Introduction to Thermodynamics and Statistical Physics phần 3 pdf

1.7. Problems Set 1 a) Sho w that α is given by α = a 2 · 1+tanh µ Fa 2τ ¶¸ . (1.135) b) Show that in the limit of high temperature the spring constant is given approximately by k ' 4τ Na 2 . (1.136) N α N α 19. A long elastic molecule can be modelled as a linear chain of N links. The state of each link is charac terized by two quantum numbers l and n.The length of a link is either l = a or l = b. The vibrational state of a link is modelled as a harmonic oscillator whose angular frequency is ω a for a link of length a and ω b for a link of length b. Thus, the energy of a link is E n,l = ½ }ω a ¡ n + 1 2 ¢ for l = a }ω b ¡ n + 1 2 ¢ for l = b , (1.137) n =0, 1, 2, The chain is held under a tension F . Show that the mean length hLi of the chain in the limit of high temperature T is given by hLi = N aω b + bω a ω b + ω a + N Fω b ω a (a −b) 2 (ω b + ω a ) 2 β + O ¡ β 2 ¢ , (1.138) where β =1/τ. 20. The elasticity of a rubber band can be described in terms of a one- dimensional model of N polymer molecules linked together end-to-end. The angle between successive links is equally likely to be 0 ◦ or 180 ◦ .The length of eac h polymer is d and the total length is L. The system is in thermal equilibrium at temperature τ. Show that the force f required to main tain a length L is given by f = τ d tanh −1 L Nd . (1.139) 21. Consider a sys tem which has two single particle states both of the same energy. When both states are unoccupied, the energy of the system is Eyal Buks Thermodynamics and Statist ical Physics 27 Chapter 1. The Principle of Largest Uncertainty zero; when one state or the other is occupied by one particle, th e energy is ε. We suppose that the energy of the system is much higher (infinitely higher) when both states are occupied. Show that in thermal equilibrium at temperature τ the av erage number of particles in the level is hNi = 2 2+exp[β (ε − µ)] , (1.140) where µ is the chemical potential and β =1/τ. 22. Consider an array of N tw o-lev el particles. Each one can be in o ne of two states, having energy E 1 and E 2 respectively. The n umbers of particles instates1and2aren 1 and n 2 respectively, where N = n 1 + n 2 (assume n 1 À 1andn 2 À 1). Consider a n energy exchange with a reservoir at temperature τ leading to population changes n 2 → n 2 −1andn 1 → n 1 +1. a) Calculate the en tropy change of the two-level system, (∆σ) 2LS . b) Calculate the entropy change of the reservoir, (∆σ) R . c) What can be said about th e relation bet ween (∆σ) 2LS and (∆σ) R in thermal equilibrium? Use your answer to express the ration n 2 /n 1 as a function of E 1 , E 2 and τ. 23. Consider a lattice containing N sites of one ty pe, which is denoted as A , and the same number of sites of another type, which is denoted as B. The lattice is occupied by N atoms. The number of atoms occupying sites of type A is denoted as N A , w hereas the number of atoms occupying atoms of type B is denoted as N B ,whereN A + N B = N.Letε be the energy necessary to remove an atom from a lattice site of type A to a lattice site of type B. The system is in thermal equilibrium at temperature τ . Assume that N,N A ,N B À 1. a) Calculate the entropy σ. b) Calculate the average number hN B i of atoms occupying sites of type B. 24. Consider a microcanonical ensem ble of N quantum harmonic oscillators in thermal equilibrium at temperature τ. The resonance frequency of all oscillators is ω. The quantum energy levels of each quantum oscillator is given by ε n = }ω µ n + 1 2 ¶ , (1.141) where n =0, 1, 2, is integer. The total energy E of the system is given by E = }ω µ m + N 2 ¶ , (1.142) where Eyal Buks Thermodynamics and Statist ical Physics 28 1.8. Solutions Set 1 m = N X l=1 n l , (1.143) and n l is state number of oscillator l. a) Calculate the number of states g (N,m) of the system with total energy }ω (m + N/2). b) Use this re sult to calculate the en tropy σ of the system with total energy }ω (m + N/2). Appro ximate the r esult by assuming t hat N À 1andm À 1. c) Use this result to calculate (in the same limit of N À 1andm À 1) the average energy of the system U as a function of the temperature τ. 25. The energy of a donor level in a semiconductor is −ε when occupied by an electron (and the energy is zero otherwise). A donor level can be either occupied by a sp in u p electron or a spin do wn electron, however, it cannot be simultaneously occupied by two electrons. Express the average occupation of a donor state hN d i as a function of ε and the chemical potential µ. 1.8 Solutions Set 1 1. Final answers: a) N! ( N 2 ) ! ( N 2 ) ! ¡ 1 2 ¢ N b) 0 2. Final answers: a) ¡ 5 6 ¢ N b) ¡ 5 6 ¢ N−1 1 6 c) In general ∞ X N=0 Nx N−1 = d dx ∞ X N=0 x N = d dx 1 1 − x = 1 (1 − x) 2 , thus ¯ N = 1 6 ∞ X N=0 N µ 5 6 ¶ N−1 = 1 6 1 ¡ 1 − 5 6 ¢ 2 =6. 3. Let W (m) be the probability for for taking n 1 steps to the right and n 2 = N − n 1 steps to the left, where m = n 1 − n 2 ,andN = n 1 + n 2 . Using Eyal Buks Thermodynamics and Statist ical Physics 29 Chapter 1. The Principle of Largest Uncertainty n 1 = N + m 2 , n 2 = N −m 2 , one finds W (m)= N! ¡ N+m 2 ¢ ! ¡ N−m 2 ¢ ! p N+m 2 q N−m 2 . It is convenient to employ the moment generating function, defined as φ (t)=  e tm ® . In general, the following holds φ (t)= ∞ X k=0 t k k!  m k ® , thus from the kth derivative of φ (t) one can calculate the kth moment of m  m k ® = φ (k) (0) . Using W (m)onefinds φ (t)= N X m=−N W (m) e tm = N X m=−N N! ¡ N+m 2 ¢ ! ¡ N−m 2 ¢ ! p N+m 2 q N−m 2 e tm , or using the summation variable n 1 = N + m 2 , one has φ (t)= N X n 1 =0 N! n 1 !(N − n 1 )! p n 1 q N−n 1 e t(2n 1 −N) = e −tN N X n 1 =0 N! n 1 !(N − n 1 )! ¡ pe 2t ¢ n 1 q N−n 1 = e −tN ¡ pe 2t + q ¢ N . Using p = q =1/2 Eyal Buks Thermodynamics and Statist ical Physics 30 1.8. Solutions Set 1 φ (t)= µ e t + e −t 2 ¶ N =(cosht) N . Thus using the expansion (cosh t) N =1+ 1 2! Nt 2 + 1 4! [N +3N (N −1)] t 4 + O ¡ t 5 ¢ , one finds hmi =0,  m 2 ® = N,  m 3 ® =0,  m 4 ® = N (3N − 2) . 4. Using the binomial distribution W (n)= N! n!(N − n)! p n (1 − p) N−n = N (N − 1)(N −1) × (N −n +1) n! p n (1 − p) N−n ∼ = (Np) n n! exp (−Np) . 5. a) ∞ X n=0 W (n)=e −λ ∞ X n=0 λ n n! =1 b) AsinEx.1-6,itisconvenienttousethemomentgeneratingfunction φ (t)=  e tn ® = ∞ X n=0 e tn W (n)=e −λ ∞ X n=0 λ n e tn n! = e −λ ∞ X n=0 (λe t ) n n! =exp £ λ ¡ e t − 1 ¢¤ . Using the expansion exp £ λ ¡ e t − 1 ¢¤ =1+λt + 1 2 λ (1 + λ) t 2 + O ¡ t 3 ¢ , one finds hni = λ. c) Using the same expansion one finds  n 2 ® = λ (1 + λ) , thus D (∆n) 2 E =  n 2 ® − hni 2 = λ (1 + λ) −λ 2 = λ. Eyal Buks Thermodynamics and Statist ical Physics 31 Chapter 1. The Principle of Largest Uncertainty 6.  R 2 ® = *à N X n=1 r n ! 2 + = N X n=1  r 2 n ® | {z} =l 2 + X n6=m hr n · r m i | {z } =0 = Nl 2 7. 20 X n=11 20! n!(20− n)! 0.2 n × 0.8 20−n =5.6 × 10 −4 . (1.144) 8. Using N + + N − = N, (1.145a) N + − N − = M m , (1.145b) one has N + = N 2 µ 1+ M mN ¶ , (1.146a) N − = N 2 µ 1 − M mN ¶ , (1.146b) or N + = N 2 (1 + x) , (1.147a) N − = N 2 (1 − x) , (1.147b) where x = M mN . The number of states having total magnetization M is giv en by Ω (M)= N! N + !N − ! = N! £ N 2 (1 + x) ¤ ! £ N 2 (1 − x) ¤ ! . (1.148) Since all states have equal proba bility one has f (M)= Ω (M) 2 N . (1.149) Taking the natural logarithm of Stirling’s formula one finds log N!=N log N − N + O µ 1 N ¶ , (1.150) Eyal Buks Thermodynamics and Statist ical Physics 32 1.8. Solutions Set 1 th us in the limit N À 1 one has log f = −log 2 N + N log N − N − · N 2 (1 + x) ¸ log · N 2 (1 + x) ¸ + · N 2 (1 + x) ¸ − · N 2 (1 − x) ¸ log · N 2 (1 − x) ¸ + · N 2 (1 − x) ¸ = −N log 2 + N log N − · N 2 (1 + x) ¸ log · N 2 (1 + x) ¸ − · N 2 (1 − x) ¸ log · N 2 (1 − x) ¸ = µ − N 2 ¶½ −2log N 2 +(1+x)log · N 2 (1 + x) ¸ +(1− x)log · N 2 (1 − x) ¸¾ = µ − N 2 ¶· −2log N 2 +(1+x) µ log N 2 +log(1+x) ¶ +(1− x) µ log N 2 +log(1− x) ¶¸ = µ − N 2 ¶µ log ¡ 1 − x 2 ¢ + x log 1+x 1 − x ¶ . (1.151) The function log f (x) has a sharp peak near x = 0, thus we can approx- imate it by assuming x ¿ 1. To lowest order log ¡ 1 − x 2 ¢ + x log 1+x 1 − x = x 2 + O ¡ x 3 ¢ , (1.152) thus f (M)=A exp µ − M 2 2m 2 N ¶ , (1.153) where A is a normalizatio n constant, which is determined by requiring that 1= ∞ Z −∞ f (M)dM. (1.154) Using the iden tity ∞ Z −∞ exp ¡ −ay 2 ¢ dy = r π a , (1.155) one finds 1 A = ∞ Z −∞ exp µ − M 2 2m 2 N ¶ dM = m √ 2πN , (1.156) Eyal Buks Thermodynamics and Statist ical Physics 33 Chapter 1. The Principle of Largest Uncertainty thus f (M)= 1 m √ 2πN exp µ − M 2 2m 2 N ¶ , (1.157) The expectation value is giving by hMi = ∞ Z −∞ Mf (M)dM =0, (1.158) and the v ariance is giv en b y D (M −hMi) 2 E =  M 2 ® = ∞ Z −∞ M 2 f (M)dM = m 2 N. (1.159) 9. The probability to have n steps to the righ t is given by W (n)= N! n!(N − n)! p n q N−n . (1.160) a) hni = N X n=0 N!n n!(N − n)! p n q N−n (1.161) = p ∂ ∂p N X n=0 N! n!(N − n)! p n q N−n = p ∂ ∂p (p + q) N = pN (p + q) N−1 = pN . Since X = an − a (N −n)=a (2n − N) (1.162) we find hXi = aN (2p − 1) = aN (p − q) . (1.163) b)  n 2 ® = N X n=0 N!n 2 n!(N − n)! p n q N−n = N X n=0 N!n (n −1) n!(N −n)! p n q N−n + N X n=0 N!n n!(N −n)! p n q N−n = p 2 ∂ 2 ∂p 2 N X n=0 N! n!(N −n)! p n q N−n + hni = p 2 ∂ 2 ∂p 2 (p + q) N + hni = p 2 N (N − 1) + pN . Eyal Buks Thermodynamics and Statist ical Physics 34 1.8. Solutions Set 1 Thus D (n −hni) 2 E = p 2 N (N − 1) + pN −p 2 N 2 = Npq , (1.164) and D (X − hXi) 2 E =4a 2 Npq . (1.165) 10. The total energy is given by E = kx 2 2 + m ˙x 2 2 = ka 2 2 , (1.166) where a is the amplitude of oscillations. The time period T is given by T =2 Z a −a dx ˙x =2 r m k Z a −a dx √ a 2 − x 2 =2π r m k , (1.167) thus f(x)= 2 T | ˙x| = 1 π √ a 2 − x 2 . (1.168) 11. The six experiments are independen t, thus σ =6× µ − 2 3 ln 2 3 − 1 3 ln 1 3 ¶ =3.8191 . 12. The random variable X obtains t he value n with proba b ility p n = q n , where n =1, 2, 3, ,andq =1/2. a) The entropy is given by σ = − ∞ X n=1 p n log 2 p n = − ∞ X n=1 q n log 2 q n = ∞ X n=1 nq n . This can be rewritten as σ = q ∂ ∂q ∞ X n=1 q n = q ∂ ∂q µ 1 1 − q − 1 ¶ = q (1 − q) 2 =2. b) A series of questions is of the form: ”Was a head obtained in the 1st time ?”, ”Was a h ea d obtain ed in the 2nd time ?”, etc. The expected number of questioned required to find X is 1 2 × 1+ 1 4 × 2+ 1 8 × 3+ =2, which is exactly the entropy σ. 13. Eyal Buks Thermodynamics and Statist ical Physics 35 Chapter 1. The Principle of Largest Uncertainty a) Consider an infinitesimal change in the va riable z = z (x, y) δz = µ ∂z ∂x ¶ y δx + µ ∂z ∂y ¶ x δy . (1.169) For a process for which z is a constant δz =0,thus 0= µ ∂z ∂x ¶ y (δx) z + µ ∂z ∂y ¶ x (δy) z . (1.170) Dividing by (δx) z yields µ ∂z ∂x ¶ y = − µ ∂z ∂y ¶ x (δy) z (δx) z = − µ ∂z ∂y ¶ x µ ∂y ∂x ¶ z = − ³ ∂y ∂x ´ z ³ ∂y ∂z ´ x . (1.171) b) Consider a process for which the variable w is k ept constant. An infinitesimal change in the va riable z = z (x, y) is expressed as (δz) w = µ ∂z ∂x ¶ y (δx) w + µ ∂z ∂y ¶ x (δy) w . (1.172) Dividing by (δx) w yields (δz) w (δx) w = µ ∂z ∂x ¶ y + µ ∂z ∂y ¶ x (δy) w (δx) w . (1.173) or µ ∂z ∂x ¶ w = µ ∂z ∂x ¶ y + µ ∂z ∂y ¶ x µ ∂y ∂x ¶ w . (1.174) 14. We have found in class that hUi = − µ ∂ log Z gc ∂β ¶ η , (1.175) hNi = − µ ∂ log Z gc ∂η ¶ β . (1.176) Eyal Buks Thermodynamics and Statist ical Physics 36 [...]... sinh 2 b n=0 (1.196) To rst order in hli = Ă Â a b + b a F b a (a b)2 + + O 2 2 b + a (b + a ) (1.197) The average total length is hLi = n hli 20 The length L is given by L = N hli , where hli is the average contribution of a single molecule to the total length, which can be either +d or d The probability of each possibility Eyal Buks Thermodynamics and Statistical Physics 39 Chapter 1 The Principle... n2 = exp (1.202) n1 23 The number of ways to select NB occupied sites of type B out of N sites is N !/n! (N n)! Similarly the number of ways to select NB empty sites of type A out of N sites is N!/n! (N n)! Eyal Buks Thermodynamics and Statistical Physics 40 1.8 Solutions Set 1 a) Thus = log à N! NB ! (N NB )! ả2 ' 2 [N log N NB log NB (N NB ) log (N NB )] (1.2 03) b) The energy of the system... (1.208a) (1.208b) (1.209) thus hNB i = Eyal Buks N Ă 1 + exp 2 Thermodynamics and Statistical Physics (1.210) 41 Chapter 1 The Principle of Largest Uncertainty 24 In general, g (N, m) = {# os ways to distribute m identical balls in N boxes} Moreover {# os ways to distribute m identical balls in N boxes} = {# os ways to arrange m identical balls and N 1 identical partitions in a line} a) Therefore g (N,... average energy is hU i = 2N sinh () log Zc = , 1 + 2 cosh (1.190) b) and the variance is D E 2 log Z hU i cosh () + 2 c (U hUi)2 = = 2N2 = 2 [1 + 2 cosh ()]2 (1.191) Eyal Buks Thermodynamics and Statistical Physics 38 1.8 Solutions Set 1 18 Each section can be in one of two possible sates with corresponding energies 0 and F a a) By denition, is the mean length of each segment, which is given... canonical partition function is given by N Zc = Z1 , (1.181) where Z1 = exp Eyal Buks à 2 ả à ả à ả + exp = 2 cosh 2 2 Thermodynamics and Statistical Physics (1.182) 37 Chapter 1 The Principle of Largest Uncertainty Thus hU i = log Z1 N log Zc = N = tanh , 2 2 (1.1 83) and = 2hU i 2 tanh N 1 (1.184) The negative temperature is originated by our assumption that the energy of a single magnet... or in terms of the total energy E = } (m + N/2) ã à ảá ã à ảá E N N E = N+ log N + } 2 } 2 à ả à ả N N E E N log N log } 2 } 2 (1.2 13) c) The temperature is given by 1 = E 2} 1 ln 2E+N} ln 2} 1 = + 2EN} } à ả } 1 2E + N } ln = } 2E N } Eyal Buks Thermodynamics and Statistical Physics (1.214) 42 1.8 Solutions Set 1 In the thermodynamical limit (N 1, m 1) the energy E and its average U... indistinguishable, thus à ả } 2U + N } , (1.215) exp = 2U N } or U= } N} coth 2 2 (1.216) 25 The grand canonical partition function is given by = 1 + 2 exp () , (1.217) where = exp (à) is the fugacity, thus hNd i = Eyal Buks 2e log 1 = = 1 + 2e 1 + 1 e(+à) 2 Thermodynamics and Statistical Physics (1.218) 43 ... Statistical Physics 39 Chapter 1 The Principle of Largest Uncertainty is determined by the Boltzmann factor The energy change due to ipping of one link from 0 to 180 is 2fd, therefore hli = d ef d ef d = d tanh (fd) , ef d + ef d where = 1/ Thus L = Nd tanh (f d) , or f= L tanh1 d Nd 21 The grand partition function Zgc is given by Zgc = 1 + 2 exp [ (à )] , (1.198) thus hN i = 2 1 log Zgc = ... given by ã à ảá a a exp (F a) F a = = 1 + tanh , (1.192) 1 + exp (F a) 2 2 where = 1/ b) At high temperature F a 1 the length of the chain L = N is given by ã à ảá à ả F a Na F a Na 1 + tanh ' 1+ , (1.1 93) L= 2 2 2 2 or à Na F = k L 2 ả , (1.194) where the spring constant k is given by k= 4 Na2 (1.195) 19 The average length of a single link is given by a exp (F a) hli = X n=0 exp (F a) X n=0 = a exp(F . independen t, thus σ =6× µ − 2 3 ln 2 3 − 1 3 ln 1 3 ¶ =3. 8191 . 12. The random variable X obtains t he value n with proba b ility p n = q n , where n =1, 2, 3, ,andq =1/2. a) The entropy is given. − 5 6 ¢ 2 =6. 3. Let W (m) be the probability for for taking n 1 steps to the right and n 2 = N − n 1 steps to the left, where m = n 1 − n 2 ,andN = n 1 + n 2 . Using Eyal Buks Thermodynamics and Statist. expected number of questioned required to find X is 1 2 × 1+ 1 4 × 2+ 1 8 × 3+ =2, which is exactly the entropy σ. 13. Eyal Buks Thermodynamics and Statist ical Physics 35 Chapter 1. The Principle of