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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 Thisinstructor’smanualforthefourtheditionofStatistical Mechanics isbased 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. Theoriginalideaofproducinganinstructor’smanualfirstcamefromRKP’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 oth- erwise spend on solving these problems oneself. Secondly, before one selects problems either for homework or for an exam, one can consult the manual to determinethelevelofdifficultyofthevariousproblemsandmakeone’sselection accordingly. Thirdly,onemayevenusesomeofthesesolvedproblems,especially theonesappearinginlaterchapters,aslecturematerial,therebysupplementing 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 learnbestwhentheyworktogetherandstruggleoverdifficultproblems. Readily availablesolutionsinterferewiththiscrucialaspectofgraduatephysicstraining. R.K.P. San Diego, CA P.D.B. Boulder, CO 3 4 Chapter 1 1.1. (a) WeexpandthequantitylnΩ(0)(E )asaTaylorseriesinthevariable 1 (E −E¯ ) and get 1 1 lnΩ(0)(E )≡lnΩ (E )+lnΩ (E ) (E =E(0)−E ) 1 1 1 2 2 2 1 ={lnΩ (E¯ )+lnΩ (E¯ )}+ 1 1 2 2 (cid:26) (cid:27) ∂lnΩ (E ) ∂lnΩ (E )∂E 1 1 + 2 2 2 (E −E¯ )+ ∂E ∂E ∂E 1 1 1 2 1 E1=E¯1 1(cid:40)∂2lnΩ (E ) ∂2lnΩ (E )(cid:18)∂E (cid:19)2(cid:41) 1 1 + 2 2 2 (E −E¯ )2+··· . 2 ∂E2 ∂E2 ∂E 1 1 1 2 1 E1=E¯1 The first term of this expansion is a constant, the second term van- ishes as a result of equilibrium (β = β ), while the third term may 1 2 be written as (cid:26) (cid:27) (cid:26) (cid:27) 1 ∂β1 + ∂B2 (cid:0)E −E¯ (cid:1)2 =−1 1 + 1 (E −E¯ )2, 2 ∂E ∂E 1 1 2 kT2(C ) kT2(C ) 1 1 1 2 eq. 1 v 1 2 v 2 with T = T . Ignoring the subsequent terms (which is justified 1 2 if the systems involved are large) and taking the exponentials, we readily see that the function Ω0(E ) is a Gaussian in the variable 1 (E −E¯ ),withvariancekT2(C ) (C ) /{(C ) +(C ) }. Notethat 1 1 v 1 v 2 v 1 v 2 if(C ) >>(C ) —correspondingtosystem1beinginthermalcon- v 2 v 1 tact with a very large reservoir — then the variance becomes simply kT2(C ) , regardless of the nature of the reservoir; cf. eqn. (3.6.3). v 1 (b) If the systems involved are ideal classical gases, then (C ) = 3N k v 1 2 1 and (C ) = 3N k; the variance then becomes 3k2T2·N N /(N + v 2 2 2 2 1 2 1 N ). Again, if N >> N , we obtain the simplified expression 2 2 1 3N k2T2; cf. Problem 3.18. 2 1 1.2. Since S is additive and Ω multiplicative, the function f(Ω) must satisfy the condition f(Ω Ω )=f(Ω )+f(Ω ). (1) 1 2 1 2 5 6 CHAPTER 1. Differentiating (1) with respect to Ω (and with respect to Ω ), we get 1 2 Ω f(cid:48)(Ω Ω )=f(cid:48)(Ω ) and Ω f(cid:48)(Ω Ω )=f(cid:48)(Ω ), 2 1 2 1 1 1 2 2 so that Ω f(cid:48)(Ω )=Ω f(cid:48)(Ω ). (2) 1 1 2 2 Sincetheleft-handsideof(2)isindependentofΩ andtheright-handside 2 isindependentofΩ ,eachsidemustbeequaltoaconstant,k,independent 1 of both Ω and Ω . It follows that f(cid:48)(Ω)=k/Ω and hence 1 2 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 −v )(V −2v )...(V −N −1v ), 0 0 0 so that ln Ω=C+ln V +ln (V −v )+ln (V −2v )+...+ln (V −N −1v ), 0 0 0 whereC isindependentofV. Theexpressionontherightmaybewritten as C+Nln V+N(cid:88)−1ln(cid:18)1− jv0(cid:19)(cid:39)C+Nln V+N(cid:88)−1(cid:18)−jv0(cid:19)(cid:39)C+Nln V−N2v0. V V 2V j=1 j=1 Equation (1.4.2) is then replaced by P N N2v N (cid:18) Nv (cid:19) = + 0 = 1+ 0 , i.e. kT V 2V2 V 2V (cid:18) Nv (cid:19)−1 PV 1+ 0 =NkT. 2V Since Nv << V, (1+Nv /2V)−1 (cid:39) 1−Nv /2V. Our last result then 0 0 0 takes the form: P(V −b)=NkT, where b= 1Nv . 2 0 A little reflection shows that v =(4π/3)σ3, with the result that 0 1 4π 4π (cid:18)1 (cid:19)3 b= N · σ3 =4N · σ . 2 3 3 2 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 (cid:88) (πε∗1/2/L)3 (πε∗1/2/L)2 (ε∗)= V ∓ S. 1 6π2 16π 7 Substituting V =L3 and S =6L2, we obtain eqns. (1.4.15 and 16). The expression for T now follows straightforwardly; we get (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) 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 (cid:30) (cid:18) (cid:19) hν Nhν T = ln 1+ . k E For E >>Nhν, we recover the classical result: T =E/Nk. 1.9. Since the function S(N,V,E) of a given thermodynamic system is an ex- tensive quantity, we may write (cid:18) (cid:19) (cid:18) (cid:19) V E V E S(N,V,E)=Nf , =Nf(v,ε) v= ,ε= . N N N N It follows that (cid:18) (cid:19) (cid:20) (cid:18) (cid:19) (cid:18) (cid:19) (cid:21) ∂S ∂f −V ∂f −E N =N f +N · +N · , ∂N ∂v N2 ∂ε N2 V,E ε v (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) ∂S ∂f ∂S ∂f 1 V =VN · . =EN · . ∂V ∂v ∂E ∂ε N N,E ε N,V v Adding these expressions, we obtain the desired result. 1.11. Clearly,theinitialtemperaturesandtheinitialparticledensitiesofthetwo gases(andhenceofthemixture)arethesame. Theentropyofmixingmay, therefore, be obtained from eqn. (1.5.4), with N = 4N and N = N . 1 A 2 A We get (∆S)∗ =k[4N ln(5/4)+N ln 5] A A =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 N , N and V fixed and vary only V . The 1 2 1 2 first and second derivatives of this expression are then given by (cid:20) (cid:21) (cid:20) (cid:21) N +N N N +N N k 1 2 − 2 and k − 1 2 + 2 (1a,b) V +V V (V +V )2 V2 1 2 2 1 2 2 respectively. Equating (1a) to zero gives the desired condition, viz. N V = N V , i.e. N /V = N /V = n, say. Expression (1b) then 1 2 2 1 1 1 2 2 reduces to (cid:20) (cid:21) n n knV k − + = 1 >0. V +V V V (V +V ) 1 2 2 2 1 2 Clearly, (∆S) is at its minimum when N /V =N /V , and it is 1≡2 1 1 2 2 straightforward to check that the value at the minimum is zero. 8 CHAPTER 1. (b) The expression now in question is given by eqn. (1.5.4). With N = 1 αN and N = (1−α)N, where N = N +N (which is fixed), the 2 1 2 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 (cid:20) (cid:21) 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 N = 1 N = (1/2)N, and it is straightforward to check that the value at the 2 maximum is Nln2. 1.13. Proceeding with eqn. (1.5.1), with T replaced by T , it is straightforward i to see that the extra contribution to ∆S, owing to the fact that T (cid:54)=T , 1 2 is given by the expression 3 3 N k ln (T /T )+ N k ln(T /T ), 2 1 f 1 2 2 f 2 where T = (N T +N T )/(N +N ). It is worth checking that this f 1 1 2 2 1 2 expression is always greater than or equal to zero, the equality holding if andonlyifT =T . Furthermore,theresultquotedheredoesnotdepend 1 2 on whether the two gases were different or identical. 1.14. By eqn. (1.5.1a), given on page 24 of the text, we get 3 (∆S) = Nk ln(T /T ). v 2 f i Now, since PV =NkT, the same equation may also be written as (cid:18) (cid:19) (cid:26) (cid:18) (cid:19)(cid:27) kT 3 5 2πmkT S =Nk ln + Nk +ln . (1b) P 2 3 h2 It follows that 5 5 (∆S) = Nk ln(T / T )= (∆S) . P 2 f i 3 V A numerical verification of this result is straightforward. It should be noted that, quite generally, (∆S) T(∂S / ∂T) C P = P = P =γ (∆S) T(∂S / ∂T) C V V V which, in the present case, happens to be 5/3. 9 1.15. For an ideal gas, C −C = nR, where n is the number of moles of the P V gas. With C /C =γ, one gets P V C =γnR / (γ−1) and C =nR / (γ−1). P V For a mixture of two ideal gases, (cid:18) (cid:19) n R n R f f C = 1 + 2 = 1 + 2 (n +n )R. V γ −1 γ −1 γ −1 γ −1 1 2 1 2 1 2 Equating this to the conventional expression (n +n )R/(γ−1), we get 1 2 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 +Ndµ. Combining this with eqn. (1.3.4), we get −SdT +VdP =Ndµ, i.e. dP =(N / V)dµ+(S / V)dT. Clearly, then, (∂P / ∂µ) =N / V and (∂P / ∂T) =S / V. T µ Now, for the ideal gas NkT (cid:40)N (cid:18) h2 (cid:19)3/2(cid:41) P = and µ=kT ln ; V V 2πmkT see eqn. (1.5.7). Eliminating (N/V), we get (cid:18)2πmkT(cid:19)3/2 P =kT eµ/kT, h2 which is the desired expression. It follows quite readily now that for this system (cid:18) (cid:19) ∂P 1 = P. ∂µ kT T which is indeed equal to N/V, whereas (cid:18)∂P(cid:19) 5 µ (cid:34)5 (cid:40)N (cid:18) h2 (cid:19)3/2(cid:41)(cid:35)Nk = P − P = −ln ∂T 2T kT2 2 V 2πmkT V µ 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 p is then mρ2ϕ˙, where the various symbols ϕ have their usual meanings. The energy of rotation is given by 1 E = m(ρϕ˙)2 =p2 / 2mρ2. 2 ϕ Lines of constant energy in the (ϕ,p )-plane are “straight lines, running ϕ 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 ∆p =h/2π”. Clearly, ϕ the eigenvalues of p , starting with p = 0, are n(cid:126) and those of E are ϕ ϕ n2(cid:126)2/2I, where I =mρ2 and n=0,±1,±2,... The eigenvalues of E obtained here are precisely the ones given by quan- tum mechanics for the energy “associated with the z-component of 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 spacemaybedenotedbytheanglesθandϕ,theconjugatevariablesbeing p =mr2θ˙ and p =mr2sin2θϕ˙. The energy of rotation is given by θ ϕ 1 1 p2 p2 M2 E = m(rθ˙)2+ m(rsin θϕ˙)2 = θ + ϕ = , 2 2 2mr2 2mr2 sin2θ 2I where I =mr2 and M2 =p2+(cid:0)p2/sin2θ(cid:1). θ ϕ The “volume” of the relevant region of the phase space is given by the (cid:82)(cid:48) integral dp dp dθ dϕ, where the region of integration is constrained θ ϕ by the value of M. A little reflection shows that in the subspace of p θ and p we are restricted by an elliptical boundary with semi-axes M and ϕ Msinθ, the enclosed area being πM2sinθ. The “volume” of the relevant region, therefore, is π 2π (cid:90) (cid:90) (πM2sinθ)dθ dϕ=4π2M2. θ=0 ϕ=0 1