Statistical Mechanics (Instructor's Solution Manual) - 4th ed [4 ed.] 0081026935, 9780081026939

Citation preview

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

1

2

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

3

4

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    
2 1 ∂B2 1 1 1 ∂β1 ¯ ¯ 1 )2 , + E1 − E1 = − + (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 see that the function Ω0 (E1 ) is a Gaussian in the variable ¯1 ), with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that (E1 − E 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

6

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 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 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 ε¯ =

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. ≡ ε¯

(ε2 − ε¯2 ) ε¯

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∞

u F (u)du

0 R∞

·

R∞

F (u)du

0

≥ 1, i.e. F (u)du

0

Z∞

Z∞ u F (u)du ·

0

u−1 F (u)du

0

∞ 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 > 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  3 1− 2n(kT ln z)  3 = 1− 2n(kT ln z)

κT =

 −1 π2 π2 (ln z)−2 + . . . 1+ (ln z)−2 + . . . 24 8  π2 (ln z)−2 + . . . . 6

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 ˚ A, while m0 = me — giving εF = 5.49 eV and TF = 6.37 × 104 K. For lead, ne = 4, na = 4, a = 4.95 ˚ A, 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 ˚ A, 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 π2 εF = kT ln z 1 + (ln z)−2 + 8  π2 = kT ln z 1 + (ln z)−2 + 12 

2/3 7π 4 −4 (ln z) + . . . 640  π4 −4 (ln z) + . . . . 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 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 Since µ → εF as T → 0, the leading term is cµ ≈

π2 2 3 k 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 /(4π0 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 /(4π0 ~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 uu¯ = 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 uν¯e = 8 2   7 uγ uν¯µ = 8 2   7 uγ uν¯τ = 8 2   7 uγ uµ+ = 2 8 2   7 uγ uu¯ = 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=1 j! m kT A (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=1 j! m kT A0 (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

3

a2 λ = −2π

r0 2

Z

−1 · r dr + 0

=

r1

(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

r13 . − 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 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  ¯  ¯ ∂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 (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| 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 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

P

s

s+

X

ln ρi (si )

i

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 ¯ = (N ¯+ − N ¯− )µ and N ¯+ + N ¯− = N , we readily see that 13.1. Since M  ¯  1 P (β, B) ± sinh x M 1 ¯ = N (x = βµB). N± = 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 −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 (11.3.17) 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 (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 N 2α =e = e2α+2α q/(q−1) 0 ¯ cosh(α + α − γ) N− and to eqn. (11.6.21), 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. (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 Heff

X 1 = − 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  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 βN ∂J1 e−4βJ2 + sinh2 (βJ1 )

1 ∂ ln Q = 1− βN ∂J2 2e−4βJ2



e−4βJ2 + sinh2 (βJ1 )

1/2 h

 1/2 i . 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 > a, coth(a/2ξ) ≈ (2ξ/a), making χ0 ∝ ξ 1 — consistent with the fact that for this system (2 − η) = 1. For ξ