Introduction to Thermodynamics and Statistical Physics (114016) - Lecture Notes

Eyal Buks

September 24, 2017

Technion

Preface

to be written...

Contents

1. The Principle of Largest Uncertainty ...... 1 1.1 Entropy in Information Theory ...... 1 1.1.1 Example - Two States System ...... 1 1.1.2 Smallest and Largest Entropy ...... 2 1.1.3 The composition property ...... 5 1.1.4 Alternative Deﬁnition of Entropy ...... 8 1.2 Largest Uncertainty Estimator ...... 9 1.2.1 Useful Relations...... 11 1.2.2 The FreeEntropy ...... 13 1.3 The Principle of Largest Uncertainty in Statistical Mechanics 14 1.3.1 Microcanonical Distribution ...... 14 1.3.2 Canonical Distribution ...... 15 1.3.3 Grandcanonical Distribution ...... 16 1.3.4 Temperature and Chemical Potential ...... 17 1.4 Time Evolution of Entropy of an Isolated System ...... 18 1.5 Thermal Equilibrium ...... 19 1.5.1 Externally Applied Potential Energy ...... 20 1.6 Free Entropy and Free Energies ...... 21 1.7 ProblemsSet1...... 21 1.8 SolutionsSet1 ...... 30

2. Ideal Gas ...... 47 2.1 AParticleinaBox ...... 47 2.2 GibbsParadox ...... 50 2.3 FermionsandBosons ...... 52 2.3.1 Fermi-Dirac Distribution ...... 53 2.3.2 Bose-Einstein Distribution ...... 54 2.3.3 ClassicalLimit ...... 54 2.4 Ideal Gas in the Classical Limit ...... 55 2.4.1 Pressure ...... 57 2.4.2 Useful Relations...... 58 2.4.3 HeatCapacity ...... 59 2.4.4 Internal Degrees of Freedom ...... 59 2.5 Processes inIdeal Gas ...... 62 Contents

2.5.1 Isothermal Process ...... 64 2.5.2 IsobaricProcess ...... 64 2.5.3 IsochoricProcess ...... 65 2.5.4 Isentropic Process ...... 65 2.6 CarnotHeatEngine ...... 66 2.7 Limits Imposed Upon the Eﬃciency ...... 68 2.8 ProblemsSet2...... 73 2.9 SolutionsSet2...... 81

3. Bosonic and Fermionic Systems ...... 99 3.1 Electromagnetic Radiation ...... 99 3.1.1 Electromagnetic Cavity ...... 99 3.1.2 Partition Function...... 102 3.1.3 CubeCavity...... 102 3.1.4 AverageEnergy ...... 104 3.1.5 Stefan-Boltzmann Radiation Law ...... 105 3.2 PhononsinSolids ...... 107 3.2.1 One Dimensional Example ...... 107 3.2.2 The3DCase ...... 109 3.3 FermiGas...... 112 3.3.1 Orbital Partition Function ...... 112 3.3.2 Partition Function of the Gas ...... 112 3.3.3 Energy and Number of Particles ...... 114 3.3.4 Example: Electrons in Metal ...... 114 3.4 Semiconductor Statistics ...... 116 3.5 ProblemsSet3...... 117 3.6 SolutionsSet3...... 119

4. Classical Limit of Statistical Mechanics ...... 129 4.1 Classical Hamiltonian ...... 129 4.1.1 Hamilton-Jacobi Equations ...... 130 4.1.2 Example ...... 130 4.1.3 Example ...... 131 4.2 Density Function ...... 132 4.2.1 Equipartition Theorem ...... 132 4.2.2 Example ...... 133 4.3 NyquistNoise ...... 134 4.4 Thermal Equilibrium From Stochastic Processes ...... 138 4.4.1 Langevin Equation ...... 138 4.4.2 The Smoluchowski-Chapman-Kolmogorov Relation . . . 139 4.4.3 The Fokker-Planck Equation ...... 139 4.4.4 The Potential Condition...... 141 4.4.5 FreeEnergy ...... 142 4.4.6 Fokker-Planck Equation in One Dimension ...... 143 4.4.7 Ornstein—Uhlenbeck Process in One Dimension ...... 145

Eyal Buks Thermodynamics and Statistical Physics 6 Contents

4.5 ProblemsSet4...... 147 4.6 SolutionsSet4...... 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 e , where m = 1, 2, 3, , m ··· 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 p the entropy is deﬁned as { m}

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 e with probability 1 p, where 0 p 1. The entropy is given by 2 − ≤ ≤ 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 0.4x 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 − →

d log p lim ( p log p) = lim dp = 0 . (1.4) p 0 p 0 d 1 → − → − dp p 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

0.2 -p*log(p) 0.15

0.1

0.05

0 0.2 0.4p 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 = g (¯p) = p 1 , (1.5) 0 m − m where p¯ denotes the vector of probabilities p¯ = (p , p , ) . (1.6) 1 2 ··· A small change in σ (denoted as δσ) due to a small change in p¯ (denoted as δp¯ = (δp , δp , )) can be expressed as 1 2 ··· ∂σ δσ = δp , (1.7) ∂p m m m or in terms of the gradient of σ (denoted as ¯ σ) as ∇ δσ = ¯ σ δp¯ . (1.8) ∇ · In addition the variables (p1, p2, ) are subjected to the constrain (1.5). Similarly to Eq. (1.8) we have ··· δg = ¯ g δp¯ . (1.9) 0 ∇ 0 · Both vectors ¯ σ and δp¯ can be decomposed as ∇ ¯ σ = ¯ σ + ¯ σ , (1.10) ∇ ∇ ∇ ⊥ δp¯ = (δp¯) + (δp¯) , (1.11) ⊥ where ¯ σ and (δp¯) are parallel to ¯ g0, and where ¯ σ and (δp¯) are ∇ ∇ ∇ ⊥ ⊥ orthogonal to ¯ g . Using this notation Eq. (1.8) can be expressed as ∇ 0

Eyal Buks Thermodynamics and Statistical Physics 3 Chapter 1. The Principle of Largest Uncertainty

δσ = ¯ σ (δp¯) + ¯ σ (δp¯) . (1.12) ∇ · ∇ ⊥ · ⊥ Given that the constrain g0 (¯p) = 0 is satisfied at a given point p¯, one has g0 (¯p + δp¯) = 0 to first order in δp¯ provided that δp¯, is orthogonal to ¯ g0, namely, provided that (δp¯) = 0. Thus, a stationary (maximum or minimum∇ or saddle point) point of σ occurs iff for every small change δp¯, which is orthogonal to ¯ g0 (namely, δp¯ ¯ g0 = 0) one has 0 = δσ = ¯ σ δp¯. As can be seen from∇ Eq. (1.12), this· condition∇ is fulfilled only when ∇¯ σ· = 0, ∇ namely only when the vectors ¯ σ and ¯ g are parallel to each other. In⊥ other 0 words, only when ∇ ∇

¯ σ = ξ ¯ g , (1.13) ∇ 0∇ 0

where ξ0 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 = ξ0 . (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 1 p = p = = . (1.15) 1 2 ··· M 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 δp¯ σ (¯p + δp¯) = exp δp¯ ¯ σ (¯p) · ∇ 2 δp¯ ¯ = 1 + δp¯ ¯ + · ∇ + σ (¯p) 2! · ∇ ··· 2 δp¯ ¯ = σ (¯p) + δp¯ ¯ σ + · ∇ σ + 2! · ∇ ··· ∂σ 1 ∂2σ = σ (¯p) + δpm + δpmδpm′ + ∂pm 2 ∂pm∂pm′ ··· m m,m′ (1.16) Using Eq. (1.2) one finds that

∂2σ 1 = δm,m′ . (1.17) ∂pm∂pm′ −pm

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

1 1 1 Λ (M) = σ , , , = log M, (1.18) M M ··· M 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 (p , p , p ) = p log p p log p p log p . (1.19) 1 2 3 − 1 1 − 2 2 − 3 3 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 e , it can either occupy− state e with probability p / (1 p ) or occupy 1 2 2 − 1 state e3 with probability p3/ (1 p1). Assuming that uncertainty (entropy) is additive, the total uncertainty− (entropy) is given by p p σ = σ (p , 1 p ) + (1 p ) σ 2 , 3 . (1.20) i 1 − 1 − 1 1 p 1 p − 1 − 1 The factor (1 p ) in the second term is included since the uncertainty asso- − 1 ciated 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

σ = p log p (1 p ) log (1 p ) i − 1 1 − − 1 − 1 p p p p + (1 p ) 2 log 2 3 log 3 − 1 −1 p 1 p − 1 p 1 p − 1 − 1 − 1 − 1 = p log p p log p p log p − 1 1 − 2 2 − 3 3 (1 p p p ) log (1 p ) − − 1 − 2 − 3 − 1 = σ (p1, p2, p3) , (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 p3 − − 1 p1 1 p1

e2 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 e , e , , e with probabilities q , q , , q respectively. This set of { 1 2 ··· M0 } 1 2 ··· M0 states is grouped as follows. The first group includes the first M1 states e , e , , e ; the second group includes the next M states e , e , , e , { 1 2 ··· M1 } 2 { M1+1 M1+2 ··· M1+M2 } etc., where M1 +M2 + = M0. The probability that one of the states in the first group is occupied is···p = q +q + +q , the probability that one of the 1 1 2 ··· M1 states in the second group is occupied is p2 = qM1+1 + qM1+2 + +qM1+M2 , etc., where ···

p + p + = 1 . (1.23) 1 2 ··· The composition property requires that the following holds [see Fig. 1.2]

σ (q , q , , q ) = σ (p , p , ) 1 2 ··· M0 1 2 ··· q q q + p σ 1 , 2 , , M1 1 p p ··· p 1 1 1 q q q + p σ M1+1 , M1+2 , , M1+M2 + 2 p p ··· p ··· 2 2 2 (1.24)

Using the deﬁnition (1.2) the following holds

σ (p , p , ) = p log p p log p , (1.25) 1 2 ··· − 1 1 − 2 2 − · · ·

Eyal Buks Thermodynamics and Statistical Physics 6 1.1. Entropy in Information Theory

p p 1 2 …

e ,e ,..., e e + ,..., e + 1 2 M1 M1 1 M1 M 2

q q qM q q + qM +M 1 2 1 M1 M1 2 1 2 p p p p p p 1 1 … 1 2 2 … 2 e e eM e + e + e + 1 2 1 M1 1 M1 2 M1 M 2 = + + + p q q ... q p = q + +... + q + 1 1 2 M1 2 M1 1 M1 M 2

Fig. 1.2. The composition property - the general case.

q q q p σ 1 , 2 , , M1 1 p p ··· p 1 1 1 q q q q q q = p 1 log 1 2 log 2 M1 log M1 1 −p p − p p − · · · − p p 1 1 1 1 1 1 = q log q q log q q log q − 1 1 − 2 2 − · · · − M1 M1 + p1 log p1 , (1.26)

q q q p σ M1+1 , M1+2 , , M1+M2 2 p p ··· p 2 2 2 = q log q q log q − M1+1 M1+1 − M1+2 M1+2 q log q − · · · − M1+M2 M1+M2 + p2 log p2 , (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 (p , p , , p ) is a continuous function of its arguments p , p , , p . 1 2 ··· N 1 2 ··· N 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 p , p , , p using rational numbers, namely p = M /M , p = M /M , 1 2 ··· N 1 1 0 2 2 0 etc., where M1,M2, are integers and M0 = M1 + M2 + + MN . Using the composition property··· (1.24) one ﬁnds ···

Λ (M ) = σ (p , p , , p ) + p Λ (M ) + p Λ (M ) + (1.28) 0 1 2 ··· N 1 1 2 2 ···

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) − = . (1.31) Kx K x Taking the limit x 0 yields → dΛ C = , (1.32) dK K where Λ (1 + x) C = lim . (1.33) x 0 → x Integrating Eq. (1.32) and using the initial condition (1.30) yields

Λ (K) = C log K. (1.34)

Eyal Buks Thermodynamics and Statistical Physics 8 1.2. Largest Uncertainty Estimator

Moreover, the second property requires that C > 0. Choosing C = 1 and using Eq. (1.28) yields

σ (p , p , , p ) = Λ (M ) p Λ (M ) p Λ (M ) 1 2 ··· N 0 − 1 1 − 2 2 − · · · M1 M2 MN = p1 log p2 log pM log − M0 − M0 − · · · − M0 = p log p p log p p log p , − 1 1 − 2 2 − · · · − N N (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 p . Let X ···(l = 1, 2, ,L) be a set of variables characterizing the system m l ··· (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

X = p X (m) , (1.36) l m l m where l = 1, 2, ,L. However, the probability distribution p is not given. ··· { m} 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 = g (¯p) = p X (m) X , (1.37) l m l − l m where l = 1, 2, L. To ﬁrst order one has ··· δσ = ¯ σ δp¯ , (1.38a) ∇ · δg = ¯ g δp¯ , (1.38b) l ∇ l ·

Eyal Buks Thermodynamics and Statistical Physics 9 Chapter 1. The Principle of Largest Uncertainty where l = 0, 1, 2, L. A stationary point of σ occurs iﬀ for every small ··· change δp¯, which is orthogonal to all vectors ¯ g0, ¯ g1, ¯ g2, , ¯ gL one has ∇ ∇ ∇ ··· ∇

0 = δσ = ¯ σ δp¯ . (1.39) ∇ · This condition is fulﬁlled only when the vector ¯ σ belongs to the subspace ∇ spanned by the vectors ¯ g0, ¯ g1, ¯ g2, , ¯ gL [see also the discussion below Eq. (1.12) above]. In∇ other∇ words,∇ only··· ∇ when ¯ σ = ξ ¯ g + ξ ¯ g + ξ ¯ g + + ξ ¯ g , (1.40) ∇ 0∇ 0 1∇ 1 2∇ 2 ··· L∇ L

where the numbers ξ0, ξ1, , ξL, 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 p 1 = ξ + ξ X (m) . (1.41) − m − 0 l l l=1 From Eq. (1.41) one obtains

L p = exp ( 1 ξ ) exp ξ X (m) . (1.42) m − − 0 − l l l=1

The Lagrange multipliers ξ0, ξ1, , ξL can be determined from Eqs. (1.5) and (1.37) ···

L 1 = p = exp ( 1 ξ ) exp ξ X (m) , (1.43) m − − 0 − l l m m l=1

Xl = pmXl (m) m L = exp ( 1 ξ ) exp ξ X (m) X (m) . − − 0 − l l l m l=1 (1.44)

Using Eqs. (1.42) and (1.43) one ﬁnds

L exp ξ X (m) − l l p = l=1 . (1.45) m L exp ξ X (m) − l l m l=1 In terms of the partition function , which is deﬁned as Z

Eyal Buks Thermodynamics and Statistical Physics 10 1.2. Largest Uncertainty Estimator

L = exp ξ X (m) , (1.46) Z − l l m l=1 one ﬁnds

1 L p = exp ξ X (m) . (1.47) m − l l l=1 Z 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 X can be expressed as l

Xl = pmXl (m) m 1 L = exp ξ Xl (m) Xl (m) m − l l=1 Z 1 ∂ = Z − ∂ξ Z l ∂ log = Z . − ∂ξl (1.48)

2 Similarly, Xl can be expressed as

2 2 Xl = pmXl (m) m 1 L = exp ξ X (m) X2 (m) − l l l m l=1 Z 1 ∂2 = Z . ∂ξ2 Z l (1.49)

Using Eqs. (1.48) and (1.49) one ﬁnds that the variance of the variable Xl is given by

2 2 2 2 1 ∂ 1 ∂ (∆Xl) = (Xl Xl ) = Z2 Z . (1.50) − ∂ξl − ∂ξl Z Z However, using the following identity

Eyal Buks Thermodynamics and Statistical Physics 11 Chapter 1. The Principle of Largest Uncertainty

∂2 log ∂ 1 ∂ 1 ∂2 1 ∂ 2 Z = Z = Z Z , (1.51) ∂ξ2 ∂ξ ∂ξ ∂ξ2 − ∂ξ l l Z l Z l Z l one finds 2 2 ∂ log (∆Xl) = 2 Z . (1.52) ∂ξl Note that the above results Eqs. (1.48) and (1.52) are valid only when is expressed as a function of the the Lagrange multipliers, namely Z = (ξ , ξ , , ξ ) . (1.53) Z Z 1 2 ··· L Using the definition of entropy (1.2) and Eq. (1.47) one finds

σ = pm log pm − m 1 L = p log exp ξ X (m) − m − l l m l=1 Z L = p log + ξ X (m) m Z l l m l=1 L = log + ξ p X (m) , Z l m l l=1 m (1.54) thus L σ = log + ξ X . (1.55) Z l l 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 ( X , X , , X ) . (1.56) 1 2 ··· L Using Eq. (1.55) one has L L ∂σ ∂ log ∂ξl′ ∂ Xl′ = Z + Xl′ + ξl′ ∂ Xl ∂ Xl ∂ Xl ∂ Xl l′=1 l′=1 L ∂ log ∂ξl′ = Z + Xl′ + ξl ∂ Xl ∂ Xl l′=1 L L ∂ log ∂ξl′ ∂ξl′ = Z + Xl′ + ξl , ∂ξ ′ ∂ Xl ∂ Xl l′=1 l l′=1 (1.57)

Eyal Buks Thermodynamics and Statistical Physics 12 1.2. Largest Uncertainty Estimator thus using Eq. (1.48) one ﬁnds ∂σ = ξ . (1.58) ∂ X l l 1.2.2 The Free Entropy

The free entropy σ is deﬁned as the term log in Eq. (1.54) F Z σ = log F Z L = σ ξ p X (m) − l m l l=1 m L = p log p ξ p X (m) . − m m − l m l m l=1 m (1.59) The free entropy is commonly expressed as a function of the Lagrange multipliers

σ = σ (ξ , ξ , , ξ ) . (1.60) F F 1 2 ··· L We have seen above that the LUE maximizes σ for given values of expectation values X , X , , X . We show below that a similar result can be 1 2 ··· L 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 ξ , ξ , , ξ . 1 2 ··· L Proof. As before, the normalization condition is expressed as

0 = g0 (¯p) = pm 1 . (1.61) m − At a stationary point of σF, as we have seen previously, the following holds ¯ σ = η ¯ g , (1.62) ∇ F ∇ 0 where η is a Lagrange multiplier. Thus

L (log p + 1) ξ X (m) = η , (1.63) − m − l l l=1 or

L p = exp ( η 1) exp ξ X (m) . (1.64) m − − − l l l=1 Eyal Buks Thermodynamics and Statistical Physics 13 Chapter 1. The Principle of Largest Uncertainty

Table 1.1. The microcanonical, canonical and grandcanonical distributions.

energy numberofparticles 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 σF corresponds to the LUE probability distribution. Since 2 ∂ σF 1 = δm,m′ < 0 , (1.65) ∂pm∂pm′ −pm 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

p = p = = 1/M , (1.66) 1 2 ··· 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 1 p = exp ( βU (m)) , (1.68) m − Zc where β is the Lagrange multiplier associated with the given expectation value U , and the partition function is given by = exp ( βU (m)) . (1.69) Zc − m The term exp ( βU (m)) is called Boltzmann factor. Moreover, Eq.− (1.48) yields ∂ log U = Zc , (1.70) − ∂β Eq. (1.52) yields

∂2 log (∆U)2 = Zc , (1.71) ∂β2 and Eq. (1.55) yields

σ = log + β U . (1.72) Zc Using Eq. (1.58) one can expressed the Lagrange multiplier β as ∂σ β = . (1.73a) ∂U The temperature τ = 1/β is deﬁned as 1 = β . (1.74) τ 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 βε βε = exp + exp = 2 cosh , (1.75) Zc 2 − 2 2 thus using Eqs. (1.70) and (1.71) one ﬁnds

ε βε U = tanh , (1.76) −2 2 and ε 2 1 (∆U)2 = , (1.77) 2 cosh2 βε 2 where β = 1/τ.

1 2x 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 1 p = exp ( βU (m) ηN (m)) , (1.78) m − − Zgc 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

gc = exp ( βU (m) ηN (m)) . (1.79) Z m − − 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

∂ log U = Zgc , (1.80) − ∂β η ∂ log N = Zgc (1.81) − ∂η β Eq. (1.52) yields

∂2 log (∆U)2 = Zgc , (1.82) ∂β2 η 2 2 ∂ log gc (∆N) = 2Z , (1.83) ∂η β and Eq. (1.55) yields

σ = log + β U + η N . (1.84) Zgc

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 + βU , (1.85) Zc and the one in Eq. (1.84) as

σ = log + βU + ηN . (1.86) Zgc Using Eq. (1.58) one can expressed the Lagrange multipliers β and η as

∂σ β = , (1.87) ∂U N ∂σ η = . (1.88) ∂N U The chemical potential µ is deﬁned as

µ = τη . (1.89) − In the deﬁnition (1.2) the entropy σ is dimensionless. Historically, the entropy was deﬁned as

S = kBσ , (1.90)

where

Eyal Buks Thermodynamics and Statistical Physics 17 Chapter 1. The Principle of Largest Uncertainty

23 1 k = 1.38 10− JK− (1.91) B × is the Boltzmann constant. Moreover, the historical deﬁnition of the temperature is τ T = . (1.92) 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 ∂ log U = Zgc + τµ Zgc , (1.93) − ∂β ∂µ µ β ∂ log N = λ Zgc , (1.94) ∂λ where λ is the fugacity , which is deﬁned by

η λ = exp (βµ) = e− . (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 . (1.97) dt ≥ Moreover, equality holds iff p = p for all pairs of states for which Γ = 0. s r sr To prove this theorem we express the rate of change in the probability ps in terms of these transition rates dp r = p Γ p Γ . (1.98) dt s sr r rs s − 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 dp r = Γ (p p ) . (1.99) dt sr s r s − Eyal Buks Thermodynamics and Statistical Physics 18 1.5. Thermal Equilibrium

The last result and the deﬁnition (1.2) allows calculating the rate of change of entropy dσ d = p log p dt dt r r − r dp = r (log p + 1) − dt r r = Γsr (ps pr) (log pr + 1) . − r s − (1.100)

One the other hand, using Eq. (1.96) and exchanging the summation indices allow rewriting the last result as dσ = Γ (p p ) (log p + 1) . (1.101) dt sr s r s r s − Thus, using both expressions (1.100) and (1.101) yields dσ 1 = Γ (p p ) (log p log p ) . (1.102) dt 2 sr s r s r r s − − In general, since log x is a monotonic increasing function

(p p ) (log p log p ) 0 , (1.103) s − r s − r ≥

and equality holds iff ps = pr. Thus, in general dσ 0 , (1.104) dt ≥ 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 = σ1 (U1,N1) and σ2 = 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 = σ1 + 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

∂σ ∂σ δσ = 1 δU + 2 δU − ∂U1 ∂U2 N1 N2 ∂σ ∂σ 1 δN + 2 δN − ∂N1 ∂N2 U1 U2 1 1 µ µ = + δU 1 + 2 δN . −τ τ − − τ τ 1 2 1 2 (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

τ 1 = τ 2 (1.108)

and

µ1 = µ2 . (1.109)

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 , (1.111)

Eyal Buks 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)]

σ = σ βU , (1.114) F,c − whereas for the grandcanonical case it is given by [see Eq. (1.86)]

σ = σ βU ηN . (1.115) F,gc − − We deﬁne below two corresponding free energies, the canonical free energy (known also as the Helmholtz free energy )

F = τσ = U τσ , (1.116) − F,c − and the grandcanonical free energy

G = τσ = U τσ + τηN = U τσ µN . − F,gc − − −

In section 1.2.2 above we have shown that the LUE maximizes σF for given values of the Lagrange multipliers ξ1, ξ2, , ξL. 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 grand- • canonical 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 (M states) given Xl where expectation U U , N l = 1, 2, ..., L values Z = Zc = Zgc = partition L − ξlXl(m) e−βU(m) e−βU(m)−ηN(m) function e l=1 m m m pm = pm = pm = p L p = 1 m − ξ X (m) m M 1 −βU(m) 1 −βU(m)−ηN(m) 1 l l e e e l=1 Zc Zgc Z ∂ log Zgc U = − ∂β ∂ log Z ∂ log Zc η Xl Xl = − U = − ∂ξl ∂β ∂ log Zgc N = − ∂η β 2 2 ∂ log Zgc 2 2 (∆U) = ∂β2 2 2 ∂ log Z 2 ∂ log Zc η (∆Xl) (∆Xl) = ∂ξ2 (∆U) = ∂β2 ∂2 log Z l 2 gc (∆N) = ∂η2 β σ = σ = σ = σ L σ = log M log Z + ξl Xl log Zc + β U log Zgc + β U + η N l=1 Lagrange β = ∂σ ξ = ∂σ β = ∂σ ∂U N l ∂Xl ∂U ∂σ multipliers {Xn}n= l η = ∂N U max min / max min F (τ) min G (τ, µ) σF (ξ1, ξ2, ..., ξL) max σ principle L F = U − τσ G = U − τσ − µN σF = σ − ξl Xl 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 22 1.7. Problems Set 1

b) What is the probability of surviving (N 1) turns in this game and then being shot the Nth 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

N! n N n W (n) = p (1 p) − . (1.117) n!(N n)! − − 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. N n Np a) Using the result ln(1 p) p, show that (1 p) − e− . b) Show that N!/(N n−)! ≃N −n. − ≃ c) Hence show that Eq.− (1.117)≃ reduces to

n λ λ W (n) = 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 approxima tion, 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→ probab ∞ ility 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 1 N! = (2πN)1/2 N N exp N + + (1.119) − 2N ··· to show that 1 M 2 f (M) = exp . (1.120) m√2πN −2m2N 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) . (1.121) − − b) Show that the variance (X X )2 is given by − (X X )2 = 4a2Npq . (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) = . (1.123) π√a2 x2 − 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:

∂y ∂z ∂x = z . (1.124) ∂y ∂x y − ∂z x

Eyal Buks Thermodynamics and Statistical Physics 25 Chapter 1. The Principle of Largest Uncertainty

b) Show that:

∂z ∂z ∂z ∂y = + . (1.125) ∂x ∂x ∂y ∂x w y x w 18. Let be a grandcanonical partition function. Zgc a) Show that:

∂ log ∂ log U = Zgc + τµ Zgc . (1.126) − ∂β ∂µ µ β where τ is the temperature, β = 1/τ, and µ is the chemical potential. b) Show that:

∂ log N = λ Zgc , (1.127) ∂λ 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 = , (1.129) 1 2 U 2 tanh− − Nε 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

1 ε = ℏω n + , (1.130) n 2 where n = 0, 1, 2, is integer. ··· a) Show that the average energy of the system is given by

Nℏω βℏω U = coth , (1.131) 2 2 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 N ℏω (∆U)2 = 2 . (1.132) sinh 2 βℏω 2 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

2Nε sinh (βε) U = , (1.133) − 1 + 2 cosh βε where β = 1/τ. b) Show the variance of the energy of the system is given by

cosh (βε) + 2 (U U )2 = 2Nε2 . (1.134) − [1 + 2 cosh (βε)]2 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 Nα, 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

a F a α = 1 + tanh . (1.135) 2 2τ b) Show that in the limit of high temperature the spring constant is given approximately by 4τ k . (1.136) ≃ Na2

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 ωa for a link of length a and ωb for a link of length b. Thus, the energy of a link is ℏ 1 ωa n + 2 for l = a En,l = 1 , (1.137) ℏωb n + for l = b 2 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 aωb + bωa F ωbωa (a b) 2 L = N + N −2 β + O β , (1.138) ωb + ωa (ωb + ωa) where β = 1/τ. 24. The elasticity of a rubber band can be described in terms of a one- dimensional model of N polymer molecules linked together end-to-end. The angle between successive links is equally likely to be 0◦ or 180◦. The length of 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

τ 1 L f = tanh− . (1.139) d Nd 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 2 N = , (1.140) 2 + exp [β (ε µ)] − 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 n n 1 and n n +1. 2 → 2− 1 → 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,N ,N 1. A B ≫ 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

1 ε = ℏω n + , (1.141) n 2 where n = 0, 1, 2, is integer. The total energy E of the system is given by ···

N E = ℏω m + , (1.142) 2 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: N! 1 N a) N N 2 ( 2 )!( 2 )! b) 0 2. Final answers: 5 N a) 6 N 1 5 − 1 b) 6 6 c) In general

∞ N 1 d ∞ N d 1 1 Nx − = x = = , dx dx 1 x 2 N=0 N=0 (1 x) − − thus

N 1 1 ∞ 5 − 1 1 N¯ = N = = 6 . 6 6 6 5 2 N=0 1 − 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 n = , 1 2 N m n = − , 2 2 one ﬁnds

N! N+m N−m 2 2 W (m) = N+m N m p q . 2 ! −2 ! It is convenient to employ the moment generating function, deﬁned as

φ (t) = etm .

In general, the following holds

∞ tk φ (t) = mk , k! k=0 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 φ (t) = W (m) etm m= N − N N! N+m N−m tm = p 2 q 2 e , N+m ! N m ! m= N 2 −2 − or using the summation variable N + m n = , 1 2 one has N N! n N n t(2n N) φ (t) = p 1 q − 1 e 1− n !(N n )! n =0 1 1 1 − N tN N! 2t n1 N n = e− pe q − 1 n1!(N n1)! n1=0 − tN 2t N = e− pe + q . Using p = q = 1/2

et + e t N φ (t) = − = (cosh t)N . 2 Thus using the expansion 1 1 (cosh t)N = 1 + Nt2 + [N + 3N (N 1)] t4 + O t5 , 2! 4! − one ﬁnds m = 0 , m2 = N, 3 m = 0 , m4 = N (3N 2) . − 4. Using the binomial distribution

N! n N n W (n) = p (1 p) − n!(N n)! − − N (N 1) (N 1) (N n + 1) n N n = − − × · · · − p (1 p) − n! − (Np)n = exp ( Np) . ∼ n! −

Eyal Buks Thermodynamics and Statistical Physics 31 Chapter 1. The Principle of Largest Uncertainty

5. ∞ ∞ λ λn a) W (n) = e− n! = 1 n=0 n=0 b) As in Ex. 1-6, it is convenient to use the moment generating function

n tn t n tn ∞ tn λ ∞ λ e λ ∞ (λe ) t φ (t) = e = e W (n) = e− = e− = exp λ e 1 . n! n! − n=0 n=0 n=0 Using the expansion 1 exp λ et 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.

N 2 N R2 = r = r2 + r r = Nl2 n n n · m n=1 n=1 n=m =l2 =0 7. 20 20! n 20 n 4 0.2 0.8 − = 5.6 10− . (1.144) n! (20 n)! × × n=11 − 8. Let σAn = 1 (σBn = 1) if object A (B) makes a move to the right (left) at step n, and σAn = 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 (x , y ) = (S , m S ) , (1.145) Am Am Am − Am (x , y ) = (N S ,N m + S ) , (1.146) Bm Bm − Bm − Bm where m SAm = σAn , (1.147) n=1 m SBm = σBn . (1.148) n=1 Eyal Buks Thermodynamics and Statistical Physics 32 1.8. Solutions Set 1

A meeting occurs if for some m S = N S , (1.149) Am − Bm m S = N m + S , (1.150) − Am − Bm i.e. if

S + S = N = 2m N. (1.151) Am Bm − Thus, a meeting is possible only after m = N steps, and it occurs if SAm + SBm = N. Therefore, the probability is given by

2N N (2N)! pN = = . (1.152) 22N N 2 (N!2 ) With the help of the Stirling’s formula (1.119) one ﬁnds that 1 lim pN = . (1.153) N √ →∞ Nπ 9. The following holds

p0 = 0 , 5 1 p = p + p , 1 6 0 6 2 4 2 p = p + p , 2 6 1 6 3 3 3 p = p + p , 3 6 2 6 4 2 4 p = p + p , 4 6 3 6 5 1 5 p = p + p , 5 6 4 6 6 p6 = 1 , and thus 1 3 1 13 31 p = , p = , p = , p = , p = . 1 32 2 16 3 2 4 16 5 32 10. For the case of a round table (and N > 2)

N 2 (N 2)! 2 p = × × − = , (1.154) C N! N 1 − and for the case of a row (2 + (N 2) 2) (N 2)! 2 p = − × × − = . (1.155) R N! N

Eyal Buks Thermodynamics and Statistical Physics 33 Chapter 1. The Principle of Largest Uncertainty

11. Let the binary representation of p be given by

m ∞ 1 p = σ , (1.156) m 2 m=1 where σ 0, 1 . Let Σ be a sequence of random variables generated m ∈ { } m 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 σm = Σm the value of m is increased by 1, i.e. m m + 1, and the loop continues. If σ = Σ the program stops and the→ value 1 is m m returned if σm > Σm and the value 0 is returned if σm < 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, (1.157a) − M N+ N = , (1.157b) − − m one has N M N = 1 + , (1.158a) + 2 mN N M N = 1 , (1.158b) − 2 − mN or N N = (1 + x) , (1.159a) + 2 N N = (1 x) , (1.159b) − 2 − where M x = . mN The number of states having total magnetization M is given by N! N! Ω (M) = = N N . (1.160) N+!N ! (1 + x) ! (1 x) ! − 2 2 − Since all states have equal probability one has Ω (M) f (M) = . (1.161) 2N

Eyal Buks Thermodynamics and Statistical Physics 34 1.8. Solutions Set 1

Taking the natural logarithm of Stirling’s formula one ﬁnds 1 log N! = N log N N + O , (1.162) − N 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 ≪ 1 + x log 1 x2 + x log = x2 + O x3 , (1.164) − 1 x − thus M 2 f (M) = A exp , (1.165) −2m2N where A is a normalization constant, which is determined by requiring that

∞ 1 = f (M) dM. (1.166) −∞ Using the identity

∞ π exp ay2 dy = , (1.167) − a −∞

Eyal Buks Thermodynamics and Statistical Physics 35 Chapter 1. The Principle of Largest Uncertainty

one ﬁnds

1 ∞ M 2 = exp dM = m√2πN , (1.168) A −2m2N −∞ thus 1 M 2 f (M) = exp , (1.169) m√2πN −2m2N The expectation value is giving by

∞ M = Mf (M) dM = 0 , (1.170) −∞ and the variance is given by

∞ (M M )2 = M 2 = M 2f (M) dM = m2N. (1.171) − −∞ 13. The probability to have n steps to the right is given by

N! n N n W (n) = p q − . (1.172) n!(N n)! − a)

N N!n n N n n = p q − (1.173) n!(N n)! n=0 − N ∂ N! n N n = p p q − ∂p n!(N n)! n=0 − ∂ N N 1 = p (p + q) = pN (p + q) − = 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 2 2 N!n n N n n = p q − n!(N n)! n=0 − N N N!n (n 1) n N n N!n n N n = − p q − + p q − n!(N n)! n!(N n)! n=0 n=0 − − 2 N 2 ∂ N! n N n = p p q − + n ∂p2 n!(N n)! n=0 − ∂2 = p2 (p + q)N + n = p2N (N 1) + pN . ∂p2 − Thus (n n )2 = p2N (N 1) + pN p2N 2 = Npq , (1.176) − − − and (X X )2 = 4a2Npq . (1.177) − 14. The total energy is given by kx2 mx˙ 2 ka2 E = + = , (1.178) 2 2 2 where a is the amplitude of oscillations. The time period T is given by a dx m a dx m T = 2 = 2 = 2π , (1.179) 2 2 a x˙ k a √a x k − − − thus 2 1 f(x) = = . (1.180) T x˙ π√a2 x2 | | − 15. The six experiments are independent, thus 2 2 1 1 σ = 6 ln ln = 3.8191 . × −3 3 − 3 3 n 16. The random variable X obtains the value n with probability pn = q , where n = 1, 2, 3, , and q = 1/2. ··· a) The entropy is given by

∞ ∞ ∞ σ = p log p = qn log qn = nqn . − n 2 n − 2 n=1 n=1 n=1 This can be rewritten as

∂ ∞ ∂ 1 q σ = q qn = q 1 = = 2 . ∂q ∂q 1 q − 2 n=1 (1 q) − −

Eyal Buks 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)

∂z ∂z δz = δx + δy . (1.181) ∂x ∂y y x For a process for which z is a constant δz = 0, thus

∂z ∂z 0 = (δx) + (δy) . (1.182) ∂x z ∂y z y x

Dividing by (δx)z yields ∂z ∂z (δy) = z ∂x − ∂y (δx) y x z ∂z ∂y = − ∂y ∂x x z ∂y ∂x = z . − ∂y ∂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 ∂z (δz) = (δx) + (δy) . (1.184) w ∂x w ∂y w y x

Dividing by (δx)w yields (δz) ∂z ∂z (δy) w = + w . (1.185) (δx) ∂x ∂y (δx) w y x w or ∂z ∂z ∂z ∂y = + . (1.186) ∂x ∂x ∂y ∂x w y x w

Eyal Buks Thermodynamics and Statistical Physics 38 1.8. Solutions Set 1

18. We have found in class that ∂ log U = Zgc , (1.187) − ∂β η ∂ log N = Zgc . (1.188) − ∂η β a) Using relation (1.125) one has

∂ log U = Zgc − ∂β η ∂ log ∂ log ∂µ = Zgc Zgc − ∂β − ∂µ ∂β µ β η ∂ log η ∂ log = Zgc Zgc − ∂β − β2 ∂µ µ β ∂ log ∂ log = Zgc + τµ Zgc . − ∂β ∂µ µ β (1.189)

b) Using Eq. (1.188) one has

∂ log N = Zgc − ∂η β ∂µ ∂ log = Zgc − ∂η ∂µ β β ∂ log = τ Zgc , ∂µ β (1.190)

or in terms of the fugacity λ, which is deﬁned by

η λ = exp (βµ) = e− , (1.191)

one has ∂ log N = τ Zgc ∂µ β ∂λ ∂ log = τ Zgc ∂µ ∂λ ∂ log = λ Zgc . ∂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

= N , (1.193) Zc Z1 where βε βε βε = exp + exp = 2 cosh . (1.194) Z1 2 − 2 2 Thus ∂ log ∂ log Nε βε U = Zc = N Z1 = tanh , (1.195) − ∂β − ∂β − 2 2 and ε τ = . (1.196) 1 2 U 2 tanh− − Nε 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

= N , (1.197) Zc Z1 where

βℏω ∞ = exp exp ( βℏωn) (1.198) Z1 − 2 − n=0 βℏω exp 2 1 = − = . 1 exp ( βℏω) 2 sinh βℏω − − 2 a) ∂ log ∂ log Nℏω βℏω U = Zc = N Z1 = coth (1.199) − ∂β − ∂β 2 2 b)

2 ∂2 log ∂2 log N ℏω (∆U)2 = Zc = N Z1 = 2 (1.200) ∂β2 ∂β2 sinh 2 βℏω 2 21. The canonical partition function is given by

= [exp (βε) + 1 + exp ( βε)]N = [1 + 2 cosh (βε)]N , (1.201) Zc − where β = 1/τ.

Eyal Buks Thermodynamics and Statistical Physics 40 1.8. Solutions Set 1

a) Thus the average energy is

∂ log 2Nε sinh (βε) U = Zc = , (1.202) − ∂β − 1 + 2 cosh βε b) and the variance is

∂2 log ∂ U cosh (βε) + 2 (U U )2 = Zc = = 2Nε2 . − ∂β2 − ∂β [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

a exp (Faβ) a F aβ α = = 1 + tanh , (1.204) 1 + exp (F aβ) 2 2 where β = 1/τ. b) At high temperature F aβ 1 the length of the chain L = Nα is given by ≪

Na F aβ Na F aβ L = 1 + tanh 1 + , (1.205) 2 2 ≃ 2 2 or Na F = k L , (1.206) − 2 where the spring constant k is given by 4τ k = . (1.207) Na2 23. The average length of a single link is given by

∞ ∞ a exp (βF a) exp βℏω n + 1 + b exp (βF b) exp βℏω n + 1 − a 2 − b 2 n=0 n=0 l = ∞ ∞ exp (βF a) exp βℏω n + 1 + exp (βF b) exp βℏω n + 1 − a 2 − b 2 n=0 n=0 (1.208) a exp(βF a) + b exp(βF b) 2 sinh βℏ ωa 2 sinh βℏ ωb = 2 2 . exp(βF a) + exp(βF b) 2 sinh βℏ ωa βℏ ωb 2 2 sinh 2

Eyal Buks Thermodynamics and Statistical Physics 41 Chapter 1. The Principle of Largest Uncertainty

To ﬁrst order in β

2 aωb + bωa F ωbωa (a b) 2 l = + −2 β + O β . (1.209) ωb + ωa (ωb + ωa) 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

βfd βfd e e− l = d βfd − βfd = d tanh (βfd) , e + e− where β = 1/τ. Thus

L = Nd tanh (βfd) ,

τ 1 L f = tanh− . d Nd 25. The grand partition function is given by Zgc

= 1 + 2 exp [β (µ ε)] , (1.210) Zgc − thus 1 ∂ log 2 N = Zgc = . (1.211) β ∂µ 2 + exp [β (ε µ)] − 26. a) N! N! (∆σ)2LS = log log (n2 1)! (n1 + 1)! − n2!n1! n − n = log 2 log 2 . n1 + 1 ≃ n1 (1.212)

b) E E (∆σ) = 2 − 1 (1.213) R τ

Eyal Buks Thermodynamics and Statistical Physics 42 1.8. Solutions Set 1

c) For a small change near thermal equilibrium one expects (∆σ)2LS + (∆σ)R = 0, thus n E E 2 = exp 2 − 1 . (1.214) n − τ 1 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

N! 2 σ = log 2 [N log N N log N (N N ) log (N N )] . N !(N N )! ≃ − B B − − B − B B − B (1.215)

b) The energy of the system is given by U = NBε. Thus, the Helmholtz free energy is given by

U U U U F = U τσ = U 2τ N log N log N log N . − − − ε ε − − ε − ε (1.216)

At thermal equilibrium (∂F/∂U)τ = 0, thus ∂F 2τ U U 0 = = 1 + log log N , (1.217) ∂U ε ε − − ε τ or N N ε − B = exp , (1.218) NB 2τ therefore N NB = ε . 1 + exp 2τ Alternatively, one can calculate the chemical potential from the requirement N N 1 = A + B , (1.219) N N where N exp (βµ) A = , (1.220a) N 1 + exp (βµ) N exp (βµ βε) B = − , (1.220b) N 1 + exp (βµ βε) −

Eyal Buks Thermodynamics and Statistical Physics 43 Chapter 1. The Principle of Largest Uncertainty

which is satisﬁed when ε µ = , (1.221) 2 thus N NB = ε . (1.222) 1 + exp 2τ 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 (N 1 + m)! g (N, m) = − . (1.223) (N 1)!m! − b) The entropy is given by

(N 1 + m)! σ = log − (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 σ = N + log N + ℏω − 2 ℏω − 2 E N E N N log N log . − − ℏω − 2 ℏω − 2 (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 1 ln 2ℏω ln 2ℏω 1 2E+Nℏω 2E Nℏω = − + − − ℏω ℏω 1 2E + Nℏω = ln . ℏω 2E Nℏω − (1.226)

In the thermodynamical limit (N 1, m 1) the energy E and its average U are indistinguishable, thus≫ ≫

ℏω 2U + Nℏω exp = , (1.227) τ 2U Nℏω − or Nℏω ℏω U = coth . (1.228) 2 2τ 29. The grand canonical partition function is given by

ζ = 1 + 2λ exp (βε) , (1.229)

where λ = exp (βµ) is the fugacity, thus

∂ log ζ 2λeβε 1 N = λ = = . (1.230) d βε 1 β(ε+µ) ∂λ 1 + 2λe 1 + 2 e−

Eyal Buks Thermodynamics and Statistical Physics 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 0 0 x, y, z L V (x, y, z) = . (2.1) ≤ else ≤ ∞ The quantum eigenstates and eigenenergies are determined by requiring that the wavefunction ψ (x, y, z) satisﬁes the Schrödinger equation 2 ∂2ψ ∂2ψ ∂2ψ + + + V ψ = Eψ . (2.2) −2M ∂x2 ∂y2 ∂z2 In addition, we require that the wavefunction ψ vanishes on the surfaces of the box. The normalized solutions are given by

2 3/2 n πx n πy n πz ψ (x, y, z) = sin x sin y sin z , (2.3) nx,ny,nz L L L L where

n , n , n = 1, 2, 3, (2.4) x y z ··· Chapter 2. Ideal Gas

The corresponding eigenenergies are given by

2 π 2 ε = n2 + n2 + n2 . (2.5) nx,ny,nz 2M L x y z 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

∞ ∞ ∞ εn ,n ,n Z = exp x y z 1 − τ n =1 n =1 n =1 x y z ∞ ∞ ∞ = exp α2 n2 + n2 + n2 , − x y z nx=1 ny=1 nz=1 (2.6) where 2π2 α2 = . (2.7) 2ML2τ The following relation can be employed to estimate the dimensionless parameter α

17 7.9 10− α2 = × , (2.8) M L 2 τ mp cm 300 K 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 α2n2 exp α2n2 dn . (2.9) − x ≃ − x x nx=1 0

By changing the integration variable x = αnx one ﬁnds

∞ 1 ∞ √π exp α2n2 dn = exp x2 dx = , (2.10) − x x α − 2α 0 0 thus

√π 3 ML2τ 3/2 Z = = = n V, (2.11) 1 2α 2π2 Q where we have introduced the quantum density

Eyal Buks Thermodynamics and Statistical Physics 48 2.1. A Particle in a Box

Mτ 3/2 n = . (2.12) Q 2π2 The partition function (2.11) together with Eq. (1.70) allows evaluating the average energy (recall that β = 1/τ) ∂ log Z ε = 1 − ∂β

2 3/2 ∂ log ML 2π2β = − ∂β 3/2 ∂ log β− = − ∂β 3 ∂ log β = 2 ∂β 3τ = . 2 (2.13) This result can be written as τ ε = d , (2.14) 2 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 Z (∆ε)2 = 1 ∂β2 ∂ ε = − ∂β ∂ 3 = −∂β 2β 3 = 2β2 3τ 2 = . 2 (2.15) Thus, using Eq. (2.13) the standard deviation is given by

2 (∆ε)2 = ε . (2.16) 3 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

h λ = . (2.17) Mv For a particle having energy equals to the average energy ε = 3τ/2 one has Mv2 3τ = , (2.18) 2 2

thus in this case the de-Broglie wavelength, which is denoted as λT (the thermal wavelength)

h λT = , (2.19) √3Mτ and therefore one has (recall that = h/2π)

3/2 3 π 1 2π n = = . (2.20) Q h2 λ 3 2Mτ T 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 ? N c = Z1 = exp ( βεn) . (2.22) Z 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 =? log ZN = N log (n V ) , (2.23) − Zc 1 Q or in terms of the gas density N n = , (2.24) V we find n σ βU =? N log N Q . (2.25) − n 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 , (2.26)

V = V1 + V2 . (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 N N n = = 1 = 2 . (2.28) V V1 V2 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 σ2 − − n n n =? N log N Q N log N Q N log N Q n − 1 1 n − 2 2 n = N log N N log N N log N . − 1 1 − 2 2 (2.29) Using the Stirling’s formula (1.162)

log N! N log N N, (2.30) ≃ −

Eyal Buks Thermodynamics and Statistical Physics 51 Chapter 2. Ideal Gas

2 … 3 1

orbital 1 orbital 2 orbital 3 orbital 4

N1=0 N2=1 N3=2 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 N! ∆σ log > 0 . (2.31) ≃ N1!N2! 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)? Ex- panding this partition function yields a sum of terms each having the form exp β N ε , where N is the number of particles occupying orbital n. − n n n n Let g (N1,N2, ) be the number of terms in such an expansion associated with a given set··· of occupation numbers N ,N , . Since the partition { 1 2 ···} 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 ··· N! g (N ,N , ) = . (2.32) 1 2 ··· N !N ! 1 2 × · · · 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 N ,N , . { 1 2 ···}

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 oc-

Eyal 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: N can be any integer. • n 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 f (ε) = N (ε) , is found using Eq. (1.94) FD ∂ log ζ f (ε) = λ FD ∂λ λ exp ( βε) = − 1 + λ exp ( βε) − 1 = . exp [β (ε µ)] + 1 − (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 ( Nβε) − 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 f (ε) = λ BE ∂λ exp ( βε) = λ − 1 λ exp ( βε) − − 1 = . exp [β (ε µ)] 1 − − (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

f (ε) f (ε) exp [β (µ ε)] 1 , (2.39) FD ≃ BE ≃ − ≪ and

ζ 1 + λ exp ( βε) . (2.40) ≃ − 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, Bose- Einstein 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 1 Fermions 1 + λ exp (−βε) exp[β(ε−µ)]+1 1 1 Bosons 1−λ exp(−βε) exp[β(ε−µ)]−1 classical limit 1 + λ exp (−βε) exp [β (µ − ε)]

1.5

0.5

0 -2 2 4 6 8energy 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 ζn of orbital n having energy εn is given by Eq. (2.40) above. The grandcanonical partition function of the entire system is Zgc found by multiplying ζn of all orbitals

= (1 + λ exp ( βε )) . (2.41) Zgc − n 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 = log (1 + λ exp ( βε )) Zgc − n n λ exp ( βεn) ≃ n − = λZ1 , (2.44) where Mτ 3/2 Z = V (2.45) 1 2π2 [see Eq. (2.11)] is the single particle partition function. In terms of the La- grange multipliers η = µ/τ and β = 1/τ the last result can be rewritten as − 3/2 η M log = e− V . (2.46) Zgc 2π2β The average energy and average number of particle are calculated using Eqs. (1.80) and (1.81) respectively ∂ log 3 U = Zgc = log , (2.47) − ∂β 2β Zgc η ∂ log N = Zgc = log . (2.48) − ∂η Zgc β 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 3Nτ U = . (2.49) 2 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 + βU + ηN Zgc 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 µβ = log , (2.51) nQ where n = N/V is the density. This allows expressing the entropy as 5 n σ = N + log Q . (2.52) 2 n Using the definition (1.116) and Eqs. (2.49) and (2.52) one finds that the Helmholtz free energy is given by n F = Nτ log 1 . (2.53) n − Q 2.4.1 Pressure The pressure p is defined by ∂F p = . (2.54) − ∂V τ,N Using Eq. (2.53) and keeping in mind that n = N/V one finds Nτ p = . (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 p momentum in the x direction− from the particle to the wall. | x| 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 { } px 2 px | | = | | × 2mL L2 p2 = x . mV (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)

3τ p2 + p2 + p2 = ε = x y z , (2.57) 2 2m thus one ﬁnds that p2 x = τ . (2.58) m 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

Nτ p = . (2.59) V

2.4.2 Useful Relations

In this section we derive some useful relations between thermodynamical quantities.

Claim. The following holds

∂U p = − ∂V σ,N Proof. Using the deﬁnition (2.54) and recalling that F = U τσ one ﬁnds − ∂F ∂U ∂ (τσ) p = = . (2.60) − ∂V ∂V − ∂V τ,N τ,N τ,N Using identity (1.125), which is given by

∂z ∂z ∂z ∂y = + , (2.61) ∂x ∂x ∂y ∂x w y x w one ﬁnds ∂U ∂U ∂U ∂σ = + , (2.62) ∂V ∂V ∂σ ∂V τ,N σ,N V,N τ,N τ thus ∂U ∂σ ∂σ ∂U p = + τ τ = . (2.63) − ∂V ∂V − ∂V ∂V σ,N τ,N τ,N σ,N

Eyal Buks Thermodynamics and Statistical Physics 58 2.4. Ideal Gas in the Classical Limit

In a similar way the following relations can be obtained ∂σ ∂U ∂F p = τ = = , (2.64) ∂V − ∂V − ∂V U,N σ,N τ,N ∂σ ∂U ∂F µ = τ = = . (2.65) − ∂N ∂N ∂N U,V σ,V τ,V Another useful relation is given below. Claim. The following holds ∂σ ∂p = . (2.66) ∂V ∂τ τ V Proof. See problem 2 of set 2.

2.4.3 Heat Capacity

The heat capacity at constant volume is defined by ∂σ c = τ , (2.67) V ∂τ V whereas the heat capacity at constant pressure is defined by ∂σ c = τ . (2.68) p ∂τ p 3/2 Using Eq. (2.52) and recalling that nQ ∝ τ one finds 3N 3N c = τ = . (2.69) V 2τ 2 Claim. The following holds 5N c = . (2.70) p 2 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 = (1 + λ exp ( βε ) exp ( βE )) , (2.71) FD,n − n − l 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 = λm exp ( βmε ) exp ( βmE ) . (2.72) BE,n − n − l l m=0 In the classical limit the average occupation of an orbital is close to zero. In this limit, namely when

λ exp ( βε ) 1 , (2.73) − n ≪ [see Eq. (2.38)] the following holds

ζ ζ ζ , (2.74) FD,n ≃ BE,n ≃ n where

ζ = 1 + λ exp ( βε ) Z , (2.75) n − n int and where

Z = exp ( βE ) , (2.76) int − l 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 f = λ n n ∂λ λZ exp ( βε ) = int − n 1 + λZ exp ( βε ) int − n λZ exp ( βε ) . ≃ int − n (2.77) The total grandcanonical partition function is given by

gc = ζn , (2.78) Z n thus

Eyal Buks Thermodynamics and Statistical Physics 60 2.4. Ideal Gas in the Classical Limit

log = log ζ Zgc n n = log [1 + λZint exp ( βεn)] n − λZint exp ( βεn) ≃ n − = λZintZ1 , (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− ≪ − 3/2 η M log = e− Z V . (2.80) Zgc int 2π2β This result together with Eqs. (1.80) and (1.81) yield ∂ log 3 U = Zgc = log + E log , (2.81) − ∂β 2β Zgc l Zgc η ∂ log N = Zgc = log , (2.82) − ∂η Zgc β where

El exp ( βEl) l − ∂ log Zint El = = . (2.83) exp ( βEl) − ∂β l − Claim. The following hold

n µ = τ log log Z , (2.84) n − int Q 3τ ∂ log Z U = N int , (2.85) 2 − ∂β n F = Nτ log log Z 1 , (2.86) n − int − Q 5 n ∂ (τ log Z ) σ = N + log Q + int , (2.87) 2 n ∂τ 3 ∂2 (τ log Z ) c = N + τ int , (2.88) V 2 ∂τ 2 cp = cV + N. (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 ∂σ ∂σ dσ = dU + dV + dN. (2.90) ∂U ∂V ∂N V,N U,N U,V Using Eqs. (1.87), (2.64) and (2.65) 1 p µ dσ = dU + dV dN, (2.91) τ τ − τ or

dU = τdσ pdV + µdN. (2.92) − This relation expresses the change in the energy of the system dU in terms of dQ = τdσ 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 isobaric

c i i r sot o her h m

c al

s i i s en tr op ic 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

∂U c = . (2.95) V ∂τ 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 . (2.96)

Eyal Buks 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

V2 dV = Nτ V V1 V = Nτ log 2 . V1 (2.97)

2.5.2 Isobaric Process

Integrating the relation dW = pdV for this case where the pressure is constant yields

V2 W = pdV = p (V V ) . (2.98) 2 − 1 V1

The change in energy ∆U can be found by integrating Eq. (2.96)

τ 2

∆U = cV dτ . (2.99)

τ1

The heat added to the system Q can be found using Eq. (2.94) Q = W + ∆U τ 2 = p (V V ) + c dτ . 2 − 1 V τ1 (2.100)

Note that if the temperature dependence of cV can be ignored to a good approximation one has

∆U = c (τ τ ) . (2.101) V 2 − 1

Eyal Buks Thermodynamics and Statistical Physics 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

τ 2

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 = c (τ τ ) . V 2 − 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 − c dτ = pdV, (2.103) V − or using Eq. (2.55)

dτ dV c = N . (2.104) V τ − V This relation can be rewritten using Eq. (2.89) as

dτ dV = (1 γ) . (2.105) τ − V where c γ = p . (2.106) cV 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

1 γ τ V − log 2 = log 2 . (2.107) τ V 1 1 Thus

γ 1 γ 1 τ 1V1 − = τ 2V2 − , (2.108) or using Eq. (2.55)

γ γ p1V1 = p2V2 . (2.109)

Eyal Buks Thermodynamics and Statistical Physics 65 Chapter 2. Ideal Gas

W Q Q τ h l τ h l

Fig. 2.3. Carnot heat engine.

γ 1 γ In other words both quantities τV − and pV remain unchanged during this process. Using the last result allows integrating the relation dW = pdV

V2 ∆U = W = pdV − V1

V2 γ γ = p1V1 V − dV

V1 γ+1 γ+1 V1− V2− = p V γ − 1 1 γ 1 − p V p V = 2 2 − 1 1 1 γ − N (τ τ ) = 2 − 1 1 γ − = c (τ τ ) . − V 2 − 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. Isothermal expansion at temperature τ h (a b) 2. Isentropic expansion from temperature τ to→τ (b c) h l → 3. Isothermal compression at temperature τ l (c d) 4. Isentropic compression from temperature τ to→τ (d a) l h → 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 τ h b

pd d c p τ c l V

Va VbVd 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 → Vb Qh = Nτ h log , (2.111) Va and the heat extracted from the cold thermal reservoir Ql at temperature τ l during step 3 (c d) is given by → Vd Ql = Nτ l log , (2.112) Vc where V is the volume at point n a, b, c, d . Note that Q > 0 since n ∈ { } h 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 = τdσ). The total work done by the system per cycle is given by

W = Wab + Wcd + Wbc + Wda Vb Vd = Nτ h log + Nτ l log Va Vc N (τ τ ) N (τ τ ) + l − h + h − l 1 γ 1 γ − − V V = N τ log b + τ log d , h V l V a c (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 W Q η = = 1 + l . (2.115) Qh Qh Using Eqs. (2.111) and (2.113) one finds

Vd τ l log η = 1 + Vc . (2.116) τ log Vb h Va Employing Eq. (2.108) for both isentropic processes yields γ 1 γ 1 τ hVb − = τ lVc − , (2.117) γ 1 γ 1 τ hVa − = τ lVd − , (2.118) thus by dividing these equations one ﬁnds

γ 1 γ 1 Vb − Vc − γ 1 = γ 1 , (2.119) Va − Vd − or V V b = c . (2.120) Va Vd

Using this result one ﬁnds that the eﬃciency of Carnot heat engine ηC is given by

τ l ηC = 1 . (2.121) − τ h

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.

Q ∆σ = , (2.123) − τ thus since Q = W (energy conservation) we ﬁnd that

W 0 . (2.124) τ ≤ 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. Ac- cording 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)

τ τ h l

-Q0 Q0 M

(b) W=Q-Q0

Q -Q0 M τ τ h l

-Q0 Q0 M

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

τ l ηC = 1 . (2.125) − τ h 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 W ηI = , (2.126) Qh′ 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)

τ τ h l W= Q 0 Q 0 M

(b)

-Q0- Q Q M τ τ h l W= Q 0 Q 0 M

Fig. 2.8. The assumption that Kelvin’s principle does not hold.

Q’h Q’l I τ τ h W l Q Q h C l

Fig. 2.9. Limit imposed upon engine eﬃciency.

W τ l ηC = = 1 . (2.127) Qh − τ h For the combined system, the Clausius’s principle requires that

Q′ Q > 0 . (2.128) h − h thus τ l ηI ηC = 1 . ≤ − τ h

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 C = τ . (2.129) p ∂τ p

Calculate Cp of an classical ideal gas having no internal degrees of freedom. 2. Show that ∂σ ∂p = , (2.130) ∂V ∂τ τ 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

n µ = τ log log Z , (2.131) n − int Q

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:

3τ ∂ log Z U = N int , (2.132) 2 − ∂β where β = bl/τ. c) Show that the Helmholtz free energy is given by

n F = Nτ log log Z 1 . (2.133) n − int − Q d) Show that the entropy is given by

5 n ∂ (τ log Z ) σ = N + log Q + int . (2.134) 2 n ∂τ

Eyal Buks Thermodynamics and Statistical Physics 73 Chapter 2. Ideal Gas

e) Show that the heat capacity at constant volume is given by

3 ∂2 (τ log Z ) c = N + τ int . (2.135) V 2 ∂τ 2 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 ∂σ c = τ , (2.137) ∂τ where τ is the temperature. Show that

(∆U)2 c = , (2.138) τ 2 where U is the energy of the body and where ∆U = U U . 5. Consider an ideal classical gas made of diatomic molecule−s. 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

1 ε = n + ω , (2.139) n 2 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

n λ λ p = e− , (2.140) n n! where λ = Nv/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 = µ0H and ε+ = µ0H, where the mag- − − netic 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 M = , (2.141) − ∂H τ where F is the Helmholtz free energy. b) Calculate the heat capacity

∂σ C = τ . (2.142) ∂τ 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 = ∆ µ0H, − − − ε0 = 0 , ε = ∆ + µ H, 1 − 0 where the magnetic moment µ0 is a constant. The system is in thermal equilibrium at temperature τ. Calculate the magnetic susceptibility M χ = lim , (2.143) H 0 H → where ∂F M = , (2.144) − ∂H 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 ∂F M = , (2.145) − ∂H τ is given by Nµ µH µH M = (2J + 1) coth (2J + 1) coth . (2.146) 2 2τ − 2τ 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

λ + 2λ2e βε N = N − , (2.147) a 0 2 βε 1 + λ + λ e− where β = 1/τ and

3/2 M − 5/2 λ = τ − p . (2.148) 2πℏ2 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

2 ε 3 ε g1g2 exp τ cV = N + − , (2.149) ε 2 2 τ g1 + g2 exp − τ 2 ε 5 ε g1g2 exp τ cp = N + − . (2.150) ε 2 2 τ g1 + g2 exp − τ 14. A classical gas is described by the following equation of state

p (V b) = Nτ , (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 a p + (V b) = Nτ , (2.152) V 2 − 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 a p + (V b) = Nτ , (2.153) V 2 − 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 V b V V W = Nτ log 2 − a 2 − 1 , (2.154) 0 V b − V V 1 − 2 1 V b Q = ∆U + W = Nτ log 2 − . (2.155) 0 V b 1 − 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

τ τ ∆σ = C a 1 log a . (2.156) τ − − τ b b 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 V , the second one b c is → 2 →

Eyal Buks Thermodynamics and Statistical Physics 77 Chapter 2. Ideal Gas p b p1

p c 2 a V V2 V1

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

p (V V ) η = 1 γ 2 1 − 2 , (2.157) − V (p p ) 2 1 − 2

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 Q γ = l . (2.158) W Show that the second law of thermodynamics imposes an upper bound on γ τ γ l . (2.159) ≤ τ τ h − l 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

Aτ 3 p = , (2.160) V 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 V U = Bτ n log + f (τ) , (2.161) V0

where B and V0 are constants, f (τ) only depends on the temperature. Find B and n. 23. 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. 24. 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. 25. 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. 26. 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 c as a function of N, M, V and τ. 27. Consider an engine working in aZ 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. 28. 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 V and c d at volume V , joined by → 1 → 2

Eyal Buks Thermodynamics and Statistical Physics 79 Chapter 2. Ideal Gas p a b p1

p 2 d c V

Fig. 2.11.

c a

d V

V1 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 σ . ﬁnal − initial

2.9 Solutions Set 2

1. The entropy is given by Mτ 3/2 V 5 σ = N log + , (2.162) 2πℏ2 N 2 or using pV = Nτ M 3/2 τ 5/2 5 σ = N log + , (2.163) 2πℏ2 p 2 thus ∂σ 5 C = τ = N. (2.164) p ∂τ 2 p 2. Since ∂2F ∂2F = , (2.165) ∂V ∂τ ∂τ∂V where F is Helmholtz free energy, one has ∂ ∂F ∂ ∂F = . (2.166) ∂V ∂τ ∂τ ∂V V τ τ V By deﬁnition ∂F = p . ∂V − τ Moreover, using F = U τσ one ﬁnds − ∂F ∂U ∂σ = τ σ ∂τ ∂τ − ∂τ − V V V ∂U ∂U ∂σ = σ ∂τ − ∂σ ∂τ − V V V = σ , − (2.167)

Eyal Buks Thermodynamics and Statistical Physics 81 Chapter 2. Ideal Gas

thus ∂σ ∂p = . (2.168) ∂V ∂τ τ V 3. We have found in class the following relations

Mτ 3/2 n = , (2.169) Q 2π2 µ η = , (2.170) − τ η log = e− Z V n , (2.171) Zgc int Q 3τ ∂ log Z U = int log , (2.172) 2 − ∂β Zgc N = log . (2.173) Zgc a) Using Eqs. (2.171) and (2.173) one ﬁnds n µ log = , (2.174) nQZint τ thus n µ = τ log log Z . (2.175) n − int Q b) Using Eqs. (2.172) and (2.173)

3τ ∂ log Z U = N int . (2.176) 2 − ∂β c) Using the relations

F = U τσ , (2.177) − σ = log + βU + ηN , (2.178) Zgc one ﬁnds

F = U τσ (2.179) − = Nτ ( η 1) (2.180) − − µ = Nτ 1 (2.181) τ − n = Nτ log log Z 1 . (2.182) n − int − Q (2.183)

Eyal Buks Thermodynamics and Statistical Physics 82 2.9. Solutions Set 2

d) Using the relation

∂F σ = , (2.184) − ∂τ V one ﬁnds ∂F σ = − ∂τ V ∂ τ log n nQ ∂ (τ log Z ) = N + int + 1 −  ∂τ  ∂τ  V     ∂ log n nQ n ∂ (τ log Z ) = N τ log + int + 1 −  ∂τ  − nQ ∂τ  V 3  n ∂ (τlog Z )  = N log + int + 1 2 − n ∂τ Q 5 n ∂ (τ log Z ) = N + log Q + int . 2 n ∂τ (2.185)

e) By deﬁnition

∂σ c = τ V ∂τ V 3 ∂2 (τ log Z ) = N + τ int . 2 ∂τ 2 (2.186)

f) The following holds

∂σ c = τ (2.187) p ∂τ p ∂σ ∂σ ∂V = τ + τ ∂τ ∂V ∂τ V τ p ∂σ ∂V = c + τ . V ∂V ∂τ τ p

Using V p = Nτ and Eq. (2.185) one ﬁnds N N c = c + τ = c + N. (2.188) p V V p V

Eyal Buks Thermodynamics and Statistical Physics 83 Chapter 2. Ideal Gas

4. With the help of Eqs. (1.87), (1.70) and (1.71) together with the following relation ∂ 1 ∂ = , (2.189) ∂τ −τ 2 ∂β one ﬁnds that

2 ∂σ ∂U 1 ∂U (∆U) c = τ = = = . (2.190) ∂τ ∂τ −τ 2 ∂β τ 2 5. The internal partition function is given by 1 τ Z = , (2.191) int ω 2 sinh 2τ ≃ ω thus using Eqs. (2.186) and (2.188)

3 ∂2 τ log τ 5N c = N + τ ω = , (2.192) V 2 ∂τ 2 2 7N c = . (2.193) p 2 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 n 5 σ (N, n) = N log Q + n 2 thus the change in entropy is given by ∆σ = σ σ σ mix − A − B N N N N = σ N , A + σ N , B σ N , A σ N , B A V + V B V + V − A V − B V A B A B A B VA + VB VA + VB = NA log + NB log VA VB 7. The probability to ﬁnd a molecule in the volume v is given by p = v/V . Thus, pn is given by

N! n N n p = p (1 p) − . (2.194) n n!(N n)! − − Using the solution of problem 4 of set 1 n λ λ p = e− , n n! where λ = Nv/V .

Eyal Buks Thermodynamics and Statistical Physics 84 2.9. Solutions Set 2

8. The partition function of a single atom is given by

Z = exp (µ Hβ) + exp ( µ Hβ) = 2 cosh (µ Hβ) , 1 0 − 0 0 where β = 1/τ, thus the partition function of the entire system is

N Z = (2 cosh (µ0Hβ)) ,

a) The free energy is given by

F = τ log Z = Nτ log (2 cosh (µ Hβ)) . (2.195) − − 0 The magnetization is given by

∂F M = = Nµ tanh (µ Hβ) . (2.196) − ∂H 0 0 τ b) The energy U is given by

∂ log Z U = = Nµ H tanh (µ Hβ) , (2.197) − ∂β − 0 0 thus ∂σ C = τ ∂τ H ∂U = ∂τ H ∂ tanh µ0H = Nµ H τ − 0 ∂τ H 2 µ H 1 = N 0 . τ µ0H cosh τ (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

H2 τ 2 = τ 1 . (2.200) H1 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 ( βε ) − m m= 1 − = 1 + 2 exp (β∆) cosh (βµ0H) , (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 M = − ∂H τ 2Nµ exp (β∆) sinh (βµ H) = 0 0 . 1 + 2 exp (β∆) cosh (βµ0H) (2.203) and the magnetic susceptibility is given by Nµ2 χ = 0 , (2.204) τ 1 + 1 exp ( β∆) 2 − 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

µHβ 1 µHβ µHβ J sinh Z = exp exp exp (mµHβ) 2 2 2 − − 2 m= J − 1 1 1 = exp J + µHβ exp J + µHβ , 2 2 − − 2 (2.205)

thus sinh J + 1 µHβ Z = 2 . (2.206) µHβ sinh 2 a) The free energy is given by

sinh J + 1 µHβ F = Nτ log Z = Nτ log 2 . (2.207) − −  µHβ  sinh 2   Eyal Buks Thermodynamics and Statistical Physics 86 2.9. Solutions Set 2

b) The magnetization is given by

∂F Nµ µH µH M = = (2J + 1) coth (2J + 1) coth . − ∂H 2 2τ − 2τ τ (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 mgz n (z) = n (0) exp . (2.210) −k T B The condition n (z) = 0.5 n (0) yields × 23 1 k T ln 2 1.3806568 10 JK− 300 K ln 2 z = B = × − × × = 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

Mτ 3/2 F = τN log V + τN log N τN , (2.212) − 2πℏ2 − thus the chemical potential is

∂F Mτ 3/2 µ = = τ log V + τ log N, (2.213) ∂N − 2πℏ2 τ,V and the pressure is ∂F Nτ p = = . (2.214) − ∂V V τ,V Using these results the fugacity λ = exp (βµ) can be expressed in terms of p

3/2 3/2 βµ Mτ − N M − 5/2 λ = e = = τ − p . (2.215) 2πℏ2 V 2πℏ2 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

βµ β(2µ ε) = 1 + e + e − , (2.216) Z or in terms of the fugacity λ = exp (βµ)

2 βε = 1 + λ + λ e− . (2.217) Z Thus ∂ log λ + 2λ2e βε N = N λ Z = N − , (2.218) a 0 0 2 βε ∂λ 1 + λ + λ e− where λ is given by Eq. (2.215). 13. The internal partition function is given by ε Z = g + g exp . (2.219) int 1 2 −τ Using Eq. (2.135)

3 ∂2 c = N + Nτ (τ log Z ) V 2 ∂τ 2 int V ε 3 ε 2 g1g2 exp = N + − τ . ε 2 2 τ g1 + g2 exp − τ (2.220) Using Eq. (2.136)

2 ε 5 ε g1g2 exp τ cp = N + − . (2.221) ε 2 2 τ g1 + g2 exp − τ 14. Using Maxwell’s relation

∂σ ∂p = , (2.222) ∂V ∂τ τ,N V,N and the equation of state one ﬁnds that

∂σ N = . (2.223) ∂V V b τ,N − a) Using the deﬁnitions

∂σ c = τ , (2.224) V ∂τ V,N ∂σ c = τ , (2.225) p ∂τ p,N

Eyal Buks Thermodynamics and Statistical Physics 88 2.9. Solutions Set 2

and the general identity

∂z ∂z ∂z ∂y = + , (2.226) ∂x ∂x ∂y ∂x α y x α one ﬁnds ∂σ ∂V c c = τ , (2.227) p − V ∂V ∂τ τ,N p,N or with the help of Eq. (2.223) and the equation of state

Nτ c c = N = N. (2.228) p − V p (V b) − b) Using the identity

∂U ∂U ∂U ∂σ = + , (2.229) ∂V ∂V ∂σ ∂V τ,N σ,N V,N τ,N together with Eq. (2.223) yields

∂U Nτ = p + = 0 . (2.230) ∂V − V b τ,N − 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 c is independent on temperature one has − V W = ∆U = c (τ τ ) . (2.232) − − V 2 − 1 15. Using the deﬁnitions

∂σ c = τ , (2.233) V ∂τ V,N ∂σ c = τ , (2.234) p ∂τ p,N and the general identity

∂z ∂z ∂z ∂y = + , (2.235) ∂x ∂x ∂y ∂x α y x α

Eyal Buks Thermodynamics and Statistical Physics 89 Chapter 2. Ideal Gas

one ﬁnds ∂σ ∂V c c = τ . (2.236) p − V ∂V ∂τ τ,N p,N Using Maxwell’s relation ∂σ ∂p = , (2.237) ∂V ∂τ τ,N V,N and the equation of state a p + (V b) = Nτ , (2.238) V 2 − one ﬁnds ∂p ∂V c c = τ p − V ∂τ ∂τ V,N p,N N τ (V b) − = aV +2ab+pV 3 − NV 3 N = 2aV +2ab , 1 + − V 3 p+ a ( V 2 ) (2.239) or N cp cV = 2 . (2.240) − 2a(1 b ) 1 − V − V Nτ 16. The work W is given by

V2 W = pdV . (2.241) V1 Using a p + (V b) = Nτ , (2.242) V 2 − one ﬁnds V2 Nτ a V b V V W = 0 dV = Nτ log 2 − a 2 − 1 . (2.243) V b − V 2 0 V b − V V V1 − 1 − 2 1 Using the identity

∂U ∂U ∂U ∂σ ∂σ = + = p + τ , ∂V ∂V ∂σ ∂V − ∂V τ,N σ,N V,N τ,N τ,N

Eyal Buks Thermodynamics and Statistical Physics 90 2.9. Solutions Set 2

(2.244)

and Maxwell’s relation ∂σ ∂p = , (2.245) ∂V ∂τ τ,N V,N one ﬁnds ∂U ∂p = τ p . (2.246) ∂V ∂τ − τ,N V,N In the present case

∂U Nτ a = p = , (2.247) ∂V V b − V 2 τ,N − thus

V2 ∂U V2 dV V V ∆U = dV = a = a 2 − 1 , (2.248) ∂V V 2 V V V1 τ,N V1 2 1 and V b Q = ∆U + W = Nτ log 2 − . (2.249) 0 V b 1 − 17. In general the following holds

1 ∂n En = Zgc , (2.250) − ∂βn Zgc η and ∂ log U = Zgc . (2.251) − ∂β η Thus the variance is given by

(∆E)2 = E2 E 2 − 2 2 1 ∂ gc 1 ∂ gc = Z2 2 Z gc ∂β − ∂β Z η Zgc η ∂2 log = Zgc . ∂β2 η (2.252)

Furthermore, the following holds:

Eyal Buks Thermodynamics and Statistical Physics 91 Chapter 2. Ideal Gas

(∆E)3 = E3 3E2U + 3EU 2 U 3 − − = E3 3U E2 + 2U 3 − 3 2 3 1 ∂ gc 3 ∂ gc ∂ gc 2 ∂ gc = Z3 2 Z Z2 + 3 Z − gc ∂β − ∂β ∂β ∂β Z η Zgc η η Zgc η 2 2 ∂ 1 ∂ gc 1 ∂ gc = Z2 2 Z −∂β gc ∂β − ∂β Z η Zgc η ∂3 log = Zgc . − ∂β3 η (2.253) For classical gas having no internal degrees of freedom one has

3/2 η M N = log = e− V , (2.254) Zgc 2π2β thus ∂ log 3Nτ U = Zgc = . (2.255) − ∂β 2 η a) Using Eq. (2.252) ∂ 3N 3 N 2U 2 (∆E)2 = = = . (2.256) −∂β 2β 2 β2 3N b) Using Eq. (2.253) ∂ 3N 3N 8U 3 (∆E)3 = = = . (2.257) −∂β 2β2 β3 9N 2 18. The entropy change of the body is τ b dτ τ ∆σ = C = C log b , (2.258) 1 τ τ τ a a and that of the bath is

∆Q C (τ a τ b) ∆σ2 = = − , (2.259) τ b τ b thus τ τ ∆σ = C a 1 log a . (2.260) τ − − τ b b 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 2.9. Solutions Set 2

19. The eﬃciency is given by

Q Q C (τ τ ) p (V V ) η = 1+ l = 1+ ca = 1+ p a − c = 1 γ 2 1 − 2 , (2.261) Q Q C (τ τ ) − V (p p ) h ab v b − a 2 1 − 2

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 Qh′ from the hot bath per cycle and delivered Ql′ heat to the cold one per cycle, where W = Ql′ + Qh′ and −

W τ l ηc = = 1 . (2.262) Qh′ − τ h According to Clausius principle

Q + Q′ 0 , (2.263) l l ≤ thus Q Q Q W τ τ γ = l l′ = h′ − = h 1 = l . (2.264) W ≤ −W W τ τ − τ τ h − l h − l 21. Using Eq. (2.264)

A (τ τ ) τ h − l = l , (2.265) P τ τ h − l thus P τ 2 2τ τ + + τ 2 = 0 , (2.266) l − l h 2A h or

2 P P 2 τ l = τ h + τ h + τ . (2.267) 2A ± 2A − h The solution for which τ τ is l ≤ h P P 2 τ = τ + τ + τ 2 . (2.268) l h 2A − h 2A − h 22. Using Eq. (2.244)

∂U ∂p = τ p , (2.269) ∂V ∂τ − τ,N V,N

Eyal Buks Thermodynamics and Statistical Physics 93 Chapter 2. Ideal Gas

thus Bτ n 3Aτ 3 2Aτ 3 = p = , (2.270) V V − V therefore

B = 2A, (2.271) n = 3 . (2.272)

23. In general the following holds

3 ∂2 (τ log Z ) C = N + τ int , (2.273) V 2 ∂τ 2 Cp = cv + N, (2.274)

where in our case ∆ Z = 1 + 3 exp , (2.275) int − τ thus

a) CV is given by

2 ∆ ∆ τ 3 3 τ e− CV = N + 2 , (2.276) 2 ∆  1 + 3e− τ     b) and Cp is given by

2 ∆ ∆ τ 5 3 τ e− Cp = N + 2 , (2.277) 2 ∆  1 + 3e− τ     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

dQ dQ dτ dτ 0 = dσ = dσ + dσ = 1 + 2 = C 1 + 2 . (2.278) 1 2 τ τ τ τ 1 2 1 2 a) Thus, by integration the equation

dτ dτ 1 = 2 , (2.279) τ 1 − τ 2 one ﬁnds

Eyal Buks Thermodynamics and Statistical Physics 94 2.9. Solutions Set 2

τ f dτ τ f dτ 1 = 2 , (2.280) τ − τ τ 1 1 τ 2 2 or τ τ log f = log 2 , (2.281) τ 1 τ f thus

τ f = √τ 1τ 2. (2.282) b) Employing energy conservation law yields

W = ∆U + ∆U = C (τ τ ) + C (τ τ ) = C (√τ √τ )2 . 1 2 1 − f 2 − f 1 − 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 n 5 σ (N, n) = N log Q + , (2.284) n 2 where Mτ 3/2 n = , (2.285) Q 2π2 N n = . (2.286) V Using the relation

pV = Nτ , (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

2p1p2 pﬁnal = . p1 + p2 Thus, the change in entropy is given by

∆σ = σ σ σ ﬁnal − 1 − 2 (p + p ) τn 5 τn 5 τn 5 = 2N log 1 2 Q + N log Q + N log Q + 2p p 2 − p 2 − p 2 1 2 1 2 (p + p )2 = N log 1 2 . 4p1p2 (2.288)

Eyal Buks Thermodynamics and Statistical Physics 95 Chapter 2. Ideal Gas

26. Using

σ = log + βU = log + βU + ηN Zc Zgc one ﬁnds

log = log + ηN . (2.289) Zc Zgc The following holds for classical ideal gas having no internal degrees of freedom

log = N Zgc n V η = βµ = log Q , − N (2.290)

where

Mτ 3/2 n = , (2.291) Q 2π2 thus n V log = N 1 + log Q Zc N = N log (n V ) + N N log N Q − N log (n V ) log N! ≃ Q − (n V )N = log Q , N! (2.292)

N 1 Mτ 3/2 = V . (2.293) Zc N! 2π2

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 → → Q C (τ τ ) η = 1 + l = 1 + p d − c . (2.294) Q C (τ τ ) h p b − a

Eyal Buks Thermodynamics and Statistical Physics 96 2.9. Solutions Set 2

Using pV = Nτ yields

τ τ p (V V ) η = 1 + d − c = 1 + 2 d − c . (2.295) τ τ p (V V ) b − a 1 b − a γ Along the isentropic process pV is constant, where γ = Cp/Cv, thus

1 γ p1 γ−1 (Va Vb) γ p p2 p η = 1 + 2 − = 1 2 . (2.296) p V V − p 1 b − a 1 28. No heat is exchanged in the isentropic processes, thus the eﬃciency is given by Q η = 1 + l Qh Qc d = 1 + → Qa b → c (τ τ ) = 1 + V d − c . c (τ τ ) V b − a (2.297) γ 1 Since τV − remains unchanged in an isentropic process, where c γ = p , (2.298) cV one ﬁnds that γ 1 γ 1 τ bV1 − = τ cV2 − , (2.299) γ 1 γ 1 τ dV2 − = τ aV1 − , (2.300) or

1 γ τ τ V − c = d = 2 , (2.301) τ τ V b a 1 thus

1 γ V − η = 1 2 . (2.302) − V 1

29. Let VA1 = Nτ A/p (VB1 = Nτ 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 Nτ V = V = f . (2.303) A2 B2 p

Eyal Buks Thermodynamics and Statistical Physics 97 Chapter 2. Ideal Gas

The entropy of an ideal gas of density n = N/V , which contains N particles, is given by n 5 σ = N log Q + , (2.304) n 2 where Mτ 3/2 n = , (2.305) Q 2π2 or as a function of τ and p

M 3/2 5/2 2 τ 5 σ = N log 2π + . (2.306) p 2 Thus the change in entropy is given by ∆σ = σ σ ﬁnal − initial 5N τ 2 = log f . 2 τ Aτ B (2.307) In general, for an isobaric process the following holds Q = W + ∆U = p (V V ) + c (τ τ ) , (2.308) 2 − 1 V 2 − 1 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 = Nτ this can be written as Q = (N + c )(τ τ ) . (2.309) V 2 − 1 Since no heat is exchanged with the environment during this process the following holds

QA + QB = 0 , where Q = (N + c )(τ τ ) , (2.310) A V f − A Q = (N + c )(τ τ ) , (2.311) B V f − B thus τ + τ τ = A B , (2.312) f 2 and therefore 5N (τ + τ )2 ∆σ = log A B . (2.313) 2 4τ Aτ B

Eyal Buks Thermodynamics and Statistical Physics 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 ∂E H = ǫ , (3.1) ∇ × 0 ∂t ∂H E = µ , (3.2) ∇ × − 0 ∂t E = 0 , (3.3) ∇ · and

H = 0 , (3.4) ∇ · 12 1 6 2 where ǫ0 = 8.85 10− F m− and µ0 = 1.26 10− NA− are the permittivity and permeability× respectively of free space,× and the following holds 1 ǫ µ = , (3.5) 0 0 c2 8 1 where c = 2.99 10 m s− 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 ∂A E = , (3.7) − ∂t and

µ H = A . (3.8) 0 ∇ × 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 ∂ ( A) ∂H E = ∇ × = µ , (3.9) ∇ × − ∂t − 0 ∂t ∂ ( A) E = ∇ · = 0 , (3.10) ∇ · − ∂t 1 H = ( A) = 0 , (3.11) ∇ · µ0 ∇ · ∇ × where in the last equation the general vector identity ( A) = 0 has been employed. Substituting Eqs. (3.7) and (3.8) into∇ the · ∇ on ×ly remaining nontrivial equation, namely into Eq. (3.1), leads to

1 ∂2A ( A) = . (3.12) ∇ × ∇ × −c2 ∂t2 Using the vector identity

( A) = ( A) 2A , (3.13) ∇ × ∇ × ∇ ∇ · − ∇ and the gauge condition (3.6) one ﬁnds that

1 ∂2A 2A = . (3.14) ∇ c2 ∂t2 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

1 d2q q 2u = u . (3.17) ∇ c2 dt2

Eyal Buks Thermodynamics and Statistical Physics 100 3.1. Electromagnetic Radiation

Multiplying by an arbitrary unit vector ˆn leads to

2u ˆn 1 d2q ∇ · = . (3.18) u ˆn c2q dt2 · 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 − 2u+κ2u = 0 , (3.19) ∇ and d2q +ω2 q = 0 , (3.20) dt2 κ where

ωκ = cκ . (3.21)

Equation (3.19) should be solved with the boundary conditions of a per- fectly conductive surface. Namely, on the surface S enclosing the cavity we have H ˆs = 0 and E ˆs = 0, where ˆs 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 = ˆs (u ˆs) 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 q (H ˆs) dS = [( u) ˆs] dS ′ · µ ′ ∇ × · S 0 S q = u dl , µ · 0 C (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

(H ˆs) dS = 0 . (3.23) ′ · S

Since S′ is arbitrary we conclude that H ˆs =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 ωκ = 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 ωκ? We have seen that the mode amplitude has the dynamics of an harmonic oscillator having angular frequency ωκ. Thus, the quantum eigenenergies of the mode are

εs = sωκ , (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 ( sβωκ ) − s=0 1 = . 1 exp ( βωκ ) − − (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 Z ε = κ κ − ∂β ω = κ . eβωκ 1 − (3.26) The partition function of the entire system is given by

Z = Zκ , (3.27) κ and the average total energy by ∂ log Z U = = ε . (3.28) ∂β κ − 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) ωκ , one obtains Zκ = 1/2 sinh (βℏωκ /2) and εn = (ωκ /2) coth (βℏωκ /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 ux = ax cos (kxx) sin (kyy) sin (kzz) , (3.29) V 8 uy = ay sin (kxx) cos (kyy) sin (kzz) , (3.30) V 8 uz = az sin (kxx) sin (kyy) cos (kzz) . (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 n π k = x , (3.32) x L n π k = y , (3.33) y L n π k = z , (3.34) z L

where nx, ny and nz are integers. This solution clearly satisﬁes Eq. (3.19) where the eigen value κ is given by

2 2 2 κ = kx + ky + kz . (3.35) Alternatively, using the notation

n = (nx, ny, nz) , (3.36)

one has π κ = n , (3.37) L where

2 2 2 n = nx + ny + nz . (3.38) Using Eq. (3.21) one ﬁnds that the angular frequency of a mode characterized by the vector of integers n is given by πc ω = n . (3.39) n L 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 , (3.40) ·

Eyal Buks Thermodynamics and Statistical Physics 103 Chapter 3. Bosonic and Fermionic Systems where

a = (ax, ay, az) . (3.41) Thus, for each set of integers n , n , n there are two orthogonal modes { x y z} (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) ω U = n eβωn 1 n − ∞ ∞ ∞ αn = 2τ , eαn 1 n =0 n =0 n =0 x y z − (3.42) where the dimensionless parameter α is given by βπc α = . (3.43) L The following relation can be employed to estimate the dimensionless parameter α 3 2.4 10− α = L× τ . (3.44) cm 300 K In the limit where α 1 (3.45) ≪ the sum can be approximated by the integral

4π ∞ αn U 2τ dn n2 . (3.46) ≃ 8 eαn 1 0 − Employing the integration variable transformation [see Eq. (3.39)] L n = ω , (3.47) πc allows expressing the energy per unit volume U/V as

U ∞ = dω u , (3.48) V ω 0

Eyal Buks Thermodynamics and Statistical Physics 104 3.1. Electromagnetic Radiation where ω3 uω = . (3.49) c3π2 eβω 1 − 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 βω0 = 2.82. In terms of the wavelength λ0 = 2πc/ω0 one has

1 λ T − 0 = 5.1 . (3.50) µm 1000 K

1.4

1.2

0.8

0.6

0.4

0.2

0 2 4x 6 8 10 The function x3/ (ex 1). The total energy is found by integrating Eq.− (3.48) and by employing the variable transformation x = βω

U τ 4 ∞ x3dx = V c3π23 ex 1 0 −

π4 15 π2τ 4 = . 153c3 (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

uˆ θ cdt

dAcos θ

Fig. 3.1. Radiation emitted through a small hole in the cavity wall.

V = dA cos θ cdt . (3.52) θ × 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 2π 1 1 U J = dθ sin θ dϕ V dAdt 4π V θ 0 0 π/2 2π π2τ 4 1 = dθ sin θ cos θ dϕ 153c2 4π 0 0

1/4 π2τ 4 = 603c2 (3.53)

In terms of the historical deﬁnition of temperature T = τ/kB [see Eq. (1.92)] one has

4 J = σBT , (3.54)

where σB, which is given by

2 4 π kB 8 W σB = = 5.67 10− , (3.55) 603c2 × m2 K4 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 one- dimensional 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 mω2. 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

k a ω = ω 2 (1 cos k a) = 2ω sin n , (3.56) n − n 2 where a is the lattice spacing, 2πn k = , (3.57) n aN and n is integer ranging from N/2 to N/2. −

0.8

0.6

0.4

0.2

0 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x 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 ωn. Thus, as we had in the previous section, where we have discussed EM modes, the quantum eigenenergies of the mode are

εs = sωn , (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 ( sβωκ ) − s=0 1 = . 1 exp ( βωκ ) − − (3.59) Similarly to the EM case, the average total energy is given by

N/2 ℏω U = n , (3.60) exp (βℏωn) 1 n= N/2 − − where β = 1/τ, and the total heat capacity is given by

N/2 ∂U (βℏω )2 exp (βℏω ) C = = n n . (3.61) V ∂τ ℏ 2 n= N/2 [exp (β ωn) 1] − − High Temperature Limit. In the high temperature limit βℏω 1 ≪ (βℏω )2 exp (βℏω ) n n 1 , (3.62) [exp (βℏω ) 1]2 ≃ n − 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 fo≫r which n N/βℏω. Thus, to a good approximation the dispersion relation can |be| approximated by k a k a 2π ω = 2ω sin n 2ω n = ω n . (3.64) n 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 )2 exp (βℏω ) C = n n V ℏ 2 n= N/2 [exp (β ωn) 1] − − 2π 2 2π ∞ βℏω n exp βℏω n 2 N N ≃ ℏ 2π 2 n=0 exp β ω n 1 N − 2π 2 2π ∞ βℏω n exp βℏω n 2 dn N N 2π 2 ≃ 0 exp βℏω n 1 N − 2 N τ ∞ x exp (x) = ℏ dx 2 π ω 0 (exp (x) 1) − 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

nD ∞ ∞ ∞ 3 4π dn n2 , (3.66) → 8 n =0 n =0 n =0 x y z 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

nD 3 4π dn n2 = 3N, (3.67) 8 0 thus 6N 1/3 n = . (3.68) D π

Eyal Buks Thermodynamics and Statistical Physics 109 Chapter 3. Bosonic and Fermionic Systems

Similarly to the EM case, the average total energy is given by ℏω U = n (3.69) exp (βℏω ) 1 n n − nD 3π ℏω = dn n2 n . 2 exp (βℏωn) 1 0 − (3.70)

To proceed with the calculation the dispersion relation ωn (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

ωn = vkn , (3.71)

where v is the sound velocity. The wave vector kn is related to n = 2 2 2 nx + ny + nz by πn k = , (3.72) n L where L = V 1/3 and V is the volume. In this approximation one ﬁnds using the variable transformation βℏvπn x = , (3.73) L that

nD 3π ℏvπn U = dn n2 L 2 exp βℏvπn 1 0 L − xD 3V τ 4 x3 = dx . 2ℏ3v3π2 exp x 1 0 − (3.74) where

1/3 βℏvπn βℏvπ 6N x = D = π . D L L Alternatively, in terms of the Debye temperature, which is deﬁned as

6π2N 1/3 Θ = ℏv , (3.75) V one has

Eyal Buks Thermodynamics and Statistical Physics 110 3.2. Phonons in Solids

Θ x = , (3.76) D τ and

xD τ 3 x3 U = 9Nτ dx . (3.77) Θ exp x 1 0 −

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 ≪

xD τ 3 x3 U = 9Nτ dx Θ exp x 1 0 − xD τ 3 9Nτ dx x2 ≃ Θ 0 τ 3 x3 = 9Nτ D Θ 3 = 3Nτ , (3.78) and therefore ∂U C = = 3N. (3.79) V ∂τ 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 ≫

xD τ 3 x3 U = 9Nτ dx Θ exp x 1 0 − τ 3 ∞ x3 9Nτ dx ≃ Θ exp x 1 0 − π4/15 3π4 τ 3 = Nτ , 5 Θ (3.80)

Eyal Buks Thermodynamics and Statistical Physics 111 Chapter 3. Bosonic and Fermionic Systems and therefore ∂U 12π4 τ 3 C = = N . (3.81) V ∂τ 5 Θ Note that Eq. (3.80) together with Eq. (3.75) yield

U 3 π2τ 4 = . (3.82) V 2 15ℏ3v3 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 = 1 + λ exp ( βεn) , (3.83) n − where

η λ = exp (βµ) = e− , (3.84)

is the fugacity and β = 1/τ. Taking into account internal degrees of freedom the grandcanonical Fermionic partition function becomes

ζ = (1 + λ exp ( βεn) exp ( βE )) , (3.85) n − − l 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

gc = ζn . (3.86) Z n Eyal Buks 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 π 2 ε = n2 , (3.87) n 2m L where

n = (nx, ny, nz) , (3.88)

2 2 2 n = nx + ny + nz , (3.89) n , n, n = 1, 2, 3, , (3.90) x y z ··· and L3 = V is the volume of the box. Thus, log can be written as Zgc

∞ ∞ ∞ log = log ζ . (3.91) Zgc n n =1 n =1 n =1 x y z Alternatively, using the notation

β2π2 α2 = , (3.92) 2mL2 and Eq. (3.85) one has

∞ ∞ ∞ log = log 1 + λ exp α2n2 exp ( βE ) . (3.93) Zgc − − l l nx=1 ny=1 nz=1 For a macroscopic system α 1, and consequently the sum over n can be approximately replaced by an≪ integral

∞ ∞ ∞ 1 ∞ 4π dn n2 , (3.94) → 8 n =0 n =0 n =0 x y z 0 thus, one has

π ∞ log = dn n2 log 1 + λ exp α2n2 exp ( βE ) . (3.95) Zgc 2 − − l l 0 This can be further simpliﬁed by employing the variable transformation

βε = α2n2 . (3.96)

The following holds

Eyal Buks Thermodynamics and Statistical Physics 113 Chapter 3. Bosonic and Fermionic Systems

√ε β 3/2 dε = n2dn . 2 α2 Thus, by introducing the density of states

3/2 V 2m ε1/2 ε 0 D (ε) = 2π2 2 , (3.97) 0 ε≥ < 0 one has

1 ∞ log = dε D (ε) log (1 + λ exp ( β (ε + E ))) . (3.98) Zgc 2 − l 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 U = Zgc , (3.99) − ∂β η ∂ log N = λ Zgc , (3.100) ∂λ one ﬁnds that 1 ∞ U = dε D (ε)(ε + E ) f (ε + E ) , (3.101) 2 l FD l l −∞ 1 ∞ N = dε D (ε) f (ε + E ) , (3.102) 2 FD l l −∞ where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)] 1 f (ǫ) = . (3.103) FD exp [β (ǫ µ)] + 1 − 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) −∞ ∞ N = dε D (ε) fFD (ε) , (3.105) −∞ Typically for metals at room temperature or below the following holds τ µ. Thus, it is convenient to employ the following theorem (Sommerfeld expan-≪ sion) to evaluate these integrals. Theorem 3.3.1. Let g (ε) be a function that vanishes in the limit ε , and that diverges no more rapidly than some power of ε as ε →. Then, −∞ the following holds → ∞

∞ dε g (ε) fFD (ε) −∞ µ π2g (µ) 1 4 = dε g (ε) + ′ + O . 6β2 βµ −∞

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 π2τ 2D (µ) N = dε D (ε) + ′ . (3.106) 6 −∞ 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 ε one has − F µ εF dε D (ε) = dε D (ε) + (µ ε ) D (ε ) + O (µ ε )2 , (3.108) − F F − F −∞ −∞ and therefore π2τ 2D (ε ) N = N + (µ ε ) D (ε ) + ′ F , (3.109) 0 − F F 6 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

2 π D′ (εF) µ = εF 2 . (3.111) − 6β D (εF) Similarly, the energy U at low temperatures is given approximately by

∞ U = dε D (ε) εfFD (ε) −∞ εF π2τ 2 = dε D (ε) ε + (µ ε ) D (ε ) ε + (D′ (ε ) ε + D (ε )) − F F F 6 F F F −∞

U0 2 2 2 2 π τ D′ (εF) π τ = U0 D (εF) εF + (D′ (εF) εF + D (εF)) − 6D (εF) 6 π2τ 2 = U + D (ε ) , 0 6 F (3.112) where

εF

U0 = dε D (ε) ε . (3.113) −∞ From this result one ﬁnds that the electronic heat capacity is given by

∂U π2τ C = = D (ε ) . (3.114) V ∂τ 3 F 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 26 10 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 temper- 8 ature over the day-night cycle. Use TSun = 5800 K; RSun = 6.96 10 m; 11 × and the Mars-Sun distance of DM S = 2.28 10 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 U p = . (3.115) 3V

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

ω = ω 2 (1 cos k a) , (3.116) n − n 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

exp [n (µ ε) β] p (n) = − , (3.117) F 1 + exp [(µ ε) β] − for the case of Fermions, where n 0, 1 , and by ∈ { } p (n) = 1 exp [(µ ε) β] exp [n (µ ε) β] , (3.118) B { − − } − where n 0, 1, 2, , for the case of Bosons. ∈ { ···}

Eyal Buks Thermodynamics and Statistical Physics 117 Chapter 3. Bosonic and Fermionic Systems

b) Show that the variance (∆n)2 = (n n )2 is given by − (∆n)2 = n (1 n ) , (3.119) F F − F for the case of Fermions, and by

(∆n)2 = n (1 + n ) , (3.120) B B B 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 π2g (µ) 1 4 = dε g (ε) + ′ + O . 6β2 βµ −∞

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

v2 α = . (3.121) v 2 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) = Nε (3.122) 4 F 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

nπ2 µ = τ log exp 1 . (3.123) m τ − e

Eyal Buks Thermodynamics and Statistical Physics 118 3.6. Solutions Set 3

3.6 Solutions Set 3

1. The density of states of the photon gas is given by

V ε2 dg = dε . (3.124) π2ℏ3c3 Thus

V ∞ ε2 N = dε π2ℏ3c3 eε/τ 1 0 − τ 3 = V α , (3.125) ℏc where

1 ∞ x2 α = dx . (3.126) π2 ex 1 0 − The number α is calculated numerically

α = 0.24359 . (3.127)

For the universe

23 1 3 4π 3 1.3806568 10 JK− 3 K N = 1026 m × − 0.24359 3 1.05457266 10 34 J s2.99792458 108 m s 1 × × − × − 2.29 1087. (3.128) ≃ × 2. The energy emitted by the Sun is

2 4 ESun = 4πRSunσBTSun , (3.129)

and the energy emitted by a planet is

2 4 Eplanet = 4πRplanetσBTplanet . (3.130)

The fraction of Sun energy that planet receives is

2 πRplanet 2 ESun , (3.131) 4πDM S − and this equals to the energy it reradiates. Therefore

πR2 planet E = E , 4πD2 Sun planet

Eyal Buks Thermodynamics and Statistical Physics 119 Chapter 3. Bosonic and Fermionic Systems

thus

RSun Tplanet = TSun , 2D and for Mars

6.96 108 m T = × 5800 K = 226 K . Mars 2 2.28 1011 m × × 3. The partition function is given by

∞ 1 Z = exp (sβℏω ) = , n 1 exp ( βℏω ) n s=0 n n − − thus the free energy is given by

F = τ log Z = τ log [1 exp ( βℏω )] . − − − n n Transforming the sum over modes into integral yields

∞ F = τπ dnn2 log [1 exp ( βℏω )] (3.132) − − n 0 ∞ βℏπcn = τπ dnn2 log 1 exp , − − L 0 or, by integrating by parts

2 3 1 ℏπ c ∞ n 1 F = dn = U, −3 L 0 exp βℏπcn 1 −3 L − where π2τ 4V U = . (3.133) 15ℏ3c3 Thus ∂F U p = = . (3.134) − ∂V 3V τ 4. Using the expression for Helmholtz free energy, which was derived in the previous problem,

U π2τ 4V F = = , − 3 −45ℏ3c3 one ﬁnds that the entropy is given by

Eyal Buks Thermodynamics and Statistical Physics 120 3.6. Solutions Set 3

∂F 4π2τ 3V σ = = . − ∂τ 45ℏ3c3 V Thus, for an isentropic process, for which σ is a constant, one has

V 1/3 τ = τ 1 . 2 1 V 2 Using again the previous problem, the pressure p is given by U p = , 3V thus π2τ 4 p = , 45ℏ3c3 and

π2τ 4 V 4/3 p = 1 1 . 2 45ℏ3c3 V 2 p1

5. Let u (na)be the displacement of point particle number n. The equations of motion are given by

mu¨ (na) = mω2 2u (na) u [(n 1) a] u [(n + 1) a] . (3.135) − { − − − } Consider a solution of the form

i(kna ω t) u (na, t) = e − n . (3.136)

Periodic boundary condition requires that

eikNa = 1 , (3.137)

thus 2πn k = . (3.138) n aN Substituting in Eq. 3.135 yields

2 2 ika ika mω u (na) = mω 2u (na) u (na) e− u (na) e , (3.139) − n − − − or k a ω = ω 2 (1 cos k a) = 2ω sin n . (3.140) n − n 2 Eyal Buks Thermodynamics and Statistical Physics 121 Chapter 3. Bosonic and Fermionic Systems

6. In general using Gibbs factor exp [n (µ ε) β] p (n) = − , (3.141) exp [n (µ ε) β] ′ − n′ where β = 1/τ, one ﬁnds for Fermions exp [n (µ ε) β] p (n) = − , (3.142) F 1 + exp [(µ ε) β] − where n 0, 1 , and for Bosons ∈ { } exp [n (µ ε) β] pB (n) = − = 1 exp [(µ ε) β] exp [n (µ ε) β] , ∞ { − − } − exp [n (µ ε) β] ′ − n′=0 (3.143) where n 0, 1, 2, . The expectation value of n in general is given by ∈ { ···} n exp [n (µ ε) β] ′ − n′ n = n′p (n′) = , (3.144) ′ exp [n (µ ε) β] n ′ − n′ thus for Fermions 1 n = , (3.145) F exp [(ε µ) β] + 1 − and for Bosons

∞ n = 1 exp [(µ ε) β] n′ exp [n′ (µ ε) β] B { − − } − n′=0 exp [(µ ε) β] = 1 exp [(µ ε) β] − { − − } (1 exp [(µ ε) β])2 − − 1 = . exp [(ε µ) β] 1 − − (3.146) In general, the following holds 2 (n )2 exp [n (µ ε) β] n exp [n (µ ε) β] ′′ − ′ ′ − ∂ n n′′ n′ τ =   ∂µ − τ exp [n′ (µ ε) β] exp [n′ (µ ε) β] ′ −  ′ −  n  n    = n2 n 2 = (n n )2 . − − (3.147)

Eyal Buks Thermodynamics and Statistical Physics 122 3.6. Solutions Set 3

Thus exp [(ε µ) β] (∆n)2 = − = n (1 n ) , (3.148) F (exp [(ε µ) β] + 1)2 F − F − exp [(ε µ) β] (∆n)2 = − = n (1 + n ) . (3.149) B (exp [(ε µ) β] 1)2 B B − − 7. Let ε

G (ε) = dε′ g (ε′) . (3.150) −∞ Integration by parts yields

∞ I = dε g (ε) fFD (ε) −∞ ∞ ∂fFD = [G (ε) fFD (ε) ∞ + dε G (ε) , |−∞ − ∂ε =0 −∞ (3.151) where the following holds

∂f βeβ(ε µ) β FD = − = . (3.152) − ∂ε β(ε µ) 2 2 β e − + 1 4 cosh (ε µ) 2 − Using the Taylor expansion of G (ε) about ε µ, which has the form − ∞ G(n) (µ) G (ε) = (ε µ)n , (3.153) n! − n=0 yields

n ∞ G(n) (µ) ∞ β (ε µ) dε I = − . (3.154) n! 2 β n=0 4 cosh 2 (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) ∞ x2ndx I = . (3.156) (2n)!β2n 4 cosh2 x n=0 2 −∞ With the help of the identities

∞ dx = 1 , (3.157) 4 cosh2 x 2 −∞ ∞ x2dx π2 = , (3.158) 4 cosh2 x 3 2 −∞ one ﬁnds π2G(2) (µ) 1 4 I = G (µ) + + O 6β2 βµ µ π2g (µ) 1 4 = dε g (ε) + ′ + O . 6β2 βµ −∞ (3.159) 8. In general, at zero temperature the average of the energy ε to the power n is given by

εF dε D (ε) εn εn = 0 , (3.160) εF dε D (ε) 0 where D (ε) is the density of states

V 2m 3/2 D (ε) = ε1/2 , (3.161) 2π2 2 thus n n εF ε = 2n . (3.162) 3 + 1 a) Using Eq. (3.162) one ﬁnds that 3ε ε = F . (3.163) 5 b) The speed v is related to the energy by

2ε v = , (3.164) m

Eyal Buks Thermodynamics and Statistical Physics 124 3.6. Solutions Set 3

thus

εF 2 ε 3 +1 16 α = 2 = 2 = . (3.165) 1/2 −1/2 15 ε εF 2 1 3 2 +1 c) The number of electrons N is given by

εF εF D (εF) 1/2 D (εF) 2 3/2 N = dε D (ε) = dε ε = εF , ε1/2 ε1/2 3 0 F 0 F thus 2 3π2N 2/3 ε = , (3.166) F 2m V and therefore 3N 2 3π2N 2/3 U = . (3.167) 5 2m V 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 τ,N 3N 2 3π2N 2/3 2 = 5 2m V 3V 2Nε = F . 5V (3.168) 9. The energy of the particles are

εn, = pc (3.169) ± where p = k, and k = πn , n = n2 + n2 + n2 and n = 1, 2, . L x y z i ··· Therefore 1 4 π N = 2 πn3 = n3 (3.170) × 8 × 3 F 3 F 3N 1/3 n = F π cπ 3N 1/3 3N 1/3 ε = = cπ F L π πV

Eyal Buks Thermodynamics and Statistical Physics 125 Chapter 3. Bosonic and Fermionic Systems

The energy is given by

εF L3 εF E (T = 0) = εg (ε) dε = ε3dε = (3.171) 23 3 π c 0 0 L3 ε4 1 L3 = F = ε3 ε = π23c3 × 4 4 π23c3 F × F 1 L3 3π23c3N 3 = ε = Nε 4 π23c3 L3 F 4 F

10. The energy of an electron having a wave function proportional to exp (ikxx) exp (ikyy) is 2 2 2 kx + ky . (3.172) 2me For periodic boundary conditions one has

2πnx kx = , (3.173) Lx 2πny ky = , (3.174) Ly

where the sample is of area LxLy, and nx and ny are both integers. The number of states having energy smaller than E′ is given by (including both spin directions) 2m E L L 2π e ′ x y , (3.175) 2 4π2 thus, the density of state per unit area is given by

me E > 0 D (E) = π2 . (3.176) 0 E < 0 Using Fermi-Dirac function 1 f (E) = , (3.177) 1 + exp [β (E µ)] − where β = 1/τ, the density n is given by

∞ n = D (E) f (E) dE −∞ m ∞ dE = e 2 π 0 1 + exp [β (E µ)] m τ − = e log 1 + eβµ , π2 (3.178)

Eyal Buks Thermodynamics and Statistical Physics 126 3.6. Solutions Set 3

thus nπ2 µ = τ log exp 1 . (3.179) m τ − e

Eyal Buks Thermodynamics and Statistical Physics 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 Hamilton- Jacobi 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 q¯ = (q , q , , q ) . (4.1) 1 2 ··· d 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 q¯· , and where V depends only on coordinates, namely V = V (¯q). The notation overdot is used to express time derivative, namely dq dq dq q¯· = 1 , 2 , , d . (4.2) dt dt ··· dt 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 ∂T pi = . (4.3) ∂q˙i The classical Hamiltonian of the system is expressed as a function of the vector of coordinates q¯ andH as a function of the vector of canonical conjugate momentum variables Chapter 4. Classical Limit of Statistical Mechanics

p¯ = (p , p , , p ) , (4.4) 1 2 ··· d namely

= (¯q, p¯) , (4.5) H H and it is deﬁned by

d = p q˙ T + V. (4.6) H i i − i=1 4.1.1 Hamilton-Jacobi Equations

The equations of motion of the system are given by ∂ q˙i = H (4.7) ∂pi ∂ p˙i = H , (4.8) − ∂qi where i = 1, 2, d. ···

4.1.2 Example V(q)

q Consider a particle having mass m in a one dimensional potential V (q). The kinetic energy is given by T = mq˙2/2, thus the canonical conjugate momentum is given by [see Eq. (4.3)] p = mq˙. 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

mq˙2 = mq˙2 + V (q) H − 2 p2 = + V (q) . 2m (4.9) Hamilton-Jacobi equations (4.7) and (4.8) read p q˙ = (4.10) m ∂V p˙ = . (4.11) − ∂q The second equation, which can be rewritten as ∂V mq¨ = , (4.12) − ∂q 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 = Lq˙2/2 is the energy stored in the inductor, and the potential energy V = q2/2C is the energy stored in the capacitor. The canonical conjugate momentum is given by [see Eq. (4.3)] p = Lq˙, and the Hamiltonian (4.6) is given by p2 q2 = + . (4.13) H 2L 2C Hamilton-Jacobi equations (4.7) and (4.8) read p q˙ = (4.14) L q p˙ = . (4.15) −C The second equation, which can be rewritten as q Lq¨+ = 0 , (4.16) C expresses the requirement that the voltage across the capacitor is the same as the one across the inductor.

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 (¯q, p¯) is the probability distribution to ﬁnd the system in the point (¯q, p¯). The following theorem is given without a proof. Let (¯q, p¯) be an Hamil- tonian of a system, and assume that has the followingH form H d = A p2 + V (¯q) , (4.17) H i i 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 (¯q, p¯) = N exp ( β (¯q, p¯)) , (4.18) − H where 1 N = (4.19) d¯q d¯p exp ( β (¯q, p¯)) − H is a normalization constant, β = 1/τ, and τ is the temperature. The notation d¯q indicates integration over all coordinates, namely d¯q = dq dq 1 1 · dqd, and similarly d¯p = dp1 dp1 dpd. ···· Let A (¯q, p¯) be a variable which depends····· on the coordinates q¯ and their canonical conjugate momentum variables p¯. Using the above theorem the average value of A can be calculates as:

A (¯q, p¯) = d¯q d¯p A (¯q, p¯) ρ (¯q, p¯) d¯q d¯p A (¯q, p¯) exp ( β (¯q, p¯)) = − H . d¯q d¯p exp ( β (¯q, p¯)) − H (4.20)

4.2.1 Equipartition Theorem

Assume that the Hamiltonian has the following form

= B q2 + ˜ , (4.21) H i i H

where Bi is a constant and where ˜ is independent of qi. Then the following holds H τ B q2 = . (4.22) i i 2 Similarly, assume that the Hamiltonian has the following form

Eyal Buks Thermodynamics and Statistical Physics 132 4.2. Density Function

= A p2 + ˜ , (4.23) H i i H

where Ai is a constant and where ˜ is independent of pi. Then the following holds H τ A p2 = . (4.24) i i 2 To prove the theorem for the ﬁrst case we use Eq. (4.20)

d¯q d¯p B q2 exp ( β (¯q, p¯)) B q2 = i i − H i i d¯q d¯p exp ( β (¯q, p¯)) − H 2 2 dqi Biq exp βBiq = i − i dq exp ( βB q2) i − i i ∂ = log dq exp βB q2 −∂β i − i i ∂ π = log −∂β βB i 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) kq2, 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

p2 kq2 = + . (4.26) H 2m 2 In the classical limit the average energy of the system can be easily calculated using the equipartition theorem

U = = τ . (4.27) H In the quantum treatment the system has energy levels given by

Es = sω , 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

∞ 1 Z = exp ( sβω) = , (4.28) − 1 exp ( βω) s=0 − − thus the average energy U is given by ∂ log Z ω U = = . (4.29) − ∂β eβω 1 − Using the expansion

1 U = β− + O (β) , (4.30) one ﬁnds that in the limit of high temperatures, namely when βω 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 + Lq¨ + Rq˙ = V (t) , (4.31) C 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 − q(t) T/2 < t < T/2 q (t) = . (4.32) T 0− else The energy stored in the capacitor is given by q2/2C. Using the equipartition theorem one ﬁnds q2 τ = , (4.33) 2C 2 where q2 is obtained by averaging q2 (t), namely

+ 2 1 ∞ 2 q lim dt qT (t) . (4.34) ≡ T T →∞ −∞ Introducing the Fourier transform:

1 ∞ iωt qT (t) = dω qT (ω)e− , (4.35) √2π −∞ one ﬁnds

+ 2 1 ∞ 1 ∞ iωt 1 ∞ iω′t q = lim dt dω qT (ω)e− dω′ qT (ω′)e− T T √2π √2π →∞ −∞ −∞ −∞ + 1 ∞ ∞ 1 ∞ i(ω+ω′)t = dω qT (ω) dω′ qT (ω′) lim dte− 2π T T −∞ −∞ →∞ −∞ 2πδ(ω+ω′) 1 ∞ = lim dω qT (ω)qT ( ω) . T T − →∞ −∞ (4.36)

Moreover, using the fact that q(t) is real one ﬁnds

2 1 ∞ 2 q = lim dω qT (ω) . (4.37) T T | | →∞ −∞ In terms of the power spectrum Sq(ω) of q (t), which is deﬁned as

1 2 Sq(ω) = lim qT (ω) , (4.38) T T | | →∞ one ﬁnds

2 ∞ q = 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) , (4.40) C − − where V (ω) is the Fourier transform of V (t), namely

1 ∞ iωt V (t) = dω V (ω)e− . (4.41) √2π −∞ In terms of the resonance frequency

1 ω0 = , (4.42) LC one has

L ω2 ω2 iωR q(ω) = V (ω) . (4.43) 0 − − Taking the absolute value squared yields

SV (ω) Sq(ω) = , (4.44) L2 (ω2 ω2)2 + ω2R2 0 −

where SV (ω) is the power spectrum of V (t). Integrating the last result yields

∞ ∞ SV (ω) dω Sq(ω) = dω L2 (ω2 ω2)2 + ω2R2 −∞ −∞ 0 − 1 ∞ S (ω) = dω V . 2 2 2 ω2R2 L (ω0 + ω) (ω0 ω) + 2 −∞ − L (4.45)

The integrand has a peak near ω0, having a width R/2L. The Quality factor Q is deﬁned as ≃ ω R 0 = . (4.46) Q 2L

Assuming SV (ω) is a smooth function near ω0 on the scale ω0/Q, and assuming Q 1 yield ≫

Eyal Buks Thermodynamics and Statistical Physics 136 4.3. Nyquist Noise

∞ SV (ω0) ∞ dω dω Sq(ω) 2 2 ≃ L 2 2 2ωω0 −∞ −∞ (ω0 + ω) (ω0 ω) + − Q SV (ω0) ∞ dω 4 2 2 2 ≃ 4ω0L ω0 ω 1 − + −∞ ω0 Q

SV (ω0) ∞ dx = 3 2 2 4ω0L 2 1 −∞ x + Q

πQ SV (ω0)πQ = 3 2 . 4ω0L (4.47)

On the other hand, using Eqs. (4.33) and (4.39) one ﬁnds

∞ 2 dω Sq(ω) = q = Cτ , (4.48) −∞ therefore 4Cω3L2 S (ω ) = 0 τ , (4.49) V 0 πQ

or using Eqs. (4.42) and (4.46)

2Rτ S (ω ) = . (4.50) V 0 π Thus, Eq. (4.44) can be rewritten as

2Rτ 1 Sq(ω) = . (4.51) π L2 (ω2 ω2)2 + ω2R2 0 − 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

2 ∞ V = dω SV (ω) , (4.52) −∞ 2 one ﬁnds that the variance in such a measurement V ∆f is given by

2 V ∆f = 4Rτ∆f . (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 ˙x = 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) and

q (t) q (t′) = g δ (t t′) . (4.56) i j ij − Let δx = x (t + δt) x (t) . (4.57) − To ﬁrst order in δt one ﬁnds by integrating Eq. (4.54) that

t+δt δ 2 ( x)i = Ai (x, t) δt + dt′ qi (t′) + O (δt) . (4.58) t With the help of Eqs. (4.55) and (4.56) one ﬁnds that

(δx) = A (x, t) δt + O (δt)2 , (4.59) i i and δ δ ( x)i ( x)j 2 = Ai (x, t) Aj (x, t)(δt) t+δt t+δt + dt′ dt′′ q (t′) q (t′′) + i j ··· t t = A (x, t) A (x, t)(δt)2 + g δt + , i j ij ··· (4.60) thus to ﬁrst order in δt δ δ 2 ( x)i ( x)j = gijδt + O (δt) . (4.61) In a similar way one can show that all higher order moments (e.g. third order

moments (δx) ′ (δx) ′′ (δx) ′′′ ) vanish to ﬁrst order in δt. i i i

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 (x, t x , t ) be the conditional probability density to ﬁnd the system at P | ′ ′ point x at time t given that it was (or will be) at point x′ at time t′. The following holds

p (x , t ; x , t ) = (x , t x , t ) p (x , t ) . (4.63) 2 3 3 1 1 P 3 3| 1 1 1 1 1 Moreover, by assuming that t t t and by assuming the case of a 1 ≤ 2 ≤ 3 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) = (x , t x , t ) (x , t x , t ) p (x , t ) . P 3 3| 2 2 P 2 2| 1 1 1 1 1

With the help of these relations Eq. (4.62) becomes

(x , t x , t ) p (x , t ) P 3 3| 1 1 1 1 1 = dx (x , t x , t ) (x , t x , t ) p (x , t ) , 2 P 3 3| 2 2 P 2 2| 1 1 1 1 1 (4.64)

thus by dividing by p1 (x1, t1) one ﬁnds that

(x , t x , t ) = dx (x , t x , t ) (x , t x , t ) . (4.65) P 3 3| 1 1 2 P 3 3| 2 2 P 2 2| 1 1 4.4.3 The Fokker-Planck Equation

Equation (4.65) can be written as

(x, t + δt x , t ) = dx′ (x, t + δt x′, t′) (x′, t′ x , t ) . (4.66) P | 0 0 P | P | 0 0 On the other hand

(x, t + δt x′, t′) = δ (x δx x′) , (4.67) P | − −

Eyal Buks Thermodynamics and Statistical Physics 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 · (δx) (δx) 2 δ dF i j d F = F (x′) + ( x)i + + , dxi′ 2! dxi′dxj′ ··· (4.69) thus

δ (x δx x′) = δ (x x′) − − − δ dδ (x x′) + ( x)i − dxi′ δ δ 2 ( x) ( x) d δ (x x′) + i j − + . 2! dxi′ dxj′ ··· (4.70) Inserting this result into Eq. (4.66) (x, t + δt x , t ) P | 0 0 = dx′ δ (x δx x′) (x′, t′ x , t ) − − P | 0 0 = dx′ δ (x x′) (x′, t′ x , t ) − P | 0 0 δ dδ (x x′) + dx′ ( x)i − (x′, t′ x0, t0) dxi′ P | 2 1 δ δ d δ (x x′) + dx′ ( x)i ( x)j − (x′, t′ x0, t0) 2 dxi′ dxj′ P | + , ··· (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 → ∂ dδ (x x′) P = dx′ Ai (x′, t) − (x′, t′ x0, t0) ∂t dxi′ P | 2 1 d δ (x x′) + dx′ g − (x′, t′ x , t ) , 2 ij dx dx P | 0 0 i′ j′ (4.72)

Eyal Buks Thermodynamics and Statistical Physics 140 4.4. Thermal Equilibrium From Stochastic Processes and ﬁnally integrating by parts and assuming that 0 in the limit x yield P → → ±∞ ∂ d 1 d2 P = (Ai ) + (gij ) . (4.73) ∂t −dxi′ P 2 dxi′ dxj′ P This result, which is known as the Fokker-Planck equation, can also be written as ∂ P + J = 0 , (4.74) ∂t ∇ · where the probability current density J is given by 1 d Ji = Ai (gij ) . (4.75) P − 2 dxj′ P

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’ as (the potential condition) H A (x, t) = . (4.77) −∇H Moreover, for simplicity assume that

gij = 2τ δij , (4.78)

where the ’temperature’ τ is a constant. For this case one has d d Ji = H τ P , (4.79) −P dxi′ − dxi′ thus

J = τ −P∇H − ∇P = ( + τ log ) . −P∇ H P (4.80)

Substituting a solution having the form

H = Ne− τ (4.81) P yields

Eyal Buks Thermodynamics and Statistical Physics 141 Chapter 4. Classical Limit of Statistical Mechanics

H H J = τNe− τ (log N) = τe− τ N, (4.82) − ∇ − ∇ and thus

H J = [ N τ ( N)] e− τ . (4.83) ∇ · ∇H · ∇ − ∇ · ∇ In terms of N the Fokker-Planck equation (4.74) becomes ∂N + [ N τ ( N)] = 0 . (4.84) ∂t ∇H · ∇ − ∇ · ∇ In steady state Eq. (4.84) becomes

N = τ ( N) . (4.85) ∇H · ∇ ∇ · ∇ 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

1 H = e− τ , (4.86) P Z where

Z = dx′ . (4.87) P 4.4.5 Free Energy

The free energy functional is deﬁned by

F ( ) = U τσ , (4.88) P − where U is the energy

U = dx′ , (4.89) HP and σ is the entropy

σ = dx′ log . (4.90) − P P The distribution is constrained to satisfy the normalization constrain P

G ( ) = dx′ 1 = 0 . (4.91) P P − Minimizing F ( ) under the constrain (4.91) is done by introducing the La- grange multiplierP η

Eyal Buks Thermodynamics and Statistical Physics 142 4.4. Thermal Equilibrium From Stochastic Processes

δ (F ( ) + ηG ( )) P P = dx′ [ + τ (1 + log ) + η] δ . H P P (4.92)

At a stationary point the following holds

+ τ (1 + log ) + η = 0 , (4.93) H P thus

η 1 H = e− τ − e− τ , (4.94) P η 1 where the constant e− τ − 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 = H + τ P , (4.95) ∂t ∂x P ∂x ∂x or ∂ P = , (4.96) ∂t LP where the operator is given by L ∂ ∂ ∂2 = H + τ . (4.97) L ∂x ∂x ∂x2 It is convenient to deﬁne the operator ˆ, which is given by L H H ˆ = e 2τ e− 2τ . (4.98) L L The following holds

Eyal Buks Thermodynamics and Statistical Physics 143 Chapter 4. Classical Limit of Statistical Mechanics

2 H ∂ ∂ H H ∂ H ˆ = e 2τ He− 2τ + τe 2τ e− 2τ L ∂x ∂x ∂x2 H ∂ ∂ H ∂ ∂ = e 2τ He− 2τ + H ∂x ∂x ∂x ∂x 2 H 2 H ∂ H H ∂e− 2τ ∂ ∂ + τe 2τ e− 2τ + 2τe 2τ + τ ∂x2 ∂x ∂x ∂x2 ∂2 1 ∂ 2 ∂ ∂ = H H + H ∂x2 − 2τ ∂x ∂x ∂x 1 ∂2 1 ∂ 2 ∂ ∂ ∂2 H + H H + τ − 2 ∂x2 4τ ∂x − ∂x ∂x ∂x2 1 ∂2 1 ∂ 2 ∂2 = H H + τ , 2 ∂x2 − 4τ ∂x ∂x2 (4.99)

thus ∂2 ˆ = τ V,ˆ (4.100) L ∂x2 − where the potential Vˆ is given by

1 ∂ 2 1 ∂2 Vˆ = H H . (4.101) 4τ ∂x − 2 ∂x2 Note that while the operator is not Hermitian, the operator ˆ is since L L H H ˆ† = e− 2τ †e 2τ L H L H H = e− 2τ †e τ e− 2τ H LH H = e− 2τ e τ e− 2τ H HL = e 2τ e− 2τ = ˆ . L L (4.102)

In terms of ˆ Eq. (4.96) becomes (it is assume that is time independent) L H ∂ ˆ P = ˆ ˆ , (4.103) ∂t LP where

H H ˆ = e 2τ e− 2τ . (4.104) P P Let ψ (x) be a set of eigenvectors of ˆ n L ˆψ = λ ψ . (4.105) L n n n

Eyal Buks Thermodynamics and Statistical Physics 144 4.4. Thermal Equilibrium From Stochastic Processes

The following holds [see Eq. (4.98)]

ϕ = λ ϕ , (4.106) L n n n where

H 2τ ϕn = e− ψn . (4.107)

The conditional probability distribution (x, t x′, t′) is given by [see Eq. (4.96)] P |

(t t′) (x, t x′, t′) = eL − δ (x x′) . (4.108) P | − With the help of the closure relation (recall that ˆ is Hermitian) L

δ (x x′) = ψ∗ (x′) ψ (x) − n n n H(x)+H(x′) 2τ = e ϕn∗ (x′) ϕn (x) n ′ H(x ) τ = e ϕn∗ (x′) ϕn (x) , n (4.109) one ﬁnds that

′ ′ H(x ) (x)(t t ) τ (x, t x′, t′) = eL − e ϕn∗ (x′) ϕn (x) P | n ′ H(x ) ′ τ (x)(t t ) = e ϕn∗ (x′) eL − ϕn (x) n ′ H(x ) ′ τ λn(t t ) = e ϕn∗ (x′) e − ϕn (x) n ′ ′ H(x ) H(x ) ′ H(x) τ 2τ λn(t t ) 2τ = e e− ψn∗ (x′) e − e− ψn (x) , n (4.110)

thus

′ H(x )−H(x) ′ 2τ λn(t t ) (x, t x′, t′) = e ψn∗ (x′) ψn (x) e − . (4.111) P | 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

x˙ + Γx = 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 (x) = , (4.114) H 2 the Fokker-Planck equation for the conditional probability distribution is given by [see Eq. (4.74)] P

∂ ∂ ∂ P = Γ x + τ P , (4.115) ∂t ∂x P ∂x and the operator ˆ is given by [see Eq. (4.100)] L 2 2 ˆ Γ ∂ x =  + 1 , L − 2 − ∂ x  x0 − x0       where

2τ x0 = . (4.116) Γ The eigenvectors of ˆ are given by [see Eq. (4.131)] L 2 1 x 2 x x e− 0 Hn x0 ψn (x) = , (4.117) 1/4 1/2 n π x0 √2 n! and the corresponding eigenvalues by

1 1 λ = Γ n + , (4.118) n − 2 − 2 where n = 0, 1, 2, . Using these results··· one ﬁnds that is given by [see Eq. (4.111)] P x′ x Γ n t t′ 2 H H e− ( − ) x n x0 n x0 x (x, t x′, t′) = e− 0 . (4.119) √πx 2nn! P | n 0

Eyal Buks Thermodynamics and Statistical Physics 146 4.5. Problems Set 4

With the help of the general identity (4.138) one ﬁnds that the following holds

2 Y Xα−1 2 n X α exp √−−2 αe ∞ Hn (Y ) Hn (X) − α 1 − 2 = − , (4.120) √π n! 2 n=0 π (α− 1) − thus Eq. (4.119) becomes

′ 2 x x′e−Γ (t−t ) exp − − δ (x, t x′, t′) = , (4.121) P | √πδ where

2τ 1 e 2Γ (t t′) δ = x2 1 e 2Γ (t t′) = − − − . (4.122) 0 − − Γ −

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

2 2 mc (λ λ0) I (λ) ∝ exp 2− , (4.123) − 2λ0τ 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 I2 , where I is the current in the RCL

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 − q(t) T/2 < t < T/2 q (t) = . (4.124) T 0− else The Fourier transform is given by

1 ∞ iωt qT (t) = dω qT (ω)e− , (4.125) √2π −∞ and the power spectrum is given by

1 2 Sq(ω) = lim qT (ω) . T T →∞ | | a) Show that

+ 2 1 ∞ 2 ∞ q lim dt qT (t) = dω Sq(ω) . (4.126) ≡ T T →∞ −∞ −∞ b) Wiener-Khinchine Theorem - show that the correlation function of the random signal q(t) is given by

+ 1 ∞ ∞ iωt′ q (t) q (t + t′) lim dt qT (t) qT (t + t′) = dω e Sq(ω) . ≡ T T →∞ −∞ −∞ (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

2 L Q = 1 , (4.128) RC ≫ 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 q , q , , q { 1 2 ··· N } and the corresponding canonically conjugate variables p1, p2, , pN . The Hamiltonian of the system is given by { ··· }

N s t H = Anpn + Bnqn , (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

1 1 U = Nτ + , (4.130) s t 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 degrees of freedom at temperature τ. The velocity of a particle is denoted 2 2 2 as v = vx + vy + vz . Calculate the quantity 1 v v (the symbol denoted averaging) in the: a) classical limit (high temperatures). b) zero temperature. 8. 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) . 9. Consider an ideal classical gas of particles having≤ mass≤ M and having no 2 2 2 internal degrees of freedom at temperature τ. Let v = vx + vy + vz . be the velocity of a particle. Calculate a) v b) v2 10. A mixture of two classical ideal gases, consisting of N and N particles 1 2 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. 11. The Hermite polynomial Hn (X) of order n is defined by X2 d n X2 H (X) = exp X exp . (4.131) n 2 − dX − 2 For some low values of n the Hermite polynomials are given by

H0 (X) = 1 , (4.132)

H1 (X) = 2X, (4.133) H (X) = 4X2 2 , (4.134) 2 − H (X) = 8X3 12X, (4.135) 3 − H (X) = 16X4 48X2 + 12 . (4.136) 4 −

Eyal Buks Thermodynamics and Statistical Physics 149 Chapter 4. Classical Limit of Statistical Mechanics

Show that

∞ tn exp 2Xt t2 = H (X) . (4.137) − n n! n=0 12. Show that

α 2XY αX2 αY 2 ( − − ) α n exp 1 α2 ∞ Hn (X) Hn (Y ) 2 = − . (4.138) n! √ 2 n=0 1 α −

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

mv2 f (v ) ∝ exp x , (4.140) x − 2τ thus using

c (λ λ0) vx = − , (4.141) λ0 the probability distribution I (λ) is proportional to

2 2 mc (λ λ0) I (λ) ∝ exp 2− . (4.142) − 2λ0τ

2 2. The energy stored in the inductor is UL = LI /2. Using the equipartition theorem U = τ/2, thus L τ I2 = . (4.143) L 3. a)

Eyal Buks Thermodynamics and Statistical Physics 150 4.6. Solutions Set 4

+ 2 1 ∞ 1 ∞ iωt 1 ∞ iω′t q = lim dt dω qT (ω)e− dω′ qT (ω′)e− T T √2π √2π →∞ −∞ −∞ −∞ + 1 ∞ ∞ 1 ∞ i(ω+ω′)t = dω qT (ω) dω′ qT (ω′) lim dt e− 2π T T −∞ −∞ →∞ −∞ 2πδ(ω+ω′) 1 ∞ = lim dω qT (ω)qT ( ω) . T T − →∞ −∞ (4.144)

Since q(t) is real one has q ( ω) = q (ω), thus T − T∗

2 ∞ q = dω Sq(ω) . (4.145) −∞ b)

+ 1 ∞ q (t) q (t + t′) lim dt qT (t) qT (t + t′) ≡ T T →∞ −∞ + 1 1 ∞ ∞ iωt ∞ iω′(t+t′) = lim dt dω qT (ω)e− dω′ qT (ω′)e− 2π T T →∞ −∞ −∞ −∞ + 1 1 ∞ iω′t′ ∞ ∞ i(ω+ω′)t = lim dω e− qT (ω) dω′ qT (ω′) dt e− 2π T T →∞ −∞ −∞ −∞ 2πδ(ω+ω′)

1 ∞ iωt′ = lim dω e qT (ω)qT ( ω) T T − →∞ −∞ ∞ iωt′ = dω e Sf (ω) . −∞ (4.146)

4. Using I (ω) = iωq (ω) and q2 = Cτ one ﬁnds for the case Q 1 − ≫

∞ ∞ τ I2 = dω S (ω) ω2 dω S (ω) = ω2 q2 = , (4.147) I ≃ 0 q 0 L −∞ −∞ in agreement with the equipartition theorem for the energy stored in the inductor LI2/2. 5. Calculate for example

Eyal Buks Thermodynamics and Statistical Physics 151 Chapter 4. Classical Limit of Statistical Mechanics

∞ t t t Bnqn exp ( βBnqn) dqn B q = −∞ − n n t ∞ exp ( βBnqxn) dqn −∞ − t t ∞ Bnq exp ( βBnq ) dqn = 0 n − n ∞ exp ( βB qt ) dq 0 − n xn n d ∞ = log exp βB qt dq . −dβ − n n n 0 (4.148)

where β = 1/τ. Changing integration variable

t x = βBnqn, (4.149) t 1 dx = tβBnqn− dqn , (4.150) leads to

t d 1 1 ∞ 1 1 x τ B q = log (βB )− t t− x t − e− dx = . (4.151) n n −dβ n t 0 Thus

N 1 1 U = A ps + B qt = Nτ + . (4.152) n n n n s t n=1 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) ≤ ≤ 1 2π ∞ N dN = dv d (cos θ) dϕv2v (dt) A cos θ f (v) . (4.154) V 0 0 0 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 2π ∞ ∞ 1 = dv d (cos θ) dϕv2f (v) = 4π dvv2f (v) , (4.155) 0 0 0 0 thus

dN πNA ∞ NA v = dvv2vf (v) = . (4.156) dt V 4V 0 Eyal Buks Thermodynamics and Statistical Physics 152 4.6. Solutions Set 4

In the classical limit Mv2 f (v) exp , (4.157) ∝ − 2τ thus, by changing the integration variable

Mv2 x = (4.158) 2τ one ﬁnds

2 ∞ dvv3 exp Mv 0 − 2τ v = 2 ∞ dvv2 exp Mv 0 − 2τ 1/2 2τ ∞ dxx exp ( x) = 0 − M ∞ dxx1/2 exp ( x) 0 − 1/2 8τ = , πM (4.159)

and

dN NA 8τ 1/2 = . (4.160) dt 4V πM 7. The probability that an orbital having energy ε is occupied is given by 1 f (ε) = , (4.161) F 1 + exp [(ε µ) β] − where β = 1/τ and µ is the chemical potential. The velocity v of such an orbital is related to the energy ε by

Mv2 ε = . (4.162) 2 The 3D density of state per unit volume is given by

1 2M 3/2 g (ε) = ε1/2 . (4.163) 2π2 2 Thus

Eyal Buks Thermodynamics and Statistical Physics 153 Chapter 4. Classical Limit of Statistical Mechanics

∞dε g (ε) f (ε) v ∞dε g (ε) f (ε) 1 1 F F v v = 0 0 v ∞ ∞ dε g (ε) fF (ε) dε g (ε) fF (ε) 0 0

∞ ∞ dε εfF (ε) dε fF (ε) = 0 0 . 2 ∞ 1/2 dε ε fF (ε) 0 (4.164)

a) In the classical limit

f (ε) exp ( βε) , (4.165) F ∝ − thus using the identities

∞ n n n dε ε exp ( βε) = Γ (n) β− − β 0 ∞ 1 dε exp ( βε) = − β 0 Γ (1) = 1 1 Γ = √π 2 one ﬁnds

∞dε ε exp ( βε) ∞dε exp ( βε) 1 − − v = 0 0 v 2 ∞dε ε1/2 exp ( βε) − 0 1 1 1 Γ (1) β− β β = 2 1 1/2 1 Γ 2 β− 2β 4 = . π (4.166)

b) Using the identity

εF εn+1 dε εn = F n + 1 0

Eyal Buks Thermodynamics and Statistical Physics 154 4.6. Solutions Set 4

one ﬁnds

εF εF dε ε dε 1 0 0 v = 2 . v εF dε ε1/2 0 1 ε2ε = 2 F F 3 2 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) = Mω2r2 , (4.168) −2 thus βMω2 n (r) = A exp [ βU (r)] = A exp r2 , (4.169) − 2 where the normalization constant A is found from the condition

R N = 2πL n (r) rdr 0 R βMω2 = 2πLA exp r2 rdr 2 0 2πLA βMω2 = exp R2 1 , βMω2 2 − (4.170)

thus NβMω2 β n (r) = exp Mω2r2 . (4.171) 2 2πL exp βMω R2 1 2 2 − 9. In the classical limit the probability distribution of the velocity vector v satisﬁes Mv2 f (v) exp , (4.172) ∝ − 2τ where v = v . | |

Eyal Buks Thermodynamics and Statistical Physics 155 Chapter 4. Classical Limit of Statistical Mechanics

a) By changing the integration variable Mv2 x = (4.173) 2τ one ﬁnds

2 ∞ dvv3 exp Mv 0 − 2τ v = 2 (4.174) ∞ dvv2 exp Mv 0 − 2τ 1/2 2τ ∞ dxx exp ( x) = 0 − M ∞ dxx1/2 exp ( x) 0 − 1/2 8τ = . πM (4.175) b) Similarly

2 ∞ dv v4 exp Mv 2 0 − 2τ v = 2 ∞ 2 Mv 0 dv v exp 2τ − 2τ ∞ dx x3/2 exp ( x) = 0 − M ∞ dx x1/2 exp ( x) 0 − 2τ 3 = , M 2 (4.176) thus 3τ 1/2 3π 1/2 v2 = = v . (4.177) M 8 10. For each gas the density is given by n (z) = n (0) exp ( βM gz) , l l − l where l 1, 2 , 0 z h and the normalization constant is found from the requirement∈ { } ≤ ≤

S n (z) dz = Nl , (4.178) 0 therefore

Nl βMlgNl nl (0) = = . (4.179) h βMlgh S (1 e− ) S exp ( βMlgz) dz − 0 − Eyal Buks Thermodynamics and Statistical Physics 156 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 M N M N g = 1 1 + 2 2 . exp (βM gh) 1 exp (βM gh) 1 S 1 − 2 − (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 − dn H (X) = exp 2Xt t2 . (4.181) n dtn − t=0 2 The identity 2Xt t2 = X2 (X t) yields − − − dn H (X) = exp X2 exp (X t)2 . (4.182) n dtn − − t=0 Moreover, using the relation d d exp (X t)2 = exp (X t)2 , (4.183) dt − − −dX − − one ﬁnds that dn H (X) = exp X2 ( 1)n exp (X t)2 n − dXn − − t=0 dn = exp X2 ( 1)n exp X2 . − dXn − (4.184)

Note that for an arbitrary function g (X) the following holds

d d exp X2 exp X2 g = 2X g , (4.185) − dX − − dX and X2 d X2 d exp X exp g = 2X g , (4.186) 2 − dX − 2 − dX thus X2 d n X2 H (X) = exp X exp . (4.187) n 2 − dX − 2

Eyal Buks Thermodynamics and Statistical Physics 157 Chapter 4. Classical Limit of Statistical Mechanics

12. With the help of Eq. (4.184) and the general identity

∞ 2 2 π 1 4ca+b exp ax + bx + c dx = e 4 a , (4.188) − a −∞ according to which the following holds (for the case a = 1, b = 2iX and c = 0)

1 ∞ exp X2 = exp x2 + 2iXx dx , (4.189) − √π − −∞ one ﬁnds that exp X2 d n ∞ H (X) = exp x2 + 2iXx dx n √π −dX − −∞ 1 ∞ = ( 2ix)n exp X2 x2 + 2iXx dx √π − − −∞ ∞ 1 2 = ( 2ix)n e(X+ix) dx , √π − −∞ (4.190) thus the following holds [see Eq. (4.188)] α n ∞ 2 Hn (X) Hn (Y ) n! n=0 ∞ ∞ n 1 2 2 ∞ ( 2αxy) = dx dy e(X+ix) e(Y +iy) − π n! n=0 −∞ −∞ e−2αxy ∞ ∞ 1 (X+ix)2 (Y +iy)2 2αxy = dx e dy e e− π −∞ −∞ √πeαx(αx−2iY ) ∞ 1 1 α2 x2+2i(X Y α)x+X2 = dx e−( − ) − √π −∞ α 2XY αX2 αY 2 ( − − ) exp 1 α2 = − . √1 α2 − (4.191)

Eyal Buks Thermodynamics and Statistical Physics 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) = Mω2x2 , (5.1) 2 where the angular frequency ω is a constant.

a) Calculate the canonical partition function of the system c,B for the case where the particles are Bosons. Z b) Calculate the canonical partition function of the system c,F for the case where the particles are Fermions. Z 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 = µ0H and ε+ = µ0H, where the − − magnetic moment µ0 is a constant. Calculate the magnetization of the system, which is defined by

∂F M = , (5.2) − ∂H 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

2 3/2 ψ (x, y, z) = sin (k x) sin (k y) sin (k z) , (5.3) L x y z Chapter 5. Exercises

is given by

ε (k) = kc , (5.4)

2 2 2 where c is the speed of light and where k = kx + ky + kz (contrary to the non-relativistic case where it is given byε (k) = 2k2/2M). 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 one- dimensional potential with coordinates x1 and x2 respectively. The potential energy is given by mω2x2 mω2x2 V (x , x ) = 1 + 2 + mΩ2 (x x )2 , (5.5) 1 2 2 2 1 − 2 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 2 a) x1 for the case Ω = 0. b) x2 , however without assuming that Ω = 0. 1 c) (x x )2 , again without assuming that Ω = 0. 1 − 2 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

2 3/2 ψ (x, y, z) = sin (k x) sin (k y) sin (k z) , (5.6) L x y z is given by

ε (k) = kc , (5.7)

2 2 2 where c is the speed of light and where k = kx + ky + kz (contrary to the non-relativistic case where it is given byε (k) = 2k2/2M). 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 ε = µ H (σ + σ ) + Jσ σ , (5.8) σ1,σ2 − 0 1 2 1 2 where both σ1 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 ∂M χ = lim , (5.9) H 0 ∂H → where ∂F M = (5.10) − ∂H 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) = mω2x2 , (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 p = , (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 vol−ume.

5.2 Solutions

1. The single particle eigen energies are given by 1 ǫ = ω n + , (5.13) n 2 where n = 0, 1, 2, . ···

Eyal Buks Thermodynamics and Statistical Physics 161 Chapter 5. Exercises

a) For Bosons 1 ∞ ∞ 1 ∞ = exp [ β (ǫ + ǫ )] + exp ( 2βǫ ) Zc,B 2 − n m 2 − n n=0 m=0 n=0 2 1 ∞ 1 ∞ = exp ( βǫ ) + exp ( 2βǫ ) 2 − n 2 − n n=0 n=0 exp ( βω) exp ( βω) = − + − . 2 (1 exp ( βω))2 2 (1 exp ( 2βω)) − − − − (5.14) Note that the average energy UB is given by

2βω βω ∂ log c,B 1 + 2e− + e− UB = Z = ω . (5.15) − ∂β 1 e 2βω − − b) For Fermions 1 ∞ ∞ 1 ∞ = exp [ β (ǫ + ǫ )] exp ( 2βǫ ) Zc,F 2 − n m − 2 − n n=0 m=0 n=0 exp ( βω) exp ( βω) = − − . 2 (1 exp ( βω))2 − 2 (1 exp ( 2βω)) − − − − (5.16) Note that for this case the average energy UF is given by

2βω βω ∂ log c,F 2 + e− + e− UF = Z = ω . (5.17) − ∂β 1 e 2βω − − 2. The orbital eigenenergies are given by 2 π 2 ε = n2 , (5.18) n 2m L where n = 1, 2, 3, . The grandcanonical partition function of the gas is given by ···

gc = ζn , (5.19) Z n where

ζ = (1 + λ exp ( βε ) exp ( βE )) (5.20) n − n − l l is the orbital grandcanonical Fermionic partition function where,

η λ = exp (βµ) = e− , (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 can be written as Zgc ∞ log = log ζ Zgc n n=1 ∞ = 2 log (1 + λ exp ( βε )) − n n=1 ∞ 2 π 2 2 dn log 1 + λ exp β n2 . ≃ − 2m L 0 (5.22) By employing the variable transformation 2 π 2 ε = n2 , (5.23) 2m L one has

1 ∞ log = dε D (ε) log (1 + λ exp ( βε)) , (5.24) Zgc 2 − 0 where

2L 2m ε 1/2 ε 0 D (ε) = π 2 − ≥ (5.25) 0 ε < 0 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 U = Zgc , (5.26) − ∂β η ∂ log N = λ Zgc , (5.27) ∂λ one ﬁnds that

∞ U = dε D (ε) εfFD (ε) , (5.28) −∞ ∞ N = dε D (ε) fFD (ε) , (5.29) −∞ where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)] 1 f (ǫ) = . (5.30) FD exp [β (ǫ µ)] + 1 −

Eyal Buks Thermodynamics and Statistical Physics 163 Chapter 5. Exercises

At zero temperature, where µ = εF one has

εF D (εF) 1/2 2D (εF) 2 U = 1/2 dε ε = εF , (5.31) ε− 3 F 0 εF D (εF) 1/2 N = 1/2 dε ε− = 2D (εF) εF , (5.32) ε− F 0 thus U N = . (5.33) εF 3 3. The Helmholtz free energy is given by n F = Nτ log log Z 1 , (5.34) n − int − Q where

Z = exp (βµ H) + exp ( βµ H) = 2 cosh (βµ H) (5.35) int 0 − 0 0 is the internal partition function. Thus the magnetization is given by ∂F M = = Nµ tanh (βµ H) . (5.36) − ∂H 0 0 τ 4. The grandcanonical partition function of the gas is given by

gc = ζn , (5.37) Z n where

ζ = (1 + λ exp ( βεn) exp ( βE )) (5.38) n − − l l is a grandcanonical Fermionic partition function of an orbital having energy εn given by πcn ε = , (5.39) n L

where n = n2 + n2 + n2, n , n , n = 1, 2, 3, , x y z x y z ··· η λ = exp (βµ) = e− (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 can be written as Zgc ∞ ∞ ∞ log = log (1 + λ exp ( βεn) exp ( βE )) . (5.41) Zgc − − l l n =1 n =1 n =1 x y z For a macroscopic system the sum over n can be approximately replaced by an integral

∞ ∞ ∞ 4π ∞ dn n2 , (5.42) → 8 n =0 n =0 n =0 x y z 0 thus, one has

4π ∞ πcn log = 2 dn n2 log 1 + λ exp β . (5.43) Zgc 8 − L 0 By employing the variable transformation πcn ε = . (5.44) L one has

∞ V ε2 log = dε log (1 + λ exp ( βε)) . (5.45) Zgc π23c3 − 0 The energy U and the number of particles N are given by

∂ log ∞ V ε3 U = gc = dε f (ǫ) , (5.46) Z 23 3 FD − ∂β η π c 0 ∂ log ∞ V ε2 N = λ Zgc = dε f (ǫ) , (5.47) ∂λ π23c3 FD 0

where fFD is the Fermi-Dirac distribution function [see Eq. (2.35)] 1 f (ǫ) = . (5.48) FD exp [β (ǫ µ)] + 1 − At zero temperature

εF V ε3 V ε4 U = dε = F , (5.49) π23c3 π23c3 4 0 εF V ε2 V ε3 N = dε = F , (5.50) π23c3 π23c3 3 0

Eyal Buks Thermodynamics and Statistical Physics 165 Chapter 5. Exercises

and therefore 3N U = ε . (5.51) 4 F The energy U can be expressed as a function of V and N as

4/3 2 3 3 1/3 1/3 (3N) π c V − U = . 4 At zero temperature the Helmholtz free energy F equals the energy U, thus the pressure p is given by

4/3 1/3 ∂F ∂U 1 3N π23c3 p = = = V , (5.52) − ∂V − ∂V 3 4 τ,N τ,N thus p 1 = . (5.53) U 3V 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)]

5 n ∂ (τ log Z ) σ = N + log Q + int , (5.54) 2 n ∂τ 3 c = N + h , (5.55) V 2 int cp = cV + N, (5.56) where n = N/V is the density,

Mτ 3/2 n = (5.57) Q 2π2 is the quantum density, M is the mass of a particle in the gas, and ∂2 (τ log Z ) c 3 h = τ int = V . (5.58) int ∂τ 2 N − 2

The requirement that hint is temperature independent leads to

∂ (τ log Zint) τ = gint + hint log , (5.59) ∂τ τ 0

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 V2τ 2 cV 3 τ 2 ∆σ = σ2 σ1 = N log 3/2 + log − N − 2 τ 1 V1τ 1 cV V τ N = N log 2 2 . V τ 1 1 (5.60) Thus the total change in the entropy of the mixture is given by

∆σ = ∆σA + ∆σB cV,A cV,B V τ NA V τ NB = N log 2 2 + N log 2 2 (5.61) A V τ B V τ 1 1 1 1 cV,A+cV,B N +N V2 τ 2 A B = (NA + NB) log , (5.62)  V τ  1 1 and the requirement ∆σ = 0 leads to 

cV,A+cV,B V τ NA+NB 2 2 = 1 . (5.63) V τ 1 1 Alternatively, by employing the equation of state

pV = Nτ , (5.64)

this can be rewritten as

c +c c +c V,A V,B +1 V,A V,B V NA+NB p NA+NB 2 2 = 1 , (5.65) V p 1 1 or with the help of Eq. (5.56) as

cV,A+cV,B +1 NA+NB c +c V2 − V,A V,B p = p NA+NB 2 1 V 1 cp,A+cp,B V cV,A+cV,B = p 1 . 1 V 2 (5.66) 6. It is convenient to employ the coordinate transformation

x1 + x2 x+ = , (5.67) √2 x x x = 1 − 2 . (5.68) − √2 The inverse transformation is given by

Eyal Buks Thermodynamics and Statistical Physics 167 Chapter 5. Exercises

x+ + x x1 = − , (5.69) √2 x+ x x2 = − − . (5.70) √2 The following holds

2 2 2 2 x1 + x2 = x+ + x , (5.71) − and

2 2 2 2 x˙ 1 +x ˙ 2 =x ˙ + +x ˙ . (5.72) − Thus, the kinetic energy T of the system is given by

2 2 2 2 m x˙ 1 +x ˙ 2 m x˙ + +x ˙ T = = − , (5.73) 2 2 and the potential energy V is given by mω2x2 mω2x2 V (x , x ) = 1 + 2 + mΩ2 (x x )2 1 2 2 2 1 − 2 2 2 2 2 2 mω x+ m ω + 4Ω x = + − . 2 2 (5.74) The equipartition theorem yields

2 2 2 2 2 mω x+ m ω + 4Ω x τ = − = , (5.75) 2 2 2 thus 2τ (x + x )2 = , (5.76) 1 2 mω2 and 2τ (x x )2 = . (5.77) 1 − 2 m (ω2 + 4Ω2) Furthermore, since by symmetry x+x = 0 one has − 2 1 2 x1 = (x+ + x ) 2 − 1 2 2 = x+ + x 2 − 4Ω2 τ 1 2 = 1 ω . mω2 2 4Ω2 − 1 + ω2 (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 πn k = , (5.79) L where

n = (nx, ny, nz) , (5.80)

and nx, ny, nz = 1, 2, 3, . The single particle partition function is given by ···

∞ ∞ ∞ ε (k) Z = exp . (5.81) 1 − τ n =1 n =1 n =1 x y z Approximating the discrete sum by a continuous integral according to

∞ ∞ ∞ 4π ∞ dn n2 , (5.82) → 8 n =0 n =0 n =0 x y z 0 one has 4π ∞ nπc Z = dn n2 exp 1 8 − Lτ 0 4V τ 3 ∞ = dx x2 exp ( x) 8π23c3 − 0

2 V τ 3 = . π23c3 (5.83)

In the classical limit the grandcanonical partition function gc is given by [see Eq. (2.44)] Z

log = λZ , (5.84) Zgc 1 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 −

η V log gc = e− . (5.85) Z π23c3β3

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

∂ log 3 U = Zgc = log , (5.86) − ∂β β Zgc η ∂ log N = Zgc = log , (5.87) − ∂η Zgc β thus

U = 3Nτ , (5.88)

and V τ 3 η = log . (5.89) π2N3c3 b) The entropy is evaluate using Eq. (1.86) σ = log + βU + ηN Zgc = N (1 + 3 + η) V τ 3 = N 4 + log , π2N3c3 (5.90) and the Helmholtz free energy by the deﬁnition (1.116)

V τ 3 F = U τσ = Nτ 1 + log . (5.91) − − π2N3c3 Thus the pressure p is given by ∂F Nτ p = = . (5.92) − ∂V V τ,N 8. The partition function is given by Z = exp ( βε ) − σ1,σ2 σ ,σ = 1 1 2 ± = exp ( βJ) [exp ( 2βµ H) + exp (2βµ H)] + 2 exp (βJ) , − − 0 0 (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 M = − ∂H τ 2µ exp ( βJ)[ exp ( 2βµ H) + exp (2βµ H)] = 0 − − − 0 0 , exp ( βJ) [exp ( 2βµ H) + exp (2βµ H)] + 2 exp (βJ) − − 0 0 (5.95)

Eyal Buks Thermodynamics and Statistical Physics 170 5.2. Solutions

and the magnetic susceptibility is given by 4βµ2 χ = 0 . (5.96) 1 + e2βJ Note that in the high temperature limit βJ 1 ≪ 2µ2 χ 0 . (5.97) ≃ τ + J 9. The orbital eigenenergies in this case are given by 1 ε = ω n + , (5.98) n 2 where n = 0, 1, 2, . The grandcanonical partition function of the gas is given by ···

gc = ζn , (5.99) Z n where ζ = (1 + λ exp ( βε ) exp ( βE )) (5.100) n − n − l l is the orbital grandcanonical Fermionic partition function where,

η λ = exp (βµ) = e− (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 can be written as Zgc ∞ log = log ζ Zgc n n=0 ∞ = 2 log (1 + λ exp ( βε )) . − n n=0 (5.102) The number of particles N is given by

∂ log ∞ N = λ Zgc = 2 f (ε ) , (5.103) ∂λ FD n n=0 where fFD is the Fermi-Dirac distribution function 1 f (ε) = . (5.104) FD exp [β (ε µ)] + 1 −

Eyal Buks Thermodynamics and Statistical Physics 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 2ε N = F , (5.105) ω thus Nω µ = ε = . (5.106) F 2 b) Using the approximation

f (ε) exp [ β (ε µ)] , (5.107) FD ≃ − − 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 dn exp ( βωn) − 2 − 0 2 exp β µ ω = − 2 , βω (5.108) thus Nβω βω µ = τ log + 2 2 Nβω τ log . ≃ 2 (5.109) 10. Using Eq. (2.239), which is given by

∂p ∂V c c = τ , (5.110) p − V ∂τ ∂τ V,N p,N one ﬁnds that

2 A τ 2 2(n 1) 2 n 1 c c = n τ − = n Aτ − . (5.111) p − V pV

Eyal Buks Thermodynamics and Statistical Physics 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 information theory, 1 Bose-Einstein function, 54 internal degrees of freedom, 59 Boson, 52 isentropic process, 65 isobaric process, 64 canonical conjugate momentum, 129 isochoric process, 65 canonical distribution, 15 isothermal process, 64 Carnot, 66 cavity, 99 Kelvin’s principle, 70 chemical potential, 17, 20 classical limit, 54 Lagrange multiplier, 4 Clausius’s principle, 70 Langevin equation, 138 composition property, 5 largest uncertainty estimator, 9 Coulomb gauge, 99 Maxwell’s equations, 99 Debye temperature, 110 microcanonical distribution, 14 density function, 132 density of states, 114 Nyquist noise, 134 eﬃciency, 68, 71 orbitals, 50 electromagnetic radiation, 99 Ornstein—Uhlenbeck process, 145 entropy, 1 equipartition theorem, 132 partition function, 10 permeability, 99 Fermi energy, 115 permittivity, 99 Fermi-Dirac distribution, 53 phonon, 107 Fermi-Dirac function, 53 photon, 102 Fermion, 52 Plank’s radiation law, 105 Fokker-Planck equation, 139 pressure, 57 free energy, 142 fugacity, 18, 53 quantum density, 48 grandcanonical distribution, 16 Shannon, 1 Sommerfeld expansion, 115 H theorem, 18 Stefan-Boltzmann constant, 106 Hamilton-Jacobi equations, 130 Stefan-Boltzmann radiation law, 105 Hamiltonian, 129 heat capacity, 59 temperature, 15, 17 Helmholtz free energy, 21, 57 thermal equilibrium, 19 ideal gas, 47 vector potential, 99