Eyal Buks Introduction to Thermodynamics and Statistical Physics (114016) - Lecture Notes September 24, 2017 Technion Preface to be written... Contents 1. The Principle of Largest Uncertainty . . . . . . . . . . . . . . . . . . . . . 1.1 Entropy in Information Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Example - Two States System . . . . . . . . . . . . . . . . . . . . . 1.1.2 Smallest and Largest Entropy . . . . . . . . . . . . . . . . . . . . . . 1.1.3 The composition property . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.4 Alternative Deﬁnition of Entropy . . . . . . . . . . . . . . . . . . . 1.2 Largest Uncertainty Estimator . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 The Free Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 The Principle of Largest Uncertainty in Statistical Mechanics 1.3.1 Microcanonical Distribution . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Canonical Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Grandcanonical Distribution . . . . . . . . . . . . . . . . . . . . . . 1.3.4 Temperature and Chemical Potential . . . . . . . . . . . . . . . 1.4 Time Evolution of Entropy of an Isolated System . . . . . . . . . . . 1.5 Thermal Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.1 Externally Applied Potential Energy . . . . . . . . . . . . . . . . 1.6 Free Entropy and Free Energies . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Problems Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Solutions Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 1 2 5 8 9 11 13 14 14 15 16 17 18 19 20 21 21 30 2. Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 A Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Gibbs Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Fermions and Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Fermi-Dirac Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Bose-Einstein Distribution . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Classical Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Ideal Gas in the Classical Limit . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.4 Internal Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . 2.5 Processes in Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 47 50 52 53 54 54 55 57 58 59 59 62 Contents 2.5.1 Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Isobaric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Isochoric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.4 Isentropic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Carnot Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limits Imposed Upon the Eﬃciency . . . . . . . . . . . . . . . . . . . . . . Problems Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 64 65 65 66 68 73 81 3. Bosonic and Fermionic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Electromagnetic Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Electromagnetic Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Cube Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Average Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Stefan-Boltzmann Radiation Law . . . . . . . . . . . . . . . . . . . 3.2 Phonons in Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 One Dimensional Example . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 The 3D Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Fermi Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Orbital Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Partition Function of the Gas . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Energy and Number of Particles . . . . . . . . . . . . . . . . . . . . 3.3.4 Example: Electrons in Metal . . . . . . . . . . . . . . . . . . . . . . . 3.4 Semiconductor Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Problems Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Solutions Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 99 99 102 102 104 105 107 107 109 112 112 112 114 114 116 117 119 4. Classical Limit of Statistical Mechanics . . . . . . . . . . . . . . . . . . . 4.1 Classical Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Hamilton-Jacobi Equations . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Density Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Equipartition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Nyquist Noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Thermal Equilibrium From Stochastic Processes . . . . . . . . . . . . 4.4.1 Langevin Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 The Smoluchowski-Chapman-Kolmogorov Relation . . . 4.4.3 The Fokker-Planck Equation . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 The Potential Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.5 Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.6 Fokker-Planck Equation in One Dimension . . . . . . . . . . 4.4.7 Ornstein—Uhlenbeck Process in One Dimension . . . . . . . 129 129 130 130 131 132 132 133 134 138 138 139 139 141 142 143 145 2.6 2.7 2.8 2.9 Eyal Buks Thermodynamics and Statistical Physics 6 Contents 4.5 Problems Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 4.6 Solutions Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 5. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 5.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 5.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Eyal Buks Thermodynamics and Statistical Physics 7 1. The Principle of Largest Uncertainty In this chapter we discuss relations between information theory and statistical mechanics. We show that the canonical and grand canonical distributions can be obtained from Shannon’s principle of maximum uncertainty [1, 2, 3]. Moreover, the time evolution of the entropy of an isolated system and the H theorem are discussed. 1.1 Entropy in Information Theory The possible states of a given system are denoted as em , where m = 1, 2, 3, · · · , and the probability that state em is occupied is denoted by pm . The normalization condition reads pm = 1 . (1.1) m For a given probability distribution {pm } the entropy is deﬁned as σ=− pm log pm . (1.2) m Below we show that this quantity characterizes the uncertainty in the knowledge of the state of the system. 1.1.1 Example - Two States System Consider a system which can occupy either state e1 with probability p, or state e2 with probability 1 − p, where 0 ≤ p ≤ 1. The entropy is given by σ = −p log p − (1 − p) log (1 − p) . (1.3) Chapter 1. The Principle of Largest Uncertainty 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.2 0.4 x 0.6 0.8 1 −p log p − (1 − p) log (1 − p) As expected, the entropy vanishes at p = 0 and p = 1, since in both cases there is no uncertainty in what is the state which is occupied by the system. The largest uncertainty is obtained at p = 0.5, for which σ = log 2 = 0.69. 1.1.2 Smallest and Largest Entropy Smallest value. The term −p log p in the range 0 ≤ p ≤ 1 is plotted in the ﬁgure below. Note that the value of −p log p in the limit p → 0 can be calculated using L’Hospital’s rule lim (−p log p) = lim p→0 p→0 − d log p dp d 1 dp p =0. (1.4) From this ﬁgure, which shows that −p log p ≥ 0 in the range 0 ≤ p ≤ 1, it is easy to infer that the smallest possible value of the entropy is zero. Moreover, since −p log p = 0 iﬀ p = 0 or p = 1, it is evident that σ = 0 iﬀ the system occupies one of the states with probability one and all the other states with probability zero. In this case there is no uncertainty in what is the state which is occupied by the system. Eyal Buks Thermodynamics and Statistical Physics 2 1.1. Entropy in Information Theory 0.35 0.3 0.25 -p*log(p) 0.2 0.15 0.1 0.05 0 0.2 0.4 p 0.6 0.8 1 −p log p Largest value. We seek a maximum point of the entropy σ with respect to all probability distributions {pm } which satisfy the normalization condition. This constrain, which is given by Eq. (1.1), is expressed as 0 = g0 (p̄) = m pm − 1 , (1.5) where p̄ denotes the vector of probabilities p̄ = (p1 , p2 , · · · ) . (1.6) A small change in σ (denoted as δσ) due to a small change in p̄ (denoted as δ p̄ = (δp1 , δp2 , · · · )) can be expressed as δσ = m ∂σ δpm , ∂pm (1.7) ¯ as or in terms of the gradient of σ (denoted as ∇σ) ¯ · δ p̄ . δσ = ∇σ (1.8) In addition the variables (p1 , p2 , · · · ) are subjected to the constrain (1.5). Similarly to Eq. (1.8) we have ¯ 0 · δ p̄ . δg0 = ∇g ¯ and δ p̄ can be decomposed as Both vectors ∇σ ¯ = ∇σ ¯ ¯ ∇σ + ∇σ , ⊥ δ p̄ = (δ p̄) + (δ p̄)⊥ , (1.9) (1.10) (1.11) ¯ ¯ 0 , and where ∇σ ¯ where ∇σ and (δ p̄) are parallel to ∇g and (δ p̄)⊥ are ⊥ ¯ 0 . Using this notation Eq. (1.8) can be expressed as orthogonal to ∇g Eyal Buks Thermodynamics and Statistical Physics 3 Chapter 1. The Principle of Largest Uncertainty ¯ δσ = ∇σ ¯ · (δ p̄) + ∇σ ⊥ · (δ p̄)⊥ . (1.12) Given that the constrain g0 (p̄) = 0 is satisﬁed at a given point p̄, one has ¯ 0, g0 (p̄ + δ p̄) = 0 to ﬁrst order in δ p̄ provided that δ p̄, is orthogonal to ∇g namely, provided that (δ p̄) = 0. Thus, a stationary (maximum or minimum or saddle point) point of σ occurs iﬀ for every small change δ p̄, which is ¯ 0 (namely, δ p̄ · ∇g ¯ 0 = 0) one has 0 = δσ = ∇σ ¯ · δ p̄. As orthogonal to ∇g ¯ can be seen from Eq. (1.12), this condition is fulﬁlled only when ∇σ = 0, ⊥ ¯ and ∇g ¯ 0 are parallel to each other. In other namely only when the vectors ∇σ words, only when ¯ = ξ 0 ∇g ¯ 0, ∇σ (1.13) where ξ 0 is a constant. This constant is called Lagrange multiplier . Using Eqs. (1.2) and (1.5) the condition (1.13) is expressed as log pm + 1 = ξ 0 . (1.14) Let M be the number of available states. From Eq. (1.14) we ﬁnd that all probabilities are equal. Thus using Eq. (1.5), one ﬁnds that p1 = p2 = · · · = 1 . M (1.15) After ﬁnding this stationary point it is necessary to determine whether it is a maximum or minimum or saddle point. To do this we expand σ to second order in δ p̄ ¯ σ (p̄) σ (p̄ + δ p̄) = exp δ p̄ · ∇ = ¯+ 1 + δ p̄ · ∇ ¯ δ p̄ · ∇ 2! 2 +··· σ (p̄) ¯ 2 ¯ + δ p̄ · ∇ σ + · · · = σ (p̄) + δ p̄ · ∇σ 2! 1 ∂σ ∂2σ = σ (p̄) + δpm + δpm δpm′ +··· ∂pm 2 m,m′ ∂pm ∂pm′ m (1.16) Using Eq. (1.2) one ﬁnds that ∂2σ 1 = − δ m,m′ . ∂pm ∂pm′ pm (1.17) Since the probabilities pm are non-negative one concludes that any stationary point of σ is a local maximum point. Moreover, since only a single stationary point was found, one concludes that the entropy σ obtains its largest value, which is denoted as Λ (M), and which is given by Eyal Buks Thermodynamics and Statistical Physics 4 1.1. Entropy in Information Theory Λ (M) = σ 1 1 1 , ,··· , M M M = log M , (1.18) for the probability distribution given by Eq. (1.15). For this probability distribution that maximizes σ, as expected, the state which is occupied by the system is most uncertain. 1.1.3 The composition property The composition property is best illustrated using an example. Example - A Three States System. A system can occupy one of the states e1 , e2 or e3 with probabilities p1 , p2 and p3 respectively. The uncertainty associated with this probability distribution can be estimated in two ways, directly and indirectly. Directly, it is simply given by the deﬁnition of entropy in Eq. (1.2) σ (p1 , p2 , p3 ) = −p1 log p1 − p2 log p2 − p3 log p3 . (1.19) Alternatively [see Fig. 1.1], the uncertainty can be decomposed as follows: (a) the system can either occupy state e1 with probability p1 , or not occupy state e1 with probability 1 − p1 ; (b) given that the system does not occupy state e1 , it can either occupy state e2 with probability p2 / (1 − p1 ) or occupy state e3 with probability p3 / (1 − p1 ). Assuming that uncertainty (entropy) is additive, the total uncertainty (entropy) is given by σi = σ (p1 , 1 − p1 ) + (1 − p1 ) σ p2 p3 , 1 − p1 1 − p1 . (1.20) The factor (1 − p1 ) in the second term is included since the uncertainty associated with distinction between states e2 and e3 contributes only when state e1 is not occupied, an event which occurs with probability 1 − p1 . Using the deﬁnition (1.2) and the normalization condition p1 + p2 + p3 = 1 , (1.21) one ﬁnds σi = −p1 log p1 − (1 − p1 ) log (1 − p1 ) p2 p2 p3 p3 log − log 1 − p1 1 − p1 1 − p1 1 − p1 = −p1 log p1 − p2 log p2 − p3 log p3 − (1 − p1 − p2 − p3 ) log (1 − p1 ) = σ (p1 , p2 , p3 ) , + (1 − p1 ) − (1.22) that is, for this example the entropy satisﬁes the decomposition property. Eyal Buks Thermodynamics and Statistical Physics 5 Chapter 1. The Principle of Largest Uncertainty p1 1 − p1 e1 e2 , e3 p2 1 − p1 e2 p3 1 − p1 e3 Fig. 1.1. The composition property - three states system. The general case. The composition property in the general case can be deﬁned as follows. Consider a system which can occupy one of the states {e1 , e2 , · · · , eM0 } with probabilities q1 , q2 , · · · , qM0 respectively. This set of states is grouped as follows. The ﬁrst group includes the ﬁrst M1 states {e1 , e2 , · · · , eM1 }; the second group includes the next M2 states {eM1 +1 , eM1 +2 , · · · , eM1 +M2 }, etc., where M1 + M2 + · · · = M0 . The probability that one of the states in the ﬁrst group is occupied is p1 = q1 +q2 +· · ·+qM1 , the probability that one of the states in the second group is occupied is p2 = qM1 +1 + qM1 +2 + · · · + qM1 +M2 , etc., where p1 + p2 + · · · = 1 . (1.23) The composition property requires that the following holds [see Fig. 1.2] σ (q1 , q2 , · · · , qM0 ) = σ (p1 , p2 , · · · ) + p1 σ + p2 σ q1 q2 qM , ,··· , 1 p1 p1 p1 qM1 +1 qM1 +2 qM +M , ,··· , 1 2 p2 p2 p2 + ··· (1.24) Using the deﬁnition (1.2) the following holds σ (p1 , p2 , · · · ) = −p1 log p1 − p2 log p2 − · · · , Eyal Buks Thermodynamics and Statistical Physics (1.25) 6 1.1. Entropy in Information Theory p1 e1 , e2 ,..., eM 1 q1 p1 e1 q2 p1 p2 … eM 1 +1 ,..., eM 1 + M 2 qM 1 qM1 qM 1 + 2 qM 1 + M 2 p1 p2 p2 p2 … e e2 M 1 p1 = q1 + q2 + ... + qM 1 … eM 1 +1 eM1 + 2 eM 1 + M 2 p2 = qM 1 +1 + ... + qM 1 + M 2 Fig. 1.2. The composition property - the general case. q1 q2 qM , ,··· , 1 p1 p1 p1 q1 q2 q2 qM1 qM q1 − log − ··· − log 1 = p1 − log p1 p1 p1 p1 p1 p1 = −q1 log q1 − q2 log q2 − · · · − qM1 log qM1 + p1 log p1 , p1 σ (1.26) qM1 +1 qM1 +2 qM +M , ,··· , 1 2 p2 p2 p2 = −qM1 +1 log qM1 +1 − qM1 +2 log qM1 +2 − · · · − qM1 +M2 log qM1 +M2 + p2 log p2 , p2 σ (1.27) etc., thus it is evident that condition (1.24) is indeed satisﬁed. Eyal Buks Thermodynamics and Statistical Physics 7 Chapter 1. The Principle of Largest Uncertainty 1.1.4 Alternative Deﬁnition of Entropy Following Shannon [1, 2], the entropy function σ (p1 , p2 , · · · , pN ) can be alternatively deﬁned as follows: 1. σ (p1 , p2 , · · · , pN ) is a continuous function of its arguments p1 , p2 , · · · , pN . 2. If all probabilities are equal, namely if p1 = p2 = · · · = pN = 1/N , then the quantity Λ (N ) = σ (1/N, 1/N, · · · , 1/N ) is a monotonic increasing function of N. 3. The function σ (p1 , p2 , · · · , pN ) satisﬁes the composition property given by Eq. (1.24). Exercise 1.1.1. Show that the above deﬁnition leads to the entropy given by Eq. (1.2) up to multiplication by a positive constant. Solution 1.1.1. The 1st property allows approximating the probabilities p1 , p2 , · · · , pN using rational numbers, namely p1 = M1 /M0 , p2 = M2 /M0 , etc., where M1 , M2 , · · · are integers and M0 = M1 + M2 + · · · + MN . Using the composition property (1.24) one ﬁnds Λ (M0 ) = σ (p1 , p2 , · · · , pN ) + p1 Λ (M1 ) + p2 Λ (M2 ) + · · · (1.28) In particular, consider the case were M1 = M2 = · · · = MN = K. For this case one ﬁnds Λ (NK) = Λ (N) + Λ (K) . (1.29) Taking K = N = 1 yields Λ (1) = 0 . (1.30) Taking N = 1 + x yields Λ (K + Kx) − Λ (K) 1 Λ (1 + x) = . Kx K x (1.31) Taking the limit x → 0 yields dΛ C = , dK K (1.32) where C = lim x→0 Λ (1 + x) . x (1.33) Integrating Eq. (1.32) and using the initial condition (1.30) yields Λ (K) = C log K . Eyal Buks Thermodynamics and Statistical Physics (1.34) 8 1.2. Largest Uncertainty Estimator Moreover, the second property requires that C > 0. Choosing C = 1 and using Eq. (1.28) yields σ (p1 , p2 , · · · , pN ) = Λ (M0 ) − p1 Λ (M1 ) − p2 Λ (M2 ) − · · · M1 M2 MN − p2 log − · · · − pM log = −p1 log M0 M0 M0 = −p1 log p1 − p2 log p2 − · · · − pN log pN , (1.35) in agreement with the deﬁnition (1.2). 1.2 Largest Uncertainty Estimator As before, the possible states of a given system are denoted as em , where m = 1, 2, 3, · · · , and the probability that state em is occupied is denoted by pm . Let Xl (l = 1, 2, · · · , L) be a set of variables characterizing the system (e.g., energy, number of particles, etc.). Let Xl (m) be the value which the variable Xl takes when the system is in state em . Consider the case where the expectation values of the variables Xl are given Xl = pm Xl (m) , (1.36) m where l = 1, 2, · · · , L. However, the probability distribution {pm } is not given. Clearly, in the general case the knowledge of X1 , X2 , · · · , XL is not suﬃcient to obtain the probability distribution because there are in general many diﬀerent possibilities for choosing a probability distribution which is consistent with the contrarians (1.36) and the normalization condition (1.1). For each such probability distribution the entropy can be calculated according to the deﬁnition (1.2). The probability distribution {pm }, which is consistent with these conditions, and has the largest possible entropy is called the largest uncertainty estimator (LUE). The LUE is found by seeking a stationary point of the entropy σ with respect to all probability distributions {pm } which satisfy the normalization constrain (1.5) in addition to the constrains (1.36), which can be expressed as 0 = gl (p̄) = m pm Xl (m) − Xl , (1.37) where l = 1, 2, · · · L. To ﬁrst order one has ¯ · δ p̄ , δσ = ∇σ ¯ l · δ p̄ , δgl = ∇g Eyal Buks Thermodynamics and Statistical Physics (1.38a) (1.38b) 9 Chapter 1. The Principle of Largest Uncertainty where l = 0, 1, 2, · · · L. A stationary point of σ occurs iﬀ for every small ¯ 0 , ∇g ¯ 1 , ∇g ¯ 2 , · · · , ∇g ¯ L one change δ p̄, which is orthogonal to all vectors ∇g has ¯ · δ p̄ . 0 = δσ = ∇σ (1.39) ¯ belongs to the subspace This condition is fulﬁlled only when the vector ∇σ ¯ ¯ ¯ ¯ spanned by the vectors ∇g0 , ∇g1 , ∇g2 , · · · , ∇gL [see also the discussion below Eq. (1.12) above]. In other words, only when ¯ = ξ 0 ∇g ¯ 0 + ξ 1 ∇g ¯ 1 + ξ 2 ∇g ¯ 2 + · · · + ξ L ∇g ¯ L, ∇σ (1.40) where the numbers ξ 0 , ξ 1 , · · · , ξ L , which are called Lagrange multipliers, are constants. Using Eqs. (1.2), (1.5) and (1.37) the condition (1.40) can be expressed as L − log pm − 1 = ξ 0 + ξ l Xl (m) . (1.41) l=1 From Eq. (1.41) one obtains L pm = exp (−1 − ξ 0 ) exp − ξ l Xl (m) . (1.42) l=1 The Lagrange multipliers ξ 0 , ξ 1 , · · · , ξ L can be determined from Eqs. (1.5) and (1.37) L 1= m pm = exp (−1 − ξ 0 ) Xl = m m exp − ξ l Xl (m) , (1.43) l=1 pm Xl (m) L = exp (−1 − ξ 0 ) m exp − ξ l Xl (m) Xl (m) . l=1 (1.44) Using Eqs. (1.42) and (1.43) one ﬁnds L pm = exp − m ξ l Xl (m) l=1 L exp − l=1 . (1.45) ξ l Xl (m) In terms of the partition function Z, which is deﬁned as Eyal Buks Thermodynamics and Statistical Physics 10 1.2. Largest Uncertainty Estimator L Z= exp − m ξ l Xl (m) , (1.46) ξ l Xl (m) . (1.47) l=1 one ﬁnds 1 exp − Z pm = L l=1 Using the same arguments as in section 1.1.2 above [see Eq. (1.16)] it is easy to show that at the stationary point that occurs for the probability distribution given by Eq. (1.47) the entropy obtains its largest value. 1.2.1 Useful Relations The expectation value Xl can be expressed as Xl = pm Xl (m) m = 1 Z L m exp − ξ l Xl (m) Xl (m) l=1 1 ∂Z Z ∂ξ l ∂ log Z =− . ∂ξ l =− (1.48) Similarly, Xl2 can be expressed as Xl2 = pm Xl2 (m) m = 1 Z L m exp − ξ l Xl (m) Xl2 (m) l=1 1 ∂2Z = . Z ∂ξ 2l (1.49) Using Eqs. (1.48) and (1.49) one ﬁnds that the variance of the variable Xl is given by (∆Xl )2 = (Xl − Xl )2 = 1 ∂2Z − Z ∂ξ 2l 1 ∂Z Z ∂ξ l 2 . (1.50) However, using the following identity Eyal Buks Thermodynamics and Statistical Physics 11 Chapter 1. The Principle of Largest Uncertainty ∂ 2 log Z ∂ 1 ∂Z 1 ∂2Z = = − 2 ∂ξ l Z ∂ξ l Z ∂ξ 2l ∂ξ l one ﬁnds 1 ∂Z Z ∂ξ l 2 , ∂ 2 log Z . ∂ξ 2l (∆Xl )2 = (1.51) (1.52) Note that the above results Eqs. (1.48) and (1.52) are valid only when Z is expressed as a function of the the Lagrange multipliers, namely Z = Z (ξ 1 , ξ 2 , · · · , ξ L ) . (1.53) Using the deﬁnition of entropy (1.2) and Eq. (1.47) one ﬁnds σ=− =− pm log pm m pm log m 1 exp − Z L ξ l Xl (m) l=1 L = m pm log Z + ξ l Xl (m) l=1 L = log Z + ξl l=1 pm Xl (m) , m (1.54) thus L σ = log Z + ξ l Xl . (1.55) l=1 Using the above relations one can also evaluate the partial derivative of the entropy σ when it is expressed as a function of the expectation values, namely σ = σ ( X1 , X2 , · · · , XL ) . (1.56) Using Eq. (1.55) one has ∂σ ∂ log Z = + ∂ Xl ∂ Xl = ∂ log Z + ∂ Xl L = l′ =1 L X l′ l′ =1 L Xl′ l′ =1 ∂ξ l′ + ∂ Xl L ξ l′ l′ =1 ∂ Xl′ ∂ Xl ∂ξ l′ + ξl ∂ Xl ∂ log Z ∂ξ l′ + ∂ξ l′ ∂ Xl L Xl′ l′ =1 ∂ξ l′ + ξl , ∂ Xl (1.57) Eyal Buks Thermodynamics and Statistical Physics 12 1.2. Largest Uncertainty Estimator thus using Eq. (1.48) one ﬁnds ∂σ = ξl . ∂ Xl (1.58) 1.2.2 The Free Entropy The free entropy σF is deﬁned as the term log Z in Eq. (1.54) σF = log Z =σ− L ξl pm Xl (m) m l=1 L =− m pm log pm − ξl pm Xl (m) . m l=1 (1.59) The free entropy is commonly expressed as a function of the Lagrange multipliers σF = σF (ξ 1 , ξ 2 , · · · , ξ L ) . (1.60) We have seen above that the LUE maximizes σ for given values of expectation values X1 , X2 , · · · , XL . We show below that a similar result can be obtained for the free energy σF with respect to given values of the Lagrange multipliers. Claim. The LUE maximizes σF for given values of the Lagrange multipliers ξ1 , ξ2, · · · , ξL. Proof. As before, the normalization condition is expressed as 0 = g0 (p̄) = m pm − 1 . (1.61) At a stationary point of σF , as we have seen previously, the following holds ¯ F = η∇g ¯ 0, ∇σ (1.62) where η is a Lagrange multiplier. Thus L − (log pm + 1) − ξ l Xl (m) = η , (1.63) l=1 or L pm = exp (−η − 1) exp − Eyal Buks ξ l Xl (m) . (1.64) l=1 Thermodynamics and Statistical Physics 13 Chapter 1. The Principle of Largest Uncertainty Table 1.1. The microcanonical, canonical and grandcanonical distributions. energy number of particles microcanonical distribution constrained U (m) = U constrained N (m) = N canonical distribution average is given U constrained N (m) = N grandcanonical distribution average is given U average is given N This result is the same as the one given by Eq. (1.42). Taking into account the normalization condition (1.61) one obtains the same expression for pm as the one given by Eq. (1.47). Namely, the stationary point of σ F corresponds to the LUE probability distribution. Since ∂ 2 σF 1 = − δ m,m′ < 0 , ∂pm ∂pm′ pm (1.65) one concludes that this stationary point is a maximum point [see Eq. (1.16)]. 1.3 The Principle of Largest Uncertainty in Statistical Mechanics The energy and number of particles of state em are denoted by U (m) and N (m) respectively. The probability that state em is occupied is denoted as pm . We consider below three cases (see table 1.1). In the ﬁrst case (microcanonical distribution) the system is isolated and its total energy U and number of particles N are constrained , that is for all accessible states U (m) = U and N (m) = N. In the second case (canonical distribution) the system is allowed to exchange energy with the environment, and we assume that its average energy U is given. However, its number of particles is constrained , that is N (m) = N . In the third case (grandcanonical distribution) the system is allowed to exchange both energy and particles with the environment, and we assume that both the average energy U and the average number of particles N are given. However, in all cases, the probability distribution {pm } is not given. According to the principle of largest uncertainty in statistical mechanics the LUE is employed to estimate the probability distribution {pm }, namely, we will seek a probability distribution which is consistent with the normalization condition (1.1) and with the given expectation values (energy, in the second case, and both energy and number of particles, in the third case), which maximizes the entropy. 1.3.1 Microcanonical Distribution In this case no expectation values are given. Thus we seek a probability distribution which is consistent with the normalization condition (1.1), and Eyal Buks Thermodynamics and Statistical Physics 14 1.3. The Principle of Largest Uncertainty in Statistical Mechanics which maximizes the entropy. The desired probability distribution is p1 = p2 = · · · = 1/M , (1.66) where M is the number of accessible states of the system [see also Eq. (1.18)]. Using Eq. (1.2) the entropy for this case is given by σ = log M . (1.67) 1.3.2 Canonical Distribution Using Eq. (1.47) one ﬁnds that the probability distribution is given by pm = 1 exp (−βU (m)) , Zc (1.68) where β is the Lagrange multiplier associated with the given expectation value U , and the partition function is given by Zc = exp (−βU (m)) . (1.69) m The term exp (−βU (m)) is called Boltzmann factor. Moreover, Eq. (1.48) yields U =− ∂ log Zc , ∂β (1.70) Eq. (1.52) yields (∆U )2 = ∂ 2 log Zc , ∂β 2 (1.71) and Eq. (1.55) yields σ = log Zc + β U . (1.72) Using Eq. (1.58) one can expressed the Lagrange multiplier β as β= ∂σ . ∂U (1.73a) The temperature τ = 1/β is deﬁned as 1 =β. τ (1.74) Exercise 1.3.1. Consider a system that can be in one of two states having energies ±ε/2. Calculate the average energy U and the variance (∆U )2 in thermal equilibrium at temperature τ. Eyal Buks Thermodynamics and Statistical Physics 15 Chapter 1. The Principle of Largest Uncertainty Solution: The partition function is given by Eq. (1.69) βε 2 Zc = exp + exp − βε 2 = 2 cosh βε 2 , (1.75) thus using Eqs. (1.70) and (1.71) one ﬁnds ε U = − tanh 2 βε 2 , (1.76) and 2 (∆U ) = ε 2 2 1 cosh2 βε 2 , (1.77) where β = 1/τ . 1 2 x 3 4 5 0 -0.2 -0.4 -tanh(1/x) -0.6 -0.8 -1 1.3.3 Grandcanonical Distribution Using Eq. (1.47) one ﬁnds that the probability distribution is given by pm = 1 exp (−βU (m) − ηN (m)) , Zgc (1.78) where β and η are the Lagrange multipliers associated with the given expectation values U and N respectively, and the partition function is given by Zgc = m exp (−βU (m) − ηN (m)) . (1.79) The term exp (−βU (m) − ηN (m)) is called Gibbs factor. Moreover, Eq. (1.48) yields Eyal Buks Thermodynamics and Statistical Physics 16 1.3. The Principle of Largest Uncertainty in Statistical Mechanics U =− ∂ log Zgc ∂β η N =− ∂ log Zgc ∂η β , (1.80) (1.81) Eq. (1.52) yields 2 η ∂ 2 log Zgc ∂η2 β (1.82) , (1.83) σ = log Zgc + β U + η N . (1.84) 2 (∆N ) = ∂ 2 log Zgc ∂β 2 , (∆U ) = and Eq. (1.55) yields 1.3.4 Temperature and Chemical Potential Probability distributions in statistical mechanics of macroscopic parameters are typically extremely sharp and narrow. Consequently, in many cases no distinction is made between a parameter and its expectation value. That is, the expression for the entropy in Eq. (1.72) can be rewritten as σ = log Zc + βU , (1.85) and the one in Eq. (1.84) as σ = log Zgc + βU + ηN . (1.86) Using Eq. (1.58) one can expressed the Lagrange multipliers β and η as β= η= ∂σ ∂U ∂σ ∂N , (1.87) . (1.88) N U The chemical potential µ is deﬁned as µ = −τ η . (1.89) In the deﬁnition (1.2) the entropy σ is dimensionless. Historically, the entropy was deﬁned as S = kB σ , (1.90) where Eyal Buks Thermodynamics and Statistical Physics 17 Chapter 1. The Principle of Largest Uncertainty kB = 1.38 × 10−23 J K−1 (1.91) is the Boltzmann constant. Moreover, the historical deﬁnition of the temperature is τ . (1.92) T = kB When the grandcanonical partition function is expressed in terms of β and µ (instead of in terms of β and η), it is convenient to rewrite Eqs. (1.80) and (1.81) as (see homework exercises 14 of chapter 1). ∂ log Zgc ∂β U =− + τµ µ ∂ log Zgc ∂µ , (1.93) β ∂ log Zgc , ∂λ where λ is the fugacity , which is deﬁned by N =λ λ = exp (βµ) = e−η . (1.94) (1.95) 1.4 Time Evolution of Entropy of an Isolated System Consider a perturbation which results in transitions between the states of an isolated system. Let Γrs denotes the resulting rate of transition from state r to state s. The probability that state s is occupied is denoted as ps . The following theorem (H theorem) states that if for every pair of states r and s Γrs = Γsr , (1.96) then dσ ≥0. dt (1.97) Moreover, equality holds iﬀ ps = pr for all pairs of states for which Γsr = 0. To prove this theorem we express the rate of change in the probability ps in terms of these transition rates dpr = dt s ps Γsr − pr Γrs . (1.98) s The ﬁrst term represents the transitions to state r, whereas the second one represents transitions from state r. Using property (1.96) one ﬁnds dpr = dt Eyal Buks s Γsr (ps − pr ) . Thermodynamics and Statistical Physics (1.99) 18 1.5. Thermal Equilibrium The last result and the deﬁnition (1.2) allows calculating the rate of change of entropy dσ d =− dt dt =− =− pr log pr r dpr (log pr + 1) dt r r s Γsr (ps − pr ) (log pr + 1) . (1.100) One the other hand, using Eq. (1.96) and exchanging the summation indices allow rewriting the last result as dσ = dt r s Γsr (ps − pr ) (log ps + 1) . (1.101) Thus, using both expressions (1.100) and (1.101) yields dσ 1 = dt 2 r s Γsr (ps − pr ) (log ps − log pr ) . (1.102) In general, since log x is a monotonic increasing function (ps − pr ) (log ps − log pr ) ≥ 0 , (1.103) and equality holds iﬀ ps = pr . Thus, in general dσ ≥0, dt (1.104) and equality holds iﬀ ps = pr holds for all pairs is states satisfying Γsr = 0. When σ becomes time independent the system is said to be in thermal equilibrium. In thermal equilibrium, when all accessible states have the same probability, one ﬁnds using the deﬁnition (1.2) σ = log M , (1.105) where M is the number of accessible states of the system. Note that the rates Γrs , which can be calculated using quantum mechanics, indeed satisfy the property (1.96) for the case of an isolated system. 1.5 Thermal Equilibrium Consider two isolated systems denoted as S1 and S2 . Let σ1 = σ 1 (U1 , N1 ) and σ 2 = σ2 (U2 , N2 ) be the entropy of the ﬁrst and second system respectively Eyal Buks Thermodynamics and Statistical Physics 19 Chapter 1. The Principle of Largest Uncertainty and let σ = σ 1 + σ2 be the total entropy. The systems are brought to contact and now both energy and particles can be exchanged between the systems. Let δU be an inﬁnitesimal energy, and let δN be an inﬁnitesimal number of particles, which are transferred from system 1 to system 2. The corresponding change in the total entropy is given by δσ = − − ∂σ 1 ∂U1 ∂σ1 ∂N1 = − δU + N1 δN + U1 1 1 + τ1 τ2 ∂σ 2 ∂U2 ∂σ 2 ∂N2 δU − − δU N2 δN U2 µ1 µ2 + τ1 τ2 δN . (1.106) The change δσ in the total entropy is obtained by removing a constrain. Thus, at the end of this process more states are accessible, and therefore, according to the principle of largest uncertainty it is expected that δσ ≥ 0 . (1.107) For the case where no particles can be exchanged (δN = 0) this implies that energy ﬂows from the system of higher temperature to the system of lower temperature. Another important case is the case where τ 1 = τ 2 , for which we conclude that particles ﬂow from the system of higher chemical potential to the system of lower chemical potential. In thermal equilibrium the entropy of the total system obtains its largest possible value. This occurs when τ1 = τ2 (1.108) µ1 = µ2 . (1.109) and 1.5.1 Externally Applied Potential Energy In the presence of externally applied potential energy µex the total chemical potential µtot is given by µtot = µint + µex , (1.110) where µint is the internal chemical potential . For example, for particles having charge q in the presence of electric potential V one has µex = qV , Eyal Buks (1.111) Thermodynamics and Statistical Physics 20 1.7. Problems Set 1 whereas, for particles having mass m in a constant gravitational ﬁeld g one has µex = mgz , (1.112) where z is the height. The thermal equilibrium relation (1.109) is generalized in the presence of externally applied potential energy as µtot,1 = µtot,2 . (1.113) 1.6 Free Entropy and Free Energies The free entropy [see Eq. (1.59)] for the canonical distribution is given by [see Eq. (1.85)] σF,c = σ − βU , (1.114) whereas for the grandcanonical case it is given by [see Eq. (1.86)] σF,gc = σ − βU − ηN . (1.115) We deﬁne below two corresponding free energies, the canonical free energy (known also as the Helmholtz free energy ) F = −τ σF,c = U − τ σ , (1.116) and the grandcanonical free energy G = −τ σ F,gc = U − τ σ + τ ηN = U − τ σ − µN . In section 1.2.2 above we have shown that the LUE maximizes σF for given values of the Lagrange multipliers ξ 1 , ξ 2 , · · · , ξ L . This principle can be implemented to show that: • In equilibrium at a given temperature τ the Helmholtz free energy obtains its smallest possible value. • In equilibrium at a given temperature τ and chemical potential µ the grandcanonical free energy obtains its smallest possible value. Our main results are summarized in table 1.2 below 1.7 Problems Set 1 Note: Problems 1-6 are taken from the book by Reif, chapter 1. Eyal Buks Thermodynamics and Statistical Physics 21 Chapter 1. The Principle of Largest Uncertainty Table 1.2. Summary of main results. micro general −canonical canonical grandcanonical U U , N (M states) given Xl where expectation l = 1, 2, ..., L values Z= partition function e L − l=1 Zc = ξl Xl (m) m m pm = pm − 1 e Z l=1 ξl Xl (m) pm = 1 M log Z Xl = − ∂ ∂ξ Xl (∆Xl )2 L ∂ 2 log Z ∂ξ2 l σ L ξ l Xl log Z + pm = pm = 1 −βU (m) Zc e 1 e−βU (m)−ηN (m) Zgc ∂ 2 log Zc ∂β 2 (∆U )2 = σ= σ = log M e−βU (m)−ηN (m) m Zc U = − ∂ log ∂β l (∆Xl )2 = Zgc = e−βU (m) U =− ∂ log Zgc ∂β N =− ∂ log Zgc ∂η (∆U )2 = β ∂ 2 log Zgc ∂β 2 (∆N )2 = ∂ 2 log Zgc ∂η 2 σ= σ= log Zc + β U log Zgc + β U + η N l=1 Lagrange multipliers ξl = ∂σ ∂ Xl β= { Xn }n=l ∂σ ∂U β= η= ∂σ ∂U N ∂σ ∂N U max min / max principle σ F (ξ 1 , ξ 2 , ..., ξ L ) L σF = σ − max σ ξ l Xl min F (τ ) min G (τ , µ) F = U − τσ G = U − τ σ − µN l=1 1. A drunk starts out from a lamppost in the middle of a street, taking steps of equal length either to the right or to the left with equal probability. What is the probability that the man will again be at the lamppost after taking N steps a) if N is even? b) if N is odd? 2. In the game of Russian roulette, one inserts a single cartridge into the drum of a revolver, leaving the other ﬁve chambers of the drum empty. One then spins the drum, aims at one’s head, and pulls the trigger. a) What is the probability of being still alive after playing the game N times? Eyal Buks Thermodynamics and Statistical Physics η 22 η β 1.7. Problems Set 1 b) What is the probability of surviving (N − 1) turns in this game and then being shot the N th time one pulls the trigger? c) What is the mean number of times a player gets the opportunity of pulling the trigger in this macabre game? 3. Consider the random walk problem with p = q and let m = n1 − n2 , denote the net displacement to the right. After a total of N steps, calculate the following mean values: m , m2 , m3 , and m4 . Hint: Calculate the moment generating function. 4. The probability W (n), that an event characterized by a probability p occurs n times in N trials was shown to be given by the binomial distribution W (n) = N! pn (1 − p)N−n . n! (N − n)! (1.117) Consider a situation where the probability p is small (p << 1) and where one is interested in the case n << N . (Note that if N is large, W (n) becomes very small if n → N because of the smallness of the factor pn when p << 1. Hence W (n) is indeed only appreciable when n << N .) Several approximations can then be made to reduce Eq. (1.117) to simpler form. a) Using the result ln(1 − p) ≃ −p, show that (1 − p)N−n ≃ e−Np . b) Show that N !/(N − n)! ≃ N n . c) Hence show that Eq. (1.117) reduces to W (n) = λn −λ e , n! where λ ≡ Np is the mean number of events. This distribution is called the "Poisson distribution." 5. Consider the Poisson distribution of the preceding problem. a) Show that it is properly normalized in the sense that ∞ W (n) = 1. n=0 (The sum can be extended to inﬁnity to an excellent approximation, since W (n) is negligibly small when n N .) b) Use the Poisson distribution to calculate n . c) Use the Poisson distribution to calculate (∆n)2 = (n − n )2 . 6. A molecule in a gas moves equal distances l between collisions with equal probability in any direction. After a total of N such displacements, what is the mean square displacement R2 of the molecule from its starting point ? 7. A multiple choice test contains 20 problems. The correct answer for each problem has to be chosen out of 5 options. Show that the probability to pass the test (namely to have at least 11 correct answers) using guessing only, is 5.6 × 10−4 . Eyal Buks Thermodynamics and Statistical Physics 23 Chapter 1. The Principle of Largest Uncertainty 8. Consider two objects traveling in the xy plane. Object A starts from the point (0, 0) and object B starts from the point (N, N ), where N is an integer. At each step both objects A and B simultaneously make a single move of length unity. Object A makes either a move to the right (x, y) → (x + 1, y) with probability 1/2 or an upward move (x, y) → (x, y + 1) with probability 1/2. On the other hand, object B makes either a move to the left (x, y) → (x − 1, y) with probability 1/2 or a downward move (x, y) → (x, y − 1) with probability 1/2. What is the probability that objects A and B meet along the way in the limit N → ∞? 9. Consider A dice having 6 faces. All faces have equal probability of outcome. Initially, n faces are colored white and 6−n faces are colored black, where n ∈ {0, 1, 2, · · · , 6}. Each time the outcome is white (black) one black (white) face is turned into a white (black) face before the next roll. The process continues until all faces have the same color. What is the probability pn that all faces will become white? 10. Alice, Bob and other N − 2 people are randomly seated at a round table. What is the probability pC that Alice and Bob will be seated next to each other? What is the probability pR that Alice and Bob will be seated next to each other for the case where the group is randomly seated in a row. 11. Write a computer function returning the value 1 with probability p and the value 0 with probability 1 − p for any given 0 ≤ p ≤ 1. The function can use another given function, which returns the value 1 with probability 1/2 and the value 0 with probability 1/2. Make sure the running time is ﬁnite. 12. Consider a system of N spins. Each spin can be in one of two possible states: in state ’up’ the magnetic moment of each spin is +m, and in state ’down’ it is −m. Let N+ (N− ) be the number of spins in state ’up’ (’down’), where N = N+ +N− . The total magnetic moment of the system is given by M = m (N+ − N− ) . (1.118) Assume that the probability that the system occupies any of its 2N possible states is equal. Moreover, assume that N ≫ 1. Let f (M) be the probability distribution of the random variable M (that is, M is considered in this approach as a continuous random variable). Use the Stirling’s formula N! = (2πN )1/2 N N exp −N + 1 +··· 2N (1.119) to show that f (M ) = 1 M2 √ exp − 2 2m N m 2πN . (1.120) Use this result to evaluate the expectation value and the variance of M . Eyal Buks Thermodynamics and Statistical Physics 24 1.7. Problems Set 1 13. Consider a one dimensional random walk. The probabilities of transiting to the right and left are p and q = 1 − p respectively. The step size for both cases is a. a) Show that the average displacement X after N steps is given by X = aN (2p − 1) = aN (p − q) . b) Show that the variance (X − X ) 2 (1.121) is given by (X − X )2 = 4a2 N pq . (1.122) 14. A classical harmonic oscillator of mass m, and spring constant k oscillates with amplitude a. Show that the probability density function f(x), where f (x)dx is the probability that the mass would be found in the interval dx at x, is given by 1 f(x) = √ 2 . π a − x2 (1.123) 15. A coin having probability p = 2/3 of coming up heads is ﬂipped 6 times. Show that the entropy of the outcome of this experiment is σ = 3.8191 (use log in natural base in the deﬁnition of the entropy). 16. A fair coin is ﬂipped until the ﬁrst head occurs. Let X denote the number of ﬂips required. a) Find the entropy σ. In this exercise use log in base 2 in the deﬁnition of the entropy, namely σ = − i pi log2 pi . b) A random variable X is drawn according to this distribution. Find an “eﬃcient” sequence of yes-no questions of the form, “Is X contained in the set S?” Compare σ to the expected number of questions required to determine X. 17. In general the notation ∂z ∂x y is used to denote the partial derivative of z with respect to x, where the variable y is kept constant. That is, to correctly calculate this derivative the variable z has to be expressed as a function of x and y, namely, z = z (x, y). a) Show that: ∂z ∂x Eyal Buks y =− ∂y ∂x z ∂y ∂z x . Thermodynamics and Statistical Physics (1.124) 25 Chapter 1. The Principle of Largest Uncertainty b) Show that: ∂z ∂x ∂z ∂x = w ∂z ∂y + y x ∂y ∂x . (1.125) w 18. Let Zgc be a grandcanonical partition function. a) Show that: ∂ log Zgc ∂β U =− + τµ µ ∂ log Zgc ∂µ . (1.126) β where τ is the temperature, β = 1/τ, and µ is the chemical potential. b) Show that: N =λ ∂ log Zgc , ∂λ (1.127) where λ = exp (βµ) (1.128) is the fugacity. 19. Consider an array on N distinguishable two-level (binary) systems. The two-level energies of each system are ±ε/2. Show that the temperature τ of the system is given by ε τ= −1 2 tanh U − 2Nε , (1.129) where U is the average total energy of the array. Note that the temperature can become negative if U > 0. Why a negative temperature is possible for this system ? 20. Consider an array of N distinguishable 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 εn = ℏω n + 1 2 , (1.130) where n = 0, 1, 2, · · · is integer. a) Show that the average energy of the system is given by U = N ℏω βℏω coth , 2 2 (1.131) where β = 1/τ. Eyal Buks Thermodynamics and Statistical Physics 26 1.7. Problems Set 1 b) Show that the variance of the energy of the system is given by 2 (∆U ) = ℏω 2 2 sinh 2 βℏω 2 N . (1.132) 21. Consider a lattice containing N non-interacting atoms. Each atom has 3 non-degenerate energy levels E1 = −ε, E2 = 0, E3 = ε. The system is at thermal equilibrium at temperature τ . a) Show that the average energy of the system is U =− 2N ε sinh (βε) , 1 + 2 cosh βε (1.133) where β = 1/τ. b) Show the variance of the energy of the system is given by (U − U )2 = 2N ε2 cosh (βε) + 2 . [1 + 2 cosh (βε)]2 (1.134) 22. Consider a one dimensional chain containing N ≫ 1 sections (see ﬁgure). Each section can be in one of two possible sates. In the ﬁrst one the section contributes a length a to the total length of the chain, whereas in the other state the section has no contribution to the total length of the chain. The total length of the chain in N α, and the tension applied to the end points of the chain is F . The system is in thermal equilibrium at temperature τ . a) Show that α is given by α= a 1 + tanh 2 Fa 2τ . (1.135) b) Show that in the limit of high temperature the spring constant is given approximately by k≃ 4τ . Na2 (1.136) Nα 23. A long elastic molecule can be modelled as a linear chain of N links. The state of each link is characterized 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 Eyal Buks Thermodynamics and Statistical Physics 27 Chapter 1. The Principle of Largest Uncertainty is modelled as a harmonic oscillator whose angular frequency is ω a for a link of length a and ωb for a link of length b. Thus, the energy of a link is ℏω a n + 12 for l = a , ℏωb n + 12 for l = b En,l = (1.137) n = 0, 1, 2, · · · The chain is held under a tension F . Show that the mean length L of the chain in the limit of high temperature T is given by 2 L =N aω b + bω a F ω b ω a (a − b) +N β + O β2 , 2 ωb + ω a (ω b + ω a ) (1.138) where β = 1/τ . 24. The elasticity of a rubber band can be described in terms of a onedimensional 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 each 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 maintain a length L is given by f= τ L tanh−1 . d Nd (1.139) 25. Consider a system which has two single particle states both of the same energy. When both states are unoccupied, the energy of the system is zero; when one state or the other is occupied by one particle, the 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 average number of particles in the level is N = 2 , 2 + exp [β (ε − µ)] (1.140) where µ is the chemical potential and β = 1/τ . 26. Consider an array of N two-level particles. Each one can be in one of two states, having energy E1 and E2 respectively. The numbers of particles in states 1 and 2 are n1 and n2 respectively, where N = n1 + n2 (assume n1 ≫ 1 and n2 ≫ 1). Consider an energy exchange with a reservoir at temperature τ leading to population changes n2 → n2 −1 and n1 → n1 +1. a) Calculate the entropy change of the two-level system, (∆σ)2LS . b) Calculate the entropy change of the reservoir, (∆σ)R . c) What can be said about the relation between (∆σ)2LS and (∆σ)R in thermal equilibrium? Use your answer to express the ration n2 /n1 as a function of E1 , E2 and τ . Eyal Buks Thermodynamics and Statistical Physics 28 1.7. Problems Set 1 27. Consider a lattice containing N sites of one type, 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 NA , whereas the number of atoms occupying atoms of type B is denoted as NB , where NA + NB = N . Let ε 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, NA , NB ≫ 1. a) Calculate the entropy σ. b) Calculate the average number NB of atoms occupying sites of type B. 28. Consider a microcanonical ensemble 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 εn = ℏω n + 1 2 , (1.141) where n = 0, 1, 2, · · · is integer. The total energy E of the system is given by E = ℏω m + N 2 , (1.142) where N m= nl , (1.143) l=1 and nl is state number of oscillator l. a) Calculate the number of states g (N, m) of the system with total energy ℏω (m + N/2). b) Use this result to calculate the entropy σ of the system with total energy ℏω (m + N/2). Approximate the result by assuming that N ≫ 1 and m ≫ 1. c) Use this result to calculate (in the same limit of N ≫ 1 and m ≫ 1) the average energy of the system U as a function of the temperature τ. 29. 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 spin up electron or a spin down electron, however, it cannot be simultaneously occupied by two electrons. Express the average occupation of a donor state Nd as a function of ε and the chemical potential µ. Eyal Buks Thermodynamics and Statistical Physics 29 Chapter 1. The Principle of Largest Uncertainty 1.8 Solutions Set 1 1. Final answers: 1 a) N N! ( 2 )!( N2 )! 2 b) 0 2. Final answers: N a) 56 N N−1 1 b) 56 6 c) In general ∞ N xN−1 = N=0 ∞ d d 1 1 xN = = 2 , dx N=0 dx 1 − x (1 − x) thus 1 N̄ = 6 ∞ N N=0 5 6 N−1 = 1 1 6 1− 5 6 2 =6. 3. Let W (m) be the probability for for taking n1 steps to the right and n2 = N − n1 steps to the left, where m = n1 − n2 , and N = n1 + n2 . Using N +m , 2 N −m n2 = , 2 n1 = one ﬁnds W (m) = N+m 2 N +m N −m N! p 2 q 2 . ! N−m ! 2 It is convenient to employ the moment generating function, deﬁned as φ (t) = etm . In general, the following holds φ (t) = ∞ k=0 tk mk , k! thus from the kth derivative of φ (t) one can calculate the kth moment of m mk = φ(k) (0) . Eyal Buks Thermodynamics and Statistical Physics 30 1.8. Solutions Set 1 Using W (m) one ﬁnds N W (m) etm φ (t) = m=−N N = N+m 2 m=−N N+m N −m N! p 2 q 2 etm , ! N−m ! 2 or using the summation variable n1 = N +m , 2 one has N φ (t) = n1 N! pn1 q N−n1 et(2n1 −N) n ! (N − n )! 1 1 =0 N = e−tN n1 −tN =e N! pe2t n ! (N − n )! 1 1 =0 pe2t + q N n1 q N−n1 . Using p = q = 1/2 φ (t) = et + e−t 2 N = (cosh t)N . Thus using the expansion (cosh t)N = 1 + 1 1 N t2 + [N + 3N (N − 1)] t4 + O t5 , 2! 4! one ﬁnds m =0, m2 = N , m3 = 0 , m4 = N (3N − 2) . 4. Using the binomial distribution N! pn (1 − p)N−n n! (N − n)! N (N − 1) (N − 1) × · · · (N − n + 1) n = p (1 − p)N−n n! (Np)n ∼ exp (−N p) . = n! W (n) = Eyal Buks Thermodynamics and Statistical Physics 31 Chapter 1. The Principle of Largest Uncertainty 5. a) ∞ ∞ W (n) = e−λ n=0 λn n! =1 n=0 b) As in Ex. 1-6, it is convenient to use the moment generating function φ (t) = etn = ∞ etn W (n) = e−λ n=0 ∞ ∞ n λn etn (λet ) = e−λ = exp λ et − 1 n! n! n=0 n=0 Using the expansion exp λ et − 1 1 = 1 + λt + λ (1 + λ) t2 + O t3 , 2 one ﬁnds n =λ. c) Using the same expansion one ﬁnds n2 = λ (1 + λ) , thus (∆n)2 = n2 − n 2 = λ (1 + λ) − λ2 = λ . 6. 2 N 2 R = rn n=1 N r2n + = n=1 =l2 n=m rn · rm = N l2 =0 7. 20 20! 0.2n × 0.820−n = 5.6 × 10−4 . n! (20 − n)! n=11 (1.144) 8. Let σAn = 1 (σBn = 1) if object A (B) makes a move to the right (left) at step n, and σ An = 0 (σBn = 0) if object A (B) makes an upward (downward) move at step n. The location (xAm , yAm ) of object A and the location (xBm , yBm ) of object B after m steps is given by (xAm , yAm ) = (SAm , m − SAm ) , (xBm , yBm ) = (N − SBm , N − m + SBm ) , (1.145) (1.146) where m SAm = σAn , (1.147) σBn . (1.148) n=1 m SBm = n=1 Eyal Buks Thermodynamics and Statistical Physics 32 . 1.8. Solutions Set 1 A meeting occurs if for some m SAm = N − SBm , m − SAm = N − m + SBm , (1.149) (1.150) i.e. if SAm + SBm = N = 2m − N . (1.151) Thus, a meeting is possible only after m = N steps, and it occurs if SAm + SBm = N . Therefore, the probability is given by pN = 2N N 22N = (2N )! 2 (N !2N ) . (1.152) With the help of the Stirling’s formula (1.119) one ﬁnds that 1 lim pN = √ . N→∞ Nπ (1.153) 9. The following holds p0 = 0 , 5 p1 = p0 + 6 4 p2 = p1 + 6 3 p3 = p2 + 6 2 p4 = p3 + 6 1 p5 = p4 + 6 p6 = 1 , 1 p2 6 2 p3 6 3 p4 6 4 p5 6 5 p6 6 , , , , , and thus p1 = 1 3 1 13 31 , p2 = , p3 = , p4 = , p5 = . 32 16 2 16 32 10. For the case of a round table (and N > 2) pC = N × 2 × (N − 2)! 2 = , N! N −1 (1.154) and for the case of a row pR = Eyal Buks (2 + (N − 2) × 2) × (N − 2)! 2 = . N! N Thermodynamics and Statistical Physics (1.155) 33 Chapter 1. The Principle of Largest Uncertainty 11. Let the binary representation of p be given by p= ∞ σm m=1 1 2 m , (1.156) where σ m ∈ {0, 1}. Let Σm be a sequence of random variables generated by the given computer function (i.e. Σm = 1 with probability 1/2 and Σm = 0 with probability 1/2). The proposed function has a while loop running over integer values of the variable m starting from the value m = 1. At each iteration the random variable Σm is generated and compared with σm . If σ m = Σm the value of m is increased by 1, i.e. m → m + 1, and the loop continues. If σm = Σm the program stops and the value 1 is returned if σ m > Σm and the value 0 is returned if σ m < Σm . Note that the probability that the program will never stop vanishes even when p is irrational and/or the number of nonzero binary digits σm is inﬁnite. 12. Using N+ + N− = N , M N+ − N− = , m (1.157a) (1.157b) one has N 2 N N− = 2 M mN M 1− mN N+ = 1+ , (1.158a) , (1.158b) or N (1 + x) , 2 N N− = (1 − x) , 2 N+ = (1.159a) (1.159b) where x= M . mN The number of states having total magnetization M is given by Ω (M ) = N! = N+ !N− ! N 2 N! . (1 + x) ! N2 (1 − x) ! (1.160) Since all states have equal probability one has f (M ) = Eyal Buks Ω (M ) . 2N Thermodynamics and Statistical Physics (1.161) 34 1.8. Solutions Set 1 Taking the natural logarithm of Stirling’s formula one ﬁnds 1 N log N! = N log N − N + O , (1.162) thus in the limit N ≫ 1 one has log f = − log 2N + N log N − N N N N (1 + x) log (1 + x) + (1 + x) 2 2 2 N N N − (1 − x) log (1 − x) + (1 − x) 2 2 2 = −N log 2 + N log N − N N N N (1 + x) log (1 + x) − (1 − x) log (1 − x) 2 2 2 2 N N N N = − −2 log + (1 + x) log (1 + x) + (1 − x) log (1 − x) 2 2 2 2 N N N N = − −2 log + (1 + x) log + log (1 + x) + (1 − x) log + log (1 − x) 2 2 2 2 N 1+x = − log 1 − x2 + x log . 2 1−x (1.163) − The function log f (x) has a sharp peak near x = 0, thus we can approximate it by assuming x ≪ 1. To lowest order log 1 − x2 + x log 1+x = x2 + O x3 , 1−x (1.164) thus f (M ) = A exp − M2 2m2 N , (1.165) where A is a normalization constant, which is determined by requiring that ∞ 1= f (M ) dM . (1.166) −∞ Using the identity ∞ −∞ Eyal Buks exp −ay2 dy = π , a Thermodynamics and Statistical Physics (1.167) 35 Chapter 1. The Principle of Largest Uncertainty one ﬁnds 1 = A ∞ −∞ exp − M2 2m2 N √ dM = m 2πN , (1.168) thus f (M ) = 1 M2 √ exp − 2 2m N m 2πN , (1.169) The expectation value is giving by ∞ M = M f (M) dM = 0 , (1.170) −∞ and the variance is given by 2 (M − M ) = M 2 ∞ = M 2 f (M ) dM = m2 N . (1.171) −∞ 13. The probability to have n steps to the right is given by W (n) = N! pn q N−n . n! (N − n)! (1.172) a) N n = N!n pn q N−n n! (N − n)! n=0 (1.173) N =p =p ∂ N! pn q N−n ∂p n=0 n! (N − n)! ∂ (p + q)N = pN (p + q)N−1 = pN . ∂p Since X = an − a (N − n) = a (2n − N ) (1.174) we ﬁnd X = aN (2p − 1) = aN (p − q) . (1.175) b) Eyal Buks Thermodynamics and Statistical Physics 36 1.8. Solutions Set 1 N n2 = N !n2 pn q N−n n! (N − n)! n=0 N = N N !n (n − 1) n N−n N !n p q + pn q N−n n! (N − n)! n! (N − n)! n=0 n=0 N = p2 = p2 ∂2 N! pn q N−n + n ∂p2 n=0 n! (N − n)! ∂2 N (p + q) + n = p2 N (N − 1) + pN . ∂p2 Thus (n − n )2 = p2 N (N − 1) + pN − p2 N 2 = N pq , (1.176) (X − X )2 = 4a2 N pq . (1.177) and 14. The total energy is given by E= kx2 mẋ2 ka2 + = , 2 2 2 (1.178) where a is the amplitude of oscillations. The time period T is given by a T =2 −a dx =2 ẋ m k a −a thus f(x) = dx √ = 2π 2 a − x2 m , k (1.179) 2 1 . = √ 2 T |ẋ| π a − x2 (1.180) 15. The six experiments are independent, thus 2 2 1 1 σ = 6 × − ln − ln 3 3 3 3 = 3.8191 . 16. The random variable X obtains the value n with probability pn = q n , where n = 1, 2, 3, · · · , and q = 1/2. a) The entropy is given by σ=− ∞ n=1 pn log2 pn = − ∞ n=1 q n log2 q n = ∞ nq n . n=1 This can be rewritten as σ=q Eyal Buks ∂ ∂q ∞ n=1 qn = q ∂ ∂q 1 q −1 = =2. 1−q (1 − q)2 Thermodynamics and Statistical Physics 37 Chapter 1. The Principle of Largest Uncertainty b) A series of questions is of the form: "Was a head obtained in the 1st time ?", "Was a head obtained in the 2nd time ?", etc. The expected number of questioned required to ﬁnd X is 1 1 1 ×1 + ×2 + ×3 +··· = 2 , 2 4 8 which is exactly the entropy σ. 17. a) Consider an inﬁnitesimal change in the variable z = z (x, y) δz = ∂z ∂x ∂z ∂y δx + y δy . (1.181) x For a process for which z is a constant δz = 0, thus 0= ∂z ∂x y ∂z ∂y (δx)z + x (δy)z . (1.182) Dividing by (δx)z yields ∂z ∂x y ∂z ∂y =− ∂z =− ∂y ∂y ∂x =− z ∂y ∂z x (δy)z (δx)z x ∂y ∂x z . x (1.183) b) Consider a process for which the variable w is kept constant. An inﬁnitesimal change in the variable z = z (x, y) is expressed as ∂z ∂x (δz)w = y (δx)w + ∂z ∂y x (δy)w . (1.184) Dividing by (δx)w yields (δz)w = (δx)w ∂z ∂x ∂z ∂y + y x (δy)w . (δx)w (1.185) or ∂z ∂x Eyal Buks = w ∂z ∂x + y ∂z ∂y x ∂y ∂x . (1.186) w Thermodynamics and Statistical Physics 38 1.8. Solutions Set 1 18. We have found in class that U =− ∂ log Zgc ∂β η N =− ∂ log Zgc ∂η β , (1.187) . (1.188) a) Using relation (1.125) one has U =− ∂ log Zgc ∂β =− ∂ log Zgc ∂β µ =− ∂ log Zgc ∂β η − 2 β µ ∂ log Zgc ∂µ β =− ∂ log Zgc ∂β + τµ ∂ log Zgc ∂µ β η − ∂ log Zgc ∂µ µ β ∂µ ∂β η . (1.189) b) Using Eq. (1.188) one has N =− ∂ log Zgc ∂η =− ∂µ ∂η =τ ∂ log Zgc ∂µ β β ∂ log Zgc ∂µ β , β (1.190) or in terms of the fugacity λ, which is deﬁned by λ = exp (βµ) = e−η , (1.191) one has N =τ ∂ log Zgc ∂µ β ∂λ ∂ log Zgc =τ ∂µ ∂λ ∂ log Zgc =λ . ∂λ (1.192) Eyal Buks Thermodynamics and Statistical Physics 39 Chapter 1. The Principle of Largest Uncertainty 19. The canonical partition function is given by Zc = Z1N , (1.193) where βε 2 Z1 = exp + exp − βε 2 = 2 cosh βε 2 . (1.194) Thus U =− ∂ log Zc ∂ log Z1 Nε βε = −N =− tanh , ∂β ∂β 2 2 (1.195) and τ= ε . U 2 tanh−1 − 2Nε (1.196) The negative temperature is originated by our assumption that the energy of a single magnet has an upper bound. In reality this is never the case. 20. The canonical partition function is given by Zc = Z1N , (1.197) where Z1 = exp − = βℏω 2 exp − βℏω 2 ∞ exp (−βℏωn) (1.198) n=0 1 − exp (−βℏω) = 1 . 2 sinh βℏω 2 a) U =− ∂ log Zc ∂ log Z1 N ℏω βℏω = −N = coth ∂β ∂β 2 2 (1.199) b) 2 2 (∆U ) N ℏω ∂ 2 log Zc ∂ 2 log Z1 2 = = N = ∂β 2 ∂β 2 sinh 2 βℏω 2 (1.200) 21. The canonical partition function is given by Zc = [exp (βε) + 1 + exp (−βε)]N = [1 + 2 cosh (βε)]N , (1.201) where β = 1/τ . Eyal Buks Thermodynamics and Statistical Physics 40 1.8. Solutions Set 1 a) Thus the average energy is U =− ∂ log Zc 2N ε sinh (βε) =− , ∂β 1 + 2 cosh βε (1.202) b) and the variance is 2 = (U − U ) ∂ U ∂ 2 log Zc cosh (βε) + 2 =− = 2Nε2 . ∂β ∂β 2 [1 + 2 cosh (βε)]2 (1.203) 22. Each section can be in one of two possible sates with corresponding energies 0 and −F a. a) By deﬁnition, α is the mean length of each segment, which is given by α= a exp (F aβ) a = 1 + tanh 1 + exp (F aβ) 2 F aβ 2 , (1.204) where β = 1/τ. b) At high temperature F aβ ≪ 1 the length of the chain L = Nα is given by L= F aβ 2 Na 1 + tanh 2 ≃ Na 2 1+ F aβ 2 , (1.205) or F = k L− Na 2 , (1.206) where the spring constant k is given by k= 4τ . Na2 (1.207) 23. The average length of a single link is given by ∞ a exp (βF a) l = exp (βF a) n=0 ∞ n=0 exp −βℏω a n + 1 2 exp −βℏω a n + 1 2 + b exp (βF b) + exp (βF b) ∞ n=0 ∞ n=0 exp −βℏωb n + exp −βℏω b n + (1.208) = Eyal Buks a exp(βF a) 2 sinh βℏ2ωa + exp(βF a) 2 sinh βℏ2ωa + b exp(βF b) βℏ ω 2 sinh 2 b exp(βF b) βℏ ω 2 sinh 2 b . Thermodynamics and Statistical Physics 41 1 2 1 2 Chapter 1. The Principle of Largest Uncertainty To ﬁrst order in β l = aω b + bω a F ω b ω a (a − b)2 + β + O β2 . 2 ωb + ωa (ωb + ω a ) (1.209) The average total length is L = n l . 24. The length L is given by L=N l , where l is the average contribution of a single molecule to the total length, which can be either +d or −d. The probability of each possibility is determined by the Boltzmann factor. The energy change due to ﬂipping of one link from 0◦ to 180◦ is 2fd, therefore l =d eβf d − e−βf d = d tanh (βfd) , eβf d + e−βf d where β = 1/τ . Thus L = Nd tanh (βf d) , or f= τ L tanh−1 . d Nd 25. The grand partition function Zgc is given by Zgc = 1 + 2 exp [β (µ − ε)] , (1.210) thus N = 1 ∂ log Zgc 2 = . β ∂µ 2 + exp [β (ε − µ)] (1.211) 26. a) N! N! − log (n2 − 1)! (n1 + 1)! n2 !n1 ! n2 n2 = log ≃ log . n1 + 1 n1 (∆σ)2LS = log (1.212) b) (∆σ)R = Eyal Buks E2 − E1 τ Thermodynamics and Statistical Physics (1.213) 42 1.8. Solutions Set 1 c) For a small change near thermal equilibrium one expects (∆σ)2LS + (∆σ)R = 0, thus E2 − E1 n2 = exp − n1 τ . (1.214) 27. 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)!. a) Thus σ = log N! NB ! (N − NB )! 2 ≃ 2 [N log N − NB log NB − (N − NB ) log (N − NB )] . (1.215) b) The energy of the system is given by U = NB ε. Thus, the Helmholtz free energy is given by F = U −τ σ = U −2τ N log N − U U U log − N − ε ε ε log N − U . ε (1.216) At thermal equilibrium (∂F/∂U)τ = 0, thus ∂F ∂U 0= =1+ τ 2τ U U log − log N − ε ε ε , (1.217) or ε N − NB = exp NB 2τ , (1.218) therefore NB = N 1 + exp ε 2τ . Alternatively, one can calculate the chemical potential from the requirement 1= NA NB + , N N (1.219) where NA exp (βµ) = , N 1 + exp (βµ) NB exp (βµ − βε) = , N 1 + exp (βµ − βε) Eyal Buks Thermodynamics and Statistical Physics (1.220a) (1.220b) 43 Chapter 1. The Principle of Largest Uncertainty which is satisﬁed when ε µ= , 2 (1.221) thus NB = N 1 + exp ε 2τ . (1.222) 28. 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, m) = (N − 1 + m)! . (N − 1)!m! (1.223) b) The entropy is given by σ = log (N − 1 + m)! ≃ (N + m) log (N + m) − N log N − m log m , (N − 1)!m! (1.224) or in terms of the total energy E = ℏω (m + N/2) E N E N − log N + − ℏω 2 ℏω 2 E N E N − N log N − − log − . ℏω 2 ℏω 2 σ= N+ (1.225) Eyal Buks Thermodynamics and Statistical Physics 44 1.8. Solutions Set 1 c) The temperature τ is given by 1 ∂σ = τ ∂E 2ℏω 1 − ln 2E+Nℏω ln 2ℏω − 1 = + 2E−Nℏω ℏω ℏω 1 2E + N ℏω ln . = ℏω 2E − N ℏω (1.226) In the thermodynamical limit (N ≫ 1, m ≫ 1) the energy E and its average U are indistinguishable, thus 2U + N ℏω , 2U − N ℏω exp ℏω τ U= Nℏω ℏω coth . 2 2τ = (1.227) or (1.228) 29. The grand canonical partition function is given by ζ = 1 + 2λ exp (βε) , (1.229) where λ = exp (βµ) is the fugacity, thus Nd = λ Eyal Buks ∂ log ζ 1 2λeβε = . = ∂λ 1 + 2λeβε 1 + 12 e−β(ε+µ) Thermodynamics and Statistical Physics (1.230) 45 2. Ideal Gas In this chapter we study some basic properties of ideal gas of massive identical particles. We start by considering a single particle in a box. We then discuss the statistical properties of an ensemble of identical indistinguishable particles and introduce the concepts of Fermions and Bosons. In the rest of this chapter we mainly focus on the classical limit. For this case we derive expressions for the pressure, heat capacity, energy and entropy and discuss how internal degrees of freedom may modify these results. In the last part of this chapter we discuss an example of an heat engine based on ideal gas (Carnot heat engine). We show that the eﬃciency of such a heat engine, which employs a reversible process, obtains the largest possible value that is allowed by the second law of thermodynamics. 2.1 A Particle in a Box Consider a particle having mass M in a box. For simplicity the box is assumed to have a cube shape with a volume V = L3 . The corresponding potential energy is given by V (x, y, z) = 0 0 ≤ x, y, z ≤ L . ∞ else (2.1) The quantum eigenstates and eigenenergies are determined by requiring that the wavefunction ψ (x, y, z) satisﬁes the Schrödinger equation 2 − 2M ∂2ψ ∂2ψ ∂2ψ + + ∂x2 ∂y 2 ∂z 2 + V ψ = Eψ . (2.2) In addition, we require that the wavefunction ψ vanishes on the surfaces of the box. The normalized solutions are given by ψnx ,ny ,nz (x, y, z) = 2 L 3/2 sin nx πx ny πy nz πz sin sin , L L L (2.3) where nx , ny , nz = 1, 2, 3, · · · (2.4) Chapter 2. Ideal Gas The corresponding eigenenergies are given by 2 εnx ,ny ,nz = π L 2M 2 n2x + n2y + n2z . (2.5) For simplicity we consider the case where the particle doesn’t have any internal degree of freedom (such as spin). Later we will release this assumption and generalize the results for particles having internal degrees of freedom. The partition function is given by ∞ Z1 = ∞ ∞ nx =1 ny =1 nz =1 ∞ = ∞ ∞ nx =1 ny =1 nz =1 exp − εnx ,ny ,nz τ exp −α2 n2x + n2y + n2z , (2.6) where α2 = 2 2 π . 2ML2 τ (2.7) The following relation can be employed to estimate the dimensionless parameter α α2 = 7.9 × 10−17 , τ L 2 cm 300 K M mp (2.8) where mp is the proton mass. As can be seen from the last result, it is often the case that α2 ≪ 1. In this limit the sum can be approximated by an integral ∞ exp nx =1 −α2 n2x ∞ ≃ 0 exp −α2 n2x dnx . (2.9) By changing the integration variable x = αnx one ﬁnds ∞ exp 0 −α2 n2x 1 dnx = α ∞ 0 exp −x2 dx = √ π , 2α (2.10) thus Z1 = √ π 2α 3 = M L2 τ 2π 2 3/2 = nQ V , (2.11) where we have introduced the quantum density Eyal Buks Thermodynamics and Statistical Physics 48 2.1. A Particle in a Box nQ = Mτ 2π 2 3/2 . (2.12) The partition function (2.11) together with Eq. (1.70) allows evaluating the average energy (recall that β = 1/τ ) ε =− ∂ log Z1 ∂β ML2 2π 2 β ∂ log =− 3/2 ∂β −3/2 ∂ log β ∂β 3 ∂ log β = 2 ∂β 3τ = . 2 =− (2.13) This result can be written as τ ε =d , 2 (2.14) where d = 3 is the number of degrees of freedom of the particle. As we will see later, this is an example of the equipartition theorem of statistical mechanics. Similarly, the energy variance can be evaluated using Eq. (1.71) ∂ 2 log Z1 ∂β 2 ∂ ε =− ∂β ∂ 3 =− ∂β 2β 3 = 2β 2 3τ 2 = . 2 (∆ε)2 = (2.15) Thus, using Eq. (2.13) the standard deviation is given by (∆ε)2 = 2 ε . 3 (2.16) What is the physical meaning of the quantum density? The de Broglie wavelength λ of a particle having mass M and velocity v is given by Eyal Buks Thermodynamics and Statistical Physics 49 Chapter 2. Ideal Gas λ= h . Mv (2.17) For a particle having energy equals to the average energy ε = 3τ /2 one has Mv 2 3τ = , 2 2 (2.18) thus in this case the de-Broglie wavelength, which is denoted as λT (the thermal wavelength) h λT = √ , 3M τ (2.19) and therefore one has (recall that π nQ = h2 2Mτ 3/2 = 1 λT 2π 3 = h/2π) 3 . (2.20) Thus the quantum density is inversely proportional to the thermal wavelength cubed. 2.2 Gibbs Paradox In the previous section we have studied the case of a single particle. Let us now consider the case where the box is occupied by N particles of the same type. For simplicity, we consider the case where the density n = N/V is suﬃciently small to safely allowing to neglect any interaction between the particles. In this case the gas is said to be ideal. Deﬁnition 2.2.1. Ideal gas is an ensemble of non-interacting identical particles. What is the partition function of the ideal gas? Recall that for the single particle case we have found that the partition function is given by [see Eq. (2.6)] Z1 = exp (−βεn ) . (2.21) n In this expression Z1 is obtained by summing over all single particle orbital states, which are denoted by the vector of quantum numbers n = (nx , ny , nz ). These states are called orbitals . Since the total number of particles N is constrained we need to calculate the canonical partition function. For the case of distinguishable particles one may argue that the canonical partition function is given by Eyal Buks Thermodynamics and Statistical Physics 50 2.2. Gibbs Paradox N ? Zc = Z1N = exp (−βεn ) . (2.22) n However, as was demonstrated by Gibbs in his famous paradox, this answer is wrong. To see this, we employ Eqs. (1.72) and (2.11) and assume that the partition function is given by Eq. (2.22) above, thus ? σ − βU = log Zc = log Z1N = N log (nQ V ) , (2.23) or in terms of the gas density n= N , V (2.24) we ﬁnd ? σ − βU = N log N nQ n . (2.25) What is wrong with this result? It suggests that the quantity σ − βU is not simply proportional to the size of the system. In other words, for a given n and a given nQ , σ −βU is not proportional to N. As we will see below, such a behavior may lead to a violation of the second law of thermodynamics. To see this consider a box containing N identical particles having volume V . What happens when we divide the box into two sections by introducing a partition? Let the number of particles in the ﬁrst (second) section be N1 (N2 ) whereas the volume in the ﬁrst (second) section be V1 (V2 ). The following hold N = N1 + N2 , V = V1 + V2 . (2.26) (2.27) The density in each section is expected to be the same as the density in the box before the partition was introduced n= N2 N N1 = . = V V1 V2 (2.28) Now we use Eq. (2.25) to evaluate the change in entropy ∆σ due to the process of dividing the box. Since no energy is required to add (or to remove) the partition one has ∆σ = σtot − σ1 − σ 2 nQ nQ nQ ? = N log N − N1 log N1 − N2 log N2 n n n = N log N − N1 log N1 − N2 log N2 . (2.29) Using the Stirling’s formula (1.162) log N ! ≃ N log N − N , Eyal Buks Thermodynamics and Statistical Physics (2.30) 51 Chapter 2. Ideal Gas 2 … orbital 1 N1=0 3 1 orbital 2 N2=1 orbital 3 N3=2 orbital 4 N4=0 Fig. 2.1. A ﬁgure describing an example term in the expansion (2.22) for a gas containing N = 3 identical particles. one ﬁnds ∆σ ≃ log N! >0. N1 !N2 ! (2.31) Thus we came to the conclusion that the process of dividing the box leads to reduction in the total entropy! This paradoxical result violates the second law of thermodynamics. According to this law we expect no change in the entropy since the process of dividing the box is a reversible one. What is wrong with the partition function given by Eq. (2.22)? Expanding this partition function yields a sum of terms each having the form exp −β Nn εn , where Nn is the number of particles occupying orbital n. n Let g (N1 , N2 , · · · ) be the number of terms in such an expansion associated with a given set of occupation numbers {N1 , N2 , · · · }. Since the partition function (2.22) treats the particles as being distinguishable, g (N1 , N2 , · · · ) may in general be larger than unity. In fact, it is easy to see that g (N1 , N2 , · · · ) = N! . N1 !N2 ! × · · · (2.32) For example, consider the state that is described by Fig. 2.1 below for a gas containing N = 3 particles. The expansion (2.22) contains 3!/1!/2! = 3 terms having the same occupation numbers (Nn = 1 if n = 2,Nn = 2 if n = 3, and Nn = 0 for all other values of n). However, for identical particles these 3 states are indistinguishable. Therefore, only a single term in the partition function should represent such a conﬁguration. In general, the partition function should include a single term only for each given set of occupation numbers {N1 , N2 , · · · }. 2.3 Fermions and Bosons As we saw in the previous section the canonical partition function given by Eq. (2.22) is incorrect. For indistinguishable particles each set of orbital ocEyal Buks Thermodynamics and Statistical Physics 52 2.3. Fermions and Bosons cupation numbers {N1 , N2 , · · · } should be counted only once. In this section we take another approach and instead of evaluating the canonical partition function of the system we consider the grandcanonical partition function. This is done by considering each orbital as a subsystem and by evaluating its grandcanonical partition function, which we denote below as ζ. To do this correctly, however, it is important to take into account the exclusion rules imposes by quantum mechanics upon the possible values of the occupation numbers Nn . The particles in nature are divided into two type: Fermion and Bosons. While Fermions have half integer spin Bosons have integer spin. According to quantum mechanics the orbital occupation numbers Nn can take the following values: • For Fermions: Nn = 0 or 1 • For Bosons: Nn can be any integer. These rules are employed below to evaluate the grandcanonical partition function of an orbital. 2.3.1 Fermi-Dirac Distribution In this case the occupation number can take only two possible values: 0 or 1. Thus, by Eq. (1.79) the grandcanonical partition function of an orbital having energy is ε is given by ζ = 1 + λ exp (−βε) , (2.33) where λ = exp (βµ) (2.34) is the fugacity [see Eq. (1.95)]. The average occupation of the orbital, which is denoted by fFD (ε) = N (ε) , is found using Eq. (1.94) ∂ log ζ ∂λ λ exp (−βε) = 1 + λ exp (−βε) 1 = . exp [β (ε − µ)] + 1 fFD (ε) = λ (2.35) The function fFD (ε) is called the Fermi-Dirac function . Eyal Buks Thermodynamics and Statistical Physics 53 Chapter 2. Ideal Gas 2.3.2 Bose-Einstein Distribution In this case the occupation number can take any integer value. Thus, by Eq. (1.79) the grandcanonical partition function of an orbital having energy is ε is given by ζ= = ∞ λN exp (−N βε) N=0 ∞ [λ exp (−βε)]N N=0 = 1 . 1 − λ exp (−βε) (2.36) The average occupation of the orbital, which is denoted by fBE (ε) = N (ε) , is found using Eq. (1.94) ∂ log ζ ∂λ exp (−βε) =λ 1 − λ exp (−βε) 1 = . exp [β (ε − µ)] − 1 fBE (ε) = λ (2.37) The function fBE (ε) is called the Bose-Einstein function . 2.3.3 Classical Limit The classical limit occurs when exp [β (ε − µ)] ≫ 1 . (2.38) As can be seen from Eqs. (2.35) and (2.37) the following holds fFD (ε) ≃ fBE (ε) ≃ exp [β (µ − ε)] ≪ 1 , (2.39) ζ ≃ 1 + λ exp (−βε) . (2.40) and Thus the classical limit corresponds to the case where the average occupation of an orbital is close to zero, namely the orbital is on average almost empty. The main results of the above discussed cases (Fermi-Dirac distribution, BoseEinstein distribution and the classical limit) are summarized in table 2.1 below. Eyal Buks Thermodynamics and Statistical Physics 54 2.4. Ideal Gas in the Classical Limit Table 2.1. Fermi-Dirac, Bose-Einstein and classical distributions. orbital partition function average occupation Fermions 1 + λ exp (−βε) Bosons 1 1−λ exp(−βε) 1 exp[β(ε−µ)]+1 1 exp[β(ε−µ)]−1 classical limit 1 + λ exp (−βε) exp [β (µ − ε)] 2 1.5 1 0.5 -2 0 2 4 6 8 energy 10 12 14 16 18 20 2.4 Ideal Gas in the Classical Limit The rest of this chapter is devoted to the classical limit. The grandcanonical partition function ζ n of orbital n having energy εn is given by Eq. (2.40) above. The grandcanonical partition function of the entire system Zgc is found by multiplying ζ n of all orbitals Zgc = (1 + λ exp (−βεn )) . (2.41) n Each term in the expansion of the above expression represents a set of orbital occupation numbers, where each occupation number can take one of the possible values: 0 or 1. We exploit the fact that in the classical limit λ exp (−βε) ≪ 1 (2.42) and employ the ﬁrst order expansion log (1 + x) = x + O x2 (2.43) to obtain Eyal Buks Thermodynamics and Statistical Physics 55 Chapter 2. Ideal Gas log Zgc = log (1 + λ exp (−βεn )) n ≃λ exp (−βεn ) n = λZ1 , (2.44) where Z1 = V Mτ 2π 2 3/2 (2.45) [see Eq. (2.11)] is the single particle partition function. In terms of the Lagrange multipliers η = −µ/τ and β = 1/τ the last result can be rewritten as log Zgc = e−η V M 2π 2 β 3/2 . (2.46) The average energy and average number of particle are calculated using Eqs. (1.80) and (1.81) respectively U =− ∂ log Zgc ∂β η N =− ∂ log Zgc ∂η β = 3 log Zgc , 2β = log Zgc . (2.47) (2.48) In what follows, to simplify the notation we remove the diagonal brackets and denote U and N by U and N respectively. As was already pointed out earlier, probability distributions in statistical mechanics of macroscopic parameters are typically extremely sharp and narrow. Consequently, in many cases no distinction is made between a parameter and its expectation value. Using this simpliﬁed notation and employing Eqs. (2.47) and (2.48) one ﬁnds that U= 3N τ . 2 (2.49) Namely, the total energy is N ε , where ε is the average single particle energy that is given by Eq. (2.13). The entropy is evaluate using Eq. (1.86) σ = log Zgc + βU + ηN 3 µ = N 1+ − 2 τ 5 =N − µβ . 2 (2.50) Eyal Buks Thermodynamics and Statistical Physics 56 2.4. Ideal Gas in the Classical Limit Furthermore, using Eqs. (2.44), (2.48), (2.11) and (1.95) one ﬁnds that n , (2.51) µβ = log nQ where n = N/V is the density. This allows expressing the entropy as σ=N 5 nQ + log 2 n . (2.52) Using the deﬁnition (1.116) and Eqs. (2.49) and (2.52) one ﬁnds that the Helmholtz free energy is given by F = Nτ log n −1 nQ . (2.53) 2.4.1 Pressure The pressure p is deﬁned by p=− ∂F ∂V . (2.54) τ ,N Using Eq. (2.53) and keeping in mind that n = N/V one ﬁnds Nτ . (2.55) V The pressure represents the force per unit area acting on the walls of the box containing the gas due to collisions between the particles and the walls. To see that this is indeed the case consider a gas of N particles contained in a box having cube shape and volume V = L3 . One of the walls is chosen to lie on the x = 0 plane. Consider and elastic collision between this wall and a particle having momentum p = (px , py , pz ). After the collision py and pz remain unchanged, however, px becomes −px . Thus each collision results in transferring 2 |px | momentum in the x direction from the particle to the wall. The rate at which a particle collides with the wall x = 0 is |px | /2mL. Thus the pressure acting on the wall due to a single particle is {force} {pressure} = {area} rate of momentum change = {area} p= 2 |px | × L2 2 p = x . mV = |px | 2mL (2.56) Eyal Buks Thermodynamics and Statistical Physics 57 Chapter 2. Ideal Gas The average energy of a particle is given by Eq. (2.13) p2x + p2y + p2z 3τ = ε = , 2 2m (2.57) thus one ﬁnds that p2x =τ . m (2.58) Using this result and Eq. (2.56) one ﬁnds that the pressure due to a single particle is p = τ /V , thus the total pressure is p= Nτ . V (2.59) 2.4.2 Useful Relations In this section we derive some useful relations between thermodynamical quantities. Claim. The following holds ∂U ∂V p=− σ,N Proof. Using the deﬁnition (2.54) and recalling that F = U − τσ one ﬁnds ∂F ∂V −p = ∂U ∂V = τ ,N τ ,N ∂ (τσ) ∂V − . (2.60) τ ,N Using identity (1.125), which is given by ∂z ∂x ∂z ∂x = w + y ∂z ∂y x ∂y ∂x , (2.61) w one ﬁnds ∂U ∂V = τ ,N ∂U ∂V + σ,N ∂U ∂σ V,N ∂σ ∂V , (2.62) τ ,N τ thus −p = Eyal Buks ∂U ∂V +τ σ,N ∂σ ∂V τ ,N −τ ∂σ ∂V = τ ,N ∂U ∂V Thermodynamics and Statistical Physics . (2.63) σ,N 58 2.4. Ideal Gas in the Classical Limit In a similar way the following relations can be obtained ∂σ ∂V p=τ µ = −τ U,N ∂σ ∂N =− ∂U ∂V = ∂U ∂N U,V σ,N =− = σ,V ∂F ∂V ∂F ∂N , (2.64) τ ,N . (2.65) τ ,V Another useful relation is given below. Claim. The following holds ∂σ ∂V ∂p ∂τ = τ . (2.66) V Proof. See problem 2 of set 2. 2.4.3 Heat Capacity The heat capacity at constant volume is deﬁned by ∂σ ∂τ cV = τ , (2.67) V whereas the heat capacity at constant pressure is deﬁned by ∂σ ∂τ cp = τ . (2.68) p Using Eq. (2.52) and recalling that nQ ∝ τ 3/2 one ﬁnds cV = τ 3N 3N = . 2τ 2 (2.69) Claim. The following holds cp = 5N . 2 (2.70) Proof. See problem 1 of set 2. 2.4.4 Internal Degrees of Freedom In this section we generalize our results for the case where the particles in the gas have internal degrees of freedom. Taking into account internal degrees of freedom the grandcanonical partition function of an orbital having orbital energy εn for the case of Fermions becomes ζ FD,n = (1 + λ exp (−βεn ) exp (−βEl )) , (2.71) l Eyal Buks Thermodynamics and Statistical Physics 59 Chapter 2. Ideal Gas where {El } are the eigenenergies of a particle due to internal degrees of freedom, and where λ = exp (βµ) and β = 1/τ . As is required by the Pauli exclusion principle, no more than one Fermion can occupy a given internal eigenstate. Similarly, for the Bosonic case, where each state can be occupied by any integer number of Bosons, one has ∞ ζ BE,n = λm exp (−βmεn ) exp (−βmEl ) . (2.72) m=0 l In the classical limit the average occupation of an orbital is close to zero. In this limit, namely when λ exp (−βεn ) ≪ 1 , (2.73) [see Eq. (2.38)] the following holds ζ FD,n ≃ ζ BE,n ≃ ζ n , (2.74) where ζ n = 1 + λ exp (−βεn ) Zint , (2.75) and where Zint = exp (−βEl ) , (2.76) l is the internal partition function. Using Eq. (1.94) one ﬁnds that the average occupation of the orbital fn in the classical limit is given by ∂ log ζ n ∂λ λZint exp (−βεn ) = 1 + λZint exp (−βεn ) ≃ λZint exp (−βεn ) . fn = λ (2.77) The total grandcanonical partition function is given by Zgc = ζn , (2.78) n thus Eyal Buks Thermodynamics and Statistical Physics 60 2.4. Ideal Gas in the Classical Limit log Zgc = log ζ n n = log [1 + λZint exp (−βεn )] n ≃ λZint exp (−βεn ) n = λZint Z1 , (2.79) where we have used the fact that in the classical limit λZint exp (−βεn ) ≪ 1. Furthermore, using Eq. (2.11) and recalling that η = −µ/τ one ﬁnds log Zgc = e−η Zint V M 2π 2 β 3/2 . (2.80) This result together with Eqs. (1.80) and (1.81) yield U =− ∂ log Zgc ∂β η N =− ∂ log Zgc ∂η β = 3 log Zgc + El log Zgc , 2β = log Zgc , (2.81) (2.82) where El exp (−βEl ) El = l exp (−βEl ) =− ∂ log Zint . ∂β (2.83) l Claim. The following hold n − log Zint nQ , (2.84) 3τ ∂ log Zint − 2 ∂β , (2.85) µ = τ log U =N F = Nτ σ=N cV = N n − log Zint − 1 nQ log , 5 nQ ∂ (τ log Zint ) + log + 2 n ∂τ 3 ∂ 2 (τ log Zint ) +τ 2 ∂τ 2 (2.86) , , cp = cV + N . (2.87) (2.88) (2.89) Proof. See problem 3 of set 2. Eyal Buks Thermodynamics and Statistical Physics 61 Chapter 2. Ideal Gas 2.5 Processes in Ideal Gas The state of an ideal gas is characterized by extensive parameters (by definition, parameters that are proportional to the system size) such as U , V , N and σ and by intensive parameters (parameters that are independent on the system size) such as τ , µ and p. In this section we discuss some examples of processes that occur by externally changing some of these parameters. We will use these processes in the next section to demonstrate how one can construct a heat engine based on an ideal gas. In general, the entropy is commonly expresses as a function of the energy, volume and number of particles σ = σ (U, V, N ). A small change in σ is expressed in terms of the partial derivatives ∂σ ∂U dσ = dU + V,N ∂σ ∂V dV + U,N ∂σ ∂N dN . (2.90) U,V Using Eqs. (1.87), (2.64) and (2.65) dσ = 1 p µ dU + dV − dN , τ τ τ (2.91) or dU = τ dσ − pdV + µdN . (2.92) This relation expresses the change in the energy of the system dU in terms of dQ = τ dσ heat added to the system dW = pdV work done by the system µdN energy change due to added particles For processes that keep the number of particles unchanged dN = 0 , one has dU = dQ − dW . (2.93) Integrating this relation for the general case (not necessarily an inﬁnitesimal process) yields ∆U = Q − W , (2.94) We discuss below some speciﬁc examples for processes for which dN = 0. The initial values of the pressure, volume and temperature are denoted as p1 , V1 and τ 1 respectively, whereas the ﬁnal values are denoted as p2 , V2 and τ 2 respectively. In all these processes we assume that the gas remains in Eyal Buks Thermodynamics and Statistical Physics 62 2.5. Processes in Ideal Gas p isochoric isobaric i so is e the ntr op rma ic l V Fig. 2.2. Four processes for which dN = 0. thermal equilibrium throughout the entire process. This can be achieved by varying the external parameters at a rate that is suﬃciently slow to allow the system to remain very close to thermal equilibrium at any moment during the process. The four example to be analyzed below are (see ﬁg. 2.2): • • • • Isothermal process - temperature is constant Isobaric process - pressure is constant Isochoric process - volume is constant Isentropic process - entropy is constant Note that in general, using the deﬁnition of the heat capacity at constant volume given by Eq. (2.67) together with Eq. (1.87) one ﬁnds that cV = ∂U ∂τ . (2.95) N,V Furthermore, as can be seen from Eq. (2.85), the energy U of an ideal gas in the classical limit is independent on the volume V (it can be expressed as a function of τ and N only). Thus we conclude that for processes for which dN = 0 the change in energy dU can be expressed as dU = cV dτ . Eyal Buks (2.96) Thermodynamics and Statistical Physics 63 Chapter 2. Ideal Gas 2.5.1 Isothermal Process Since τ is constant one ﬁnds using Eq. (2.96) that ∆U = 0. Integrating the relation dW = pdV and using Eq. (2.55) one ﬁnds V2 Q=W = pdV V1 V2 = Nτ dV V V1 = N τ log V2 . V1 (2.97) 2.5.2 Isobaric Process Integrating the relation dW = pdV for this case where the pressure is constant yields V2 pdV = p (V2 − V1 ) . W = (2.98) V1 The change in energy ∆U can be found by integrating Eq. (2.96) τ2 cV dτ . ∆U = (2.99) τ1 The heat added to the system Q can be found using Eq. (2.94) Q = W + ∆U τ2 = p (V2 − V1 ) + cV dτ . τ1 (2.100) Note that if the temperature dependence of cV can be ignored to a good approximation one has ∆U = cV (τ 2 − τ 1 ) . Eyal Buks Thermodynamics and Statistical Physics (2.101) 64 2.5. Processes in Ideal Gas 2.5.3 Isochoric Process In this case the volume is constant, thus W = 0. By integrating Eq. (2.96) one ﬁnds that τ2 Q = ∆U = cV dτ . (2.102) τ1 Also in this case, if the temperature dependence of cV can be ignored to a good approximation one has Q = ∆U = cV (τ 2 − τ 1 ) . 2.5.4 Isentropic Process In this case the entropy is constant, thus dQ = τdσ = 0, and therefore dU = −dW , thus using the relation dW = pdV and Eq. (2.96) one has cV dτ = −pdV , (2.103) or using Eq. (2.55) cV dτ dV = −N . τ V (2.104) This relation can be rewritten using Eq. (2.89) as dV dτ = (1 − γ) . τ V (2.105) where γ= cp . cV (2.106) The last result can be easily integrated if the temperature dependence of the factor γ can be ignored to a good approximation. For that case one has log τ2 = log τ1 V2 V1 1−γ . (2.107) Thus τ 1 V1γ−1 = τ 2 V2γ−1 , (2.108) or using Eq. (2.55) p1 V1γ = p2 V2γ . Eyal Buks Thermodynamics and Statistical Physics (2.109) 65 Chapter 2. Ideal Gas W τh Qh Ql τl Fig. 2.3. Carnot heat engine. In other words both quantities τV γ−1 and pV γ remain unchanged during this process. Using the last result allows integrating the relation dW = pdV V2 −∆U = W = pdV V1 V2 = p1 V1γ V −γ dV V1 = p1 V1γ V1−γ+1 − V2−γ+1 γ−1 p2 V2 − p1 V1 = 1−γ N (τ 2 − τ 1 ) = 1−γ = −cV (τ 2 − τ 1 ) . (2.110) 2.6 Carnot Heat Engine In this section we discuss an example of a heat engine proposed by Carnot that is based on an ideal classical gas. Each cycle is made of four steps (see Figs. 2.3 and 2.4) 1. 2. 3. 4. Isothermal expansion at temperature τ h (a → b) Isentropic expansion from temperature τ h to τ l (b → c) Isothermal compression at temperature τ l (c → d) Isentropic compression from temperature τ l to τ h (d → a) All four steps are assumed to be suﬃciently slow to maintain the gas in thermal equilibrium throughout the entire cycle. The engine exchanges heat Eyal Buks Thermodynamics and Statistical Physics 66 2.6. Carnot Heat Engine p a pa τh b pb pd d pc Va Vb Vd τl c V Vc Fig. 2.4. A cycle of Carnot heat engine. with the environment during both isothermal processes. Using Eq. (2.97) one ﬁnds that the heat extracted from the hot reservoir Qh at temperature τ h during step 1 (a → b) is given by Qh = N τ h log Vb , Va (2.111) and the heat extracted from the cold thermal reservoir Ql at temperature τ l during step 3 (c → d) is given by Ql = N τ l log Vd , Vc (2.112) where Vn is the volume at point n ∈ {a, b, c, d}. Note that Qh > 0 since the system undergoes expansion in step 1 whereas Ql < 0 since the system undergoes compression during step 3. Both thermal reservoirs are assumed to be very large systems that can exchange heat with the engine without changing their temperature. No heat is exchanged during the isentropic steps 2 and 4 (since dQ = τ dσ). The total work done by the system per cycle is given by W = Wab + Wcd + Wbc + Wda Vb Vd = N τ h log + N τ l log Va Vc N (τ l − τ h ) N (τ h − τ l ) + + 1−γ 1−γ Vb Vd = N τ h log + τ l log , Va Vc (2.113) Eyal Buks Thermodynamics and Statistical Physics 67 Chapter 2. Ideal Gas where the work in both isothermal processes Wab and Wcd is calculated using Eq. (2.97), whereas the work in both isentropic processes Wbc and Wda is calculate using Eq. (2.110). Note that the following holds W = Qh + Ql . (2.114) This is expected in view of Eq. (2.94) since the gas returns after a full cycle to its initial state and therefore ∆U = 0. The eﬃciency of the heat engine is deﬁned as the ratio between the work done by the system and the heat extracted from the hot reservoir per cycle η= W Ql =1+ . Qh Qh (2.115) Using Eqs. (2.111) and (2.113) one ﬁnds η =1+ τ l log VVdc τ h log VVab . (2.116) Employing Eq. (2.108) for both isentropic processes yields τ h Vbγ−1 = τ l Vcγ−1 , τ h Vaγ−1 = τ l Vdγ−1 , (2.117) (2.118) thus by dividing these equations one ﬁnds Vbγ−1 Vcγ−1 = , Vaγ−1 Vdγ−1 (2.119) Vb Vc = . Va Vd (2.120) or Using this result one ﬁnds that the eﬃciency of Carnot heat engine ηC is given by ηC = 1 − τl . τh (2.121) 2.7 Limits Imposed Upon the Eﬃciency Is it possible to construct a heat engine that operates between the same heat reservoirs at temperatures τ h and τ l that will have eﬃciency larger than the value given by Eq. (2.121)? As we will see below the answer is no. This conclusion is obtained by noticing that the total entropy remains unchanged in each of the four steps that constructs the Carnot’s cycle. Consequently, Eyal Buks Thermodynamics and Statistical Physics 68 2.7. Limits Imposed Upon the Eﬃciency W Q τ M Fig. 2.5. An idle heat engine (Perpetuum Mobile of the second kind). the entire process is reversible, namely, by varying the external parameters in the opposite direction, the process can be reversed. We consider below a general model of a heat engine. In a continuos operation the heat engine repeats a basic cycle one after another. We make the following assumptions: • At the end of each cycle the heat engine returns to the same macroscopic state that it was in initially (otherwise, continuous operation is impossible). • The work W done per cycle by the heat engine does not change the entropy of the environment (this is the case when, for example, the work is used to lift a weight - a process that only changes the center of mass of the weight, and therefore causes no entropy change). Figure (2.5) shows an ideal heat engine that fully transforms the heat Q extracted from a thermal reservoir into work W , namely Q = W . Such an idle engine has a unity eﬃciency η = 1. Is it possible to realized such an idle engine? Such a process does not violate the law of energy conservation (ﬁrst law of thermodynamics). However, as we will see below it violates the second law of thermodynamics. Note also that the opposite process, namely a process that transforms work into heat without losses is possible, as can be seen from the example seen in Fig. (2.6). In this system the weigh normally goes down and consequently the blender rotates and heats the liquid in the container. In principle, the opposite process at which the weigh goes up and the liquid cools down doesn’t violate the law of energy conservation, however, it violates the second law (Perpetuum Mobile of the second kind), as we will see below. To show that the idle heat engine shown in Fig. (2.5) can not be realized we employ the second law and require that the total change in entropy ∆σ per cycle is non-negative ∆σ ≥ 0 . (2.122) The only change in entropy per cycle is due to the heat that is subtracted from the heat bath Eyal Buks Thermodynamics and Statistical Physics 69 Chapter 2. Ideal Gas Fig. 2.6. Transforming work into heat. ∆σ = − Q , τ (2.123) thus since Q = W (energy conservation) we ﬁnd that W ≤0. τ (2.124) Namely, the work done by the heat engine is non-positive W ≤ 0. This result is known as Kelvin’s principle. Kelvin’s principle: In a cycle process, it is impossible to extract heat from a heat reservoir and convert it all into work. As we will see below, Kelvin’s principle is equivalent to Clausius’s principle that states: Clausius’s principle: It is impossible that at the end of a cycle process, heat has been transferred from a colder to a hotter thermal reservoirs without applying any work in the process. A refrigerator and an air conditioner (in cooling mode) are examples of systems that transfer heat from a colder to a hotter thermal reservoirs. According to Clausius’s principle such systems require that work is consumed for their operation. Theorem 2.7.1. Kelvin’s principle is equivalent to Clausius’s principle. Proof. Assume that Clausius’s principle does not hold. Thus the system shown in Fig. 2.7(a) that transfers heat Q0 > 0 from a cold thermal reservoir at temperature τ l to a hotter one at temperature τ h > τ l is possible. In Fig. 2.7(b) a heat engine is added that extracts heat Q > Q0 from the hot thermal reservoir, delivers heat Q0 to the cold one, and performs work W = Q − Q0 . The combination of both systems extracts heat Q − Q0 from the hot thermal reservoir and converts it all into work, in contradiction with Kelvin’s principle. Assume that Kelvin’s principle does not hold. Thus the system shown in Fig. 2.8(a) that extracts heat Q0 from a thermal reservoir at temperature τ h and converts it all into work is possible. In Fig. 2.8(b) a refrigerator is added Eyal Buks Thermodynamics and Statistical Physics 70 2.7. Limits Imposed Upon the Eﬃciency (a) τh τl -Q0 (b) M Q0 W=Q-Q0 Q M -Q0 τh τl -Q0 M Q0 Fig. 2.7. The assumption that Clausius’s principle does not hold. that employs the work W = Q0 to remove heat Q from a colder thermal reservoir at temperature τ l < τ h and to deliver heat Q0 + Q to the hot thermal reservoir. The combination of both systems transfers heat Q from a colder to a hotter thermal reservoirs without consuming any work in the process, in contradiction with Clausius’s principle. As we have seen unity eﬃciency is impossible. What is the largest possible eﬃciency of an heat engine? Theorem 2.7.2. The eﬃciency η of a heat engine operating between a hotter and colder heat reservoirs at temperature τ h and τ l respectively can not exceed the value τl . (2.125) ηC = 1 − τh Proof. A heat engine (labeled as ’I’) is seen in Fig. 2.9. A Carnot heat engine operated in the reverse direction (labeled as ’C’) is added. Here we exploit the fact the Carnot’s cycle is reversible. The eﬃciency η of the heat engine ’I’ is given by ηI = W , Q′h (2.126) whereas the eﬃciency of the reversed Carnot heat engine ’C’ is given by Eq. (2.121) Eyal Buks Thermodynamics and Statistical Physics 71 Chapter 2. Ideal Gas (a) τh W= Q0 Q0 τl M (b) -Q0- Q τh Q M W= Q0 Q0 τl M Fig. 2.8. The assumption that Kelvin’s principle does not hold. Q’h τh Q’l I τl W Qh C Ql Fig. 2.9. Limit imposed upon engine eﬃciency. ηC = W τl =1− . Qh τh (2.127) For the combined system, the Clausius’s principle requires that Q′h − Qh > 0 . (2.128) thus ηI ≤ ηC = 1 − τl . τh The same argument that was employed in the proof above can be used to deduce the following corollary: Eyal Buks Thermodynamics and Statistical Physics 72 2.8. Problems Set 2 Corollary 2.7.1. All reversible heat engines operating between a hotter heat reservoir and a colder one at temperatures τ h and τ l respectively have the same eﬃciency. Note that a similar bound is imposed upon the eﬃciency of refrigerators (see problem 16 of set 2). 2.8 Problems Set 2 1. The heat capacity at constant pressure is deﬁned as ∂σ ∂τ Cp = τ . (2.129) p Calculate Cp of an classical ideal gas having no internal degrees of freedom. 2. Show that ∂σ ∂V = τ ∂p ∂τ , (2.130) V where σ is entropy, V is volume, and p is pressure. 3. Consider a classical ideal gas having internal partition function Zint . a) Show that the chemical potential µ is given by µ = τ log n − log Zint nQ , (2.131) where τ is the temperature, n = N/V , V is the volume, and nQ is the quantum density. b) Show that the energy U is related to the number of particles by the following relation: U =N ∂ log Zint 3τ − 2 ∂β , (2.132) where β = bl/τ . c) Show that the Helmholtz free energy is given by F = Nτ log n − log Zint − 1 nQ . (2.133) d) Show that the entropy is given by σ=N Eyal Buks 5 nQ ∂ (τ log Zint ) + log + 2 n ∂τ . Thermodynamics and Statistical Physics (2.134) 73 Chapter 2. Ideal Gas e) Show that the heat capacity at constant volume is given by cV = N 3 ∂ 2 (τ log Zint ) +τ 2 ∂τ 2 . (2.135) f) Show that the heat capacity at constant pressure is given by cp = cV + N . (2.136) 4. The heat capacity c of a body having entropy σ is given by c=τ ∂σ , ∂τ (2.137) where τ is the temperature. Show that (∆U )2 c= , τ2 (2.138) where U is the energy of the body and where ∆U = U − U . 5. Consider an ideal classical gas made of diatomic molecules. The internal vibrational degree of freedom is described using a model of a one dimensional harmonic oscillator with angular frequency ω. That is, the eigen energies associated with the internal degree of freedom are given by εn = n + 1 2 ω, (2.139) where n = 0, 1, 2, · · · . The system is in thermal equilibrium at temperature τ , which is assumed to be much larger than ω. Calculate the heat capacities cV and cp . 6. A thermally isolated container is divided into two chambers, the ﬁrst containing NA particles of classical ideal gas of type A, and the second one contains NB particles of classical ideal gas of type B. Both gases have no internal degrees of freedom. The volume of ﬁrst chamber is VA , and the volume of the second one is VB . Both gases are initially in thermal equilibrium at temperature τ. An opening is made in the wall separating the two chambers, allowing thus mixing of the two gases. Calculate the change in entropy during the process of mixing. 7. Consider an ideal gas of N molecules in a vessel of volume V . Show that the probability pn to ﬁnd n molecules in a small volume v (namely, v << V ) contained in the vessel is given by pn = λn −λ e , n! (2.140) where λ = N v/V . Eyal Buks Thermodynamics and Statistical Physics 74 2.8. Problems Set 2 8. A lattice contains N sites, each is occupied by a single atom. The set of eigenstates of each atom, when a magnetic ﬁeld H is applied, contains 2 states having energies ε− = −µ0 H and ε+ = µ0 H, where the magnetic moment µ0 is a constant. The system is in thermal equilibrium at temperature τ . a) Calculate the magnetization of the system, which is deﬁned by ∂F ∂H M =− , (2.141) τ where F is the Helmholtz free energy. b) Calculate the heat capacity C=τ ∂σ ∂τ . (2.142) H where σ is the entropy of the system. c) Consider the case where initially the magnetic ﬁeld is H1 and the temperature is τ 1 . The magnetic ﬁeld is then varied slowly in an isentropic process from H1 to H2 . Calculate the ﬁnal temperature of the system τ 2 . 9. A lattice contains N sites, each occupied by a single atom. The set of eigenstates of each atom, when a magnetic ﬁeld H is applied, contains 3 states with energies ε−1 = −∆ − µ0 H , ε0 = 0 , ε1 = −∆ + µ0 H , where the magnetic moment µ0 is a constant. The system is in thermal equilibrium at temperature τ . Calculate the magnetic susceptibility χ = lim H→0 M , H (2.143) where M =− ∂F ∂H , (2.144) τ is the magnetization of the system, and where F is the Helmholtz free energy. 10. A lattice contains N sites, each occupied by a single atom. The set of eigenstates of each atom, when a magnetic ﬁeld H is applied, contains 2J + 1 states with energies εm = −mµH, where J is integer, m = −J, −J + 1, · · · J − 1, J, and the magnetic moment µ is a constant. The system is in thermal equilibrium at temperature τ . Eyal Buks Thermodynamics and Statistical Physics 75 Chapter 2. Ideal Gas a) Calculate the free energy F of the system. b) Show that the average magnetization, which is deﬁned as M =− ∂F ∂H , (2.145) τ is given by M= Nµ 2 (2J + 1) coth (2J + 1) µH − coth 2τ µH 2τ . (2.146) 11. Assume the earth’s atmosphere is pure nitrogen in thermodynamic equilibrium at a temperature of 300 K. Calculate the height above sea level at which the density of the atmosphere is one-half its sea-level value (answer: 12.6 km). 12. Consider a box containing an ideal classical gas made of atoms of mass M having no internal degrees of freedom at pressure p and temperature τ . The walls of the box have N0 absorbing sites, each of which can absorb 0, 1, or 2 atoms of the gas. The energy of an unoccupied site and the energy of a site occupying one atom is zero. The energy of a site occupying two atoms is ε. Show that the mean number of absorbed atoms is given by Na = N0 λ + 2λ2 e−βε , 1 + λ + λ2 e−βε (2.147) where β = 1/τ and λ= M 2πℏ2 −3/2 τ −5/2 p . (2.148) 13. An ideal gas containing N atoms is in equilibrium at temperature τ . The internal degrees of freedom have two energy levels, the ﬁrst one has energy zero and degeneracy g1 , and the second one energy ε and degeneracy g2 . Show that the heat capacities at constant volume and at constant pressure are given by ! g1 g2 exp − τε ε 2 3 cV = N + , (2.149) 2 2 τ g1 + g2 exp − τε ! g1 g2 exp − τε ε 2 5 cp = N + . (2.150) 2 2 τ g1 + g2 exp − ε τ 14. A classical gas is described by the following equation of state p (V − b) = N τ , (2.151) where p is the pressure, V is the volume, τ is the temperature, N is the number of particles and b is a constant. Eyal Buks Thermodynamics and Statistical Physics 76 2.8. Problems Set 2 a) Calculate the diﬀerence cp − cV between the heat capacities at constant pressure and at constant volume. b) Consider an isentropic expansion of the gas from volume V1 and temperature τ 1 to volume V2 and temperature τ 2 . The number of particles N is kept constant. Assume that cV is independent on temperature. Calculate the work W done by the gas during this process. 15. A classical gas is described by the following equation of state p+ a (V − b) = N τ , V2 (2.152) where p is the pressure, V is the volume, τ is the temperature, and a and b are constants. Calculate the diﬀerence cp − cV between the heat capacities at constant pressure and at constant volume. 16. A classical gas is described by the following equation of state p+ a (V − b) = N τ , V2 (2.153) where p is the pressure, V is the volume, τ is the temperature, and a and b are constants. The gas undergoes a reversible isothermal expansion at a ﬁxed temperature τ 0 from volume V1 to volume V2 . Show that the work W done by the gas in this process, and the heat Q which is supplied to the gas during this process are given by V2 − b V2 − V1 , −a V1 − b V2 V1 V2 − b Q = ∆U + W = N τ 0 log . V1 − b W = N τ 0 log (2.154) (2.155) 17. The energy of a classical ideal gas having no internal degrees of freedom is denoted as E, the deviation from the average value U = E as ∆E = E − U . The gas, which contains N particles and has volume is V , is in thermal equilibrium at temperature τ . a) Calculate (∆E)2 . b) Calculate (∆E)3 . 18. A body having a constant heat capacity C and a temperature τ a is put into contact with a thermal bath at temperature τ b . Show that the total change in entropy after equilibrium is establishes is given by ∆σ = C τa τa − 1 − log τb τb . (2.156) Use this result to show that ∆σ ≥ 0. 19. The ﬁgure below shows a cycle of an engine made of an ideal gas. In the ﬁrst step a → b the volume is constant V2 , the second one b → c is Eyal Buks Thermodynamics and Statistical Physics 77 Chapter 2. Ideal Gas p p1 b c p2 a V2 V1 V Fig. 2.10. an isentropic process, and in the third one the pressure is constant p2 . Assume that the heat capacities CV and Cp are temperature independent. Show that the eﬃciency of this engine is given by η =1−γ p2 (V1 − V2 ) , V2 (p1 − p2 ) (2.157) where γ = Cp /Cv . 20. Consider a refrigerator consuming work W per cycle to extract heat from a cold thermal bath at temperature τ l to another thermal bath at higher temperature τ h . Let Ql be the heat extracted from the cold bath per cycle and −Qh the heat delivered to the hot one per cycle. The coeﬃcient of refrigerator performance is deﬁned as γ= Ql . W (2.158) Show that the second law of thermodynamics imposes an upper bound on γ γ≤ τl . τh − τl (2.159) 21. A room air conditioner operates as a Carnot cycle refrigerator between an outside temperature τ h and a room at a lower temperature τ l . The room gains heat from the outdoors at a rate A (τ h − τ l ); this heat is removed by the air conditioner. The power supplied to the cooling unit is P . Calculate the steady state temperature of the room. 22. The state equation of a given matter is Eyal Buks Thermodynamics and Statistical Physics 78 2.8. Problems Set 2 p= Aτ 3 , V (2.160) where p, V and τ are the pressure, volume and temperature, respectively, A is a constant. The internal energy of the matter is written as U = Bτ n log 23. 24. 25. 26. 27. 28. V + f (τ) , V0 (2.161) where B and V0 are constants, f (τ) only depends on the temperature. Find B and n. An ideal classical gas is made of N identical molecules each having mass M . The volume of the gas is V and the temperature is τ . The energy spectrum due to internal degrees of freedom of each molecule has a ground state, which is nondegenerate state (singlet state), and a ﬁrst excited energy state, which has degeneracy 3 (triplet state). The energy gap between the ground state and the ﬁrst excited state is ∆ and all other states have a much higher energy. Calculate: a) the heat capacity at constant volume CV . b) the heat capacity at constant pressure Cp . Two identical bodies have internal energy U = Cτ , with a constant heat capacity C. The initial temperature of the ﬁrst body is τ 1 and that of the second one is τ 2 . The two bodies are used to produce work by connecting them to a reversible heat engine and bringing them to a common ﬁnal temperature τ f . a) Calculate τ f . b) Calculate the total work W , which is delivered by the process. An ideal classical gas having no internal degrees of freedom is contained in a vessel having two parts separated by a partition. Each part contains the same number of molecules, however, while the pressure in the ﬁrst one is p1 , the pressure in the second one is p2 . The system is initially in thermal equilibrium at temperature τ . Calculate the change of entropy caused by a fast removal of the partition. A classical ideal gas contains N particles having mass M and no internal degrees of freedom is in a vessel of volume V at temperature τ . Express the canonical partition function Zc as a function of N , M , V and τ. Consider an engine working in a reversible cycle and using an ideal classical gas as the working substance. The cycle consists of two processes at constant pressure (a → b and c → d), joined by two isentropic processes (b → c and d → a), as show in Fig. 2.11. Assume that the heat capacities CV and Cp are temperature independent. Calculate the eﬃciency of this engine. Consider an engine working in a cycle and using an ideal classical gas as the working substance. The cycle consists of two isochoric processes (constant volume) a→b at volume V1 and c→d at volume V2 , joined by Eyal Buks Thermodynamics and Statistical Physics 79 Chapter 2. Ideal Gas p p1 a p2 b d c V Fig. 2.11. p b c a d V1 V V2 Fig. 2.12. two isentropic processes (constant entropy) b→c and d→a, as shown in Fig. 2.12. Assume that the heat capacities CV and Cp are temperature independent. Calculate the eﬃciency η of this engine. Eyal Buks Thermodynamics and Statistical Physics 80 2.9. Solutions Set 2 29. Consider two vessels A and B each containing ideal classical gas of particles having no internal degrees of freedom . The pressure and number of particles in both vessels are p and N respectively, and the temperature is τ A in vessel A and τ B in vessel B. The two vessels are brought into thermal contact. No heat is exchanged with the environment during this process. Moreover, the pressure is kept constant at the value p in both vessels during this process. Find the change in the total entropy ∆σ = σ ﬁnal − σ initial . 2.9 Solutions Set 2 1. The entropy is given by " Mτ σ = N log 2πℏ2 or using pV = N τ " σ=N 3/2 3/2 log M 2πℏ2 ∂σ ∂τ 5 = N. 2 V N # 5 + 2 ! , (2.162) # ! τ 5/2 5 + , p 2 (2.163) thus Cp = τ p (2.164) 2. Since ∂2F ∂2F = , ∂V ∂τ ∂τ∂V where F is Helmholtz free energy, one has ∂ ∂V ∂F ∂τ = V τ ∂ ∂τ ∂F ∂V (2.165) . τ (2.166) V By deﬁnition ∂F ∂V τ = −p . Moreover, using F = U − τσ one ﬁnds ∂F ∂τ = V = ∂U ∂τ ∂U ∂τ V −τ V − ∂σ ∂τ ∂U ∂σ V V −σ ∂σ ∂τ V −σ = −σ , Eyal Buks Thermodynamics and Statistical Physics (2.167) 81 Chapter 2. Ideal Gas thus ∂σ ∂V = τ ∂p ∂τ . (2.168) V 3. We have found in class the following relations 3/2 Mτ , 2π 2 µ η=− , τ log Zgc = e−η Zint V nQ , nQ = 3τ ∂ log Zint − 2 ∂β N = log Zgc . U= (2.169) (2.170) (2.171) log Zgc , (2.172) (2.173) a) Using Eqs. (2.171) and (2.173) one ﬁnds log µ n = , nQ Zint τ (2.174) thus µ = τ log n − log Zint nQ . (2.175) . (2.176) b) Using Eqs. (2.172) and (2.173) U =N 3τ ∂ log Zint − 2 ∂β c) Using the relations F = U − τσ , σ = log Zgc + βU + ηN , (2.177) (2.178) one ﬁnds F = U − τσ = Nτ (−η − 1) µ −1 = Nτ τ n = Nτ log − log Zint − 1 nQ (2.179) (2.180) (2.181) . (2.182) (2.183) Eyal Buks Thermodynamics and Statistical Physics 82 2.9. Solutions Set 2 d) Using the relation σ=− ∂F ∂τ , (2.184) V one ﬁnds ∂F ∂τ V ∂ τ log nnQ =N − ∂τ ∂ log nnQ = N −τ ∂τ σ=− =N =N ∂ (τ log Z ) int + + 1 ∂τ V n ∂ (τ log Z ) int − log + + 1 nQ ∂τ V ∂ (τ log Zint ) n 3 + − log +1 2 nQ ∂τ 5 nQ ∂ (τ log Zint ) + log + . 2 n ∂τ (2.185) e) By deﬁnition cV = τ =N ∂σ ∂τ V 3 ∂ 2 (τ log Zint ) +τ 2 ∂τ 2 . (2.186) f) The following holds cp = τ =τ ∂σ ∂τ ∂σ ∂τ = cV + τ (2.187) p ∂σ ∂V +τ V ∂σ ∂V τ ∂V ∂τ ∂V ∂τ τ p . p Using V p = N τ and Eq. (2.185) one ﬁnds cp = cV + τ Eyal Buks NN = cV + N . V p Thermodynamics and Statistical Physics (2.188) 83 Chapter 2. Ideal Gas 4. With the help of Eqs. (1.87), (1.70) and (1.71) together with the following relation ∂ 1 ∂ =− 2 , ∂τ τ ∂β (2.189) one ﬁnds that c=τ (∆U )2 ∂σ ∂U 1 ∂U = =− 2 = ∂τ ∂τ τ ∂β τ2 . (2.190) 5. The internal partition function is given by Zint = 1 2 sinh ω 2τ ≃ τ , ω (2.191) thus using Eqs. (2.186) and (2.188) cV = N cp = ∂ 2 τ log 3 +τ 2 ∂τ 2 τ ω = 5N , 2 (2.192) 7N . 2 (2.193) 6. Energy conservation requires that the temperature of the mixture will remain τ . The entropy of an ideal gas of density n, which contains N particles, is given by σ (N, n) = N log nQ 5 + n 2 thus the change in entropy is given by ∆σ = σmix − σ A − σB NA NB + σ NB , VA + VB VA + VB VA + VB VA + VB = NA log + NB log VA VB = σ NA , − σ NA , NA VA − σ NB , 7. The probability to ﬁnd a molecule in the volume v is given by p = v/V . Thus, pn is given by pn = N! pn (1 − p)N−n . n! (N − n)! (2.194) Using the solution of problem 4 of set 1 pn = λn −λ e , n! where λ = N v/V . Eyal Buks Thermodynamics and Statistical Physics 84 NB VB 2.9. Solutions Set 2 8. The partition function of a single atom is given by Z1 = exp (µ0 Hβ) + exp (−µ0 Hβ) = 2 cosh (µ0 Hβ) , where β = 1/τ , thus the partition function of the entire system is N Z = (2 cosh (µ0 Hβ)) , a) The free energy is given by F = −τ log Z = −N τ log (2 cosh (µ0 Hβ)) . (2.195) The magnetization is given by M =− ∂F ∂H = Nµ0 tanh (µ0 Hβ) . (2.196) τ b) The energy U is given by U =− thus C=τ = ∂ log Z = −Nµ0 H tanh (µ0 Hβ) , ∂β ∂σ ∂τ ∂U ∂τ H H = −N µ0 H =N (2.197) ∂ tanh µ0τH ∂τ µ0 H 1 τ cosh µ0 H τ H 2 . (2.198) c) The entropy σ, which is given by σ = β (U − F ) µ H µ H µ H = N log 2 cosh 0 − 0 tanh 0 τ τ τ , (2.199) and which remains constant, is a function of the ratio H/τ, therefore τ2 = τ1 H2 . H1 (2.200) 9. The partition function of a single atom is given by Eyal Buks Thermodynamics and Statistical Physics 85 Chapter 2. Ideal Gas 1 Z= exp (−βεm ) m=−1 = 1 + 2 exp (β∆) cosh (βµ0 H) , (2.201) where β = 1/τ . The free energy is given by F = −Nτ log Z , (2.202) thus the magnetization is given by ∂F ∂H τ 2N µ0 exp (β∆) sinh (βµ0 H) = . 1 + 2 exp (β∆) cosh (βµ0 H) M =− (2.203) and the magnetic susceptibility is given by Nµ20 , τ 1 + 12 exp (−β∆) χ= (2.204) 10. The partition function of a single atom is given by J Z= exp (mµHβ) , m=−J where β = 1/τ . By multiplying by a factor sinh (µHβ/2) one ﬁnds sinh µHβ 2 Z= = 1 exp 2 µHβ 2 1 exp 2 J+ − exp − 1 2 µHβ 2 J exp (mµHβ) m=−J µHβ − exp − J + 1 2 µHβ (2.205) thus Z= sinh 1 2 µHβ µHβ 2 J+ sinh . (2.206) a) The free energy is given by F = −N τ log Z = −Nτ log Eyal Buks sinh 1 2 µHβ µHβ 2 J+ sinh Thermodynamics and Statistical Physics . (2.207) 86 , 2.9. Solutions Set 2 b) The magnetization is given by M =− ∂F ∂H = τ Nµ 2 (2J + 1) coth (2J + 1) µH − coth 2τ µH . 2τ (2.208) 11. The internal chemical potential µg is given by Eq. (2.131). In thermal equilibrium the total chemical potential µtot = µg + mgz , (2.209) (m is the mass of the each diatomic molecule N2 , g is the gravity acceleration constant, and z is the height) is z independent. Thus, the density n as a function of height above see level z can be expressed as n (z) = n (0) exp − mgz kB T . (2.210) The condition n (z) = 0.5 × n (0) yields z= kB T ln 2 1.3806568 × 10−23 J K−1 × 300 K × ln 2 = 12.6 km . = mg 14 × 1.6605402 × 10−27 kg × 9.8 m s−2 (2.211) 12. The Helmholtz free energy of an ideal gas of N particles is given by " # 3/2 Mτ F = −τN log V + τN log N − τ N , (2.212) 2πℏ2 thus the chemical potential is µ= ∂F ∂N τ ,V Mτ 2πℏ2 = −τ log 3/2 V + τ log N , (2.213) and the pressure is p=− ∂F ∂V = τ ,V Nτ . V (2.214) Using these results the fugacity λ = exp (βµ) can be expressed in terms of p λ = eβµ = Mτ 2πℏ2 −3/2 N = V M 2πℏ2 −3/2 τ −5/2 p . (2.215) At equilibrium the fugacity of the gas and that of the system of absorbing sites is the same. The grand canonical partition function of a single absorption site is given by Eyal Buks Thermodynamics and Statistical Physics 87 Chapter 2. Ideal Gas Z = 1 + eβµ + eβ(2µ−ε) , (2.216) or in terms of the fugacity λ = exp (βµ) Z = 1 + λ + λ2 e−βε . (2.217) Thus Na = N0 λ ∂ log Z λ + 2λ2 e−βε = N0 , ∂λ 1 + λ + λ2 e−βε (2.218) where λ is given by Eq. (2.215). 13. The internal partition function is given by Zint = g1 + g2 exp − ε τ . (2.219) Using Eq. (2.135) ∂2 3 (τ log Zint ) cV = N + N τ 2 ∂τ 2 3 ε + 2 τ =N V g1 g2 exp − τε 2 g1 + g2 exp − τε 2 ! . (2.220) Using Eq. (2.136) cp = N g1 g2 exp − τε 2 5 ε + 2 τ g1 + g2 exp − τε 2 ! . (2.221) 14. Using Maxwell’s relation ∂σ ∂V = τ ,N ∂p ∂τ , (2.222) V,N and the equation of state one ﬁnds that ∂σ ∂V = τ ,N N . V −b (2.223) a) Using the deﬁnitions cV = τ cp = τ Eyal Buks ∂σ ∂τ V,N ∂σ ∂τ p,N , (2.224) , (2.225) Thermodynamics and Statistical Physics 88 2.9. Solutions Set 2 and the general identity ∂z ∂x ∂z ∂x = α + y ∂z ∂y ∂y ∂x x , (2.226) α one ﬁnds ∂σ ∂V cp − cV = τ ∂V ∂τ τ ,N , (2.227) p,N or with the help of Eq. (2.223) and the equation of state cp − cV = N Nτ =N . p (V − b) (2.228) b) Using the identity ∂U ∂V ∂U ∂V = τ ,N ∂U ∂σ + σ,N V,N ∂σ ∂V , (2.229) τ ,N together with Eq. (2.223) yields ∂U ∂V τ ,N = −p + Nτ =0. V −b (2.230) Thus, the energy U is independent on the volume V (it can be expressed as a function of τ and N only), and therefore for processes for which dN = 0 the change in energy dU can be expressed as dU = cV dτ . (2.231) For an isentropic process no heat is exchanged, and therefore dW = −dU , thus since cV is independent on temperature one has W = −∆U = −cV (τ 2 − τ 1 ) . (2.232) 15. Using the deﬁnitions cV = τ cp = τ ∂σ ∂τ V,N ∂σ ∂τ p,N , (2.233) , (2.234) and the general identity ∂z ∂x Eyal Buks = α ∂z ∂x + y ∂z ∂y x ∂y ∂x , (2.235) α Thermodynamics and Statistical Physics 89 Chapter 2. Ideal Gas one ﬁnds ∂σ ∂V cp − cV = τ ∂V ∂τ τ ,N . (2.236) p,N Using Maxwell’s relation ∂σ ∂V ∂p ∂τ = , (2.237) a (V − b) = N τ , V2 (2.238) τ ,N V,N and the equation of state p+ one ﬁnds ∂p ∂τ cp − cV = τ ∂V ∂τ V,N p,N τ (VN−b) = −aV +2ab+pV 3 NV 3 N = 1+ −2aV +2ab V 3 (p+ Va2 ) , (2.239) or cp − cV = 1− N 2 . 2a(1− Vb ) (2.240) V Nτ 16. The work W is given by V2 W = pdV . (2.241) V1 Using p+ a (V − b) = N τ , V2 (2.242) one ﬁnds V2 Nτ0 a − 2 V −b V W = V1 dV = N τ 0 log V2 − b V2 − V1 −a . (2.243) V1 − b V2 V1 Using the identity ∂U ∂V Eyal Buks = τ ,N ∂U ∂V + σ,N ∂U ∂σ V,N ∂σ ∂V τ ,N = −p + τ Thermodynamics and Statistical Physics ∂σ ∂V , τ ,N 90 2.9. Solutions Set 2 (2.244) and Maxwell’s relation ∂σ ∂V ∂p ∂τ = τ ,N , (2.245) V,N one ﬁnds ∂U ∂V ∂p ∂τ =τ τ ,N −p. V,N (2.246) In the present case ∂U ∂V = τ ,N Nτ a −p= 2 , V −b V (2.247) thus V2 ∆U = V1 ∂U ∂V V2 dV = a V1 τ ,N V2 − V1 dV =a , V2 V2 V1 (2.248) and Q = ∆U + W = N τ 0 log V2 − b . V1 − b (2.249) 17. In general the following holds En = 1 Zgc − ∂ n Zgc ∂β n , (2.250) η and U =− ∂ log Zgc ∂β . (2.251) η Thus the variance is given by (∆E)2 = E 2 − E = = 1 Zgc 2 ∂ 2 Zgc ∂β 2 ∂ 2 log Zgc ∂β 2 η − 1 2 Zgc ∂Zgc ∂β 2 η . η (2.252) Furthermore, the following holds: Eyal Buks Thermodynamics and Statistical Physics 91 Chapter 2. Ideal Gas 3 = E 3 − 3E 2 U + 3EU 2 − U 3 (∆E) = E 3 − 3U E 2 + 2U 3 " 1 ∂ 3 Zgc 3 ∂Zgc ∂ 2 Zgc =− − 2 3 Zgc Zgc ∂β η ∂β ∂β 2 η " # 2 ∂ 1 1 ∂ 2 Zgc ∂Zgc =− − 2 ∂β Zgc ∂β η ∂β 2 η Zgc =− ∂ 3 log Zgc ∂β 3 2 + 3 Z gc η . η (2.253) For classical gas having no internal degrees of freedom one has N = log Zgc = e−η V M 2π 2 β 3/2 , (2.254) thus ∂ log Zgc ∂β U =− = η 3N τ . 2 (2.255) a) Using Eq. (2.252) (∆E)2 = − ∂ 3N 2U 2 3N = = . 2 ∂β 2β 2β 3N (2.256) ∂ 3N 3N 8U 3 = = . ∂β 2β 2 9N 2 β3 (2.257) b) Using Eq. (2.253) (∆E)3 = − 18. The entropy change of the body is τb ∆σ1 = C τa dτ τb , = C log τ τa (2.258) and that of the bath is ∆σ2 = ∆Q C (τ a − τ b ) = , τb τb (2.259) thus ∆σ = C τa τa − 1 − log τb τb . (2.260) The function f (x) = x − 1 − log x in the range 0 < x < ∞ satisfy f (x) ≥ 0, where f (x) > 0 unless x = 1. Eyal Buks Thermodynamics and Statistical Physics 92 ∂Zgc ∂β 3 η # 2.9. Solutions Set 2 19. The eﬃciency is given by η = 1+ Ql Qca Cp (τ a − τ c ) p2 (V1 − V2 ) = 1−γ , (2.261) = 1+ = 1+ Qh Qab Cv (τ b − τ a ) V2 (p1 − p2 ) where γ = Cp /CV . 20. Energy conservation requires that W = Ql + Qh . Consider a Carnot heat engine operating between the same thermal baths producing work W per cycle. The Carnot engine consumes heat Q′h from the hot bath per cycle and delivered −Q′l heat to the cold one per cycle, where W = Q′l + Q′h and W τl = 1− . Q′h τh ηc = (2.262) According to Clausius principle Ql + Q′l ≤ 0 , (2.263) thus γ= τl Ql Q′ Q′ − W τh −1= . ≤− l = h = W W W τh − τl τh − τl (2.264) 21. Using Eq. (2.264) A (τ h − τ l ) τl = , P τh − τl (2.265) thus τ 2l − 2τ l τ h + P 2A + τ 2h = 0 , (2.266) or P τl = τh + ± 2A , τh + P 2A The solution for which τ l ≤ τ h is , P P τl = τh + − τh + 2A 2A 2 − τ 2h . (2.267) − τ 2h . (2.268) 2 22. Using Eq. (2.244) ∂U ∂V Eyal Buks =τ τ ,N ∂p ∂τ V,N −p, Thermodynamics and Statistical Physics (2.269) 93 Chapter 2. Ideal Gas thus Bτ n 3Aτ 3 2Aτ 3 = −p= , V V V (2.270) therefore B = 2A , n=3. (2.271) (2.272) 23. In general the following holds 3 ∂ 2 (τ log Zint ) +τ 2 ∂τ 2 Cp = cv + N , , CV = N (2.273) (2.274) where in our case Zint = 1 + 3 exp − ∆ τ thus a) CV is given by , (2.275) ∆ 2 −∆ e τ τ 3 3 CV = N + ∆ 2 1 + 3e− τ b) and Cp is given by 5 Cp = N + 2 3 ∆ 2 −∆ e τ τ −∆ τ 1 + 3e , (2.276) , (2.277) 2 2 24. Consider an inﬁnitesimal change in the temperatures of both bodies dτ 1 and dτ 2 . The total change in entropy associated with the reversible process employed by the heat engine vanishes, thus 0 = dσ = dσ 1 + dσ 2 = dQ1 dQ2 + =C τ1 τ2 dτ 1 dτ 2 + τ1 τ2 . (2.278) a) Thus, by integration the equation dτ 1 dτ 2 =− , τ1 τ2 (2.279) one ﬁnds Eyal Buks Thermodynamics and Statistical Physics 94 2.9. Solutions Set 2 τf τ1 τf dτ 1 =− τ1 τ2 dτ 2 , τ2 (2.280) or log τ2 τf = log , τ1 τf (2.281) thus τf = √ τ 1τ 2. (2.282) b) Employing energy conservation law yields √ √ 2 W = ∆U1 + ∆U2 = C (τ 1 − τ f ) + C (τ 2 − τ f ) = C ( τ 1 − τ 2 ) . (2.283) 25. Energy conservation requires that the temperature of the mixture will remain τ . The entropy of an ideal gas of density n, which contains N particles, is given by σ (N, n) = N log nQ 5 + n 2 , (2.284) where Mτ 2π 2 N n= . V 3/2 nQ = , (2.285) (2.286) Using the relation pV = N τ , (2.287) one ﬁnds that the ﬁnal pressure of the gas after the partition has been removed and the system has reached thermal equilibrium is given by pﬁnal = 2p1 p2 . p1 + p2 Thus, the change in entropy is given by ∆σ = σﬁnal − σ1 − σ 2 = 2N log = N log (p1 + p2 ) τ nQ 5 + 2p1 p2 2 − N log τ nQ 5 + p1 2 − N log τ nQ 5 + p2 2 (p1 + p2 )2 . 4p1 p2 (2.288) Eyal Buks Thermodynamics and Statistical Physics 95 Chapter 2. Ideal Gas 26. Using σ = log Zc + βU = log Zgc + βU + ηN one ﬁnds log Zc = log Zgc + ηN . (2.289) The following holds for classical ideal gas having no internal degrees of freedom log Zgc = N η = −βµ = log nQ V , N (2.290) where Mτ 2π 2 nQ = 3/2 , (2.291) thus nQ V N = N log (nQ V ) + N − N log N ≃ N log (nQ V ) − log N ! log Zc = N 1 + log = log (nQ V )N , N! (2.292) or 1 Zc = N! Mτ 2π 2 N 3/2 V . (2.293) 27. The eﬃciency is deﬁned as η = W/Qh , where W is the total work, and Qh is the heat extracted from the heat bath at higher temperature. Energy conservation requires that W = Qh + Ql , where Ql is the heat extracted from the heat bath at lower temperature, thus η = 1 + Ql /Qh . In the present case Qh is associated with process a → b, while Ql is associated with process c → d. In both isentropic processes (b → c and d → a) no heat is exchanged. Thus η =1+ Eyal Buks Ql Cp (τ d − τ c ) =1+ . Qh Cp (τ b − τ a ) Thermodynamics and Statistical Physics (2.294) 96 2.9. Solutions Set 2 Using pV = Nτ yields η =1+ τd − τc p2 (Vd − Vc ) . =1+ τb − τa p1 (Vb − Va ) (2.295) Along the isentropic process pV γ is constant, where γ = Cp /Cv , thus η =1+ p2 p1 1 γ p1 p2 (Va − Vb ) Vb − Va =1− p2 p1 γ−1 γ . (2.296) 28. No heat is exchanged in the isentropic processes, thus the eﬃciency is given by Ql Qh Qc→d = 1+ Qa→b cV (τ d − τ c ) = 1+ . cV (τ b − τ a ) η = 1+ (2.297) Since τV γ−1 remains unchanged in an isentropic process, where γ= cp , cV (2.298) one ﬁnds that τ b V1γ−1 = τ c V2γ−1 , τ d V2γ−1 = τ a V1γ−1 , (2.299) (2.300) or τc τd = = τb τa V2 V1 1−γ , (2.301) thus η =1− V2 V1 1−γ . (2.302) 29. Let VA1 = Nτ A /p (VB1 = N τ B /p) be the initial volume of vessel A (B) and let VA2 (VB2 ) be the ﬁnal volume of vessel A (B). In terms of the ﬁnal temperature of both vessels, which is denoted as τ f , one has VA2 = VB2 = Eyal Buks Nτ f . p Thermodynamics and Statistical Physics (2.303) 97 Chapter 2. Ideal Gas The entropy of an ideal gas of density n = N/V , which contains N particles, is given by σ = N log nQ 5 + n 2 , (2.304) where Mτ 2π 2 nQ = 3/2 , (2.305) or as a function of τ and p σ=N log 3/2 5/2 M τ 2π 2 p + 5 2 . (2.306) Thus the change in entropy is given by ∆σ = σﬁnal − σinitial τ 2f 5N log = . 2 τ Aτ B (2.307) In general, for an isobaric process the following holds Q = W + ∆U = p (V2 − V1 ) + cV (τ 2 − τ 1 ) , (2.308) where Q is the heat that was added to the gas, W the work done by the gas and ∆U the change in internal energy of the gas. Using the equation of state pV = N τ this can be written as Q = (N + cV ) (τ 2 − τ 1 ) . (2.309) Since no heat is exchanged with the environment during this process the following holds QA + QB = 0 , where QA = (N + cV ) (τ f − τ A ) , QB = (N + cV ) (τ f − τ B ) , (2.310) (2.311) thus τf = τA + τB , 2 (2.312) and therefore ∆σ = Eyal Buks 5N (τ A + τ B )2 log . 2 4τ A τ B Thermodynamics and Statistical Physics (2.313) 98 3. Bosonic and Fermionic Systems In the ﬁrst part of this chapter we study two Bosonic systems, namely photons and phonons. A photon is the quanta of electromagnetic waves whereas a phonon is the quanta of acoustic waves. In the second part we study two Fermionic systems, namely electrons in metals and electrons and holes in semiconductors. 3.1 Electromagnetic Radiation In this section we study an electromagnetic cavity in thermal equilibrium. 3.1.1 Electromagnetic Cavity Consider an empty volume surrounded by conductive walls having inﬁnite conductivity. The Maxwell’s equations in SI units are given by ∇ × H = ǫ0 ∂E , ∂t ∇ × E = −µ0 ∂H , ∂t (3.1) (3.2) ∇·E=0, (3.3) ∇·H=0, (3.4) and where ǫ0 = 8.85 × 10−12 F m−1 and µ0 = 1.26 × 10−6 N A−2 are the permittivity and permeability respectively of free space, and the following holds ǫ0 µ0 = 1 , c2 (3.5) where c = 2.99 × 108 m s−1 is the speed of light in vacuum. In the Coulomb gauge, where the vector potential A is chosen such that ∇·A=0 , (3.6) Chapter 3. Bosonic and Fermionic Systems the scalar potential φ vanishes in the absence of sources (charge and current), and consequently both ﬁelds E and H can be expressed in terms of A only as E=− ∂A , ∂t (3.7) and µ0 H =∇ × A . (3.8) The gauge condition (3.6) and Eqs. (3.7) and (3.8) guarantee that Maxwell’s equations (3.2), (3.3), and (3.4) are satisﬁed ∇×E=− ∂ (∇ × A) ∂H = −µ0 , ∂t ∂t ∂ (∇ · A) =0, ∂t 1 ∇ · (∇ × A) = 0 , ∇·H= µ0 ∇·E=− (3.9) (3.10) (3.11) where in the last equation the general vector identity ∇ · (∇ × A) = 0 has been employed. Substituting Eqs. (3.7) and (3.8) into the only remaining nontrivial equation, namely into Eq. (3.1), leads to ∇ × (∇ × A) = − 1 ∂2A . c2 ∂t2 (3.12) Using the vector identity ∇ × (∇ × A) = ∇ (∇ · A) − ∇2 A , (3.13) and the gauge condition (3.6) one ﬁnds that ∇2 A = 1 ∂2A . c2 ∂t2 (3.14) Consider a solution in the form A = q (t) u (r) , (3.15) where q (t) is independent on position r and u (r) is independent on time t. The gauge condition (3.6) leads to ∇·u=0. (3.16) From Eq. (3.14) one ﬁnds that q∇2 u = Eyal Buks 1 d2 q u . c2 dt2 Thermodynamics and Statistical Physics (3.17) 100 3.1. Electromagnetic Radiation Multiplying by an arbitrary unit vector n̂ leads to ∇2 u · n̂ 1 d2 q = 2 . u · n̂ c q dt2 (3.18) The left hand side of Eq. (3.18) is a function of r only while the right hand side is a function of t only. Therefore, both should equal a constant, which is denoted as −κ2 , thus ∇2 u+κ2 u = 0 , (3.19) d2 q +ω 2κ q = 0 , dt2 (3.20) and where ω κ = cκ . (3.21) Equation (3.19) should be solved with the boundary conditions of a perfectly conductive surface. Namely, on the surface S enclosing the cavity we have H · ŝ = 0 and E × ŝ = 0, where ŝ is a unit vector normal to the surface. To satisfy the boundary condition for E we require that u be normal to the surface, namely, u = ŝ (u · ŝ) on S. This condition guarantees also that the boundary condition for H is satisﬁed. To see this we calculate the integral of the normal component of H over some arbitrary portion S ′ of S. Using Eq. (3.8) and Stoke’s’ theorem one ﬁnds that S′ q [(∇ × u) · ŝ] dS µ0 S ′ / q = u·dl , µ0 C (H · ŝ) dS = (3.22) where the close curve C encloses the surface S ′ . Thus, since u is normal to the surface one ﬁnds that the integral along the close curve C vanishes, and therefore S′ (H · ŝ) dS = 0 . (3.23) Since S ′ is arbitrary we conclude that H · ŝ =0 on S. Each solution of Eq. (3.19) that satisﬁes the boundary conditions is called an eigen mode. As can be seen from Eq. (3.20), the dynamics of a mode amplitude q is the same as the dynamics of an harmonic oscillator having angular frequency ω κ = cκ. Eyal Buks Thermodynamics and Statistical Physics 101 Chapter 3. Bosonic and Fermionic Systems 3.1.2 Partition Function What is the partition function of a mode having eigen angular frequency ω κ ? We have seen that the mode amplitude has the dynamics of an harmonic oscillator having angular frequency ω κ . Thus, the quantum eigenenergies of the mode are εs = s ωκ , (3.24) where s = 0, 1, 2, · · · is an integer1 . When the mode is in the eigenstate having energy εs the mode is said to occupy s photons. The canonical partition function of the mode is found using Eq. (1.69) ∞ Zκ = exp (−sβ ωκ ) s=0 1 . 1 − exp (−β ω κ ) = (3.25) Note the similarity between this result and the orbital partition function ζ of Bosons given by Eq. (2.36). The average energy is found using ∂ log Zκ ∂β ωκ = β ωκ . e −1 εκ = − (3.26) The partition function of the entire system is given by Z= Zκ , (3.27) κ and the average total energy by U =− ∂ log Z = ∂β εκ . (3.28) κ 3.1.3 Cube Cavity For simplicity, consider the case of a cavity shaped as a cube of volume V = L3 . We seek solutions of Eq. (3.19) satisfying the boundary condition 1 In Eq. (3.24) above the ground state energy was taken to be zero. Note that by taking instead εs = (s + 1/2) ω κ , one obtains Zκ = 1/2 sinh (βℏωκ /2) and εn = ( ωκ /2) coth (βℏω κ /2). In some cases the oﬀset energy term ωκ /2 is very important (e.g., the Casimir force), however, in what follows we disregard it. Eyal Buks Thermodynamics and Statistical Physics 102 3.1. Electromagnetic Radiation that the tangential component of u vanishes on the walls. Consider a solution having the form 8 ax cos (kx x) sin (ky y) sin (kz z) , (3.29) V 8 ay sin (kx x) cos (ky y) sin (kz z) , (3.30) uy = V 8 az sin (kx x) sin (ky y) cos (kz z) . uz = (3.31) V While the boundary condition on the walls x = 0, y = 0, and z = 0 is guaranteed to be satisﬁed, the boundary condition on the walls x = L, y = L, and z = L yields nx π , (3.32) kx = L ny π ky = , (3.33) L nz π kz = , (3.34) L where nx , ny and nz are integers. This solution clearly satisﬁes Eq. (3.19) where the eigen value κ is given by 0 κ = kx2 + ky2 + kz2 . (3.35) ux = Alternatively, using the notation n = (nx , ny , nz ) , (3.36) one has κ= π n, L (3.37) 0 n2x + n2y + n2z . (3.38) where n= Using Eq. (3.21) one ﬁnds that the angular frequency of a mode characterized by the vector of integers n is given by ωn = πc n. L (3.39) In addition to Eq. (3.19) and the boundary condition, each solution has to satisfy also the transversality condition ∇ · u = 0 (3.16), which in the present case reads n·a= 0 , Eyal Buks (3.40) Thermodynamics and Statistical Physics 103 Chapter 3. Bosonic and Fermionic Systems where a = (ax , ay , az ) . (3.41) Thus, for each set of integers {nx , ny , nz } there are two orthogonal modes (polarizations), unless nx = 0 or ny = 0 or nz = 0. In the latter case, only a single solution exists. 3.1.4 Average Energy The average energy U of the system is found using Eqs. (3.26), (3.28) and (3.39) U = ωn −1 eβ n ωn ∞ = 2τ ∞ ∞ nx =0 ny =0 nz αn , αn − 1 e =0 (3.42) where the dimensionless parameter α is given by α= β πc . L (3.43) The following relation can be employed to estimate the dimensionless parameter α α= 2.4 × 10−3 τ L cm 300 K . (3.44) In the limit where α≪1 (3.45) the sum can be approximated by the integral 4π U ≃ 2τ 8 ∞ dn n2 0 αn . eαn − 1 (3.46) Employing the integration variable transformation [see Eq. (3.39)] n= L ω, πc (3.47) allows expressing the energy per unit volume U/V as U = V ∞ dω uω , (3.48) 0 Eyal Buks Thermodynamics and Statistical Physics 104 3.1. Electromagnetic Radiation where uω = c3 π 2 ω3 . −1 (3.49) eβ ω This result is know as Plank’s radiation law. The factor uω represents the spectral distribution of the radiation. The peak in uω is obtained at β ω0 = 2.82. In terms of the wavelength λ0 = 2πc/ω0 one has λ0 = 5.1 µm T 1000 K −1 . (3.50) 1.4 1.2 1 0.8 0.6 0.4 0.2 0 2 4 6 x 3 8 10 x The function x / (e − 1). The total energy is found by integrating Eq. (3.48) and by employing the variable transformation x = β ω U τ4 = 3 2 V c π ∞ 3 0 x3 dx ex − 1 π4 15 = π2 τ 4 . 15 3 c3 (3.51) 3.1.5 Stefan-Boltzmann Radiation Law Consider a small hole having area dA drilled into the conductive wall of an electromagnetic (EM) cavity. What is the rate of energy radiation emitted from the hole? We employ below a kinetic approach to answer this question. Consider radiation emitted in a time interval dt in the direction of the unit vector û. Let θ be the angle between û and the normal to the surface of the hole. Photons emitted during that time interval dt in the direction û came from the region in the cavity that is indicated in Fig. 3.1, which has volume Eyal Buks Thermodynamics and Statistical Physics 105 Chapter 3. Bosonic and Fermionic Systems û cdt θ dAcosθ Fig. 3.1. Radiation emitted through a small hole in the cavity wall. Vθ = dA cos θ × cdt . (3.52) The average energy in that region can be found using Eq. (3.51). Integrating over all possible directions yields the total rate of energy radiation emitted from the hole per unit area π/2 1 1 J= dAdt 4π 2π dθ sin θ 0 = π2τ 4 1 15 3 c2 4π dϕ U Vθ V 0 π/2 2π dθ sin θ cos θ 0 dϕ 0 1/4 2 4 = π τ 60 3 c2 (3.53) In terms of the historical deﬁnition of temperature T = τ/kB [see Eq. (1.92)] one has J = σB T 4 , (3.54) where σ B , which is given by σB = 4 π2 kB W = 5.67 × 10−8 2 4 , 3 2 60 c m K (3.55) is the Stefan-Boltzmann constant. Eyal Buks Thermodynamics and Statistical Physics 106 3.2. Phonons in Solids m mω2 m mω2 m mω2 m mω2 m Fig. 3.2. 1D lattice. 3.2 Phonons in Solids In this section we study elastic waves in solids. We start with a onedimensional example, and then generalize some of the results for the case of a 3D lattice. 3.2.1 One Dimensional Example Consider the 1D lattice shown in Fig. 3.2 below, which contains N ’atoms’ having mass m each that are attached to each other by springs having spring constant mω 2 . The lattice spacing is a. The atoms are allowed to move in one dimension along the array axis. In problem 5 of set 3 one ﬁnds that the normal mode angular eigen-frequencies are given by 2 2 1 2 kn a 2 2 , ω n = ω 2 (1 − cos kn a) = 2ω 22sin (3.56) 2 2 where a is the lattice spacing, kn = 2πn , aN (3.57) and n is integer ranging from −N/2 to N/2. 0.8 0.6 0.4 0.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 x 0.6 0.8 1 The function |sin (πx/2)|. Eyal Buks Thermodynamics and Statistical Physics 107 Chapter 3. Bosonic and Fermionic Systems What is the partition function of an eigen-mode having eigen angular frequency ωn ? The mode amplitude has the dynamics of an harmonic oscillator having angular frequency ω n . Thus, as we had in the previous section, where we have discussed EM modes, the quantum eigenenergies of the mode are εs = s ωn , (3.58) where s = 0, 1, 2, · · · is an integer. When the mode is in an eigenstate having energy εs the mode is said to occupy s phonons. The canonical partition function of the mode is found using Eq. (1.69) ∞ Zκ = exp (−sβ ωκ ) s=0 = 1 . 1 − exp (−β ω κ ) (3.59) Similarly to the EM case, the average total energy is given by N/2 U= n=−N/2 ℏω n , exp (βℏωn ) − 1 (3.60) where β = 1/τ , and the total heat capacity is given by CV = ∂U = ∂τ N/2 (βℏωn )2 exp (βℏω n ) . [exp (βℏω n ) − 1]2 n=−N/2 (3.61) High Temperature Limit. In the high temperature limit βℏω ≪ 1 (βℏω n )2 exp (βℏωn ) [exp (βℏωn ) − 1]2 ≃1, (3.62) therefore CV = N . (3.63) Low Temperature Limit. In the low temperature limit βℏω ≫ 1 the main contribution to the sum in Eq. (3.61) comes from terms for which |n| N/βℏω. Thus, to a good approximation the dispersion relation can be approximated by 2 2 2 2 2 2 kn a 2 kn a 22 2 2 2 = ω 2π |n| . ω n = 2ω 2sin ≃ 2ω 2 (3.64) 2 2 2 2 N Moreover, in the limit N ≫ 1 the sum in Eq. (3.61) can be approximated by an integral, and to a good approximation the upper limit N/2 can be substituted by inﬁnity, thus Eyal Buks Thermodynamics and Statistical Physics 108 3.2. Phonons in Solids N/2 (βℏω n )2 exp (βℏωn ) CV = 2 [exp (βℏω n ) − 1] n=−N/2 ≃2 ∞ βℏω 2π N n ∞ dn 2 βℏω 2π Nn ∞ 2 exp βℏω 2π N n exp βℏω 2π N n −1 0 N τ π ℏω = exp βℏω 2π Nn exp βℏω 2π Nn −1 n=0 ≃2 2 dx 0 2 x2 exp (x) (exp (x) − 1)2 π2 /3 Nπ τ . 3 ℏω = (3.65) 3.2.2 The 3D Case The case of a 3D lattice is similar to the case of EM cavity that we have studied in the previous section. However, there are 3 important distinctions: 1. The number of modes of a lattice containing N atoms that can move in 3D is ﬁnite, 3N instead of inﬁnity as in the EM case. 2. For any given vector k there are 3, instead of only 2, orthogonal modes (polarizations). 3. Dispersion: contrary to the EM case, the dispersion relation (namely, the function ω (k)) is in general nonlinear. Due to distinctions 1 and 2, the sum over all modes is substituted by an integral according to ∞ ∞ ∞ nx =0 ny =0 nz 3 → 4π 8 =0 nD dn n2 , (3.66) 0 where the factor of 3 replaces the factor of 2 we had in the EM case. Moreover, the upper limit is nD instead of inﬁnity, where nD is determined from the requirement 3 4π 8 nD dn n2 = 3N , (3.67) 0 thus nD = Eyal Buks 6N π 1/3 . Thermodynamics and Statistical Physics (3.68) 109 Chapter 3. Bosonic and Fermionic Systems Similarly to the EM case, the average total energy is given by U = n 3π = 2 ℏω n exp (βℏω n ) − 1 nD dn n2 0 (3.69) ℏω n . exp (βℏωn ) − 1 (3.70) To proceed with the calculation the dispersion relation ω n (kn ) is needed. Here we assume for simplicity that dispersion can be disregarded to a good approximation, and consequently the dispersion relation can be assumed to be linear ω n = vkn , (3.71) where v is the sound velocity. The wave vector kn is related to n = 0 n2x + n2y + n2z by kn = πn , L (3.72) where L = V 1/3 and V is the volume. In this approximation one ﬁnds using the variable transformation x= βℏvπn , L (3.73) that 3π U = 2 = nD ℏvπn L βℏvπn L dn n2 exp 0 3V τ 4 2ℏ3 v3 π2 xD dx 0 −1 x3 . exp x − 1 (3.74) where βℏvπ 6N βℏvπnD π xD = = L L 1/3 . Alternatively, in terms of the Debye temperature, which is deﬁned as Θ = ℏv 6π2 N V 1/3 , (3.75) one has Eyal Buks Thermodynamics and Statistical Physics 110 3.2. Phonons in Solids xD = Θ , τ (3.76) and τ U = 9N τ Θ 3 xD dx 0 x3 . exp x − 1 (3.77) As an example Θ/kB = 88 K for Pb, while Θ/kB = 1860 K for diamond. Below we calculate the heat capacity CV = ∂U/∂τ in two limits. High Temperature Limit. In the high temperature limit xD = Θ/τ ≪ 1, thus τ U = 9Nτ Θ 3 τ Θ 3 τ = 9Nτ Θ = 3Nτ , 3 ≃ 9Nτ xD dx 0 xD x3 exp x − 1 dx x2 0 x3D 3 (3.78) and therefore CV = ∂U = 3N . ∂τ (3.79) Note that in this limit the average energy of each mode is τ and consequently U = 3Nτ . This result demonstrates the equal partition theorem of classical statistical mechanics that will be discussed in the next chapter. Low Temperature Limit. In the low temperature limit xD = Θ/τ ≫ 1, thus τ U = 9Nτ Θ ≃ 9Nτ τ Θ 3 3 xD dx x3 exp x − 1 dx x3 exp x − 1 0 ∞ 0 π4 /15 = 3π4 τ Nτ 5 Θ 3 , (3.80) Eyal Buks Thermodynamics and Statistical Physics 111 Chapter 3. Bosonic and Fermionic Systems and therefore CV = ∂U 12π4 τ = N ∂τ 5 Θ 3 . (3.81) Note that Eq. (3.80) together with Eq. (3.75) yield U 3 π2τ 4 = . V 2 15ℏ3 v3 (3.82) Note the similarity between this result and Eq. (3.51) for the EM case. 3.3 Fermi Gas In this section we study an ideal gas of Fermions of mass m. While only the classical limit was considered in chapter 2, here we consider the more general case. 3.3.1 Orbital Partition Function Consider an orbital having energy εn . Disregarding internal degrees of freedom, its grandcanonical Fermionic partition function is given by [see Eq. (2.33)] ζ n = 1 + λ exp (−βεn ) , (3.83) where λ = exp (βµ) = e−η , (3.84) is the fugacity and β = 1/τ . Taking into account internal degrees of freedom the grandcanonical Fermionic partition function becomes ζn = (1 + λ exp (−βεn ) exp (−βEl )) , (3.85) l where {El } are the eigenenergies of a particle due to internal degrees of freedom [see Eq. (2.71)]. As is required by the Pauli exclusion principle, no more than one Fermion can occupy a given internal eigenstate and a given orbital. 3.3.2 Partition Function of the Gas The grandcanonical partition function of the gas is given by Zgc = Eyal Buks ζn . (3.86) n Thermodynamics and Statistical Physics 112 3.3. Fermi Gas As we have seen in chapter 2, the orbital eigenenergies of a particle of mass m in a box are given by Eq. (2.5) 2 εn = 2 π 2m L n2 , (3.87) where n = (nx , ny , nz ) , 0 n = n2x + n2y + n2z , (3.88) (3.89) nx , ny , nz = 1, 2, 3, · · · , (3.90) and L3 = V is the volume of the box. Thus, log Zgc can be written as log Zgc = ∞ ∞ ∞ log ζ n . (3.91) nx =1 ny =1 nz =1 Alternatively, using the notation α2 = β 2 π2 , 2mL2 (3.92) and Eq. (3.85) one has log Zgc = ∞ l ∞ ∞ nx =1 ny =1 nz =1 log 1 + λ exp −α2 n2 exp (−βEl ) . (3.93) For a macroscopic system α ≪ 1, and consequently the sum over n can be approximately replaced by an integral ∞ ∞ ∞ nx =0 ny =0 nz =0 → 1 4π 8 ∞ dn n2 , (3.94) 0 thus, one has log Zgc π = 2 ∞ l 0 dn n2 log 1 + λ exp −α2 n2 exp (−βEl ) . (3.95) This can be further simpliﬁed by employing the variable transformation βε = α2 n2 . (3.96) The following holds Eyal Buks Thermodynamics and Statistical Physics 113 Chapter 3. Bosonic and Fermionic Systems √ ε 2 3/2 β α2 dε = n2 dn . Thus, by introducing the density of states 2m 3/2 1/2 V 2π2 D (ε) = ε 2 0 ε≥0 , ε<0 (3.97) one has log Zgc = 1 2 ∞ dε D (ε) log (1 + λ exp (−β (ε + El ))) . (3.98) l −∞ 3.3.3 Energy and Number of Particles Using Eqs. (1.80) and (1.94) for the energy U and the number of particles N , namely using ∂ log Zgc ∂β U =− , ∂ log Zgc , ∂λ one ﬁnds that N =λ 1 U = 2 N = 1 2 (3.99) η (3.100) ∞ dε D (ε) (ε + El ) fFD (ε + El ) , (3.101) dε D (ε) fFD (ε + El ) , (3.102) l −∞ ∞ l −∞ where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)] fFD (ǫ) = 1 . exp [β (ǫ − 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 orthogonal 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 114 3.3. Fermi Gas ∞ U = dε D (ε) εfFD (ε) , (3.104) dε D (ε) fFD (ε) , (3.105) −∞ ∞ N = −∞ Typically for metals at room temperature or below the following holds τ ≪ µ. Thus, it is convenient to employ the following theorem (Sommerfeld expansion) 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 power of ε as ε → ∞. Then, the following holds ∞ dε g (ε) fFD (ε) −∞ µ = dε g (ε) + −∞ π 2 g′ (µ) +O 6β 2 1 βµ 4 . Proof. See problem 7 of set 3. With the help of this theorem the number of particles N to second order in τ is given by µ N= dε D (ε) + −∞ π2 τ 2 D′ (µ) . 6 (3.106) Moreover, at low temperatures, the chemical potential is expected to be close the the Fermi energy εF , which is deﬁned by εF = lim µ . (3.107) τ →0 Thus, to lowest order in µ − εF one has µ εF dε D (ε) = −∞ −∞ dε D (ε) + (µ − εF ) D (εF ) + O (µ − εF )2 , (3.108) and therefore N = N0 + (µ − εF ) D (εF ) + π2 τ 2 D ′ (εF ) , 6 (3.109) where Eyal Buks Thermodynamics and Statistical Physics 115 Chapter 3. Bosonic and Fermionic Systems εF N0 = dε D (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 = N0 and consequently π 2 D′ (εF ) . 6β 2 D (εF ) µ = εF − (3.111) Similarly, the energy U at low temperatures is given approximately by ∞ U = dε D (ε) εfFD (ε) −∞ εF = −∞ dε D (ε) ε + (µ − εF ) D (εF ) εF + π2 τ 2 (D′ (εF ) εF + D (εF )) 6 U0 2 2 π τ D′ (εF ) π2τ 2 D (εF ) εF + (D′ (εF ) εF + D (εF )) 6D (εF ) 6 π2τ 2 = U0 + D (εF ) , 6 = U0 − (3.112) where εF U0 = dε D (ε) ε . (3.113) −∞ From this result one ﬁnds that the electronic heat capacity is given by CV = ∂U π2 τ = D (εF ) . ∂τ 3 (3.114) Comparing this result with Eq. (3.81) for 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 temperatures the phonons’ contribution becomes dominant. 3.4 Semiconductor Statistics To be written... Eyal Buks Thermodynamics and Statistical Physics 116 3.5. Problems Set 3 3.5 Problems Set 3 1. Calculate the average number of photons N in equilibrium at temperature τ in a cavity of volume V . Use this result to estimate the number of photons in the universe assuming it to be a spherical cavity of radius 1026 m and at temperature τ = kB × 3 K. 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 reradiates as much thermal radiation, as it receives from the Sun. Assume also, that the surface of the planet is at constant temperature over the day-night cycle. Use TSun = 5800 K; RSun = 6.96 × 108 m; and the Mars-Sun distance of DM−S = 2.28 × 1011 m and calculate the temperature of Mars surface. 3. Calculate the Helmholtz free energy F of photon gas having total energy 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 V1 . The gas is adiabatically compressed from volume V1 to volume V2 in an isentropic process. Calculate the ﬁnal temperature τ 2 and ﬁnal pressure p2 . 5. Consider a one-dimensional lattice of N identical point particles of mass m, interacting via nearest-neighbor spring-like forces with spring constant mω2 (see Fig. 3.2). Denote the lattice spacing by a. Show that the normal mode eigen-frequencies are given by 1 ωn = ω 2 (1 − cos kn a) , (3.116) where kn = 2πn/aN , and n is integer ranging from −N/2 to N/2 (assume N ≫ 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 that the orbital is occupied by n particles is given by pF (n) = exp [n (µ − ε) β] , 1 + exp [(µ − ε) β] (3.117) for the case of Fermions, where n ∈ {0, 1}, and by pB (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 117 Chapter 3. Bosonic and Fermionic Systems 2 2 b) Show that the variance (∆n) = (n − n ) (∆n)2F = n F (1 − n F) is given by , (3.119) for the case of Fermions, and by (∆n)2B = n B (1 + n B) , (3.120) for the case of Bosons. 7. Let g (ε) be a function that vanishes in the limit ε → −∞, and that diverges no more rapidly than some power of ε as ε → ∞. Show that the following holds (Sommerfeld expansion) ∞ I= dε g (ε) fFD (ε) −∞ µ = dε g (ε) + −∞ π2 g ′ (µ) +O 6β 2 1 βµ 4 . 8. Consider a metal at zero temperature having Fermi energy εF , number of electrons N and volume V . a) Calculate the mean energy of electrons. b) Calculate the ratio α of the mean-square-speed of electrons to the square of the mean speed α= v2 v 2 . (3.121) c) Calculate the pressure exerted by an electron gas at zero temperature. 9. For electrons with energy ε ≫ mc2 (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 3 E (T = 0) = N εF 4 (3.122) 10. A gas of two dimensional electrons is free to move in a plane. The mass of each electron is me , the density (number of electrons per unit area) is n, and the temperature is τ. Show that the chemical potential µ is given by µ = τ log exp Eyal Buks nπ 2 me τ −1 . Thermodynamics and Statistical Physics (3.123) 118 3.6. Solutions Set 3 3.6 Solutions Set 3 1. The density of states of the photon gas is given by dg = V ε2 dε . π2 ℏ3 c3 (3.124) Thus V N= 2 3 3 π ℏ c =V τ ℏc ∞ 0 3 ε2 dε eε/τ − 1 α, (3.125) x2 dx . ex − 1 (3.126) where 1 α= 2 π ∞ 0 The number α is calculated numerically α = 0.24359 . (3.127) For the universe N= 4π 1026 m 3 3 ≃ 2.29 × 1087 . 1.3806568 × 10−23 J K−1 3 K 1.05457266 × 10−34 J s2.99792458 × 108 m s−1 3 (3.128) 2. The energy emitted by the Sun is 4 ESun = 4πR2Sun σB TSun , (3.129) and the energy emitted by a planet is 2 4 Eplanet = 4πRplanet σB Tplanet . (3.130) The fraction of Sun energy that planet receives is 2 πRplanet ESun , 2 4πDM−S (3.131) and this equals to the energy it reradiates. Therefore πR2planet ESun = Eplanet , 4πD2 Eyal Buks Thermodynamics and Statistical Physics 119 × 0.24359 Chapter 3. Bosonic and Fermionic Systems thus RSun TSun , 2D Tplanet = and for Mars 6.96 × 108 m 5800 K = 226 K . 2 × 2.28 × 1011 m TMars = 3. The partition function is given by ∞ Z= exp (sβℏωn ) = n s=0 n 1 , 1 − exp (−βℏω n ) thus the free energy is given by F = −τ log Z = τ n log [1 − exp (−βℏωn )] . Transforming the sum over modes into integral yields ∞ F = τπ 0 ∞ = τπ 0 dnn2 log [1 − exp (−βℏω n )] dnn2 log 1 − exp − βℏπcn L (3.132) , or, by integrating by parts F =− ∞ 1 ℏπ2 c 3 L 0 n3 dn exp βℏπcn L 1 =− U , 3 −1 where U= π2τ 4V . 15ℏ3 c3 (3.133) Thus p=− ∂F ∂V = τ U . 3V (3.134) 4. Using the expression for Helmholtz free energy, which was derived in the previous problem, F =− U π2 τ 4 V =− , 3 45ℏ3 c3 one ﬁnds that the entropy is given by Eyal Buks Thermodynamics and Statistical Physics 120 3.6. Solutions Set 3 σ=− ∂F ∂τ = V 4π2 τ 3 V . 45ℏ3 c3 Thus, for an isentropic process, for which σ is a constant, one has V1 V2 τ2 = τ1 1/3 . Using again the previous problem, the pressure p is given by p= U , 3V thus p= π2 τ 4 , 45ℏ3 c3 and p2 = π2 τ 41 45ℏ3 c3 V1 V2 4/3 . p1 5. Let u (na) be the displacement of point particle number n. The equations of motion are given by mü (na) = −mω 2 {2u (na) − u [(n − 1) a] − u [(n + 1) a]} . (3.135) Consider a solution of the form u (na, t) = ei(kna−ωn t) . (3.136) Periodic boundary condition requires that eikNa = 1 , (3.137) thus kn = 2πn . aN (3.138) Substituting in Eq. 3.135 yields −mω 2n u (na) = −mω 2 2u (na) − u (na) e−ika − u (na) eika , (3.139) or 2 2 1 2 kn a 2 2 2 . ωn = ω 2 (1 − cos kn a) = 2ω 2sin 2 2 Eyal Buks Thermodynamics and Statistical Physics (3.140) 121 Chapter 3. Bosonic and Fermionic Systems 6. In general using Gibbs factor exp [n (µ − ε) β] p (n) = , exp [n′ (µ − ε) β] n′ (3.141) where β = 1/τ , one ﬁnds for Fermions pF (n) = exp [n (µ − ε) β] , 1 + exp [(µ − ε) β] (3.142) where n ∈ {0, 1}, and for Bosons exp [n (µ − ε) β] pB (n) = ∞ n′ =0 = {1 − exp [(µ − ε) β]} exp [n (µ − ε) β] , exp [n′ (µ − ε) β] (3.143) where n ∈ {0, 1, 2, · · · }. The expectation value of n in general is given by n′ p (n′ ) = n = n′ n′ exp [n (µ − ε) β] n′ n′ exp [n′ (µ − ε) β] , (3.144) thus for Fermions n F = 1 , exp [(ε − µ) β] + 1 (3.145) and for Bosons n B = {1 − exp [(µ − ε) β]} = {1 − exp [(µ − ε) β]} = 1 . exp [(ε − µ) β] − 1 ∞ n′ =0 n′ exp [n′ (µ − ε) β] exp [(µ − ε) β] (1 − exp [(µ − ε) β])2 (3.146) In general, the following holds τ ∂ n ∂µ = n′′ (n′′ )2 exp [n (µ − ε) β] τ n′ exp [n′ (µ − ε) β] = n2 − n 2 = (n − n )2 − n′ n′ n′ exp [n′ (µ − ε) β] exp [n′ (µ − ε) β] . (3.147) Eyal Buks Thermodynamics and Statistical Physics 122 2 3.6. Solutions Set 3 Thus 2 (∆n)F = exp [(ε − µ) β] 2 = n (exp [(ε − µ) β] + 1) exp [(ε − µ) β] 2 (∆n)B = 7. Let 2 (exp [(ε − µ) β] − 1) = n F B (1 − n F) , (3.148) (1 + n B) . (3.149) ε dε′ g (ε′ ) . G (ε) = (3.150) −∞ Integration by parts yields ∞ I= dε g (ε) fFD (ε) −∞ = [G (ε) fFD (ε)|∞ −∞ ∞ + =0 dε G (ε) − −∞ ∂fFD ∂ε , (3.151) where the following holds − ∂fFD ∂ε = βeβ(ε−µ) eβ(ε−µ) +1 2 β = 2 4 cosh β 2 . (3.152) (ε − µ) Using the Taylor expansion of G (ε) about ε − µ, which has the form G (ε) = ∞ G(n) (µ) (ε − µ)n , n! n=0 (3.153) yields ∞ G(n) (µ) I= n! n=0 ∞ −∞ β (ε − µ)n dε 4 cosh2 β 2 . (3.154) (ε − µ) Employing the variable transformation x = β (ε − µ) , (3.155) and exploiting the fact that (−∂fFD /∂ε) is an even function of ε−µ leads to Eyal Buks Thermodynamics and Statistical Physics 123 Chapter 3. Bosonic and Fermionic Systems ∞ ∞ G(2n) (µ) I= (2n)!β 2n n=0 −∞ x2n dx . 4 cosh2 x2 (3.156) With the help of the identities ∞ −∞ ∞ −∞ dx 4 cosh2 x 2 =1, (3.157) x2 dx π2 = , 3 4 cosh2 x2 (3.158) one ﬁnds I = G (µ) + π2 G(2) (µ) +O 6β 2 µ = dε g (ε) + −∞ 1 βµ π2 g ′ (µ) +O 6β 2 4 1 βµ 4 . (3.159) 8. In general, at zero temperature the average of the energy ε to the power n is given by εn = 3εF dε D (ε) εn 0 3εF , (3.160) dε D (ε) 0 where D (ε) is the density of states D (ε) = V 2π2 2m 2 3/2 ε1/2 , (3.161) thus εn = εnF . 2n 3 +1 (3.162) a) Using Eq. (3.162) one ﬁnds that ε = 3εF . 5 (3.163) b) The speed v is related to the energy by v= Eyal Buks 2ε , m Thermodynamics and Statistical Physics (3.164) 124 3.6. Solutions Set 3 thus ε α= 2 ε1/2 =4 εF 2 3 +1 −1/2 εF 2 1 3 2 +1 52 = 16 . 15 (3.165) c) The number of electrons N is given by εF N= dε D (ε) = εF D (εF ) 1/2 εF 0 dε ε1/2 = D (εF ) 2 3/2 ε , 1/2 3 F ε F 0 thus 2 εF = 2m 3π 2 N V 2/3 , (3.166) and therefore U= 3N 2 5 2m 3π 2 N V 2/3 . (3.167) Moreover, at zero temperature the Helmholtz free energy F = U − τσ = U , thus the pressure is given by ∂F p=− ∂V τ ,N =− ∂U ∂V 3N 2 5 2m 2N εF = . 5V = τ ,N 3π 2 N V 2/3 2 3V (3.168) 9. The energy of the particles are εn,± = pc (3.169) 0 k, and k = πn n2x + n2y + n2z and ni = 1, 2, · · · . L , n = where p = Therefore 1 4 3 π × πn = n3F 8 3 F 3 1/3 3N π N =2× nF = εF = Eyal Buks cπ L 3N π 1/3 = cπ (3.170) 3N πV 1/3 Thermodynamics and Statistical Physics 125 Chapter 3. Bosonic and Fermionic Systems The energy is given by εF E (T = 0) = εg (ε) dε = π2 3 c3 0 L3 εF L3 ε3 dε = (3.171) 0 ε4F 1 L3 = ε3 × εF = 4 4 π2 3 c3 F 3 1 L3 3π2 3 c3 N = εF = N εF 4 π 2 3 c3 L3 4 = π2 3 c3 × 10. The energy of an electron having a wave function proportional to exp (ikx x) exp (iky y) is 2 2me kx2 + ky2 . (3.172) For periodic boundary conditions one has 2πnx , Lx 2πny 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 ′ is given by (including both spin directions) 2π 2me E ′ Lx Ly , 2 4π2 (3.175) thus, the density of state per unit area is given by me π 2 E>0 . 0 E<0 D (E) = (3.176) Using Fermi-Dirac function f (E) = 1 , 1 + exp [β (E − µ)] (3.177) where β = 1/τ , the density n is given by n= ∞ D (E) f (E) dE −∞ me ∞ dE π 2 0 1 + exp [β (E − µ)] me τ log 1 + eβµ , = π 2 = (3.178) Eyal Buks Thermodynamics and Statistical Physics 126 3.6. Solutions Set 3 thus µ = τ log exp Eyal Buks nπ 2 me τ −1 . Thermodynamics and Statistical Physics (3.179) 127 4. Classical Limit of Statistical Mechanics In this chapter we discuss the classical limit of statistical mechanics. We discuss Hamilton’s formalism, deﬁne the Hamiltonian and present the HamiltonJacobi equations of motion. 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 brieﬂy review Hamilton’s formalism, which is analogous to Newton’s laws of classical mechanics . Consider a classical system having d degrees of freedom. The system is described using the vector of coordinates q̄ = (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 , · where T depends only on velocities, namely T = T q̄ , and where V depends only on coordinates, namely V = V (q̄). The notation overdot is used to express time derivative, namely · q̄ = dq1 dq2 dqd , ,··· , dt dt dt . (4.2) We refer to the ﬁrst term T as kinetic energy and to the second one V as potential energy. The canonical conjugate momentum pi of the coordinate qi is deﬁned as pi = ∂T . ∂ q̇i (4.3) The classical Hamiltonian H of the system is expressed as a function of the vector of coordinates q̄ and as a function of the vector of canonical conjugate momentum variables Chapter 4. Classical Limit of Statistical Mechanics p̄ = (p1 , p2 , · · · , pd ) , (4.4) namely H = H (q̄, p̄) , (4.5) and it is deﬁned by d H= i=1 pi q̇i − T + V . (4.6) 4.1.1 Hamilton-Jacobi Equations The equations of motion of the system are given by ∂H ∂pi ∂H ṗi = − , ∂qi q̇i = (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 kinetic energy is given by T = mq̇ 2 /2, thus the canonical conjugate momentum is given by [see Eq. (4.3)] p = mq̇. Thus for this example the canonical conjugate momentum equals the mechanical momentum. Note, however, that this is not necessarily always the case. Using the deﬁnition (4.6) one ﬁnds that the Hamiltonian is given by Eyal Buks Thermodynamics and Statistical Physics 130 4.1. Classical Hamiltonian H = mq̇ 2 − = mq̇ 2 + V (q) 2 p2 + V (q) . 2m (4.9) Hamilton-Jacobi equations (4.7) and (4.8) read p q̇ = m ∂V ṗ = − . ∂q (4.10) (4.11) The second equation, which can be rewritten as mq̈ = − ∂V , ∂q (4.12) expresses Newton’s second law. 4.1.3 Example L C Consider a capacitor having capacitance C connected in parallel to an inductor having inductance L. Let q be the charge stored in the capacitor. The kinetic energy in this case T = Lq̇ 2 /2 is the energy stored in the inductor, and the potential energy V = q 2 /2C is the energy stored in the capacitor. The canonical conjugate momentum is given by [see Eq. (4.3)] p = Lq̇, and the Hamiltonian (4.6) is given by p2 q2 + . (4.13) 2L 2C Hamilton-Jacobi equations (4.7) and (4.8) read p q̇ = (4.14) L q ṗ = − . (4.15) C The second equation, which can be rewritten as q Lq̈ + = 0 , (4.16) C expresses the requirement that the voltage across the capacitor is the same as the one across the inductor. H= Eyal Buks Thermodynamics and Statistical Physics 131 Chapter 4. Classical Limit of Statistical Mechanics 4.2 Density Function Consider a classical system in thermal equilibrium. The density function ρ (q̄, p̄) is the probability distribution to ﬁnd the system in the point (q̄, p̄). The following theorem is given without a proof. Let H (q̄, p̄) be an Hamiltonian of a system, and assume that H has the following form d H= Ai p2i + V (q̄) , (4.17) i=1 where Ai are constants. Then in the classical limit, namely in the limit where Plank’s constant approaches zero h → 0, the density function is given by ρ (q̄, p̄) = N exp (−βH (q̄, p̄)) , (4.18) where N=3 dq̄ 3 1 dp̄ exp (−βH (q̄, p̄)) (4.19) The 3 3is a normalization constant, β = 1/τ , and τ is the temperature. 3 3 notation dq̄ 3 indicates integration over all coordinates, namely dq̄ = dq1 dq1 · 3 3 3 3 · · · · dqd , and similarly dp̄ = dp1 dp1 · · · · · dpd . Let A (q̄, p̄) be a variable which depends on the coordinates q̄ and their canonical conjugate momentum variables p̄. Using the above theorem the average value of A can be calculates as: A (q̄, p̄) = = 3 dq̄ dp̄ A (q̄, p̄) ρ (q̄, p̄) 3 dq̄ dp̄ A (q̄, p̄) exp (−βH (q̄, p̄)) 3 3 . dq̄ dp̄ exp (−βH (q̄, p̄)) (4.20) 4.2.1 Equipartition Theorem Assume that the Hamiltonian has the following form H = Bi qi2 + H̃ , (4.21) where Bi is a constant and where H̃ is independent of qi . Then the following holds Bi qi2 = τ . 2 (4.22) Similarly, assume that the Hamiltonian has the following form Eyal Buks Thermodynamics and Statistical Physics 132 4.2. Density Function H = Ai p2i + H̃ , (4.23) where Ai is a constant and where H̃ is independent of pi . Then the following holds τ (4.24) Ai p2i = . 2 To prove the theorem for the ﬁrst case we use Eq. (4.20) 3 3 dq̄ dp̄ Bi qi2 exp (−βH (q̄, p̄)) 2 3 Bi qi = 3 dq̄ dp̄ exp (−βH (q̄, p̄)) 3 dqi Bi qi2 exp −βBi qi2 = 3 dqi exp (−βBi qi2 ) ∂ =− log dqi exp −βBi qi2 ∂β ∂ π =− log ∂β βBi 1 = . 2β (4.25) The proof for the second case is similar. 4.2.2 Example Here we calculate the average energy of an harmonic oscillator using both, classical and quantum approaches. Consider a particle having mass m in a one dimensional parabolic potential given by V (q) = (1/2) kq 2 , where k is the spring constant. The kinetic energy is given by p2 /2m, where p is the canonical momentum variable conjugate to q. The Hamiltonian is given by H= p2 kq 2 + . 2m 2 (4.26) In the classical limit the average energy of the system can be easily calculated using the equipartition theorem U = H =τ . (4.27) In the quantum treatment the system has energy levels given by Es = s ω , 1 where s = 0, 1, 2, · · · , and where ω = k/m is the angular resonance frequency. The partition function is given by Eyal Buks Thermodynamics and Statistical Physics 133 Chapter 4. Classical Limit of Statistical Mechanics Z= ∞ exp (−sβ ω) = s=0 1 , 1 − exp (−β ω) (4.28) thus the average energy U is given by U =− ∂ log Z ω = β ω . ∂β e −1 (4.29) Using the expansion U = β −1 + O (β) , (4.30) one ﬁnds that in the limit of high temperatures, namely when β ω ≪ 1, the quantum result [Eq. (4.30)] coincides with the classical limit [Eq. (4.27)]. 4.3 Nyquist Noise Here we employ the equipartition theorem in order to evaluate voltage noise across a resistor. Consider the circuit shown in the ﬁgure below, which consists of a capacitor having capacitance C, an inductor having inductance L, and a resistor having resistance R, all serially connected. The system is assumed to be in thermal equilibrium at temperature τ. To model the eﬀect of thermal ﬂuctuations we add a ﬁctitious voltage source, which produces a random ﬂuctuating voltage V (t). Let q (t) be the charge stored in the capacitor at time t. The classical equation of motion, which is given by q + Lq̈ + Rq̇ = V (t) , C (4.31) represents Kirchhoﬀ’s voltage law. L R V(t) ~ C Fig. 4.1. Eyal Buks Thermodynamics and Statistical Physics 134 4.3. Nyquist Noise Consider a sampling of the ﬂuctuating function q (t) in the time interval (−T /2, T /2), namely qT (t) = q(t) −T /2 < t < T /2 . 0 else (4.32) The energy stored in the capacitor is given by q 2 /2C. Using the equipartition theorem one ﬁnds q2 τ = , 2C 2 (4.33) where q 2 is obtained by averaging q 2 (t), namely 1 T →∞ T q 2 ≡ lim +∞ −∞ dt qT2 (t) . (4.34) Introducing the Fourier transform: ∞ 1 qT (t) = √ 2π dω qT (ω)e−iωt , (4.35) −∞ one ﬁnds 1 T →∞ T +∞ ∞ ′ 1 dω qT (ω)e−iωt √ dω′ qT (ω ′ )e−iω t 2π −∞ −∞ −∞ ∞ ∞ +∞ ′ 1 1 = dω qT (ω) dω ′ qT (ω′ ) lim dte−i(ω+ω )t T →∞ T −∞ 2π −∞ −∞ 1 T →∞ T ∞ q 2 = lim 1 dt √ 2π ∞ 2πδ(ω+ω′ ) = lim dω qT (ω)qT (−ω) . −∞ (4.36) Moreover, using the fact that q(t) is real one ﬁnds 1 T →∞ T q 2 = lim ∞ −∞ 2 dω |qT (ω)| . (4.37) In terms of the power spectrum Sq (ω) of q (t), which is deﬁned as Sq (ω) = lim T →∞ 1 |qT (ω)|2 , T (4.38) one ﬁnds q2 = ∞ dω Sq (ω) . (4.39) −∞ Eyal Buks Thermodynamics and Statistical Physics 135 Chapter 4. Classical Limit of Statistical Mechanics Taking the Fourier transform of Eq. (4.31) yields 1 − iωR − Lω2 q(ω) = V (ω) , C (4.40) where V (ω) is the Fourier transform of V (t), namely 1 V (t) = √ 2π ∞ dω V (ω)e−iωt . (4.41) −∞ In terms of the resonance frequency 1 , LC ω0 = (4.42) one has L ω 20 − ω 2 − iωR q(ω) = V (ω) . (4.43) Taking the absolute value squared yields Sq (ω) = SV (ω) 2 L2 (ω 20 − ω2 ) + ω2 R2 , (4.44) where SV (ω) is the power spectrum of V (t). Integrating the last result yields ∞ dω Sq (ω) = −∞ ∞ −∞ = 1 L2 dω ∞ −∞ SV (ω) L2 (ω20 dω 2 − ω 2 ) + ω2 R2 SV (ω) (ω 0 + ω)2 (ω 0 − ω)2 + ω 2 R2 L2 . (4.45) The integrand has a peak near ω 0 , having a width ≃ R/2L. The Quality factor Q is deﬁned as ω0 R = . Q 2L (4.46) Assuming SV (ω) is a smooth function near ω0 on the scale ω 0 /Q, and assuming Q ≫ 1 yield Eyal Buks Thermodynamics and Statistical Physics 136 4.3. Nyquist Noise ∞ −∞ dω Sq (ω) ≃ SV (ω 0 ) L2 ≃ SV (ω 0 ) 4ω40 L2 = SV (ω 0 ) 4ω30 L2 ∞ −∞ ∞ dω (ω 0 + ω)2 (ω 0 − ω)2 + 2 dω 2 ω0 −ω ω0 −∞ ∞ −∞ 2ωω0 Q + 1 Q 2 dx x2 + 1 Q 2 πQ SV (ω 0 )πQ = . 4ω30 L2 (4.47) On the other hand, using Eqs. (4.33) and (4.39) one ﬁnds ∞ dω Sq (ω) = q 2 = Cτ , (4.48) −∞ therefore SV (ω0 ) = 4Cω 30 L2 τ, πQ (4.49) or using Eqs. (4.42) and (4.46) SV (ω0 ) = 2Rτ . π (4.50) Thus, Eq. (4.44) can be rewritten as Sq (ω) = 2Rτ 1 . π L2 (ω 20 − ω 2 )2 + ω 2 R2 (4.51) Note that the spectral density of V given by Eq. (4.50) is frequency independent. Consider a measurement of the ﬂuctuating voltage V (t) in a frequency band having width ∆f. Using the relation V2 = ∞ dω SV (ω) , (4.52) −∞ one ﬁnds that the variance in such a measurement V 2 V2 ∆f ∆f = 4Rτ ∆f . is given by (4.53) The last result is the Nyquist’s noise formula. Eyal Buks Thermodynamics and Statistical Physics 137 Chapter 4. Classical Limit of Statistical Mechanics 4.4 Thermal Equilibrium From Stochastic Processes This section demonstrates that under appropriate conditions a stochastic process can lead to thermal equilibrium in steady state. 4.4.1 Langevin Equation Consider the Langevin equation ẋ = A (x, t) + q (t) , (4.54) where x is a vector of coordinates that depends on the time t, overdot denotes time derivative, the vector A (x, t) is a deterministic function of x and t, and the vector q (t) represents random noise that satisﬁes q (t) = 0 , (4.55) qi (t) qj (t′ ) = gij δ (t − t′ ) . (4.56) and Let δx = x (t + δt) − x (t) . (4.57) To ﬁrst order in δt one ﬁnds by integrating Eq. (4.54) that t+δt (δx)i = Ai (x, t) δt + dt′ qi (t′ ) + O (δt)2 . (4.58) t With the help of Eqs. (4.55) and (4.56) one ﬁnds that (δx)i = Ai (x, t) δt + O (δt)2 , (4.59) and (δx)i (δx)j = Ai (x, t) Aj (x, t) (δt)2 t+δt + t t+δt dt′ t dt′′ qi (t′ ) qj (t′′ ) + · · · = Ai (x, t) Aj (x, t) (δt)2 + gij δt + · · · , (4.60) thus to ﬁrst order in δt (δx)i (δx)j = gij δt + O (δt)2 . (4.61) In a similar way one can show that all higher order moments (e.g. third order moments (δx)i′ (δx)i′′ (δx)i′′′ ) vanish to ﬁrst order in δt. Eyal Buks Thermodynamics and Statistical Physics 138 4.4. Thermal Equilibrium From Stochastic Processes 4.4.2 The Smoluchowski-Chapman-Kolmogorov Relation Let p1 (x, t) be the probability density to ﬁnd the system at point x at time t, let p2 (x′′ , t′′ ; x′ , t′ ) be the probability density to ﬁnd the system at point x′ at time t′ and at point x′′ at time t′′ , and similarly let p3 (x′′′ , t′′′ ; x′′ , t′′ ; x′ , t′ ) be the probability density to ﬁnd the system at point x′ at time t′ , at point x′′ at time t′′ and at point x′′′ at time t′′′ . The following holds p2 (x3 , t3 ; x1 , t1 ) = dx2 p3 (x3 , t3 ; x2 , t2 ; x1 , t1 ) . (4.62) Let P (x, t|x′ , t′ ) be the conditional probability density to ﬁnd the system at point x at time t given that it was (or will be) at point x′ at time t′ . The following holds p2 (x3 , t3 ; x1 , t1 ) = P (x3 , t3 |x1 , t1 ) p1 (x1 , t1 ) . (4.63) Moreover, by assuming that t1 ≤ t2 ≤ t3 and by assuming the case of a Markov process, i.e. the case where the future (t3 ) depends on the present (t2 ), but not on the past (t1 ), one ﬁnds that p3 (x3 , t3 ; x2 , t2 ; x1 , t1 ) = P (x3 , t3 |x2 , t2 ) P (x2 , t2 |x1 , t1 ) p1 (x1 , t1 ) . With the help of these relations Eq. (4.62) becomes P (x3 , t3 |x1 , t1 ) p1 (x1 , t1 ) = dx2 P (x3 , t3 |x2 , t2 ) P (x2 , t2 |x1 , t1 ) p1 (x1 , t1 ) , (4.64) thus by dividing by p1 (x1 , t1 ) one ﬁnds that P (x3 , t3 |x1 , t1 ) = dx2 P (x3 , t3 |x2 , t2 ) P (x2 , t2 |x1 , t1 ) . (4.65) 4.4.3 The Fokker-Planck Equation Equation (4.65) can be written as P (x, t + δt|x0 , t0 ) = dx′ P (x, t + δt|x′ , t′ ) P (x′ , t′ |x0 , t0 ) . (4.66) On the other hand P (x, t + δt|x′ , t′ ) = δ (x − δx − x′ ) , Eyal Buks Thermodynamics and Statistical Physics (4.67) 139 Chapter 4. Classical Limit of Statistical Mechanics where δx = x (t + δt) − x (t) . (4.68) For a general scalar function F of x′ the following holds F (x′ + δx) = exp δx · ∇′ F = F (x′ ) + (δx)i (δx)i (δx)j d2 F dF + +··· , ′ dxi 2! dx′i dx′j (4.69) thus δ (x − δx − x′ ) = δ (x − x′ ) dδ (x − x′ ) + (δx)i dx′i (δx)i (δx)j d2 δ (x − x′ ) + +··· . 2! dx′i dx′j (4.70) Inserting this result into Eq. (4.66) P (x, t + δt|x0 , t0 ) = dx′ δ (x − δx − x′ ) P (x′ , t′ |x0 , t0 ) = dx′ δ (x − x′ ) P (x′ , t′ |x0 , t0 ) + dx′ (δx)i + 1 2 dx′ dδ (x − x′ ) P (x′ , t′ |x0 , t0 ) dx′i d2 δ (x − x′ ) (δx)i (δx)j P (x′ , t′ |x0 , t0 ) dx′i dx′j +··· , (4.71) employing Eqs. (4.59) and (4.61) (recall that higher order moments vanish to ﬁrst order in δt), dividing by δt, taking the limit δt → 0 ∂P = ∂t + dδ (x − x′ ) P (x′ , t′ |x0 , t0 ) dx′i d2 δ (x − x′ ) dx′ gij P (x′ , t′ |x0 , t0 ) , dx′i dx′j dx′ Ai (x′ , t) 1 2 (4.72) Eyal Buks Thermodynamics and Statistical Physics 140 4.4. Thermal Equilibrium From Stochastic Processes and ﬁnally integrating by parts and assuming that P → 0 in the limit x → ±∞ yield 1 d2 ∂P d (gij P) . = − ′ (Ai P) + ∂t dxi 2 dx′i dx′j (4.73) This result, which is known as the Fokker-Planck equation, can also be written as ∂P +∇·J= 0, ∂t (4.74) where the probability current density J is given by Ji = Ai P − 1 d (gij P) . 2 dx′j (4.75) In steady state the Fokker-Planck equation (4.74) becomes ∇·J=0 . (4.76) 4.4.4 The Potential Condition Consider the case where A can be expressed in terms of a scalar ’Hamiltonian’ H as (the potential condition) A (x, t) = −∇H . (4.77) Moreover, for simplicity assume that gij = 2τ δ ij , (4.78) where the ’temperature’ τ is a constant. For this case one has Ji = −P dH dP −τ ′ , dx′i dxi (4.79) thus J = −P∇H − τ ∇P = −P∇ (H + τ log P) . (4.80) Substituting a solution having the form H P = N e− τ (4.81) yields Eyal Buks Thermodynamics and Statistical Physics 141 Chapter 4. Classical Limit of Statistical Mechanics H H J = −τ N e− τ ∇ (log N ) = −τ e− τ ∇N , (4.82) and thus H ∇ · J = [∇H · ∇N − τ ∇ · (∇N)] e− τ . (4.83) In terms of N the Fokker-Planck equation (4.74) becomes ∂N + [∇H · ∇N − τ ∇ · (∇N )] = 0 . ∂t (4.84) In steady state Eq. (4.84) becomes ∇H · ∇N = τ ∇ · (∇N) . (4.85) This equation can be solved by choosing N to be a constant, which can be determined by the normalization condition. In terms of the partition function Z, where Z = 1/N , the steady state solution is expressed as P= 1 −H e τ , Z (4.86) where Z= dx′ P . (4.87) 4.4.5 Free Energy The free energy functional is deﬁned by F (P) = U − τ σ , (4.88) where U is the energy U= dx′ HP , (4.89) and σ is the entropy σ=− dx′ P log P . (4.90) The distribution P is constrained to satisfy the normalization constrain G (P) = dx′ P − 1 = 0 . (4.91) Minimizing F (P) under the constrain (4.91) is done by introducing the Lagrange multiplier η Eyal Buks Thermodynamics and Statistical Physics 142 4.4. Thermal Equilibrium From Stochastic Processes δ (F (P) + ηG (P)) = dx′ [H + τ (1 + log P) + η] δP . (4.92) At a stationary point the following holds H + τ (1 + log P) + η = 0 , (4.93) thus η H P = e− τ −1 e− τ , (4.94) η where the constant e− τ −1 is determined by the normalization condition, i.e. the obtained distribution is identical to the steady state solution of the Fokker-Planck equation (4.86). 4.4.6 Fokker-Planck Equation in One Dimension In one dimension the Fokker-Planck equation (4.74) becomes [see Eq. (4.79)] ∂P ∂ = ∂t ∂x ∂H ∂P +τ ∂x ∂x P , (4.95) or ∂P = LP , ∂t (4.96) where the operator L is given by L= ∂ ∂H ∂2 +τ 2 . ∂x ∂x ∂x (4.97) It is convenient to deﬁne the operator L̂, which is given by H H L̂ = e 2τ Le− 2τ . (4.98) The following holds Eyal Buks Thermodynamics and Statistical Physics 143 Chapter 4. Classical Limit of Statistical Mechanics H L̂ = e 2τ H = e 2τ 2 H H ∂ H ∂ ∂H − 2τ e + τ e 2τ 2 e− 2τ ∂x ∂x ∂x ∂ ∂H − 2τ H ∂H ∂ e + ∂x ∂x ∂x ∂x ∂2 − H e 2τ ∂x2 H + τe 2τ H H + 2τe 2τ 2 = ∂2H 1 − ∂x2 2τ 1 1 ∂2H + 2 2 ∂x 4τ ∂H ∂x 2 − 1 ∂2H 1 − 2 ∂x2 4τ ∂H ∂x 2 = ∂H ∂x + ∂e− 2τ ∂ ∂2 +τ 2 ∂x ∂x ∂x ∂H ∂ ∂x ∂x − ∂H ∂ ∂2 +τ 2 ∂x ∂x ∂x +τ ∂2 , ∂x2 (4.99) thus L̂ = τ ∂2 − V̂ , ∂x2 (4.100) where the potential V̂ is given by V̂ = 1 4τ 2 ∂H ∂x 1 ∂2H . 2 ∂x2 − (4.101) Note that while the operator L is not Hermitian, the operator L̂ is since H H H H L̂† = e− 2τ L† e 2τ H = e− 2τ L† e τ e− 2τ H H H = e− 2τ e τ Le− 2τ H H = e 2τ Le− 2τ = L̂ . (4.102) In terms of L̂ Eq. (4.96) becomes (it is assume that H is time independent) ∂ P̂ = L̂P̂ , ∂t (4.103) where H H P̂ = e 2τ Pe− 2τ . (4.104) Let ψn (x) be a set of eigenvectors of L̂ L̂ψn = λn ψn . Eyal Buks Thermodynamics and Statistical Physics (4.105) 144 4.4. Thermal Equilibrium From Stochastic Processes The following holds [see Eq. (4.98)] Lϕn = λn ϕn , (4.106) where H ϕn = e− 2τ ψn . (4.107) The conditional probability distribution P (x, t|x′ , t′ ) is given by [see Eq. (4.96)] P (x, t|x′ , t′ ) = eL(t−t ) δ (x − x′ ) . ′ (4.108) With the help of the closure relation (recall that L̂ is Hermitian) δ (x − x′ ) = ψ∗n (x′ ) ψn (x) n =e H(x)+H(x′ ) 2τ ϕ∗n (x′ ) ϕn (x) n =e H(x′ ) τ ϕ∗n (x′ ) ϕn (x) , n (4.109) one ﬁnds that P (x, t|x′ , t′ ) = eL(x)(t−t ) e ′ =e H(x′ ) τ H(x′ ) τ ϕ∗n (x′ ) ϕn (x) n ϕ∗n (x′ ) eL(x)(t−t ) ϕn (x) ′ n =e H(x′ ) τ ϕ∗n (x′ ) eλn (t−t ) ϕn (x) ′ n =e H(x′ ) τ e− H(x′ ) 2τ ψ ∗n (x′ ) eλn (t−t ) e− ′ H(x) 2τ ψn (x) , n (4.110) thus P (x, t|x′ , t′ ) = e H(x′ )−H(x) 2τ ψ∗n (x′ ) ψn (x) eλn (t−t ) . ′ (4.111) n 4.4.7 Ornstein—Uhlenbeck Process in One Dimension Consider the following Langevin equation Eyal Buks Thermodynamics and Statistical Physics 145 Chapter 4. Classical Limit of Statistical Mechanics ẋ + Γ x = q (t) , (4.112) where x can take any real value, Γ is a positive constant and where the real noise term q (t) satisﬁes q (t) = 0 and q (t) q (t′ ) = 2τδ (t − t′ ) , (4.113) where τ is positive. For this case [see Eq. (4.77)] Γ x2 , 2 H (x) = (4.114) the Fokker-Planck equation for the conditional probability distribution P is given by [see Eq. (4.74)] ∂ ∂P = ∂t ∂x PΓ x + τ ∂P ∂x , (4.115) and the operator L̂ is given by [see Eq. (4.100)] 2 2 Γ ∂ x L̂ = − − + − 1 , x 2 x 0 ∂ x0 where x0 = 2τ . Γ (4.116) The eigenvectors of L̂ are given by [see Eq. (4.131)] − 12 e ψn (x) = x x0 2 Hn xx0 1/2 √ n π1/4 x0 2 n! , (4.117) and the corresponding eigenvalues by λn = −Γ n+ 1 1 − 2 2 , (4.118) where n = 0, 1, 2, · · · . Using these results one ﬁnds that P is given by [see Eq. (4.111)] − P (x, t|x′ , t′ ) = e Eyal Buks x x0 Hn 2 n x′ x0 Hn xx0 e−Γ n(t−t ) √ . πx0 2n n! Thermodynamics and Statistical Physics ′ (4.119) 146 4.5. Problems Set 4 With the help of the general identity (4.138) one ﬁnds that the following holds 2 −1 −X 2 αe √ π ∞ α n Hn (Y ) Hn (X) 2 n! n=0 thus Eq. (4.119) becomes P (x, t|x′ , t′ ) = exp − exp − Y√−Xα α−2 −1 1 = π (α−2 − 1) ′ x−x′ e−Γ (t−t ) δ √ πδ , (4.120) 2 , (4.121) where 0 δ = x20 1 − e−2Γ (t−t′ ) = , 2τ 1 − e−2Γ (t−t′ ) . Γ (4.122) 4.5 Problems Set 4 1. A gas at temperature τ emits a spectral line at wavelength λ0 . The width of the observed spectral line is broadened due to motion of the molecules (this is called Doppler broadening). Show that the relation between spectral line intensity I and wavelength is given by " # mc2 (λ − λ0 )2 , (4.123) I (λ) ∝ exp − 2λ20 τ where c is velocity of ﬁght, and m is mass of a molecule. 2. The circuit seen in the ﬁgure below, which contains a resistor R, capacitor C, and an inductor L, is at thermal equilibrium at temperature τ . Calculate the average value I 2 , where I is the current in the R C L inductor. Eyal Buks Thermodynamics and Statistical Physics 147 Chapter 4. Classical Limit of Statistical Mechanics 3. Consider a random real signal q(t) varying in time. Let qT (t) be a sampling of the signal q(t) in the time interval (−T /2, T /2), namely qT (t) = q(t) −T /2 < t < T /2 . 0 else (4.124) The Fourier transform is given by ∞ 1 qT (t) = √ 2π dω qT (ω)e−iωt , (4.125) −∞ and the power spectrum is given by Sq (ω) = lim T →∞ 1 |qT (ω)|2 . T a) Show that 1 T →∞ T q 2 ≡ lim +∞ −∞ dt qT2 (t) = ∞ dω Sq (ω) . (4.126) −∞ b) Wiener-Khinchine Theorem - show that the correlation function of the random signal q(t) is given by 1 T →∞ T q (t) q (t + t′ ) ≡ lim +∞ −∞ ∞ dt qT (t) qT (t + t′ ) = −∞ ′ dω eiωt Sq (ω) . (4.127) 4. Consider a resonator made of a capacitor C, an inductor L, and a resistor R connected in series, as was done in class. Let I (t) be the current in the circuit. Using the results obtained in class calculate the spectral density SI (ω) of I at thermal equilibrium. Show that in the limit of high quality factor, namely when Q= 2 R L ≫1, C (4.128) your result is consistent with the equipartition theorem applied for the energy stored in the inductor. 5. A classical system is described using a set of coordinates {q1 , q2 , · · · , qN } and the corresponding canonically conjugate variables {p1 , p2 , · · · , pN }. The Hamiltonian of the system is given by N An psn + Bn qnt , H= (4.129) n=1 where An and Bn are positive constants and s and t are even positive integers. Show that the average energy of the system in equilibrium at temperature τ is given by Eyal Buks Thermodynamics and Statistical Physics 148 4.5. Problems Set 4 U = Nτ 1 1 + s t , (4.130) 6. A small hole of area A is made in the walls of a vessel of volume V containing a classical ideal gas of N particles of mass M each in equilibrium at temperature τ. Calculate the number of particles dN , which escape through the opening during the inﬁnitesimal time interval dt. 7. Consider an ideal gas of Fermions having mass M and having no internal degrees0 of freedom at temperature τ . The velocity of a particle is denoted as v = vx2 + vy2 + vz2 . Calculate the quantity 4 5 1 v v 8. 9. 10. 11. (the symbol denoted averaging) in the: a) classical limit (high temperatures). b) zero temperature. Consider an ideal gas of N molecules, each of mass M, contained in a centrifuge of radius R and length L rotating with angular velocity ω about its axis. Neglect the eﬀect of gravity. The system is in equilibrium at temperature τ = 1/β. Calculate the particle density n (r) as a function of the radial distance from the axis r ( where 0 ≤ r ≤ R) . Consider an ideal classical gas of particles having mass M0and having no internal degrees of freedom at temperature τ . Let v = vx2 + vy2 + vz2 . be the velocity of a particle. Calculate a) 1 v b) v2 A mixture of two classical ideal gases, consisting of N1 and N2 particles of mass M1 and M2 , respectively, is enclosed in a cylindrical vessel of height h and area of bottom and top side S. The vessel is placed in a gravitational ﬁeld having acceleration g. The system is in thermal equilibrium at temperature τ. Find the pressure exerted on the upper wall of the cylinder. The Hermite polynomial Hn (X) of order n is deﬁned by Hn (X) = exp X2 2 X− d dX n exp − X2 2 . (4.131) For some low values of n the Hermite polynomials are given by H0 (X) = 1 , H1 (X) = 2X , (4.132) (4.133) H2 (X) = 4X 2 − 2 , (4.134) H4 (X) = 16X 4 − 48X 2 + 12 . (4.136) 3 H3 (X) = 8X − 12X , Eyal Buks Thermodynamics and Statistical Physics (4.135) 149 Chapter 4. Classical Limit of Statistical Mechanics Show that exp 2Xt − t2 = ∞ Hn (X) n=0 tn . n! (4.137) 12. Show that ∞ α n Hn (X) Hn (Y ) 2 n! n=0 exp = α(2XY −αX 2 −αY 2 ) 1−α2 √ 1 − α2 . (4.138) 4.6 Solutions Set 4 1. Let λ be the wavelength measured by an observer, and let λ0 be the wavelength of the emitted light in the reference frame where the molecule is at rest. Let vx be the velocity of the molecule in the direction of the light ray from the molecule to the observer. Due to Doppler eﬀect λ = λ0 (1 + vx /c) . (4.139) The probability distribution f (vx ) is proportional to f (vx ) ∝ exp − mvx2 2τ , (4.140) thus using vx = c (λ − λ0 ) , λ0 the probability distribution I (λ) is proportional to " # mc2 (λ − λ0 )2 I (λ) ∝ exp − . 2λ20 τ (4.141) (4.142) 2. The energy stored in the inductor is UL = LI 2 /2. Using the equipartition theorem UL = τ/2, thus I2 = τ . L (4.143) 3. a) Eyal Buks Thermodynamics and Statistical Physics 150 4.6. Solutions Set 4 +∞ ∞ ′ 1 dω qT (ω)e−iωt √ dω ′ qT (ω ′ )e−iω t 2π −∞ −∞ −∞ ∞ ∞ +∞ ′ 1 1 = dω qT (ω) dω ′ qT (ω′ ) lim dt e−i(ω+ω )t T →∞ 2π −∞ T −∞ −∞ 1 T →∞ T q 2 = lim ∞ 1 dt √ 2π 2πδ(ω+ω′ ) = lim T →∞ 1 T ∞ dω qT (ω)qT (−ω) . −∞ (4.144) Since q(t) is real one has qT (−ω) = qT∗ (ω), thus q2 = ∞ dω Sq (ω) . (4.145) −∞ b) 1 T →∞ T q (t) q (t + t′ ) ≡ lim = +∞ −∞ 1 1 lim 2π T →∞ T dt qT (t) qT (t + t′ ) +∞ ∞ dt −∞ ∞ 1 1 = lim T →∞ 2π T ∞ dω qT (ω)e−iωt −∞ ∞ ′ ′ dω e−iω t qT (ω) −∞ dω ′ qT (ω ′ )e−iω (t+t ) −∞ dω ′ qT (ω ′ ) −∞ ′ +∞ dt e−i(ω+ω )t −∞ 2πδ(ω+ω′ ) 1 T →∞ T ∞ = lim = ∞ ′ dω eiωt qT (ω)qT (−ω) −∞ ′ dω eiωt Sf (ω) . −∞ (4.146) 4. Using I (ω) = −iωq (ω) and q 2 = Cτ one ﬁnds for the case Q ≫ 1 I2 = ∞ −∞ dω SI (ω) ≃ ω 20 ∞ −∞ dω Sq (ω) = ω 20 q 2 = τ , L (4.147) in agreement with the equipartition theorem for the energy stored in the inductor LI 2 /2. 5. Calculate for example Eyal Buks Thermodynamics and Statistical Physics 151 ′ ′ Chapter 4. Classical Limit of Statistical Mechanics Bn qnt = = = 3∞ t t −∞ Bn qn exp (−βBn qn ) dqn 3∞ t −∞ exp (−βBn qxn ) dqn 3∞ t t 0 3 Bn qn exp (−βBn qn ) dqn ∞ t 0 exp (−βBn qxn ) dqn ∞ d log − exp −βBn qnt dqn dβ 0 . (4.148) where β = 1/τ . Changing integration variable x = βBn qnt , dx = (4.149) tβBn qnt−1 dqn , (4.150) leads to Bn qnt = − d −1 log (βBn ) t t−1 dβ ∞ 1 x t −1 e−x dx = 0 τ . t (4.151) Thus N An psn + Bn qnt = N τ U = n=1 1 1 + s t . (4.152) 6. Let f (v) be the probability distribution of velocity v of particles in the gas. The vector u is expressed in spherical coordinates, where the z axis is chosen in the direction of the normal outward direction v = v (sin θ cos ϕ, sin θ sin ϕ, cos θ) . (4.153) By symmetry, f (v) is independent of θ and ϕ. The number dN is calculated by integrating over all possible values of the velocity of the leaving particles (note that θ can be only in the range 0 ≤ θ ≤ π/2) ∞ dN = 1 dv 2π d (cos θ) 0 0 dϕv2 v (dt) A cos θ 0 N f (v) . V (4.154) Note that v (dt) A cos θ represents the volume of a cylinder from which particles of velocity v can escape during the time interval dt. Since f (v) is normalized ∞ 1= 1 dv 0 2π d (cos θ) 0 ∞ dϕv 2 f (v) = 4π 0 dvv 2 f (v) , (4.155) 0 thus dN πN A = dt V Eyal Buks ∞ 0 dvv2 vf (v) = NA v . 4V Thermodynamics and Statistical Physics (4.156) 152 4.6. Solutions Set 4 In the classical limit f (v) ∝ exp − Mv 2 2τ , (4.157) thus, by changing the integration variable x= Mv2 2τ (4.158) one ﬁnds v = = = 3∞ 2 dvv3 exp − Mv 2τ 0 3∞ 2 dvv 2 exp − Mv 2τ 0 1/2 3 ∞ dxx exp (−x) 2τ 3 ∞0 1/2 exp (−x) M 0 dxx 1/2 8τ πM , (4.159) and dN NA = dt 4V 8τ πM 1/2 . (4.160) 7. The probability that an orbital having energy ε is occupied is given by fF (ε) = 1 , 1 + exp [(ε − µ) β] (4.161) where β = 1/τ and µ is the chemical potential. The velocity v of such an orbital is related to the energy ε by ε= Mv 2 . 2 (4.162) The 3D density of state per unit volume is given by g (ε) = 1 2π 2 2M 2 3/2 ε1/2 . (4.163) Thus Eyal Buks Thermodynamics and Statistical Physics 153 Chapter 4. Classical Limit of Statistical Mechanics 3∞ 3∞ dε g (ε) fF (ε) v dε g (ε) fF (ε) v1 4 5 1 0 = 03∞ v 3∞ v dε g (ε) fF (ε) dε g (ε) fF (ε) = 0 3∞ 0 3∞ dε εfF (ε) 0 dε fF (ε) 0 3∞ . 2 dε ε1/2 fF (ε) 0 (4.164) a) In the classical limit fF (ε) ∝ exp (−βε) , (4.165) thus using the identities ∞ dε εn exp (−βε) = Γ (n) β −n n β 0 ∞ dε exp (−βε) = 1 β 0 Γ (1) = 1 √ 1 Γ = π 2 one ﬁnds 4 5 1 v = v 3∞ dε ε exp (−βε) 0 Γ = Γ = 3∞ dε exp (−βε) 0 3∞ 2 dε ε1/2 exp (−βε) 0 −1 1 1 (1) β β β 1 2 1 β −1/2 2β 2 4 . π (4.166) b) Using the identity εF dε εn = εn+1 F n+1 0 Eyal Buks Thermodynamics and Statistical Physics 154 4.6. Solutions Set 4 one ﬁnds 3εF 4 5 1 v = v dε ε 0 3εF dε 0 ε 3F dε 2 . ε1/2 0 = = 1 2 2 εF εF 2 3 2 2 3 εF 9 . 8 (4.167) 8. The eﬀect of rotation is the same as an additional external ﬁeld with potential energy given by 1 U (r) = − M ω2 r2 , 2 (4.168) thus n (r) = A exp [−βU (r)] = A exp βMω 2 2 r 2 , (4.169) where the normalization constant A is found from the condition R N = 2πL n (r) rdr 0 R = 2πLA exp 0 2πLA = exp βM ω 2 βMω 2 2 r rdr 2 βM ω 2 2 R −1 , 2 (4.170) thus n (r) = 8 N βM ω 2 2πL exp βMω2 2 2 R −1 9 exp β Mω 2 r2 2 . (4.171) 9. In the classical limit the probability distribution of the velocity vector v satisﬁes f (v) ∝ exp − M v2 2τ , (4.172) where v = |v|. Eyal Buks Thermodynamics and Statistical Physics 155 Chapter 4. Classical Limit of Statistical Mechanics a) By changing the integration variable x= M v2 2τ (4.173) one ﬁnds 3∞ 0 3∞ 0 v = dvv 3 exp − Mv 2τ 2 2 dvv2 exp − Mv 2τ 1/2 3 ∞ dxx exp (−x) 2τ 3 ∞0 1/2 M exp (−x) 0 dxx = 8τ πM = (4.174) 1/2 . (4.175) b) Similarly v2 = = = 3∞ 2 dv v4 exp − Mv 2τ 0 3∞ 2 dv v2 exp − Mv 2τ 0 3∞ 2τ 0 dx x3/2 exp (−x) 3∞ M 0 dx x1/2 exp (−x) 2τ 3 , M2 (4.176) thus 1 v2 = 1/2 3τ M = 1/2 3π 8 v . (4.177) 10. For each gas the density is given by nl (z) = nl (0) exp (−βMl gz) , where l ∈ {1, 2}, 0 ≤ z ≤ h and the normalization constant is found from the requirement h S n (z) dz = Nl , (4.178) 0 therefore Nl βMl gNl . S (1 − e−βMl gh ) (4.179) Thermodynamics and Statistical Physics 156 nl (0) = S 3h exp (−βMl gz) dz = 0 Eyal Buks 4.6. Solutions Set 4 Using the equation of state p = nτ, where n = N/V is the density, one ﬁnds that the pressure on the upper wall of the cylinder is given by p = (n1 (h) + n2 (h)) τ M1 N1 M2 N2 = + exp (βM1 gh) − 1 exp (βM2 gh) − 1 g . S (4.180) 11. The relation (4.137), which is a Taylor expansion of the function f (t) = exp 2Xt − t2 around the point t = 0, implies that 2 2 dn 2 2 . (4.181) Hn (X) = n exp 2Xt − t 2 dt t=0 The identity 2Xt − t2 = X 2 − (X − t)2 yields Hn (X) = exp X 2 2 dn 2 22 exp − (X − t) . 2 dtn t=0 (4.182) Moreover, using the relation d d exp − (X − t)2 = − exp − (X − t)2 dt dX , (4.183) one ﬁnds that 2 dn 2 22 exp − (X − t) 2 dX n t=0 n n d 2 exp −X . (−1) dX n Hn (X) = exp X 2 (−1)n = exp X 2 (4.184) Note that for an arbitrary function g (X) the following holds − exp X 2 d d exp −X 2 g = 2X − dX dX g, (4.185) and exp X2 2 X− d dX exp − X2 2 g = 2X − d dX g, (4.186) thus Hn (X) = exp Eyal Buks X2 2 X− d dX n exp − X2 2 . (4.187) Thermodynamics and Statistical Physics 157 Chapter 4. Classical Limit of Statistical Mechanics 12. With the help of Eq. (4.184) and the general identity ∞ −∞ π 1 4ca+b2 e4 a , a exp −ax2 + bx + c dx = (4.188) according to which the following holds (for the case a = 1, b = 2iX and c = 0) exp −X 2 1 =√ π ∞ −∞ exp −x2 + 2iXx dx , (4.189) one ﬁnds that exp X 2 √ Hn (X) = π 1 =√ π 1 =√ π ∞ −∞ ∞ d − dX n ∞ −∞ exp −x2 + 2iXx dx (−2ix)n exp X 2 − x2 + 2iXx dx 2 (−2ix)n e(X+ix) dx , −∞ (4.190) thus the following holds [see Eq. (4.188)] α n Hn 2 ∞ n=0 1 = π = 1 π ∞ −∞ ∞ 2 2 dy e(X+ix) e(Y +iy) −∞ 2 dx e(X+ix) ∞ −∞ ∞ ∞ (−2αxy)n n! n=0 e−2αxy −∞ 2 dy e(Y +iy) e−2αxy √ αx(αx−2iY ) πe 2 2 2 dx e−(1−α )x +2i(X−Y α)x+X −∞ exp Eyal Buks ∞ dx 1 =√ π = (X) Hn (Y ) n! α(2XY −αX 2 −αY 2 ) 1−α2 √ 1 − α2 . Thermodynamics and Statistical Physics (4.191) 158 5. Exercises 5.1 Problems 1. Two identical non-interacting particles each having mass M are conﬁned in a one dimensional parabolic potential given by 1 V (x) = M ω 2 x2 , 2 (5.1) where the angular frequency ω is a constant. a) Calculate the canonical partition function of the system Zc,B for the case where the particles are Bosons. b) Calculate the canonical partition function of the system Zc,F for the case where the particles are Fermions. 2. Consider a one dimensional gas containing N non-interacting electrons moving along the x direction. The electrons are conﬁned to a section of length L. At zero temperature τ = 0 calculate the ratio U/εF between the total energy of the system U and the Fermi energy εF . 3. Consider an ideal classical gas at temperature τ. The set of internal eigenstates of each particle in the gas, when a magnetic ﬁeld H is applied, contains 2 states having energies ε− = −µ0 H and ε+ = µ0 H, where the magnetic moment µ0 is a constant. Calculate the magnetization of the system, which is deﬁned by M =− ∂F ∂H , (5.2) τ where F is the Helmholtz free energy. 4. (Note: replace this with Ex. 3.9 in the lecture notes) Consider an ideal gas made of N electrons in the extreme relativistic limit. The gas is contained in a box having a cube shape with a volume V = L3 . In the extreme relativistic limit the dispersion relation ε (k) is modiﬁed: the energy ε of a single particle quantum state having a wavefunction ψ given by ψ (x, y, z) = 2 L 3/2 sin (kx x) sin (ky y) sin (kz z) , (5.3) Chapter 5. Exercises is given by ε (k) = kc , (5.4) where c is the speed of light and where k = 0 kx2 + ky2 + kz2 (contrary to the non-relativistic case where it is given by ε (k) = 2 k2 /2M ). The system is in thermal equilibrium at zero temperature τ = 0. Calculate the ratio p/U between the pressure p and the total energy of the system U. 5. Consider a mixture of two classical ideal gases, consisting of NA particles of type A and NB particles of type B. The heat capacities cp,A and cV,A (cp,B and cV,B ) at constant pressure and at constant volume respectively of gas A (B) are assumed to be temperature independent. The volume of the mixture is initially V1 and the pressure is initially p1 . The mixture undergoes an adiabatic (slow) and isentropic (at a constant entropy) process leading to a ﬁnal volume V2 . Calculate the ﬁnal pressure p2 . 6. Consider two particles, both having the same mass m, moving in a onedimensional potential with coordinates x1 and x2 respectively. The potential energy is given by V (x1 , x2 ) = mω2 x21 mω2 x22 + + mΩ 2 (x1 − x2 )2 , 2 2 (5.5) where the angular frequencies ω and Ω are real constants Assume that the temperature τ of the system is suﬃciently high to allow treating it classically. Calculate the following average values a) x21 for the case Ω = 0. b) x21 , however without assuming that Ω = 0. c) (x1 − x2 )2 , again without assuming that Ω = 0. 7. Consider an ideal classical gas containing N identical particles having each mass M in the extreme relativistic limit. The gas is contained in a vessel having a cube shape with a volume V = L3 . In the extreme relativistic limit the dispersion relation ε (k) is modiﬁed: the energy ε of a single particle quantum state having a wavefunction ψ given by ψ (x, y, z) = 2 L 3/2 sin (kx x) sin (ky y) sin (kz z) , (5.6) is given by ε (k) = kc , where c is the speed of light and where k = (5.7) 0 kx2 + ky2 + kz2 (contrary to the non-relativistic case where it is given by ε (k) = 2 k2 /2M ). The system is in thermal equilibrium at temperature τ . Calculate: Eyal Buks Thermodynamics and Statistical Physics 160 5.2. Solutions a) the total energy U of the system. b) the pressure p. 8. Consider a system made of two localized spin 1/2 particles whose energy is given by εσ1 ,σ2 = −µ0 H (σ 1 + σ2 ) + Jσ1 σ2 , (5.8) where both σ 1 and σ2 can take one of two possible values σn = ±1 (n = 1, 2). While H is the externally applied magnetic ﬁeld, J is the coupling constant between both spins. The system is in thermal equilibrium at temperature τ . Calculate the magnetic susceptibility χ = lim H→0 ∂M , ∂H (5.9) where M =− ∂F ∂H (5.10) τ is the magnetization of the system, and where F is the Helmholtz free energy. 9. Consider a one dimensional gas containing N non-interacting electrons moving along the x direction. The electrons are conﬁned by a potential given by 1 V (x) = mω 2 x2 , (5.11) 2 where m is the electron mass and where ω is the angular frequency of oscillations. Calculate the chemical potential µ a) in the limit of zero temperature τ = 0. b) in the limit of high temperatures τ ≫ ω. 10. The state equation of a given matter is Aτ n , (5.12) V where p, V and τ are the pressure, volume and temperature, respectively, and A and n are both constants. Calculate the diﬀerence cp −cV between the heat capacities at constant pressure and at constant volume. p= 5.2 Solutions 1. The single particle eigen energies are given by ǫn = ω n + 1 2 , (5.13) where n = 0, 1, 2, · · · . Eyal Buks Thermodynamics and Statistical Physics 161 Chapter 5. Exercises a) For Bosons ∞ ∞ ∞ 1 1 exp [−β (ǫn + ǫm )] + exp (−2βǫn ) 2 n=0 m=0 2 n=0 Zc,B = ∞ 1 = 2 2 exp (−βǫn ) + n=0 ∞ 1 exp (−2βǫn ) 2 n=0 exp (−β ω) exp (−β ω) 2 + 2 (1 − exp (−2β ω)) . 2 (1 − exp (−β ω)) = (5.14) Note that the average energy UB is given by UB = − ∂ log Zc,B 1 + 2e−2β ω + e−β = ω ∂β 1 − e−2β ω ω . (5.15) b) For Fermions ∞ ∞ ∞ 1 1 exp [−β (ǫn + ǫm )] − exp (−2βǫn ) Zc,F = 2 n=0 m=0 2 n=0 = exp (−β ω) exp (−β ω) 2 − 2 (1 − exp (−2β ω)) . 2 (1 − exp (−β ω)) (5.16) Note that for this case the average energy UF is given by UF = − ∂ log Zc,F 2 + e−2β ω + e−β = ω ∂β 1 − e−2β ω ω . (5.17) 2. The orbital eigenenergies are given by 2 εn = π 2m L 2 n2 , (5.18) where n = 1, 2, 3, · · · . The grandcanonical partition function of the gas is given by Zgc = ζn , (5.19) n where ζn = (1 + λ exp (−βεn ) exp (−βEl )) (5.20) l is the orbital grandcanonical Fermionic partition function where, λ = exp (βµ) = e−η , (5.21) is the fugacity, β = 1/τ and {El } are the eigenenergies of a particle due to internal degrees of freedom. For electrons, in the absence of magnetic Eyal Buks Thermodynamics and Statistical Physics 162 5.2. Solutions ﬁeld both spin states have the same energy, which is taken to be zero. Thus, log Zgc can be written as log Zgc = ∞ log ζ n n=1 ∞ =2 log (1 + λ exp (−βεn )) n=1 ∞ ≃2 2 dn log 1 + λ exp −β 0 π 2m L 2 n2 . (5.22) By employing the variable transformation 2 ε= 2 π 2m L n2 , (5.23) one has log Zgc 1 = 2 ∞ dε D (ε) log (1 + λ exp (−βε)) , (5.24) 0 where 2L π D (ε) = 0 2m −1/2 2 ε 0 ε≥0 ε<0 (5.25) is the 1D density of states. Using Eqs. (1.80) and (1.94) for the energy U and the number of particles N , namely using ∂ log Zgc ∂β U =− , (5.26) η ∂ log Zgc , ∂λ one ﬁnds that N =λ (5.27) ∞ U= dε D (ε) εfFD (ε) , (5.28) dε D (ε) fFD (ε) , (5.29) −∞ ∞ N= −∞ where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)] fFD (ǫ) = Eyal Buks 1 . exp [β (ǫ − µ)] + 1 Thermodynamics and Statistical Physics (5.30) 163 Chapter 5. Exercises At zero temperature, where µ = εF one has U= N= D (εF ) εF dε ε1/2 = −1/2 εF 0 εF D (εF ) dε −1/2 εF 0 2D (εF ) 2 εF , 3 ε−1/2 = 2D (εF ) εF , (5.31) (5.32) thus U N = . εF 3 (5.33) 3. The Helmholtz free energy is given by F = Nτ log n − log Zint − 1 nQ , (5.34) where Zint = exp (βµ0 H) + exp (−βµ0 H) = 2 cosh (βµ0 H) (5.35) is the internal partition function. Thus the magnetization is given by ∂F ∂H M =− = N µ0 tanh (βµ0 H) . (5.36) τ 4. The grandcanonical partition function of the gas is given by Zgc = ζn , (5.37) n where ζn = (1 + λ exp (−βεn ) exp (−βEl )) (5.38) l is a grandcanonical Fermionic partition function of an orbital having energy εn given by π cn , L 0 where n = n2x + n2y + n2z , nx , ny , nz = 1, 2, 3, · · · , εn = (5.39) λ = exp (βµ) = e−η (5.40) is the fugacity, β = 1/τ and {El } are the eigenenergies of a particle due to internal degrees of freedom. For electrons, in the absence of magnetic Eyal Buks Thermodynamics and Statistical Physics 164 5.2. Solutions ﬁeld both spin states have the same energy, which is taken to be zero. Thus, log Zgc can be written as ∞ log Zgc = ∞ ∞ log (1 + λ exp (−βεn ) exp (−βEl )) . (5.41) nx =1 ny =1 nz =1 l For a macroscopic system the sum over n can be approximately replaced by an integral ∞ ∞ ∞ ∞ nx =0 ny =0 nz 4π → 8 =0 dn n2 , (5.42) 0 thus, one has log Zgc 4π =2 8 ∞ 0 dn n2 log 1 + λ exp −β π cn L . (5.43) By employing the variable transformation ε= π cn . L (5.44) one has ∞ log Zgc = dε V ε2 log (1 + λ exp (−βε)) . π 2 3 c3 (5.45) 0 The energy U and the number of particles N are given by ∞ ∂ log Zgc U =− ∂β N =λ ∂ log Zgc = ∂λ = η dε V ε3 fFD (ǫ) , π2 3 c3 (5.46) 0 ∞ dε V ε2 fFD (ǫ) , π2 3 c3 (5.47) 0 where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)] fFD (ǫ) = 1 . exp [β (ǫ − µ)] + 1 (5.48) At zero temperature εF U= dε V ε3 V ε4F = , π 2 3 c3 π2 3 c3 4 (5.49) dε V ε2 V ε3F = , π 2 3 c3 π2 3 c3 3 (5.50) 0 εF N= 0 Eyal Buks Thermodynamics and Statistical Physics 165 Chapter 5. Exercises and therefore U= 3N εF . 4 (5.51) The energy U can be expressed as a function of V and N as U= 3 3 1/3 (3N)4/3 π2 c V −1/3 4 . At zero temperature the Helmholtz free energy F equals the energy U , thus the pressure p is given by p=− ∂F ∂V τ ,N =− ∂U ∂V = τ ,N 1 3 3N 4/3 V π2 4 3 3 1/3 c , (5.52) thus 1 p = . U 3V (5.53) 5. First, consider the case of an ideal gas made of a unique type of particles. Recall that the entropy σ, cV and cp are given by [see Eqs. (2.87), (2.88) and (2.89)] σ=N cV = N 5 nQ ∂ (τ log Zint ) + log + 2 n ∂τ 3 + hint 2 , , cp = cV + N , (5.54) (5.55) (5.56) where n = N/V is the density, nQ = Mτ 2π 2 3/2 (5.57) is the quantum density, M is the mass of a particle in the gas, and hint = τ ∂ 2 (τ log Zint ) cV 3 = − . ∂τ 2 N 2 (5.58) The requirement that hint is temperature independent leads to ∂ (τ log Zint ) τ = gint + hint log , ∂τ τ0 (5.59) where both gint and τ 0 are constants. Using this notation, the change in entropy due to a change in V from V1 to V2 and a change in τ from τ 1 to τ 2 is given by Eyal Buks Thermodynamics and Statistical Physics 166 5.2. Solutions 3/2 ∆σ = σ2 − σ1 = N = N log V2 V1 log V2 τ 2 + 3/2 V1 τ 1 cV 3 − N 2 log τ2 τ1 cV N τ2 τ1 . (5.60) Thus the total change in the entropy of the mixture is given by ∆σ = ∆σA + ∆σB V2 V1 = NA log cV,A NA τ2 τ1 V2 = (NA + NB ) log V1 + NB log τ2 τ1 cV,A +cV,B NA +NB and the requirement ∆σ = 0 leads to τ2 τ1 V2 V1 cV,A +cV,B NA +NB V2 V1 τ2 τ1 cV,B NB , =1. (5.61) (5.62) (5.63) Alternatively, by employing the equation of state pV = N τ , (5.64) this can be rewritten as V2 V1 cV,A +cV,B NA +NB +1 p2 p1 cV,A +cV,B NA +NB =1, (5.65) or with the help of Eq. (5.56) as p2 = p1 V2 V1 = p1 V1 V2 − cV,A +cV,B +1 NA +NB cV,A +cV,B NA +NB cp,A +cp,B cV,A +cV,B . (5.66) 6. It is convenient to employ the coordinate transformation x1 + x2 x+ = √ , 2 x1 − x2 x− = √ . 2 The inverse transformation is given by Eyal Buks Thermodynamics and Statistical Physics (5.67) (5.68) 167 Chapter 5. Exercises x+ + x− √ , 2 x+ − x− √ x2 = . 2 The following holds x1 = (5.69) (5.70) x21 + x22 = x2+ + x2− , (5.71) ẋ21 + ẋ22 = ẋ2+ + ẋ2− . (5.72) and Thus, the kinetic energy T of the system is given by T = m ẋ21 + ẋ22 m ẋ2+ + ẋ2− = , 2 2 (5.73) and the potential energy V is given by mω 2 x21 mω 2 x22 + + mΩ 2 (x1 − x2 )2 2 2 mω 2 x2+ m ω 2 + 4Ω 2 x2− + . = 2 2 V (x1 , x2 ) = (5.74) The equipartition theorem yields mω 2 x2+ m ω2 + 4Ω 2 = 2 2 x2− = τ , 2 (5.75) thus (x1 + x2 )2 = 2τ , mω 2 (5.76) (x1 − x2 )2 = 2τ . m (ω 2 + 4Ω 2 ) (5.77) and Furthermore, since by symmetry x+ x− = 0 one has 1 (x+ + x− )2 2 1 = x2+ + x2− 2 2 τ 1 4Ω ω2 = 1− 2 mω 2 2 1 + 4Ω ω2 x21 = . (5.78) Eyal Buks Thermodynamics and Statistical Physics 168 5.2. Solutions 7. The k vector is restricted due to boundary conditions to the values k= πn , L (5.79) where n = (nx , ny , nz ) , (5.80) and nx , ny , nz = 1, 2, 3, · · · . The single particle partition function is given by Z1 = ∞ ∞ ∞ nx =1 ny =1 nz =1 exp − ε (k) τ . (5.81) Approximating the discrete sum by a continuous integral according to ∞ ∞ ∞ nx =0 ny =0 nz 4π → 8 =0 ∞ dn n2 , (5.82) 0 one has 4π Z1 = 8 ∞ 0 dn n2 exp − ∞ 4V τ 3 = 8π 2 3 c3 n πc Lτ dx x2 exp (−x) 0 2 V τ3 = 2 3 3 . π c (5.83) In the classical limit the grandcanonical partition function Zgc is given by [see Eq. (2.44)] log Zgc = λZ1 , (5.84) where λ = exp (βµ) is the fugacity. In terms of the Lagrange multipliers η = −µ/τ and β = 1/τ the last result can be rewritten as log Zgc = e−η π2 V . 3 c3 β 3 (5.85) a) The average energy U and average number of particle N are calculated using Eqs. (1.80) and (1.81) respectively Eyal Buks Thermodynamics and Statistical Physics 169 Chapter 5. Exercises U =− ∂ log Zgc ∂β η N =− ∂ log Zgc ∂η β thus = 3 log Zgc , β (5.86) = log Zgc , (5.87) U = 3N τ , (5.88) and V τ3 π2 N 3 c3 η = log . (5.89) b) The entropy is evaluate using Eq. (1.86) σ = log Zgc + βU + ηN = N (1 + 3 + η) V τ3 = N 4 + log , π 2 N 3 c3 (5.90) and the Helmholtz free energy by the deﬁnition (1.116) F = U − τ σ = −N τ 1 + log V τ3 π2 N 3 c3 . (5.91) Thus the pressure p is given by p=− ∂F ∂V = τ ,N Nτ . V (5.92) 8. The partition function is given by Z= exp (−βεσ1 ,σ2 ) σ1 ,σ 2 =±1 = exp (−βJ) [exp (−2βµ0 H) + exp (2βµ0 H)] + 2 exp (βJ) , (5.93) where β = 1/τ . The free energy is given by F = −τ log Z , (5.94) thus the magnetization is given by ∂F ∂H τ 2µ0 exp (−βJ) [− exp (−2βµ0 H) + exp (2βµ0 H)] = , exp (−βJ) [exp (−2βµ0 H) + exp (2βµ0 H)] + 2 exp (βJ) (5.95) M =− Eyal Buks Thermodynamics and Statistical Physics 170 5.2. Solutions and the magnetic susceptibility is given by χ= 4βµ20 . 1 + e2βJ (5.96) Note that in the high temperature limit βJ ≪ 1 χ≃ 2µ20 . τ +J (5.97) 9. The orbital eigenenergies in this case are given by εn = ω n + 1 2 , (5.98) where n = 0, 1, 2, · · · . The grandcanonical partition function of the gas is given by Zgc = ζn , (5.99) n where ζn = (1 + λ exp (−βεn ) exp (−βEl )) (5.100) l is the orbital grandcanonical Fermionic partition function where, λ = exp (βµ) = e−η (5.101) is the fugacity, β = 1/τ and {El } are the eigenenergies of a particle due to internal degrees of freedom. For electrons, in the absence of magnetic ﬁeld both spin states have the same energy, which is taken to be zero. Thus, log Zgc can be written as log Zgc = ∞ log ζ n n=0 ∞ =2 log (1 + λ exp (−βεn )) . n=0 (5.102) The number of particles N is given by N =λ ∞ ∂ log Zgc =2 fFD (εn ) , ∂λ n=0 (5.103) where fFD is the Fermi-Dirac distribution function fFD (ε) = Eyal Buks 1 . exp [β (ε − µ)] + 1 Thermodynamics and Statistical Physics (5.104) 171 Chapter 5. Exercises a) At zero temperature the chemical potential µ is the Fermi energy εF , and the Fermi-Dirac distribution function becomes a step function, thus with the help of Eq. (5.103) one ﬁnds that N= 2εF , ω (5.105) thus µ = εF = N ω . 2 (5.106) b) Using the approximation fFD (ε) ≃ exp [−β (ε − µ)] , (5.107) for the the limit of high temperatures and approximating the sum by an integral one has ∞ 1 N =2 exp −β ω n + −µ 2 n=0 ∞ ω = 2 exp β µ − 2 2 exp β µ − = β ω ω 2 dn exp (−β ωn) 0 , (5.108) thus µ = τ log ≃ τ log Nβ ω β ω + 2 2 Nβ ω . 2 (5.109) 10. Using Eq. (2.239), which is given by cp − cV = τ ∂p ∂τ V,N ∂V ∂τ , (5.110) p,N one ﬁnds that cp − cV = Eyal Buks A2 τ 2 2(n−1) n τ = n2 Aτ n−1 . pV Thermodynamics and Statistical Physics (5.111) 172 References 1. C. E. Shannon, Bell System Tech. J. 27, 379 (1948). 2. C. E. Shannon, Bell System Tech. J. 27, 623 (1948). 3. E. T. Jaynes, Phys. rev. Lett. 106, 620 (1957). Index Bose-Einstein distribution, 54 Bose-Einstein function, 54 Boson, 52 canonical conjugate momentum, 129 canonical distribution, 15 Carnot, 66 cavity, 99 chemical potential, 17, 20 classical limit, 54 Clausius’s principle, 70 composition property, 5 Coulomb gauge, 99 Debye temperature, 110 density function, 132 density of states, 114 eﬃciency, 68, 71 electromagnetic radiation, 99 entropy, 1 equipartition theorem, 132 Fermi energy, 115 Fermi-Dirac distribution, 53 Fermi-Dirac function, 53 Fermion, 52 Fokker-Planck equation, 139 free energy, 142 fugacity, 18, 53 grandcanonical distribution, 16 H theorem, 18 Hamilton-Jacobi equations, 130 Hamiltonian, 129 heat capacity, 59 Helmholtz free energy, 21, 57 ideal gas, 47 information theory, 1 internal degrees of freedom, 59 isentropic process, 65 isobaric process, 64 isochoric process, 65 isothermal process, 64 Kelvin’s principle, 70 Lagrange multiplier, 4 Langevin equation, 138 largest uncertainty estimator, 9 Maxwell’s equations, 99 microcanonical distribution, 14 Nyquist noise, 134 orbitals, 50 Ornstein—Uhlenbeck process, 145 partition function, 10 permeability, 99 permittivity, 99 phonon, 107 photon, 102 Plank’s radiation law, 105 pressure, 57 quantum density, 48 Shannon, 1 Sommerfeld expansion, 115 Stefan-Boltzmann constant, 106 Stefan-Boltzmann radiation law, 105 temperature, 15, 17 thermal equilibrium, 19 vector potential, 99