NOTES AND SOLUTIONS TO THERMAL PHYSICS BY CHARLES KITTLE .
NOTES AND SOLUTIONS TO THERMAL PHYSICS BY CHARLES KITTLE AND HERBERT KROEMERERNEST YEUNG - LOS ANGELESA BSTRACT. These are notes and solutions to Kittle and Kroemer’s Thermal Physics. The solutions are (almost) complete: I willcontinuously add to subsections, before the problems in each chapter, my notes that I write down as I read (and continuously reread).I am attempting a manifold formulation of the equilibrium states in the style of Schutz’s Geometrical Methods of MathematicalPhysics and will point out how it applies directly to Thermal Physics. Other useful references along this avenue of investigation isprovided at the very bottom in the references.Any and all feedback, including negative feedback, is welcomed and you can reach me by email or my wordpress.com blog.You are free to copy, edit, paste, and add onto the pdf and LaTeX files as you like in the spirit of open-source software. You areresponsible adults to use these notes and solutions as governed by the Caltech Honor Code: “No member of the Caltech communityshall take unfair advantage of any other member of the Caltech community” and follow the Honor Code in spirit.S ECOND E DITION. Thermal Physics. Charles Kittel. Herbert Kroemer. W. H. Freeman and Company. New York.QC311.5.K52 1980 536’.7 ISBN 0-7167-1088-91. S TATES OF A M ODEL S YSTEM2. E NTROPY AND T EMPERATUREThermal Equilibrium. EY : 20150821 Based on considering the physical setup of two systems that can only exchangeenergy between each other, that are in thermal contact, this is a derivation of temperature.U U1 U2 is constant total energy of 2 systems 1, 2 in thermal contactmultiplicity g(N, U ) of combined system isXg(N, U ) g1 (N1 , U1 )g2 (N2 , U U1 )U1 UThe “differential” of g(N, U ) is dg g1 U1 g2 dU g1N1 g2 U2 dU2 0N2EY : 20150821 This step can be made mathematically sensible by considering the exterior derivative d of g C (Σ), whereΣ is the manifold of states of the system, with local coordinates N, U , where U happens to be a global coordinate. Then, consider a curve in Σ s.t. it has no component in N, N, and this curve is a “null curve” so that the vector field X X(Σ)1generated by this curve is s.t. dg(X) 0.With dU1 dU2 , 1 g2 ln g1 ln g21 g1 g1 U1 N1g2 U2 N2 U1 N1 U2 N2Defineσ(N, U ) : ln g(N, U )Then σ1 U1 N1 σ2 U2 N2Date: Fall 2008.I am crowdfunding on Tilt/Open to support basic sciences research: ernestyalumni.tilt.com. If you find this pdf (and LaTeX file) valuableand/or helpful, please consider donating to my crowdfunding campaign. There is also a Paypal button there and it’s easy to donate with PayPal, as I hadrecently to the Memorial Fund for the creator of Python’s matplotlib. But the most important thing is for anyone, anywhere, and at anytime be able tolearn thermodynamics and use it for research and application and so I want to keep this as openly public as possible, in the spirit of open-source software.Tilt/Open is an open-source crowdfunding platform that is unique in that it offers open-source tools for building a crowdfunding campaign. Tilt/Open hasbeen used by Microsoft and Dicks Sporting Goods to crowdfund their respective charity causes.1
Temperature. T1 T2 - temperaturesof 2 systems in thermal equilibrium are equal. σ[?].T “must be a function of UN 1 σ kBT U NExperimentally, kB 1.381 10 23 J/K 1.381 10 16 ergs/K.Now 1 σ τ kB Tτ U NProblems. Solution 1. Entropy and temperature. σand σ(N, U ) log g(N, U ). Given g(U ) CU 3N/2 ,(a) Recall that τ1 UN,Vσ(N, U ) log CU 3N/2 log C 3Nlog U2 σ3N 113N U τ U2 Uτ2(b) 2σ U 2 N 0? 2σ U 2 N3N2 1U2 0Solution 2. Paramagnetism.U (s) U1 (s1 ) U2 (s2 ) 2mB(s1 s2 ) 2mBs or s U 2mBi.e. potential energy U (s) 2s · mB.For s N , then2 g(N, s) ' g(N, 0) exp 2s /N g(N, 0) expσ(N, U ) ln g(N, U ) σ0 U 22(mB)2 N U2 1where σ0 ln g(N, 0)2m2 B 2 N σ U 11 2 2τ U Nm B NWhat is the thermal equilibrium value of this N -spin system of fractional magnetization? If U denotes hU i, thermal averageenergy, we also get the thermal average spin excess.hU i h 2mBsi 2mBhsi τ m2 B 2 NmBN U2hsiSolution 3. Quantum harmonic oscillator.(a) Result from Ch. 1: g(N, n) Let N 1 N g(N(N n 1)!n!(N 1)! . n)! 1, n) (Nn!N! .(N n)! ln (N n)! ln (n!) ln (N !)n!N ! (N n) ln (N n) N n n ln n n N ln N N (N n) ln (N n) n ln n N ln Nσ(N 1, n) ln g(N 1, n) ln(b) Let U denote total energy n ω of oscillators.UU n ω or n ωσ(N, U ) (N At τ ,1τ UUUU) ln (N ) ln N ln Nωωωω σ U N . 11U1 U1 ln (N ) ln lnτωωωωω2 NωNω 1 or U Uexp (ω/τ ) 1
Solution 4. The meaning of “never.”Suppose 1010 monkeys.(a) Hamlet represents one specific ordering of 1015 with 44 possibilities for each character. The probability of hitting 11 100000 upon Hamlet from a given, random sequence is 44. Given that log10 44 1.64345, then44100000101.64345 44 or 10 1.64345 44 1 so then 1000001 10 16434544(b) (age of universe) ·10 keyssecond 10 keyssecond 1 hamlet510 characters 1018 s 1019 keys · 1010 monkeys 1029 keys typed out 1019 keys 1024 possible “Hamlets”From part (a), the probability that a given, random sequence is Hamlet, 10 164345(1029 characters)(10 164345 ) 10 164316 164345Note, I think that the probability should be (1029 characters) 1015 Hamlet) 10 164321characters (105Since we are considering the number of “Hamlet”, 10 character sequences.3. B OLTZMANN D ISTRIBUTION AND H ELMHOLTZ F REE E NERGYcf. Example: Energy and heat capacity of a two state system, pp. 62 of Kittel and Kroemer [1]. Kittel and Kroemerintroduces the heat capacity very early, specific to this example.Definition 1. heat capacity CV at constant volume is defined as σ(1)CV : τ τ VRecall the thermodynamic identity (which is introduced many equations later):dU τ dσ pdV Ω1 (Σ)where Σ is a manifold of states of all systems. i.e. no componentConsider local coordinates of Σ, (σ, V ). Consider curve c : R Σ s.t. c generates a vector field ċ σ̇ σc(τ ) Σin the V direction. Notice the prescient choice of parameter τ .Now for internal energy U C (Σ), taking the exterior derivative d results indU U Udσ dV σ V Then applying dU onto vector field σ̇ σ, σdUσ̇ σ τ Uσ̇ σ̇τ 0 σNow, U σV V U τ V σ τ τVHence, (2)CV : τ σ τ V U τ VEY: 20150825 Why do we need differential geometry? It’s because I always wondered why you could do this: σ U?CV : τ with τ dσ dU τ σ U τ V τ Vand talk of “differentials.”Definition: Reversible process. EY : 20150824 Mathematically, 1-forms are exact.3
Pressure. Consider coordinates (σ, V ) Σ of manifold of thermodynamic states Σ.Imagine a reversible compression of a cube system (so imagine dV 0; cube’s volume get smaller).σ constant, i.e. dσ 0 (on this curve in Σ) because as particles in cube gets squeezed, less positions particles could sit in,but they get more kinetic energy, more energetic (more momentum squared).Now U U (σ, V ) C (Σ). U dU U σ V dσ V σ dV Again, imagine a curve c : R Σ, connecting 1 state (σ, V ) Σ to another state (σ, V dV ) Σ s.t. ċ V̇ V. U dU (ċ) V̇ V σIntroduce 1-form W Ω1 (Σ) of work done on the cube system from some external agentW pdVso W 0 when dV 0.Then W (ċ) pV̇ dU (ċ) p (3) U V U V V̇σ σConsider 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 ), so that σ σdσ UdU VdVVU Consider curve c (U, V ) Σ. Then ċ U̇ U V̇ V. For this curve c, σ is constant, meaning dσ(ċ) 0 (it’s a “nullcurve” of dσ σ σU̇ V̇dσ(ċ) 0 U V V UNow defineDefinition 2.1: τ(4) σ U VSo then we have τ1 U̇ V U V̇ 0. For the parameter of curve c, choose the parameter to be V , knowing that σ is constanton this curve, or thermodynamic process. Thus 1 U σ σ p orτ V σ V Uτ V U σ (5)p τ σ V UThermodynamicLet σ σ(U, V ) C (Σ). Then Identity. σ σdσ U V dU V U dV Ω1 (Σ). σ σNow recall the quantities we’ve recently used: τ1 : U(this is a definition) and τp V(it comes from the physics,VUof doing work on the system, by some external agent). Then the thermodynamic identity is obtained:Theorem 1.(6)τ dσ dU pdVIdeal Gas: A First Look.4
One atom in a box. one atom of mass M in cubical box of volume V L3 2 2p i p2 ψ ψ2M ψ ψ ny πy nx πx ψ(x) A sin L sin L sin nzLπz π 2 2(n2x n2y n2z ) n 2MLThen the partition function Z1 is XX n 2 π 2 222Z1 exp(nx ny nz ) expτ2M τ Ln(nx ,ny ,nz )Letα2 2 π 2 πor α 22M L τ(2M τ )1/2 V 2/dThen Z ZdnxZ1 0Zdny0In general, Z1 Now dnz exp [ α2(n2x n2y Zn2z )] 3 dnx exp ( α2 n2x )00 3 1/2 3 1/2 31ππ α22α d1/2π2αZ1 π 1/2 V 1/d2 (2M πτ )1/2!dnQV nQ V d/22 πn Mτin terms of concentration n 1/V . M τ d/2nQ : 2π is the quantum concentration.2Problems. Solution 1. Free energy of a two state system.(a)Z 1 e /τF τ ln Z τ ln (1 e /τ )(b) e /τ (F/τ ) τ1 e /τ /τe Fσ ln (1 e /τ ) τ /τ τ(1 e)U τ 2Solution 2. Magnetic susceptibility(a) Remember to calculate the multiplicity in the N -spin system (it’s not enough to sum up exp ( s /τ ) factors).Us M B 2smBM 2smN N N 2s N N N (N N ) 2N NN/2Z X N 2smB expN τs N/2 e N mBτ(1 e2mBτN/2XN2s N/2)N 2N coshN mBτN! exp s ! N2 s ! 2mBsτ NXs 0N!exps!(N s)! 2mBτ Ns 2 where it was crucial to use (1 x)N neglect dropping the s 0 term.PNj 1Nj xj . Note, in changing the sum index, since N is large, we can mBmB m) sinhτττ2 mBM τ 2ln Z N m tanh ττ 2 MNmmBχ sech2 Bττ τ Z 2N (N )(coshN 1 5
(b)For x Mnm mB NmBF τ ln Z τ ln (2 cosh) N τ ln (2 cosh)ττ 2 N τ42 2 tanh mB ln.Now1 tanhy sechy.F Nτln2τ21 x1 x2m2 NτSolution 3. Free energy of a harmonic oscillator(a) X s ω01 (1 e ω0 /τ ) 1Z exp ω0 /ττ1 es 0 ω0 ω0 ω0 /τF τ ln Z τ ln (1 e) ' τ lnfor 1ττ(b) ω0 Fτ ω0 ω0 /τ ω0 /τσ ( e ln (1 e τ ) {ln (1 e ω0 /τ ) )} ω /τ ω/τ00 ττ1 ee 1Solution 4. Energy fluctuations.Xe s βZ 1sPX β s e s /ττ β Z e s βU s β ln Z β 1 Zs 2X τ τ βτ β 2 Z 2 e s β(c) FormBτ1, cosh2mBτ 1. χ βss U β Z β 2 U β 2 β β2 τ βZ( β2 Z)Z ( β Z)2Z2! β2 β2 Z Z β ZZ 2 ! U h 2 i h i2 τSolution 5. Overhauser effect. System S in energy eigenstate En n .P (E) (1)gR (E)dURdUtotSNote UR (α 1) . US . dUd d 1 (α 1) α d in a specific energy eigenstate; gS (n ) 1While gR (UR ) multiplicity of reservoir R with UR energy.Now σR1 EsτandgR (UR ) exp (σR (UR ))RIf dU (α 1) UsmallcomparedtoURR.d τ 2dUR UR (n ) (α 1) d 1σR (UR ((n 1) )) ' σR (UR (n )) (α 1) τ exp (σR (UR (n )) τ1 (α 1) )P (ES (n 1) ) exp (1 α)P (ES n )exp (σR (UR (n )))τSolution 6. Rotation of diatomic molecules.(a) (j) j(j 1) 0 . g(j) 2j 1Remember that Z is a sum over all states, not over all levels. 2XX τd (j 2 j) 0 /τ τ X d 0 /τ j j j(j 1) 0 /τ(e) eZ (2j 1)e dj 0 0 j 0 djj 0j 0UR (ES (n 1) ) UR (n ) (b) For 1 0ττZR (τ ) 0Z0 d 0 /τ xedx2 xdx 6 τ 0 /τ x2 xτ(e) (e 0 /τ )0 0 0
(c) Forτ 01ZR (τ ) 1 3e 2 0 /τ(d) 0τ2U τfor 1U τ τ ln ττ 0 6 0 e 2 0 /ττ12 0 2 0 /τ for1, U τ 2(3e)2 2 /τ 0τ1 3e 01 3e 2 0 /τ U 0CV 1 when 1 τ Vτ 2 /τ ! 2 0 2 0 /τ 2 0 /τ0 2 0 /τ(1 3e) (3e)e 0e 2 0 /τ 2 1222eττ 12 0τ2(1 3e 2 0 /τ )2(1 3e 2 0 /τ )22 CV 6 0For very smallτ 0ln Z τ 1, CV 12e 2 0 /τ(τ / 0 )2 (e) See sketch.Solution 7. Zipper problem.(a) N links. s 0 closed, open.Z NXexp ( s /τ ) s 01 e (N 1) /τ1 e /τ(b) 1 τ .PNZ s 0 exp ( s B)PN 1 1 s 0 ( s )e s β β ln Z hsi Z 1 e /τ e (N 1) /τ ( (N 1) )(1 e /τ ) ( e /τ )( )(1 e (N 1)/ /τ ) 1 e (N 1) /τ(1 e /τ )2 e /τ (e N β (N 1)(1 e β ) 1) e /τ (e N β (N 1)(1 e β ) (1 e (N 1) /τ ))'(1 e β )(1 e (N 1) /τ )(1 e /τ )This still does not give the desired approximation. Consider the following:e β e N β1 e (N 1) β 1 e βe β 1 β N β ((e N βe)(e β 1) (e β )(e β e N β )) (e β (e β N e N β e β e N β ) e β N e N β ) β Z (e β 1)2(e β 1)2Z (e β (N 1)e N β e β N e N β ) (e β (N 1)e N β e β N e N β ) β Z ' β β N βZ(e 1)(e e)e2 β (N e (N 1) β e (N 1) β e β N e N β ) ((N 1)e (N 1) β (e β N e N β )) 2 βe2 β β βe β N β β β Ne (N 1)e N e eeN e (Ne 1)e eNn β e β e βN β βeN βN eN β 'e2 βe2 βe2 βeN β e β ' ( eN β ) e βeN β e β hsi e /τ n πy Solution 8. Quantum concentration. Now Ψ(x, y, z) A sin nxLπx sin yLsinGround orbital: nx ny nz 1.3 π 23 π 2T hψ0 ψ0 i 2m L2m L7nz πzL . p 1i ,p22m1 2m 2 .
where hψ0 ψ0 i 1, ψ normalized. It was normalized in this way:Z0 Z L2nx πx n πx nx sin1 cosxLdnx sin2dnx L202nx πxL L2nx π2 L L20 3LA2 L382hψ ψi A 1 or A2 328L mτ 3/2Recall that nQ 2π .2Consider the condition that there will be a concentration for which the zero-point quantum kinetic energy is equal to thetemperature τ : 3/2 3 2 2/32mτ3 π22mτ2/3 orn πn τorn 2m L22m3π 2 23π 2 2 3/244mτ 3/2 n () nQ23π 2π 3πSolution 9.Partition function for two systems. XX E1 2 E1 E2g(E1 2 ) expZ(1 2) g(E1 E2 ) exp exp τττE1 2E1 E2 E0 XX E2 E1 g(E1 )g(E2 ) expexp Z(1)Z(2)ττE1 E2since systems are independent.Solution 10. Elasticity of polymers.(a) Consider 2s N N ; N N N , 2s N (N N ) 2N N .For 2s, consider 2s N (N N ) 2N N .N 2s N2 g(N, s) g(N, s) N2N 2s N2 .2N ! s ! N2 s !(b) s N l l2(N!) σ(l) ln g N, g N, ln l lNN2ρ2ρ ! 22ρ22ρ ! LNlNlNlNlNlN ln ln } ln (2N !) {22ρ22ρ22ρ22ρ22ρ22ρwhere we used ln(x x) ' ln x x1 x. lN1 lNNlN1 lN ln ln } σ(l) ln (2N !) {22ρ2N/2 2ρ222ρ2N/2 2ρ 2 NNNNNNlNl2l llNl2 ln (2N !) { ln ln ln ln }2222222ρ22ρ N 2ρ2ρ2ρ2N 2ρ2!2N !l2 σ(l) ln NNN ρ22 ! 2 !(c) σ 2l lN ρ2 f 2lτN ρ2Solution 11. One-dimensional gas. 2 π 2 2 n nin one dimension2m L Z X 2 π 2 2π α2 n2Z1 expn /τ dne 2mL2α0n 18
2 π 22mτ L.where π/Lα 2mτ 1/2 Recall that σ F τ , F U τ σ, so thatα2 π2α2 π/LF τ ln Z τ N ln τ N ln τ N ln 2αππ 2mτ 1/2! 12π 2π 2 τ N ln τ N ln 2mL2 τmLτ 1/2! 12π 2τN1 2π 2 F N ln 2π 2 τ2mL2 τ2mL2 τ 2mL2 τ 12π 2 τ NN2π 2 N ln ln 12mL2 τ2τ2mL2 τ4. T HERMAL R ADIATION AND P LANCK D ISTRIBUTIONProblems. Solution 1. Number of thermal photons.We consider a cavity of volume V , and of edge length L (so V L3 ). So then ωn nπc/L.1is the thermal average number of photons in a single mode frequency ω. So thenNow exp ω( τ n ) 1XX1 hsn i ωnexp τ 1Consider (nx , ny , nz ) on positive octant, and 2 independent polarization s of em field.Z ZX(2) n2 dn112 π 4πndn8 0 1 1 1exp nπcexp nπcexp nπc0LτLτLτn 3 Z 2 Lτx dxV τ 3 π N 2(2.404)x πce 1π c0where I used the substitutions πcnLτLτ x n πcLτ dn(dx) πcx Now σ(τ ) (4π 2 V /45)(τ / c)3 , so thenσ1 N2.404 4π 445 ' 3.602R 2dxNow how was 0 dx exx 1evaluated?Solution 2. Surface temperature of a Sun. Given the solar constant of the Earth, the total radiant energy flux density atthe Earth from the Sun normal to the incident rays, integrated over all emission wavelengths,solar constant 0.136 J s 1 cm 2 ,(49)(a)4π(1.49 10112 1m) · 0.136 J scm 2 102 cm1m 2 (4π)(1.49)2 1022 · 0.136 104 3.8 1026 J · s 1Note that I had used 1.49 1011 m as the distance of the Earth from the Sun.(b) Jν energy flux density or rate of energy emission per unit area.π 2 k4σB 60 3Bc2 5.670 10 8 W m 2 K 4 .Note that W 1sJ . I will use R 6.9599 1010 cm as the radius of the Sun.4 1026 J · s 1 J ν σB T 44π(6.9599 1010 cm)29
2 102 cm1m 2 4 1026 J · s 1 1m 44 k T 1024π(6.9599 10 cm)5.670 10 8 J/sT ' 5830 KSolution 3. Average temperature of the interior of the Sun.(a)RZU G433 πρr0 3GM 2U 5RZ1 RGU 2 0 (4πr2 ρdr)16 π 2 Gρ2r3433 πρrZRr4 dr 0 16 2 24π ρ GR5 ; M πR3 ρ153Z 8 2 2 8 2 2 R 4r dr (4πr ρdr) π Gρπ ρ GR5 r3150 2 8 2 3 GM 2M π GR5 4 3 1.14 1041 J1510RπR3(b) Using the virial theorem of mechanics, note that 1 kgm2 1133kg · sm2 · kgg 102 · 2 103g 13 GM 23 6.67 10 U 5.72 1040 J220 R207 1010 cm 1012mcm 21Now hsi exp ( ω/τ) 1 , is the Planck distribution function, giving the thermal average number of photons in asingle mode frequency ω. ωthermal average energy h i hsi ω exp ( ω/τ) 1 for τ ω, h i ' τ40So then 5.72 10 J N h i N τ .τ 5.72 1040 J 4.14 106 K1 1057 (1.381 10 23 J/K)Solution 4. Age of the Sun.(a) Consider 4H 42 He. Then 4(1.0078) 4.0026 0.0286 amu. Then 1.6726 10 27 kg(3 108 m/s)2 4.27 10 12 J(0.0286 amu)1.00727647 uGiven M 2 103 3 g, 11.00727647 u30(2 10 kg)(0.10)(4.27 10 12 J) 1.28 1044 J4 (1.0078 amu)1.6726 10 27 kgSo 1.28 1044 J energy is available.(b)1.15 1045 J4 1026 J · s 1 1 hr3600 s 1 day24 hr 1 yr365 days 1.02 1010 yearsSolution 5. Surface temperature of the Earth. JS σb T 4 is the radiant power per unit area.Total emitted radiation energy of the sun is JS 4πR2 .4πR2 JS24πRESR2 R2 JS radiation energy hitting 1 cm2 of Earth’s surface in one secondESSince the Earth is considered a black-body, the rate of absorption must equal the rate of emission:rR27 1010 cm4444 2σb T σb Te or Te T R Te 5800 K 396.2 K 123 C2RES1.5 1013 cmSolution 6. Pressure of thermal radiation.(a) sj number of photons in that mode. Suppose modes of ωj , j 0, 1, 2, . . . . j sj ωj total energy in jth mode, sj photons in jth mode.U Xj j Xsj ωjP j10X U ωj sj V Vj
(b) ωj jπc/L, V L3 . So then ωj jπcV 1/3 . dωj 1 1 ωj jπcV 4/3 dV33 VU(c) p 3V(d) We want the Kinetic pressure at a concentration of (1 mol/cm3 ). Recalling P N kVB T , 2 3 6.022 102310 cmN1 mol 23 J1.381 10(2 107 K) 1.663 1014 2P cm31 mol1mKmNow for the thermal radiation pressure,p U1 π2Nτ 4 4.03 1013 2 3V3 15 3 c3mwhere t 2 107 K.For the pressures to be equal,1 π2N kB45( c)3 N kB4 4kBT T or T 3 so that T 3.2 107 K4V333 15 cVπ 2 kBFree energy of a photon gasP Q11Consider Z s 0 e s ω/τ 1 e ω/τfor a single mode.(a) Z n 1 e ωn /τPnπc ωn /τ(b) F τ ln Z τ n ln (1 e)ωn L .Z Z X4πn2 dn(2) ln (1 e nπc/τ L ) πτdnn2 ln (1 e nπc/τ L ) F τln (1 e ωn /τ ) τ800n Z Z3 nπc/τ L n e πcn3 π 2 c πτ (n3 ln (1 e nπc/τ L )) dndn nπ/τ L nπc/τLτLL01 ee 100 3 Z x3(τ L)3 π 2τL πcex 1( 2 c3 ) 450Solution 7.where I used x nπcτL .Solution 8. Heat shields. For Ju σB (Tu4 Tl4 ), the thermal flux without the heat shield, in the middle region. Plane m4absorbs σB Tu4 σB Tl4 and emits Jm σB (Tu4 Tl4 ) σTm.44 1/4Tm (Tu Tl ) . The key point is that, by symmetry, plane m emits J2m flux on each side. 4 4Tu Tl4JuTu Tl44Jnet σB Tu (σB) σB 222Jnet is the same for the other side of the heat shield:Jnet σB Tl4 σBTu4 Tl42 σBTu4 Tl42 Solution 9. Photon gas in one dimension. E E0 sin (kx) cos (ωt) is the form of a solution with kL nπ or k sincenπL ,nπω v, ωn vkL1 1 e ω/τv 2 Ex x Et t v 2 k 2 ω 2 orZj Xe s ω/τs 0where Zj is the partition function for a particular mode frequency ω.P n s ωτd ωn 1 ω 2 ωn 1 1s 0 se Z(1 exp) Z (1 exp) exp hsi Zd( ωn /τ )τττexp ( ωn /τ ) 1 exp ( ωn /τ )So vnπ/L exp vnπ 1exp τ 1Lτnn 2XX vπ n2 exp vnπX vπ 2 n2 exp vπn U vnπ vnπ vnπ/LLτLτLτLτ 2 exp2 τLτLτ 2(exp nvπ 1)2(exp vπexp vnπ 1LτLτ n 1)nnnLτhsi ωn h n i ωnU ωn11Xh n i X
R PNow n 0 dn for one-dimensional photon. Let α vπLτ .Letting x αn,ZZ nvπ 2 Zn2 exp (αn)1x2 ex1
ABSTRACT.These are notes and solutions to Kittle and Kroemer’s Thermal Physics. The solutions are (almost) complete: I will The solutions are (almost) complete: I will continuously add to subsections, before the problems in each chapter, my notes that I write down as I read (and continuously reread).
Energies 2018, 11, 1879 3 of 14 R3 Thermal resistance of the air space between a panel and the roof surface. R4 Thermal resistance of roof material (tiles or metal sheet). R5 Thermal resistance of the air gap between the roof material and a sarking sheet. R6 Thermal resistance of a gabled roof space. R7 Thermal resistance of the insulation above the ceiling. R8 Thermal resistance of ceiling .
thermal models is presented for electronic parts. The thermal model of an electronic part is extracted from its detailed geometry configuration and material properties, so multiple thermal models can form a thermal network for complex steady-state and transient analyses of a system design. The extracted thermal model has the following .
Thermal Transfer Overprinting is a printing process that applies a code to a flexible film or label by using a thermal printhead and a thermal ribbon. TTO uses a thermal printhead and thermal transfer ribbon. The printhead comprises a ceramic coating, covering a row of thermal pixels at a resolution of 12 printing dots per mm
Thermal Control System for High Watt Density - Low thermal resistance is needed to minimize temperature rise in die-level testing Rapid Setting Temperature Change - High response thermal control for high power die - Reducing die-level test time Thermal Model for New Thermal Control System - Predict thermal performance for variety die conditions
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for evaluating thermal shock resistance of ceramic composites. Jin and Batra [9], Jin and Luo [10], and Jin and Feng [11] developed theo-retical thermo-fracture mechanics models to evaluate the critical thermal shock and thermal shock residual strength of FGMs. This entry introduces concepts of the critical thermal shock and the thermal shock .
Transient Thermal Measurements and thermal equivalent circuit models Title_continued 2 Thermal equivalent circuit models 2.1 ntroduction The thermal behavior of semiconductor components can be described using various equivalent circuit models: Figure 6 Continued-fraction circuit, also known as Cauer model, T-model or ladder network
ANSI A300 (Part 1)-2001 Pruning Glossary of Terms . I. Executive Summary Trees within Macon State College grounds were inventoried to assist in managing tree health and safety. 500 trees or tree groupings were identified of 40 different species. Trees inventoried were 6 inches at DBH or greater. The attributes that were collected include tree Latitude and Longitude, and a visual assessment of .
