate is
n
gi !
w( N1, N2  N n )  
{What are the Ni
N
!
(
g

N
)!
i 1
i
i
i
at equilibrium?}
Again we will work with ln(w)
n
n
n
i 1
i 1
i 1
ln( w)   ln( gi ! )   ln( Ni ! )   ln(( gi  Ni )! )
The first term is constant. We use Stirling’s approximation for the
n
n
n
other terms.
ln( w)   ln( g i ! )   N i ln( N i )   N i
(gi constant!)
i 1
n
i 1
i 1
n
  ( g i  N i ) ln( g i  N i )   ( g i  N i )
i 1
i 1
29
n
n
n
Ni
d (ln( w))   dN i ln( N i )   dN i   (dN i ) ln( g i  N i )
i 1
i 1 N i
i 1
( gi  Ni )

(dN i )
i 1 ( g i  N i )
n
n
n
i 1
i 1
d (ln( w))   dNi ln( Ni )   (dNi ) ln( gi  Ni )
 gi  Ni 
(dN i )
d (ln( w))   ln
i 1
 Ni 
As before, we have the constraints:
n
N
i 1
i
 N  constant
n
n
N 
i 1
i i
We introduce the Lagrange multipliers
 U  constant
 and  
n
n
 gi  Ni 
(dNi )    (dNi )    i (dNi )  0
ln

i 1
i 1
i 1
 Ni 
n
30
  gi  Ni 

    i (dN i )  0
ln 

Ni 
i 1  

n
With the constraints included in the equation, the N i are independent
The coefficient of each dN i  0
 gi  Ni 
 gi  Ni 
    i  0 ln
    i
ln
 Ni 
 Ni 
gi
   i
1  e
Ni
Ni
1
   
i
gi
e
1
It is not trivial to determine the Lagrange multipliers. It turns out, as
1
before, that
 
kT
The other multiplier is related to the chemical potential
 

kT

With these assignments we have
31
Ni
1
fi 
 (   ) / kT
gi
e i
1
Fermi-Dirac distribution
For a continuous energy distribution, this becomes
f ( ) 
1
e
(   ) / kT
1
Later in the course we will use the Fermi-Dirac Distribution.
32
Bose-Einstein Statistics
The particles are again indistinguishable, but now any number
of particles can be in a particular energy state. (The Pauli Exclusion
Principle does not apply.)
Example: 13 particles in the ith energy level which has a degeneracy
of 8
partition
xxx x
xxx x
xx xx x
1
2
xx xx
3
4
5
x xx x
6
7
8 State
x x xx x
We can obtain different distributions by moving particles and
partitions around. How many ways can we arrange N i particles
and ( g i  1) partitions to form different distributions given that
the particles and partitions are identical? The answer is
Ni  gi  1! (Students: convince yourselves.)
Number of ways=
Ni ! gi  1!
33
6!
 20
Example: 3 particles in 4 degenerate states
3! 3!
Students: show these 20 distributions
(BEEx.ppt)
The thermodynamic probability for a macrostate is then:
n
w( N1, N2 , N n )  
i 1
N
 g i  1!
N i ! g i  1!
i
We proceed as before:
n
n
n
i 1
i 1
i 1
ln( w)   ln(( Ni  gi  1)! )   ln( Ni ! )   ln(( gi  1)! )
Using Stirling’s formula
ln( w) 
n
 (N
i 1
i
n
n
i 1
i 1
 g i  1) ln( N i  g i  1)   ( N i  g i  1)   N i ln( N i )
n
n
n
i 1
i 1
i 1
  N i   ( g i  1) ln( g i  1)   ( g i  1)
34
n
n
n
ln( w)   ( Ni  gi  1) ln( Ni  gi  1)   Ni ln( Ni )   ( gi  1) ln( gi  1)
i 1
i 1
n
n
i 1
n
n
i 1
i 1
d (ln( w))   (dNi ) ln( Ni  gi  1)   dNi   (dNi ) ln( Ni )   dNi
i 1
i 1
n
n
i 1
i 1
d (ln( w))   (dNi ) ln( Ni  gi  1)   (dNi ) ln( Ni )
 N  gi  1 
dNi
d (ln( w))   ln i
Ni
i 1


n
Introducing Lagrange multipliers with the constraint conditions:
n
n
n
 Ni  gi  1 
dN i    dN i    i dN i  0
ln

Ni
i 1
i 1
i 1


Proceeding as before: ln Ni  gi  1     
i


N
i


Neglecting the 1 (gi large!)
 Ni  gi 
    i
ln
 Ni 
35

gi 
    i
ln1 
Ni 

gi
   i
e
1
Ni
gi
   i
1
e
Ni
Ni
1
   
i
gi
e
1
Again, a more detailed analysis gives the same values as before
for the Lagrange multipliers
Ni
1
fi 
 (   ) / kT
gi
e i
1
Bose-Einstein Distribution
This distribution will be used later in the course.
36
Maxwell-Boltzmann Statistics and Review
Let us consider an assembly (system) of N indistinguishable
particles. A macrostate is a given distribution of particles in the
various energy levels. A microstate is a given distribution of particles
in the energy states.
Basic postulate of statistical mechanics: All accessible microstates
of an isolated system are equally probable of occurring.
We have considered some general assembly with:
N1 particles in any of the g1 states of 1
N 2 particles in any of the g2 states of 2

N i particles in any of the g i states of  i

We now impose the restriction that g i  N i for all i. This gives the
Maxwell-Boltzmann Statistics. This condition holds for all real gases
except at very low temperatures. At low temperatures one must use
either Bose-Einstein or Fermi-Dirac Statistics depending on the
37
spin of the molecules. This restriction means that it is very unlikely
that more than one particle will be in a given state. Subject to this
restriction we first consider the number of ways that N i
distinguishable particles can be distributed among the g i states.
The first particle can be placed in any one of the g i states. The
second particle can then be placed in any of the remaining ( g i  1)
states, and so forth. The total number of different ways is then:
g i ( g i  1)( g i  2)( g i  N i1)
Since g i  N i this is approximately g iN i
At present we are interested in indistinguishable particles and so
many of these distributions will be the same when the condition of
distinguishability is removed. We can start with one particular
distribution and then obtain identical distributions by permuting the
indistinguishable particles among themselves. The number of such
permutations is N i ! . Hence the number of ways, subject to the
restriction, that the N i indistinguishable particles can be
distributed among the states is (g iNi / N i ! )(1)
38
A fundamental problem in statistical mechanics is to determine the
particular macrostate of a system when at equilibrium.
The U of the isolated system is fixed and so each microstate has this
value. The laws of mechanics do not lead one to expect that the
system will be found in one microstate rather than in any other. This
is consistent with with our postulate that all microstates are equally
probable. Of course, all such postulates must be verified by
comparing the calculations based on the postulates with experimental
results.
Now let us return to a consideration of a particular macrostate,
that is, a particular set ( i , N i ) As we stated earlier, the number of
ways this particular macrostate can be achieved is called the
thermodynamic probability w. We have calculated this probability
(unnormalized) above for one energy level. The total thermodynamic
probability is the product of the individual probabilities for all the
N
n
accessible levels. From equation (1):
gi i
w( N1, N2 , N n )  
i 1 N i !
39
As shown in the textbook, the Fermi-Dirac and Bose-Einstein
probabilities reduce to this equation when one makes the
approximation g i  N i
The thermodynamic probability for Boltzmann Statistics and
Maxwell-Boltzmann Statistics is
w(Boltzmann )  N! w(Maxwell - Boltzmann )
Distinguishable
Indistinguishable
These two probabilities differ by a constant and hence their
derivatives (dw) will be the same. In addition the constraint
conditions (constant N and U) are the same. Hence, in using the
method of Lagrange Undetermined Multipliers, we will obtain the
same result.
Z 
n
 gie
i 1

i
kT
Ni
N
fi 
 e
gi
Z
g i  N i

i
kT
Maxwell-Boltzmann
so fi<<1
40
Statistical Interpretation of Heat and Work
For concreteness we return to the quantum mechanical formula
for the energy levels available for molecules of a gas in a container
2
of volume V.
h2 2 / 3 2
h
2
2
2 / 3 2
i 
8m
V
(n x  n y  n z ) 
8m
V
n
For a given quantum level (given n) the energy depends only upon
the volume.
d i
 i   i (V ) d i 
dV
dV
The internal energy of a gas is U 
n
N
i 1
i i
n
d i 
dU    i dNi   Ni d i    i dNi   Ni
 dV
dV 
i 1
i 1
i 1
 i 1
n
n
d i 
dU    i dN i   N i
 dV
dV 
i 1
 i 1
n
n
Macroscopically
n
dU= đ Q-PdV
41
Now suppose that the volume of the gas does not change. (No work
is done on the gas.) Comparison of the two equations yields
n
đ Q    i dN i
i 1
When energy is added by the heating process, the energies do not
change, but the distribution of particles among the energy levels
changes. (See Fig. 13.4). In heating the gas you are promoting
molecules from lower levels to higher levels.
Now suppose that, instead of heating the gas, one does work on the
gas, so that its volume decreases. In this case, the above formulae
shows that:
n
d i 
 n
dV   N i d i
đ W   N i

dV 
i 1
 i 1
In this case the distribution of particles among the energy levels does
not change, but the energy of the levels change. Doing work on the
gas raises the energy levels. (See Fig. 13.5).
42



 

W

 
Q
43
The entropy, Helmholtz function and the chemical potential in
terms of the partition function.
We obtain formulae using the Maxwell-Boltzmann distribution.
n
Ni
i
g
w( N1, N2 , N n )  
i 1 N i !
Ni
N
 e
gi
Z

i
kT
n
n

S  k ln w  k  N i ln g i   ln( N i ! ) Using the Stirling approx.
i 1
 i 1

n
n
n

n

Ni 
S  k  Ni ln gi   Ni ln N i   N i   k  N   N i ln 
gi 
i 1
i 1
i 1
 i 1


Ni
i
But ln
 ln N  ln Z 
n

gi
kT S  k N  N  ln N  ln Z   i  



i 1

1 
S  k  N  (ln N )( N )  (ln Z )( N ) 
U
kT 

i

kT  
44
U
S   Nk ln Z  ln N  1
T
(M-B Dist.)
Since we can write the Helmholtz function in terms of U and S we
can easily obtain an expression for F in terms of Z
U

F=U-TS
F  U  T   Nk ln Z  ln N  1
T

F   NkTln Z  ln N  1
{Note: Z=Z(T,V) so F(T,V)}
We have already determined the chemical potential in terms of F:
 F 
   
 N T ,V
Differentiating F with respect to N
 1
  kT ln Z  ln N  1  NkT  
 N
N
  kT ln  
(M-B Dist.)
Z
45
Finally we write the Maxwell-Boltzmann distribution in terms of
the chemical potential. This will permit a comparison of the three
i

distributions.
N
N
fi 
i
gi

Z
e
kT

Using the above expression for the chemical potential:

Substituting into the distribution:
N
 e kT
Z

 i
Ni
fi 
 e kT e kT
gi
Ni
1
fi 
 (   ) / kT
gi
e i
46
Comparison of the distributions
The chemical potential, which enters these distributions, will
be discussed when we use the distributions. At this stage we will
simply plot the distributions, which can all be represented by
fi 
a=
+1 Fermi-Dirac
-1 Bose-Einstein
0 Maxwell-Boltzman
1
e
( i   ) / kT
We plot these distributions as a function of
(MAPLE plot distrib.mws)
i   x  0
i   x  0
a
x
( i   )
kT
i   x  0
47
limit of validity (M-B)
48
Consider the BE distribution:
For  i   the distribution is infinite and for  i  
the distribution is negative and hence meaningless. The particles
tend to condense into the lower energy states.
Consider the FD distribution.
For x=0 f i  1 / 2
For x  
fi  1
The low-energy levels are very nearly uniformly populated
with one particle per state.
Consider the MB distribution
This distribution is only valid for f i  1
In this limit this distribution is an approximation for the BE and FD
distributions.
49
We have introduced four distributions. These give the distribution
of particles in the accessible states.
Boltzmann: Distinguishable particles. Any number can go into
an energy state.
Fermi-Dirac: Indistinguishable particles which obey the PEP.
Bose-Einstein: Indistinguishable particles. Any number can go
into an energy state.
Maxwell-Boltzmann: Indistinguishable particles. Any number
can go into an energy state. Valid when g i  N i
An extremely useful function was introduced: The partition function.
n
Z(T, V )   g i e

 i
kT
{Boltzmann Factor}
i 1
Many thermodynamic variables and functions can be written
in terms of this function. This function contains all the statistical
information about the system.
50
We also introduced the canonical distribution, a probability
i
distribution, N i

1
N

Z
e
kT
which has wide applicability.
51