Download Introduction to Thermodynamics and Statistical Physics

Eyal Buks
Introduction to
Thermodynamics and Statistical
Physics (114016) - Lecture Notes
September 15, 2013
Technion
Preface
to be written...
Contents
1.
The Principle of Largest Uncertainty . . . . . . . . . . . . . . . . . . . . .
1.1 Entropy in Information Theory . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Example - Two States System . . . . . . . . . . . . . . . . . . . . .
1.1.2 Smallest and Largest Entropy . . . . . . . . . . . . . . . . . . . . . .
1.1.3 The composition property . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.4 Alternative Definition of Entropy . . . . . . . . . . . . . . . . . . .
1.2 Largest Uncertainty Estimator . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 The Free Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 The Principle of Largest Uncertainty in Statistical Mechanics
1.3.1 Microcanonical Distribution . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Canonical Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.3 Grandcanonical Distribution . . . . . . . . . . . . . . . . . . . . . .
1.3.4 Temperature and Chemical Potential . . . . . . . . . . . . . . .
1.4 Time Evolution of Entropy of an Isolated System . . . . . . . . . . .
1.5 Thermal Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Externally Applied Potential Energy . . . . . . . . . . . . . . . .
1.6 Free Entropy and Free Energies . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Problems Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 Solutions Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1
2
5
8
9
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21
29
2.
Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 A Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Gibbs Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Fermions and Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Fermi-Dirac Distribution . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Bose-Einstein Distribution . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 Classical Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Ideal Gas in the Classical Limit . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.4 Internal Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . .
2.5 Processes in Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Contents
2.5.1 Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.2 Isobaric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.3 Isochoric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.4 Isentropic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Carnot Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Limits Imposed Upon the Efficiency . . . . . . . . . . . . . . . . . . . . . .
Problems Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
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66
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79
3.
Bosonic and Fermionic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Electromagnetic Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Electromagnetic Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.2 Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.3 Cube Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.4 Average Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.5 Stefan-Boltzmann Radiation Law . . . . . . . . . . . . . . . . . . .
3.2 Phonons in Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 One Dimensional Example . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 The 3D Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Fermi Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Orbital Partition Function . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2 Partition Function of the Gas . . . . . . . . . . . . . . . . . . . . . .
3.3.3 Energy and Number of Particles . . . . . . . . . . . . . . . . . . . .
3.3.4 Example: Electrons in Metal . . . . . . . . . . . . . . . . . . . . . . .
3.4 Semiconductor Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Problems Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Solutions Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.
Classical Limit of Statistical Mechanics . . . . . . . . . . . . . . . . . . .
4.1 Classical Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Hamilton-Jacobi Equations . . . . . . . . . . . . . . . . . . . . . . . .
4.1.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Density Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 Equipartition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Nyquist Noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Thermal Equilibrium From Stochastic Processes . . . . . . . . . . . .
4.4.1 Langevin Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.2 The Smoluchowski-Chapman-Kolmogorov Relation . . .
4.4.3 The Fokker-Planck Equation . . . . . . . . . . . . . . . . . . . . . . .
4.4.4 The Potential Condition . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.5 Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.6 Fokker-Planck Equation in One Dimension . . . . . . . . . .
4.4.7 Ornstein—Uhlenbeck Process in One Dimension . . . . . . .
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2.6
2.7
2.8
2.9
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Contents
4.5 Problems Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
4.6 Solutions Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
5.
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
5.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
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1. The Principle of Largest Uncertainty
In this chapter we discuss relations between information theory and statistical
mechanics. We show that the canonical and grand canonical distributions
can be obtained from Shannon’s principle of maximum uncertainty [1, 2, 3].
Moreover, the time evolution of the entropy of an isolated system and the H
theorem are discussed.
1.1 Entropy in Information Theory
The possible states of a given system are denoted as em , where m = 1, 2, 3, ...,
and the probability that state em is occupied is denoted by pm . The normalization condition reads
pm = 1 .
(1.1)
m
For a given probability distribution {pm } the entropy is defined as
pm log pm .
σ=−
(1.2)
m
Below we show that this quantity characterizes the uncertainty in the knowledge of the state of the system.
1.1.1 Example - Two States System
Consider a system which can occupy either state e1 with probability p, or
state e2 with probability 1 − p, where 0 ≤ p ≤ 1. The entropy is given by
σ = −p log p − (1 − p) log (1 − p) .
(1.3)
Chapter 1. The Principle of Largest Uncertainty
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.2
0.4
x
0.6
0.8
1
−p log p − (1 − p) log (1 − p)
As expected, the entropy vanishes at p = 0 and p = 1, since in both cases
there is no uncertainty in what is the state which is occupied by the system.
The largest uncertainty is obtained at p = 0.5, for which σ = log 2 = 0.69.
1.1.2 Smallest and Largest Entropy
Smallest value. The term −p log p in the range 0 ≤ p ≤ 1 is plotted in
the figure below. Note that the value of −p log p in the limit p → 0 can be
calculated using L’Hospital’s rule
d log p lim (−p log p) = lim
p→0
p→0
−
dp
d 1
dp p
=0.
(1.4)
From this figure, which shows that −p log p ≥ 0 in the range 0 ≤ p ≤ 1, it is
easy to infer that the smallest possible value of the entropy is zero. Moreover,
since −p log p = 0 iff p = 0 or p = 1, it is evident that σ = 0 iff the system
occupies one of the states with probability one and all the other states with
probability zero. In this case there is no uncertainty in what is the state which
is occupied by the system.
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Thermodynamics and Statistical Physics
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1.1. Entropy in Information Theory
0.35
0.3
0.25
-p*log(p)
0.2
0.15
0.1
0.05
0
0.2
0.4
p
0.6
0.8
1
−p log p
Largest value. We seek a maximum point of the entropy σ with respect to
all probability distributions {pm } which satisfy the normalization condition.
This constrain, which is given by Eq. (1.1), is expressed as
pm − 1 ,
(1.5)
0 = g0 (p̄) =
m
where p̄ denotes the vector of probabilities
p̄ = (p1 , p2 , ...) .
(1.6)
A small change in σ (denoted as δσ) due to a small change in p̄ (denoted as
δ p̄ = (δp1 , δp2 , ...)) can be expressed as
δσ =
∂σ
δpm ,
∂pm
m
(1.7)
¯ as
or in terms of the gradient of σ (denoted as ∇σ)
¯ · δ p̄ .
δσ = ∇σ
(1.8)
In addition the variables (p1 , p2 , ...) are subjected to the constrain (1.5). Similarly to Eq. (1.8) we have
¯ 0 · δ p̄ .
δg0 = ∇g
¯ and δ p̄ can be decomposed as
Both vectors ∇σ
¯ = ∇σ
¯
¯
∇σ
+ ∇σ
,
⊥
(1.9)
(1.10)
δ p̄ = (δ p̄) + (δ p̄)⊥ ,
(1.11)
¯
¯ 0 , and where ∇σ
¯
where ∇σ
and (δ p̄) are parallel to ∇g
and (δ p̄)⊥ are
⊥
¯ 0 . Using this notation Eq. (1.8) can be expressed as
orthogonal to ∇g
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Thermodynamics and Statistical Physics
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Chapter 1. The Principle of Largest Uncertainty
¯
¯
δσ = ∇σ
· (δ p̄) + ∇σ
· (δ p̄)⊥ .
⊥
(1.12)
Given that the constrain g0 (p̄) = 0 is satisfied at a given point p̄, one has
¯ 0,
g0 (p̄ + δ p̄) = 0 to first order in δ p̄ provided that δ p̄, is orthogonal to ∇g
namely, provided that (δ p̄) = 0. Thus, a stationary (maximum or minimum
or saddle point) point of σ occurs iff for every small change δ p̄, which is
¯ 0 (namely, δ p̄ · ∇g
¯ 0 = 0) one has 0 = δσ = ∇σ
¯
orthogonal to ∇g
·δ p̄. As
¯
can be seen from Eq. (1.12), this condition is fulfilled only when ∇σ
= 0,
⊥
¯ and ∇g
¯ 0 are parallel to each other. In other
namely only when the vectors ∇σ
words, only when
¯ = ξ 0 ∇g
¯ 0,
∇σ
(1.13)
where ξ 0 is a constant. This constant is called Lagrange multiplier . Using
Eqs. (1.2) and (1.5) the condition (1.13) is expressed as
log pm + 1 = ξ 0 .
(1.14)
Let M be the number of available states. From Eq. (1.14) we find that all
probabilities are equal. Thus using Eq. (1.5), one finds that
p1 = p2 = ... =
1
.
M
(1.15)
After finding this stationary point it is necessary to determine whether it
is a maximum or minimum or saddle point. To do this we expand σ to second
order in δ p̄
¯ σ (p̄)
σ (p̄ + δ p̄) = exp δ p̄ · ∇
¯ 2
δ
p̄
·
∇
¯+
= 1 + δ p̄ · ∇
+ ... σ (p̄)
2!
¯ 2
δ p̄ · ∇
¯
= σ (p̄) + δ p̄ · ∇σ +
σ + ...
2!
∂σ
1 ∂2σ
= σ (p̄) +
δpm +
δpm δpm′
+ ...
∂pm
2 m,m′
∂pm ∂pm′
m
(1.16)
Using Eq. (1.2) one finds that
∂2σ
1
= − δ m,m′ .
∂pm ∂pm′
pm
(1.17)
Since the probabilities pm are non-negative one concludes that any stationary
point of σ is a local maximum point. Moreover, since only a single stationary
point was found, one concludes that the entropy σ obtains its largest value,
which is denoted as Λ (M), and which is given by
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Thermodynamics and Statistical Physics
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1.1. Entropy in Information Theory
Λ (M) = σ
1 1
1
, , ...,
M M
M
= log M ,
(1.18)
for the probability distribution given by Eq. (1.15). For this probability distribution that maximizes σ, as expected, the state which is occupied by the
system is most uncertain.
1.1.3 The composition property
The composition property is best illustrated using an example.
Example - A Three States System. A system can occupy one of the
states e1 , e2 or e3 with probabilities p1 , p2 and p3 respectively. The uncertainty associated with this probability distribution can be estimated in two
ways, directly and indirectly. Directly, it is simply given by the definition of
entropy in Eq. (1.2)
σ (p1 , p2 , p3 ) = −p1 log p1 − p2 log p2 − p3 log p3 .
(1.19)
Alternatively [see Fig. 1.1], the uncertainty can be decomposed as follows:
(a) the system can either occupy state e1 with probability p1 , or not occupy
state e1 with probability 1 − p1 ; (b) given that the system does not occupy
state e1 , it can either occupy state e2 with probability p2 / (1 − p1 ) or occupy
state e3 with probability p3 / (1 − p1 ). Assuming that uncertainty (entropy)
is additive, the total uncertainty (entropy) is given by
p2
p3
σi = σ (p1 , 1 − p1 ) + (1 − p1 ) σ
,
.
(1.20)
1 − p1 1 − p1
The factor (1 − p1 ) in the second term is included since the uncertainty associated with distinction between states e2 and e3 contributes only when state
e1 is not occupied, an event which occurs with probability 1 − p1 . Using the
definition (1.2) and the normalization condition
p1 + p2 + p3 = 1 ,
(1.21)
one finds
σi = −p1 log p1 − (1 − p1 ) log (1 − p1 )
p2
p2
p3
p3
+ (1 − p1 ) −
log
−
log
1 − p1
1 − p1 1 − p1
1 − p1
= −p1 log p1 − p2 log p2 − p3 log p3
− (1 − p1 − p2 − p3 ) log (1 − p1 )
= σ (p1 , p2 , p3 ) ,
(1.22)
that is, for this example the entropy satisfies the decomposition property.
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Thermodynamics and Statistical Physics
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Chapter 1. The Principle of Largest Uncertainty
p1
1 − p1
e1
e2 , e3
p2
1 − p1
e2
p3
1 − p1
e3
Fig. 1.1. The composition property - three states system.
The general case. The composition property in the general case can
be defined as follows. Consider a system which can occupy one of the
states {e1 , e2 , ..., eM0 } with probabilities q1 , q2 , ..., qM0 respectively. This set
of states is grouped as follows. The first group includes the first M1 states
{e1 , e2 , ..., eM1 }; the second group includes the next M2 states {eM1 +1 , eM1 +2 , ..., eM1 +M2 },
etc., where M1 + M2 + ... = M0 . The probability that one of the states in the
first group is occupied is p1 = q1 +q2 +...+qM1 , the probability that one of the
states in the second group is occupied is p2 = qM1 +1 + qM1 +2 + ... + qM1 +M2 ,
etc., where
p1 + p2 + ... = 1 .
(1.23)
The composition property requires that the following holds [see Fig. 1.2]
σ (q1 , q2 , ..., qM0 ) = σ (p1 , p2 , ...)
q1 q2
qM
+ p1 σ
, , ..., 1
p1 p1
p1
qM1 +1 qM1 +2
qM1 +M2
+ p2 σ
,
, ...,
+ ...
p2
p2
p2
(1.24)
Using the definition (1.2) the following holds
σ (p1 , p2 , ...) = −p1 log p1 − p2 log p2 − ... ,
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Thermodynamics and Statistical Physics
(1.25)
6
1.1. Entropy in Information Theory
p1
e1 , e2 ,..., eM 1
q1
p1
e1
q2
p1
p2
…
eM 1 +1 ,..., eM 1 + M 2
qM 1
qM1
qM 1 + 2
qM 1 + M 2
p1
p2
p2
p2
… e
e2
M
1
p1 = q1 + q2 + ... + qM 1
…
eM 1 +1 eM1 + 2 eM 1 + M 2
p2 = qM 1 +1 + ... + qM 1 + M 2
Fig. 1.2. The composition property - the general case.
q1 q2
qM
, , ..., 1
p1 p1
p1
q1
q2
q2
qM1
qM
q1
−
log
− ... −
log 1
= p1 − log
p1
p1 p1
p1
p1
p1
= −q1 log q1 − q2 log q2 − ... − qM1 log qM1
+ p1 log p1 ,
p1 σ
(1.26)
qM1 +1 qM1 +2
qM1 +M2
p2 σ
,
, ...,
p2
p2
p2
= −qM1 +1 log qM1 +1 − qM1 +2 log qM1 +2
− ... − qM1 +M2 log qM1 +M2
+ p2 log p2 ,
(1.27)
etc., thus it is evident that condition (1.24) is indeed satisfied.
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Chapter 1. The Principle of Largest Uncertainty
1.1.4 Alternative Definition of Entropy
Following Shannon [1, 2], the entropy function σ (p1 , p2 , ..., pN ) can be alternatively defined as follows:
1. σ (p1 , p2 , ..., pN ) is a continuous function of its arguments p1 , p2 , ..., pN .
2. If all probabilities are equal, namely if p1 = p2 = ... = pN = 1/N, then
the quantity Λ (N ) = σ (1/N, 1/N, ..., 1/N ) is a monotonic increasing
function of N.
3. The function σ (p1 , p2 , ..., pN ) satisfies the composition property given by
Eq. (1.24).
Exercise 1.1.1. Show that the above definition leads to the entropy given
by Eq. (1.2) up to multiplication by a positive constant.
Solution 1.1.1. The 1st property allows approximating the probabilities
p1 , p2 , ..., pN using rational numbers, namely p1 = M1 /M0 , p2 = M2 /M0 ,
etc., where M1 , M2 , ... are integers and M0 = M1 + M2 + ... + MN . Using the
composition property (1.24) one finds
Λ (M0 ) = σ (p1 , p2 , ..., pN ) + p1 Λ (M1 ) + p2 Λ (M2 ) + ...
(1.28)
In particular, consider the case were M1 = M2 = ... = MN = K. For this
case one finds
Λ (NK) = Λ (N) + Λ (K) .
(1.29)
Taking K = N = 1 yields
Λ (1) = 0 .
(1.30)
Taking N = 1 + x yields
Λ (K + Kx) − Λ (K)
1 Λ (1 + x)
=
.
Kx
K
x
(1.31)
Taking the limit x → 0 yields
dΛ
C
=
,
dK
K
(1.32)
where
C = lim
x→0
Λ (1 + x)
.
x
(1.33)
Integrating Eq. (1.32) and using the initial condition (1.30) yields
Λ (K) = C log K .
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Thermodynamics and Statistical Physics
(1.34)
8
1.2. Largest Uncertainty Estimator
Moreover, the second property requires that C > 0. Choosing C = 1 and
using Eq. (1.28) yields
σ (p1 , p2 , ..., pN ) = Λ (M0 ) − p1 Λ (M1 ) − p2 Λ (M2 ) − ...
M1
M2
MN
− p2 log
− ... − pM log
= −p1 log
M0
M0
M0
= −p1 log p1 − p2 log p2 − ... − pN log pN ,
(1.35)
in agreement with the definition (1.2).
1.2 Largest Uncertainty Estimator
As before, the possible states of a given system are denoted as em , where
m = 1, 2, 3, ..., and the probability that state em is occupied is denoted by
pm . Let Xl (l = 1, 2, ..., L) be a set of variables characterizing the system
(e.g., energy, number of particles, etc.). Let Xl (m) be the value which the
variable Xl takes when the system is in state em . Consider the case where
the expectation values of the variables Xl are given
Xl =
pm Xl (m) ,
(1.36)
m
where l = 1, 2, ..., L. However, the probability distribution {pm } is not given.
Clearly, in the general case the knowledge of X1 , X2 , ..., XL is not
sufficient to obtain the probability distribution because there are in general
many different possibilities for choosing a probability distribution which is
consistent with the contrarians (1.36) and the normalization condition (1.1).
For each such probability distribution the entropy can be calculated according
to the definition (1.2). The probability distribution {pm }, which is consistent
with these conditions, and has the largest possible entropy is called the largest
uncertainty estimator (LUE).
The LUE is found by seeking a stationary point of the entropy σ with
respect to all probability distributions {pm } which satisfy the normalization
constrain (1.5) in addition to the constrains (1.36), which can be expressed
as
0 = gl (p̄) =
pm Xl (m) − Xl ,
(1.37)
m
where l = 1, 2, ...L. To first order one has
¯ · δ p̄ ,
δσ = ∇σ
¯ l · δ p̄ ,
δgl = ∇g
Eyal Buks
Thermodynamics and Statistical Physics
(1.38a)
(1.38b)
9
Chapter 1. The Principle of Largest Uncertainty
where l = 0, 1, 2, ...L. A stationary point of σ occurs iff for every small change
¯ 0 , ∇g
¯ 1 , ∇g
¯ 2 , ..., ∇g
¯ L one has
δ p̄, which is orthogonal to all vectors ∇g
¯ · δ p̄ .
0 = δσ = ∇σ
(1.39)
¯
This condition is fulfilled
only when the vector ∇σ belongs to the subspace
¯
¯
¯
¯
spanned by the vectors ∇g0 , ∇g1 , ∇g2 , ..., ∇gL [see also the discussion below Eq. (1.12) above]. In other words, only when
¯ = ξ 0 ∇g
¯ 0 + ξ 1 ∇g
¯ 1 + ξ 2 ∇g
¯ 2 + ... + ξ L ∇g
¯ L,
∇σ
(1.40)
where the numbers ξ 0 , ξ 1 , ..., ξ L , which are called Lagrange multipliers, are
constants. Using Eqs. (1.2), (1.5) and (1.37) the condition (1.40) can be
expressed as
− log pm − 1 = ξ 0 +
L
ξ l Xl (m) .
(1.41)
l=1
From Eq. (1.41) one obtains
pm = exp (−1 − ξ 0 ) exp −
L
ξ l Xl (m)
l=1
.
(1.42)
The Lagrange multipliers ξ 0 , ξ 1 , ..., ξ L can be determined from Eqs. (1.5) and
(1.37)
L
1=
pm = exp (−1 − ξ 0 )
exp −
ξ l Xl (m) ,
(1.43)
m
Xl =
m
m
l=1
pm Xl (m)
= exp (−1 − ξ 0 )
m
exp −
L
l=1
ξ l Xl (m) Xl (m) .
Using Eqs. (1.42) and (1.43) one finds
L
exp −
ξ l Xl (m)
pm =
l=1L
.
exp −
ξ l Xl (m)
m
(1.44)
(1.45)
l=1
In terms of the partition function Z, which is defined as
Eyal Buks
Thermodynamics and Statistical Physics
10
1.2. Largest Uncertainty Estimator
Z=
,
(1.46)
L
1
= exp −
ξ l Xl (m) .
Z
(1.47)
m
exp −
L
ξ l Xl (m)
l=1
one finds
pm
l=1
Using the same arguments as in section 1.1.2 above [see Eq. (1.16)] it is easy to
show that at the stationary point that occurs for the probability distribution
given by Eq. (1.47) the entropy obtains its largest value.
1.2.1 Useful Relations
The expectation value Xl can be expressed as
pm Xl (m)
Xl =
m
L
1 =
exp −
ξ l Xl (m) Xl (m)
m
Z
l=1
1 ∂Z
=−
Z ∂ξ l
∂ log Z
=−
.
∂ξ l
(1.48)
Similarly, Xl2 can be expressed as
2 Xl =
pm Xl2 (m)
m
L
1 exp −
ξ l Xl (m) Xl2 (m)
=
Z m
l=1
1 ∂2Z
=
.
Z ∂ξ 2l
(1.49)
Using Eqs. (1.48) and (1.49) one finds that the variance of the variable Xl is
given by
2
1 ∂2Z
1 ∂Z
2
2
(∆Xl ) = (Xl − Xl ) =
−
.
(1.50)
Z ∂ξ 2l
Z ∂ξ l
However, using the following identity
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Thermodynamics and Statistical Physics
11
Chapter 1. The Principle of Largest Uncertainty
∂ 2 log Z
∂ 1 ∂Z
1 ∂2Z
=
=
−
2
∂ξ l Z ∂ξ l
Z ∂ξ 2l
∂ξ l
one finds
∂ 2 log Z
.
(∆Xl )2 =
∂ξ 2l
1 ∂Z
Z ∂ξ l
2
,
(1.51)
(1.52)
Note that the above results Eqs. (1.48) and (1.52) are valid only when Z is
expressed as a function of the the Lagrange multipliers, namely
Z = Z (ξ 1 , ξ 2 , ..., ξ L ) .
(1.53)
Using the definition of entropy (1.2) and Eq. (1.47) one finds
pm log pm
σ=−
m
L
1
=−
pm log
ξ l Xl (m)
exp −
Z
m
l=1
L
=
pm log Z +
ξ l Xl (m)
m
l=1
= log Z +
L
ξl
L
ξ l Xl .
l=1
pm Xl (m) ,
m
(1.54)
thus
σ = log Z +
l=1
(1.55)
Using the above relations one can also evaluate the partial derivative of
the entropy σ when it is expressed as a function of the expectation values,
namely
σ = σ (X1 , X2 , ..., XL ) .
(1.56)
Using Eq. (1.55) one has
L
L
l =1
l =1
∂ Xl′ ∂σ
∂ log Z ∂ξ l′
=
+
Xl′ +
ξ l′
∂ Xl ∂ Xl ∂ Xl ∂ Xl ′
′
=
=
∂ log Z
+
∂ Xl L
l′ =1
Eyal Buks
L
l′ =1
Xl′ ∂ξ l′
+ ξl
∂ Xl L
∂ log Z ∂ξ l′
∂ξ l′
+
+ ξl ,
Xl′ ∂ξ l′ ∂ Xl ∂
Xl ′
l =1
Thermodynamics and Statistical Physics
(1.57)
12
1.2. Largest Uncertainty Estimator
thus using Eq. (1.48) one finds
∂σ
= ξl .
∂ Xl (1.58)
1.2.2 The Free Entropy
The free entropy σF is defined as the term log Z in Eq. (1.54)
σF = log Z
L
=σ−
ξl
pm Xl (m)
m
l=1
=−
m
pm log pm −
L
l=1
ξl
pm Xl (m) .
m
(1.59)
The free entropy is commonly expressed as a function of the Lagrange multipliers
σF = σF (ξ 1 , ξ 2 , ..., ξ L ) .
(1.60)
We have seen above that the LUE maximizes σ for given values of expectation values X1 , X2 , ..., XL . We show below that a similar result can be
obtained for the free energy σF with respect to given values of the Lagrange
multipliers.
Claim. The LUE maximizes σF for given values of the Lagrange multipliers
ξ 1 , ξ 2 , ..., ξ L .
Proof. As before, the normalization condition is expressed as
0 = g0 (p̄) =
pm − 1 .
(1.61)
m
At a stationary point of σF , as we have seen previously, the following holds
¯ F = η∇g
¯ 0,
∇σ
(1.62)
where η is a Lagrange multiplier. Thus
− (log pm + 1) −
or
L
ξ l Xl (m) = η ,
pm = exp (−η − 1) exp −
Eyal Buks
(1.63)
l=1
L
l=1
ξ l Xl (m)
.
Thermodynamics and Statistical Physics
(1.64)
13
Chapter 1. The Principle of Largest Uncertainty
Table 1.1. The microcanonical, canonical and grandcanonical distributions.
energy
number of particles
microcanonical distribution
constrained U (m) = U
constrained N (m) = N
canonical distribution
average is given U constrained N (m) = N
grandcanonical distribution
average is given U average is given N This result is the same as the one given by Eq. (1.42). Taking into account
the normalization condition (1.61) one obtains the same expression for pm as
the one given by Eq. (1.47). Namely, the stationary point of σ F corresponds
to the LUE probability distribution. Since
∂ 2 σF
1
= − δ m,m′ < 0 ,
∂pm ∂pm′
pm
(1.65)
one concludes that this stationary point is a maximum point [see Eq. (1.16)].
1.3 The Principle of Largest Uncertainty in Statistical
Mechanics
The energy and number of particles of state em are denoted by U (m) and
N (m) respectively. The probability that state em is occupied is denoted as
pm . We consider below three cases (see table 1.1). In the first case (microcanonical distribution) the system is isolated and its total energy U and number of particles N are constrained , that is for all accessible states U (m) = U
and N (m) = N. In the second case (canonical distribution) the system is
allowed to exchange energy with the environment, and we assume that its
average energy U is given. However, its number of particles is constrained
, that is N (m) = N . In the third case (grandcanonical distribution) the system is allowed to exchange both energy and particles with the environment,
and we assume that both the average energy U and the average number
of particles N are given. However, in all cases, the probability distribution
{pm } is not given.
According to the principle of largest uncertainty in statistical mechanics
the LUE is employed to estimate the probability distribution {pm }, namely,
we will seek a probability distribution which is consistent with the normalization condition (1.1) and with the given expectation values (energy, in the
second case, and both energy and number of particles, in the third case),
which maximizes the entropy.
1.3.1 Microcanonical Distribution
In this case no expectation values are given. Thus we seek a probability
distribution which is consistent with the normalization condition (1.1), and
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Thermodynamics and Statistical Physics
14
1.3. The Principle of Largest Uncertainty in Statistical Mechanics
which maximizes the entropy. The desired probability distribution is
p1 = p2 = ... = 1/M ,
(1.66)
where M is the number of accessible states of the system [see also Eq. (1.18)].
Using Eq. (1.2) the entropy for this case is given by
σ = log M .
(1.67)
1.3.2 Canonical Distribution
Using Eq. (1.47) one finds that the probability distribution is given by
pm =
1
exp (−βU (m)) ,
Zc
(1.68)
where β is the Lagrange multiplier associated with the given expectation
value U , and the partition function is given by
Zc =
exp (−βU (m)) .
(1.69)
m
The term exp (−βU (m)) is called Boltzmann factor.
Moreover, Eq. (1.48) yields
U = −
∂ log Zc
,
∂β
(1.70)
Eq. (1.52) yields
∂ 2 log Z
c
(∆U )2 =
,
∂β 2
(1.71)
and Eq. (1.55) yields
σ = log Zc + β U .
(1.72)
Using Eq. (1.58) one can expressed the Lagrange multiplier β as
β=
∂σ
.
∂U
(1.73a)
The temperature τ = 1/β is defined as
1
=β.
τ
(1.74)
Exercise 1.3.1. Consider a system that can be in one of two states
having
energies ±ε/2. Calculate the average energy U and the variance (∆U )2
in thermal equilibrium at temperature τ.
Eyal Buks
Thermodynamics and Statistical Physics
15
Chapter 1. The Principle of Largest Uncertainty
Solution: The partition function is given by Eq. (1.69)
βε
βε
βε
+ exp −
= 2 cosh
,
Zc = exp
2
2
2
thus using Eqs. (1.70) and (1.71) one finds
ε
βε
U = − tanh
,
2
2
(1.75)
(1.76)
and
ε 2
1
2
(∆U ) =
2 cosh2
βε
2
,
(1.77)
where β = 1/τ .
1
2
x
3
4
5
0
-0.2
-0.4
-tanh(1/x)
-0.6
-0.8
-1
1.3.3 Grandcanonical Distribution
Using Eq. (1.47) one finds that the probability distribution is given by
pm =
1
exp (−βU (m) − ηN (m)) ,
Zgc
(1.78)
where β and η are the Lagrange multipliers associated with the given expectation values U
and N respectively, and the partition function is given
by
exp (−βU (m) − ηN (m)) .
(1.79)
Zgc =
m
The term exp (−βU (m) − ηN (m)) is called Gibbs factor.
Moreover, Eq. (1.48) yields
Eyal Buks
Thermodynamics and Statistical Physics
16
1.3. The Principle of Largest Uncertainty in Statistical Mechanics
U = −
∂ log Zgc
∂β
∂ log Zgc
N = −
∂η
,
(1.80)
η
(1.81)
β
Eq. (1.52) yields
∂ 2 log Z gc
2
,
(∆U ) =
∂β 2
η
∂ 2 log Z gc
2
(∆N ) =
,
∂η2
β
(1.82)
(1.83)
and Eq. (1.55) yields
σ = log Zgc + β U
+ η N .
(1.84)
1.3.4 Temperature and Chemical Potential
Probability distributions in statistical mechanics of macroscopic parameters
are typically extremely sharp and narrow. Consequently, in many cases no
distinction is made between a parameter and its expectation value. That is,
the expression for the entropy in Eq. (1.72) can be rewritten as
σ = log Zc + βU ,
(1.85)
and the one in Eq. (1.84) as
σ = log Zgc + βU + ηN .
(1.86)
Using Eq. (1.58) one can expressed the Lagrange multipliers β and η as
∂σ
β=
,
(1.87)
∂U N
∂σ
η=
.
(1.88)
∂N U
The chemical potential µ is defined as
µ = −τ η .
(1.89)
In the definition (1.2) the entropy σ is dimensionless. Historically, the entropy
was defined as
S = kB σ ,
(1.90)
where
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Thermodynamics and Statistical Physics
17
Chapter 1. The Principle of Largest Uncertainty
kB = 1.38 × 10−23 J K−1
(1.91)
is the Boltzmann constant. Moreover, the historical definition of the temperature is
τ
.
(1.92)
T =
kB
When the grandcanonical partition function is expressed in terms of β
and µ (instead of in terms of β and η), it is convenient to rewrite Eqs. (1.80)
and (1.81) as (see homework exercises 14 of chapter 1).
∂ log Zgc
∂ log Zgc
U = −
+ τµ
,
(1.93)
∂β
∂µ
µ
β
∂ log Zgc
,
∂λ
where λ is the fugacity , which is defined by
N = λ
(1.94)
λ = exp (βµ) = e−η .
(1.95)
1.4 Time Evolution of Entropy of an Isolated System
Consider a perturbation which results in transitions between the states of an
isolated system. Let Γrs denotes the resulting rate of transition from state r
to state s. The probability that state s is occupied is denoted as ps .
The following theorem (H theorem) states that if for every pair of states
r and s
Γrs = Γsr ,
(1.96)
then
dσ
≥0.
dt
(1.97)
Moreover, equality holds iff ps = pr for all pairs of states for which Γsr = 0.
To prove this theorem we express the rate of change in the probability ps
in terms of these transition rates
dpr =
ps Γsr −
pr Γrs .
(1.98)
dt
s
s
The first term represents the transitions to state r, whereas the second one
represents transitions from state r. Using property (1.96) one finds
dpr =
Γsr (ps − pr ) .
dt
s
Eyal Buks
Thermodynamics and Statistical Physics
(1.99)
18
1.5. Thermal Equilibrium
The last result and the definition (1.2) allows calculating the rate of change
of entropy
dσ
d =−
pr log pr
dt
dt r
dpr
(log pr + 1)
=−
dt
r
=−
Γsr (ps − pr ) (log pr + 1) .
r
s
(1.100)
One the other hand, using Eq. (1.96) and exchanging the summation indices
allow rewriting the last result as
dσ Γsr (ps − pr ) (log ps + 1) .
=
dt
r
s
(1.101)
Thus, using both expressions (1.100) and (1.101) yields
dσ
1 =
Γsr (ps − pr ) (log ps − log pr ) .
dt
2 r s
(1.102)
In general, since log x is a monotonic increasing function
(ps − pr ) (log ps − log pr ) ≥ 0 ,
(1.103)
and equality holds iff ps = pr . Thus, in general
dσ
≥0,
dt
(1.104)
and equality holds iff ps = pr holds for all pairs is states satisfying Γsr =
0. When σ becomes time independent the system is said to be in thermal
equilibrium. In thermal equilibrium, when all accessible states have the same
probability, one finds using the definition (1.2)
σ = log M ,
(1.105)
where M is the number of accessible states of the system.
Note that the rates Γrs , which can be calculated using quantum mechanics, indeed satisfy the property (1.96) for the case of an isolated system.
1.5 Thermal Equilibrium
Consider two isolated systems denoted as S1 and S2 . Let σ1 = σ 1 (U1 , N1 ) and
σ 2 = σ2 (U2 , N2 ) be the entropy of the first and second system respectively
Eyal Buks
Thermodynamics and Statistical Physics
19
Chapter 1. The Principle of Largest Uncertainty
and let σ = σ 1 + σ2 be the total entropy. The systems are brought to contact
and now both energy and particles can be exchanged between the systems.
Let δU be an infinitesimal energy, and let δN be an infinitesimal number of
particles, which are transferred from system 1 to system 2. The corresponding
change in the total entropy is given by
∂σ 1
∂σ 2
δσ = −
δU +
δU
∂U1 N1
∂U2 N2
∂σ1
∂σ 2
−
δN +
δN
∂N1 U1
∂N2 U2
1
1
µ
µ
δU − − 1 + 2 δN .
= − +
τ1 τ2
τ1
τ2
(1.106)
The change δσ in the total entropy is obtained by removing a constrain.
Thus, at the end of this process more states are accessible, and therefore,
according to the principle of largest uncertainty it is expected that
δσ ≥ 0 .
(1.107)
For the case where no particles can be exchanged (δN = 0) this implies that
energy flows from the system of higher temperature to the system of lower
temperature. Another important case is the case where τ 1 = τ 2 , for which
we conclude that particles flow from the system of higher chemical potential
to the system of lower chemical potential.
In thermal equilibrium the entropy of the total system obtains its largest
possible value. This occurs when
τ1 = τ2
(1.108)
µ1 = µ2 .
(1.109)
and
1.5.1 Externally Applied Potential Energy
In the presence of externally applied potential energy µex the total chemical
potential µtot is given by
µtot = µint + µex ,
(1.110)
where µint is the internal chemical potential . For example, for particles having
charge q in the presence of electric potential V one has
µex = qV ,
Eyal Buks
(1.111)
Thermodynamics and Statistical Physics
20
1.7. Problems Set 1
whereas, for particles having mass m in a constant gravitational field g one
has
µex = mgz ,
(1.112)
where z is the height. The thermal equilibrium relation (1.109) is generalized
in the presence of externally applied potential energy as
µtot,1 = µtot,2 .
(1.113)
1.6 Free Entropy and Free Energies
The free entropy [see Eq. (1.59)] for the canonical distribution is given by
[see Eq. (1.85)]
σF,c = σ − βU ,
(1.114)
whereas for the grandcanonical case it is given by [see Eq. (1.86)]
σF,gc = σ − βU − ηN .
(1.115)
We define below two corresponding free energies, the canonical free energy
(known also as the Helmholtz free energy )
F = −τ σF,c = U − τ σ ,
(1.116)
and the grandcanonical free energy
G = −τ σ F,gc = U − τ σ + τ ηN = U − τ σ − µN .
In section 1.2.2 above we have shown that the LUE maximizes σF for
given values of the Lagrange multipliers ξ 1 , ξ 2 , ..., ξ L . This principle can be
implemented to show that:
• In equilibrium at a given temperature τ the Helmholtz free energy obtains
its smallest possible value.
• In equilibrium at a given temperature τ and chemical potential µ the grandcanonical free energy obtains its smallest possible value.
Our main results are summarized in table 1.2 below
1.7 Problems Set 1
Note: Problems 1-6 are taken from the book by Reif, chapter 1.
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Thermodynamics and Statistical Physics
21
Chapter 1. The Principle of Largest Uncertainty
Table 1.2. Summary of main results.
micro
general
−canonical
canonical
grandcanonical
U U , N
(M states)
given
Xl where
expectation
l = 1, 2, ..., L
values
Z=
partition
function
e
−
m
pm
1
e
Z
(∆Xl )2
l=1
L
σ
Zc =
e−βU (m)
m
ξl Xl (m)
l=1
Xl =
ξl Xl (m)
pm =
−
Xl L
pm =
1
M
log Z
− ∂ ∂ξ
l
(∆Xl )2 =
σ=
L
ξ l Xl log Z +
σ = log M
pm =
pm =
1 −βU (m)
Zc e
1
e−βU (m)−ηN (m)
Zgc
Zc
− ∂ log
∂β
(∆U )2 =
∂ 2 log Zc
∂β 2
Lagrange
multipliers
min / max
principle
ξl =
∂σ
∂Xl σ=
log Zc + β U β=
{Xn }n=l
∂σ
∂U
max
σ F (ξ 1 , ξ 2 , ..., ξ L )
L
σF = σ −
ξ l Xl max σ
min F (τ )
F = U − τσ
l=1
1. A drunk starts out from a lamppost in the middle of a street, taking steps
of equal length either to the right or to the left with equal probability.
What is the probability that the man will again be at the lamppost after
taking N steps
a) if N is even?
b) if N is odd?
2. In the game of Russian roulette, one inserts a single cartridge into the
drum of a revolver, leaving the other five chambers of the drum empty.
One then spins the drum, aims at one’s head, and pulls the trigger.
a) What is the probability of being still alive after playing the game N
times?
Eyal Buks
Thermodynamics and Statistical Physics
U = −
∂ log Zgc
∂β
η
∂ log Zgc
N = −
∂η
β
∂ 2 log Zgc (∆U )2 =
∂β 2
2
η
∂ log Zgc
2
(∆N ) =
∂η 2
β
l=1
Zgc =
e−βU (m)−ηN (m)
m
U =
∂ 2 log Z
∂ξ2
l
22
σ=
log Zgc + β U + η N
∂σ β = ∂U
∂σ N
η = ∂N
U
min G (τ , µ)
G = U − τ σ − µN
1.7. Problems Set 1
b) What is the probability of surviving (N − 1) turns in this game and
then being shot the N th time one pulls the trigger?
c) What is the mean number of times a player gets the opportunity of
pulling the trigger in this macabre game?
3. Consider the random walk problem with p = q and let m = n1 − n2 , denote the net displacement to the right.
After
a total of N
steps, calculate
the following mean values: m
, m2 , m3 , and m4 . Hint: Calculate
the moment generating function.
4. The probability W (n), that an event characterized by a probability p
occurs n times in N trials was shown to be given by the binomial distribution
W (n) =
N!
pn (1 − p)N−n .
n! (N − n)!
(1.117)
Consider a situation where the probability p is small (p << 1) and where
one is interested in the case n << N . (Note that if N is large, W (n)
becomes very small if n → N because of the smallness of the factor pn
when p << 1. Hence W (n) is indeed only appreciable when n << N .)
Several approximations can then be made to reduce Eq. (1.117) to simpler
form.
a) Using the result ln(1 − p) ≃ −p, show that (1 − p)N−n ≃ e−Np .
b) Show that N !/(N − n)! ≃ N n .
c) Hence show that Eq. (1.117) reduces to
W (n) =
λn −λ
e ,
n!
where λ ≡ Np is the mean number of events. This distribution is
called the "Poisson distribution."
5. Consider the Poisson distribution of the preceding problem.
∞
a) Show that it is properly normalized in the sense that
W (n) = 1.
n=0
(The sum can be extended to infinity to an excellent approximation,
since W (n) is negligibly small when n N .)
b) Use the Poisson distribution to calculate n
.
c) Use the Poisson distribution to calculate (∆n)2 = (n − n
)2 .
6. A molecule in a gas moves equal distances l between collisions with equal
probability in any direction. Aftera total
of N such displacements, what
is the mean square displacement R2 of the molecule from its starting
point ?
7. A multiple choice test contains 20 problems. The correct answer for each
problem has to be chosen out of 5 options. Show that the probability to
pass the test (namely to have at least 11 correct answers) using guessing
only, is 5.6 × 10−4 .
Eyal Buks
Thermodynamics and Statistical Physics
23
Chapter 1. The Principle of Largest Uncertainty
8. Consider a system of N spins. Each spin can be in one of two possible
states: in state ’up’ the magnetic moment of each spin is +m, and in
state ’down’ it is −m. Let N+ (N− ) be the number of spins in state ’up’
(’down’), where N = N+ +N− . The total magnetic moment of the system
is given by
M = m (N+ − N− ) .
(1.118)
Assume that the probability that the system occupies any of its 2N possible states is equal. Moreover, assume that N ≫ 1. Let f (M) be the
probability distribution of the random variable M (that is, M is considered in this approach as a continuous random variable). Use the Stirling’s
formula
1
1/2
N
+ ...
(1.119)
N! = (2πN ) N exp −N +
2N
to show that
1
M2
exp − 2
f (M ) = √
.
2m N
m 2πN
(1.120)
Use this result to evaluate the expectation value and the variance of M .
9. Consider a one dimensional random walk. The probabilities of transiting
to the right and left are p and q = 1 − p respectively. The step size for
both cases is a.
a) Show that the average displacement X
after N steps is given by
X
= aN (2p − 1) = aN (p − q) .
b) Show that the variance (X − X
)2 is given by
(X − X
)2 = 4a2 N pq .
(1.121)
(1.122)
10. A classical harmonic oscillator of mass m, and spring constant k oscillates
with amplitude a. Show that the probability density function f(x), where
f (x)dx is the probability that the mass would be found in the interval
dx at x, is given by
1
f(x) = √
.
2
π a − x2
(1.123)
11. A coin having probability p = 2/3 of coming up heads is flipped 6 times.
Show that the entropy of the outcome of this experiment is σ = 3.8191
(use log in natural base in the definition of the entropy).
12. A fair coin is flipped until the first head occurs. Let X denote the number
of flips required.
Eyal Buks
Thermodynamics and Statistical Physics
24
1.7. Problems Set 1
a) Find the entropy σ. In this exercise
use log in base 2 in the definition
of the entropy, namely σ = − i pi log2 pi .
b) A random variable X is drawn according to this distribution. Find
an “efficient” sequence of yes-no questions of the form, “Is X contained in the set S?” Compare σ to the expected number of questions
required to determine X.
13. In general the notation
∂z
∂x y
is used to denote the partial derivative of z with respect to x, where the
variable y is kept constant. That is, to correctly calculate this derivative
the variable z has to be expressed as a function of x and y, namely,
z = z (x, y).
a) Show that:
∂y
∂x
∂z
= − z .
(1.124)
∂y
∂x y
∂z
x
b) Show that:
∂z
∂z
∂z
∂y
=
+
.
∂x w
∂x y
∂y x ∂x w
14. Let Zgc be a grandcanonical partition function.
a) Show that:
∂ log Zgc
∂ log Zgc
+ τµ
.
U = −
∂β
∂µ
µ
β
(1.125)
(1.126)
where τ is the temperature, β = 1/τ, and µ is the chemical potential.
b) Show that:
N = λ
∂ log Zgc
,
∂λ
(1.127)
where
λ = exp (βµ)
(1.128)
is the fugacity.
15. Consider an array on N distinguishable two-level (binary) systems. The
two-level energies of each system are ±ε/2. Show that the temperature
τ of the system is given by
Eyal Buks
Thermodynamics and Statistical Physics
25
Chapter 1. The Principle of Largest Uncertainty
τ=
ε
,
2 tanh
− 2U
Nε
−1
(1.129)
where U
is the average total energy of the array. Note that the temperature can become negative if U > 0. Why a negative temperature
is possible for this system ?
16. Consider an array of N distinguishable quantum harmonic oscillators in
thermal equilibrium at temperature τ . The resonance frequency of all
oscillators is ω. The quantum energy levels of each quantum oscillator
is given by
1
εn = ℏω n +
,
(1.130)
2
where n = 0, 1, 2, ... is integer.
a) Show that the average energy of the system is given by
U =
βℏω
N ℏω
coth
,
2
2
(1.131)
where β = 1/τ.
b) Show that the variance of the energy of the system is given by
2
N ℏω
2
2
(∆U ) =
.
sinh 2 βℏω
2
(1.132)
17. Consider a lattice containing N non-interacting atoms. Each atom has 3
non-degenerate energy levels E1 = −ε, E2 = 0, E3 = ε. The system is at
thermal equilibrium at temperature τ .
a) Show that the average energy of the system is
U = −
2N ε sinh (βε)
,
1 + 2 cosh βε
(1.133)
where β = 1/τ.
b) Show the variance of the energy of the system is given by
cosh (βε) + 2
(U − U
)2 = 2N ε2
.
[1 + 2 cosh (βε)]2
(1.134)
18. Consider a one dimensional chain containing N ≫ 1 sections (see figure).
Each section can be in one of two possible sates. In the first one the
section contributes a length a to the total length of the chain, whereas
in the other state the section has no contribution to the total length of
the chain. The total length of the chain in N α, and the tension applied
to the end points of the chain is F . The system is in thermal equilibrium
at temperature τ .
Eyal Buks
Thermodynamics and Statistical Physics
26
1.7. Problems Set 1
a) Show that α is given by
a
Fa
1 + tanh
.
α=
2
2τ
(1.135)
b) Show that in the limit of high temperature the spring constant is
given approximately by
k≃
4τ
.
Na2
(1.136)
Nα
19. A long elastic molecule can be modelled as a linear chain of N links. The
state of each link is characterized by two quantum numbers l and n. The
length of a link is either l = a or l = b. The vibrational state of a link
is modelled as a harmonic oscillator whose angular frequency is ω a for a
link of length a and ωb for a link of length b. Thus, the energy of a link
is
ℏω a n + 12 for l = a
En,l =
,
(1.137)
ℏωb n + 12 for l = b
n = 0, 1, 2, ...
The chain is held under a tension F . Show that the mean length L
of
the chain in the limit of high temperature T is given by
L
= N
F ω b ω a (a − b)2
aω b + bω a
+N
β + O β2 ,
2
ωb + ω a
(ω b + ω a )
(1.138)
where β = 1/τ .
20. The elasticity of a rubber band can be described in terms of a onedimensional model of N polymer molecules linked together end-to-end.
The angle between successive links is equally likely to be 0◦ or 180◦ . The
length of each polymer is d and the total length is L. The system is in
thermal equilibrium at temperature τ. Show that the force f required to
maintain a length L is given by
f=
τ
L
tanh−1
.
d
Nd
(1.139)
21. Consider a system which has two single particle states both of the same
energy. When both states are unoccupied, the energy of the system is
Eyal Buks
Thermodynamics and Statistical Physics
27
Chapter 1. The Principle of Largest Uncertainty
zero; when one state or the other is occupied by one particle, the energy
is ε. We suppose that the energy of the system is much higher (infinitely
higher) when both states are occupied. Show that in thermal equilibrium
at temperature τ the average number of particles in the level is
N =
2
,
2 + exp [β (ε − µ)]
(1.140)
where µ is the chemical potential and β = 1/τ .
22. Consider an array of N two-level particles. Each one can be in one of two
states, having energy E1 and E2 respectively. The numbers of particles
in states 1 and 2 are n1 and n2 respectively, where N = n1 + n2 (assume
n1 ≫ 1 and n2 ≫ 1). Consider an energy exchange with a reservoir at
temperature τ leading to population changes n2 → n2 −1 and n1 → n1 +1.
a) Calculate the entropy change of the two-level system, (∆σ)2LS .
b) Calculate the entropy change of the reservoir, (∆σ)R .
c) What can be said about the relation between (∆σ)2LS and (∆σ)R in
thermal equilibrium? Use your answer to express the ration n2 /n1 as
a function of E1 , E2 and τ .
23. Consider a lattice containing N sites of one type, which is denoted as A,
and the same number of sites of another type, which is denoted as B. The
lattice is occupied by N atoms. The number of atoms occupying sites of
type A is denoted as NA , whereas the number of atoms occupying atoms
of type B is denoted as NB , where NA + NB = N . Let ε be the energy
necessary to remove an atom from a lattice site of type A to a lattice
site of type B. The system is in thermal equilibrium at temperature τ .
Assume that N, NA , NB ≫ 1.
a) Calculate the entropy σ.
b) Calculate the average number NB of atoms occupying sites of type
B.
24. Consider a microcanonical ensemble of N quantum harmonic oscillators
in thermal equilibrium at temperature τ. The resonance frequency of all
oscillators is ω. The quantum energy levels of each quantum oscillator is
given by
1
εn = ℏω n +
,
(1.141)
2
where n = 0, 1, 2, ... is integer. The total energy E of the system is given
by
N
E = ℏω m +
,
(1.142)
2
where
Eyal Buks
Thermodynamics and Statistical Physics
28
1.8.
m=
N
Solutions Set 1
nl ,
(1.143)
l=1
and nl is state number of oscillator l.
a) Calculate the number of states g (N, m) of the system with total
energy ℏω (m + N/2).
b) Use this result to calculate the entropy σ of the system with total
energy ℏω (m + N/2). Approximate the result by assuming that N ≫
1 and m ≫ 1.
c) Use this result to calculate (in the same limit of N ≫ 1 and m ≫ 1)
the average energy of the system U as a function of the temperature
τ.
25. The energy of a donor level in a semiconductor is −ε when occupied
by an electron (and the energy is zero otherwise). A donor level can be
either occupied by a spin up electron or a spin down electron, however, it
cannot be simultaneously occupied by two electrons. Express the average
occupation of a donor state Nd as a function of ε and the chemical
potential µ.
1.8 Solutions Set 1
1. Final answers:
1 N
a) N N!
( 2 )!( N2 )! 2
b) 0
2. Final answers:
N
a) 56
N−1 1
b) 56
6
c) In general
∞
N=0
N xN−1 =
∞
d N
d 1
1
x =
=
,
dx N=0
dx 1 − x
(1 − x)2
thus
N−1
∞
1 5
1
1
N̄ =
N
= =6.
6 N=0
6
6 1− 5 2
6
3. Let W (m) be the probability for for taking n1 steps to the right and
n2 = N − n1 steps to the left, where m = n1 − n2 , and N = n1 + n2 .
Using
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Thermodynamics and Statistical Physics
29
Chapter 1. The Principle of Largest Uncertainty
N +m
,
2
N −m
n2 =
,
2
n1 =
one finds
N +m
N −m
N!
W (m) = N+m N−m p 2 q 2 .
!
!
2
2
It is convenient to employ the moment generating function, defined as
φ (t) = etm .
In general, the following holds
φ (t) =
∞ k
t k
m ,
k!
k=0
thus from the kth derivative of φ (t) one can calculate the kth moment
of m
k
m = φ(k) (0) .
Using W (m) one finds
φ (t) =
N
W (m) etm
m=−N
=
N
m=−N
N+m
N −m
N!
N+m N−m p 2 q 2 etm ,
!
!
2
2
or using the summation variable
n1 =
N +m
,
2
one has
φ (t) =
N
n1
N!
pn1 q N−n1 et(2n1 −N)
n
!
(N
−
n
)!
1
1
=0
N
2t n1 N−n1
N!
pe
q
n ! (N − n1 )!
n1 =0 1
N
= e−tN pe2t + q
.
= e−tN
Using p = q = 1/2
Eyal Buks
Thermodynamics and Statistical Physics
30
1.8.
φ (t) =
et + e−t
2
N
Solutions Set 1
= (cosh t)N .
Thus using the expansion
(cosh t)N = 1 +
one finds
1
1
N t2 + [N + 3N (N − 1)] t4 + O t5 ,
2!
4!
m
= 0 ,
2
m =N ,
3
m =0,
4
m = N (3N − 2) .
4. Using the binomial distribution
N!
pn (1 − p)N−n
n! (N − n)!
N (N − 1) (N − 1) × ... (N − n + 1) n
=
p (1 − p)N−n
n!
(Np)n
∼
exp (−N p) .
=
n!
W (n) =
5.
a)
∞
−λ
W (n) = e
n=0
∞
λn
n!
=1
n=0
b) As in Ex. 1-6, it is convenient to use the moment generating function
∞
∞
∞
n
λn etn
(λet )
φ (t) = etn =
etn W (n) = e−λ
= e−λ
= exp λ et − 1 .
n!
n!
n=0
n=0
n=0
Using the expansion
1
exp λ et − 1 = 1 + λt + λ (1 + λ) t2 + O t3 ,
2
one finds
n
= λ .
c) Using the same expansion one finds
2
n = λ (1 + λ) ,
thus
(∆n)2 = n2 − n
2 = λ (1 + λ) − λ2 = λ .
Eyal Buks
Thermodynamics and Statistical Physics
31
Chapter 1. The Principle of Largest Uncertainty
6.
2
R =
N
rn
n=1
2 7.
=
N
2 rn +
rn · rm = N l2
n=1
n=m
=l2
=0
20
20!
0.2n × 0.820−n = 5.6 × 10−4 .
n!
(20
−
n)!
n=11
(1.144)
8. Using
N+ + N− = N ,
M
,
N+ − N− =
m
(1.145a)
(1.145b)
one has
N
N+ =
1+
2
N
N− =
1−
2
M
,
mN
M
,
mN
(1.146a)
(1.146b)
or
N
(1 + x) ,
2
N
N− =
(1 − x) ,
2
N+ =
(1.147a)
(1.147b)
where
x=
M
.
mN
The number of states having total magnetization M is given by
Ω (M ) =
N!
N!
N
.
= N
N+ !N− !
2 (1 + x) ! 2 (1 − x) !
(1.148)
Since all states have equal probability one has
f (M ) =
Ω (M )
.
2N
Taking the natural logarithm of Stirling’s formula one finds
1
log N! = N log N − N + O
,
N
Eyal Buks
Thermodynamics and Statistical Physics
(1.149)
(1.150)
32
1.8.
Solutions Set 1
thus in the limit N ≫ 1 one has
log f = − log 2N + N log N − N
N
N
N
(1 + x) log
(1 + x) +
(1 + x)
−
2
2
2
N
N
N
−
(1 − x) log
(1 − x) +
(1 − x)
2
2
2
= −N log 2 + N log N
N
N
N
N
−
(1 + x) log
(1 + x) −
(1 − x) log
(1 − x)
2
2
2
2
N
N
N
N
−2 log
+ (1 + x) log
(1 + x) + (1 − x) log
(1 − x)
= −
2
2
2
2
N
N
N
N
= −
−2 log
+ (1 + x) log
+ log (1 + x) + (1 − x) log
+ log (1 − x)
2
2
2
2
N
1+x
= −
log 1 − x2 + x log
.
2
1−x
(1.151)
The function log f (x) has a sharp peak near x = 0, thus we can approximate it by assuming x ≪ 1. To lowest order
1+x
log 1 − x2 + x log
= x2 + O x3 ,
1−x
(1.152)
thus
M2
f (M ) = A exp − 2
,
2m N
(1.153)
where A is a normalization constant, which is determined by requiring
that
1=
∞
f (M ) dM .
(1.154)
−∞
Using the identity
∞
−∞
exp −ay2 dy =
π
,
a
(1.155)
√
M2
dM = m 2πN ,
exp − 2
2m N
(1.156)
one finds
1
=
A
Eyal Buks
∞
−∞
Thermodynamics and Statistical Physics
33
Chapter 1. The Principle of Largest Uncertainty
thus
1
M2
√
exp − 2
,
2m N
m 2πN
The expectation value is giving by
∞
M
=
M f (M) dM = 0 ,
f (M ) =
(1.157)
(1.158)
−∞
and the variance is given by
∞
2
(M − M
) = M 2 =
M 2 f (M ) dM = m2 N .
(1.159)
−∞
9. The probability to have n steps to the right is given by
W (n) =
a)
N!
pn q N−n .
n! (N − n)!
n
=
N
(1.160)
N!n
pn q N−n
n!
(N
−
n)!
n=0
=p
=p
(1.161)
N
∂ N!
pn q N−n
∂p n=0 n! (N − n)!
∂
(p + q)N = pN (p + q)N−1 = pN .
∂p
Since
X = an − a (N − n) = a (2n − N )
(1.162)
we find
X
= aN (2p − 1) = aN (p − q) .
(1.163)
b)
N
2 n =
N !n2
pn q N−n
n!
(N
−
n)!
n=0
=
N
N !n (n − 1)
n! (N − n)!
n=0
= p2
= p2
Eyal Buks
pn q N−n +
N
N !n
pn q N−n
n!
(N
−
n)!
n=0
N
∂2 N!
pn q N−n + n
∂p2 n=0 n! (N − n)!
∂2
(p + q)N + n
= p2 N (N − 1) + pN .
∂p2
Thermodynamics and Statistical Physics
34
1.8.
Solutions Set 1
Thus
(n − n
)2 = p2 N (N − 1) + pN − p2 N 2 = N pq ,
(1.164)
and
(X − X
)2 = 4a2 N pq .
(1.165)
10. The total energy is given by
E=
kx2 mẋ2
ka2
+
=
,
2
2
2
(1.166)
where a is the amplitude of oscillations. The time period T is given by
a
a
dx
m
dx
m
√
T =2
=2
= 2π
,
(1.167)
2
2
k −a a − x
k
−a ẋ
thus
f(x) =
2
1
.
= √
T |ẋ|
π a2 − x2
(1.168)
11. The six experiments are independent, thus
2 2 1 1
σ = 6 × − ln − ln
= 3.8191 .
3 3 3 3
12. The random variable X obtains the value n with probability pn = q n ,
where n = 1, 2, 3, ..., and q = 1/2.
a) The entropy is given by
σ=−
∞
n=1
pn log2 pn = −
∞
q n log2 q n =
n=1
∞
nq n .
n=1
This can be rewritten as
σ=q
∞
∂ n
∂
1
q
q =q
−1 =
=2.
∂q n=1
∂q 1 − q
(1 − q)2
b) A series of questions is of the form: "Was a head obtained in the 1st
time ?", "Was a head obtained in the 2nd time ?", etc. The expected
number of questioned required to find X is
1
1
1
× 1 + × 2 + × 3 + ... = 2 ,
2
4
8
which is exactly the entropy σ.
13.
Eyal Buks
Thermodynamics and Statistical Physics
35
Chapter 1. The Principle of Largest Uncertainty
a) Consider an infinitesimal change in the variable z = z (x, y)
∂z
∂z
δx +
δy .
(1.169)
δz =
∂x y
∂y x
For a process for which z is a constant δz = 0, thus
∂z
∂z
0=
(δx)z +
(δy)z .
∂x y
∂y x
(1.170)
Dividing by (δx)z yields
(δy)z
∂z
∂z
=−
∂x y
∂y x (δx)z
∂z
∂y
=−
∂y
∂x z
x
=
∂y
∂x z
− ∂y
∂z x
.
(1.171)
b) Consider a process for which the variable w is kept constant. An
infinitesimal change in the variable z = z (x, y) is expressed as
∂z
∂z
(δx)w +
(δy)w .
(1.172)
(δz)w =
∂x y
∂y x
Dividing by (δx)w yields
(δy)w
(δz)w
∂z
∂z
=
+
.
(δx)w
∂x y
∂y x (δx)w
(1.173)
or
∂z
∂x
w
=
∂z
∂x
+
y
∂z
∂y
x
∂y
∂x
.
14. We have found in class that
∂ log Zgc
U = −
,
∂β
η
∂ log Zgc
N = −
.
∂η
β
Eyal Buks
(1.174)
w
Thermodynamics and Statistical Physics
(1.175)
(1.176)
36
1.8.
Solutions Set 1
a) Using relation (1.125) one has
∂ log Zgc
U = −
∂β
η
∂ log Zgc
∂ log Zgc
∂µ
=−
−
∂β
∂µ
∂β η
µ
β
∂ log Zgc
η ∂ log Zgc
=−
− 2
∂β
∂µ
β
µ
β
∂ log Zgc
∂ log Zgc
=−
+ τµ
.
∂β
∂µ
µ
β
(1.177)
b) Using Eq. (1.176) one has
∂ log Zgc
N = −
∂η
β
∂µ
∂ log Zgc
=−
∂η β
∂µ
β
∂ log Zgc
=τ
,
∂µ
β
(1.178)
or in terms of the fugacity λ, which is defined by
λ = exp (βµ) = e−η ,
(1.179)
one has
N = τ
∂ log Zgc
∂µ
β
∂λ ∂ log Zgc
=τ
∂µ
∂λ
∂ log Zgc
=λ
.
∂λ
(1.180)
15. The canonical partition function is given by
Zc = Z1N ,
(1.181)
where
Z1 = exp
Eyal Buks
βε
2
βε
βε
+ exp −
= 2 cosh
.
2
2
Thermodynamics and Statistical Physics
(1.182)
37
Chapter 1. The Principle of Largest Uncertainty
Thus
U = −
∂ log Zc
∂ log Z1
Nε
βε
= −N
=−
tanh
,
∂β
∂β
2
2
(1.183)
and
τ=
ε
.
− 2U
2 tanh
Nε
−1
(1.184)
The negative temperature is originated by our assumption that the energy of a single magnet has an upper bound. In reality this is never the
case.
16. The canonical partition function is given by
Zc = Z1N ,
(1.185)
where
∞
βℏω Z1 = exp −
exp (−βℏωn)
2
n=0
exp − βℏω
2
1
=
.
=
1 − exp (−βℏω)
2 sinh βℏω
2
(1.186)
a)
U = −
∂ log Z1
N ℏω
βℏω
∂ log Zc
= −N
=
coth
∂β
∂β
2
2
(1.187)
b)
2
∂ 2 log Z
N ℏω
∂ 2 log Z1
c
2
2
(∆U ) =
=N
=
∂β 2
∂β 2
sinh 2 βℏω
2
(1.188)
17. The canonical partition function is given by
Zc = [exp (βε) + 1 + exp (−βε)]N = [1 + 2 cosh (βε)]N ,
(1.189)
where β = 1/τ .
a) Thus the average energy is
U = −
∂ log Zc
2N ε sinh (βε)
=−
,
∂β
1 + 2 cosh βε
(1.190)
b) and the variance is
∂ 2 log Z
∂ U cosh (βε) + 2
c
(U − U
)2 =
=−
= 2Nε2
.
∂β
∂β 2
[1 + 2 cosh (βε)]2
(1.191)
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Thermodynamics and Statistical Physics
38
1.8.
Solutions Set 1
18. Each section can be in one of two possible sates with corresponding energies 0 and −F a.
a) By definition, α is the mean length of each segment, which is given
by
a exp (F aβ)
a
F aβ
α=
=
1 + tanh
,
(1.192)
1 + exp (F aβ)
2
2
where β = 1/τ.
b) At high temperature F aβ ≪ 1 the length of the chain L = Nα is
given by
Na
F aβ
Na
F aβ
1 + tanh
≃
1+
,
(1.193)
L=
2
2
2
2
or
Na
F = k L−
2
,
(1.194)
where the spring constant k is given by
k=
4τ
.
Na2
(1.195)
19. The average length of a single link is given by
a exp (βF a)
l
=
∞
n=0
exp (βF a)
∞
n=0
=
a exp(βF a)
2 sinh βℏ2ωa
+
exp(βF a)
2 sinh βℏ2ωa
+
∞
exp −βℏω a n + 12 + b exp (βF b)
exp −βℏωb n + 12
n=0
∞
exp −βℏω a n + 12 + exp (βF b)
exp −βℏω b n + 12
b exp(βF b)
βℏ ω
2 sinh 2 b
exp(βF b)
βℏ ω
2 sinh 2 b
n=0
(1.196)
.
To first order in β
l
=
aω b + bω a F ω b ω a (a − b)2
+
β + O β2 .
2
ωb + ωa
(ωb + ω a )
(1.197)
The average total length is L
= n l
.
20. The length L is given by
L = N l
,
where l
is the average contribution of a single molecule to the total
length, which can be either +d or −d. The probability of each possibility
Eyal Buks
Thermodynamics and Statistical Physics
39
Chapter 1. The Principle of Largest Uncertainty
is determined by the Boltzmann factor. The energy change due to flipping
of one link from 0◦ to 180◦ is 2fd, therefore
l
= d
eβf d − e−βf d
= d tanh (βfd) ,
eβf d + e−βf d
where β = 1/τ . Thus
L = Nd tanh (βf d) ,
or
f=
τ
L
tanh−1
.
d
Nd
21. The grand partition function Zgc is given by
Zgc = 1 + 2 exp [β (µ − ε)] ,
(1.198)
thus
N =
1 ∂ log Zgc
2
=
.
β
∂µ
2 + exp [β (ε − µ)]
(1.199)
22.
a)
N!
N!
− log
(n2 − 1)! (n1 + 1)!
n2 !n1 !
n2
n2
= log
≃ log
.
n1 + 1
n1
(∆σ)2LS = log
(1.200)
b)
(∆σ)R =
E2 − E1
τ
(1.201)
c) For a small change near thermal equilibrium one expects (∆σ)2LS +
(∆σ)R = 0, thus
n2
E2 − E1
= exp −
.
(1.202)
n1
τ
23. The number of ways to select NB occupied sites of type B out of N sites
is N !/n! (N − n)!. Similarly the number of ways to select NB empty sites
of type A out of N sites is N !/n! (N − n)!.
Eyal Buks
Thermodynamics and Statistical Physics
40
1.8.
Solutions Set 1
a) Thus
σ = log
N!
NB ! (N − NB )!
2
≃ 2 [N log N − NB log NB − (N − NB ) log (N − NB )] .
(1.203)
b) The energy of the system is given by U = NB ε. Thus, the Helmholtz
free energy is given by
U
U
U
U
F = U −τ σ = U −2τ N log N − log − N −
log N −
.
ε
ε
ε
ε
(1.204)
At thermal equilibrium (∂F/∂U)τ = 0, thus
2τ
U
U
∂F
=1+
log − log N −
,
0=
∂U τ
ε
ε
ε
(1.205)
or
ε N − NB
= exp
,
NB
2τ
(1.206)
therefore
NB =
N
ε.
1 + exp 2τ
Alternatively, one can calculate the chemical potential from the requirement
1=
NA NB
+
,
N
N
(1.207)
where
NA
exp (βµ)
=
,
N
1 + exp (βµ)
exp (βµ − βε)
NB
=
,
N
1 + exp (βµ − βε)
which is satisfied when
ε
µ= ,
2
(1.208a)
(1.208b)
(1.209)
thus
NB =
Eyal Buks
N
ε.
1 + exp 2τ
Thermodynamics and Statistical Physics
(1.210)
41
Chapter 1. The Principle of Largest Uncertainty
24. In general,
g (N, m) = {# os ways to distribute m identical balls in N boxes}
Moreover
{# os ways to distribute m identical balls in N boxes}
= {# os ways to arrange m identical balls and N − 1 identical partitions in a line}
…
…
a) Therefore
g (N, m) =
(N − 1 + m)!
.
(N − 1)!m!
(1.211)
b) The entropy is given by
σ = log
(N − 1 + m)!
≃ (N + m) log (N + m) − N log N − m log m ,
(N − 1)!m!
(1.212)
or in terms of the total energy E = ℏω (m + N/2)
E
N
E
N
σ= N+
−
log N +
−
ℏω
2
ℏω
2
E
N
E
N
− N log N −
−
log
−
.
ℏω
2
ℏω
2
(1.213)
c) The temperature τ is given by
1
∂σ
=
τ
∂E
2ℏω
1 − ln 2E+Nℏω
ln 2ℏω − 1
=
+ 2E−Nℏω
ℏω
ℏω
1
2E + N ℏω
=
ln
.
ℏω
2E − N ℏω
Eyal Buks
Thermodynamics and Statistical Physics
(1.214)
42
1.8.
Solutions Set 1
In the thermodynamical limit (N ≫ 1, m ≫ 1) the energy E and its
average U are indistinguishable, thus
ℏω
2U + N ℏω
exp
=
,
(1.215)
τ
2U − N ℏω
or
U=
Nℏω
ℏω
coth
.
2
2τ
(1.216)
25. The grand canonical partition function is given by
ζ = 1 + 2λ exp (βε) ,
(1.217)
where λ = exp (βµ) is the fugacity, thus
Nd = λ
Eyal Buks
∂ log ζ
2λeβε
1
=
=
.
∂λ
1 + 2λeβε
1 + 12 e−β(ε+µ)
Thermodynamics and Statistical Physics
(1.218)
43
2. Ideal Gas
In this chapter we study some basic properties of ideal gas of massive identical
particles. We start by considering a single particle in a box. We then discuss
the statistical properties of an ensemble of identical indistinguishable particles
and introduce the concepts of Fermions and Bosons. In the rest of this chapter
we mainly focus on the classical limit. For this case we derive expressions for
the pressure, heat capacity, energy and entropy and discuss how internal
degrees of freedom may modify these results. In the last part of this chapter
we discuss an example of an heat engine based on ideal gas (Carnot heat
engine). We show that the efficiency of such a heat engine, which employs a
reversible process, obtains the largest possible value that is allowed by the
second law of thermodynamics.
2.1 A Particle in a Box
Consider a particle having mass M in a box. For simplicity the box is assumed
to have a cube shape with a volume V = L3 . The corresponding potential
energy is given by
0 0 ≤ x, y, z ≤ L
V (x, y, z) =
.
(2.1)
∞
else
The quantum eigenstates and eigenenergies are determined by requiring that
the wavefunction ψ (x, y, z) satisfies the Schrödinger equation
2 ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ
−
+
+
+ V ψ = Eψ .
(2.2)
2M ∂x2
∂y 2
∂z 2
In addition, we require that the wavefunction ψ vanishes on the surfaces of
the box. The normalized solutions are given by
3/2
2
nx πx
ny πy
nz πz
ψnx ,ny ,nz (x, y, z) =
sin
sin
sin
,
(2.3)
L
L
L
L
where
nx , ny , nz = 1, 2, 3, ...
(2.4)
Chapter 2. Ideal Gas
The corresponding eigenenergies are given by
εnx ,ny ,nz =
2 π 2 2
nx + n2y + n2z .
2M L
(2.5)
For simplicity we consider the case where the particle doesn’t have any internal degree of freedom (such as spin). Later we will release this assumption
and generalize the results for particles having internal degrees of freedom.
The partition function is given by
∞ ∞ ∞
ε
n ,n ,n
Z1 =
exp − x y z
τ
n =1 n =1 n =1
x
y
z
∞ ∞ ∞
=
nx =1 ny =1 nz =1
exp −α2 n2x + n2y + n2z ,
(2.6)
where
α2 =
2 π2
.
2ML2 τ
(2.7)
The following relation can be employed to estimate the dimensionless parameter α
α2 =
7.9 × 10−17
L 2 τ
M
mp
cm
,
(2.8)
300 K
where mp is the proton mass. As can be seen from the last result, it is often
the case that α2 ≪ 1. In this limit the sum can be approximated by an
integral
∞
nx =1
exp −α2 n2x ≃
∞
0
exp −α2 n2x dnx .
(2.9)
By changing the integration variable x = αnx one finds
∞
exp
0
−α2 n2x
1
dnx =
α
∞
0
exp −x2 dx =
√
π
,
2α
(2.10)
thus
√ 3 3/2
π
M L2 τ
Z1 =
=
= nQ V ,
2α
2π2
(2.11)
where we have introduced the quantum density
Eyal Buks
Thermodynamics and Statistical Physics
46
2.1. A Particle in a Box
nQ =
Mτ
2π2
3/2
.
(2.12)
The partition function (2.11) together with Eq. (1.70) allows evaluating
the average energy (recall that β = 1/τ )
ε
= −
=−
∂ log Z1
∂β
3/2 ML2
∂ log
2π 2 β
∂β
−3/2
∂ log β
∂β
3 ∂ log β
=
2 ∂β
3τ
=
.
2
=−
(2.13)
This result can be written as
τ
ε
= d ,
2
(2.14)
where d = 3 is the number of degrees of freedom of the particle. As we will see
later, this is an example of the equipartition theorem of statistical mechanics.
Similarly, the energy variance can be evaluated using Eq. (1.71)
∂ 2 log Z
1
(∆ε)2 =
∂β 2
∂ ε
=−
∂β
∂ 3
=−
∂β 2β
3
=
2β 2
3τ 2
=
.
2
(2.15)
Thus, using Eq. (2.13) the standard deviation is given by
2
2
(∆ε) =
ε
.
3
(2.16)
What is the physical meaning of the quantum density? The de Broglie
wavelength λ of a particle having mass M and velocity v is given by
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Thermodynamics and Statistical Physics
47
Chapter 2. Ideal Gas
λ=
h
.
Mv
(2.17)
For a particle having energy equals to the average energy ε
= 3τ /2 one has
Mv 2
3τ
=
,
2
2
(2.18)
thus in this case the de-Broglie wavelength, which is denoted as λT (the
thermal wavelength)
h
λT = √
,
3M τ
(2.19)
and therefore one has (recall that = h/2π)
nQ =
π
h2
2Mτ
3/2
=
1
λT
2π
3
3
.
(2.20)
Thus the quantum density is inversely proportional to the thermal wavelength
cubed.
2.2 Gibbs Paradox
In the previous section we have studied the case of a single particle. Let us
now consider the case where the box is occupied by N particles of the same
type. For simplicity, we consider the case where the density n = N/V is
sufficiently small to safely allowing to neglect any interaction between the
particles. In this case the gas is said to be ideal.
Definition 2.2.1. Ideal gas is an ensemble of non-interacting identical particles.
What is the partition function of the ideal gas? Recall that for the single
particle case we have found that the partition function is given by [see Eq.
(2.6)]
exp (−βεn ) .
(2.21)
Z1 =
n
In this expression Z1 is obtained by summing over all single particle orbital
states, which are denoted by the vector of quantum numbers n = (nx , ny , nz ).
These states are called orbitals .
Since the total number of particles N is constrained we need to calculate
the canonical partition function. For the case of distinguishable particles one
may argue that the canonical partition function is given by
Eyal Buks
Thermodynamics and Statistical Physics
48
2.2. Gibbs Paradox
?
Zc =
Z1N
=
N
exp (−βεn )
n
.
(2.22)
However, as was demonstrated by Gibbs in his famous paradox, this answer
is wrong. To see this, we employ Eqs. (1.72) and (2.11) and assume that the
partition function is given by Eq. (2.22) above, thus
?
σ − βU = log Zc = log Z1N = N log (nQ V ) ,
(2.23)
or in terms of the gas density
n=
N
,
V
(2.24)
we find
n ?
Q
σ − βU = N log N
.
n
(2.25)
N = N1 + N2 ,
V = V1 + V2 .
(2.26)
(2.27)
What is wrong with this result? It suggests that the quantity σ − βU is not
simply proportional to the size of the system. In other words, for a given n
and a given nQ , σ −βU is not proportional to N. As we will see below, such a
behavior may lead to a violation of the second law of thermodynamics. To see
this consider a box containing N identical particles having volume V . What
happens when we divide the box into two sections by introducing a partition?
Let the number of particles in the first (second) section be N1 (N2 ) whereas
the volume in the first (second) section be V1 (V2 ). The following hold
The density in each section is expected to be the same as the density in the
box before the partition was introduced
n=
N2
N
N1
=
.
=
V
V1
V2
(2.28)
Now we use Eq. (2.25) to evaluate the change in entropy ∆σ due to the
process of dividing the box. Since no energy is required to add (or to remove)
the partition one has
∆σ = σtot − σ1 − σ 2
n n n ?
Q
Q
Q
= N log N
− N1 log N1
− N2 log N2
n
n
n
= N log N − N1 log N1 − N2 log N2 .
(2.29)
Using the Stirling’s formula (1.150)
log N ! ≃ N log N − N ,
Eyal Buks
Thermodynamics and Statistical Physics
(2.30)
49
Chapter 2. Ideal Gas
2
…
orbital 1
N1=0
3
1
orbital 2
N2=1
orbital 3
N3=2
orbital 4
N4=0
Fig. 2.1. A figure describing an example term in the expansion (2.22) for a gas
containing N = 3 identical particles.
one finds
∆σ ≃ log
N!
>0.
N1 !N2 !
(2.31)
Thus we came to the conclusion that the process of dividing the box leads
to reduction in the total entropy! This paradoxical result violates the second
law of thermodynamics. According to this law we expect no change in the
entropy since the process of dividing the box is a reversible one.
What is wrong with the partition function given by Eq. (2.22)? Expanding
function yields a sum of terms each having the form
this partition
exp −β Nn εn , where Nn is the number of particles occupying orbital
n
n. Let g (N1 , N2 , ...) be the number of terms in such an expansion associated
with a given set of occupation numbers {N1 , N2 , ...}. Since the partition function (2.22) treats the particles as being distinguishable, g (N1 , N2 , ...) may in
general be larger than unity. In fact, it is easy to see that
g (N1 , N2 , ...) =
N!
.
N1 !N2 ! × ...
(2.32)
For example, consider the state that is described by Fig. 2.1 below for a
gas containing N = 3 particles. The expansion (2.22) contains 3!/1!/2! =
3 terms having the same occupation numbers (Nn = 1 if n = 2,Nn = 2
if n = 3, and Nn = 0 for all other values of n). However, for identical
particles these 3 states are indistinguishable. Therefore, only a single term
in the partition function should represent such a configuration. In general,
the partition function should include a single term only for each given set of
occupation numbers {N1 , N2 , ...}.
2.3 Fermions and Bosons
As we saw in the previous section the canonical partition function given
by Eq. (2.22) is incorrect. For indistinguishable particles each set of orbital
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Thermodynamics and Statistical Physics
50
2.3. Fermions and Bosons
occupation numbers {N1 , N2 , ...} should be counted only once. In this section
we take another approach and instead of evaluating the canonical partition
function of the system we consider the grandcanonical partition function.
This is done by considering each orbital as a subsystem and by evaluating
its grandcanonical partition function, which we denote below as ζ. To do this
correctly, however, it is important to take into account the exclusion rules
imposes by quantum mechanics upon the possible values of the occupation
numbers Nn .
The particles in nature are divided into two type: Fermion and Bosons.
While Fermions have half integer spin Bosons have integer spin. According to
quantum mechanics the orbital occupation numbers Nn can take the following
values:
• For Fermions: Nn = 0 or 1
• For Bosons: Nn can be any integer.
These rules are employed below to evaluate the grandcanonical partition
function of an orbital.
2.3.1 Fermi-Dirac Distribution
In this case the occupation number can take only two possible values: 0 or
1. Thus, by Eq. (1.79) the grandcanonical partition function of an orbital
having energy is ε is given by
ζ = 1 + λ exp (−βε) ,
(2.33)
where
λ = exp (βµ)
(2.34)
is the fugacity [see Eq. (1.95)]. The average occupation of the orbital, which
is denoted by fFD (ε) = N (ε)
, is found using Eq. (1.94)
∂ log ζ
∂λ
λ exp (−βε)
=
1 + λ exp (−βε)
1
=
.
exp [β (ε − µ)] + 1
fFD (ε) = λ
(2.35)
The function fFD (ε) is called the Fermi-Dirac function .
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Thermodynamics and Statistical Physics
51
Chapter 2. Ideal Gas
2.3.2 Bose-Einstein Distribution
In this case the occupation number can take any integer value. Thus, by Eq.
(1.79) the grandcanonical partition function of an orbital having energy is ε
is given by
ζ=
=
∞
N=0
∞
λN exp (−N βε)
[λ exp (−βε)]N
N=0
=
1
.
1 − λ exp (−βε)
(2.36)
The average occupation of the orbital, which is denoted by fBE (ε) = N (ε)
,
is found using Eq. (1.94)
∂ log ζ
∂λ
exp (−βε)
=λ
1 − λ exp (−βε)
1
=
.
exp [β (ε − µ)] − 1
fBE (ε) = λ
(2.37)
The function fBE (ε) is called the Bose-Einstein function .
2.3.3 Classical Limit
The classical limit occurs when
exp [β (ε − µ)] ≫ 1 .
(2.38)
As can be seen from Eqs. (2.35) and (2.37) the following holds
fFD (ε) ≃ fBE (ε) ≃ exp [β (µ − ε)] ≪ 1 ,
(2.39)
ζ ≃ 1 + λ exp (−βε) .
(2.40)
and
Thus the classical limit corresponds to the case where the average occupation
of an orbital is close to zero, namely the orbital is on average almost empty.
The main results of the above discussed cases (Fermi-Dirac distribution, BoseEinstein distribution and the classical limit) are summarized in table 2.1
below.
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Thermodynamics and Statistical Physics
52
2.4. Ideal Gas in the Classical Limit
Table 2.1. Fermi-Dirac, Bose-Einstein and classical distributions.
orbital partition function
average occupation
Fermions
1 + λ exp (−βε)
Bosons
1
1−λ exp(−βε)
1
exp[β(ε−µ)]+1
1
exp[β(ε−µ)]−1
classical limit
1 + λ exp (−βε)
exp [β (µ − ε)]
2
1.5
1
0.5
-2 0
2
4
6
8 energy
10 12
14
16
18
20
2.4 Ideal Gas in the Classical Limit
The rest of this chapter is devoted to the classical limit. The grandcanonical
partition function ζ n of orbital n having energy εn is given by Eq. (2.40)
above. The grandcanonical partition function of the entire system Zgc is
found by multiplying ζ n of all orbitals
Zgc =
(1 + λ exp (−βεn )) .
(2.41)
n
Each term in the expansion of the above expression represents a set of orbital occupation numbers, where each occupation number can take one of the
possible values: 0 or 1. We exploit the fact that in the classical limit
λ exp (−βε) ≪ 1
and employ the first order expansion
log (1 + x) = x + O x2
(2.42)
(2.43)
to obtain
Eyal Buks
Thermodynamics and Statistical Physics
53
Chapter 2. Ideal Gas
log Zgc =
log (1 + λ exp (−βεn ))
n
≃λ
exp (−βεn )
n
= λZ1 ,
(2.44)
where
Z1 = V
Mτ
2π2
3/2
(2.45)
[see Eq. (2.11)] is the single particle partition function. In terms of the Lagrange multipliers η = −µ/τ and β = 1/τ the last result can be rewritten
as
3/2
M
−η
.
(2.46)
log Zgc = e V
2π2 β
The average energy and average number of particle are calculated using Eqs.
(1.80) and (1.81) respectively
∂ log Zgc
3
U = −
=
log Zgc ,
(2.47)
∂β
2β
η
∂ log Zgc
N = −
= log Zgc .
(2.48)
∂η
β
In what follows, to simplify the notation we remove the diagonal brackets
and denote U and N
by U and N respectively. As was already pointed
out earlier, probability distributions in statistical mechanics of macroscopic
parameters are typically extremely sharp and narrow. Consequently, in many
cases no distinction is made between a parameter and its expectation value.
Using this simplified notation and employing Eqs. (2.47) and (2.48) one finds
that
U=
3N τ
.
2
(2.49)
Namely, the total energy is N ε
, where ε
is the average single particle
energy that is given by Eq. (2.13).
The entropy is evaluate using Eq. (1.86)
σ = log Zgc + βU + ηN
3 µ
= N 1+ −
2 τ
5
=N
− µβ .
2
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Thermodynamics and Statistical Physics
(2.50)
54
2.4. Ideal Gas in the Classical Limit
Furthermore, using Eqs. (2.44), (2.48), (2.11) and (1.95) one finds that
n
,
(2.51)
µβ = log
nQ
where n = N/V is the density. This allows expressing the entropy as
5
nQ
σ=N
+ log
.
(2.52)
2
n
Using the definition (1.116) and Eqs. (2.49) and (2.52) one finds that the
Helmholtz free energy is given by
n
−1 .
(2.53)
F = Nτ log
nQ
2.4.1 Pressure
The pressure p is defined by
∂F
p=−
.
∂V τ ,N
(2.54)
Using Eq. (2.53) and keeping in mind that n = N/V one finds
Nτ
.
(2.55)
V
The pressure represents the force per unit area acting on the walls of the box
containing the gas due to collisions between the particles and the walls. To
see that this is indeed the case consider a gas of N particles contained in a
box having cube shape and volume V = L3 . One of the walls is chosen to
lie on the x = 0 plane. Consider and elastic collision between this wall and
a particle having momentum p = (px , py , pz ). After the collision py and pz
remain unchanged, however, px becomes −px . Thus each collision results in
transferring 2 |px | momentum in the x direction from the particle to the wall.
The rate at which a particle collides with the wall x = 0 is |px | /2mL. Thus
the pressure acting on the wall due to a single particle is
{force}
{pressure} =
{area}
rate of
momentum change
=
{area}
p=
2 |px | ×
L2
2
p
= x .
mV
=
|px |
2mL
(2.56)
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Chapter 2. Ideal Gas
The average energy of a particle is given by Eq. (2.13)
2
px + p2y + p2z
3τ
= ε
=
,
2
2m
(2.57)
thus one finds that
2
px
=τ .
m
(2.58)
Using this result and Eq. (2.56) one finds that the pressure due to a single
particle is p = τ /V , thus the total pressure is
p=
Nτ
.
V
(2.59)
2.4.2 Useful Relations
In this section we derive some useful relations between thermodynamical
quantities.
Claim. The following holds
∂U
p=−
∂V σ,N
Proof. Using the definition (2.54) and recalling that F = U − τσ one finds
∂U
∂ (τσ)
∂F
−p =
=
−
.
(2.60)
∂V τ ,N
∂V τ ,N
∂V
τ ,N
Using identity (1.125), which is given by
∂z
∂z
∂z
∂y
=
+
,
∂x w
∂x y
∂y x ∂x w
(2.61)
one finds
∂U
∂U
∂U
∂σ
=
+
,
∂V τ ,N
∂V σ,N
∂σ V,N ∂V τ ,N
(2.62)
τ
thus
−p =
Eyal Buks
∂U
∂V
+τ
σ,N
∂σ
∂V
τ ,N
−τ
∂σ
∂V
τ ,N
=
∂U
∂V
Thermodynamics and Statistical Physics
.
(2.63)
σ,N
56
2.4. Ideal Gas in the Classical Limit
In a similar way the following relations can be obtained
∂U
∂F
∂σ
p=τ
=−
=−
,
∂V U,N
∂V σ,N
∂V τ ,N
∂σ
∂U
∂F
µ = −τ
=
=
.
∂N U,V
∂N σ,V
∂N τ ,V
(2.64)
(2.65)
Another useful relation is given below.
Claim. The following holds
∂p
∂σ
=
.
∂V τ
∂τ V
(2.66)
Proof. See problem 2 of set 2.
2.4.3 Heat Capacity
The heat capacity at constant volume is defined by
∂σ
cV = τ
,
∂τ V
whereas the heat capacity at constant pressure is defined by
∂σ
cp = τ
.
∂τ p
(2.67)
(2.68)
Using Eq. (2.52) and recalling that nQ ∝ τ 3/2 one finds
cV = τ
3N
3N
=
.
2τ
2
(2.69)
Claim. The following holds
cp =
5N
.
2
(2.70)
Proof. See problem 1 of set 2.
2.4.4 Internal Degrees of Freedom
In this section we generalize our results for the case where the particles in the
gas have internal degrees of freedom. Taking into account internal degrees of
freedom the grandcanonical partition function of an orbital having orbital
energy εn for the case of Fermions becomes
ζ FD,n =
(1 + λ exp (−βεn ) exp (−βEl )) ,
(2.71)
l
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Chapter 2. Ideal Gas
where {El } are the eigenenergies of a particle due to internal degrees of
freedom, and where λ = exp (βµ) and β = 1/τ . As is required by the Pauli
exclusion principle, no more than one Fermion can occupy a given internal
eigenstate. Similarly, for the Bosonic case, where each state can be occupied
by any integer number of Bosons, one has
∞
m
ζ BE,n =
λ exp (−βmεn ) exp (−βmEl ) .
(2.72)
l
m=0
In the classical limit the average occupation of an orbital is close to zero.
In this limit, namely when
λ exp (−βεn ) ≪ 1 ,
(2.73)
[see Eq. (2.38)] the following holds
ζ FD,n ≃ ζ BE,n ≃ ζ n ,
(2.74)
where
ζ n = 1 + λ exp (−βεn ) Zint ,
(2.75)
and where
Zint =
exp (−βEl ) ,
(2.76)
l
is the internal partition function.
Using Eq. (1.94) one finds that the average occupation of the orbital fn
in the classical limit is given by
∂ log ζ n
∂λ
λZint exp (−βεn )
=
1 + λZint exp (−βεn )
≃ λZint exp (−βεn ) .
fn = λ
The total grandcanonical partition function is given by
Zgc =
ζn ,
(2.77)
(2.78)
n
thus
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58
2.4. Ideal Gas in the Classical Limit
log Zgc =
log ζ n
n
=
log [1 + λZint exp (−βεn )]
n
≃ λZint
exp (−βεn )
n
= λZint Z1 ,
(2.79)
where we have used the fact that in the classical limit λZint exp (−βεn ) ≪ 1.
Furthermore, using Eq. (2.11) and recalling that η = −µ/τ one finds
log Zgc = e−η Zint V
M
2π2 β
3/2
.
This result together with Eqs. (1.80) and (1.81) yield
3
∂ log Zgc
=
U = −
log Zgc + El log Zgc ,
∂β
2β
η
∂ log Zgc
N = −
= log Zgc ,
∂η
β
(2.80)
(2.81)
(2.82)
where
El exp (−βEl )
∂ log Zint
=−
.
El = exp (−βEl )
∂β
l
(2.83)
l
Claim. The following hold
n
µ = τ log
− log Zint ,
nQ
3τ
∂ log Zint
U =N
−
,
2
∂β
n
F = Nτ log
− log Zint − 1 ,
nQ
5
nQ ∂ (τ log Zint )
σ=N
+ log
+
,
2
n
∂τ
3
∂ 2 (τ log Zint )
cV = N
+τ
,
2
∂τ 2
cp = cV + N .
(2.84)
(2.85)
(2.86)
(2.87)
(2.88)
(2.89)
Proof. See problem 3 of set 2.
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Chapter 2. Ideal Gas
2.5 Processes in Ideal Gas
The state of an ideal gas is characterized by extensive parameters (by definition, parameters that are proportional to the system size) such as U , V ,
N and σ and by intensive parameters (parameters that are independent on
the system size) such as τ , µ and p. In this section we discuss some examples of processes that occur by externally changing some of these parameters.
We will use these processes in the next section to demonstrate how one can
construct a heat engine based on an ideal gas.
In general, the entropy is commonly expresses as a function of the energy,
volume and number of particles σ = σ (U, V, N ). A small change in σ is
expressed in terms of the partial derivatives
∂σ
∂σ
∂σ
dσ =
dU +
dV +
dN .
(2.90)
∂U V,N
∂V U,N
∂N U,V
Using Eqs. (1.87), (2.64) and (2.65)
dσ =
1
p
µ
dU + dV − dN ,
τ
τ
τ
(2.91)
or
dU = τ dσ − pdV + µdN .
(2.92)
This relation expresses the change in the energy of the system dU in terms
of
dQ = τ dσ
heat added to the system
dW = pdV
work done by the system
µdN
energy change due to added particles
For processes that keep the number of particles unchanged
dN = 0 ,
one has
dU = dQ − dW .
(2.93)
Integrating this relation for the general case (not necessarily an infinitesimal
process) yields
∆U = Q − W ,
(2.94)
We discuss below some specific examples for processes for which dN = 0.
The initial values of the pressure, volume and temperature are denoted as
p1 , V1 and τ 1 respectively, whereas the final values are denoted as p2 , V2
and τ 2 respectively. In all these processes we assume that the gas remains in
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Thermodynamics and Statistical Physics
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2.5. Processes in Ideal Gas
p
isochoric
isobaric
i so
is e
the
ntr
op
rma
ic
l
V
Fig. 2.2. Four processes for which dN = 0.
thermal equilibrium throughout the entire process. This can be achieved by
varying the external parameters at a rate that is sufficiently slow to allow the
system to remain very close to thermal equilibrium at any moment during
the process. The four example to be analyzed below are (see fig. 2.2):
•
•
•
•
Isothermal process - temperature is constant
Isobaric process - pressure is constant
Isochoric process - volume is constant
Isentropic process - entropy is constant
Note that in general, using the definition of the heat capacity at constant
volume given by Eq. (2.67) together with Eq. (1.87) one finds that
∂U
cV =
.
(2.95)
∂τ N,V
Furthermore, as can be seen from Eq. (2.85), the energy U of an ideal gas
in the classical limit is independent on the volume V (it can be expressed as
a function of τ and N only). Thus we conclude that for processes for which
dN = 0 the change in energy dU can be expressed as
dU = cV dτ .
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(2.96)
Thermodynamics and Statistical Physics
61
Chapter 2. Ideal Gas
2.5.1 Isothermal Process
Since τ is constant one finds using Eq. (2.96) that ∆U = 0. Integrating the
relation dW = pdV and using Eq. (2.55) one finds
Q=W =
V2
pdV
V1
V2
= Nτ
dV
V
V1
= N τ log
V2
.
V1
(2.97)
2.5.2 Isobaric Process
Integrating the relation dW = pdV for this case where the pressure is constant yields
W =
V2
pdV = p (V2 − V1 ) .
(2.98)
V1
The change in energy ∆U can be found by integrating Eq. (2.96)
∆U =
τ 2
cV dτ .
(2.99)
τ1
The heat added to the system Q can be found using Eq. (2.94)
Q = W + ∆U
= p (V2 − V1 ) +
τ 2
cV dτ .
τ1
(2.100)
Note that if the temperature dependence of cV can be ignored to a good
approximation one has
∆U = cV (τ 2 − τ 1 ) .
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Thermodynamics and Statistical Physics
(2.101)
62
2.5. Processes in Ideal Gas
2.5.3 Isochoric Process
In this case the volume is constant, thus W = 0. By integrating Eq. (2.96)
one finds that
Q = ∆U =
τ 2
cV dτ .
(2.102)
τ1
Also in this case, if the temperature dependence of cV can be ignored to a
good approximation one has
Q = ∆U = cV (τ 2 − τ 1 ) .
2.5.4 Isentropic Process
In this case the entropy is constant, thus dQ = τdσ = 0, and therefore
dU = −dW , thus using the relation dW = pdV and Eq. (2.96) one has
cV dτ = −pdV ,
(2.103)
or using Eq. (2.55)
cV
dτ
dV
= −N
.
τ
V
(2.104)
This relation can be rewritten using Eq. (2.89) as
dV
dτ
= (1 − γ)
.
τ
V
(2.105)
where
γ=
cp
.
cV
(2.106)
The last result can be easily integrated if the temperature dependence of the
factor γ can be ignored to a good approximation. For that case one has
log
τ2
= log
τ1
V2
V1
1−γ
.
(2.107)
Thus
τ 1 V1γ−1 = τ 2 V2γ−1 ,
(2.108)
or using Eq. (2.55)
p1 V1γ = p2 V2γ .
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(2.109)
63
Chapter 2. Ideal Gas
W
τh
Qh
Ql
τl
Fig. 2.3. Carnot heat engine.
In other words both quantities τV γ−1 and pV γ remain unchanged during
this process. Using the last result allows integrating the relation dW = pdV
−∆U = W =
V2
pdV
V2
V −γ dV
V1
=
p1 V1γ
V1
= p1 V1γ
V1−γ+1 − V2−γ+1
γ−1
p2 V2 − p1 V1
=
1−γ
N (τ 2 − τ 1 )
=
1−γ
= −cV (τ 2 − τ 1 ) .
(2.110)
2.6 Carnot Heat Engine
In this section we discuss an example of a heat engine proposed by Carnot
that is based on an ideal classical gas. Each cycle is made of four steps (see
Figs. 2.3 and 2.4)
1.
2.
3.
4.
Isothermal expansion at temperature τ h (a → b)
Isentropic expansion from temperature τ h to τ l (b → c)
Isothermal compression at temperature τ l (c → d)
Isentropic compression from temperature τ l to τ h (d → a)
All four steps are assumed to be sufficiently slow to maintain the gas in
thermal equilibrium throughout the entire cycle. The engine exchanges heat
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2.6. Carnot Heat Engine
p
a
pa
τh
b
pb
pd
d
pc
Va
Vb Vd
τl
c
V
Vc
Fig. 2.4. A cycle of Carnot heat engine.
with the environment during both isothermal processes. Using Eq. (2.97) one
finds that the heat extracted from the hot reservoir Qh at temperature τ h
during step 1 (a → b) is given by
Qh = N τ h log
Vb
,
Va
(2.111)
and the heat extracted from the cold thermal reservoir Ql at temperature τ l
during step 3 (c → d) is given by
Ql = N τ l log
Vd
,
Vc
(2.112)
where Vn is the volume at point n ∈ {a, b, c, d}. Note that Qh > 0 since
the system undergoes expansion in step 1 whereas Ql < 0 since the system
undergoes compression during step 3. Both thermal reservoirs are assumed
to be very large systems that can exchange heat with the engine without
changing their temperature. No heat is exchanged during the isentropic steps
2 and 4 (since dQ = τ dσ).
The total work done by the system per cycle is given by
W = Wab + Wcd + Wbc + Wda
Vb
Vd
= N τ h log
+ N τ l log
Va
Vc
N (τ l − τ h ) N (τ h − τ l )
+
+
1−γ
1−γ
Vb
Vd
= N τ h log
+ τ l log
,
Va
Vc
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Thermodynamics and Statistical Physics
(2.113)
65
Chapter 2. Ideal Gas
where the work in both isothermal processes Wab and Wcd is calculated using
Eq. (2.97), whereas the work in both isentropic processes Wbc and Wda is
calculate using Eq. (2.110). Note that the following holds
W = Qh + Ql .
(2.114)
This is expected in view of Eq. (2.94) since the gas returns after a full cycle
to its initial state and therefore ∆U = 0.
The efficiency of the heat engine is defined as the ratio between the work
done by the system and the heat extracted from the hot reservoir per cycle
η=
W
Ql
=1+
.
Qh
Qh
(2.115)
Using Eqs. (2.111) and (2.113) one finds
η =1+
τ l log VVdc
τ h log VVab
.
(2.116)
Employing Eq. (2.108) for both isentropic processes yields
τ h Vbγ−1 = τ l Vcγ−1 ,
τ h Vaγ−1
=
τ l Vdγ−1
,
(2.117)
(2.118)
thus by dividing these equations one finds
Vbγ−1
Vcγ−1
=
,
Vaγ−1
Vdγ−1
(2.119)
Vb
Vc
=
.
Va
Vd
(2.120)
or
Using this result one finds that the efficiency of Carnot heat engine ηC is
given by
ηC = 1 −
τl
.
τh
(2.121)
2.7 Limits Imposed Upon the Efficiency
Is it possible to construct a heat engine that operates between the same heat
reservoirs at temperatures τ h and τ l that will have efficiency larger than
the value given by Eq. (2.121)? As we will see below the answer is no. This
conclusion is obtained by noticing that the total entropy remains unchanged
in each of the four steps that constructs the Carnot’s cycle. Consequently,
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2.7. Limits Imposed Upon the Efficiency
W
Q
τ
M
Fig. 2.5. An idle heat engine (Perpetuum Mobile of the second kind).
the entire process is reversible, namely, by varying the external parameters
in the opposite direction, the process can be reversed.
We consider below a general model of a heat engine. In a continuos operation the heat engine repeats a basic cycle one after another. We make the
following assumptions:
• At the end of each cycle the heat engine returns to the same macroscopic
state that it was in initially (otherwise, continuous operation is impossible).
• The work W done per cycle by the heat engine does not change the entropy
of the environment (this is the case when, for example, the work is used
to lift a weight - a process that only changes the center of mass of the
weight, and therefore causes no entropy change).
Figure (2.5) shows an ideal heat engine that fully transforms the heat Q
extracted from a thermal reservoir into work W , namely Q = W . Such an
idle engine has a unity efficiency η = 1. Is it possible to realized such an
idle engine? Such a process does not violate the law of energy conservation
(first law of thermodynamics). However, as we will see below it violates the
second law of thermodynamics. Note also that the opposite process, namely
a process that transforms work into heat without losses is possible, as can be
seen from the example seen in Fig. (2.6). In this system the weigh normally
goes down and consequently the blender rotates and heats the liquid in the
container. In principle, the opposite process at which the weigh goes up and
the liquid cools down doesn’t violate the law of energy conservation, however,
it violates the second law (Perpetuum Mobile of the second kind), as we will
see below.
To show that the idle heat engine shown in Fig. (2.5) can not be realized
we employ the second law and require that the total change in entropy ∆σ
per cycle is non-negative
∆σ ≥ 0 .
(2.122)
The only change in entropy per cycle is due to the heat that is subtracted
from the heat bath
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Chapter 2. Ideal Gas
Fig. 2.6. Transforming work into heat.
∆σ = −
Q
,
τ
(2.123)
thus since Q = W (energy conservation) we find that
W
≤0.
τ
(2.124)
Namely, the work done by the heat engine is non-positive W ≤ 0. This result
is known as Kelvin’s principle.
Kelvin’s principle: In a cycle process, it is impossible to extract heat
from a heat reservoir and convert it all into work.
As we will see below, Kelvin’s principle is equivalent to Clausius’s principle
that states:
Clausius’s principle: It is impossible that at the end of a cycle process,
heat has been transferred from a colder to a hotter thermal reservoirs without
applying any work in the process.
A refrigerator and an air conditioner (in cooling mode) are examples of
systems that transfer heat from a colder to a hotter thermal reservoirs. According to Clausius’s principle such systems require that work is consumed
for their operation.
Theorem 2.7.1. Kelvin’s principle is equivalent to Clausius’s principle.
Proof. Assume that Clausius’s principle does not hold. Thus the system
shown in Fig. 2.7(a) that transfers heat Q0 > 0 from a cold thermal reservoir at temperature τ l to a hotter one at temperature τ h > τ l is possible.
In Fig. 2.7(b) a heat engine is added that extracts heat Q > Q0 from the
hot thermal reservoir, delivers heat Q0 to the cold one, and performs work
W = Q − Q0 . The combination of both systems extracts heat Q − Q0 from
the hot thermal reservoir and converts it all into work, in contradiction with
Kelvin’s principle.
Assume that Kelvin’s principle does not hold. Thus the system shown in
Fig. 2.8(a) that extracts heat Q0 from a thermal reservoir at temperature τ h
and converts it all into work is possible. In Fig. 2.8(b) a refrigerator is added
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2.7. Limits Imposed Upon the Efficiency
(a)
τh
τl
-Q0
(b)
M
Q0
W=Q-Q0
Q
M
-Q0
τh
τl
-Q0
M
Q0
Fig. 2.7. The assumption that Clausius’s principle does not hold.
that employs the work W = Q0 to remove heat Q from a colder thermal
reservoir at temperature τ l < τ h and to deliver heat Q0 + Q to the hot
thermal reservoir. The combination of both systems transfers heat Q from
a colder to a hotter thermal reservoirs without consuming any work in the
process, in contradiction with Clausius’s principle.
As we have seen unity efficiency is impossible. What is the largest possible
efficiency of an heat engine?
Theorem 2.7.2. The efficiency η of a heat engine operating between a hotter
and colder heat reservoirs at temperature τ h and τ l respectively can not exceed
the value
τl
.
(2.125)
ηC = 1 −
τh
Proof. A heat engine (labeled as ’I’) is seen in Fig. 2.9. A Carnot heat engine
operated in the reverse direction (labeled as ’C’) is added. Here we exploit
the fact the Carnot’s cycle is reversible. The efficiency η of the heat engine
’I’ is given by
ηI =
W
,
Q′h
(2.126)
whereas the efficiency of the reversed Carnot heat engine ’C’ is given by Eq.
(2.121)
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Chapter 2. Ideal Gas
(a)
τh
W= Q0
Q0
τl
M
(b)
-Q0- Q
τh
Q
M
W= Q0
Q0
τl
M
Fig. 2.8. The assumption that Kelvin’s principle does not hold.
Q’h
τh
Q’l
I
τl
W
Qh
C
Ql
Fig. 2.9. Limit imposed upon engine efficiency.
ηC =
W
τl
=1−
.
Qh
τh
(2.127)
For the combined system, the Clausius’s principle requires that
Q′h − Qh > 0 .
(2.128)
thus
ηI ≤ ηC = 1 −
τl
.
τh
The same argument that was employed in the proof above can be used to
deduce the following corollary:
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2.8. Problems Set 2
Corollary 2.7.1. All reversible heat engines operating between a hotter heat
reservoir and a colder one at temperatures τ h and τ l respectively have the
same efficiency.
Note that a similar bound is imposed upon the efficiency of refrigerators
(see problem 16 of set 2).
2.8 Problems Set 2
1. The heat capacity at constant pressure is defined as
∂σ
Cp = τ
.
∂τ p
(2.129)
Calculate Cp of an classical ideal gas having no internal degrees of freedom.
2. Show that
∂σ
∂p
=
,
(2.130)
∂V τ
∂τ V
where σ is entropy, V is volume, and p is pressure.
3. Consider a classical ideal gas having internal partition function Zint .
a) Show that the chemical potential µ is given by
n
− log Zint ,
(2.131)
µ = τ log
nQ
where τ is the temperature, n = N/V , V is the volume, and nQ is
the quantum density.
b) Show that the energy U is related to the number of particles by the
following relation:
∂ log Zint
3τ
U =N
−
,
(2.132)
2
∂β
where β = bl/τ .
c) Show that the Helmholtz free energy is given by
n
F = Nτ log
− log Zint − 1 .
nQ
d) Show that the entropy is given by
5
nQ ∂ (τ log Zint )
σ=N
+ log
+
.
2
n
∂τ
Eyal Buks
Thermodynamics and Statistical Physics
(2.133)
(2.134)
71
Chapter 2. Ideal Gas
e) Show that the heat capacity at constant volume is given by
3
∂ 2 (τ log Zint )
+τ
.
(2.135)
cV = N
2
∂τ 2
f) Show that the heat capacity at constant pressure is given by
cp = cV + N .
(2.136)
4. The heat capacity c of a body having entropy σ is given by
c=τ
∂σ
,
∂τ
where τ is the temperature. Show that
(∆U )2
c=
,
τ2
(2.137)
(2.138)
where U is the energy of the body and where ∆U = U − U .
5. Consider an ideal classical gas made of diatomic molecules. The internal
vibrational degree of freedom is described using a model of a one dimensional harmonic oscillator with angular frequency ω. That is, the eigen
energies associated with the internal degree of freedom are given by
1
εn = n +
ω ,
(2.139)
2
where n = 0, 1, 2, .... The system is in thermal equilibrium at temperature τ , which is assumed to be much larger than ω. Calculate the heat
capacities cV and cp .
6. A thermally isolated container is divided into two chambers, the first
containing NA particles of classical ideal gas of type A, and the second
one contains NB particles of classical ideal gas of type B. Both gases have
no internal degrees of freedom. The volume of first chamber is VA , and
the volume of the second one is VB . Both gases are initially in thermal
equilibrium at temperature τ. An opening is made in the wall separating
the two chambers, allowing thus mixing of the two gases. Calculate the
change in entropy during the process of mixing.
7. Consider an ideal gas of N molecules in a vessel of volume V . Show
that the probability pn to find n molecules in a small volume v (namely,
v << V ) contained in the vessel is given by
pn =
λn −λ
e ,
n!
(2.140)
where λ = N v/V .
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Thermodynamics and Statistical Physics
72
2.8. Problems Set 2
8. A lattice contains N sites, each is occupied by a single atom. The set of
eigenstates of each atom, when a magnetic field H is applied, contains
2 states having energies ε− = −µ0 H and ε+ = µ0 H, where the magnetic moment µ0 is a constant. The system is in thermal equilibrium at
temperature τ .
a) Calculate the magnetization of the system, which is defined by
∂F
M =−
,
(2.141)
∂H τ
where F is the Helmholtz free energy.
b) Calculate the heat capacity
∂σ
C=τ
.
∂τ H
(2.142)
where σ is the entropy of the system.
c) Consider the case where initially the magnetic field is H1 and the
temperature is τ 1 . The magnetic field is then varied slowly in an
isentropic process from H1 to H2 . Calculate the final temperature of
the system τ 2 .
9. A lattice contains N sites, each occupied by a single atom. The set of
eigenstates of each atom, when a magnetic field H is applied, contains 3
states with energies
ε−1 = −∆ − µ0 H ,
ε0 = 0 ,
ε1 = −∆ + µ0 H ,
where the magnetic moment µ0 is a constant. The system is in thermal
equilibrium at temperature τ . Calculate the magnetic susceptibility
χ = lim
H→0
M
,
H
(2.143)
where
M =−
∂F
∂H
,
(2.144)
τ
is the magnetization of the system, and where F is the Helmholtz free
energy.
10. A lattice contains N sites, each occupied by a single atom. The set
of eigenstates of each atom, when a magnetic field H is applied, contains 2J + 1 states with energies εm = −mµH, where J is integer,
m = −J, −J + 1, ...J − 1, J, and the magnetic moment µ is a constant.
The system is in thermal equilibrium at temperature τ .
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73
Chapter 2. Ideal Gas
a) Calculate the free energy F of the system.
b) Show that the average magnetization, which is defined as
∂F
M =−
,
∂H τ
(2.145)
is given by
µH
µH
(2J + 1) coth (2J + 1)
− coth
. (2.146)
2τ
2τ
Nµ
M=
2
11. Assume the earth’s atmosphere is pure nitrogen in thermodynamic equilibrium at a temperature of 300 K. Calculate the height above sea level
at which the density of the atmosphere is one-half its sea-level value
(answer: 12.6 km).
12. Consider a box containing an ideal classical gas made of atoms of mass M
having no internal degrees of freedom at pressure p and temperature τ .
The walls of the box have N0 absorbing sites, each of which can absorb 0,
1, or 2 atoms of the gas. The energy of an unoccupied site and the energy
of a site occupying one atom is zero. The energy of a site occupying two
atoms is ε. Show that the mean number of absorbed atoms is given by
Na = N0
λ + 2λ2 e−βε
,
1 + λ + λ2 e−βε
(2.147)
where β = 1/τ and
λ=
M
2πℏ2
−3/2
τ −5/2 p .
(2.148)
13. An ideal gas containing N atoms is in equilibrium at temperature τ .
The internal degrees of freedom have two energy levels, the first one
has energy zero and degeneracy g1 , and the second one energy ε and
degeneracy g2 . Show that the heat capacities at constant volume and at
constant pressure are given by
!
g1 g2 exp − τε
3 ε 2
cV = N
+
(2.149)
2 ,
2
τ
g1 + g2 exp − τε
!
g1 g2 exp − τε
5 ε 2
cp = N
+
(2.150)
2 .
2
τ
g1 + g2 exp − ε
τ
14. A classical gas is described by the following equation of state
p (V − b) = N τ ,
(2.151)
where p is the pressure, V is the volume, τ is the temperature, N is the
number of particles and b is a constant.
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74
2.8. Problems Set 2
a) Calculate the difference cp − cV between the heat capacities at constant pressure and at constant volume.
b) Consider an isentropic expansion of the gas from volume V1 and
temperature τ 1 to volume V2 and temperature τ 2 . The number of
particles N is kept constant. Assume that cV is independent on temperature. Calculate the work W done by the gas during this process.
15. A classical gas is described by the following equation of state
a p + 2 (V − b) = N τ ,
(2.152)
V
where p is the pressure, V is the volume, τ is the temperature, and a
and b are constants. Calculate the difference cp − cV between the heat
capacities at constant pressure and at constant volume.
16. A classical gas is described by the following equation of state
a (2.153)
p + 2 (V − b) = N τ ,
V
where p is the pressure, V is the volume, τ is the temperature, and a and
b are constants. The gas undergoes a reversible isothermal expansion at a
fixed temperature τ 0 from volume V1 to volume V2 . Show that the work
W done by the gas in this process, and the heat Q which is supplied to
the gas during this process are given by
V2 − b
V2 − V1
,
−a
V1 − b
V2 V1
V2 − b
Q = ∆U + W = N τ 0 log
.
V1 − b
W = N τ 0 log
(2.154)
(2.155)
17. The energy of a classical ideal gas having no internal degrees of freedom
is denoted as E, the deviation from the average value U = E
as ∆E =
E − U . The gas, which contains N particles and has volume is V , is in
thermal equilibrium at temperature τ .
a) Calculate (∆E)2 .
b) Calculate (∆E)3 .
18. A body having a constant heat capacity C and a temperature τ a is put
into contact with a thermal bath at temperature τ b . Show that the total
change in entropy after equilibrium is establishes is given by
τa
τa
∆σ = C
− 1 − log
.
(2.156)
τb
τb
Use this result to show that ∆σ ≥ 0.
19. The figure below shows a cycle of an engine made of an ideal gas. In
the first step a → b the volume is constant V2 , the second one b → c is
Eyal Buks
Thermodynamics and Statistical Physics
75
Chapter 2. Ideal Gas
p
p1
b
c
p2
a
V2
V1
V
Fig. 2.10.
an isentropic process, and in the third one the pressure is constant p2 .
Assume that the heat capacities CV and Cp are temperature independent.
Show that the efficiency of this engine is given by
η =1−γ
p2 (V1 − V2 )
,
V2 (p1 − p2 )
(2.157)
where γ = Cp /Cv .
20. Consider a refrigerator consuming work W per cycle to extract heat from
a cold thermal bath at temperature τ l to another thermal bath at higher
temperature τ h . Let Ql be the heat extracted from the cold bath per cycle
and −Qh the heat delivered to the hot one per cycle. The coefficient of
refrigerator performance is defined as
γ=
Ql
.
W
(2.158)
Show that the second law of thermodynamics imposes an upper bound
on γ
γ≤
τl
.
τh − τl
(2.159)
21. A room air conditioner operates as a Carnot cycle refrigerator between
an outside temperature τ h and a room at a lower temperature τ l . The
room gains heat from the outdoors at a rate A (τ h − τ l ); this heat is
removed by the air conditioner. The power supplied to the cooling unit
is P . Calculate the steady state temperature of the room.
22. The state equation of a given matter is
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Thermodynamics and Statistical Physics
76
2.8. Problems Set 2
p=
Aτ 3
,
V
(2.160)
where p, V and τ are the pressure, volume and temperature, respectively,
A is a constant. The internal energy of the matter is written as
U = Bτ n log
23.
24.
25.
26.
27.
28.
V
+ f (τ) ,
V0
(2.161)
where B and V0 are constants, f (τ) only depends on the temperature.
Find B and n.
An ideal classical gas is made of N identical molecules each having mass
M . The volume of the gas is V and the temperature is τ . The energy
spectrum due to internal degrees of freedom of each molecule has a ground
state, which is nondegenerate state (singlet state), and a first excited
energy state, which has degeneracy 3 (triplet state). The energy gap
between the ground state and the first excited state is ∆ and all other
states have a much higher energy. Calculate:
a) the heat capacity at constant volume CV .
b) the heat capacity at constant pressure Cp .
Two identical bodies have internal energy U = Cτ , with a constant heat
capacity C. The initial temperature of the first body is τ 1 and that of the
second one is τ 2 . The two bodies are used to produce work by connecting
them to a reversible heat engine and bringing them to a common final
temperature τ f .
a) Calculate τ f .
b) Calculate the total work W , which is delivered by the process.
An ideal classical gas having no internal degrees of freedom is contained
in a vessel having two parts separated by a partition. Each part contains
the same number of molecules, however, while the pressure in the first
one is p1 , the pressure in the second one is p2 . The system is initially in
thermal equilibrium at temperature τ . Calculate the change of entropy
caused by a fast removal of the partition.
A classical ideal gas contains N particles having mass M and no internal
degrees of freedom is in a vessel of volume V at temperature τ . Express
the canonical partition function Zc as a function of N , M , V and τ.
Consider an engine working in a reversible cycle and using an ideal classical gas as the working substance. The cycle consists of two processes at
constant pressure (a → b and c → d), joined by two isentropic processes
(b → c and d → a), as show in Fig. 2.11. Assume that the heat capacities
CV and Cp are temperature independent. Calculate the efficiency of this
engine.
Consider an engine working in a cycle and using an ideal classical gas
as the working substance. The cycle consists of two isochoric processes
(constant volume) a→b at volume V1 and c→d at volume V2 , joined by
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Thermodynamics and Statistical Physics
77
Chapter 2. Ideal Gas
p
p1
a
p2
b
d
c
V
Fig. 2.11.
p
b
c
a
d
V1
V
V2
Fig. 2.12.
two isentropic processes (constant entropy) b→c and d→a, as shown in
Fig. 2.12. Assume that the heat capacities CV and Cp are temperature
independent. Calculate the efficiency η of this engine.
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Thermodynamics and Statistical Physics
78
2.9. Solutions Set 2
29. Consider two vessels A and B each containing ideal classical gas of particles having no internal degrees of freedom . The pressure and number
of particles in both vessels are p and N respectively, and the temperature is τ A in vessel A and τ B in vessel B. The two vessels are brought
into thermal contact. No heat is exchanged with the environment during
this process. Moreover, the pressure is kept constant at the value p in
both vessels during this process. Find the change in the total entropy
∆σ = σ final − σ initial .
2.9 Solutions Set 2
1. The entropy is given by
!
"
3/2 #
Mτ
5
V
σ = N log
+
,
2πℏ2
N
2
or using pV = N τ
"
!
3/2 5/2 #
M
τ
5
σ = N log
+
,
2πℏ2
p
2
thus
Cp = τ
∂σ
∂τ
p
5
= N.
2
(2.162)
(2.163)
(2.164)
2. Since
∂2F
∂2F
=
,
∂V ∂τ
∂τ∂V
where F is Helmholtz free energy, one has
∂
∂F
∂ ∂F
=
.
∂V ∂τ V τ
∂τ ∂V τ V
(2.165)
(2.166)
By definition
∂F
= −p .
∂V τ
Moreover, using F = U − τσ one finds
∂F
∂U
∂σ
=
−τ
−σ
∂τ V
∂τ V
∂τ V
∂U
∂U
∂σ
=
−
−σ
∂τ V
∂σ V ∂τ V
= −σ ,
Eyal Buks
Thermodynamics and Statistical Physics
(2.167)
79
Chapter 2. Ideal Gas
thus
∂σ
∂V
=
τ
∂p
∂τ
.
(2.168)
V
3. We have found in class the following relations
nQ
η
log Zgc
U
N
3/2
Mτ
=
,
2π2
µ
=− ,
τ
= e−η Zint V nQ ,
3τ
∂ log Zint
=
−
log Zgc ,
2
∂β
= log Zgc .
(2.169)
(2.170)
(2.171)
(2.172)
(2.173)
a) Using Eqs. (2.171) and (2.173) one finds
log
µ
n
= ,
nQ Zint
τ
(2.174)
thus
n
µ = τ log
− log Zint .
nQ
b) Using Eqs. (2.172) and (2.173)
3τ
∂ log Zint
U =N
−
.
2
∂β
(2.175)
(2.176)
c) Using the relations
F = U − τσ ,
σ = log Zgc + βU + ηN ,
(2.177)
(2.178)
one finds
F = U − τσ
= Nτ (−η − 1)
µ
−1
= Nτ
τ
n
= Nτ log
− log Zint − 1 .
nQ
Eyal Buks
Thermodynamics and Statistical Physics
(2.179)
(2.180)
(2.181)
(2.182)
(2.183)
80
2.9. Solutions Set 2
d) Using the relation
∂F
σ=−
,
∂τ V
(2.184)
one finds
∂F
σ=−
∂τ V


  ∂ τ log nnQ
∂
(τ
log
Z
)
int
 +
+ 1
= N − 
∂τ
∂τ
V
  
n
∂ log nQ
n
∂
(τ
log
Z
)
int
 − log
= N −τ 
+
+ 1
∂τ
nQ
∂τ
V
∂ (τ log Zint )
n
3
+
− log
+1
=N
2
nQ
∂τ
5
nQ ∂ (τ log Zint )
=N
+ log
+
.
2
n
∂τ
(2.185)
e) By definition
∂σ
cV = τ
∂τ V
3
∂ 2 (τ log Zint )
=N
.
+τ
2
∂τ 2
f) The following holds
∂σ
cp = τ
∂τ p
∂σ
∂σ
∂V
=τ
+τ
∂τ V
∂V τ ∂τ p
∂σ
∂V
= cV + τ
.
∂V τ ∂τ p
(2.186)
(2.187)
Using V p = N τ and Eq. (2.185) one finds
cp = cV + τ
Eyal Buks
NN
= cV + N .
V p
Thermodynamics and Statistical Physics
(2.188)
81
Chapter 2. Ideal Gas
4. With the help of Eqs. (1.87), (1.70) and (1.71) together with the following
relation
∂
1 ∂
=− 2
,
∂τ
τ ∂β
(2.189)
one finds that
c=τ
∂σ
∂U
1 ∂U
=
=− 2
=
∂τ
∂τ
τ ∂β
(∆U )2
τ2
.
(2.190)
5. The internal partition function is given by
Zint =
1
τ
,
≃
ω
ω
2 sinh 2τ
thus using Eqs. (2.186) and (2.188)
τ
∂ 2 τ log ω
3
5N
cV = N
+τ
=
,
2
∂τ 2
2
cp =
7N
.
2
(2.191)
(2.192)
(2.193)
6. Energy conservation requires that the temperature of the mixture will
remain τ . The entropy of an ideal gas of density n, which contains N
particles, is given by
nQ 5
σ (N, n) = N log
+
n
2
thus the change in entropy is given by
∆σ = σmix − σ A − σB
NA
NB
NA
NB
= σ NA ,
+ σ NB ,
− σ NA ,
− σ NB ,
VA + VB
VA + VB
VA
VB
VA + VB
VA + VB
= NA log
+ NB log
VA
VB
7. The probability to find a molecule in the volume v is given by p = v/V .
Thus, pn is given by
pn =
N!
pn (1 − p)N−n .
n! (N − n)!
(2.194)
Using the solution of problem 4 of set 1
pn =
λn −λ
e ,
n!
where λ = N v/V .
Eyal Buks
Thermodynamics and Statistical Physics
82
2.9. Solutions Set 2
8. The partition function of a single atom is given by
Z1 = exp (µ0 Hβ) + exp (−µ0 Hβ) = 2 cosh (µ0 Hβ) ,
where β = 1/τ , thus the partition function of the entire system is
N
Z = (2 cosh (µ0 Hβ))
,
a) The free energy is given by
F = −τ log Z = −N τ log (2 cosh (µ0 Hβ)) .
The magnetization is given by
∂F
M =−
= Nµ0 tanh (µ0 Hβ) .
∂H τ
(2.195)
(2.196)
b) The energy U is given by
U =−
∂ log Z
= −Nµ0 H tanh (µ0 Hβ) ,
∂β
(2.197)
thus
∂σ
C=τ
∂τ H
∂U
=
∂τ H
∂ tanh µ0τH
= −N µ0 H
∂τ
H
2
µ0 H
1
=N
.
τ cosh µ0 H
τ
(2.198)
c) The entropy σ, which is given by
σ = β (U − F )
µ0 H
µ0 H
µ0 H
= N log 2 cosh
−
tanh
,
τ
τ
τ
(2.199)
and which remains constant, is a function of the ratio H/τ, therefore
τ2 = τ1
H2
.
H1
(2.200)
9. The partition function of a single atom is given by
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Thermodynamics and Statistical Physics
83
Chapter 2. Ideal Gas
Z=
1
exp (−βεm )
m=−1
= 1 + 2 exp (β∆) cosh (βµ0 H) ,
(2.201)
where β = 1/τ . The free energy is given by
F = −Nτ log Z ,
(2.202)
thus the magnetization is given by
∂F
M =−
∂H τ
2N µ0 exp (β∆) sinh (βµ0 H)
=
.
1 + 2 exp (β∆) cosh (βµ0 H)
(2.203)
and the magnetic susceptibility is given by
χ=
τ 1+
Nµ20
,
1
2 exp (−β∆)
(2.204)
10. The partition function of a single atom is given by
Z=
J
exp (mµHβ) ,
m=−J
where β = 1/τ . By multiplying by a factor sinh (µHβ/2) one finds
sinh
µHβ
2
J
1
µHβ
µHβ
exp
− exp −
exp (mµHβ)
2
2
2
m=−J
1
1
1
=
exp J +
µHβ − exp − J +
µHβ
,
2
2
2
(2.205)
Z=
thus
Z=
J + 12 µHβ
.
sinh µHβ
2
sinh
(2.206)
a) The free energy is given by

F = −N τ log Z = −Nτ log 
Eyal Buks

J + 12 µHβ
 .
sinh µHβ
2
sinh
Thermodynamics and Statistical Physics
(2.207)
84
2.9. Solutions Set 2
b) The magnetization is given by
Nµ
µH
µH
∂F
(2J + 1) coth (2J + 1)
− coth
.
M =−
=
∂H τ
2
2τ
2τ
(2.208)
11. The internal chemical potential µg is given by Eq. (2.131). In thermal
equilibrium the total chemical potential
µtot = µg + mgz ,
(2.209)
(m is the mass of the each diatomic molecule N2 , g is the gravity acceleration constant, and z is the height) is z independent. Thus, the density
n as a function of height above see level z can be expressed as
mgz
n (z) = n (0) exp −
.
(2.210)
kB T
The condition n (z) = 0.5 × n (0) yields
z=
kB T ln 2
1.3806568 × 10−23 J K−1 × 300 K × ln 2
= 12.6 km .
=
mg
14 × 1.6605402 × 10−27 kg × 9.8 m s−2
(2.211)
12. The Helmholtz free energy of an ideal gas of N particles is given by
"
3/2 #
Mτ
F = −τN log
V + τN log N − τ N ,
(2.212)
2πℏ2
thus the chemical potential is
3/2 ∂F
Mτ
µ=
= −τ log
V + τ log N ,
∂N τ ,V
2πℏ2
and the pressure is
∂F
Nτ
p=−
=
.
∂V τ ,V
V
(2.213)
(2.214)
Using these results the fugacity λ = exp (βµ) can be expressed in terms
of p
−3/2
−3/2
Mτ
N
M
=
λ = eβµ =
τ −5/2 p .
(2.215)
2πℏ2
V
2πℏ2
At equilibrium the fugacity of the gas and that of the system of absorbing sites is the same. The grand canonical partition function of a single
absorption site is given by
Eyal Buks
Thermodynamics and Statistical Physics
85
Chapter 2. Ideal Gas
Z = 1 + eβµ + eβ(2µ−ε) ,
(2.216)
or in terms of the fugacity λ = exp (βµ)
Z = 1 + λ + λ2 e−βε .
(2.217)
Thus
Na = N0 λ
∂ log Z
λ + 2λ2 e−βε
= N0
,
∂λ
1 + λ + λ2 e−βε
where λ is given by Eq. (2.215).
13. The internal partition function is given by
ε
Zint = g1 + g2 exp −
.
τ
(2.218)
(2.219)
Using Eq. (2.135)
2
∂
3
(τ log Zint )
cV = N + N τ
2
∂τ 2
V
ε !
2
g1 g2 exp − τ
3
ε
=N
+
2 .
2
τ
g1 + g2 exp − ε
τ
(2.220)
Using Eq. (2.136)
cp = N
!
g1 g2 exp − τε
5 ε 2
+
2 .
2
τ
g1 + g2 exp − τε
14. Using Maxwell’s relation
∂p
∂σ
=
,
∂V τ ,N
∂τ V,N
and the equation of state one finds that
∂σ
N
=
.
∂V τ ,N
V −b
a) Using the definitions
∂σ
cV = τ
,
∂τ V,N
∂σ
cp = τ
,
∂τ p,N
Eyal Buks
Thermodynamics and Statistical Physics
(2.221)
(2.222)
(2.223)
(2.224)
(2.225)
86
2.9. Solutions Set 2
and the general identity
∂z
∂z
∂z
∂y
=
+
,
∂x α
∂x y
∂y x ∂x α
(2.226)
one finds
cp − cV = τ
∂σ
∂V
τ ,N
∂V
∂τ
,
(2.227)
p,N
or with the help of Eq. (2.223) and the equation of state
cp − cV = N
Nτ
=N .
p (V − b)
b) Using the identity
∂U
∂U
∂U
∂σ
=
+
,
∂V τ ,N
∂V σ,N
∂σ V,N ∂V τ ,N
together with Eq. (2.223) yields
Nτ
∂U
= −p +
=0.
∂V τ ,N
V −b
(2.228)
(2.229)
(2.230)
Thus, the energy U is independent on the volume V (it can be expressed as a function of τ and N only), and therefore for processes
for which dN = 0 the change in energy dU can be expressed as
dU = cV dτ .
(2.231)
For an isentropic process no heat is exchanged, and therefore dW =
−dU , thus since cV is independent on temperature one has
W = −∆U = −cV (τ 2 − τ 1 ) .
15. Using the definitions
∂σ
cV = τ
,
∂τ V,N
∂σ
cp = τ
,
∂τ p,N
and the general identity
∂z
∂z
∂z
∂y
=
+
,
∂x α
∂x y
∂y x ∂x α
Eyal Buks
Thermodynamics and Statistical Physics
(2.232)
(2.233)
(2.234)
(2.235)
87
Chapter 2. Ideal Gas
one finds
cp − cV = τ
∂σ
∂V
τ ,N
∂V
∂τ
.
(2.236)
p,N
Using Maxwell’s relation
∂σ
∂p
=
,
∂V τ ,N
∂τ V,N
(2.237)
and the equation of state
a p + 2 (V − b) = N τ ,
V
one finds
cp − cV = τ
=
=
∂p
∂τ
V,N
τ (VN−b)
(2.238)
∂V
∂τ
p,N
−aV +2ab+pV 3
NV 3
N
1+
−2aV +2ab
V 3 (p+ Va2 )
,
(2.239)
or
cp − cV =
1−
N
2 .
2a(1− Vb )
(2.240)
V Nτ
16. The work W is given by
V2
W =
pdV .
(2.241)
V1
Using
a p + 2 (V − b) = N τ ,
V
(2.242)
one finds
W =
V2
V1
Nτ0
a
− 2
V −b V
dV = N τ 0 log
V2 − b
V2 − V1
−a
. (2.243)
V1 − b
V2 V1
Using the identity
∂U
∂U
∂U
∂σ
∂σ
=
+
= −p + τ
,
∂V τ ,N
∂V σ,N
∂σ V,N ∂V τ ,N
∂V τ ,N
Eyal Buks
Thermodynamics and Statistical Physics
88
2.9. Solutions Set 2
(2.244)
and Maxwell’s relation
∂p
∂σ
=
,
∂V τ ,N
∂τ V,N
(2.245)
one finds
∂U
∂p
=τ
−p.
∂V τ ,N
∂τ V,N
(2.246)
In the present case
Nτ
∂U
a
=
−p= 2 ,
∂V τ ,N
V −b
V
(2.247)
thus
∆U =
V2
V1
∂U
∂V
dV = a
V2
V1
τ ,N
V2 − V1
dV
=a
,
V2
V2 V1
(2.248)
and
Q = ∆U + W = N τ 0 log
V2 − b
.
V1 − b
17. In general the following holds
n
1
∂ Zgc
E n =
−
,
Zgc
∂β n η
(2.249)
(2.250)
and
∂ log Zgc
U =−
∂β
.
(2.251)
η
Thus the variance is given by
(∆E)2 = E 2 − E
2
2
2
∂ Zgc
∂Zgc
1
1
=
−
2
Zgc
∂β η
∂β 2 η Zgc
2
∂ log Zgc
=
.
∂β 2
η
(2.252)
Furthermore, the following holds:
Eyal Buks
Thermodynamics and Statistical Physics
89
Chapter 2. Ideal Gas
3
(∆E) = E 3 − 3E 2 U + 3EU 2 − U 3
= E 3 − 3U E 2 + 2U 3
"
3
2
3 #
1
∂ Zgc
3
∂Zgc
∂ Zgc
2
∂Zgc
=−
− 2
+ 3
Zgc
∂β η
∂β η
∂β 3 η Zgc
∂β 2 η Zgc
"
2
2 #
∂
1
1
∂ Zgc
∂Zgc
=−
− 2
2
∂β Zgc
Zgc
∂β η
∂β
η
3
∂ log Zgc
=−
.
∂β 3
η
(2.253)
For classical gas having no internal degrees of freedom one has
3/2
M
−η
,
N = log Zgc = e V
2π2 β
(2.254)
thus
U =−
∂ log Zgc
∂β
=
η
3N τ
.
2
(2.255)
a) Using Eq. (2.252)
∂ 3N
2U 2
3N
(∆E)2 = −
=
=
.
2
∂β 2β
2β
3N
(2.256)
b) Using Eq. (2.253)
∂ 3N
3N
8U 3
(∆E)3 = −
=
=
.
∂β 2β 2
9N 2
β3
18. The entropy change of the body is
τb
dτ
τb
∆σ1 = C
,
= C log
τ
τ
a
τa
(2.257)
(2.258)
and that of the bath is
∆σ2 =
∆Q
C (τ a − τ b )
=
,
τb
τb
(2.259)
thus
∆σ = C
τa
τa
− 1 − log
τb
τb
.
(2.260)
The function f (x) = x − 1 − log x in the range 0 < x < ∞ satisfy
f (x) ≥ 0, where f (x) > 0 unless x = 1.
Eyal Buks
Thermodynamics and Statistical Physics
90
2.9. Solutions Set 2
19. The efficiency is given by
η = 1+
Ql
Qca
Cp (τ a − τ c )
p2 (V1 − V2 )
= 1−γ
, (2.261)
= 1+
= 1+
Qh
Qab
Cv (τ b − τ a )
V2 (p1 − p2 )
where γ = Cp /CV .
20. Energy conservation requires that W = Ql + Qh . Consider a Carnot heat
engine operating between the same thermal baths producing work W per
cycle. The Carnot engine consumes heat Q′h from the hot bath per cycle
and delivered −Q′l heat to the cold one per cycle, where W = Q′l + Q′h
and
ηc =
W
τl
= 1−
.
Q′h
τh
(2.262)
According to Clausius principle
Ql + Q′l ≤ 0 ,
(2.263)
thus
γ=
τl
Ql
Q′
Q′ − W
τh
−1=
.
≤− l = h
=
W
W
W
τh − τl
τh − τl
(2.264)
21. Using Eq. (2.264)
A (τ h − τ l )
τl
=
,
P
τh − τl
(2.265)
P
τ 2l − 2τ l τ h +
+ τ 2h = 0 ,
2A
(2.266)
thus
or
P
τl = τh +
±
2A
,
2
P
τh +
− τ 2h .
2A
The solution for which τ l ≤ τ h is
,
2
P
P
τl = τh +
−
τh +
− τ 2h .
2A
2A
22. Using Eq. (2.244)
∂U
∂p
=τ
−p,
∂V τ ,N
∂τ V,N
Eyal Buks
Thermodynamics and Statistical Physics
(2.267)
(2.268)
(2.269)
91
Chapter 2. Ideal Gas
thus
Bτ n
3Aτ 3
2Aτ 3
=
−p=
,
V
V
V
(2.270)
therefore
B = 2A ,
n=3.
(2.271)
(2.272)
23. In general the following holds
3
∂ 2 (τ log Zint )
+τ
,
CV = N
2
∂τ 2
Cp = cv + N ,
(2.273)
(2.274)
where in our case
∆
Zint = 1 + 3 exp −
,
τ
thus
a) CV is given by

∆ 2
(2.275)
−∆
τ

3
e

3
CV = N  + τ
2  ,
∆
2
1 + 3e− τ
b) and Cp is given by


∆ 2 − ∆
τ
3
e

5
Cp = N  + τ
2  ,
∆
2
1 + 3e− τ
(2.276)
(2.277)
24. Consider an infinitesimal change in the temperatures of both bodies
dτ 1 and dτ 2 . The total change in entropy associated with the reversible
process employed by the heat engine vanishes, thus
dQ1 dQ2
dτ 1 dτ 2
+
=C
+
.
(2.278)
0 = dσ = dσ 1 + dσ 2 =
τ1
τ2
τ1
τ2
a) Thus, by integration the equation
dτ 1
dτ 2
=−
,
τ1
τ2
(2.279)
one finds
Eyal Buks
Thermodynamics and Statistical Physics
92
2.9. Solutions Set 2
τf
τ1
or
log
dτ 1
=−
τ1
τf
τ2
dτ 2
,
τ2
τ2
τf
= log
,
τ1
τf
(2.280)
(2.281)
thus
τf =
√
τ 1τ 2.
(2.282)
b) Employing energy conservation law yields
√
√ 2
W = ∆U1 + ∆U2 = C (τ 1 − τ f ) + C (τ 2 − τ f ) = C ( τ 1 − τ 2 ) .
(2.283)
25. Energy conservation requires that the temperature of the mixture will
remain τ . The entropy of an ideal gas of density n, which contains N
particles, is given by
nQ 5
σ (N, n) = N log
+
,
(2.284)
n
2
where
Mτ
2π2
N
n=
.
V
nQ =
3/2
,
(2.285)
(2.286)
Using the relation
pV = N τ ,
(2.287)
one finds that the final pressure of the gas after the partition has been
removed and the system has reached thermal equilibrium is given by
pfinal =
2p1 p2
.
p1 + p2
Thus, the change in entropy is given by
∆σ = σfinal − σ1 − σ 2
(p1 + p2 ) τ nQ 5
τ nQ 5
τ nQ 5
= 2N log
+
− N log
+
− N log
+
2p1 p2
2
p1
2
p2
2
= N log
(p1 + p2 )2
.
4p1 p2
(2.288)
Eyal Buks
Thermodynamics and Statistical Physics
93
Chapter 2. Ideal Gas
26. Using
σ = log Zc + βU = log Zgc + βU + ηN
one finds
log Zc = log Zgc + ηN .
(2.289)
The following holds for classical ideal gas having no internal degrees of
freedom
log Zgc = N
η = −βµ = log
nQ V
,
N
(2.290)
where
nQ =
Mτ
2π2
3/2
,
(2.291)
thus
nQ V
log Zc = N 1 + log
N
= N log (nQ V ) + N − N log N
≃ N log (nQ V ) − log N !
= log
(nQ V )N
,
N!
(2.292)
or
1
Zc =
N!
Mτ
2π2
3/2
V
N
.
(2.293)
27. The efficiency is defined as η = W/Qh , where W is the total work, and Qh
is the heat extracted from the heat bath at higher temperature. Energy
conservation requires that W = Qh + Ql , where Ql is the heat extracted
from the heat bath at lower temperature, thus η = 1 + Ql /Qh . In the
present case Qh is associated with process a → b, while Ql is associated
with process c → d. In both isentropic processes (b → c and d → a) no
heat is exchanged. Thus
η =1+
Eyal Buks
Ql
Cp (τ d − τ c )
=1+
.
Qh
Cp (τ b − τ a )
Thermodynamics and Statistical Physics
(2.294)
94
2.9. Solutions Set 2
Using pV = Nτ yields
η =1+
τd − τc
p2 (Vd − Vc )
.
=1+
τb − τa
p1 (Vb − Va )
(2.295)
Along the isentropic process pV γ is constant, where γ = Cp /Cv , thus
η =1+
p2
p1
p1
p2
γ1
(Va − Vb )
Vb − Va
=1−
p2
p1
γ−1
γ
.
(2.296)
28. No heat is exchanged in the isentropic processes, thus the efficiency is
given by
Ql
Qh
Qc→d
= 1+
Qa→b
cV (τ d − τ c )
= 1+
.
cV (τ b − τ a )
η = 1+
(2.297)
Since τV γ−1 remains unchanged in an isentropic process, where
γ=
cp
,
cV
(2.298)
one finds that
τ b V1γ−1 = τ c V2γ−1 ,
τ d V2γ−1 = τ a V1γ−1 ,
(2.299)
(2.300)
or
τc
τd
=
=
τb
τa
V2
V1
1−γ
,
(2.301)
thus
η =1−
V2
V1
1−γ
.
(2.302)
29. Let VA1 = Nτ A /p (VB1 = N τ B /p) be the initial volume of vessel A (B)
and let VA2 (VB2 ) be the final volume of vessel A (B). In terms of the
final temperature of both vessels, which is denoted as τ f , one has
VA2 = VB2 =
Eyal Buks
Nτ f
.
p
Thermodynamics and Statistical Physics
(2.303)
95
Chapter 2. Ideal Gas
The entropy of an ideal gas of density n = N/V , which contains N
particles, is given by
nQ 5
+
,
(2.304)
σ = N log
n
2
where
nQ =
Mτ
2π2
3/2
,
or as a function of τ and p
M 3/2 5/2
τ
5
2π 2
σ = N log
+
.
p
2
(2.305)
(2.306)
Thus the change in entropy is given by
∆σ = σfinal − σinitial
τ 2f
5N
log
=
.
2
τ Aτ B
(2.307)
In general, for an isobaric process the following holds
Q = W + ∆U = p (V2 − V1 ) + cV (τ 2 − τ 1 ) ,
(2.308)
where Q is the heat that was added to the gas, W the work done by the
gas and ∆U the change in internal energy of the gas. Using the equation
of state pV = N τ this can be written as
Q = (N + cV ) (τ 2 − τ 1 ) .
(2.309)
Since no heat is exchanged with the environment during this process the
following holds
QA + QB = 0 ,
where
QA = (N + cV ) (τ f − τ A ) ,
QB = (N + cV ) (τ f − τ B ) ,
(2.310)
(2.311)
thus
τf =
τA + τB
,
2
(2.312)
and therefore
∆σ =
Eyal Buks
5N
(τ A + τ B )2
log
.
2
4τ A τ B
Thermodynamics and Statistical Physics
(2.313)
96
3. Bosonic and Fermionic Systems
In the first part of this chapter we study two Bosonic systems, namely photons
and phonons. A photon is the quanta of electromagnetic waves whereas a
phonon is the quanta of acoustic waves. In the second part we study two
Fermionic systems, namely electrons in metals and electrons and holes in
semiconductors.
3.1 Electromagnetic Radiation
In this section we study an electromagnetic cavity in thermal equilibrium.
3.1.1 Electromagnetic Cavity
Consider an empty volume surrounded by conductive walls having infinite
conductivity. The Maxwell’s equations in SI units are given by
∇ × H = ǫ0
∂E
,
∂t
∇ × E = −µ0
∂H
,
∂t
(3.1)
(3.2)
∇·E=0,
(3.3)
∇·H=0,
(3.4)
and
where ǫ0 = 8.85 × 10−12 F m−1 and µ0 = 1.26 × 10−6 N A−2 are the permittivity and permeability respectively of free space, and the following holds
ǫ0 µ0 =
1
,
c2
(3.5)
where c = 2.99 × 108 m s−1 is the speed of light in vacuum.
In the Coulomb gauge, where the vector potential A is chosen such that
∇·A=0 ,
(3.6)
Chapter 3. Bosonic and Fermionic Systems
the scalar potential φ vanishes in the absence of sources (charge and current),
and consequently both fields E and H can be expressed in terms of A only
as
E=−
∂A
,
∂t
(3.7)
and
µ0 H =∇ × A .
(3.8)
The gauge condition (3.6) and Eqs. (3.7) and (3.8) guarantee that
Maxwell’s equations (3.2), (3.3), and (3.4) are satisfied
∇×E=−
∂ (∇ × A)
∂H
= −µ0
,
∂t
∂t
∂ (∇ · A)
=0,
∂t
1
∇ · (∇ × A) = 0 ,
∇·H=
µ0
∇·E=−
(3.9)
(3.10)
(3.11)
where in the last equation the general vector identity ∇ · (∇ × A) = 0 has
been employed. Substituting Eqs. (3.7) and (3.8) into the only remaining
nontrivial equation, namely into Eq. (3.1), leads to
∇ × (∇ × A) = −
1 ∂2A
.
c2 ∂t2
(3.12)
Using the vector identity
∇ × (∇ × A) = ∇ (∇ · A) − ∇2 A ,
(3.13)
and the gauge condition (3.6) one finds that
∇2 A =
1 ∂2A
.
c2 ∂t2
(3.14)
Consider a solution in the form
A = q (t) u (r) ,
(3.15)
where q (t) is independent on position r and u (r) is independent on time t.
The gauge condition (3.6) leads to
∇·u=0.
(3.16)
From Eq. (3.14) one finds that
q∇2 u =
Eyal Buks
1 d2 q
u
.
c2 dt2
Thermodynamics and Statistical Physics
(3.17)
98
3.1. Electromagnetic Radiation
Multiplying by an arbitrary unit vector n̂ leads to
2 ∇ u · n̂
1 d2 q
= 2
.
u · n̂
c q dt2
(3.18)
The left hand side of Eq. (3.18) is a function of r only while the right hand
side is a function of t only. Therefore, both should equal a constant, which is
denoted as −κ2 , thus
∇2 u+κ2 u = 0 ,
(3.19)
d2 q
+ω 2κ q = 0 ,
dt2
(3.20)
and
where
ω κ = cκ .
(3.21)
Equation (3.19) should be solved with the boundary conditions of a perfectly conductive surface. Namely, on the surface S enclosing the cavity we
have H · ŝ = 0 and E × ŝ = 0, where ŝ is a unit vector normal to the surface.
To satisfy the boundary condition for E we require that u be normal to the
surface, namely, u = ŝ (u · ŝ) on S. This condition guarantees also that the
boundary condition for H is satisfied. To see this we calculate the integral of
the normal component of H over some arbitrary portion S ′ of S. Using Eq.
(3.8) and Stoke’s’ theorem one finds that
q
(H · ŝ) dS =
[(∇ × u) · ŝ] dS
µ0 S ′
S′
/
q
=
u·dl ,
µ0 C
(3.22)
where the close curve C encloses the surface S ′ . Thus, since u is normal to
the surface one finds that the integral along the close curve C vanishes, and
therefore
(H · ŝ) dS = 0 .
(3.23)
S′
Since S ′ is arbitrary we conclude that H · ŝ =0 on S.
Each solution of Eq. (3.19) that satisfies the boundary conditions is called
an eigen mode. As can be seen from Eq. (3.20), the dynamics of a mode
amplitude q is the same as the dynamics of an harmonic oscillator having
angular frequency ω κ = cκ.
Eyal Buks
Thermodynamics and Statistical Physics
99
Chapter 3. Bosonic and Fermionic Systems
3.1.2 Partition Function
What is the partition function of a mode having eigen angular frequency ω κ ?
We have seen that the mode amplitude has the dynamics of an harmonic
oscillator having angular frequency ω κ . Thus, the quantum eigenenergies of
the mode are
εs = sωκ ,
(3.24)
where s = 0, 1, 2, ... is an integer1 . When the mode is in the eigenstate having
energy εs the mode is said to occupy s photons. The canonical partition
function of the mode is found using Eq. (1.69)
Zκ =
∞
exp (−sβωκ )
s=0
1
.
1 − exp (−βω κ )
=
(3.25)
Note the similarity between this result and the orbital partition function ζ
of Bosons given by Eq. (2.36). The average energy is found using
∂ log Zκ
∂β
ω κ
= βωκ
.
e
−1
εκ = −
The partition function of the entire system is given by
Z=
Zκ ,
(3.26)
(3.27)
κ
and the average total energy by
U =−
∂ log Z εκ .
=
∂β
κ
(3.28)
3.1.3 Cube Cavity
For simplicity, consider the case of a cavity shaped as a cube of volume
V = L3 . We seek solutions of Eq. (3.19) satisfying the boundary condition
1
In Eq. (3.24) above the ground state energy was taken to be zero. Note that
by taking instead εs = (s + 1/2) ω κ , one obtains Zκ = 1/2 sinh (βℏωκ /2) and
εn = (ωκ /2) coth (βℏω κ /2). In some cases the offset energy term ωκ /2 is
very important (e.g., the Casimir force), however, in what follows we disregard
it.
Eyal Buks
Thermodynamics and Statistical Physics
100
3.1. Electromagnetic Radiation
that the tangential component of u vanishes on the walls. Consider a solution
having the form
8
ax cos (kx x) sin (ky y) sin (kz z) ,
ux =
(3.29)
V
8
ay sin (kx x) cos (ky y) sin (kz z) ,
(3.30)
uy =
V
8
az sin (kx x) sin (ky y) cos (kz z) .
uz =
(3.31)
V
While the boundary condition on the walls x = 0, y = 0, and z = 0 is
guaranteed to be satisfied, the boundary condition on the walls x = L, y = L,
and z = L yields
nx π
,
(3.32)
kx =
L
ny π
ky =
,
(3.33)
L
nz π
kz =
,
(3.34)
L
where nx , ny and nz are integers. This solution clearly satisfies Eq. (3.19)
where the eigen value κ is given by
0
κ = kx2 + ky2 + kz2 .
(3.35)
Alternatively, using the notation
n = (nx , ny , nz ) ,
(3.36)
one has
κ=
π
n,
L
(3.37)
0
n2x + n2y + n2z .
(3.38)
where
n=
Using Eq. (3.21) one finds that the angular frequency of a mode characterized
by the vector of integers n is given by
ωn =
πc
n.
L
(3.39)
In addition to Eq. (3.19) and the boundary condition, each solution has to
satisfy also the transversality condition ∇ · u = 0 (3.16), which in the present
case reads
n·a= 0 ,
Eyal Buks
(3.40)
Thermodynamics and Statistical Physics
101
Chapter 3. Bosonic and Fermionic Systems
where
a = (ax , ay , az ) .
(3.41)
Thus, for each set of integers {nx , ny , nz } there are two orthogonal modes
(polarizations), unless nx = 0 or ny = 0 or nz = 0. In the latter case, only a
single solution exists.
3.1.4 Average Energy
The average energy U of the system is found using Eqs. (3.26), (3.28) and
(3.39)
ω n
U =
eβωn − 1
n
= 2τ
∞ ∞ ∞
nx =0 ny =0 nz
αn
,
αn − 1
e
=0
(3.42)
where the dimensionless parameter α is given by
α=
βπc
.
L
(3.43)
The following relation can be employed to estimate the dimensionless parameter α
α=
2.4 × 10−3
τ
L
cm 300 K
.
(3.44)
In the limit where
α≪1
(3.45)
the sum can be approximated by the integral
4π
U ≃ 2τ
8
∞
dn n2
0
αn
.
eαn − 1
(3.46)
Employing the integration variable transformation [see Eq. (3.39)]
n=
L
ω,
πc
(3.47)
allows expressing the energy per unit volume U/V as
U
=
V
∞
dω uω ,
(3.48)
0
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Thermodynamics and Statistical Physics
102
3.1. Electromagnetic Radiation
where
uω =
c3 π 2
ω3
.
−1
(3.49)
eβω
This result is know as Plank’s radiation law. The factor uω represents the
spectral distribution of the radiation. The peak in uω is obtained at βω0 =
2.82. In terms of the wavelength λ0 = 2πc/ω0 one has
−1
λ0
T
= 5.1
.
(3.50)
µm
1000 K
1.4
1.2
1
0.8
0.6
0.4
0.2
0
2
4
6
x
3
8
10
x
The function x / (e − 1).
The total energy is found by integrating Eq. (3.48) and by employing the
variable transformation x = βω
∞ 3
U
τ4
x dx
= 3 2 3
V
c π ex − 1
0
π4
15
=
π2 τ 4
.
153 c3
(3.51)
3.1.5 Stefan-Boltzmann Radiation Law
Consider a small hole having area dA drilled into the conductive wall of an
electromagnetic (EM) cavity. What is the rate of energy radiation emitted
from the hole? We employ below a kinetic approach to answer this question.
Consider radiation emitted in a time interval dt in the direction of the unit
vector û. Let θ be the angle between û and the normal to the surface of the
hole. Photons emitted during that time interval dt in the direction û came
from the region in the cavity that is indicated in Fig. 3.1, which has volume
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Thermodynamics and Statistical Physics
103
Chapter 3. Bosonic and Fermionic Systems
û
cdt
θ
dAcosθ
Fig. 3.1. Radiation emitted through a small hole in the cavity wall.
Vθ = dA cos θ × cdt .
(3.52)
The average energy in that region can be found using Eq. (3.51). Integrating
over all possible directions yields the total rate of energy radiation emitted
from the hole per unit area
1 1
J=
dAdt 4π
π/2
2π
U
dθ sin θ dϕ Vθ
V
0
0
π/2
2π
π2τ 4 1
=
dθ
sin
θ
cos
θ
dϕ
153 c2 4π
0
0
1/4
2 4
=
π τ
603 c2
(3.53)
In terms of the historical definition of temperature T = τ/kB [see Eq. (1.92)]
one has
J = σB T 4 ,
(3.54)
where σ B , which is given by
σB =
4
π2 kB
W
= 5.67 × 10−8 2 4 ,
3
2
60 c
m K
(3.55)
is the Stefan-Boltzmann constant.
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104
3.2. Phonons in Solids
m mω2 m mω2 m mω2 m mω2 m
Fig. 3.2. 1D lattice.
3.2 Phonons in Solids
In this section we study elastic waves in solids. We start with a onedimensional example, and then generalize some of the results for the case
of a 3D lattice.
3.2.1 One Dimensional Example
Consider the 1D lattice shown in Fig. 3.2 below, which contains N ’atoms’
having mass m each that are attached to each other by springs having spring
constant mω 2 . The lattice spacing is a. The atoms are allowed to move in
one dimension along the array axis. In problem 5 of set 3 one finds that the
normal mode angular eigen-frequencies are given by
2
2
1
2 kn a 2
2 ,
ω n = ω 2 (1 − cos kn a) = 2ω 22sin
(3.56)
2 2
where a is the lattice spacing,
kn =
2πn
,
aN
(3.57)
and n is integer ranging from −N/2 to N/2.
0.8
0.6
0.4
0.2
-1
-0.8 -0.6 -0.4 -0.2 0
0.2
0.4 x 0.6
0.8
1
The function |sin (πx/2)|.
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Chapter 3. Bosonic and Fermionic Systems
What is the partition function of an eigen-mode having eigen angular frequency ωn ? The mode amplitude has the dynamics of an harmonic oscillator
having angular frequency ω n . Thus, as we had in the previous section, where
we have discussed EM modes, the quantum eigenenergies of the mode are
εs = sωn ,
(3.58)
where s = 0, 1, 2, ... is an integer. When the mode is in an eigenstate having
energy εs the mode is said to occupy s phonons. The canonical partition
function of the mode is found using Eq. (1.69)
Zκ =
∞
exp (−sβωκ )
s=0
=
1
.
1 − exp (−βω κ )
(3.59)
Similarly to the EM case, the average total energy is given by
N/2
U=
n=−N/2
ℏω n
,
exp (βℏωn ) − 1
(3.60)
where β = 1/τ , and the total heat capacity is given by
CV =
∂U
=
∂τ
N/2
(βℏωn )2 exp (βℏω n )
.
[exp (βℏω n ) − 1]2
n=−N/2
(3.61)
High Temperature Limit. In the high temperature limit βℏω ≪ 1
(βℏω n )2 exp (βℏωn )
[exp (βℏωn ) − 1]2
≃1,
(3.62)
therefore
CV = N .
(3.63)
Low Temperature Limit. In the low temperature limit βℏω ≫ 1 the
main contribution to the sum in Eq. (3.61) comes from terms for which
|n| N/βℏω. Thus, to a good approximation the dispersion relation can
be approximated by
2
2
2
2
2
2 kn a 2
kn a 22
2
2
2 = ω 2π |n| .
ω n = 2ω 2sin
≃ 2ω 2
(3.64)
2
2
2 2
N
Moreover, in the limit N ≫ 1 the sum in Eq. (3.61) can be approximated
by an integral, and to a good approximation the upper limit N/2 can be
substituted by infinity, thus
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3.2. Phonons in Solids
N/2
(βℏω n )2 exp (βℏωn )
CV =
2
[exp (βℏω n ) − 1]
2
∞ βℏω 2π
exp βℏω 2π
N n
Nn
≃2
2
exp βℏω 2π
n=0
Nn −1
2
∞ βℏω 2π
exp βℏω 2π
Nn
N n
≃2
dn 2
0
exp βℏω 2π
N n −1
∞
N τ
x2 exp (x)
=
dx
π ℏω 0
(exp (x) − 1)2
n=−N/2
π2 /3
Nπ τ
.
3 ℏω
=
(3.65)
3.2.2 The 3D Case
The case of a 3D lattice is similar to the case of EM cavity that we have
studied in the previous section. However, there are 3 important distinctions:
1. The number of modes of a lattice containing N atoms that can move in
3D is finite, 3N instead of infinity as in the EM case.
2. For any given vector k there are 3, instead of only 2, orthogonal modes
(polarizations).
3. Dispersion: contrary to the EM case, the dispersion relation (namely, the
function ω (k)) is in general nonlinear.
Due to distinctions 1 and 2, the sum over all modes is substituted by an
integral according to
∞ ∞
∞ nx =0 ny =0 nz
3
→ 4π
8
=0
nD
dn n2 ,
(3.66)
0
where the factor of 3 replaces the factor of 2 we had in the EM case. Moreover,
the upper limit is nD instead of infinity, where nD is determined from the
requirement
3
4π
8
nD
dn n2 = 3N ,
(3.67)
0
thus
nD =
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6N
π
1/3
.
Thermodynamics and Statistical Physics
(3.68)
107
Chapter 3. Bosonic and Fermionic Systems
Similarly to the EM case, the average total energy is given by
ℏω n
U =
exp
(βℏω
n) − 1
n
3π
=
2
nD
dn n2
0
(3.69)
ℏω n
.
exp (βℏωn ) − 1
(3.70)
To proceed with the calculation the dispersion relation ω n (kn ) is needed.
Here we assume for simplicity that dispersion can be disregarded to a good
approximation, and consequently the dispersion relation can be assumed to
be linear
ω n = vkn ,
(3.71)
where v is the sound velocity. The wave vector kn is related to n =
0
n2x + n2y + n2z by
kn =
πn
,
L
(3.72)
where L = V 1/3 and V is the volume. In this approximation one finds using
the variable transformation
x=
βℏvπn
,
L
(3.73)
that
3π
U =
2
=
nD
dn n2
0
3V τ 4
2ℏ3 v3 π2
exp
xD
dx
0
ℏvπn
L βℏvπn
L
−1
x3
.
exp x − 1
(3.74)
where
1/3
βℏvπ 6N
βℏvπnD
π
xD =
=
.
L
L
Alternatively, in terms of the Debye temperature, which is defined as
Θ = ℏv
6π2 N
V
1/3
,
(3.75)
one has
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Thermodynamics and Statistical Physics
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3.2. Phonons in Solids
xD =
Θ
,
τ
(3.76)
and
τ 3 xD
x3
U = 9N τ
dx
.
Θ
exp x − 1
(3.77)
0
As an example Θ/kB = 88 K for Pb, while Θ/kB = 1860 K for diamond.
Below we calculate the heat capacity CV = ∂U/∂τ in two limits.
High Temperature Limit. In the high temperature limit xD = Θ/τ ≪ 1,
thus
τ 3 xD
x3
U = 9Nτ
dx
Θ
exp x − 1
0
τ 3 xD
≃ 9Nτ
dx x2
Θ
0
τ 3 x3
D
= 9Nτ
Θ
3
= 3Nτ ,
(3.78)
and therefore
CV =
∂U
= 3N .
∂τ
(3.79)
Note that in this limit the average energy of each mode is τ and consequently
U = 3Nτ . This result demonstrates the equal partition theorem of classical
statistical mechanics that will be discussed in the next chapter.
Low Temperature Limit. In the low temperature limit xD = Θ/τ ≫ 1,
thus
τ 3 xD
x3
U = 9Nτ
dx
Θ
exp x − 1
0
τ 3 ∞
x3
≃ 9Nτ
dx
Θ
exp x − 1
0
π4 /15
=
τ 3
3π4
Nτ
,
5
Θ
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(3.80)
109
Chapter 3. Bosonic and Fermionic Systems
and therefore
CV =
∂U
12π4 τ 3
=
N
.
∂τ
5
Θ
(3.81)
Note that Eq. (3.80) together with Eq. (3.75) yield
U
3 π2τ 4
=
.
V
2 15ℏ3 v3
(3.82)
Note the similarity between this result and Eq. (3.51) for the EM case.
3.3 Fermi Gas
In this section we study an ideal gas of Fermions of mass m. While only the
classical limit was considered in chapter 2, here we consider the more general
case.
3.3.1 Orbital Partition Function
Consider an orbital having energy εn . Disregarding internal degrees of freedom, its grandcanonical Fermionic partition function is given by [see Eq.
(2.33)]
ζ n = 1 + λ exp (−βεn ) ,
(3.83)
where
λ = exp (βµ) = e−η ,
(3.84)
is the fugacity and β = 1/τ . Taking into account internal degrees of freedom
the grandcanonical Fermionic partition function becomes
ζn =
(1 + λ exp (−βεn ) exp (−βEl )) ,
(3.85)
l
where {El } are the eigenenergies of a particle due to internal degrees of
freedom [see Eq. (2.71)]. As is required by the Pauli exclusion principle, no
more than one Fermion can occupy a given internal eigenstate and a given
orbital.
3.3.2 Partition Function of the Gas
The grandcanonical partition function of the gas is given by
Zgc =
ζn .
(3.86)
n
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3.3. Fermi Gas
As we have seen in chapter 2, the orbital eigenenergies of a particle of mass
m in a box are given by Eq. (2.5)
εn =
where
2 π 2 2
n ,
2m L
(3.87)
n = (nx , ny , nz ) ,
0
n = n2x + n2y + n2z ,
(3.88)
(3.89)
nx , ny , nz = 1, 2, 3, ... ,
(3.90)
and L3 = V is the volume of the box. Thus, log Zgc can be written as
log Zgc =
∞ ∞
∞ log ζ n .
(3.91)
nx =1 ny =1 nz =1
Alternatively, using the notation
α2 =
β2 π2
,
2mL2
(3.92)
and Eq. (3.85) one has
log Zgc =
∞ ∞ ∞
l
nx =1 ny =1 nz =1
log 1 + λ exp −α2 n2 exp (−βEl ) . (3.93)
For a macroscopic system α ≪ 1, and consequently the sum over n can
be approximately replaced by an integral
∞ ∞
∞ nx =0 ny =0 nz =0
→
1
4π
8
∞
dn n2 ,
(3.94)
0
thus, one has
∞
log Zgc
π
=
dn n2 log 1 + λ exp −α2 n2 exp (−βEl ) .
2
l
(3.95)
0
This can be further simplified by employing the variable transformation
βε = α2 n2 .
(3.96)
The following holds
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Chapter 3. Bosonic and Fermionic Systems
√ 3/2
ε β
dε = n2 dn .
2
α2
Thus, by introducing the density of states
2m 3/2
V
2π2
D (ε) =
2
0
ε1/2 ε ≥ 0 ,
ε<0
(3.97)
one has
log Zgc =
1
2
∞
dε D (ε) log (1 + λ exp (−β (ε + El ))) .
(3.98)
l −∞
3.3.3 Energy and Number of Particles
Using Eqs. (1.80) and (1.94) for the energy U and the number of particles
N , namely using
∂ log Zgc
U =−
,
(3.99)
∂β
η
∂ log Zgc
,
∂λ
one finds that
∞
1
U =
dε D (ε) (ε + El ) fFD (ε + El ) ,
2 l
N =λ
N =
−∞
∞
1
2
dε D (ε) fFD (ε + El ) ,
(3.100)
(3.101)
(3.102)
l −∞
where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)]
fFD (ǫ) =
1
.
exp [β (ǫ − µ)] + 1
(3.103)
3.3.4 Example: Electrons in Metal
Electrons are Fermions having spin 1/2. The spin degree of freedom gives rise
to two orthogonal eigenstates having energies E+ and E− respectively. In the
absent of any external magnetic field these states are degenerate, namely
E+ = E− . For simplicity we take E+ = E− = 0. Thus, Eqs. (3.101) and
(3.102) become
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Thermodynamics and Statistical Physics
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3.3. Fermi Gas
U =
N =
∞
−∞
∞
dε D (ε) εfFD (ε) ,
(3.104)
dε D (ε) fFD (ε) ,
(3.105)
−∞
Typically for metals at room temperature or below the following holds τ ≪ µ.
Thus, it is convenient to employ the following theorem (Sommerfeld expansion) to evaluate these integrals.
Theorem 3.3.1. Let g (ε) be a function that vanishes in the limit ε → −∞,
and that diverges no more rapidly than some power of ε as ε → ∞. Then,
the following holds
∞
=
−∞
µ
dε g (ε) fFD (ε)
dε g (ε) +
−∞
π 2 g′ (µ)
+O
6β 2
1
βµ
4
.
Proof. See problem 7 of set 3.
With the help of this theorem the number of particles N to second order
in τ is given by
N=
µ
dε D (ε) +
−∞
π2 τ 2 D′ (µ)
.
6
(3.106)
Moreover, at low temperatures, the chemical potential is expected to be close
the the Fermi energy εF , which is defined by
εF = lim µ .
(3.107)
τ →0
Thus, to lowest order in µ − εF one has
µ
dε D (ε) =
−∞
εF
−∞
dε D (ε) + (µ − εF ) D (εF ) + O (µ − εF )2 ,
(3.108)
and therefore
N = N0 + (µ − εF ) D (εF ) +
π2 τ 2 D ′ (εF )
,
6
(3.109)
where
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Thermodynamics and Statistical Physics
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Chapter 3. Bosonic and Fermionic Systems
εF
N0 =
dε D (ε) ,
(3.110)
−∞
is the number of electrons at zero temperature. The number of electrons N
in metals is expected to be temperature independent, namely N = N0 and
consequently
π 2 D′ (εF )
.
6β 2 D (εF )
µ = εF −
(3.111)
Similarly, the energy U at low temperatures is given approximately by
∞
U =
dε D (ε) εfFD (ε)
=
−∞
εF
−∞
dε D (ε) ε + (µ − εF ) D (εF ) εF +
U0
2 2
π2 τ 2
(D′ (εF ) εF + D (εF ))
6
π τ D′ (εF )
π2τ 2
D (εF ) εF +
(D′ (εF ) εF + D (εF ))
6D (εF )
6
π2τ 2
= U0 +
D (εF ) ,
6
= U0 −
(3.112)
where
U0 =
εF
dε D (ε) ε .
(3.113)
−∞
From this result one finds that the electronic heat capacity is given by
CV =
∂U
π2 τ
=
D (εF ) .
∂τ
3
(3.114)
Comparing this result with Eq. (3.81) for the phonons heat capacity, which
is proportional to τ 3 at low temperatures, suggests that typically, while the
electronic contribution is the dominant one at very low temperatures, at
higher temperatures the phonons’ contribution becomes dominant.
3.4 Semiconductor Statistics
To be written...
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Thermodynamics and Statistical Physics
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3.5. Problems Set 3
3.5 Problems Set 3
1. Calculate the average number of photons N in equilibrium at temperature
τ in a cavity of volume V . Use this result to estimate the number of
photons in the universe assuming it to be a spherical cavity of radius
1026 m and at temperature τ = kB × 3 K.
2. Write a relation between the temperature of the surface of a planet and its
distance from the Sun, on the assumption that as a black body in thermal
equilibrium, it reradiates as much thermal radiation, as it receives from
the Sun. Assume also, that the surface of the planet is at constant temperature over the day-night cycle. Use TSun = 5800 K; RSun = 6.96 × 108 m;
and the Mars-Sun distance of DM−S = 2.28 × 1011 m and calculate the
temperature of Mars surface.
3. Calculate the Helmholtz free energy F of photon gas having total energy
U and volume V and use your result to show that the pressure is given
by
p=
U
.
3V
(3.115)
4. Consider a photon gas initially at temperature τ 1 and volume V1 . The gas
is adiabatically compressed from volume V1 to volume V2 in an isentropic
process. Calculate the final temperature τ 2 and final pressure p2 .
5. Consider a one-dimensional lattice of N identical point particles of mass
m, interacting via nearest-neighbor spring-like forces with spring constant
mω2 (see Fig. 3.2). Denote the lattice spacing by a. Show that the normal
mode eigen-frequencies are given by
1
ωn = ω 2 (1 − cos kn a) ,
(3.116)
where kn = 2πn/aN , and n is integer ranging from −N/2 to N/2 (assume
N ≫ 1).
6. Consider an orbital with energy ε in an ideal gas. The system is in thermal
equilibrium at temperature τ and chemical potential µ.
a) Show that the probability that the orbital is occupied by n particles
is given by
pF (n) =
exp [n (µ − ε) β]
,
1 + exp [(µ − ε) β]
(3.117)
for the case of Fermions, where n ∈ {0, 1}, and by
pB (n) = {1 − exp [(µ − ε) β]} exp [n (µ − ε) β] ,
(3.118)
where n ∈ {0, 1, 2, ...}, for the case of Bosons.
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Thermodynamics and Statistical Physics
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Chapter 3. Bosonic and Fermionic Systems
2
2
b) Show that the variance (∆n) = (n − n
) is given by
(∆n)2F = n
F (1 − n
F ) ,
(3.119)
for the case of Fermions, and by
(∆n)2B = n
B (1 + n
B ) ,
(3.120)
for the case of Bosons.
7. Let g (ε) be a function that vanishes in the limit ε → −∞, and that
diverges no more rapidly than some power of ε as ε → ∞. Show that the
following holds (Sommerfeld expansion)
I=
=
∞
−∞
µ
−∞
dε g (ε) fFD (ε)
dε g (ε) +
π2 g ′ (µ)
+O
6β 2
1
βµ
4
.
8. Consider a metal at zero temperature having Fermi energy εF , number
of electrons N and volume V .
a) Calculate the mean energy of electrons.
b) Calculate the ratio α of the mean-square-speed of electrons to the
square of the mean speed
2
v
.
(3.121)
α=
v
2
c) Calculate the pressure exerted by an electron gas at zero temperature.
9. For electrons with energy ε ≫ mc2 (relativistic fermi gas), the energy
is given by ε = pc. Find the fermi energy of this gas and show that the
ground state energy is
3
E (T = 0) = N εF
4
(3.122)
10. A gas of two dimensional electrons is free to move in a plane. The mass
of each electron is me , the density (number of electrons per unit area) is
n, and the temperature is τ. Show that the chemical potential µ is given
by
nπ2
µ = τ log exp
−1 .
(3.123)
me τ
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Thermodynamics and Statistical Physics
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3.6. Solutions Set 3
3.6 Solutions Set 3
1. The density of states of the photon gas is given by
dg =
V ε2
dε .
π2 ℏ3 c3
(3.124)
Thus
V
N= 2 3 3
π ℏ c
∞
0
where
ε2
dε
eε/τ − 1
τ 3
=V
α,
ℏc
1
α= 2
π
∞
0
(3.125)
x2
dx .
ex − 1
(3.126)
The number α is calculated numerically
α = 0.24359 .
(3.127)
For the universe
N=
4π 26 3
10 m
3
≃ 2.29 × 1087 .
1.3806568 × 10−23 J K−1 3 K
1.05457266 × 10−34 J s2.99792458 × 108 m s−1
3
(3.128)
2. The energy emitted by the Sun is
4
ESun = 4πR2Sun σB TSun
,
(3.129)
and the energy emitted by a planet is
2
4
Eplanet = 4πRplanet
σB Tplanet
.
(3.130)
The fraction of Sun energy that planet receives is
2
πRplanet
ESun ,
2
4πDM−S
(3.131)
and this equals to the energy it reradiates. Therefore
πR2planet
ESun = Eplanet ,
4πD2
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× 0.24359
Chapter 3. Bosonic and Fermionic Systems
thus
Tplanet =
and for Mars
TMars =
RSun
TSun ,
2D
6.96 × 108 m
5800 K = 226 K .
2 × 2.28 × 1011 m
3. The partition function is given by
Z=
∞
exp (sβℏωn ) =
n s=0
n
1
,
1 − exp (−βℏω n )
thus the free energy is given by
F = −τ log Z = τ
log [1 − exp (−βℏωn )] .
n
Transforming the sum over modes into integral yields
∞
F = τπ
dnn2 log [1 − exp (−βℏω n )]
0
∞
βℏπcn
2
= τπ
dnn log 1 − exp −
,
L
0
(3.132)
or, by integrating by parts
1 ℏπ2 c ∞
n3
1
F =−
dn
=− U ,
βℏπcn
3 L 0
3
exp
−1
L
where
U=
π2τ 4V
.
15ℏ3 c3
(3.133)
Thus
∂F
p=−
∂V
τ
=
U
.
3V
(3.134)
4. Using the expression for Helmholtz free energy, which was derived in the
previous problem,
F =−
U
π2 τ 4 V
=−
,
3
45ℏ3 c3
one finds that the entropy is given by
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Thermodynamics and Statistical Physics
118
3.6. Solutions Set 3
σ=−
∂F
∂τ
=
V
4π2 τ 3 V
.
45ℏ3 c3
Thus, for an isentropic process, for which σ is a constant, one has
τ2 = τ1
V1
V2
1/3
.
Using again the previous problem, the pressure p is given by
p=
U
,
3V
thus
p=
π2 τ 4
,
45ℏ3 c3
and
p2 =
π2 τ 41
3 3
45ℏ
c p1
V1
V2
4/3
.
5. Let u (na) be the displacement of point particle number n. The equations
of motion are given by
mü (na) = −mω 2 {2u (na) − u [(n − 1) a] − u [(n + 1) a]} .
(3.135)
Consider a solution of the form
u (na, t) = ei(kna−ωn t) .
(3.136)
Periodic boundary condition requires that
eikNa = 1 ,
(3.137)
thus
kn =
2πn
.
aN
(3.138)
Substituting in Eq. 3.135 yields
−mω 2n u (na) = −mω 2 2u (na) − u (na) e−ika − u (na) eika , (3.139)
or
2
2
1
2 kn a 2
2
2 .
ωn = ω 2 (1 − cos kn a) = 2ω 2sin
2 2
Eyal Buks
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(3.140)
119
Chapter 3. Bosonic and Fermionic Systems
6. In general using Gibbs factor
exp [n (µ − ε) β]
,
p (n) = exp [n′ (µ − ε) β]
(3.141)
n′
where β = 1/τ , one finds for Fermions
pF (n) =
exp [n (µ − ε) β]
,
1 + exp [(µ − ε) β]
(3.142)
where n ∈ {0, 1}, and for Bosons
pB (n) =
exp [n (µ − ε) β]
= {1 − exp [(µ − ε) β]} exp [n (µ − ε) β] ,
∞
exp [n′ (µ − ε) β]
n′ =0
(3.143)
where n ∈ {0, 1, 2, ...}. The expectation value of n
in general is given
by
n′ exp [n (µ − ε) β]
′
n′ p (n′ ) = n
,
(3.144)
n
=
exp [n′ (µ − ε) β]
n′
n′
thus for Fermions
n
F =
1
,
exp [(ε − µ) β] + 1
(3.145)
and for Bosons
n
B = {1 − exp [(µ − ε) β]}
= {1 − exp [(µ − ε) β]}
=
1
.
exp [(ε − µ) β] − 1
∞
n′ =0
n′ exp [n′ (µ − ε) β]
exp [(µ − ε) β]
(1 − exp [(µ − ε) β])2
(3.146)
In general, the following holds

2
(n′′ )2 exp [n (µ − ε) β]
n′ exp [n′ (µ − ε) β]
 n′

′′
∂ n

−
τ
= n


∂µ τ
exp [n′ (µ − ε) β]
exp [n′ (µ − ε) β]
n′
= n2 − n
2 = (n − n
)2 .
n′
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Thermodynamics and Statistical Physics
(3.147)
120
3.6. Solutions Set 3
Thus
2
(∆n)F =
exp [(ε − µ) β]
2 = n
F (1 − n
F ) ,
(exp [(ε − µ) β] + 1)
exp [(ε − µ) β]
2
(∆n)B =
7. Let
G (ε) =
2
(exp [(ε − µ) β] − 1)
ε
= n
B (1 + n
B ) .
dε′ g (ε′ ) .
(3.148)
(3.149)
(3.150)
−∞
Integration by parts yields
∞
I=
dε g (ε) fFD (ε)
−∞
=
[G (ε) fFD (ε)|∞
−∞
=0
+
∞
−∞
∂fFD
dε G (ε) −
,
∂ε
where the following holds
βeβ(ε−µ)
∂fFD
β
.
−
=
2 =
2
β(ε−µ)
∂ε
e
+1
4 cosh β2 (ε − µ)
(3.151)
(3.152)
Using the Taylor expansion of G (ε) about ε − µ, which has the form
G (ε) =
∞
G(n) (µ)
(ε − µ)n ,
n!
n=0
(3.153)
yields
I=
∞
∞
G(n) (µ)
n=0
n!
−∞
β (ε − µ)n dε
.
4 cosh2 β2 (ε − µ)
(3.154)
Employing the variable transformation
x = β (ε − µ) ,
(3.155)
and exploiting the fact that (−∂fFD /∂ε) is an even function of ε−µ leads
to
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Thermodynamics and Statistical Physics
121
Chapter 3. Bosonic and Fermionic Systems
∞
∞
G(2n) (µ)
x2n dx
I=
.
2n
(2n)!β
4 cosh2 x2
n=0
(3.156)
−∞
With the help of the identities
∞
dx
=1,
4 cosh2 x2
−∞
∞
−∞
x2 dx
π2
=
,
3
4 cosh2 x2
(3.157)
(3.158)
one finds
4
1
π2 G(2) (µ)
+O
I = G (µ) +
2
βµ
6β
4
µ
π2 g ′ (µ)
1
=
dε g (ε) +
+
O
.
βµ
6β 2
−∞
(3.159)
8. In general, at zero temperature the average of the energy ε to the power
n is given by
εn =
3εF
dε D (ε) εn
0
3εF
,
(3.160)
dε D (ε)
0
where D (ε) is the density of states
D (ε) =
V
2π2
2m
2
3/2
ε1/2 ,
(3.161)
thus
εn =
εnF
.
2n
3 +1
(3.162)
a) Using Eq. (3.162) one finds that
ε
=
3εF
.
5
b) The speed v is related to the energy by
2ε
,
v=
m
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Thermodynamics and Statistical Physics
(3.163)
(3.164)
122
3.6. Solutions Set 3
thus
εF
2
16
ε
3 +1
α= .
2 = 4
52 =
−1/2
1/2
15
ε
εF
(3.165)
2 1
3 2 +1
c) The number of electrons N is given by
N=
εF
dε D (ε) =
D (εF )
0
1/2
εF
εF
0
dε ε1/2 =
D (εF ) 2 3/2
ε
,
1/2 3 F
ε
F
thus
εF =
2
2m
3π 2 N
V
2/3
,
(3.166)
and therefore
3N 2
U=
5 2m
3π 2 N
V
2/3
.
(3.167)
Moreover, at zero temperature the Helmholtz free energy F = U −
τσ = U , thus
the
pressure is given by
∂F
p=−
∂V τ ,N
∂U
=−
∂V τ ,N
2/3
3N 2 3π 2 N
2
=
5 2m
V
3V
2N εF
=
.
5V
(3.168)
9. The energy of the particles are
εn,± = pc
(3.169)
0
where p = k, and k = πn
n2x + n2y + n2z and ni = 1, 2, ...
L , n=
Therefore
1 4
π
N = 2 × × πn3F = n3F
(3.170)
8 3
3
1/3
3N
nF =
π
1/3
1/3
cπ 3N
3N
εF =
= cπ
L
π
πV
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123
Chapter 3. Bosonic and Fermionic Systems
The energy is given by
E (T = 0) =
εF
L3
εg (ε) dε = 2 3 3
π c
0
εF
ε3 dε =
(3.171)
0
ε4F
1 L3
=
ε3 × εF =
4
4 π2 3 c3 F
3
1 L3 3π2 3 c3 N
=
εF = N εF
4 π 2 3 c3
L3
4
=
L3
π2 3 c3
×
10. The energy of an electron having a wave function proportional to exp (ikx x) exp (iky y)
is
2 2
k + ky2 .
2me x
(3.172)
For periodic boundary conditions one has
2πnx
,
Lx
2πny
ky =
,
Ly
kx =
(3.173)
(3.174)
where the sample is of area Lx Ly , and nx and ny are both integers. The
number of states having energy smaller than E ′ is given by (including
both spin directions)
2π
2me E ′ Lx Ly
,
2
4π2
thus, the density of state per unit area is given by
me
E>0
D (E) = π 2
.
0 E<0
(3.175)
(3.176)
Using Fermi-Dirac function
f (E) =
1
,
1 + exp [β (E − µ)]
where β = 1/τ , the density n is given by
∞
n=
D (E) f (E) dE
−∞
∞
me
dE
=
π2 0 1 + exp [β (E − µ)]
me τ
log 1 + eβµ ,
=
π2
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Thermodynamics and Statistical Physics
(3.177)
(3.178)
124
3.6. Solutions Set 3
thus
nπ2
µ = τ log exp
−1 .
me τ
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Thermodynamics and Statistical Physics
(3.179)
125
4. Classical Limit of Statistical Mechanics
In this chapter we discuss the classical limit of statistical mechanics. We discuss Hamilton’s formalism, define the Hamiltonian and present the HamiltonJacobi equations of motion. The density function in thermal equilibrium is
used to prove the equipartition theorem. This theorem is then employed to
analyze an electrical circuit in thermal equilibrium, and to calculate voltage
noise across a resistor (Nyquist noise formula).
4.1 Classical Hamiltonian
In this section we briefly review Hamilton’s formalism, which is analogous to
Newton’s laws of classical mechanics . Consider a classical system having d
degrees of freedom. The system is described using the vector of coordinates
q̄ = (q1 , q2 , ..., qd ) .
(4.1)
Let E be the total energy of the system. For simplicity we restrict the discussion to a special case where E is a sum of two terms
E =T +V ,
·
where T depends only on velocities, namely T = T q̄ , and where V depends
only on coordinates, namely V = V (q̄). The notation overdot is used to
express time derivative, namely
·
dq1 dq2
dqd
q̄ =
,
, ...,
.
(4.2)
dt dt
dt
We refer to the first term T as kinetic energy and to the second one V as
potential energy.
The canonical conjugate momentum pi of the coordinate qi is defined as
pi =
∂T
.
∂ q̇i
(4.3)
The classical Hamiltonian H of the system is expressed as a function of the
vector of coordinates q̄ and as a function of the vector of canonical conjugate
momentum variables
Chapter 4. Classical Limit of Statistical Mechanics
p̄ = (p1 , p2 , ..., pd ) ,
(4.4)
namely
H = H (q̄, p̄) ,
(4.5)
and it is defined by
H=
d
i=1
pi q̇i − T + V .
(4.6)
4.1.1 Hamilton-Jacobi Equations
The equations of motion of the system are given by
∂H
∂pi
∂H
ṗi = −
,
∂qi
q̇i =
(4.7)
(4.8)
where i = 1, 2, ...d.
4.1.2 Example
V(q)
m
q
Consider a particle having mass m in a one dimensional potential V (q).
The kinetic energy is given by T = mq̇ 2 /2, thus the canonical conjugate momentum is given by [see Eq. (4.3)] p = mq̇. Thus for this example the canonical conjugate momentum equals the mechanical momentum. Note, however,
that this is not necessarily always the case. Using the definition (4.6) one
finds that the Hamiltonian is given by
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Thermodynamics and Statistical Physics
128
4.1. Classical Hamiltonian
H = mq̇ 2 −
=
mq̇ 2
+ V (q)
2
p2
+ V (q) .
2m
(4.9)
Hamilton-Jacobi equations (4.7) and (4.8) read
p
q̇ =
m
∂V
ṗ = −
.
∂q
(4.10)
(4.11)
The second equation, which can be rewritten as
mq̈ = −
∂V
,
∂q
(4.12)
expresses Newton’s second law.
4.1.3 Example
L
C
Consider a capacitor having capacitance C connected in parallel to an
inductor having inductance L. Let q be the charge stored in the capacitor.
The kinetic energy in this case T = Lq̇ 2 /2 is the energy stored in the inductor,
and the potential energy V = q 2 /2C is the energy stored in the capacitor.
The canonical conjugate momentum is given by [see Eq. (4.3)] p = Lq̇, and
the Hamiltonian (4.6) is given by
p2
q2
+
.
(4.13)
2L 2C
Hamilton-Jacobi equations (4.7) and (4.8) read
p
q̇ =
(4.14)
L
q
ṗ = − .
(4.15)
C
The second equation, which can be rewritten as
q
Lq̈ + = 0 ,
(4.16)
C
expresses the requirement that the voltage across the capacitor is the same
as the one across the inductor.
H=
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Thermodynamics and Statistical Physics
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Chapter 4. Classical Limit of Statistical Mechanics
4.2 Density Function
Consider a classical system in thermal equilibrium. The density function
ρ (q̄, p̄) is the probability distribution to find the system in the point (q̄, p̄).
The following theorem is given without a proof. Let H (q̄, p̄) be an Hamiltonian of a system, and assume that H has the following form
H=
d
Ai p2i + V (q̄) ,
(4.17)
i=1
where Ai are constants. Then in the classical limit, namely in the limit where
Plank’s constant approaches zero h → 0, the density function is given by
ρ (q̄, p̄) = N exp (−βH (q̄, p̄)) ,
(4.18)
where
N=3
dq̄
3
1
dp̄ exp (−βH (q̄, p̄))
(4.19)
The
3
3is a normalization constant, β = 1/τ , and τ is the temperature.
3
3 notation
dq̄3 indicates integration
over
all
coordinates,
namely
dq̄
=
dq1 dq1 ·
3
3
3
3
... · dqd , and similarly dp̄ = dp1 dp1 · ... · dpd .
Let A (q̄, p̄) be a variable which depends on the coordinates q̄ and their
canonical conjugate momentum variables p̄. Using the above theorem the
average value of A can be calculates as:
A (q̄, p̄)
= dq̄ dp̄ A (q̄, p̄) ρ (q̄, p̄)
3
3
dq̄ dp̄ A (q̄, p̄) exp (−βH (q̄, p̄))
3
3
=
.
dq̄ dp̄ exp (−βH (q̄, p̄))
(4.20)
4.2.1 Equipartition Theorem
Assume that the Hamiltonian has the following form
H = Bi qi2 + H̃ ,
(4.21)
where Bi is a constant and where H̃ is independent of qi . Then the following
holds
τ
(4.22)
Bi qi2 = .
2
Similarly, assume that the Hamiltonian has the following form
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Thermodynamics and Statistical Physics
130
4.2. Density Function
H = Ai p2i + H̃ ,
(4.23)
where Ai is a constant and where H̃ is independent of pi . Then the following
holds
τ
(4.24)
Ai p2i = .
2
To prove the theorem for the first case we use Eq. (4.20)
3
3
dq̄ dp̄ Bi qi2 exp (−βH (q̄, p̄))
2
3
Bi qi = 3
dq̄ dp̄ exp (−βH (q̄, p̄))
3
dqi Bi qi2 exp −βBi qi2
3
=
dqi exp (−βBi qi2 )
∂
2
=−
log
dqi exp −βBi qi
∂β
∂
π
=−
log
∂β
βBi
1
=
.
2β
(4.25)
The proof for the second case is similar.
4.2.2 Example
Here we calculate the average energy of an harmonic oscillator using both,
classical and quantum approaches. Consider a particle having mass m in a
one dimensional parabolic potential given by V (q) = (1/2) kq 2 , where k is
the spring constant. The kinetic energy is given by p2 /2m, where p is the
canonical momentum variable conjugate to q. The Hamiltonian is given by
H=
p2
kq 2
+
.
2m
2
(4.26)
In the classical limit the average energy of the system can be easily calculated
using the equipartition theorem
U = H
= τ .
(4.27)
In the quantum treatment the system has energy levels given by
Es = sω ,
1
where s = 0, 1, 2, ..., and where ω = k/m is the angular resonance frequency. The partition function is given by
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131
Chapter 4. Classical Limit of Statistical Mechanics
Z=
∞
exp (−sβω) =
s=0
1
,
1 − exp (−βω)
(4.28)
thus the average energy U is given by
U =−
∂ log Z
ω
= βω
.
∂β
e
−1
(4.29)
Using the expansion
U = β −1 + O (β) ,
(4.30)
one finds that in the limit of high temperatures, namely when βω ≪ 1, the
quantum result [Eq. (4.30)] coincides with the classical limit [Eq. (4.27)].
4.3 Nyquist Noise
Here we employ the equipartition theorem in order to evaluate voltage noise
across a resistor. Consider the circuit shown in the figure below, which consists
of a capacitor having capacitance C, an inductor having inductance L, and a
resistor having resistance R, all serially connected. The system is assumed to
be in thermal equilibrium at temperature τ. To model the effect of thermal
fluctuations we add a fictitious voltage source, which produces a random
fluctuating voltage V (t). Let q (t) be the charge stored in the capacitor at
time t. The classical equation of motion, which is given by
q
+ Lq̈ + Rq̇ = V (t) ,
C
(4.31)
represents Kirchhoff’s voltage law.
L
R
V(t)
~
C
Fig. 4.1.
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4.3. Nyquist Noise
Consider a sampling of the fluctuating function q (t) in the time interval
(−T /2, T /2), namely
q(t) −T /2 < t < T /2
qT (t) =
.
(4.32)
0
else
The energy stored in the capacitor is given by q 2 /2C. Using the equipartition
theorem one finds
2
q
τ
= ,
(4.33)
2C
2
where q 2 is obtained by averaging q 2 (t), namely
2
1
q ≡ lim
T →∞ T
+∞
−∞
dt qT2 (t) .
(4.34)
Introducing the Fourier transform:
∞
1
dω qT (ω)e−iωt ,
qT (t) = √
2π −∞
(4.35)
one finds
∞
∞
2
′
1
1 +∞
−iωt 1
√
q = lim
dt √
dω qT (ω)e
dω′ qT (ω ′ )e−iω t
T →∞ T −∞
2π −∞
2π −∞
∞
∞
+∞
′
1
1
=
dω qT (ω)
dω ′ qT (ω′ ) lim
dte−i(ω+ω )t
T →∞ T −∞
2π −∞
−∞
1
T →∞ T
= lim
2πδ(ω+ω′ )
∞
dω qT (ω)qT (−ω) .
−∞
(4.36)
Moreover, using the fact that q(t) is real one finds
2
1 ∞
2
q = lim
dω |qT (ω)| .
T →∞ T −∞
(4.37)
1
|qT (ω)|2 ,
T
(4.38)
In terms of the power spectrum Sq (ω) of q (t), which is defined as
Sq (ω) = lim
T →∞
one finds
2
q =
Eyal Buks
∞
dω Sq (ω) .
(4.39)
−∞
Thermodynamics and Statistical Physics
133
Chapter 4. Classical Limit of Statistical Mechanics
Taking the Fourier transform of Eq. (4.31) yields
1
2
− iωR − Lω q(ω) = V (ω) ,
C
where V (ω) is the Fourier transform of V (t), namely
∞
1
V (t) = √
dω V (ω)e−iωt .
2π −∞
In terms of the resonance frequency
1
ω0 =
,
LC
one has
2
L ω 0 − ω 2 − iωR q(ω) = V (ω) .
(4.40)
(4.41)
(4.42)
(4.43)
Taking the absolute value squared yields
Sq (ω) =
SV (ω)
2
L2 (ω 20 − ω2 ) + ω2 R2
,
(4.44)
where SV (ω) is the power spectrum of V (t). Integrating the last result yields
∞
∞
SV (ω)
dω Sq (ω) =
dω
2
2
2
L (ω0 − ω 2 ) + ω2 R2
−∞
−∞
∞
1
SV (ω)
= 2
dω
2 2 .
2
L −∞
(ω 0 + ω) (ω 0 − ω)2 + ωLR2
(4.45)
The integrand has a peak near ω 0 , having a width ≃ R/2L. The Quality
factor Q is defined as
ω0
R
=
.
Q
2L
(4.46)
Assuming SV (ω) is a smooth function near ω0 on the scale ω 0 /Q, and assuming Q ≫ 1 yield
Eyal Buks
Thermodynamics and Statistical Physics
134
4.3. Nyquist Noise
∞
−∞
dω Sq (ω) ≃
SV (ω 0 )
L2
∞
≃
SV (ω 0 )
4ω40 L2
∞
SV (ω 0 )
4ω30 L2
∞
=
−∞
−∞
−∞
SV (ω 0 )πQ
=
.
4ω30 L2
dω
(ω 0 + ω)2 (ω 0 − ω)2 +
ω0 −ω
ω0
dω
2 2
1
+ Q
2ωω0
Q
2
dx
2
1
x2 + Q
πQ
(4.47)
On the other hand, using Eqs. (4.33) and (4.39) one finds
∞
dω Sq (ω) = q 2 = Cτ ,
(4.48)
−∞
therefore
SV (ω0 ) =
4Cω 30 L2
τ,
πQ
(4.49)
or using Eqs. (4.42) and (4.46)
SV (ω0 ) =
2Rτ
.
π
(4.50)
Thus, Eq. (4.44) can be rewritten as
Sq (ω) =
2Rτ
1
.
π L2 (ω 20 − ω 2 )2 + ω 2 R2
(4.51)
Note that the spectral density of V given by Eq. (4.50) is frequency independent. Consider a measurement of the fluctuating voltage V (t) in a
frequency band having width ∆f. Using the relation
∞
2
V =
dω SV (ω) ,
(4.52)
−∞
one finds that the variance in such a measurement V 2 ∆f is given by
2
V ∆f = 4Rτ ∆f .
(4.53)
The last result is the Nyquist’s noise formula.
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135
Chapter 4. Classical Limit of Statistical Mechanics
4.4 Thermal Equilibrium From Stochastic Processes
This section demonstrates that under appropriate conditions a stochastic
process can lead to thermal equilibrium in steady state.
4.4.1 Langevin Equation
Consider the Langevin equation
ẋ = A (x, t) + q (t) ,
(4.54)
where x is a vector of coordinates that depends on the time t, overdot denotes
time derivative, the vector A (x, t) is a deterministic function of x and t, and
the vector q (t) represents random noise that satisfies
q (t)
= 0 ,
(4.55)
qi (t) qj (t′ )
= gij δ (t − t′ ) .
(4.56)
and
Let
δx = x (t + δt) − x (t) .
To first order in δt one finds by integrating Eq. (4.54) that
t+δt
(δx)i = Ai (x, t) δt +
dt′ qi (t′ ) + O (δt)2 .
(4.57)
(4.58)
t
With the help of Eqs. (4.55) and (4.56) one finds that
(δx)i = Ai (x, t) δt + O (δt)2 ,
and
(4.59)
(δx)i (δx)j
= Ai (x, t) Aj (x, t) (δt)2
t+δt
t+δt
+
dt′
dt′′ qi (t′ ) qj (t′′ )
+ · · ·
t
t
= Ai (x, t) Aj (x, t) (δt)2 + gij δt + · · · ,
thus to first order in δt
(δx)i (δx)j = gij δt + O (δt)2 .
(4.60)
(4.61)
In a similar way one can show that all higher order moments (e.g. third order
moments (δx)i′ (δx)i′′ (δx)i′′′ ) vanish to first order in δt.
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4.4. Thermal Equilibrium From Stochastic Processes
4.4.2 The Smoluchowski-Chapman-Kolmogorov Relation
Let p1 (x, t) be the probability density to find the system at point x at time
t, let p2 (x′′ , t′′ ; x′ , t′ ) be the probability density to find the system at point x′
at time t′ and at point x′′ at time t′′ , and similarly let p3 (x′′′ , t′′′ ; x′′ , t′′ ; x′ , t′ )
be the probability density to find the system at point x′ at time t′ , at point
x′′ at time t′′ and at point x′′′ at time t′′′ . The following holds
p2 (x3 , t3 ; x1 , t1 ) = dx2 p3 (x3 , t3 ; x2 , t2 ; x1 , t1 ) .
(4.62)
Let P (x, t|x′ , t′ ) be the conditional probability density to find the system at
point x at time t given that it was (or will be) at point x′ at time t′ . The
following holds
p2 (x3 , t3 ; x1 , t1 ) = P (x3 , t3 |x1 , t1 ) p1 (x1 , t1 ) .
(4.63)
Moreover, by assuming that t1 ≤ t2 ≤ t3 and by assuming the case of a
Markov process, i.e. the case where the future (t3 ) depends on the present
(t2 ), but not on the past (t1 ), one finds that
p3 (x3 , t3 ; x2 , t2 ; x1 , t1 )
= P (x3 , t3 |x2 , t2 ) P (x2 , t2 |x1 , t1 ) p1 (x1 , t1 ) .
With the help of these relations Eq. (4.62) becomes
P (x3 , t3 |x1 , t1 ) p1 (x1 , t1 )
= dx2 P (x3 , t3 |x2 , t2 ) P (x2 , t2 |x1 , t1 ) p1 (x1 , t1 ) ,
thus by dividing by p1 (x1 , t1 ) one finds that
P (x3 , t3 |x1 , t1 ) = dx2 P (x3 , t3 |x2 , t2 ) P (x2 , t2 |x1 , t1 ) .
(4.64)
(4.65)
4.4.3 The Fokker-Planck Equation
Equation (4.65) can be written as
P (x, t + δt|x0 , t0 ) = dx′ P (x, t + δt|x′ , t′ ) P (x′ , t′ |x0 , t0 ) .
(4.66)
On the other hand
P (x, t + δt|x′ , t′ ) = δ (x − δx − x′ )
,
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Thermodynamics and Statistical Physics
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137
Chapter 4. Classical Limit of Statistical Mechanics
where
δx = x (t + δt) − x (t) .
(4.68)
For a general scalar function F of x′ the following holds
F (x′ + δx)
= exp δx · ∇′ F
= F (x′ ) + (δx)i
(δx)i (δx)j d2 F
dF
+
+··· ,
′
dxi
2!
dx′i dx′j
(4.69)
thus
δ (x − δx − x′ ) = δ (x − x′ )
dδ (x − x′ )
+ (δx)i
dx′i
(δx)i (δx)j d2 δ (x − x′ )
+
+··· .
2!
dx′i dx′j
(4.70)
Inserting this result into Eq. (4.66)
P (x, t + δt|x0 , t0 )
= dx′ δ (x − δx − x′ )
P (x′ , t′ |x0 , t0 )
= dx′ δ (x − x′ ) P (x′ , t′ |x0 , t0 )
dδ (x − x′ )
+ dx′ (δx)i P (x′ , t′ |x0 , t0 )
dx′i
d2 δ (x − x′ )
1
+
P (x′ , t′ |x0 , t0 )
dx′ (δx)i (δx)j
2
dx′i dx′j
+··· ,
(4.71)
employing Eqs. (4.59) and (4.61) (recall that higher order moments vanish
to first order in δt), dividing by δt, taking the limit δt → 0
∂P
dδ (x − x′ )
= dx′ Ai (x′ , t)
P (x′ , t′ |x0 , t0 )
∂t
dx′i
1
d2 δ (x − x′ )
+
dx′ gij
P (x′ , t′ |x0 , t0 ) ,
2
dx′i dx′j
(4.72)
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4.4. Thermal Equilibrium From Stochastic Processes
and finally integrating by parts and assuming that P → 0 in the limit x →
±∞ yield
1 d2
∂P
d
(gij P) .
= − ′ (Ai P) +
∂t
dxi
2 dx′i dx′j
(4.73)
This result, which is known as the Fokker-Planck equation, can also be written
as
∂P
+∇·J= 0,
∂t
(4.74)
where the probability current density J is given by
Ji = Ai P −
1 d
(gij P) .
2 dx′j
(4.75)
In steady state the Fokker-Planck equation (4.74) becomes
∇·J=0 .
(4.76)
4.4.4 The Potential Condition
Consider the case where A can be expressed in terms of a scalar ’Hamiltonian’
H as (the potential condition)
A (x, t) = −∇H .
(4.77)
Moreover, for simplicity assume that
gij = 2τ δ ij ,
(4.78)
where the ’temperature’ τ is a constant. For this case one has
Ji = −P
dH
dP
−τ ′ ,
dx′i
dxi
(4.79)
thus
J = −P∇H − τ ∇P
= −P∇ (H + τ log P) .
(4.80)
Substituting a solution having the form
H
P = N e− τ
(4.81)
yields
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Thermodynamics and Statistical Physics
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Chapter 4. Classical Limit of Statistical Mechanics
H
H
J = −τ N e− τ ∇ (log N ) = −τ e− τ ∇N ,
(4.82)
and thus
H
∇ · J = [∇H · ∇N − τ ∇ · (∇N)] e− τ .
(4.83)
In terms of N the Fokker-Planck equation (4.74) becomes
∂N
+ [∇H · ∇N − τ ∇ · (∇N )] = 0 .
∂t
(4.84)
In steady state Eq. (4.84) becomes
∇H · ∇N = τ ∇ · (∇N) .
(4.85)
This equation can be solved by choosing N to be a constant, which can be
determined by the normalization condition. In terms of the partition function
Z, where Z = 1/N , the steady state solution is expressed as
P=
1 −H
e τ ,
Z
(4.86)
(4.87)
where
Z=
dx′ P .
4.4.5 Free Energy
The free energy functional is defined by
F (P) = U − τ σ ,
where U is the energy
U = dx′ HP ,
and σ is the entropy
σ = − dx′ P log P .
(4.88)
(4.89)
(4.90)
The distribution P is constrained to satisfy the normalization constrain
(4.91)
G (P) = dx′ P − 1 = 0 .
Minimizing F (P) under the constrain (4.91) is done by introducing the Lagrange multiplier η
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4.4. Thermal Equilibrium From Stochastic Processes
δ (F (P) + ηG (P))
= dx′ [H + τ (1 + log P) + η] δP .
(4.92)
At a stationary point the following holds
H + τ (1 + log P) + η = 0 ,
(4.93)
thus
η
H
P = e− τ −1 e− τ ,
(4.94)
η
where the constant e− τ −1 is determined by the normalization condition,
i.e. the obtained distribution is identical to the steady state solution of the
Fokker-Planck equation (4.86).
4.4.6 Fokker-Planck Equation in One Dimension
In one dimension the Fokker-Planck equation (4.74) becomes [see Eq. (4.79)]
∂P
∂
∂H
∂P
=
P
+τ
,
(4.95)
∂t
∂x
∂x
∂x
or
∂P
= LP ,
∂t
(4.96)
where the operator L is given by
L=
∂ ∂H
∂2
+τ 2 .
∂x ∂x
∂x
(4.97)
It is convenient to define the operator L̂, which is given by
H
H
L̂ = e 2τ Le− 2τ .
(4.98)
The following holds
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Thermodynamics and Statistical Physics
141
Chapter 4. Classical Limit of Statistical Mechanics
2
H
H ∂
H
∂ ∂H − 2τ
e
+ τ e 2τ 2 e− 2τ
∂x ∂x
∂x
H ∂
∂H
H
∂H
∂
−
e 2τ +
= e 2τ
∂x ∂x
∂x ∂x
2
−H
H
∂ −H
H ∂e 2τ ∂
∂2
2τ
2τ
+ τe 2τ
+
2τe
+
τ
e
∂x2
∂x ∂x
∂x2
2
∂2H
1 ∂H
∂H ∂
=
−
+
∂x2
2τ ∂x
∂x ∂x
2
2
1 ∂H
∂H ∂
1∂ H
∂2
+
−
−
+
τ
2 ∂x2
4τ ∂x
∂x ∂x
∂x2
2
1 ∂2H
1 ∂H
∂2
=
−
+τ 2 ,
2
2 ∂x
4τ ∂x
∂x
H
L̂ = e 2τ
(4.99)
thus
L̂ = τ
∂2
− V̂ ,
∂x2
(4.100)
where the potential V̂ is given by
2
1 ∂2H
1 ∂H
V̂ =
−
.
4τ ∂x
2 ∂x2
(4.101)
Note that while the operator L is not Hermitian, the operator L̂ is since
H
H
H
H
L̂† = e− 2τ L† e 2τ
H
= e− 2τ L† e τ e− 2τ
H
H
H
= e− 2τ e τ Le− 2τ
H
H
= e 2τ Le− 2τ = L̂ .
(4.102)
In terms of L̂ Eq. (4.96) becomes (it is assume that H is time independent)
∂ P̂
= L̂P̂ ,
∂t
(4.103)
where
H
H
P̂ = e 2τ Pe− 2τ .
(4.104)
Let ψn (x) be a set of eigenvectors of L̂
L̂ψn = λn ψn .
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142
4.4. Thermal Equilibrium From Stochastic Processes
The following holds [see Eq. (4.98)]
Lϕn = λn ϕn ,
(4.106)
where
H
ϕn = e− 2τ ψn .
(4.107)
The conditional probability distribution P (x, t|x′ , t′ ) is given by [see Eq.
(4.96)]
P (x, t|x′ , t′ ) = eL(t−t ) δ (x − x′ ) .
′
(4.108)
With the help of the closure relation (recall that L̂ is Hermitian)
∗
ψn (x′ ) ψn (x)
δ (x − x′ ) =
n
=e
H(x)+H(x′ )
2τ
ϕ∗n (x′ ) ϕn (x)
n
=e
H(x′ )
τ
ϕ∗n (x′ ) ϕn (x)
,
n
(4.109)
one finds that
P (x, t|x′ , t′ ) = eL(x)(t−t ) e
′
=e
H(x′ )
τ
H(x′ )
τ
ϕ∗n (x′ ) ϕn (x)
n
ϕ∗n (x′ ) eL(x)(t−t ) ϕn (x)
′
n
=e
H(x′ )
τ
ϕ∗n (x′ ) eλn (t−t ) ϕn (x)
′
n
=e
H(x′ )
τ
e−
H(x′ )
2τ
ψ ∗n (x′ ) eλn (t−t ) e−
′
H(x)
2τ
ψn (x) ,
n
(4.110)
thus
P (x, t|x′ , t′ ) = e
H(x′ )−H(x)
2τ
ψ∗n (x′ ) ψn (x) eλn (t−t ) .
′
(4.111)
n
4.4.7 Ornstein—Uhlenbeck Process in One Dimension
Consider the following Langevin equation
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Thermodynamics and Statistical Physics
143
Chapter 4. Classical Limit of Statistical Mechanics
ẋ + Γ x = q (t) ,
(4.112)
where x can take any real value, Γ is a positive constant and where the real
noise term q (t) satisfies q (t)
= 0 and
q (t) q (t′ )
= 2τδ (t − t′ ) ,
(4.113)
where τ is positive.
For this case [see Eq. (4.77)]
Γ x2
,
2
H (x) =
(4.114)
the Fokker-Planck equation for the conditional probability distribution P is
given by [see Eq. (4.74)]
∂
∂P
∂P
=
PΓ x + τ
,
(4.115)
∂t
∂x
∂x
and the operator L̂ is given by [see Eq. (4.100)]
 

2
2
Γ 
∂
x

L̂ = − −   +
− 1 ,
x
2
x
0
∂
x0
where
x0 =
2τ
.
Γ
(4.116)
The eigenvectors of L̂ are given by [see Eq. (4.131)]
ψn (x) =
2
Hn xx0
,
1/2 √ n
π1/4 x0
2 n!
− 12
e
x
x0
and the corresponding eigenvalues by
1 1
λn = −Γ n + −
,
2 2
(4.117)
(4.118)
where n = 0, 1, 2, · · · .
Using these results one finds that P is given by [see Eq. (4.111)]
′
−Γ n(t−t′ )
x
x
2
H
H
n x0
n x0 e
x
−
√
P (x, t|x′ , t′ ) = e x0
.
(4.119)
πx0 2n n!
n
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144
4.5. Problems Set 4
With the help of the general identity (4.138) one finds that the following
holds
2 Y√−Xα−1
2 ∞ α n
exp
−
α−2 −1
αe−X 2 Hn (Y ) Hn (X)
√
1
=
,
(4.120)
π n=0
n!
π (α−2 − 1)
thus Eq. (4.119) becomes
′ −Γ (t−t′ ) 2
exp − x−x e δ
√
,
P (x, t|x′ , t′ ) =
πδ
(4.121)
where
0 δ = x20 1 − e−2Γ (t−t′ ) =
,
2τ 1 − e−2Γ (t−t′ )
.
Γ
(4.122)
4.5 Problems Set 4
1. A gas at temperature τ emits a spectral line at wavelength λ0 . The
width of the observed spectral line is broadened due to motion of the
molecules (this is called Doppler broadening). Show that the relation
between spectral line intensity I and wavelength is given by
"
#
mc2 (λ − λ0 )2
,
(4.123)
I (λ) ∝ exp −
2λ20 τ
where c is velocity of fight, and m is mass of a molecule.
2. The circuit seen in the figure below, which contains a resistor R, capacitor C, and an inductor L, is at thermal
equilibrium at tempera
ture τ . Calculate the average value I 2 , where I is the current in the
R
C
L
inductor.
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Chapter 4. Classical Limit of Statistical Mechanics
3. Consider a random real signal q(t) varying in time. Let qT (t) be a sampling of the signal q(t) in the time interval (−T /2, T /2), namely
q(t) −T /2 < t < T /2
qT (t) =
.
(4.124)
0
else
The Fourier transform is given by
∞
1
dω qT (ω)e−iωt ,
qT (t) = √
2π −∞
(4.125)
and the power spectrum is given by
Sq (ω) = lim
T →∞
1
|qT (ω)|2 .
T
a) Show that
2
1
q ≡ lim
T →∞ T
+∞
−∞
dt qT2 (t) =
∞
dω Sq (ω) .
(4.126)
−∞
b) Wiener-Khinchine Theorem - show that the correlation function
of the random signal q(t) is given by
1
T →∞ T
q (t) q (t + t′ )
≡ lim
+∞
−∞
dt qT (t) qT (t + t′ ) =
∞
−∞
′
dω eiωt Sq (ω) .
(4.127)
4. Consider a resonator made of a capacitor C, an inductor L, and a resistor
R connected in series, as was done in class. Let I (t) be the current in the
circuit. Using the results obtained in class calculate the spectral density
SI (ω) of I at thermal equilibrium. Show that in the limit of high quality
factor, namely when
2 L
Q=
≫1,
(4.128)
R C
your result is consistent with the equipartition theorem applied for the
energy stored in the inductor.
5. A classical system is described using a set of coordinates {q1 , q2 , ..., qN }
and the corresponding canonically conjugate variables {p1 , p2 , ..., pN }.
The Hamiltonian of the system is given by
H=
N
An psn + Bn qnt ,
(4.129)
n=1
where An and Bn are positive constants and s and t are even positive
integers. Show that the average energy of the system in equilibrium at
temperature τ is given by
Eyal Buks
Thermodynamics and Statistical Physics
146
4.5. Problems Set 4
U = Nτ
1 1
+
s
t
,
(4.130)
6. A small hole of area A is made in the walls of a vessel of volume V containing a classical ideal gas of N particles of mass M each in equilibrium
at temperature τ. Calculate the number of particles dN , which escape
through the opening during the infinitesimal time interval dt.
7. Consider an ideal gas of Fermions having mass M and having no internal
degrees0
of freedom at temperature τ . The velocity of a particle is denoted
as v = vx2 + vy2 + vz2 . Calculate the quantity
4 5
1
v
v
8.
9.
10.
11.
(the symbol denoted averaging) in the:
a) classical limit (high temperatures).
b) zero temperature.
Consider an ideal gas of N molecules, each of mass M, contained in a
centrifuge of radius R and length L rotating with angular velocity ω
about its axis. Neglect the effect of gravity. The system is in equilibrium
at temperature τ = 1/β. Calculate the particle density n (r) as a function
of the radial distance from the axis r ( where 0 ≤ r ≤ R) .
Consider an ideal classical gas of particles having mass M0and having no
internal degrees of freedom at temperature τ . Let v = vx2 + vy2 + vz2 .
be the velocity of a particle. Calculate
a) 1
v
b) v2 A mixture of two classical ideal gases, consisting of N1 and N2 particles
of mass M1 and M2 , respectively, is enclosed in a cylindrical vessel of
height h and area of bottom and top side S. The vessel is placed in a
gravitational field having acceleration g. The system is in thermal equilibrium at temperature τ. Find the pressure exerted on the upper wall of
the cylinder.
The Hermite polynomial Hn (X) of order n is defined by
2
n
X
d
X2
Hn (X) = exp
X−
exp −
.
(4.131)
2
dX
2
For some low values of n the Hermite polynomials are given by
H0 (X) = 1 ,
H1 (X) = 2X ,
(4.132)
(4.133)
H2 (X) = 4X 2 − 2 ,
(4.134)
H4 (X) = 16X 4 − 48X 2 + 12 .
(4.136)
3
H3 (X) = 8X − 12X ,
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(4.135)
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Chapter 4. Classical Limit of Statistical Mechanics
Show that
∞
tn
Hn (X)
exp 2Xt − t2 =
.
n!
n=0
(4.137)
12. Show that
∞
n=0
α n
2
Hn (X) Hn (Y )
=
n!
exp
α(2XY −αX 2 −αY 2 )
1−α2
√
1 − α2
.
(4.138)
4.6 Solutions Set 4
1. Let λ be the wavelength measured by an observer, and let λ0 be the
wavelength of the emitted light in the reference frame where the molecule
is at rest. Let vx be the velocity of the molecule in the direction of the
light ray from the molecule to the observer. Due to Doppler effect
λ = λ0 (1 + vx /c) .
The probability distribution f (vx ) is proportional to
mvx2
f (vx ) ∝ exp −
,
2τ
(4.139)
(4.140)
thus using
vx =
c (λ − λ0 )
,
λ0
the probability distribution I (λ) is proportional to
"
#
mc2 (λ − λ0 )2
I (λ) ∝ exp −
.
2λ20 τ
(4.141)
(4.142)
2. The energy stored in the inductor is UL = LI 2 /2. Using the equipartition
theorem UL = τ/2, thus
2 τ
I = .
L
3.
(4.143)
a)
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148
4.6. Solutions Set 4
∞
∞
2
′
1
1 +∞
1
q = lim
dt √
dω qT (ω)e−iωt √
dω ′ qT (ω ′ )e−iω t
T →∞ T −∞
2π −∞
2π −∞
∞
∞
′
1 +∞
1
′
′
=
dω qT (ω)
dω qT (ω ) lim
dt e−i(ω+ω )t
T
→∞
2π −∞
T
−∞
−∞
= lim
T →∞
1
T
2πδ(ω+ω′ )
∞
dω qT (ω)qT (−ω) .
−∞
(4.144)
Since q(t) is real one has qT (−ω) = qT∗ (ω), thus
∞
2
dω Sq (ω) .
q =
(4.145)
−∞
b)
1
T →∞ T
q (t) q (t + t′ )
≡ lim
=
+∞
−∞
1
1
lim
2π T →∞ T
1
1
=
lim
T
→∞
2π
T
dt qT (t) qT (t + t′ )
+∞
−∞
∞
dt
∞
dω qT (ω)e−iωt
−∞
′ ′
−iω t
dω e
qT (ω)
−∞
∞
∞
dω ′ qT (ω ′ )e−iω (t+t )
−∞
′
′
dω qT (ω )
−∞
∞
−∞
(4.146)
2
4. Using I (ω) = −iωq (ω) and q = Cτ one finds for the case Q ≫ 1
∞
∞
2
τ
dω SI (ω) ≃ ω 20
dω Sq (ω) = ω 20 q 2 = , (4.147)
I =
L
−∞
−∞
in agreement with the equipartition theorem for the energy stored in the
inductor LI 2 /2.
5. Calculate for example
Thermodynamics and Statistical Physics
+∞
−∞
′
1
= lim
dω eiωt qT (ω)qT (−ω)
T →∞ T −∞
∞
′
=
dω eiωt Sf (ω) .
Eyal Buks
′
149
′
dt e−i(ω+ω )t
2πδ(ω+ω′ )
′
Chapter 4. Classical Limit of Statistical Mechanics
Bn qnt =
=
=
3∞
t
t
−∞ Bn qn exp (−βBn qn ) dqn
3∞
t
−∞ exp (−βBn qxn ) dqn
3∞
t
t
0 3 Bn qn exp (−βBn qn ) dqn
∞
t
0 exp (−βBn qxn ) dqn
∞
d
log
−
exp −βBn qnt dqn
dβ
0
.
(4.148)
where β = 1/τ . Changing integration variable
x = βBn qnt ,
dx =
(4.149)
tβBn qnt−1 dqn
,
(4.150)
leads to
∞
1
τ
d
−1
x t −1 e−x dx = .
Bn qnt = −
log (βBn ) t t−1
dβ
t
0
(4.151)
Thus
U =
N
n=1
An psn + Bn qnt = N τ
1 1
+
s
t
.
(4.152)
6. Let f (v) be the probability distribution of velocity v of particles in the
gas. The vector u is expressed in spherical coordinates, where the z axis
is chosen in the direction of the normal outward direction
v = v (sin θ cos ϕ, sin θ sin ϕ, cos θ) .
(4.153)
By symmetry, f (v) is independent of θ and ϕ. The number dN is calculated by integrating over all possible values of the velocity of the leaving
particles (note that θ can be only in the range 0 ≤ θ ≤ π/2)
∞ 1
2π
N
dN =
dv
d (cos θ)
dϕv2 v (dt) A cos θ f (v) .
(4.154)
V
0
0
0
Note that v (dt) A cos θ represents the volume of a cylinder from which
particles of velocity v can escape during the time interval dt. Since f (v)
is normalized
2π
∞
∞ 1
2
dv
d (cos θ)
dϕv f (v) = 4π
dvv 2 f (v) , (4.155)
1=
0
0
0
0
thus
dN
πN A
=
dt
V
Eyal Buks
0
∞
dvv2 vf (v) =
N A v
.
4V
Thermodynamics and Statistical Physics
(4.156)
150
4.6. Solutions Set 4
In the classical limit
Mv 2
f (v) ∝ exp −
,
2τ
(4.157)
thus, by changing the integration variable
x=
Mv2
2τ
(4.158)
one finds
Mv 2
3
dvv
exp
−
2τ
0
v
= 3 ∞
Mv2 2
dvv exp − 2τ
0
1/2 3 ∞
dxx exp (−x)
2τ
3 ∞0
=
1/2 exp (−x)
M
0 dxx
1/2
8τ
=
,
πM
3∞
(4.159)
and
dN
NA
=
dt
4V
8τ
πM
1/2
.
(4.160)
7. The probability that an orbital having energy ε is occupied is given by
fF (ε) =
1
,
1 + exp [(ε − µ) β]
(4.161)
where β = 1/τ and µ is the chemical potential. The velocity v of such an
orbital is related to the energy ε by
ε=
Mv 2
.
2
(4.162)
The 3D density of state per unit volume is given by
1
g (ε) = 2
2π
2M
2
3/2
ε1/2 .
(4.163)
Thus
Eyal Buks
Thermodynamics and Statistical Physics
151
Chapter 4. Classical Limit of Statistical Mechanics
3∞
3∞
dε g (ε) fF (ε) v dε g (ε) fF (ε) v1
4 5
1
0
= 03∞
v
3∞
v
dε g (ε) fF (ε)
dε g (ε) fF (ε)
0
0
∞
∞
3
3
dε εfF (ε)
dε fF (ε)
0
= 0 .
2
3∞
dε ε1/2 fF (ε)
0
(4.164)
a) In the classical limit
fF (ε) ∝ exp (−βε) ,
(4.165)
thus using the identities
∞
dε εn exp (−βε) = Γ (n) β −n
n
β
0
∞
dε exp (−βε) =
1
β
0
Γ (1) = 1
√
1
Γ
= π
2
one finds
4 5
1
v
=
v
∞
3
dε ε exp (−βε)
0
∞
3
0
dε
dε exp (−βε)
2
ε1/2 exp (−βε)
0
−1 1 1
(1) β β β
Γ
= 2
1
Γ 12 β −1/2 2β
=
∞
3
4
.
π
(4.166)
b) Using the identity
εF
dε εn =
εn+1
F
n+1
0
Eyal Buks
Thermodynamics and Statistical Physics
152
4.6. Solutions Set 4
one finds
4 5
1
v
=
v
εF
3
0
εF 3
dε ε
dε
εF
3
0
dε
0
1 2
εF εF
3 2
2 2
ε
3 F
= 2
=
ε1/2
2
.
9
.
8
(4.167)
8. The effect of rotation is the same as an additional external field with
potential energy given by
1
U (r) = − M ω2 r2 ,
2
(4.168)
thus
n (r) = A exp [−βU (r)] = A exp
βMω 2 2
r
2
,
(4.169)
where the normalization constant A is found from the condition
R
N = 2πL
n (r) rdr
0
R
βMω 2 2
= 2πLA
exp
r rdr
2
0
2πLA
βM ω 2 2
=
exp
R −1 ,
βM ω 2
2
(4.170)
thus
N βM ω 2
8
9 exp
n (r) =
2
2 −1
2πL exp βMω
2 R
β
Mω 2 r2
2
.
(4.171)
9. In the classical limit the probability distribution of the velocity vector v
satisfies
M v2
f (v) ∝ exp −
,
(4.172)
2τ
where v = |v|.
Eyal Buks
Thermodynamics and Statistical Physics
153
Chapter 4. Classical Limit of Statistical Mechanics
a) By changing the integration variable
x=
M v2
2τ
(4.173)
one finds
Mv2
3
dvv
exp
−
2τ
0
Mv2 v
= 3 ∞
2
0 dvv exp − 2τ
1/2 3 ∞
dxx exp (−x)
2τ
3 ∞0
=
1/2 exp (−x)
M
0 dxx
1/2
8τ
=
.
πM
3∞
(4.174)
(4.175)
b) Similarly
Mv 2
4
dv
v
exp
−
2
2τ
0
v = 3∞
Mv2 2
dv v exp − 2τ
0
3∞
2τ 0 dx x3/2 exp (−x)
3
=
M 0∞ dx x1/2 exp (−x)
2τ 3
=
,
M2
3∞
(4.176)
thus
1/2 1/2
1
3τ
3π
2
v =
=
v
.
M
8
(4.177)
10. For each gas the density is given by
nl (z) = nl (0) exp (−βMl gz) ,
where l ∈ {1, 2}, 0 ≤ z ≤ h and the normalization constant is found from
the requirement
S
h
n (z) dz = Nl ,
(4.178)
0
therefore
Nl
βMl gNl
.
S (1 − e−βMl gh )
(4.179)
Thermodynamics and Statistical Physics
154
nl (0) =
S
3h
exp (−βMl gz) dz
=
0
Eyal Buks
4.6. Solutions Set 4
Using the equation of state p = nτ, where n = N/V is the density, one
finds that the pressure on the upper wall of the cylinder is given by
p = (n1 (h) + n2 (h)) τ
M1 N1
M2 N2
g
=
+
.
exp (βM1 gh) − 1 exp (βM2 gh) − 1 S
(4.180)
11. Therelation (4.137),
which is a Taylor expansion of the function f (t) =
exp 2Xt − t2 around the point t = 0, implies that
2
2
dn
2 2
.
(4.181)
Hn (X) = n exp 2Xt − t 2
dt
t=0
The identity 2Xt − t2 = X 2 − (X − t)2 yields
2
dn
2 22
Hn (X) = exp X 2
exp
−
(X
−
t)
.
2
dtn
t=0
(4.182)
Moreover, using the relation
d
d
exp − (X − t)2 = −
exp − (X − t)2 ,
dt
dX
(4.183)
one finds that
2
dn
2 22
Hn (X) = exp X 2 (−1)n
exp
−
(X
−
t)
2
dX n
t=0
n
2
n d
2
= exp X (−1)
exp −X .
dX n
Note that for an arbitrary function g (X) the following holds
d
d
− exp X 2
exp −X 2 g = 2X −
g,
dX
dX
(4.184)
(4.185)
and
exp
X2
2
d
X2
d
X−
exp −
g = 2X −
g,
dX
2
dX
(4.186)
thus
n
d
X2
X−
exp −
.
dX
2
(4.187)
Thermodynamics and Statistical Physics
155
Hn (X) = exp
Eyal Buks
X2
2
Chapter 4. Classical Limit of Statistical Mechanics
12. With the help of Eq. (4.184) and the general identity
∞
π 1 4ca+b2
exp −ax2 + bx + c dx =
e4 a ,
a
(4.188)
−∞
according to which the following holds (for the case a = 1, b = 2iX and
c = 0)
1
exp −X 2 = √
π
∞
−∞
exp −x2 + 2iXx dx ,
(4.189)
one finds that
n ∞
exp X 2
d
√
Hn (X) =
−
exp −x2 + 2iXx dx
π
dX
1
=√
π
1
=√
π
∞
−∞
∞
−∞
(−2ix)n exp X 2 − x2 + 2iXx dx
2
(−2ix)n e(X+ix) dx ,
−∞
(4.190)
thus the following holds [see Eq. (4.188)]
∞ α n
Hn (X) Hn (Y )
2
n!
n=0
1
=
π
1
=
π
∞
−∞
∞
dx e
∞
(−2αxy)n
n=0
(X+ix)2
∞
−∞
∞
2
−∞
e−2αxy
2
√ αx(αx−2iY )
πe
2
2
2
dx e−(1−α )x +2i(X−Y α)x+X
α(2XY −αX 2 −αY 2 )
1−α2
√
1 − α2
n!
dy e(Y +iy) e−2αxy
−∞
exp
Eyal Buks
2
dy e(X+ix) e(Y +iy)
−∞
1
=√
π
=
dx
∞
.
Thermodynamics and Statistical Physics
(4.191)
156
5. Exercises
5.1 Problems
1. Two identical non-interacting particles each having mass M are confined
in a one dimensional parabolic potential given by
1
V (x) = M ω 2 x2 ,
2
(5.1)
where the angular frequency ω is a constant.
a) Calculate the canonical partition function of the system Zc,B for the
case where the particles are Bosons.
b) Calculate the canonical partition function of the system Zc,F for the
case where the particles are Fermions.
2. Consider a one dimensional gas containing N non-interacting electrons
moving along the x direction. The electrons are confined to a section of
length L. At zero temperature τ = 0 calculate the ratio U/εF between
the total energy of the system U and the Fermi energy εF .
3. Consider an ideal classical gas at temperature τ. The set of internal
eigenstates of each particle in the gas, when a magnetic field H is applied,
contains 2 states having energies ε− = −µ0 H and ε+ = µ0 H, where the
magnetic moment µ0 is a constant. Calculate the magnetization of the
system, which is defined by
∂F
M =−
,
(5.2)
∂H τ
where F is the Helmholtz free energy.
4. (Note: replace this with Ex. 3.9 in the lecture notes) Consider an ideal gas
made of N electrons in the extreme relativistic limit. The gas is contained
in a box having a cube shape with a volume V = L3 . In the extreme
relativistic limit the dispersion relation ε (k) is modified: the energy ε of
a single particle quantum state having a wavefunction ψ given by
ψ (x, y, z) =
3/2
2
sin (kx x) sin (ky y) sin (kz z) ,
L
(5.3)
Chapter 5. Exercises
is given by
ε (k) = kc ,
(5.4)
where c is the speed of light and where k =
0
kx2 + ky2 + kz2 (contrary
to the non-relativistic case where it is given by ε (k) = 2 k2 /2M ). The
system is in thermal equilibrium at zero temperature τ = 0. Calculate
the ratio p/U between the pressure p and the total energy of the system
U.
5. Consider a mixture of two classical ideal gases, consisting of NA particles
of type A and NB particles of type B. The heat capacities cp,A and cV,A
(cp,B and cV,B ) at constant pressure and at constant volume respectively
of gas A (B) are assumed to be temperature independent. The volume of
the mixture is initially V1 and the pressure is initially p1 . The mixture
undergoes an adiabatic (slow) and isentropic (at a constant entropy)
process leading to a final volume V2 . Calculate the final pressure p2 .
6. Consider two particles, both having the same mass m, moving in a onedimensional potential with coordinates x1 and x2 respectively. The potential energy is given by
V (x1 , x2 ) =
mω2 x21 mω2 x22
+
+ mΩ 2 (x1 − x2 )2 ,
2
2
(5.5)
where the angular frequencies ω and Ω are real constants Assume that
the temperature τ of the system is sufficiently high to allow treating it
classically. Calculate the following average values
a) x21 for the case Ω = 0.
b) x21 , however
without assuming that Ω = 0.
c)
(x1 − x2 )2 , again without assuming that Ω = 0.
7. Consider an ideal classical gas containing N identical particles having
each mass M in the extreme relativistic limit. The gas is contained in
a vessel having a cube shape with a volume V = L3 . In the extreme
relativistic limit the dispersion relation ε (k) is modified: the energy ε of
a single particle quantum state having a wavefunction ψ given by
3/2
2
ψ (x, y, z) =
sin (kx x) sin (ky y) sin (kz z) ,
(5.6)
L
is given by
ε (k) = kc ,
where c is the speed of light and where k =
(5.7)
0
kx2 + ky2 + kz2 (contrary
to the non-relativistic case where it is given by ε (k) = 2 k2 /2M ). The
system is in thermal equilibrium at temperature τ . Calculate:
Eyal Buks
Thermodynamics and Statistical Physics
158
5.2. Solutions
a) the total energy U of the system.
b) the pressure p.
8. Consider a system made of two localized spin 1/2 particles whose energy
is given by
εσ1 ,σ2 = −µ0 H (σ 1 + σ2 ) + Jσ1 σ2 ,
(5.8)
where both σ 1 and σ2 can take one of two possible values σn = ±1 (n =
1, 2). While H is the externally applied magnetic field, J is the coupling
constant between both spins. The system is in thermal equilibrium at
temperature τ . Calculate the magnetic susceptibility
χ = lim
H→0
where
∂M
,
∂H
∂F
M =−
∂H
(5.9)
(5.10)
τ
is the magnetization of the system, and where F is the Helmholtz free
energy.
9. Consider a one dimensional gas containing N non-interacting electrons
moving along the x direction. The electrons are confined by a potential
given by
1
V (x) = mω 2 x2 ,
(5.11)
2
where m is the electron mass and where ω is the angular frequency of
oscillations. Calculate the chemical potential µ
a) in the limit of zero temperature τ = 0.
b) in the limit of high temperatures τ ≫ ω.
10. The state equation of a given matter is
Aτ n
,
(5.12)
V
where p, V and τ are the pressure, volume and temperature, respectively,
and A and n are both constants. Calculate the difference cp −cV between
the heat capacities at constant pressure and at constant volume.
p=
5.2 Solutions
1. The single particle eigen energies are given by
1
ǫn = ω n +
,
2
(5.13)
where n = 0, 1, 2, · · · .
Eyal Buks
Thermodynamics and Statistical Physics
159
Chapter 5. Exercises
a) For Bosons
∞ ∞
∞
1
1
exp [−β (ǫn + ǫm )] +
exp (−2βǫn )
2 n=0 m=0
2 n=0
∞
2
∞
1 1
=
exp (−βǫn ) +
exp (−2βǫn )
2 n=0
2 n=0
Zc,B =
exp (−βω)
exp (−βω)
2 + 2 (1 − exp (−2βω)) .
2 (1 − exp (−βω))
=
(5.14)
Note that the average energy UB is given by
UB = −
∂ log Zc,B
1 + 2e−2βω + e−βω
= ω
.
∂β
1 − e−2βω
(5.15)
b) For Fermions
∞ ∞
∞
1
1
exp [−β (ǫn + ǫm )] −
exp (−2βǫn )
Zc,F =
2 n=0 m=0
2 n=0
=
exp (−βω)
exp (−βω)
2 − 2 (1 − exp (−2βω)) .
2 (1 − exp (−βω))
(5.16)
Note that for this case the average energy UF is given by
UF = −
∂ log Zc,F
2 + e−2βω + e−βω
.
= ω
∂β
1 − e−2βω
(5.17)
2. The orbital eigenenergies are given by
εn =
2 π 2 2
n ,
2m L
(5.18)
where n = 1, 2, 3, · · · . The grandcanonical partition function of the gas
is given by
Zgc =
ζn ,
(5.19)
n
where
ζn =
(1 + λ exp (−βεn ) exp (−βEl ))
(5.20)
l
is the orbital grandcanonical Fermionic partition function where,
λ = exp (βµ) = e−η ,
(5.21)
is the fugacity, β = 1/τ and {El } are the eigenenergies of a particle due
to internal degrees of freedom. For electrons, in the absence of magnetic
Eyal Buks
Thermodynamics and Statistical Physics
160
5.2. Solutions
field both spin states have the same energy, which is taken to be zero.
Thus, log Zgc can be written as
log Zgc =
∞
log ζ n
n=1
∞
=2
≃2
log (1 + λ exp (−βεn ))
n=1
∞
0
2 π 2 2
.
dn log 1 + λ exp −β
n
2m L
(5.22)
By employing the variable transformation
2 π 2 2
ε=
n ,
2m L
(5.23)
one has
log Zgc
1
=
2
∞
dε D (ε) log (1 + λ exp (−βε)) ,
(5.24)
0
where
2L
π
D (ε) =
0
2m −1/2
2 ε
0
ε≥0
ε<0
(5.25)
is the 1D density of states. Using Eqs. (1.80) and (1.94) for the energy
U and the number of particles N , namely using
∂ log Zgc
U =−
,
(5.26)
∂β
η
∂ log Zgc
,
∂λ
one finds that
∞
U=
dε D (ε) εfFD (ε) ,
N =λ
N=
−∞
∞
dε D (ε) fFD (ε) ,
(5.27)
(5.28)
(5.29)
−∞
where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)]
fFD (ǫ) =
Eyal Buks
1
.
exp [β (ǫ − µ)] + 1
Thermodynamics and Statistical Physics
(5.30)
161
Chapter 5. Exercises
At zero temperature, where µ = εF one has
εF
D (εF )
2D (εF ) 2
εF ,
U = −1/2
dε ε1/2 =
3
εF
(5.31)
0
N=
D (εF )
−1/2
εF
εF
dε ε−1/2 = 2D (εF ) εF ,
(5.32)
0
thus
U
N
=
.
εF
3
3. The Helmholtz free energy is given by
n
− log Zint − 1 ,
F = N τ log
nQ
(5.33)
(5.34)
where
Zint = exp (βµ0 H) + exp (−βµ0 H) = 2 cosh (βµ0 H)
(5.35)
is the internal partition function. Thus the magnetization is given by
∂F
M =−
= N µ0 tanh (βµ0 H) .
(5.36)
∂H τ
4. The grandcanonical partition function of the gas is given by
ζn ,
Zgc =
(5.37)
n
where
ζn =
(1 + λ exp (−βεn ) exp (−βEl ))
(5.38)
l
is a grandcanonical Fermionic partition function of an orbital having
energy εn given by
πcn
,
L
0
where n = n2x + n2y + n2z , nx , ny , nz = 1, 2, 3, · · · ,
εn =
(5.39)
λ = exp (βµ) = e−η
(5.40)
is the fugacity, β = 1/τ and {El } are the eigenenergies of a particle due
to internal degrees of freedom. For electrons, in the absence of magnetic
Eyal Buks
Thermodynamics and Statistical Physics
162
5.2. Solutions
field both spin states have the same energy, which is taken to be zero.
Thus, log Zgc can be written as
log Zgc =
∞ ∞ ∞
log (1 + λ exp (−βεn ) exp (−βEl )) . (5.41)
nx =1 ny =1 nz =1
l
For a macroscopic system the sum over n can be approximately replaced
by an integral
∞ ∞ ∞
nx =0 ny =0 nz
4π
→
8
=0
∞
dn n2 ,
(5.42)
0
thus, one has
log Zgc
4π
=2
8
∞
0
πcn
dn n log 1 + λ exp −β
.
L
2
(5.43)
By employing the variable transformation
ε=
πcn
.
L
(5.44)
one has
log Zgc =
∞
dε
V ε2
log (1 + λ exp (−βε)) .
π 2 3 c3
(5.45)
0
The energy U and the number of particles N are given by
∞
∂ log Zgc
V ε3
U =−
= dε 2 3 3 fFD (ǫ) ,
∂β
π c
η
(5.46)
0
N =λ
∂ log Zgc
=
∂λ
∞
dε
V ε2
fFD (ǫ) ,
π2 3 c3
(5.47)
0
where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)]
fFD (ǫ) =
1
.
exp [β (ǫ − µ)] + 1
At zero temperature
εF
V ε3
V
ε4
U = dε 2 3 3 = 2 3 3 F ,
π c
π c 4
(5.48)
(5.49)
0
N=
εF
dε
V ε2
V
ε3F
=
,
π 2 3 c3
π2 3 c3 3
(5.50)
0
Eyal Buks
Thermodynamics and Statistical Physics
163
Chapter 5. Exercises
and therefore
U=
3N
εF .
4
(5.51)
The energy U can be expressed as a function of V and N as
1/3 −1/3
(3N)4/3 π2 3 c3
V
.
U=
4
At zero temperature the Helmholtz free energy F equals the energy U ,
thus the pressure p is given by
∂F
p=−
∂V
τ ,N
∂U
=−
∂V
τ ,N
1
=
3
3N 4/3 2 3 3 1/3
π c
V
,
4
(5.52)
thus
1
p
=
.
U
3V
(5.53)
5. First, consider the case of an ideal gas made of a unique type of particles.
Recall that the entropy σ, cV and cp are given by [see Eqs. (2.87), (2.88)
and (2.89)]
5
nQ ∂ (τ log Zint )
σ=N
+ log
+
,
(5.54)
2
n
∂τ
3
cV = N
(5.55)
+ hint ,
2
cp = cV + N ,
(5.56)
where n = N/V is the density,
nQ =
Mτ
2π2
3/2
(5.57)
is the quantum density, M is the mass of a particle in the gas, and
hint = τ
∂ 2 (τ log Zint )
cV
3
=
− .
∂τ 2
N
2
(5.58)
The requirement that hint is temperature independent leads to
∂ (τ log Zint )
τ
= gint + hint log
,
∂τ
τ0
(5.59)
where both gint and τ 0 are constants. Using this notation, the change in
entropy due to a change in V from V1 to V2 and a change in τ from τ 1
to τ 2 is given by
Eyal Buks
Thermodynamics and Statistical Physics
164
5.2. Solutions
∆σ = σ2 − σ1 = N
= N log
V2
V1
3/2
log
τ2
τ1
V2 τ 2
3/2
V1 τ 1
cV +
cV
3
−
N
2
τ2
log
τ1
N
.
(5.60)
Thus the total change in the entropy of the mixture is given by
∆σ = ∆σA + ∆σB
cV,A cV,B V2 τ 2 NA
V2 τ 2 NB
+ NB log
= NA log
V1 τ 1
V1 τ 1


cV,A +cV,B
N +N
A
B
V
τ
2
2
 ,
= (NA + NB ) log 
V1
τ1
(5.61)
(5.62)
and the requirement ∆σ = 0 leads to
V2
V1
τ2
τ1
+cV,B
cV,A
N +N
A
B
=1.
(5.63)
Alternatively, by employing the equation of state
pV = N τ ,
(5.64)
this can be rewritten as
V2
V1
+cV,B
cV,A
+1 N +N
A
B
p2
p1
+cV,B
cV,A
N +N
A
B
=1,
(5.65)
or with the help of Eq. (5.56) as
p2 = p1
= p1
V2
V1
V1
V2
−
cV,A +cV,B
+1
NA +NB
cV,A +cV,B
NA +NB
p,B
ccp,A +c
+c
V,A
V,B
.
6. It is convenient to employ the coordinate transformation
x1 + x2
x+ = √
,
2
x1 − x2
x− = √
.
2
The inverse transformation is given by
Eyal Buks
Thermodynamics and Statistical Physics
(5.66)
(5.67)
(5.68)
165
Chapter 5. Exercises
x+ + x−
√
,
2
x+ − x−
√
x2 =
.
2
The following holds
x1 =
(5.69)
(5.70)
x21 + x22 = x2+ + x2− ,
(5.71)
ẋ21 + ẋ22 = ẋ2+ + ẋ2− .
(5.72)
and
Thus, the kinetic energy T of the system is given by
m ẋ21 + ẋ22
m ẋ2+ + ẋ2−
T =
=
,
2
2
(5.73)
and the potential energy V is given by
mω 2 x21 mω 2 x22
+
+ mΩ 2 (x1 − x2 )2
2
2
mω 2 x2+ m ω 2 + 4Ω 2 x2−
+
.
=
2
2
V (x1 , x2 ) =
The equipartition theorem yields
mω 2 x2+
m ω2 + 4Ω 2 x2−
τ
=
= ,
2
2
2
thus
2τ
(x1 + x2 )2 =
,
mω 2
(5.74)
(5.75)
(5.76)
and
(x1 − x2 )2 =
2τ
.
m (ω 2 + 4Ω 2 )
(5.77)
Furthermore, since by symmetry x+ x− = 0 one has
2 1 x1 =
(x+ + x− )2
2
1 2 2 =
x+ + x−
2
2
τ
1 4Ω
ω2
=
1−
.
2
mω 2
2 1 + 4Ω
ω2
(5.78)
Eyal Buks
Thermodynamics and Statistical Physics
166
5.2. Solutions
7. The k vector is restricted due to boundary conditions to the values
k=
πn
,
L
(5.79)
where
n = (nx , ny , nz ) ,
(5.80)
and nx , ny , nz = 1, 2, 3, · · · . The single particle partition function is given
by
Z1 =
ε (k)
exp −
.
τ
=1
∞ ∞ ∞
nx =1 ny =1 nz
(5.81)
Approximating the discrete sum by a continuous integral according to
∞ ∞ ∞
nx =0 ny =0 nz
4π
→
8
=0
∞
dn n2 ,
(5.82)
0
one has
4π
Z1 =
8
∞
0
nπc
dn n exp −
Lτ
4V τ 3
=
8π 2 3 c3
2
∞
0
=
V τ3
.
π2 3 c3
dx x2 exp (−x)
2
(5.83)
In the classical limit the grandcanonical partition function Zgc is given
by [see Eq. (2.44)]
log Zgc = λZ1 ,
(5.84)
where λ = exp (βµ) is the fugacity. In terms of the Lagrange multipliers
η = −µ/τ and β = 1/τ the last result can be rewritten as
log Zgc = e−η
V
.
π2 3 c3 β 3
(5.85)
a) The average energy U and average number of particle N are calculated using Eqs. (1.80) and (1.81) respectively
Eyal Buks
Thermodynamics and Statistical Physics
167
Chapter 5. Exercises
U =−
N =−
thus
∂ log Zgc
∂β
∂ log Zgc
∂η
=
η
β
3
log Zgc ,
β
= log Zgc ,
U = 3N τ ,
(5.86)
(5.87)
(5.88)
and
η = log
V τ3
π2 N3 c3
.
b) The entropy is evaluate using Eq. (1.86)
σ = log Zgc + βU + ηN
= N (1 + 3 + η)
V τ3
= N 4 + log
,
π 2 N 3 c3
and the Helmholtz free energy by the definition (1.116)
V τ3
F = U − τ σ = −N τ 1 + log
.
π2 N 3 c3
Thus the pressure p is given by
∂F
Nτ
p=−
=
.
∂V τ ,N
V
(5.89)
(5.90)
(5.91)
(5.92)
8. The partition function is given by
Z=
exp (−βεσ1 ,σ2 )
σ1 ,σ 2 =±1
= exp (−βJ) [exp (−2βµ0 H) + exp (2βµ0 H)] + 2 exp (βJ) ,
(5.93)
where β = 1/τ . The free energy is given by
F = −τ log Z ,
(5.94)
thus the magnetization is given by
∂F
M =−
∂H τ
2µ0 exp (−βJ) [− exp (−2βµ0 H) + exp (2βµ0 H)]
=
,
exp (−βJ) [exp (−2βµ0 H) + exp (2βµ0 H)] + 2 exp (βJ)
(5.95)
Eyal Buks
Thermodynamics and Statistical Physics
168
5.2. Solutions
and the magnetic susceptibility is given by
χ=
4βµ20
.
1 + e2βJ
(5.96)
Note that in the high temperature limit βJ ≪ 1
χ≃
2µ20
.
τ +J
(5.97)
9. The orbital eigenenergies in this case are given by
1
,
εn = ω n +
2
(5.98)
where n = 0, 1, 2, · · · . The grandcanonical partition function of the gas
is given by
Zgc =
ζn ,
(5.99)
n
where
ζn =
(1 + λ exp (−βεn ) exp (−βEl ))
(5.100)
l
is the orbital grandcanonical Fermionic partition function where,
λ = exp (βµ) = e−η
(5.101)
is the fugacity, β = 1/τ and {El } are the eigenenergies of a particle due
to internal degrees of freedom. For electrons, in the absence of magnetic
field both spin states have the same energy, which is taken to be zero.
Thus, log Zgc can be written as
log Zgc =
∞
log ζ n
n=0
∞
=2
log (1 + λ exp (−βεn )) .
n=0
(5.102)
The number of particles N is given by
N =λ
∞
∂ log Zgc
=2
fFD (εn ) ,
∂λ
n=0
(5.103)
where fFD is the Fermi-Dirac distribution function
fFD (ε) =
Eyal Buks
1
.
exp [β (ε − µ)] + 1
Thermodynamics and Statistical Physics
(5.104)
169
Chapter 5. Exercises
a) At zero temperature the chemical potential µ is the Fermi energy εF ,
and the Fermi-Dirac distribution function becomes a step function,
thus with the help of Eq. (5.103) one finds that
N=
2εF
,
ω
(5.105)
thus
µ = εF =
N ω
.
2
(5.106)
b) Using the approximation
fFD (ε) ≃ exp [−β (ε − µ)] ,
(5.107)
for the the limit of high temperatures and approximating the sum by
an integral one has
∞
1
N =2
exp −β ω n +
−µ
2
n=0
∞
ω
= 2 exp β µ −
dn exp (−βωn)
2
0
2 exp β µ − ω
2
=
,
βω
(5.108)
thus
N βω
βω
µ = τ log
+
2
2
N βω
≃ τ log
.
2
10. Using Eq. (2.239), which is given by
∂p
∂V
,
cp − cV = τ
∂τ V,N ∂τ p,N
(5.109)
(5.110)
one finds that
cp − cV =
Eyal Buks
A2 τ 2 2(n−1)
n τ
= n2 Aτ n−1 .
pV
Thermodynamics and Statistical Physics
(5.111)
170
References
1. C. E. Shannon, Bell System Tech. J. 27, 379 (1948).
2. C. E. Shannon, Bell System Tech. J. 27, 623 (1948).
3. E. T. Jaynes, Phys. rev. Lett. 106, 620 (1957).
Index
Bose-Einstein distribution, 52
Bose-Einstein function, 52
Boson, 50
canonical conjugate momentum, 127
canonical distribution, 15
Carnot, 64
cavity, 97
chemical potential, 17, 20
classical limit, 52
Clausius’s principle, 68
composition property, 5
Coulomb gauge, 97
Debye temperature, 108
density function, 130
density of states, 112
efficiency, 66, 69
electromagnetic radiation, 97
entropy, 1
equipartition theorem, 130
Fermi energy, 113
Fermi-Dirac distribution, 51
Fermi-Dirac function, 51
Fermion, 50
Fokker-Planck equation, 137
free energy, 140
fugacity, 18, 51
grandcanonical distribution, 16
H theorem, 18
Hamilton-Jacobi equations, 128
Hamiltonian, 127
heat capacity, 57
Helmholtz free energy, 21, 55
ideal gas, 45
information theory, 1
internal degrees of freedom, 57
isentropic process, 63
isobaric process, 62
isochoric process, 63
isothermal process, 62
Kelvin’s principle, 68
Lagrange multiplier, 4
Langevin equation, 136
largest uncertainty estimator, 9
Maxwell’s equations, 97
microcanonical distribution, 14
Nyquist noise, 132
orbitals, 48
Ornstein—Uhlenbeck process, 143
partition function, 10
permeability, 97
permittivity, 97
phonon, 105
photon, 100
Plank’s radiation law, 103
pressure, 55
quantum density, 46
Shannon, 1
Sommerfeld expansion, 113
Stefan-Boltzmann constant, 104
Stefan-Boltzmann radiation law, 103
temperature, 15, 17
thermal equilibrium, 19
vector potential, 97