Instructor’s Manual

Containing Solutions toOver 280 Problems

Selected from

Statistical Mechanics

Third Edition

By

R. K. Pathria and Paul D. Beale

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Preface

This instructor’s manual for the third edition of Statistical Mechanics is basedon RKP’s instructor’s manual for the second edition. Most of the solutionshere were retypeset into TeX from that manual. PDB is responsible for thesolutions of the new problems added in the third edition. The result is a manualcontaining solutions to some 280 problems selected from the third edition.

The original idea of producing an instructor’s manual first came from RKP’sfriend and colleague Wing-Ki Liu in the 1990’s when RKP had just embarkedon the task of preparing the second edition of Statistical Mechanics.

This should provide several benefits to the statistical mechanics instructor.First of all, there is the obvious advantage of saving time that one would oth-erwise spend on solving these problems oneself. Secondly, before one selectsproblems either for homework or for an exam, one can consult the manual todetermine the level of difficulty of the various problems and make one’s selectionaccordingly. Thirdly, one may even use some of these solved problems, especiallythe ones appearing in later chapters, as lecture material, thereby supplementingthe text. We hope that this manual will enhance the usefulness of the text –both for the instructors and (indirectly) for the students.

We implore that instructors not share copies of any of the material in thismanual with students or post any part of this manual on the web. Studentslearn best when they work together and struggle over difficult problems. Readilyavailable solutions interfere with this crucial aspect of graduate physics training.

R.K.P. San Diego, CAP.D.B. Boulder, CO

iii

Chapter 1

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

ln Ω(0)(E1) ≡ lnΩ1(E1) + ln Ω2(E2) (E2 = E(0) − E1)

= ln Ω1(E1) + ln Ω2(E2)+∂ ln Ω1(E1)

∂E1+∂ ln Ω2(E2)

∂E2

∂E2

∂E1

E1=E1

(E1 − E1)+

1

2

∂2 ln Ω1(E1)

∂E21

+∂2 ln Ω2(E2)

∂E22

(∂E2

∂E1

)2E1=E1

(E1 − E1)2 + · · · .

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 maybe written as

1

2

∂β1

∂E1+∂B2

∂E2

eq.

(E1 − E1

)2= −1

2

1

kT 21 (Cv)1

+1

kT 22 (Cv)2

(E1−E1)2,

with T1 = T2. Ignoring the subsequent terms (which is justified if thesystems involved are large) and taking the exponentials, we readilysee that the function Ω0(E1) is a Gaussian in the variable (E1− E1),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 avery large reservoir — then the variance becomes simply kT 2(Cv)1,regardless of the nature of the reservoir; cf. eqn. (3.6.3).

(b) If the systems involved are ideal classical gases, then (Cv)1 = 32N1k

and (Cv)2 = 32N2k; the variance then becomes 3

2k2T 2 ·N1N2/(N1 +

N2). Again, ifN2 N1, we obtain the simplified expression 32N1k

2T 2;cf. Problem 3.18.

1.2. Since S is additive and Ω multiplicative, the function f(Ω) must satisfythe condition

f(Ω1Ω2) = f(Ω1) + f(Ω2). (1)

1

2

Differentiating (1) with respect to Ω1 (and with respect to Ω2), we get

Ω2f′(Ω1Ω2) = f ′(Ω1) and Ω1f

′(Ω1Ω2) = f ′(Ω2),

so thatΩ1f

′(Ω1) = Ω2f′(Ω2). (2)

Since the left-hand side of (2) is independent of Ω2 and the right-hand sideis independent of Ω1, each side must be equal to a constant, k, independentof 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.

1.4. Instead of eqn. (1.4.1), we now have

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

so that

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 writtenas

C+N ln V+

N−1∑j=1

ln

(1− jv0

V

)' C+N ln V+

N−1∑j=1

(−jv0

V

)' C+N ln V−N

2v0

2V.

Equation (1.4.2) is then replaced by

P

kT=N

V+N2v0

2V 2=N

V

(1 +

Nv0

2V

), i.e.

PV

(1 +

Nv0

2V

)−1

= NkT .

Since Nv0 V, (1 + Nv0/2V )−1 ' 1 − Nv0/2V . Our last result thentakes the form: P (V − b) = NkT , where b = 1

2Nv0.

A little reflection shows that v0 = (4π/3)σ3, with the result that

b =1

2N · 4π

3σ3 = 4N · 4π

3

(1

2σ

)3

.

1.5. This problem is essentially solved in Appendix A; all that remains to bedone is to substitute from eqn. (B.12) into (B.11), to get

Σ1(ε∗) =(πε∗1/2/L)3

6π2V ∓ (πε∗1/2/L)2

16πS.

3

Substituting V = L3 and S = 6L2, we obtain eqns. (1.4.15 and 16).

The expression for T now follows straightforwardly; we get

1

T= k

(∂ ln Ω

∂E

)N

=k

hν

(∂ ln Ω

∂R

)N

=k

hνln

(R+N

R

)=

k

hνln

(1 +

Nhν

E

),

so that

T =hν

k

/ln

(1 +

Nhν

E

).

For E Nhν, we recover the classical result: T = E/Nk .

1.9. Since the function S(N,V,E) of a given thermodynamic system is an ex-tensive quantity, we may write

S(N,V,E) = Nf

(V

N,E

N

)= Nf (v, ε)

(v =

V

N, ε =

E

N

).

It follows that

N

(∂S

∂N

)V,E

= N

[f +N

(∂f

∂v

)ε

· −VN2

+N

(∂f

∂ε

)v

· −EN2

],

V

(∂S

∂V

)N,E

= VN

(∂f

∂v

)ε

· 1

Nand E

(∂S.∂E

)N,V

= EN

(∂f

∂ε

)v

· 1

N.

Adding these expressions, we obtain the desired result.

1.11. Clearly, the initial temperatures and the initial particle densities of the twogases (and hence of the mixture) are the same. The entropy of mixing may,therefore, be obtained from eqn. (1.5.4), with N1 = 4NA and N2 = NA.We get

(∆S)∗ = k[4NA ln(5/4) +NA ln 5]

= R[4 ln(5/4) + ln 5] = 2.502 R,

which is equivalent to about 0.5 R per mole of the mixture.

1.12. (a) The expression in question is given by eqn. (1.5.3a). Without loss ofgenerality, we may keep N1, N2 and V1 fixed and vary only V2 . Thefirst and second derivatives of this expression are then given by

k

[N1 +N2

V1 + V2− N2

V2

]and k

[− N1 +N2

(V1 + V2)2+N2

V 22

](1a,b)

respectively. Equating (1a) to zero gives the desired condition, viz.N1V2 = N2V1, i.e. N1/V1 = N2/V2 = n, say. Expression (1b) thenreduces to

k

[− n

V1 + V2+

n

V2

]=

knV1

V2(V1 + V2)> 0.

Clearly, (∆S)1≡2 is at its minimum when N1/V1 = N2/V2, and it isstraightforward to check that the value at the minimum is zero.

4

(b) The expression now in question is given by eqn. (1.5.4). With N1 =αN and N2 = (1 − α)N , where N = N1 + N2 (which is fixed), theexpression for (∆S)∗/k takes the form

−αN ln α− (1− α)N ln (1− α).

The first and second derivatives of this expression with respect to α are

[−N ln α+N ln(1− α)] and

[−Nα− N

1− α

](2a,b)

respectively. Equating (2a) to zero gives the condition α = 1/2, whichreduces (2b) to −4N . Clearly, (∆S)∗/k is at its maximum when N1 =N2 = (1/2)N , and it is straightforward to check that the value at themaximum is N ln 2.

1.13. Proceeding with eqn. (1.5.1), with T replaced by Ti, it is straightforwardto see that the extra contribution to ∆S, owing to the fact that T1 6= T2,is given by the expression

3

2N1k ln (Tf/T1) +

3

2N2k ln(Tf/T2),

where Tf = (N1T1 + N2T2)/(N1 + N2). It is worth checking that thisexpression is always greater than or equal to zero, the equality holding ifand only if T1 = T2. Furthermore, the result quoted here does not dependon whether the two gases were different or identical.

1.14. By eqn. (1.5.1a), given on page 19 of the text, we get

(∆S)v =3

2Nk ln(Tf/Ti).

Now, since PV = NkT , the same equation may also be written as

S = Nk ln

(kT

P

)+

3

2Nk

5

3+ ln

(2πmkT

h2

). (1b)

It follows that

(∆S)P =5

2Nk ln(Tf / Ti) =

5

3(∆S)V.

A numerical verification of this result is straightforward.

It should be noted that, quite generally,

(∆S)P(∆S)V

=T (∂S / ∂T )PT (∂S / ∂T )V

=CPCV

= γ

which, in the present case, happens to be 5/3.

5

1.15. For an ideal gas, CP − CV = nR, where n is the number of moles of thegas. With CP /CV = γ, one gets

CP = γnR / (γ − 1) and CV = nR / (γ − 1).

For a mixture of two ideal gases,

CV =n1R

γ1 − 1+

n2R

γ2 − 1=

(f1

γ1 − 1+

f2

γ2 − 1

)(n1 + n2)R.

Equating this to the conventional expression (n1 + n2)R/(γ − 1), we getthe desired result.

1.16. In view of eqn. (1.3.15), E − TS + PV = µN . It follows that

dE − TdS − SdT + PdV + VdP = µdN + Ndµ.

Combining this with eqn. (1.3.4), we get

−SdT + VdP = Ndµ, i.e. dP = (N / V )dµ+ (S / V )dT .

Clearly, then,

(∂P / ∂µ)T = N / V and (∂P / ∂T )µ = S / V.

Now, for the ideal gas

P =NkT

Vand µ = kT ln

N

V

(h2

2πmkT

)3/2

;

see eqn. (1.5.7). Eliminating (N/V ), we get

P = kT

(2πmkT

h2

)3/2

eµ/kT ,

which is the desired expression. It follows quite readily now that for thissystem (

∂P

∂µ

)T

=1

kTP.

which is indeed equal to N/V , whereas(∂P

∂T

)µ

=5

2TP − µ

kT 2P =

[5

2− ln

N

V

(h2

2πmkT

)3/2]

Nk

V

which, by eqn. (1.5.1a), is precisely equal to S/V .

Chapter 2

2.3. The rotator in this problem may be regarded as confined to the (z = 0)-plane and its position at time t may be denoted by the azimuthal angleϕ. The conjugate variable pϕ is then mρ2ϕ, where the various symbolshave their usual meanings. The energy of rotation is given by

E =1

2m(ρϕ)2 = p2

ϕ / 2mρ2.

Lines of constant energy in the (ϕ, pϕ)-plane are “straight lines, runningparallel to the ϕ-axis from ϕ = 0 to ϕ = 2π”. The basic cell of area h inthis plane is a “rectangle with sides ∆ϕ = 2π and ∆pϕ = h/2π”. Clearly,the eigenvalues of pϕ, starting with pϕ = 0, are n~ and those of E aren2~2/2I, where I = mρ2 and n = 0,±1,±2, . . .

The eigenvalues of E obtained here are precisely the ones given by quan-tum mechanics for the energy “associated with the z-component of therotational motion”.

2.4. The rigid rotator is a model for a diatomic molecule whose internucleardistance r may be regarded as fixed. The orientation of the molecule in

6

7

space may be denoted by the angles θ and ϕ, the conjugate variables beingpθ = mr2θ and pϕ = mr2 sin2 θϕ. The energy of rotation is given by

E =1

2m(rθ)2 +

1

2m(r sin θϕ)2 =

p2θ

2mr2+

p2ϕ

2mr2 sin2 θ=M2

2I,

where I = mr2 and M2 = p2θ +

(p2ϕ/ sin2 θ

).

The “volume” of the relevant region of the phase space is given by theintegral

∫ ′dpθdpϕdθ dϕ, where the region of integration is constrained

by the value of M . A little reflection shows that in the subspace of pθand pϕ we are restricted by an elliptical boundary with semi-axes M andM sin θ, the enclosed area being πM2 sin θ. The “volume” of the relevantregion, therefore, is

π∫θ=0

2π∫ϕ=0

(πM2 sin θ)dθ dϕ = 4π2M2.

The number of microstates available to the rotator is then given by 4π2M2/h2,which is precisely (M/~)2. At the same time, the number of microstatesassociated with the quantized value M2

j = j(j+ 1)~2 may be estimated as

1

~2

[M2j+ 1

2−M2

j− 12

]=

(j +

1

2

)(j +

3

2

)−(j − 1

2

)(j +

1

2

)= 2j + 1.

This is precisely the degeneracy arising from the eigenvalues that the az-imuthal quantum number m has, viz. j, j − 1, . . . ,−j + 1,−j.

2.6. In terms of the variables θ and L(= m`2θ), the state of the simple pendu-lum is given by, see eqns. (2.4.9),

θ = (A/`) cos(ωt+ ϕ), L = −m`ωA sin(ωt+ ϕ),

with E = 12mω

2A2 and τ = 2π/ω. The trajectory in the (θ, L)-plane isgiven by the equation

θ2

(A/`)2+

L2

(m` ωA)2= 1,

which is an ellipse — just like in Fig. 2.2. The enclosed area turns out tobe πmωA2, which is precisely equal to the product Eτ .

2.7. Following the argument developed on page 68–69 of the text, the numberof microstates for a given energy E turns out to be

Ω(E) = (R+N − 1)!/ R!(N − 1)!, R =

(E − 1

2N~ω

)/~ω. (1)

8

For R N , we obtain the asymptotic result

Ω(E) ≈ RN−1 / (N − 1)!, where R ≈ E / ~ω. (3.8.25a)

The corresponding expression for Γ(E; ∆) would be

Γ(E; ∆) ≈ (E / ~ω)N−1

(N − 1)!· ∆

~ω=

EN−1∆

(N − 1)!(~ω)N. (1)

The “volume” of the relevant region of the phase space may be derivedfrom the integral∫ ′ N∏

i=1

(dq idpi), with

N∑i=1

(1

2kq2i +

1

2mp2i

)≤ E.

This is equal to, see eqn. (7a) of Appendix C,(2

k

) 12N

(2m)12N · π

N

N !EN =

(2π

ω

)NEN

N !,

where ω =√k/m. The “volume” of the shell in question is then given by(

2π

ω

)NNEN−1

N !·∆ =

(2π

ω

)NEN−1∆

(N − 1)!. (2)

Dividing (2) by (1), we see that the conversion factor ω0 is precisely hN .

2.8. We write V3N = AR3N , so that dV 3N = A · 3NR3N−1dR. At the sametime, we have

∞∫0

. . .

∞∫0

e−

N∑i=1

riN∏i=1

r2i dr i =

N∏i=1

∞∫0

e−rir2i dr i = 2N . (1)

The integral on the left may be written as

∞∫0

e−R(4π)−N dV 3N =

∞∫0

e−R(4π)−N A·3NR3N−1 dR = (4π)−N A·3N Γ(3N).

(2)Equating (1) and (2), we get: A = (8π)N/(3N)!, which yields the desiredresult for V3N .

The “volume” of the relevant region of the phase space is given by∫ ′ 3N∏i=1

dq idpi = V N∫ ′ N∏

i=1

(4πp2

i dpi)

= V N (8π E3 / c3)N / (3N)!,

9

so thatΣ(n, V, E) = V N (8π E3 / h3 c3)N / (3N)!,

which is a function of N and VE 3. An isentropic process then impliesthat VE 3 = const .

The temperature of the system is given by

1

T=

(∂(k ln Σ)

∂E

)N,V

=3Nk

E, i.e. E = 3NkT .

The equation for the isentropic process then becomes VT 3 = const ., i.e.T ∝ V −1/3; this implies that γ = 4/3. The rest of the thermodynamicsfollows straightforwardly. See also Problems 1.7 and 3.15.

Chapter 3

3.4. For the first part, we use eqn. (3.2.31) with all ωr = 1. We get

k

Nln Γ = k ln

∑r

e−βEr

+ kβU,

which is indeed equal to −(A/T ) + (U/T ) = S.

For the second part, we use eqn. (3.2.5), with the result that

k

Nln W n∗r =

k

N

[N ln N −

∑r

n∗r ln n∗r

]

= −k∑r

n∗rN

lnn∗rN

= −k⟨

lnn∗rN

⟩.

Substituting for n∗r from eqn. (3.2.10), we get

k

Nln W n∗r = kβ〈Er〉+ k ln

∑r

e−βEr

,

which is precisely the result obtained in the first part.

3.5. Since the function A(N,V, T ) of a given thermodynamic system is anextensive quantity, we may write

A(N,V, T ) = Nf (v, T ) (v = V / N).

It follows that

N

(∂A

∂N

)V,T

= N

[f +N

(∂f

∂v

)T

· −VN2

], and V

(∂A

∂V

)N,T

= VN

(∂f

∂v

)T

· 1

N.

Adding these expressions, we obtain the desired result.

3.6. Let’s go to part (c) right away. Our problem here is to maximize theexpression S/k = −

∑r,sPr,s ln Pr,s, subject to the constraints

∑r,sPr,s =

10

11

1,∑r,sEsPr,s = E and

∑r,sNrPr,s = N . Varying P ’s and using the method

of Lagrange’s undetermined multipliers, we are led to the condition∑r,s

−(1 + ln Pr,s)− γ − βEs − αNr δPr,s = 0.

In view of the arbitrariness of the δP ’s in this expression, we require that

−(1 + ln Pr,s)− γ − βEs − αNr = 0

for all r and s. It follows that

Pr,s ∝ exp(−βEs − αNr).

The parameters α and β are to be determined by the given values of Nand E.

In the absence of the constraint imposed by N , the parameter α does noteven figure in the calculation, and we obtain

Pr ∝ exp(−βEr),

as desired in part (b). And if the constraint imposed by E is also absent,we obtain

Pr = const .,

as desired in part (a).

3.7. From thermodynamics,

CP − CV = T

(∂P

∂T

)V

(∂V

∂T

)P

= −T(∂P

∂T

)2

V

/(∂P

∂V

)T

> 0. (1)

From Sec. 3.3,

P = −(∂A

∂V

)N,T

= kT

(∂ ln Q

∂V

)N,T

. (2)

Substituting (2) into (1), we obtain the desired result.

For the ideal gas, Q ∝ V NT 3N/2. Therefore, (∂ lnQ/∂V )T = N/V . Wethen get

CP − CV = −k (N/V )2

−N/V 2= Nk .

3.8. For an ideal gas,

Q1 = V(2πmkT )3/2

h3=

NkT

P

(2πmkT )3/2

h3.

12

It follows that T (∂ ln Q1/∂T )P = 5/2; the expression on the right-handside of the given equation then is

ln

V

N

(2πmkT )3/2

h3

+

5

2

which, by eqn. (3.5.13), is indeed equal to the quantity S/Nk.

3.12. We start with eqn. (3.5.5), substitute H(q,p) =∑i

(p2i /2m

)+ U(q) and

integrate over the pi′s, to get

QN (V, T ) =1

N !

(2πmkT

h2

)3N/2

ZN (V, T ), where ZN (V, T ) =

∫e−U(q)/kT d3N q.

It follows that, for N 1,

A = NkT

[ln

N

(h2

2πmkT

)3/2− 1

]− kT ln Z, whence

S = Nk

[ln

1

N

(2πmkT

h2

)3/2

+5

2

]+ k ln Z + kT

(∂ ln Z

∂T

)N,V

.

Now

kT

(∂ ln Z

∂T

)N,V

=kT∫e−U/kT (U/kT 2)d3N q∫e−U/kT d3N q

=U

T, while

k ln Z = k lnV N e−U/kT

= Nk ln V − U

T.

Substituting these results into the above expression for S, we obtain thedesired result for S. In passing, we note that 〈H〉 ≡ A+ TS = 3

2NkT + U .

For the second part of the question, we write U(q) =∑i<j

u(rij ), so that

e−βU(q) =∏i<j

e−βu(rij ) =∏i<j

(1 + fij ) ,

and follow Problems 3.23 and 1.4. The quantity V then appears to be inthe nature of a “free volume” for the molecules of the system.

3.14. a) The Lagrangian is given by

L = K − V =∑iα

1

2mr2

iα −∑i<j

u(rij)−∑iα

[uw (riα) + uw (L− riα)],

where i = 1, · · · , N denotes the particle number, α = x, y, z denotes thecartesian directions, and r2

ij =∑α (riα − rjα)2. The canonical momenta

are

piα =∂L

∂riα= mriα.

13

The Hamiltonian is given by

H =∑iα

piαriα −L

=∑iα

p2iα

2m+∑i<j

u(rij) +∑iα

[uw (riα) + uw (L− riα)].

The canonical pressure can be written

P = −∂H

∂V= − 1

3L2

∂H

∂L= − 1

3L2

∑iα

u′w(L− riα) =1

3L2(Fx + Fy + Fz) .

This is clearly the instantaneous force per unit area on the right, back,and top walls.

b) The cartesian coordinates for the scaled position inside the box aresiα = riα/L so the Lagrangian becomes

L =∑iα

1

2mL2s2

iα −∑i<j

u(Lsij)−∑iα

[uw (Lsiα) + uw (L− Lsiα)].

In this case the canonical momenta are

piα =∂L

∂siα= mL2siα.

This leads to a Hamiltonian of the form

H =∑iα

p2iα

2mL2+∑i<j

u(Lsij) +∑iα

[uw (Lsiα) + uw (L− Lsiα)],

with canonical pressure is

P = +1

3L2

∑iα

p2iα

mL3

− 1

3L2

∑i<j

u′(Lsij)sij

− 1

3L2

∑iα

[u′w(Lsiα)siα + u′w(L− Lsiα)(1− siα)].

Converting back to normal cartesian coordinates and momenta gives

P = +2

3V

∑iα

p2iα

2m

− 1

3V

∑i<j

u′(rij)rij

− 1

3V

∑iα

[u′w(riα)riα + u′w(L− riα)(L− riα)].

14

The first term is (2/3)(N/V ) times the kinetic energy per particle so isO(N). The second term is (1/3)(N/V ) times the virial per particle so isalso O(N). On the other hand, third term is proportional to the force onthe walls divided by the volume so is O(N2/3) which is negligible in thethermodynamic limit.

Comparing to equation (3.7.15) for the average pressure we see that

P

nkT= 1− 1

3NkT

⟨∑i<j

u′w(rij)rij

⟩.

3.15. Here, QN (V, T ) = (1/N !)QN1 (V, T ), while

Q1(V, T ) =

∞∫0

e−βpc V · 4πp2dp

h3=

8πV

h3

1

β3c3,

which yields the desired result for QN . The thermodynamics of the systemnow follows straightforwardly.

As regards the density of states, the expression

Q1(V, T ) =

∞∫0

e−βεg(ε)dε =8πV

h3

1

β3c3

leads to

g(ε) =4πV

h3c3ε2

for a single particle, while the expression for QN (V, T ) leads to

g(E) =1

N !

(8πV

h3c3

)NE3N−1

Γ(3N)

for the N -particle system; cf. the expression for Σ(E) derived in Prob-lem 2.8.

3.17. Differentiate the stated result with respect to β, to get∫ ∂U

∂β−H(U −H)

e−βH dω = 0.

This means that ⟨∂U

∂β−HU +H2

⟩= 0,

which amounts to the desired result: 〈H2〉 − 〈H〉2 = −(∂U/∂β).

15

3.18. We start with eqn. (3.6.2), viz.

∂U

∂β= −

∑rE2re−βEr∑

re−βEr

+ U2, (1)

and differentiate it with respect to β, keeping the Er fixed. We get

∂2U

∂β2= 〈E3〉 − 〈E2〉〈E〉+ 2U

∂U

∂β.

Substituting for (∂U/∂β) from eqn. (1), we get

∂2U

∂β2= 〈E3〉 − 3〈E2〉U + 2U3,

which is precisely equal to 〈(E−U)3〉. As for ∂2U/∂β2, we note that, since(∂U

∂β

)Er

= −kT 2

(∂U

∂T

)V

= −kT 2CV,(∂2U

∂β2

)Er

= −kT 2

[∂

∂T

(−kT 2CV

)]V

= k2T 2

[2TC V + T 2

(∂CV

∂T

)V

].

Hence the desired result.

For the ideal classical gas, U = 32NkT and CV = 3

2Nk , which readily yieldthe stated results.

3.19. Since G =∑i

qipi, G =∑i

(qipi + qipi). Averaging over a time interval τ ,

we get

1

τ

t+τ∫t

∑i

(qipi + qipi)dt =1

τ

t+τ∫t

Gdt =G(t+ τ)−G(t)

τ. (1)

For a finite V and finite E, the quantity G is bounded ; therefore, in thelimit τ → ∞, the right-hand side of (1) vanishes. The left-hand sidethen gives ⟨∑

i

(qipi + qipi)

⟩= 0.

which leads to the desired result.

3.20. The virial of the noninteracting system, by eqn. (3.7.12), is −3PV . Thecontribution from interparticle interactions, by eqn. (3.7.15), is given bythe “expectation value of the sum of the quantity −r(∂u/∂r) over all pairsof particles in the system”. If u(r) is a homogeneous function (of degreen) of the particle coordinates, this contribution will be −nU , where U is

16

the mean potential energy (not the internal energy) of the system. Thetotal virial is then given by

V = −3PV − nU .

The relation K = − 12V still holds, and the rest of the results follow

straightforwardly.

3.21. All systems considered here are localized. The pressure term, therefore,drops out, and we are left with the result

K =n

2U =

n

n+ 2E.

Example (a) pertains to n = 2, while examples (b) and (c) pertain ton = −1. In the former case, K = U = 1

2E; in the latter, K = − 12U = −E.

The next problem pertains to n = 4.

3.22. Note that a force proportional to q3 implies a potential energy proportionalto q4. Thus

H =1

2mp2 + cq4 (c > 0).

It follows that

⟨1

2mp2

⟩=

∞∫−∞

e−βp2/2m(p2/2m)dp

∞∫−∞

e−βp2/2m dp

=1

2β;

for the values of these integrals, see eqns. (13a) of Appendix B. Next,

〈cq4〉 =

∞∫−∞

e−βcq4

(cq4)dq

∞∫−∞

e−βcq4 dq

= − ∂

∂βln I(B),

where I(β) denotes the integral in the denominator. It is straightforwardto see that I(β) is proportional to β−1/4, whence 〈cq4〉 = 1/4β, whichproves the desired result.

3.23. The key to this derivation is writing the partition function in terms ofposition integrals over scaled coordinates. Assume a cubic box of size Land volume V = L3. The scaled position for particle i is si = ri/L. The

17

partition function is

QN (V, T ) =1

λ3NN !

∫exp

−β∑i<j

u(rij)

dNr

=V N

λ3NN !

∫exp

−β∑i<j

u(V 1/dsij)

dNs.

Now the pressure is

P = −(∂A

∂V

)N,T

=kT

QN

NQNV− βV N

dV λ3NN !

∫ ∑i<j

V 1/dsiju′(V 1/dsij)

exp

−β∑i<j

u(V 1/dsij)

dNs

.This can be simplified by going back to integrals over the normal positionvariables to give equation (3.7.15).

3.24. By eqn. (3.7.5), we have, for a single particle,⟨3∑i=1

piqi

⟩= 3kT . (1)

The left-hand side of (1) is the expectation value of the quantity p ·u, i.e.pu which, for a relativistic particle, is equal to m0u

2(1− u2/c2)−1/2. Thedesired result follows readily.

In the non-relativistic limit (u c), one obtains:⟨

12m0u

2⟩≈ 3

2kT ; in theextreme relativistic limit (u → c), one obtains: 〈mc2〉 ≈ 3kT . Note that,in the latter case, m0c

2 is negligible in comparison with mc2, so there isno significant difference between the kinetic energy and the total energyof the particle.

3.25. For the first part of this problem, see Sec. 6.4 — especially the derivationof the formula (6.4.9). For the second part, equate the result obtained inthe first part with the one stated in eqn. (3.7.5).

3.26. The multiplicity w(j)= (j + s − 1)!/j!(s − 1)! arises from the varietyof ways in which j indistinguishable quanta can be divided among thes dimensions of the oscillator: j = j1 + . . . + js; this is similar to thecalculation done on page 68–69 of the text.

As for the partition function, Q(s)N (β) =

[Q

(s)1 (β)

]N, where

Q(s)1 (β) =

∞∑j=0

(j + s− 1)!

j!(s− 1)!e−β(j+ 1

2 s)~ω

= e−12 sβ~ω (1− e−β~ω)−s.

18

Calculation of the various thermodynamic quantities is now straightfor-ward. The results are found to be essentially the same as for a system ofsN one-dimensional oscillators. However, since

Q(s)N (β) = Q

(1)Ns (β),

the chemical potential µs will turn out to be s times µ1.

3.28. (a) When one of the oscillators is in the quantum state n, the energy leftfor the remaining (N − 1) oscillators is E −

(n+ 1

2

)~ω; the corre-

sponding number of quanta to be distributed among these oscillatorsis R−n; see eqn. (3.8.24). The relevant number of microstates is thengiven by the expression (R−n+N−2)!/(R−n)!(N−2)!. Combinedwith expression (3.8.25), this gives

pn =(R− n+N − 2)!

(R− n)!(N − 2)!÷ (R+N − 1)!

R!(N − 1)!. (1)

It follows that

pn+1

pn=

R− nR− n+N − 2

' R

R+N=

n

n+ 1.

By iteration, pn = p0n/(n+ 1)n.

Going back to eqn. (1), we note that

p0 =N − 1

R+N − 1' N

R+N= − 1

n+ 1,

which completes the desired calculation.

(b) The probability in question is proportional to gN−1(E − ε), i.e. to

(E − ε)32 (N−1)−1. For 1 N , this is essentially proportional to

(1− ε/E)32N and, for ε −E, to e−3Nε/2E .

3.29. The partition function of the anharmonic oscillator is given by

Q1(β) =1

h

∞∫−∞

∞∫−∞

e−βHdq dq

H =

p2

2m+ cq2 − gq3 − fq4

.

The integration over p gives a factor of√

2πm/β. For integration over q,we write

e−βcq2

eβ(gq3+fq4) = e−βcq2

[1 + β(gq3 + fq4) +

1

2β2(gq3 + fq4)2 + . . .

];

the integration then gives√π

βc+ βf · 3

4

√π

β5c5+

1

2β2 g2 · 15

8

√π

β7c7+ . . . .

19

It follows that

Q1(β) =π

βh

√2m

c

[1 +

3f

4βc2+

15g2

16βc3+ . . .

],

so that

lnQ1(β) = const .− lnβ +3f

4βc2+

15g2

16βc3+ . . . ,

whence

U(β) =1

β+

3f

4β2c2+

15g2

16β2c3+ . . .

and

C(β) = k +3f k2T

2c2+

15g2 k2T

8c3+ . . . .

Next, the mean value of the displacement q is given by

〈q〉 =

∫ ∞−∞

∫ ∞−∞

exp(−βH)q dp dq /

∫ ∞−∞

∫ ∞−∞

exp(−βH)dpdq .

In the desired approximation, we get

〈q〉 ' βg

∞∫−∞

e−βcq2

q4 dq

/ ∞∫−∞

e−βcq2

dq

= βg · 3

4

√π

β5c5/

√π

βc=

3g

4βc2.

3.30. The single-oscillator partition function is now given by

Q1(β) =

∞∑n=0

e−β(n+ 12 )~ω+βx(n+ 1

2 )2~ω.

For x 1, we may write

Q1(β) =

∞∑n=0

e−β(n+ 12 )~ω

[1 + βx

(n+

1

2

)2

~ω + . . .

].

With u = β~ω, the sums involved are

S1(u) =

∞∑n=0

e−u(n+ 12 ) =

[2 sinh

(1

2u

)]−1

, and

S2(u) =

∞∑n=0

e−u(n+ 12 )(n+

1

2

)2

=d2

du2S1(u) =

[4 sinh

(1

2u

)]−1coth2

(1

2u

)− 1

2

.

20

It follows that

ln Q1 = ln[S1 + xuS2 + . . .] ' lnS1 + xu(S2/S1)

= − ln

[2 sinh

(1

2u

)]+

1

2xu

coth2

(1

2u

)− 1

2

.

The first part of this expression leads to the standard results (3.8.20 and21). The second part may, for simplification, be expressed as a powerseries in u, viz.

x

(2

u+

u

12+

u3

120+ . . .

).

The resulting contribution to the internal energy per oscillator turns outto be

x~ω(

2

u2− 1

12− u2

40− . . .

)and the corresponding contribution to the specific heat is given by

xk

(4

u+u3

20+ . . .

).

3.31. This problem is essentially the same as Problem 3.32, with g1 = g2 =1, ε1 = 0 and ε2 = ε.

3.32. We use formula (3.3.13), with Pr = p1/g1 for each of the states in group1 and p2/g2 for each of the states in group 2. We get

S = −k[g1

(p1

g1lnp1

g1

)+ g2

(p2

g2lnp2

g2

)]. (1)

(a) In thermal equilibrium,

pi =gie−βεi∑

i

gie−βεi(i = 1, 2).

With x = β(ε2 − ε1), we have: p1 = g1/(g1 + g2e−x) and p2 =

g2/(g1ex + g2). Substituting these results into (1), we obtain

S = k

[g1

g1 + g2e−xln(g1 + g2e

−x)+g2

g1ex + g2ln (g1e

x + g2)

].

Writing the first log as ln g1 + ln1 + (g2/g1)e−x and the second logas ln g1 + x + ln1 + (g2/g1)e−x, we obtain the stated expressionfor S.

(b) With Q = g1e−βε1 + g2e

−βε2 , it is straightforward to see that

A = −kT lng1e−βε1 + g2e

−βε2 and

U = g1ε1e−βε1 + g2ε2e

−βε2/g1e−βε1 + g2e

−βε2.

The formula S = (U −A)/T then leads to the desired result.

21

(c) As T → 0, x → ∞ and S indeed tends to the value k ln g1. Thiscorresponds to the fact that the probabilities p1 and p2 in this limittend to the values 1 and 0, respectively.

3.35. The partition function of the system is given by

QN =1

N !QN1 , where Q1 =

V

λ3· Z,

Z being the factor that arises from the rotational/orientational degrees offreedom of the molecule:

Z =

∫exp

[−β

p2θ

2I+

p2ϕ

2I sin2 θ− µE cos θ

]dpθdpϕdθdϕ

h2

=

π∫0

(2πI

β

)1/2(2πI sin2 θ

β

)1/2

eβµE cos θ 2πdθ

h2

=I

β~2· 2 sinh(βµE)

βµE.

The study of the various thermodynamical quantities of the system is nowstraightforward.

Concentrating on the electrical quantities alone, we obtain for the netdipole moment of the system

Mz = N〈µ cos θ〉 =N

β

∂ ln Z

∂E= Nµ

[coth (βµE)− 1

βµE

];

cf. eqns. (3.9.4 and 6). For βµE 1,

Mz ≈ Nµ ·1

3βµE.

The polarization P , per unit volume, of the system is then given by

P ≈ nµ2 E / 3kT (n = N/V ),

and the dielectric constant ε by

ε =E + 4πP

E≈ 1 +

4πnµ2

3kT.

The numerical part of the problem is straightforward.

3.36. The mean force 〈F〉 between the two dipoles is given by

〈F〉 =

⟨−∂U∂R

⟩=

∫e−βU (−∂U / ∂R) sin θd θdϕ · sin θ′ dθ′ dϕ′∫

e−βU sin θdθdϕ · sin θ′ dθ′ dϕ′(1)

=1

β

∂

∂Rln Z . (2)

22

where Z denotes the integral in the denominator of (1). At high temper-atures, we may write

Z =

∫ [1− βU +

1

2β2U2 − . . .

]sin θ dθdϕ · sin θ′ dθ′ dϕ′.

The linear term vanishes on integration and we are left with

Z =

∫ [1 +

1

2β2 (µµ′)2

R62 cos θ cos θ′ − sin θ sin θ′ cos(ϕ− ϕ′)2 − . . .

]sin θd θdϕ · sin θ′ dθ′ dϕ′

= 16π2

[1 +

1

2β2 (µµ′)2

R6

4 · 1

3· 1

3− 0 +

2

3· 2

3· 1

2

− . . .

].

It follows that

ln Z = const .+1

3β2 (µµ′)2

R6− . . .

and hence, at high temperatures,

〈F〉 ≈ −2β(µµ′)2

R7R.

3.37. By eqns. (3.9.17 and 18), we have, for a single dipole,

µz =

J∑m=−J

(gµBm) exp(βgµBmH )

J∑m=−J

exp(βgµBmH )

.

At high temperatures, the exponential may be approximated by (1 +βgµBmH ) which yields, to the leading order in H,

µz = βg2 µ2BHm

2.

One readily obtains for the Curie constant (per unit volume) of the system

CJ = N0

(g2 µ2

B / k)m2.

Writing m = J cos θ, one obtains the desired result.

For the second part, we simply note that, for a given J ,

m2 =

J∑m=−J

m2

2J + 1=J(J + 1)(2J + 1)/3

2J + 1=

1

3J(J + 1).

23

3.38. Treating m as a continuous variable, the partition function of a magneticdipole assumes the form

Q1(β) =

∫ J

−JeβgµBHm dm =

2

βgµBHsin h (βgµBJH );

cf. eqn. (3.9.5). It is clear that this approximation will lead essentially tothe same results as the ones following from the Langevin theory — exceptfor the fact that the role of µ will be played by gµBJ , which should becontrasted with the expression (3.9.16) of the quantum theory.

3.40. (a) By definition, CH = T (∂S/∂T )H and CM = T (∂S/∂T )M . Now(∂S

∂T

)H

=

(∂S

∂T

)M

+

(∂S

∂M

)T

(∂M

∂T

)H

; (1)

at the same time, dA ≡ dU − TdS − SdT = HdM − SdT , with theresult that (∂H/∂T )M = −(∂S/∂M)T . Equation (1) then becomes(

∂S

∂T

)H

=

(∂S

∂T

)M

−(∂H

∂T

)M

(∂M

∂T

)H

. (2)

Multiplying (2) by T , we obtain the desired result for CH − CM .

(b) The Curie law implies thatM = CH /T . This means that (∂H/∂T )M =H/T , while (∂M/∂T )H = −CH /T 2. It follows that CH − CM =CH 2/T 2.

3.42. Let N1(N2) be the number of dipoles aligned parallel (opposite) to thefield. Then

N1 +N2 = N, while −N1ε+N2ε = E.

It follows that

N1 =1

2(N − E/ε), N2 =

1

2(N + E/ε).

The number of microstates associated with this macrostate is given by

Ω(N, E) =N !

12 (N − E/ε)

!

12 (N + E/ε)

!

The entropy of the system is then given by the expression

S = k ln Ω ≈ k[N ln N − 1

2

(N − E

ε

)ln

1

2

(N − E

ε

)− 1

2

(N +

E

ε

)ln

1

2

(N +

E

ε

)],

which is essentially the same as eqn. (3.10.9).

24

For the temperature of the system, we get

1

T=

(∂S

∂E

)N

=k

2εln

N − E/εN + E/ε

,

which agrees with eqn. (3.10.8).

3.43. The partition function of this system is given by the usual expression(3.5.5), except for the fact that the Hamiltonian of the system is now afunction of the quantities pj + (ej / c)A(rj), and not of the pj as such.However, on integration over any component of pj , from −∞ to +∞, we

obtain the same standard factor√

2πmkT — regardless of the value ofthe corresponding component of A. The partition function is, therefore,independent of the applied field and hence the net magnetization of thesystem is zero.

3.44. The Shannon information for a single message is given by I1 = −∑r Pr lnPr

where Pr is the a priori probability of message r from among all Ω pos-sible messages. The maximum information is obtained from varying theprobabilities, using a Lagrange multiplier µ to maintain the normalization∑r Pr = 1, and demanding the solution is stationary.

0 = δI1 − µδ

(∑r

Pr

)= −

∑r

δPr [ln pr − 1− µ].

This implies the Pr = const, i.e. all messages are equally likely. ThereforePr = 1/Ω , which gives I1 = ln Ω. Any other set of probabilities givessmaller information per message.

Keeping to the general cases in which probabilities of individual mes-sages messages do not need to be equal, consider a sequence of two mes-sages. The a priori probability of message r followed by message r′ isPrr′ = PrPr′Grr′ . The quantity Grr′ is the correlation between the twomessages. A value of Grr′ greater than unit implies that the first message rincreases the probability of finding the second message r′ above Pr′ . Thetwo message probabilities have the following properties:

∑r Prr′ = Pr′

and∑r′ Prr′ = Pr, i.e.

∑r PrGrr′ = 1 and

∑r′ Pr′Grr′ = 1. The infor-

mation contained in two messages is given by

I2 =∑rr′

Prr′ lnPrr′ =∑rr′

PrPr′Grr′ ln (PrPr′Grr′).

Expanding the logarithm and using the above summation properties gives

I2 = 2I1 −∑rr′

PrPr′Grr′ lnGrr′ = 2I1 +∑rr′

PrPr′Grr′ ln

(1

Grr′

).

25

Now, using lnx ≤ x− 1 for all x > 0, we get

I2 ≤ 2I1 +∑rr′

PrPr′ [1−Grr′ ] = 2I1.

The information contained in two correlated messages is reduced comparedto sum of the information contained in two uncorrelated messages. Anal-ysis of the first 65536 digits of π results in an information per characterof I1 ≈ 2.3 = ln 10. That makes sense because the characters 0, · · · , 9 areevenly distributed in the digital representation of π. Furthermore, sincethe digits of π are uncorrelated, the information per pair of characters isI2 ≈ 4.6 = 2I1. Analysis of the first 15, 000 characters of A ChrismasCarol by Charles Dickens gives I1 ≈ 3.08 ≈ ln 21.75. This value is rea-sonable since most of the characters are lower case letters of the alphabetand blanks. The nonuniformity of the distribution of letters reduces theinformation below ln 27 . When analyzed two characters at a time, theinformation is I2 ≈ 5.45 ≈ 2 ln 15.25. The strong correlations betweencharacters in English text reduces the information well below 2I1.

Chapter 4

4.1. By eqns. (4.1.9), (4.3.10) and (4.1.8), we get∑r,s

Pr,s ln Pr,s ≡ 〈ln Pr,s〉 = −αN − βE − βPV

= (µN − U − PV )/kT . (1)

Since µN = G = U + PV − TS , the right-hand side of (1) equals −S/k.Hence the result.

4.2. According to the grand canonical ensemble theory,

PV = kT ln

∑Nr

zNrQNr (V, T )

. (1)

Now, the largest term in the sum pertains to the value N∗, of Nr, whichis determined by the condition

∂

∂NrNr ln z − ln QNrNr=N∗ = 0.

By Sec. 3.3, this is equivalent to the statement: z = exp(µ∗/kT ), whereµ∗ is the chemical potential of the given system in a canonical ensemble(with N = N∗). If we replace the sum in (1) by its largest term, wewould get

PV ≈ N∗µ∗ −A∗ = P ∗V,

where P ∗ is the pressure of the system in the canonical ensemble (withN = N∗). How different would P be from P ∗ depends essentially onhow different the particle density n is from n∗ — a question thoroughlydiscussed in Sec. 4.5.

4.3. The probability distribution in question is the binomial distribution

P (N, V ) =N (0)!

N !(N (0) −N)!pN qN

(0)−N(p =

V

V (0), q = 1− V

V (0)

).

26

27

We note thatN(0)∑N=0

P (N, V ) = (q + p)N(0)

= 1.

For part (i), we have

N =

N(0)∑N=0

N P (N, V ) = N (0)p(q + p)N(0)−1 = N (0)p, while

N(N − 1) =

N(0)∑N=0

N(N − 1)P (N, V ) = N (0)(N (0) − 1)p2(q + p)N (0)−2 = N (0)(N (0) − 1)p2.

It follows that

N2 =N(N − 1) + N = (N (0)p)2 −N (0)p2 +N (0)p, whence

(∆N)2 ≡ N2 − N2 = N (0)p(1− p), etc.

For part (ii), we shift the origin to N = N (0)p, write

N = N (0)p+ x, N (0) −N = N (0)q − x

and examine the function

ln P (x) = ln N (0)!−ln(N (0)p+x)!−ln(N (0)q−x)!+(N (0)p+x) ln p+(N (0)q−x) ln q.

Since N (0)p and N (0)q are both 1, we apply Stirling’s formula, ln v ! ≈v ln v − v, and get (after some reduction)

ln P (x) ≈ −(N (0)p+ x) ln

(1 +

x

N (0)p

)− (N (0)q − x) ln

(1− x

N (0)q

).

For x N (0)p and N (0)q, we expand this expression in powers of x, withthe result that ln P(x) ≈ −x2/2N (0)pq . It follows that the distributionP (x), under the stated conditions, is a Gaussian, with (∆N)2 = N (0)pq .

For part (iii), we write

P (N) =N (0)(N (0) − 1) . . . (N (0) −N + 1)

N !pN (1− p)N

(0)−N .

Now, if p 1 and N N (0), we obtain the Poisson distribution

P (N) ≈ [N (0)]N

N !pNe−N

(0)p =(N)N

N !e−N ,

with (∆N)2 = N .

28

4.4. For obvious reasons,

P (Nr) =∑s

Pr,s =

e−αNr∑se−βEs

Q(α, β, V )=zNrQNr (V, T )

Q(z, V, T ). (1)

For an ideal classical gas, see Sec. 4.4,

z =Nλ3

V, QN =

1

N !

(V

λ3

)N, Q ≡ ePV /kT = eN . (2)

Substituting (2) into (1), we get

P (N) =(N)N

N !e−N ,

which is a Poisson distribution, with (∆N)2 = N .

We note that the variance of N , calculated from the general formula(4.5.3), also turns out to be the same:

(∆N)2 = −(∂N

∂α

)T,V

= z

(∂N

∂z

)T,V

= z

[∂

∂z

(zV

λ3

)]T,V

= zV

λ3= N .

4.5. The first term on the right-hand side of (4.3.20) may be written as

kT

(∂q

∂T

)z,V

= kT

[(∂q

∂T

)µ,V

+

(∂q

∂µ

)T,V

(∂µ

∂T

)z,V

]

= kT

(∂q

∂T

)µ,V

+ kT · NkT· k ln z (for µ = kT ln z ).

Equation (4.3.20) then reduces to

S = kT

(∂q

∂T

)µ,V

+ kq = k

[∂

∂T(Tq)

]µ,V

.

Note that this result is directly related to the formula, see Problem 1.16,

d(PV ) = PdV + Ndµ+ SdT ,

whence

S =

[∂

∂T(PV )

]µ,V

.

4.6. The Gibbs free energy is

G(N,P, T ) = −kT ln (YN (P, T )) .

29

For example(∂G

∂P

)N,T

=1

YN (P, T )

∫ ∞0

V Q(N,V, T )e−βPV = 〈V 〉.

The ideal gas gives

YN (P, T ) =1

(βPλ3)N+1

,

G(N,P, T ) ≈ NkT ln(βPλ3

),

V =

(∂G

∂P

)N,T

=NkT

P.

4.10. The partition function of the adsorbed molecules, assumed noninteracting,is given by

QN (N0, T ) = g(N)aN =N0!

N !(N0 −N)!aN [a = a(T )]. (1)

Using Stirling’s formula (B.29), we get

ln QN ≈ N0 ln N0 −N ln N − (N0 −N) ln(N0 −N) +N ln a,

with the result that

µ = −kT∂ ln QN

∂N= kT ln

N

(N0 −N)a. (2)

Alternatively, the grand partition function of the system consisting of allN0 sites (of which some are empty while others are occupied by a singlemolecule) is given by

Q(z,N0, T ) = [Q(z, 1, T )]N0 = [1 + za(T )]N0 ; (3)

see eqn. (4.4.15), with Nr = 0 or 1. Note that expression (3) could also be

obtained by using the standard definition Q(z, N0, T ) =N0∑N=0

zNQN (N0, T )

and employing expression (1) for QN . The mean value of N now turnsout to be

N = z∂

∂zln Q = N0

za

1 + za,whence z =

1

a

N

N0 − N, (4)

which agrees with (2).

4.11. By eqn. (4) of the preceding problem, the fraction θ of the adsorption sitesthat are occupied is given by

θ =N

N0=

za

1 + za, whence z =

1

a

θ

1− θ. (1)

30

Now, if the molecules in the adsorbed phase are in equilibrium with thosein the gaseous phase, then their fugacity z would be equal to the fugacityzg of the gaseous phase. The latter is given by eqns. (4.4.5 and 29),whereby

zg =PgkT

h3

(2πmkT )3/2. (2)

Equating (1) and (2), we obtain the desired result

Pg =θ

1− θ× 1

a(T )kT

(2πmkT )3/2

h3.

4.12. From eqn. (4.5.1), we get(∂N

∂β

)α,V

= −NE + NE.

The left-hand side here is equal to, see eqns. (4.5.3 and 12),

−kT 2

(∂N

∂T

)z,V

= −kT

(∂U

∂µ

)T,V

= −kT

(∂U

∂N

)T,V

(∂N

∂µ

)T,V

= −(∂U

∂N

)T,V

(∆N)2.

Hence the result.

4.13. With µ fixed (as it is in the grand canonical ensemble),

(∆J)2 = (∆E)2 − 2µ(∆E)(∆N) + µ2(∆N)2.

Substituting from eqn. (4.5.14) and from the previous problem, we get

(∆J)2 = 〈(∆E)2〉can +

(∂U

∂N

)2

T,V

− 2µ

(∂U

∂N

)T,V

+ µ2

(∆N)2,

which is the desired result.

4.14. The Clausius–Clapeyron equation (4.7.7) can be integrated to give

Pσ(T ) = Pσ(T0) exp

[L

k

(1

T0− 1

T

)],

where T0 = 373 K, Pσ(T0) = 1 atm and

L

k=

(2260 kJ/kg)(18 kg/kmol)

(6.02× 1026 kmol−1)(1.38× 10−23 J/K)= 4890 K.

31

This gives Pσ(273 K) ' 0.0082 atm and Pσ(473 K) ' 16 atm. The experi-mental values are 0.006 atm and 15.3 atm respectively.

4.15. The correct value for the latent heat of sublimation near the triple pointis 2833 kJ/kg. Following the solution to problem 4.14,

Pσ(T ) = Pσ(T0) exp

[L

k

(1

T0− 1

T

)],

where T0 = 273 K, Pσ(T0) = 612 Pa and

L

k=

(2833 kJ/kg)(18 kg/kmol)

(6.02× 1026 kmol−1)(1.38× 10−23 J/K)= 6138 K.

This gives Pσ(193 K) ' 0.055 Pa which corresponds nearly exactly withthe experimental value.

4.16. The slope of the melting line is

dPmdT

=Lm

T (∆v)' (80 cal/g)(4.18 J/cal)(106 cm3/m3)

(273 K)(−0.09 cm3/g)' −1.3× 107 Pa/K.

This gives Tm(100 atm) = −0.77C.

4.17. The slopes at the triple point are of the form(dPσdT

)ij

=si − sjvi − vj

=∆y

∆x,

so the vectors

[(v1 − v2)x+ (s1 − s2)y] + [(v2 − v3)x+ (s2 − s3)y] + [(v3 − v1)x+ (s3 − s1)y] = 0

sum to zero. This makes the third vector the negative of the sum of thefirst two vectors in each case, guaranteeing the stated geometry.

32

4.18. The liquid–vapor lines will appear much like in figure 6.2 but the liquidbranch will extend to P = 0. The upper end of the solid–liquid lines willappear as in figure 6.2 but the lines will end at Ps.

4.19. Since p1(µσ(T ), T ) = p2(µσ(T ), T ) on the coexistence line(∂p1

∂µ

)T

dµσdT

+

(∂p1

∂T

)µ

=

(∂p2

∂µ

)T

dµσdT

+

(∂p2

∂T

)µ

,

which gives

dµσdT

= − s1 − s2

n1 − n2=−LT∆n

.

4.20. The liquid–vapor lines will appear much like in figure 6.2 but with theliquid branch ending abruptly at Pt. The liquid side of solid–liquid lineswill start at Pt and extend upward as in figure 6.2 but the solid side ofthe solid–liquid transition will be to the right of the liquid line (since thesolid has lower density) and will extend to P = 0.

Chapter 5

5.1. On transformation, a given operator A would become

A′ = U AU−1 =

(1/√

2 1/√

2

−1/√

2 1/√

2

)(a11 a12

a21 a22

)(1/√

2 −1/√

2

1/√

2 1/√

2

).

Equations (2), (3) and (4) of Sec. 5.3 would then be replaced by

σ′x =

(1 00 −1

), σ′y =

(0 −ii 0

), σ′z =

(0 −1−1 0

),

ρ′ =1

eβµBB + e−βµBB

(cosh(βµBB) − sinh(βµBB)− sinh(βµBB) cosh(βµBB)

)and

〈σ′z〉 = Tr (ρ′σ′z) =1

eβµBB + e−βµBB· 2 sinh(βµBB) = tanh(βµBB),

with no change in the final result.

5.2. For a formal solution to this problem, see Kubo (1965), problem 2.32, pp.178–80.

5.4. If we use the unsymmetrized wave function (5.4.3), rather than the sym-metrized wave function (5.5.7), the density matrix of the system turns outto be, cf. eqn. (5.5.11),

〈1, . . . , N |e−βH |1′, . . . , N ′〉 =∑K

e−β~2K2/2muk1(1) . . . ukN (N)

u∗k1(1′) . . . u∗kN (N ′)

=

∑k1,...,kN

e−β~2(k21+...+k2N)/2m [uk1(1)u∗k1(1′)

. . .ukN (N)u∗kN (N ′)

]

=

N∏j=1

∑kj

e−β~2k2j/2m

ukj (j)u

∗kj (j

′) .

33

34

Replacing the summation over kj by an integration, one gets [see thecorresponding passage from eqn. (5.5.12) to (5.5.14)]

〈1, . . . , N |e−βH |1′, . . . , N ′〉 =

(m

2πβ~2

)3N/2

exp

− m

2β~2

(ξ21 + . . .+ ξ2

N

),

(1)

where ξj =∣∣rj − r′j

∣∣. The diagonal elements of the density matrix then are

〈1, . . . , N |e−βH |1, . . . , N〉 = (m/2πβ~2)3N/2 = 1/λ3N , (2)

where λ is the mean thermal wavelength of the particles. The structure ofexpressions (1) and (2) shows that there is no spatial correlation amongthe particles of this system.

The partition function now turns out to be

QN (V, T ) ≡ Tr(e−βH) =

∫1

λ3Nd3Nr =

vN

λ3N,

with no Gibbs’ correction factor.

5.5. By eqn. (5.5.17), we have

QN (V, T ) ≡ Tr(e−βH

)=

1

N !λ3NZN (V, T ),

where

ZN (V, T ) =

∫ ∑P

. . .d3Nr. (1)

In the zeroth approximation,∑P

= 1; see eqn. (5.5.19). So, ZN (V, T ) =

V N . In the first approximation,∑P

= 1±∑i<j

fij fji = 1±∑i<j

e−2πr2ij/λ2

. (2)

If λ is much smaller than the mean interparticle distance, we may write

∑P

≈∏i<j

(1± e−2πr2ij/λ

2)

=∏i<j

e−βvs(rij ) = exp

−β∑i<j

vs(rij )

,

which leads to the desired result.

For the second part, we substitute (2) into (1) and integrate over the posi-tion coordinates of the particles. We obtain, on assembling contributionsfrom all pairs of particles,

ZN (V, T ) = V N ± N(N − 1)

2· V N−2V · λ3

23/2.

35

The case N = 2 corresponds to eqn. (5.5.25) for Q2(V, T ). For N 1and Nλ3 V , we may write

ZN (V, T ) = V N[1±N2 λ3

25/2V

]≈ V N

[1± Nλ3

25/2V

]N.

It follows that

ln QN (V, T ) ≈ −N ln N +N +N ln

(V

λ3

)+N

(± Nλ3

25/2V

),

whence

P

kT≡(∂ ln QN

∂V

)N,T

≈ N

V∓ N2λ3

25/2V 2=

1

v∓ 1

25/2

λ3

v2,

where v = V/N ; cf. eqns. (7.1.13) and (8.1.17).

5.7 and 8. For solutions to these problems, consult the references cited in Notes 10and 11.

Chapter 6

6.1. We start with eqn. (6.1.19) and write it in the form

S = k∑i

[n∗i ln

(gin∗i

)+(n∗i −

gia

)ln

(1− an

∗i

gi

)]. (1)

Now, setting all gi = 1 and identifying (n∗i /gi) with 〈nε〉, see eqns. (6.1.18a)and (6.2.22), we get

S = k∑ε

[−〈nε〉 ln〈nε〉+

(〈nε〉 −

1

a

)ln(1− a〈nε〉)

]. (2)

Choosing a = −1 or +1, we obtain the desired results.

Next we have to verify that

S = −k∑ε

∑n

pε(n) ln pε(n)

= −k

∑ε

〈ln pε(n)〉. (3)

Substituting for pε(n) from eqn. (6.3.10) into (3) leads to the desiredresult (2), with a = −1; substituting from eqn. (6.3.11) instead leads tothe desired result (2), with a = +1.

6.2. In the B.E. case, see eqn. (6.3.10),

pε(n) = (1− r)rn [r = 〈nε〉/(〈nε〉+ 1); n = 0, 1, 2, . . .].

It follows that

〈nε〉 = (1− r)∞∑n=0

nrn = r/(1− r),

⟨n2ε

⟩= (1− r)

∞∑n=0

n2rn = r(1 + r)/(1− r)2, so that⟨n2ε

⟩− 〈nε〉2 = r/(1− r)2 = 〈nε〉+ 〈nε〉2. (1)

36

37

In the F.D. case, see eqn. (6.3.11),

⟨n2ε

⟩=

1∑n=0

n2pε(n) = pε(1) = 〈nε〉, so that⟨n2ε

⟩− 〈nε〉2 = 〈nε〉 − 〈nε〉2 (2)

In the M.B. case, see eqn. (6.3.12), one can readily see that

〈nε(nε − 1)〉 =∑n

n(n− 1)〈nε〉n

n!e−〈nε〉 = 〈nε〉2

∑n

〈nε〉n−2

(n− 2)!e−〈nε〉 = 〈nε〉2, so that⟨

n2ε

⟩− 〈nε〉2 = 〈nε〉. (3)

For the second part, we note, from eqn. 6.2.22, that

〈nε〉−1 = e(ε−µ)/kT + a.

Differentiating this result with respect to µ, we get

−〈nε〉−2

[∂〈nε〉∂µ

]T

= − 1

kTe(ε−µ)/kT = − 1

kT[〈nε〉−1 − a].

It follows that

kT

[∂〈nε〉∂µ

]T

= 〈nε〉 − a〈nε〉2. (4)

Comparing (4) with our previous results (1)–(3), and with formula (6.3.9),we infer that, quite generally,⟨

n2ε

⟩− 〈nε〉2 = kT [∂〈nε〉/∂µ]T .

6.3. Starting with eqn. (6.2.15), we now have

Q(z, V, T ) =∏ε

[ ∑nε=0

(ze−βε)nε

]=∏ε

[1− (ze−βε)`+1

1− ze−βε

],

so that

q(z, V, T ) =∑ε

[ln1− (ze−βε)`+1 − ln1− ze−βε];

cf. eqn. (6.2.17). It follows that

〈nε〉 = − 1

β

(∂q

∂ε

)z,T,all other ε

= − (`+ 1)(ze−βε)`(ze−βε)

1− (ze−βε)`+1+

ze−βε

1− ze−βε

=1

z−1eβε − 1− `+ 1

(z−1eβε)`+1 − 1

For ` = 1, we obtain the Fermi-Dirac result; for ` → ∞ and z−1eβε > 1[see eqn. (6.2.16a)], we obtain the Bose-Einstein result.

38

6.4. To determine the state of equilibrium of the given system, we minimizeits free energy, U − TS , under the constraint that the total number ofparticles, N , is fixed. For this, we vary the particle distribution from n(r)to n(r) + δn(r) and require that the resulting variation

δ(U − TS ) =e2

2

∫ ∫n(r)δn(r′) + n(r′)δn(r)

|r− r′|drdr′ + e

∫δn(r)ϕext(r)dr

+ kT

∫[1 + ln n(r)]δn(r)dr = 0,

while δN =∫δn(r)dr is, of necessity, zero. Introducing the Lagrange

multiplier λ, our requirement takes the form∫ [e2

∫n(r′)

|r− r′|dr′ + eϕext(r) + kT [1 + ln n(r)]− λ

]δn(r)dr = 0.

Since the variation δn(r) in this expression is arbitrary, the condition forequilibrium turns out to be

e2

∫n(r′)

|r− r|′dr′ + eϕext(r) + kT ln n(r)− µ = 0, (1)

where µ = λ− kT .

Introducing the total potential ϕ(r), viz.

ϕ(r) = ϕext(r) + e

∫n(r)′

|r− r′|dr′, (2)

condition (1) takes the Boltzmannian form

n(r) = exp[µ− eϕ(r)/kT ]. (3)

Choosing n(r) to be n0 at the point where ϕ(r) = 0, eqn. (3) may bewritten as

n(r) = n0 exp[−eϕ(r)/kT ]. (4)

With ϕext(r) given, the coupled equations (2) and (4) together determinethe desired functions n(r) and ϕ(r).

6.5. The (un-normalized) distribution function for the variable ε in this prob-lem is given by

f(ε)dε ∼ e−βεε1/2dε,

where use has been made of expression (2.4.7) for the density of states ofa free particle. It is now straightforward to show that

ε =β−5/2Γ(5/2)

β−3/2Γ(3/2)=

3

2βand ε2 =

β−7/2Γ(7/2)

β−3/2Γ(3/2)=

15

4β2.

39

It follows that

(∆ε)r.m.s.ε

≡

√(ε2 − ε2)

ε=

√(3/2β2)

(3/2β)=√

(2/3).

6.6. We have to show that, for any law of distribution of molecular speeds [say,F (u)du],

∞∫0

uF (u)du

∞∫0

F (u)du

·

∞∫0

u−1 F (u)du

∞∫0

F (u)du

≥ 1, i.e.

∞∫0

uF (u)du ·∞∫

0

u−1 F (u)du ≥

∞∫0

F (u)du

2

.

For this, we employ Schwarz’s inequality (see Abramowitz and Stegun,1964), b∫

a

f(x)g(x)dx

2

≤b∫a

[f(x)]2dx ·b∫a

[g(x)]2dx ,

which holds for arbitrary functions f(x) and g(x) — so long as the integralsexist; the equality holds if and only if f(x) = c g(x), where c is a constant.Now, with f(u) =

√uF (u) and g(u) =

√u−1F (u), we obtain the desired

result.

For the Maxwellian distribution,

F (u)du ∼ e− 12βmu2

u2du.

It is then straightforward to see, with the help of the formulae (B.13), that

〈u〉 =I3I2

=

(8

πβm

)1/2

and 〈u−1〉 =I1I2

=

(2βm

π

)1/2

,

whence 〈u〉〈u−1〉 = 4/π, in conformity with the inequality stated.

6.7. For light emitted in the x-direction, only the x-component of the molecularvelocity u will contribute to the Doppler effect. Moreover, for ux c, (ν − ν0)/ν0 ' ux/c, which means that (λ − λ0)/λ0 ' −ux/c. Now,the distribution of ux among the molecules of the gas is governed bythe Boltzmann factor exp

(− 1

2mu2x/kT

); the distribution of λ in the light

emerging from the window will, therefore, be determined by the factorexp

− 1

2mc2(λ− λ0)2/λ20kT

.

40

6.8. The partition function QN (β) = (1/N !)QN1 (β), where Q1(β) is given by

Q1(β) =1

h3

∫exp

−β(p2

2m+ mgz

)dpxdpydpzdxdydz

=

(2πm

βh2

)3/2

·A1− e−βmgL

βmg, (1)

A being the area of cross-section of the cylinder. In the limit L→∞,

Q1(β) =

(2πm

βh2

)3/2A

βmg∝ T 5/2. (2)

The thermodynamic properties of the system now follow straightforwardly.In particular, U turns out to be 5

2NkT and hence Cv = 52Nk . The extra

contribution comes from the potential energy of the system, which alsorises with T . Note, from eqns. (1) and (2), that the effective height of thegas molecules is (1 − e−βmgL)/βmg which for small heights is essentiallyL itself but for large heights is essentially kT/mg — making the totalpotential energy of the gas equal to NkT.

6.9. Correction to the first printing of third edition: the correct Hamiltonianis

H (pr, pθ, pz, r, θ, z) =p2r

2m+

(p2θ −mr2ω)2

2mr2+

p2z

2m− mr2ω2

2.

This gives for the partition function

Q1(V, T ) =2πH

λ3

∫ R

0

exp

(βmr2ω2

2

)rdr =

2πHkT

λ3mω2

[exp

(βmR2ω2

2

)− 1

]In the limit of small rotation rate, this becomes Q1 = πHR2/λ3 = V/λ3

as expected.

The density is determined from 〈δ(z − z1)δ(θ − θ1)δ(r − r1)/r〉. This gives

n(r) = n(0) exp

(βmω2r2

2

).

Since the 238UF6 molecules are heaver, their concentration is enhanced atr = R, while the concentration of the 235UF6 is enhanced near r = 0. Theratio at r = 0 is given by

n235(0)

n238(0)=m235N235

m238N238

[exp

(12βm238ω

2R2)− 1

exp(

12βm235ω2R2

)− 1

]

≈ m235N235

m238N238exp

[1

2β (m238 −m235)ω2R2

].

41

A value of ωR = 500 m/s gives a 16% enhancement compared to the inputfraction. Drawing the uranium hexafluoride gas from near the center ofthe cylinder results in a sample that is isotopically enhanced with 235Ucompared to the input concentration. This process may be repeated asoften as needed to achieve the isotopic fraction needed.

6.10. Consider a layer of the gas confined between heights z and z + dz . Forhydrostatic equilibrium, we must have

P (z + dz ) + ρgdz = P (z),

where ρ is the mass density of the gas. In differential form, one gets

dP/dz = −ρg = (−mg/kT )P. (1)

(a) If T is uniform, eqn. (1) can be readily integrated, with the result

ln P = −(mg/kT )z + const ., (2)

which yields the desired formula: P (z) = P (0) exp(−mgz/kT ).

(b) If, on the other hand, the equilibrium is attained adiabatically, thenT is related to P ; in fact, T ∝ P (γ−1)/γ . We now get

dT

T=γ − 1

γ

dP

P= −γ − 1

γ

mg

kTdz . (3)

This means that T now decreases essentially linearly with height.The pressure P and the density ρ go hand in hand with T — varyingas T γ/γ−1 and T 1/γ−1, respectively.

6.11. (a) For the given system,

f(p)dp = const .e−βε(p)(4πp2dp) = C e−βc(p2+m2

0c2)

1/2

p2dp.

The normalization constant C is determined by the condition∫f(p)dp = C

∞∫0

e−βc(p2+m2

0c2)

1/2

p2 dp = 1.

Substituting p = m0c sinh θ, we get for the left-hand side of thisequation

C

∞∫0

e−βm0c2 cosh θm3

0c3 sinh2 θ cosh θ dθ

= Cm30c

3

e−βm0c2 cosh θ

−βm0c2sinh θ cosh θ|∞0 +

∞∫0

e−βm0c2 cosh θ

βm0c3cosh(2θ)dθ

= Cm3

0c3 · (βm0c

2)−1K2(βm0c2).

Equating this result with 1, we obtain the desired expression for C.

42

(b) Using the limiting forms

K2(x) ≈

(π/2x)1/2e−x (x 1)

2/x2 (x 1),

we obtain, rather straightforwardly, the nonrelativistic and the ex-treme relativistic limits of the distribution.

(c) Since

u =dε

dp=m0c

2d(cosh θ)

m0cd(sinh θ)= c tanh θ,

〈pu〉 = C

∞∫0

m0c2 sinh θ tanh θ e−βm0c

2 cosh θm30c

3 sinh2 θ cosh θ dθ

= Cm40c

5

∞∫0

e−βm0c2 cosh θ sinh4 θ dθ.

Once again, integrating by parts (this time twice), we obtain

〈pu〉 = Cm40c

5 · 3(βm0c2)−2K2(βm0c

2).

Substituting for C, we obtain: 〈pu〉 = 3/β — regardless of the sever-ity of the relativistic effects and in conformity with the results ofSecs. 3.7 and 6.4.

6.12. Ordinarily, when a molecule is reflected from a stationary wall that isperpendicular to the z-direction, the z-component of its velocity u simplychanges sign, i.e. u′z = −uz. If the wall is receding at velocity v in thedirection of its normal, the above result changes to (u′z − v) = −(uz − v),so that u′z = −(uz − 2v). This results in a change in the translationalenergy of the molecule which, for small v, is given by

∆ε =1

2mu ′2z −

1

2mu2

z ' −2muzv.

If A is the area of the wall, the net change in the energy of the gas, in

43

time δt, is then given by, cf. eqn. (6.4.10),

δE = Aδt · n∞∫

ux=−∞

∞∫uy=−∞

∞∫uz=−∞

(−2muzv)uzf(u)duxduyduz

= −Avδt · n2π∫

φ=0

π/2∫θ=0

∞∫u=0

2mu2 cos2 θf(u)(u2 sin θ du dθ dφ)

= −δV · n1

3

∞∫0

mu2 f(u)(4π u2 du)

= −δV · 2

3nεk = −δV · 2

3

EkV, (1)

where Ek is the total kinetic energy of the gas. Note that, since the gascontinues to be in a state of (quasi-static) equilibrium, the change δE (eventhough it originates in the translational motion of the molecules) becomeseventually a change in the internal energy U of the gas (which may wellhave contributions from degrees of freedom other than translational). IfU = aEk, we may write

δEk =1

aδU = − 2

3aEk

δV

V. (2)

Next, since PV = (2/3)Ek, we get

δP

P+δV

V=δEkEk

= − 2

3a

δV

V. (3)

Re-arranging (3) and integrating it, we obtain the desired result.

In the extreme relativistic case, the factor 2/3 is replaced throughout by1/3, leading to the alternate value of γ.

6.13. We refer to expression (6.4.11) of the text. For part (a) of the question,we integrate only over u and ϕ, to get

dRθ = n(u/4π) · 2π sin θ cos θdθ =1

2nu sin θ cos θdθ.

For part (b), we integrate only over θ and ϕ, to get

dRu = nπ · f(u)u3du, where f(u)|M.B. =( m

2πkT

)3/2

e−mu2/2kT .

For part (c), we refer to expression (6.4.10) instead and get

RE = n( m

2πkT

)3/2∞∫

ux=−∞

∞∫uy=−∞

∞∫uz=√

2E/m

e−m(u2x+u2

y+u2z)/2kT uzduxduyduz

= n

(kT

2πm

)1/2

e−E/kT .

44

It follows that

RE(T2)

RE(T1)=

(T2

T1

)1/2

exp

−Ek

(1

T2− 1

T1

).

With T1 = 300K, T2 = 310K and E = 10−19 J , this ratio turns out tobe about 2.2.

6.14. (a) We start by calculating the kinetic energy associated with the z-component of the motion of the effused molecules. Proceeding asSection 6.4 of the text, we get [see eqn. (6.4.11)]

⟨1

2mu2

z

⟩=

1

2m〈u2 cos2 θ〉 =

1

2m

π/2∫0

∞∫0

(u3 cos3 θ)f(u)u2 sin θdudθ

π/2∫0

∞∫0

(u cos θ)f(u)u2 sin θdudθ

=1

4m〈u3〉〈u〉

;

note that the averages on the right-hand side are taken over the gasinside the vessel. It is not difficult to show, see the correspondingcalculation in Problem 6.6 and the formulae (B.13b), that

〈u3〉〈u〉

=I5I3

=4

βm,

so that⟨

12mu2

z

⟩, for the effused molecules, = 1/β = kT . The kinetic

energy associated with the x- and y-components of the molecularmotion will be the same as inside the vessel, viz. 1

2kT each. Itfollows that the mean energy ε of an effused molecule is 2 kT.

(b) Assuming quasi-static equilibrium, the relations E = (3/2)NkT andP = NkT/V will continue to hold for the gas inside the vessel. How-ever, in view of the result obtained in part (a), we shall also have

dE

dt≡ d

dt

(3

2NkT

)= 2kT

dN

dt.

It follows thatdT

T=

1

3

dN

Nand hence T ∝ N1/3;

it further follows that P ∝ N4/3.

As for explicit variations with t, we make use of eqn. (6.4.13) and write

dN

dt= −1

4an〈u〉 = −1

4

aN

V

(8kT

πm

)1/2

.

45

Combining the last two results, we get

dT

T= −1

3

a

V

(kT

2πm

)1/2

dt ,

so that T = T0(1+ct)−2, where c = (a/6V )(kT 0/2πm)1/2. The variationsof N and P with t follow straightforwardly.

6.15. If nH is the number of holes per unit area of the surface of the balloon (ofradius r), a the area of each hole and t the duration of the leak, then thetotal number of molecules leaking is given by

∆N =1

4nu · nH(4πr2)at

[u = (8kT/πm)1/2

].

The fraction of the molecules leaking is thus given by

∆N

N=

1

4V

(8kT

πm

)1/2

· nH(4πr2)at .

Since V = (4π/3)r3, we get

nH =∆N

N· r

3at

(2πm

kT

)1/2

.

Substituting the data given, we obtain: nH ' 187 holes/m2.

6.16. The rate of effusion of molecules from side A to side B, through a hole ofcross-section S, is given by the expression

RA→B =1

4nA

√8kTA

πmAS =

PA√2πmAkTA

S;

the same from side B to side A is given by

RB→A =1

4nB

√8kTB

πmBS =

PB√2πmBkTB

S.

In the stationary state, these two expressions will be equal — which leadsto the condition of dynamic equilibrium

PA/PB = (mATA/mBTB)1/2.

If the two gases are samples of the same gas, the condition simplifies to

PA/PB = (TA/TB)1/2.

46

6.18. The (un-normalized) velocity distribution for a pair of molecules is given by

F (u1,u2)d3u1d3u2 ∼ e−

12βm(u2

1+u22)d3u1 d

3u2.

We define the relative velocity, v, and the velocity of the centre-of-mass,V, in the usual manner, viz.

v = u2 − u1, V =1

2(u1 + u2).

This results in a new distribution for the variables v and V:

F (v,V)d3vd3V ∼ e− 14βmv2

d3v · e−βmV2

d3V.

It is now straightforward to show that 〈v〉 = (16/πβm)1/2 =√

2〈u〉, while〈v2〉 = (6/βm) = 2〈u2〉. The latter result implies that vr.m.s. =

√2 ur.m.s..

We note that, since

v2 = u21 + u2

2 + u22 − 2u1 · u2,

〈v2〉 = 2〈u2〉, regardless of the law of distribution of velocities — so longas it is isotropic, making 〈u1 · u2〉 = 0.

6.19. The (un-normalized) joint distribution for the molecular energies ε1 andε2 is

f(ε1, ε2)dε1dε2 ∼ e−β(ε1+ε2) ε1/21 ε

1/22 dε1dε2.

To obtain the desired distribution, we set ε2 = E − ε1 and integrate overall relevant values of ε1, with the result that

P (E)dE ∼ e−βEE∫

0

ε1(E − ε1)1/2dε1dE

∼ e−βEE2 dE ;

cf. eqns. (3.4.3) and (3.5.16), with N = 2. It is now straightforward tocheck that

〈E〉 =β−4Γ(4)

β−3Γ(3)=

3

β.

6.20. The relative fraction of the excited atoms in the given sample of the heliumgas would be 3e−βε1 , where

βε1 =hc

kTλ1' 38.22.

The desired fraction turns out to be extremely small — about 7× 10−17.

47

6.21. We extend the treatment of Problem 3.14 to the reaction AB + CD ↔AD + CB and obtain, in equilibrium,

nADnCB

nABnCD=fADfCB

fABnCD= K(T ).

For the given reaction,

K(T ) =f2

HD

fHH fDD,

where each f is a product of three factors — the translational, the rota-tional and the vibrational.

Now, for a heteronuclear molecule like HD we have, at high temperatures,

fHD ≈ V(mHDkT

2π~2

)3/2

· 2IHDkT

~2· kT

~ωHD,

while for a homonuclear molecule like HH we have instead

fHH ≈ V(mHH kT

2π~2

)3/2

· IHH kT

~2· kT

~ωHH;

see Note 11 of the text. It follows that, at high temperatures,

K(T ) ≈ 4m3

HD

m3/2HHm

3/2DD

· I2HD

IHH IDD· I2

HD

IHH IDD· ωHHωDD

ω2HD

. (1)

Assuming the internuclear distances to be the same, the I’s here will beproportional to the reduced masses of the molecules; the ω’s, on the otherhand, are inversely proportional to the square roots of the reduced masses.Accordingly,

I2HD

IHH IDD· ωHHωDD

ω2HD

=µ3

HD

µ3/2HHµ

3/2DD

=mHmD/(mH +mD)3(

12mH

)3/2 ( 12mD

)3/2 . (2)

At the same time,

m3HD

m3/2HHm

3/2DD

=(mH +mD)3

(2mH)3/2(2mD)3/2. (3)

Substituting (2) and (3) into (1), we see that K(T ) ≈ 4.

6.23. The potential V (r) is minimum at r = r0, which determines the equilib-rium value of r. Accordingly, the quantum of the rotational motion ofthe molecule is ~2/2I, where I = µr2

0. This gives for Θr the expression~2/2µr2

0k = ~2/mr20k because the reduced mass µ in this case is equal to

m/2. Substituting the given data, Θr turns out to be about 75 K. Thisgives a fairly clear idea of the “temperature range” where the rotational

48

motion of the hydrogen molecules begins to contribute towards the specificheat of the gas.

Next we expand V (r) in the neighborhood of r = r0 and write

V (r) = −V0 + (V0/a2)(r − r0)2 + . . . .

This gives an ω equal to (2V0/µa2)1/2 = (4V0/ma2)1/2 and hence a Θv

equal to ~(4V0/ma2)1/2/k. Substituting the given data, Θv turns out tobe about 6260 K. Again, this gives a fairly clear idea of the “temperaturerange” where the vibrational motion of the hydrogen molecules begins tocontribute towards the specific heat of the gas.

6.24. The effective potential of a diatomic molecule (including both rotationand vibration) is given by

V (r) = −V0 +1

2µω2(r − r0)2 +

~2

2µr2J(J + 1).

The equilibrium value of r is obtained by minimizing V (r), with the result

(req − r0) =~2

µ2ω2r3eq

J(J + 1) ' ~2

µ2ω2r30

J(J + 1).

It follows that

∆r0

r0' ~2

µ2ω2r40

J(J + 1) = 4

(Θr

Θv

)2

J(J + 1).

Using data from the preceding problem, we find that for a hydrogenmolecule the fractional change in r0 is O(10−3).

6.25. The occupation number NJ is proportional to (2J + 1) e−εJ/kT . It fol-lows that

N0

N2=

1

5e−(ε0−ε2)/kT ,

N1

N2=

3

5e−(ε1−ε2)/kT .

Substituting the given data, we get

N0

N2=

1

5e−1.086 ' 0.0675,

N1

N2=

3

5e−0.760 ' 0.2806;

in other words,

N0 : N1 : N2 :: 0.050 : 0.208 : 0.742.

6.29. The various contributions to the molar specific heat of the gas at 300 K are:

(i) translational — the amount being (3/2)R.

49

(ii) rotational — since the characteristic values of the parameter Θr inthis case are of the order of 10 K, these degrees of freedom maybe treated classically, which yields a contribution of (3/2)R; seeeqn. (6.5.42).

(iii) vibrational — here, the parameters Θv are such that the various con-tributions have to be calculated quantum-mechanically, using formula(6.5.44). We find that

Θ1,2

T' 16.00,

Θ3,4

T' 4.56,

Θ5

T' 16.37,

Θ6

T' 7.80,

with the result that only modes 3 and 4 make appreciable contributions tothe specific heat of the gas; it turns out that each of these contributions isabout 0.22R. The contribution from mode 6 is about 0.02R, while thosefrom modes 1,2 and 5 are entirely negligible.

The net result is: 3.46R.

6.30. Equation (6.6.3) can be written∑α ναµα = 0 where the stioichiometric

coefficients are understood to be positive if they appear on the right handside of equation (6.6.1) and negative if on the left. Using equation (6.6.5)gives ∑

α

να[εα + kT ln

(nαλ

3α

)− kT ln jα

]= 0,

which gives∑α

να

[εα + kT ln

(n0λ

3α

)− kT ln jα + kT ln

(nαn0

)]= 0.

Rearranging gives∑α

να ln

(nαn0

)= −β

∑α

να[εα + kT ln

(n0λ

3α

)− kT ln jα

]= −β

∑α

ναµ(0)α ,

where µ(0)α is the chemical potential of species α at temperature T and

standard density n0. Equation (6.6.6) follows from exponentiating bothsides.

6.31. Equation (6.6.11) gives

[CO]

[CO2]=

√1

K(T ) [O2]

50

where

K(T ) = exp(−2βµ

(0)CO2

+ 2βµ(0)CO + βµ

(0)O2

).

For the parameters given in the problem K(1500 K) = 4 × 1010 whichyields [CO] / [CO2] ' 5 × 10−5 = 50 ppm, while K(600 K) = 1.7 × 1040

which yields [CO] / [CO2] ' 7× 10−20 which yields a negligible [CO] con-centration.

6.32. The equilibrium constant for N2 + O2 → 2NO is

[NO]2

[N2] [O2]= K(T ) = e−β∆ε j2

NO

jN2jO2

λ3N2λ3

O2

λ6NO

.

The internal partition functions are of the form

j =

(TΘr

)for kT ~ω(

TΘr

) (kT~ω)

for kT ~ω

which leads to

K(T ) =

e−β∆ε(

ΘN2ΘO2

Θ2NO

)(303

283/2 323/2

)for kT ~ω

e−β∆ε(

ΘN2ΘO2

Θ2NO

)(303

283/2 323/2

)(ωN2

ωO2

ω2NO

)for kT ~ω

6.33. The equilibrium relation is

[CO2]H2O]2

[CH4][O2]2= K

Let [excess] be the initial excess amount of O2 above stoichiometry, and[unburned] be the unburned amount of CH4. Then

[CO2]H2O]2

[CH4][O2]2=

4 ([CH4]0 − [unburned])3

[unburned] ([excess] + 2[unburned])2 = K

Since K 1, at the stoichiometric point [excess] = 0 so

[unburned] ≈ [CH4]0K1/3

.

On the lean side of the stoichiometric point [excess] > 0 so

[unburned] ≈ 4[CH4]20[excess]K

.

Finally, on the rich side of the stoichiometric point [excess] < 0 so

[unburned] ≈ − [excess]

2.

51

6.34. Equation (6.6.3) gives µNa = µNa+ + µe where

µNa = −εb − kT ln 2 + kT ln(nNaλ

3Na

),

µNa+ = kT ln(nNa+λ3

Na

),

µe = −kT ln 2 + kT ln(neλ

3e

),

where εb is the ionization energy of Na. These lead to

nNa

nNa+ne= eβεbλ3

e.

If the total density is n0 = nNa +nNa+ , the ionized fraction f = nNa+/n0,and the system is charge neutral, then

1− ff2

= eβεbn0λ3e = s,

which has solution

f =

√1 + 4s− 1

2s.

Chapter 7

7.2. With N0 N , eqn. (7.1.8) reads

nλ3 = g3/2(z) = z + 2−3/2z2 + 3−3/2z3 + 4−3/2z4 + . . . , (1)

where n is the particle density. To invert this series, we write

z = c1(nλ3) + c2(nλ3)2 + c3(nλ3)3 + c4(nλ3)4 + . . . (2)

and substitute into (1). Equating coefficients of like powers of (nλ3) onthe two sides of the resulting equation, we get

1 = c1, 0 = c2 + 2−3/2c21, 0 = c3 + 2−3/2 · 2c1c2 + 3−3/2c31,

0 = c4 + 2−3/2(c22 + 2c1c3

)+ 3−3/2 · 3c21c2 + 4−3/2c41, . . . .

It follows that

c1 = 1, c2 = −2−3/2, c3 = (1/4)− 3−3/2,

c4 = 5.6−3/2 − 5.2−9/2 − (1/8), . . .

We now write eqn. (7.1.7) in the form

PV

NkT=

1

nλ3(z + 2−5/2z2 + 3−5/2z3 + 4−5/2z4 + . . .)

and substitute expression (2) into it. This leads to the desired result(7.1.13), with

a1 = c1 = 1, a2 = c2 + 2−5/2c21 = −2−5/2,

a3 = c3 + 2−5/2 · 2c1c2 + 3−5/2c31 = (1/8)− 2.3−5/2,

a4 = c4 + 2−5/2(c22 + 2c1c3

)+ 3−5/2 · 3c21c2 + 4−5/2c41

= 3.6−3/2 − 5.2−11/2 − (3/32),

in agreement with the values quoted in expressions (7.1.14).

52

53

7.3. By eqns. (7.1.24) and (7.1.26), nλ3 = g3/2(z) while nλ3c = ζ(3/2). It

follows that

T

Tc≡(λ

λc

)−2

=

(g3/2(z)

ζ(3/2)

)−2/3

.

The right-hand side of this equation may be approximated with the helpof formula (D.9), with the result that

T

Tc=

(ζ(3/2)− 2π1/2α1/2 + . . .

ζ(3/2)

)−2/3

≈ 1 +4π1/2α1/2

3ζ(3/2),

valid for α 1 and hence for T & Tc. The desired result now followsreadily.

7.4. By eqn. (7.1.7), P = cT 5/2g5/2(z), where c is a constant. Differentiatingthis result with respect to T at constant P, we get

0 =5

2cT 3/2g5/2(z) + cT 5/2 ∂g5/2(z)

∂z

(∂z

∂T

)P

,

so that (∂z

∂T

)P

= − 5

2T

g5/2(z)

∂g5/2(z)/∂z.

Using the recurrence relation (D.10), we get the desired result

1

z

(∂z

∂T

)P

= − 5

2T

g5/2(z)

g3/2(z). (1)

Now, CP = T (∂S/∂T )P,N and CV = T (∂S/∂T )V,N . In view of thefact that S, at constant N, is a function of z only, see eqn. (7.1.44a),we may write

CP = T

(∂S

∂z

)N

(∂z

∂T

)P

and CV = T

(∂S

∂z

)N

(∂z

∂T

)v

.

It follows that

γ =CPCV

=(∂z/∂T )P(∂z/∂T )v

.

Substituting from eqn. (1) above and from eqn. (7.1.36), we obtain thedesired result

CP /CV = (5/3)[g5/2(z)g1/2(z)/g3/2(z)2

].

For T Tc, which implies z 1, we recover the classical result: γ =5/3. As T → Tc, z → 1 and the function g1/2(z) diverges as α−1/2; seeeqn. (D.8). Along with it, both γ and CP diverge as (T − Tc)−1; see therelation established in Problem 7.3.

54

7.5. (a) We have to evaluate the quantities

κT =1

n

(∂n

∂P

)T

and κS =1

n

(∂n

∂P

)z

,

where n = N/V . For N0 N, n(T, z) = aT 3/2g3/2(z), where a is aconstant; see eqn. (7.1.8). It follows that

dn =3

2aT 1/2g3/2(z)dT + aT 3/2

1

zg1/2(z)

dz .

Similarly, since P = cT 5/2g5/2(z), where c is a constant,

dP =5

2cT 3/2g5/2(z)dT + cT 5/2

1

zg3/2(z)

dz .

The quantities κT and κS are then given by

κT =1

n

a

cT

g1/2(z)

g3/2(z)and κS =

1

n

3a

5cT

g3/2(z)

g5/2(z).

Since c = ak , the desired results follow readily.

Note that, as z → 1, κT diverges in the same manner as γ and CP .

(b) Since P = 2U/3V, (∂P/∂T )V = 2CV/3V . It follows that

CP − CV = TV · 1

nkT

g1/2(z)

g3/2(z)· 4C2

V

9V 2=

4C2V

9Nk

g1/2(z)

g3/2(z),

in agreement with eqn. (7.1.48a). The other result follows straight-forwardly.

7.6. For T > Tc, we employ expression (7.1.37) and write

1

Nk

(∂CV

∂T

)V

=∂

∂ ln z

(CV

Nk

)·(∂ ln z

∂T

)v

.

The first factor turns out to be

15

4

g3/2(z)g3/2(z)− g5/2(z)g1/2(z)

g3/2(z)2− 9

4

g1/2(z)g1/2(z)− g3/2(z)g−1/2(z)

g1/2(z)2

=3

2− 15

4

g5/2(z)g1/2(z)

g3/2(z)2+

9

4

g3/2(z)g−1/2(z)

g1/2(z)2.

The second factor is given by eqn. (7.1.36). Multiplying the two, we obtainthe desired result.

For T < Tc, we employ expression (7.1.31) instead. Since CV is nowproportional to T 3/2,

1

Nk

(∂CV

∂T

)V

=3

2T

CV

Nk,

55

which leads to the result quoted in the problem.

As T → Tc from above, the quantity under study approaches the limitingvalue

1

Tc

[45

8

ζ(5/2)

ζ(3/2)− 0− 27

8

ζ(3/2)2 · Γ(3/2)α−3/2

Γ(1/2)α−1/23

];

on the other hand, as T → Tc from below, we obtain simply

1

Tc· 45

8

ζ(5/2)

ζ(3/2).

The discontinuity in the slope of the specific heat curve at T = Tc is,therefore, given by

Nk

Tc· 27

8

ζ

(3

2

)2

· (π1/2/2)

π3/2=

27Nk

16πTc

ζ

(3

2

)2

.

7.7. Since P = 2U/3V, (∂2P/∂T 2)v = (2/3V )(∂CV/∂T )V. An explicit ex-pression for this quantity can be written down using the result quoted inProblem 7.6.

Next, since µ = kT ln z, we obtain using eqn. (7.1.36)(∂µ

∂T

)v

= k ln z +kT

z·(∂z

∂T

)v

= k ln z − 3

2kg3/2(z)

g1/2(z),(

∂2µ

∂T 2

)v

=

[k − 3

2kg1/2(z)g1/2(z)− g3/2(z)g−1/2(z)

g1/2(z)2

](∂ ln z

∂T

)v

=3k

4T

g3/2(z)

g1/2(z)− 9k

4T

g3/2(z)2g−1/2(z)

g1/2(z)3

Similarly, using a result from Problem 7.4, we obtain(∂2µ

∂T 2

)P

=15k

4T

g5/2(z)

g3/2(z)− 25k

4T

g5/2(z)2g1/2(z)

g3/2(z)3.

We also note, see eqns. (7.1.37) and (7.1.48b), that

CPNk

=25

4

g5/2(z)2g1/2(z)

g3/2(z)3− 15

4

g5/2(z)

g3/2(z).

It is now straightforward to see that the stated thermodynamic rela-tions are indeed satisfied. The critical behavior of these quantities is alsostraightforward to check.

7.8. One readily sees that

w2 =

(∂P

∂(nm)

)S

=1

mnκS,

56

where κS is the adiabatic compressibility of the fluid. Using a result fromProblem 7.5, we get for the ideal Bose gas

w2 =5kT

3m

g5/2(z)

g3/2(z).

Next,

〈u2〉 =

⟨2ε

m

⟩=

2

m

U

N=

3kT

m

g5/2(z)

g3/2(z);

see eqns. (7.1.8) and (7.1.11). Clearly, w2 = (5/9) < u2 >.

7.9. We start by calculating the expectation values of the quantities ε1/2 andε−1/2:

〈ε1/2〉 =

∞∫0

〈nε〉ε1/2a(ε)dε

∞∫0

〈nε〉a(ε)dε

, 〈ε−1/2〉 =

∞∫0

〈nε〉ε−1/2a(ε)dε

∞∫0

〈nε〉a(ε)dε

.

The integral in the denominator has been evaluated in Section 7.1; thosein the numerator can be evaluated like-wise, with the results

〈ε1/2〉 = (kT )1/2 Γ(2)g2(z)

Γ(3/2)g3/2(z), 〈ε−1/2〉 = (kT )−1/2 Γ(1)g1(z)

Γ(3/2)g3/2(z).

It follows that

〈u〉 =

√2

m〈ε1/2〉 =

√8kT

πm

g2(z)

g3/2(z), while

〈u−1〉 =

√m

2〈ε−1/2〉 =

√2m

πkT

g1(z)

g3/2(z).

Multiplying the last two expressions, we obtain the desired result.

For z → 0, we recover the classical result stated in Problem 6.6. Forz → 1, we encounter divergence of the quantity 〈u−1〉, which arises fromthe contribution made by the particles in the condensate (for which u = 0).

7.11. Under the conditions of this problem, the summation in eqn. (7.1.2) hasto be carried out over the states of the internal spectrum as well as overthe translational states. Expression (7.1.16) is then replaced by

Ne = (Ne)0 + (Ne)1 =V

λ3g3/2

exp

( µ

kT

)+V

λ3g3/2

exp

(µ− ε1

kT

).

The critical temperature Tc is then determined by the condition

V

λ3c

g3/2(1) +V

λ3c

g3/2(x) = N, where x = e−ε1/kTc . (1)

57

For x 1, g3/2(x) ' x and eqn. (1) gives

λ3c ' (V/N)[ζ(3/2) + x].

Comparing this with the standard result(λ0c

)3= (V/N)ζ(3/2), we get

TcT 0c

≡(λ0c

λc

)2

'[1 +

x

ζ(3/2)

]−2/3

' 1− 2/3

ζ(3/2)x ' 1− 2/3

ζ(3/2)e−ε1/kT0

c .

For x . 1, on the other hand, g3/2(x) ' ζ(3/2)−2π1/2(− ln x)1/2; eqn. (1)now gives

λ3c ' (2V/N)[ζ(3/2)− π1/2(ε1/kT c)

1/2], whence

TcT 0c

'

2

[1− π1/2

ζ(3/2)

(ε1

kT c

)1/2]−2/3

' 2−2/3

[1 +

2

3

π1/2

ζ(3/2)21/3

(ε1

kT 0c

)1/2].

7.12. The relative mean-square fluctuation in N is given by the general for-mula (4.5.7),

(∆N)2

N2=

kT

VκT , (1)

while κT for the ideal Bose gas is given in Problem 7.5. As T → Tc fromabove, the function g1/2(z) and, along with it, both κT and the relativefluctuation in N diverge!

The mean-square fluctuation in E is given by the general formula (4.5.14),viz.

(∆E)2 = kT 2CV + (∂U/∂N)T,V2(∆N)2. (2)

The first term in (2), for the ideal Bose gas, is determined by eqn. (7.1.37)and stays finite at all T. The second term can be evaluated with the helpof eqns. (7.1.8 and 11), whereby(

∂U

∂N

)T,V

=

(∂g5/2(z)

∂g3/2(z)

)T,V

=g3/2(z)

g3/2(z). (3)

The second term in (2) is, therefore, inversely proportional to g1/2(z) andhence vanishes as T → Tc; this happens because the energy associatedwith the Bose condensate (which is, in fact, the component responsiblefor the dramatic rise in the fluctuation of N) is zero. Thus, all in all, therelative fluctuation in E is negligible at all T.

7.13. It is straightforward to see that for a Bose gas in two dimensions

Ne =

∞∫0

1

z−1eβε−1

A · 2πp dp

h2=A · 2πmkT

h2

∞∫0

dx

z−1ex − 1=A

λ

2

g1(z),

58

whileN0 =

z

1− z.

Since Bose-Einstein condensation requires that z → 1, the critical tem-perature Tc, by the usual argument, is given by(

N

A

)λ2c = g1(1) =∞ [for g1(z) = − ln(1− z)].

It follows that Tc = 0.

More accurately, the phenomenon of condensation requires that both Neand N0 be of order N . This means that, while z ' 1, (1 − z) be oforder N−1 and hence λ2 be of order (A ln N /N). Since the ratio (A/N) ∼`2, the condition for condensation takes the form (λ2/`2) = O(ln N ). Itfollows that

T ≡ h2

2πmkλ2∼ h2

mk`21

ln N.

7.14. With energy spectrum ε = Aps, the density of states in the system isgiven by, see formula (C.7b),

a(ε)dε =V

hn2πn/2

Γ(n− 2)pn−1dp =

V

hn2πn/2

sAn/sΓ(n/2)ε(n/s)−1dε. (1)

This leads to the expression

N −N0 =V

hn2πn/2

sAn/sΓ(n/2)

∞∫0

ε(n/2)−1

z−1eβε − 1dε

=V

hn2πn/2Γ(n/s)

sAn/sΓ(n/2)(kT )n/sgn/s(z), (2)

while N0 = z/(1− z). Similarly,

P =1

hn2πn/2Γ(n/s)

sAn/sΓ(n/2)(kT )(n/s)+1g(n+s)+1(z). (3)

Next, following the derivation of eqn. (7.1.11), we get

U = kT 2

∂

∂T

(PV

kT

)z,v

=n

sPV , (4)

so that P = sU /nV .

The onset of Bose-Einstein condensation requires that z → 1 at a finitetemperature Tc. A glance at eqn. (2) tells us that this will happen onlyif n > s and that the critical temperature Tc will then be determined bythe equation

N =V

hn2πn/2Γ(n/s)

sAn/sΓ(n/2)(kT c)

n/sζ(ns

). (5)

59

For T < Tc, Ne will be equal to N(T/Tc)n/s while N0 will be given by

the balance (N −Ne).To study the specific heats we first observe, from eqns. (2)–(4), that forT > Tc (when N0 N)

U =n

sNkT · g(n/s)+1(z)/gn/s(z) (6)

Next, using eqns. (2) and (3), and the recurrence relation (D.10), we get

1

2

(∂z

∂T

)v

= −ns

1

T

gn/s(z)

g(n/s)−1(z)and

1

z

(∂z

∂T

)P

= −(ns

+ 1) 1

T

g(n/s)+1(z)

gn/s(z).

(7)It is now straightforward to show that

Cv

Nk=n

s

(ns

+ 1) g(n/s)+1(z)

gn/s(z)−(ns

)2 gn/s(z)

g(n/s)−1(z)(8)

and

CPNk

=(ns

+ 1)2 g(n/s)+1(z)2g(n/s−1)(z)

gn/s(z)3− ns

(ns

+ 1) g(n/s)+1(z)

gn/s(z). (9)

The limiting cases suggested in the problem follow quite easily.

7.15. The position and momentum representations of the Schrodinger equationafter the potential is turned off at time t = 0 is

− ~2

2m

∂2ψ

∂x2= i~

∂ψ

∂t,

p2

2mψ = i~

∂ψ

∂t.

The momentum representation is easily solved

ψ(p, t) = exp

(p2t

2i~m

)ψ(p, 0),

where

ψ(p, 0) =1√2π~

∫eipx/~ψ(x, 0)dx.

This leads to (suppressing the normalization factor)

ψ(p, t) ∼ exp

[− p2

2~2

(a2 +

i~tm

)].

Inverse Fourier transforming gives

ψ(x, t) =

√a

π1/4√a2 + i~t/m

exp

(−1

2

x2

a2 + i~t/m

).

60

This solves the Schrodinger equation and leads to the one-dimensionaldensity

|ψ(x, t)|2 =1

π1/2a

1√1 + (~t/ma2)2

exp

(− x2

a2(1 + (~t/ma2)2)

).

This gives the spatial distribution for one cartesian direction once younote that ~/ma2 = ω0. At long-time, the width of the distribution growslinearly in time.

7.16. The one-dimensional normalized joint momentum–position density at timet = 0 is given by

f(p, x, 0) =ω

2πkTexp

(−βp

2

2m− βmω2x2

2

).

After the potential is turned off at t = 0, the particles move ballisticallyso the density becomes

f(p, x, t) = f(p, x+ pt/m, 0) =ω

2πkTexp

(−βp

2

2m− βmω2(x+ pt/m)2

2

).

The spatial density is then given by

n(x, t) =

∫f(p, x+ pt/m, 0)dp =

ω

2πkT

√2πmkT

1 + ω2t2exp

(−1

2

βmω2x2

1 + ω2t2

).

The high-temperature limit of equation (7.2.15) is given by the first termin the series since at high temperature the chemical potential is large andnegative.

7.17. The ground state density at the center of the trap is N0/(π3/2a3); see

problem 7.15. Using N0/N = 1−(T/Tc)3, a =

√~/(mω), and kTc/(~ω) =

(N/ζ(3)1/3), we get

n(0)λ3 = 7ζ(3)1/2N1/2 1.

7.18. Integrating equation (7.2.15) gives∫nex(r)dr =

1

λ3

∞∑j=1

eβµj(kT )3/2

j3m3/2ω30

=

(kT

~ω0

)3 ∞∑j=1

eβµj

j3.

The excited particles can be counted using the density of states and theBose-Einstein factor,

Nex =

∫a(ε)

1

eβ(ε−µ) − 1dε =

(kT )3

2(~ω)3

∫x2∞∑j=1

e−xeβµjdx =

(kT

~ω0

)3 ∞∑j=1

eβµj

j3.

Above Tc when µ < 0 this counts all of the particles. Below Tc whenµ = 0, this counts the particles that are not in the ground state.

61

7.19. The density of states for a two-dimensional harmonic oscillator isa(ε) = ε/(~ω0)2 so the number particles in the trap is given by

N(T, µ) =

∫dε

ε

(~ω)2

1

eβ(ε−µ) − 1.

As T → Tc, µ→ 0 so

N =

∫dε

ε

(~ω)2

1

eβc(ε) − 1=

(kTc~ω0

)2 ∫xdx

ex − 1= ζ(2)

(kTc~ω0

)2

=π2

6

(kTc~ω0

)2

.

so kTc = ~ω√

6N/π2. The condensate fraction for T ≤ Tc isN0/N = 1 − (T/Tc)

2. For this two-dimensional theory to be valid, theoccupancy of the first excited z-state must be negligible which requires~ωz kTc ∼

√N~ω0, i.e. ωz

√Nω0.

7.20. By eqn. (3.8.14),

−kT ln Q1 = kT ln(eβ~ω/2 − e−β~ω/2) =~ω2

+ kT ln(1− e−β~ω).

Now, concentrating on the thermal part alone and utilizing eqn. (7.3.2),we get

A(V, T ) ≡ −kT ln Q(V, T ) =VkT

π2c3

∞∫0

ln(1− e−β~ω)ω2dω.

After an integration by parts, we obtain

A(V, T ) = − V ~3π2c3

∞∫0

ω3dω

eβ~ω−1= −π

2Vk4T 4

45~3c3;

cf. eqns. (7.3.17 and 18). We also get

S = −(∂A

∂T

)V

= −4A

Tand U = A+ TS = −3A = 3PV .

Other results of Sec. 7.3 follow straightforwardly.

7.21. Using expressions (7.3.12) and (7.3.23), we readily get

U

N=

π4

30ζ(3)kT ' 2.7 kT .

Note that the numerical factor appearing here is actually Γ(4)ζ(4)/Γ(3)ζ(3).

62

7.22. Since ω = 2πc/λ, the characteristic frequencies of the vibrational modes ofa radiation cavity (and hence the energy eigenvalues of these modes) areproportional to L−1, i.e. to V −1/3. Just as in Problem 1.7, we infer thatthe entropy of this system is a function of the combination (V 1/3U). Itthen follows that during an isentropic process the quantity (V 1/3U) staysconstant, i.e. (

1

3V −2/3dV

)U + V 1/3dU = 0.

Consequently, the pressure of the system is given by

P ≡ −(∂U

∂V

)S

=1

3

U

V.

7.24. The number density of photons in the cosmic microwave background(CMB) follows from equation (7.3.23)

n =2ζ(3)

π2

(kT

~c

)3

' 4.10× 108 m−3 ' 410. cm−3

The energy density is

u =π2

15

(kT )4

(~c)3' 4.17× 10−14 J/m3.

The entropy density is

s =4π2k

45

(kT

~c

)3

' 1.48× 109km−3 ' 2.04× 10−14 J/m3K.

7.25. According to Sec. 7.4,

CV(T ) =

∫ω

∂

∂T

~ω

e~ω/kT − 1

g(ω)dω, while CV(∞) =

∫ω

kg(ω)dω.

It follows that

∞∫0

CV(∞)− CV(T )dT =

∫ω

[kT − ~ω

e~ω/kT − 1

]∞0

g(ω)dω.

It is easy to show that

limT→∞

~ωe~ω/kT − 1

≈ kT − 1

2~ω;

see Section 3.8 as well as Fig. 3.4 of the text. The integral on the right-hand side then becomes ∫

ω

1

2~ω · g(ω)dω,

63

which is indeed equal to the zero-point energy of the solid.

The physical interpretation of this result lies in noting that the actualamount of heat required to raise the temperature of a solid is less thanthe value predicted classically because the solid already possesses a finiteamount of energy even at T = 0K.

7.26. Using the Debye spectrum (7.4.15), we have for the zero-point energy ofthe solid

ωD∫0

1

2~ω · 9N

ω3D

ω2dω =9

8N~ωD =

9

8NkΘD.

Indeed,

ω =

ωD∫0

ω · ω2dω

ωD∫0

ω2dω

=3

4ωD

and hence the mean energy per mode is equal to 12~ω = 3

8~ωD = 38kΘD.

7.27. We’ll show that if the entropy of a system is given by S = aVTn, where ais a constant, then the quantity (CP −CV) of that system is proportionalto T 2n+1. For the Debye solid, at T ΘD, this indeed is the case, theparameter n being equal to 3. Hence the stated result.

We know that

CP − CV = T

(∂P

∂T

)V

(∂V

∂T

)P

= −T(∂V

∂P

)T

(∂P

∂T

)2

V

.

Since

S ≡ −(∂A

∂T

)V

= aVTn,

we must haveA = −aVTn+1/(n+ 1) + f(V ),

where f(V ) is a function of V alone. It follows that

P ≡ −(∂A

∂V

)T

= aTn+1

n+ 1− f ′(V ),

so that (∂P

∂V

)T

= −f ′′(V ),

(∂P

∂T

)V

= aTn;

clearly, f ′′(V ) must be non-negative. We thus get

CP − CV = −T · −1

f ′′(V )(aTn)2 =

a2

f ′′(V )T 2n+1.

64

7.33. The specific heat of the system is given by the general expression (7.4.8),which may in the present case be written as

CV(T ) = k

ωD∫0

(~ω/kT )2e~ω/kT

(e~ω/kT − 1)2g(ω)dω. (1)

The mode density, g(ω), is given by the relation

g(ω)dω = 3 · V (4πp2dp)/h3,

where p = ~k = ~(A−1ω)1/s. It follows that

g(ω)dω = Cω(3/s)−1)dω [C = 3V/(2sπ2A3/s)]. (2)

Substituting (2) into (1) and introducing the variable x = ~ω/kT , we get

CV(T ) ∼ T 3/s

∫ x0

0

x(3/s)+1ex

(ex − 1)2dx

(x0 =

~ωDkT

).

At low temperatures, the upper limit of this integral may be replaced byinfinity — making the integral essentially T -independent; this leads to thedesired result CV ∼ T 3/s.

7.34. The mode density in this case is given by, see eqn. (C.7b),

g(ω)dω ∼ kn−1dk ∼ ωn−1dω.

The rest of the argument is similar to the one made in the previous prob-lem; the net result is that the specific heat of the given system, at lowtemperatures, is proportional to Tn.

It is not difficult to see that if the dispersion relation were ω ∼ ks and thedimensionality of the system were n, then the low-temperature specificheat of the system would be proportional to Tn/s.

7.35. The Hamiltonian of this system is given by eqn. (7.4.6); the partitionfunction then turns out to be, see eqn. (3.8.14),

Q = e−βΦ0

∏i

2 sinh

(1

2β~ωi

)−1

,

with the result that

A = −kT ln Q = Φ0 + kT∑i

ln2 sinh(~ωi/2kT ), and hence

P = −(∂A

∂V

)T

= −∂Φ0

∂V− 1

2~∑i

coth

(~ωi2kT

)· ∂ωi∂V

.

65

Recognizing that (i) the total vibrational energy U ′ of this system is givenby the expression

∑i

(12~ωi

)coth(~ωi/2kT ), see eqn. (3.8.20), and (ii) the

coefficient ∂ωi/∂V = −γωi/V , the expression for P may be written as

P =∂Φ0

∂V+ γ

U ′

V(U ′ = U − Φ0); (1)

see eqn. (7.4.7). With Φ0(V ) = (V − V0)2/2κ0V0, eqn. (1) takes the form

P = −V − V0

κ0V0+ γ

U ′

V. (2)

Now, the coefficient of thermal expansion of any thermodynamic systemis given by

α ≡ 1

V

(∂V

∂T

)P

= − 1

V

(∂V

∂P

)T

(∂P

∂T

)V

= κT

(∂P

∂T

)V

, (3)

where κT is the isothermal compressibility. In the present case, eqn. (2)gives

κ−1T ≡ −V

(∂P

∂V

)T

=V

κ0V0+ γ

U ′

V− γ

(∂U ′

∂V

)T

;

using the thermodynamic formula (∂U/∂V )T = T (∂P/∂T )v − P , whereU = Φ0 + U ′, we get

κ−1T =

V

κ0V0+ γ

U ′

V− γT

(∂P

∂T

)V

+ γP + γ∂Φ0

∂V.

Next, since (∂P

∂V

)V

= γCV

V, (4)

we get

κ−1T =

V

κ0V0+ (1 + γ)

[P +

V − V0

κ0V0

]− γ2TC V

V(5)

Under the conditions of the problem, all terms on the right-hand side of(5), except the first one, can be neglected; the term retained may also beapproximated by κ−1

0 — with the result that κT ≈ κ0. Equations (3) and(4) then lead to the desired result for α.

Finally, the quantity (CP − CV) is given by

CP − CV = T

(∂P

∂T

)V

(∂V

∂T

)P

= T

(∂P

∂T

)V

· αV ≈ γ2κ0TC 2V

V0.

Note that, at low temperatures, (Cp − Cv) ∼ T 7 — as in Problem 7.27.

66

7.36. For rotons, ε = ∆ + (p − p0)2/2µ. Therefore, u ≡ dε/dp = (p − p0)/µ.Consequently,

P =1

3n〈p(p− p0)/µ〉

=1

3n

∫e−(p−p0)2/2µkTp(p− p0)/µp2dp∫

e−(p−p0)2/2µkT p2dp.

Substituting p = p0 +√

2µkTx, we get

P =1

3n

∫e−x

2 (p0 +

√2µkT x

)3(2kT/µ)1/2 x dx∫

e−x2(p0 +

√2µkT x

)2dx

.

As explained in Section 7.6, these integrals are well-approximated byletting the range of x extend from −∞ to +∞; also remembering that√

2µkT p0, we get

P ' 1

3n

(2kT

µ

)1/23p2

0

√2µkT · (

√π/2)

p20 ·√π

= nkT .

7.37. Following Secs. 7.5 and 7.6, the free energy A(v) of a roton gas in massmotion is given by

A(v) = −NkT = −kT · Vh3

∫n(ε− v · p)2πp2 · sin θdpdθ.

As explained in Section 7.6, though rotons obey Bose-Einstein statistics,their distribution function is practically Boltzmannian; see eqns. (7.6.6and 7). We may, therefore, write

A(v) = −kT · Vh3

∫e−βε+βvp cos θ2πp2 sin θ dpdθ.

Integrating over θ, we get

A(v) = −kT · Vh3

∞∫0

e−βεsinh(βvp)

βvp4πp2 dp.

Integration over p is now carried out the same way as in eqn. (7.6.9); withappropriate approximation, we end up with the result

A(v) = A(0) sinh(βvp0)/(βvp0).

Next, the inertial density of the roton gas is given by

ρ(v) =1

v· 1

h3

∫n(ε− v · p)p cos θ 2πp2 sin θdpdθ

' 1

vh3

∫e−βε+βvp cos θ2πp3 cos θ sin θdpdθ

67

Integration over θ now gives

ρ(v) =1

vh3

∞∫0

e−βε

cosh(βvp)

βvp− sinh(βvp)

(βvp)2

4πp3 dp

=β

h3

∞∫0

e−βε(βvp) cosh(βvp)− sinh(βvp)

(βvp)34πp4dp.

Finally, integrating over p (under appropriate approximation) and com-paring the resulting expression with eqn. (7.6.19), we obtain

ρ(v) = ρ(0)3(βvp0) cosh(βvp0)− sin(βvp0)

(βvp0)3.

7.38. We write eqn. (7.6.17) in the form

ρ0 = − 4π

3h3

∞∫0

∂n(p)

∂p

(p4 dp

dε

)dp

and integrate it by parts, to get

ρ0 = − 4π

3h3

n(p)p4 dp

dε

∣∣∣∣∞0

−∞∫

0

n(p)d

dp

(p4 dp

dε

)dp

.The integrated part vanishes at both limits, and we are left with

ρ0 =4π

3h3

∞∫0

n(p)d

dp

(p4 dp

dε

)dp.

Comparing this with the standard result for the equilibrium number ofexcitations in the system, viz.

N =4πV

h3

∞∫0

n(p)p2 dp,

we obtain for the effective mass of an excitation

meff =ρ0V

N=

1

3

⟨1

p2

d

dp

(p4 dp

dε

)⟩.

For ideal-gas particles, ε = p2/2m; the effective mass then turns out to beprecisely equal to m. For phonons, ε = pc; we then get

(meff )ph = 4 < ε > /3c2,

in agreement with eqn. (7.5.15). Unfortunately, in the case of rotons thisexpression presents certain problems of analyticity at the point p = p0;we then resort to direct calculation — leading to eqn. (7.6.19), whereby(meff )rot ' p2

0/3kT .

Chapter 8

8.1. Referring to Fig. 8.11 and noting that the slope of the tangent at the pointx = ξ is −1/4, the approximate distribution is given by

f(x) =

1 0 ≤ x ≤ (ξ − 2)(ξ + 2− x)/4 (ξ − 2) ≤ x ≤ (ξ + 2)

0 (ξ + 0) ≤ x,

where x = ε/kT and ξ = µ/kT . Accordingly,

N = g · 2πV

h3(2m)3/2

∞∫0

n(ε)ε1/2dε = C

∞∫0

f(x)x1/2dx ,

where C = g(2πV/h3)(2mkT )3/2. After some algebra, one gets

N =1

5C(ξ+2)5/2−(ξ−2)5/2 =

2

3Cξ3/2

1 +

1

2ξ−2 + . . .

(ξ 1).

(1)Comparing (1) with eqn. (8.1.24), which may be written as

N =2

3C( εF

kT

)3/2

,

we get

ξ =εFkT

1− 1

3

(kT

εF

)2

+ . . .

. (2)

Similarly,

U = CkT

∞∫0

f(x)x3/2dx =1

35CkT(ξ + 2)7/2 − (ξ − 2)7/2

=2

5CkT ξ5/2

1 +

5

2ξ−2 + . . .

. (3)

Combining (1) and (3), and then making use of (2), we get

U =3

5NkT ξ

1 + 2ξ−2 + . . .

=

3

5NεF

1 +

5

3(kT/εF )2 + . . .

.

68

69

It follows that, at temperatures much less than εF /k,

CV = 2Nk(kT/εF ),

which is “correct” insofar as the dependence on T is concerned but isnumerically less than the true value, given by eqn. (8.1.39), by a factor of4/π2.

The reason for the numerical discrepancy lies in the fact that the presentapproximation takes into account only a fraction of the particles that arethermally excited; see Fig. 8.11. In fact, the ones that are not taken intoaccount have a higher ∆ε than the ones that are, which explains why themagnitude of the discrepancy is so large.

8.2. By eqns. (8.1.4) and (8.1.5), the temperature T0 is given by

T0 =

(N

gV f3/2(1)

)2/3(h2

2πmk

). (1)

At the same time, the Fermi temperature TF is given by, see eqn. (8.1.24),

TF ≡εFk

=

(3N

4πgV

)2/3h2

2mk. (2)

It follows thatT0

TF=

(4π

3 f3/2(1)

)2/31

π. (3)

Now, by eqn. (E. 16), f3/2(1) = (1–2−1/2)ζ(3/2) ' 0.765. Substitutingthis into (3), we get: T0/TF ' 0.989.

8.3. This problem is similar to Problem 7.4 of the Bose gas and can be donethe same way — only the functions gv(z) get replaced by fv(z).

To obtain the low-temperature expression for γ, we make use of expansions(8.1.30–32), with the result

γ =

1 +

5π2

8(ln z)−2 + . . .

1− π2

24(ln z)−2 + . . .

1 +

π2

8(ln z)−2 + . . .

−2

= 1 +π2

3(ln z)−2 + . . . ' 1 +

π2

3

(kT

εF

)2

.

8.4. This problem is similar to Problem 7.5 of the Bose gas and can be done thesame way. To obtain the various low-temperature expressions, we makeuse of expansions (8.1.30–32). Thus

κT =3

2n(kT ln z)

1− π2

24(ln z)−2 + . . .

1 +

π2

8(ln z)−2 + . . .

−1

=3

2n(kT ln z)

1− π2

6(ln z)−2 + . . .

.

70

We now employ eqn. (8.1.35) and get

κT =3

2nεF

1− π2

12

(kT

εF

)2

+ . . .

−11− π2

6

(kT

εF

)2

+ . . .

' 3

2nεF

1− π2

12

(kT

εF

)2, (1)

which is the desired result.

Similarly, using appropriate expansions, we get

κs =3

2n(kT ln z)

1− π2

2(ln z)−2 + . . .

' 3

2nεF

1− 5π2

12

(kT

εF

)2. (2)

Dividing (1) by (2), we obtain the low-temperature expression for γ, thesame as the one quoted in the previous problem; this also yields the desiredresult for (CP − CV)/CV, which is simply (γ − 1).

8.6. This problem is similar to Problem 7.8 of the Bose gas and can be donethe same way. In the limit z →∞, which corresponds to T → 0K,

w2 ≈ 2kT ln z/3m,

which tends to the limiting value 2εF /3m. Thus

w0 = (2εF /3m)1/2.

For comparison, the Fermi velocity uF = (2εF /m)1/2. It follows thatw0 = uF /

√3.

8.7. This problem is similar to Problem 7.9 of the Bose gas and can be donethe same way. At low temperatures, using formula (E. 17), we get

〈u〉〈u−1〉 =9

8

1 +

π2

3(ln z)−2 + . . .

1 +

π2

8(ln z)−2 + . . .

−2

=9

8

1 +

π2

12(ln z)−2 + . . .

' 9

8

1 +

π2

12

(kT

εF

)2

;

cf. Problem 6.6.

8.8. (i) Refer to eqns. (8.3.1 and 2) of the text and note that for silver ne =1, na = 4, a = 4.09 A, while m′ = me — giving εF = 5.49 eV andTF = 6.37 × 104 K. For lead, ne = 4, na = 4, a = 4.95 A, whilem′ = 2.1me — giving εF = 9.45 eV and TF = 10.96 × 104 K. Foraluminum, ne = 3, na = 4, a = 4.05 A, while m′ = 1.6me — givingεF = 11.63 eV and TF = 13.50× 104 K.

71

(ii) The nuclear radius for 80Hg200 is about 8.4 × 10−13 cm. Taking allthe nucleons together, this gives a particle density of about 8.06 ×1037 cm−3. Substituting this into eqn. (8.1.34), we get: εF = 3.7 ×107 eV and TF = 4.3× 1011 K.

(iii) For liquid He3, the particle density is about 1.59 × 1022 cm−3. Thisyields an εF of about 4.1× 10−4 eV and a TF of about 4.8 K.

8.9. By eqns. (8.1.4, 5 and 24), the Fermi energy εF is given by

εF =

3

4πf3/2(z)

2/3h2

2mλ2=

3π1/2

4f3/2(z)

2/3

kT .

With the help of Sommerfeld’s lemma (E.17), this becomes

εF = kT ln z

1 +

π2

8(ln z)−2 +

7π4

640(ln z)−4 + . . .

2/3

= kT ln z

1 +

π2

12(ln z)−2 +

π4

180(ln z)−4 + . . .

. (1)

To invert this series, we write

kT ln z ≡ µ = εF

1 + a2

(kT

εF

)2

+ a4

(kT

εF

)4

+ . . .

(2)

and substitute into (1), to get

1−a2

(kT

εF

)2

+(a2

2 − a4

)(kT

εF

)4

+. . . = 1+π2

12

(kT

εF

)2

+

(π4

180− π2

6a2

)(kT

εF

)4

+. . . .

Equating coefficients on the two sides of this equality, we get: a2 =−π2/12, a4 = −π4/80, . . .. Equation (2) then gives the desired result(8.1.35a).

Next, we have from eqns. (8.1.7) and (E.17)

U

N=

3

5kT ln z

1 +

5π2

8(ln z)−2 − 7π4

384(ln z)−4 + . . .

1 +π2

8(ln z)−2 +

7π4

640(ln z)−4 + . . .

−1

=3

5kT ln z

1 +

π2

2(ln z)−2 − 11π4

120(ln z)−4 + . . .

. (3)

72

Substituting from eqn. (8.1.35a) into (3), we get

U

N=

3

5εF

1− π2

12

(kT

εF

)2

− π4

80

(kT

εF

)4

+ . . .+

1 +π2

2

(kT

εF

)2

− π4

120

(kT

εF

)4

+ . . .

=3

5εF

1 +

5π2

12

(kT

εF

)2

− π4

16

(kT

εF

)4

+ . . .

. (4)

The specific heat of the gas is then given by

CV

Nk=π2

2

kT

εF− 3π4

20

(kT

εF

)3

+ . . . . (5)

We note that the ratio of the T 3-term here to the Debye expression (7.3.23)is (1/16)(ΘD/TF )3. For a typical metal, this is O(10−8–10−9).

8.10. This problem is similar to Problem 7.14 of the Bose gas and can be donethe same way.

Parts (i) and (ii) are straightforward. For part (iii), we have to show that

CPCV

= 1 +( sn

)2 CV

Nk

f(n/s)−1(z)

fn/s(z)=(

1 +s

n

) f(n/s)+1(z)f(n/s)−1(z)

fn/s(z)2, (1)

which can be done quite easily; see eqns. (7)–(9) of the solution to Prob-lem 7.14. For part (iv), we observe that, since the quantity S/N is afunction of z only, an isentropic process implies that z = const . Accord-ingly, for such a process,

VTn/s = const . and P/T (n/s)+1 = const .;

see eqns. (2) and (3) of the solution to Problem 7.14. Eliminating T amongthese relations, we obtain the desired equation of an adiabat. For part (v),we proceed as follows.

In this limit z → 0, eqn. (1) gives

CP /CV → 1 + (s/n). (1a)

For z 1, on the other hand, we obtain [see formula (E.17)]

CPCV

=

1 +

(ns

+ 1) ns

π2

6(ln z)−2 + . . .

1 +

(ns− 1)(n

s− 2) π2

6(ln z)−2 + . . .

×

1 +n

s

(ns− 1) π2

6(ln z)−2 + . . .

−2

= 1 +π2

3(ln z)−2 + . . . ' 1 +

π2

3(kT/εF )2,

regardless of the values of s and n.

73

8.11. For T TF , we get

CV

Nk' n

s,CP − CV

Nk' 1, so that

CPNk'(ns

+ 1).

For T TF , we obtain [see formula (E.17)]

CV

Nk=n

sln z

1 +

n

s

π2

3(ln z)−2 + . . .

− n

sln z

1 +

(ns− 1) π2

3(ln z)−2 + . . .

=n

s

π2

3(ln z)−1 + . . . ' n

s

π2

3

(kT

εF

).

To this order of accuracy, the quantity CP /Nk has the same value asCv/Nk . As for the difference between the two, we obtain

CP − CV

Nk' n

s

π4

9

(kT

εF

)3

,

consistent with the corresponding value of γ quoted in the previous prob-lem. The non-relativistic case pertains to s = 2 while the extreme rela-tivistic one pertains to s = 1.

8.12. For a Fermi gas confined to a two-dimensional region of area A,

N =A

λ2f1(zF ) =

A

λ2ln(1 + zF ), EF =

AkT

λ2f2(zF ), (1a,b)

while the corresponding results for the Bose gas are

N =A

λ2g1(zB) =

A

λ2ln(1− zB), EB =

AkT

λ2g2(zB). (2a,b)

Equating (la) and (2a), we get

1 + zF =1

1− zB, i.e. zF =

zB1− zB

or zB =zF

1 + zF.

Next, since z∂f2(z)/∂z = f1(z),

f2(zF ) =

∫ zF

0

1

zln(1 + z)dz =

∫ zF

0

1

1 + z+

1

z(1 + z)

ln(1 + z)dz .

The first part of this integral is readily evaluated; in the second part, wesubstitute z = z′/(1− z′), to get

f2(zF ) =1

2ln2(1+zF )−

∫ zF /(1+zF )

0

1

z′ln(1−z′)dz ′ =

1

2ln2(1+zF )+g2(zB).

Equations (1b) and (2b) then yield the desired result, viz.

EF (N,T ) =N2h2

4πmA+ EB(N,T ), whence CV(N,T )F = CV(N,T )B .

74

Letting T → 0, we recognize that the constant appearing in the aboveresult must be equal to EF (N, 0). To verify this, we note that, since theFermi momentum of the gas in two dimensions is given by the equationN = A ·πp2

F /h2, the Fermi energy is given by εF = p2

F /2m = Nh2/2πmA.The ground-state energy of the gas then follows readily:

EF (N, 0) =

∫ pF

0

p2

2m

A · 2πpdp

h2=A · πp4

F

4mh2 =N2h2

4πmA=

1

2NεF .

8.13. The Fermi energy of the gas is given by the obvious relation

N =

∫ εF

0

a(ε)dε. (1)

At the same time, the quantities N and U , as functions of µ and T , aregiven by the standard integrals

N =

∫ ∞0

a(ε)dε

eβ(ε−µ) + 1and U =

∫ ∞0

εa(ε)dε

eβ(ε−µ) + 1.

At low temperatures we employ formula (E.18), with x = βε and ξ = βµ,to obtain

N =

∫ µ

0

a(ε)dε+π2

6(kT )2

da(ε)

dε

ε=µ

+ . . .

'∫ εF

0

a(ε)dε+ (µ− εF )a(εF ) +π2

6(kT )2

da(ε)

dε

ε=εF

, (2)

U =

∫ µ

0

εa(ε)dε+π2

6(kT )2

a(ε) + ε

da(ε)

dε

ε=µ

+ . . .

'∫ εF

0

εa(ε)dε+ (µ− εF )εFa(εF ) +π2

6(kT )2

a(εF ) + εF

[da(ε)

dε

]ε=εF

.

(3)

Comparing (1) and (2), we obtain for the chemical potential of the gas

µ ' εF −π2

6

(kT )2

a(εF )

da(ε)

dε

ε=εF

, (4)

which leads to the desired result for µ.

Next, substituting (4) into (3), we obtain the remarkably simple expression

U ' U0 + (π2/6)k2T 2a(εF ),

whenceCV ' (π2/3)k2T a(εF ). (5)

75

It follows that

S =

∫ T

0

CVdT

T' (π2/3)k2T a(εF ). (6)

For a gas with energy spectrum ε ∝ ps, confined to a space of n dimensions,

a(ε)dε ∼ pn−1dp ∼ ε(n/s)−1dε.

By eqn. (1), the Fermi energy of the gas is given by

N =

∫ εF

0

Aε(n/s)−1dε =sA

nεn/sF =

sεFna(εF ).

Substituting this result into (5), we get

CV

Nk' n

s· π

2

3

(kT

εF

); (7)

cf. eqn. (8.1.39), which pertains to the case n = 3, s = 2. See alsoProblem 8.11.

8.14. In the notation of Sec. 3.9, the potential energy of a magnetic dipole inthe presence of a magnetic field B = (0 , 0 , B) is given by the expression−(gµBm)B, where m = −J, . . . ,+J . The total energy ε of the dipole isthen given by ε = (p2/2m′) − gµBmB , m′ being the (effective) mass ofthe particle; the momentum of the particle may then be written as

p = 2m′(ε+ gµBmB)1/2.

At T = 0, the number of such particles in the gas will be

Nm =4πV

3h32m′(εF + gµBmB)3/2

and hence the net magnetic moment of the gas will be given by

M =∑m

(gµBm)Nm =4πgµBV

3h3(2m′)3/2

∑m

m(εF + gµBmB)3/2.

We thus obtain for the low-field susceptibility (per unit volume) of thesystem

χ0 = LimB→0

(M

VB

)=

4πgµB3h3

(2m′)3/2 · 3

2gµBε

1/2F

J∑m=−J

m2

=2πg2µ2

B

3h3(2m′)3/2ε

1/2F J(J + 1)(2J + 1). (1)

By eqn. (8.1.24),

ε3/2F =

3n

4π(2J + 1)

h3

(2m′)3/2

(n =

N

V

). (2)

76

Substituting (2) into (1), we obtain the desired result

χ0 =1

2nµ∗2/εF

µ∗2 = g2µ2

BJ(J + 1).

With g = 2 and J = 1/2, we obtain: χ0 = (3/2)nµ2B/εF , in agreement

with eqn. (8.2.6).

The corresponding result in the limit T →∞ is given by

χ∞ =1

2nµ∗2/kT ;

see eqn. (3.9.26). We note that the ratio χ0/χ∞ = 3kT/2εF , valid for allJ .

8.15. We note that the symbol µ0(xN ) denotes the chemical potential (≡ kT ln z)of an ideal gas of xN “spinless” (g = 1) fermions. The corresponding fu-gacity z is determined by the equation

f3/2(z) = xNλ3/V. (1)

Differentiating (1) with respect to x, we get

∂f3/2(z)

∂ ln z

∂ ln z

∂x=Nλ3

V=

1

xf3/2(z).

It follows that∂µ0

∂x=

kT

x

f3/2(z)

f1/2(z).

Equation (8.2.20) then assumes the form stated in the problem.

At low temperatures, we get

χ =nµ∗2

kT· 3

2 ln z

1− π2

6(ln z)−2 + . . .

=3nµ∗2

2εF

1− π2

12

(kT

εF

)2

+ . . .

−11− π2

6

(kT

εF

)2

+ . . .

' χ0

1− π2

12

(kT

εF

)2. (8.2.24)

At high temperatures, on the other hand,

χ =nµ∗2

kT

z − 2−1/2z2 + . . .

z − 2−3/2z2 + . . .=nµ∗2

kT(1− 2−3/2z + . . .)

' χ∞(1− 2−5/2nλ3), (8.2.27)

where use has been made of eqn. (1), with f3/2(z) ' z and x = 1/2.

77

8.18. The ground-state energy of a relativistic gas of electrons is given by

E0 =8πV

h3

∫ pF

0

mc2[1 + (p/mc)21/2 − 1]p2 dp.

Making the substitution (8.5.9), we get

E0 =8πm4c5V

h3

∫ θF

0

(cosh θ − 1) sinh2 θ cosh θ dθ. (1)

Now the integral∫ θF

0

sinh2 θ cosh2 θ dθ =1

3sinh3 θ cosh θ

∣∣∣∣θF0

− 1

3

∫ θF

0

sinh4 θ dθ. (2)

Substituting (2) into (1) and making use of eqn. (8.5.12), we get

E0 =8πm4c5V

3h3sinh3 θF cosh θF − P0V −

8πm4c5V

3h3sinh3 θF ; (3)

note that the last term is simply Nmc2. Finally, using the definitionx = sinh θF , we obtain the desired result.

We observe that eqn. (3) can also be written as

E0 + P0V = Nmc2(cosh θF − 1) = NεF ≡ Nµ0.

To verify that the derivative (∂E0/∂V )N is equal to −P0, we have toshow that

[∂VB(x)/∂V ](Vx3) = −A(x), i.e. ∂x−3B(x)/∂x−3 = −A(x), i.e.

x4 ∂

∂x[8(x2 + 1)1/2 − 1 − x−3A(x)] = 3A(x), i.e.

∂A(x)/∂x = 8x4(x2 + 1)−1/2,

which can be readily verified with the help of expression (8.5.13).

8.19. Utilizing the result obtained in Problem 8.13, we have for a Fermi gas atlow temperatures

CV

Nk=π2

3

a(εF )

NkT . (1)

Now, the density of states for the relativistic gas is given by, see eqn. (8.5.7),

a(ε) =8πV

h3p2 dp

dε=

8πmV

h3p

1 +

( p

mc

)21/2

,

where p = p(ε). Substituting this result into (1) and making use ofeqn. (8.5.4), we get

CV

Nk=π2m

p2F

1 +

( pFmc

)1/2

kT ,

78

which leads to the desired result.

In the non-relativistic case(pF mc and εF = p2

F /2m), we obtain the fa-

miliar expression (8.1.39); in the extreme relativistic case (pF mc and ε =pc), we obtain

CV

Nk= π2

(kT

εF

),

consistent with expression (7) of the solution to Problem 8.13.

8.22. The number of fermions in the trap is

N(T, µ) =

∫dε ε2

2(~ω)3

1

eβ(ε−µ) − 1=

∫ εF

0

dε ε2

2(~ω)3=

ε3F

6(~ω)3.

Using kTF = εF this gives the following relation for the fugacity z = e−βµ,

3

(T

TF

)3 ∫x2dx

exe−βµ + 1= 1.

The internal energy is

U(T, µ) =

∫dε ε3

2(~ω)3

1

eβ(ε−µ) − 1=

(kT )4

2(~ω)3

∫x3

exe−βµ − 1.

When compared to the ground state energy U0 = (kTF )4/[8(~ω)3], we get

U

U0= 4

(T

TF

)4 ∫x3

exe−βµ − 1.

Chapter 9

9.1. Using the Friedmann equation (9.1.1)

da

dt=

√8πGu

3c2a,

and the connection between scale factor a and blackbody temperature T ,Ta = T0a0, along with (9.3.4b) we get

dT

dt= −

√8πGu

3c2T = −

√8π3Ggk4

45~3c5T 3,

where g = 43/8 is the effective number of relativistic species from equation(9.3.6b). The solution of the differential equation is

T (t) = T0

√t0T,

where

t0 =1

2

√45~3c5

8π3Gg(kT0)4' 0.99 s

for the case of T0 = 1010 K.

9.2. Just use equations (9.3.4) and (9.3.6) with T = 1010 K. The pressure

and energy density are of order 1025 J/m3, and the number density and

entropy divided by k are of order 1038 m−3.

9.3. The average kinetic energy per relativistic electron/positron is of the orderof ue/ne ∼ kT . The Coulomb energy per electron/positron is of the orderof uc ≈ e2/(4πε0a) where a ≈ (1/ne)

1/3 is of the order of the averagedistance between the charged particles. Using ne ∼ (kT/~c)3 we getuc/ue ∼ e2/(4πε0~c) ≈ 1/137. This is the justification for treating therelativistic electrons and positrons as noninteracting.

9.4. Correction to the first printing of third edition: The exponent in theresult should be −3/2. For βmc2 1 but before the time when the

79

80

electron density approaches the protron density, the density of electronsand positrons are almost identical so µ ≈ 0. Equation (9.5.6) gives

n−nγ≈ n+

nγ≈ 1

ζ(3)

∫ ∞βmc2

x√x+ βmc2

√x− βmc2dx

ex + 1

≈e−βmc

2 (βmc2

)3/22ζ(3)

∫ ∞0

√ye−ydy.

9.5. Correction to the first printing of third edition: The exponent in theresult should be −3/2. After the density of electrons levels off at thenearly the proton density, you can use equation (9.5.8) to show that thechemical potential µ− ≈ mc2. Then the positron number density is givenby equation (9.5.7),

n+

nγ≈ 1

ζ(3)

∫ ∞βmc2

x√x+ βmc2

√x− βmc2dx

ex+βmc2 + 1

≈e−2βmc2

(βmc2

)3/22ζ(3)

∫ ∞0

√ye−ydy

≈e−2βmc2

(βmc2

)3/2√π

4ζ(3).

9.6. Correction to the first printing of third edition: the energy density in thestatement of the problem should read

utotal = (1 + (21/8)(4/11)4/3)uγ .

After the electron–positron annihilation, the only relativistic species leftare the photons and the neutrinos. The factor 21/8 = (3)(1)(7/8) in theenergy is because there are three families of neutrinos, the spin degeneracyfactor is 1 (all left handed), and 7/8 is the Fermi-Dirac factor. The factor(4/11)4/3 is due to the lower temperature of the neutrinos compared tothe photons; see equation (9.6.4). Following the solution to problem 9.1,we get

t0 =1

2

√√√√ 45~3c5

8π3G(

1 +(

218

) (411

)4/3)(kT0)4

' 1.79 s.

9.7. If the current CMB temperature was 27K rather than 2.7K, the baryon-to-photon ratio would be 103 times smaller. Equation (9.7.8) implies thatthe nucleosynthesis temperature would have been about 20% lower whichwould have delayed the nucleosynthesis by an extra two minutes. Thiswould have given the neutrons a longer time to decay leading to q ≈ 0.10rather than 0.12, leading to a helium content in the universe of about 20%

81

by weight. If the current CMB temperature were 0.27K, that would haveincreased the baryon-to-photon ratio by a factor of 103. Fewer photons perbaryon would have led to an earlier nucleosynthesis, less time for neutronsto decay and an increase of the neutron fraction to q ≈ 0.135 leading toabout 27% helium content.

9.8. The strong interaction exhibits asymptotic freedom at high energies jus-tifying treating the quarks an gluons as noninteracting. The effectivenumber of species in equilibrium in these tiny quark–gluon plasmas isaccounted for using only the up and down quarks and the gluons. Pho-tons, and leptons, for example, easily escape without interacting with theplasma.

uu = 2

(7

8

)uγ2

uu = 2

(7

8

)uγ2

up quarks and antiquarks

ud = 2

(7

8

)uγ2

ud = 2

(7

8

)uγ2

down quarks and antiquarks

ug = (8)2uγ2

gluons

Therefore, the effective number of species is g = 8 + 28/8 = 23/2 and

uQGP = guγ . The energy density is 4 GeV/fm3

= 6.4× 1035 J/m3, so

kT '(

15(~c)3

gπ2

[4

GeV

fm3

])1/4

' 4× 10−11 J ' 250 MeV,

and T ' 3 × 1012 K. This is the record hottest temperature for mattercreated in the laboratory.

9.9. The strong interaction exhibits asymptotic freedom at high energies jus-tifying treating the quarks an gluons as noninteracting. The effectivenumber of species is much larger than during the time near t = 1s due to

82

the muons, quarks and gluons.

uγ = 2uγ2

photons

ue− = 2

(7

8

)uγ2

ue+ = 2

(7

8

)uγ2

electrons/positrons

uνe =

(7

8

)uγ2

uνe =

(7

8

)uγ2

electron neutrinos/antineutrinos

uνµ =

(7

8

)uγ2

uνµ =

(7

8

)uγ2

muon neutrinos/antineutrinos

uντ =

(7

8

)uγ2

uντ =

(7

8

)uγ2

tau neutrinos/antineutrinos

uµ− = 2

(7

8

)uγ2

uµ+ = 2

(7

8

)uγ2

muons/antimuons

uu = 2

(7

8

)uγ2

uu = 2

(7

8

)uγ2

up quarks/antiquarks

ud = 2

(7

8

)uγ2

ud = 2

(7

8

)uγ2

down quarks/antiquarks

ug = (8)2uγ2

gluons

The result is u = (149/8)uγ . Proceeding as in problem 9.1 we get

T (t) = 1010 K

√0.53 s

T.

Therefore at kT = 300 MeV (T ' 3.5 × 1012 K), the age of the universewas about 4× 10−6 s.

Chapter 10

10.1. By eqn. (10.2.3), the second virial coefficient of the gas with the giveninterparticle interaction would be

a2 = −2π

λ3

[∫ D

0

−1 · r2dr +

∫ ∞D

eε(σ/r)6/kT − 1r2dr

]

=2π

λ3

1

3D3 −

∫ ∞r0

∞∑j=1

1

j!

(εσ6

kT r6

)jr2dr

=

2πD3

3λ3

1−∞∑j=1

1

(2j − 1)j!

(εσ6

kT D6

)j ;

cf. eqn. (10.3.6). For the rest of the question, follow the solution toProblem 10.7.

10.2. For this problem, we integrate (10.2.3) by parts and write

a2λ3 = − 2π

3kT

∫ ∞0

e−u(r)/kT ∂u(r)

∂rr3 dr ;

cf. eqn. (3.7.17) and Problem 3.23. With the given u(r), we get

a2λ3 =

2π

3kT

∫ ∞0

e−A/kTrm eB/kTrn(

mA

rm−2− nB

rn−2

)dr

=2π

3kT

∫ ∞0

e−A/kTrm∞∑j=0

1

j!

(B

kT

)j (mA

rm−2+nj− nB

rn−2+nj

)dr

=2π

3kT

∞∑j=0

1

j!

(B

kT

)j AΓ

(m− 3 + nj

m

)(kT

A

)(m−3+nj )/m

− n

mBΓ

(n− 3 + nj

m

)(kT

A

)(n−3+nj )/m.

From the first sum we take the (j = 0)-term out and combine the remain-ing terms with the second sum (in which the index j is changed to j − 1);

83

84

after considerable simplification, we get

a2λ3 =

2π

3

(A

kT

)3/mΓ

(m− 3

m

)− 3

m

∞∑j=1

1

j!Γ

(nj − 3

m

)[B

kT

(kT

A

)n/m]j .

(1)

For comparison with other cases, we set A = A′rm0 and B = B′rn0 (sothat A′ and B′ become direct measures of the energy of interaction).Expression (1) then becomes

a2λ3 =

2π

3r30

(A′

kT

)3/mΓ

(m− 3

m

)− 3

m

∞∑j=1

1

j!Γ

(nj − 3

m

)[B′

kT

(kT

A′

)n/m]j .

(2)Now, to simulate a hard-core repulsive interaction, we let m → ∞, withthe result that

a2λ3 =

2π

3r30

1− 3

∞∑j=1

1

(nj − 3)j!

(B′

kT

)j . (2a)

With n = 6, expression (2a) reduces to the one derived in the precedingproblem. Furthermore, if terms with j > 1 are neglected, we recover thevan der Waals approximation (10.3.8).

For further comparison, we look at the behavior of the coefficient B2(≡a2λ

3) at high temperatures. While the hard-core expression (2a) predictsa constant B2 as T → ∞, the soft-core expression (2) predicts a B2 thatultimately vanishes, as T−3/m, which agrees qualitatively with the datashown in Fig. 10.2.

10.3. (a) Using the thermodynamic relation

CP − CV = T (∂P/∂T )V(∂V/∂T )P = −T (∂P/∂T )2V/(∂P/∂V )T

and the equation of state (10.3.9), we get

CP − CV

Nk= −T (∂P/∂T )2

v

k(∂P/∂v)T= − Tk/(v − b)2

k−kT/(v − b2) + 2a/v3=

1

1− 2a(v − b)2/kTv3.

(b) In view of the thermodynamic relation

TdS = CVdT + T (∂P/∂T )VdV

and the equation of state (10.3.9), an adiabatic process is character-ized by the fact that

CVdT + NkT (v − b)−1dv = 0.

Integrating this result, under the assumption that CV = const ., weget

TCV/Nk (v − b) = const .

85

(c) For this process we evaluate the Joule coefficient(∂T

∂V

)U

=(∂U/∂V )T(∂U/∂T )V

= −T (∂P/∂T )V − PCV

= −a/v2

CV= − N2a

CVV 2.

Now integrating from state 1 to state 2, we readily obtain the desiredresult.

10.4. Since, by definition,

α = v−1(∂v/∂T )P and B−1 ≡ κT = −v−1(∂v/∂P )T ,

we must have:[∂(αv)/∂P ]T = −[∂(vB−1)/∂T ]P . (1)

Using the given empirical expressions, we obtain for the left-hand side of(1) (

∂(αv)

∂P

)T

=1

T

(∂v

∂P

)T

= −vB−1

T= − 1

PT

(v +

a′

T 2

)and for the right-hand side

−(∂(vB−1)

∂T

)P

= − 1

P

[(∂v

∂T

)P

− 2a′

T 3

]= − 1

P

(αv − 2a′

T 3

)= − 1

P

(v

T+a′

T 3

).

The compatibility of the given expressions is thus established.

To determine the equation of state of the gas, we note from the givenexpression for α that(

∂v

∂T

)P

=v

T+

3a′

T 3, i.e.

[∂

∂T

( v

T

)]P

=3a′

T 4,

whencev

T= − a′

T 3+ f(P ), i.e. v = − a′

T 2+ Tf (P ), (2)

where f is a function of P only. We then obtain for B

vB−1 = −Tf ′(P ). (3)

Combining (2) and (3), we get

f ′(P )

f(P )= − vB−1

(v + a′/T )2= − 1

P.

It follows that f(P ) is proportional to 1/P and hence, by (2),

P = const . T (v + a′/T 2)−1.

86

10.5. The Joule-Thomson coefficient of a gas is given by(∂T

∂P

)H

= − (∂H/∂P )T(∂H/∂T )P

=1

CP

[T

(∂V

∂T

)P

− V]

=N

CP

[(∂v

∂T

)P

− v

].

By eqn. (10.2.1),

Pv

kT= 1 +

a2λ3

v+ . . . , so that

v =kT

P

(1 +

a2λ3P

kT+ . . .

)=

kT

P+ a2λ

3 + . . . .

It follows that

T

(∂v

∂T

)P

− v =

[T∂(a2λ

3)

∂T− a2λ

3

]+ . . .

and hence the quoted result for (∂T/∂P )H .

With the given interparticle interaction, eqn. (10.2.3) gives

a2λ3 = −2π

[∫ D

0

−1 · r2dr +

∫ r1

D

(eu0/kT − 1)r2dr

]

=2π

3

[D3 −

(r31 −D3

)eu0/kT

],

whence

T∂(a2λ

3)

∂T− a2λ

3 =2π

3

[(r31 −D3

) (1 +

u0

kT

)eu0/kT − r3

1

].

The desired result for (∂T/∂P )H now follows readily.

We note that the Joule-Thomson coefficient obtained here vanishes ata temperature T0, known as the temperature of inversion, given by theimplicit relationship (

1 +u0

kT 0

)eu0/kT0 =

r31

r31 −D3

.

For T < T0, (∂T/∂P )H > 0, which means that the Joule-Thomson ex-pansion causes a cooling of the gas. For T > T0, (∂T/∂P )H < 0; theexpansion now causes a heating instead.

10.7. To the desired approximation,

P

kT≡ 1

Vln Q =

1

λ3(z − a2z

2), n =N

V=

1

λ3(z − 2a2z

2), (1a,b)

where a2 is the second virial coefficient of the gas. It follows that

z = nλ3(1 + 2a2 · nλ3), whence P = nkT (1 + a2 · nλ3). (2a,b)

87

Next

A = NkT ln z − PV = NkTln(nλ3)− 1 + a2 · nλ3,G = NkT ln z = NkTln(nλ3) + 2a2 · nλ3,

S = −(∂A

∂T

)N,V

= Nk

5

2− ln(nλ3)− n ∂

∂T(Ta2λ

3)

;

remember that the coefficient a2 is a function of T . Furthermore,

U = A+ TS = NkT

3

2− nT

∂

∂T(a2λ

3)

,

H = U + PV = NkT

5

2− nT 2 ∂

∂T

(a2λ

3

T

),

CV =

(∂U

∂T

)N,V

= Nk

3

2− n ∂

∂T

(T 2 ∂

∂T(a2λ

3)

), and

CP − CV = −T(∂P/∂T )2

N,V

(∂P/∂V )N,T= Nk

1 + 2nT

∂

∂T(a2λ

3)

.

For the second part, use the expression for a2λ3 derived in Problem 10.5

and examine the temperature dependence of the various thermodynamicquantities.

10.8. We consider a volume element dx 1dy1dz 1 around the point P (x1, 0, 0) insolid 1 and a volume element dx 2dy2dz 2 around the point Q(x2, y2, z2) insolid 2. The force of attraction between these elements will be

−α(n dx 1dy1dz 1)(n dx 2dy2dz 2)`5

(x2 − x1)2 + y22 + z2

25/2

,

directed along the line joining the points P and Q. The normal componentof this force will be

−αn2(dy1dz 1)`5 (x2 − x1)

(x2 − x1)2 + y22 + z2

23 dx 1dx 2dy2dz 2.

The net force (per unit area) experienced by solid 1, because of attractionby all the molecules of solid 2, will thus be

− αn2`5∫ 0

x1=−∞

∫ ∞x2=d

∫ ∞ρ=0

(x2 − x1)

(x2 − x1)2 + ρ23dx 1dx 2 · 2πρdρ

= −παn2`5

2

∫ 0

x1=−∞

∫ ∞x2=d

1

(x2 − x1)3dx 1dx 2 =

παn2`5

4d,

i.e. inversely proportional to d.

88

10.9. For x 1, the spherical Bessel function j`(x) behaves like x`/1.3 . . . (2`+1) while the spherical Neumann function behaves like−1.3 . . . (2`−1)/x`+1;see Abramowitz and Stegun (1964). Substituting these results into eqn. (10.5.31),we readily obtain the desired result.

10.10. The symmetrized wave functions for a pair of non-interacting bosons/fermionsare given by

Ψα(r1, r2) =1√2V

(eik1·r1eik2·r2 ± eik1·r2eik2·r1).

The probability density operator W2 of the pair is then given through thematrix elements

〈1′, 2′|W2|1, 2〉 = 2λ6∑α

Ψα(1′, 2′)Ψ∗α(1, 2)e−βEα

=λ6

V 2

∑α

(eik1·r′1eik2·r′2 ± eik1·r′2eik2·r′1)×

(e−ik1·r1e−ik2·r2 ± e−ik1·r2e−ik2·r1)e−β~2(k21+k22)/2m

=λ6

2V 2

∑k1

∑k2

[eik1·(r′1−r1)eik2·(r′2−r2) + eik1·(r′2−r2)eik2·(r′1−r1)±eik1·(r′2−r1)eik2·(r′1−r2) ± eik1·(r′1−r2)eik2·(r′2−r1)

]e−β~

2k21/2me−β~2k22/2m

=1

2[〈1′|W1|1〉〈2′|W1|2〉+ 〈2′|W1|2〉〈1′|W1|1〉±

〈2′|W1|1〉〈1′|W1|2〉 ± 〈1′|W1|2〉〈2′|W1|1〉]= 〈1′|W1|1〉〈2′|W1|2〉 ± 〈2′|W1|1〉〈1′|W1|2〉.

Comparing this with eqn. (10.6.18), we obtain the desired result.

10.11. A particle with spin J can be in any one of the (2J + 1) spin statescharacterized by the spin functions χm(m = −J, . . . , J). For a pair ofsuch particles, we will have (2J + 1)2 spin states characterized by thesymmetrized spin functions

χm1(1)χm2

(2)± χm1(2)χm2

(1) (m1,2 = −J, . . . , J).

Of these, (2J + 1) functions, for which m1 = m2, can only be symmetric,for the corresponding antisymmetric combinations vanish identically. Theremaining 2J(2J + 1) functions, for which m1 6= m2, can be symmetricor antisymmetric; however, only half of them are linearly independentfunctions (because an interchange of the suffices m1 and m2 does notproduce anything new). Thus, in all, we have J(2J + 1) antisymmetricspin functions, and (J + 1)(2J + 1) symmetric spin functions, that arelinearly independent.

89

Now the total wave function of the pair will be the product of a sym-metrized space function (like the ones considered in the previous problem)and a symmetrized spin function (like the ones discussed above). For thetotal wave function to be symmetric, as required for a pair of bosons, wemay associate any of the (J + 1)(2J + 1) symmetric spin functions witha symmetric space function or any of the J(2J + 1) antisymmetric spinfunctions with an antisymmetric space function. This will lead to thequoted expression for the coefficient bs2. On the other hand, for the totalwave function to be antisymmetric, as required for a pair of fermions, wemay associate any of the J(2J + 1) antisymmetric spin functions with asymmetric space function or any of the (J + 1)(2J + 1) symmetric spinfunctions with an antisymmetric space function. This will lead to thequoted expression for the coefficient bA2 .

10.12. To derive the desired results, we make the following observations:

(i) Since a pair of particles with spin J has (2J+1)2 possible spin stateswhile a pair of spinless particles has only one, we have to divide theexpression for b2 pertaining to the former by (2J+1)2 so that we aretalking of the average contribution per state.

(ii) To make a transition from discreteness in orientation (that is associ-ated with a finite value of J) to continuity in orientation, we shouldtake the limit J →∞.

(iii) In view of the foregoing, the distinction between the original systembeing symmetric or antisymmetric is completely lost, and we are ledto the results quoted in the problem.

Next, using eqns. (10.5.28, 36 and 37), we obtain for the quantum-mechanicalBoltzmannian gas

b2 = −(D

λ

)1

− 3π

(D

λ

)3

+22π2

3

(D

λ

)5

+ . . . ,

which differs significantly from the corresponding classical result.

10.13 Expand the definition of the pair density n2(r, r′) in powers of the fugacityz using the grand canonical partition function and the Mayer functionsfij = exp (−βu(rij))− 1.

n2(r12) =1

Q(µ, V, T )

∞∑N=2

zN

(N − 2)!

∫dr3 · · · drN exp (−βu(r12)− βu(r13)− · · · )

= e−βu(r12)

[z2 + z3

∫(1 + f13 + f23 + f13f23) dr3 − z3Q1 + · · ·

]= e−βu(r12)

[(z2 + 2z3

∫f(r)dr

)+ z3

∫f13f23 dr3 + · · ·

]

90

Note every term includes the factor e−βu(r12). The coefficients of thoseterms are integrals over the Mayer functions that are continuous functionsof r12 even for the infinite step function potential; see equation (10.3.19)and discussion in Hansen and McDonald (1986) Chapter 5.

10.14 The pressure is given by

P

nkT= 1− n

2dkT

∫rg(r)

du

drdr,

where g(r) = y(r)e−βu(r). This gives

P

nkT= 1− n

2dkT

∫ry(r)

du

dre−βu(r)dr = 1 +

n

2d

∫ry(r)

d

dr

(e−βu(r)

)dr,

For the case of hard spheres,

d

dr

(e−βu(r)

)= δ(r −D),

so

P

nkT= 1 +

nDd

2dΩdy(D).

where Ωd is the area of the d-dimensional unit sphere. For hard spheresy(D) = g(D+). In three dimensions η = πnD3/6 and Ω3 = 4π, soP/(nkT ) = 1 + 4ηg(D+).

10.15 Let P (r) be the cumulative probability that no particles are closer thanr to a given particle. Breaking up the interval between zero and r intosmall intervals starting ar rk = k∆r with width ∆r gives

P (r) =

r/∆r∏k=0

(1− 4πng(rk)r2

k∆r),

since each factor represents the probability there are no neighbors in in-terval k. This gives

ln (P (r)) ≈r/∆r∑k=0

ln(1− 4πng(rk)r2

k∆r)≈ −

r/∆r∑k=0

4πng(rk)r2k∆r.

Therefore

P (r) = exp

(−4πn

∫ r

0

r2g(r)dr

).

Finally

w(r) = −dPdr.

For an ideal gas g(r) = 1, so the integrals are easily evaluated.

91

10.17 & 10.18. For a complete solution to these problems, see Landau and Lifshitz (1958),sec. 117, pp. 369–74.

10.19. (a) In this problem we are concerned with the integral

I =

∞∫0

∂u

dre−βur3dr .

Integrating by parts, we get

I = − 1

β

[e−βu + c

]r3 |∞0 +

1

β

∞∫0

(e−βu + c

)3r2 dr .

An arbitrary constant c has been introduced here to secure “properbehavior” at r = ∞. Since exp(−βu) → 1 as r → ∞, we choosec = −1. The integrated part then vanishes [assuming that u(r)→ 0faster than 1/r3], and we are left with the result

I =3

β

∞∫0

[e−βu − 1

]r2 dr .

This reduces eqn. (10.7.11) to the desired form.

(b) In the case of hard-sphere potential, the function f(r) = −1 for r ≤ σand 0 for r > σ. We then get

PV

NkT' 1 +

2πnσ3

3

(n =

N

V

).

For nσ3 1, we may write this result in the approximate form

PV

(1− 2πnσ3

3

)= NkT .

Comparison with Problem 1.4 shows that the parameter b of that prob-lem is equal to (2π/3)Nσ3, which is indeed four times the actual spaceoccupied by the particles.

10.20 Use

[κT (n, T )]−1

= n

(∂p

∂n

)T

P (n, T ) = n2

(∂f

∂n

)T

92

where f = A/N is the Helmholtz free energy per particle. Then

P (n, T ) = p(n0, T ) +

∫ n

n0

dn′

n′κ(n′, T ),

f(n, T ) = f(n0, T ) +

∫ n

n0

P (n′, T )

(n′)2dn′.

10.21 The most general Gaussian distribution of variables u1, · · · , uN is of theform

P (u1, · · · , uN ) ∼ exp(−1

2uTAu

)where A is a symmetric positive definite matrix. The matrix has only pos-itive eigenvalues and can be diagonalized into diagonal matrix (A = UT =U−1 using the orthogonal the matrix U . The eigenvalues λ1, · · · , λNare all positive and detU = 1. The normalization is

N =

∫dNu exp

(−1

2uTAu

)=

∫dNy exp

(−1

2yT Ay

)

=

√√√√(2π)NN∏i=1

λi =√

(2π)NdetA .

The transformed variables are y = UTu so the Jacobian is unity. Theintegral of the average of exp

(aTu

)can be determined from completing

the square inside the exponential,∫dNu exp

(aTu− 1

2uTAu

)= N exp

(1

2aTA−1a

).

Averaging the quantity(aTu

)2is accomplished by transforming to the y

variables which, using∫dxx2 exp(−λx2/2) = (2π)1/2/λ3/2,

gives∫dNu

(aTu

)2exp

(−1

2uTAu

)= N aTUA−1UTa = N aTA−1a.

10.22 The pressure is given by

p = −(∂A

∂V

)N,T

= n2

(∂A/N

∂n

)T

,

93

and the excess pressure is given by

P ex = Pcs − Pideal = nkT

(1 + η + η2 − η3

(1− η)3− 1

)= nkT

4η − 2η2

(1− η)3= n2

(∂Aex/N

∂n

)T

.

This can be integrated to give

βAex

N=

∫ η

0

4− 2η′

(1− η′)2dη′ =

3− 2η

(1− η)2− 3 =

4η − 3η2

(1− η)2.

10.23 The simplest rational approximations are

P

nkT=

1 + η/8

(1− η)2≈ 1 + 2η + 3.125η2 + 4.25η3 + 5.375η4 + 6.5η5

+ 7.625η6 + 8.75η7 + 9.875η8 + 11.000η9 + · · · ,

and

P

nkT=

1 + 0.128018 η

(1− η)2≈ 1 + 2η + 3.128018η2 + 4.256036η3 + 5.384054η4 + 6.512072η5

+ 7.64009η6 + 8.768108η7 + 9.896126η8 + 11.024144η9 + · · ·

The later gets the first two orders exactly correct, and the third and fourthorder coefficients correct to better that 1%.

Chapter 11

11.4. The relevant results for T < Tc are given in eqns. (11.2.13–15). Thecorresponding results for T > Tc follow from eqn. (11.2.10) by neglectingn0 altogether; we get, to the first order in a,

1

NA(N,V, T ) =

1

NAid(N,V, T ) +

4πa~2

mv, (13a)

P = Pid +4πa~2

mv2, (14a)

µ = µid +8πa~2

mv. (15a)

Remembering that vc ∝ T−3/2, the various quantities of interest turn outto be

CV = −T(∂2A

∂T 2

)N,V

= (CV)id +N2πa~2

mT

0 (T > Tc)(

− 32vc

+ 6vv2c

)(T < Tc)

K = −v

(∂P

∂v

)T

= Kid +2πa~2

m

4/v2 (T > Tc)

2/v2 (T < Tc)(∂2P

∂T 2

)v

=

(∂2P

∂T 2

)v,id

+2πa~2

mT 2

0 (T > Tc)

6/v2c (T < Tc)(

∂2µ

∂T 2

)v

=

(∂2µ

∂T 2

)v,id

+4πa~2

mT 2

0 (T > Tc)

3/4vc (T < Tc)

The thermodynamic relationship quoted in part (b) of the problem isreadily verified.

As for the discontinuities at T = Tc, we get (setting v = vc)

∆CV = N9πa~2

mT c

1

vc= Nk

9aλ2c

2

ζ(3/2)

λ3c

= Nk9a

2λcζ(3/2),

∆K = −4πa~2

m

1

v2c

,

∆

(∂2P

∂T 2

)v

=12πa~2

mT 2cv

2c

, ∆

(∂2µ

∂T 2

)v

=3πa~2

mT 2cvc

.

94

95

11.5. (a) We replace the sum over p appearing in eqn. (11.3.14) by an integral,viz.∫ ∞

0

ε(p)− p2

2m− 4πa~2N

mV+

(4πa~2N

mV

)2m

p2

V · 4πp2dp

h3.

Substituting p = (8πa~2N/V )1/2x, we get∫ ∞0

(4πa~2N

mV

)x(x2 + 2)1/2 − x2 − 1 +

1

2x2

4πV

h3

(8πa~2N

V

)3/2

x2dx ,

which readily leads us from eqn. (11.3.14) to (11.3.15). The resultingintegral over x can be done by elementary means, giving∫ ∞

0

x(x2 + 2)1/2 − x2 − 1 +

1

2x2

x2dx =

1

15(3x2 − 4)(x2 + 2)3/2 − 1

5x5 − 1

3x3 +

1

2x

∣∣∣∣∞0

.

For x 1,

1

15(3x2 − 4)(x2 + 2)3/2 =

1

15(3x5 − 4x3)

1 +

3

x2+

3

2x4+O

(1

x6

)=

1

5x5 +

1

3x3 − 1

2x+O

(1

x

).

The contribution from the upper limit is, therefore, zero. From thelower limit we get

√128/15, which leads to eqn. (11.3.16).

(b) Noting that

4πVp2dp

h3=

4πV

h3

(8πa~2N

V

)3/2

x2 dx = N

128

π(na3)

1/2

x2dx ,

we readily obtain eqn. (11.3.23). Now the integral∫ ∞0

x(x2 + 1)

(x2 + 2)1/2− x2

dx =

1

3(x2 − 1)(x2 + 2)1/2 − 1

3x3|∞0 .

Again, for x 1,

1

3(x2−1)(x2+2)1/2 =

1

3(x3−x)

1 +

1

x2+O

(1

x4

)=

1

3x3+O

(1

x

),

with the result that the contribution from the upper limit vanishes.From the lower limit we get

√2/3, which leads to eqn. (11.3.24).

11.6. We invert the given equation for n and write

µ0 =4πa~2n

m

[1 +

32

3π1/2(na3)1/2 + . . .

].

96

Substituting this into the given expressions for E0 and P0, we get

E0

V=

2πa~2n2

m

[1 +

32

3π1/2(na3)1/2 + . . .

]2 [1− 64

5π1/2(na3)1/2 + . . .

]=

2πa~2n2

m

[1 +

128

15π1/2(na3)1/2 + . . .

], and

P0 =2πa~2n2

m

[1 +

32

3π1/2(na3)1/2 + . . .

]2 [1− 128

15π1/2(na3)1/2 + . . .

]=

2πa~2n2

m

[1 +

64

5π1/2(na3)1/2 + . . .

],

in complete agreement with eqns. (11.3.16 and 17).

11.7. By eqns. (11.3.11), the number operator np for the real particles is given by

np = a+p ap =

1

1 + α2p

b+p bp − αp

(b−pbp + b+p b

+−p

)+ α2

pb−pb+−p

.

The terms linear in αp do not contribute to the expectation value of np

(with p 6= 0) because of the absence of the diagonal matrix elements inb−pbp and b+p b

+−p. Further, since b+p bp = Np and b−pb

+−p = N−p +1, we get

np =1

1− α2p

Np + α2

p(N−p + 1)

(p 6= 0).

Finally, in view of the isotropy of the problem, N−p = Np and we get thedesired result.

11.8. For a solution to this problem, see Feynman (1954).

11.10 and 11. For solutions to these problems, see Fetter (1963, 1965).

11.14. We set x = 1 + ε, where |ε| 1, and find that

2x4 lnx2

|x2 − 1|' 2(1 + 4ε) ln

1 + 2ε

2|ε|(1 + 1

2ε) ' 2(1 + 4ε)

− ln |ε| − ln 2 +

3

2ε

,

10

(x− 1

x

)ln

∣∣∣∣x+ 1

x− 1

∣∣∣∣ ' 20ε ln2 + ε

|ε|' 20ε− ln |ε|+ ln 2,

(2− x2)5/2

xln

(1 + x

√(2− x2)

1− x√

((2− x2)

)=

2(2− x2)5/2

xln

∣∣∣∣∣x+√

2− x2

x−√

2− x2

∣∣∣∣∣' 2(1− 2ε)5/2

1 + εln

∣∣∣∣ (1 + ε) + (1− ε− ε2)

(1 + ε)− (1− ε− ε2)

∣∣∣∣' 2(1− 6ε) ln

∣∣∣∣ 2

2ε+ ε2

∣∣∣∣ ' 2(1− 6ε)

− ln |ε| − 1

2ε

.

97

Substituting these results into the square bracket appearing in the formulafor ε(p), we get, to the desired degree of approximation,

11 + −2 ln |ε| − 2 ln 2 + 3ε− 8ε ln |ε| − 8ε ln 2− −20ε ln |ε|+ 20ε ln 2 − −2 ln |ε| − ε+ 12ε ln |ε|= (11− 2 ln 2)− 4(7 ln 2− 1)ε,

which yields the stated result.

Comparing this result with eqn. (11.8.10), we find that

V (p) ' const .− 8

15π2(7 ln 2− 1)

p3Fa

2

m~2(p− pF ).

Equation (11.8.11) then gives

1

m∗' 1

m

1− 8

15π2(7 ln 2− 1)(kFa)2

,

which leads to the desired result for the ratio m∗/m.

11.15. At T = 0K, the chemical potential of a thermodynamic system is given by

µ =

(∂E

∂N

)v

=∂(E/V )

∂(N/V ).

It follows that, in the ground state of the given system,

E = V

∫ n

0

µ(n)dn =N

n

∫ n

0

µ(n)dn

(n =

N

V

).

Now, since pF = (3π2n)1/3~, the given expression for µ may be written as

µ(n) ' (3π2n)2/3 ~2

2m+ (2πn)

~2a

m+ (3π2n)4/3 2

15π2(11− 2 ln 2)

~2a2

m.

It follows that

E

N' 3

5(3π2n)2/3 ~2

2m+

1

2(2πn)

~2a

m+

3

7(3π2n)4/3 2

15π2(11− 2 ln 2)

~2a2

m,

which agrees with eqn. (11.7.31).

11.16. For a complete solution to this problem, see the first edition of this book —Sec. 10.3, pages 311-5.

11.17. Correction to the first printing of third edition: In line 3, the definition of

the dimensionless wavefunction should read: ψ = a3/2osc Ψ/

√N . Using that

substitution and aosc =√~/(mω0) gives

−1

2∇2ψ +

1

2s2ψ +

4πNa

aosc|ψ|2ψ = µψ,

where s = r/aosc, ∇ = ∂/∂sx + ... and µ = µ/(~ω0).

98

11.18. The solution for the case V = 0 is Ψ =√N/V which gives µ = Nu0/V

and E = (2πa~2N2)/(mV ).

11.19. For the case a→ 0 the dimensionless G-P equation is

−1

2∇2ψ +

1

2s2ψ+ = µψ,

which has solution ψ = 1π3/4 exp

(− 1

2s2)

with E/(N~ω0) = 3/2, i.e. thezero point energy for N particles in the trap.

11.20. Use the dimensionless form from problem 11.17. Ignoring the kinetic en-ergy term

ψ =

√µ− s2

2√4πNa/a0

.

The normalization is

1 =4πa0

4πNa

∫ √2µ

0

(µ− s2

2

)ds

which gives

N =a0

15πa(2µ)5/2.

Using the definitions for u0, µ, and a0 gives equations (11.2.25) and(11.2.26). Equation (11.2.28) follows from the definition of the dimension-less length scale, and (11.2.27) comes from integrating the dimensionlessenergy in problem 11.17, again ignoring the kinetic energy term.

Chapter 12

12.1. We assume the equation of state to be

P =kT

v

(1− 1

2β1λ3

v− 2

3β2λ6

v2

), (1)

where β1 and β2 are certain functions of T . It follows that(∂P

∂v

)T

= −kT

v2+ β1

kTλ3

v3+ 2β2

kTλ6

v4, and(

∂2P

∂v2

)T

= 2kT

v3− 3β1

kTλ3

v4− 8β2

kTλ6

v5.

At the critical point, both these derivatives vanish — with the result that

(β1)cλ3c

vc+ 2(β2)c

λ6c

v2c

= 1 and 3(β1)cλ3c

vc+ 8(β2)c

λ6c

v2c

= 2,

whence(β1)c = 2vc/λ

3c and (β2)c = −v2

c/2λ6c . (2)

We infer that, at the critical point, β21 = −8β2.

Finally, substituting (2) into (1), we get(Pv

kT

)c

= 1− 1

2(β1)c

λ3c

vc− 2

3(β2)c

λ6c

v2c

= 1− 1 +1

3=

1

3.

12.2. The given equation of state is

P =kT

v − be−a/kTv. (1)

It follows that(∂P

∂v

)T

= kTe−a/kTv

− 1

(v − b)2+

1

v − b· a

kTv2

= P

− 1

(v − b)+

a

kTv2

,(

∂2P

∂v2

)T

=

(∂P

∂v

)T

− 1

(v − b)+

a

kTv2

+ P

1

(v − b)2− 2a

kTv3

.

99

100

At the critical point, both these derivatives vanish — with the result that

a

kT c=

v2c

vc − band

2a

kT c=

v3c

(vc − b)2,

whence vc = 2b and kT c = a/4b. Equation (1) then gives: Pc = (a/4b2)e−2

and hence kT c/Pcvc = e2/2 ' 3.695.

(a) For large v, the given equation of state may be approximated as

P =kT

v

(1− b

v

)−1

e−a/kTv ' kT

v

1 +

b

v− a

kTv

.

Comparing this with eqns. (10.3.7–10), we see that the coefficient B2

in the present case is formally the same as the one for the van derWaals gas, viz. b− (a/kT ).

(b) We note that the derivative (∂P/∂v)T for the Dietrici gas can bewritten as(

∂P

∂v

)T

= − kT

v2(v − b)2e−a/kTv

v2 − a

kT(v − b)

= − kT

v2(v − b)2e−a/kTv

(v − a

2kT

)2

+ab

kT 2 (T − Tc).

Clearly, if T > Tc, then (∂P/∂v)T is definitely negative; the same istrue at T = Tc — except for the special case v = a/2kT c = 2b when(∂P/∂v)T is zero. In any case, for all T ≥ Tc, P is a monotonicallydecreasing function of v — with the result that, for any given T andP , we have a unique v.

(c) For T < Tc, P is a non-monotonic function of v — generally decreas-ing with v but increasing between the values

vmin =a

2kT−√

ab

kT 2 (Tc − T ) and vmax =a

2kT+

√ab

kT 2 (Tc − T ).

For any given T , we now have (for a certain range of P ) three possiblevalues of v such that

v1 > vmax > v2 > vmin > v3;

see Figs. 12.2 and 12.3. We further note that

vmin

vc=

1

1 + (1− T/Tc)1/2and

vmax

vc=

1

1− (1− T/Tc)1/2.

Clearly, vmin < vc < vmax and, hence, v3 < vc < v1.

101

(d) To examine the critical behavior of the Dietrici gas, we write

P =a

4e2b2(1 + π), v = 2b(1 + ψ), T =

a

4bk(1 + t).

The equation of state then takes the form

1 + π =1 + t

1 + 2ψexp

[2

(1− 1

(1 + t)(1 + ψ)

)].

Taking logarithms, carrying out expansions and retaining the most impor-tant terms, we get

π ≈ t−(

2ψ − 2ψ2 +8

3ψ3

)+21−(1−t)(1−ψ+ψ2−ψ3) ≈ 3t−2

3ψ3−2tψ.

We now observe that

(i) at t = 0, π ≈ − 23ψ

3, while at ψ = 0, π ≈ 3t,

(ii) for t < 0, we obtain three values of ψ: while |ψ2| |ψ1,3|, implyingonce again π ≈ 3t, ψ1,3 ≈ ±(3|t|)1/2,

(iii) the quantity −(∂ψ∂π

)t≈ 1

2ψ2+2t ≈

1/2t (t > 0, ψ = 0)

1/4|t| (t < 0, ψ = ψ1,3).

Comparing these results with the ones derived in Sec. 12.2, we infer thatthe critical exponents of this gas are precisely the same as those of thevan der Waals gas; the amplitudes, however, are different.

12.3. The given equation of state (for one mole) of the gas is

P = RT/(v − b)− a/vn (n > 1). (1)

Equating (∂P/∂v)T and (∂2P/∂v2)T to zero, we get

vc =n+ 1

n− 1b and Tc =

4n(n− 1)n−1

(n+ 1)n+1

a

bn−1R.

Equation (1) then gives

Pc =

(n− 1

n+ 1

)n+1a

bn,

whence RT c/Pcvc = 4n/(n2 − 1).

To determine the critical behaviour of this gas, we write

P = Pc(1 + π), v = vc(1 + ψ), T = Tc(1 + t).

The equation of state then takes the form

1 + π =4n(1 + t)

(n2 − 1)(1 + ψ)− (n− 1)2− n+ 1

(n− 1)(1 + ψ)n.

102

Carrying out the usual expansions and retaining only the most importantterms, we get

π ≈ 2n

n− 1t− n(n+ 1)2

12ψ3 − n(n+ 1)

n− 1tψ.

It follows that

(i) at t = 0, π ≈ −n(n+ 1)2/12

ψ3, while at ψ = 0, π ≈ 2n/(n −

1)t,(ii) for t < 0, we obtain three values of ψ; while |ψ2| |ψ1,3|, implying

once again that π ≈ 2n/(n− 1)t, ψ1,3 ≈ ±12/(n2 − 1)1/2|t|1/2,

(iii) the quantity

−(∂ψ∂π

)t≈ 4(n−1)

n(n+1)(n2−1)ψ2+4t

≈

(n− 1)/n(n+ 1)t (t > 0)

(n− 1)/2n(n+ 1)|t| (t < 0).

Clearly, the critical exponents of this gas are the same as those of the vander Waals gas — regardless of the value of n. The critical amplitudes (aswell as the critical constants Pc, vc and Tc), however, do vary with n andhence are model-dependent.

12.4. The partition function of the system may be written as∑L

exp f(L), where

f(L) = lnN !− ln(Np)!− ln(Nq)! + βN

(1

2qJL2 + µBL

).

Using the Stirling approximation (B.29), we get

f(L) ≈ −Np ln p −Nq ln q + βN

(1

2qJL2 + µBL

).

With p and q given by eqn. (1) of the problem, the function f(L) ismaximum when

−1

2N(1 + ln p) +

1

2N(1 + ln q) + βN(qJL + µB) = 0.

Substituting for p and q, the above condition takes the form

1

2ln

1 + L

1− L= β(qJL + µB).

Comparing this with eqn. (12.5.10), we see that the value, L∗, of L, thatmaximizes the function f(L) is identical with L.

The free energy and the internal energy of the system are now given by

A ≈− kTf(L∗) ≈ NkT (p∗ ln p∗ + q∗ ln q∗)−N(

1

2qJL∗2 + µBL∗

),

U ≈− 1

2NqJL∗2 −NµBL∗,

103

whenceS ≈ −Nk(p∗ ln p∗ + q∗ ln q∗).

12.5. The relevant results of the preceding problem are

A

N= kT

1 + L∗

2ln

1 + L∗

2+

1− L∗

2ln

1− L∗

2

− 1

2qJL∗2 − µBL∗,

(1)

N+ =1

2N(1 + L∗), N− =

1

2N(1− L∗), (2)

where L∗ satisfies the maximization condition

1

2ln(1 + L∗)− ln(1− L∗) = β(qJL∗ + µB). (3)

Combining (1) and (3), we get

A

N=

1

2kT ln

1− L∗2

4+

1

2qJL∗2. (4)

Now, using the correspondence given in Section 11.4 and rememberingthat L∗ is identical with L, we obtain from eqns. (2) and (4) the desiredresults for the quantities P and v pertaining to a lattice gas.

For the critical constants of the gas, we first note from eqn. (12.5.13) thatTc = qJ/k, i.e. qε0/4k; the other constants then follow from the statedresults for P and v, with B = 0 and L = 0.

12.6. The Hamiltonian of this model may be written as

H = −1

2c∑i 6=j

σiσj − µB∑i

σi.

The double sum here is equal to∑i

σi∑j

σj −∑i

σ2i = (NL)2 −N which,

for N 1, is essentially equal to N2L2. It follows that asymptotically ourHamiltonian is of the form

H = −1

2(cN )NL2 − µBNL.

Now, this is precisely the Hamiltonian of the model studied in Prob-lem 12.4, except for the fact that the quantity qJ there is replaced bythe quantity cN here. We, therefore, infer that, in the limit N →∞ andc→ 0 (such that the product cN is held fixed), the mean-field approach ofProblem 12.4 would be exact for the present model — provided that thefixed value of the product cN is identified with the quantity qJ. It followsthat the critical temperature of this model would be cN/k.

104

12.7 & 8. Let us concentrate on one particular spin, s0, in the lattice and look at the

part of the energy E that involves this spin, viz. −2Jq∑j=1

s0 ·sj−gµBs0 ·H;

for notation, see Secs. 3.9 and 12.3. In the spirit of the mean field theory,we replace each of the sj by s, which modifies the foregoing expression to−gµBs0 ·Heff , where

Heff = H + H′ (H ′ = 2qJ s/gµB). (1)

We now apply the theory of Sec. 3.9. Taking H (and hence s) to be in thedirection of the positive z-axis, we get from eqn. (3.9.22)

µz = gµBsBs(x) [x = β(gµBs)Heff ], (2)

where Bs(x) is the Brillouin function of order s. At high temperatures(where x 1), the function Bs(x) may be approximated by (s+1)/3sx,with the result that

µz ≈g2µ2

Bs(s+ 1)

3kT

H +

2qJµzg2µ2

B

. (3)

The net magnetization, per unit volume, of the system is now given bythe formula M = nµz, where n(= N/V ) is the spin density in the lattice.We thus get from (3)

M

(1− Tc

T

)≈ CH

T, i.e. M ≈ CH

T − Tc, where (4)

Tc =2s(s+ 1)qJ

3k, C =

ng2µ2Bs(s+ 1)

3k. (5)

The Curie-Weiss law (4) signals the possibility of a phase transition as T →Tc from above. However, this is only a high-temperature approximation,so no firm conclusion about a phase transition can be drawn from it. Forthat, we must look into the possibility of spontaneous magnetization inthe system.

To study the possibility of spontaneous magnetization, we let H → 0 andwrite from (2)

µz = gµBsBs(x0) [x0 = β(gµBs)H′ = 2sqJµz/gµBkT ]. (6)

In the close vicinity of the transition temperature, we expect µz to bemuch less than the saturation value gµBs, so once again we approximatethe function Bs(x0) for x0 1. However, this time we need a betterapproximation than the one employed above; this can be obtained byutilizing the series expansion

cothx =1

x+x

3− x3

45+ . . . (x 1),

105

which yields the desired result:

Bs(x) =s+ 1

3sx− (s+ 1)s2 + (s+ 1)2

90s3x3 + . . . (x 1). (7)

Substituting (7) into (6), we get

µz =TcTµz −

b

g2µ2Bs

2

(TcTµz

)3

+ . . . , (8)

where Tc is the same as defined in (5), while b is a positive number givenby

b =3

10

1 +

s2

(s+ 1)2

. (9)

Clearly, for T < Tc, a non-zero solution for µz is possible; in fact, forT . Tc,

µz ≈ (gµBs/√b)(1− T/Tc)1/2. (10)

The long-range order L0 is then given by

L0 ≡ µz/(gµBs) ≈ (1/√b)(1− T/Tc)1/2. (11)

For T Tc, we employ the approximation

cothx ≈ 1 + 2e−2x,whence Bs(x) ≈ 1− s−1e−x/s (x 1). (12)

Equation (6) now gives

L0 ≈ 1− s−1 exp

(−2sqJ

kT

)= 1− s−1 exp

− 3Tc

(s+ 1)T

. (13)

For s = 1/2, expressions (11) and (13) reduce precisely to eqns. (12.5.14and 15) of the Ising model; the expression for Tc is different though.

12.9 & 10. We shall consider only the Heisenberg model; the study of the Ising modelis somewhat simpler. Following the procedure of Problem 12.7, we findthat the “effective field” Ha experienced by any given spin s on the sub-lattice a would be

Ha = H− 2q′J ′

gµBsb −

2qJ

gµBsa, (1a)

where q′ and q are, respectively, the number of nearest neighbors on theother and on the same sub-lattice, while J ′ and J are the magnitudes ofthe corresponding interaction energies. Similarly,

Hb = H− 2q′J ′

gµBsa −

2qJ

gµBsb. (1b)

106

The net magnetization, per unit volume, of the sub-lattices a and b athigh temperatures is then given by, see eqns. (2) and (3) of the precedingproblem,

Ma ≡1

2n(gµB sa) ≈ 1

2n · g

2µ2Bs(s+ 1)

3kT

H− 4q′J ′

ng2µ2B

Mb −4qJ

ng2µ2B

Ma

,

Mb ≡1

2n(gµB sb) ≈

1

2n · g

2µ2Bs(s+ 1)

3kT

H− 4q′J ′

ng2µ2B

Ma −4qJ

ng2µ2B

Mb

.

Adding these two results, we obtain for the total magnetization of thelattice

M ≈ C

TH− (γ′ + γ)M

[γ′ =

2q′J ′

ng2µ2B

, γ =2qJ

ng2µ2B

]; (2)

the parameter C here is the same as defined in eqn. (5) of the precedingproblem. Equation (2) may be written in the form

M ≈ C

T + θH, where θ = (γ′ + γ)C; (3)

this yields the desired result for the paramagnetic susceptibility of thelattice. Note that the parameter θ here has no direct bearing on theonset of a phase transition in the system; for that, we must examine thepossibility of spontaneous magnetization in the two sub-lattices.

To study the possibility of spontaneous magnetization, we let H → 0; inthat limit, the vectors Ma and Mb are equal in magnitude but oppositein direction. We may then write: Ma = M∗, Mb = −M∗, and studyonly the former. In analogy with eqn. (6) of the preceding problem, wenow have

M∗ =1

2n(gµBs)Bs(x0)

[x0 =

4s(q′J ′ − qJ )M∗

n gµBkT

]. (4)

We now employ expansion (7) of the preceding problem and get

M∗ =TNTM∗ − b(

12n gµBs

)2 (TNT M∗)3

+ . . . , (5)

where

TN =2s(s+ 1)

3k(q′J ′ − qJ ), (6)

while the number b is the same as given by eqn. (9) of the precedingproblem. Clearly, for T < TN , a nonzero solution for M∗ is possible. Notethat the Neel temperature TN = (γ′ − γ)C, which should be contrastedwith the parameter θ of eqn. (3); moreover, for the antiferromagnetictransition to take place in the given system, we must have q′J ′ > qJ , thephysical reason for which is not difficult to understand.

107

12.11. To determine the equilibrium distribution f(σ), we minimize the free en-ergy (E − TS ) of the system under the obvious constraint

∑σf(σ) = 1.

For this, we vary the function f(σ) to f(σ) + δf(σ) and require that theresulting variation

δ(E − TS ) =1

2qN

∑σ′,σ′′

u(σ′, σ′′)f(σ′)δf(σ′′) + f(σ′′)δf(σ′)

+ NkT∑σ′

1 + ln f(σ′)δf(σ′) = 0,

while∑σ′δf(σ′) is, of necessity, zero. Introducing the Lagrange multiplier

λ and remembering that the function u(σ′, σ′′) is symmetric in σ′ and σ′′,our requirement takes the form

∑σ′

qN∑σ′′

u(σ′, σ′′)f(σ′′) + NkT1 + ln f (σ′) − λ

δf(σ′) = 0.

Since the variation δf(σ′) in this expression is arbitrary, the condition forequilibrium becomes

qN∑σ′′

u(σ′, σ′′)f(σ′′) + NkT ln f (σ′)−Nµ = 0,

where µ = (λ − NkT )/N . By a change of notation, we get the desiredresult

f(σ) = C exp

[−βq

∑σ′

u(σ, σ′)f(σ′)

], (1)

where C is a constant to be determined by the normalization condition∑σf(σ) = 1.

For the special case u(σ, σ′) = −Jσσ′, where the σ’s can be either +1 or−1, eqn. (1) becomes

f(σ) = C exp[βqJσf(1)− f(−1)]. (2)

Writing f(σ) = 12 (1 + L0σ), the quantity f(1)− f(−1) becomes precisely

equal to L0, and eqn. (2) takes the form

f(σ) = C exp[βqJσL0]. (3)

From equation (3), we obtain

f(1) + f(−1) = 2C cosh(βqJL0) = 1, while (4)

f(1)− f(−1) = 2C sinh(βqJL0) = L0. (5)

Dividing (5) by (4), we obtain the Weiss eqn. (12.5.11) for L0.

108

12.12. The configurational energy of the lattice is given by

E =1

2qN [ε11 · xA(1 +X) · xA(1−X) + ε12xA(1 +X)(xB + xAX)

+ xA(1−X)(xB − xAX)+ ε22(xB − xAX)(xB + xAX)]

=1

2qN[(ε11x

2A + 2ε12xAxB + ε22x

2B

)− 2εx2

AX2] [

ε =1

2(ε11 + ε22)− ε12

].

The entropy, on the other hand, is given by

S = k

[ln

(1

2N

)!− ln

1

2NxA(1 +X)

!− ln

1

2N(xB − xAX)

!+

ln

(1

2N

)!− ln

1

2NxA(1−X)

!− ln

1

2N(xB + xAX)

!

]≈ 1

2Nk [−xA(1 +X) lnxA(1 +X) − (xB − xAX) ln(xB − xAX)

− xA(1−X) lnxA(1−X) − (xB + xAX) ln(xB + xAX)].

To determine the equilibrium value of X, we minimize the free energy ofthe system and obtain

∂(E − TS )

∂X= −2qN εx2

AX +1

2NkTxA ln

(1 +X)(xB + xAX)

(1−X)(xB − xAX)

= 0.

(1)Now, since xA + xB = 1, the argument of the logarithm can be writtenas (1 + z)/(1 − z), where z = X/(xB + xAX

2). Equation (1) then takesthe form

2qεxAX

kT=

1

2ln

1 + z

1− z= tanh−1 z, (2)

which is identical with the result quoted in the problem. For xA = xB = 12 ,

eqn. (2) reduces to the more familiar result

qεX

kT= tanh−1 2X

1 +X2= 2 tanh−1X, (2a)

leading to a phase transition at the critical temperature T 0c = qε/2k.

To determine the transition temperature Tc in the general case when xA 6=xB , we go back to eqn. (2) and write it in the form

X

xB + xAX2= tanh

(4xAT

0c

TX

). (3)

For small X, we get

1

xBX − xA

x2B

X3 + . . . =4xAT

0c

TX − 1

3

(4xAT

0c

T

)3

X3 + . . . , i.e.

1

xB

(4xAxBT

0c

T− 1

)X − 1

3x3B

[(4xAxBT

0c

T

)3

− 3xAxB

]X3 + . . . = 0.

109

It is now straightforward to see that for T < 4xAxBT0c , a non-zero solution

for X is possible whereas for T ≥ 4xAxBT0c , X = 0 is the only possibility.

The transition temperature Tc is, therefore, given by 4xA(1− xA)T 0c .

12.13. For a complete solution to this problem, see Kubo (1965), problem 5.6,pp. 335–7.

12.14. (a) Setting N++ + N−− + N+− = 12qN , we find that, in equilibrium,

γ = 1/(1 + sL2). So, in general, it may be written as 1/(1 + sL2).

(b) As in Problem 12.4, we write Q(B, T ) =∑L

exp f(L), where

f(L) = ln N !− ln N+!− ln N−!− βE, with

N+ =1

2N(1 + L), N− =

1

2N(1− L),

and E = −J(N++ +N−− −N+−)− µB(N+ −N−)

= −J · 1

2qN (L2 + s)/(1 + sL2)− µBNL.

The condition that maximizes f(L) now reads:

1

2ln

1 + L

1− L= β

[qJ

(1− s2)L

(1 + sL2)2+ µB

].

In the close vicinity of the critical point, L 1 — with the resultthat

1

2ln

1 + L

1− L' β

[qJ (1− s2)L+ µB

](T ' Tc).

Comparing this with the corresponding equation in the solution toProblem 12.4, we infer that the critical behavior of this model is qual-itatively the same as one encounters in the Bragg-Williams approxi-mation. Quantitatively, though, the effective spin-spin interaction isreduced by the factor (1 − s2) — leading to a critical temperatureTc = (1− s2)qJ/k, instead of qJ/k.

(c) As for the specific-heat singularity, the limit T → Tc− would be iden-tical with the one obtained in Sec. 12.5; see the derivation leadingto eqn. (12.5.18) and note that the replacement of J by (1 − s2)Jdoes not affect the final result 3

2Nk . For T > Tc, L0 is identi-cally zero. We are then left with a finite configurational energy,− 1

2qJNs, that arises from the (assumed) short-range order in thesystem; however, unlike in the Bethe approximation, this energy istemperature-independent and hence does not entail any specific heat.The singularity in question is, therefore, precisely the same as the oneencountered in Sec. 12.5 and depicted in Fig. 12.8.

110

12.15. Using eqn. (12.6.30), we get

S∞ − ScNk

=1

Nk

∫ ∞Tc

C(T )dT

T=

1

2q

∫ γc

0

γ sech 2γdγ

=1

2q(γc tanh γc − ln cosh γc).

Next, we use eqn. (12.6.11) and obtain

S∞ − ScNk

=1

2q

[1

2ln

(q

q − 2

)1

q − 1+

1

2ln

1−

(1

q − 1

)2]

=1

4

q

q − 1ln q − ln(q − 2)+

1

4qln q + ln(q − 2)− 2 ln(q − 1).

In view of the fact that S∞ = Nk ln 2, we finally get

ScNk

= ln 2− 1

4

q2

q − 1ln q +

1

2q ln(q − 1)− 1

4

q(q − 2)

q − 1ln(q − 2),

which leads to the desired result.

For q 1, the stated expression for Sc/Nk reduces to

ln 2 +q

2

−1

q+O

(1

q2

)− q

4

1 +O

(1

q

)−2

q+O

(1

q2

)= ln 2 +O

(1

q

),

which tends to the limit ln 2 as q →∞.

12.16. Using eqn. (12.6.14), we get

χ ≡(∂M

∂B

)T

=Nµ2

kT

(∂L

∂α

)T

=2Nµ2

kT

1 + exp(−2γ) cosh(2α+ 2α′)

cosh(2α+ 2α′) + exp(−2γ)21 +

(∂α′

∂α

)T

.

(1)

To determine (∂α′/∂α)T , we differentiate (12.6.8) logarithmically and ob-tain after some simplification(

∂α′

∂α

)T

=(q − 1)tanh(α+ α′ + γ)− tanh(α+ α′ − γ)

2− (q − 1)tanh(α+ α′ + γ)− tanh(α+ α′ + γ). (2)

Substituting (2) into (1) and letting α→ 0, we get

χ0 =4Nµ2

kT

1 + exp(−2γ) cosh(2α′)

cosh(2α′) + exp(−2γ)21

2− (q − 1)tanh(α′ + γ)− tanh(α′ − γ). (3)

111

To study the critical behavior of χ0, we let α′ → 0 and γ → γc. Usingeqn. (12.6.11), we see that, while the first two factors of expression (3)reduce to

4Nµ2

kT c

1

1 + exp(−2γc)=

2Nµ2

kT c

q

q − 1, (4)

the last factor diverges. To determine the nature of the divergence, wewrite γ = γc(1− t) and carry out expansions in powers of t and α′. Thus

tanh(γ ± α′) = tanh γc + sech 2γc (−γct± α′)

− sech 2γc tanh γc (−γct± α′)2

+ . . . ,

so that

tanh(γ + α′) + tanh(γ − α′) ≈ 2 tanh γc − 2 sech2γc · γct− 2 sech2γc tanh γc · α′2;

note that we have dropped terms of order t2 and higher. It now followsthat

2− (q − 1)tanh(γ + α′) + tanh(γ − α′)≈ 2(q − 1)sech2γc

(γct+ tanh γc · α′2

). (5)

Substituting (4) and (5) into (3), we finally obtain

χ0≈ Nµ2

kT c

1

(q − 2) (γct+ tanh γc · α′2). (6)

For t > 0, α′ = 0; eqn. (6) then gives

χ0≈ Nµ2

kT c

1

(q − 2)γct. (7a)

For t < 0, α′ is given by eqn. (12.6.13), whence α′2 ' −3(q − 1)γct; wenow get

χ0 ≈Nµ2

kT c

1

2(q − 2)γc|t|. (7b)

Note that, for large q, the quantity

(q − 2)γc =1

2(q − 2) ln

(q

q − 2

)= 1 +O

(1

q

);

eqns. (7) then reduce to eqns. (12.5.22) of the Bragg-Williams approxima-tion.

112

12.19. We refer to the solutions to Problems 12.4 and 12.5, whereby

ψ0 ≡(

A

NkT

)B=0

=1 +m0

2ln

1 +m0

2+

1−m0

2ln

1−m0

2− 1

2

qJ

kTm2

0

=1

2ln

1−m20

4+m0

2ln

1 +m0

1−m0− 1

2

TcTm2

0

= − ln 2 +1

2

(−m2

0 −m4

0

2− . . .

)+m0

(m0 +

m30

3+ . . .

)− 1

2(1− t+ . . .)m2

0

≈ − ln 2 +1

2tm2

0 +1

12m4

0.

Comparing this expression with eqn. (12.9.5), we infer that in the Bragg-Williams approximation r1 = 1/2 and s0 = 1/12. Now, substitutingthese values of r1 and s0 into eqns. (12.9.4, 9–11 and 15), we see that thecorresponding eqns. (12.5.14, 22, 24 and 18) are readily verified.

A similar calculation under the Bethe approximation is somewhat tedious;the answer, nevertheless, is

r1 =q − 2

4ln

q

q − 2, s0 =

(q − 1)(q − 2)

12q2.

12.20. The equilibrium values of m in this case are given by the equation

ψ′h = −h+ 2 rm + 4 sm3 + 6 um5 = 0 (u > 0). (1)

With h = 0, we get: m0 = 0,±√A+ or ±

√A−, where

A± =−s±

√s2 − 3 ur

3u. (2)

First of all, we note that, for A± to be real, s2 must be ≥ 3ur . Thispresents no problem if r ≤ 0; however, if r > 0, then s must be either≥√

3ur or ≤ −√

3ur . We also observe that

A+A− =r

3uand A+ +A− = − 2s

3u.

It follows that (i) if r < 0, then one of the A’s will be positive, the othernegative (in fact, since A− < A+, A− will be negative and A+ positive),(ii) if r = 0, then for s > 0, A− will be negative and A+ = 0, for s = 0both A− and A+ will be zero whereas for s < 0, A− will be zero while A+

will be positive (and equal to 2|s|/3u), (iii) if r > 0, then for s ≥√

3urboth A+ and A− will be negative whereas for s ≤ −

√3ur both A+ and A−

will be positive. We must, in this context, remember that only a positiveA will yield a real m0. Finally, since

ψ′′0 = 2r + 12 sm20 + 30 um4

0, (3)

113

the extremum at m0 = 0 is a maximum if r < 0, a minimum if r > 0. Itfollows that for r < 0 the function ψ0 is minimum at m0 = ±

√A+ and

for r > 0 (and s ≤ −√

3ur) it is maximum at m0 = ±√A− and minimum

at m0 = ±√A+. We, therefore, have to contend only with A+.

The foregoing observations should suffice to prove statements (a), (e), (f)and (g) of this problem. For the rest, we note that the function ψ0(m0)may be written as

ψ0(m0) = q +(rm2

0 + sm40 + um6

0

)− 1

4m0

(2rm0 + 4sm3

0 + 6um50

)(4a)

= q +1

2m2

0

(r − um4

0

); (4b)

note that in writing (4a) we have added an expression which, by the min-imization condition, is identically zero. It now follows from eqn. (4b) thatψ0(m0 = 0) is less than, equal to or greater than ψ0(m0 6= 0) accrod-ing as m2

0 is less than, equal to or greater than√r/u. The dividing line

corresponds to A+ =√r/u, i.e.

−s+√s2 − 3ur

3u=

√r

u, i.e. s = −

√4ur ;

see the accompanying figure. We also note that, in reference to the di-viding line, m2

0 decreases monotonically towards the limiting value√r/3u

as s → −√

3ur and increases montonically as s decreases below −√

4ur .These observations should suffice to prove statements (b), (c) and (d).

12.21. With s = 0, the order parameter m is given by the equation

ψ′h = −h+ 2rm + 6um5 = 0.

114

For |t| 1, we set r ≈ r1t; the equation of state then takes the form

h ≈ 2r1tm + 6um5.

With t < 0 and h → 0, we get: m0 ≈ (r1/3u)1/4|t|1/4, giving β = 1/4.With t = 0, we get: h ≈ 6um5, giving δ = 5. For susceptibility, we have

χ ∼(∂m

∂h

)t

≈ 1

2r1t+ 30um4.

It follows that

χ0 ≈

1/2 r1t (t > 0, m→ 0)1/8 r1|t| (t < 0, m→ m0),

giving γ = γ′ = 1. Finally, using the scaling relation α + 2β + γ = 2, weget: α = 1/2.

12.22. (a) We introduce the variable ψ[= (vg−vc)/vc ' (vc−v`)/vc] and obtain

ψ ∼(∂G(s)

∂π

)t

∼ |t|2−α−∆ g′(

π

|t|∆

). (2)

With t < 0 and π → 0, we get: ψ ∼ |t|β , where β = 2 − α −∆. Itfollows that the quantities (ρ` − ρc), (ρc − ρg) and (ρ` − ρg) all varyas |t|β .

(b) Writing g′(x) as xβ/∆f(x), eqn. (1) takes the form

ψ ∼ πβ/∆ f(π/|t|∆). (2)

It follows that the quantity |t|∆/π is a universal function of the quan-tity πβ/∆/ψ, i.e.

|t| ∼ π1/∆ × a universal function of (π/ψ∆/β).

It is now clear that, at t = 0, π ∼ ψδ, where δ = ∆/β.

(c) For the isothermal compressibility of the system, we have from (1)

κT ≡ −1

v

(∂v

∂P

)T

∼(∂ψ

∂π

)t

∼ |t|β−∆ g′′(

π

|t|∆

).

It follows that, in the limit π → 0, κT ∼ |t|−γ , where γ = ∆−β = β(δ−1).As for the coefficient of volume expansion, we have the relationship

αP =1

v

(∂v

∂T

)P

= −1

v

(∂v

∂P

)T

(∂P

∂T

)v

= κT

(∂P

∂T

)v

.

In the region of phase transition, (∂P/∂T )v is simply (dP/dT ), which isnon-singular. Accordingly, αP ∼ κT ∼ |t|−γ . Similarly, in view of the

115

relation CP = VT (dP/dT )2κT , established in Problem 13.25, we inferthat CP ∼ κT ∼ |t|−γ .

For CV , we go back to the given expression for G(s) and write

S(s) = −(∂G(s)

∂T

)P

∼ |t|1−α × a universal function of

(π

|t|∆

)∼ |t|1−α × a universal function of

(ψ

|t|β

).

It then follows that

C(s)V = T

(∂S(s)

∂T

)V

∼ |t|−α × a universal function of

(ψ

|t|β

).

Now, letting ψ → 0, we obtain: C(s)V ∼ |t|−α. And, in view of the relation

CV = VT (dP/dT )2κS , also established in Problem 13.25, we infer thatκS ∼ CV ∼ |t|−α.

Finally, for the latent heat of vaporization `, we invoke the Clapeyronequation,

dPσdT

=`

T (vg − v`),

and conclude that ` ∼ |t|β .

12.23. We make the following observations:

(i) With h = 0 and t < 0, m0 = 0 or ±(b−1|t|)1/2, giving β = 1/2.

(ii) With t = 0, h = abm2Θ+1, giving δ = 2Θ + 1.

(iii) The quantity (∂m

∂h

)t

=1

a(1 + 3bm2)(t+ bm2)Θ−1.

With t > 0 and h→ 0, m→ 0 and we are left with(∂m

∂h

)t

≈ 1

atΘ, giving γ = Θ.

The scaling relation (12.10.22) is readily verified.

12.24. For ri 6= rj , eqns. (12.11.22 and 26) give

(∇2 − ξ−2)g(r) = 0. (1)

116

Now, if g(r) is a function of r only then

∇g =dg

dr∇r =

dg

dr

r

r=

dg

dr

(x1

r, . . . ,

xdr

), whence

∇ · ∇g =

d∑i=1

∂

∂xi

(dg

dr

xir

)=

d∑i=1

dg

dr

1

r+

(d2g

dr2

1

r− dg

dr

1

r2

)x2i

r

=d2g

dr2 +dg

dr

1

r(d− 1). (2)

Substituting (2) into (1), we obtain the desired differential equation — ofwhich (12.11.26) is the exact solution.

Substituting (12.11.27) into the left-hand side of the given differentialequation, we get

const .e−r/ξ

r(d−1)/2

[(1

ξ2+ . . .

)+ (. . .)− 1

ξ2

];

the ratio of the terms omitted to the ones retained is O(ξ/r). Clearly, theequation is satisfied for r ξ. Similarly, substituting (12.11.28) instead,we get

const .1

rd−2

[(2− d)(1− d)

r2+d− 1

r

2− dr− (. . .)

];

the ratio of the term omitted to the ones retained is now O(r/ξ)2. Clearly,the equation is again satisfied but this time for r ξ.

12.25. By the scaling hypothesis of Sec. 12.10, we expect that, for h > 0 andt > 0,

ξ(t, h) = ξ(t, 0)× a universal function of (h/t∆)

= ξ(t, 0)× a universal function of (t/h1/∆).

Now, in view of eqn. (11.12.1), we may write

ξ(t, h) ∼ t−ν(t/h1/∆)v × a universal function of (t/h∆)

= h−ν/∆ × a universal function of (t/h1/∆).

At t = 0, we obtain: ξ(0, h) ∼ h−νc , where νc = ν/∆.

By a similar argument, χ(0, h) ∼ h−γc , where

γc = γ/∆ = β(δ − 1)/βδ = (δ − 1)/δ.

12.26. Clearly, the ratio (ρ0/ρs) ∼ |t|2β−ν . In view of the scaling relations α +2β + γ = 2, γ = (2− η)ν and dν = 2− α, we get

2β − ν = (2− α− γ)− ν = dν − (2− η)ν − ν = (d− 3 + η)ν.

Setting d = 3, we obtain the desired result.

117

12.27. By definition,σ ≡ ψA(t) ∼ |t|µ (t . 0). (1)

By an argument similar to the one that led to eqn. (12.12.14), we get

ψA(t) ∼ A−1 ∼ ξ−(d−1), (2)

where A is the “area of a typical domain in the liquid-vapor interface”.Now, for a scalar model (n = 1), to which the liquid-vapor transitionbelongs, ξ ∼ |t|−ν . Equations (1) and (2) then lead to the desired result

µ = (d− 1)ν = (2− α)(d− 1)/d.

Chapter 13

13.1. Since M = (N+ − N−)µ and N+ + N− = N , we readily see that

N± =1

2N

(1± M

Nµ

)=

1

2NP (β,B)± sinh x

P (β,B)(x = βµB). (1)

Next, comparing eqns. (12.3.19) and (13.2.12), and keeping in mind thatq = 2, we get

N+ − N++ = Ne−4βJ

2D(β,B)= N

P 2(β,B)− sinh2 x

2D(β,B). (2)

It follows from eqns. (1) and (2) that

N++ = (N/2D)(P + sinhx)(P + coshx)− (P 2 − sinh2 x)= (N/2D)(P + sinhx)(coshx+ sinhx), (3)

which is the desired result. Equation (12.3.17) now gives

N+− = 2(N+ − N++) and N−− = N − 2N+ + N++ = N++ − (M/µ).

The number N+− is just twice the expression (2). The number N−− isgiven by

N−− = (N/2D)(P + sinhx)(coshx+ sinhx)− sinhx · 2(P + coshx)= (N/2D)(P − sinhx)(coshx− sinhx), (4)

which is the desired result.

It is straightforward to check that the sum

N+++N−− = (N/D)(P coshx+sinh2 x) = (N/D)P coshx+(P 2−e−4βJ);

adding N+−, one obtains the expected result N . Finally, the product

N++N−− = (N/2D)2P 2 − sinh2 x = (N/2D)2e−4βJ .

Equation (12.6.22) is now readily verified.

118

119

13.2. (a) In view of eqn. (12.3.18), the quantity (N++ + N−− − N+−) thatappears in the Hamiltonian (12.3.19) of the lattice may be written as(

12qN − 2N+−

). The partition function (12.3.20) then assumes the

form stated here.

(b) A complete solution to this problem can be found in the first editionof this book — Sec. 12.9A, pp. 414–8. In any case, this problem isa special case, q = 2, of the next problem which is treated here atsufficient length.

13.3. In the notation of Problem 13.2, we now have

ln gN (N+, N+−) ≈ 1

2qN ln

(1

2qN

)−N++ ln N++ −N−− ln N−−

−N+− ln

(1

2N+−

)+ (q − 1)N+ ln N+ +N− ln N− −N ln N ,

(1)

where, by virtue of eqn. (12.3.17),

N++ =1

2qN +−

1

2N+−, N−− =

1

2qN− 1

2qN +−

1

2N+− and N− = N−N+.

(2)Now, as usual, the logarithm of the partition function may be approxi-mated by the logarithm of the largest term in the sum over N+ and N+−,with the result that

ln QN (B, T ) ≈ ln gN(N∗+, N

∗+−)+βJ

(1

2qN − 2N∗+−

)+βµB

(2N∗+ −N

),

(3)where N∗+ and N∗+− are the values of the variables N+ and N+− thatmaximize the summand (or the log of it). The maximizing conditionsturn out to be

∂

∂N+(. . .) = −(1 + ln N++)

(1

2q

)− (1 + ln N−−)

(−1

2q

)+

(q − 1)(1 + ln N+) + (1 + ln N−)(−1)+ 2βµB

= ln

(N−−N++

)q/2(N+

N−

)q−1

+ 2βµB = 0, and (4)

∂

∂N+−(. . .) = −(1 + ln N++)

(−1

2

)− (1 + ln N−−)

(−1

2

)−

1 + ln

(1

2N+−

)− 2βJ

= ln

N

1/2++N

1/2−−(

12N+−

) − 2βJ = 0. (5)

120

Equations (4) and (5), with the help of eqns. (2), determine the equilibriumvalues of all the numbers involved in the problem; eqn. (3) then determinesthe rest of the properties of the system.

To compare these results with the ones following from the Bethe approx-imation, we first observe that eqn. (5) here is identical with the corre-sponding eqn. (12.6.22) of that treatment. As for eqn. (4), we go back toeqns. (12.6.4 and 8) of the Bethe approximation, whereby

N+

N−= e2α

[cosh(α+ α′ + γ)

cosh(α+ α′ − γ)

]q= e2α+2α′q/(q−1)

and to eqn. (12.6.21), whereby

N−−N++

= e−4(α+α′).

It follows that(N−−

N++

)q/2(N+

N−

)q−1

= e−2q(α+α′)+2α(q−1)+2α′q = e−2α (α = βµB),

in complete agreement with eqn. (4). Hence the equivalence of the twotreatments.

13.4. By eqns. (13.2.8 and 37),

ξ−1(B, T ) = ln

[coshx+ e−4βJ + sinh2 x1/2

coshx− e−4βJ + sinh2 x1/2

](x = βµB).

As T → Tc (which, in this case, is 0),

ξ−1(B, Tc) ≈ ln

[coshx+ sinhx

coshx− sinhx

]= 2x.

It follows that ξ(B, Tc) ≈ (1/2x) ∼ B−1, which means that the criticalexponent νc = 1. Now, a reference to eqns. (13.2.21 and 35) tells us thatν = ∆. The relation νc = ν/∆ is thus verified.

13.5. On integration over B, our partition function takes the form

QN (s, T ) ∼(

2πs

βN

)1/2 ∑σi

exp

βµ2s

2N

(∑i

σi

)2

+ βJ∑i

σiσi+1

,

which implies an effective Hamiltonian given by the expression

Heff = −1

2µ2sNL2 − J

∑i

σiσi+1

(L = N−1

∑i

σi

).

121

The first term here is equivalent to an infinite-range interaction of the typeconsidered in Problem 12.6, with µ2s playing the role of the quantity cNof that model. This is also equivalent to the Bragg-Williams model, withµ2s ↔ qJ ; see Problem 12.4. It follows that, by virtue of this term, thesystem will undergo an order-disorder transition at a critical temperatureTc = µ2s/k. The second term in the Hamiltonian will contribute towardsthe short-range order in the system.

We also note that the root-mean-square value ofB in this model is (s/βN)1/2

which, for a given value of s, is negligibly small when N is large. Theorder-disorder transition is made possible by the fact that the resultinginteraction is of an infinite range.

13.6. Since the Hamiltonian Hτi of the present model is formally similar tothe Hamiltonian Hσi of Sec. 13.2, the partition function of this systemcan be written down in analogy with eqn. (13.2.10):

1

Nln Q ≈ ln

[eβJ2 cosh(βJ1) +

e−2βJ2 + e2βJ2 sinh2(βJ1)

1/2],

from which the various thermodynamic properties of the system can bederived. In particular, we get

σiσi+1 =1

βN

∂

∂J1ln Q =

sinh(βJ1)e−4βJ2 + sinh2(βJ1)

1/2, and

σiσi+2 =1

βN

∂

∂J2ln Q = 1−

2e−4βJ2e−4βJ2 + sinh2(βJ1)

1/2[cosh(βJ1) +

e−4βJ2 + sinh2(βJ1)

1/2] .

As a check, we see that in the limit J2 → 0 these expressions reduce totanh(βJ1) and tanh2(βJ1), respectively; on the other hand, if J1 → 0,they reduce to 0 and tanh(βJ2) instead.

13.7. In the symmetrized version of this problem, the transfer matrix P isgiven by

⟨σi, σ

′i|P|σi+1, σ

′i+1

⟩= exp

K1

(σiσi+1 + σ′iσ

′i+1

)+

1

2K2

(σiσ′i + σi+1σ

′i+1

),

where K1 = βJ1 and K2 = βJ2. Since (σi, σ′i) = (1, 1), (1,−1), (−1, 1) or

(−1,−1), we get

(P) =

e2K1+K2 1 1 e−2K1+K2

1 e2K1−K2 e−2K1−K2 11 e−2K1−K2 e2K1−K2 1

e−2K1+K2 1 1 e2K1+K2

122

The eigenvalues of this matrix are(λ1

λ2

)=

1

2

[(A+B + C +D)± (A−B + C −D)2 + 161/2

],

λ3 = A− C, λ4 = B −D,

where

A = e2K1+K2 , B = e2K1−K2 , C = e−2K1+K2 , D = e−2K1−K2 .

Since λ1 is the largest eigenvalue of P,

1

Nln Q ≈ lnλ1 = ln

1

2

[(A+B + C +D) + (A−B + C −D)2 + 161/2

].

Substituting for A, B, C and D, we obtain the quoted result. The studyof the various thermodynamic properties of the system is now straightfor-ward.

13.8. In the notation of Sec. 13.2, the transfer matrix of this model is

〈σi|P|σi+1〉 = exp(βJ σiσi+1) (σi = −1, 0, 1).

It follows that

(P) =

eK 1 e−K

1 1 1e−K 1 eK

(K = βJ),

with eigenvalues(λ1

λ2

)=

1

2

[(1 + 2 coshK)± 8 + (2 coshK − 1)21/2

], λ3 = 2 sinhK.

Since λ1 is the largest eigenvalue of P,

1

Nln Q ≈ lnλ1 = ln

1

2

[(1 + 2 coshK) + 8 + (2 coshK − 1)21/2

],

which leads to the quoted expression for the free energy A.

In the limit T → 0, K →∞, with the result that coshK ≈ 12eK and hence

A ≈ −NJ ; this corresponds to a state of perfect order in the system, withU = −NJ and S = 0. On the other hand, when T →∞, coshK → 1 andhence A→ −NkT ln 3; this corresponds to a state of complete randomnessin a system with 3N microstates.

13.9. (a) Making use of the correspondence established in Sec. 12.4, we obtainfor a one-dimensional lattice gas (q = 2).

(i) The fugacity z = e−4βJ+2βµB = η2y, where η = e−2βJ andy ↔ e2βµB .

123

(ii) The pressure P = −(A/N) − J + µB; using eqn. (13.2.11), thisbecomes

P = kT ln[cosh(βµB) + e−4βJ + sinh2(βµB)1/2

]+ µB. (1)

(iii) The density (1/v) = 121 + (M/Nµ); using eqn. (13.2.13), this

becomes

1

v=

1

2

[1 +

sinh(βµB)

e−βJ + sinh2(βµB)1/2

]. (2)

Our next step consists in eliminating the magnetic variable (βµB) infavor of the fluid variable y. For this, we note that

cosh(βµB) =1

2(y1/2 + y−1/2) = (y + 1)/2y1/2,

sinh(βµB) =1

2(y1/2 − y−1/2) = (y − 1)/2y1/2,

while βµB = 12 ln y. Substituting these results into eqns. (1) and (2),

we obtain the quoted expressions for P/kT and 1/v. It may beverified that these expressions satisfy the thermodynamic relation

1

v= z

[∂

∂z

(P

kT

)]T

= y

[∂

∂y

(P

kT

)]η

At high temperatures, η → 1 and we get

P

kT≈ ln(y + 1),

1

v≈ y

y + 1.

Moreover, in this limit y ' z 1. One may, therefore, write

P

kT≈ y, 1

v≈ y and hence

Pv

kT≈ 1.

At low temperatures, η becomes very small and y very large — withthe result that

P

kT≈ ln y +

η2

y,

1

v≈ 1− η2

y.

(b) For a hard-core lattice gas (y → 0, η →∞), we get

P

kT= ln

[1 +√

1 + 4z

2

], ρ =

√1 + 4z − 1

2√

1 + 4z.

From the second equation, it follows that√

1 + 4z = 1/(1 − 2ρ). Sub-stituting this into the first equation, we obtain the quoted expression forP (ρ).

124

13.10. Applying eqn. (12.11.11) to a one-dimensional system, we get

χ0

Nβµ2=

∞∑x=−∞

e−(a/ξ)|x| = 1 + 2

∞∑x=1

e−(a/ξ)x

= 1 + 2e−a/ξ

1− e−a/ξ=

1 + e−a/ξ

1− e−a/ξ= coth

(a

2ξ

).

For ξ a, coth(a/2ξ) ≈ (2ξ/a), making χ0 ∝ ξ1 — consistent with thefact that for this system (2 − η) = 1. For ξ a, we recover the familiarresult: χ0 ≈ Nµ2/kT .

For an n-vector model, eqn. (13.3.17) leads to the result

χ0

Nβµ2=

1 + In/2(βJ)/I(n−2)/2(βJ)

1− In/2(βJ)/I(n−2)/2(βJ),

which agrees with the expression quoted in the problem. For the specialcase n = 1, the ratio I1/2(x)/I−1/2(x) = tanhx, whence

χ0

Nβµ2=

1 + tanh(βJ)

1− tanh(βJ)=

cosh(βJ) + sinh(βJ)

cosh(βJ)− sinh(βJ)= e2βJ ,

in agreement with eqn. (13.2.14).

13.11. We introduce an extra factor σ2k+1 . . . σ

2`−1σ

2m+1 . . . σ

2n−1, which is identi-

cally equal to 1, and write

σkσ`σmσn = (σkσk+1) . . . (σ`−1σ`)(σmσm+1) . . . (σn−1σn).

Following the same procedure that led to eqn. (13.2.31), we now get

σkσ`σmσn =

`−1∏i=k

tanh(βJi)

n−1∏i=m

tanh(βJi).

Employing a common J , we obtain the desired result

σkσ`σmσn = tanh(βJ)`−ktanh(βJ)n−m = tanh(βJ)n−m+`−k

13.12. For a complete solution to this problem, see Thompson (1972b), sec. 6.1,pp. 147–9.

13.13. With J ′ = 0, we get a much simpler result, viz.

1

Nln Q = ln 2 +

1

2π2

π∫0

π∫0

lncosh(2γ)− sinh(2γ)− cosωdωdω′.

125

The integration over ω′ is straightforward; the one over ω can be donewith the help of the formula

π∫0

ln(a− b cosω)dω = π ln

[a+√a2 − b22

](|b| < a), (1)

which yields the expected result

1

Nln Q = ln 2 +

1

2ln

[cosh(2γ) + 1

2

]= ln(2 cosh γ).

With J ′ = J , we have

1

Nln Q = ln 2 +

1

2π2

π∫0

π∫0

lncosh2(2γ)− sinh 2γ(cosω + cosω′)dωdω′.

(2)We substitute ω = θ + ϕ and ω′ = θ − ϕ; this will replace the sum(cosω + cosω′) by the product 2 cos θ cosϕ and the element dωdω′ by2dθdϕ. As for the limits of integration, the periodicity of the integrandallows us to choose the rectangle [0 ≤ θ ≤ π, 0 ≤ ϕ ≤ π/2] withoutaffecting the value of the integral. We thus have

1

Nln Q = ln 2 +

1

π2

π∫0

π/2∫0

lncosh2(2γ)− 2 sinh(2γ) cosϕ cos θdθdϕ.

Integration over θ may now be carried out using formula (1), with theresult

1

Nln Q = ln 2 +

1

π

π/2∫0

ln

[1

2

cosh2(2γ) +

√cosh4(2γ)− 4 sinh2(2γ) cos2 ϕ

]dϕ

= ln

2 · 1√

2cosh(2γ)

+

1

π

π/2∫0

ln

1 +√

1− κ2 cos2 ϕdϕ,

where κ is given by eqn. (13.4.23). Finally, we replace cos2 ϕ by sin2 ϕ(without affecting the value of the integral) and recover eqn. (13.4.22).

We know that this model is singular at κ = 1. A close look at the integralin (2) shows that the singularity arises when contributions in the neigh-borhood of the point ω = ω′ = 0 pile up. Since cosω and cosω′ are almostunity there, the situation becomes catastrophic when cosh2 2γ = 2 sinh 2γ,i.e. when κ = 1. In the anisotropic case, a similar observation suggeststhat the singularity will arise when

cosh(2γ) cosh(2γ′) = sinh(2γ) + sinh(2γ′).

126

Squaring both sides of this equation and simplifying, we get

sinh(2γ) sinh(2γ′) = 1 (3)

as the criterion for the onset of phase transition in this system; cf. eqn. (13.4.18).The study of the thermodynamic behavior of the system in the neighbor-hood of the critical point in the general case, when J ′ 6= J , is highlycomplicated; the fact, however, remains that the internal energy U0 iscontinuous and the specific heat C0 displays a logarithmic divergence ata critical temperature Tc given by eqn. (3).

13.14. As κ→ 1, the first integral tends to the limit

π/2∫0

1− sinϕ

cosϕdϕ =

π/2∫0

cosϕ

1 + sinϕdϕ = ln(1 + sinϕ)|π/20 = ln 2.

The second integral diverges as κ→ 1, so we need to examine it carefully.Setting cosϕ = x, this integral takes the form

1∫0

κ dx√(1− κ2) + κ2x2

'1∫

0

dx√κ′2 + x2

= lnx+√κ′2 + x2

∣∣∣10.

Since κ is close to 1, κ′2 1; we, therefore, get for this integral theasymptotic result ln 2 − ln |κ′|. It follows that K1(κ) ≈ 2 ln 2 − ln |κ′| =ln(4/|κ′|).

13.15. The quantity to be evaluated here is

ScNk

=

(U

NkT

)c

−(

A

NkT

)c

.

The first term, by eqn. (13.4.28), is −Kc coth(2Kc) = −√

2Kc. The sec-ond, by eqn. (13.4.22), is

ln 2 +1

π

π/2∫0

ln(1 + cosϕ)dϕ

= ln 2 +1

π

ϕ ln(1 + cosϕ)|π/20 +

π/2∫0

ϕ sinϕ

1 + cosϕdϕ

.The integrated part vanishes while the remaining integral has the value−(π/2) ln 2+2G; see Gradshteyn and Ryzhik (1965), p. 435. Thus, finally,

ScNk

= −√

2Kc +1

2ln 2 + 2G/π ' 0.3065.

The corresponding result under the Bethe approximation is 2 ln 3−(7/3) ln 2 '0.5799 and that under the Bragg-Williams approximation is ln 2 ' 0.6931.

127

13.16. Expanding around K = Kc, we get

sinh(2K) = sinh(2Kc) + 2 cosh(2Kc)(K −Kc) + 2 sinh(2Kc)(K −Kc)2 + . . .

= 1 + 2√

2(K −Kc) + 2(K −Kc)2 + . . . . (1)

Since (K−Kc) = Kc(1 + t)−1−1 = Kc(−t+ t2− . . .), eqn. (1) becomes

sinh(2K) = 1− 2√

2Kct+(

2√

2Kc + 2K2c

)t2 + . . . . (2)

Raising expression (2) to the power −4, we get

sinh(2K)−4 = 1 + 8√

2Kct−(

8√

2Kc + 8K2c − 80K2

c

)t2 + . . . ,

so that

1− sinh(2K)−4 = −8√

2Kct+(

8√

2Kc − 72K2c

)t2 + . . .

= 8√

2Kc|t|

1 +

(1− 9√

2Kc

)|t|+ . . .

. (3)

Taking the eighth-root of (3), we obtain the desired result.

13.17. Making use of the correspondence established in Sec. 12.4 and utilizing aresult obtained in Problem 13.15, we have for a two-dimensional latticegas (q = 4)

PckT c

= −(

A

NkT

)c

− 2J

kT c=

(1

2ln 2 +

2G

π

)− 2Kc.

We also have: vc = 2 (because, at T = Tc, the spontaneous magnetizationof the corresponding ferromagnet is zero). It follows that

PcvckT c

= ln 2 +4G

π− 4Kc ' 0.09659.

Taking the reciprocal of this result, we obtain the one stated in the prob-lem.

13.18. In one dimension, expression (13.5.31) assumes the form

W1(ϕ) =

∞∫0

e−(1+ϕ)xI0(x)dx =1

(1 + ϕ)2 − 11/2=

1

(λ2 − J2)1/2.

The constraint equation (13.5.19) then becomes

N

2β

1

(λ2 − J2)1/2+

Nµ2 β2

4(λ− J)2= N, (1)

128

which agrees with the quoted result. Comparing (1) with the formal con-straint equation (13.5.13), we conclude that

∂Aλ∂λ

=N

2β

1

(λ2 − J2)1/2+

Nµ2B2

4(λ− J)2, (2)

It follows that

Aλ =N

2βlnλ+ (λ2 − J2)1/2

+Nµ2B2

4(λ− J)+ C, (2)

where C is a constant of integration. To determine C, we observe fromeqn. (13.5.12a) that, for B = 0 and J = 0, the partition function QN =(π/βλ)N/2 — with the result that Aλ = (N/2β) ln(βλ/π). It follows thatC = (N/2β) ln(β/2π), which leads to the quoted result for Aλ.

With B = 0 but J 6= 0, eqn. (1) gives: (λ2 − J2)1/2 = 1/2β, and henceλ = (1 + 4β2J2)1/2/2β. Equation (13.5.15) then gives

βAS

N=βAλN− βλ =

1

2ln

(1 + 4β2 J2)1/2 + 1

4π

− 1

2(1 + 4β2 J2)1/2.

13.19. With βJi = β · nJ ′ = nK , eqn. (13.3.8) becomes

QN (nK ) =

∣∣∣∣∣ Γ(n/2)(12nK

)(n−2)/2I(n−2)/2(nK )

∣∣∣∣∣N−1

.

For n, N 1, we get

1

nNln QN (nK ) ≈ 1

n

[ln Γ

(n2

)− n

2ln

(1

2nK

)+ ln In/2

(n2· 2K

)]=

1

n

[n

2lnn

2− n

2+ . . .− n

2ln

(1

2nK

)+n

2

(4K2 + 1)1/2 − ln

((4K2 + 1)1/2 + 1

2K

)+ . . .

]≈ 1

2

[(4K2 + 1)1/2 − 1− ln

((4K2 + 1)1/2 + 1

2

)].

13.20. By eqn. (13.5.69), we have for the spherical model at T < Tc

χ0 =Nµ2

2 Jϕ≈ N2 µ2 (K −Kc)

J.

Replacing (K − Kc) by m20K, see eqn. (13.5.44), and remembering that

K = J/kBT , we obtain the desired result.

129

13.21. Before employing the suggested approximation, we observe that the majorcontribution to the integral over θj in expression (13.5.58) comes fromthose values of θj that are either close to the lower limit 0 or close to theupper limit 2π. We, therefore, write this integral in the form

2

π∫0

cos(Rjθj / a)e−x+x cos θjNj2π

dθj ,

so that the major contribution now comes only from those values of θ thatare close to 0. We may, therefore, replace (1− cos θj) by θ2

j/2 and, at thesame time, replace the upper limit of the integral by ∞. This yields theasymptotic result

2

π∫0

cos(Rjθj / a)e−xθ2j/2

Nj2π

dθj =Nj√2πx

e−R2j/2a

2x,

where use has been made of formula (B.41). Equation (13.5.57) thenbecomes

G(R) ≈ 1

2NβJ

∞∫0

N

(2πx)d/2e−ϕx−R

2/2a2x dx ,

which is precisely the expression that led to eqns. (13.5.61 and 62).

In the case of the Bose gas, we are concerned with the expression

1

(2π)d

∫exp(ik ·R)

eα+β~2k2/2m − 1dd k,

see Section 13.6, which may now be approximated by

1

(2π)d

∫exp(ik ·R)

α+ β~2k2 /2mdd k.

Using the representation (13.5.27), this may be written as

∞∫0

e−αx

d∏n=1

∞∫−∞

1

2πeiknRn−β~2k2nx/2m dkn

dx

=

∞∫0

e−αx

[d∏

n=1

(1

λ√xe−πR

2n/λ

2x

)]dx

[λ = ~

(2πβ

m

)1/2]

=1

λd

∞∫0

e−αx−πR2/λ2x 1

xd/2dx ,

which is precisely the expression that led to eqns. (13.6.35 and 36).

130

13.22. The constraint equation (13.5.21) now takes the form

2Nβ(1−m2) =∑k

J

(ϕ+

1

2kσ aσ

)−1

,

which may as well be written as

2K(1−m2) = F (ϕ), whereK = βJ and F (ϕ) = N−1∑k

(ϕ+

1

2kσ aσ

)−1

.

To determine the behavior of the function F (ϕ) at small ϕ, we look at thederivative

F ′(ϕ) = −N−1∑k

(ϕ+

1

2kσ aσ

)−2

.

In view of eqns. (13.5.17) and (C.7b), we obtain

F ′(ϕ) ≈ −N−1d∏j=1

(Nja

2π

) ∞∫0

(ϕ+

1

2kσ aσ

)−22πd/2

Γ(d/2)kd−1 dk

= − ad

2d−1 πd/2Γ(d/2)ϕ2

∞∫0

(1 +

1

2ϕkσ aσ

)−2

kd−1 dk .

We substitute kσ = (2ϕ/aσ)x and get

F ′(ϕ) ≈ − (2ϕ)d/σ

2d−1 πd/2Γ(d/2)σϕ2

∞∫0

x(d/σ)−1 dx

(1 + x)2

= −2d/σΓ(d/σ)Γ(2− d / σ)

2d−1 πd/2Γ(d/2)σϕ(d−2σ)/σ (d < 2σ).

Integrating over ϕ, we obtain

F (ϕ) ≈ F (0)− 2d/σΓ(d/σ)/σΓ(2σ − d)/σ2d−1 πd/2Γ(d/2)σ

ϕ(d−2σ)/σ;

cf. eqn. (G.7c). The function F (0) exists for all d > σ and may beidentified with 2Kc, leading to the constraint equation

2K(1−m2) = 2Kc − const . ϕ(d−σ)/σ (σ < d < 2σ).

The critical exponents of the model follow straightforwardly from thisequation. The first one to emerge is β = 1/2, as before. Next, γ =σ/(d−σ), whence α = 2−2β−γ = (d−2σ)/(d−σ), while δ = 1+(γ/β) =(d+ σ)/(d− σ). Next, from the very starting form of the function F (ϕ),we infer that the correlation length ξ ∼ ϕ−1/σ and hence ∼ t−1/(d−σ); itfollows that ν = 1/(d− σ). We then get: η = 2− (γ/ν) = 2− σ.

131

For d > 2σ, the derivative F ′(0) exists — with the result that

F (ϕ) ≈ F (0)− |F ′(0)|ϕ.

This leads to mean-field results for the exponents β, γ, α and δ. Thecorrelation length is, once again, given by ξ ∼ ϕ−1/σ but now this is∼ t−1/σ, so that ν = 1/σ (and not 1/2); accordingly, η is, once again,2− σ (and not 0).

13.23. The derivation of eqns. (13.6.9 and 11) proceeds exactly as of eqns. (7.1.36and 37). The derivation of eqns. (13.6.10 and 13) proceeds exactly asin Problem 7.4; eqn. (13.6.12) then follows as a product of expressions(13.6.11 and 13): The derivation of eqns. (13.6.14 and 15) proceeds exactlyas in Problem 7.5; note that one may first obtain here

κT =1

nkBT

g(d−2)/2 (z)

gd/2 (z), κS =

d

(d+ 2)nkBT

gd/2 (z)

g(d+2)/2 (z),

and then use eqn. (13.6.7) to express these quantities in terms of P . Fi-nally, the derivation of eqn. (13.6.23) proceeds exactly as in Problem 7.6.

13.24. The derivations here proceed exactly as in Problem 7.7. The singularityof these quantities arises from the last term of the two expressions, andis qualitatively similar to the singularity of the quantity (∂CV /∂T ). Notethat the singularity of the combination

v(∂2P/∂T 2)v − (∂2µ/∂T 2)v

is

relatively mild.

13.25. By definition,

CP ≡ T(∂S

∂T

)P

= T

(∂S

∂V

)P

(∂V

∂T

)P

.

Using the Maxwell relation (∂S/∂V )P = (∂P/∂T )S , we get

CP = T

(∂P

∂T

)S

(∂V

∂T

)P

= T

(∂P

∂T

)S

−(∂V

∂P

)T

(∂P

∂T

)V

=

T

(∂P

∂T

)S

V κT

(∂P

∂T

)V

.

Next

CV ≡ T(∂S

∂T

)V

= T

(∂S

∂P

)V

(∂P

∂T

)V

.

Now, using the Maxwell relation (∂S/∂P )V = −(∂V/∂T )S , we get

CV = −T(∂V

∂T

)S

(∂P

∂T

)V

= −T(∂V

∂P

)S

(∂P

∂T

)S

(∂P

∂T

)V

=

T (V κS)

(∂P

∂T

)S

(∂P

∂T

)V

.

132

In the two-phase region, (∂P/∂T )S = (∂P/∂T )V = dP/dT , with theresult that

CP = VT (dP / dT )2 κT , CV = VT (dP / dT )2 κS .

Now, by eqn. (13.6.28), dP/dT, at T < Tc,= (d + 2)P/2T . Using thisresult and eqn. (13.6.15), we get

CV = VT

(d+ 2)P

2T

2d

(d+ 2)P=d(d+ 2)

4

PV

T.

Substituting for P from eqn. (13.6.28), we recover expression (13.6.30) forCV.

13.26. As shown in Problem 1.16, dµ = −sdT + vdP . It follows that

κT ≡ −1

v

(∂v

∂P

)T

= −(∂v

∂µ

)T

.

Since v = 1/ρ, we readily obtain the desired result for κT .

For the ideal Bose gas at T < Tc, the particle numbers N0 and Ne aregiven by eqns. (13.6.24) and (13.6.27). Since α ≡ −µ/kBT , we have

ρ =N

V=N0

V+NeV

= −kBTV µ

+ζ(d / 2)

λd.

It follows that

κT =1

ρ2

(kBT

V µ2

)=

1

ρ2

kBT

V

(− N0

kBT

)2

=N2

0

ρ2VkBT=

V ρ20

ρ2kBT.

Incidently, using the relationship between CP and κT , as developed inProblem 13.25, we can show that, in the two-phase region of the Bose gas,

CPNkB

= N

d+ 2

2

ζ(d+ 2)/2ζ(d/2)

2(T

Tc

)d(ρ0

ρ

)2

.

Comparing this with eqn. (13.6.30), we find that, in this region, the ratioCP /CV = O(N).

13.28. Use the relations N/(L−ND) = βP and

limn→∞

(1 +

z

n

)n= exp(z),

and collect factors.

133

13.29. Using equation (10.7.20a) and ignoring the delta–function contribution toS(k) gives

S(k) = 1 +n

βP (1− nD)

∞∑j=1

1

(j − 1)!

∫ ∞jD

(βP (x− jD))j exp (−βP (x− jD) + ikx) + c.c.

.Using n/(βP (1− nD)) = 1 we get

S(k) = 1 +

(βPeikD

βP − ik − βPeikD+ c.c.

)=

k2

k2 + 2(βP )2(1− cos(kD)) + 2βPk sin(kD)

13.30. The isobaric partition function is

YN (P, T ) =

[1

λ

∫ ∞−∞

exp

(−βPy − βmω2

2(y − a)2

)dx

],

G(N,P, T ) = NkT ln

(kT

~ω

)+NPa− NP 2

2mω2,

which gives L = (∂G/∂P )T = N(a− P/mω2

). As P → 0 the length

goes to Na, i.e. N times the equilibrium length of one spring. However,the masses and springs do not form a long-range-ordered lattice since thevariance of the neighbor distances grows with n.

〈xn − x0〉 = n〈y〉 = a− P

mω2

〈(xn − x0)2〉 − 〈xn − x0〉2 = n(〈y2〉 − 〈y〉2

)= na2 kT

mω2a2.

The heat capacity is CP = Nk.

13.31. Here is a C code snippet that performs the calculation.

int L=4;

int n=L*L;

int* s=new int[n]; //spins: 0 or 1

int* i1 = new int[n]; //neighbors to the right

int* i3 = new int[n]; //neighbors above

for (int i=0;i<n;i++)

i1[i] = i+1; // site to right

i3[i] = i+L; // site above

if ((i1[i] % L) == 0) i1[i] -= L; //implement periodic boundary conditions

if (i3[i] >= n) i3[i] -= n; //implement periodic boundary conditions

134

double* hist=new double[n+1]; // histogram: double since would overflow integers for L>5;

for (int e=0;e<=n;e++) hist[e]=0.0;

double nconfig=pow(2.0,n); // number of configurations is 2 ∧ n

for (double iconfig=0.0; iconfig < nconfig; iconfig += 1.0)

double state = iconfig;

for (int i=0;i<n;i++) // determine spins (0 or 1) for each site

s[i]=(int) fmod(state,2.0);

state = floor(0.5*state);

int e=0;

for (int i=0;i<n;i++) // count energy above ground state (unequal neighbors)

e += (s[i] != s[i1[i]]);

e += (s[i] != s[i3[i]]);

hist[e/2] += 1.0; //increment histogram of energies, only even values needed

for (int e=0;e<=n;e++) cout << e << ” ” << hist[e] << endl; //output the results

This code gives the coefficients in the problem. Since the L = 6 caseinvolves 236 configurations, you might try a bitwise calculation with eachrow represented by an integer between 0 and 2L − 1 and the spins repre-sented by the bits. This is computationally much more efficient since theenergy can be determined easily using simple bit rotations and exclusiveors.

13.32. Separating off the ground state energy gives

T =

(1 xx 1

)which has eigenvalues λ = 1± x. Therefore

QN = (1 + x)N + (1− x)N ,

which can be expanded

QN =

N∑j=0

(N !xj

j!(N − j)!+

N !(−x)j

j!(N − j)!

)=

N/2∑k=0

(2

n!

(2k)!(N − 2k)!x2k

)

135

13.33. The 8× 8 partition function coefficients are

g = 2, 0, 128, 256, 4672, 17920, 145408, 712960, 4274576, 22128384, 118551552, 610683392,

3150447680, 16043381504, 80748258688, 396915938304, 1887270677624, 8582140066816,

36967268348032, 149536933509376, 564033837424064, 1971511029384704,

6350698012553216, 18752030727310592, 50483110303426544, 123229776338119424,

271209458049836032, 535138987032308224, 941564975390477248, 1469940812209435392,

2027486077172296064, 2462494093546483712, 2627978003957146636,

2462494093546483712, 2027486077172296064, 1469940812209435392,

941564975390477248, 535138987032308224, 271209458049836032,

123229776338119424, 50483110303426544, 18752030727310592, 6350698012553216,

1971511029384704, 564033837424064, 149536933509376, 36967268348032,

8582140066816, 1887270677624, 396915938304, 80748258688, 16043381504, 3150447680,

610683392, 118551552, 22128384, 4274576, 712960, 145408, 17920, 4672, 256, 128, 0, 2

The specific heats for the 8×8, 16×16 and 32×32 Ising model are shownbelow.

0 1 2 3 4

kT

J0.0

0.5

1.0

1.5

2.0

2.5

3.0

C

Nk

Notice that the height of the peak grows linearly with ln(L).

136

13.34. The specific heat for the 64× 64 Ising model is shown below.

0 1 2 3 4

kT

J0.0

0.5

1.0

1.5

2.0

2.5

3.0

C

Nk

Chapter 14

14.1. We start with expression (13.2.3) for the partition function QN and carryout summation over σ2, σ4, . . .. The resulting expression will consist of12N factors such as∑

σ2

〈σ1|P|σ2〉〈σ2|P|σ3〉 = 〈σ1|P2|σ3〉,

and will be formally similar to the expression we started with. Calling thenew transfer operator P′K′, we clearly get eqn. (1) of the problem.

From the given expression for PK, we readily get

P′K′ = e2K0

(e2(K1+K2) + e−2K1 eK2 + e−K2

eK2 + e−K2 e2(K1−K2) + e−2K1

).

Expressing this in a form similar to eqn. (2), we obtain

eK′0+K′1+K′2 = e2K0e2(K1+K2) + e−2K1

eK′0+K′1−K

′2 = e2K0e2(K1−K2) + e−2K1, and

eK′0−K

′1 = e2K0eK2 + e−K22

which are identical with eqns. (14.2.7) and will lead precisely to eqns. (14.2.8).

14.2. We’ll do the second part only, for it includes the first as a special case.For this, we have to show that the given function f(K1,K2) satisfies thefunctional equation (14.2.11). Now, the right-hand side of this equation is

−1

2ln

eK′0eK′1+K′2 + eK

′1−K

′2

2+

e−2K′1 +

(eK′1+K′2 − eK′1−K′2

2

)2

1/2 .

Substituting from eqns. (14.2.7) with K0 = 0, this becomes

− 1

2ln[eK2 cosh(2K1 +K2) + e−K2 cosh(2K1 −K2) +

4 cosh2K2+

(eK2 cosh(2K1 +K2)− e−K2 cosh(2K1 −K2))21/2

]

= −1

2ln[e2K1 cosh(2K2) + e−2K1 + 4 cosh2K2 + (e2K1 sinh(2K2))21/2].

137

138

The left-hand side of the same equation is

− ln[eK1 coshK2 + e−2K1 + e2K1 sinh2K21/2]

= −1

2ln[e2K1 cosh2 K2 + e−2K1 + e2K1 sinh2K2

+2eK1 coshK2e−2K1 + e2K1 sinh2K21/2]

= −1

2ln[e2K1 cosh(2K2) + e−2K1 + 4 cosh2K2 + 4e4K1 cosh2K2 sinh2K21/2],

which is precisely the same as the right-hand side.

14.3. We’ll do the second part only, for it includes the first as a special case. Forthis, we have to show that the given function f(K1,K2,Λ) satisfies thefunctional equation (14.2.27). Now, the right-hand side of this equation is

−1

2K ′0 +

1

4ln

[Λ′ +

√Λ′2 −K ′212π

]− K ′22

8 (Λ′ −K ′1).

Substituting from eqns. (14.2.25 and 28) with K0 = 0, this becomes

−1

4ln(π

Λ

)− K2

2

8Λ+

1

4ln

[(Λ−K2

1/2Λ)

+√

Λ2 −K21

2π

]

− K22

8(Λ−K1)

(1 +

K1

Λ

)=

1

4ln

[2Λ

2π.

(Λ−K2

1/2Λ)

+√

Λ2 −K21

2π

]

− K22

8(Λ−K1)

[Λ−K1

Λ+

Λ +K1

Λ

]=

1

2ln

[Λ +

√Λ2 −K2

1

2π

]− K2

2

4(Λ−K1).

which is precisely f(K1,K2,Λ).

14.4. Making the suggested substitution into eqn. (14.2.24), we get

QN =

∫· · ·∫

exp

N ′∑j=1

K ′0 +K ′1 ·

2Λ

K1s′js′j+1 − Λ′ · 2Λ

K1s′2j

( 2Λ

K1

)N ′/2ds ′1 . . . ds ′N .

In view of eqns. (14.2.25), with K0 = K2 = 0, we now have

eK′0 ·(

2Λ

K1

)1/2

=(π

Λ

)1/2

·(

2Λ

K1

)1/2

=

(2π

K1

)1/2

= eK′′0 , say,

K ′1 ·2Λ

K1= K1, and Λ′ · 2Λ

K1=

2Λ2

K1−K1 = Λ′′, say.

139

The resulting expression for QN , when compared with eqn. (14.2.19), leadsto the functional equation

f(K1,Λ) = −1

2K ′′0 +

1

2f(K1,Λ

′′), (1)

where

K ′′0 =1

2ln

(2π

K1

)and Λ′′ =

2Λ2

K1−K1. (2)

To verify that the function (14.2.32) satisfies the functional equation (1),we note that the right-hand side of this equation is

− 1

4ln

(2π

K1

)+

1

4ln

[Λ′′ +

√Λ′′2 −K2

1

2π

]

=1

4ln

[K1

2π· (2Λ2/K1 −K1) +

√4Λ4/K2

1 − 4Λ2

2π

]

=1

4ln

[2Λ2 −K2

1 + 2Λ√

Λ2 −K21

4π2

]=

1

2ln

[Λ +

√Λ2 −K2

1

2π

],

which is precisely f(K1,Λ).

14.5. For a solution to this problem, see Kadanoff (1976a).

14.6. The eigenvalues λ1 and λ2 of the matrixA∗` are determined by the equation∣∣∣∣a11 − λ a12

a21 a22 − λ

∣∣∣∣ = 0.

Clearly,

λ1 + λ2 = a11 + a22, while λ1λ2 = a11a22 − a12a21. (1)

The eigenfunctions ϕ1 and ϕ2 are given by

ϕ1 = const

(x1

y1

), where

y1

x1=λ1 − a11

a12=

a21

λ1 − a22, and (2)

ϕ2 = const

(x2

y2

), where

y2

x2=λ2 − a11

a12=

a21

λ2 − a22, (3)

(a) Now, by eqn. (14.3.13a),

k1 = u1x1 + u2x2, k2 = u1y1 + u2y2, where

u1 =k1y2 − k2x2

x1y2 − y1x2, u2 =

k1y1 − k2x1

x2y1 − y2x1.

It follows that the slope of the line u1 = 0 in the (k1, k2)-plane ism1 = y2/x2, which is given by eqn. (3), while the slope of the line

140

u2 = 0 is m2 = y1/x1, which is given by eqn. (2). We readily seethat the product

m1m2 =λ2 − a11

a12· λ1 − a11

a12=λ2λ1 − a11(λ2 + λ1) + a2

11

a212

Substituting from eqns. (1), we get

m1m2 =(a11a22 − a12a21)− a11(a11 + a22) + a2

11

a212

= −a21

a12.

It follows that the two lines will be mutually perpendicular if andonly if a12 = a21.

(b) If a12 or a21 = 0, then by eqns. (1), λ1 = a11 and λ2 = a22. Thestated results then follow straightforwardly.

(c) If a11 = 0, then m1 = −a21/λ1 = λ2/a12 and m2 = −a21/λ2 =λ1/a12. On the other hand, if a22 = 0, then m1 = a21/λ2 = −λ1/a12

and m2 = a21/λ1 = −λ2/a12.

14.7. In the limit n→∞, we obtain from eqns. (14.4.38–40)

ν ≈ 1

2+

1

4ε, ∆ ≈ 3

2+

1

2ε, α ≈ −1

2ε,

β ≈ 1

2, γ ≈ 1 +

1

2ε, δ ≈ 3 + ε, η ≈ 0.

At the same time, we obtain directly from eqns. (13.5.47, 66 and 67), withd = 4− ε where 0 < ε 1,

α ' −ε2, β =

1

2, γ =

2

2− ε' 1 +

1

2ε,

δ =6− ε2− ε

= 31− 1

6ε

1− 12ε' 3

(1 +

1

3ε

)= 3 + ε, η = 0,

ν =1

2− ε' 1

2

(1 +

1

2ε

)=

1

2+

1

4ε.

To the given order in ε, the two sets of results are in complete agreement.

14.8 & 9. For d = 4− ε where 0 < ε 1, eqn. (14.4.46) gives

Sd ≈sin(π − πε/2)Γ(3)

2πΓ(2)2≈ 1

2ε.

Equations (14.4.43–45) then give

η '4ε · 1

2ε

4

1

n=ε2

2n,

γ ' 2

2− ε

(1− 3ε

n

)'(

1 +1

2ε

)1− 3ε

n

' 1 +

(1

2− 3

n

)ε,

α ' − ε

2− ε

(1− 12ε

ε

1

n

)' −ε

2

(1− 12

n

).

141

Next, we obtain

β =1

2(2− α− γ) =

1

2− 2(2d− 5)Sd

d− 2

1

n+O

(1

n2

)'1

2− 3

2nε,

δ = 1 +γ

β= 1 +

4

d− 2

1− 6Sd/n+O(1/n2)

1− 4(2d− 5)/(d− 2) · (Sd/n) +O(1/n2)

= 1 +4

2− ε

[1 +O

(1

n2

)]' 1 + 2

(1 +

1

2ε

)= 3 + ε,

ν =γ

2− η=

1

d− 2

1− 6Sd/n+O(1 + n2)

1− 2(4− d)/d · (Sd/n) +O(1/n2)

' 1

2− ε

(1− 3ε

n

)' 1

2

(1 +

1

2ε

)(1− 3ε

n

)' 1

2

[1 +

(1

2− 3

n

)ε

]=

1

2+

1

4

(1− 6

n

)ε.

All these results agree with the corresponding ones following from eqns. (14.4.38–40).

Chapter 15

15.1. (i) We multiply expression (15.1.11) by ∆T , take its average and utilizerelations (15.1.14), to obtain (∆T∆S) = kT .

(ii) We multiply expression (15.1.12) by ∆V , take its average and utilizerelations (15.1.14), to obtain (∆P∆V ) = −kT .

(iii) We multiply expression (15.1.11) by ∆V , take its average and utilizerelations (15.1.14), to obtain

∆S∆V =

(∂P

∂T

)V

kT

[− 1

V

(∂V

∂P

)T

]V = kT

(∂V

∂T

)P

.

(iv) We multiply expression (15.1.12) by ∆T , take its average and utilizerelations (15.1.14), to obtain (∆P∆T ) = kT 2C−1

V (∂P/∂T )V.

15.2. If we choose ∆S and ∆P as our independent variables, then

∆T =

(∂T

∂S

)P

∆S +

(∂T

∂P

)S

∆P =T

CP∆S +

(∂T

∂P

)S

∆P, and

∆V =

(∂V

∂S

)P

∆S +

(∂V

∂P

)S

∆P =

(∂T

∂P

)S

∆S − V κS∆P.

It follows that

−∆T∆S + ∆P∆V = − T

CP(∆S)2 − V κS(∆P )2,

which converts expression (15.1.8) into (15.1.15), leading directly to ex-pressions (15.1.16) for (∆S)2, (∆P )2 and (∆S∆P ).

For an independent evaluation of these averages, we proceed as in Prob-

142

143

lem 15.1. From eqns. (15.1.11, 12 and 14), we readily obtain

(∆S)2 =C2

V

T 2(∆T )2 +

(∂P

∂T

)2

V

(∆V )2 = k

[CV + TV κT

(∂P

∂T

)2

V

]= kCP ,

(∆P )2 =

(∂P

∂T

)2

V

(∆T )2 +1

κ2TV

2(∆V )2 =

kT

CV

[T

(∂P

∂T

)2

V

+CV

κTV

]

=kT

CV· CPκTV

=kT

κSV, and

(∆S∆P ) =CV

T

(∂P

∂T

)V

(∆T )2 −(∂P

∂T

)V

1

κTV(∆V )2 = 0.

15.3. We start with expression (15.1.6) and eliminate ∆S by writing

∆S =

(∂S

∂E

)0

∆E +

(∂S

∂V

)0

∆V +1

2

[(∂2S

∂E2

)0

(∆E)2

+2

(∂2S

∂E∂V

)0

∆E∆V +

(∂2S

∂V 2

)0

(∆V 2)

]+ . . . .

Replacing (∂S/∂E)0 by 1/T and (∂S/∂V )0 by P/T, and retaining termsup to second order only, expression (15.1.6) takes the form

p ∝ exp

[1

2k

(∂θ

∂E

)0

(∆E)2 + 2

(∂θ

∂Vor

∂π

∂E

)0

∆E∆V +

(∂π

∂V

)0

(∆V )2

],

where θ = 1/T and π = P/T . The covariance matrix of this distributionis given by (

(∆E)2 (∆E∆V )

(∆V∆E) (∆V )2

)= k

(− ∂θ∂E − ∂θ

∂V

− ∂π∂E − ∂π

∂V

)−1

.

The evaluation of the inverse here is rather tricky; the interested readermay consult Kubo (1965), problem 6.2, pp. 382–5, where a complete so-lution, along with the desired results for (∆E)2, (∆V )2 and (∆E∆V ), isgiven.

In passing, we note that two of the aforementioned results are also givenin eqns. (15.1.14 and 18); the third may be obtained as follows: multiply(15.1.17) by ∆V , take its average and utilize relations (15.1.14), to get

(∆E∆V ) =

(∂E

∂V

)T

(∆V )2 =

[T

(∂P

∂T

)V

− P]

kTκTV

= kT

[T

(∂V

∂T

)P

+ P

(∂V

∂P

)T

].

144

15.4. With a given displacement y(x), the overall shape of the string would, onan average, be as shown in Fig. 1. This amounts to a strain, ∆`, in thestring given by the expression

∆` =√x2 + y2 +

√(`− x)2 + y2 − `;

the energy Φ associated with this strain is obviously F∆`. For small y,

Φ(y) ≈ F[y2

2x+

y2

2(`− x)

]=

F`

2x(`− x)y2,

which leads to a probability distribution for y that is Gaussian, with vari-ance

(∆y)2 =kT

F`x(`− x).

For the second part, we refer to Fig. 2 for which

∆` =√x2

1 + y21 +

√(x2 − x1)2 + (y1 − y2)2 +

√(`− x2)2 + y2

2 − `

≈ y21

2x1+

(y1 − y2)2

2(x2 − x1)+

y22

2(`− x2), and hence

Φ(y1, y2) ≈ F

2x1(x2 − x1)(`− x2)

[x2(`− x2)y2

1 − 2x1(`− x2)y1y2 + x1(`− x1)y22

].

This leads to a bivariate Gaussian distribution in the variables y1 and y2,with the covariance matrix(y2

1 y1y2

y2y1 y22

)=

kTx1(x2 − x1)(`− x2)

F

(x2(`− x2) −x1(`− x2)−x1(`− x2) x1(`− x1)

)−1

=kT

F`

(x1(`− x1) x1(`− x2)x1(`− x2) x2(`− x2)

).

15.5. The quantity in question here is

(∆NA)21/2/NA = (kT κT /VA)1/2; (1)

see eqn. (15.1.20). Assuming the gas to be an ideal one, the compressibilityκT may be taken as 1/(nkT ), where n is the particle density in the system;see eqn. (15.2.12). This reduces (1) to the simple expression (1/NA)1/2.

145

For this fraction to be 1 per cent, the volume VA of the subsystem must besuch that it contains, on an average, 104 particles. At normal temperatureand pressure, this volume would be about 3.7× 10−22m3 — for instance,a cube of side 7.2× 10−8m.

15.6. By eqns. (15.2.23) and (15.3.11), and by Note 5, we have

x2 = 2Dt = 2BkTt = kTt/3πηa.

It follows thatk = 3πηax2/Tt .

Substituting the given data, we get: k = 1.18×10−16 erg K−1, which maybe compared with the accepted value of 1.38× 10−16 erg K−1.

15.7. By eqn. (15.3.2), we have

〈v · F〉 = M

⟨v · dv

dt

⟩+

1

B〈v · v〉 =

1

2M

d

dt〈v2〉+

1

B〈v2〉.

Substituting for 〈v2〉 from eqn. (15.3.29) and remembering that B = τ/M ,we get

〈v · F〉 =M

τ

3kT

M− v2(0)

e−2t/τ +

1

B

[3kT

M−

3kT

M− v2(0)

e−2t/τ

]=

3kT

BM=

3kT

τ,

which holds at all t. By tacit assumption, the statement 〈r · F〉 = 0 alsoholds at all t. On the other hand, the quantities 〈v ·F 〉 and 〈r ·F 〉 behavesomewhat differently.

First of all,

〈v ·F 〉 = M

⟨v · dv

dt

⟩=

1

2M

d

dt〈v2〉

If the Brownian particle has already attained thermal equilibrium, then〈v2〉 = 3kT/M and hence 〈v ·F 〉 = 0; if it hasn’t, then

〈v ·F 〉 =M

τ

3kT

M− v2(0)

e−2t/τ ,

which decays exponentially with t. Next, by eqns. (15.3.1 and 5),

〈r ·F 〉 = M

⟨r · dv

dt

⟩= −M

τ〈r · v〉 = −M

2τ

d

dt〈r2〉.

Once again, if the particle has already attained thermal equilibrium, then,by eqn. (15.3.7),

〈r ·F 〉 = −3kT (1− e−t/τ )−→tτ−3kT ;

146

if it hasn’t, then, by eqn. (15.3.31),

〈r ·F 〉 = [−3kT + 3kT −Mv2(0)e−t/τ ](1− e−t/τ )

which too approaches −3kT when t becomes much larger than τ .

15.8. Integrating eqn. (15.3.14) over t, we get

r(t) =

t∫0

v(t′)dt ′ = v(0)[−τe−t

′/τ]t

0+

t∫0

e−t′/τ t′∫0

eu/τA(u)du

dt ′.

(1)The remaining integration may be carried out by parts, with the result(−τe−t

′/τ )

t′∫0

ru/τA(u)du

t0

−t∫

0

(−τe−t′/τ )et

′/τA(t′)dt ′

= −τe−t/τt∫

0

eu/τA(u)du + τ

t∫0

A(t′)dt ′. (2)

Substituting (2) into (1), we obtain the desired result

r(t) = v(0)τ(1− e−t/τ ) + τ

t∫0

1− e(u−t)/τA(u)du. (3)

To obtain an expression for 〈r2(t)〉, we take the square of (3) and averageit over an ensemble. The cross-term vanishes on averaging, and we areleft with

〈r2(t)〉 = v2(0)τ2(1− e−t/τ )2 + τ2

t∫0

t∫0

1− e(u1−t)/τ1− e(u2−t)/τ

〈A(u1) ·A(u2)〉du1du2. (4)

Noting that the autocorrelation function 〈A(u1) · A(u2)〉, which is thesame as the function K(s) of Sec. 15.3, may be treated as a delta function,see the passage from eqn. (15.3.24) to (15.3.25) along with eqns. (15.3.26and 28), we may write

〈A(u1) ·A(u2)〉 = Cδ(u2 − u1), where C = 6kT/Mτ.

The second term in (4) then takes the form

6kT τ

M

t∫0

1− e(u−t)/τ2du

=6kT τ

M

[t− 1

2τ(1− e−t/τ )(3− e−t/τ )

]. (5)

Substituting (5) into (4), we obtain eqn. (15.3.31).

147

15.13. By eqn. (15.5.14), we have in the first case

w(f) = 4

∞∫0

K(0)e−αs2

cos(2πf∗s) cos(2πfs)ds

= 2K(0)

∞∫0

e−αs2

[cos2π(f − f∗)s+ cos2π(f + f∗)s]ds.

Using formula (B.41), we get the desired result

w(f) = K(0)(πα

)1/2

[e−π2(f−f∗)2/α + e−π

2(f+f∗)2/α].

In the limit α→ 0 (with f∗ > 0), w(f)→ K(0)δ(f − f∗); see eqn. (B.43).In the limit f∗ → 0 (with α > 0), w(f)→ 2K(0)(π/α)1/2 exp−(π2f2/α).On the other hand, if both α and f∗ → 0, w(f) tends to be 2K(0)δ(f).In either case, eqn. (15.5.16) is satisfied.

In the second case, we get

w(f) = 2K(0)

[α

α2 + 4π2(f − f∗)2+

α

α2 + 4π2(f + f∗)2

].

Now, in the limit α → 0 (with f∗ > 0), w(f) → 2πK(0)δ2π(f − f∗) =K(0)δ(f −f∗); see eqn. (B.36). In the limit f∗ → 0 (with α > 0), w(f)→4K(0)α/(α2 + 4π2f2). On the other hand, if both α and f∗ → 0, w(f)again tends to be 2K(0)δ(f).

15.14. By eqn. (15.5.14), we get

w(f) = 4

∞∫0

K(0)sin(as) sin(bs)

abs2 cos(2πfs)ds

=2K(0)

ab

∞∫0

sin(as)[sin(b− 2πf)s+ sin(b+ 2πf)s]ds

s2. (1)

To evaluate the integral in (1), we use the formula, see Gradshteyn andRyzhik (1965),

∞∫0

sin(px ) sin(qx )dx

x2=

pπ/2 if p ≤ qqπ/2 if q ≤ p

.

It follows that if 0 < f ≤ (a− b)/2π, then the integral in (1) is equal to

(b− 2πf)π/2 + (b+ 2πf)π/2 = bπ. (2)

148

If (a− b)/2π ≤ f ≤ (a+ b)/2π, then our integral is equal to

(b− 2πf)π/2 + aπ/2 = (a+ b− 2πf)π/2. (3)

If f ≥ (a+ b)/2π, then we have

−aπ/2 + aπ/2 = 0. (4)

Substituting (2)–(4) into (1), we obtain the desired result for w(f).

It is quite straightforward to check that the function w(f) obtained heresatisfies eqn. (15.5.16).

15.15. (a) From the defining equation of the variable Y (t), we get

〈Y 2(t)〉 =

u+t∫u

u+t∫u

〈y(u1)y(u2)〉du1du2 (1)

Since y(u) is statistically stationary, we may write

〈y(u1)y(u2)〉 =

∞∫0

w(f) cos(2πfs)df (s = u2 − u1); (2)

see eqn. (15.5.15). Substituting (2) into (1), we get

〈Y 2(t)〉 =

∞∫0

w(f)I(f, t)df , where (3)

I(f, t) =

u+t∫u

u+t∫u

cos(2πfu2) cos(2πfu1) + sin(2πfu2) sin(2πfu1)du1du2

=

u+t∫u

cos(2πfu)du

2

+

u+t∫u

sin(2πfu)du

2

=1

4π2f2

[[sin2πf(u+ t) − sin(2πfu)]2+

[cos(2πfu)− cos2πf(u+ t)]2]

=1

2π2f2[1− cos(2πft)], regardess of the initial instant u.

(4)

Substituting (4) into (3), we obtain the desired result for 〈Y 2(t)〉.

149

Next, it follows that

∂

∂t〈Y 2(t)〉 =

1

π

∞∫0

w(f)

fsin(2πft)df , and (5)

∂2

∂t2〈Y 2(t)〉 = 2

∞∫0

w(f) cos(2πft)df . (6)

Taking the sine transform of (5) and the cosine transform of (6), weobtain the other quoted results. Finally, a comparison of eqns. (2)and (6) shows that

Ky(s) =1

2

∂2

ds2 〈Y2(s)〉. (7)

(b) If the variable y(u) is the x-component of the velocity of a Brownianparticle, with power spectrum (15.5.21), then eqns. (3) and (4) give

〈x2(t)〉 =2kT τ

π2M

∞∫0

1− cos(2πft)

f21 + (2πfτ)2df

=4kT τ2

πM

∞∫0

1− cos(xt/τ)

x2(1 + x2)dx

=2kT τ2

M

[t

τ− (1− e−t/τ )

], (8)

in complete agreement with eqn. (15.3.7) for the quantity 〈r2(t)〉. Wealso note that

1

2

∂2

∂s2〈x2(s)〉 =

kT

Me−s/τ (s > 0), (9)

which indeed is equal to the autocorrelation function K(s) of thevariable vx; see eqn. (15.6.20).

15.16. First we’ll prove the following lemma.

Lemma:

For a given variable x(t), define a complementary function

yx(f, T ) =1√T

T/2∫−T/2

x(t)e−2πiftdt . (1)

The power spectrum of the variable x(t) is then given by

wx(f) = 2 limT→∞

|yx(f, T )|2. (2)

150

Proof:

From (1), it readily follows that

|yx(f, T )|2 =1

T−

T/2∫−T/2

T/2∫−T/2

x(t1)x(t2)e2πif (t2−t1)dt1dt2.

Changing over to the variables S and s, as defined in eqns. (15.3.23), weget

|yx(f, T )|2 =1

T

∫ ∫x

(S − 1

2s

)x

(S +

1

2s

)cos(2πfs)dSds.

Integrating over S and letting T → ∞ amounts to taking an ensembleaverage of the quantity x

(S − 1

2s)x(S + 1

2s); this reduces the above ex-

pression to∞∫−∞

Kx(s) cos(2πfs)ds

which, by eqn. (15.5.14), is equal to 12wx(f). Hence the lemma.

We now proceed to establish the stated relation between the power spectrawv(f) and wA(f). For this we refer to eqn. (15.3.5) for the variable A(t)and construct its complementary function

yA(f, T ) =1√T

T/2∫−T/2

(dv

dt+

v

τ

)e−2πiftdt . (3)

The first part here gives

1√T

ve−2πift∣∣∣T/2−T/2

−T/2∫−T/2

v(−2πif )e−2πiftdt

.Equation (3) then becomes

yA(f, T ) =1√T

[v

(T

2

)e−πift − v

(−T

2

)eπifT

]+

(1

τ+ 2πif

)yv(f, T ).

Since the variable v(t) is bounded, the limit T →∞ gives

|yA(f, T )|2 ≈∣∣∣∣1τ + 2πif

∣∣∣∣2 |yv(f, T )|2. (4)

Using lemma (2), we finally get

wA(f) =

[1

τ2+ (2πf)2

]wv(f), (5)

151

which is the desired result.

Now, by eqn. (15.5.21),

wv(f) =12kT τ

M

1

1 + (2πfτ)2. (6)

Substituting (6) into (5), we readily obtain the stated result for wA(f).Note that this result is consistent with the assertion that, for most practi-cal purposes, the autocorrelation function KA(s) may be taken as Cδ(s),with C = 6kT/Mτ ; see eqns. (15.3.26 and 28).

15.17. (a) Using eqn. (15.3.14), we construct the quantity v(t) · v(t + s) andaverage it over an ensemble. The cross-term vanishes on averaging,and we are left with

〈v(t)·v(t+s)〉 = e−2(t+s)/τ

v2(0) +

t∫0

t+s∫0

e(u1+u2)τ 〈A(u1) ·A(u2)〉du1du2

.(1)

In view of the argument leading from eqn. (15.3.24) to (15.3.25), wemay replace the function 〈A(u1) ·A(u2)〉 by the singular expressionCδ(u2 − u1), where C = 6kT/Mτ . At the same time, we observethat the integral

t∫0

e(u1+u2)/τδ(u1 − u2)du1 =

e2u2/τ if 0 < u2 < t

0 otherwise.

The double integral in (1) is then equal to

t∫0

Ce2u2/τdu2 = Cτ

2(e2t/τ − 1) if s > 0, and

t+s∫0

Ce2u2/τdu2 = Cτ

2(e2(t+s)/τ − 1) if s < 0.

Substituting these results into (1), we obtain eqns. (15.6.7) and (15.6.8).Equation (15.6.9) follows straightforwardly.

(b) To evaluate 〈r2(t)〉, we write eqn. (15.6.9) in the form

Kv(s) =

(v2(0)− 3kT

M

)e−2S/τ +

3kT

Me−|s|/τ , (2)

where S = 12 (u1 + u2) and s = (u2 − u1). Substituting (2) into

152

eqn. (15.6.6), we get

〈r2(t)〉 =

(v2(0)− 3kT

M

) t/2∫0

e−2S/τ · 4SdS +

t∫t/2

e−2S/τ · 4(t− S)dS

+

3kT

M

t/2∫0

2

2S∫0

e−2s/τdsdS +

t∫t/2

2

2(t−S)∫0

e−2s/τdsdS

=

(v2(0)− 3kT

M

)· τ2(1− e−t/τ )2 +

3kT

M· 2τt− τ(1− e−t/r),

which is the same as expression (15.3.31). Note that the second partof this result is identical with expression (15.3.7) that pertains to astationary ensemble.

15.18. Using equation (15.3.37), the response function is

χvx(ω) =

∫ ∞0

χvx(s)eiωsds

which has imaginary part

χ′′vx(ω) =1

M

ω(ω2 − ω2

0

)(ω2 − ω2

0)2

+ γ2ω2.

The correlation function is

Gvx(t− t′) =

∫ t

−∞ds

∫ t′

−∞ds′χvx(t− s)χxx(t′ − s′) 〈F (s)F (s′)〉

Using 〈F (s)F (s′)〉 = 2γMkTδ(s− s′) and Fourier transforming gives

Svx(ω) =2kT

M

(ω2 − ω2

0

)(ω2 − ω2

0)2

+ γ2ω2.

15.19. Just differentiate equation (15.6.29) with respect to t and equation (15.6.28)drops out.

Correction to the first printing of third edition: 15.20 and 15.21: Thecorrelation function relation should read:

GAB(t) = GBA(−t− iβ~).

153

15.20. Since we need to evaluate 〈[A(t), B(0)]〉, we need to relate 〈B(0)A(t)〉 to〈A(t)B(0)〉.

〈B(0)A(t)〉 =1

QTr(BeiHt/~Ae−iHt/~e−βH

)=

1

QTr(eiHt/~+βHAe−iHt/~−βHBe−βH

)= 〈A(t− iβ~)B(0)〉

Now equation (15.6.34) can be evaluated as

χ′′AB(ω) =1

2~

∫(〈A(t)B(0)〉 − 〈A(t− iβ~)B(0)〉) eiωtdt

=1

2~

∫ (〈A(t)B(0)〉eiωt − 〈A(t− iβ~)B(0)〉eiω(t−iβ~)e−β~ω

)dt

=1

2~(1− e−β~ω

)SAB(ω)

15.21. Since 〈B(0)A(t)〉 = 〈A(t− iβ~)B(0)〉,

〈A(t)B(0)−B(0)A(t)〉 ≈ iβ~⟨dA

dtB

⟩as ~→ 0. Therefore,

χ′′AB(ω) =1

2~

∫〈A(t)B(0)−B(0)A(t)〉eiωtdt ≈ iβ~

2~

∫ ⟨dA

dtB(0)

⟩eiωtdt

=βω

2

∫〈A(t)B(0)〉 eiωtdt =

ω

2kTSAB(ω)

15.22. The self–diffusion term can be written

Sself(ω) =

∫〈e−ik·(r(t)−r(0))〉eiωtdt

=

∫eiωt−Dk

2|t|dt,

using the diffusion relation 〈(x(t)− x(0))2〉 = 2D|t|. Integrating gives

Sself(ω) =2Dk2

ω2 + (Dk2)2;

compare to the heat diffusion term in equation (15.6.45).

15.23. The magnitude of the wavevector transfer is k =√

2k0 = 7× 106 m−1 andthe width of the Rayleigh peak is ∆ω = DT k

2 = 7×106 s−1. The locationof the sound peak is at ω = ck = 2.4× 109 m−1 is well separated from theRayleigh peak.

15.24. The Raman peak has ~ω = 0.05 eV ' 2kT at room temperature so thepeaks are not symmetric. Since Γ ∼ 1012 s−1 and ω ∼ 8 × 1013 s−1, theRaman peak is well resolved.

Chapter 16

16.1. Here is a C code snippet for a pseudorandom number generator basedon the L’Ecuyer prime number linear congruential generator discussed inAppendix I.

double rand(double seed[])

seed[0] = fmod(seed[0] * 40014., 2147483563.);

seed[1] = fmod(seed[1] * 40692., 2147483399.);

double r=seed[0]- seed[1];

if (r<= 0.0) r += 2147483562.;

return r/2147483563.;

For a sequence of N numbers, one should test that 〈x〉 ≈ 0.5± 1/(12√N)

and 〈x2〉 − 〈x〉2 ≈ 1/12.

16.2 Here is a code snippet for generating gaussian pseudorandom numbersbased on the Box-Muller algorithm in Appendix I.

double s,w;

do

double x=2.0*rand(seed)-1.0;

double y=2.0*rand(seed)-1.0;

s=x*x+y*y;

while( s >= 1.0);

w=sqrt(-2.0*log(s)/s);

gaussrand = x*w;

For efficiency, one can also use y*w as an independent gaussian pseudo-random number. For a sequence of N numbers, one should test that〈x〉 ≈ 0.0 ± 1/

√N and 〈x2〉 ≈ 1.0. The reader should also determine the

expected uncertainty in the value of the variance for N numbers. The his-togram of points for pairs of gaussian random numbers should be centeredat 0, be isotropic, and have variance 〈x2 + y2〉 = 2.

154

155

16.3 First note that the sum of two gaussian random distrubutions is alsogaussian,

P1(s1) =exp

(− s21

2σ21

)√

2πσ21

, P2(s2) =exp

(− s22

2σ22

)√

2πσ22

,

P (S = s1 + s2) =

∫ ∫δ(S − s1 − s2)P1(s1)P2(s2)ds1ds2,

P (S) =exp

(− S2

2(σ21+σ2

2)

)√

2π(σ21 + σ2

2).

Iterating the equation defining the correlated random numbers gives

sk = (1− α)

∞∑j=0

αjrk−j .

This implies that the s’s are also gaussian. The averages 〈sk〉 are clearlyzero and the variance is given by

〈s2k〉 = (1− α)2

∞∑i=0

∞∑j=0

αi+j〈rk−irk−j〉,

which is easily evaluated to give

〈s2k〉 = (1− α)2

∞∑j=0

α2j =(1− α)2

1− α2=

1− α1 + α

.

The correlations are then given by

〈sksk−l〉 = α|l|1− α1 + α

.

16.4 A Monte Carlo Sweep of an ordered list of N particlesx0 < x2 < x3 < · · · < xN−1 is done with the following C code snippet.

xtrial = x[0] + dx*(rand(seed)-0.5);

if (xtrial - (x[n-1] - L) > 1.0 && x[1] - xtrial > 1.0) x[i]=xtrial;

for (int i=1;i<n-1;i++)

xtrial=x[i] + dx*(rand(seed)-0.5);

if (xtrial - x[i-1] > 1.0 && x[i+1]-xtrial > 1.0) x[i]=xtrial;

xtrial = x[n-1] + dx*(rand(seed)-0.5);

if (xtrial - x[n-2] > 1.0 && (x[0] + L) - xtrial > 1.0) x[i]=xtrial;

Note the periodic boundary conditions treat particle to the left of particleparticle 0 as particle N − 1 shifted left by L, and particle to the right of

156

particle N − 1 as particle 0 shifted to the right by L. The random stepsize dx is typically chosen on the order of L/(DN) − 1 but must be lessthan 2 to avoid particles getting out of order.

16.5 Here is a C code snippet for a Monte Carlo of a two-dimensional systemof hard spheres in a LX × LY periodic box.

int i = n*rand(seed); // choose a particle randomly

double xtrial = x[i] + dx*(rand(seed) -0.5);

double ytrial = y[i] + dx*(rand(seed) -0.5);

int collision = 0;

for (int j=0;j<n;j++)

if( j != i)

double dx = fabs(x[j]-x[i]);

double dy = fabs(y[j]-y[i]);

if (dx > halfLX) dx = LX-dx; // use periodic boundary conditions

if (dy > halfLY) dy = LY-dy; // halfLX=0.5*LX and halfLY=0.5*LY

if (dx*dx+dy*dy < 1.0) collision=1; // test for collision

if (collision == 1) break;

if (collision == 0) //accept trial position if no collision

x[i] = xtrial;

y[i] = ytrial;

if (x[i] > LX) x[i] -= LX; // impose periodic boundary conditions

if (y[i] > LY) y[i] -= LY;

if (x[i] < 0.0) x[i] += LX;

if (y[i] < 0.0) y[i] += LY;

Here is a C code snippet to collect correlation function information.

for (int i=0;i<n;i++) for (int j=i+1;j<n;j++)

double dx = fabs(x[j]-x[i]);

double dy = fabs(y[j]-y[i]);

if (dx > halfLX) dx = LX-dx; // use periodic boundary conditions

if (dy > halfLY) dy = LY-dy; // halfLX=0.5*LX and halfLY=0.5*LY

int ir=sqrt(dx*dx+dy*dy)/dr; // dr is the binsize

histogram[ir] ++; //increment the histogram

16.6 The new additions to the code accept moves ∆y > 0 with probabilityexp(−βmg∆y), and to reject moves that go outside the vertical bound-aries. At low density and small βmgLy, the density will be proportional toexp(−βmgy). Large βmgLy will result in an interface with a low densityphase above a high density phase.

157

16.7 Here is a C code snippet for one time step for Lennard-Jones particles ina two dimensional LX×LY box with periodic boundary conditions. Thearrays x1 and x0 store the current and previous positions of the n particlesrespectively.

// calculate forces

for (int i=0;i<n;i++)

fx[i]=0.0;

fy[i]=0.0;

for (int i=0;i<n;i++) for (int j=i+1;j<n;j++)

double dx=x1[j]-x1[i];

double dy=y1[j]-y1[i];

if (dx > halfLX) dx -= LX; // use periodic boundary conditions

if (dy > halfLY) dy -= LY; // halfLX=0.5*LX and halfLY=0.5*LY

if (dx < -halfLX) dx += LX;

if (dy < -halfLY) dy += LY;

double r2=1.0/(dx*dx+dy*dy);

double r4=r2*r2;

double r6=r2*r4;

double r8=r4*r4;

double r14=r8*r6;

double f0=48.0*r14 - 24.0*r8; // see equation (16.3.5)

double fx0=f0*dx;

double fy0=f0*dy;

fx[j] += fx0; // Use Newtons’s third law to update forces on each particle

fy[j] += fy0;

fx[i] -= fx0;

fy[i] -= fy0;

//update positions using Verlet

for (int i=0;i<n;i++)

double xnew=2*x1[i]-x0[i]+dtsqr*fx[i];

double ynew=2*y1[i]-y0[i]+dtsqr*fy[i];

x0[i]=x1[i];

x1[i]=xnew;

y0[i]=y1[i];

y1[i]=ynew;

// impose periodic boundary conditions

for (int i=0;i<n;i++)

if (x1[i] > LX) x1[i] -= LX; x0[i] -= LX;

158

if (y1[i] > LY) y1[i] -= LY; y0[i] -= LY;

if (x1[i] < 0.0) x1[i] += LX; x0[i] += LX;

if (y1[i] < 0.0) y1[i] += LY; y0[i] += LY;

16.8 The new additions are to generate a one–body force Fy = −mg and re-pulsive forces with the top and bottom walls. The average kinetic energyper particle will be independent of the position in the box. At low densityand small βmgLy, the density will be proportional to exp(−βmgy). LargeβmgLy will result in an interface with a low density phase above a highdensity phase.

16.9 Each Monte Carlo step involves determining the energy change of a spinflip with is proportional to ∆ = si(si+1 + si−1) using periodic boundaryconditions. If ∆ ≤ 0 flip the spin. Otherwise ∆ = +2, so flip the spinwith probability exp(−4K). Due to the periodic boundary conditionsthe correlation function will also be periodic. You can generalize thecalculation of the correlation function in section 13.2 for a finite periodiclattice to show that the zero field correlation function is of form

〈sisj〉 =1

1 +(λ2

λ1

)N[(

λ2

λ1

)|i−j|+

(λ2

λ1

)N−|i−j|],

so the correlations are minimized halfway across the lattice.

159

16.10 Here is a C code snippet for the two dimensional Ising model

int L=32; // size of lattice

int n=L*L; // number of sites

double K=-0.5*log(sqrt(2.0)-1); // K=critical value

int nstat = 1000000; // number of Monte Carlo Sweeps

int neq=100000; // number of equilibration sweeps

int* s=new int[n]; // spins: +1 or -1

for (int i=0;i<n;i++) s[i]=1; // all spins initially up (+1)

int* i1 = new int[n]; // arrays i1[], i2[], i3[], i4[]: neighbor sites

int* i2 = new int[n];

int* i3 = new int[n];

int* i4 = new int[n];

for (int i=0;i<n;i++)

i1[i] = i+1; // site to right

i2[i] = i-1; // site to left

i3[i] = i+L; // site above

i4[i] = i-L; // site below

if ((i1[i] % L) == 0) i1[i] -= L; //implement periodic boundary conditions

if (((i2[i]+L) % L) == L-1) i2[i] += L; //implement periodic boundary conditions

if (i3[i] >= n) i3[i] -= n; //implement periodic boundary conditions

if (i4[i] < 0) i4[i] += n; //implement periodic boundary conditions

double* boltz=new double[5]; //precompute spin flip Boltzmann factors for efficiency

boltz[2]=exp(-4.0*K); // energy increase = 4

boltz[4]=exp(-8.0*K); // energy increase = 8

int* e=new int[nstat]; //stored energy after each pass

int* m=new int[nstat]; //stored magnetization after each pass

int energy;

int mag;

for (int iter=0; iter<(nstat+neq);iter++) // perform nstat Monte Carlo Sweeps

//after neq equilibration steps

for (int ii=0;ii<n;ii++)

int i=n*rand(seed); //choose a random site

int neighborsum=s[i1[i]]+s[i2[i]]+s[i3[i]]+s[i4[i]]; //sum of spins on

//neighboring sites

int de = s[i]*neighborsum; // energy change of spin flip is 2*de

if (de <= 0) s[i]=-s[i]; // accept if energy change is not positive

else if (rand(seed) < boltz[de]) s[i]=-s[i]; //if energy increase,

//accept with Boltzmann factor probability

if (iter >= neq)

160

mag=0;

energy=0;

for (int i=0;i<n;i++)

mag += s[i];

energy -= s[i]*(s[i1[i]]+s[i3[i]]);

e[iter-neq] = energy; //store energy for later analysis

m[iter-neq] = mag; //store magnetization for later analysis

// collect other statistics here, especially for correlations

16.11 Use the code snippet to collect a histogram of energies. Use the codeposted at www.elsevierdirect.com to calculate the energy distribution atK = 0.4,Kc = 0.4406868, 0.5. Here is a plot. The horizontal axis is theenergy above the ground state in units of 4J and the vertical axis is theprobability for each energy.

0 100 200 300 400 5000.000

0.005

0.010

0.015

0.020

0.025

0.030

16.12 Each Monte Carlo step will involve involve creating a trial state(θtrial = θi + ∆θ(rand(seed)− 0.5)) and calculating the change in energy,

∆ε = cos(θi + θi+1)− cos(θi − θi−1)

− cos(θtrial − θi+1)− cos(θtrial − θi−1).

Accept (i.e. set θi = θtrial) if ∆ε < 0 or rand(seed) < exp(−β∆ε).

161

16.13 Here is is C code snippet for the two dimensional XY model

int L=32; // size of lattice

int n=L*L; // number of sites

double K=1.12; // K=critical value

int nstat = 1000000; // number of Monte Carlo Sweeps

int neq=100000; // number of equilibration sweeps

double* theta=new double[n]; // angles of spins

double dtheta=1.0; // range for random angle changes

double twopi=8.0*atan(1.0);

for (int i=0;i<n;i++) theta[i]=0.0; // all spins initially along x direction

int* i1 = new int[n]; // arrays i1[], i2[], i3[], i4[]: neighbor sites

int* i2 = new int[n];

int* i3 = new int[n];

int* i4 = new int[n];

for (int i=0;i<n;i++)

i1[i] = i+1; // site to right

i2[i] = i-1; // site to left

i3[i] = i+L; // site above

i4[i] = i-L; // site below

if ((i1[i] % L) == 0) i1[i] -= L; //implement periodic boundary conditions

if (((i2[i]+L) % L) == L-1) i2[i] += L; //implement periodic boundary conditions

if (i3[i] >= n) i3[i] -= n; //implement periodic boundary conditions

if (i4[i] < 0) i4[i] += n; //implement periodic boundary conditions

double* e=new double[nstat]; //stored energy after each pass

double* mx=new double[nstat]; //stored x component of magnetization after each pass

for (int iter=0; iter<(nstat+neq);iter++) // perform nstat Monte Carlo Sweeps

//after neq equilibration steps

for (int ii=0;ii<n;ii++)

int i=n*rand(seed); //choose a random site

double thetatrial = fmod(theta[i] + dtheta*(rand(seed)-0.5) + twopi , twopi);

double de=cos(theta[i]-theta[i1[i]]) - cos(thetatrial-theta[i1[i]])

+ cos(theta[i]-theta[i2[i]]) - cos(thetatrial-theta[i2[i]])

+ cos(theta[i]-theta[i3[i]]) - cos(thetatrial-theta[i3[i]])

+ cos(theta[i]-theta[i4[i]]) - cos(thetatrial-theta[i4[i]]);

if (de<0.0) theta[i]=thetatrial; //accept if energy decreases

else if (rand(seed) < exp(K*de) theta[i]=thetatrial; //or with Boltzmann factor

if (iter >= neq)

double magx=0.0;

double magy=0.0;

162

double energy=0.0;

for (int i=0;i¡n;i++)

magx += cos(theta[i]);

magy += sin(theta[i]);

energy -= (cos(theta[i]-theta[i1[i]])+cos(theta[i]-theta[i3[i]]));

e[iter-neq] = energy; //store energy for later analysis

mx[iter-neq] = magx; //store x magnetization for later analysis

my[iter-neq] = magy; //store y magnetization for later analysis

// collect other statistics here, especially for correlations