Introduction to Thermodynamics and Statistical Physics phần 8 ppt

Chapter 3. Boso nic and Fermionic Systems √ ε 2 µ β α 2 ¶ 3/2 dε = n 2 dn. Thus, b y introducing the densityofstates D (ε)= ( V 2π 2 ¡ 2m ~ 2 ¢ 3/2 ε 1/2 ε ≥ 0 0 ε<0 , (3.97) one has log Z gc = 1 2 X l ∞ Z −∞ dεD(ε)log(1+λ exp (−β (ε + E l ))) . (3.98) 3.3.3 Energy and Number of Particles Using Eqs. (1.80) a nd (1.94) for the energy U and the number of particles N,namelyusing U = − µ ∂ log Z gc ∂β ¶ η , (3.99) N = λ ∂ logZ gc ∂λ , (3.100) one finds that U = 1 2 X l ∞ Z −∞ dεD(ε)(ε + E l ) f FD (ε + E l ) , (3.101) N = 1 2 X l ∞ Z −∞ dεD(ε) f FD (ε + E l ) , (3.102) where f FD is the Fermi-Dirac distribution function [see Eq. (2.35)] f FD ()= 1 exp [β ( − µ)] + 1 . (3.103) 3.3.4 Example: Electrons in Metal Electrons are Fermions having spin 1/2. The spin degree of freedom gives rise to two orthogo nal eigenstates having energies E + and E − respectively. In the absent of any external magnetic field these states are degenerate, namely E + = E − . For simplicity we take E + = E − = 0. Thus, Eqs. (3.101) and (3.102) become Eyal Buks Thermodynamics and Statistical Physics 112 3.3. Fermi Gas U = ∞ Z −∞ dεD(ε) εf FD (ε) , (3.104) N = ∞ Z −∞ dεD(ε) f FD (ε) , (3.105) T ypically f or metals at room temperature or below the following holds τ ¿ µ. Thus, it is c onvenien t to employ the following theorem ( Sommerfeld expan- sion) to evaluate these integrals. Theorem 3.3.1. Let g (ε) be a function that vanishes in the limit ε →−∞, and that diverges no more rapidly than some p ower of ε as ε →∞.Then, the following holds ∞ Z −∞ dεg(ε) f FD (ε) = µ Z −∞ dεg(ε)+ π 2 g 0 (µ) 6β 2 + O µ 1 βµ ¶ 4 . Proof. See problem 7 of set 3. With the help of this theorem the n umber of particles N to second order in τ is given by N = µ Z −∞ dεD(ε)+ π 2 τ 2 D 0 (µ) 6 . (3.106) Moreover, at low temperatures, the chemical potential is expected to be close the the Fermi energy ε F , which is defined by ε F = lim τ→0 µ. (3.107) Thus, to lowest order in µ − ε F one has µ Z −∞ dεD(ε)= ε F Z −∞ dεD(ε)+(µ − ε F ) D (ε F )+O (µ − ε F ) 2 , (3.108) and therefore N = N 0 +(µ −ε F ) D (ε F )+ π 2 τ 2 D 0 (ε F ) 6 , (3.109) where Eyal Buks Thermodynamics and Statistical Physics 113 Chapter 3. Boso nic and Fermionic Systems N 0 = ε F Z −∞ dεD(ε) , (3.110) is the number of electrons at zero temperature. The number of electrons N in metals is expected to be temperature independent, namely N = N 0 and consequently µ = ε F − π 2 D 0 (ε F ) 6β 2 D (ε F ) . (3.111) Similarly, the energy U at low temperatures is given approximat ely by U = ∞ Z −∞ dεD(ε) εf FD (ε) = ε F Z −∞ dεD(ε) ε | {z } U 0 +(µ −ε F ) D (ε F ) ε F + π 2 τ 2 6 (D 0 (ε F ) ε F + D (ε F )) = U 0 − π 2 τ 2 D 0 (ε F ) 6D (ε F ) D (ε F ) ε F + π 2 τ 2 6 (D 0 (ε F ) ε F + D (ε F )) = U 0 + π 2 τ 2 6 D (ε F ) , (3.112) where U 0 = ε F Z −∞ dεD(ε) ε. (3.113) From this result one finds that the e lectronic heat capacity is given by C V = ∂U ∂τ = π 2 τ 3 D (ε F ) . (3.114) Comparing this result with Eq. (3.81) fo r the phonons heat capacity, which is proportional to τ 3 at low temperatures, suggests that typically, while the electronic contribution is the dominant one at very low temperatures, at higher te mperatures the phonons’ contribution becomes dominant. 3.4 Semiconductor Statistics To be written Eyal Buks Thermodynamics and Statistical Physics 114 3.5. Problems Set 3 3.5 Problems Set 3 1. Calculate the average number of photons N in equilibrium at tempera- ture τ in a cavity of volum e V . Use this result to estimate the number of photons in the universe assuming it to be a spherical cavity of radius 10 26 m and at temperature τ = k B × 3K. 2. Write a relation between the temperature of the surface of a planet and its distance from the Sun, on the assumption that as a black body in thermal equilibrium, it rera d iates as much thermal radiation, as it receives from the Sun. Assume also, that the surface of the planet is at constant temper- ature over the day-nigh t cycle. Use T Sun =5800K;R Sun =6.96 × 10 8 m; and the Mars-Sun distance of D M−S =2.28 × 10 11 m and calculate the temperature of M ars surface. 3. Calculate the Helmholtz free energy F of photon gas having total e nergy U and volume V and use your result to show that the pressure is given by p = U 3V . (3.115) 4. Consider a photon gas initially at temperature τ 1 and volume V 1 .Thegas is adiabatically compressed from volume V 1 to volume V 2 in an isentropic process. Calculate the final temperature τ 2 and final p ressure p 2 . 5. Consider a one-dimensional lattice of N iden tical point particles of mass m, int eracting via nearest-neighbor spring-like forces with spring constant mω 2 (see Fig. 3.2). Denote the lattice spacing by a. Show that the normal mode eigen-frequencies are given b y ω n = ω p 2(1− cos k n a) , (3.116) where k n =2πn/aN,andn is in teger r anging from −N/2toN/2 (assume N À 1). 6. Consider an orbital with energy ε in an ideal gas. The system is in thermal equilibrium at temperature τ and chemical potential µ. a) Show that the probability t hat the orbital is occupied by n particles is given by p F (n)= exp [n (µ − ε) β] 1+exp[(µ − ε) β] , (3.117) for the c ase of Fermions, w here n ∈ {0, 1},andby p B (n)={1 − exp [(µ − ε) β]} exp [n (µ − ε) β] , (3.118) where n ∈ {0, 1, 2, }, for the case of Bosons. Eyal Buks Thermodynamics and Statistical Physics 115 Chapter 3. Boso nic and Fermionic Systems b) Sho w that the variance (∆n) 2 = D (n − hni) 2 E is given by (∆n) 2 F = hni F (1 − hni F ) , (3.119) for the c ase of Fermions, and by (∆n) 2 B = hni B (1 + hni B ) , (3.120) for the case of Bosons. 7. Let g (ε) be a function that vanishes in the limit ε →−∞,andthat diverges no more rapidly than some pow er of ε as ε →∞. Show that the following holds (Sommerfeld expansion) I = ∞ Z −∞ dεg(ε) f FD (ε) = µ Z −∞ dεg(ε)+ π 2 g 0 (µ) 6β 2 + O µ 1 βµ ¶ 4 . 8. Cons ider a metal at zero temperature having Ferm i e nergy ε F ,number of electrons N and v olume V . a) Calculate the mean en ergy of electrons. b) Calculate the ratio α of the mean-square- speed of electrons to the square of the mean s peed α =  v 2 ® hvi 2 . (3.121) c) Calculate the pressure exerted by an electr on gas at zero tempera- ture. 9. For electrons with energy ε À mc 2 (relativistic fermi gas), the energy is given by ε = pc. Find the fermi energy of this gas and show that the ground state energy is E (T =0)= 3 4 Nε F (3.122) 10. A gas of two dimensional electrons is free to mo ve in a p lane. The mass of each electron is m e , the density (number of electrons per unit area) is n,andthetemperatureisτ. Show that the chemical potential µ is given by µ = τ log · exp µ nπ~ 2 m e τ ¶ − 1 ¸ . (3.123) Eyal Buks Thermodynamics and Statistical Physics 116 3.6. Solutions Set 3 3.6 Solutions Set 3 1. The density of states of the photon gas is giv en b y dg = Vε 2 π 2 } 3 c 3 dε. (3.124) Thus N = V π 2 } 3 c 3 ∞ Z 0 ε 2 e ε/τ − 1 dε = V ³ τ }c ´ 3 α, (3.125) where α = 1 π 2 ∞ Z 0 x 2 e x − 1 dx. (3.126) The number α is calculated numerically α =0.24359 . (3.127) For the universe N = 4π 3 ¡ 10 26 m ¢ 3 µ 1.3806568 × 10 −23 JK −1 3K 1.05457266 × 10 −34 Js2.99792458 × 10 8 ms −1 ¶ 3 × 0.24359 ' 2.29 × 10 87 . (3.128) 2. The energy emitted by the Sun is E Sun =4πR 2 Sun σ B T 4 Sun , (3.129) and the energy emitte d by a planet is E planet =4πR 2 planet σ B T 4 planet . (3.130) The fraction of Sun energy that pla net receives is πR 2 planet 4πD 2 M−S E Sun , (3.131) and this e quals to the energy it reradiates. Therefore πR 2 planet 4πD 2 E Sun = E planet , Eyal Buks Thermodynamics and Statistical Physics 117 Chapter 3. Boso nic and Fermionic Systems thus T planet = r R Sun 2D T Sun , and for Mars T Mars = r 6.96 × 10 8 m 2 × 2.28 × 10 11 m 5800 K = 226 K . 3. The partition function is giv en by Z = Y n ∞ X s=0 exp (sβ}ω n )= Y n 1 1 − exp (−β}ω n ) , th us the free energy is given by F = −τ log Z = τ X n log [1 − exp (−β}ω n )] . Transforming the sum over modes into integral yields F = τπ Z ∞ 0 dnn 2 log [1 − exp (−β}ω n )] (3.132) = τπ Z ∞ 0 dnn 2 log · 1 − exp µ − β}πcn L ¶¸ , or, by integrating by parts F = − 1 3 }π 2 c L Z ∞ 0 dn n 3 exp ³ β}πcn L ´ − 1 = − 1 3 U, where U = π 2 τ 4 V 15} 3 c 3 . (3.133) Thus p = − µ ∂F ∂V ¶ τ = U 3V . (3.134) 4. Using the expression for Helmholtz free energy, which was derived in the previous problem, F = − U 3 = − π 2 τ 4 V 45} 3 c 3 , one finds that the en tropy is given by Eyal Buks Thermodynamics and Statistical Physics 118 3.6. Solutions Set 3 σ = − µ ∂F ∂τ ¶ V = 4π 2 τ 3 V 45} 3 c 3 . Thus, for an isentropic process, for which σ is a constant, one has τ 2 = τ 1 µ V 1 V 2 ¶ 1/3 . Using again the previous problem, the pressure p is given by p = U 3V , thus p = π 2 τ 4 45} 3 c 3 , and p 2 = π 2 τ 4 1 45} 3 c 3 | {z } p 1 µ V 1 V 2 ¶ 4/3 . 5. Let u (na) be the displacement of poin t p article number n. The equations of motion are given by m¨u (na)=−mω 2 {2u (na) − u [(n − 1) a] − u [(n +1)a]} . (3.135) Consider a s olution of the form u (na, t)=e i(kna−ω n t) . (3.136) P eriodic boundary condition requires that e ikNa =1, (3.137) thus k n = 2πn aN . (3.138) Substituting in Eq. 3.135 yields −mω 2 n u (na)=−mω 2 £ 2u (na) − u (na) e −ika − u (na) e ika ¤ , (3.139) or ω n = ω p 2(1− cos k n a)=2ω ¯ ¯ ¯ ¯ sin k n a 2 ¯ ¯ ¯ ¯ . (3.140) Eyal Buks Thermodynamics and Statistical Physics 119 Chapter 3. Boso nic and Fermionic Systems 6. In general using Gibbs factor p (n)= exp [n (µ − ε) β] X n 0 exp [n 0 (µ − ε) β] , (3.141) where β =1/τ ,onefinds for Fermions p F (n)= exp [n (µ − ε) β] 1+exp[(µ − ε) β] , (3.142) where n ∈ {0, 1},andforBosons p B (n)= exp [n (µ − ε) β] ∞ X n 0 =0 exp [n 0 (µ − ε) β] = {1 − exp [(µ − ε) β]} exp [ n (µ − ε) β] , (3.143) where n ∈ {0, 1, 2, }. The expectation value of hni in general is given by hni = X n 0 n 0 p (n 0 )= X n 0 n 0 exp [n (µ − ε) β] X n 0 exp [n 0 (µ − ε) β] , (3.144) thus for Fermions hni F = 1 exp [(ε − µ) β]+1 , (3.145) and for Bosons hni B = {1 − exp [(µ − ε) β]} ∞ X n 0 =0 n 0 exp [n 0 (µ − ε) β] = {1 − exp [(µ − ε) β]} exp [(µ − ε) β] (1 − exp [(µ − ε) β]) 2 = 1 exp [(ε −µ) β] − 1 . (3.146) In general, the following holds τ µ ∂ hni ∂µ ¶ τ = X n 00 (n 00 ) 2 exp [n (µ − ε) β] X n 0 exp [n 0 (µ − ε) β] −     X n 0 n 0 exp [n 0 (µ − ε) β] X n 0 exp [n 0 (µ − ε) β]     2 =  n 2 ® − hni 2 = D (n − hni) 2 E . (3.147) Eyal Buks Thermodynamics and Statistical Physics 120 3.6. Solutions Set 3 Thus (∆n) 2 F = exp [(ε − µ) β] (exp [( ε − µ) β]+1) 2 = hni F (1 − hni F ) , (3.148) (∆n) 2 B = exp [(ε − µ) β] (exp [( ε − µ) β] − 1) 2 = hni B (1 + hni B ) . (3.149) 7. Let G (ε)= ε Z −∞ dε 0 g (ε 0 ) . (3.150) Integration by parts yields I = ∞ Z −∞ dεg(ε) f FD (ε) =[G (ε) f FD (ε)| ∞ −∞ | {z } =0 + ∞ Z −∞ dεG(ε) µ − ∂f FD ∂ε ¶ , (3.151) where the fo llowing holds µ − ∂f FD ∂ε ¶ = βe β(ε−µ) ¡ e β(ε−µ) +1 ¢ 2 = β 4cosh 2 ³ β 2 (ε − µ) ´ . (3.152) Using the Taylor expansion of G ( ε)aboutε −µ, which has the form G (ε)= ∞ X n=0 G (n) (µ) n! (ε − µ) n , (3.153) yields I = ∞ X n=0 G (n) (µ) n! ∞ Z −∞ β (ε − µ) n dε 4cosh 2 ³ β 2 (ε − µ) ´ . (3.154) Employing the variable transformation x = β (ε − µ) , (3.155) and exploiting the fact that (−∂f FD /∂ε)isanevenfunctionofε−µ leads to Eyal Buks Thermodynamics and Statistical Physics 121 [...]... , = ~2 Eyal Buks Thermodynamics and Statistical Physics (3.177) (3.1 78) 124 3.6 Solutions Set 3 thus ã à ả á n~2 à = log exp 1 me Eyal Buks Thermodynamics and Statistical Physics (3.179) 125 4 Classical Limit of Statistical Mechanics In this chapter we discuss the classical limit of statistical mechanics We discuss Hamiltons formalism, dene the Hamiltonian and present the HamiltonJacobi equations... ,N à ả2/3 3N ~2 3 2 N 2 = 5 2m V 3V 2N F = 5V (3.1 68) 9 The energy of the particles are n, = pc (3.169) q where p = ~k, and k = n , n = n2 + n2 + n2 and ni = 1, 2, x y z L Therefore 1 4 N = 2 ì ì n3 = n3 (3.170) 8 3 F 3 F à ả1/3 3N nF = à à ả1/3 ả1/3 ~c 3N 3N F = = ~c L V Eyal Buks Thermodynamics and Statistical Physics 123 Chapter 3 Bosonic and Fermionic Systems The energy is given by E (T =... The density function in thermal equilibrium is used to prove the equipartition theorem This theorem is then employed to analyze an electrical circuit in thermal equilibrium, and to calculate voltage noise across a resistor (Nyquist noise formula) 4.1 Classical Hamiltonian In this section we briey review Hamiltons formalism, which is analogous to Newtons laws of classical mechanics Consider a classical... dt dt We refer to the rst term T as kinetic energy and to the second one V as potential energy The canonical conjugate momentum pi of the coordinate qi is dened as pi = T qi (4.3) The classical Hamiltonian H of the system is expressed as a function of the vector of coordinates q and as a function of the vector of canonical conjugate momentum variables Chapter 4 Classical Limit of Statistical Mechanics... 3F 5 b) The speed v is related to the energy by r 2 , v= m Eyal Buks Thermodynamics and Statistical Physics (3.163) (3.164) 122 3.6 Solutions Set 3 thus F 2 hi 16 3 +1 = ư đ2 = 1/2 2 = 1/2 15 F (3.165) 2 1 3 2 +1 c) The number of electrons N is given by N= ZF d D () = D (F ) 0 1/2 F ZF 0 d 1/2 = D (F ) 2 3/2 , 1/2 3 F F thus F = ~2 2m à 3 2 N V ả2/3 , (3.166) and therefore 3N ~2 U= 5 2m à 3... canon ical conjugate momentum equals the mechanical momentum Note, however, that this is not necessarily always the case Using the denition (4.6) one nds that the Hamiltonian is given by Eyal Buks Thermodynamics and Statistical Physics 1 28 ... The system is described using the vector of coordinates q = (q1 , q2 , , qd ) (4.1) Let E be the total energy of the system For simplicity we restrict the discussion to a special case where E is a sum of two terms E =T +V , ã where T depends only on velocities, namely T = T q , and where V depends only on coordinates, namely V = V () The notation overdot is used to q express time derivative, namely...Chapter 3 Bosonic and Fermionic Systems X G(2n) (à) Z x2n dx I= 2n (2n)! 4 cosh2 x 2 n=0 (3.156) With the help of the identities Z dx =1, 4 cosh2 x 2 Z 2 x2 dx , = 3 4 cosh2 x 2 (3.157) (3.1 58) one nds à ả4 2 G(2) (à) 1 I = G (à) + +O 2 à 6 à ả4 Zà 2 g 0 (à) 1 = d g () + +O à 6 2 (3.159) 8 In general, at zero temperature the average of the energy to the power n is given by hn... variables Chapter 4 Classical Limit of Statistical Mechanics p = (p1 , p2 , , pd ) , (4.4) namely H = H (, p) , q (4.5) and it is dened by H= d X i=1 pi qi T + V (4.6) 4.1.1 Hamilton-Jacobi Equations The equations of motion of the system are given by H pi H , pi = qi qi = (4.7) (4 .8) where i = 1, 2, d 4.1.2 Example V(q) m q Consider a particle having mass m in a one dimensional potential V (q) The... F = N F 4 2 ~3 c3 L3 4 = L3 Z 2 ~3 c3 ì 10 The energy of an electron having a wave function proportional to exp (ikx x) exp (iky y) is  ~2 Ă 2 2 k + ky 2me x (3.172) For periodic boundary conditions one has 2nx , Lx 2ny ky = , Ly kx = (3.173) (3.174) where the sample is of area Lx Ly , and nx and ny are both integers The number of states having energy smaller than E 0 is given by (including both . µ)] = m e τ π~ 2 log ¡ 1+e βµ ¢ , (3.1 78) Eyal Buks Thermodynamics and Statistical Physics 124 3.6. Solutions Set 3 thus µ = τ log · exp µ nπ~ 2 m e τ ¶ − 1 ¸ . (3.179) Eyal Buks Thermodynamics and Statistical Physics 125 4 T Sun = 580 0K;R Sun =6.96 × 10 8 m; and the Mars-Sun distance of D M−S =2. 28 × 10 11 m and calculate the temperature of M ars surface. 3. Calculate the Helmholtz free energy F of photon gas having total. 1},andby p B (n)={1 − exp [(µ − ε) β]} exp [n (µ − ε) β] , (3.1 18) where n ∈ {0, 1, 2, }, for the case of Bosons. Eyal Buks Thermodynamics and Statistical Physics 115 Chapter 3. Boso nic and