NOTES AND SOLUTIONS TO THERMAL PHYSICS BY CHARLES KITTLE AND HERBERT KROEMER

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by  ERNEST YEUNG

NOTES AND SOLUTIONS TO THERMAL PHYSICS BY CHARLES KITTLE AND HERBERT KROEMER
ERNESTYEUNG-LOSANGELES
ABSTRACT. ThesearenotesandsolutionstoKittleandKroemer’sThermalPhysics. Thesolutionsare(almost)complete: Iwill
continuouslyaddtosubsections,beforetheproblemsineachchapter,mynotesthatIwritedownasIread(andcontinuouslyreread).
IamattemptingamanifoldformulationoftheequilibriumstatesinthestyleofSchutz’sGeometricalMethodsofMathematical
PhysicsandwillpointouthowitappliesdirectlytoThermalPhysics. Otherusefulreferencesalongthisavenueofinvestigationis
providedattheverybottominthereferences.
Anyandallfeedback,includingnegativefeedback,iswelcomedandyoucanreachmebyemailormywordpress.comblog.
Youarefreetocopy,edit,paste,andaddontothepdfandLaTeXfilesasyoulikeinthespiritofopen-sourcesoftware. Youare
responsibleadultstousethesenotesandsolutionsasgovernedbytheCaltechHonorCode:“NomemberoftheCaltechcommunity
shalltakeunfairadvantageofanyothermemberoftheCaltechcommunity”andfollowtheHonorCodeinspirit.
SECOND EDITION. Thermal Physics. Charles Kittel. Herbert Kroemer. W. H. Freeman and Company. New York.
QC311.5.K521980536’.7ISBN0-7167-1088-9
1. STATESOFAMODELSYSTEM
2. ENTROPYANDTEMPERATURE
Thermal Equilibrium. EY : 20150821 Based on considering the physical setup of two systems that can only exchange
energybetweeneachother,thatareinthermalcontact,thisisaderivationoftemperature.
U =U +U isconstanttotalenergyof2systems1,2inthermalcontact
1 2
multiplicityg(N,U)ofcombinedsystemis
(cid:88)
g(N,U)= g (N ,U )g (N ,U −U )
1 1 1 2 2 1
U1≤U
The“differential”ofg(N,U)is
(cid:18) (cid:19) (cid:18) (cid:19)
∂g ∂g
dg = 1 g dU +g 2 dU =0
∂U 2 1 ∂U 2
1 N1 2 N2
EY:20150821Thisstepcanbemademathematicallysensiblebyconsideringtheexteriorderivativedofg ∈C∞(Σ),where
Σ is the manifold of states of the system, with local coordinates N,U, where U happens to be a global coordinate. Then,
consideracurveinΣs.t. ithasnocomponentin ∂ , ∂ ,andthiscurveisa“nullcurve”sothatthevectorfieldX ∈X(Σ)
∂N ∂N1
generatedbythiscurveiss.t. dg(X)=0.
With−dU =dU ,
1 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
1 ∂g 1 ∂g ∂lng ∂lng
1 = 2 =⇒ 1 = 2
g ∂U g ∂U ∂U ∂U
1 1 N1 2 2 N2 1 N1 2 N2
Define
σ(N,U):=lng(N,U)
Then
(cid:18) (cid:19) (cid:18) (cid:19)
∂σ ∂σ
=⇒ 1 = 2
∂U ∂U
1 N1 2 N2
Date:Fall2008.
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1
Temperature. T =T -temperaturesof2systemsinthermalequilibriumareequal.
1 2
T “mustbeafunctionof(cid:0)∂σ(cid:1) [?].
∂U N
(cid:18) (cid:19)
1 ∂σ
=⇒ =k
T B ∂U
N
Experimentally,k =1.381×10−23J/K =1.381×10−16 ergs/K.
B
Now
(cid:18) (cid:19)
1 ∂σ
= τ =k T
τ ∂U B
N
Problems. Solution1. Entropyandtemperature.
(a) Recallthat 1 ≡(cid:0)∂σ(cid:1) andσ(N,U)≡logg(N,U). Giveng(U)=CU3N/2,
τ ∂U N,V
3N
σ(N,U)=logCU3N/2 =logC+ logU
2
∂σ 3N 1 1 3N
= = =⇒ U = τ
∂U 2 U τ 2
(cid:16) (cid:17)
(b) ∂2σ <0?
∂U2
N
(cid:18)∂2σ(cid:19) 3N (cid:18) 1 (cid:19)
=− <0
∂U2 2 U2
N
Solution2. Paramagnetism.
U
U(s)=U (s )+U (s )=−2mB(s +s )=−2mBsors=
1 1 2 2 1 2 −2mB
i.e. potentialenergyU(s)=−2s·mB.
For|s|(cid:28)N,then
g(N,s)(cid:39)g(N,0)exp(cid:0)−2s2/N(cid:1)=g(N,0)exp(cid:18) −U2 (cid:19)
2(mB)2N
U2 1
σ(N,U)=lng(N,U)=σ − whereσ =lng(N,0)
0 2m2B2N 0
(cid:18) (cid:19)
1 ∂σ −U 1
= =
τ ∂U m2B2N
N
WhatisthethermalequilibriumvalueofthisN-spinsystemoffractionalmagnetization? IfU denotes(cid:104)U(cid:105),thermalaverage
energy,wealsogetthethermalaveragespinexcess.
(cid:104)U(cid:105)=(cid:104)−2mBs(cid:105)=−2mB(cid:104)s(cid:105)
m2B2N mBN
=⇒τ = =
−U 2(cid:104)s(cid:105)
Solution3. Quantumharmonicoscillator.
(a) ResultfromCh. 1: g(N,n)= (N+n−1)!.
n!(N−1)!
LetN −1→N =⇒g(N +1,n)= (N+n)!.
n!N!
(N +n)!
σ(N +1,n)≡lng(N +1,n)=ln =ln(N +n)!−ln(n!)−ln(N!)
n!N!
≈(N +n)ln(N +n)−N −n−nlnn+n−NlnN +N = (N +n)ln(N +n)−nlnn−NlnN
(b) LetU denotetotalenergyn(cid:126)ωofoscillators.
U =n(cid:126)ωorn= U
(cid:126)ω
U U U U
σ(N,U)=(N + )ln(N + )− ln −NlnN
ω ω ω ω
Atτ, 1 =(cid:0)∂σ(cid:1) .
τ ∂U N
(cid:18) (cid:19)
1 1 U 1 U 1 Nω Nω
= ln(N + )− ln = ln +1 or U =
τ ω ω ω ω ω U exp(ω/τ)−1
2
Solution4. Themeaningof“never.”
Suppose1010monkeys.
(a) Hamlet represents one specific ordering of 1015 with 44 possibilities for each character. The probability of hitting
upon Hamlet from a given, random sequence is (cid:0) 1 (cid:1)100000 = 1 . Given that log 44 = 1.64345, then
44 44100000 10
101.64345 =44or10−1.64345 =44−1sothen
(cid:18) 1 (cid:19)100000
=10−164345
44
(b)
(cid:18) (cid:19) (cid:18) (cid:19)
10keys 10keys
(ageofuniverse)· =1018s =1019keys
second second
(cid:18) (cid:19)
1hamlet
1019keys·1010monkeys=1029keystypedout =1024possible“Hamlets”
105characters
Frompart(a),theprobabilitythatagiven,randomsequenceisHamlet,10−164345
(1029characters)(10−164345)=10−164316
Note,Ithinkthattheprobabilityshouldbe(1029characters)(cid:0) 1Hamlet (cid:1)(10−164345)=10−164321
105characters
Sinceweareconsideringthenumberof“Hamlet”,105charactersequences.
3. BOLTZMANNDISTRIBUTIONANDHELMHOLTZFREEENERGY
cf. Example: Energy and heat capacity of a two state system, pp. 62 of Kittel and Kroemer [1]. Kittel and Kroemer
introducestheheatcapacityveryearly,specifictothisexample.
Definition1. heatcapacityC atconstantvolumeisdefinedas
V
(cid:18) (cid:19)
∂σ
(1) C :=τ
V ∂τ
V
Recallthethermodynamicidentity(whichisintroducedmanyequationslater):
dU =τdσ−pdV ∈Ω1(Σ)
whereΣisamanifoldofstatesofallsystems.
ConsiderlocalcoordinatesofΣ,(σ,V). Considercurve c:R→Σs.t. cgeneratesavectorfieldc˙=σ˙ ∂ i.e. nocomponent
∂σ
c(τ)∈Σ
intheV direction. Noticetheprescientchoiceofparameterτ.
NowforinternalenergyU ∈C∞(Σ),takingtheexteriorderivativedresultsin
∂U ∂U
dU = dσ+ dV
∂σ ∂V
ThenapplyingdU ontovectorfieldσ˙ ∂ ,
∂σ
(cid:18) (cid:19)
∂ ∂U
dU σ˙ = σ˙ =σ˙τ +0
∂σ ∂σ
Now,
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
∂U ∂σ ∂U ∂σ
= = τ
∂σ ∂τ ∂τ ∂τ
V V V V
Hence,
(cid:18) (cid:19) (cid:18) (cid:19)
∂σ ∂U
(2) C :=τ =
V ∂τ ∂τ
V V
EY:20150825Whydoweneeddifferentialgeometry? It’sbecauseIalwayswonderedwhyyoucoulddothis:
(cid:18) (cid:19) (cid:18) (cid:19)
∂σ ∂U
? ?
C :=τ = withτdσ =dU =⇒τ∂σ =∂U
V ∂τ ∂τ
V V
andtalkof“differentials.”
Definition: Reversibleprocess. EY:20150824Mathematically,1-formsareexact.
3
Pressure. Considercoordinates(σ,V)∈ΣofmanifoldofthermodynamicstatesΣ.
Imagineareversiblecompressionofacubesystem(soimaginedV <0;cube’svolumegetsmaller).
σ constant,i.e. dσ = 0(onthiscurveinΣ)becauseasparticlesincubegetssqueezed,lesspositionsparticlescouldsitin,
buttheygetmorekineticenergy,moreenergetic(moremomentumsquared).
NowU =U(σ,V)∈C∞(Σ).
=⇒dU =(cid:0)∂U(cid:1) dσ+(cid:0)∂U(cid:1) dV
∂σ V ∂V σ
Again,imagineacurvec:R→Σ,connecting1state(σ,V)∈Σtoanotherstate(σ,V +dV)∈Σs.t. c˙=V˙ ∂ .
∂V
(cid:18) (cid:19)
∂U
=⇒dU(c˙)= V˙
∂V
σ
Introduce1-formW ∈Ω1(Σ)ofworkdoneonthecubesystemfromsomeexternalagent
W =−pdV
soW >0whendV <0.
Then
(cid:18) (cid:19)
∂U
W(c˙)=−pV˙ =dU(c˙)= V˙
∂V
σ
(cid:18) (cid:19)
∂U
(3) =⇒ p=−
∂V
σ
Consider another set of coordinates (U,V) ∈ Σ for manifold Σ. Now entropy σ is a function of U,V, as σ = σ(U,V) ∈
C∞(M),sothat
dσ =(cid:0)∂σ(cid:1) dU +(cid:0)∂σ(cid:1) dV
∂U V ∂V U
Considercurvec = (U,V) ∈ Σ. Thenc˙ = U˙ ∂ +V˙ ∂ . Forthiscurvec,σ isconstant,meaningdσ(c˙) = 0(it’sa“null
∂U ∂V
curve”ofdσ
(cid:18) (cid:19) (cid:18) (cid:19)
∂σ ∂σ
dσ(c˙)=0= U˙ + V˙
∂U ∂V
V U
Nowdefine
Definition2.
(cid:18) (cid:19)
1 ∂σ
(4) :=
τ ∂U
V
Sothenwehave 1U˙ +(cid:0)∂σ(cid:1) V˙ =0. Fortheparameterofcurvec,choosetheparametertobeV,knowingthatσisconstant
τ ∂V U
onthiscurve,orthermodynamicprocess. Thus
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
1 ∂U ∂σ −p ∂σ
=− =⇒ =− or
τ ∂V ∂V τ ∂V
σ U U
(cid:18) (cid:19)
∂σ
(5) p=τ
∂V
U
ThermodynamicIdentity. Letσ =σ(U,V)∈C∞(Σ). Then
dσ =(cid:0)∂σ(cid:1) dU +(cid:0)∂σ(cid:1) dV ∈Ω1(Σ).
∂U V ∂V U
Nowrecallthequantitieswe’verecentlyused: 1 :=(cid:0)∂σ(cid:1) (thisisadefinition)and p =(cid:0)∂σ(cid:1) (itcomesfromthephysics,
τ ∂U V τ ∂V U
ofdoingworkonthesystem,bysomeexternalagent). Thenthethermodynamicidentityisobtained:
Theorem1.
(6) τdσ =dU +pdV
IdealGas: AFirstLook.
4
Oneatominabox. oneatomofmassM incubicalboxofvolumeV =L3
−(cid:126)2∇2ψ =(cid:15)ψ p= (cid:126)∇ p2ψ =(cid:15)ψ
2M i
=⇒ψ(x)=Asin(cid:0)nxπx(cid:1)sin(cid:0)nyπy(cid:1)sin(cid:0)nzπz(cid:1)
L L L
(cid:15) = (cid:126)2 (cid:0)π(cid:1)2(n2 +n2 +n2)
n 2M L x y z
ThenthepartitionfunctionZ is
1
Z =(cid:88)exp(cid:18)−(cid:15)n(cid:19)= (cid:88) exp(cid:18) −(cid:126)2 (cid:16)π(cid:17)2(n2 +n2 +n2)(cid:19)
1 τ 2Mτ L x y z
n (nx,ny,nz)
Let
(cid:126)2π2 (cid:126)π
α2 = orα=
2ML2τ (2Mτ)1/2V2/d
Then
(cid:90) ∞ (cid:90) ∞ (cid:90) ∞ (cid:18)(cid:90) ∞ (cid:19)3 (cid:18)1(cid:19)3(cid:18)π1/2(cid:19)3 (cid:18)π1/2(cid:19)3
Z = dn dn dn exp[−α2(n2 +n2 +n2)]= dn exp(−α2n2) = =
1 x y z x y z x x α 2 2α
0 0 0 0
(cid:16) (cid:17)d
Ingeneral,Z = π1/2
1 2α
Now
(cid:32) (cid:33)d
π1/2V1/d V n
Z = = =n V = Q
1 2 (cid:126)π (cid:0)2(cid:126)π(cid:1)d/2 Q n
(2Mτ)1/2 Mτ
intermsofconcentrationn=1/V.
n :=(cid:0) Mτ (cid:1)d/2isthequantumconcentration.
Q 2π(cid:126)2
Problems. Solution1. Freeenergyofatwostatesystem.
(a)
Z =1+e−(cid:15)/τ
F =−τlnZ =−τln(1+e−(cid:15)/τ)
(b)
∂(F/τ) (cid:15)e−(cid:15)/τ
U =−τ2 =
∂τ 1+e−(cid:15)/τ
∂F (cid:15)e−(cid:15)/τ
σ =− =ln(1+e−(cid:15)/τ)+ τ
∂τ (1+e−(cid:15)/τ)
Solution2. Magneticsusceptibility
(a) RemembertocalculatethemultiplicityintheN-spinsystem(it’snotenoughtosumupexp(−(cid:15) /τ)factors).
s
M =2sm U =−MB =−2smB N =N +N
s + −
2s=N −N =N −(N −N )=2N −N
+ − + + +
N/2 (cid:18) (cid:19) (cid:18) (cid:19) N/2 (cid:18) (cid:19) N (cid:18) (cid:18) (cid:19)(cid:19)
(cid:88) N 2smB (cid:88) N! 2mBs (cid:88) N! 2mB N
Z = exp = exp = exp s− =
N τ (cid:0)N +s(cid:1)!(cid:0)N −s(cid:1)! τ s!(N −s)! τ 2
s=−N/2 + s=−N/2 2 2 s=0
(cid:18) (cid:19)
mB
=e−NmτB(1+e2mτB)N =2NcoshN
τ
whereitwascrucialtouse(1+x)N = (cid:80)N (cid:0)N(cid:1)xj. Note, inchangingthesumindex, sinceN islarge, wecan
j=1 j
neglectdroppingthes=0term.
(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
mB mB −m
∂ Z =2N(N)(coshN−1 )sinh
τ τ τ τ2
(cid:18) (cid:19)
∂ mB
M =−τ2 lnZ =Nmtanh
∂τ τ
∂M Nm2 (cid:18)mB(cid:19)
χ= = sech2
∂B τ τ
5
(b)
(cid:18) (cid:18) (cid:19) (cid:19) (cid:18) (cid:19)
mB mB
F =−τlnZ =−τln (2cosh )N =−Nτln(2cosh )
τ τ
Forx≡ M =tanh(cid:0)mB(cid:1). Now1−tanh2y =sech2y. F =−Nτln(cid:16)√ 2 (cid:17)= −Nτ ln(cid:16) 4 (cid:17).
nm τ 1−x2 2 1−x2
(c) For mB (cid:28)1,cosh2(cid:0)mB(cid:1)→1. χ= m2N
τ τ τ
Solution3. Freeenergyofaharmonicoscillator
(a)
Z =(cid:88)∞ exp(cid:18)−s(cid:126)ω0(cid:19)= 1 =(1−e−(cid:126)ω0/τ)−1
τ 1−e−(cid:126)ω0/τ
s=0
(cid:18)(cid:126)ω (cid:19) (cid:126)ω
F =−τlnZ =τln(1−e−(cid:126)ω0/τ)(cid:39)τln 0 for1(cid:29) 0
τ τ
(b)
σ =−∂∂Fτ =−{ln(1−e−(cid:126)ω0/τ)+ 1−e−τ(cid:126)ω0/τ(−e−(cid:126)ω0/τ)(cid:18)(cid:126)τω0(cid:19)}= e(cid:126)−ω(cid:126)0/ωτ0/−τ1 −ln(1−e−(cid:126)ωτ0)
Solution4. Energyfluctuations.
(cid:88)
Z = e−(cid:15)sβ
1 s
τ =β ∂βZ =(cid:88)−(cid:15)e−(cid:15)sβ U = (cid:80)s(cid:15)se−(cid:15)s/τ =−∂ lnZ
∂ ∂β ∂ 1 ∂ Z β
= =− s
∂τ ∂τ ∂β τ2∂β ∂2Z =(cid:88)(cid:15)2e−(cid:15)sβ
β s
s
∂U =−β2 ∂ U =β2(cid:18)∂ (cid:18)∂βZ(cid:19)(cid:19)=β2(cid:32)(∂β2Z)Z−(∂βZ)2(cid:33)=β2(cid:32)∂β2Z −(cid:18)∂βZ(cid:19)2(cid:33)=
∂τ ∂β β Z Z2 Z Z
∂U
=⇒τ2 =(cid:104)(cid:15)2(cid:105)−(cid:104)(cid:15)(cid:105)2
∂τ
Solution5. Overhausereffect. SystemS inenergyeigenstateEn =n(cid:15).
P(E)=(1)g (E)
R
Note∆UR =(α−1)(cid:15). ∆US =(cid:15). ddU(cid:15)S + dUd(cid:15)R =1+(α−1)=α= dUdt(cid:15)ot inaspecificenergyeigenstate;gS(n(cid:15))=1
Whileg (U )=multiplicityofreservoirRwithU energy.
R R R
Now
∂σ 1
R =
∂E τ
s
and
g (U )=exp(σ (U ))
R R R R
If(cid:15)dUd(cid:15)R =(α−1)(cid:15)=∆URsmallcomparedtoUR.
dU
U (E =(n+1)(cid:15))=U (n(cid:15))+ R(cid:15)=U (n(cid:15))+(α−1)(cid:15)
R S R d(cid:15) R
1
σ (U ((n+1)(cid:15)))(cid:39)σ (U (n(cid:15)))+ (α−1)(cid:15)
R R R R τ
P(E =(n+1)(cid:15)) exp(σ (U (n(cid:15)))+ 1(α−1)(cid:15)) (cid:16) (cid:15) (cid:17)
S = R R τ = exp − (1−α)
P(E =n(cid:15)) exp(σ (U (n(cid:15)))) τ
S R R
Solution6. Rotationofdiatomicmolecules.
(a) (cid:15)(j)=j(j+1)(cid:15) . g(j)=2j+1
0
RememberthatZ isasumoverallstates,notoveralllevels.
Z =(cid:88)∞ (2j+1)e−j(j+1)(cid:15)0/τ =(cid:88)∞ d(e−(j2+j)(cid:15)0/τ)(cid:18)−τ(cid:19)= −τ (cid:88)∞ d (cid:16)e−(cid:15)0/τ(cid:17)j2+j
dj (cid:15) (cid:15) dj
0 0
j=0 j=0 j=0
(b) For1(cid:29) (cid:15)0
τ
τ (cid:90) ∞ d (cid:16) (cid:17)x2+x τ (cid:16) (cid:17) τ
Z (τ)=− e−(cid:15)0/τ dx=− (e−(cid:15)0/τ)x2+x−(e−(cid:15)0/τ)0 =
R (cid:15) dx (cid:15) (cid:15)
0 0 0 0
6
(c) For τ (cid:28)1
(cid:15)0
Z (τ)=1+3e−2(cid:15)0/τ
R
(d)
(cid:18) (cid:19)
∂lnZ (cid:15) τ
U =τ2 for1(cid:29) 0 U =τ2∂ ln =τ
∂τ τ τ (cid:15)
0
for τ (cid:28)1, U =τ2(cid:18) 1 (cid:19)(3e−2(cid:15)0/τ)(cid:18)2(cid:15)0(cid:19)= 6(cid:15)0e−2(cid:15)0/τ
(cid:15)0 1+3e−2(cid:15)0/τ τ2 1+3e−2(cid:15)0/τ
(cid:18) (cid:19)
∂U (cid:15)
C = =1when1(cid:29) 0
V ∂τ τ
V
(cid:32)e−2(cid:15)0/τ(cid:0)2(cid:15)0(cid:1)(1+3e−2(cid:15)0/τ)−(3e−2(cid:15)0/τ)(cid:0)2(cid:15)0(cid:1)e−2(cid:15)0/τ(cid:33) e−2(cid:15)0/τ (cid:18) 1 (cid:19)
C =6(cid:15) τ2 τ2 =12(cid:15)2
V 0 (1+3e−2(cid:15)0/τ)2 0 τ2 (1+3e−2(cid:15)0/τ)2
(cid:18)e−2(cid:15)0/τ(cid:19)
Forverysmall τ (cid:28)1, C ≈12
(cid:15)0 V (τ/(cid:15)0)2
(e) Seesketch.
Solution7. Zipperproblem.
(a) N links. (cid:15) =0closed,(cid:15)open.
s
(cid:88)N 1−e−(N+1)(cid:15)/τ
Z = exp(−s(cid:15)/τ)=
1−e−(cid:15)/τ
s=0
(b) 1(cid:29) τ.
(cid:15)
Z =(cid:80)N exp(−s(cid:15)B)
s=0
−1 −1(cid:80)N (−s(cid:15))e−s(cid:15)β
∂ lnZ = s=0 =(cid:104)s(cid:105)=
(cid:15) β (cid:15) Z
(cid:18) 1−e−(cid:15)/τ (cid:19)(cid:18)−e−(N+1)(cid:15)/τ(−(N +1)(cid:15))(1−e−(cid:15)/τ)−(−e−(cid:15)/τ)(−(cid:15))(1−e−(N+1)/(cid:15)/τ)(cid:19)
= =
1−e−(N+1)(cid:15)/τ (1−e−(cid:15)/τ)2
(cid:15)e−(cid:15)/τ(e−N(cid:15)β(N +1)(1−e−(cid:15)β)−(1−e−(N+1)(cid:15)/τ)) (cid:15)e−(cid:15)/τ(e−N(cid:15)β(N +1)(1−e−(cid:15)β)−1)
= (cid:39)
(1−e−(N+1)(cid:15)/τ)(1−e−(cid:15)/τ) (1−e−(cid:15)β)
Thisstilldoesnotgivethedesiredapproximation. Considerthefollowing:
1−e−(N+1)(cid:15)β e(cid:15)β −e−N(cid:15)β
Z = =
1−e−(cid:15)β e(cid:15)β −1
(cid:15)((e(cid:15)β +Nβe−N(cid:15)β)(e(cid:15)β −1)−(e(cid:15)β)(e(cid:15)β −e−N(cid:15)β)) (cid:15)(e(cid:15)β(e(cid:15)β +Ne−N(cid:15)β −e(cid:15)β +e−N(cid:15)β)−e(cid:15)β −Ne−N(cid:15)β)
∂ Z = =
β (e(cid:15)β −1)2 (e(cid:15)β −1)2
∂ Z (cid:15)(e(cid:15)β(N +1)e−N(cid:15)β −e(cid:15)β −Ne−N(cid:15)β) (cid:15)(e(cid:15)β(N +1)e−N(cid:15)β −e+(cid:15)β −Ne−N(cid:15)β)
β = (cid:39) =
Z (e(cid:15)β −1)(e(cid:15)β −e−N(cid:15)β) e2(cid:15)β
(cid:15)(Ne−(N−1)(cid:15)β +e−(N−1)(cid:15)β −e(cid:15)β −Ne−N(cid:15)β) (cid:15)((N +1)e−(N−1)(cid:15)β −(e(cid:15)β +Ne−N(cid:15)β))
= = =
e2(cid:15)β e2(cid:15)β
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
(cid:15) (Ne+N1(cid:15))βe(cid:15)β − eNn(cid:15)β −e(cid:15)β (cid:15) (N+e1N)e(cid:15)(cid:15)ββ−N −e(cid:15)β (cid:15) Ne(cid:15)β+ee(cid:15)Nβ−(cid:15)βe(cid:15)βeN(cid:15)β (cid:18)N −eN(cid:15)β(cid:19)
= = = =(cid:15) (cid:39)
e2(cid:15)β e2(cid:15)β e2(cid:15)β eN(cid:15)βe(cid:15)β
(cid:15)(−eN(cid:15)β)
(cid:39) =−(cid:15)e−(cid:15)β
eN(cid:15)βe(cid:15)β
=⇒ (cid:104)s(cid:105)=e−(cid:15)/τ
Solution8. Quantumconcentration. NowΨ(x,y,z)=Asin(cid:0)nxπx(cid:1)sin(cid:0)nyπy(cid:1)sin(cid:0)nzπz(cid:1). p= 1∇, p2 =− 1 ∇2.
L L L i 2m 2m
Groundorbital: n =n =n =1.
x y z
3 (cid:16)π(cid:17)2 3 (cid:16)π(cid:17)2
T = (cid:104)ψ |ψ (cid:105)=
2m L 0 0 2m L
7
where(cid:104)ψ |ψ (cid:105)=1,ψnormalized. Itwasnormalizedinthisway:
0 0
(cid:90) ∞sin2(cid:16)nxLπx(cid:17)dnx =(cid:90) L 1−cos2(cid:0)2nLxπx(cid:1)dnx = nx−sin(cid:0)2n2Lxπx(cid:1)(cid:16)2nLxπ(cid:17)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)L = L2
0 0 (cid:12)
0
(cid:18)L(cid:19)3 A2L3 8
(cid:104)ψ|ψ(cid:105)=A2 = =1orA2 =
2 8 L3
Recallthatn =(cid:0) mτ (cid:1)3/2.
Q 2π(cid:126)2
Consider the condition that there will be a concentration for which the zero-point quantum kinetic energy is equal to the
temperatureτ:
3 π2 3 2mτ (cid:18) 2mτ (cid:19)3/2
=⇒ = π2n2/3 =τ orn2/3 = orn=
2mL2 2m 3π2(cid:126)2 3π2(cid:126)2
(cid:18) 4 (cid:19) mτ (cid:18) 4 (cid:19)3/2
=⇒ n=( )3/2 = n
3π 2π(cid:126)2 3π Q
Solution9. Partitionfunctionfortwosystems.
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Z(1+2)= (cid:88) g(E )exp −E1+2 = (cid:88) g(E +E )exp −E1 exp −E2 =
1+2 τ 1 2 τ τ
E1+2 E1+E2=E0
(cid:18) (cid:19) (cid:18) (cid:19)
=(cid:88)(cid:88)g(E )g(E )exp −E1 exp −E2 =Z(1)Z(2)
1 2 τ τ
E1 E2
sincesystemsareindependent.
Solution10. Elasticityofpolymers.
(a) Consider2s=N −N ; N =N +N , 2s=N −(N −N )=2N −N. N = 2s+N.
+ − + − + + + + 2
For2s,consider−2s=N −(N −N )=2N −N. N = −2s+N.
+ + + + 2
2N!
=⇒ g(N,−s)+g(N,s)=
(cid:0)N +s(cid:1)!(cid:0)N −s(cid:1)!
2 2
(b) |s|(cid:28)N
 
(cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
l −l 2(N!)
σ(l)=ln g N, +g N, =ln(cid:16) (cid:17) (cid:16) (cid:17) =
2ρ 2ρ N + l ! N + −l !
2 2ρ 2 2ρ
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
N L N l N l N l N l N l
=ln(2N!)={ + ln + − + + − ln − − − }
2 2ρ 2 2ρ 2 2ρ 2 2ρ 2 2ρ 2 2ρ
whereweusedln(x+∆x)(cid:39)lnx+ 1∆x.
x
(cid:18) (cid:19)(cid:18) (cid:18) (cid:19) (cid:19) (cid:18) (cid:19)(cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
N l N 1 l N N l N 1 −l
σ(l)=ln(2N!)−{ + ln + − + − ln + }=
2 2ρ 2 N/22ρ 2 2 2ρ 2 N/2 2ρ
N N N N N N l (cid:18)N(cid:19) l 2 (cid:18) l (cid:19)2 −l l (cid:18)N(cid:19) l2
=ln(2N!)−{ ln − + ln − + ln + + + − ln + }
2 2 2 2 2 2 2ρ 2 2ρ N 2ρ 2ρ 2ρ 2 N2ρ2
(cid:32) (cid:33)
2N! l2
σ(l)=ln −
(cid:0)N(cid:1)!(cid:0)N(cid:1)! Nρ2
2 2
(c)
∂σ −2l 2lτ
= =⇒ f =
∂l Nρ2 Nρ2
Solution11. One-dimensionalgas.
(cid:126)2 (cid:16)π(cid:17)2
(cid:15) = n2 inonedimension
n 2m L
√
Z = (cid:88)∞ exp(cid:18)−(cid:126)2 (cid:16)π(cid:17)2n2/τ(cid:19)=⇒(cid:90) ∞dne−α2n2 = π
1 2m L 2α
n=1 0
8
(cid:126)2 (cid:16)π(cid:17)2
α2 =
2mτ L
where .
(cid:126)π/L
α= √
2mτ1/2
Recallthatσ =(cid:0)∂F(cid:1),F =U −τσ,sothat
∂τ
√
π (cid:18)2α(cid:19) (cid:18) 2(cid:126)π/L (cid:19)
F =−τlnZ =−τNln =τNln √ =τNln √ √ =
2α π π 2mτ1/2
(cid:32) √2π(cid:126) (cid:33) 1 (cid:18) 2π(cid:126)2 (cid:19)
=τNln √ = τNln
mLτ1/2 2 mL2τ
∂F 1 (cid:18) 2π(cid:126)2 (cid:19) τN (cid:32) 1 (cid:33)(cid:18)−2π(cid:126)2(cid:19)
= Nln + =
∂τ 2 mL2τ 2 (cid:0) 2π(cid:126)2 (cid:1) mL2τ2
mL2τ
1 (cid:18) 2π(cid:126)2 (cid:19) −τN N (cid:18) (cid:18) 2π(cid:126)2 (cid:19) (cid:19)
= Nln + = ln −1
2 mL2τ 2τ 2 mL2τ
4. THERMALRADIATIONANDPLANCKDISTRIBUTION
Problems. Solution1. Numberofthermalphotons.
WeconsideracavityofvolumeV,andofedgelengthL(soV L3). Sothenω =nπc/L.
n
Now 1 isthethermalaveragenumberofphotonsinasinglemodefrequencyω. Sothen
exp((cid:126)ωn)−1
τ
(cid:88) (cid:88) 1
(cid:104)s (cid:105)=
n exp(cid:0)(cid:126)ωn(cid:1)−1
τ
Consider(n ,n ,n )onpositiveoctant,and2independentpolarizationsofemfield.
x y z
(cid:88) 1 (2)(cid:90) ∞ 1 (cid:90) ∞ n2dn
= 4πn2dn =π =
exp(cid:0)(cid:126)nπc(cid:1)−1 8 exp(cid:0)(cid:126)nπc(cid:1)−1 exp(cid:0)(cid:126)nπc(cid:1)−1
n Lτ 0 Lτ 0 Lτ
(cid:18)Lτ (cid:19)3(cid:90) ∞(cid:18) x2dx (cid:19) V (cid:16) τ (cid:17)3
=π = N = (2.404)
(cid:126)πc ex−1 π2 (cid:126)c
0
whereIusedthesubstitutions
(cid:126)πcn
x=
Lτ
Lτx
=n
(cid:126)πc
Lτ
(dx) =dn
(cid:126)πc
Nowσ(τ)=(4π2V/45)(τ/(cid:126)c)3,sothen
σ 1 (cid:18)4π4(cid:19)
= (cid:39) 3.602
N 2.404 45
Nowhowwas(cid:82)∞dxx2dx evaluated?
0 ex−1
Solution2. SurfacetemperatureofaSun. GiventhesolarconstantoftheEarth,thetotalradiantenergyfluxdensityat
theEarthfromtheSunnormaltotheincidentrays,integratedoverallemissionwavelengths,
solarconstant =0.136Js−1cm−2, (49)
(a)
(cid:18)102cm(cid:19)2
4π(1.49×1011m)2·0.136Js−1cm−2 =(4π)(1.49)21022·0.136×104 = 3.8×1026J ·s−1
1m
NotethatIhadused1.49×1011masthedistanceoftheEarthfromtheSun.
(b) J =energyfluxdensityorrateofenergyemissionperunitarea.
ν
σB = 6π02(cid:126)k3B4c2 =5.670×10−8W m−2K−4.
NotethatW = 1J. IwilluseR =6.9599×1010cmastheradiusoftheSun.
s ◦
4×1026J ·s−1
=J =σ T4
4π(6.9599×1010cm)2 ν B
9
 (cid:16) (cid:17)2 
4×1026J ·s−1  1m2 1012mcm k4 =T4
4π(6.9599×1010cm)2 5.670×10−8J/s
T (cid:39)5830K
Solution3. AveragetemperatureoftheinterioroftheSun.
(a)
(cid:90) R G(cid:0)4πρr3(cid:1)(4πr2ρdr) 16 (cid:90) R −16 4
U =− 3 =− π2Gρ2 r4dr = π2ρ2GR5; M = πR3ρ
r 3 15 3
0 0
−3GM2
U =
5R
1(cid:90) R G(cid:0)4πρr3(cid:1) −8 (cid:90) R −8
U =− 3 (4πr2ρdr)= π2Gρ2 r4dr = π2ρ2GR5 =
2 r 3 15
0 0
−8 (cid:18) M (cid:19)2 −3GM2
= π2GR5 = =1.14×1041J
15 4πR3 10 R
3
(b) Usingthevirialtheoremofmechanics,notethat
(cid:16) (cid:17)
−1U = 3 GM2 = 3 6.67×10−11kg· sm2 · kmg22 ·2×1033g 110k3gg =5.72×1040J
2 20 R 20 7×1010cm(cid:0) 1m (cid:1)
102cm
Now (cid:104)s(cid:105) = 1 , is the Planck distribution function, giving the thermal average number of photons in a
exp((cid:126)ω/τ)−1
singlemodefrequencyω.
thermalaverageenergy(cid:104)(cid:15)(cid:105)=(cid:104)s(cid:105)(cid:126)ω = (cid:126)ω forτ (cid:28)(cid:126)ω,(cid:104)(cid:15)(cid:105)(cid:39)τ
exp((cid:126)ω/τ)−1
Sothen5.72×1040J =N(cid:104)(cid:15)(cid:105)=Nτ.
5.72×1040J
τ = = 4.14×106K
1×1057(1.381×10−23J/K)
Solution4. AgeoftheSun.
(a) Consider4H →4 He. Then4(1.0078)−4.0026=0.0286amu. Then
2
(cid:18)1.6726×10−27kg(cid:19)
(0.0286amu) (3×108m/s)2 =4.27×10−12J
1.00727647u
GivenM =2×1033g,
(cid:12)
(cid:18) (cid:19)(cid:18) (cid:19)
1 1.00727647u
(2×1030kg)(0.10) (4.27×10−12J)=1.28×1044J
4×(1.0078amu) 1.6726×10−27kg
So 1.28×1044J energyisavailable.
(b)
1.15×1045J (cid:18) 1hr (cid:19)(cid:18)1day(cid:19)(cid:18) 1yr (cid:19)
=1.02×1010years
4×1026J ·s−1 3600s 24hr 365days
Solution5. SurfacetemperatureoftheEarth. JS =σbT(cid:12)4 istheradiantpowerperunitarea.
TotalemittedradiationenergyofthesunisJ 4πR2.
S (cid:12)
4πR(cid:12)2JS = R(cid:12)2 J = radiationenergyhitting1cm2ofEarth’ssurfaceinonesecond
4πR2 R2 S
ES ES
SincetheEarthisconsideredablack-body,therateofabsorptionmustequaltherateofemission:
(cid:114)
R2 7×1010cm
(cid:12) σ T4 =σ T4orT4 =T4R2 =⇒T =5800K =396.2K =123C
R2 b (cid:12) b e e (cid:12) (cid:12) e 1.5×1013cm
ES
Solution6. Pressureofthermalradiation.
(a) s =numberofphotonsinthatmode. Supposemodesofω ,j =0,1,2,....
j j
(cid:15) =s (cid:126)ω =totalenergyinjthmode,s photonsinjthmode.
j j j j
U =(cid:88)(cid:15) =(cid:88)s (cid:126)ω P =−∂U =−(cid:88)s (cid:126)∂ωj
j j j ∂V j ∂V
j j j
10