10.40 Appendix
Connection to Thermodynamics and Derivation of
Boltzmann Distribution
Bernhardt L. Trout
Outline
• Cannonical ensemble
• Maximum term method
• Most probable distribution
• Ensembles continued: Canonical, Microcanonical, Grand Canonical, etc.
• Connection to thermodynamics
• Relation of thermodynamic quantities to Q
0.1
Canonical ensemble
In the Canonical ensemble, each system has constant N, V ,and T.
After equilibration, remove all of the systems from the bath, and put them all
together:
Apply postulate 2 to the ensemble of systems, also called a supersystem.
P
P Let nj = number of systems with energy E j . Also, N = j nj and E tot =
j nj E j .
If we know all E j ’s, then the state of the entire ensemble would be well-defined.
For example, let’s analyze an ensemble with 4 systems, labeled A, B, C, and D,
where
A
E2
B
E3
C
E2
D
E1
Then, E tot =E 1 + 2E 2 +E 3
→
Also, the distribution of the systems, −
n = (n1 , n2, n3 , ...) = (1, 2, 1).
But there are many different supersystems consistent with this distribution. In
fact, the number of supersystems consistent with this distribution is
N!
4!
→
Ωtot (−
n) = Q
=
= 12
n
!
1!2!1!
j j
What is the probability of observing a given quantum state, e.g. E j ? In other
words, what is the fraction of systems in the ensemble in the state E j ?
1
10.40: Fall 2003
Appendix
2
(Ensemble of systems)
N,V,EA
N,V,EB
...
N,V,EC
...
N,V,Ej
(Bath at temperature T)
Figure 1: Canonical ensemble
The answer is
nj
N.
However, it may be the case that many distributions fulfill the conditions of the
ensemble, (N, V , E tot ).
For example, assume that there are two:
n1 = 1, n2 = 2, n3 = 1; Ωtot = 12
and
n1 = 2, n2 = 0, n3 = 2; Ωtot = 6
Then the probability of observing, for example, E 3 is 14 in the first distribution
and 12 in the second.
The probability in the case where both distributions make up the ensemble is:
µ ¶
1 1 × 12 + 2 × 6
1
p3 =
=
4
12 + 6
3
In general,
pj =
µ
1
N
¶P
→
−
→
−
→
−
Ωtot ( n )nj ( n )
nP
−
→
→
−
n Ωtot ( n )
where the sum is over all distributions satisfying the conditions of (N, V , E tot ).
Then, for example, we could compute ensemble averages of mechanical quantities:
X
E = hEi =
Pj E j
j
2
10.40: Fall 2003
Appendix
3
N,V,EA N,V,EB N,V,EC
...
...
N,V,Ej
Figure 2: Canonical ensemble forming its own bath
and
P = hP i =
X
pj Pj ,
j
where p is the pressure.
0.2
Maximum term method
Recall:
pj =
where
µ
1
N
¶P
→
−
→
−
→
−
Ωtot ( n )nj ( n )
nP
−
→
→
−
n Ωtot ( n )
N!
→
Ωtot (−
n) = Q
.
j nj !
As N → ∞, nj → ∞, for each j.
Thus, the most probable distribution becomes dominant. We can call this dis→
tribution, −
n ∗.
→
∗
Let nj = nj in the −
n ∗ distribution. Then
pj =
0.3
→
−
n∗j
1 Ωtot ( n ∗ )n∗j
=
→
N Ωtot (−
n ∗)
N
Most probable distribution
Which distribution gives the largest Ωtot ?
Solve via method of undetermined multipliers:
Take natural log of Ωtot .
µ
¶ ÃX ! ÃX ! X
N!
→
−
ln (Ωtot ( n )) = ln Q
ni ln
ni −
ni ln ni ,
=
i ni !
i
i
i
where we have switched the index from j to i and used Stirling’s approximation,
which becomes exact as ni → ∞ :
ln y! ≈ y ln y − y.
→
→
We wish to find the set of nj ’s, which maximize Ωtot (−
n ) and hence ln(Ωtot (−
n )) :
#
"
X
X
∂
→
n )) − α
ni − β
ni E i = 0 , j = 1, 2, 3, ...
ln (Ωtot (−
∂nj
i
i
3
10.40: Fall 2003
Appendix
4
where α and β are the undetermined multipliers. Carrying out the differentiation
yields
Ã
!
X
ln
ni − ln n∗j − α − βE j = 0, j = 1, 2, 3, ...
i
or
n∗j = N e−α e−βE j , j = 1, 2, 3, ...
Recalling that
X
N =
yields
X
n∗j
j
e−α e−βE j = 1
j
or
eα =
X
e−βE j .
j
Also,
hEi =
P
j
n∗j E j
N
and
pj =
where
=
P
j
N e−α e−βE j E j
N
P
−βE j
Ej
je
= P −βE
j
je
n∗j
e−βEj
= e−α e−βEj = P −βEj ,
N
je
Q=
X
e−βE j
j
and, as we discussed in the last lecture, is the partition function, the normalization
factor.
0.4
Canonical ensemble continued and connection to thermodynamics
Recall from last time, via the maximum-term method in the canonical ensemble:
P ∗
P
P −βE
−α −βE j
jE
e
Ej
j
j nj E j
j Ne
je
hEi =
=
= P −βE
j
N
N
e
j
and
pj =
where,
n∗j
e−βE j
= e−α e−βE j = P −βE ,
j
N
je
Q=
X
e−βE j ,
j
as we discussed in the last lecture, is the partition function, the normalization factor.
In addition, as we have shown:
E = hEi =
4
X
j
pj E j
10.40: Fall 2003
Appendix
and
P = hP i =
where P is the pressure.
X
5
pj Pj ,
j
If we differentiate the equation for hEi,
X
X
dhEi =
E j dpj +
pj dE j
j
=−
j
X µ ∂E j ¶
1X
(ln pj + ln Q) dpj +
pj
dV .
β j
∂V N
j
Recall that the pressure,
µ
P =
or
µ
Pj =
This yields for equation 1
dhEi = −
(1)
¶
∂E
∂V
N
∂E j
∂V
¶
.
N
X
1X
1X
ln pj dpj −
ln Qdpj +
pj Pj dV .
β j
β j
j
[Note that


X
d
pj ln pj 
j
=
X
ln pj dpj +
j
=
pj d (ln pj )
j
X
ln pj dpj +
j
Since,
X
X
pj
j
X
dpj
.
pj
(2)
pj = 1,
j
X
dpj = 0.]
j
Thus, the right term in equation 2 is equal to 0. This yields


1 X
pj ln pj  − hP idV .
dhEi = − d
β
j
Recalling from the combined first and second laws (in intensive form, noting that
since N is a constant, intensive and extensive forms are equivalent):
dU = T dS − P dV
Since U ↔ hEi and p ↔ hpi,


1 X
T dS ↔ − d
pj ln pj  .
β
j
5
10.40: Fall 2003
Appendix
Let
X =−
X
6
pj ln pj .
j
Then,
dS =
1
dX.
βT
(3)
We know that the left side of the equation is an exact differential, so the right side
1
must be a function of X. This means that
must be too, and thus, βT
dS = φ(X)dX = df (X).
Integrating,
S = f (X) + const,
(4)
where we can set the arbitrary constant, const, equal to 0 for convenience.
Now we can make use of the additive property of S, and we can divide a system
into two parts, A and B. This yields:
S = S A + S B = f (X A ) + f (X B ).
Note that
X A+B = −
X
(5)
pi,j ln pi,j ,
i,j
where i is the index for the possible states of A and j is the index for the possible
states of B. Then
X
B
A
B
X A+B = −
pA
i pj (ln pi + ln pj )
i,j
=−
X
i
A
pA
i ln pi −
X
B
pB
j ln pj
j
= XA + XB.
Thus, from equation 5,
S = f (X A ) + f (X B ) = f (X A + X B ).
For this to be so,
f (X) = kX,
where k is a constant. Thus,
S = −k
X
pj ln pj .
j
From equations 3 and 4,
1
= k,
βT
and thus,
1
.
kT
We designate k as Boltzmann’s constant, a universal constant.
β=
6
(6)
10.40: Fall 2003
0.5
Appendix
7
Microcanonical, Grand Canonical, and other ensembles
Recalling the formulation for S from equation 6, and noting that in the microcanonical ensemble,
1
pj = ,
Ω
where we recall that Ω is the total number of states with the same energy, then
S = −k
X
j
pj ln pj = −k
X1
1
ln
Ω
Ω
j
= k ln Ω(N, V , E).
This is Boltzmann’s famous formula for the entropy.
In the Grand Canonical ensemble, the number of particles in each system is
allowed to fluctuate, but µ is kept constant. This is called the (V , T, µ) ensemble.
Also, there are other ensembles, such as (N, P, T ), etc. Note that from an analysis
of fluctuations (Lecture 27), we shall see that in the macroscopic limit of a large
number of systems, all of these ensembles are equivalent.
0.6
Relation of thermodynamic quantities to Q
Recall that
S
= −k
X
pj ln pj
j
e−βE j
Q
X
Q =
e−βE j
pj
=
j
Plugging in the formula for pj into that for S yields
S
X e−βE j
= −k
j
e−βE j
Q
¶
X e−βE j µ E j
−
− ln Q
Q
kT
j
= −k
=
Q
ln
hEi
+ k ln Q
T
Recalling our definitions from macroscopic thermodynamics and the fact that U ↔
hEi yields
A= −kT ln Q
Similarly,
S
P
U
= −
µ
∂A
∂T
¶
= kT
V ,Ni
∂ ln Q
∂T
¶
V ,Ni
¶
∂ ln Q
∂V
T,Ni
T,Ni
µ
¶
∂
ln
Q
= A + T S = kT 2
.
∂T
V ,Ni
= −
µ
∂A
∂V
¶
µ
= kT
7
µ
+ k ln Q
10.40: Fall 2003
Appendix
8
Thus, all thermodynamic properties can be written in terms of the partition function, Q(N, V , T )!
In order to compute Q, all we need are the possible energy levels of the system.
We can obtain these from solving the equations of quantum mechanics.
8