Solution Manual for Statistical Mechanics, Edition Fourth R.K. Pathria, Paul D. Beale

SOLUTION MANUAL

Instructor’s Manual Containing Solutions to Over 300 Problems Selected From STATISTICAL MECHANICS (FOURTH EDITION) By R. K. PATHRIA and PAUL. D. BEALE

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Preface This instructor’s manual for the fourth edition of Statistical Mechanics is based on RKP’s instructor’s manual for the second edition. Most of the solutions here were retypeset into TeX from that manual. PDB is responsible for the solutions of the new problems added in the third and fourth editions. The result is a manual containing solutions to over 300 problems selected from the fourth edition.

The original idea of producing an instructor’s manual first came from RKP’s friend and colleague Wing-Ki Liu in the 1990’s when RKP had just embarked on 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 otherwise spend on solving these problems oneself. Secondly, before one selects problems either for homework or for an exam, one can consult the manual to determine the level of difficulty of the various problems and make one’s selection accordingly. Thirdly, one may even use some of these solved problems, especially the ones appearing in later chapters, as lecture material, thereby supplementing the 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 this manual with students or post any part of this manual on the web. Students learn best when they work together and struggle over difficult problems. Readily available solutions interfere with this crucial aspect of graduate physics training.

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

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Chapter 1

1.1. (a) We expand the quantity ln Ω(0) (E1 ) as a Taylor series in the variable (E1 − Ē1 ) and get ln Ω(0) (E1 ) ≡ lnΩ1 (E1 ) + ln Ω2 (E2 )

(E2 = E (0) − E1 )

= {ln Ω1 (Ē1 ) + ln Ω2 (Ē2 )}+

∂ ln Ω1 (E1 ) ∂ ln Ω2 (E2 ) ∂E2 + (E1 − Ē1 )+ ∂E1 ∂E2 ∂E1 E1 =Ē1 (

2 ) 1 ∂ 2 ln Ω1 (E1 ) ∂ 2 ln Ω2 (E2 ) ∂E2 + (E1 − Ē1 )2 + · · · .

2 ∂E12 ∂E22 ∂E1 E1 =Ē1

The first term of this expansion is a constant, the second term vanishes as a result of equilibrium (β1 = β2 ), while the third term may be written as

2 1 ∂B2 1 1 1 ∂β1 + E1 − Ē1 = − + (E1 −Ē1 )2 , 2 ∂E1 ∂E2 eq.

2 kT12 (Cv )1 kT22 (Cv )2 with T1 = T2 . Ignoring the subsequent terms (which is justified if the systems involved are large) and taking the exponentials, we readily see that the function Ω0 (E1 ) is a Gaussian in the variable (E1 − Ē1 ), with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that if (Cv )2 >> (Cv )1 — corresponding to system 1 being in thermal contact with a very large reservoir — then the variance becomes simply kT 2 (Cv )1 , regardless of the nature of the reservoir; cf. eqn. (3.6.3). (b) If the systems involved are ideal classical gases, then (Cv )1 = 23 N1 k and (Cv )2 = 32 N2 k; the variance then becomes 32 k 2 T 2 · N1 N2 /(N1 + N2 ). Again, if N2 >> N1 , we obtain the simplified expression 3 2 2 2 N1 k T ; cf. Problem 3.18.

1.2. Since S is additive and Ω multiplicative, the function f (Ω) must satisfy the condition f (Ω1 Ω2 ) = f (Ω1 ) + f (Ω2 ).

(1) 5

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CHAPTER 1.

Differentiating (1) with respect to Ω1 (and with respect to Ω2 ), we get Ω2 f 0 (Ω1 Ω2 ) = f 0 (Ω1 )

and Ω1 f 0 (Ω1 Ω2 ) = f 0 (Ω2 ),

so that Ω1 f 0 (Ω1 ) = Ω2 f 0 (Ω2 ).

(2)

Since the left-hand side of (2) is independent of Ω2 and the right-hand side is independent of Ω1 , each side must be equal to a constant, k, independent of both Ω1 and Ω2 . It follows that f 0 (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 written as

N −1 N −1 X X jv0 jv0 N 2 v0 ln 1 − − C+N ln V + ' C+N ln V + ' C+N ln V − .

V V 2V j=1 j=1 Equation (1.4.2) is then replaced by

P N N 2 v0 N N v0 = + = 1 + , i.e. kT V 2V 2 V 2V

−1 N v0 PV 1 + = NkT .

2V Since N v0 << V, (1 + N v0 /2V )−1 ' 1 − N v0 /2V . Our last result then takes the form: P (V − b) = NkT , where b = 12 N v0 .

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

1 4π 3 4π N· σ = 4N · 2 3 3

1 σ 2

3 .

1.5. This problem is essentially solved in Appendix A; all that remains to be done is to substitute from eqn. (B.12) into (B.11), to get X 1

(ε∗ ) =

(πε∗1/2 /L)3 (πε∗1/2 /L)2 V ∓ S. 6π 2 16π

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

The expression for T now follows straightforwardly; we get

1 ∂ ln Ω k ∂ ln Ω k R+N k Nhν =k = = ln = ln 1 + , T ∂E hν ∂R hν R hν E N N so that

hν T = k

Nhν ln 1 + .

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 extensive quantity, we may write

E V E V , S(N, V, E) = Nf = Nf (v, ε) v = ,ε = .

N N N N It follows that

∂f −V ∂f −E ∂S =N f +N · 2 +N · 2 , N ∂N V,E ∂v ε N ∂ε v N

∂f ∂S. ∂f 1 ∂S = VN · = EN · .

V ∂V N,E ∂v ε ∂E N,V ∂ε v N Adding these expressions, we obtain the desired result.

1.11. Clearly, the initial temperatures and the initial particle densities of the two gases (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 of generality, we may keep N1 , N2 and V1 fixed and vary only V2 . The first and second derivatives of this expression are then given by

N1 + N2 N2 N1 + N2 N2 k − and k − + 2 (1a,b) V1 + V2 V2 (V1 + V2 )2 V2 respectively. Equating (1a) to zero gives the desired condition, viz.

N1 V2 = N2 V1 , i.e. N1 /V1 = N2 /V2 = n, say. Expression (1b) then reduces to

n n knV1 k − + = > 0.

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

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CHAPTER 1. (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), the expression 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 N [−N ln α + N ln(1 − α)] and − − (2a,b) α 1−α respectively. Equating (2a) to zero gives the condition α = 1/2, which reduces (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 the maximum is N ln 2.

1.13. Proceeding with eqn. (1.5.1), with T replaced by Ti , it is straightforward to see that the extra contribution to ∆S, owing to the fact that T1 6= T2 , is given by the expression 3 3 N1 k ln (Tf /T1 ) + N2 k ln(Tf /T2 ), 2 2 where Tf = (N1 T1 + N2 T2 )/(N1 + N2 ). It is worth checking that this expression is always greater than or equal to zero, the equality holding if and only if T1 = T2 . Furthermore, the result quoted here does not depend on whether the two gases were different or identical.

1.14. By eqn. (1.5.1a), given on page 24 of the text, we get (∆S)v =

3 Nk ln(Tf /Ti ).

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

kT 3 5 2πmkT S = Nk ln + Nk + ln .

P 2 3 h2 It follows that (∆S)P =

5 5 Nk ln(Tf / Ti ) = (∆S)V .

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A numerical verification of this result is straightforward.

It should be noted that, quite generally, (∆S)P T (∂S / ∂T )P CP = = =γ (∆S)V T (∂S / ∂T )V CV which, in the present case, happens to be 5/3.

(1b)

9 1.15. For an ideal gas, CP − CV = nR, where n is the number of moles of the gas. With CP /CV = γ, one gets CP = γnR / (γ − 1)

and CV = nR / (γ − 1).

For a mixture of two ideal gases, CV =

n1 R n2 R + = γ1 − 1 γ2 − 1

f1 f2 + γ1 − 1 γ2 − 1

(n1 + n2 )R.

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

1.16. In view of eqn. (1.3.15), E − TS + PV = µN . It follows that dE − TdS − SdT + PdV + VdP = µdN + Nd µ.

Combining this with eqn. (1.3.4), we get −SdT + VdP = Nd µ, 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 NkT P = and µ = kT ln V

(

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 this system

1 ∂P = P. ∂µ T kT which is indeed equal to N/V , whereas " (

3/2 )# ∂P 5 µ 5 N h2 Nk = P− 2 P = 2 − ln ∂T µ 2T V 2πmkT V kT 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 angle ϕ. The conjugate variable pϕ is then mρ2 ϕ̇, where the various symbols have their usual meanings. The energy of rotation is given by E=

1 m(ρϕ̇)2 = p2ϕ / 2mρ2 .

2

Lines of constant energy in the (ϕ, pϕ )-plane are “straight lines, running parallel to the ϕ-axis from ϕ = 0 to ϕ = 2π”. The basic cell of area h in this 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 are n2 ~2 /2I, where I = mρ2 and n = 0, ±1, ±2, . . .

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

2.4. The rigid rotator is a model for a diatomic molecule whose internuclear distance r may be regarded as fixed. The orientation of the molecule in space may be denoted by the angles θ and ϕ, the conjugate variables being pθ = mr 2 θ̇ and pϕ = mr 2 sin2 θϕ̇. The energy of rotation is given by p2ϕ 1 1 p2θ M2 m(rθ̇)2 + m(r sin θϕ̇)2 = + = , 2 2 2mr 2 2I 2mr 2 sin2 θ

where I = mr 2 and M 2 = p2θ + p2ϕ / sin2 θ .

E=

The “volume” of the relevant region of the phase space is given by the R0 integral dp θ dp ϕ dθ dϕ, where the region of integration is constrained by the value of M . A little reflection shows that in the subspace of pθ and pϕ we are restricted by an elliptical boundary with semi-axes M and M sin θ, the enclosed area being πM 2 sin θ. The “volume” of the relevant region, therefore, is Zπ

Z2π

θ=0

ϕ=0

(πM 2 sin θ)dθ dϕ = 4π 2 M 2 .

1

2 The number of microstates available to the rotator is then given by 4π 2 M 2 /h2 , which is precisely (M/~)2 . At the same time, the number of microstates associated with the quantized value Mj2 = j(j + 1)~2 may be estimated as

i 1 h 2 1 3 1 1 2 Mj+ 1 − Mj− 1 = j + j+ − j− j+ = 2j + 1.

2 2 ~2 2 2 2 2 This is precisely the degeneracy arising from the eigenvalues that the azimuthal quantum number m has, viz. j, j − 1, . . . , −j + 1, −j.

2.6. In terms of the variables θ and L(= m`2 θ), the state of the simple pendulum is given by, see eqns. (2.4.9), θ = (A/`) cos(ωt + ϕ), L = −m`ωA sin(ωt + ϕ), with E = 12 mω 2 A2 and τ = 2π/ω.

The trajectory in the (θ, L)-plane is

given by the equation θ2 L2 + = 1, (A/`)2 (m` ωA)2 which is an ellipse — just like in Fig. 2.2. The enclosed area turns out to be πmωA2 , which is precisely equal to the product Eτ .

2.7. Following the argument developed on page 70 of the text, the number of microstates for a given energy E turns out to be

1 R = E − N ~ω /~ω.

2

Ω(E) = (R + N − 1)!/ R!(N − 1)!,

(1)

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 ∆ E N −1 ∆ · = .

(N − 1)! ~ω (N − 1)!(~ω)N

(1)

The “volume” of the relevant region of the phase space may be derived from the integral Z 0Y N i=1

(dq i dp i ), with

N X 1 i=1

2

kq 2i +

1 2 p 2m i

≤ E.

This is equal to, see eqn. (7a) of Appendix C, 12 N N N 1 2 πN N 2π E (2m) 2 N · E = , k N! ω N!

3 where ω =

p

k/m. The “volume” of the shell in question is then given by

2π ω

N

NE N −1 ·∆= N!

2π ω

N

E N −1 ∆ .

(N − 1)!

(2)

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

2.8. We write V3N = AR 3N , so that dV 3N = A · 3NR 3N −1 dR. At the same time, we have Z∞

Z∞ ...

0

N P

−

e

i=1

N ri Y i=1

0

∞

ri2 dr i =

N Z Y

e−ri ri2 dr i = 2N .

(1)

i=1 0

The integral on the left may be written as Z∞

e−R (4π)−N dV 3N =

0

Z∞

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

0

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

The “volume” of the relevant region of the phase space is given by Z 0 Y 3N i=1

dq i dp i = V N

Z 0 Y N

4πp2i dp i = V N (8π E 3 / c3 )N / (3N )!,

i=1

so that Σ(n, V, E) = V N (8π E 3 / h3 c3 )N / (3N )!, which is a function of N and VE 3 . An isentropic process then implies that VE 3 = const.

The temperature of the system is given by

∂(k ln Σ) 3Nk 1 = = , i.e. E = 3NkT .

T ∂E E N,V 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 thermodynamics follows 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 ( ) X k −βEr ln Γ = k ln e + kβU, N r which is indeed equal to −(A/T ) + (U/T ) = S. For the second part, we use eqn. (3.2.5), with the result that " # X k k ln W {n∗r } = N ln N − n∗r ln n∗r N N r

X n∗ n∗r n∗ r = −k ln = −k ln r .

N N N r Substituting for n∗r from eqn. (3.2.10), we get ( ) X k ∗ −βEr ln W {nr } = kβhEr i + k ln e , N r 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 an extensive quantity, we may write A(N, V, T ) = Nf (v, T )

(v = V / N ).

It follows that

∂A ∂f −V ∂A ∂f 1 N =N f +N · 2 , and V = VN · . ∂N V,T ∂v T N ∂V N,T ∂v T N Adding these expressions, we obtain the desired result.

3.6. Let’s go to part (c) P right away. Our problem here is to maximize the P expression S/k = − Pr,s ln Pr,s , subject to the constraints Pr,s = r,s

r,s

1

2 1,

P

Es Pr,s = E and

r,s

P

Nr Pr,s = N . Varying P ’s and using the method

r,s

of Lagrange’s undetermined multipliers, we are led to the condition X {−(1 + ln Pr,s ) − γ − βEs − αNr } δPr,s = 0. r,s

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 N̄ and Ē.

In the absence of the constraint imposed by N̄ , the parameter α does not even figure in the calculation, and we obtain Pr ∝ exp(−βEr ), as desired in part (b). And if the constraint imposed by Ē is also absent, we obtain Pr = const., as desired in part (a).

3.7. From thermodynamics,

CP − CV = T

∂P ∂T

V

∂V ∂T

= −T

P

∂P ∂T

2 V

∂P ∂V

> 0.

(1)

T

From Sec. 3.3,

P =−

∂A ∂V

= kT

N,T

∂ ln Q ∂V

.

(2)

N,T

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

For the ideal gas, Q ∝ V N T 3N/2 . Therefore, (∂ ln Q/∂V )T = N/V . We then get (N/V )2 CP − CV = −k = Nk . −N/V 2 3.8. For an ideal gas, Q1 = V

(2πmkT )3/2 NkT (2πmkT )3/2 = . h3 P h3

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

5 V (2πmkT )3/2 + ln N h3 2 which, by eqn. (3.5.13), is indeed equal to the quantity S/Nk.

P 2 3.12. We start with eqn. (3.5.5), substitute H(q,p) = pi /2m + U (q) and i

integrate over the pi 0 s, to get 3N/2

Z 1 2πmkT Z (V, T ), where Z (V, T ) = e−U (q)/kT d3N q.

QN (V, T ) = N N N! h2 It follows that, for N >> 1, " ( # 3/2 ) h2 A = NkT ln N − 1 − kT ln Z, whence 2πmkT " ( # 3/2 )

5 ∂ ln Z 1 2πmkT S = Nk ln + + k ln Z + kT .

N h2 2 ∂T N,V Now e−U/kT (U/kT 2 )d3N q Ū R = , while −U/kT d3N q T e N,V n o Ū k ln Z = k ln V̄ N e−Ū /kT = Nk ln V̄ − .

T Substituting these results into the above expression for S, we obtain the desired result for S. In passing, we note that hHi ≡ A + TS = 23 NkT + Ū .

P For the second part of the question, we write U (q) = u(rij ), so that

kT

∂ ln Z ∂T

=

kT

R

i<j

e−βU (q) =

Y i<j

e−βu(rij ) =

Y

(1 + fij ) ,

i<j

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

3.14. a) The Lagrangian is given by X1 X X 2 L =K −V = mṙiα − u(rij ) − [uw (riα ) + uw (L − riα )], 2 iα i<j iα where i = 1, · · · , N denotes thePparticle number, α = x, y, z denotes the 2 cartesian directions, and rij = α (riα − rjα )2 . The canonical momenta are ∂L piα = = mṙiα . ∂ ṙiα

4 The Hamiltonian is given by X H = piα ṙiα − L iα

=

X p2

iα

2m

iα

+

X

u(rij ) +

i<j

X

[uw (riα ) + uw (L − riα )].

iα

The canonical pressure can be written ∂H 1 ∂H 1 X 0 1 (Fx + Fy + Fz ) .

P =− =− 2 =− 2 u (L − riα ) = ∂V 3L ∂L 3L iα w 3L2 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 are siα = riα /L so the Lagrangian becomes X X X1 mL2 ṡ2iα − u(Lsij ) − [uw (Lsiα ) + uw (L − Lsiα )].

L = 2 i<j iα iα In this case the canonical momenta are p̃iα =

∂L = mL2 ṡiα . ∂ ṡiα

This leads to a Hamiltonian of the form X p̃2 X X iα H = + u(Ls ) + [uw (Lsiα ) + uw (L − Lsiα )], ij 2mL2 i<j iα iα with canonical pressure is 1 X p̃2iα 3L2 iα mL3 1 X 0 − u (Lsij )sij 3L2 i<j

P =+

−

1 X 0 [u (Lsiα )siα + u0w (L − Lsiα )(1 − siα )].

3L2 iα w

Converting back to normal cartesian coordinates and momenta gives 2 X p2iα 3V iα 2m 1 X 0 u (rij )rij − 3V i<j

P =+

−

1 X 0 [u (riα )riα + u0w (L − riα )(L − riα )].

3V iα w

5 The first term is (2/3)(N/V ) times the kinetic energy per particle so is O(N ). The second term is (1/3)(N/V ) times the virial per particle so is also O(N ). On the other hand, third term is proportional to the shortranged virial on the walls divided by the volume so is O(N 2/3 ) which is negligible in the thermodynamic limit.

Comparing to equation (3.7.15) for the average pressure we see that P 1 =1− nkT 3N kT

*

+ X

u0w (rij )rij

.

i<j

3.15. Here, QN (V, T ) = (1/N !)QN 1 (V, T ), while Z∞ Q1 (V, T ) =

e−βpc

8πV 1 V · 4πp2 dp = 3 3 3, h3 h β c

0

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

As regards the density of states, the expression Z∞ Q1 (V, T ) =

e−βε g(ε)dε =

8πV 1 h3 β 3 c3

0

leads to g(ε) =

4πV 2 ε h3 c3

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

1 N!

8πV h3 c3

N

E 3N −1 Γ(3N )

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

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

Z ∂U − H(U − H) e−βH dω = 0. ∂β This means that

∂U − HU + H 2 ∂β

= 0,

which amounts to the desired result: hH 2 i − hHi2 = −(∂U/∂β).

6 3.18. We start with eqn. (3.6.2), viz.

P 2 −βEr Er e ∂U r = − P −βEr + U 2 , ∂β e

(1)

r

and differentiate it with respect to β, keeping the Er fixed. We get ∂U ∂2U = hE 3 i − hE 2 ihEi + 2U . ∂β 2 ∂β Substituting for (∂U/∂β) from eqn. (1), we get ∂2U = hE 3 i − 3hE 2 iU + 2U 3 , ∂β 2 which is precisely equal to h(E −U )3 i. As for ∂ 2 U/∂β 2 , we note that, since

∂U ∂U = −kT 2 = −kT 2 CV , ∂β Er ∂T V 2

∂ U ∂CV ∂ 2 2 2 2 2 = −kT . −kT CV = k T 2TC V + T ∂β 2 Er ∂T ∂T V V Hence the desired result.

For the ideal classical gas, U = 32 NkT and CV = 32 Nk , which readily yield the stated results.

P P 3.19. Since G = qi pi , Ġ = (q̇i pi + qi ṗi ). Averaging over a time interval τ , i

we get 1 τ

i

t+τ Z X t

(q̇i pi + qi ṗi )dt =

i

1 τ

t+τ Z

Ġdt =

G(t + τ ) − G(t) . τ

(1)

t

For a finite V and finite E, the quantity G is bounded ; therefore, in the limit τ → ∞, the right-hand side of (1) vanishes. The left-hand side then gives * + X (q̇i pi + qi ṗi ) = 0. i

which leads to the desired result.

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

7 the mean potential energy (not the internal energy) of the system. The total virial is then given by v = −3PV − nU .

The relation K = − 12 v 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 n n U= E. 2 n+2

K=

Example (a) pertains to n = 2, while examples (b) and (c) pertain to n = −1. In the former case, K = U = 12 E; in the latter, K = − 21 U = −E. The next problem pertains to n = 4.

3.22. Note that a force proportional to q 3 implies a potential energy proportional to q 4 . Thus

H=

1 2 p + cq 4 2m

(c > 0).

It follows that R∞

1 2 p 2m

=

2

e−βp /2m (p2 /2m)dp

−∞

R∞

= e

−βp2 /2m

dp

1 ; 2β

−∞

for the values of these integrals, see eqns. (13a) of Appendix B. Next, R∞ hcq 4 i =

4

e−βcq (cq 4 )dq

−∞

R∞

=− e

−βcq 4

dq

∂ ln I(B), ∂β

−∞

where I(β) denotes the integral in the denominator. It is straightforward to see that I(β) is proportional to β −1/4 , whence hcq 4 i = 1/4β, which proves the desired result.

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

8 partition function is  QN (V, T ) =

1 λ3N N !

N

=

Z exp −β

 X

u(rij ) dN r

i<j



V λ3N N !

Z exp −β

 X

u(V 1/d sij ) dN s.

i<j

Now the pressure is

∂A P =− ∂V N,T       Z X N X kT  N QN βV  = V 1/d sij u0 (V 1/d sij ) exp −β u(V 1/d sij ) dN s . − QN V dV λ3N N ! i<j i<j This can be simplified by going back to integrals over the normal position variables to give equation (3.7.15).

3.24. By eqn. (3.7.5), we have, for a single particle, * 3 + X pi q̇i = 3kT .

(1)

i=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 m0 u2 (1 − u2 /c2 )−1/2 . The desired result follows readily.

In the non-relativistic limit (u << c), one obtains: 12 m0 u2 ≈ 32 kT ; in the extreme relativistic limit (u → c), one obtains: hmc 2 i ≈ 3kT . Note that, in the latter case, m0 c2 is negligible in comparison with mc 2 , so there is no significant difference between the kinetic energy and the total energy of the particle.

3.25. For the first part of this problem, see Sec. 6.4 — especially the derivation of the formula (6.4.9). For the second part, equate the result obtained in the 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 variety of ways in which j indistinguishable quanta can be divided among the s dimensions of the oscillator: j = j1 + . . . + js ; this is similar to the calculation done on page 70 of the text. h iN (s) (s) As for the partition function, QN (β) = Q1 (β) , where (s) Q1 (β) =

∞ X (j + s − 1)! j=0 1

j!(s − 1)!

e−β (j+ 2 s)~ω 1

= e− 2 sβ~ω (1 − e−β~ω )−s .

9 Calculation of the various thermodynamic quantities is now straightforward. The results are found to be essentially the same as for a system of sN one-dimensional oscillators. However, since (s)

(1)

QN (β) = QNs (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 left for the remaining (N − 1) oscillators is E − n + 21 ~ω; the corresponding number of quanta to be distributed among these oscillators is R−n; see eqn. (3.8.24). The relevant number of microstates is then given by the expression (R − n + N − 2)!/(R − n)!(N − 2)!. Combined with expression (3.8.25), this gives pn =

(R − n + N − 2)! (R + N − 1)! ÷ .

(R − n)!(N − 2)!

R!(N − 1)!

(1)

It follows that R n̄ pn+1 R−n ' = . = pn R−n+N −2 R+N n̄ + 1 By iteration, pn = p0 {n̄/(n̄ + 1)}n .

Going back to eqn. (1), we note that p0 =

N −1 N 1 ' =− , R+N −1 R+N n̄ + 1

which completes the desired calculation. (b) The probability in question is proportional to gN −1 (E − ε), i.e. to 3 (E − ε) 2 (N −1)−1 . For 1 << N , this is essentially proportional to 3 (1 − ε/E) 2 N and, for ε << −E, to e−3N ε/2E .

3.29. The partition function of the anharmonic oscillator is given by 1 Q1 (β) = h

Z∞ Z∞ e

−βH

dq dq

p2 4 2 3 H= + cq − gq − fq .

2m

−∞ −∞

p The integration over p gives a factor of 2πm/β. For integration over q, we write

1 2 3 4 4 2 −βcq 2 β(gq 3 +fq 4 ) −βcq 2 3 e e =e 1 + β(gq + fq ) + β (gq + fq ) + . . . ; 2 the integration then gives r r r π 3 π 1 2 2 15 π + βf · + β g · + .... βc 4 β 5 c5 2 8 β 7 c7

10 It follows that π Q1 (β) = βh

r

15g 2 2m 3f + + . . . , 1+ c 4βc2 16βc3

so that ln Q1 (β) = const. − ln β + whence U (β) =

3f 15g 2 + + ..., 2 4βc 16βc3

15g 2 1 3f + ... + 2 2+ β 4β c 16β 2 c3

and C(β) = k +

3f k 2 T 15g 2 k 2 T + + ....

2c2 8c3

Next, the mean value of the displacement q is given by Z ∞ Z ∞ Z ∞ Z ∞ hqi = exp(−βH)q dp dq / exp(−βH)dpdq. −∞

−∞

−∞

−∞

In the desired approximation, we get Z∞ hqi ' βg

e

, Z∞

−βcq 2 4

q dq

−∞

3 = βg · 4

2

e−βcq dq

−∞

r

r 3g π π / = . β 5 c5 βc 4βc2

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

Q1 (β) =

∞ X

2

e−β (n+ 2 )~ω+βx(n+ 2 ) ~ω .

1

1

n=0

For x << 1, we may write Q1 (β) =

∞ X n=0

" −β (n+ 12 )~ω

e

1 1 + βx n + 2

#

2

~ω + . . . .

With u = β~ω, the sums involved are ∞ X

−1 1 S1 (u) = e = 2 sinh u , and 2 n=0

2

−1 ∞ X 1 d2 1 1 1 1 2 S2 (u) = e−u(n+ 2 ) n + = S (u) = 4 sinh u coth u − .

1 2 2 2 2 du 2 n=0 −u(n+ 12 )

11 It follows that ln Q1 = ln[S1 + xuS2 + . . .] ' ln S1 + xu(S2 /S1 )

1 1 1 1 = − ln 2 sinh u + xu coth2 u − .

2 2 2 2 The first part of this expression leads to the standard results (3.8.20 and 21). The second part may, for simplification, be expressed as a power series in u, viz.

u u3 2 + + + ... . x u 12 120 The resulting contribution to the internal energy per oscillator turns out to be

2 1 u2 − − − . . . x~ω u2 12 40 and the corresponding contribution to the specific heat is given by

4 u3 + + ... . xk u 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 group 1 and p2 /g2 for each of the states in group 2. We get

p1 p1 p2 p2 S = −k g1 ln + g2 ln .

(1) g1 g1 g2 g2 (a) In thermal equilibrium, gi e−βεi pi = P −βεi gi e

(i = 1, 2).

i

With x = β(ε2 − ε1 ), we have: p1 = g1 /(g1 + g2 e−x ) and p2 = g2 /(g1 ex + g2 ). Substituting these results into (1), we obtain

g1 g2 x −x S=k ln g1 + g2 e + ln (g1 e + g2 ) . g1 + g2 e−x g1 ex + g2 Writing the first log as ln g1 + ln{1 + (g2 /g1 )e−x } and the second log as ln g1 + x + ln{1 + (g2 /g1 )e−x }, we obtain the stated expression for S. (b) With Q = g1 e−βε1 + g2 e−βε2 , it is straightforward to see that A = −kT ln{g1 e−βε1 + g2 e−βε2 } and U = {g1 ε1 e−βε1 + g2 ε2 e−βε2 }/{g1 e−βε1 + g2 e−βε2 }.

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

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

3.34. The Curie paramagnetic susceptibility per unit volume of a gas is given by χ N µ2 µ0 P µ2 µ0 . = = V V kT (kT )2 Using χ/V = 1.8 × 10−6 at STP gives µ = 1.6µB , because of the two unpaired valence electrons in O2 .

3.35. The partition function of the system is given by 1 N V Q , where Q1 = 3 · Z, N! 1 λ

QN =

Z being the factor that arises from the rotational/orientational degrees of freedom of the molecule: " ( )# Z p2ϕ dp θ dp ϕ dθdϕ p2θ Z = exp −β + − µE cos θ 2I h2 2I sin2 θ Zπ =

2πI β

1/2

2πI sin2 θ β

1/2

eβµE cos θ

2πdθ h2

0

=

I 2 sinh(βµE) · .

2 β~ βµE

The study of the various thermodynamical quantities of the system is now straightforward.

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

N ∂ ln Z 1 Mz = N hµ cos θi = = N µ coth (βµE) − ; β ∂E βµE cf. eqns. (3.9.4 and 6). For βµE << 1, 1 Mz ≈ N µ · βµE. 3 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 4πnµ2 ≈1+ .

E 3kT

The numerical part of the problem is straightforward.

13 3.36. The mean force hFi between the two dipoles is given by

R −βU e (−∂U / ∂R) sin θd θdϕ · sin θ0 dθ0 dϕ0 ∂U R hFi = − = ∂R e−βU sin θdθdϕ · sin θ0 dθ0 dϕ0 1 ∂ = ln Z . β ∂R

(1) (2)

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

Z 1 Z= 1 − βU + β 2 U 2 − . . . sin θ dθdϕ · sin θ0 dθ0 dϕ0 .

2 The linear term vanishes on integration and we are left with

Z 1 (µµ0 )2 0 0 0 2 {2 cos θ cos θ − sin θ sin θ cos(ϕ − ϕ )} − . . .

Z= 1 + β2 2 R6 sin θd θd ϕ · sin θ0 dθ0 dϕ0

1 (µµ0 )2 2 2 1 1 1 = 16π 2 1 + β 2 · − 0 + · · − . . . .

4 · 2 R6 3 3 3 3 2 It follows that

1 (µµ0 )2 − ... ln Z = const. + β 2 3 R6 and hence, at high temperatures, hFi ≈ −2β

(µµ0 )2 R̂.

R7

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

µ̄z =

(gµB m) exp(βgµB mH )

m=−J J P

. exp(βgµB mH )

m=−J

At high temperatures, the exponential may be approximated by (1 + βgµB mH ) which yields, to the leading order in H, µ̄z = βg 2 µ2B Hm2 .

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

CJ = N0 g 2 µ2B / k m2 .

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

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

m2 =

m2

m=−J

2J + 1

=

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

2J + 1 3

14 3.38. Treating m as a continuous variable, the partition function of a magnetic dipole assumes the form Z J Q1 (β) =

eβgµB Hm dm =

−J

2 sin h (βgµB JH ); βgµB H

cf. eqn. (3.9.5). It is clear that this approximation will lead essentially to the same results as the ones following from the Langevin theory — except for the fact that the role of µ will be played by gµB J, which should be contrasted 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 ∂S ∂S ∂M = + ; (1) ∂T H ∂T M ∂M T ∂T H at the same time, dA ≡ dU − TdS − SdT = HdM − SdT , with the result that (∂H/∂T )M = −(∂S/∂M )T . Equation (1) then becomes

∂S ∂H ∂M ∂S = − .

(2) ∂T H ∂T M ∂T M ∂T H Multiplying (2) by T , we obtain the desired result for CH − CM . (b) The Curie law implies that M = 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 the field. Then N1 + N2 = N, while − N1 ε + N2 ε = E. It follows that N1 =

1 1 (N − E/ε), N2 = (N + E/ε).

2 2

The number of microstates associated with this macrostate is given by N!

(N − E/ε) ! 21 (N + E/ε) !

2

Ω(N, E) = 1

The entropy of the system is then given by the expression

1 E 1 E S = k ln Ω ≈ k N ln N − N− ln N− 2 ε 2 ε

1 E 1 E − N+ ln N+ , 2 ε 2 ε which is essentially the same as eqn. (3.10.9).

15 For the temperature of the system, we get

1 ∂S k N − E/ε = = ln , T ∂E N 2ε N + 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 a function 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 of the corresponding component of A. The partition function is, therefore, independent of the applied field and hence the net magnetization of the system is zero.

P 3.44. The Shannon information for a single message is given by I1 = − r Pr ln Pr where Pr is the a priori probability of message r from among all Ω possible messages. The maximum information is obtained from varying the probabilities, using a Lagrange multiplier µ to maintain the normalization P P = 1, and demanding the solution is stationary. r r !

0 = δI1 − µδ

X

Pr

=−

r

X

δPr [ln pr − 1 − µ].

r

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

Keeping to the general cases in which probabilities of individual messages messages do not need to be equal, consider a sequence of two messages. The a priori probability of message r followed by message r0 is Prr0 = Pr Pr0 Grr0 . The quantity Grr0 is the correlation between the two messages. A value of Grr0 greater than unit implies that the first message r increases the probability of finding the second message r0 above Pr0 . The P 0 = Pr 0 two P message probabilities have the following properties: P rr r P P 0 = 1 and 0 Grr 0 = 1. The inforand r0 Prr0 = Pr , i.e.

P G P 0 r rr r r r mation contained in two messages is given by I2 =

X rr 0

Prr0 ln Prr0 =

X

Pr Pr0 Grr0 ln (Pr Pr0 Grr0 ).

rr 0

Expanding the logarithm and using the above summation properties gives

X X 1 .

I2 = 2I1 − Pr Pr0 Grr0 ln Grr0 = 2I1 + Pr Pr0 Grr0 ln Grr0 0 0 rr

rr

16 Now, using ln x ≤ x − 1 for all x > 0, we get X I2 ≤ 2I1 + Pr Pr0 [1 − Grr0 ] = 2I1 . rr 0

The information contained in two correlated messages is reduced compared to sum of the information contained in two uncorrelated messages.

Analysis of the digits of the first 65536 digits of π results in an information per character of I1 ≈ 2.3 = ln 10. That makes sense because the characters 0, · · · , 9 are evenly distributed in the digital representation of π. Furthermore, since the digits of π are uncorrelated, the information per pair of characters is I2 ≈ 4.6 = 2I1 . Analysis of the first 15, 000 characters of A Chrismas Carol by Charles Dickens gives I1 ≈ 3.08 ≈ ln 21.75. This value is reasonable since most of the characters are lower case letters of the alphabet and blanks. The nonuniformity of the distribution of letters reduces the information below ln 27 . When analyzed two characters at a time, the information is I2 ≈ 5.45 ≈ 2 ln 15.25. The strong correlations between characters in English text reduces the information well below 2I1 .

R d B · da, 3.45. Starting from Faraday’s Law, the back-EMF is given by E = − dt so the external work needed to change the magnetic flux is Z dW = I δB · da ZS =I ∇ × δA · da using B = ∇ × A IS =I δA · dl using curl theorem L Z = δA · J dr expanding current into physical wire Z = δA · (∇ × H )dr using Ampére’s law The vector identity ∇ · (X × Y ) = Y · (∇ × X ) − X · (∇ × Y ), gives Z dW =

Z δH · (∇ × δA)dr =

H · δBdr,

since the integral of the divergence term over all space vanishes for localized sources. For fixed applied field, this becomes Z dW = µ0 H · δM dr.

17 If the field and magentization density are uniform, this gives dW = µ0 H · dM . Here is a simpler argument for a special geometry.

Assume a very long solenoid has N turns, length L, and uniform crosssectional area A, and the magnetic material inside the solenoid has uniform cross-sectional area A0 and has uniform magnetization per unit volume M along the axis of the solenoid. Therefore the magnetization is M = A0 LM. Faraday’s law gives gives the external work needed to change the uniform field inside the solenoid by amount dB as dW = N IA0 dB. The uniform H field along the axis of the solenoid is given by H = N I/L, so the external work can be written as dW = A0 LH dB. The fields are related

by B = µ0 (H + M), so changing the magnetic dipole moment while keeping H constant gives dW = µ0 HA0 LdM = µ0 HdM .

3.46. The work done to quasistatically change the distribution of space charge by amount δρ(r), on capacitor plates for example, is given by Z dW = V (r)δρ(r)dr, where V (r) is the potential already present. Using ∇ · D = ρ and E = −∇V , and integrating by parts gives Z dW = V (r)∇ · δDdr, Z = (−∇V ) · δDdr, Z = E · δD dr.

R Note that the integration by parts surface term V δD · da ∼ 1/R tends to zero as the size of the integration volume R goes to infinity since for localized sources near the origin the voltage and field scale as V ∼ 1/r and δD ∼ 1/r2 . For constant uniform field and uniform polarization density the work is dW = EdP . Here is a simpler argument for a special geometry.

Assume a thin plate capacitor has area A and thickness d, and is held at constant voltage V . The volume between the plates contains a material with uniform polarization density P oriented perpendicular to the plates.

Changing the polarization density by δP leads to a charge transfer by the battery of dQ = AδP. The external work done by the battery is dW = V Q = (Ed)(AδP) = EdP .

3.47. The Hamiltonian takes the form X H({qk , pk }) = pk q̇k − L k

=

p2φ p2θ + − mgL cos θ, 2mL2 2mL2 sin2 θ

18 since pθ = mLθ̇ and pφ = mLφ̇ sin θ. The one-particle partition function is then

Z 1 dθ dpθ dφ dpφ e−βH h2 Z 2πL2 π dθsinθeβmgL cos θ = λ2 0 Z 2πL2 1 d(cos θ)eβmgL cos θ = λ2 −1

Q1 (T ) =

=

4πL2 sinh(βmgL) , λ2 βmgL

√ where λ = h/ 2πmkT . The one-particle internal energy is

∂ ln Q1 ∂βF1 =− ∂β ∂β mgL

, = 2kT − tanh mgL kT

U1 (T ) =

and the one-particle heat capacity is

C1 (T ) =

∂U = 2k − k ∂T

mgL kT

2 2

sinh

1

mgL kT

.

The plot of C1 /k vs. kT /(mgL) is shown in the figure. The behavior can be explained using the the equipartition theorem since there are four squared degrees of freedom at low temperature where the pendulum acts like a two-dimensional harmonic oscillator, but only two degrees of freedom at high temperature where the only squared degrees of freedom are pθ and pφ .

19

C/k 2.5

2.0

1.5

1.0

0.5

0.0 0.0

0.5

1.0

1.5

2.0

kT/(mgL)

CHAPTER 4 4.1. By eqns. (4.1.9), (4.3.10) and (4.1.8), we get X Pr,s ln Pr,s ≡ hln Pr,s i = −αN̄ − β Ē − βPV r,s

= (µN̄ − U − PV )/kT .

(1)

Since µN̄ = 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, ( ) X PV = kT ln z Nr QNr (V, T ) .

(1)

Nr

Now, the largest term in the sum pertains to the value N ∗ , of Nr , which is determined by the condition ∂ {Nr ln z − ln QNr }Nr =N ∗ = 0. ∂Nr

By Sec. 3.3, this is equivalent to the statement: z = exp(µ∗ /kT ), where µ∗ 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, we would get PV ≈ N ∗ µ∗ − A∗ = P ∗ V, where P ∗ is the pressure of the system in the canonical ensemble (with N = N ∗ ). How different would P be from P ∗ depends essentially on how different the particle density n̄ is from n∗ — a question thoroughly discussed in Sec. 4.5.

4.3. The probability distribution in question is the binomial distribution

N (0) !

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

N !(N (0) − N )!

V (0) V (0) We note that (0) N X

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

N =0

1

(0)

= 1.

2 For part (i), we have

N̄ =

(0) N X

(

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

N =0

N (N − 1) =

(0) N X

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

N =0

It follows that N 2 =N (N − 1) + N̄ = (N (0) p)2 − N (0) p2 + N (0) p, whence (∆N )2 ≡ N 2 − N̄ 2 = 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)

x x − (N (0) q − x) ln 1 − (0) . ln P (x) ≈ −(N (0) p + x) ln 1 + (0) N p N q For x << N (0) p and N (0) q, we expand this expression in powers of x, with the result that ln P (x) ≈ −x2 /2N (0) pq. It follows that the distribution P (x), under the stated conditions, is a Gaussian, with (∆N )2 = N (0) pq.

For part (iii), we write P (N ) =

(0) N (0) (N (0) − 1) . . . (N (0) − N + 1) N p (1 − p)N −N .

N!

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

[N (0) ]N N −N (0) p (N̄ )N −N̄ p e = e , N! N!

with (∆N )2 = N̄ .

4.4. For obvious reasons, e−αNr P (Nr ) =

X s

Pr,s =

P −βEs e s

L(α, β, V )

=

z Nr QNr (V, T ) .

L(z, V, T )

(1)

3 For an ideal classical gas, see Sec. 4.4, N̄ λ3 1 z= , QN = V N!

V λ3

N

, L ≡ ePV /kT = eN̄ .

(2)

Substituting (2) into (1), we get P (N ) =

(N̄ )N −N̄ e , N!

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

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

(∆N )2 = −

∂ N̄ ∂α

=z T,V

∂ N̄ ∂z

=z T,V

∂ ∂z

zV λ3

=z T,V

V = N̄ . λ3

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

# ∂q ∂q ∂q ∂µ kT = kT + ∂T z,V ∂T µ,V ∂µ T,V ∂T z,V

∂q N̄ = kT + kT · · k ln z (for µ = kT ln z ). ∂T µ,V kT Equation (4.3.20) then reduces to

∂q ∂ S = kT + kq = k (Tq) . ∂T µ,V ∂T µ,V Note that this result is directly related to the formula, see Problem 1.16, d(PV ) = PdV + Nd µ + SdT , whence

∂ S= (PV ) . ∂T µ,V

4.6. The Gibbs free energy is G(N, P, T ) = −kT ln (YN (P, T )) .

For example

Z ∞ ∂G 1 = V Q(N, V, T )e−βP V = hV i. ∂P N,T YN (P, T ) 0

4 The ideal gas gives YN (P, T ) =

1 (βP λ3 )

N +1

,

G(N, P, T ) ≈ N kT ln βP λ3 ,

∂G N kT V = = . ∂P N,T p 4.10. The partition function of the adsorbed molecules, assumed noninteracting, is given by QN (N0 , T ) = g(N )aN =

N0 ! aN N !(N0 − N )!

[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

N ∂ ln QN = kT ln . ∂N (N0 − N )a

(2)

Alternatively, the grand partition function of the system consisting of all N0 sites (of which some are empty while others are occupied by a single molecule) is given by L(z, N0 , T ) = [L(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 N0 P obtained by using the standard definition L(z, N0 , T ) = z N QN (N0 , T ) N =0

and employing expression (1) for QN . The mean value of N now turns out to be N̄ = z

za 1 N̄ ∂ ln L = N0 , whence z = , ∂z 1 + za a N0 − N̄

(4)

which agrees with (2).

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

N̄ za 1 θ = , whence z = .

N0 1 + za a1−θ

(1)

Now, if the molecules in the adsorbed phase are in equilibrium with those in the gaseous phase, then their fugacity z would be equal to the fugacity

5 zg of the gaseous phase. The latter is given by eqns. (4.4.5 and 29), whereby Pg h3 .

(2) zg = kT (2πmkT )3/2 Equating (1) and (2), we obtain the desired result Pg =

θ 1 (2πmkT )3/2 . × kT 1 − θ a(T ) h3

4.12. From eqn. (4.5.1), we get

∂ N̄ ∂β

= −NE + N̄ Ē. α,V

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

∂U ∂U ∂ N̄ ∂ N̄ = −kT = −kT −kT 2 ∂T z,V ∂µ T,V ∂ N̄ T,V ∂µ T,V

∂U =− (∆N )2 . ∂ N̄ T,V 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 ( )

2

∂U ∂U 2 2 (∆J)2 = h(∆E) ican + − 2µ + µ (∆N )2 , ∂ N̄ T,V ∂ N̄ T,V which is, in fact, the desired result.

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

L 1 1 Pσ (T ) = Pσ (T0 ) exp − , k T0 T where T0 = 373 K, Pσ (T0 ) = 1 atm and L (2260 kJ/kg)(18 kg/kmol) = = 4890 K. k (6.02 × 1026 kmol−1 )(1.38 × 10−23 J/K) This gives Pσ (273 K) ' 0.0082 atm and Pσ (473 K) ' 16 atm. The experimental values are 0.006 atm and 15.3 atm respectively.

6 4.15. The correct value for the latent heat of sublimation near the triple point is 2833 kJ/kg. Following the solution to problem 4.14,

L 1 1 Pσ (T ) = Pσ (T0 ) exp − , k T0 T where T0 = 273 K, Pσ (T0 ) = 612 Pa and (2833 kJ/kg)(18 kg/kmol) L = 6138 K. = k (6.02 × 1026 kmol−1 )(1.38 × 10−23 J/K) This gives Pσ (193 K) ' 0.055 Pa which corresponds nearly exactly with the experimental value.

4.16. The slope of the melting line is dPm Lm (80 cal/g)(4.18 J/cal)(106 cm3 /m3 ) = ' ' −1.3 × 107 Pa/K. dT T (∆v) (273 K)(−0.09 cm3 /g) This gives Tm (100 atm) = −0.77◦ C. 4.17. The slopes at the triple point are of the form

dPσ si − sj ∆y = , = dT ij vi − vj ∆x so the vectors [(v1 − v2 )x̂ + (s1 − s2 )ŷ] + [(v2 − v3 )x̂ + (s2 − s3 )ŷ] + [(v3 − v1 )x̂ + (s3 − s1 )ŷ] = 0 sum to zero. This makes the third vector the negative of the sum of the first two vectors in each case, guaranteeing the stated geometry.

4.18. The liquid–vapor lines will appear much like in figure 6.2 but the liquid branch will extend to P = 0. The upper end of the solid–liquid lines will appear 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 dµσ ∂p1 ∂p2 dµσ ∂p2 + + = , ∂µ T dT ∂T µ ∂µ T dT ∂T µ which gives s1 − s2 −L dµσ =− = . dT n1 − n2 T ∆n 4.20. The liquid–vapor lines will appear much like in figure 6.2 but with the liquid branch ending abruptly at Pt . The liquid side of solid–liquid lines will start at Pt and extend upward as in figure 6.2 but the solid side of the solid–liquid transition will be to the right of the liquid line (since the solid has lower density) and will extend to P = 0.

7 4.21. The bulk and surface chemical potentials are equal at equilibrium, µb = µs where

2IkT ∂A V , µb = = −kT ln ∂N V,T N λ3 ~2 ! r

∂As 2πIkT A µs = , = −ε − kT ln ∂Ns A,T Ns λ 2 ~2 and the ”pressures” are given by

∂A N kT P =− = , ∂V N,T V

Ps =

∂As Ns kT = . ∂A Ns ,T A

Solving for the equilibrium surface density Ns /A gives Pλ 1 ~ Ns √ √ = eβε .

A kT π 2πIkT For this set of parameters, we get N ' 2.5 × 1022 , Ns ' 4 × 1013 .

If you pump down to 10−7 torr, the number of molecules inside the chamber is reduced to N ' 3 × 1012 , so if the walls did not outgas, the number of molecules remaining on the wall is larger than the number inside the vacuum chamber. Those molecules could slowly outgas for months, preventing reaching much lower pressures. To mitigate this, experimenters increase the outgassing rate of the walls by heating up the chamber to above 100◦ C during the pump-down using electric heat tape. That is why UHV chambers are typically covered with aluminum foil to keep the heat in during the bake-out.

Chapter 5 5.1. On transformation, a given operator  would become

Â0 = Û ÂÛ −1 =

√ 1/ √2 −1/ 2

√ a11 1/√2 a21 1/ 2

a12 a22

√ 1/√2 1/ 2

√ −1/√ 2 .

1/ 2

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

0 0 −i 0 −1 , σˆy0 = , σ̂z0 = , −1 i 0 −1 0

1 cosh(βµB B) − sinh(βµB B) ρ̂0 = βµB B e + e−βµB B − sinh(βµB B) cosh(βµB B) σ̂x0 =

1 0

and hσz0 i = Tr (ρ̂0 σ̂z0 ) =

1 eβµB B + e−βµB B

· 2 sinh(βµB B) = tanh(βµB B),

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 symmetrized wave function (5.5.7), the density matrix of the system turns out to be, cf. eqn. (5.5.11), h1, . . . , N |e−β Ĥ |10 , . . . , N 0 i =

X

2 2 e−β~ K /2m {uk1 (1) . . . ukN (N )} u∗k1 (10 ) . . . u∗kN (N 0 )

K

=

X

e

2 −β~2 (k12 +...+kN )/2m

uk1 (1)u∗k1 (10 ) . . . ukN (N )u∗kN (N 0 )

k1 ,...,kN

  N n o Y X 2 2  = e−β~ kj /2m ukj (j)u∗k (j 0 )  . j

j=1

kj

1

2 Replacing the summation over kj by an integration, one gets [see the corresponding passage from eqn. (5.5.12) to (5.5.14)] 3N/2

m m −β Ĥ 0 0 2 2 exp − h1, . . . , N |e |1 , . . . , N i = ξ + . . . + ξN , 2πβ~2 2β~2 1 (1) where ξj = rj − r0j . The diagonal elements of the density matrix then are h1, . . . , N |e−β Ĥ |1, . . . , N i = (m/2πβ~2 )3N/2 = 1/λ3N ,

(2)

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

The partition function now turns out to be Z vN 1 3N d r = , QN (V, T ) ≡ Tr (e−β Ĥ ) = λ3N λ3N with no Gibbs’ correction factor.

5.5. By eqn. (5.5.17), we have

QN (V, T ) ≡ Tr e−β Ĥ =

1 ZN (V, T ), N !λ3N

where ZN (V, T ) =

Z X {. . .}d3N r.

(1)

P

In the zeroth approximation,

P

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

P

V N . In the first approximation, X X X 2 2 =1± fij fji = 1 ± e−2πrij /λ .

P

i<j

(2)

i<j

If λ is much smaller than the mean interparticle distance, we may write     Y X Y X 2 2 ≈ 1 ± e−2πrij /λ = e−βvs (rij ) = exp −β vs (rij ) ,   P

i<j

i<j

i<j

which leads to the desired result.

For the second part, we substitute (2) into (1) and integrate over the position coordinates of the particles. We obtain, on assembling contributions from all pairs of particles, ZN (V, T ) = V N ±

V · λ3 N (N − 1) · V N −2 3/2 .

2 2

3 The case N = 2 corresponds to eqn. (5.5.25) for Q2 (V, T ). For N >> 1 and N λ3 << V , we may write

N

λ3 N λ3 . ≈ V N 1 ± 5/2 ZN (V, T ) = V N 1 ± N 2 5/2 2 V 2 V It follows that

`nQ N (V, T ) ≈ −N `nN + N + N `n

V λ3

N λ3 + N ± 5/2 , 2 V

whence P ≡ kT

∂`nQ N ∂V

≈ N,T

N N 2 λ3 1 1 λ3 ∓ 5/2 2 = ∓ 5/2 2 , V v 2 v 2 V

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 10 and 11 on page 126 of the text.

5.9.

Tr ρ̂(t) =

XX α

pk hα| ψk (t)i hψk (t)| αi

k

! =

X

pk hψk (t)|

X

|αi hα| |ψk (t)i

α

k

=

X

pk hψk (t)| ψk (t)i =

k

X

pk = 1.

k

Likewise, Tr ρ̂2 (t) =

XX

pk pk0 hα| ψk (t)i hψk (t)| ψk0 (t)i hψk0 (t)| αi

α

kk0

=

X

pk pk0 |hψk (t)| ψk0 (t)i|

≤

X

2

kk0

pk pk0 = 1.

kk0

For any normalized wavefunction |φi X 2 hφ| ρ̂(t) |φi = pk |hφ| ψk (t)i| k

≤

X

pk = 1.

k

Since 0 ≤ hφ|ρ̂|φi ≤ 1, all of the eigenvalues of ρ̂ must be in the range [0, 1]. For a pure state density matrix ρ̂ = |ψihψ|, the square of the

4 density P matrix is given P by ρ̂2 = |ψihψ|ψihψ = |ψihψ| = ρ̂. Therefore 2 2 Tr ρ̂ = i λi = Tr ρ̂ = i λi = 1. Since 0 ≤ λi ≤ 1, the only possibility is λi = 0 for every eigenvalue P except one, which must have the value unity.

Then S[ρ̂] = −Tr ρ̂ ln ρ̂ = − i λi ln λi = 0. For a mixed state, at least two Pof the eigenvalues are in the range 0 < λi < 1 so S[ρ̂] = −Tr ρ̂ ln ρ̂ = − i λi ln λi > 0.

5.10. For equilibrium density matrices with HAB = H = HA + HB obeying the principle of the independence of noninteracting subsystems, the density matrices must have the property ρAB [H] = ρA [HA ]ρB [HB ].

Differentiating both sides gives ∂ρAB ∂ρAB ∂ρAB ∂ρA ∂ρB = = = ρB [HB ] = ρA [HA ] . ∂H ∂HA ∂HB ∂HA ∂HB Therefore ∂ρA ∂ρB 1 1 = = −β, ρA [HA ] ∂HA ρB [HB ] ∂HB

(1)

for constant β since the two sides are independent of each other and if they are equal, they must be equal to a constant. Solving the differential equation and normalizing gives ρA [HA ] = e−βHA /Tr e−βHA , and likewise for systems B and AB. If the Hamiltonian is unbounded above, the constant β must be positive for the density matrix to be normalizable. This corresponds to some positive temperature. If the Hamiltonian is bounded, no normalization problem arises so β can be negative or zero, which corresponds to negative or infinite temperature, respectively.

5.11. The eigenvalues λ1 and λ2 are the two solutions of the determinate x−λ z

z = 0, y−λ

The quadratic equation gives x+y λ= ± 2

s

x−y 2

2 + z2,

so the average of the two eigenvalues is (x + y)/2. (a) Since x and y are selected from normal distributions with zero mean, hxi = hyi = 0, and the two eigenvalues are symmetric about (x + y)/2, the average is given by + * 2

1X x+y λi = = 0.

2 i=1 2

5 (b) The average of the squares of the eigenvalues is given by + * 2 + * 2 2 x−y 1X 2 x+y 2 + +z λ = 2 i=1 i 2 2 2

x y2 = + + z 2 = 1 + 1 + 1 = 3.

2 2 Parts (c)-(e) all involve integrals of the following form: Z

1 dxdydz √ exp −(x2 + y 2 )/4 − z 2 /2 F (·).

4π 2π Let u = (x + y)/2, v = (x − y)/2, with Jacobian J = 2 so the integrals become Z

2 dudvdz √ exp −(u2 + v 2 + z 2 )/2 F (·).

4π 2π √ All of the integrands will depend on the quantity r = v 2 + z 2 so we can transform to 2D polar coordinates, which gives Z Z ∞

1 du dr r √ exp −(u2 + r2 )/2 F (·).

2π 0 In these coordinates λ = u ± r. (c) For calculating the distribution of eigenvalues P (λ) use F = (δ(λ − u + r) + δ(λ − u − r))/2 which gives

2 1 −λ2 /2 √ P (λ) = √ 2e + πλe−λ /4 erf(λ/2) , 4 2π R R which can confirm has the properties P (λ)dλ = 1, λP (λ)dλ = 0, R you and λ2 P (λ)dλ = 3, which agrees with parts √ (a) and (b). This function is shown in the next problem with x = λ/ 2 compared with the results of many samplings of the eigenvalues of random 2 × 2 matrices. (d) For calculating the joint distribution of eigenvalues ρ(λ1 , λ2 ) use F = (δ(λ1 − u + r) + δ(λ2 − u − r))/2 which gives

2 1 λ + λ22 , ρ(λ1 , λ2 ) = √ |λ1 − λ2 | exp − 1 4 8 2π R which you can confirm has the property ρ(λ1 , λ2 )dλ2 = P (λ1 ) in patt (c). (e) For calculating the distribution of the difference between the two eigenvalues p(t) where t = |λ1 − λ2 | = 2r, use F = δ(t − 2r).

6 This gives 2 t t p(t) = exp − .

4 8 Integrating to get the average separations gives Z √ hti = tp(t)dt = 2π.

Expressing the eigenvalue spacing in terms of s = t/hti gives the Wigner surmise

πs πs2 p(s) = exp − .

2 r 5.12. The numerical distribution of the eigenvalues of 2 × 2 GOE matrices compared with the result in problem 5.11.

GOE 2 √ × 2 random matrix eigenvalue distribution using scaled variable x = λ/ 2 (solid line), along with the results from sampling eigenvalues from 105 random matrices (histogram). The dashed line is the asymptotic semicircular distribution which results from Γ × Γ GOE random matrices in the limit Γ → ∞.

5.13. (a) The average of the eigenvalues of R can be written as the expectation value of the trace of R: * +

1X 1 1X λi = Tr R = Rii = 0. Γ i Γ Γ i

7 (b) The average of the square of the eigenvalues of R can be written as the expectation value of the trace of R2 :

*

1X 2 λ Γ i i

+

1 Tr R2 Γ

* =

1X Rij Rji Γ ij

=.

1 (Γ(Γ − 1) + 2Γ) = Γ + 1. Γ

=

+ =.

1X (Rij )2 Γ ij

(c) For 2 × 2 GOE random matrices, the average of the eigenvalues and average of the square of the eigenvalues are given by

*

*

1X λi Γ i

+

1X 2 λ Γ i i

= 0, + = 2 + 1 = 3.

The large Γ limiting semicircular elgenvalue distribution gives

Z hλi = dλ λρ(λ) = 0, Γ Z hλ2 i = dλ λ2 ρ(λ) = 1 Γ

5.14. The figure shows GOE 100 × 100 random matrix eigenvalue distribution 4 results from sampling eigenvalues √ from 10 random matrices (histogram) using scaled variable x = λ/ 100. The dashed line is the asymptotic semicircular distribution which results from Γ × Γ GOE random matrices in the limit Γ → ∞.

8

5.15. The infinite time-average of he square of δhAit is given by Z 1 ∞ 2 (δ hAi) = lim (δ hAit ) dt, T →∞ T 0 Z n0 1 ∞ X X ∗ cm cn c∗m0 cn0 ei(Em +Em0 −En −E )t/~ Amn Am0 n0 = lim T →∞ T 0 m6=n m0 6=n0 X X = c∗m cn c∗m0 cn0 Amn Am0 n0 δmn0 δm0 n 2

m6=n m0 6=n0

=

X

2

2

2

|cn | |cm | |Anm | .

n6=m

√ Now, if the off-diagonal terms are of the form A2 / ΓRmn then 2

(δ hAi) =

X

2

2

|cn | |cm | |Anm |

2

n6=m

=

A22 X 2 2 2 |cn | |cm | |Rnm | Γ n6=m

A2 ≈ 2, Γ where the approximation involves neglecting the diagonal terms.

5.16. If the matrix elements of A are of the form given by equation (5.6.15),

9 then the time varying portion of the expectation value of A is given by A2 X ∗ δhAit ≈ √ c̃m c̃n Rmn , Γ mn where the coefficients c̃n = cn e−iEn t/~ are the original coefficients with a P time-dependent phase. In order to estimate δhAit , note that mn Rmn = O(Γ) since the Γ2 random matrix elements have zero mean and unit √ variance. At most times the coefficients are of the order of c̃n = O(1/ Γ) and are uncorrelated with the random matrix elements so 1 1 1 X Rmn ∼ 3/2 Γ ∼ √ . δhAit ∼ 3/2 Γ Γ Γ mn However, is at some time, say t = 0, c̃n is highly correlated with one on (α) the eigenvectors of R, i.e. c̃n ≈ φn , then A2 X (α) δhAit ≈ √ φ ∗m Rmn φn(α) , Γ mn A2 λα ≈ √ λα ≈ A2 √ . Γ Γ Since the eigenvalue distribution is given by equation (5.6.17), this gives a result in the range ±2A2 .

5.17. The density matrix ρ̂1 (t) has the form

3/4 ρ̂1 (t) = √ 3/4e−iωt

√

3/4eiωt , 1/4

so Tr ρ̂1 = 3/4 + 1/4 = 1. Matrix multiplication shows ρ̂21 = ρ̂1 so Tr ρ̂21 = 3/4 + 1/4 = 1. The eigenvalues satisfy √ λ 3/4eiωt √3/4 −−iωt = (3/4 − λ)(1/4 − λ) − 3/16 = λ(1 − λ) = 0, 3/4e 1/4 − λ so the eigenvalues are λ = {0, 1}, which gives Tr ρ̂ = Tr ρ̂2 = 1. Since the eigenvalues are 0 and 1, the von Neumann entropy is S[ρ̂1 ] = −Trρ̂1 ln(ρ̂1 ) = P − i λi ln λi = 0.

The density matrix ρ̂2 (t) has the form

5/6 √ ρ̂2 (t) = 1/(2 3)e−iωt

√

1/(2 3)eiωt , 1/6

so Tr ρ̂2 = 1, while ρ̂22 (t) =

7/9 √ 1/(2 3)e−iωt

√

1/(2 3)eiωt , 1/9

10 √ so Tr ρ̂22 = 8/9 < 1. The eigenvalues of ρ̂2 are λ = (3 ± 7)/6 6= {0, 1}.

These are both signs of a mixed state.

P The mixed state von Neumann entropy is S[ρ̂2 ] = −Trρ̂2 ln(ρ̂2 ) = − i λi ln λi ' 0.2243 > 0.

5.18.

QN (V, T ) =

1 N!

X {nxj ,nyj ,nzj }N j=1

1 N exp −βE({nxj , nyj , nzj }N Q (V, T ), j=1 ) = N! 1

where β~2 π 2 Q1 (V, T ) = exp − 2mL2 n ,n ,n X

x

y

z

n2y n2 n2x + + z2 2 2 Lx Ly Lz

!!

If λ is small compared to each dimension, then the summand is slowly varying with respect to each summation so !!

Z n2y n2z β~2 π 2 n2x + 2 + 2 Q1 (V, T ) ≈ dnx dny dnz exp − 2m L2x Ly Lz √ √ √ Lx 2πmkT Ly 2πmkT Lz 2πmkT V = 3. = 2π~ 2π~ 2π~ λ If λ Lz then the summation over nz reduces to the single term from the ground state with nz = 1 so that √ √

Lx 2πmkT Ly 2πmkT β~2 π 2 Q1 (Lx , Ly , Lz , T ) = exp − 2π~ 2π~ 2mL2z

A β~2 π 2 = 2 exp − , λ 2mL2z where A = Lx Ly is the area of that face, i.e. the partition function reduces to the case of a classical two-dimensional system. Likewise, if λ Ly and λ Lz , then the partition function becomes √

β~2 π 2 β~2 π 2 Lx 2πmkT exp − exp − Q1 (Lx , Ly , Lz , T ) = 2π~ 2mL2y 2mL2z

2 2 2 2 Lx β~ π β~ π = exp − exp − , 2 λ 2mLy 2mL2z i.e. the partition function reduces to the case of a classical one-dimensional system. The exponential factors result in constant additions to the free energy because every particle is in the ground state for motions in the other directions.

Chapter 6 6.1. We start with eqn. (6.1.19) and write it in the form

X gi n∗i gi ∗ S=k n∗i ln ln 1 − a + n − . i n∗i a gi i

(1)

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

X 1 ln(1 − ahnε i) .

(2) S=k −hnε i lnhnε i + hnε i − a ε Choosing a = −1 or +1, we obtain the desired results.

Next we have to verify that ( ) X X X S = −k pε (n) ln pε (n) = −k hln pε (n)i. ε

n

(3)

ε

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

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

pε (n) = (1 − r)rn

[r = hnε i/(hnε i + 1); n = 0, 1, 2, . . .].

It follows that hnε i = (1 − r) n2ε = (1 − r)

∞ X n=0 ∞ X

nr n = r/(1 − r), n2 rn = r(1 + r)/(1 − r)2 , so that

n=0

n2ε

2

− hnε i = r/(1 − r)2 = hnε i + hnε i2 .

1

(1)

2 In the F.D. case, see eqn. (6.3.11), n2ε =

1 X

n2 pε (n) = pε (1) = hnε i, so that

n=0

n2ε − hnε i2 = hnε i − hnε i2

(2)

In the M.B. case, see eqn. (6.3.12), one can readily see that hnε (nε − 1)i =

X

n(n − 1)

n

X hnε in−2 hnε in −hnε i e = hnε i2 e−hnε i = hnε i2 , so that n! (n − 2)! n

n2ε − hnε i2 = hnε i.

(3)

For the second part, we note, from eqn. 6.2.22, that hnε i−1 = e(ε−µ)/kT + a.

Differentiating this result with respect to µ, we get

1 (ε−µ)/kT 1 ∂hnε i =− e =− [hnε i−1 − a]. −hnε i−2 ∂µ T kT kT It follows that

kT

∂hnε i = hnε i − ahnε i2 . ∂µ T

(4)

Comparing (4) with our previous results (1)–(3), and with formula (6.3.9), we infer that, quite generally, n2ε − hnε i2 = kT [∂hnε i/∂µ]T .

6.3. Starting with eqn. (6.2.15), we now have " ` # Y X Y 1 − (ze −βε )`+1 −βε nε L(z, V, T ) = (ze ) = , 1 − ze −βε ε ε n =0 ε

so that q(z, V, T ) =

X

[ln{1 − (ze −βε )`+1 } − ln{1 − ze −βε }];

ε

cf. eqn. (6.2.17). It follows that

1 ∂q hnε i = − β ∂ε z,T,all other ε (` + 1)(ze −βε )` (ze −βε ) ze −βε + −βε `+1 1 − (ze ) 1 − ze −βε 1 `+1 = −1 βε − z e − 1 (z −1 eβε )`+1 − 1 =−

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

3 6.4. To determine the state of equilibrium of the given system, we minimize its free energy, U − TS , under the constraint that the total number of particles, N , is fixed. For this, we vary the particle distribution from n(r) to n(r) + δn(r) and require that the resulting variation Z Z Z n(r)δn(r0 ) + n(r0 )δn(r) e2 0 drdr + e δn(r)ϕext (r)dr δ(U − TS ) = 2 |r − r0 | Z + kT [1 + ln n(r)]δn(r)dr = 0, R while δN = δn(r)dr is, of necessity, zero. Introducing the Lagrange multiplier λ, our requirement takes the form

Z Z n(r0 ) 0 2 dr + eϕext (r) + kT [1 + ln n(r)] − λ δn(r)dr = 0. e |r − r0 | Since the variation δn(r) in this expression is arbitrary, the condition for equilibrium turns out to be Z n(r0 ) e2 dr0 + eϕext (r) + kT ln n(r) − µ = 0, (1) |r − r|0 where µ = λ − kT .

Introducing the total potential ϕ(r), viz.

Z ϕ(r) = ϕext (r) + e

n(r)0 dr0 , |r − r0 |

(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 be written as n(r) = n0 exp[−eϕ(r)/kT ].

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

6.5. The (un-normalized) distribution function for the variable ε in this problem is given by f (ε)dε ∼ e−βε ε1/2 dε, where use has been made of expression (2.4.7) for the density of states of a free particle. It is now straightforward to show that ε̄ =

β −5/2 Γ(5/2) 3 β −7/2 Γ(7/2) 15 = and ε2 = −3/2 = . −3/2 2β 4β 2 β Γ(3/2) β Γ(3/2)

It follows that q (∆ε)r.m.s. ≡ ε̄

(ε2 − ε̄2 ) ε̄

p =

(3/2β 2 ) p = (2/3).

(3/2β)

4 6.6. We have to show that, for any law of distribution of molecular speeds [say, F (u)du], R∞

R∞ −1 u F (u)du

u F (u)du

0 R∞

F (u)du

· 0 R∞

0

0

Z∞

Z∞ u F (u)du ·

0

≥ 1, i.e.

F (u)du

∞ 2 Z u−1 F (u)du ≥  F (u)du  .

0

0

For this, we employ Schwarz’s inequality (see Abramowitz and Stegun, 1964),  b 2 Z Zb Zb 2  f (x)g(x)dx  ≤ [f (x)] dx · [g(x)]2 dx , a

a

a

which holds for arbitrary functions f (x) and g(x) — so long as the integrals exist; the equality holds p = c g(x), where c is a constant. p if and only if f (x) Now, with f (u) = uF (u) and g(u) = u−1 F (u), we obtain the desired result.

For the Maxwellian distribution, 1

2

F (u)du ∼ e− 2 βmu u2 du.

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

1/2

1/2 I3 8 I1 2βm −1 hui = = and hu i = = , I2 πβm I2 π whence huihu−1 i = 4/π, in conformity with the inequality stated.

6.7. For light emitted in the x-direction, only the x-component of the molecular velocity 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 by

the Boltzmann factor exp − 12 mu 2x /kT ; the distribution of λ in the light emerging from the window will, therefore, be determined by the factor

exp − 21 mc 2 (λ − λ0 )2 /λ20 kT .

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

2

Z 1 p Q1 (β) = 3 exp −β + mgz dp x dp y dp z dxdydz h 2m

3/2 2πm 1 − e−βmgL = · A , (1) βh2 βmg

5 A being the area of cross-section of the cylinder. In the limit L → ∞, 3/2

A 2πm Q1 (β) = ∝ T 5/2 .

(2) 2 βh βmg The thermodynamic properties of the system now follow straightforwardly.

In particular, U turns out to be 52 NkT and hence Cv = 52 Nk . The extra contribution comes from the potential energy of the system, which too rises with T . Note, from eqns. (1) and (2), that the effective height of the gas molecules is (1 − e−βmgL )/βmg which for small heights is essentially L itself but for large heights is essentially kT/mg — making the total potential energy of the gas equal to NkT. 6.9. Correction to the first printing of third edition: the correct Hamiltonian is H (pr , pθ , pz , r, θ, z) =

p2r (p2 − mr2 ω)2 p2z mr2 ω 2 + θ + − .

2m 2mr2 2m 2

This gives for the partition function

Z βmr2 ω 2 2πH R 2πHkT βmR2 ω 2 exp Q1 (V, T ) = 3 rdr = 3 exp −1 λ 2 λ mω 2 2 0 In the limit of small rotation rate, this becomes Q1 = πHR2 /λ3 = V /λ3 as expected.

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

βmω 2 r2 n(r) = n(0) exp .

2 Since the 238 UF6 molecules are heaver, their concentration is enhanced at r = R, while the concentration of the 235 UF6 is enhanced near r = 0. The ratio at r = 0 is given by " #

m235 N235 exp 12 βm238 ω 2 R2 − 1 n235 (0)

= n238 (0) m238 N238 exp 12 βm235 ω 2 R2 − 1

m235 N235 1 2 2 ≈ exp β (m238 − m235 ) ω R . m238 N238 2 A value of ωR = 500 m/s gives a 16% enhancement compared to the input fraction. Drawing the uranium hexafluoride gas from near the center of the cylinder results in a sample that is isotopically enhanced with 235 U compared to the input concentration. This process may be repeated as often as needed to achieve the isotopic fraction needed.

6.10. Consider a layer of the gas confined between heights z and z + dz . For hydrostatic equilibrium, we must have P (z + dz ) + ρgdz = P (z),

6 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, then T is related to P ; in fact, T ∝ P (γ−1)/γ . We now get dT γ − 1 dP γ − 1 mg = =− dz .

T γ P γ kT

(3)

This means that T now decreases essentially linearly with height.

The pressure P and the density ρ go hand in hand with T — varying as T γ/γ−1 and T 1/γ−1 , respectively.

6.11. (a) For the given system, 2

2 2 1/2

f (p)dp = const.e−βε(p) (4πp2 dp) = C e−βc(p +m0 c )

p2 dp.

The normalization constant C is determined by the condition Z∞

Z f (p)dp = C

2

2 2 1/2

e−βc(p +m0 c )

p2 dp = 1.

0

Substituting p = m0 c sinh θ, we get for the left-hand side of this equation Z∞ C

2

e−βm0 c cosh θ m30 c3 sinh2 θ cosh θ dθ

0

 = C m30 c3 

e

−βm0 c2 cosh θ

−βm0 c2

 Z∞ −βm0 c2 cosh θ e sinh θ cosh θ|∞ cosh(2θ)dθ 0 + βm0 c3 0

= Cm 30 c3 · (βm0 c2 )−1 K2 (βm0 c2 ).

Equating this result with 1, we obtain the desired expression for C. (b) Using the limiting forms ( (π/2x)1/2 e−x K2 (x) ≈ 2/x2

(x >> 1) , (x << 1)

we obtain, rather straightforwardly, the nonrelativistic and the extreme relativistic limits of the distribution.

7 (c) Since dε m0 c2 d(cosh θ) = = c tanh θ, dp m0 cd (sinh θ) Z∞ 2 hpui = C {m0 c2 sinh θ tanh θ} e−βm0 c cosh θ m30 c3 sinh2 θ cosh θ dθ

u=

0

= C m40 c5

Z∞

2

e−βm0 c cosh θ sinh4 θ dθ.

0

Once again, integrating by parts (this time twice), we obtain hpui = C m40 c5 · 3(βm0 c2 )−2 K2 (βm0 c2 ).

Substituting for C, we obtain: hpui = 3/β — regardless of the severity of the relativistic effects and in conformity with the results of Secs. 3.7 and 6.4.

6.12. Ordinarily, when a molecule is reflected from a stationary wall that is perpendicular to the z-direction, the z-component of its velocity u simply changes sign, i.e. u0z = −uz . If the wall is receding at velocity v in the direction of its normal, the above result changes to (u0z − v) = −(uz − v), so that u0z = −(uz − 2v). This results in a change in the translational energy of the molecule which, for small v, is given by ∆ε =

1 1 2 mu 02 z − mu z ' −2mu z v.

2 2

If A is the area of the wall, the net change in the energy of the gas, in time δt, is then given by, cf. eqn. (6.4.10), Z∞

Z∞

Z∞

δE = Aδt · n

(−2mu z v)uz f (u)du x du y du z

ux =−∞ uy =−∞ uz =−∞

Z2π Zπ/2 Z∞ = −Avδt · n

= −δV · n

1 3

{2mu 2 cos2 θf (u)}(u2 sin θ du dθ dφ)

φ=0 θ=0 u=0 Z∞

{mu 2 f (u)}(4π u2 du)

0

2 2 Ek = −δV · nε̄k = −δV · , 3 3 V

(1)

where Ek is the total kinetic energy of the gas. Note that, since the gas continues to be in a state of (quasi-static) equilibrium, the change δE (even though it originates in the translational motion of the molecules) becomes

8 eventually a change in the internal energy U of the gas (which may well have contributions from degrees of freedom other than translational). If U = aE k , we may write δEk =

2 δV 1 δU = − Ek . a 3a V

(2)

Next, since PV = (2/3)Ek , we get δP δV δEk 2 δV + = =− .

P V Ek 3a 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 by 1/3, 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 1 nū sin θ cos θdθ.

2

dR θ = n(ū/4π) · 2π sin θ cos θdθ =

For part (b), we integrate only over θ and ϕ, to get dR u = nπ · f (u)u3 du, where f (u)|M.B. =

m 3/2 2 e−mu /2kT .

2πkT

For part (c), we refer to expression (6.4.10) instead and get m 3/2 RE = n 2πkT

Z∞

Z∞

Z∞ √

ux =−∞ uy =−∞ u = z

=n

kT 2πm

1/2

2

2

1 1 − T2 T1

2

e−m(ux +uy +uz )/2kT uz du x du y du z

2E/m

e−E/kT .

It follows that RE (T2 ) = RE (T1 )

T2 T1

1/2

exp

−E k

.

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

6.14. (a) We start by calculating the kinetic energy associated with the zcomponent of the motion of the effused molecules. Proceeding as on

9 page 139 of the text, we get [see eqn. (6.4.11)] π/2 R R∞

1 mu 2z 2

(u3 cos3 θ)f (u)u2 sin θdudθ 1 1 0 0 2 2 = mhu cos θi = m π/2 2 2 R R∞ (u cos θ)f (u)u2 sin θdudθ 0

0

1 hu3 i ; = m 4 hui note that the averages on the right-hand side are taken over the gas inside the vessel. It is not difficult to show, see the corresponding calculation in Problem 6.6 and the formulae (B.13b), that I5 4 hu3 i = = , hui I3 βm so that 21 mu 2z , for the effused molecules, = 1/β = kT . The kinetic energy associated with the x- and y-components of the molecular motion will be the same as inside the vessel, viz. 12 kT each. It follows that the mean energy ε of an effused molecule is 2 kT. (b) Assuming quasi-static equilibrium, the relations E = (3/2)NkT and P = NkT /V will continue to hold for the gas inside the vessel. However, in view of the result obtained in part (a), we shall also have dE d ≡ dt dt

3 NkT 2

= 2kT

dN . dt

It follows that 1 dN dT = and hence T ∝ N 1/3 ; T 3 N it further follows that P ∝ N 4/3 .

As for explicit variations with t, we make use of eqn. (6.4.13) and write dN 1 1 aN = − anhui = − dt 4 4 V

8kT πm

1/2 .

Combining the last two results, we get dT 1a =− T 3V

kT 2πm

1/2 dt,

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

10 6.15. If nH is the number of holes per unit area of the surface of the balloon (of radius r), a the area of each hole and t the duration of the leak, then the total number of molecules leaking is given by ∆N =

1 nū · nH (4πr2 )at 4

i h ū = (8kT /πm)1/2 .

The fraction of the molecules leaking is thus given by ∆N 1 = N 4V

8kT πm

1/2

· nH (4πr2 )at.

Since V = (4π/3)r3 , we get ∆N r nH = · N 3at

2πm kT

1/2 .

2

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

6.16. The rate of effusion of molecules from side A to side B, through a hole of cross-section S, is given by the expression r 1 8kT A PA RA→B = nA S=√ S; 4 πmA 2πmA kT A the same from side B to side A is given by r 1 8kT B PB RB→A = nB S=√ S. 4 πmB 2πmB kT B In the stationary state, these two expressions will be equal — which leads to the condition of dynamic equilibrium PA /PB = (mA TA /mB TB )1/2 .

If the two gases are samples of the same gas, the condition simplifies to PA /PB = (TA /TB )1/2 .

6.18. The (un-normalized) velocity distribution for a pair of molecules is given by 2

2

F (u1 , u2 )d3 u1 d3 u2 ∼ e− 2 βm(u1 +u2 ) d3 u1 d3 u2 .

1

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 (u1 + u2 ).

2

11 This results in a new distribution for the variables v and V: 2

1

2

F (v,V)d3 vd 3 V ∼ e− 4 βmv d3 v · e−βmV d3 V. √ while It is now straightforward to show that hvi = (16/πβm)1/2 = 2hui, √ hv2 i = (6/βm) = 2hu2 i. The latter result implies that vr.m.s. = 2 ur.m.s ..

We note that, since v2 = u21 + u22 + u22 − 2u1 · u2 , hv2 i = 2hu2 i, regardless of the law of distribution of velocities — so long as it is isotropic, making hu1 · u2 i = 0.

6.19. The (un-normalized) joint distribution for the molecular energies ε1 and ε2 is 1/2 1/2

f (ε1 , ε2 )dε1 dε2 ∼ e−β(ε1 +ε2 ) ε1 ε2 dε1 dε2 .

To obtain the desired distribution, we set ε2 = E − ε1 and integrate over all relevant values of ε1 , with the result that p(E)dE ∼ e

−βE

ZE

{ε1 (E − ε1 )}1/2 dε1 dE

0

∼ e−βE E 2 dE ; cf. eqns. (3.4.3) and (3.5.16), with N = 2. It is now straightforward to check that B −4 Γ(4) 3 hEi = −3 = . β Γ(3) β 6.20. The relative fraction of the excited atoms in the given sample of the helium gas would be 3e−βε1 , where βε1 =

hc ' 38.22. kT λ1

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

6.21. We extend the treatment of Problem 3.14 to the reaction AB + CD ↔ AD + CB and obtain, in equilibrium, nAD nCB fAD fCB = = K(T ). nAB nCD fAB nCD For the given reaction, K(T ) =

2 fHD , fHH fDD

12 where each f is a product of three factors — the translational, the rotational and the vibrational.

Now, for a heteronuclear molecule like HD we have, at high temperatures, 3/2

2IHD kT kT mHD kT · · fHD ≈ V , 2 2 2π~ ~ ~ωHD while for a homonuclear molecule like HH we have instead

3/2 mHH kT IHH kT kT · fHH ≈ V · ; 2π~2 ~2 ~ωHH see Note 11 on page 156 of the text. It follows that, at high temperatures, K(T ) ≈ 4

2 2 IHD IHD ωHH ωDD . · · · 2 3/2 3/2 I I I I ωHD HH DD HH DD mHH mDD

m3HD

(1)

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

Accordingly, 2 {mH mD /(mH + mD )}3 IHD ωHH ωDD µ3HD = · = 3/2 1 3/2 .

2 3/2 3/2 1 IHH IDD ωHD µ µ mH mD HH

DD

2

(2)

2

At the same time, m3HD 3/2

= 3/2

mHH mDD

(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 equilibrium value of r. Accordingly, the quantum of the rotational motion of the molecule is ~2 /2I, where I = µr02 . This gives for Θr the expression ~2 /2µr02 k = ~2 /mr 20 k 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. This gives a fairly clear idea of the “temperature range” where the rotational motion of the hydrogen molecules begins to contribute towards the specific heat 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 (2 V0 /µa2 )1/2 = (4V0 /ma 2 )1/2 and hence a Θv equal to ~(4V0 /ma 2 )1/2 /k. Substituting the given data, Θv turns out to be about 6260 K. Again, this gives a fairly clear idea of the “temperature range” where the vibrational motion of the hydrogen molecules begins to contribute towards the specific heat of the gas.

13 6.24. The effective potential of a diatomic molecule (including both rotation and vibration) is given by 1 ~2 J(J + 1).

V (r) = −V0 + µω 2 (r − r0 )2 + 2 2µr2 The equilibrium value of r is obtained by minimizing V (r), with the result (req − r0 ) =

~2 ~2 J(J + 1) ' J(J + 1).

3 µ2 ω 2 req µ2 ω 2 r03

It follows that ∆r0 ~2 ' 2 2 4 J(J + 1) = 4 r0 µ ω r0

Θr Θv

2 J(J + 1).

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

6.25. The occupation number NJ is proportional to (2J + 1) e−εJ /kT . It follows that 1 N1 3 N0 = e−(ε0 −ε2 )/kT , = e−(ε1 −ε2 )/kT .

N2 5 N2 5 Substituting the given data, we get N0 1 N1 3 = e−1.086 ' 0.0675, = e−0.760 ' 0.2806; N2 5 N2 5 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. (ii) rotational — since the characteristic values of the parameter Θr in this case are of the order of 10 K, these degrees of freedom may be treated classically, which yields a contribution of (3/2)R; see eqn. (6.5.42). (iii) vibrational — here, the parameters Θv are such that the various contributions have to be calculated quantum-mechanically, using formula (6.5.44). We find that Θ3,4 Θ5 Θ6 Θ1,2 ' 16.00, ' 4.56, ' 16.37, ' 7.80, T T T T

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

The net result is: 3.46R. P 6.30. Equation (6.6.3) can be written α να µα = 0 where the stioichiometric coefficients are understood to be positive if they appear on the right hand side of equation (6.6.1) and negative if on the left. Using equation (6.6.5) gives X

να α + kT ln nα λ3α − kT ln jα = 0, α

which gives

nα 3 = 0. να α + kT ln n0 λα − kT ln jα + kT ln n0 α

X

Rearranging gives

X X

nα να ln = −β να α + kT ln n0 λ3α − kT ln jα n 0 α α X = −β να µ(0) α , α (0)

where µα is the chemical potential of species α at temperature T and standard density n0 . Equation (6.6.6) follows from exponentiating both sides.

6.31. Equation (6.6.11) gives [CO] = [CO2 ]

s

1 K(T ) [O2 ]

where

(0) (0) (0) K(T ) = exp −2βµCO2 + 2βµCO + βµO2 .

For the parameters given in the problem K(1500 K) = 4 × 1010 which yields [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] concentration.

6.32. The equilibrium constant for N2 + O2 → 2NO is 2 λ3 λ3 [NO] j2 = K(T ) = e−β∆ε NO N26 O2 .

[N2 ] [O2 ] jN2 jO2 λNO

15 The internal partition functions are of the form   T for kT ~ω Θr j=

kT  T for kT ~ω Θr ~ω which leads to 

303  e−β∆ε ΘN22ΘO2 3/2 323/2 Θ 28 NO

K(T ) = ωN2 ωO2 303  e−β∆ε ΘN22ΘO2 2 3/2 3/2 Θ ω 28 32 NO

NO

for kT ~ω for kT ~ω

6.33. The equilibrium relation is [CO2 ]H2 O]2 =K [CH4 ][O2 ]2 Let [excess] be the initial excess amount of O2 above stoichiometry, and [unburned] be the unburned amount of CH4 . Then 3

[CO2 ]H2 O]2 4 ([CH4 ]0 − [unburned]) = 2 =K [CH4 ][O2 ]2 [unburned] ([excess] + 2[unburned]) Since K 1, at the stoichiometric point [excess] = 0 so [unburned] ≈

[CH4 ]0 .

K 1/3

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

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

µNa = −εb − kT ln 2 + kT ln nNa λ3Na ,

µNa+ = kT ln nNa+ λ3Na ,

µe = −kT ln 2 + kT ln ne λ3e , where b is the ionization energy of Na. These lead to nNa = eβ b λ3e . nNa+ ne

16 If the total density is n0 = nNa + nNa+ , the ionized fraction f = nNa+ /n0 , and the system is charge neutral, then 1−f = eβ b n0 λ3e = s, f2 which has solution √ f=

1 + 4s − 1 .

2s

s 6.35. The partition function QNs = N1s ! QN 1 where one-particle partition function is of the form Q1 = jrot Qtrans where Z 2 2 Ae−βε e−βε dpx dpy dxdye−β(px +py )/(2m) = Qtrans = 2 h λ2 √ Z X 2 2 2 2 2πIkT jrot = e−β~ m /(2I) ≈ dme−β~ m /(2I) = , ~ m

√ λ = h/ 2πmkT . The Helmholtz free energy is A(Ns , A, T ) = −kT ln QNs = −Ns ε − Ns kT ln

A Ns λ 2

r

2πIkT ~2

! − Ns kT,

where we have used ln N ! ≈ N ln N − N . The internal energy is

∂βA 3Ns kT U= = , ∂β Ns ,V 2 so the heat capacity is

C=

∂βU 3Ns k = , ∂T Ns ,V 2

i.e. k/2 for each squared degree of freedom in the Hamiltonian (p2x /2m, p2y /2m, L2z /2I).

The pressure and chemical potential are

Ns kT ∂A = , P =− ∂V Ns ,T A

∂A A µ= = −ε − kT ln ∂N V,T Ns λ2 6.36. The single-particle rotational partition function is jrot =

∞ X l=0

(2l + 1)2 e−l(l+1)θ/T = 1 + 9e−2θ/T + 25e−6θ/T + · · ·

17 At low temperature the successive terms get exponentially smaller so the contribution to the rotational free energy, internal energy and heat capacity per particle is jrot (T ) ≈ −9kT e−2θ/T , Urot (T ) ≈ 18kθe−2θ/T , 2 θ e−2θ/T .

Crot (T ) = 36k T The rotational heat capacity is 2

CV θ = Z2 /Z0 − (Z1 /Z0 )2 , Nk T where Zn =

∞ X

n

(2l + 1)2 (l(l + 1)) e−l(l+1)θ/T

l=0

By approximating the summation jrot as an integral, you can show that the heat capacity at high temperature is C ≈ 32 N k, and evaluating the summations numerically gives the following figure.

6.37. The canonical momenta are given by pX = ∂L /∂ Ẋ, etc., and the Legendre transformation H = −L + PX Ẋ + · · · + pφ φ̇ gives

1 H (pX , X, . . . , pφ , φ) = p2X + p2Y + p2Z 2M 1 2 1 2 1 + pr + pθ + p2φ .

2 2 2µ 2µr 2µr sin2 φ

18 The partition function is given by

Z Q1 (T, V ) =

e−βH dΓ =

4πV λ3M λ3µ

Z ∞

2

e−βks (r−a) r2 dr,

0

√ √ where λp M = h/ 2πM kT and λµ = h/ 2πµkT . For small T the r-integral gives a2 2πkT /ks , while at large T the r-integral gives (kT /ks )3/2 /2. The low temperature partition function is proportional to T 7/2 so the specific heat is CV /N = 7k/2, while the high temperature partition function is proportional to T 9/2 so the specific heat is CV /N = 9k/2. There are six squared momentum degrees of freedom in both cases. At low temperature (kT ks a2 ) the vibrations of the spring only stretch the spring a small amount along one-dimension, while at high temperature (kT ks a2 ) the spring acts like a three-dimensional harmonic oscillator. In terms of integrals, the specific heat is given by

C T =3+ Nk 4θ

Z2 − Z0

Z2 Z0

2 ! ,

where θ = ks a2 /k,

Z ∞ Z0 = Z0 ∞ Z1 = Z0 ∞ Z2 =

2

x2 e−(θ/T )(x−1) /2 , 2

x2 (x − 1)2 e−(θ/T (x−1) /2 , 2

x2 (x − 1)4 e−(θ/T (x−1) /2 .

0

The result as a function of T /θ is shown in this figure.

19

Chapter 7 7.2. With N0 << N , eqn. (7.1.8) reads nλ3 = g3/2 (z) = z + 2−3/2 z 2 + 3−3/2 z 3 + 4−3/2 z 4 + . . . ,

(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 ) on the two sides of the resulting equation, we get 1 = c1 , 0 = c2 + 2−3/2 c21 , 0 = c3 + 2−3/2 · 2c1 c2 + 3−3/2 c31 ,

0 = c4 + 2−3/2 c22 + 2c1 c3 + 3−3/2 · 3c21 c2 + 4−3/2 c41 , . . . .

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 1 = (z + 2−5/2 z 2 + 3−5/2 z 3 + 4−5/2 z 4 + . . .)

NkT nλ3 and substitute expression (2) into it. This leads to the desired result (7.1.13), with a1 = c1 = 1, a2 = c2 + 2−5/2 c21 = −2−5/2 , a3 = c3 + 2−5/2 · 2c1 c2 + 3−5/2 c31 = (1/8) − 2.3−5/2 ,

a4 = c4 + 2−5/2 c22 + 2c1 c3 + 3−5/2 · 3c21 c2 + 4−5/2 c41 = 3.6−3/2 − 5.2−11/2 − (3/32), in perfect agreement with the values quoted in expressions (7.1.14).

1

2 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 help of formula (D.9), with the result that T = Tc

ζ(3/2) − 2π 1/2 α1/2 + . . . ζ(3/2)

−2/3 ≈1+

4π 1/2 α1/2 , 3ζ(3/2)

valid for α << 1 and hence for T & Tc . The desired result now follows readily.

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

5 3/2 ∂z 5/2 ∂g5/2 (z) 0 = cT g5/2 (z) + cT , 2 ∂z ∂T P so that

∂z ∂T

=− P

g5/2 (z) 5 .

2T {∂g5/2 (z)/∂z}

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

5 g5/2 (z) 1 ∂z =− . z ∂T P 2T g3/2 (z)

(1)

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

∂S ∂S ∂z ∂z CP = T and CV = T . ∂z N ∂T P ∂z N ∂T v It follows that γ=

CP (∂z/∂T )P = .

CV (∂z/∂T )v

Substituting from eqn. (1) above and from eqn. (7.1.36), we obtain the desired 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 ; see eqn. (D.8). Along with it, both γ and CP diverge as (T − Tc )−1 ; see the relation established in Problem 7.3.

3 7.5. (a) We have to evaluate the quantities

1 ∂n 1 ∂n κT = and κS = , n ∂P T n ∂P z where n = N/V . For N0 << N, n(T, z) = aT 3/2 g3/2 (z), where a is a constant; see eqn. (7.1.8). It follows that

3 1/2 1 3/2 dn = aT g3/2 (z)dT + aT g1/2 (z) dz .

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

1 5 3/2 5/2 g3/2 (z) dz . dP = cT g5/2 (z)dT + cT 2 z The quantities κT and κS are then given by κT =

1 a g1/2 (z) n cT g3/2 (z)

and κS =

1 3a g3/2 (z) . n 5cT 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 ·

2 2 g 4CV 1 g1/2 (z) 4CV 1/2 (z) · = , nkT g3/2 (z) 9V 2 9Nk g3/2 (z)

in agreement with eqn. (7.1.48a). The other result follows straightforwardly.

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

∂ ∂CV CV ∂ ln z 1 = . · Nk ∂T V ∂ ln z Nk ∂T v The first factor turns out to be 15 g3/2 (z)g3/2 (z) − g5/2 (z)g1/2 (z) 9 g1/2 (z)g1/2 (z) − g3/2 (z)g−1/2 (z) − 4 {g3/2 (z)}2 4 {g1/2 (z)}2 =

3 15 g5/2 (z)g1/2 (z) 9 g3/2 (z)g−1/2 (z) − + .

2 4 {g3/2 (z)}2 4 {g1/2 (z)}2

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

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

∂CV 3 CV 1 = , Nk ∂T V 2T Nk

4 which leads to the result quoted in the problem.

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

1 45 ζ(5/2) 27 {ζ(3/2)}2 · Γ(3/2)α−3/2 ; −0− Tc 8 ζ(3/2) 8 {Γ(1/2)α−1/2 }3 on the other hand, as T → Tc from below, we obtain simply 1 45 ζ(5/2) · .

Tc 8 ζ(3/2) The discontinuity in the slope of the specific heat curve at T = Tc is, therefore, given by Nk 27 · Tc 8

2 2 3 (π 1/2 /2) 27Nk 3 ζ · = . ζ 2 16πTc 2 π 3/2

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

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

∂µ kT ∂z 3 g3/2 (z) = k ln z + · = k ln z − k , ∂T v z ∂T v 2 g1/2 (z) 2

∂ µ 3 g1/2 (z)g1/2 (z) − g3/2 (z)g−1/2 (z) ∂ ln z = k− k ∂T 2 v 2 {g1/2 (z)}2 ∂T v =

9k {g3/2 (z)}2 g−1/2 (z) 3k g3/2 (z) − 4T g1/2 (z) 4T {g1/2 (z)}3

Similarly, using a result from Problem 7.4, we obtain 2 15k g5/2 (z) 25k {g5/2 (z)}2 g1/2 (z) ∂ µ = − .

2 ∂T P 4T g3/2 (z) 4T {g3/2 (z)}3 We also note, see eqns. (7.1.37) and (7.1.48b), that 25 {g5/2 (z)}2 g1/2 (z) 15 g5/2 (z) CP = − .

Nk 4 {g3/2 (z)}3 4 g3/2 (z) It is now straightforward to see that the stated thermodynamic relations are indeed satisfied. The critical behavior of these quantities is also straightforward to check.

7.8. One readily sees that w2 =

∂P ∂(nm)

= S

1 , mnκS

5 where κS is the adiabatic compressibility of the fluid. Using a result from Problem 7.5, we get for the ideal Bose gas w2 =

5kT g5/2 (z) .

3m g3/2 (z)

Next,

2

hu i =

2ε m

=

2 U 3kT g5/2 (z) = ; mN m 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 and ε−1/2 : R∞ hnε iε1/2 a(ε)dε

R∞ hnε iε−1/2 a(ε)dε

, hε−1/2 i = 0 R∞ hε1/2 i = 0 R∞ hnε ia(ε)dε hnε ia(ε)dε 0

.

0

The integral in the denominator has been evaluated on pp. 158–9 of the text; those in the numerator can be evaluated like-wise, with the results hε1/2 i = (kT )1/2

Γ(2)g2 (z) Γ(1)g1 (z) , hε−1/2 i = (kT )−1/2 . Γ(3/2)g3/2 (z) Γ(3/2)g3/2 (z)

It follows that r 2 1/2 8kT g2 (z) hui = hε i = , while m πm g3/2 (z) r r m −1/2 2m g1 (z) −1 hu i = hε i= .

2 πkT g3/2 (z) r

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

For z → 0, we recover the classical result stated in Problem 6.6. For z → 1, we encounter divergence of the quantity hu−1 i, which arises from the 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) has to be carried out over the states of the internal spectrum as well as over the translational states. Expression (7.1.16) is then replaced by

n µ o V V µ − ε1 Ne = (Ne )0 + (Ne )1 = 3 g3/2 exp + 3 g3/2 exp . λ kT λ kT The critical temperature Tc is then determined by the condition V V g3/2 (1) + 3 g3/2 (x) = N, where x = e−ε1 /kT c . λ3c λc

(1)

6 For x << 1, g3/2 (x) ' x and eqn. (1) gives λ3c ' (V /N )[ζ(3/2) + x].

3 Comparing this with the standard result λ0c = (V /N )ζ(3/2), we get Tc ≡ Tc0

λ0c λc

2

' 1+

x ζ(3/2)

−2/3 '1−

2/3 2/3 −ε1 /kT 0c x'1− e . ζ(3/2) ζ(3/2)

For x . 1, on the other hand, g3/2 (x) ' ζ(3/2)−2π 1/2 (− ln x)1/2 ; eqn. (1) now gives λ3c ' (2V /N )[ζ(3/2) − π 1/2 (ε1 /kT c )1/2 ], whence ( "

1/2 #)−2/3 π 1/2 ε1 Tc ' 2 1− Tc0 ζ(3/2) kT c "

1/2 # 1/2 2 ε π 1 ' 2−2/3 1 + 21/3 .

3 ζ(3/2) kT 0c 7.12. The relative mean-square fluctuation in N is given by the general formula (4.5.7), (∆N )2 kT = κT , (1) V N̄ 2 while κT for the ideal Bose gas is given in Problem 7.5. As T → Tc from above, the function g1/2 (z) and, along with it, both κT and the relative fluctuation in N diverge!

The mean-square fluctuation in E is given by the general formula (4.5.14), viz. (∆E)2 = kT 2 CV + {(∂U/∂N )T,V }2 (∆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 help of eqns. (7.1.8 and 11), whereby

∂g5/2 (z) g3/2 (z) ∂U = = .

(3) ∂N T,V ∂g3/2 (z) T,V g3/2 (z) The second term in (2) is, therefore, inversely proportional to g1/2 (z) and hence vanishes as T → Tc ; this happens because the energy associated with the Bose condensate (which is, in fact, the component responsible for the dramatic rise in the fluctuation of N ) is zero. Thus, all in all, the relative fluctuation in E is negligible at all T. 7.13. It is straightforward to see that for a Bose gas in two dimensions Z∞ Ne = 0

1 A · 2πp dp A · 2πmkT = z −1 eβε−1 h2 h2

Z∞ 0

dx A2 = g1 (z), z −1 ex − 1 λ

7 while

z .

1−z Since Bose-Einstein condensation requires that z → 1, the critical temperature Tc , by the usual argument, is given by

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

A N0 =

It follows that Tc = 0.

More accurately, the phenomenon of condensation requires that both Ne and N0 be of order N . This means that, while z ' 1, (1 − z) be of order 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 ). It follows that h2 1 h2 . ∼ T ≡ 2 2πmk λ mk `2 ln N 7.14. With energy spectrum ε = Ap s , the density of states in the system is given by, see formula (C.7b), a(ε)dε =

V 2π n/2 n−1 V 2π n/2 ε(n/s)−1 dε. p dp = hn Γ(n − 2) hn sAn/s Γ(n/2)

(1)

This leads to the expression V 2π n/2 N − N0 = n n/s h sA Γ(n/2)

Z∞

ε(n/2)−1 z −1 eβε − 1

dε

0

V 2π n/2 Γ(n/s) = n n/s (kT )n/s gn/s (z), h sA Γ(n/2)

(2)

while N0 = z/(1 − z). Similarly, P =

1 2π n/2 Γ(n/s) (kT )(n/s)+1 g(n+s)+1 (z). hn sAn/s Γ(n/2)

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

PV n ∂ 2 = PV , U = kT ∂T kT s z,v

(3)

(4)

so that P = sU /nV .

The onset of Bose-Einstein condensation requires that z → 1 at a finite temperature Tc . A glance at eqn. (2) tells us that this will happen only if n > s and that the critical temperature Tc will then be determined by the equation n V 2π n/2 Γ(n/s) .

(5) N = n n/s (kT c )n/sζ h sA Γ(n/2) s

8 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 for T > Tc (when N0 << N ) U=

n NkT · g(n/s)+1 (z)/gn/s (z) s

(6)

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

n 1g 1 ∂z n 1 gn/s(z) 1 ∂z (n/s)+1 (z) =− and =− +1 .

2 ∂T v s T g(n/s)−1 (z) z ∂T P s T gn/s (z) (7) It is now straightforward to show that g n 2 g (z) n n Cv (n/s)+1 (z) n/s = +1 − (8) Nk s s gn/s (z) s g(n/s)−1 (z) and n 2 {g g 2 CP n n (n/s)+1 (z) (n/s)+1 (z)} g(n/s−1) (z) = +1 − + 1 . (9) Nk s {gn/s (z)}3 s s gn/s (z) The limiting cases suggested in the problem follow quite easily.

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

~2 ∂ 2 ψ ∂ψ = i~ , 2m ∂x2 ∂t p2 ∂ ψ̂ ψ̂ = i~ .

2m ∂t

The momentum representation is easily solved 2 p t ψ̂(p, t) = exp ψ̂(p, 0), 2i~m where ψ̂(p, 0) = √

1 2π~

Z

eipx/~ ψ(x, 0)dx.

This leads to (suppressing the normalization factor)

p2 i~t 2 ψ̂(p, t) ∼ exp − 2 a + .

2~ m Inverse Fourier transforming gives

√ a 1 x2 p ψ(x, t) = exp − 2 .

2 a + i~t/m π 1/4 a2 + i~t/m

9 This solves the Schrodinger equation and leads to the one-dimensional density

x2 1 1 2 exp − 2 |ψ(x, t)| = 1/2 p . a (1 + (~t/ma2 )2 ) π a 1 + (~t/ma2 )2 This gives the spatial distribution for one cartesian direction once you note that ~/ma2 = ω0 . At long-time, the width of the distribution grows linearly in time.

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

βp2 βmω 2 x2 ω exp − − . f (p, x, 0) = 2πkT 2m 2 After the potential is turned off at t = 0, the particles move ballistically so the density becomes

ω βp2 βmω 2 (x + pt/m)2 f (p, x, t) = f (p, x + pt/m, 0) = exp − − .

2πkT 2m 2 The spatial density is then given by r

Z ω 2πmkT 1 βmω 2 x2 n(x, t) = f (p, x + pt/m, 0)dp = exp − .

2πkT 1 + ω 2 t2 2 1 + ω 2 t2 The high-temperature limit of equation (7.2.15) is given by the first term in the series since at high temperature the chemical potential is large and negative.

7.17. The ground state density at the center of theptrap is N0 /(π 3 /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)1/2 N 1/2 1.

7.18. Integrating equation (7.2.15) gives 3 X

Z ∞ ∞ eβµj 1 X eβµj (kT )3/2 kT . nex (r)dr = 3 = 3 λ j=1 j 3 m3/2 ω0 ~ω0 j3 j=1 The excited particles can be counted using the density of states and the Bose-Einstein factor, 3 X

Z Z ∞ ∞ X (kT )3 kT eβµj 1 2 −x βµj dε = x e e dx = .

Nex = a(ε) β(ε−µ) 2(~ω)3 ~ω0 j3 e −1 j=1 j=1 Above Tc when µ < 0 this counts all of the particles. Below Tc when µ = 0, this counts the particles that are not in the ground state.

10 7.19. The density of states for a two-dimensional harmonic oscillator is a(ε) = ε/(~ω0 )2 so the number particles in the trap is given by Z N (T, µ) =

dε

1 ε .

2 β(ε−µ) (~ω) e −1

As T → Tc , µ → 0 so Z N=

1 ε = dε 2 β (ε) c (~ω) e −1

kTc ~ω0

2 Z

xdx = ζ(2) x e −1

kTc ~ω0

2

π2 = 6

kTc ~ω0

p so kTc = ~ω 6N/π 2 . The condensate fraction for T ≤ Tc is N0 /N = 1 − (T /Tc )2 . For this two-dimensional theory to be valid, the occupancy of √ the first excited z-state must be negligible which requires √ ~ωz kTc ∼ N ~ω0 , i.e. ωz N ω0 .

7.20. By eqn. (3.8.14), −kT ln Q1 = kT ln(eβ~ω/2 − e−β~ω/2 ) =

~ω + kT ln(1 − e−β~ω ).

2

Now, concentrating on the thermal part alone and utilizing eqn. (7.2.2), we get VkT A(V, T ) ≡ −kT ln Q(V, T ) = 2 3 π c

Z∞

ln(1 − e−β~ω )ω 2 dω.

0

After an integration by parts, we obtain V~ A(V, T ) = − 2 3 3π c

Z∞

ω 3 dω π 2 Vk 4 T 4 = − ; eβ~ω−1 45~3 c3

0

cf. eqns. (7.2.17 and 18). We also get

S=−

∂A ∂T

=− V

4A T

and U = A + TS = −3A = 3PV .

Other results of Sec. 7.2 follow straightforwardly.

7.21. Using expressions (7.2.12) and (7.2.23), we readily get U π4 = kT ' 2.7 kT .

30ζ(3) N̄ Note that the numerical factor appearing here is actually Γ(4)ζ(4)/Γ(3)ζ(3).

2 .

11 7.22. Since ω = 2πc/λ, the characteristic frequencies of the vibrational modes of a radiation cavity (and hence the energy eigenvalues of these modes) are proportional to L−1 , i.e. to V −1/3 . Just as in Problem 1.7, we infer that the entropy of this system is a function of the combination (V 1/3 U ). It then follows that during an isentropic process the quantity (V 1/3 U ) stays cosntant, i.e.

1 −2/3 V dV U + V 1/3 dU = 0.

3 Consequently, the pressure of the system is given by

∂U 1U P ≡− = . ∂V S 3V 7.24. The number density of photons in the cosmic microwave background (CMB) follows from equation (7.3.23)

3 2ζ(3) kT n= ' 4.10 × 108 m−3 ' 410. cm−3 π2 ~c The energy density is u=

π 2 (kT )4 ' 4.17 × 10−14 J/m3 .

15 (~c)3

The entropy density is

3 4π 2 k kT s= ' 1.48 × 109 k m−3 ' 2.04 × 10−14 J/m3 K. 45 ~c 7.25. According to Sec. 7.3,

Z Z ∂ ~ω CV (T ) = g(ω)dω, while C (∞) = kg(ω)dω.

V ∂T e~ω/kT − 1 ω

ω

It follows that Z∞ Z kT − {CV (∞) − CV (T )}dT = ω

0

~ω

∞

e~ω/kT − 1 0

g(ω)dω.

It is easy to show that ~ω

lim

T →∞ e~ω/kT − 1

1 ≈ kT − ~ω; 2

see page 69 as well as Fig. 3.4 of the text. The intgral on the right-hand side then becomes Z 1 ~ω · g(ω)dω, 2 ω

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

The physical interpretation of this result lies in noting that the actual amount of heat required to raise the temperature of a solid is less than the value predicted classically because the solid already possesses a finite amount of energy even at T = 0K. 7.26. Using the Debye spectrum (7.3.15), we have for the zero-point energy of the solid ZωD 1 9N 9 9 ~ω · 3 ω 2 dω = N ~ωD = Nk ΘD .

2 ωD 8 8 0

Indeed, ω RD

ω · ω 2 dω

0

ω̄ =

ω RD

= ω 2 dω

3 ωD 4

0

and hence the mean energy per mode is equal to 12 ~ω̄ = 83 ~ωD = 38 kΘD .

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

We know that

CP − CV = T

∂P ∂T

V

Since

S≡−

∂V ∂T

∂A ∂T

= −T P

∂V ∂P

T

∂P ∂T

= aVT n , V

we must have A = −aVT n+1 /(n + 1) + f (V ), where f (V ) is a function of V alone. It follows that

∂A T n+1 P ≡− =a − f 0 (V ), ∂V T n+1 so that

∂P ∂V

00

= −f (V ), T

∂P ∂T

= aT n ; V

clearly, f 00 (V ) must be non-negative. We thus get CP − CV = −T ·

−1 f 00 (V )

(aT n )2 =

a2 f 00(V )

T 2n+1 .

2 .

V

13 7.33. The specific heat of the system is given by the general expression (7.3.8), which may in the present case be written as ZωD CV (T ) = k

(~ω/kT )2 e~ω/kT g(ω)dω. (e~ω/kT − 1)2

(1)

0

The mode density, g(ω), is given by the relation g(ω)dω = 3 · V (4πp2 dp)/h3 , where p = ~k = ~(A−1 ω)1/s . It follows that g(ω)dω = Cω (3/s)−1) dω

[C = 3V /(2sπ 2 A3/s )].

(2)

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

Z x0 (3/s)+1 x ~ωD x e dx x = .

CV (T ) ∼ T 3/s 0 (ex − 1)2 kT 0 At low temperatures, the upper limit of this integral may be replaced by infinity — making the integral essentially T -independent; this leads to the desired result CV ∼ T 3/s .

7.34. The mode density in this case is given by, see eqn. (C.7b), g(ω)dω ∼ k n−1 dk ∼ ω n−1 dω.

The rest of the argument is similar to the one made in the previous problem; the net result is that the specific heat of the given system, at low temperatures, is proportional to T n .

It is not difficult to see that if the dispersion relation were ω ∼ k s and the dimensionality of the system were n, then the low-temperature specific heat of the system would be proportional to T n/s .

7.35. The Hamiltonian of this system is given by eqn. (7.3.6); the partition function then turns out to be, see eqn. (3.8.14), Q = e−βΦ0

Y

2 sinh

i

1 β~ωi 2

−1 ,

with the result that A = −kT ln Q = Φ0 + kT

X

ln{2 sinh(~ωi /2kT )}, and hence

i

P =−

∂A ∂V

=− T

∂Φ0 1 X − ~ coth ∂V 2 i

~ωi 2kT

·

∂ωi . ∂V

14 0 Recognizing that (i) total vibrational energy U of this system is given P the 1 by the expression 2 ~ωi coth(~ωi /2kT ), see eqn. (3.8.20), and (ii) the i

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

∂Φ0 U0 +γ ∂V V

(U 0 = U − Φ0 );

(1)

see eqn. (7.3.7). With Φ0 (V ) = (V − V0 )2 /2κ0 V0 , eqn. (1) takes the form P =−

U0 V − V0 +γ . κ0 V0 V

(2)

Now, the coefficient of thermal expansion of any thermodynamic system is given by

1 ∂V ∂P 1 ∂V ∂P =− = κT , (3) α≡ V ∂T P V ∂P T ∂T V ∂T V where κT is the isothermal compressibility. In the present case, eqn. (2) gives

0 ∂P V ∂U U0 −1 κT ≡ −V = −γ ; +γ ∂V T κ0 V0 V ∂V T using the thermodynamic formula (∂U/∂V )T = T (∂P/∂T )v − P , where U = Φ0 + U 0 , we get

V ∂P ∂Φ0 U0 κ−1 = − γT + γP + γ . + γ T κ0 V0 V ∂T V ∂V Next, since

∂P ∂V

=γ V

CV , V

(4)

we get κ−1 T =

V V − V0 γ 2 TC V + (1 + γ) P + − κ0 V0 κ0 V0 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 be approximated 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

=T P

∂P ∂T

· αV ≈ V

γ 2 κ0 TC 2V .

V0

Note that, at low temperatures, (Cp − Cv ) ∼ T 7 — as in Problem 7.21.

15 7.36. For rotons, ε = ∆ + (p − p0 )2 /2µ. Therefore, u ≡ dε/dp = (p − p0 )/µ.

Consequently, 1 nhp(p − p0 )/µi 3 R −(p−p )2 /2µkT 0 e {p(p − p0 )/µ}p2 dp 1 R = n .

3 e−(p−p0 )2 /2µkT p2 dp √ Substituting p = p0 + 2µkT x, we get 3 R −x2 √ e p0 + 2µkT x (2kT /µ)1/2 x dx 1 P = n .

2 R √ 3 e−x2 p0 + 2µkT x dx P =

As explained on page 185 of the text, these integrals are well-approximated by √ letting the range of x extend from −∞ to +∞; also remembering that 2µkT << p0 , we get √

1/2 √ 2kT 1 3p20 2µkT · ( π/2) √ P ' n 3 µ p20 · π = nkT .

7.37. Following Secs. 7.4 and 7.5, the free energy A(v) of a roton gas in mass motion is given by Z V A(v) = −N̄ kT = −kT · 3 n(ε − v · p)2πp2 · sin θdpd θ. h As explained on page 185 of the text, though rotons obey Bose-Einstein statistics, their distribution function is practically Boltzmannian; see eqns. (7.5.6 and 7). We may, therefore, write Z V A(v) = −kT · 3 e−βε+βvp cos θ 2πp2 sin θ dpd θ. h Integrating over θ, we get V A(v) = −kT · 3 h

Z∞

e−βε

sinh(βvp) 4πp2 dp. βvp

0

Integration over p is now carried out the same way as in eqn. (7.5.9); with appropriate 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 Z 1 1 ρ(v) = · 3 n(ε − v · p)p cos θ 2πp2 sin θdpd θ v h Z 1 ' e−βε+βvp cos θ 2πp3 cos θ sin θdpd θ vh3

16 Integration over θ now gives Z∞

1 ρ(v) = vh3

−βε

e

cosh(βvp) sinh(βvp) − βvp (βvp)2

4πp3 dp

0

=

β h3

Z∞

e−βε

(βvp) cosh(βvp) − sinh(βvp) 4πp4 dp. (βvp)3

0

Finally, integrating over p (under appropriate approximation) and comparing the resulting expression with eqn. (7.5.19), we obtain ρv = ρ(0)

3{(βvp0 ) cosh(βvp0 ) − sin(βvp0 )} . (βvp0 )3

7.38. We write eqn. (7.5.17) in the form 4π ρ0 = − 3 3h

Z∞

∂n(p) ∂p

4 dp p dp dε

0

and integrate it by parts, to get  

Z∞ ∞ 4π d dp dp ρ0 = − 3  n(p)p4 − n(p) p4 dp  .

3h dε 0 dp dε 0

The integrated part vanishes at both limits, and we are left with ρ0 =

4π 3h3

Z∞ n(p)

d dp

dp p4 dp. dε

0

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

4πV N̄ = 3 h

Z∞

n(p)p2 dp,

0

we obtain for the effective mass of an excitation

ρ0 V 1 1 d 4 dp meff = = − p .

3 p2 dp dε N̄ For ideal-gas particles, ε = p2 /2m; the effective mass then turns out to be precisely equal to m. For phonons, ε = pc; we then get (meff )ph = 4 < ε > /3c2 ,

17 in agreement with eqn. (7.4.15). Unfortunately, in the case of rotons this expression presents certain problems of analyticity at the point p = p0 ; we then resort to direct calculation — leading to eqn. (7.5.19), whereby (meff )rot ' p20 /3kT .

7.39. The occupancy of the first excited state below the condensation temperature is given by

kT 1 T ≈ hN1 i = β~ω0 ∼ N 1/3 since kTc ~ω0 . −1 e ~ω0 Tc

hN1 i 1 T ∼ 2/3 → 0 in the thermodynamic limit.

N Tc N 7.40. The general solution for for power-law density of states εσ (d − 1)!( 1 )σ+1

σ+1 X ∞ zj kT

g(ε) = α hN i =

1 j σ+1 j=1

σ+1 kT hN i = α gσ+1 (z) ~ω0

σ+1 kT N =α ζσ+1 ~ω0 Z ∞ ∞ X 1 1 ds for σ 1. ≈ = ζσ+1 = σ+1 σ+1 j s σ 1 j=1 1/(σ+1)

1/(σ+1) kTc N σN → 0 as σ → 0. = ≈ 1 αζσ+1 α Applying this to a d-dimensional harmonic oscillator gives g(ε) = hN i =

εd−1 (d − 1)!(~ω0 )d

d X ∞ kT zd ~ω0 kT ~ω0

d

kTc ~ω0 ∞ X 1

d

hN i =

N=

j=1

jd

gd (z) ζd Z ∞

ds 1 ≈ = for d − 1 1. d d j s d − 1 1 j=1 1/d kTc N 1/d = ≈ ((d − 1)N ) → 0 as d → 1. ~ω0 ζd ζd =

Chapter 8 8.1. Referring to Fig. 8.11 and noting that the slope of the tangent at the point x = ξ is −1/4, the approximate distribution is given by  1 0 ≤ x ≤ (ξ − 2)  f (x) = (ξ + 2 − x)/4 (ξ − 2) ≤ x ≤ (ξ + 2)  0 (ξ + 0) ≤ x, where x = ε/kT and ξ = µ/kT . Accordingly, 2πV N = g · 3 (2m)3/2 h

Z∞ n(ε)ε 0

1/2

Z∞ dε = C

f (x)x1/2 dx ,

0

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

2 3/2 1 1 −2 5/2 5/2 (ξ >> 1).

N = C{(ξ+2) −(ξ−2) } = Cξ 1 + ξ + ...

5 3 2 (1) Comparing (1) with eqn. (8.1.24), which may be written as 2 εF 3/2 N= C , 3 kT we get ( )

2 1 kT εF ξ= 1− + ... .

(2) kT 3 εF Similarly, Z∞

1 CkT {(ξ + 2)7/2 − (ξ − 2)7/2 } 35 0

2 5 −2 5/2 = CkT ξ 1 + ξ + ... .

5 2

U = CkT

f (x)x3/2 dx =

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

3 3 5 U = NkT ξ 1 + 2ξ −2 + . . . = N εF 1 + (kT /εF )2 + . . . .

5 5 3 1

(3)

2 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 is numerically less than the true value, given by eqn. (8.1.39), by a factor of 4/π 2 .

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

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

2/3 2 N h T0 = . gV f3/2 (1) 2πmk

(1)

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

2/3 3N h2 εF = .

(2) TF ≡ k 4πgV 2mk It follows that T0 = TF

4π 3 f3/2 (1)

2/3

1 . π

(3)

Now, by eqn. (E. 14), f3/2 (1) = (1–2−1/2 )ζ(3/2) ' 0.765. Substituting this 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 done the 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

−2

π2 π2 5π 2 (ln z)−2 + . . .

1− (ln z)−2 + . . .

1+ (ln z)−2 + . . . γ = 1+ 8 24 8

2 2 2 π π kT =1+ (ln z)−2 + . . . ' 1 + .

3 3 εF 8.4. This problem is similar to Problem 7.5 of the Bose gas and can be done the same way. To obtain the various low-temperature expressions, we make use of expansions (8.1.30–32). Thus

−1 3 π2 π2 1− (ln z)−2 + . . .

1+ (ln z)−2 + . . .

2n(kT ln z) 24 8

3 π2 = 1− (ln z)−2 + . . . .

2n(kT ln z) 6

κT =

3 We now employ eqn. (8.1.35) and get ( )−1 ( )

2

2 3 π 2 kT π 2 kT κT = 1− + ...

1− + ...

2nεF 12 εF 6 εF (

2 ) 3 π 2 kT ' 1− , (1) 2nεF 12 εF which is the desired result.

Similarly, using appropriate expansions, we get

3 π2 −2 κs = 1− (ln z) + . . .

2n(kT ln z) 2 (

2 ) 3 5π 2 kT ' 1− .

2nεF 12 εF

(2)

Dividing (1) by (2), we obtain the low-temperature expression for γ, the same as the one quoted in the previous problem; this also yields the desired result 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 done the 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 that √ w0 = uF / 3.

8.7. This problem is similar to Problem 7.9 of the Bose gas and can be done the same way. At low temperatures, using formula (E. 15), we get

−2

9 π2 π2 −1 −2 −2 huihu i = (ln z) + . . .

1+ (ln z) + . . .

1+ 8 3 8 (

2 ) 2 9 π2 9 π kT = 1+ (ln z)−2 + . . . ' 1+ ; 8 12 8 12 εF 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 Å, while m0 = me — giving εF = 5.49 eV and TF = 6.37 × 104 K. For lead, ne = 4, na = 4, a = 4.95 Å, while m0 = 2.1 me — giving εF = 9.45 eV and TF = 10.96 × 104 K. For aluminum, ne = 3, na = 4, a = 4.05 Å, while m0 = 1.6 me — giving εF = 11.63 eV and TF = 13.50 × 104 K.

4 (ii) The nuclear radius for 80 Hg200 is about 8.4 × 10−13 cm. Taking all the 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 . This yields 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 =

1/2 2/3 2/3 3π 3 h2 = kT . f3/2 (z) f (z) 3/2 4π 2mλ2 4

With the help of Sommerfeld’s lemma (E.15), this becomes 2/3 π2 7π 4 −2 −4 εF = kT ln z 1 + (ln z) + (ln z) + . . .

8 640

π2 π4 −2 −4 = kT ln z 1 + (ln z) + (ln z) + . . . .

12 180

(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

+ a22 − a4

kT εF

4

π2 +. . . = 1+ 12

kT εF

2 4

4 π π2 kT − a2 + +. . . .

180 6 εF

Equating coefficients on the two sides of this equality, we get: a2 = −π 2 /12, a4 = −π 4 /80, . . .. Equation (2) then gives the desired result (8.1.35a).

Next, we have from eqns. (8.1.7) and (E.15)

3 5π 2 7π 4 U = kT ln z 1 + (ln z)−2 − (ln z)−4 + . . .

N 5 8 384

−1 2 4 π 7π 1+ (ln z)−2 + (ln z)−4 + . . .

8 640

3 π2 11π 4 = kT ln z 1 + (ln z)−2 − (ln z)−4 + . . . .

5 2 120

(3)

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

2

4 U 3 π 4 kT π 2 kT = εF 1 − − + ...+ N 5 12 εF 80 εF ) (

2

4 π 4 kT π 2 kT − + ...

1+ 2 εF 120 εF ) (

2

4 3 π 4 kT 5π 2 kT = εF 1 + − + ... .

5 12 εF 16 εF The specific heat of the gas is then given by

3 CV π 2 kT 3π 4 kT = − + ....

Nk 2 εF 20 εF

(4)

(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 done the same way.

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

s 2 C f s f(n/s)+1 (z)f(n/s)−1 (z) CP V (n/s)−1 (z) = 1+ , (1) =1+ CV n Nk fn/s (z) n {fn/s (z)}2 which can be done quite easily; see eqns. (7)–(9) of the solution to Problem 7.14. For part (iv), we observe that, since the quantity S/N is a function of z only, an isentropic process implies that z = const. Accordingly, for such a process, VT n/s = const. and

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

see eqns. (2) and (3) of the solution to Problem 7.14. Eliminating T among these 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.15)]

n n π2 n n π2 CP = 1+ +1 (ln z)−2 + . . .

1+ −1 −2 (ln z)−2 + . . .

CV s s 6 s s 6

−2

2 n n π × 1+ −1 (ln z)−2 + . . . s s 6 π2 π2 =1+ (ln z)−2 + . . . ' 1 + (kT /εF )2 , 3 3 regardless of the values of s and n.

6 8.11. For T >> TF , we get CV n CP − CV ' , ' 1, Nk s Nk

so that

n

CP ' +1 .

Nk s

For T << TF , we obtain [see formula (E.15)]

n π2 n n π2 n CV −2 −2 = ln z 1 + (ln z) + . . . − ln z 1 + −1 (ln z) + . . .

Nk s s 3 s s 3

n π2 n π 2 kT = (ln z)−1 + . . . ' . s 3 s 3 εF To this order of accuracy, the quantity CP /Nk has the same value as Cv /Nk . As for the difference between the two, we obtain

3 n π 4 kT CP − CV ' , Nk s 9 εF consistent with the corresponding value of γ quoted in the previous problem. The non-relativistic case pertains to s = 2 while the extreme relativistic one pertains to s = 1.

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

A A AkT f1 (zF ) = 2 ln(1 + zF ), EF = 2 f2 (zF ), λ2 λ λ

(1a,b)

while the corresponding results for the Bose gas are N=

A A AkT g1 (zB ) = 2 ln(1 − zB ), EB = 2 g2 (zB ). λ2 λ λ

(2a,b)

Equating (la) and (2a), we get 1 + zF =

1 zB zF , i.e. zF = or zB = .

1 − zB 1 − zB 1 + zF

Next, since z∂f2 (z)/∂z = f1 (z),

Z zF Z zF 1 1 1 ln(1 + z)dz = + ln(1 + z)dz . f2 (zF ) = z 1+z z(1 + z) 0 0 The first part of this integral is readily evaluated; in the second part, we substitute z = z 0 /(1 − z 0 ), to get Z zF /(1+zF ) 1 1 1 ln(1−z 0 )dz 0 = ln2 (1+zF )+g2 (zB ). f2 (zF ) = ln2 (1+zF )− 0 2 z 2 0 Equations (1b) and (2b) then yield the desired result, viz.

EF (N, T ) =

N 2 h2 + EB (N, T ), whence {CV (N, T )}F = {CV (N, T )}B .

4πmA

7 Letting T → 0, we recognize that the constant appearing in the above result must be equal to EF (N, 0). To verify this, we note that, since the Fermi momentum of the gas in two dimensions is given by the equation N = A · πp2F /h2 , the Fermi energy is given by εF = p2F /2m = Nh 2 /2πmA. The ground-state energy of the gas then follows readily: Z pF 2 A · πp4F N 2 h2 1 p A · 2πpdp = = = N εF .

EF (N, 0) = 2 2 2m h 4πmA 2 4mh 0 8.13. The Fermi energy of the gas is given by the obvious relation Z εF a(ε)dε.

N=

(1)

0

At the same time, the quantities N and U , as functions of µ and T , are given by the standard integrals Z ∞ Z ∞ a(ε)dε εa(ε)dε N= and U = . β(ε−µ) + 1 β(ε−µ) + 1 e e 0 0 At low temperatures we employ formula (E.16), with x = βε and ξ = βµ, to obtain

Z µ π2 da(ε) N= a(ε)dε + (kT )2 + ...

6 dε 0 ε=µ

Z εF da(ε) π2 2 (kT ) , (2) ' a(ε)dε + (µ − εF )a(εF ) + 6 dε 0 ε=εF

Z µ π2 da(ε) U= εa(ε)dε + (kT )2 a(ε) + ε + ...

6 dε 0 ε=µ ( )

Z εF da(ε) π2 2 (kT ) a(εF ) + εF . ' εa(ε)dε + (µ − εF )εF a(εF ) + 6 dε ε=εF 0 (3) Comparing (1) and (2), we obtain for the chemical potential of the gas

π 2 (kT )2 da(ε) µ ' εF − , (4) 6 a(εF ) dε ε=εF which leads to the desired result for µ.

Next, substituting (4) into (3), we obtain the remarkably simple expression U ' U0 + (π 2 /6)k 2 T 2 a(εF ), whence CV ' (π 2 /3)k 2 T a(εF ).

(5)

8 It follows that

Z T S= 0

CV dT ' (π 2 /3)k 2 T a(εF ).

T

(6)

For a gas with energy spectrum ε ∝ ps , confined to a space of n dimensions, a(ε)dε ∼ pn−1 dp ∼ ε(n/s)−1 dε.

By eqn. (1), the Fermi energy of the gas is given by Z εF sεF sA n/s N= Aε(n/s)−1 dε = εF = a(εF ). n n 0 Substituting this result into (5), we get CV n π2 ' · Nk s 3

kT εF

;

(7)

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

8.14. In the notation of Sec. 3.9, the potential energy of a magnetic dipole in the presence of a magnetic field B = (0 , 0 , B) is given by the expression −(gµB m)B, where m = −J, . . . , +J. The total energy ε of the dipole is then given by ε = (p2 /2m0 ) − gµB mB , m0 being the (effective) mass of the particle; the momentum of the particle may then be written as p = {2m0 (ε + gµB mB )}1/2 .

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

4πV {2m0 (εF + gµB mB )}3/2 3h3

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

X

(gµB m)Nm =

m

X 4πgµB V 0 3/2 (2m ) m(εF + gµB mB )3/2 .

3h3 m

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

χ0 = Lim

B→0

M VB

J X 4πgµB 1/2 0 3/2 3 = (2m ) · gµB εF m2 3h3 2 m=−J

2πg 2 µ2B 1/2 = (2m0 )3/2 εF J(J + 1)(2J + 1).

3h3

(1)

By eqn. (8.1.24), 3/2

εF

=

3n h3 4π(2J + 1) (2m0 )3/2

n=

N V

.

(2)

9 Substituting (2) into (1), we obtain the desired result χ0 =

1 ∗2 nµ /εF 2

∗2 µ = g 2 µ2B J(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 ∗2 nµ /kT ; 2

see eqn. (3.9.26). We note that the ratio χ0 /χ∞ = 3kT /2εF , valid for all J. 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 fugacity z is determined by the equation f3/2 (z) = xN λ3 /V.

(1)

Differentiating (1) with respect to x, we get ∂f3/2 (z) ∂ ln z N λ3 1 = = f3/2 (z). ∂ ln z ∂x V x It follows that

∂µ0 kT f3/2 (z) = . ∂x x f1/2 (z)

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

At low temperatures, we get

3 nµ∗2 π2 · (ln z)−2 + . . . ξ= 1− kT 2 ln z 6 ( )−1 ( )

2

2 3nµ∗2 π 2 kT π 2 kT = 1− + ...

1− + ...

2εF 12 εF 6 εF (

2 ) π 2 kT ' χ0 1 − .

(8.2.24) 12 εF At high temperatures, on the other hand, nµ∗2 z − 2−1/2 z 2 + . . . nµ∗2 = (1 − 2−3/2 z + . . .) kT z − 2−3/2 z 2 + . . . kT ' χ∞ (1 − 2−5/2 nλ3 ), (8.2.27)

χ=

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

10 8.18. The ground-state energy of a relativistic gas of electrons is given by Z pF 8πV E0 = 3 mc 2 [{1 + (p/mc)2 }1/2 − 1]p2 dp. h 0 Making the substitution (8.4.9), we get Z θF 8πm4 c5 V E0 = (cosh θ − 1) sinh2 θ cosh θ dθ. h3 0 Now the integral Z Z θF θF 1 1 θF 2 2 3 sinh θ cosh θ dθ = sinh θ cosh θ − sinh4 θ dθ.

3 3 0 0 0

(1)

(2)

Substituting (2) into (1) and making use of eqn. (8.4.12), we get E0 =

8πm4 c5 V 8πm4 c5 V sinh3 θF cosh θF − P0 V − sinh3 θF ; 3 3h 3h3

(3)

note that the last term is simply Nmc 2 . Finally, using the definition x = sinh θF , we obtain the desired result.

We observe that eqn. (3) can also be written as E0 + P0 V = Nmc 2 (cosh θF − 1) = N εF ≡ N µ0 .

To verify that the derivative (∂E0 /∂V )N is equal to −P0 , we have to show that [∂{VB (x)}/∂V ](Vx 3 ) = −A(x), i.e. ∂{x−3 B(x)}/∂x−3 = −A(x), i.e. ∂ x4 [8{(x2 + 1)1/2 − 1} − x−3 A(x)] = 3A(x), i.e. ∂x ∂A(x)/∂x = 8x4 (x2 + 1)−1/2 , which can be readily verified with the help of expression (8.4.13).

8.19. Utilizing the result obtained in Problem 8.13, we have for a Fermi gas at low temperatures CV π 2 a(εF ) = kT .

(1) Nk 3 N Now, the density of states for the relativistic gas is given by, see eqn. (8.4.7),

p 2 1/2 8πmV 8πV dp = , a(ε) = 3 p2 p 1 + h dε h3 mc where p = p(ε). Substituting this result into (1) and making use of eqn. (8.4.4), we get p o1/2 CV π2 m n F = 2 1+ kT , Nk pF mc

11 which leads to the desired result.

In the non-relativistic case pF << mc and εF = p2F /2m , we obtain the familiar expression (8.1.39); in the extreme relativistic case (pF >> mc and ε = pc), we obtain

kT CV = π2 , Nk εF consistent with expression (7) of the solution to Problem 8.13.

8.22. The number of fermions in the trap is Z Z εF 1 dε ε2 ε3F dε ε2 N (T, µ) = = = .

3 3 β(ε−µ) 2(~ω) e 2(~ω) 6(~ω)3 −1 0 Using kTF = εF this gives the following relation for the fugacity z = e−βµ ,

3

T TF

3 Z

x2 dx = 1. ex e−βµ + 1

The internal energy is Z Z dε ε3 1 (kT )4 x3 U (T, µ) = = .

2(~ω)3 eβ(ε−µ) − 1 2(~ω)3 ex e−βµ − 1 When compared to the ground state energy U0 = (kTF )4 /[8(~ω)3 ], we get U =4 U0

T TF

4 Z

x3 ex e−βµ − 1

.

8.23. For large and negative µ we can treat e−β(ε−µ) 1 and e+β(ε−µ) 1 for all ε so Z √ 3 2m ε −β(ε−µ) P (µ, T ) ≈ kT e dε = nkT where π 2 ~3 Z √ 3 2m ε −β(ε−µ) eβµ n(µ, T ) ≈ e dε = 2 and, π 2 ~3 λ3 h λ= √ .

2πmkT Taking two temperature derivatives of the pressure gives the specific heat as

2 Z ∂s ∂ P k eβ(ε−µ) 2 cµ = T =T = a(ε)(ε − µ) 2 dε. ∂T µ ∂T 2 µ (kT )2 eβ(ε−µ) + 1 Expand the density of states in a Taylor series centered at µ and let x = β(ε − µ). The even order terms in the expansion, starting with

12 constant term, will give nonzero results by symmetry.

Z k 1 00 eβ(ε−µ) 0 2 2 cµ = a(µ) + a a (µ)(ε − µ) + (µ)(ε − µ) + · · · (ε − µ) 2 dε, (kT )2 2 eβ(ε−µ) + 1 Z Z ex ex 2 3 00 4 dx + k(kT ) = k(kT )a(µ) x a (µ) x 2 2 dx (ex + 1) (ex + 1) π 2 k(kT )a(µ) 7π 4 k(kT )3 a00 (µ) = + + ··· 3 15 2

Since µ → εF as T → 0, the leading term is cµ ≈ π3 k 2 T a(εF ).

1

Chapter 9 9.1 Using the Friedmann equation (9.1.1) r da 8πGu a, = dt 3c2 and the connection between scale factor a and blackbody temperature T , T a = T0 a0 , along with (9.3.4b) we get r r dT 8πGu 8π 3 Ggk 4 3 =− T = − T , dt 3c2 45~3 c5 where g = 43/8 is the effective number of relativistic species from equation (9.3.6b). The solution of the differential equation is r t0 T (t) = T0 , T where 1 t0 = 2

s

45~3 c5 ' 0.99 s 8π 3 Gg(kT0 )4

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 3 and energy density are of order 1025 J/m , 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 order of ue /ne ∼ kT . The Coulomb energy per electron/positron is of the order of uc ≈ e2 /(4π 0 a) where a ≈ (1/ne )1/3 is of the order of the average distance between the charged particles. Using ne ∼ (kT /~c)3 we get uc /ue ∼ e2 /(4π 0 ~c) ≈ 1/137. This is the justification for treating the relativistic electrons and positrons as noninteracting.

9.4 For βmc2 >> 1 but before the time when the electron density approaches the proton density, the density of electrons and positrons are almost identical so µ ≈ 0. Equation (9.5.6) gives p p Z ∞ n+ 1 x x + βmc2 x − βmc2 dx n− ≈ ≈ nγ nγ ζ(3) βmc2 ex + 1 3/2 Z ∞ 2 e−βmc βmc2 √ −y ≈ ye dy.

2ζ(3) 0

2 9.5 After the density of electrons levels off at the nearly the proton density, you can use equation (9.5.8) to show that the chemical potential µ− ≈ mc2 .

Then the positron number density is given by equation (9.5.7), p p Z ∞ x x + βmc2 x − βmc2 dx 1 n+ ≈ nγ ζ(3) βmc2 ex+βmc2 + 1 3/2 Z ∞ 2 e−2βmc βmc2 √ −y ≈ ye dy 2ζ(3) 0 3/2 √ 2 π e−2βmc βmc2 . ≈ 4ζ(3) 9.6 After the electron–positron annihilation, the only relativistic species left are the photons and the neutrinos. The factor 21/8 = (3)(1)(7/8) in the energy is because there are three families of neutrinos, the spin degeneracy factor 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 to the photons; see equation (9.6.4). Following the solution to problem 9.1, we get v 1u 45~3 c5 u

t0 = t ' 1.79 s.

2 8π 3 G 1 + 21 4 4/3 (kT )4 0 8 11 9.7 If the current CMB temperature was 27K rather than 2.7K, the baryonto-photon ratio would be 103 times smaller. Equation (9.7.8) implies that the nucleosynthesis temperature would have been about 20% lower which would have delayed the nucleosynthesis by an extra two minutes. This would have given the neutrons a longer time to decay leading to q ≈ 0.10 rather than 0.12, leading to a helium content in the universe of about 20% by weight. If the current CMB temperature were 0.27K, that would have increased the baryon-to-photon ratio by a factor of 103 . Fewer photons per baryon would have led to an earlier nucleosynthesis, less time for neutrons to decay and an increase of the neutron fraction to q ≈ 0.135 leading to about 27% helium content.

9.8 The strong interaction exhibits asymptotic freedom at high energies justifying treating the quarks an gluons as noninteracting. The effective number of species in equilibrium in these tiny quark–gluon plasmas is accounted for using only the up and down quarks and the gluons. Photons, and leptons for example easily escape without interacting with the plasma.

7 uγ 7 uγ uu = 2 uū = 2 up quarks and antiquarks 8 2 8 2

7 uγ 7 uγ ud = 2 ud¯ = 2 down quarks and antiquarks 8 2 8 2 uγ ug = (8)2 gluons 2

3 Therefore, the effective number of species is g = 8 + 28/8 = 23/2 and 3 3 uQGP = guγ . The energy density is 4 GeV/fm = 6.4 × 1035 J/m , so

kT '

1/4

15(~c)3 GeV 4 ' 4 × 10−11 J ' 250 MeV, gπ 2 fm3

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

9.9 The strong interaction exhibits asymptotic freedom at high energies justifying treating the quarks an gluons as noninteracting. The effective number of species is much larger than during the time near t = 1s due to the muons, quarks and gluons. uγ 2 7 uγ u e− = 2 8 2

7 uγ uνe = 8 2

7 uγ uνµ = 8 2

7 uγ uντ = 8 2

7 uγ uµ− = 2 8 2

7 uγ uu = 2 8 2

7 uγ ud = 2 8 2 uγ ug = (8)2 2

photons

uγ = 2

7 uγ u e+ = 2 8 2

7 uγ uν̄e = 8 2

7 uγ uν̄µ = 8 2

7 uγ uν̄τ = 8 2

7 uγ uµ+ = 2 8 2

7 uγ uū = 2 8 2

7 uγ ud¯ = 2 8 2

electrons/positrons electron neutrinos/antineutrinos muon neutrinos/antineutrinos tau neutrinos/antineutrinos muons/antimuons up quarks/antiquarks down quarks/antiquarks gluons

The result is u = (149/8)uγ . Proceeding as in problem 9.1 we get r 0.53 s 10 T (t) = 10 K .

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

Chapter 10 10.1. By eqn. (9.2.3), the second virial coefficient of the gas with the given interparticle interaction would be Z r0

Z ∞ 2π 2 ε(σ/r)6 /kT 2 −1 · r dr + {e − 1}r dr a2 = − 3 λ 0 r0  

j Z ∞X ∞ 1 2π 1 εσ 6 = 3  r03 − r2 dr  6 λ 3 j! kT r r0 j=1  

j ∞ 3 6 X 2πr0  1 εσ ; = 1− 3λ3 (2j − 1)j! kT r06 j=1 cf. eqn. (9.3.6). For the rest of the question, follow the solution to Problem 9.7.

10.2. For this problem, we integrate (9.2.3) by parts and write Z ∞ 2π ∂u(r) 3 3 r dr ; a2 λ = − e−u(r)/kT 3kT 0 ∂r

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

Z ∞ m n 2π mA nB a2 λ3 = − e−A/kT r eB/kT r dr 3kT 0 rm−2 rn−2

j

Z ∞ ∞ X 1 B mA nB 2π −A/kT r m e − n−2+nj dr = 3kT 0 j! kT rm−2+nj r j=0 (

j

(m−3+nj )/m

(n−3+nj )/m ) ∞ B m − 3 + nj kT n n − 3 + nj kT 2π X 1 = AΓ − BΓ 3kT j=0 j! kT m A m m A From the first sum we take the (j = 0)-term out and combine the remaining terms with the second sum (in which the index j is changed to j − 1); after considerable simplification, we get  

3/m 

"

n/m #j  ∞ X 2π A m−3 3 1 nj − 3 B kT a2 λ3 = Γ − Γ .   3 kT m m j! m kT A j=1

(1) 1

2 For comparison with other cases, we set A = A0 r0m and B = B 0 r0n (so that A0 and B 0 become direct measures of the energy of interaction).

Expression (1) then becomes   n/m #j  0 3/m 

" 0 ∞ X m−3 2π 3 A 3 1 nj − 3 B kT . Γ a2 λ3 = r − Γ   3 0 kT m m j! m kT A0 j=1

(2) Now, to simulate a hard-core repulsive interaction, we let m → ∞, with the result that   0 j  ∞  X 2π 3 B 1 .

(2a) a2 λ3 = r 1−3 3 0 (nj − 3)j! kT  j=1

With n = 6, expression (2a) reduces to the one derived in the preceding problem. Furthermore, if terms with j > 1 are neglected, we recover the van der Waals approximation (9.3.8).

For further comparison, we look at the behavior of the coefficient B2 (≡ a2 λ3 ) at high temperatures. While the hard-core expression (2a) predicts a constant B2 as T → ∞, the soft-core expression (2) predicts a B2 that ultimately vanishes, as T −3/m , which agrees qualitatively with the data shown in Fig. 9.2.

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 (9.3.9), we get T (∂P/∂T )2v T {k/(v − b)}2 1 CP − CV =− =− = .

2 3 Nk k(∂P/∂v)T k{−kT /(v − b ) + 2a/v } 1 − 2a(v − b)2 /kT v3 (b) In view of the thermodynamic relation TdS = CV dT + T (∂P + ∂T )V dV and the equation of state (9.3.9), an adiabatic process is characterized by the fact that CV dT + NkT (v − b)−1 dv = 0.

Integrating this result, under the assumption that CV = const., we get T CV /Nk (v − b) = const. (c) For this process we evaluate the Joule coefficient

∂T (∂U/∂V )T T (∂P/∂T )V − P a/v2 N 2a = =− =− =− . ∂V U (∂U/∂T )V CV CV CV V 2 Now integrating from state 1 to state 2, we readily obtain the desired result.

3 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) 1 ∂v vB −1 1 a0 = =− =− v+ 2 ∂P T ∂P T T PT T T and for the right-hand side

1 ∂v 2a0 1 2a0 1 v a0 ∂(vB −1 ) =− − 3 =− αv − 3 = − + 3 . − ∂T P ∂T P T P T P T T P The compatibility of the given expressions is thus established.

To determine the equation of state of the gas, we note from the given expression for α that

∂v v 3a0 ∂ v 3a0 = + 3 , i.e. = 4, ∂T P T T ∂T T P T whence

v a0 a0 = − 3 + f (P ), i.e. v = − 2 + Tf (P ), T T T where f is a function of P only. We then obtain for B vB −1 = −Tf 0 (P ).

(2)

(3)

Combining (2) and (3), we get f 0 (P ) vB −1 1 =− =− . f (P ) (v + a0 /T )2 P It follows that f (P ) is proportional to 1 /P and hence, by (2), P = const. T (v + a0 /T 2 )−1 .

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

∂T (∂H/∂P )T 1 ∂V N ∂v =− = T −V = −v . ∂P H (∂H/∂T )P CP ∂T P CP ∂T P By eqn. (9.2.1), Pv a2 λ3 =1+ + . . . , so that kT v

kT a2 λ3 P kT v= 1+ + ... = + a2 λ3 + . . . .

P kT P

4 It follows that

T

∂v ∂T

∂(a2 λ3 ) 3 −v= T − a2 λ + . . . ∂T P

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

With the given interparticle interaction, eqn. (9.2.3) gives Z r0

3

a2 λ = −2π

2

Z r1

−1 · r dr + 0

=

(e

u0 /kT

2

− 1)r dr

r0

i

2π h 3 r1 − r13 − r03 eu0 /kT , 3

whence T

i

∂(a2 λ3 ) 2π h 3 u0 u0 /kT − a2 λ3 = r1 − r03 1 + e − r13 . ∂T 3 kT

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

We note that the Joule-Thomson coefficient obtained here vanishes at a temperature T0 , known as the temperature of inversion, given by the implicit relationship

u0 1+ kT 0

eu0 /kT 0 =

r13 .

3 r1 − r03

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

10.7. To the desired approximation, P 1 1 ≡ ln L = 3 (z − a2 z 2 ), kT V λ

n=

N 1 = 3 (z − 2a2 z 2 ), V λ

(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 ).

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

∂A 5 ∂ 3 3 S=− = Nk − ln(nλ ) − n (Ta 2 λ ) ; ∂T N,V 2 ∂T

(2a,b)

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

3 ∂ U = A + TS = NkT − nT (a2 λ3 ) , 2 ∂T

5 ∂ a2 λ3 H = U + PV = NkT − nT 2 , 2 ∂T T

∂U 3 ∂ ∂ CV = = Nk −n T2 (a2 λ3 ) , and ∂T N,V 2 ∂T ∂T

(∂P/∂T )2N,V ∂ CP − CV = −T = Nk 1 + 2nT (a2 λ3 ) . (∂P/∂V )N,T ∂T For the second part, use the expression for a2 λ3 derived in Problem 9.5 and examine the temperature dependence of the various thermodynamic quantities.

10.8. We consider a volume element dx 1 dy 1 dz 1 around the point P (x1 , 0, 0) in solid 1 and a volume element dx 2 dy 2 dz 2 around the point Q(x2 , y2 , z2 ) in solid 2. The force of attraction between these elements will be −α(n dx 1 dy 1 dz 1 )(n dx 2 dy 2 dz 2 )

`5 5/2

{(x2 − x1 )2 + y22 + z22 }

,

directed along the line joining the points P and Q. The normal component of this force will be −αn2 (dy 1 dz 1 )`

(x2 − x1 )

5

3 dx 1 dx 2 dy 2 dz 2 .

{(x2 − x1 )2 + y22 + z22 }

The net force (per unit area) experienced by solid 1, because of attraction by all the molecules of solid 2, will thus be Z 0 Z ∞ Z ∞ (x2 − x1 ) 2 5 − αn ` dx 1 dx 2 · 2πρdρ 2 2 3 x1 =−∞ x2 =d ρ=0 {(x2 − x1 ) + ρ } Z Z ∞ παn2 `5 0 παn2 `5 1 =− dx 1 dx 2 = , 3 2 4d x1 =−∞ x2 =d (x2 − x1 ) i.e. inversely proportional to d.

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. (9.5.31), we readily obtain the desired result.

10.10. The symmetrized wave functions for a pair of non-interacting bosons/fermions are given by 1 Ψα (r1 , r2 ) = √ (eik1 ·r1 eik2 ·r2 ± eik1 ·r2 eik2 ·r1 ).

2V

6 The probability density operator Ŵ2 of the pair is then given through the matrix elements X h10 , 20 |Ŵ2 |1, 2i = 2λ6 Ψα (10 , 20 )Ψ∗α (1, 2)e−βEα α 0 0 λ6 X ik1 ·r01 ik2 ·r02 = 2 (e e ± eik1 ·r2 eik2 ·r1 )× V α 2

2

2

(e−ik1 ·r1 e−ik2 ·r2 ± e−ik1 ·r2 e−ik2 ·r1 )e−β~ (k1 +k2 )/2m # " 0 0 0 0 λ6 X X eik1 ·(r1 −r1 ) eik2 ·(r2 −r2 ) + eik1 ·(r2 −r2 ) eik2 ·(r1 −r1 ) ± = 0 0 0 0 2V 2 eik1 ·(r1 −r1 ) eik2 ·(r1 −r2 ) ± eik1 ·(r1 −r2 ) eik2 ·(r2 −r1 ) k1 k2 2 2

2 2

e−β~ k1 /2m e−β~ k2 /2m =

1 0 [h1 |Ŵ1 |1ih20 |Ŵ1 |2i + h20 |Ŵ1 |2ih10 |Ŵ1 |1i± 2 h20 |Ŵ1 |1ih10 |Ŵ1 |2i ± h10 |Ŵ1 |2ih20 |Ŵ1 |1i]

= h10 |Ŵ1 |1ih20 |Ŵ1 |2i ± h20 |Ŵ1 |1ih10 |Ŵ1 |2i.

Comparing this with eqn. (9.6.18), we obtain the desired result.

10.11. A particle with spin J can be in any one of the (2J + 1) spin states characterized by the spin functions χm (m = −J, . . . , J). For a pair of such particles, we will have (2J + 1)2 spin states characterized by the symmetrized 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. The remaining 2J(2J + 1) functions, for which m1 6= m2 , can be symmetric or antisymmetric; however, only half of them are linearly independent functions (because an interchange of the suffices m1 and m2 does not produce anything new). Thus, in all, we have J(2J + 1) antisymmetric spin functions, and (J + 1)(2J + 1) symmetric spin functions, that are linearly independent.

Now the total wave function of the pair will be the product of a symmetrized space function (like the ones considered in the previous problem) and a symmetrized spin function (like the ones discussed above). For the total wave function to be symmetric, as required for a pair of bosons, we may associate any of the (J + 1)(2J + 1) symmetric spin functions with a symmetric space function or any of the J(2J + 1) antisymmetric spin functions with an antisymmetric space function. This will lead to the quoted expression for the coefficient bs2 . On the other hand, for the total wave function to be antisymmetric, as required for a pair of fermions, we may associate any of the J(2J + 1) antisymmetric spin functions with a symmetric space function or any of the (J + 1)(2J + 1) symmetric spin

7 functions with an antisymmetric space function. This will lead to the quoted expression for the coefficient bA 2.

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 states while a pair of spinless particles has only one, we have to divide the expression for b2 pertaining to the former by (2J + 1)2 so that we are talking of the average contribution per state. (ii) To make a transition from discreteness in orientation (that is associated with a finite value of J) to continuity in orientation, we should take the limit J → ∞. (iii) In view of the foregoing, the distinction between the original system being symmetric or antisymmetric is completely lost, and we are led to the results quoted in the problem.

Next, using eqns. (9.5.28, 36 and 37), we obtain for the quantum-mechanical Boltzmannian gas r 3 22π 2 r 5 r 1 0 0 0 − 3π + + ..., b2 = − λ λ 3 λ which differs significantly from the corresponding classical result.

10.13 Expand the definition of the pair density n2 (r, r 0 ) in powers of the fugacity z using the grand canonical partition function and the Mayer functions fij = exp (−βu(rij )) − 1.

Z ∞ X 1 zN n2 (r12 ) = dr 3 · · · dr N exp (−βu(r12 ) − βu(r13 ) − · · · ) Q(µ, V, T ) (N − 2)!

N =2

Z = e−βu(r12 ) z 2 + z 3 (1 + f13 + f23 + f13 f23 ) dr 3 − z 3 Q1 + · · ·

Z Z = e−βu(r12 ) z 2 + 2z 3 f (r)dr + z 3 f13 f23 dr 3 + · · · Note every term includes the factor e−βu(r12 ) . The coefficients of those terms are integrals over the Mayer functions that are continuous functions of 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 n =1− nkT 2dkT

Z rg(r)

du dr, dr

where g(r) = y(r)e−βu(r) . This gives Z Z P n du n d −βu(r) =1− ry(r) e−βu(r) dr = 1 + ry(r) e dr, nkT 2dkT dr 2d dr

8 For the case of hard spheres, d −βu(r) = δ(r − D), e dr so P nDd =1+ Ωd y(D). nkT 2d where Ωd is the area of the d-dimensional unit sphere. For hard spheres y(D) = g(D+ ). In three dimensions η = πnD3 /6 and Ω3 = 4π so P/(nkT ) = 1 + 4ηg(D+ ).

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

P (r) =

Y

1 − 4πng(rk )rk2 ∆r ,

k=0

since each factor represents the probability there are no neighbors in interval k. This gives r/∆r

ln (P (r)) ≈

X

r/∆r

ln

1 − 4πng(rk )rk2 ∆r

≈−

k=0

X

4πng(rk )rk2 ∆r.

k=0

Therefore

Z r 2 P (r) = exp −4πn r g(r)dr .

0

Finally w(r) = −

dP . dr

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

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 Z∞ I=

∂u −βu 3 e r dr . dr

0

Integrating by parts, we get

1 1 ∞ I = − e−βu + c r3 |0 + β β

Z∞ 0

e−βu + c 3r2 dr .

9 An arbitrary constant c has been introduced here to secure “proper behavior” at r = ∞. Since exp(−βu) → 1 as r → ∞, we choose c = −1. The integrated part then vanishes [assuming that u(r) → 0 faster than 1/r3 ], and we are left with the result 3 I= β

Z∞

−βu

e − 1 r2 dr .

0

This reduces eqn. (3.7.17) to the desired form. (b) In the case of hard-sphere potential, the function f (r) = −1 for r ≤ σ and 0 for r > σ. We then get 2πnσ 3 PV '1+ NkT 3

n=

N V

.

For nσ 3 << 1, we may write this result in the approximate form

2πnσ 3 = NkT .

PV 1 − 3 Comparison with Problem 1.4 shows that the parameter b of that problem is equal to (2π/3)N σ 3 , which is indeed four times the actual space occupied by the particles.

10.20 Use

∂p [κT (n, T )] = n ∂n T ∂f P (n, T ) = n2 ∂n T −1

where f = A/N is the Helmholtz free energy per particle. Then Z n dn0 P (n, T ) = p(n0 , T ) + , 0 0 n0 n κ(n , T ) Z n P (n0 , T ) 0 f (n, T ) = f (n0 , T ) + dn . (n0 )2 n0 10.21 The most general Gaussian distribution of variables {u1 , · · · , uN } is of the form

1 T P (u1 , · · · , uN ) ∼ exp − u Au 2 where A is a symmetric positive definite matrix. The matrix has only positive eigenvalues and can be diagonalized into diagonal matrix (Ã = U T =

10 U −1 using the orthogonal the matrix U . The eigenvalues {λ1 , · · · , λN } are all positive and detU = 1. The normalization is

Z

Z 1 1 N = dN u exp − uT Au = dN y exp − y T Ãy 2 2 v u N q u Y λi = (2π)N detA . = t(2π)N i=1

The transformed variables are y = U T u so the Jacobian is unity. The integral of the average of exp aT u can be determined from completing the square inside the exponential,

Z 1 T 1 T −1 N T d u exp a u − u Au = N exp a A a .

2 2 2 Averaging the quantity aT u is accomplished by transforming to the y variables which, using Z dx x2 exp(−λx2 /2) = (2π)1/2 /λ3/2 , gives Z

dN u aT u

2

1 exp − uT Au = N aT U Ã−1 U T a = N aT A−1 a.

2

10.22 The pressure is given by

p=−

∂A ∂V

= n2 N,T

∂A/N ∂n

, T

and the excess pressure is given by

ex

4η − 2η 2 ∂A /N 1 + η + η2 − η3 2 ex P = Pcs − Pideal = nkT − 1 = nkT =n .

(1 − η)3 (1 − η)3 ∂n T This can be integrated to give Z η βAex 4 − 2η 0 3 − 2η 4η − 3η 2 0 = dη = − 3 = .

0 2 N (1 − η)2 (1 − η)2 0 (1 − η )

10.23 The simplest rational approximations are P 1 + η/8 = ≈ 1 + 2η + 3.125η 2 + 4.25η 3 + 5.375η 4 + 6.5η 5 nkT (1 − η)2 + 7.625η 6 + 8.75η 7 + 9.875η 8 + 11.000η 9 + · · · ,

11 and P 1 + 0.128018 η ≈ 1 + 2η + 3.128018η 2 + 4.256036η 3 + 5.384054η 4 + 6.512072η 5 = nkT (1 − η)2 + 7.64009η 6 + 8.768108η 7 + 9.896126η 8 + 11.024144η 9 + · · · The later gets the first two orders exactly correct, and the third and fourth order coefficients correct to better that 1%.

Chapter 11 11.4. The relevant results for T < Tc are given in eqns. (10.2.13–15). The corresponding results for T > Tc follow from eqn. (10.2.10) by neglecting n0 altogether; we get, to the first order in a, 1 4πa~2 1 A(N, V, T ) = Aid (N, V, T ) + , N N mv 4πa~2 P = Pid + , mv2 2 8πa~ . µ = µid + mv

(13a) (14a) (15a)

Remembering that vc ∝ T−3/2 , the various quantities of interest turn out to be ( 2 0 ∂ A 2πa~2 (T > Tc ) CV = −T = (CV )id + N 3 6v 2 ∂T N,V mT − 2vc + v2 (T < Tc ) c (

∂P 2πa~2 4/v2 (T > Tc ) K = −v = Kid + ∂v T m 2/v2 (T < Tc ) ( 2 2 0 (T > Tc ) ∂ P ∂ P 2πa~2 = + 2 2 ∂T 2 v ∂T 2 v,id 6/vc (T < Tc ) mT ( 2 2 0 (T > Tc ) ∂ µ ∂ µ 4πa~2 = + 2 2 2 ∂T v ∂T v,id 3/4vc (T < Tc ) mT The thermodynamic relationship quoted in part (b) of the problem is readily verified.

As for the discontinuities at T = Tc , we get (setting v = vc ) 9πa~2 1 9aλ2c ζ(3/2) 9a = Nk = Nk ζ(3/2), 3 mT c vc 2 λc 2λc 4πa~2 1 ∆K = − , m vc2 2 2 ∂ P 12πa~2 ∂ µ 3πa~2 = , ∆ = . ∆ 2 2 2 2 ∂T v ∂T v mT c vc mT 2c vc ∆CV = N

1

2 11.5. (a) We replace the sum over p appearing in eqn. (10.3.14) by an integral, viz.

2 ) Z ∞( p2 4πa~2 N 4πa~2 N m V · 4πp2 dp ε(p) − . − + 2m mV mV p2 h3 0 Substituting p = (8πa~2 N/V )1/2 x, we get Z ∞ 0

4πa~2 N mV

3/2 4πV 8πa~2 N 1 2 1/2 2 x2 dx , x(x + 2) − x − 1 + 2 2x h3 V

which readily leads us from eqn. (10.3.14) to (10.3.15). The resulting integral over x can be done by elementary means, giving

Z ∞ ∞ 1 1 1 1 1 (3x2 − 4)(x2 + 2)3/2 − x5 − x3 + x . x(x2 + 2)1/2 − x2 − 1 + 2 x2 dx = 2x 15 5 3 2 0 0 For x >> 1,

1 1 1 3 3 2 2 3/2 5 3 (3x − 4)(x + 2) = (3x − 4x ) 1 + 2 + 4 + O 15 15 x 2x x6

1 1 1 1 = x5 + x3 − x + O .

5 3 2 x The contribution from √ the upper limit is, therefore, zero. From the lower limit we get 128/15, which leads to eqn. (10.3.16). (b) Noting that 4πVp 2 dp 4πV = 3 h3 h

8πa~2 N V

3/2

x2 dx = N

1/2 128 (na 3 ) x2 dx , π

we readily obtain eqn. (10.3.23). Now the integral

Z ∞ 1 x(x2 + 1) 1 2 − x dx = (x2 − 1)(x2 + 2)1/2 − x3 |∞ 0 .

2 + 2)1/2 3 3 (x 0 Again, for x >> 1,

1 2 1 3 1 1 1 1 3 2 1/2 (x −1)(x +2) = (x −x) 1 + 2 + O , = x +O 3 3 x x4 3 x with the result that the contribution from the upper limit vanishes. √ From the lower limit we get 2/3, which leads to eqn. (10.3.24).

11.6. We invert the given equation for n and write

4πa~2 n 32 µ0 = 1 + 1/2 (na 3 )1/2 + . . . . m 3π

3 Substituting this into the given expressions for E0 and P0 , we get 2

64 E0 2πa~2 n2 32 1 + 1/2 (na 3 )1/2 + . . . = 1 + 1/2 (na 3 )1/2 + . . .

V m 3π 5π

2 2 2πa~ n 128 = (na 3 )1/2 + . . . , and 1+ m 15π 1/2 2

2πa~2 n2 128 32 3 1/2 3 1/2 P0 = 1− (na ) + . . .

1 + 1/2 (na ) + . . . m 3π 15π 1/2

2πa~2 n2 64 = 1 + 1/2 (na 3 )1/2 + . . . , m 5π in complete agreement with eqns. (10.3.16 and 17).

11.7. By eqns. (10.3.11), the number operator n̂p for the real particles is given by

n̂p = a+ p ap =

1 + + + 2 b bp − αp b−p bp + b+ p b−p + αp b−p b−p .

1 + αp2 p

The terms linear in αp do not contribute to the expectation value of n̂p (with p 6= 0) because of the absence of the diagonal matrix elements in + + + b−p bp and b+ p b−p . Further, since bp bp = N̂p and b−p b−p = N̂−p +1, we get np =

1 N̄p + αp2 (N̄−p + 1) 1 − αp2

(p 6= 0).

Finally, in view of the isotropy of the problem, N −p = N p and we get the desired 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

x2 1 + 2ε 3 ' 2(1 + 4ε) − ln |ε| − ln 2 + ε , 2x ln 2 ' 2(1 + 4ε) ln |x − 1| 2 2|ε| 1 + 12 ε

1 x+1 2+ε 10 x − ln ' 20ε ln ' 20ε{− ln |ε| + ln 2}, x x−1 |ε| ! p √ 1 + x (2 − x2 ) 2(2 − x2 )5/2 x + 2 − x2 p √ = ln x x − 2 − x2 1 − x ((2 − x2 ) 4

(2 − x2 )5/2 ln x

2(1 − 2ε)5/2 (1 + ε) + (1 − ε − ε2 ) ln 1+ε (1 + ε) − (1 − ε − ε2 )

2 1 ' 2(1 − 6ε) ln ' 2(1 − 6ε) − ln |ε| − ε .

2ε + ε2 2 '

4 Substituting these results into the square bracket appearing in the formula for 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. (10.8.10), we find that V (p) ' const. −

p3 a2 8 (7 ln 2 − 1) F 2 (p − pF ).

2 15π m~

Equation (10.8.11) then gives

1 1 8 2 ' (7 ln 2 − 1)(kF a) , 1− m∗ m 15π 2 which leads to the desired result for the ratio m∗ /m.

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

∂E ∂(E/V ) µ= = . ∂N v ∂(N/V ) It follows that, in the ground state of the given system,

Z n Z N N n E=V µ(n)dn = µ(n)dn n= . n 0 V 0 Now, since pF = (3π 2 n)1/3 ~, the given expression for µ may be written as µ(n) ' (3π 2 n)2/3

~2 a ~2 a2 2 ~2 + (2πn) + (3π 2 n)4/3 (11 − 2 ln 2) .

2 2m m 15π m

It follows that 3 1 ~2 a 3 E ~2 2 ~2 a2 ' (3π 2 n)2/3 + (2πn) + (3π 2 n)4/3 , (11 − 2 ln 2) 2 N 5 2m 2 m 7 15π m which agrees with eqn. (10.7.31).

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 3/2 the dimensionless wavefunction should read: ψ = a Ψ/ N . Using that osc p substitution and aosc = ~/(mω0 ) gives 1 ˜2 1 4πN a 2 − ∇ ψ + s2 ψ + |ψ| ψ = µ̃ψ, 2 2 aosc ˜ = ∂/∂sx + ... and µ̃ = µ/(~ω0 ). where s = r/aosc , ∇

5 11.18. The solution for the case V = 0 is Ψ = and E = (2πa~2 N 2 /(mV ).

p

N/V which gives µ = N u0 /V

11.19. For the case a → 0 the dimensionless G-P equation is 1 ˜2 1 − ∇ ψ + s2 ψ+ = µψ, 2 2

1 which has solution ψ = π3/4 exp − 12 s2 with E/(N ~ω0 ) = 3/2, i.e. the zero point energy for N particles in the trap.

11.20. Use the dimensionless form from problem 11.17. Ignoring the kinetic energy term q 2 µ̃ − s2 . ψ=p 4πN a/a0 The normalization is 4πa0 1= 4πN a

Z √2µ̃ s2 ds µ̃ − 2 0

which gives N=

a0 (2µ̃)5/2 .

15πa

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

CHAPTER 12 12.1. We assume the equation of state to be

kT 2 λ6 1 λ3 P = − β2 2 , 1 − β1 v 2 v 3 v

(1)

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

∂P kT kT λ3 kT λ6 = − 2 + β1 3 + 2β2 4 , and ∂v v v v 2 T 3 kT kT λ kT λ6 ∂ P = 2 3 − 3β1 4 − 8β2 5 .

2 ∂v T v v v At the critical point, both these derivatives vanish — with the result that (β1 )c

λ3c λ3 λ6 λ6 + 2(β2 )c 2c = 1 and 3(β1 )c c + 8(β2 )c 2c = 2, vc vc vc vc

whence (β1 )c = 2vc /λ3c and (β2 )c = −vc2 /2λ6c .

(2)

We infer that, at the critical point, β12 = −8β2 .

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

Pv 1 1 λ3 2 λ6 1 = 1 − (β1 )c c − (β2 )c 2c = 1 − 1 + = . kT c 2 vc 3 vc 3 3 12.2. The given equation of state is P =

kT −a/kT v e . v−b

(1)

It follows that

∂P 1 1 a 1 a −a/kT v = kT e − + · =P − + , ∂v (v − b)2 v − b kT v2 (v − b) kT v2 2 T

∂ P ∂P 1 a 1 2a = − + +P − .

2 2 2 ∂v T ∂v T (v − b) kT v (v − b) kT v3 1

2 At the critical point, both these derivatives vanish — with the result that a vc2 2a vc3 , = and = kT c vc − b kT c (vc − b)2 whence vc = 2b and kT c = a/4b. Equation (1) then gives: Pc = (a/4b2 )e−2 and hence kT c /Pc vc = e2 /2 ' 3.695. (a) For large v, the given equation of state may be approximated as

−1

kT b kT b a e−a/kT v ' P = 1− 1+ − . v v v v kT v Comparing this with eqns. (9.3.7–10), we see that the coefficient B2 in the present case is formally the same as the one for the van der Waals gas, viz. b − (a/kT ). (b) We note that the derivative (∂P/∂v)T for the Dietrici gas can be written as

n o kT a ∂P =− 2 e−a/kT v v2 − (v − b) 2 ∂v T v (v − b) kT

kT a 2 ab −a/kT v =− 2 (T − T ) . e v − + c v (v − b)2 2kT kT 2 Clearly, if T > Tc , then (∂P/∂v)T is definitely negative; the same is true 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 monotonically decreasing function of v — with the result that, for any given T and P , we have a unique v. (c) For T < Tc , P is a non-monotonic function of v — generally decreasing with v but increasing between the values r r a ab a ab vmin = (Tc − T ). − 2 (Tc − T ) and vmax = 2kT + 2kT kT kT 2 For any given T , we now have (for a certain range of P ) three possible values of v such that v1 > vmax > v2 > vmin > v3 ; see Figs. 11.2 and 11.3. We further note that vmin 1 vmax 1 = and = .

1/2 vc vc 1 + (1 − T /Tc ) 1 − (1 − T /Tc )1/2 Clearly, vmin < vc < vmax and, hence, v3 < vc < v1 . (d) To examine the critical behavior of the Dietrici gas, we write P =

a a (1 + π), v = 2b(1 + ψ), T = (1 + t).

2 2 4e b 4bk

3 The equation of state then takes the form

1+t 1 1+π = exp 2 1 − .

1 + 2ψ (1 + t)(1 + ψ) Taking logarithms, carrying out expansions and retaining the most important terms, we get

8 2 π ≈ t− 2ψ − 2ψ 2 + ψ 3 +2{1−(1−t)(1−ψ+ψ 2 −ψ 3 } ≈ 3t− ψ 3 −2tψ.

3 3 We now observe that (i) at t = 0, π ≈ − 32 ψ 3 , while at ψ = 0, π ≈ 3t, (ii) for t < 0, we obtain three values of ψ: while |ψ2 | << |ψ1,3 |, implying once again π ≈ 3t, ψ1,3 ≈ ±(3|t|)1/2 , (

1/2t (t > 0, ψ = 0) ∂ψ 1 . (iii) the quantity − ∂π ≈ 2ψ2 +2t ≈ t 1/4|t| (t < 0, ψ = ψ1,3 ) Comparing these results with the ones derived in Sec. 11.2, we infer that the critical exponents of this gas are precisely the same as those of the van 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 2

(n > 1).

(1)

2

Equating (∂P/∂v)T and (∂ P/∂v )T to zero, we get vc =

4n(n − 1)n−1 a n+1 b and Tc = . n−1 (n + 1)n+1 bn−1 R

Equation (1) then gives

Pc =

n−1 n+1

n+1

a , bn

whence RT c /Pc vc = 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) n+1 − . (n2 − 1)(1 + ψ) − (n − 1)2 (n − 1)(1 + ψ)n

Carrying out the usual expansions and retaining only the most important terms, we get π≈ It follows that

n(n + 1)2 3 n(n + 1) 2n t− ψ − tψ. n−1 12 n−1

4

(i) at t = 0, π ≈ − n(n + 1)2 /12 ψ 3 , 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 ,

4(n−1) ≈ n(n+1){(n − ∂ψ 2 −1)ψ 2 +4t} ∂π t ( (iii) the quantity (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 van der Waals gas — regardless of the value of n. The critical amplitudes (as well as the critical constants Pc , vc and Tc ), however, do vary with n and hence are model-dependent.

P 12.4. The partition function of the system may be written as exp f (L), where L

f (L) = ln N ! − ln(Np)! − ln(Nq)! + βN

1 2 qJL + µBL .

2

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

1 f (L) ≈ −Np ln p − Nq ln q + βN qJL2 + µBL .

2 With p and q given by eqn. (1) of the problem, the function f (L) is maximum when 1 1 − N (1 + ln p) + N (1 + ln q) + βN (qJL + µB) = 0.

2 2 Substituting for p and q, the above condition takes the form 1 1+L ln = β(qJL + µB).

2 1−L Comparing this with eqn. (11.5.10), we see that the value, L∗ , of L, that maximizes the function f (L) is identical with L̄.

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

1 A ≈ − kT f (L∗ ) ≈ NkT (p∗ ln p ∗ + q ∗ ln q ∗ ) − N qJL∗2 + µBL∗ , 2 1 U ≈ − NqJL∗2 − N µBL∗ , 2 whence S ≈ −Nk (p∗ ln p ∗ + q ∗ ln q ∗ ).

5 12.5. The relevant results of the preceding problem are

A 1 + L∗ 1 + L∗ 1 − L∗ 1 − L∗ 1 = kT ln + ln − qJL∗2 − µBL∗ , N 2 2 2 2 2 (1) 1 1 N + = N (1 + L∗ ), N − = N (1 − L∗ ), (2) 2 2 where L∗ satisfies the maximization condition 1 {ln(1 + L∗ ) − ln(1 − L∗ )} = β(qJL∗ + µB).

2

(3)

Combining (1) and (3), we get 1 1 − L∗2 1 A = kT ln + qJL∗2 .

N 2 4 2

(4)

Now, using the correspondence given on page 320 of the text and remembering that L∗ is identical with L̄, we obtain from eqns. (2) and (4) the desired results for the quantities P and v pertaining to a lattice gas.

For the critical constants of the gas, we first note from eqn. (11.5.13) that Tc = qJ /k, i.e. qε0 /4k; the other constants then follow from the stated results for P and v, with B = 0 and L̄ = 0.

12.6. The Hamiltonian of this model may be written as X 1 X H=− c σi σj − µB σi .

2 i i6=j

The double sum here is equal to

P i

σi

P j

σj −

P i

σi2 = (NL)2 − N which,

for N >> 1, is essentially equal to N 2 L2 . It follows that asymptotically our Hamiltonian is of the form 1 H = − (cN )NL2 − µBNL. 2 Now, this is precisely the Hamiltonian of the model studied in Problem 11.4, except for the fact that the quantity qJ there is replaced by the quantity cN here. We, therefore, infer that, in the limit N → ∞ and c → 0 (such that the product cN is held fixed), the mean-field approach of Problem 11.4 would be exact for the present model — provided that the fixed value of the product cN is identified with the quantity qJ. It follows that the critical temperature of this model would be cN/k.

12.7 & 8. Let us concentrate on one particular spin, s0 , in the lattice and look at the q P part of the energy E that involves this spin, viz. −2J s0 ·sj −gµB s0 ·H; j=1

for notation, see Secs. 3.9 and 11.3. In the spirit of the mean field theory,

6 we replace each of the sj by s̄, which modifies the foregoing expression to −gµB s0 · Heff , where Heff = H + H0

(H 0 = 2qJ s̄/gµB ).

(1)

We now apply the theory of Sec. 3.9, pp. 74-5. Taking H (and hence s̄) to be in the direction of the positive z-axis, we get from eqn. (3.9.22) µz = gµB s Bs (x)

[x = β(gµB s)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)/3s}x, with the result that

2qJ µ g 2 µ2B s(s + 1) (3) H + 2 2z . µz ≈ 3kT g µB The net magnetization, per unit volume, of the system is now given by the formula M = nµ̄z , where n(= N/V ) is the spin density in the lattice.

We thus get from (3)

Tc CH CH M 1− , i.e. M ≈ , where (4) ≈ T T T − Tc ng 2 µ2B s(s + 1) 2s(s + 1)qJ , C= .

(5) Tc = 3k 3k 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. For that, we must look into the possibility of spontaneous magnetization in the system.

To study the possibility of spontaneous magnetization, we let H → 0 and write from (2) µz = gµB s Bs (x0 )

[x0 = β(gµB s)H 0 = 2sqJ µz /gµB kT ].

(6)

In the close vicinity of the transition temperature, we expect µ̄z to be much less than the saturation value gµB s, so once again we approximate the function Bs (x0 ) for x0 << 1. However, this time we need a better approximation than the one employed above; this can be obtained by utilizing the series expansion coth x =

1 x x3 + − + ... x 3 45

(x << 1),

which yields the desired result: Bs (x) =

s+1 (s + 1){s2 + (s + 1)2 } 3 x− x + ...

3s 90s3

(x << 1).

(7)

7 Substituting (7) into (6), we get Tc b µz = µ − T z g 2 µ2B s2

Tc µ T z

3 + ...,

(8)

where Tc is the same as defined in (5), while b is a positive number given by

3 s2 .

(9) b= 1+ 10 (s + 1)2 Clearly, for T < Tc , a non-zero solution for µ̄z is possible; in fact, for T . Tc , √ µz ≈ (gµB s/ b)(1 − T /Tc )1/2 .

(10) The long-range order L̄0 is then given by √ L0 ≡ µz /(gµB s) ≈ (1/ b)(1 − T /Tc )1/2 .

(11)

For T << Tc , we employ the approximation coth x ≈ 1 + 2e−2x , whence Bs (x) ≈ 1 − s−1 e−x/s

(x >> 1).

Equation (6) now gives

2sqJ 3Tc −1 −1 L0 ≈ 1 − s exp − = 1 − s exp − . kT (s + 1)T

(12)

(13)

For s = 1/2, expressions (11) and (13) reduce precisely to eqns. (11.5.14 and 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 model is somewhat simpler. Following the procedure of Problem 11.7, we find that the “effective field” Ha experienced by any given spin s on the sublattice a would be Ha = H −

2q 0 J 0 2qJ s̄b − s̄a , gµB gµB

(1a)

where q 0 and q are, respectively, the number of nearest neighbors on the other and on the same sub-lattice, while J 0 and J are the magnitudes of the corresponding interaction energies. Similarly, Hb = H −

2q 0 J 0 2qJ s̄a − s̄b . gµB gµB

(1b)

The net magnetization, per unit volume, of the sub-lattices a and b at high temperatures is then given by, see eqns. (2) and (3) of the preceding problem,

1 1 g 2 µ2B s(s + 1) 4q 0 J 0 4qJ Ma ≡ n(gµB s̄a ) ≈ n · H − 2 2 Mb − 2 2 Ma , 2 2 3kT ng µB ng µB

1 1 g 2 µ2B s(s + 1) 4q 0 J 0 4qJ Mb ≡ n(gµB s̄b ) ≈ n · H − 2 2 Ma − 2 2 Mb .

2 2 3kT ng µB ng µB

8 Adding these two results, we obtain for the total magnetization of the lattice

C 2q 0 J 0 2qJ 0 0 M ≈ {H − (γ + γ)M} γ = , γ= ; (2) T ng 2 µ2B ng 2 µ2B the parameter C here is the same as defined in eqn. (5) of the preceding problem. Equation (2) may be written in the form M≈

C H, where θ = (γ 0 + γ)C; T +θ

(3)

this yields the desired result for the paramagnetic susceptibility of the lattice. Note that the parameter θ here has no direct bearing on the onset of a phase transition in the system; for that, we must examine the possibility of spontaneous magnetization in the two sub-lattices.

To study the possibility of spontaneous magnetization, we let H → 0; in that limit, the vectors Ma and Mb are equal in magnitude but opposite in direction. We may then write: Ma = M∗ , Mb = −M∗ , and study only the former. In analogy with eqn. (6) of the preceding problem, we now have

4s(q 0 J 0 − qJ )M ∗ 1 ∗ x0 = .

(4) M = n(gµB s)Bs (x0 ) 2 n gµB kT We now employ expansion (7) of the preceding problem and get 3

TN ∗ b TN ∗ ∗ M = M − M + ..., 2 1 T T 2 n gµB s

(5)

where

2s(s + 1) 0 0 (q J − qJ ), (6) 3k while the number b is the same as given by eqn. (9) of the preceding problem. Clearly, for T < TN , a nonzero solution for M ∗ is possible. Note that the Néel temperature TN = (γ 0 − γ)C, which should be contrasted with the parameter θ of eqn. (3); moreover, for the antiferromagnetic transition to take place in the given system, we must have q 0 J 0 > qJ , the physical reason for which is not difficult to understand.

TN =

12.11. To determine the equilibrium distribution f (σ), we minimizeP the free energy (E − TS ) of the system under the obvious constraint f (σ) = 1. σ

For this, we vary the function f (σ) to f (σ) + δf (σ) and require that the resulting variation X 1 u(σ 0 , σ 00 ){f (σ 0 )δf (σ 00 ) + f (σ 00 )δf (σ 0 )} δ(E − TS ) = qN 2 σ 0 ,σ 00 X + NkT {1 + ln f (σ 0 )}δf (σ 0 ) = 0, σ0

9 while

P

δf (σ 0 ) is, of necessity, zero. Introducing the Lagrange multiplier

σ0

λ and remembering that the function u(σ 0 , σ 00 ) is symmetric in σ 0 and σ 00 , our requirement takes the form ) ( X X 0 00 00 0 qN u(σ , σ )f (σ ) + NkT {1 + ln f (σ )} − λ δf (σ 0 ) = 0. σ0

σ 00

Since the variation δf (σ 0 ) in this expression is arbitrary, the condition for equilibrium becomes qN

X

u(σ 0 , σ 00 )f (σ 00 ) + NkT ln f (σ 0 ) − N µ = 0,

σ 00

where µ = (λ − NkT )/N . By a change of notation, we get the desired result " # X 0 0 f (σ) = C exp −βq u(σ, σ )f (σ ) , (1) σ0

where C is a constant to be determined by the normalization condition P f (σ) = 1. σ

For the special case u(σ, σ 0 ) = −Jσσ 0 , where the σ’s can be either +1 or −1, eqn. (1) becomes f (σ) = C exp[βqJ σ{f (1) − f (−1)}].

(2)

Writing f (σ) = 21 (1 + L̄0 σ), the quantity f (1) − f (−1) becomes precisely equal to L̄0 , and eqn. (2) takes the form f (σ) = C exp[βqJ σL0 ].

(3)

From equation (3), we obtain f (1) + f (−1) = 2C cosh(βqJ L0 ) = 1, while

(4)

f (1) − f (−1) = 2C sinh(βqJ L0 ) = L0 .

(5)

Dividing (5) by (4), we obtain the Weiss eqn. (11.5.11) for L̄0 .

12.12. The configurational energy of the lattice is given by 1 qN [ε11 · xA (1 + X) · xA (1 − X) + ε12 {xA (1 + X)(xB + xA X) 2 + xA (1 − X)(xB − xA X)} + ε22 (xB − xA X)(xB + xA X)]

1 1 2 2 2 2 = qN ε11 xA + 2ε12 xA xB + ε22 xB − 2εxA X ε = (ε11 + ε22 ) − ε12 .

2 2

E=

10 The entropy, on the other hand, is given by

1 1 1 N ! − ln Nx A (1 + X) ! − ln N (xB − xA X) !+ S = k ln 2 2 2

1 1 1 ln N ! − ln Nx A (1 − X) ! − ln N (xB + xA X) !

2 2 2 1 ≈ Nk [−xA (1 + X) ln{xA (1 + X)} − (xB − xA X) ln(xB − xA X) 2 − xA (1 − X) ln{xA (1 − X)} − (xB + xA X) ln(xB + xA X)].

To determine the equilibrium value of X, we minimize the free energy of the system and obtain

∂(E − TS ) 1 (1 + X)(xB + xA X) 2 = −2qN εxA X + NkTx A ln = 0. ∂X 2 (1 − X)(xB − xA X) (1) Now, since xA + xB = 1, the argument of the logarithm can be written as (1 + z)/(1 − z), where z = X/(xB + xA X 2 ). Equation (1) then takes the form 2qεxA X 1 1+z = ln = tanh−1 z, (2) kT 2 1−z which is identical with the result quoted in the problem. For xA = xB = 21 , eqn. (2) reduces to the more familiar result 2X qεX = tanh−1 = 2 tanh−1 X, kT 1 + X2

(2a)

leading to a phase transition at the critical temperature Tc0 = 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 4xA Tc0 = tanh X .

(3) xB + xA X 2 T For small X, we get

3 xA 4xA Tc0 1 4xA Tc0 1 X − 2 X3 + . . . = X− X 3 + . . . , i.e. xB xB T 3 T " #

3 4xA xB Tc0 1 4xA xB Tc0 1 −1 X − 3 − 3xA xB X 3 + . . . = 0. xB T 3xB T It is now straightforward to see that for T < 4xA xB Tc0 , a non-zero solution for X is possible whereas for T ≥ 4xA xB Tc0 , X = 0 is the only possibility.

The transition temperature Tc is, therefore, given by 4xA (1 − xA )Tc0 .

12.13. For a complete solution to this problem, see Kubo (1965), problem 5.6, pp. 335–7.

11 12.14. (a) Setting N ++ + N −− + N +− = 21 qN , we find that, in equilibrium, 2

γ = 1/(1 + sL ). So, in general, it may be written as 1/(1 + sL2 ).

P (b) As in Problem 11.4, we write Q(B, T ) = exp f (L), where L

f (L) = ln N ! − ln N+ ! − ln N− ! − βE, with 1 1 N+ = N (1 + L), N− = N (1 − L), 2 2 and E = −J(N++ + N−− − N+− ) − µB(N+ − N− ) 1 = −J · qN (L2 + s)/(1 + sL2 ) − µBNL. 2 The condition that maximizes f (L) now reads:

(1 − s2 )L 1 1+L ln = β qJ + µB .

2 1−L (1 + sL2 )2 In the close vicinity of the critical point, L << 1 — with the result that

1 1+L ln ' β qJ (1 − s2 )L + µB (T ' Tc ).

2 1−L Comparing this with the corresponding equation in the solution to Problem 11.4, we infer that the critical behavior of this model is qualitatively the same as one encounters in the Bragg-Williams approximation. Quantitatively, though, the effective spin-spin interaction is reduced by the factor (1 − s2 ) — leading to a critical temperature Tc = (1 − s2 )qJ /k, instead of qJ /k. (c) As for the specific-heat singularity, the limit T → Tc− would be identical with the one obtained in Sec. 11.5; see the derivation leading to eqn. (11.5.18) and note that the replacement of J by (1 − s2 )J does not affect the final result 23 Nk . For T > Tc , L0 is identically zero. We are then left with a finite configurational energy, − 12 qJNs, that arises from the (assumed) short-range order in the system; however, unlike in the Bethe approximation, this energy is temperature-independent and hence does not entail any specific heat.

The singularity in question is, therefore, precisely the same as the one encountered in Sec. 11.5 and depicted in Fig. 11.8.

12.15. Using eqn. (11.6.30), we get Z ∞ Z γc 1 1 S∞ − Sc C(T )dT = = q γ sech 2 γdγ Nk Nk Tc T 2 0 1 = q(γc tanh γc − ln cosh γc ).

2

12 Next, we use eqn. (11.6.11) and obtain " (

2 )#

S∞ − Sc 1 1 q 1 1 1 = q ln + ln 1 − Nk 2 2 q−2 q−1 2 q−1 =

1 1 q {ln q − ln(q − 2)} + q{ln q + ln(q − 2) − 2 ln(q − 1)}.

4q−1 4

In view of the fact that S∞ = Nk ln 2, we finally get Sc 1 q2 1 1 q(q − 2) = ln 2 − ln q + q ln(q − 1) − ln(q − 2), Nk 4q−1 2 4 q−1 which leads to the desired result.

For q >> 1, the stated expression for Sc /Nk reduces to

1 1 q 1 2 1 q − +O − 1+O − +O ln 2 + 2 q q2 4 q q q2

1 = ln 2 + O , q which tends to the limit ln 2 as q → ∞.

12.16. Using eqn. (11.6.14), we get

∂ M̄ N µ2 ∂ L̄ 2N µ2 1 + exp(−2γ) cosh(2α + 2α0 ) χ≡ = = ∂B T kT ∂α T kT {cosh(2α + 2α0 ) + exp(−2γ)}2

0 ∂α 1+ . ∂α T (1) To determine (∂α0 /∂α)T , we differentiate (11.6.8) logarithmically and obtain after some simplification 0 ∂α (q − 1){tanh(α + α0 + γ) − tanh(α + α0 − γ)} = .

(2) ∂α T 2 − (q − 1){tanh(α + α0 + γ) − tanh(α + α0 + γ)} Substituting (2) into (1) and letting α → 0, we get χ0 =

4N µ2 1 + exp(−2γ) cosh(2α0 ) kT {cosh(2α0 ) + exp(−2γ)}2 1 .

2 − (q − 1){tanh(α0 + γ) − tanh(α0 − γ)}

(3)

To study the critical behavior of χ0 , we let α0 → 0 and γ → γc . Using eqn. (11.6.11), we see that, while the first two factors of expression (3) reduce to 4N µ2 1 2N µ2 q = , (4) kT c 1 + exp(−2γc ) kT c q − 1

13 the last factor diverges. To determine the nature of the divergence, we write γ = γc (1 − t) and carry out expansions in powers of t and α0 . Thus tanh(γ ± α0 ) = tanh γc + sech 2 γc (−γc t ± α0 ) − sech 2 γc tanh γc 2

(−γc t ± α0 ) + . . . , so that tanh(γ + α0 ) + tanh(γ − α0 ) ≈ 2 tanh γc − 2 sech2 γc · γc t − 2 sech2 γc tanh γc · α02 ; note that we have dropped terms of order t2 and higher. It now follows that 2 − (q − 1){tanh(γ + α0 ) + tanh(γ − α0 )}

≈ 2(q − 1)sech2 γc γc t + tanh γc · α02 .

(5)

Substituting (4) and (5) into (3), we finally obtain χ0 ≈

1 N µ2 . kT c (q − 2) (γc t + tanh γc · α02 )

(6)

For t > 0, α0 = 0; eqn. (6) then gives χ0 ≈

N µ2 1 . kT c (q − 2)γc t

(7a)

For t < 0, α0 is given by eqn. (11.6.13), whence α02 ' −3(q − 1)γc t; we now get N µ2 1 χ0 ≈ .

(7b) kT c 2(q − 2)γc |t| Note that, for large q, the quantity

q 1 1 =1+O ; (q − 2)γc = (q − 2) ln 2 q−2 q eqns. (7) then reduce to eqns. (11.5.22) of the Bragg-Williams approximation.

12.19. We refer to the solutions to Problems 11.4 and 11.5, whereby

A 1 + m0 1 + m0 1 − m0 1 − m0 1 qJ 2 ψ0 ≡ = ln + ln − m NkT B=0 2 2 2 2 2 kT 0 1 1 − m20 m0 1 + m0 1 Tc 2 ln + ln − m 2 4 2 1 − m0 2T 0

1 m4 m3 = − ln 2 + −m20 − 0 − . . . + m0 m0 + 0 + . . .

2 2 3 1 − (1 − t + . . .)m20 2 1 1 ≈ − ln 2 + t m20 + m40 .

2 12 =

14 Comparing this expression with eqn. (11.9.5), we infer that in the BraggWilliams approximation r1 = 1/2 and s0 = 1/12. Now, substituting these values of r1 and s0 into eqns. (11.9.4, 9–11 and 15), we see that the corresponding eqns. (11.5.14, 22, 24 and 18) are readily verified.

A similar calculation under the Bethe approximation is somewhat tedious; the answer, nevertheless, is r1 =

q (q − 1)(q − 2) q−2 . ln , s0 = 4 q−2 12q 2

12.20. The equilibrium values of m in this case are given by the equation ψh0 = −h + 2 rm + 4 sm 3 + 6 um 5 = 0 (u > 0). √ √ With h = 0, we get: m0 = 0, ± A+ or ± A− , where √ −s ± s2 − 3 ur A± = .

3u

(1)

(2)

First of all, we note that, for A± to be real, s2 must be ≥ 3ur . This presents no problem if r ≤ 0; however, if r > 0, then s must be either √ √ ≥ 3ur or ≤ − 3ur . We also observe that A+ A− =

r 3u

and A+ + A− = −

2s .

3u

It follows that (i) if r < 0, then one of the A’s will be positive, the other negative (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 = 0 both 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 ≥ 3ur both 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 positive A will yield a real m0 . Finally, since ψ000 = 2r + 12 sm 20 + 30 um 40 ,

(3)

the extremum at m0 = 0 is a maximum if r < 0, a minimum if√r > 0. It follows that for r < √ 0 the function ψ0 is minimum at√m0 = ± A+ and 3ur ) it is maximum at m0 = ± A− and minimum for r > 0 (and s ≤ − √ 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

1 ψ0 (m0 ) = q + rm 20 + sm 40 + um 60 − m0 2rm 0 + 4sm 30 + 6um 50 (4a) 4

1 = q + m20 r − um 40 ; (4b) 2

15 note that in writing (4a) we have added an expression which, by the minimization condition, is identically zero. It now follows from eqn. (4b) that ψ0 (m0 = 0) is less than, equal to or greater than p ψ0 (m0 6= 0) accroding as m20 is less than,pequal to or greater than r/u. The dividing line corresponds to A+ = r/u, i.e. √ r √ s2 − 3ur −s + r = , i.e. s = − 4ur ; 3u u see the accompanying figure. We also note that, in reference to p the di2 viding line, m decreases monotonically towards the limiting value √ 0 √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 ψh0 = −h + 2rm + 6um 5 = 0.

For |t| << 1, we set r ≈ r1 t; the equation of state then takes the form h ≈ 2r1 tm + 6um 5 .

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 ≈ 6um 5 , giving δ = 5. For susceptibility, we have

∂m 1 χ∼ ≈ . ∂h t 2r1 t + 30um 4 It follows that

χ0 ≈

1/2 r1 t (t > 0, m → 0) 1/8 r1 |t| (t < 0, m → m0 ),

giving γ = γ 0 = 1. Finally, using the scaling relation α + 2β + γ = 2, we get: α = 1/2.

12.22. (a) We introduce the variable ψ[= (vg −vc )/vc ' (vc −v` )/vc ] and obtain

(s)

π ∂G ∼ |t|2−α−∆ g 0 ψ∼ .

(1) ∂π |t|∆ t With t < 0 and π → 0, we get: ψ ∼ |t|β , where β = 2 − α − ∆. It follows that the quantities (ρ` − ρc ), (ρc − ρg ) and (ρ` − ρg ) all vary as |t|β . (b) Writing g 0 (x) as xβ/∆ f (x), eqn. (1) takes the form ψ ∼ π β/∆ f (π/|t|∆ ).

(2)

It follows that the quantity |t|∆ /π is a universal function of the quantity π β/∆ /ψ, i.e. |t| ∼ π 1/∆ × a universal function of (π/ψ ∆/β ).

It is now clear that, at t = 0, π ∼ ψ δ , where δ = ∆/β.

16 (c) For the isothermal compressibility of the system, we have from (1)

1 ∂v ∂ψ π . ∼ ∼ |t|β−∆ g 00 κT ≡ − v ∂P T ∂π t |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

1 ∂v 1 ∂v ∂P ∂P αP = =− = κT . v ∂T P v ∂P T ∂T v ∂T v In the region of phase transition, (∂P/∂T )v is simply (dP/dT ), which is non-singular. Accordingly, αP ∼ κT ∼ |t|−γ . Similarly, in view of the relation CP = VT (dP /dT )2 κT , established in Problem 12.25, we infer that CP ∼ κT ∼ |t|−γ .

For Cν , we go back to the given expression for G(s) and write

∂G(s) ∂T

∼ |t|1−α × a universal function of P

ψ 1−α ∼ |t| × a universal function of . |t|β

S (s) = −

π |t|∆

It then follows that

(s) ψ ∂S (s) ∼ |t|−α × a universal function of .

CV = T ∂T V |t|β (s)

Now, letting ψ → 0, we obtain: CV ∼ |t|−α . And, in view of the relation CV = VT (dP /dT )2 κS , also established in Problem 12.25, we infer that κS ∼ CV ∼ |t|−α .

Finally, for the latent heat of vaporization `, we invoke the Clapeyron equation, dP ` = , dT 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 = abm 2Θ+1 , giving δ = 2Θ + 1. (iii) The quantity

∂m ∂h

= t

1 . a(1 + 3bm )(t + bm 2 )Θ−1 2

17 With t > 0 and h → 0, m → 0 and we are left with

∂m ∂h

≈ t

1 , giving γ = Θ. at Θ

The scaling relation (11.10.22) is readily verified.

12.24. For ri 6= rj , eqns. (11.11.22 and 26) give (∇2 − ξ −2 )g(r) = 0.

(1)

Now, if g(r) is a function of r only then dg r dg x1 xd dg , whence ∇r = = ,..., dr dr r dr r r X 2

d d X ∂ dg 1 d g 1 dg 1 x2i dg xi ∇ · ∇g = = + − ∂xi dr r dr r dr r2 r dr 2 r i=1 i=1 ∇g =

=

d2 g dg 1 2 + dr r (d − 1). dr

(2)

Substituting (2) into (1), we obtain the desired differential equation — of which (11.11.26) is the exact solution.

Substituting (11.11.27) into the left-hand side of the given differential equation, we get

e−r/ξ 1 1 const. (d−1)/2 + . . . + (. . .) − 2 ; ξ2 ξ r the ratio of the terms omitted to the ones retained is O(ξ/r). Clearly, the equation is satisfied for r >> ξ. Similarly, substituting (11.11.28) instead, we get

(2 − d)(1 − d) d − 1 2 − d 1 + − (. . .) ; const. d−2 r r2 r r 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. 11.10, we expect that, for h > 0 and t > 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/∆ ).

18 c

At t = 0, we obtain: ξ(0, h) ∼ h−ν , where ν c = ν/∆. c

By a similar argument, χ(0, h) ∼ h−γ , where γ c = 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.

12.27. By definition, σ ≡ ψA (t) ∼ |t|µ

(t . 0).

(1)

By an argument similar to the one that led to eqn. (11.12.14), 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 transition belongs, ξ ∼ |t|−ν . Equations (1) and (2) then lead to the desired result µ = (d − 1)ν = (2 − α)(d − 1)/d.

12.28. The exponential has the convexity property eλφ ≥ 1 + λφ so if P (φ) is any normalized probability distribution for φ and hf (φ)i ≡ TrP (φ)f (φ) then eλφ−λhφi ≥ (1 + λ(φhφi)), eλφ ≥ (1 + λ(φ − hφi))eλhφi , heλφ i ≥ eλhφi .

Now is H = H[φ] is a function is a classical field φ, and ρ is any normalized probability distribution of φ then the canonical partition function and free energy Q = e−βF are given by Q = e−βF = Tre−βH = Trρ[φ]e−βH[φ]−ln ρ[φ] = e−βH−ln ρ ρ .

Now using the inequality above gives e−βF ≥ e−Trρ(βH+ln ρ) , so βF ≤ Fρ = Tr (ρβH + ρ ln ρ) , Now, setting the functional derivative of F ρ with respect to ρ to zero with the constraint Trρ = 1 gives the exact density matrix and free energy: ρ = e−βH /Tre−βH , βF = − ln Tre−βH .

19 12.29. Minimizing the variational free energy with respect to ρα (sα ) ! βFρ =

XY {s}

ρi (si ) −K

X

−(i, j) − h

i

with the constraint

X i

s+

X

ln ρi (si )

i

P

s ρα (s) = 1 gives

(zKhsi + h) sα + ln ρα (sα ) + 1 + ζ = 0, where z is the number of nearest neighbors on the lattice, hsi is the average spin, and ζ is a Lagrange multiplier. This gives the Ising mean field theory discussed in Section 12.5 which involves solving the self-consistency equation hsi = tanh (Kzhsi + h).

Chapter 13 13.1. Since M̄ = (N̄+ − N̄− )µ and N̄+ + N̄− = N , we readily see that

M̄ 1 P (β, B) ± sinh x 1 = N (x = βµB).

N̄± = N 1 + 2 Nµ 2 P (β, B)

(1)

Next, comparing eqns. (11.3.19) and (12.1.12), and keeping in mind that q = 2, we get N̄+ − N̄++ = N

e−4βJ P 2 (β, B) − sinh2 x =N .

2D(β, B) 2D(β, B)

(2)

It follows from eqns. (1) and (2) that N̄++ = (N/2D){(P + sinh x)(P + cosh x) − (P 2 − sinh2 x)} = (N/2D){(P + sinh x)(cosh x + sinh x)},

(3)

which is the desired result. Equation (11.3.17) now gives N̄+− = 2(N̄+ − N̄++ ) and N̄−− = N − 2N̄+ + N̄++ = N̄++ − (M̄ /µ).

The number N̄+− is just twice the expression (2). The number N̄−− is given by N̄−− = (N/2D){(P + sinh x)(cosh x + sinh x) − sinh x · 2(P + cosh x)} = (N/2D){(P − sinh x)(cosh x − sinh x)},

(4)

which is the desired result.

It is straightforward to check that the sum N̄++ +N̄−− = (N/D)(P cosh x+sinh2 x) = (N/D){P cosh x+(P 2 −e−4βJ )}; adding N̄+− , one obtains the expected result N . Finally, the product N̄++ N̄−− = (N/2D)2 {P 2 − sinh2 x} = (N/2D)2 e−4βJ .

Equation (11.6.22) is now readily verified.

1

2 13.2. (a) In view of eqn. (11.3.18), the quantity (N++ + N−− − N+− ) that appears in the Hamiltonian (11.3.19) of the lattice may be written as 1 2 qN − 2N+− . The partition function (11.3.20) then assumes the form stated here. (b) A complete solution to this problem can be found in the first edition of this book — Sec. 12.9A, pp. 414–8. In any case, this problem is a special case, q = 2, of the next problem which is treated here at sufficient length.

13.3. In the notation of Problem 12.2, we now have

1 1 ln gN (N+ , N+− ) ≈ qN ln qN − N++ ln N++ − N−− ln N−− − N+− 2 2

1 N+− ln 2 + (q − 1){N+ ln N+ + N− ln N− − N ln N }, (1) where, by virtue of eqn. (11.3.17), 1 1 1 1 1 qN − N+− , N−− = qN − qN + − N+− and N− = N −N+ .

2 + 2 2 2 2 (2) Now, as usual, the logarithm of the partition function may be approximated by the logarithm of the largest term in the sum over N+ and N+− , with the result that

1 ∗ ∗ ∗ ∗ ln QN (B, T ) ≈ ln gN N+ , N+− +βJ qN − 2N+− +βµB 2N+ −N , 2 (3) ∗ ∗ are the values of the variables N+ and N+− that and N+− where N+ maximize the summand (or the log of it). The maximizing conditions turn out to be

∂ 1 1 (. . .) = −(1 + ln N++ ) q − (1 + ln N−− ) − q + ∂N+ 2 2 N++ =

(q − 1){(1 + ln N+ ) + (1 + ln N− )(−1)} + 2βµB ( q/2 q−1 ) N+ N−− + 2βµB = 0, and (4) = ln N++ N−

∂ 1 1 (. . .) = −(1 + ln N++ ) − − (1 + ln N−− ) − − ∂N+− 2 2

1 1 + ln N+− − 2βJ 2 ( 1/2 1/2 ) N++ N−− − 2βJ = 0. = ln (5) 1 2 N+−

3 Equations (4) and (5), with the help of eqns. (2), determine the equilibrium values of all the numbers involved in the problem; eqn. (3) then determines the rest of the properties of the system.

To compare these results with the ones following from the Bethe approximation, we first observe that eqn. (5) here is identical with the corresponding eqn. (11.6.22) of that treatment. As for eqn. (4), we go back to eqns. (11.6.4 and 8) of the Bethe approximation, whereby

q 0 cosh(α + α0 + γ N̄+ 2α =e = e2α+2α q/(q−1) 0 cosh(α + α − γ) N̄− and to eqn. (11.6.21), whereby 0 N̄−− = e−4(α+α ) .

N̄++

It follows that

N −− N ++

q/2

N+ N−

q−1

0

0

= e−2q(α+α )+2α(q−1)+2α q = e−2α

(α = βµB),

in complete agreement with eqn. (4). Hence the equivalence of the two treatments.

13.4. By eqns. (12.1.8 and 37),

cosh x + {e−4βJ + sinh2 x}1/2 ξ −1 (B, T ) = ln cosh x − {e−4βJ + sinh2 x}1/2

(x = βµB).

As T → Tc (which, in this case, is 0 K),

cosh x + sinh x ξ −1 (B, Tc ) ≈ ln = 2x. cosh x − sinh x It follows that ξ(B, Tc ) ≈ (1/2x) ∼ B −1 , which means that the critical exponent ν c = 1. Now, a reference to eqns. (12.1.21 and 35) tells us that ν = ∆. The relation ν c = ν/∆ is thus verified.

13.5. On integration over B, our partition function takes the form   !2

1/2 X  βµ2 s X  X 2πs QN (s, T ) ∼ exp σi + βJ σi σi+1 ,  2N  βN i i {σi }

which implies an effective Hamiltonian given by the expression X 1 Heff = − µ2 s NL2 − J σi σi+1 2 i

!

L=N

−1

X i

σi

.

4 The first term here is equivalent to an infinite-range interaction of the type considered in Problem 11.6, with µ2 s playing the role of the quantity cN of that model. This is also equivalent to the Bragg-Williams model, with µ2 s ↔ qJ ; see Problem 11.4. It follows that, by virtue of this term, the system will undergo an order-disorder transition at a critical temperature Tc = µ2 s/k. The second term in the Hamiltonian will contribute towards the short-range order in the system.

We also note that the root-mean-square value of B in this model is (s/βN )1/2 which, for a given value of s, is negligibly small when N is large. The order-disorder transition is made possible by the fact that the resulting interaction is of an infinite range.

13.6. Since the Hamiltonian H{τi } of the present model is formally similar to the Hamiltonian H{σi } of Sec. 12.1, the partition function of this system can be written down in analogy with eqn. (12.1.10): h i

1 1/2 ln Q ≈ ln eβJ2 cosh(βJ1 ) + e−2βJ2 + e2βJ2 sinh2 (βJ1 ) , N from which the various thermodynamic properties of the system can be derived. In particular, we get σi σi+1 =

1 ∂ sinh(βJ1 ) ln Q = , and 1/2 βN ∂J1 e−4βJ2 + sinh2 (βJ ) 1

1 ∂ σi σi+2 = ln Q = 1− βN ∂J2 2e−4βJ2

e−4βJ2 + sinh2 (βJ1 )

h 1/2

i.

1/2 cosh(βJ1 ) + e−4βJ2 + sinh2 (βJ1 )

As a check, we see that in the limit J2 → 0 these expressions reduce to tanh(βJ1 ) and tanh2 (βJ1 ), respectively; on the other hand, if J1 → 0, they reduce to 0 and tanh(βJ2 ) instead. All these limiting results are indeed expected.

13.7. In the symmetrized version of this problem, the transfer matrix P is given by

1

0 0 0 0 0 0 σi , σi |P|σi+1 , σi+1 = exp K1 σi σi+1 + σi σi+1 + K2 σi σi + σi+1 σi+1 , 2 where K1 = βJ1 and K2 = βJ2 . Since (σi , σi0 ) = (1, 1), (1, −1), (−1, 1) or (−1, −1), we get  2K +K  e 1 2 1 1 e−2K1 +K2   1 e2K1 −K2 e−2K1 −K2 1  (P) =  −2K1 −K2 2K1 −K2   1 e e 1 −2K1 +K2 2K1 +K2 e 1 1 e

5 The eigenvalues of this matrix are

i 1h λ1 (A + B + C + D) ± {(A − B + C − D)2 + 16}1/2 , = λ2 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, h i 1 1 (A + B + C + D) + {(A − B + C − D)2 + 16}1/2 . ln Q ≈ ln λ1 = ln N 2 Substituting for A, B, C and D, we obtain the quoted result. The study of the various thermodynamic properties of the system is now straightforward.

13.8. In the notation of Sec. 12.1, the transfer matrix of this model is hσi |P|σi+1 i = exp(βJ σi σi+1 )

(σi = −1, 0, 1).

It follows that eK (P) =  1 e−K 

 1 e−K 1 1  1 eK

(K = βJ),

with eigenvalues

i 1h λ1 (1 + 2 cosh K) ± {8 + (2 cosh K − 1)2 }1/2 , λ3 = 2 sinh K. = λ2 2 Since λ1 is the largest eigenvalue of P, h i 1 1 ln Q ≈ ln λ1 = ln (1 + 2 cosh K) + {8 + (2 cosh K − 1)2 }1/2 , N 2 which leads to the quoted expression for the free energy A. In the limit T → 0, K → ∞, with the result that cosh K ≈ 12 ek and hence A ≈ −NJ ; this corresponds to a state of perfect order in the system, with U = −NJ and S = 0. On the other hand, when T → ∞, cosh K → 1 and hence A → −NkT ln 3; this corresponds to a state of complete randomness in a system with 3N microstates.

13.9. (a) Making use of the correspondence established in Sec. 11.4, we obtain for a one-dimensional lattice gas (q = 2). (i) The fugacity z = e−4βJ+2βµB = η 2 y, where η = e−2βJ and y ↔ e2βµB .

6 (ii) The pressure P = −(A/N ) − J + µB; using eqn. (12.1.11), this becomes i h P = kT ln cosh(βµB) + {e−4βJ + sinh2 (βµB)}1/2 + µB. (1) (iii) The density (1/v) = 12 {1 + (M /N µ)}; using eqn. (12.1.13), this becomes

1 1 sinh(βµB) .

(2) = 1 + −βJ v 2 {e + sinh2 (βµB)}1/2 Our next step consists in eliminating the magnetic variable (βµB) in favor of the fluid variable y. For this, we note that 1 1/2 (y + y −1/2 ) = (y + 1)/2y 1/2 , 2 1 sinh(βµB) = (y 1/2 − y −1/2 ) = (y − 1)/2y 1/2 , 2

cosh(βµB) =

while βµB = 21 ln y. Substituting these results into eqns. (1) and (2), we obtain the quoted expressions for P/kT and 1/v. It may be verified that these expressions satisfy the thermodynamic relation

1 ∂ ∂ P P =z =y v ∂z kT ∂y kT T η At high temperatures, η → 1 and we get P 1 y ≈ ln(y + 1), ≈ . kT v y+1 Moreover, in this limit y ' z << 1. One may, therefore, write 1 P ≈ y, ≈ y kT v

and hence

Pv ≈ 1. kT

At low temperatures, η becomes very small and y very large — with the result that η2 1 η2 P ≈ ln y + , ≈ 1 − . kT y v y (b) For a hard-core lattice gas (y → 0, η → ∞), we get √ √

P 1 + 4z − 1 1 + 1 + 4z √ = ln , ρ= . kT 2 2 1 + 4z √ From the second equation, it follows that 1 + 4z = 1/(1 − 2ρ). Substituting this into the first equation, we obtain the quoted expression for P (ρ).

7 13.10. Applying eqn. (11.11.11) to a one-dimensional system, we get ∞ ∞ X X χ0 −(a/ξ)|x| = e =1+2 e−(a/ξ)x N βµ2 x=−∞ x=1

1 + e−a/ξ e−a/ξ = = coth =1+2 1 − e−a/ξ 1 − e−a/ξ

a 2ξ

.

For ξ >> a, coth(a/2ξ) ≈ (2ξ/a), making χ0 ∝ ξ 1 — consistent with the fact that for this system (2 − η) = 1. For ξ << a, we recover the familiar result: χ0 ≈ N µ2 /kT .

For an n-vector model, eqn. (12.2.17) leads to the result 1 + In/2 (βJ)/I(n−2)/2 (βJ) χ0 = , 2 N βµ 1 − In/2 (βJ)/I(n−2)/2 (βJ) which agrees with the expression quoted in the problem. For the special case n = 1, the ratio I1/2 (x)/I−1/2 (x) = tanh x, whence cosh(βJ) + sinh(βJ) 1 + tanh(βJ) χ0 = = e2βJ , = N βµ2 1 − tanh(βJ) cosh(βJ) − sinh(βJ) in agreement with eqn. (12.1.14).

2 2 2 2 13.11. We introduce an extra factor σk+1 . . . σ`−1 σm+1 . . . σn−1 , which is identically equal to 1, and write

σk σ` σm σn = (σk σk+1 ) . . . (σ`−1 σ` )(σm σm+1 ) . . . (σn−1 σn ).

Following the same procedure that led to eqn. (12.1.31), we now get σk σ` σm σn =

`−1 Y i=k

tanh(βJi )

n−1 Y

tanh(βJi ).

i=m

Employing a common J, we obtain the desired result σk σ` σm σn = {tanh(βJ)}`−k {tanh(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 = 0, we get a much simpler result, viz.

1 1 ln Q = ln 2 + 2 N 2π

Zπ Zπ 0

0

ln{cosh(2γ) − sinh(2γ) − cos ω}dωdω 0 .

8 The integration over ω 0 is straightforward; the one over ω can be done with the help of the formula " # √ Zπ a + a2 − b2 ln(a − b cos ω)dω = π ln (|b| < a), (1) 2 0

which yields the expected result

1 1 cosh(2γ) + 1 ln Q = ln 2 + ln = ln(2 cosh γ).

N 2 2 With J 0 = J, we have 1 1 ln Q = ln 2 + 2 N 2π

Zπ Zπ 0

ln{cosh2 (2γ) − sinh 2γ(cos ω + cos ω 0 )}dωdω 0 .

0

(2) We substitute ω = θ + ϕ and ω 0 = θ − ϕ; this will replace the sum (cos ω + cos ω 0 ) by the product 2 cos θ cos ϕ and the element dωdω 0 by 2dθdϕ. As for the limits of integration, the periodicity of the integrand allows us to choose the rectangle [0 ≤ θ ≤ π, 0 ≤ ϕ ≤ π/2] without affecting the value of the integral. We thus have 1 1 ln Q = ln 2 + 2 N π

Zπ Zπ/2 ln{cosh2 (2γ) − 2 sinh(2γ) cos ϕ cos θ}dθdϕ.

0

0

Integration over θ may now be carried out using formula (1), with the result 1 1 ln Q = ln 2 + N π

Zπ/2 q 1 2 4 2 2 ln cosh (2γ) + cosh (2γ) − 4 sinh (2γ) cos ϕ dϕ 2 0

Zπ/2 n o p 1 1 = ln 2 · √ cosh(2γ) + ln 1 + 1 − κ2 cos2 ϕ dϕ, π 2

0

where κ is given by eqn. (12.3.23). Finally, we replace cos2 ϕ by sin2 ϕ (without affecting the value of the integral) and recover eqn. (12.3.22).

We know that this model is singular at κ = 1. A close look at the integral in (2) shows that the singularity arises when contributions in the neighborhood of the point ω = ω 0 = 0 pile up. Since cos ω and cos ω 0 are almost unity there, the situation becomes catastrophic when cosh2 2γ = 2 sinh 2γ, i.e. when κ = 1. In the anisotropic case, a similar observation suggests that the singularity will arise when cosh(2γ) cosh(2γ 0 ) = sinh(2γ) + sinh(2γ 0 ).

9 Squaring both sides of this equation and simplifying, we get sinh(2γ) sinh(2γ 0 ) = 1

(3)

as the criterion for the onset of phase transition in this system; cf. eqn. (12.3.18).

The study of the thermodynamic behavior of the system in the neighborhood of the critical point in the general case, when J 0 6= J, is highly complicated; the fact, however, remains that the internal energy U0 is continuous and the specific heat C0 displays a logarithmic divergence at a critical temperature Tc given by eqn. (3).

13.14. As κ → 1, the first integral tends to the limit Zπ/2

1 − sin ϕ dϕ = cos ϕ

0

Zπ/2

cos ϕ π/2 dϕ = ln(1 + sin ϕ)|0 = ln 2.

1 + sin ϕ

0

The second integral diverges as κ → 1, so we need to examine it carefully.

Setting cos ϕ = x, this integral takes the form Z1 p 0

Z1

κ dx (1 − κ2 ) + κ2 x2

'

√

0

n o 1 √ dx = ln x + κ02 + x2 .

0 κ02 + x2

02

Since κ is close to 1, κ << 1; we, therefore, get for this integral the asymptotic result ln 2 − ln |κ0 |. It follows that K1 (κ) ≈ 2 ln 2 − ln |κ0 | = ln(4/|κ0 |).

13.15. The quantity to be evaluated here is

Sc U A = − .

Nk NkT c NkT c

√ The first term, by eqn. (12.3.28), is −Kc coth(2Kc ) = − 2Kc . The second, by eqn. (12.3.22), is 1 ln 2 + π

Zπ/2 ln(1 + cos ϕ)dϕ 0

 = ln 2 +

1 π/2 ϕ ln(1 + cos ϕ)|0 + π

Zπ/2

 ϕ sin ϕ  dϕ .

1 + cos ϕ

0

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, √ Sc 1 = − 2Kc + ln 2 + 2G/π ' 0.3065.

Nk 2 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.

10 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 2Kc t + 2 2Kc + 2Kc2 t2 + . . . .

(2) Raising expression (2) to the power −4, we get

√ √ {sinh(2K)}−4 = 1 + 8 2Kc t − 8 2Kc + 8Kc2 − 80Kc2 t2 + . . . , so that √

√ 1 − {sinh(2K)}−4 = −8 2Kc t + 8 2Kc − 72Kc2 t2 + . . .

√ 9 = 8 2Kc |t| 1 + 1 − √ Kc |t| + . . . .

2

(3)

Taking the eighth-root of (3), we obtain the desired result.

13.17. Making use of the correspondence established in Sec. 11.4 and utilizing a result obtained in Problem 12.15, we have for a two-dimensional lattice gas (q = 4)

Pc 2J 2G A 1 − ln 2 + − 2Kc . =− = kT c NkT c kT c 2 π We also have: vc = 2 (because, at T = Tc , the spontaneous magnetization of the corresponding ferromagnet is zero). It follows that Pc vc 4G − 4Kc ' 0.09659. = ln 2 + kT c π Taking the reciprocal of this result, we obtain the one stated in the problem.

13.18. In one dimension, expression (12.4.31) assumes the form Z∞ W1 (ϕ) =

e−(1+ϕ)x I0 (x)dx =

1 1 = 2 . {(1 + ϕ)2 − 1}1/2 (λ − J 2 )1/2

0

The constraint equation (12.4.19) then becomes N 1 N µ2 β 2 + = N, 2 2 1/2 2β (λ − J ) 4(λ − J)2

(1)

11 which agrees with the quoted result. Comparing (1) with the formal constraint equation (12.4.13), we conclude that N µ2 B 2 ∂Aλ N 1 + , = 2 2 1/2 ∂λ 2β (λ − J ) 4(λ − J)2

(2)

N N µ2 B 2 ln λ + (λ2 − J 2 )1/2 + + C, 2β 4(λ − J)

(2)

It follows that Aλ =

where C is a constant of integration. To determine C, we observe from eqn. (12.4.12a) 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 that C = (N/2β) ln(β/2π), which leads to the quoted result for Aλ .

With B = 0 but J 6= 0, eqn. (1) gives: (λ2 − J 2 )1/2 = 1/2β, and hence λ = (1 + 4β 2 J 2 )1/2 /2β. Equation (12.4.15) then gives βAλ 1 βA& = − βλ = ln N N 2

(1 + 4β 2 J 2 )1/2 + 1 4π

1 − (1 + 4β 2 J 2 )1/2 .

2

13.19. With βJi = β · nJ 0 = nK , eqn. (12.2.8) becomes QN (nK ) =

Γ(n/2) (n−2)/2 I(n−2)/2 (nK ) 1 2 nK

N −1

.

For n, N >> 1, we get

n

n n 1 1 1 ln QN (nK ) ≈ − ln nK + ln In/2 · 2K ln Γ nN n 2 2 2 2

n 1 1 n n n ln − + . . . − ln nK = n 2 2 2 2 2

n (4K 2 + 1)1/2 + 1 + (4K 2 + 1)1/2 − ln + ...

2 2K

2 1/2 1 (4K + 1) + 1 2 1/2 ≈ (4K + 1) − 1 − ln .

2 2 13.20. By eqn. (12.4.69), we have for the spherical model at T < Tc χ0 =

N µ2 N 2 µ2 (K − Kc ) ≈ .

2 Jϕ J

Replacing (K − Kc ) by m20 K, see eqn. (12.4.44), and remembering that K = J/kB T , we obtain the desired result.

12 13.21. Before employing the suggested approximation, we observe that the major contribution to the integral over θj in expression (12.4.58) comes from those values of θj that are either close to the lower limit 0 or close to the upper limit 2π. We, therefore, write this integral in the form Zπ 2

cos(Rj θj / a)e−x+x cos θj

Nj dθj , 2π

0

so that the major contribution now comes only from those values of θ that are close to 0. We may, therefore, replace (1 − cos θj ) by θj2 /2 and, at the same time, replace the upper limit of the integral by ∞. This yields the asymptotic result Zπ 2

2

cos(Rj θj / a)e−xθj /2

Nj −Rj2 /2a2 x Nj e dθj = √ , 2π 2πx

0

where use has been made of formula (B.41). Equation (12.4.57) then becomes Z∞ 2 2 1 N G(R) ≈ e−ϕx−R /2a x dx , d/2 2N βJ (2πx) 0

which is precisely the expression that led to eqns. (12.4.61 and 62).

In the case of the Bose gas, we are concerned with the expression Z 1 exp(ik · R) dd k, (2π)d eα+β~2 k2 /2m − 1 see page 402 of the text, which may now be approximated by Z 1 exp(ik · R) d d k. d (2π) α + β~2 k 2 /2m Using the representation (12.4.27), this may be written as   Z∞ d Z∞ Y 2 2 1 eik n Rn −β~ kn x/2m dk n  dx e−αx  2π n=1 −∞

0

Z∞ =

e

−αx

0

1 = d λ

Z∞

" d # Y 1 −πRn2 /λ2 x √ e dx λ x n=1 2

2

e−αx−πR /λ x

1 xd/2

" λ=~

2πβ m

1/2 #

dx ,

0

which is precisely the expression that led to eqns. (12.5.35 and 36).

13 13.22. The constraint equation (12.4.21) now takes the form −1 X 1 2N β(1 − m2 ) = , J ϕ + k σ aσ 2 k

which may as well be written as 2

2K(1−m ) = F (ϕ), where K = βJ and F (ϕ) = N

−1

X k

1 ϕ + k σ aσ 2

−1

To determine the behavior of the function F (ϕ) at small ϕ, we look at the derivative −2 X 1 σ σ 0 −1 F (ϕ) = −N ϕ+ k a .

2 k

In view of eqns. (12.4.17) and (C.7b), we obtain ∞

0

F (ϕ) ≈ −N

−1

Z d Y Nj a 2π

j=1

ϕ+

1 σ σ k a 2

−2

2π d/2 d−1 k dk Γ(d/2)

0

a

Z∞

d

= − d−1 d/2 2 π Γ(d/2)ϕ2

1+

1 σ σ k a 2ϕ

−2

k d−1 dk .

0 σ

σ

We substitute k = (2ϕ/a )x and get (2ϕ)d/σ F (ϕ) ≈ − d−1 d/2 2 π Γ(d/2)σϕ2 0

Z∞

x(d/σ)−1 dx (1 + x)2

0

2d/σ Γ(d/σ)Γ(2 − d / σ) (d−2σ)/σ =− ϕ 2d−1 π d/2 Γ(d/2)σ

(d < 2σ).

Integrating over ϕ, we obtain F (ϕ) ≈ F (0) −

2d/σ Γ{(d/σ)/σ}Γ{(2σ − d)/σ} (d−2σ)/σ ϕ ; 2d−1 π d/2 Γ(d/2)σ

cf. eqn. (F.7c). The function F (0) exists for all d > σ and may be identified 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 this equation. 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) ; it follows that ν = 1/(d − σ). We then get: η = 2 − (γ/ν) = 2 − σ.

.

14 For d > 2σ, the derivative F 0 (0) exists — with the result that F (ϕ) ≈ F (0) − |F 0 (0)|ϕ.

This leads to mean-field results for the exponents β, γ, α and δ. The correlation 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. (12.5.9 and 11) proceeds exactly as of eqns. (7.1.36 and 37). The derivation of eqns. (12.5.10 and 13) proceeds exactly as in Problem 7.4; eqn. (12.5.12) then follows as a product of expressions (12.5.11 and 13): The derivation of eqns. (12.5.14 and 15) proceeds exactly as in Problem 7.5; note that one may first obtain here κT =

gd/2 (z) d 1 g(d−2)/2 (z) , κS = , nk B T gd/2 (z) (d + 2)nk B T g(d+2)/2 (z)

and then use eqn. (12.5.7) to express these quantities in terms of P . Finally, the derivation of eqn. (12.5.23) proceeds exactly as in Problem 7.6.

13.24. The derivations here proceed exactly as in Problem 7.7. The singularity of these quantities arises from the last term of the two expressions, and is qualitatively similar to the singularity of the quantity (∂Cν /∂T ); see eqn. (10.5.23b) and the ensuing comments. Note that the singularity of the combination v(∂ 2 P/∂T 2 )v − (∂ 2 µ/∂T 2 )v is relatively mild.

13.25. By definition,

CP ≡ T

∂S ∂T

=T P

∂S ∂V

P

∂V ∂T

.

P

Using the Maxwell relation (∂S/∂V )P = (∂P/∂T )S , we get

∂P ∂P ∂V ∂V ∂P =T CP = T − = ∂T S ∂T P ∂T S ∂P T ∂T V

∂P ∂P VκT .

T ∂T S ∂T V Next

CV ≡ T

∂S ∂T

=T V

∂S ∂P

V

∂P ∂T

.

V

Now, using the Maxwell relation (∂S/∂P )V = −(∂V /∂T )S , we get

∂V ∂P ∂V ∂P ∂P CV = −T = −T = ∂T S ∂T V ∂P S ∂T S ∂T V

∂P ∂P T (V κS ) . ∂T S ∂T V

15 In the two-phase region, (∂P/∂T )S = (∂P/∂T )V = dP /dT , with the result that CP = VT (dP / dT )2 κT , CV = VT (dP / dT )2 κS .

Now, by eqn. (12.5.28), dP/dT, at T < Tc , = (d + 2)P/2T . Using this result and eqn. (12.5.15), we get

CV = VT

(d + 2)P 2T

2

d(d + 2) PV d = . (d + 2)P 4 T

Substituting for P from eqn. (12.5.28), we recover expression (12.5.30) for CV .

13.26. As shown in Problem 1.16, dµ = −sdT + vdP . It follows that 1 κT ≡ − 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 are given by eqns. (12.5.24) and (12.5.27). Since α ≡ −µ/kB T , we have ρ=

N N0 Ne kB T ζ(d / 2) = + =− + .

V V V Vµ λd

It follows that κT =

1 ρ2

kB T V µ2

=

1 kB T ρ2 V

2 N0 N2 V ρ2 − = 2 0 = 2 0 . kB T ρ Vk B T ρ kB T

Incidently, using the relationship between CP and κT , as developed in Problem 12.25, we can show that, in the two-phase region of the Bose gas, CP =N N kB

d + 2 ζ{(d + 2)/2} 2 ζ(d/2)

2

T Tc

d

ρ0 ρ

2 .

Comparing this with eqn. (12.5.30), we find that, in this region, the ratio CP /CV = O(N ).

13.28. Use the relations N/(L − N D) = βP and

z n 1+ = exp(z), n→∞ n lim

and collect factors.

16 13.29. Using equation (10.7.20a) and ignoring the delta–function contribution to S(k) gives   Z ∞ ∞ X 1 n  (βP (x − jD))j exp (−βP (x − jD) + ikx) + c.c. .

S(k) = 1 + βP (1 − nD) j=1 (j − 1)! jD Using n/(βP (1 − nD)) = 1 we get

βP eikD S(k) = 1 + + c.c. βP − ik − βP eikD k2 = 2 k + 2(βP )2 (1 − cos(kD)) + 2βP k sin(kD) 13.30. The isobaric partition function is Z ∞

1 βmω 2 2 YN (P, T ) = (y − a) dx , exp −βP y − λ −∞ 2

kT NP2 , G(N, P, T ) = N kT ln + NPa − ~ω 2mω 2

which gives L = (∂G/∂P )T = N a − P/mω 2 . As P → 0 the length goes to N a, i.e. N times the equilibrium length of one spring. However, the masses and springs do not form a long-range-ordered lattice since the variance of the neighbor distances grows with n. hxn − x0 i = nhyi = a −

P mω 2

h(xn − x0 )2 i − hxn − x0 i2 = n hy 2 i − hyi2 = na2

kT . mω 2 a2

The heat capacity is CP = N k.

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

17 } 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 case involves 236 configurations, you might try a bitwise calculation with each row represented by an integer between 0 and 2L − 1 and the spins represented by the bits. This is computationally much more efficient since the energy can be determined easily using simple bit rotations and exclusive ors.

13.32. Separating off the ground state energy gives

T =

1 x

x 1

which has eigenvalues λ = 1 ± x. Therefore QN = (1 + x)N + (1 − x)N , which can be expanded

QN =

N X j=0

N !xj N !(−x)j + j!(N − j)! j!(N − j)!

N/2

=

X

k=0

2

n! x2k (2k)!(N − 2k)!

18 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 shown below.

Notice that the height of the peak grows linearly with ln(L).

13.34. The specific heat for the 64 × 64 Ising model is shown below.

19

C Nk 3.0 2.5 2.0 1.5 1.0 0.5 kT 0.0

0

1

2

3

4 J

Chapter 14 14.1. We start with expression (12.1.3) for the partition function QN and carry out summation over σ2 , σ4 , . . .. The resulting expression will consist of 1 2 N factors such as X hσ1 |P|σ2 ihσ2 |P|σ3 i = hσ1 |P2 |σ3 i, σ2

and will be formally similar to the expression we started with. Calling the new transfer operator P0 {K0 }, we clearly get eqn. (1) of the problem.

From the given expression for P{K}, we readily get

2(K +K ) 1 2 + e−2K1 eK2 + e−K2 0 0 2K0 e .

P {K } = e e2(K1 −K2 ) + e−2K1 eK2 + e−K2 Expressing this in a form similar to eqn. (2), we obtain 0

0

0

0

0

0

0

0

eK0 +K1 +K2 = e2K0 {e2(K1 +K2 ) + e−2K1 } eK0 +K1 −K2 = e2K0 {e2(K1 −K2 ) + e−2K1 }, and eK0 −K1 = e2K0 {eK2 + e−K2 }2 which are identical with eqns. (13.2.7) and will lead precisely to eqns. (13.2.8).

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 the functional equation (13.2.11). Now, the right-hand side of this equation is     !2 1/2   0 0 0 0 0 0 0 0     K +K K −K K +K K −K 1 2 1 2 1 2 1 2 0 e 0 1 +e e −e  − ln eK0  + e−2K1 +  .    2  2 2   Substituting from eqns. (13.2.7) with K0 = 0, this becomes −

1 ln[eK2 cosh(2K1 + K2 ) + e−K2 cosh(2K1 − K2 ) + 4 cosh2 K2 + 2 (eK2 cosh(2K1 + K2 ) − e−K2 cosh(2K1 − K2 ))2

1/2

]

1 = − ln[e2K1 cosh(2K2 ) + e−2K1 + {4 cosh2 K2 + (e2K1 sinh(2K2 ))2 }1/2 ].

2 1

2 The left-hand side of the same equation is − ln[eK1 cosh K2 + {e−2K1 + e2K1 sinh2 K2 }1/2 ] 1 = − ln e2K1 cosh2 K2 + e−2K1 + e2K1 sinh2 K2 + 2eK1 cosh K2 2 i {e−2K1 + e2K1 sinh2 K2 }1/2

1 = − ln[e2K1 cosh(2K2 ) + e−2K1 + {4 cosh2 K2 + 4e4K1 cosh2 K2 sinh2 K2 }1/2 ], 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. For this, we have to show that the given function f (K1 , K2 , Λ) satisfies the functional equation (13.2.27). Now, the right-hand side of this equation is " # p Λ0 + Λ02 − K102 K202 1 0 1 − . − K0 + ln 2 4 2π 8 (Λ0 − K10 ) Substituting from eqns. (13.2.25 and 28) with K0 = 0, this becomes " # p Λ − K12 /2Λ + Λ2 − K12 1 π K22 1 K22 − ln + ln − − 4 Λ 8Λ 4 2π 8(Λ − K1 )

K1 1+ Λ " # p 1 K22 2Λ Λ − K12 /2Λ + Λ2 − K12 = ln . − 4 2π 2π 8(Λ − K1 )

Λ + K1 Λ − K1 + Λ Λ " # p Λ + Λ2 − K12 1 K22 = ln − .

2 2π 4(Λ − K1 ) which is precisely f (K1 , K2 , Λ).

14.4. Making the suggested substitution into eqn. (13.2.24), we get   0

N 0/2 Z ∞Z N X 2Λ 2Λ 2Λ  QN = exp  K00 + K10 · s0j s0j+1 − Λ0 · s02 K1 K1 j K1 −∞ j=1 ds 01 . . . ds 0N .

In view of eqns. (13.2.25), with K0 = K2 = 0, we now have

1/2

1/2 1/2 0 00 2Λ π 1/2 2Λ 2π eK0 · = · = = eK0 , say, K1 Λ K1 K1 2Λ 2Λ 2Λ2 K10 · = K1 , and Λ0 · = − K1 = Λ00 , say.

K1 K1 K1

3 The resulting expression for QN , when compared with eqn. (13.2.19), leads to the functional equation 1 1 f (K1 , Λ) = − K000 + f (K1 , Λ00 ), 2 2 where K000 =

1 ln 2

2π K1

and Λ00 =

2Λ2 − K1 .

K1

(1)

(2)

To verify that the function (13.2.32) satisfies the functional equation (1), we note that the right-hand side of this equation is # " p

Λ00 + Λ002 − K12 1 2π 1 + ln − ln 4 K1 4 2π " # p 1 K1 (2Λ2 /K1 − K1 ) + 4Λ4 /K12 − 4Λ2 = ln · 4 2π 2π " # " # p p 2Λ2 − K12 + 2Λ Λ2 − K12 Λ + Λ2 − K12 1 1 = ln , = ln 4 4π 2 2 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 matrix A∗` are determined by the equation a11 − λ a21

a12 = 0. a22 − λ

Clearly, λ1 + λ2 = a11 + a22 ,

while

λ1 λ2 = a11 a22 − a12 a21 .

The eigenfunctions ϕ1 and ϕ2 are given by

λ1 − a11 a21 y1 x1 = = , and ϕ1 = const , where y1 x1 a12 λ1 − a22

λ2 − a11 a21 y2 x2 ϕ2 = const = = , , where y2 x2 a12 λ2 − a22

(1)

(2) (3)

(a) Now, by eqn. (13.3.13a), k1 = u1 x1 + u2 x2 , k2 = u1 y1 + u2 y2 , where k1 y2 − k2 x2 k1 y1 − k2 x1 u1 = , u2 = . x1 y2 − y1 x2 x2 y1 − y2 x1 It follows that the slope of the line u1 = 0 in the (k1 , k2 )-plane is m1 = y2 /x2 , which is given by eqn. (3), while the slope of the line

4 u2 = 0 is m2 = y1 /x1 , which is given by eqn. (2). We readily see that the product m1 m2 =

λ2 − a11 λ1 − a11 λ2 λ1 − a11 (λ2 + λ1 ) + a211 · = a12 a12 a212

Substituting from eqns. (1), we get m1 m2 =

(a11 a22 − a12 a21 ) − a11 (a11 + a22 ) + a211 a21 =− . a212 a12

It follows that the two lines will be mutually perpendicular if and only if a12 = a21 . (b) If a12 or a21 = 0, then by eqns. (1), λ1 = a11 and λ2 = a22 . The stated 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. (13.4.38–40) 1 1 3 1 1 + ε, ∆ ≈ + ε, α ≈ − ε, 2 4 2 2 2 1 1 β ≈ , γ ≈ 1 + ε, δ ≈ 3 + ε, η ≈ 0.

2 2 At the same time, we obtain directly from eqns. (12.4.47, 66 and 67), with d = 4 − ε where 0 < ε << 1, ν≈

−ε 1 2 1 , β= , γ= ' 1 + ε, 2 2 2−ε 2

1 − 16 ε 6−ε 1 δ= =3 ' 3 1 + ε = 3 + ε, η = 0, 2−ε 3 1 − 12 ε

1 1 1 1 1 ν= ' 1 + ε = + ε.

2−ε 2 2 2 4 α'

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. (13.4.46) gives sin(π − πε/2)Γ(3) 1 ≈ ε.

2π{Γ(2)}2 2 Equations (13.4.43–45) then give Sd ≈

4ε · 12 ε 1 ε2 = , 4 n 2n

2 3ε 1 3ε 1 3 γ' 1− ' 1+ ε 1− '1+ − ε, 2−ε n 2 n 2 n

ε 12ε 1 ε 12 α'− 1− '− 1− .

2−ε ε n 2 n η'

5 Next, we obtain

1 1 2(2d − 5)Sd 1 1 (2 − α − γ) = − +O 2 2 d−2 n n2 1 3 ' − ε, 2 2n 4 1 − 6Sd /n + O(1/n2 ) γ δ =1+ =1+ β d − 2 1 − 4(2d − 5)/(d − 2) · (Sd /n) + O(1/n2 )

1 4 1 ' 1 + 2 1 + =1+ 1+O ε = 3 + ε, 2−ε n2 2 γ 1 1 − 6Sd /n + O(1 + n2 ) ν= = 2−η d − 2 1 − 2(4 − d)/d · (Sd /n) + O(1/n2 )

1 3ε 1 3 3ε 1 1 1 ' 1− − 1− ' 1+ ε ' 1+ ε 2−ε n 2 2 n 2 2 n

1 1 6 = + 1− ε.

2 4 n β=

All these results agree with the corresponding ones following from eqns. (13.4.38– 40).

14.10. For zero magnetic field h = K2 = 0, the transformations in equations (12.2.7) and (12.2.8) give 0

e2K1 = cosh(2K1 ), 0

e2K0 = 4e4K0 cosh(2K1 ).

The correlation length in the 1D Ising model with spin coupling K is ξ(K1 ) = [− log tanh(K)]−1 since g(r) = exp(−r/ξ) = tanhr (K); see equations (13.2.32) and (13.2.33). To find the relation between ξ 0 and ξ let’s determine the value of 0

tanh(K10 ) = 0

where eK1 =

0

eK1 − eK1 0 0 , eK1 + eK1

p cosh(2K1 ). Plugging in gives

p −1 cosh(2K1 ) tanh(K10 ) = p p −1 cosh(2K1 ) + cosh(2K1 ) p

=

cosh(2K1 ) −

cosh(2K1 ) − 1 cosh(2K1 ) + 1

cosh2 (K1 ) + sinh2 (K1 ) − 1 = = cosh2 (K1 ) + sinh2 (K1 ) − 1 = tanh2 (K1 ).

sinh K1 cosh K1

2

6 Therefore ξ 0 = ξ/2.

The free energy per spin relation will be satisfied if the partition func0 tions are the same, i.e. eN K0 /2 QN/2 (K10 ) = eN K0 QN (K1 ) where the 1D Ising partition function is given by QN (K1 ) = (2 cosh(K))N . From the transformation relations above, we get N/2

p 0 N/2 (2 cosh(K10 )) eN K0 /2 QN/2 (K10 ) = 2e2K0 cosh(2K1 ) N/2 = 2N/2 eN K0 cosh2 (K1 ) + sinh2 (K1 ) + 1 = 2N/2 eN K0 cosh2 (K1 ) + sinh2 (K1 ) + cosh2 (K1 ) − sinh2 (K1 ) = eN K0 (2 cosh(K1 ))

N

= eN K0 QN (K1 ).

14.11. The l = 3 RG transformation formed by summing over two out of every three spins gives X 0 0 eK0 +K1 s1 s2 = e3K0 +K1 s1t1+K1 s2 t2 +K1 t1 t2 t1 ,t2

= 2e3K0 eK1 cosh(K1 (s1 + s2 )) + e−K1 cosh(K1 (s1 − s2 )) .

This gives

0 0 eK0 +K1 = e3K0 e3K1 + 3e−K1 ,

0 0 eK0 −K1 = e3K0 e−3K1 + 3eK1 .

Requiring this for s2 = s1 and s2 = −s1 gives e3K1 + 3e−K1 , e−3K1 + 3eK1

0 e2K0 = e6K0 e3K1 + 3e−K1 e−3K1 + 3eK1 .

0

e2K1 =

Since the correlation function has the form g(r) = tanhr (K1 ) (see equation (13.2.32)) the correlations in the transformed system will be given by the value of q q e3K1 +3e−K1 e−3K1 +3eK1 − e−3K1 +3eK1 e3K1 +3e−K1 0 q tanh(K1 ) = q 3K −K −3K1 +3eK1 1 1 e +3e + ee3K1 +3e −K1 e−3K1 +3eK1 =

e3K1 +3e−K1 −1 e−3K1 +3eK1 e3K1 +3e−K1 +1 e−3K1 +3eK1 3K1 K1

e − 3e + 3e−K1 − e−3K1 e3K1 + 3eK1 + 3e−K1 + e−3K1 3 K1 e − e−K1 = = tanh3 (K1 ) eK1 + e−K1 =

N/2

7 Therefore ξ 0 = ξ/3.

For the free energy relation to be correct we need to show that 0 eN K0 /3 QN/3 (K10 ) = eN K0 QN (K1 ) where QN (K) = (2 cosh(K))N , so it 0 will suffice to show that eK0 2 cosh(K10 ) = e3K0 (2 cosh(K1 ))3 . q 0 eK0 2 cosh(K10 ) = e3K0 (e3K1 + 3e−K1 ) (e−3K1 + 3eK1 ) s  s −K K 3K −3K 1 1 1 1 + 3e + 3e  e  e + e−3K1 + 3eK1 e3K1 + 3e−K1

= e3K0 e3K1 + 3eK1 + 3e−K1 + e−3K1 = e3K1 (2 cosh(K1 ))3 14.12. The renormalized couplings are 1 (ln cosh(2K1 + K2 ) + ln cosh(2K1 − K2 ) − 2 ln cosh(K2 )) , 2 0 K2 = 2K2 + ln cosh(2K1 + K2 ) − ln cosh(2K1 − K2 ) − 2 ln cosh(K2 ),

K10 =

which have fix point K1∗ = 0.609378, K20 = 0.

The exponents follow from evaluating dK10 = 2 tanh(2K1∗ ) = 1.6786 = 2yt dK1 ∗ K20 = 2 + 2 tanh(2K1∗ ) = 3.6786 = 2yh . dK2 ∗ These give yt = 0.747236, yh = 1.87915, ν = 1.33827, α = −0.67653, β = 0.161734, γ = 2.35306, and δ = 15.549. Note that the heat capacity exponent is negative so the heat capacity is bounded, but with a power law approach to the critical value. The free energy can be calculated numerically by evaluating K00 at each hierarchical level, summing and weighting by the number of spins at each level.

CHAPTER 15 15.1.

(i) We multiply expression (14.1.11) by ∆T , take its average and utilize relations (14.1.14), to obtain (∆T ∆S) = kT . (ii) We multiply expression (14.1.12) by ∆V , take its average and utilize relations (14.1.14), to obtain (∆P ∆V ) = −kT . (iii) We multiply expression (14.1.11) by ∆V , take its average and utilize relations (14.1.14), to obtain

∆S∆V =

∂P ∂T

1 ∂V ∂V kT − .

V = kT V ∂P T ∂T P V

(iv) We multiply expression (14.1.12) by ∆T , take its average and utilize relations (14.1.14), to obtain (∆P ∆T ) = kT 2 Cv−1 (∂P/∂T )v .

15.2. If we choose ∆S and ∆P as our independent variables, then

∂T T ∂T ∂T ∆S + ∆P = ∆P, and ∆T = ∆S + ∂S P ∂P S CP ∂P S

∂V ∂V ∂T ∆V = ∆S + ∆P = ∆S − V κS ∆P. ∂S P ∂P S ∂P S

It follows that −∆T ∆S + ∆P ∆V = −

T (∆S)2 − V κS (∆P )2 , CP

which converts expression (14.1.8) into (14.1.15), leading directly to expressions (14.1.16) for (∆S)2 , (∆P )2 and (∆S∆P ).

For an independent evaluation of these averages, we proceed as in Prob1

2 lem 14.1. From eqns. (14.1.11, 12 and 14), we readily obtain "

2

2 # 2 CV ∂P ∂P 2 2 2 (∆S) = 2 (∆T ) + (∆V ) = k CV + TV κT = kC P , T ∂T V ∂T V " #

2 2 ∂P ∂P 1 kT C V T (∆P )2 = (∆T )2 + 2 2 (∆V )2 = + ∂T V κT V CV ∂T V κT V kT CP kT · = , and CV κT V κS V

∂P 1 CV ∂P 2 (∆S∆P ) = (∆T ) − (∆V )2 = 0.

T ∂T V ∂T V κT V =

15.3. We start with expression (14.1.6) and eliminate ∆S by writing

∆S =

∂S ∂E

∆E + 0

∂S ∂V

2

∂2S ∂ S 2 (∆E) + 2 ∂E 2 0 ∂E∂V 0 2

∂ S 2 ∆E∆V + (∆V ) + . . . . ∂V 2 0

1 ∆V + 2 0

Replacing (∂S/∂E)0 by 1/T and (∂S/∂V )0 by P/T, and retaining terms up to second order only, expression (14.1.6) takes the form

1 ∂θ ∂θ ∂π ∂π p ∝ exp (∆E)2 + 2 or ∆E∆V + (∆V )2 , 2k ∂E 0 ∂V ∂E 0 ∂V 0 where θ = 1/T and π = P/T . The covariance matrix of this distribution is given by

∂θ (∆E)2 (∆E∆V ) − ∂E = k ∂π − ∂E (∆V ∆E) (∆V )2

∂θ − ∂V ∂π − ∂V

−1 .

The evaluation of the inverse here is rather tricky; the interested reader may consult Kubo (1965), problem 6.2, pp. 382–5, where a complete solution, along with the desired results for (∆E)2 , (∆V )2 and (∆E∆V ), is given.

In passing, we note that two of the aforementioned results are also given in eqns. (14.1.14 and 18) of the text; the third may be obtained as follows: multiply (14.1.17) by ∆V , take its average and utilize relations (14.1.14), to get

∂E ∂P (∆E∆V ) = (∆V )2 = T − P kT κT V = kT ∂V ∂T T

V ∂V ∂V T +P . ∂T P ∂P T

3

Figure 1: Figure 2: 15.4. With a given displacement y(x), the overall shape of the string would, on an average, be as shown in Fig. 1. This amounts to a strain, ∆`, in the string given by the expression p p ∆` = x2 + y 2 + (` − x)2 + y 2 − `; the energy Φ associated with this strain is obviously F ∆`. For small y, 2

y y2 F` Φ(y) ≈ F + y2 , = 2x 2(` − x) 2x(` − x) which leads to a probability distribution for y that is Gaussian, with variance kT (∆y)2 = x(` − x).

F` For the second part, we refer to Fig. 2 for which q q p ∆` = x21 + y12 + (x2 − x1 )2 + (y1 − y2 )2 + (` − x2 )2 + y22 − ` y12 (y1 − y2 )2 y22 + + , and hence 2x1 2(x2 − x1 ) 2(` − x2 )

F Φ(y1 , y2 ) ≈ x2 (` − x2 )y12 − 2x1 (` − x2 )y1 y2 + x1 (` − x1 )y22 .

2x1 (x2 − x1 )(` − x2 ) ≈

This leads to a bivariate Gaussian distribution in the variables y1 and y2 , with the covariance matrix !

−1 kT x1 (x2 − x1 )(` − x2 ) x2 (` − x2 ) −x1 (` − x2 ) y12 y1 y2 = −x1 (` − x2 ) x1 (` − x1 ) F y2 y1 y22

kT x1 (` − x1 ) x1 (` − x2 ) = .

F ` x1 (` − x2 ) x2 (` − x2 ) 15.5. The quantity in question here is {(∆NA )2 }1/2 /N̄A = (kT κT /VA )1/2 ;

(1)

see eqn. (14.1.20). Assuming the gas to be an ideal one, the compressibility κT may be taken as 1/(nkT ), where n is the particle density in the system; see eqn. (14.2.12). This reduces (1) to the simple expression (1/N̄A )1/2 .

For this fraction to be 1 per cent, the volume VA of the subsystem must be such that it contains, on an average, 104 particles. At normal temperature and pressure, this volume would be about 3.7 × 10−22 m3 — for instance, a cube of side 7.2 × 10−8 m.

4 15.6. By eqns. (14.3.23) and (14.4.11), and by Note 9 that appears on p. 493 of the text, we have x2 = 2Dt = 2BkTt = kTt/3πηa.

It follows that k = 3πηax2 /Tt. Substituting the given data, we get: k = 1.18 × 10−16 erg K −1 , which may be compared with the accepted value of 1.38 × 10−16 erg K −1 .

15.7. By eqn. (14.4.2), we have

dv 1 1 d 1 hv · Fi = M v · + hv · vi = M hv2 i + hv2 i. dt B 2 dt B Substituting for hv2 i from eqn. (14.4.29) and remembering that B = τ /M , we get

1 3kT 3kT M 3kT 2 −2t/τ 2 −2t/τ − v (0) e + − − v (0) e hv · Fi = τ M B M M 3kT 3kT = = , BM τ which holds at all t. By tacit assumption, the statement hr · Fi = 0 also holds at all t. On the other hand, the quantities hv · F i and hr · F i behave somewhat differently.

First of all, hv · F i = M

v·

dv dt

=

d 1 M hv2 i 2 dt

If the Brownian particle has already attained thermal equilibrium, then hv2 i = 3kT /M and hence hv · F i = 0; if it hasn’t, then

M 3kT − v2 (0) e−2t/τ , hv · F i = τ M which decays exponentially with t. Next, by eqns. (14.4.1 and 5),

dv M M d 2 hr · F i = M r · = − hr · vi = − hr i. dt τ 2τ dt Once again, if the particle has already attained thermal equilibrium, then, by eqn. (14.4.7), hr · F i = −3kT (1 − e−t/τ ) −→ −3kT ; t>>τ

if it hasn’t, then, by eqn. (14.4.31), hr · F i = [−3kT + {3kT − M v2 (0)}e−t/τ ](1 − e−t/τ ) which too approaches −3kT when t becomes much larger than τ .

5 15.8. Integrating eqn. (14.4.14) over t, we get Zt r(t) =

0

v(t0 )dt 0 = v(0)[−τ e−t /τ ]t0 +

Zt



Zt

0

e−t /τ

0

0

0

 eu/τ A(u)du  dt 0 . (1)

0

The remaining integration may be carried out by parts, with the result t

0

 0

(−τ e−t /τ )

Zt

r

u/τ

0

= −τ e−t/τ

Zt

Zt

A(u)du  −

0

0

0

eu/τ A(u)du + τ

0

(−τ e−t /τ ){et /τ A(t0 )}dt 0

Zt

A(t0 )dt 0 .

(2)

0

0

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

r(t) = v(0)τ (1 − e

−t/τ

Zt )+τ

{1 − e(u−t)/τ }A(u)du.

(3)

0

To obtain an expression for hr2 (t)i, we take the square of (3) and average it over an ensemble. The cross-term vanishes on averaging, and we are left with 2

2

2

hr (t)i = v (0)τ (1 − e

−t/τ 2

) +τ

2

Zt Zt 0

{1 − e(u1 −t)/τ }{1 − e(u2 −t)/τ }

0

hA(u1 ) · A(u2 )idu 1 du 2 .

(4)

Noting that the autocorrelation function hA(u1 ) · A(u2 )i, which is the same as the function K(s) of Sec. 14.4, may be treated as a delta function, see the passage from eqn. (14.4.24) to (14.4.25) along with eqns. (14.4.26 and 28), we may write hA(u1 ) · A(u2 )i = Cδ(u2 − u1 ), where C = 6kT /M τ.

The second term in (4) then takes the form 6kT τ M

Zt

{1 − e(u−t)/τ }2 du

0

=

6kT τ 1 t − τ (1 − e−t/τ )(3 − e−t/τ ) .

M 2

Substituting (5) into (4), we obtain eqn. (14.4.31).

(5)

6 15.13. By eqn. (14.6.14), we have in the first case Z∞ w(f ) = 4

2

K(0)e−αs cos(2πf ∗ s) cos(2πfs)ds

0

Z∞ = 2K(0)

2

e−αs [cos{2π(f − f ∗ )s} + cos{2π(f + f ∗ )s}]ds.

0

Using formula (B.41), we get the desired result w(f ) = K(0)

π 1/2 α

2

∗ 2

2

∗ 2

[e−π (f −f ) /α + e−π (f +f ) /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{−(π 2 f 2 /α)}.

On the other hand, if both α and f ∗ → 0, w(f ) tends to be 2K(0)δ(f ).

In either case, eqn. (14.6.16) is satisfied.

In the second case, we get

α α w(f ) = 2K(0) 2 + . α + 4π 2 (f − f ∗ )2 α2 + 4π 2 (f + f ∗ )2 Now, in the limit α → 0 (with f ∗ > 0), w(f ) → 2πK(0)δ{2π(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 + 4π 2 f 2 ). On the other hand, if both α and f ∗ → 0, w(f ) again tends to be 2K(0)δ(f ).

15.14. By eqn. (14.6.14), we get Z∞ w(f ) = 4

K(0)

sin(as) sin(bs) cos(2πfs)ds abs 2

0

2K(0) = ab

Z∞

ds sin(as)[sin{(b − 2πf )s} + sin{(b + 2πf )s}] 2 . s

(1)

0

To evaluate the integral in (1), we use the formula, see Gradshteyn and Ryzhik (1965), Z∞

dx sin(px ) sin(qx ) 2 = x

( pπ/2 qπ/2

if p ≤ q . if q ≤ p

0

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)

7 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 here satisfies eqn. (14.6.16).

15.15. (a) From the defining equation of the variable Y (t), we get u+t Z u+t Z hY (t)i = hy(u1 )y(u2 )idu 1 du 2 2

u

(1)

u

Since y(u) is statistically stationary, we may write Z∞ hy(u1 )y(u2 )i =

w(f ) cos(2πfs)df

(s = u2 − u1 );

(2)

0

see eqn. (14.6.15). Substituting (2) into (1), we get Z∞

2

hY (t)i =

w(f )I(f, t)df , where

(3)

0 u+t Z u+t Z

{cos(2πfu 2 ) cos(2πfu 1 ) + sin(2πfu 2 ) sin(2πfu 1 )}du 1 du2

I(f, t) = u

u

 u+t 2  u+t 2 Z Z = cos(2πfu)du  +  sin(2πfu)du  u

u

1 [sin{2πf (u + t)} − sin(2πfu)]2 + = 4π 2 f 2

[cos(2πfu) − cos{2πf (u + t)}]2 1 = [1 − cos(2πft)], regardess of the initial instantu.

2π 2 f 2 (4) Substituting (4) into (3), we obtain the desired result for hY 2 (t)i.

8 Next, it follows that Z∞

∂ 1 hY 2 (t)i = ∂t π ∂2 hY 2 (t)i = 2 ∂t2

w(f ) sin(2πft)df , and f

(5)

w(f ) cos(2πft)df .

(6)

0 Z∞

0

Taking the sine transform of (5) and the cosine transform of (6), we obtain the other quoted results. Finally, a comparison of eqns. (2) and (6) shows that 1 ∂2 hY 2 (s)i.

2 ds 2

Ky (s) =

(7)

(b) If the variable y(u) is the (x-component of the) velocity of a Brownian particle, with power spectrum (14.6.21), then eqns. (3) and (4) give 2kT τ hx (t)i = 2 π M 2

Z∞

1 − cos(2πft) df f 2 {1 + (2πf τ )2 }

0

4kT τ 2 πM

Z∞

1 − cos(xt/τ ) dx x2 (1 + x2 ) 0

2kT τ 2 t − (1 − e−t/τ ) , = M τ =

(8)

in complete agreement with eqn. (14.4.7) for the quantity hr2 (t)i. We also note that 1 ∂2 2 kT −s/τ hx (s)i = e 2 2 ∂s M

(s > 0),

(9)

which indeed is equal to the autocorrelation function K(s) of the variable vx ; see eqn. (14.6.20).

15.16. First we’ll prove the following lemma.

Lemma: For a given variable x(t), define a complementary function 1 yx (f, T ) = √ T

T /2 Z

x(t)e−2πift dt.

(1)

−T /2

The power spectrum of the variable x(t) is then given by wx (f ) = 2 lim |yx (f, T )|2 .

T →∞

(2)

9 Proof: From (1), it readily follows that 1 |yx (f, T )| = = T 2

T /2 Z T /2 Z

x(t1 )x(t2 )e2πif (t2 −t1 ) dt 1 dt 2 .

−T /2 −T /2

Changing over to the variables S and s, as defined in eqns. (14.4.23), we get

Z Z 1 1 1 |yx (f, T )|2 = x S − s x S + s cos(2πfs)dSds. T 2 2 Integrating over S and letting T → ∞ amounts to taking an ensemble

average of the quantity x S − 21 s x S + 12 s ; this reduces the above expression to Z∞ Kx (s) cos(2πfs)ds −∞

which, by eqn. (14.6.14), is equal to 12 wx (f ). Hence the lemma.

We now proceed to establish the stated relation between the power spectra wv (f ) and wA (f ). For this we refer to eqn. (14.4.5) for the variable A(t) and construct its complementary function T /2 Z

1 yA (f, T ) = √ T

dv v + dt τ

e−2πift dt.

(3)

−T /2

The first part here gives  1  /2 √ ve−2πift T−T /2 − T

T /2 Z

  v(−2πif )e−2πift dt  .

−T /2

Equation (3) then becomes

1 T T 1 −πift πifT e −v − e + yA (f, T ) = √ v + 2πif yv (f, T ).

2 2 τ T Since the variable v(t) is bounded, the limit T → ∞ gives |yA (f, T )|2 ≈

1 + 2πif τ

2

|yv (f, T )|2 .

Using lemma (2), we finally get

1 wA (f ) = 2 + (2πf )2 wv (f ), τ

(4)

(5)

10 which is, in fact, the desired result.

Now, by eqn. (14.6.21), wv (f ) =

1 12kT τ .

M 1 + (2πf τ )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 practical purposes, the autocorrelation function KA (s) may be taken as Cδ(s), with C = 6kT /M τ ; see eqns. (14.4.26 and 28).

15.17. (a) Using eqn. (14.4.14), we construct the quantity v(t) · v(t + s) and average it over an ensemble. The cross-term vanishes on averaging, and we are left with 

 Zt Zt+s hv(t)·v(t+s)i = e−2(t+s)/τ v2 (0) + e(u1 +u2 )τ hA(u1 ) · A(u2 )idu 1 du 2  .

0

0

(1) In view of the argument leading from eqn. (14.4.24) to (14.4.25), we may replace the function hA(u1 ) · A(u2 )i by the singular expression Cδ(u2 − u1 ), where C = 6kT /M τ . At the same time, we observe that the integral Zt e

(u1 +u2 )/τ

( e2u2 /τ δ(u1 − u2 )du 1 = 0

if 0 < u2 < t otherwise.

0

The double integral in (1) is then equal to Zt

τ Ce 2u2 /τ du 2 = C (e2t/τ − 1) 2

if s > 0, and

0 Zt+s

τ Ce 2u2 /τ du 2 = C (e2(t+s)/τ − 1) 2

if s < 0.

0

Substituting these results into (1), we obtain eqns. (14.7.7) and (14.7.8).

Equation (14.7.9) follows straightforwardly. (b) To evaluate hr2 (t)i, we write eqn. (14.7.9) in the form

3kT 3kT −|s|/τ 2 Kv (s) = v (0) − e−2S/τ + e , M M

(2)

where S = 21 (u1 + u2 ) and s = (u2 − u1 ). Substituting (2) into

11 eqn. (14.7.6), we get   Zt/2 Zt 3kT   −2S/τ · 4SdS + e−2S/τ · 4(t − S)dS  hr2 (t)i = v2 (0) −  e M

0

 +

3kT   M

Zt/2

Z2S

2 0

=

t/2

e−2s/τ dsdS +

0



 e−2s/τ dsdS 

2 t/2

3kT M

v2 (0) −

2(t−S) Z

Zt

0

· τ 2 (1 − e−t/τ )2 +

3kT · 2τ {t − τ (1 − e−t/r )}, M

which is the same as expression (14.4.31). Note that the second part of this result is identical with expression (14.4.7) that pertains to a stationary ensemble.

15.18. Using equation (15.3.37), the response function is Z ∞ χ̃vx (ω) =

χvx (s)eiωs ds

0

which has imaginary part χ00vx (ω) =

ω ω 2 − ω02 1 .

M (ω 2 − ω02 )2 + γ 2 ω 2

The correlation function is 0

Z t0

Z t

Gvx (t − t ) =

ds −∞

ds0 χvx (t − s)χxx (t0 − s0 ) hF (s)F (s0 )i

−∞

Using hF (s)F (s0 )i = 2γM kT δ(s − s0 ) and Fourier transforming gives

ω 2 − ω02 2kT .

Svx (ω) = M (ω 2 − ω02 )2 + γ 2 ω 2 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: The correlation function relation should read: GAB (t) = GBA (−t − iβ~).

12 15.20. Since we need to evaluate h[A(t), B(0)]i, we need to relate hB(0)A(t)i to hA(t)B(0)i.

1 hB(0)A(t)i = Tr BeiHt/~ Ae−iHt/~ e−βH Q

1 = Tr eiHt/~+βH Ae−iHt/~−βH Be−βH Q = hA(t − iβ~)B(0)i Now equation (15.6.34) can be evaluated as Z 1 χ̂00AB (ω) = (hA(t)B(0)i − hA(t − iβ~)B(0)i) eiωt dt 2~ Z

1 = hA(t)B(0)ieiωt − hA(t − iβ~)B(0)ieiω(t−iβ~) e−β~ω dt 2~

1 1 − e−β~ω SAB (ω) = 2~ 15.21. Since hB(0)A(t)i = hA(t − iβ~)B(0)i,

hA(t)B(0) − B(0)A(t)i ≈ iβ~

dA B dt

in the limit ~ → 0. Therefore,

Z Z 1 dA iβ~ χ00AB (ω) = hA(t)B(0) − B(0)A(t)ieiωt dt ≈ B(0) eiωt dt 2~ 2~ dt Z βω ω = hA(t)B(0)i eiωt dt = SAB (ω) 2 2kT 15.22. The self–diffusion term can be written Z Sself (ω) = he−ik·(r(t)−r(0)) ieiωt dt Z 2 = eiωt−Dk |t| dt, using the diffusion relation h(x(t) − x(0))2 i = 2D|t|. Integrating gives Sself (ω) =

2Dk 2 ; ω 2 + (Dk 2 )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 and the width of the Rayleigh peak is ∆ω = DT k 2 = 7 × 106 s−1 . The location of the sound peak is at ω = ck = 2.4 × 109 m−1 is well separated from the Rayleigh peak.

13 15.24. The Raman peak has ~ω = 0.05 eV ' 2kT at room temperature so the peaks are not symmetric. Since Γ ∼ 1012 s−1 and ω ∼ 8 × 1013 s−1 , the Raman peak is well resolved.

15.25. While the system Hamiltonian H(q, p) does not evolve deterministically due to the coupling to the reservoir, the full Hamiltonian H (q, p, q 0 , p0 ) does! Therefore the total work on the system plus reservoir done by the external force as λ evolves from 0 to 1 between t = 0 and t = τ is W = H (q(τ ), p(τ ), q 0 (τ ), p0 (τ )) − H (q(0), p(0), q 0 (0), p0 (0)). Using the same argument used in Section 15.8 for the isolated system we get e−βW =

Y1 = e−β∆F , Y0

where Yλ and Fλ are the partition function and free energy of the full system plus reservoir. Now, if we can invoke weak coupling, the partition function factorizes so ZQ1 Y1 = = e−β∆F , Y0 ZQ0 where F is the free energy change of the subsystem alone.

15.26. Note that the partition functions and free energies associated with classical Hamiltonians H0 and H1 are given by Z Q0 = exp (−βF0 ) = e−βH0 (p,q) dΓ, Z Q1 = exp (−βF1 ) = e−βH1 (p,q) dΓ, and that the equilibrium phase space density associated with H0 is given by ρ0 = exp(−βH0 )/Q0 . Therefore the ratio Q1 /Q0 can be written Z −βH0 (p,q) Q1 e = exp (−β(F1 − F0 )) = e−β(H1 (p,q)−H0 (p,q)) dΓ Q0 Q0 = hexp (−β(H1 − H0 ))i0 .

Taking a logarithm of both sides yields D E F1 − F0 = −kT ln e−β(H1 −H0 ) .

0

15.27. Let’s show that (15.8.18) and (15.8.15) give (15.8.2) for the special case that τ = 2 so that λ(0) = 0 and λ(2) = 1. The general case then follows straightforwardly. Using (15.8.14) for this special case gives X 1 e−βEλ(0) (q0 ) P1 (q 0 → q 1 )P2 (q 1 → q 2 ) e−βW = Qλ(0) q ,q ,q 0

1

2

× e−βEλ(1) (q0 ) e+βEλ(0) (q0 ) × e−βEλ(2) (q1 ) e+βEλ(1) (q1 )

14 Now using detailed balance (15.8.15) gives e−βW =

1

X

Qλ(0) q ,q ,q 0

1

e−βEλ(0) (q0 ) P1 (q 1 → q 0 )P2 (q 2 → q 1 ) 2

× e−βEλ(1) (q1 ) e+βEλ(1) (q0 ) × e−βEλ(2) (q2 ) e+βEλ(2) (q1 ) × e−βEλ(1) (q0 ) e+βEλ(0) (q0 ) × e−βEλ(2) (q1 ) e+βEλ(1) (q1 ) Now, there are a many Boltzmann factor cancellations e−βW =

1

X

Qλ(0) q ,q ,q 0

1

P1 (q 1 → q 0 )P2 (q 2 → q 1 ) 2

× e−βEλ(2) (q2 ) and the summations over q 0 and q 1 both give unity so e−βW =

1

X

Qλ(0) q

e−βEλ(2) (q2 ) =

2

Qλ(2) = e−β∆F Qλ(0)

15.28. Let’s show this for the special case that τ = 2 so that λ(0) = 0 and λ(2) = 1. The general case then follows straightforwardly. Equation (15.8.16) gives P (q 0 , q 1 , q 2 ) =

e−βEλ(0) (q0 ) P1 (q 0 → q 1 )P2 (q 1 → q 2 ) Q0

Now using detailed balance (15.8.15) gives P (q 0 , q 1 , q 2 ) =

e−βEλ(0) (q0 ) P1 (q 1 → q 0 )P2 (q 2 → q 1 ) Q0

× e−βEλ(1) (q1 ) e+βEλ(1) (q0 ) × e−βEλ(2) (q2 ) e+βEλ(2) (q1 ) Q1 P (q 2 , q 1 , q 0 )eβ(δW1 +δW2 ) = Q0 = P (q 2 , q 1 , q 0 )eβW −β∆F 15.29. The Jarzynski case is trivial since e−βW = e−β∆F so eβ∆F −βW = 1.

15 The Crooks’ case is also easily shown: Z −β∆F +βW e = dβW PF (βW )e−β∆F +βW Z

= dβW e−β∆F +βW PR (−βW )e−β∆F +βW Z = dβW PR (−βW ) = 1.

Chapter 16 16.1. Here is a C code snippet for a pseudorandom number generator based on the L’Ecuyer prime number linear congruential generator discussed in Appendix 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 hxi ≈ 0.5 ± 1/(12 N ) and hx2 i − hxi2 ≈ 1/12.

16.2 Here is a code snippet for generating gaussian pseudorandom numbers based 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 pseudorandom number. √ For a sequence of N numbers, one should test that hxi ≈ 0.0 ± 1/ N and hx2 i ≈ 1.0. The reader should also determine the expected uncertainty in the value of the variance for N numbers. The histogram of points for pairs of gaussian random numbers should be centered at 0, be isotropic, and have variance hx2 + y 2 i = 2.

1

2 16.3 First note that the sum of two gaussian random distrubutions is also gaussian,

s2 s2 exp − 2σ22 exp − 2σ12 1 2 , P2 (s2 ) = p , P1 (s1 ) = p 2πσ12 2πσ22 Z Z P (S = s1 + s2 ) = δ(S − s1 − s2 )P1 (s1 )P2 (s2 )ds1 ds2 ,

2 exp − 2(σ2S+σ2 ) 1 2 .

P (S) = p 2π(σ12 + σ22 ) Iterating the equation defining the correlated random numbers gives sk = (1 − α)

∞ X

αj rk−j .

j=0

This implies that the s’s are also gaussian. The averages hsk i are clearly zero and the variance is given by hs2k i = (1 − α)2

∞ X ∞ X

αi+j hrk−i rk−j i,

i=0 j=0

which is easily evaluated to give hs2k i = (1 − α)2

∞ X

α2j =

j=0

(1 − α)2 1−α = .

1 − α2 1+α

The correlations are then given by hsk sk−l i = α|l|

1−α .

1+α

16.4 A Monte Carlo Sweep of an ordered list of N particles x0 < 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 particle particle 0 as particle N − 1 shifted left by L, and particle to the right of

3 particle N − 1 as particle 0 shifted to the right by L. The random step size dx is typically chosen on the order of L/(DN ) − 1 but must be less than 2 to avoid particles getting out of order.

16.5 Here is a C code snippet for a Monte Carlo of a two-dimensional system of 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 probability exp(−βmg∆y), and to reject moves that go outside the vertical boundaries. At low density and small βmgLy , the density will be proportional to exp(−βmgy). Large βmgLy will result in an interface with a low density phase above a high density phase.

4 16.7 Here is a C code snippet for one time step for Lennard-Jones particles in a two dimensional LX×LY box with periodic boundary conditions. The arrays x1 and x0 store the current and previous positions of the n particles respectively. // 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; }

5 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 repulsive forces with the top and bottom walls. The average kinetic energy per particle will be independent of the position in the box. At low density and small βmgLy , the density will be proportional to exp(−βmgy). Large βmgLy will result in an interface with a low density phase above a high density phase.

16.9 Each Monte Carlo step involves determining the energy change of a spin flip with is proportional to ∆ = si (si+1 + si−1 ) using periodic boundary conditions. If ∆ ≤ 0 flip the spin. Otherwise ∆ = +2, so flip the spin with probability exp(−4K). Due to the periodic boundary conditions the correlation function will also be periodic. You can generalize the calculation of the correlation function in section 13.2 for a finite periodic lattice to show that the zero field correlation function is of form " N −|i−j| # |i−j| λ2 λ2 1 + , hsi sj i = N λ1 λ1 1 + λλ12 so the correlations are minimized halfway across the lattice.

6 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) {

7 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 code posted at www.elsevierdirect.com to calculate the energy distribution at K = 0.4, Kc = 0.4406868, 0.5. Here is a plot. The horizontal axis is the energy above the ground state in units of 4J and the vertical axis is the probability for each energy.

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).

8 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;

9 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 } }