Statistical Mechanics 3rd Edition
INSTRUCTOR’S
ST
UV
MANUAL
IA
_A
R. K. Pathria
PP
Paul D. Beale
RO
Comprehensive Instructor’s Manual for
VE
Instructors and Students
© R. K. Pathria & Paul D. Beale. All rights reserved. Reproduction or distribution without
D?
permission is prohibited.
©MedConnoisseur
, TABLE OF CONTENTS
Instructor’s Manual – Statistical Mechanics, 3rd Edition
R. K. Pathria, Paul D. Beale
ST
Chapter 1. Elements of Ensemble Theory,
Chapter 2. The Canonical Ensemble,
Chapter 3. The Grand Canonical Ensemble,
UV
Chapter 4. Formulation of Quantum Statistics,
Chapter 5. The Theory of Simple Gases,
Chapter 6. Ideal Bose Systems,
Chapter 7. Ideal Fermi Systems,
IA
Chapter 8. Thermodynamics of the Early Universe,
Chapter 9. Statistical Mechanics of Interacting Systems: The Method of Cluster
Expansions,
_A
Chapter 10. Statistical Mechanics of Interacting Systems: The Method of
Quantized Fields,
Chapter 11. Phase Transitions: Criticality, Universality, and Scaling,
Chapter 12. Phase Transitions: Exact (or Almost Exact) Results for Various
PP
Models,
Chapter 13. Phase Transitions: The Renormalization Group Approach,
Chapter 14. Fluctuations and Nonequilibrium Statistical Mechanics,
RO
Chapter 15. Computer Simulations
VE
D?
©MedConnoisseur
,ST
Chapter 1
UV
1.1. (a) We expand the quantity ln Ω(0) (E1 ) as a Taylor series in the variable
IA
(E1 − Ē1 ) and get
ln Ω(0) (E1 ) ≡ lnΩ1 (E1 ) + ln Ω2 (E2 ) (E2 = E (0) − E1 )
= {ln Ω1 (Ē1 ) + ln Ω2 (Ē2 )}+
_A
∂ ln Ω1 (E1 ) ∂ ln Ω2 (E2 ) ∂E2
+ (E1 − Ē1 )+
∂E1 ∂E2 ∂E1 E1 =Ē1
( 2 )
1 ∂ 2 ln Ω1 (E1 ) ∂ 2 ln Ω2 (E2 ) ∂E2
+ (E1 − Ē1 )2 + · · · .
2 ∂E12 ∂E22 ∂E1
E1 =Ē1
PP
The first term of this expansion is a constant, the second term van-
ishes as a result of equilibrium (β1 = β2 ), while the third term may
be written as
1 ∂β1 ∂B2 2 1 1 1
+ E1 − Ē1 = − + (E1 −Ē1 )2 ,
2 ∂E1 ∂E2 eq. 2 kT12 (Cv )1 kT22 (Cv )2
RO
with T1 = T2 . Ignoring the subsequent terms (which is justified if the
systems involved are large) and taking the exponentials, we readily
see that the function Ω0 (E1 ) is a Gaussian in the variable (E1 − Ē1 ),
with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that if (Cv )2 ≫
(Cv )1 — corresponding to system 1 being in thermal contact with a
very large reservoir — then the variance becomes simply kT 2 (Cv )1 ,
VE
regardless of the nature of the reservoir; cf. eqn. (3.6.3).
(b) If the systems involved are ideal classical gases, then (Cv )1 = 23 N1 k
and (Cv )2 = 23 N2 k; the variance then becomes 23 k 2 T 2 · N1 N2 /(N1 +
N2 ). Again, if N2 ≫ N1 , we obtain the simplified expression 23 N1 k 2 T 2 ;
cf. Problem 3.18.
D?
1.2. Since S is additive and Ω multiplicative, the function f (Ω) must satisfy
the condition
f (Ω1 Ω2 ) = f (Ω1 ) + f (Ω2 ). (1)
1
, 2
Differentiating (1) with respect to Ω1 (and with respect to Ω2 ), we get
ST
Ω2 f ′ (Ω1 Ω2 ) = f ′ (Ω1 ) and Ω1 f ′ (Ω1 Ω2 ) = f ′ (Ω2 ),
so that
Ω1 f ′ (Ω1 ) = Ω2 f ′ (Ω2 ). (2)
Since the left-hand side of (2) is independent of Ω2 and the right-hand side
UV
is independent of Ω1 , each side must be equal to a constant, k, independent
of both Ω1 and Ω2 . It follows that f ′ (Ω) = k/Ω and hence
f (Ω) = k ln Ω + const. (3)
Substituting (3) into (1), we find that the constant of integration is zero.
IA
1.4. Instead of eqn. (1.4.1), we now have
Ω ∝ V (V − v0 )(V − 2v0 ) . . . (V − N − 1v0 ),
so that
_A
ln Ω = C + ln V + ln (V − v0 ) + ln (V − 2v0 ) + . . . + ln (V − N − 1v0 ),
where C is independent of V . The expression on the right may be written
as
N −1 N −1
N 2 v0
PP
X jv0 X jv0
C+N ln V + ln 1 − ≃ C+N ln V + − ≃ C+N ln V − .
j=1
V j=1
V 2V
Equation (1.4.2) is then replaced by
N 2 v0
P N N N v0
RO
= + = 1 + , i.e.
kT V 2V 2 V 2V
−1
N v0
PV 1 + = NkT .
2V
Since N v0 ≪ V, (1 + N v0 /2V )−1 ≃ 1 − N v0 /2V . Our last result then
takes the form: P (V − b) = NkT , where b = 21 N v0 .
VE
A little reflection shows that v0 = (4π/3)σ 3 , with the result that
3
1 4π 3 4π 1
b= N· σ = 4N · σ .
2 3 3 2
1.5. This problem is essentially solved in Appendix A; all that remains to be
D?
done is to substitute from eqn. (B.12) into (B.11), to get
(πε∗1/2 /L)3 (πε∗1/2 /L)2
Σ1 (ε∗ ) = V ∓ S.
6π 2 16π
INSTRUCTOR’S
ST
UV
MANUAL
IA
_A
R. K. Pathria
PP
Paul D. Beale
RO
Comprehensive Instructor’s Manual for
VE
Instructors and Students
© R. K. Pathria & Paul D. Beale. All rights reserved. Reproduction or distribution without
D?
permission is prohibited.
©MedConnoisseur
, TABLE OF CONTENTS
Instructor’s Manual – Statistical Mechanics, 3rd Edition
R. K. Pathria, Paul D. Beale
ST
Chapter 1. Elements of Ensemble Theory,
Chapter 2. The Canonical Ensemble,
Chapter 3. The Grand Canonical Ensemble,
UV
Chapter 4. Formulation of Quantum Statistics,
Chapter 5. The Theory of Simple Gases,
Chapter 6. Ideal Bose Systems,
Chapter 7. Ideal Fermi Systems,
IA
Chapter 8. Thermodynamics of the Early Universe,
Chapter 9. Statistical Mechanics of Interacting Systems: The Method of Cluster
Expansions,
_A
Chapter 10. Statistical Mechanics of Interacting Systems: The Method of
Quantized Fields,
Chapter 11. Phase Transitions: Criticality, Universality, and Scaling,
Chapter 12. Phase Transitions: Exact (or Almost Exact) Results for Various
PP
Models,
Chapter 13. Phase Transitions: The Renormalization Group Approach,
Chapter 14. Fluctuations and Nonequilibrium Statistical Mechanics,
RO
Chapter 15. Computer Simulations
VE
D?
©MedConnoisseur
,ST
Chapter 1
UV
1.1. (a) We expand the quantity ln Ω(0) (E1 ) as a Taylor series in the variable
IA
(E1 − Ē1 ) and get
ln Ω(0) (E1 ) ≡ lnΩ1 (E1 ) + ln Ω2 (E2 ) (E2 = E (0) − E1 )
= {ln Ω1 (Ē1 ) + ln Ω2 (Ē2 )}+
_A
∂ ln Ω1 (E1 ) ∂ ln Ω2 (E2 ) ∂E2
+ (E1 − Ē1 )+
∂E1 ∂E2 ∂E1 E1 =Ē1
( 2 )
1 ∂ 2 ln Ω1 (E1 ) ∂ 2 ln Ω2 (E2 ) ∂E2
+ (E1 − Ē1 )2 + · · · .
2 ∂E12 ∂E22 ∂E1
E1 =Ē1
PP
The first term of this expansion is a constant, the second term van-
ishes as a result of equilibrium (β1 = β2 ), while the third term may
be written as
1 ∂β1 ∂B2 2 1 1 1
+ E1 − Ē1 = − + (E1 −Ē1 )2 ,
2 ∂E1 ∂E2 eq. 2 kT12 (Cv )1 kT22 (Cv )2
RO
with T1 = T2 . Ignoring the subsequent terms (which is justified if the
systems involved are large) and taking the exponentials, we readily
see that the function Ω0 (E1 ) is a Gaussian in the variable (E1 − Ē1 ),
with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that if (Cv )2 ≫
(Cv )1 — corresponding to system 1 being in thermal contact with a
very large reservoir — then the variance becomes simply kT 2 (Cv )1 ,
VE
regardless of the nature of the reservoir; cf. eqn. (3.6.3).
(b) If the systems involved are ideal classical gases, then (Cv )1 = 23 N1 k
and (Cv )2 = 23 N2 k; the variance then becomes 23 k 2 T 2 · N1 N2 /(N1 +
N2 ). Again, if N2 ≫ N1 , we obtain the simplified expression 23 N1 k 2 T 2 ;
cf. Problem 3.18.
D?
1.2. Since S is additive and Ω multiplicative, the function f (Ω) must satisfy
the condition
f (Ω1 Ω2 ) = f (Ω1 ) + f (Ω2 ). (1)
1
, 2
Differentiating (1) with respect to Ω1 (and with respect to Ω2 ), we get
ST
Ω2 f ′ (Ω1 Ω2 ) = f ′ (Ω1 ) and Ω1 f ′ (Ω1 Ω2 ) = f ′ (Ω2 ),
so that
Ω1 f ′ (Ω1 ) = Ω2 f ′ (Ω2 ). (2)
Since the left-hand side of (2) is independent of Ω2 and the right-hand side
UV
is independent of Ω1 , each side must be equal to a constant, k, independent
of both Ω1 and Ω2 . It follows that f ′ (Ω) = k/Ω and hence
f (Ω) = k ln Ω + const. (3)
Substituting (3) into (1), we find that the constant of integration is zero.
IA
1.4. Instead of eqn. (1.4.1), we now have
Ω ∝ V (V − v0 )(V − 2v0 ) . . . (V − N − 1v0 ),
so that
_A
ln Ω = C + ln V + ln (V − v0 ) + ln (V − 2v0 ) + . . . + ln (V − N − 1v0 ),
where C is independent of V . The expression on the right may be written
as
N −1 N −1
N 2 v0
PP
X jv0 X jv0
C+N ln V + ln 1 − ≃ C+N ln V + − ≃ C+N ln V − .
j=1
V j=1
V 2V
Equation (1.4.2) is then replaced by
N 2 v0
P N N N v0
RO
= + = 1 + , i.e.
kT V 2V 2 V 2V
−1
N v0
PV 1 + = NkT .
2V
Since N v0 ≪ V, (1 + N v0 /2V )−1 ≃ 1 − N v0 /2V . Our last result then
takes the form: P (V − b) = NkT , where b = 21 N v0 .
VE
A little reflection shows that v0 = (4π/3)σ 3 , with the result that
3
1 4π 3 4π 1
b= N· σ = 4N · σ .
2 3 3 2
1.5. This problem is essentially solved in Appendix A; all that remains to be
D?
done is to substitute from eqn. (B.12) into (B.11), to get
(πε∗1/2 /L)3 (πε∗1/2 /L)2
Σ1 (ε∗ ) = V ∓ S.
6π 2 16π
