Solution Manual For Thermal Physics, 2nd Edition

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CHAPTER22.1)ENTROPYANDTEMPERATURE(a)o=logg=logC+(3N/2)logU.(80/3U)y=1/1=(3N/2)(1/U),henceU=3Nt/2.(®)(2%0/00%)N=(38/2)(-1/U%)<0.2.2)PARAMAGNETISMStateswithspinexcess2shavetheenergyU=-2smB,hences=-U/2mB.Insertioninto(40)yields(41).From(41)obtain(3a/3U)yB=1/t==U/m?B2N.SolveforU:U=<U>==n2B2N/t=-MB.HenceM/Nm=mB/t.2.3)QUANTUMHARMONICOSCILLATOR(a)From(1.55):0=logg=log(N+n-1)!=logn!-log(N-1)!=log(N+n)!-logn!-logN!ReplacingN-1byNisequivalenttoneglectingaterm~log((N+n)/N},whichforlargeNissmallcomparedtothefactorialterms.WiththeStirlingapproximation:o=(N+n)log(N+n)-nlogn-NlogN=(N+n)log[(N+n)/N]-nlog(n/N)=N[(1+n/N)log(l+n/N)-(n/N)log(n/N)].(b)Setn=U/MwandNfw=Uy(N):0(UN)=N[(1+U/U))log(1+U/U)=(U/Uy)log(u/ug)].-3-[2.3]—StudyXY

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DownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com

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Withtheabbreviationx=U/U,:1/t=(30/30)=(1/04)(30/3%)=(N/Uy)[log(1+x)-1logx]=(1/Mw)log(1+Uy/U).solvingforU/NyieldsU/N=Juw/[exp(Mw/t1)-1].Alternatemethod:Forlargen:d(logn!)/dn=logn!-log(n-1)!=logn.Withthis,1/t=(1/Mw)(3a/3n)y=(1/Hw)[log(N+n)-logn]=(1/Mw)log(1+N/n),whichistheequivalenttotheresultabove.2.4)THEMEANINGOFNEVER(a)Theprobabilityacorrectkeywillbestruckis1/44.Theprobabilityasequenceof10°keyswillbecorrectis(1/44)100,000=107164,345(b)Onasingletypewriterthenumberofkeysthatcanbestruckin10%8sat10keys/sis101°.TheHamletsequencemaystartwithanyoftheseexceptthelast10%-1.Thusthereare10%-10%+1=101%possiblestarts.TheprobabilityofacorrectHamletsequenceononetypewriteris10195707164,345_|-164,326For1010monkeys:1010x7~164,326.107164,316[2.3]-4-|—StudyXY

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Comment.OneUCSBstudentsuggestedthatthemonkeysbepermittedatleastasmanytypographicalerrorsasinthelecturenotesfromwhichthemaintextwasprepared.Itisnotdifficulttoshowthateven1000errorswouldhaveincreasedtheprobabilitybylessthanafactor(43x10%)1000M106633toavaluethatisstillnegligible.2.5)ADDITIVITYOFENTROPYFORTWOSPINSYSTEMS(a)From(17),withN=N,=102%,6=101%;=2-919;=(979)ax®XP(-467/N1)=(979,)1,*exp(-4),(9195)/(979,pax=0-0183.Comparethiswith107174for6=10%2.(b)From(1.35):a(N,s)=(2/n)%x2Nxexp(-282/M).Thismaybeappliedbothtothetwoindividualsystems(N=Ny=Ny),andtothecombinedsystem(N=N+Ny).Forthetwoindividualsystemsintheirmostprobableconfigurationwehavefrom(14),§;=§,.Hence(919)dma=190481%=(27m)x2°Lxexp(-a8,2/m;)9192)max=[90,5010=1P(=45)Np)Forthecombinedsystem,s=28):29,(Npi5))g,(Ny5-51)=g(2N,281)S12N.1a2=@/mip)%x2txexp(-a3,2M)_5-[2.5]|—StudyXY

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Thetworesultsdifferbythefactor9(2M,28,)/[9(8),81%=(N)/20)%=ax10t0(c)Thetrueentropyoftheinteractingcombinedsystemiso=logg(2N),25))=2M)log2~as,2/m,-log(nN,)x1.3863x10%%=1018=25.9=1.3862x10%2Theerrormadeintheentropybyestimatingtheentropyas109(9;9,)pay=09%0,islog(4x10'%)=24.4.Thisisafractionalerrorof1.76x1072L.2.6)INTEGRATEDDEVIATIONForNy=Ny.(17)becomes2919;=(919)paySXP(-467/Np)®.me2919%(919DayJSxwi-262/mya010,©=(919)payN/0)%[exp(-thatXwherex=26/N%.Wecountbothpositiveandnegative§with[8]>6).Then®®(18156)=[exp(-t?)at/[exp(-t?)atx[Jx1(NM52=(2/n%)erfc(x)=bonyFaexp(-46,/Ny)=0.28xexp(-400"=exp(-403.6)=107273|[2.5]-6-|—StudyXY

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CHAPTER33.1)FREEENERGYFORTWOSTATESYSTEM(a)Byinserting(13)into(55):F=-1tlog2=-tlog[l+exp(-c/1)].(b)Insertedinto(49),o=-(3F/31)=log[l+exp(-e/t)]+(¢/1)/[1+exp(e/1)],(81)U=F+to=¢/[l+exp(e/1)],thesameas(14)obtaineddirectlyfromZz.Comment.Ast»0andhenceexp(-£/t)»0,thelogarithmin(S1)maybeexpanded:log[l+exp(-e/t)]+exp(-£/1).Theno»(1l+e/t)exp(-¢£/1)»0.Theexponentialfactorgoestozerofasterthananyinversepowerof1goestoinfinity:BothoandallderivativesofovanishwhentT+»0,asshowninFig.11.Thehigh-temperaturelimitofoisobtainedbylettingexp(-£/t)>1,in(Sl):o>log2+e/2t»log2.3.2)MAGNETICSUSCEPTIBILITY(a)Forasinglemagnet,with&£=mB:zy=exp(mB/t)+exp(-mB/t)=2cosh(mB/t),(s1)<m>=[mexp(+mB/1)-mexp(-mB/1)1/2;=mtanh(mB/t).ThemagnetizationMisobtainedbymultiplyingbytheparticleconcentrationn:M=n<m>=nmtanh(mB/t).Forweakfields,mB<<t:—2.M=nm°B/t(s2)_7-[3.2]|—StudyXY

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Forstrongfields,mB>>1:M=nm.ThesetwolimitsareseeninFig.3.12.(b)Inserting(S1)into(55)andmultiplyingbyn:F=-ntlog2q=-ntlog[2cosh(mB/t)].Toexpresscosh(mB/t)asafunctionofx=M/mm=tanh(mB/1)=tanhyweusetherelation1/cosh?y=(cosh?y-sinh®y)/cosh?y=1-tanh?y=1-x2.Wenextwritelog(2coshy]=-%log(1/4cosh®y)=-%log[(1-x%)/4].Withthis:F=+(n/2)logl(1-x")/4](c)Thesusceptibilityisdefinedasx=dM/dB.InthelimitmB<<t,from(S2):xX=nm?/t.3.3)FREEENERGYOFANHARMONICOSCILLATORThepartitionfunctionisageometricseries:z=poyexp(-sjw/t)=1/[1-exp(-Hw/1)].(a)Insertedinto(55):F=-1logZ=tlog[l-exp(-fuw/t)].(87)[3.2]-8-

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Athightemperatures,Mw/t<<1,sothatl-exp(-Mw/1)=Yw/t.Hencefrom(S1):F=tlog(fw/t).(s2)(b)Theexpression(88)followsdirectlybyinserting(87)into(49),o=-(3F/3T1).Comment.Thelow-temperature(1<<Mw)behavioroftheharmonicoscillatoristhesameasforthetwostatesystemwith¢=Jw,asisapparentfromcomparingFigs.3.13and3.14withFigs.3.11and3.4:Onlythetwoloweststatesmatter.Thehigh-temperaturebehavior(tv>>Mw)isquitedifferent,becausethenumberofacces-siblestatesisnotlimitedto2.Inthislimit,from(s2):0==(3F/31t)»1+log(t/Hw).Ifthisisinsertedinto(17a):Cy=1(da/31)>»1,infundamentalunits.3.4)ENERGYFLUCTUATIONSNotefirstthat<(e-<e>)2>=<6?—2g<e>+<e5%>=eZ—<g>|NextwritethepartitionfunctionasafunctionofBp=1/t:2=2exp(-Be,).2Then=9-[3.4]|—StudyXY

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==1-=-1ladzU=<g>=zeexp(-Bey)=-75w?=156?exp(-pe,)=+=ozz45%2Z452Fromthese,withdg/dt=-1/12,2(3)Ze?(&)a8(3),at),3p),dr3p“4(5%)spfz1(=fdp\zapZgg2TZ\@=<e?=<>?Comment.Manipulationsinvolvingtemperaturediffer-entialsofthepartitionfunctionsarefrequentlysimpli-fiedbyusingB=1/1ratherthan1asindependentvari-able.3.5)OVERHAUSEREFFECTLetU,betheenergyofthereservoirwhentheenergyofthesystemiszero.Then,whenthesystemhastheenergye,thereservoirhas,byoursupposition,theenergyUg-&+ag=Uy-(1-x)e.TheprobabilityP(e)tofindthesysteminaparticularstatewithenergyegisthenproportionaltothenumberofstatesofthereservoirwiththeenergyUy-(1-a)eg,Peg)=gplUg-(1-a)e],withthesameproportionalityfactorforallstates.Hence,insteadof(2)and(3):Bley)_9g[Ug-(1-a)e,}_exp{op[Uy-(1-a)e,1}(51)B(e,)=9p(Uy-(1-a)e,]=expfoglUg-(1-a)e,11°IftheentropyofthereservoirisexpandedaboutU=Uysasin(7):[3.4]-10-|—StudyXY

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0plUg=(1-a)e]=0,(Uy)=(305/8U)(1-a)e+...Ifthisisinsertedinto(S1),oneobtains,insteadof(9),P(e)expl-(l-a)e,/1]P(ey)expl-(1l-a)e,/t]whichisequivalentto(91).3.6)ROTATIONOFDIATOMICMOLECULE(a)Thereare2j+1statesatenergyj(i+l)c,,henceEN=Texploeg/)=z(23+1)expl[-i(i+l)eg/t]wherethesumoverallstateshasbeenconvertedintoasumoverallenergylevels.(b)Thesummaybeviewedasasumovertheareasofrectangleswiththewidthaj=1andwiththeheight£5=(23+1)exp[-j(i+l)eg/T],asshownbelow.Alsoshownisthecurvef(j),withjtreatedasacontinuousratherthandiscretevariable.Notethatthiscurvegoesthroughthemiddleoftheupperedgeofeachrectangle.£.,£03)3JAT]eg/T=0.1/ANN/|ANN/n<J:a3012345S11[3.6]|\—StudyXY

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When©>>£qrtheslopeofthecontinuouscurvef(j)variesonlyverylittlewithineachrectangle,andthesumcanbeapproximatedbytheintegral©z=/£(3)a3=kNotethatthelowerintegrationlimitisnotzero;thisisimportant.Substitutenewvariable:x=JG+e/T,dx=(2+1)(eo/1)dj:wz=[exp(-x)ax=I[1-exp(-x,)1(s1)EyEr0*owherex,=-e,/41isthevalueofxcorrespondingtoj=-1/2.Whent>>eg,X,<<1,andtheexponentialmaybeexpanded:exp(-x4)=z1-XoFoeIfthisisinsertedinto(S1),Z=t/ey+1/4+cee4(s2)whereallomittedtermsdecreaseatleastasrapidlyas1/twhen1>@.Amorecarefultreatmentofthesum(seebelow)yieldstheslightlydifferentresultZ2=t/e,+1/3+Lo..(s3)Thisistheformwewilluseinwhatfollows.(€)2=1+3exp(-2¢0/1)+...(54)wheretheomittedtermsvanishmorerapidlythanthesecondtermwhent>0.(d)Insertzinto(12),U(1)=t2(92/91)/2:[3.6]-12-|—StudyXY

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—2—Treg:U(r)=©£02=T/(1+ey/31)HTx(l-ey/31)=1-£o/3.(85)cv)=1.TeyUT)=begexp(-2¢0/T)/2=beyexp(-2¢,/1)<<6eB(s6)cr)=(12¢4/19)exp(-2e0/T)Notethat,asinProblems3.1and3.3,theexponentialfactorgoestozeromuchfasterthan1/t°goestoin-finity,when1>0.(e)Seedrawingsonnextpage.Theapproximation(S6)dipsbelow(S5),suggestingthatU(t)approachestheasymptote($5)frombelow.ThismeansthatC=3aU/dt>1abovesometemperature,thatis,theheatcapacitygoesthroughamaximum.Anaccuratecalculationconfirmsthisprediction.TheinstructorshouldcheckthatthestudentdrawingsshowcorrectlythevanishingslopeofbothU(r)andC(t)ast+0,ratherthanexhibitinganon-descriptbehavior.DiscussionandElaborations.Unlessforewarned,manystudentsarelikelytoreplacethesumbyanintegralwithalowerintegrationlimitofzero.Asthegraphsof£5and£(j)show,thisisinvalid.Thispitfallmaybetakenasthepointofdepartureforaclassroomdiscussiononseveralmathematicalpointsconcerningthesummationofseries,startingwithawarningagainstthepurelyformalreplacementofsumsbyintegralswithoutagraphicalvisualizationofthedifferencebetweenthetwo.Thedifferencebetween(S2)and(S3)arisesfromthenon-zerosecondderivativeofthecontinuousfunction£(j),whichintegratestoafiniteerrorin(S2).Theformaltoolforamoresystematicaccuratereplacementof-13-[3.61|—Study

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uU=rpd£9ydJSJVipyYVaI—U=1-5/3//bs£92eqAllderivativesarezeroCyLeITT.py2¢eyAllderivativesarezero[3.61“14-+StudyXy

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sumsbyintegralsistheEuler-McLaurinsummingformula.TheSupplementaryMaterialtothisChaptergivesasimplederivationthatdoesnotdrawonthepropertiesoftheEulerpolynomials,whicharecommonlyutilizedintheproofsfoundinmosttextbooks.IftheEuler-McLaurinexpansionisappliedtoourparti-tionfunction,onecanobtainanexpansionofZbypowersofn=£o/Tr1,1,02Z=g*t3+igrom)(57)whereO(n?)standsforomittedtermsofordern2orhigh-er.Toobtainallcontributionsofordern,theEuler-McLaurinexpansionmustbecarrieduptotheterm(1/720)£3)(0).From(S7)oneobtains,bystandardmethods,_22U(r)==&5d(logz)/dn=1=24/3=£,°/45T+O(1/17)or)=1+(1/85)(e/0)7+01/3)Thedetailsarelefttothereader.TheseresultsdoindeedindicatethatU(1)approachesitsasymptotefrombelowandthatC(1)approachesitslimitingvaluefromabove.Asanalternativetotheseanalyticaltreatments,itisnotdifficulttosumthepartitionfunctionnumericallyonaprogrammablecalculator.ThesummationisgreatlyspeededupbycalculatingtheBoltzmannfactorsrecursive-ly:Bj=exp[-j(i*l)eg/1]=ByjxC5C5=exp[-2jey/t]=C51xByBy=Cy=1.-15-[3.61|—StudyXY

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Inthiswayonlyoneexponentialmustbeevaluatedforeach1;theremainingexponentialsintheseriesfollowbysimplemultiplications,atalargesavingincalculationtime.If2issummednumerically,theenergyismostconve-nientlyobtainedfromthefirstequalityin(3.12),byalsosummingu=1&_exp(-e_/1)TzzseXP(eg/T-Inthiswaythedifferentiationofnumericaldataisavoided.similarly,theheatcapacitycanbeobtainedwithoutdifferentiationfrom=_1(12--v7c=52[izexp(-¢/1)v?].3.7)ZIPPERPROBLEM(a)Astateinwhichslinksareopencanberealizedinonlyoneway.ThusthepartitionfunctionisZ=1+exp(-¢/t)+exp(-2¢/t)+...+exp(~-Ne/1).NN+1=3x®=BEwherex=exp(-e/1)(93)2xs=0(b)TheaveragenumberofopenlinksisN=1Ss_da<s>=2Ysx=x3xlogz.(s1)s=0N+1If¢>>tr,thenx<<1,andwemayneglectthetermxin(93)toobtain<>=-xL1og(1-x)=X==1/[exp(e/1)-1]dxI-x:ThisisoftheformofthePlanckdistribution.[3.61-16-|—StudyXY

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Extension.Ourassumptionthateachlinkhasonlyoneopenstateisanunrealisticassumption,whichneglectsthatthetwohalvesofanopenlinkmayhavemanydiffer-entorientationsrelativetoeachother.Itisinstruc-tivetogeneralizetheproblembyassumingthateachopenlinkhasGopenstateswiththeenergy&.Thechangehasfar-reachingconsequences.AstateofthezipperwithsopenlinksisthenG°-folddegenerate,andthepartitionfunctionnowbecomesZ=1+Gexp(-£/1)+Glexp(-2e/1)+...NN+1coo+Mexp(anesr)=3xS=IEZ-xs=0whereXx=Gexp(-g/t),whichdiffersfromtheearlierformonlybythefactorG>1inthedefinitionofx.Becauseofthisfactor,valuesx>1arenowpossible.Thishasdrasticconse-quencesifthetotalnumberoflinksisverylarge,N>>1.InthiscasetheopeningofthezipperapproachesthebehaviorofanabruptphasetransitionatthesharptransitiontemperatureTy=€/logGwhichis‘thetemperatureforwhichx=1.Fortempera-turesverylittlebelowt,,onlyaverysmallfractionofthelinksareopen,fortemperaturesverylittleaboveTyalmostalllinksareopen.ThelargerN,thenarrowerthetemperatureintervaloverwhichtheopeningtakesplace.Wegivehereonlythekeypointsinthederivationofthisresult.Itisnotdifficulttoshowthattheexpression(S1)for<s>canbewrittenas_17-[3.71|—Study

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