[R.K.pathria] Solutions of Statistical Mechanics

Instructor’s Manual Containing Solutions to Over 280 Problems Selected from

Statistical Mechanics Third Edition By

R. K. Pathria and Paul D. Beale

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Preface This instructor’s manual for the third edition of Statistical Mechanics is 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 edition. The result is a manual containing solutions to some 280 problems selected from the third 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

iii

Chapter 1

1.1. (a) We expand the quantity ln Ω(0) (E1 ) as a Taylor series in the variable ¯1 ) and get (E1 − E ln Ω(0) (E1 ) ≡ lnΩ1 (E1 ) + ln Ω2 (E2 ) (E2 = E (0) − E1 ) ¯1 ) + ln Ω2 (E ¯2 )}+ = {ln Ω1 (E   ∂ ln Ω1 (E1 ) ∂ ln Ω2 (E2 ) ∂E2 ¯1 )+ + (E1 − E ∂E1 ∂E2 ∂E1 E1 =E¯1 (  2 ) 1 ∂ 2 ln Ω1 (E1 ) ∂ 2 ln Ω2 (E2 ) ∂E2 ¯1 )2 + · · · . + (E1 − E 2 ∂E12 ∂E22 ∂E1 ¯ E1 =E1

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    
1 1 ∂β1 ∂B2 1 ¯1 2 = − 1 ¯ 1 )2 , + E1 − E + (E1 −E 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 ¯1 ), see that the function Ω0 (E1 ) is a Gaussian in the variable (E1 − E 2 with variance kT (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 32 N1 k 2 T 2 ; 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) 1

2 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 = 21 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 Σ1 (ε∗ ) =

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

3 Substituting V = L3 and S = 6L2 , we obtain eqns. (1.4.15 and 16). The expression for T now follows straightforwardly; we get         1 ∂ 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 1 ∂S. ∂f 1 ∂S = VN · and E = EN · . V ∂V N,E ∂v ε N ∂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.

4 (b) The expression now in question is given by eqn. (1.5.4). With N1 = αN and N2 = (1 − α)N , where N = N1 + N2 (which is fixed), 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 19 of the text, we get (∆S)v =

3 Nk ln(Tf /Ti ). 2

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 . 2 3

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)

5 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 ϕ, ˙ where the various symbols have their usual meanings. The energy of rotation is given by E=

1 m(ρϕ) ˙ 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 6

7 space may be denoted by the angles θ and ϕ, the conjugate variables being pθ = mr 2 θ˙ and pϕ = mr 2 sin2 θϕ. ˙ The energy of rotation is given by p2ϕ M2 1 p2θ ˙ 2 + 1 m(r sin θϕ) + = m(rθ) ˙ 2= , 2 2 2 2 2 2mr 2I 2mr sin 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 .

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 3 1 1 1 h 2 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 = 21 mω 2 A2 and τ = 2π/ω. The trajectory in the (θ, L)-plane is given by the equation θ2 L2 + = 1, 2 (A/`) (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 68–69 of the text, the number of microstates for a given energy E turns out to be  Ω(E) = (R + N − 1)!/ R!(N − 1)!,

R=

 1 E − N ~ω /~ω. 2

(1)

8 For R  N , we obtain the asymptotic result Ω(E) ≈ RN −1 / (N − 1)!, where R ≈ E / ~ω.

(3.8.25a)

The corresponding expression for Γ(E; ∆) would be E N −1 ∆ (E / ~ω)N −1 ∆ . · = (N − 1)! ~ω (N − 1)!(~ω)N

Γ(E; ∆) ≈

(1)

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

(dq i dp i ), with

i=1

N  X 1 i=1

2

kq 2i

1 2 + p 2m i

 ≤ E.

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

where ω =

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

ri

i=1

N 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

−N

(4π)

Z∞ dV 3N =

0

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

3N 0 Y i=1

dq i dp i = V N

Z

0

N Y i=1


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

9 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 {nr } = N ln N − nr ln nr N N r   X n∗ n∗ n∗ r = −k ln r = −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

10

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

12 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 U R = , while T e−U/kT d3N q N,V n o ¯ U ¯ k ln Z = k ln V¯ N e−U /kT = Nk ln V¯ − . 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 + U P For the second part of the question, we write U (q) = u(rij ), so that 

kT

∂ ln Z ∂T



=

kT

R

i 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 = − (T − T ) and v = + (Tc − T ). c max 2kT 2kT kT 2 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. 12.2 and 12.3. 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 .

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

a a (1 + π), v = 2b(1 + ψ), T = (1 + t). 4e2 b2 4bk

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 3 2 2 π ≈ t− 2ψ − 2ψ + ψ +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 ≈ − 23 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. 12.2, 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

(n > 1).

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

n+1 4n(n − 1)n−1 a 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

(1)

102 Carrying out the usual expansions and retaining only the most important terms, we get π≈

2n n(n + 1)2 3 n(n + 1) t− ψ − tψ. n−1 12 n−1

It follows that  (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 qJL2 + µBL . 2

Using the Stirling approximation (B.29), we get   1 2 f (L) ≈ −Np ln p − Nq ln q + βN qJL + 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. (12.5.10), we see that the value, L∗ , of L, that ¯ maximizes the function f (L) is identical with L. 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

103 whence S ≈ −Nk (p∗ ln p ∗ + q ∗ ln q ∗ ). 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 in Section 11.4 and remembering ¯ we obtain from eqns. (2) and (4) the desired that L∗ is identical with L, results for the quantities P and v pertaining to a lattice gas. For the critical constants of the gas, we first note from eqn. (12.5.13) that Tc = qJ /k, i.e. qε0 /4k; the other constants then follow from the stated ¯ = 0. results for P and v, with B = 0 and L 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 12.4, 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 12.4 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.

104 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 12.3. In the spirit of the mean field theory, we replace each of the sj by ¯s, which modifies the foregoing expression to −gµB s0 · Heff , where Heff = H + H0

(H 0 = 2qJ ¯s/gµB ).

(1)

We now apply the theory of Sec. 3.9. Taking H (and hence ¯s) 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) H + 2 2z . (3) µz ≈ 3kT g µB The net magnetization, per unit volume, of the system is now given by the formula M = n¯ µz , 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 µ ¯z 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),

105 which yields the desired result: Bs (x) =

s+1 (s + 1){s2 + (s + 1)2 } 3 x + ... x− 3s 90s3

(x  1).

(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 µ ¯z is possible; in fact, for T . Tc , √ µz ≈ (gµB s/ b)(1 − T /Tc )1/2 . (10) ¯ 0 is then given by The long-range order L √ 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     3Tc 2sqJ −1 −1 = 1 − s exp − . L0 ≈ 1 − s exp − kT (s + 1)T

(12)

(13)

For s = 1/2, expressions (11) and (13) reduce precisely to eqns. (12.5.14 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 12.7, 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 ¯sb − ¯sa , 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 ¯sa − ¯sb . gµB gµB

(1b)

106 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,   4qJ 1 g 2 µ2B s(s + 1) 4q 0 J 0 1 H − 2 2 Mb − 2 2 Ma , Ma ≡ n(gµB ¯sa ) ≈ n · 2 2 3kT ng µB ng µB   2 2 0 0 1 4qJ 1 g µB s(s + 1) 4q J H − 2 2 Ma − 2 2 Mb . Mb ≡ n(gµB ¯sb ) ≈ n · 2 2 3kT ng µB ng µB Adding these two results, we obtain for the total magnetization of the lattice   C 2qJ 2q 0 J 0 0 0 M ≈ {H − (γ + γ)M} , γ= ; (2) γ = 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   1 4s(q 0 J 0 − qJ )M ∗ M ∗ = n(gµB s)Bs (x0 ) x0 = . (4) 2 n gµB kT We now employ expansion (7) of the preceding problem and get M∗ =

TN ∗ M − T

where

b
2 1 2 n gµB s



TN ∗ M T

3 + ...,

(5)

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 N´eel 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 =

107 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 qN u(σ 0 , σ 00 ){f (σ 0 )δf (σ 00 ) + f (σ 00 )δf (σ 0 )} 2 0 00 σ ,σ X + NkT {1 + ln f (σ 0 )}δf (σ 0 ) = 0,

δ(E − TS ) =

σ0

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 X qN 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 u(σ, σ )f (σ ) , (1) f (σ) = C exp −βq 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)

¯ 0 σ), the quantity f (1) − f (−1) becomes precisely Writing f (σ) = 21 (1 + L ¯ 0 , and eqn. (2) takes the form equal to L 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)

¯0. Dividing (5) by (4), we obtain the Weiss eqn. (12.5.11) for L

108 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 = qN ε11 x2A + 2ε12 xA xB + ε22 x2B − 2εx2A X 2 ε = (ε11 + ε22 ) − ε12 . 2 2

E=

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) = −2qN εx2A 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 qεX 2X = 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 1 xA 3 4xA Tc0 1 4xA Tc0 X − 2 X + ... = X− X 3 + . . . , i.e. xB xB T 3 T " #   3 1 4xA xB Tc0 1 4xA xB Tc0 −1 X − 3 − 3xA xB X 3 + . . . = 0. xB T 3xB T

109 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. 12.14. (a) Setting N ++ + N −− + N +− = 2

1 2 qN ,

we find that, in equilibrium,

γ = 1/(1 + sL ). So, in general, it may be written as 1/(1 + sL2 ). P (b) As in Problem 12.4, 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 1+L (1 − s2 )L + µB . ln = β qJ 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 12.4, 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. 12.5; see the derivation leading to eqn. (12.5.18) 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. 12.5 and depicted in Fig. 12.8.

110 12.15. Using eqn. (12.6.30), we get Z ∞ Z γc S∞ − Sc C(T )dT 1 1 γ sech 2 γdγ = = q Nk Nk Tc T 2 0 1 = q(γc tanh γc − ln cosh γc ). 2 Next, we use eqn. (12.6.11) and obtain " (   2 )#  1 1 q 1 1 1 S∞ − Sc = q ln + ln 1 − Nk 2 2 q−2 q−1 2 q−1 =

1 q 1 {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 1 q2 1 1 q(q − 2) Sc = 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 1 1 2 q q − +O 1 + O − + O − ln 2 + 2 2 q q 4 q q q2   1 = ln 2 + O , q which tends to the limit ln 2 as q → ∞. 12.16. Using eqn. (12.6.14), we get  ¯  ¯ ∂M N µ2 ∂ L 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 (12.6.8) 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 . 0 2 − (q − 1){tanh(α + γ) − tanh(α0 − γ)}

(3)

111 To study the critical behavior of χ0 , we let α0 → 0 and γ → γc . Using eqn. (12.6.11), we see that, while the first two factors of expression (3) reduce to 1 2N µ2 q 4N µ2 = , (4) kT c 1 + exp(−2γc ) kT c q − 1 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 ) 2

− sech 2 γc tanh γc (−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. (12.6.13), 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 − 2)γc =

1 (q − 2) ln 2



q q−2

 =1+O

  1 ; q

eqns. (7) then reduce to eqns. (12.5.22) of the Bragg-Williams approximation.

112 12.19. We refer to the solutions to Problems 12.4 and 12.5, 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 m0 1 + m0 1 Tc 2 1 1 − m20 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 2 ≈ − ln 2 + t m0 + m40 . 2 12

=

Comparing this expression with eqn. (12.9.5), we infer that in the BraggWilliams approximation r1 = 1/2 and s0 = 1/12. Now, substituting these values of r1 and s0 into eqns. (12.9.4, 9–11 and 15), we see that the corresponding eqns. (12.5.14, 22, 24 and 18) are readily verified. A similar calculation under the Bethe approximation is somewhat tedious; the answer, nevertheless, is r1 =

q (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)

113 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 for r > 0 (and s ≤ − 3ur ) it is maximum at m0 = ± A− and minimum √ at m0 = ± A+ . We, therefore, have to contend only with A+ . The foregoing observations should suffice to prove statements (a), (e), (f) and (g) of this problem. For the rest, we note that the function ψ0 (m0 ) may be written as
1
ψ0 (m0 ) = q + rm 20 + sm 40 + um 60 − m0 2rm 0 + 4sm 30 + 6um 50 (4a) 4
1 2 4 (4b) = q + m0 r − um 0 ; 2 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 √ −s + s2 − 3ur 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.

114 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   1 ∂m . ≈ 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  ψ∼

∂G(s) ∂π



2−α−∆

∼ |t|

g

0

t



π |t|∆

 .

(2)

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. (c) For the isothermal compressibility of the system, we have from (1)       ∂ψ π 1 ∂v ∼ ∼ |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 ∂P ∂P 1 ∂v =− = κT . αP = 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

115 relation CP = VT (dP /dT )2 κT , established in Problem 13.25, we infer that CP ∼ κT ∼ |t|−γ . For CV , we go back to the given expression for G(s) and write   (s)   ∂G π S (s) = − ∼ |t|1−α × a universal function of ∂T |t|∆ P   ψ ∼ |t|1−α × a universal function of . |t|β It then follows that   (s)   ∂S ψ (s) −α . CV = T ∼ |t| × a universal function of ∂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 13.25, we infer that κS ∼ CV ∼ |t|−α . Finally, for the latent heat of vaporization `, we invoke the Clapeyron equation, dPσ ` = , 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 2 )(t + bm 2 )Θ−1

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 (12.10.22) is readily verified. 12.24. For ri 6= rj , eqns. (12.11.22 and 26) give (∇2 − ξ −2 )g(r) = 0.

(1)

116 Now, if g(r) is a function of r only then dg dg r dg  x1 xd  ∇g = , whence ∇r = = ,..., dr dr r dr r r     X  2 d d  X ∂ dg xi d g 1 dg 1 x2i dg 1 ∇ · ∇g = = + − ∂xi dr r dr r dr r2 r dr 2 r i=1 i=1 =

dg 1 d2 g + (d − 1). dr r dr 2

(2)

Substituting (2) into (1), we obtain the desired differential equation — of which (12.11.26) is the exact solution. Substituting (12.11.27) into the left-hand side of the given differential equation, we get    1 1 e−r/ξ + . . . + (. . .) − ; const. (d−1)/2 ξ2 ξ2 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 (12.11.28) instead, we get   1 (2 − d)(1 − d) d − 1 2 − d 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. 12.10, 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/∆ ). 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.

117 12.27. By definition, σ ≡ ψA (t) ∼ |t|µ

(t . 0).

(1)

By an argument similar to the one that led to eqn. (12.12.14), we get ψA (t) ∼ A−1 ∼ ξ −(d−1) ,

(2)

where A is the “area of a typical domain in the liquid-vapor interface”. Now, for a scalar model (n = 1), to which the liquid-vapor transition belongs, ξ ∼ |t|−ν . Equations (1) and (2) then lead to the desired result µ = (d − 1)ν = (2 − α)(d − 1)/d.

Chapter 13 ¯ = (N ¯+ − N ¯− )µ and N ¯+ + N ¯− = N , we readily see that 13.1. Since M  ¯  ¯± = 1 N 1 ± M = 1 N P (β, B) ± sinh x N (x = βµB). 2 Nµ 2 P (β, B)

(1)

Next, comparing eqns. (12.3.19) and (13.2.12), and keeping in mind that q = 2, we get −4βJ P 2 (β, B) − sinh2 x ¯+ − N ¯++ = N e N =N . 2D(β, B) 2D(β, B)

(2)

It follows from eqns. (1) and (2) that ¯++ = (N/2D){(P + sinh x)(P + cosh x) − (P 2 − sinh2 x)} N = (N/2D){(P + sinh x)(cosh x + sinh x)},

(3)

which is the desired result. Equation (12.3.17) now gives ¯+− = 2(N ¯+ − N ¯++ ) and N ¯−− = N − 2N ¯+ + N ¯++ = N ¯++ − (M ¯ /µ). N ¯+− is just twice the expression (2). The number N ¯−− is The number N given by ¯−− = (N/2D){(P + sinh x)(cosh x + sinh x) − sinh x · 2(P + cosh x)} N = (N/2D){(P − sinh x)(cosh x − sinh x)},

(4)

which is the desired result. It is straightforward to check that the sum ¯++ +N ¯−− = (N/D)(P cosh x+sinh2 x) = (N/D){P cosh x+(P 2 −e−4βJ )}; N ¯+− , one obtains the expected result N . Finally, the product adding N ¯++ N ¯−− = (N/2D)2 {P 2 − sinh2 x} = (N/2D)2 e−4βJ . N Equation (12.6.22) is now readily verified. 118

119 13.2. (a) In view of eqn. (12.3.18), the quantity (N++ + N−− − N+− ) that appears in the
Hamiltonian (12.3.19) of the lattice may be written as 1 2 qN − 2N+− . The partition function (12.3.20) then assumes the form stated here. (b) A complete solution to this problem can be found in the first 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 13.2, we now have   1 1 ln gN (N+ , N+− ) ≈ qN ln qN − N++ ln N++ − N−− ln N−− 2 2   1 N+− + (q − 1){N+ ln N+ + N− ln N− − N ln N }, − N+− ln 2 (1) where, by virtue of eqn. (12.3.17), 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+−

120 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. (12.6.22) of that treatment. As for eqn. (4), we go back to eqns. (12.6.4 and 8) of the Bethe approximation, whereby  q 0 ¯+ 0 N 2α cosh(α + α + γ) =e = e2α+2α q/(q−1) 0 ¯ cosh(α + α − γ) N− and to eqn. (12.6.21), whereby ¯−− 0 N = e−4(α+α ) . ¯ N++ 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. (13.2.8 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),   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. (13.2.21 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 Heff

X 1 = − µ2 s NL2 − J σi σi+1 2 i

! L=N

−1

X i

σi

.

121 The first term here is equivalent to an infinite-range interaction of the type considered in Problem 12.6, 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 12.4. 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. 13.2, the partition function of this system can be written down in analogy with eqn. (13.2.10): h  1/2 i 1 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 = σi σi+2 =

1 ∂ sinh(βJ1 ) ln Q =  1/2 , and 2 βN ∂J1 −4βJ 2 + sinh (βJ ) e 1

1 ∂ ln Q = 1− βN ∂J2 

e−4βJ2

2e−4βJ2 h  1/2 i . 1/2 + sinh2 (βJ1 ) 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. 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) =    1 e−2K1 −K2 e2K1 −K2 1 e−2K1 +K2 1 1 e2K1 +K2

122 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. 13.2, 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. 12.4, 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 .

123 (ii) The pressure P = −(A/N ) − J + µB; using eqn. (13.2.11), 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. (13.2.13), 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 ≈ ln(y + 1), kT

1 y ≈ . v y+1

Moreover, in this limit y ' z  1. One may, therefore, write P 1 ≈ y, ≈ y kT v

and hence

Pv ≈ 1. kT

At low temperatures, η becomes very small and y very large — with the result that P η2 ≈ ln y + , kT y

1 η2 ≈1− . v y

(b) For a hard-core lattice gas (y → 0, η → ∞), we get √ √   P 1 + 1 + 4z 1 + 4z − 1 √ . = 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).

124 13.10. Applying eqn. (12.11.11) 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. (13.3.17) 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 χ0 1 + tanh(βJ) cosh(βJ) + sinh(βJ) = = = e2βJ , 2 N βµ 1 − tanh(βJ) cosh(βJ) − sinh(βJ) in agreement with eqn. (13.2.14). 2 2 2 2 σm+1 . . . σn−1 , which is identi. . . σ`−1 13.11. We introduce an extra factor σk+1 cally 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. (13.2.31), 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 .

125 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. (13.4.23). Finally, we replace cos2 ϕ by sin2 ϕ (without affecting the value of the integral) and recover eqn. (13.4.22). We know that this model is singular at κ = 1. A close look at the 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 ).

126 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. (13.4.18). 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. (13.4.28), is −Kc coth(2Kc ) = − 2Kc . The second, by eqn. (13.4.22), 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.

127 13.16. Expanding around K = Kc , we get sinh(2K) = sinh(2Kc ) + 2 cosh(2Kc )(K − Kc ) + 2 sinh(2Kc )(K − Kc )2 + . . . √ (1) = 1 + 2 2(K − Kc ) + 2(K − Kc )2 + . . . . 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. 12.4 and utilizing a result obtained in Problem 13.15, we have for a two-dimensional lattice gas (q = 4)     2J 2G Pc 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 (13.5.31) assumes the form Z∞ W1 (ϕ) =

e−(1+ϕ)x I0 (x)dx =

1 1 = 2 . 2 1/2 {(1 + ϕ) − 1} (λ − J 2 )1/2

0

The constraint equation (13.5.19) then becomes 1 N N µ2 β 2 + = N, 2β (λ2 − J 2 )1/2 4(λ − J)2

(1)

128 which agrees with the quoted result. Comparing (1) with the formal constraint equation (13.5.13), we conclude that N µ2 B 2 ∂Aλ N 1 + , = ∂λ 2β (λ2 − J 2 )1/2 4(λ − J)2

(2)

o N 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. (13.5.12a) that, for B = 0 and J = 0, the partition function QN = (π/βλ)N/2 — with the result that Aλ = (N/2β) ln(βλ/π). It follows 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 (13.5.15) then gives βAλ 1 βAS = − 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. (13.3.8) becomes QN (nK ) =

N −1 Γ(n/2) I (nK ) . (n−2)/2
(n−2)/2 1 2 nK

For n, N  1, we get   n n 1 n  1 1 ln QN (nK ) ≈ ln Γ nK + ln In/2 · 2K − ln nN n 2 2 2 2    1 1 n n n n = ln − + . . . − ln nK n 2 2 2 2 2     n (4K 2 + 1)1/2 + 1 2 1/2 + (4K + 1) − ln + ... 2 2K    1 (4K 2 + 1)1/2 + 1 ≈ (4K 2 + 1)1/2 − 1 − ln . 2 2 13.20. By eqn. (13.5.69), 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. (13.5.44), and remembering that K = J/kB T , we obtain the desired result.

129 13.21. Before employing the suggested approximation, we observe that the major contribution to the integral over θj in expression (13.5.58) 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 (13.5.57) 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. (13.5.61 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 Section 13.6, 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 (13.5.27), 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 2 2 1 √ e−πRn /λ x λ x n=1

e−αx−πR

2

/λ2 x

1 xd/2

#

" dx

 λ=~

2πβ m

1/2 #

dx ,

0

which is precisely the expression that led to eqns. (13.6.35 and 36).

130 13.22. The constraint equation (13.5.21) now takes the form −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. (13.5.17) and (C.7b), we obtain 0

F (ϕ) ≈ −N

−1

 Z∞  d  Y Nj a 2π

j=1

1 σ σ k a 2

−2

2π d/2 d−1 k dk Γ(d/2)

0

Z∞ 

d

=−

ϕ+

a

1+

2d−1 π d/2 Γ(d/2)ϕ2

1 σ σ k a 2ϕ

−2

k d−1 dk .

0 σ

σ

We substitute k = (2ϕ/a )x and get 0

F (ϕ) ≈ −

2d−1

(2ϕ)d/σ π d/2 Γ(d/2)σϕ2

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. (G.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.

.

131 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. (13.6.9 and 11) proceeds exactly as of eqns. (7.1.36 and 37). The derivation of eqns. (13.6.10 and 13) proceeds exactly as in Problem 7.4; eqn. (13.6.12) then follows as a product of expressions (13.6.11 and 13): The derivation of eqns. (13.6.14 and 15) proceeds 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. (13.6.7) to express these quantities in terms of P . Finally, the derivation of eqn. (13.6.23) proceeds exactly as in Problem 7.6. 13.24. The derivations here proceed exactly as in Problem 7.7. The singularity of these quantities arises from the last term of the two expressions, and is qualitatively similar to the singularity of the quantity (∂CV /∂T ). 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 V κT . 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

132 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. (13.6.28), dP/dT, at T < Tc , = (d + 2)P/2T . Using this result and eqn. (13.6.15), we get  CV = VT

(d + 2)P 2T

2

d d(d + 2) PV = . (d + 2)P 4 T

Substituting for P from eqn. (13.6.28), we recover expression (13.6.30) for CV . 13.26. As shown in Problem 1.16, dµ = −sdT + vdP . It follows that     ∂v 1 ∂v =− . κT ≡ − v ∂P T ∂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. (13.6.24) and (13.6.27). Since α ≡ −µ/kB T , we have ρ=

N0 Ne kB T ζ(d / 2) N = + =− + . V V V Vµ λd

It follows that 1 κT = 2 ρ



kB T V µ2



1 kB T = 2 ρ V



N0 − kB T

2 =

N02 V ρ20 = . ρ2 Vk B T ρ2 kB T

Incidently, using the relationship between CP and κT , as developed in Problem 13.25, 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. (13.6.30), 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 lim 1 + = exp(z), n→∞ n and collect factors.

133 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) i3[i] -= n; //implement periodic boundary conditions

134 } double* hist=new double[n+1]; // histogram: double since would overflow integers for L>5; for (int e=0;e 1.0) x[i]=xtrial; for (int i=1;i 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

156 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 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 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.

157 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