Instructor’s Manual — Statistical Mechanics, 3rd Edition — R. K. Pathria & Paul D. Beale — ISBN Verified — Latest Update 2025/2026 — (All Chapters Covered)

Statistical Mechanics 3rd Edition


INSTRUCTOR’S
ST
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MANUAL
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R. K. Pathria
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Paul D. Beale
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Comprehensive Instructor’s Manual for
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Instructors and Students
© R. K. Pathria & Paul D. Beale. All rights reserved. Reproduction or distribution without
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permission is prohibited.




©MedConnoisseur

, TABLE OF CONTENTS

Instructor’s Manual – Statistical Mechanics, 3rd Edition
R. K. Pathria, Paul D. Beale
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Chapter 1. Elements of Ensemble Theory,
Chapter 2. The Canonical Ensemble,
Chapter 3. The Grand Canonical Ensemble,
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Chapter 4. Formulation of Quantum Statistics,
Chapter 5. The Theory of Simple Gases,
Chapter 6. Ideal Bose Systems,
Chapter 7. Ideal Fermi Systems,
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Chapter 8. Thermodynamics of the Early Universe,
Chapter 9. Statistical Mechanics of Interacting Systems: The Method of Cluster
Expansions,
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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
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Models,
Chapter 13. Phase Transitions: The Renormalization Group Approach,
Chapter 14. Fluctuations and Nonequilibrium Statistical Mechanics,
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Chapter 15. Computer Simulations
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©MedConnoisseur

,ST

Chapter 1
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1.1. (a) We expand the quantity ln Ω(0) (E1 ) as a Taylor series in the variable
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(E1 − Ē1 ) and get

ln Ω(0) (E1 ) ≡ lnΩ1 (E1 ) + ln Ω2 (E2 ) (E2 = E (0) − E1 )
= {ln Ω1 (Ē1 ) + ln Ω2 (Ē2 )}+
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∂ 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
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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
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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 ,
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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.
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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
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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
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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.
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1.4. Instead of eqn. (1.4.1), we now have

Ω ∝ V (V − v0 )(V − 2v0 ) . . . (V − N − 1v0 ),

so that
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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
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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
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= + = 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 .
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A little reflection shows that v0 = (4π/3)σ 3 , with the result that
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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
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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