Chapter1
1.1 (a)Weexpandthequantityln Ω(0)(E1)asaTaylorseriesinthevariable (E1 E1)andget
ln Ω(0)(E1) ≡ lnΩ1(E1)+ln Ω
Thefirsttermofthisexpansionisaconstant,thesecondtermvanishesasaresultofequilibrium(β1 = β2),whilethethirdtermmay bewrittenas
with T1 = T2.Ignoringthesubsequentterms(whichisjustified ifthesystemsinvolvedare large)andtakingtheexponentials,we readilyseethatthefunctionΩ0(E1)isaGaussianinthevariable (E1 E1),withvariance kT 2(Cv)1(Cv)2/{(Cv)1 +(Cv)2}.Notethat if(Cv)2 >> (Cv)1 —correspondingtosystem1beinginthermalcontactwithaverylargereservoir—thenthevariancebecomessimply kT 2(Cv)1,regardlessofthenatureofthereservoir;cf.eqn.(3.6.3).
(b)Ifthesystemsinvolvedareidealclassicalgases,then(Cv)1 = 3 2 N1k and(Cv)2 = 3 2 N2k;thevariancethenbecomes 3 2 k2T 2 N1N2/(N1 + N2).Again,if N2 >>N1,weobtainthesimplifiedexpression 3 2 N1k2T 2;cf.Problem3.18.
1.2 SinceSisadditiveand Ω multiplicative,thefunction f (Ω)mustsatisfy thecondition f (Ω1Ω2)= f (Ω1)+ f (Ω2). (1)
Differentiating(1)withrespectto Ω1 (andwithrespectto Ω2),weget
Ω2f (Ω1Ω2)= f (Ω1)and Ω1f (Ω1Ω2)= f (Ω2), sothat
Ω1f (Ω1)= Ω2f (Ω2) (2)
Sincetheleft-handsideof(2)isindependentof Ω2 andtheright-handside isindependentof Ω1,eachsidemustbeequaltoaconstant, k,independent of both Ω1 and Ω2.Itfollowsthat f (Ω)= k/Ω andhence f (Ω)= k ln Ω +const (3)
Substituting(3)into(1),wefindthattheconstantofintegrationiszero.
1.4 Insteadofeqn.(1.4.1),wenowhave
Ω ∝ V (V v0)(V 2v0) (V N 1v0), sothat
ln Ω = C +ln V +ln(V v0)+ln(V 2v0)+ +ln(V N 1v0), where C isindependentof V .Theexpressionontherightmaybewritten as
Equation(1.4.2)isthenreplacedby
i e PV 1+ N v0 2V 1 = NkT
Since N v0 <<V, (1+ N v0/2V ) 1 1 N v0/2V .Ourlastresultthen takestheform: P (V b)= NkT ,where b = 1 2 N v0 Alittlereflectionshowsthatv0 =(4π/3)σ3,withtheresultthat
1.5 ThisproblemisessentiallysolvedinAppendixA;allthatremainstobe doneistosubstitutefromeqn.(B.12)into(B.11),toget
Substituting V = L3 and S =6L2,weobtaineqns.(1.4.15and16). Theexpressionfor T nowfollowsstraightforwardly;weget 1 T = k
, sothat
For E>> Nhν,werecovertheclassicalresult: T = E/Nk
1.9. Sincethefunction S(N,V,E) ofagiventhermodynamicsystemisan extensive quantity,wemaywrite S(N,V,E)= Nf V N , E N = Nf (v,ε) v= V N ,ε = E N
Itfollowsthat
Addingtheseexpressions,weobtainthedesiredresult.
1.11 Clearly,theinitialtemperaturesandtheinitialparticledensitiesofthetwo gases(andhenceofthemixture)arethesame.Theentropyofmixingmay, therefore,beobtainedfromeqn.(1.5.4),with N1 =4NA and N2 = NA Weget (∆S)∗ = k[4NA ln(5/4)+ NA ln5] = R[4ln(5/4)+ln5]=2 502 R, whichisequivalenttoabout0.5 R permoleofthemixture.
1.12 (a)Theexpressioninquestionisgivenbyeqn.(1.5.3a).Withoutlossof generality,wemaykeep N1,N2 and V1 fixedandvaryonly V2 .The firstandsecondderivativesofthisexpressionarethengivenby k N1 + N2 V1 + V2 N2 V2 and k N1 + N2 (V1 + V2)2 + N2 V 2 2 (1a,b) respectively.Equating(1a)tozerogivesthedesiredcondition,viz. N1V2 = N2V1,i.e. N1/V1 = N2/V2 = n,say.Expression(1b)then reducesto k n V1 + V2 + n V2 = knV1 V2(V1 + V2) > 0.
Clearly,(∆S)1≡2 isatits minimum when N1/V1 = N2/V2,anditis straightforwardtocheckthatthevalueattheminimumiszero.
8 CHAPTER1.
(b)Theexpressionnowinquestionisgivenbyeqn.(1.5.4).With N1 = αN and N2 =(1 α)N ,where N = N1 + N2 (whichisfixed),the expressionfor(∆S)∗/k takestheform αN ln α (1 α)N ln(1 α)
Thefirstandsecondderivativesofthisexpressionwithrespectto α are [ N ln α + N ln(1 α)]and N α N 1 α (2a,b)
respectively.Equating(2a)tozerogivesthecondition α =1/2,which reduces(2b)to 4N .Clearly,(∆S)∗/k isatits maximum when N1 = N2 =(1/2)N ,anditisstraightforwardtocheckthatthevalueatthe maximumis N ln2.
1.13. Proceedingwitheqn.(1.5.1),with T replacedby Ti,itisstraightforward toseethattheextracontributionto∆S,owingtothefactthat T1 = T2, isgivenbytheexpression 3 2 N1k ln(Tf /T1)+
ln(Tf /T2), where Tf =(N1T1 + N2T2)/(N1 + N2).Itisworthcheckingthatthis expressionisalwaysgreaterthanorequaltozero,theequalityholdingif andonlyif T1 = T2.Furthermore,theresultquotedheredoesnotdepend onwhetherthetwogasesweredifferentoridentical.
1.14 Byeqn.(1.5.1a),givenonpage24ofthetext,weget (∆S)v = 3 2 Nk ln(Tf /Ti)
Now,since PV = NkT ,thesameequationmayalsobewrittenas S = Nk
Itfollowsthat (∆S)P =
)V.
Anumericalverificationofthisresultisstraightforward. Itshouldbenotedthat,quitegenerally, (∆S)P (∆S)V = T (∂S/∂T )P T (∂S/∂T )V = CP CV = γ which,inthepresentcase,happenstobe5/3.
1.15 Foranidealgas, CP CV = nR,where n isthenumberofmolesofthe gas.With CP /CV = γ,onegets
CP = γnR / (γ 1)and CV = nR / (γ 1). Foramixtureoftwoidealgases,
Equatingthistotheconventionalexpression(n1 + n2)R/(γ 1),weget thedesiredresult.
1.16 Inviewofeqn.(1.3.15), E TS + PV = µN .Itfollowsthat
dE TdS SdT + PdV + VdP = µdN + Nd µ.
Combiningthiswitheqn.(1.3.4),weget
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,fortheidealgas P = NkT V and µ = kT ln
seeeqn.(1.5.7).Eliminating(N/V ),weget
;
, whichisthedesiredexpression.Itfollowsquitereadilynowthatforthis system
∂µ T = 1 kT P. whichisindeedequalto N/V ,whereas
∂T µ = 5 2T P µ kT 2 P = 5 2 ln
which,byeqn.(1.5.1a),ispreciselyequalto S/V
Chapter2
2.3 Therotatorinthisproblemmayberegardedasconfinedtothe(z=0)planeanditspositionattime t maybedenotedbytheazimuthalangle ϕ.Theconjugatevariable pϕ isthen mρ2ϕ,wherethevarioussymbols havetheirusualmeanings.Theenergyofrotationisgivenby E = 1 2 m(ρ ˙ ϕ)2 = p 2 ϕ / 2mρ 2 .
Linesofconstantenergyinthe(ϕ,pϕ)-planeare“straightlines,running paralleltothe ϕ-axisfrom ϕ =0to ϕ =2π”.Thebasiccellofarea h in thisplaneisa“rectanglewithsides∆ϕ =2π and∆pϕ = h/2π”.Clearly, theeigenvaluesof pϕ,startingwith pϕ =0,are n andthoseof E are n2 2/2I,where I = mρ2 and n =0, ±1, ±2,...
Theeigenvaluesof E obtainedherearepreciselytheonesgivenbyquantummechanicsfortheenergy“associatedwiththez-componentofthe rotationalmotion”.
2.4 Therigidrotatorisamodelforadiatomicmoleculewhoseinternuclear distance r mayberegardedasfixed.Theorientationofthemoleculein spacemaybedenotedbytheangles θ and ϕ,theconjugatevariablesbeing pθ = mr 2θ and pϕ = mr 2
˙
.Theenergyofrotationisgivenby
The“volume”oftherelevantregionofthephasespaceisgivenbythe integral dpθ dp ϕdθdϕ,wheretheregionofintegrationisconstrained bythevalueof M .Alittlereflectionshowsthatinthesubspaceof pθ and pϕ wearerestrictedbyanellipticalboundarywithsemi-axes M and M sin θ,theenclosedareabeing πM 2 sin θ.The“volume”oftherelevant region,therefore,is
Thenumberofmicrostatesavailabletotherotatoristhengivenby4π2M 2/h2 , whichisprecisely(M/ )2.Atthesametime,thenumberofmicrostates associatedwiththequantizedvalue M 2 j = j(j +1) 2 maybeestimatedas
.
Thisispreciselythedegeneracyarisingfromtheeigenvaluesthattheazimuthalquantumnumber m has,viz. j,j 1,..., j +1, j
2.6 Intermsofthevariables θ and L(= m 2θ),thestateofthesimplependulumisgivenby,seeeqns.(2.4.9), θ =(A/ )cos(ωt + ϕ),L = m ωA sin(ωt + ϕ), with E = 1 2 mω2A2 and τ =2π/ω.Thetrajectoryinthe(θ,L)-planeis givenbytheequation
2 (A/ )2 + L2 (m ωA)2 =1, whichisanellipse—justlikeinFig.2.2.Theenclosedareaturnsoutto be πmωA2,whichispreciselyequaltotheproduct Eτ
2.7 Followingtheargumentdevelopedonpage70ofthetext,thenumberof microstatesforagivenenergy E turnsouttobe
Ω(E)=(R + N 1)!/R!(N 1)!,R = E 1 2 N ω / ω. (1)
For R>>N ,weobtaintheasymptoticresult
Ω(E) ≈ RN 1 / (N 1)!, where R ≈ E/ ω. (3.8.25a)
ThecorrespondingexpressionforΓ(E;∆)wouldbe Γ(E;∆) ≈ (E/ ω)N 1 (N 1)! ∆ ω = EN 1∆ (N 1)!( ω)N (1)
The“volume”oftherelevantregionofthephasespacemaybederived fromtheintegral N i=1 (dq idp i), with N i=1 1 2 kq 2 i + 1 2m p 2 i ≤ E.
Thisisequalto,seeeqn.(7a)ofAppendixC,
3
where ω = k/m.The“volume”oftheshellinquestionisthengivenby
Dividing(2)by(1),weseethattheconversionfactor ω0 isprecisely hN .
2.8. Wewrite V3N = AR3N ,sothat dV 3N = A 3NR3N 1dR.Atthesame time,wehave
Theintegralontheleftmaybewrittenas
Equating(1)and(2),weget: A =(8π)N /(3N )!,whichyieldsthedesired resultfor V3N
The“volume”oftherelevantregionofthephasespaceisgivenby
(3
)!, sothat Σ(n,V,E)= V N (8πE
/h
3)N / (3N )!, whichisafunctionof N and VE 3.An isentropic processthenimplies that VE 3 = const
Thetemperatureofthesystemisgivenby
Theequationfortheisentropicprocessthenbecomes VT 3 = const .,i.e. T ∝ V 1/3;thisimpliesthat γ =4/3.Therestofthethermodynamics followsstraightforwardly.SeealsoProblems1.7and3.15.
Chapter3
3.4. Forthefirstpart,weuseeqn.(3.2.31)withall ωr =1.Weget
k N lnΓ= k ln r e βEr + kβU,
whichisindeedequalto (A/T )+(U/T )= S. Forthesecondpart,weuseeqn.(3.2.5),withtheresultthat k N ln W {n ∗ r } = k N N ln N r n ∗ r ln n ∗ r = k r n∗ r N ln n∗ r N = k ln n∗ r N .
Substitutingfor n∗ r fromeqn.(3.2.10),weget
k N ln W {n ∗ r } = kβ Er + k ln r e βEr , whichispreciselytheresultobtainedinthefirstpart.
3.5 Sincethefunction A(N,V,T )ofagiventhermodynamicsystemisan extensive quantity,wemaywrite A(N,V,T )= Nf (v,T )(v= V/N )
Itfollowsthat N ∂A ∂N V,T = N f + N ∂f ∂v T V N 2 , and V ∂A ∂V N,T = VN ∂f ∂v T 1 N
Addingtheseexpressions,weobtainthedesiredresult.
3.6 Let’sgotopart(c)rightaway.Ourproblemhereisto maximize the expression S/k = r,s Pr,s ln Pr,s,subjecttotheconstraints r,s Pr,s =
1, r,s EsPr,s = E and r,s NrPr,s = N .Varying P ’sandusingthemethod ofLagrange’sundeterminedmultipliers,weareledtothecondition
r,s {−(1+ln Pr,s) γ βEs αNr} δPr,s =0.
Inviewofthearbitrarinessofthe δP ’sinthisexpression,werequirethat (1+ln Pr,s) γ βEs αNr =0 for allr and s.Itfollowsthat
Theparameters α and β aretobedeterminedbythegivenvaluesof N and E
Intheabsenceoftheconstraintimposedby N ,theparameter α doesnot evenfigureinthecalculation,andweobtain
Pr ∝ exp( βEr), asdesiredinpart(b).Andiftheconstraintimposedby E isalsoabsent, weobtain
Pr = const ., asdesiredinpart(a).
3.7. Fromthermodynamics,
FromSec.3.3,
Substituting(2)into(1),weobtainthedesiredresult. Fortheidealgas, Q ∝ V N T 3N/2.Therefore,(∂ ln Q/∂V )T = N/V .We thenget CP CV = k (N/V )2 N/V 2 = Nk
3.8 Foranidealgas,
=
3
Itfollowsthat T (∂ ln Q1/∂T )P =5/2;theexpressionontheright-hand sideofthegivenequationthenis
ln V N (2πmkT )3/2 h3 + 5 2 which,byeqn.(3.5.13),isindeedequaltothequantity S/Nk
3.12. Westartwitheqn.(3.5.5),substitute H(q,p)= i p2 i /2m + U (q)and integrateoverthe pi s,toget
QN (V,T )= 1 N !
mkT h2 3N/2 ZN (V,T ), where ZN (V,T )= e U (q)/kT d3N q.
Itfollowsthat,for N>> 1, A = NkT ln N
N,V .
Now
ln Z ∂T N,V
Substitutingtheseresultsintotheaboveexpressionfor S,weobtainthe desiredresultfor S.Inpassing,wenotethat H ≡ A + TS = 3 2 NkT + U Forthesecondpartofthequestion,wewrite U (q)= i<j u(rij ),sothat e βU (q) = i<j
(rij ) = i<j (1+ fij ) , andfollowProblems3.23and1.4.Thequantity V thenappearstobein thenatureofa“freevolume”forthemoleculesofthesystem.
3.14. a)TheLagrangianisgivenby L = K V =
)+ uw (L riα)], where i =1, ,N denotestheparticlenumber, α = x,y,z denotesthe cartesiandirections,and r2 ij = α (riα rjα)2.Thecanonicalmomenta are
TheHamiltonianisgivenby
Thecanonicalpressurecanbewritten
Thisisclearlytheinstantaneousforceperunitareaontheright,back, andtopwalls.
b)Thecartesiancoordinatesforthescaledpositioninsidetheboxare siα = riα/L sotheLagrangianbecomes
Inthiscasethecanonicalmomentaare
ThisleadstoaHamiltonianoftheform
withcanonicalpressureis
Convertingbacktonormalcartesiancoordinatesandmomentagives
Thefirsttermis(2/3)(N/V )timesthekineticenergyperparticlesois O(N ).Thesecondtermis(1/3)(N/V )timesthevirialperparticlesois also O(N ).Ontheotherhand,thirdtermisproportionaltotheshortrangedvirialonthewallsdividedbythevolumesois O(N 2/3)whichis negligibleinthethermodynamiclimit.
Comparingtoequation(3.7.15)fortheaveragepressureweseethat
3.15 Here, QN (V,T )=(1/N !)QN 1 (V,T ),while
whichyieldsthedesiredresultfor QN .Thethermodynamicsofthesystem nowfollowsstraightforwardly.
Asregardsthedensityofstates,theexpression
leadsto
forasingleparticle,whiletheexpressionfor QN (V,T )leadsto
forthe N -particlesystem;cf.theexpressionforΣ(E)derivedinProblem2.8.
3.17. Differentiatethestatedresultwithrespectto β,toget
∂β H(U H) e βH dω =0.
Thismeansthat
HU + H 2 =0,
whichamountstothedesiredresult: H 2 − H 2 = (∂U/∂β).
3.18 Westartwitheqn.(3.6.2),viz.
anddifferentiateitwithrespectto β,keepingthe Er fixed.Weget
Substitutingfor(∂U/∂β)fromeqn.(1),weget
, whichispreciselyequalto (E U )3 .Asfor ∂2U/∂β2,wenotethat,since
Hencethedesiredresult.
Fortheidealclassicalgas, U = 3 2 NkT and CV = 3 2 Nk ,whichreadilyyield thestatedresults.
3.19 Since G =
ipi).Averagingoveratimeinterval τ , weget
Forafinite V andfinite E,thequantity G is bounded ;therefore,inthe limit τ →∞,theright-handsideof(1)vanishes.Theleft-handside thengives i (˙ qipi + qipi) =0. whichleadstothedesiredresult.
3.20 Thevirialofthenoninteractingsystem,byeqn.(3.7.12),is 3PV .The contributionfrominterparticleinteractions,byeqn.(3.7.15),isgivenby the“expectationvalueofthesumofthequantity r(∂u/∂r)over all pairs ofparticlesinthesystem”.If u(r)isahomogeneousfunction(ofdegree n)oftheparticlecoordinates,thiscontributionwillbe nU ,where U is
themeanpotentialenergy(not theinternalenergy)ofthesystem.The totalvirialisthengivenby
v = 3PV nU
Therelation K = 1 2 v stillholds,andtherestoftheresultsfollow straightforwardly.
3.21. Allsystemsconsideredherearelocalized.Thepressureterm,therefore, dropsout,andweareleftwiththeresult
Example(a)pertainsto n =2,whileexamples(b)and(c)pertainto n = 1.Intheformercase, K = U = 1 2 E;inthelatter, K = 1 2 U = E.
Thenextproblempertainsto n =4.
3.22. Notethataforceproportionalto q3 impliesapotentialenergyproportional to q4.Thus
Itfollowsthat
forthevaluesoftheseintegrals,seeeqns.(13a)ofAppendixB.Next,
where I(β)denotestheintegralinthedenominator.Itisstraightforward toseethat I(β)isproportionalto β 1/4,whence cq 4 =1/4β,which provesthedesiredresult.
3.23. Thekeytothisderivationiswritingthepartitionfunctionintermsof positionintegralsoverscaledcoordinates.Assumeacubicboxofsize L andvolume V = L3.Thescaledpositionforparticle i is si = ri/L.The
partitionfunctionis
Nowthepressureis
Thiscanbesimplifiedbygoingbacktointegralsoverthenormalposition variablestogiveequation(3.7.15).
3.24. Byeqn.(3.7.5),wehave,forasingleparticle,
Theleft-handsideof(1)istheexpectationvalueofthequantity p · u,i.e. puwhich,forarelativisticparticle,isequalto m0u2(1 u2/c2) 1/2.The desiredresultfollowsreadily.
Inthenon-relativisticlimit(u<<c),oneobtains: 1 2 m0u2 ≈ 3 2 kT ;in theextremerelativisticlimit(u → c),oneobtains: mc2 ≈ 3kT .Note that,inthelattercase, m0c2 isnegligibleincomparisonwith mc2,so thereisnosignificantdifferencebetweenthekineticenergyandthetotal energyoftheparticle.
3.25 Forthefirstpartofthisproblem,seeSec.6.4—especiallythederivation oftheformula(6.4.9).Forthesecondpart,equatetheresultobtainedin thefirstpartwiththeonestatedineqn.(3.7.5).
3.26 Themultiplicity w(j){=(j + s 1)!/j!(s 1)!} arisesfromthevariety ofwaysinwhich j indistinguishablequantacanbedividedamongthe s dimensionsoftheoscillator: j = j1 + ... + js;thisissimilartothe calculationdoneonpage70ofthetext.
Asforthepartitionfunction, Q(s) N (β)= Q(s) 1 (β) N ,where
Calculationofthevariousthermodynamicquantitiesisnowstraightforward.Theresultsarefoundtobeessentiallythesameasforasystemof sN one-dimensionaloscillators.However,since Q(s) N (β)= Q(1) Ns (β), thechemicalpotential µs willturnouttobe s times µ1.
3.28 (a) Whenoneoftheoscillatorsisinthequantumstate n,theenergyleft fortheremaining(N 1)oscillatorsis E n + 1 2 ω;thecorrespondingnumberofquantatobedistributedamongtheseoscillators is R n;seeeqn.(3.8.24).Therelevantnumberofmicrostatesisthen givenbytheexpression(R n + N 2)!/(R n)!(N 2)!.Combined withexpression(3.8.25),thisgives
Itfollowsthat
Byiteration, pn = p0{n/(¯ n +1)}n Goingbacktoeqn.(1),wenotethat
, whichcompletesthedesiredcalculation.
(b) Theprobabilityinquestionisproportionalto gN 1(E ε),i.e.to (E ε) 3 2 (N 1) 1.For1 <<N ,thisisessentiallyproportionalto (1 ε/E) 3 2 N and,for ε<< E,to e 3Nε/2E
3.29 Thepartitionfunctionofthe anharmonic oscillatorisgivenby
Theintegrationover p givesafactorof 2πm/β.Forintegrationover q, wewrite
theintegrationthengives
Itfollowsthat
sothat
whence
and
Next,themeanvalueofthedisplacement q isgivenby
Inthedesiredapproximation,weget
3.30 Thesingle-oscillatorpartitionfunctionisnowgivenby
For x<< 1,wemaywrite
With u = β ω,thesumsinvolvedare
11
Itfollowsthat
ln Q1 =ln[S1 + xuS2 + ] ln S1 + xu(S2/S1) = ln 2sinh 1 2 u + 1 2 xu coth2 1 2 u 1 2
Thefirstpartofthisexpressionleadstothestandardresults(3.8.20and 21).Thesecondpartmay,forsimplification,beexpressedasapower seriesin u,viz.
Theresultingcontributiontotheinternalenergyperoscillatorturnsout tobe
andthecorrespondingcontributiontothespecificheatisgivenby
3.31. ThisproblemisessentiallythesameasProblem3.32,with g1 = g2 = 1,ε1 =0and ε2 = ε.
3.32 Weuseformula(3.3.13),with Pr = p1/g1 foreachofthestatesingroup 1and p2/g2 foreachofthestatesingroup2.Weget
(1)
(a) Inthermalequilibrium,
With x = β(ε2 ε1),wehave: p1 = g1/(g1 + g2e x)and p2 = g2/(g1ex + g2).Substitutingtheseresultsinto(1),weobtain S = k g1
ln(g1ex + g2)
+
Writingthefirstlogasln g1 +ln{1+(g2/g1)e x} andthesecondlog asln g1 + x +ln{1+(g2/g1)e x},weobtainthestatedexpression for S
(b) With Q = g1e βε1 + g2e βε2 ,itisstraightforwardtoseethat A = kT ln{g1e βε1 + g2e βε2 } and U = {g1ε1e βε1 + g2ε2e βε2 }/{g1e βε1 + g2e βε2 }.
Theformula S =(U A)/T thenleadstothedesiredresult.
(c) As T → 0,x →∞ and S indeedtendstothevalue k ln g1.This correspondstothefactthattheprobabilities p1 and p2 inthislimit tendtothevalues1and0,respectively.
3.34 TheCurieparamagneticsusceptibilityperunitvolumeofagasisgiven by
=
Using χ/V =1 8 × 10 6 atSTPgives µ =1 6µB ,becauseofthetwo unpairedvalenceelectronsin O2
3.35. Thepartitionfunctionofthesystemisgivenby
QN = 1 N ! QN 1 , where Q1 = V λ3 · Z,
Z beingthefactorthatarisesfromtherotational/orientationaldegreesof freedomofthemolecule:
= I β 2 2sinh(βµE) βµE .
Thestudyofthevariousthermodynamicalquantitiesofthesystemisnow straightforward.
Concentratingontheelectricalquantitiesalone,weobtainforthe net dipolemomentofthesystem
Mz = N µ cos θ = N β ∂ ln Z ∂E = Nµ coth(βµE) 1 βµE ; cf.eqns.(3.9.4and6).For βµE<< 1, Mz ≈ Nµ · 1 3 βµE.
Thepolarization P ,perunitvolume,ofthesystemisthengivenby P ≈ nµ 2 E/ 3kT(n = N/V ), andthedielectricconstant ε by ε = E +4πP E ≈ 1+ 4πnµ2 3kT .
Thenumericalpartoftheproblemisstraightforward.
3.36 Themeanforce F betweenthetwodipolesisgivenby F = ∂U ∂R = e βU ( ∂U/∂R)sin θdθdϕ sin θ dθ dϕ e βU sin θdθdϕ · sin θ dθ dϕ (1) = 1 β ∂ ∂R ln Z (2)
where Z denotestheintegralinthedenominatorof(1).Athightemperatures,wemaywrite
Thelineartermvanishesonintegrationandweareleftwith
Itfollowsthat
andhence,athightemperatures,
3.37. Byeqns.(3.9.17and18),wehave,forasingledipole,
Athightemperatures,theexponentialmaybeapproximatedby(1+ βgµB mH )whichyields,totheleadingorderin H,
OnereadilyobtainsfortheCurieconstant(perunitvolume)ofthesystem
Writing m = J cos θ,oneobtainsthedesiredresult. Forthesecondpart,wesimplynotethat,foragiven J,
3.38 Treating m asacontinuousvariable,thepartitionfunctionofamagnetic dipoleassumestheform
Q1(β)= J J eβgµB Hm dm = 2 βgµB H sinh(βgµB JH ); cf.eqn.(3.9.5).Itisclearthatthisapproximationwillleadessentiallyto thesameresultsastheonesfollowingfromtheLangevintheory—except forthefactthattheroleof µ willbeplayedby gµB J,whichshouldbe contrastedwiththeexpression(3.9.16)ofthequantumtheory.
3.40 (a) Bydefinition, CH = T (∂S/∂T )H and CM = T (∂S/∂T )M .Now
H =
H ;(1)
atthesametime, dA ≡ dU TdS SdT = HdM SdT ,withthe resultthat(∂H/∂T )M = (∂S/∂M )T .Equation(1)thenbecomes
Multiplying(2)by T ,weobtainthedesiredresultfor CH CM . (b) TheCurielawimpliesthat M = CH /T .Thismeansthat(∂H/∂T )M = H/T ,while(∂M/∂T )H = CH /T 2.Itfollowsthat CH CM = CH 2/T 2
3.42. Let N1(N2)bethenumberofdipolesalignedparallel(opposite)tothe field.Then
1 + N2 = N, while N1ε + N2ε = E. Itfollowsthat
).
Thenumberofmicrostatesassociatedwiththismacrostateisgivenby
N,E)=
(N E/ε) !
2 (N + E/ε) !
Theentropyofthesystemisthengivenbytheexpression
whichisessentiallythesameaseqn.(3.10.9).
