Physics 831: Statistical Mechanics

Russell Bloomer1 University of Virginia

Note: There is no guarantee that these are correct, and they should not be copied

1 Problem Set 1 1 1.1 Kittel 8.1: Heat pump ...... 1 1.2 Kittel 8.6: Room air conditioner ...... 1 1.3 Kittel 8.9: Cooling of nonmetallic solid to T = 0 ...... 2 1.4 Sterling Heat Engine ...... 2 1.5 Unavailability for work ...... 3 1.6 Gibbs Free Energy ...... 3

2 Problem Set 2 5 2.1 Spin model ...... 5 2.2 Paramagnetism of a system of N localized spin-1/2 particles ...... 6 2.3 Kittel 2.3: Quantum harmonic oscillator ...... 7 2.4 Review Thermal mechanics ...... 7 2.5 Review Thermal mechanics ...... 8

3 Problem Set 3 9 3.1 Kittel 3.2: Magnetic susceptibility ...... 9 3.2 Kittel 3.3: Free energy of a harmonic oscillator ...... 10 3.3 Kittel 3.4: Energy ﬂuctuations ...... 10 3.4 Kittel 3.10: Elasticity of polymers ...... 11 3.5 Ising spin chain ...... 12

4 Problem Set 4 13 4.1 Application of equal partition theorem ...... 13 4.2 Thermal Equilibrium of the Sun and Earth ...... 14 4.3 Kittel 4.3: Average temperature of the interior of the Sun ...... 14 4.4 Kittel 4.6: Pressure of thermal radiation ...... 15 4.5 Kittel 4.7: Free energy of a photon gas ...... 15 4.6 Kittel 4.18: Isentropic expansion of photon gas ...... 16

5 Problem Set 5 17 5.1 5.1: Kittel 4.14: Heat capacity of liquid 4He at low temperature ...... 17 5.2 5.2: Kittel 4.15: Angular distribution of radiant energy flux ...... 17 5.3 5.3: Qualifying exam file problem ...... 17 5.4 5.4: Helmholtz free energy for the Debye model ...... 17 5.5 5.5: Kittel 5.3: Potential energy of gas in gravitational field ...... 17

i 6 Problem Set 6 19 6.1 6.1: Kittel 5.4: Active transport ...... 19 6.2 6.2: Kittel 5.7: States of positive and negative ionization ...... 19 6.3 6.3: Kittel 5.10: Concentration ﬂuctuation ...... 20 6.4 6.4: Kittel 6.3: Distribution function for double occupancy statistics ...... 20 6.5 6.5: Kittel 6.7: Relation of pressure and energy density ...... 21 6.6 6.6: Kittel 6.9: Gas of atoms with internal degree of freedom ...... 21

7 Problem Set 7 23 7.1 7.1: Pressure in types of gases ...... 23 7.2 7.2: Kittel 5.13: Isentropic expansion ...... 23 7.3 7.3: Kittel 6.8: Time for a large ﬂuctuation ...... 23 7.4 7.4: Kittel 6.10: Isentropic relations of ideal gas ...... 23 7.5 7.5: Kittel 6.12: Ideal gas in two dimensions ...... 23 7.6 7.6: Kittel 7.4: Chemical potential versus temperature ...... 23 7.7 7.7: The absorbtion of gas onto a surface ...... 23

8 Problem Set 8 25 8.1 8.1: Mixing of two distinct atoms ...... 25 8.2 8.2: Kittel 7.3: Pressure and entropy of degenerate Fermi gas ...... 25 8.3 8.3: Kittel 7.6: Mass-radius relationship for white dwarfs ...... 26 8.4 8.4: Kittel 7.10: Relativistic white dwarfs stars ...... 27 8.5 8.5: Electrons in the air oﬀ a conductor ...... 27

9 Special Problem Set (9) 29 9.1 Problem 1: Properties of “Photon Gas” ...... 29 9.2 Problem 2: Engine Cycle ...... 29 9.3 Problem 3: Vibrational Modes of a Molecule ...... 30 9.4 Problem 4: Relativistic Massless Bosons ...... 31 9.5 Problem 5: Kittel & Kroemer 7.9 and more ...... 32

10 Problem Set 10 33 10.1 10.1: Collisions with a wall for a Fermi Gas ...... 33 10.2 10.2: Free energy and pressure of a Boson gas ...... 33 10.3 10.3: Discontinuity in the slope of the heat capacity of a Bose gas ...... 34 10.4 10.4: Maximum work extracted from an ideal gas ...... 34

11 Problem Set 11 37 11.1 11.1:A review problem ...... 37 11.2 11.2: Dissociation of water ...... 37 11.3 11.3: Practice with the Jacobian ...... 38 11.4 11.4: More practice ...... 39 11.5 11.5: van der Waals Gas ...... 39

12 Problem Set 12 41 12.1 12.1: Maxwell Relations ...... 41 12.2 12.2: Equilibrium conditions ...... 41 12.3 12.3: Fluctuation in number of a Fermi gas ...... 42 12.4 12.4: Fluctuation in volume, pressure, entropy and temperature ...... 42 12.5 12.5: Kittel 10.5: Gas-solid equilibrium ...... 42

ii 13 Problem Set 13 45 13.1 13.1: Superconduction and Heat Capacity ...... 45 13.2 13.2: Kittel 10.8: First order crystal transformation ...... 45 13.3 13.3: Kittel 11.2: Mixing energy in 3He −4 He and P b − Sn mixtures ...... 46 13.4 13.4: Kittel 11.4: Solidiﬁcation range of a binary alloy ...... 46 13.5 13.5: Kittel 11.5: Alloying of gold into silicon ...... 47

iii Chapter 1

Problem Set 1

1.1 Kittel 8.1: Heat pump

(a) Show that for a reversible heat pump the energy required per unit of heat delivered inside the building is given by the Carnot eﬃciency

W τh − τl = ηC = (1.1) Qh τh What happens if the heat pump is not reversible?

Ql Qh For a reversible system: σh = σl. From the deﬁnition of entropy σl = ; σh = . Then τl τh

W Qh − Ql σhτh − σlτl ⇒ τh − τl = = σh = σl = ηC X Qh Qh σhτh τh

(b) Assume that the electricity consumed by a reversible heat pump must itself be generated by a Carnot engine operating between the temperatures τhh and τl. What is the ratio Qhh/Qh, of the beat consumed at τhh, to the heat delivered at τh? Give numerical values for Thh = 600 K; Th = 300 K; Tl = 270 K. For a Carnot engine

W τhh − τl τhh − τl = ⇒ W = Qhh Qhh τhh τhh The ratio is then

τ Q τ W/(τ − τ ) Q (τ − τ )/τ Q 1 − l hh = hh hh l ⇒ hh = h l h ⇒ hh = τh τl Qh τhW/(τh − τl) Qh (τhh − τl)/τhh Qh 1 − τhh

For τhh = 600 K, τh = 300 K and τl = 270 K,

270 1 − 300 270 = .18 X 1 − 600 (c) Draw an energy-entropy ﬂow diagram for the combination heat engine-heat pump, similar to Figures 8.1, 8.2 and 8.4, but involving no external work at all, only energy and entropy ﬂows at three temperatures.

1.2 Kittel 8.6: Room air conditioner

A room air conditioner operates as a Carnot cycle refrigerator between an outside temperature Th and a room at a lower temperature Tl. The room gains heat from the outdoors at a rate A(Th − Tl); this heat is removed by the air conditioner. The power supplied to the cooling unit is P . (a) Show that the steady state temperature of the room is

1 2 21/2 Tl = (Th + P/2A) − (Th + P/2A) − Th (1.2)

dQl The rate is dt = A(Th − Tl), so the power is

dW 1 dQ (T − T )2 P = = = A h l dt γ dt Tl

Solving for Tl

P q T 2 − (2T − P/A) T + T 2 = 0 ⇒ T = T + − (T + P/2A)2 − T 2 l h l h l h 2A h h X

(b) If the outdoors is at 37oC and the room is maintained at 17oC by a cooling power 2 kW, ﬁnd the heat loss coeﬃcient A of the room in WK−1.

s 1000 1000 2 40000 (1000)2 620000 (1000)2 290 = 310 + − 310 + − 3102 ⇒ 400 − + = 3102 + + − 3102 ⇒ A = 1450 W/K A A A A2 A A2 X 1.3 Kittel 8.9: Cooling of nonmetallic solid to T = 0

Wa saw in Chapter 4 that the heat capacity of nonmetallic solids at suﬃciently low temperatures is proportional to T 3, as C = aT 3. Assume it were possible to cool a piece of such a solid to T = 0 by means of a reversible refrigerator that uses the solid specimen as its (varying!) low-temperature reservoir, and for which the high-temperature reservoir has a ﬁxed temperature Th equal to the initial temperature Ti of the solid. Find an expression for the electrical energy required. Th−Tl 3 Rate of change in work is dW = d(Qh − Ql) = γdQl = dQl. We are given C = aT Tl

3 dQl = −dTl ⇒ dQl = −aTl dTl

Therefore the work is

a 3 dW = − (Th − Tl) Tl dTl Tl

For cooling Th → 0

Z Z 0 2 3 a 0 a 4 0 a 4 W = −a ThTl − Tl dTl ⇒ W = − ThT + T ⇒ W = Th X 3 Th 4 Th 12 Th 1.4 Sterling Heat Engine

The operation of a certain type of engine involves applying two isothermal steps and two isovolumetric steps per cycle to one mole of diatomic gas. The largest and smallest volumes shown are VL and VS , respectively. You do not need to consider the vibrational degree of freedom of the molecules.

(a) Find the eﬃciency of the engine in terms of Th, Tc, VL and VS . From 1 → 2 the process is isothermal so pV = constant, so the work is

Z Z VL cdV VL VL Q12 = pdV ⇒ Q12 = = c ln V ⇒ Q12 = nRTh ln V VS V VS S 5 VL The heat along 2 → 3: Q23 = R (Tc − Th). For 3 → 4: Q34 = nRTc ln . Finally, for 4 → 1: Q41 = 2 VS 5 2 R (Th − Tc). Then VL 5 5 VL VL VL nRTh ln + R (Th − Tc) + R (Tc − Th) − nRTc ln Th ln − Tc ln W VS 2 2 VS VS VS = ⇒ η = X Qh VL 5 VL 5 nRTh ln + R (Th − TL) Th ln + (Th − Tc) VS 2 VS 2

2 (b) Prove that this engine has a lower eﬃciency than that of Carnot engine operating at the highest and lowest temperatures in the cycle.

V (T − T ) ln L Tc h c V 1 − η η = S = Th = C 5 ηC 5 VL 5/2 Tc 1 + (Th − Tc) + Th ln 1 + 1 − 2 ln(VL/VS ) 2 VS ln(VL/VS ) Th where ηC is the Carnot’s eﬃciency. Now 5 η C > 0 2 ln (VL/VS ) because ηC > 0 and VL > VS ⇒ ln (VL/VS ) > 0. ∴ η < ηC X

1.5 Unavailability for work

A certain engine works in a cycle between reservoirs at Th and Tc. Find the diﬀerence in the work dome by an otherwise ideal engine (Carnot engine) and that done by the above engine in terms of ∆SU , where ∆SU is the change in entropy in the universe due to the operation of the engine. The change in the entropy on the Th is dUh = dQh = τhdσh. The low side, dUc = dQc = τcdσc. By conservation dQh + dQc = 0. Then dσirr = dσh + dσc = dQh/τh + dQc/τc. Then 1 1 τc − τh dσirr = − dQh = dQh τh τc τhτc

The heat ﬂow is independent of the temperature, so solving for Qh

τc − τh τhτc ∆σirr = ∆Qh ⇒ ∆Qh = ∆σirr τhτc τc − τh

τh−τc An ideal engine has W = ηC Qh where ηC = . Substituting the change in entropy to ﬁnd the change in work τh

τh − τc τhτc ∆W = ∆σirr ⇒ ∆W = −τc∆σirr τh τc − τh For the universe the sign is change, so in conventional units

∆W = Tc∆SU X

1.6 Gibbs Free Energy

The Gibbs free energy of a certain system is given by the follow formula:

aτ 5/2 G = −nτ ln (1.3) P where a is a constant. Compute (a) the entropy

∂G aτ 5/2 5 1 aτ 5/2 5 G = U + pV − τσ ⇒ = −σ ⇒ σ = N ln + Nτ ⇒ σ = N ln + N X ∂τ pV p 2 τ p 2

(b) the heat capacity at constant P , ie Cp

∂σ ∂ aτ 5/2 5 5N 5 Cp = τ = τ σ = N ln + N = τ ⇒ Cp = N X ∂τ p ∂τ p 2 2τ 2

3 (c) the equation of state relating P , V , and τ

∂G −1 V = = −Nτ ⇒ pV = Nτ ∂p p X

(d) the energy

aτ 5/2 5 aτ 5/2 3 G = U − τσ + pV ⇒ U = G + τσ − pV = Nτ ln + Nτ − Nτ − Nτ ln ⇒ U = Nτ p 2 p 2 X

(e) the chemical potential

∂G aτ 5/2 µ = ⇒ µ = −τ ln X ∂N pτ p

4 Chapter 2

Problem Set 2

2.1 Spin model

In an isolated system of a large number N of localized particles of spin 1/2, each particle has a magnetic moment of µ which can point either parallel or antiparallel to an applied magnetic ﬁeld H. The energy of the system when n ↑ and n ↓ is given by U = −(n ↑ −n ↓)µH. Consider the energy range between U and U + δU where δU is very much smaller than U but may be microscopically large so that δU µH. (a) What is the total number of states Ω(U) lying in this energy range? This problem is much like the random walk problem in that there are only two options for the spins to align along and that they are equally as likely. With that, the multiplicity is a binomial distribution. Deﬁning m ≡ n ↑ −n ↓, the energy becomes U = −mµH. From the random walk N! Ω(m) = N+m N−m 2 ! 2 ! U Solving m in terms of the energy m = − µH , the distribution becomes N! Ω(U) = U U N− µH N+ µH 2 ! 2 !

Because δU is small compared to the number of available state and that is much greater than µH, δU is the width of δU the distribution. This has to be unit of the number, so it becomes δU → 2µH . The ﬁnal solution for U + δU becomes N! δU Ω(U) = U U X N− µH N+ µH 2µH 2 ! 2 !

(b) Assume that the energy U is in a region where Ω(U) is appreciable. Apply the Gaussian approximation to part (a) to obtain a simple expression for Ω(U)dU as a function of U. To ﬁnd the Gaussian approximation the logarithm of Ω needs to be expanded. The expansion of a binomial is given by Equation 2.341. 1 log g =∼ log 2/πN + N log 2 − 2s2/N (2.1) 2 m U For this problem s = 2 = 2µH . This approximation becomes 1 log (Ω(U) + δU) =∼ log 2/πN + N log 2 2 U 2 δU − 2 /N + log 2µH 2µH δU goes to dU as the function becomes continuous. This should become a Gaussian once the removal of logarithm. 2 −2 U /N dU Ω(U)dU = (2/πN)1/2 2N e 2µH 2µH X 1Kittel and Kroemer pg 20

5 2.2 Paramagnetism of a system of N localized spin-1/2 particles

(a) Using expression for Ω(U) calculated in problem 2.1(a) and applying Stirling approximation, ﬁnd the relation between the absolute “temperature” τ = kB T and the total energy U of this system. The temperature is deﬁned by 1 ∂ (log(g)) = (2.2) kB T ∂U N The logarithm of 2.1(a) is

N − U ! N + U ! log Ω = log N! − log µH ! − log µH ! 2 2

Using Stirling’s approximation log N! ' N log N − N log Ω = N log N − N µHN − U µHN − U µHN − U − log − 2µH 2µH 2µH µHN + U µHN + U µHN + U + log − 2µH 2µH 2µH µHN − U µHN − U = N log N − log 2µH 2µH µHN + U µHN + U + log 2µH 2µH Now diﬀerentiating with respect to U ∂ (log(Ω)) 1 µHN − U = log ∂U N 2µH µH µHN + U − log µH This means with respect to temperature c (b) Under what circumstance is τ negative? For the temperature to be negative the logarithm has to be negative. The only time that happens, when the term inside is less than one. This occurs whenever there are more spins against the ﬁeld than pointing with it. In that case, U > 0, so µHN − U < 1 → µHN − U < µHN + U µHN + U X (c) Find the magnetization as a function of H and T .

µHN − U −1 k T = 2µH log B µHN + U 2µH µHN − U e kB T = µHN + U 2µH 2µH U 1 + e kB T = µHN e kB T − 1

2µH e kB T − 1 µH U = µHN 2µH → U = µHN tanh kB T 1 + e kB T The magnetization is deﬁned as the number of magnetic moments per unit volume. Deﬁne n = U/HV µH µH M = (U/HV ) µ tanh → nµ tanh X kB T kB T

6 2.3 Kittel 2.3: Quantum harmonic oscillator

(a) Find the entropy of a set of N oscillators of frequency ω as a function of the total quantum number n. Use multiplicity function (1.55)2 and make the Stirling approximation log N! ' N log N − N. Replace N − 1 by N. Equation 1.55 is (N + n − 1)! g(N, n) = (2.3) (N − 1)!(n)! Replacing N − 1 with N in equation 3 (N + n)! g(N, n) = (N)!(n)! The entropy is

σ = log g (2.4)

Then by Stirling approximation

σ = (N + n) log(N + n) − (N + n) − N log N + N − n log n + n N + n N + n = N log + n log N n X

(b) Let U denote the total energy n~ω of the oscillators. Express the entropy as σ(U, N). Show that the total energy at temperature τ is N ω U = ~ (2.5) e~ω/τ − 1

The energy is U = n ω → U , so the entropy becomes ~ ~ω ! ! N + U U N + U σ = N log ~ω + log ~ω N ~ω U ~ω Then the inverse temperature is

1 ∂ (σ) 1 = = log (N~ω/U + 1) τ ∂U N ~ω ω ~ = log (N ω/U + 1) τ ~ ω/τ N~ω N~ω e~ − 1 = → U = X U e~ω/τ − 1 2.4 Review Thermal mechanics

Focus on the two stages in the engine cycle of problem 1.4: The isothermal expansion stage followed by the isovolumetric stage. The working substance in the engine consist of N helium atoms. What is the change in the multiplicity of the gas as it traverses the two stages, with the volume increasing from VS (@τh to VL(@τc)? The entropy for an ideal gas (which helium closely approximates) is given by Equation 3.76, which is

5 σ = N log (n /n) + (2.6) Q 2

2 3/2 where n ≡ N/V and n ≡ (mτ/2π~ ) . So the multiplicity is

n N V N mτ 3N/2 g = eσ = e5N/2 Q = e5N/2 n N 2π~2 2Kittel and Kroemer pg. 25

7 The multiplicity at VS (@τh

N 3N/2 5N/2 VS mτh g1 = e N 2π~2

The multiplicity at VL(@τc

N 3N/2 5N/2 VL mτc g2 = e N 2π~2 So the diﬀerence is then

N 3N/2 5N/2 VS mτh g1 − g2 = e N 2π~2 V N mτ 3N/2 − e5N/2 L c N 2π~2 5N/2 3N/2 e m N 3N/2 N 3N/2 = VS τh − VL τc X N N 2π~2 2.5 Review Thermal mechanics

Suppose a body at temperature τ and pressure P is immersed in a medium that is at the temperature τo and pressure Po, where τ and pressure P are not necessary equal to τo and pressure Po. The body and medium form a closed system. Now an external source does work to change the state of the body. The source of work is assumed to be isolated from both the body and the system.

(a) Prove that the minimum work Wmin needed to change the state of the body is given by:

Wmin = ∆ (U + PoV + τoσ) (2.7)

The medium can be assume to be large enough that τo and Po are constants and then Uo is also constant. The heat transfer is then

dQo = −τo − dσoσodτo = dU + dUo + Wmin + PodVo

Because τo is constant and Uo is constant dτo = dUo = 0. Because the change in the volume of the medium must equal to change in the volume in the body due to the constant pressure and temperature of the medium, dVo = dV . So the heat transfer is

−τodσo = dU + Wmin + PodV

According to the second law dσall ≥ 0. The change in all entropy is dσall = dσ + dσo. Therefore dσ ≥ −dσo. When it is at its minimum the two must be equal. With this ﬁnal substitution

τodσ = dU + PodV + Wmin → Wmin = dU + PodV − τodσ Now to pull the change out front on the right

Wmin = ∆(U + PoV − τoσ) X (2.8) (b) Now consider the two special cases

(i) P = Po and τ = τo. What is Wmin equal to in this case.

In this case, ∆U = τo∆σ − Po∆V + σ∆τ − V ∆P Making the substitution in equation 8

Wmin = τo∆σ − Po∆V + σ∆τ − V ∆P + Po∆V − τo∆σ) = σ∆τ − V ∆P = −∆GX (2.9)

(ii) P = Po and the process is carried out adiabatically. What is Wmin equal to in this case.

In this case there is no change in the entropy and temperature, ∆U = −Po∆V −V ∆P Making the substitution in equation 8

Wmin = −Po∆V − V ∆P + Po∆V − τo∆σ = τo∆σ − V ∆P = −∆H X (2.10)

8 Chapter 3

Problem Set 3

3.1 Kittel 3.2: Magnetic susceptibility

(a) Use the partition function to find an exact expression for the magnetization M and the susceptibility χ = dM/dB as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result of the magnetization is M = nm tanh(mB/τ), as derived in (46) by another method. Here n is the particle concentration.

In this problem there are two diﬀerent energies, with and against the ﬁeld. For one particle the partition function is

mB Z = emB/τ + e−mB/τ = 2 cosh 1 τ

For N particles the partition function becomes

mB Z = (Z )N = 2N coshN = 2N coshN (mBβ) 1 τ where β = 1/τ. The energy is then

1 ∂Z 1 ∂ 1 U = ⇒ U = 2N coshN (mBβ) = 2N coshN−1 (mBβ) sinh (mBβ) mB Z ∂β 2N coshN (mBβ) ∂β 2N coshN (mBβ) mB U = mB tanh (mBβ) ⇒ M = nm tanh τ

Now for χ

dM d mB nm2 mB χ = = nm tanh ⇒ χ = sech2 dB dB τ τ τ X

(b) Find the free energy and express the result as a function only of τ and parameter x ≡ M/nm. The free energy is

2 sinh (mB/τ) F = −τ log Z = −τ log (2 cosh (mB/τ)) = −τ log = τ log x − τ log (2 sinh (mB/τ)) x

In this case, mB/τ → 0 ∴

nm2 mB nm2 nm2 χ = sech2 → sech20 ⇒ χ = τ τ τ τ X

9 3.2 Kittel 3.3: Free energy of a harmonic oscillator

A one-dimensional harmonic oscillator has an inﬁnite series of equally spaced energy states, with s = s~ω, where s is a positive integer or zero, and omega is the classical frequency of the oscillator. We have chosen the zero energy at the state s = 0. (a) Show that for a harmonic oscillator the free energy is h i F = τ log 1 − e−~ω/τ (3.1)

The partition function is

∞ X −s ω/τ − ω/τ −2 ω/τ Z = e ~ = 1 + e ~ + e ~ + ... s=0

This is a binomial expansion, as long as e−~ω/τ 1. Then

1 −1 Z = = 1 − e−~ω/τ 1 − e−~ω/τ From Eq. 3.55

−1 F = −τ log Z = −τ log 1 − e−~ω/τ ⇒ F = τ log 1 − e−~ω/τ

− ω/τ In the case, τ ~ω: e ~ ≈ 1 − ~ω/τ. Then

− ω/τ F = τ log 1 − e ~ ≈ τ log (1 − 1 + ~ω/τ) ⇒ F = τ log (~ω/τ) X

(b) From (87) show that the entropy is

ω/τ h i σ = ~ − log 1 − e−~ω/τ (3.2) e~ω/τ − 1

From Eq. 3.49

−~ω/τ ∂F ∂ h − ω/τ i τ~ωe − ω/τ ~ω/τ − ω/τ σ = − = − τ log 1 − e ~ = − log 1 − e ~ ⇒ σ = − log 1 − e ~ X ∂τ ∂τ τ 2 (1 − e−~ω/τ ) e~ω/τ − 1

3.3 Kittel 3.4: Energy ﬂuctuations

Consider a system of ﬁxed volume in thermal contact with a reservoir. Show that the mean square ﬂuctuation in the energy of the system is

∂U h(ε − hεi)2i = τ 2 (3.3) ∂τ V Here U is the conventional symbol for hεi.

" #2 ∂U 1 X 2 −ε /τ 1 X −ε /τ = ε e s − ε e s ∂τ Zτ 2 s Z2τ 2 s V s s " #2 2 ∂U 1 X 2 −varepsilon /τ 1 X −ε /τ τ = ε e s − ε e s ∂τ Z s Z2 s V s s 2 ∂U 2 2 τ = hε i − hεi X ∂τ V

10 3.4 Kittel 3.10: Elasticity of polymers

The thermodynamic identity for a one-dimensional system is

τdσ = dU − fdl (3.4) when f is the external force exerted on the line and dl is the extension of the line. By analogy with (32) we form the derivative to ﬁnd

f ∂σ − = (3.5) τ ∂l U

The direction of the force is opposite to the conventional direction of the pressure. We consider a polymeric chain of N links each of length ρ, with each link equally likely to be directed to the right and to the left.

(a) Show that the number of arrangements that give a head-to-tail length of l = 2|s|ρ is

2N! g(N, −s) + g(N, s) = 1 1 (3.6) 2 N + s ! 2 N − s !

The length is given by l = 2|s|ρ → |s| = l/2ρ. |s| is the “steps” of the “random walk.” So

N! N! g(N, s) = N+2s N−2s and g(N, −s) = N−2s N+2s 2 ! 2 ! 2 ! 2 !

The total is the “steps” to the “left and right”

2N! g(N, −s) + g(N, s) = N+2s N−2s X 2 ! 2 !

(b) For |s| N show that

σ(l) = log [2g(N, 0)] − l2/2Nρ2 (3.7)

Using the Gaussian approximation

2N! −2s2/N gtotal = N+2s N−2s ≈ 2g(N, 0)e 2 ! 2 !

The entropy is σ = log g = log (2g(N, 0)) − 2s2/N, because |s|2 = l2/4ρ2. Then

2 2 σ = log (2g(N, 0)) − l /2Nρ X

f = lτ/Nρ2 (3.8)

From Eq. 3.96

∂σ ∂ 2 2 τl f = −τ = −τ log (2g(N, 0)) − l /2Nρ ⇒ f = 2 X ∂l V ∂l Nρ

11 3.5 Ising spin chain

Based on the prescription of one-dimensional N spin system with only nearest-neighbor interactions J and -J for two parallel spins and two antiparallel spins, respectively, you are asked to (a) compute the partition function Z of the spin chain The partition function

X − /τ X −E/τ Z = e s = Ω(E)e s E

If there are N particles then there are N −1 pairs. The energy is E = JNp −JNa = J(Np −Na), where Na = N −1−Np ∴ E = J(2Np − (N − 1)). The multiplicity is (N − 1)! (N − 1)! Ω = ⇒ Np!(N − 1 − Np)! (N − 1 − 1/2(E/J + N − 1))!(1/2(N − 1 + E/J))! The partition function becomes

J(N−1) X (N − 1)! −E/τ Z = e (N − 1 − 1/2(E/J + N − 1))!(1/2(N − 1 + E/J))! X E=−J(N−1)

(b) determine the entropy of the system at very high and very low temperatures. The partition function for 1 pair of particles is

−Jβ Jβ Z1 = e + e where β = 1/τ. So for N − 1 pairs

N−1 Z = e−Jβ + eJβ

The entropy is given by β ∂Z σ = ln Z − Z ∂β When at high τ, β → 0, so

0 ∂Z N−1 σ = ln Z − = ln e−Jβ + eJβ ⇒ σ ≈ (N − 1) ln 2 Z ∂β For low τ, let’s consider a shift in energy from -J to J → 0 to 2J. Then

N−1 N−1 Z = e−Jβ + eJβ → Z = 1 + e−2Jβ

At low τ, β → ∞, so

N−1 ln Z = ln 1 + e−2Jβ ⇒ ln Z = ln (1 + 0)N−1 ⇒ ln Z = ln 1 = 0

σ = −τ ln Z ⇒ σ = 0 X

12 Chapter 4

Problem Set 4

4.1 Application of equal partition theorem

Each particle in a system of N particles has a mass m and is free to preform one-dimensional oscillations about its equilibrium position. Assume the temperature is suﬃciently high so that classical statistical mechanics is relevant. Calculate the heat capacity of this system of particles in each case of the following restoring forces: (a) The force is proportional to the square of the particle’s displacement x from its equilibrium position. 2 3 0 3 The force is F ∝ xi , so the energy from this force is Ui = hxi . The total energy is E = (xi) + E , where (xi) = hxi . The average energy is

R ∞ −βE e idx1dx2 . . . dxf dp1dp2 . . . dpf = −∞ i R ∞ −βE −∞ e dx1dx2 . . . dxf dp1dp2 . . . dpf

R ∞ −βi e idxi = −∞ R ∞ −β −∞ e i dxi Z ∞ ∂ −βi i = − log e dxi (4.1) ∂β −∞

3 Now i = hxi , then

∞ ∂ Z 3 −βhxi i = − log e dxi ∂β −∞

1/3 Let y = β xi, then

∞ ∞ ∞ ∂ Z 3 ∂ Z 3 ∂ Z 3 −βhxi −1/3 −hy −1/3 −hy i = − log e dxi ⇒ i = − log β e dy = − log β + log e dy ∂β −∞ ∂β −∞ ∂β −∞ Because this derivative is with respect to β only

1 1 = ⇒ = τ i 3β i 3

The total energy is then

1 1 5 E = N((p ) + (x )) ⇒ E = N τ + τ ⇒ E = Nτ i i 2 3 6 So the heat capacity is then

∂E 5 C = ⇒ C = N V ∂τ V 6 X

(b) The force is proportional to |x|3

13 3 4 4 The force is F ∝ x , so the potential energy is U = hxi , then (xi) = hxi . From equation 4.1 ∞ ∂ Z 4 −βhxi i = − log e dxi ∂β −∞ 1/4 Now, letting y = β xi, then Z ∞ Z ∞ ∂ −1/4 −hy4 ∂ −1/4 −hy4 1 1 i = − log β e dy = − log β + log e dy ⇒ i = = τ ∂β −∞ ∂β −∞ 4β 4 For the total energy of the system 1 1 3 E = N((p ) + (x )) ⇒ E = N τ + τ ⇒ E = Nτ i i 2 4 4 So the heat capacity is ∂E 3 C = ⇒ C = N V ∂τ V 4 X 4.2 Thermal Equilibrium of the Sun and Earth

8 6 The surface temperature of the sun is To; its radius is R (= 7×10 m) while the radius of the earth is r (= 6.37×10 m). The mean distance between the sun and the earth is L (= 1.5 × 1011m). Assume the earth has reached a steady state of absorbing and emitting radiation so that its temperature does not change with time. (a) Find an approximate expression for the temperature T of the earth in terms of the parameters given. The power of the Sun at its surface is 4 P = AJσTJ 4 where AJ is the surface area of the Sun. The power is constant, so the power at L is P = ALσTL. More important is the percent that the Earth receives, which is

AeAJ 4 P = σTJ AL Now at equilibrium this must equal the amount emitted by the Earth

2 2 1/2 4 AeAJ 4 2 4 πRe4πRJ 4 RJ J J J AseσTe = σT ⇒ 4πReσTe = 2 σT ⇒ T = T X AL 4πL 2L

(b) Calculate the temperature of the sun given T ≈ 290K. The temperature of the Sun is 1/2 r 2L 2(1.5 × 1011) TJ = T ⇒ 290 ⇒ TJ = 6000 K X RJ 7 × 108 4.3 Kittel 4.3: Average temperature of the interior of the Sun

(a) Estimate by a dimensional argument or otherwise the order of magnitude of the gravitational self-energy of the 33 10 −8 2 −2 Sun, MJ = 2 × 10 g and RJ = 7 × 10 cm. The gravitational constant G is 6.6 × 10 dyne cm g .

cm2 2 The self energy is negative. In cgs units energy is ergs=dyne·cm. The constant has units of dyne g2 . To remove g , there needs to be MJ squared. For the distance, only the ﬁrst power of RJ is needed. So the self-energy is

2 GMJ 48 U = − = −3.77 × 10 ergs X RJ (b) Assume that the total thermal kinetic energy of the atoms in the Sun is equal to -1/2 the gravitational self-energy. Estimate the average temperature of the Sun, assuming that there are 1 × 1057 particles.

2 3 U GMJ 6 K = −U/2 ⇒ NkB T = −U/2 ⇒ T = − ⇒ T = = 9.1 × 10 K X 2 3NkB 3NkB RJ

14 4.4 Kittel 4.6: Pressure of thermal radiation

Show for a photon gas that: (a) ∂U X dωj p = − = − s (4.2) ∂V j ~ dV σ j P From Eq. 4.39, U = hsj i~ωj . Then j

!! ∂ X X ∂sj X ∂(~ωj ) p = − sj ~ωj = − ~ωj − sj ∂V ∂V σ ∂V σ j σ j j

∂sj Because sj is the thermal average, it will not change if there is no change in entropy, so ∂V = 0. Then

X ∂ωj p = − s j ~ ∂V X j

(b) dω ω j = − j (4.3) dV 3V

Nπc dω d(V −1/3) 1 nπc 1 nπc 1 ω j = L = nπc = − = − = − j dV dV dV 3 V 4/3 3 LV 3 V X (c) U p = (4.4) 3V From parts a and b X dωj X 1 ωj 1 U p = − s = − s − ~ ⇒ p = j ~ dV j 3 V 3 V X j j

(d) Compare the pressure of the thermal radiation with the kinetic pressure of gas of H atoms at a concentration of 1 mole cm−3. At what temperature (roughly) are the two pressure equal? 2 4 nRT π kB 4 For the hydrogen gas p = , for photons p = 3 3 T . Then V 45~ c 2 4 3 3 1/3 nRT π kB 4 45~ c nR 7 = 3 3 T ⇒ T = 2 4 ⇒ T = 3.21 × 10 K X V 45~ c π kB V 4.5 Kittel 4.7: Free energy of a photon gas

(a) Show that the partition function of a photon gas given by

h i−1 −~ωn/τ Z = τΠn 1 − e (4.5)

where the product is over the modes n.

1 th 1 For one mode Z1 = . For the i mode, Zi = . The total partition function is given by 1−e−~ω1/τ 1−e−~ωi/τ h i 1 1 1 −~ωn/τ Z = Z1 ...Zi ... = ...... = Πn ⇒ Z = Πn 1 − e X 1 − e−~ω1/τ 1 − e−~ωi/τ 1 − e−~ωn/τ

15 (b) The Helmholtz free energy is found directly from above as

X h − ω /τ i F = τ log 1 − e ~ n (4.6) n Transform the sum to an integral to ﬁnd π2V τ 4 T = − (4.7) 45~3c3

h i−1 h i h i −~ωn/τ −~ωn/τ X −~ωn/τ F = −τ log Z = −τ log Πn 1 − e = τ log Πn 1 − e ⇒ F = τ log 1 − e n From Eq. 4.17

Z ∞ Z ∞ −nc~π/Lτ X h − ω /τ i 1 2 h −ncπ /Lτ i 2τ 2 ncπ e F = τ log 1 − e ~ n = 4πn τ log 1 − e ~ = 4πn dn ~ 8 8 Lτ 1 − e−nc~π/Lτ n 0 0

nc~π Letting x = Lτ and Eq. 4.18 and 4.19

π Lτ 3 Z ∞ x3dx π Lτ 3 π4 π2L3τ 4 π2V τ 4 F = − τ x = − τ = − 3 3 ⇒ F = − 3 3 X 3 ~πc 0 e − 1 3 ~πc 15 45~ c 45~ c 4.6 Kittel 4.18: Isentropic expansion of photon gas

Consider the gas of photons of the thermal equilibrium radiation in a cube of volume V at temperature τ. Let the cavity volume increase; the radiation pressure performs work during the expansion, and the temperature of the radiation will drop. From the results for the entropy we know that τV 1/3 is constant in such an expansion. (a) Assume that the temperature of the cosmic black-body radiation was decoupled from the temperature of the matter when both were at 3000 K. What was the radius of the universe at that time, compared to now? If the radius has increased linearly with time, at what time, at fraction of the present age of the universe did the decoupling take place? For the universe R = ct and τV 1/3 = constant, where R is the radius t is the time, and c is some constant. Then τV 1/3 → τR = constant. Then

Ri τf τiRi = τf Rf ⇒ = Rf τi So Ri = 2.9/3000 ≈ 1/1000 so the radius was about 1/1000 the present radius. Because the growth is linear c = Ri . Rf ti Then

Ri Ri tf 1 tf = Rf ⇒ = = present time X ti Rf ti 1000 (b) Show that the work done by the photons during the expansion is 2 π 3 W = Viτi (τi − τf ) (4.8) 15~3c3 where the subscripts i and f refer to the initial and ﬁnal states.

1 U π2τ 4 dW = pdV ⇒ dW = dV ⇒ dW = dV 3 V 45~3c3 By τV 1/3 = constant ⇒ τ 3V = constant. Now τ 3dV = −3τ 2V dτ ⇒ dV = −3τ −1V dτ

3 3 3 Remembering τ V is a constant, so τ V = τi Vi. The work is then

Z τf π2τ 3V π2τ 3V W = − i i dτ ⇒ W = i i (τ − τ ) 15 3c3 15 3c3 i f X τi ~ ~

16 Chapter 5

Problem Set 5

5.1 5.1: Kittel 4.14: Heat capacity of liquid 4He at low temperature

5.2 5.2: Kittel 4.15: Angular distribution of radiant energy ﬂux

5.3 5.3: Qualifying exam ﬁle problem

5.4 5.4: Helmholtz free energy for the Debye model

5.5 5.5: Kittel 5.3: Potential energy of gas in gravitational ﬁeld

17 18 Chapter 6

Problem Set 6

6.1 6.1: Kittel 5.4: Active transport

Consider the concentration of K+ is 104 inside the plant cell compared to the pond. Find the chemical potential diﬀerence.

Treating the ions as an ideal gas,the chemical potential diﬀerence is nin nout ∆µ = µin − µout = τ log − τ log nQ nQ n = τ log in nout Inserting the values 300 K ∆µ = log(1000) ≈ 0.24 eV 11605 K/eV X

6.2 6.2: Kittel 5.7: States of positive and negative ionization

For the following system,

State # of electrons Energy Ground 1 −∆/2 Positive ions 0 −δ/2 Negative ions 2 δ/2 Excited 1 ∆/2

Find the condition for N = 1.

The grand partition function is

Z = λe∆/2τ + eδ/2τ + λ2e−δ/2τ + λe−∆/2τ (6.1)

So the average number is

1 X N − /τ N = Nλ e s Z ASN

λ e∆/2τ + e−∆/2τ + 2λ2e−δ/2τ N = (6.2) λ (e∆/2τ + e−∆/2τ ) + λ2e−δ/2τ + eδ/2τ

δ/2τ For N = 1, then λ = e X

19 6.3 6.3: Kittel 5.10: Concentration ﬂuctuation

(a) Show that

τ 2 ∂2Z N 2 = (6.3) Z ∂µ2 Now

P 2 (Nµ−s)/τ 2 2 2 N e τ X ∂ (Nµ− )/τ N = ASN = e s Z Z ∂µ2 ASN 2 2 2 2 τ ∂ X (Nµ− )/τ 2 τ ∂ Z = e s ⇒ N = Z ∂µ2 Z ∂µ2 X ASN (b) Show that

τ∂ N (∆N)2 = (6.4) ∂µ Beginning with

∂ N ∂ τ ∂Z 1 ∂2Z 1 ∂Z ∂Z τ = τ = τ 2 − ∂µ ∂µ Z ∂µ Z ∂µ Z2 ∂µ ∂µ which reduces to ! 1 ∂2Z 1 ∂Z 2 τ 2 − = N − N 2 = (∆N)2 Z ∂µ2 Z2 ∂µ X

6.4 6.4: Kittel 6.3: Distribution function for double occupancy statistics

For a new system, which has orbitals 0,1,2 and energies 0,,2, respectively (a) Derive N , so

Z = 1 + λe−/τ + λ2e−2/τ (6.5) So N 1 N = 0 · 0 + 1 · λe−/τ + 2λ2e−2/τ Z λe−/τ + 2λ2e−2/τ N = (6.6) 1 + λe−/τ + λ2e−2/τ

(b) Now if is double generate The partition function is now 2 Z = 1 + 2λe−/τ + λ2e−2/τ = 1 + λe−/τ (6.7)

Then 1 N = 0 · 0 + 2 · λe−/τ + 2λ2e−2/τ Z −/τ −/τ 2λe 1 + λe 2λe−/τ = = (1 + λe−/τ )2 1 + λe−/τ Now 2 2 N = ⇒ N = (6.8) λ−1e/τ + 1 e(−µ)/τ + 1 X

20 6.5 6.5: Kittel 6.7: Relation of pressure and energy density

(a) Show that

P ∂s e−s/τ p = − s ∂V N (6.9) Z

P −s/τ So beginning with the partition function, Z = s e , then the Helmholtz free energy is ! X − /τ F = −τ log Z = −τ log e s s so the pressure is then

∂Z ∂F τ ∂V p = − = ∂V τ,N Z

P ∂s e−s/τ p = − s ∂V N Z X

(b) Show that for the free particles ∂ 2 s = − s (6.10) ∂V N 3 V for the boundary conditions of the system The energy levels for these boundary conditions ∂ ∂ 2π2n2 s = ~ V −2/3 ∂V N ∂V 2m 2 2π2n2 2 = − ~ V −5/3 = − s 3 2m 3 V X (c) Show that

2U p = (6.11) 3V From part 1 and 2

P ∂s −s/τ P 2 −s/τ e s e p = − s ∂V N = s 3V Z Z Then using the deﬁnition of the partition function and energies

P −s/τ ! 2 se 2U p = s ⇒ p = 3V Z 3V X

6.6 6.6: Kittel 6.9: Gas of atoms with internal degree of freedom

For an ideal gas with an internal degree of freedom, ∆. The internal partition is for one particle is Zint = e−/τ 1 + e−∆/τ (a) Find the chemical potential Because the logarithm is being taken of the partition function the internal and ideal gas can be separated. This is in fact true for all of this problem. From Equation 6.48 µ = τ [log (n/nQ) − log Zint]

µ = τ [log(n/nQ) − log Zint] h i = τ log(n/n ) + − log(1 + e−∆/τ ) Q τ X

21 (b) Free energy

From Equation 6.49, Fint = −Nτ log Zint

−/τ −∆/τ F = Nτ (log (n/nQ) − 1) − Nτ log e 1 + e

−∆/τ = Nτ (log (n/nQ) − 1) + N − Nτ log 1 + e X

From Equation 6.50, σint = − (∂F/∂τ)V

σ = σideal + σint ∂ = N [log(n/n ) + 5/2] − N − Nτ log 1 + e−∆/τ Q ∂τ −∆/τ = N [log(n/nQ) + 5/2] + N log 1 + e N∆ + τ (e∆/τ + 1) X

(d) Pressure Notice that the internal partition function is independent of the volume. From Equation 3.49 −p = ∂F , then ∂V τ Nτ −pint = 0, so the pressure is just the ideal gas, which is p = V X (d) Heat capacity at constant pressure. 5 For an ideal gas, Cp = 2 N. Now from Equation 6.37, but the volume is constant, so ∂U ∂V ∂U Cp = + p → Cp = ∂τ p ∂τ p ∂τ p

Now the energy needs to be found. So the relation Uint = Fint + σintτ yields the internal energy. N∆ U = N − Nτ log 1 + e−∆/τ + Nτ log 1 + e−∆/τ + int e∆/τ + 1 N∆ = e∆/τ + 1 So the heat capacity for the internal energy is

∂ N∆ N∆2 e∆/τ Cp = = ∂τ e∆/τ + 1 τ 2 (e∆/τ + 1)2

The total heat capacity at constant pressure

5 N∆2 e∆/τ Cp = N + 2 X 2 τ 2 (e∆/τ + 1)

22 Chapter 7

Problem Set 7

7.1 7.1: Pressure in types of gases

7.2 7.2: Kittel 5.13: Isentropic expansion

7.3 7.3: Kittel 6.8: Time for a large ﬂuctuation

7.4 7.4: Kittel 6.10: Isentropic relations of ideal gas

7.5 7.5: Kittel 6.12: Ideal gas in two dimensions

7.6 7.6: Kittel 7.4: Chemical potential versus temperature

7.7 7.7: The absorbtion of gas onto a surface

23 24 Chapter 8

Problem Set 8

8.1 8.1: Mixing of two distinct atoms

Calculate the entropy change when an impermeable partition separating two compartments each of volume V and each containing an ideal monatomic gas of N identical atoms at temperature τ is removed. However, the atoms in one compartment are distinguishable from those in the other compartment.

To identify each set of particles one side will be A and the other B. The change in entropy for A particles is given V2 by σ2 − σ1 = N log , but V2 = 2V1, so the change in entropy of A is then ∆σA = N log 2. This is identical for V1 B because there is no description of A in ﬁnding of the change in the entropy. So the change in the entropy for B is ∆σB = N log 2. The total change in entropy is then ∆σA + ∆σB = ∆σ = 2N log 2 X

8.2 8.2: Kittel 7.3: Pressure and entropy of degenerate Fermi gas

(a) Show that a Fermi electron gas in the ground state exerts a pressure

22/3 5/3 3π 2 N p = ~ (8.1) 5 m V From earlier in Kittel, the pressure is ∂U ∂ 3 3 ∂F p = − ⇒ p = − NF → p = − N ∂V σ ∂V 5 5 ∂V Now expanding the Fermi energy " # 3 ∂ 2 3 ∂ 3π2N 2/3 p = − N F → p = − ~ N 5 ∂V 2m 5 ∂V V

2 5/3 3 N 2/3 ∂ 1 = ~ 3π2 5 2m ∂V V 2/3

22/3 22/3 5/3 2N 5/3 3π 3π 2 N p = ~ → p = ~ 5m V 5/3 5 m V X

(b) Find an expression for the entropy of a Fermi electron gas in the region τ F 1 2 The heat capacity of a Fermi gas is CV = 2 π Nτ/τF . The heat capacity is related to the entropy by ∂σ CV CV = τ → ∂σ = ∂τ ∂τ V τ Using the heat capacity of Fermi gas

Z τ 1 2 1 2 dσ = π Nτ/ (ττF ) dτ ⇒ σ = π Nτ/τF X 0 2 2

25 8.3 8.3: Kittel 7.6: Mass-radius relationship for white dwarfs

Consider a white dwarf of mass M and radius R. Let the electrons be degenerate but nonrelativistic; the protons are nondegenerate. (a) Show that to the order of magnitude of the gravitational self-energy is −GM 2/R, where G is the gravitational constant. 3M The density for an uniform sphere is ρ = 4πR3 . So the potential self-energy is G∆mm(r) G4πr2∆rρm(r) ∆V = − = − r r G4πr∆rρ24πr3 16Gπ2ρ2r4∆r = − = − 3 3 3GM 2 = − r4∆r R6 Then the total potential self-energy is

Z R 2 2 2 3GM 4 3GM GM V = − 6 r dr → V = − ≈ − X 0 R 5R R (b) Show that the order of magnitude of the kinetic energy of the electrons in the state with mass of an electron, m, and mass of a proton, MH , is

2N 5/3 2M 5/3 ~ ≈ ~ (8.2) mR2 5/3 2 mMH R

2 2 2/3 ~ 3π N The energy of one electron is = 2m V . There are N electrons in the Sun. The total energy would be then 2 2 2/3 ~ N 3π N 4 3 K = 2m V . The volume of the Sun is V = 3 πR . The energy becomes " # 2 9πN 2/3 2N 5/3 1 9π 2/3 K = ~ N = ~ 2m 4R3 mR2 2 4

Because the mass of the proton is so much more than that of the electrons, so then all that has to be considered is mass of the proton. So the number of particles are N ≈ M/MH . The terms in [...] are on the order of two, so to the order of magnitude, the kinetic energy is

2 5/3 " 2/3# 2 5/3 ~ N 1 9π ~ M K = 2 ≈ 2 X mR 2 4 mR MH

2 2 5/3 2 GM ~ M 1/3 ~ = 2 ⇒ M R = 5/3 R mR MH mGMH −272 1/3 10 20 1/3 M R ≈ = 10 g cm X 10−810−27 (10−24)5/3

(d) What is the density of a white dwarf with its mass equal to the mass of the Sun (2 × 1033g) The density would be

2 −1 3M M 1/3R = 1020 ⇒ MR3 = 10−60 ⇒ ρ = 4π1060 3 2 × 10332 ρ = → ρ = 955000 g/cm3 4π1060 X

26 (e) Show for a neutron star M 1/3R ≈ 1017g1/3cm. What is the radius of the neutron star whose mass is equal to that of the Sun? 2 1/3 ~ Looking back at part c, M R = 5/3 , but the mass of a neutron is Mn ≈ 1000m and that MH ≈ Mn. Then mGMH

2 2 1/3 ~ ~ 17 1/3 M R = 5/3 = 5/3 ≈ 10 g cm MnGMH 1000mGMH The radius of a neutron star with mass of the Sun would be

M 1/3R = 1017g1/3cm 1/3 2 × 1033 R = 1017g1/3cm R = 794000cm → R = 7.94km X

8.4 8.4: Kittel 7.10: Relativistic white dwarfs stars

Consider a Fermi gas of N electrons each of rest mass m in a sphere of radius R. Treating the electrons as relativistic for a white dwarf, ﬁnd the following (a) Using the viral theorem ﬁnd the value of N, neglecting the kinetic energy of the protons.

3 1/3 From problem 7.2, the kinetic energy for the electrons is K + 4 NF and the Fermi energy is F = ~πc (3N/πV ) . 1/3 3π~c 9 4/3 Using the volume of the sphere, the kinetic energy becomes K = 4R 4π2 N . The potential energy is from 2 2 2 3GM 3GN MH Problem 7.6, V = 5R ≈ 5R . Applying the viral theorem

3GN 2M 2 3 π c 9 H = ~ N 4/3 5R 4 R 4π2 1/3 2/3 5π~c 9 N = 2 2 4GMp 4π

" 1/3#3/2 5π~c 9 N = 2 2 X 4GMp 4π

(b) Estimate the value of N Inserting the values

" 8 −34 1/3#3/2 5 π3 × 10 1.05 × 10 9 N = 4 6.6 × 10−11 (1.67 × 10−27)2 4π2 57 = 8.14 × 10 X

8.5 8.5: Electrons in the air oﬀ a conductor

The lowest possible energy of a conduction electron in a metal is −Vo below the energy of a free electron at inﬁnity. The conduction electrons have a chemical potential energy µ. The minimum energy needed to remove an electron from the metal is Φ = Vo − µ and is called the work function of the metal. Consider an electron gas outside the metal in thermal equilibrium with the electrons in the metal at the temperature τ. Assume τ Φ, ﬁnd the mean number of electrons per unit volume outside the metal.

The mean number of electrons per unit volume outside the metal can be found by relating the chemical potentials. The chemical potential inside the conductor is µ. Outside the conductor, the chemical potential is µideal + Φ + µ. The chemical potential for an ideal gas is µ = τ log (n/nQ). Therefore

−Φ/τ µideal + Φ + µ = µ → τ log (n/nQ) + Φ + µ = µ → n = nQe

27 mτ 3/2 Now n = N/V , and nQ = 2 . Then 2π~ mτ 3/2 N/V = e−Φ/τ 2π~2 Lastly, there is a spin 2 degeneracy, so

3/2 N mτ −Φ/τ = 2 e X V 2π~2

28 Chapter 9

Special Problem Set (9)

9.1 Problem 1: Properties of “Photon Gas”

The Helmholtz free energy of a certain system is given by F = bV τ 4, where b is a constant and all other symbols have their usual meaning. (a) Compute the chemical potential of the system. The chemical potential is related to the free energy by ∂F ∂ 4 µ = = bV τ ⇒ µ = 0 X ∂N τ,V ∂N

(b) What is the energy of the system? The energy of the system is related to the free energy by

∂(F/τ) ∂ U = −τ 2 = −τ 2 bV τ 3 ⇒ U = −3bV τ 4 ∂τ ∂τ X (c) Find the equation of state that describes the system. The equation of state is related to the pressure, so

∂F 4 U p = − ⇒ p = −bτ or p = X ∂V τ 3V

(d) What is the work done during an isentropic expansion from τ1 to τ2? 3 So ﬁrst the entropy has to be found σ = −(∂F/∂τ)V = −4bV τ .

1 U dW = pdV 1⇒.c dW = dV ⇒1.b dW = −bτ 4dV 3 V

An isentropic expansion σ =constant, so τ 3V =constant. Then dV = −3τ −1V dτ. The work is then dW = 3bV τ 3dτ, but because the expansion is isentropic V τ 3 =constant. With that choose it to be the initial state 1. Finally

τ Z 2 3 3 W = 3bV1τ1 dτ ⇒ W = 3bV1τ1 (τ2 − τ1) X

τ1

9.2 Problem 2: Engine Cycle

For an ideal-gas engine cycle the processes AB and CD are isentropic, process BC is isobaric, and process DA is isovolumetric.

(a) Calculate the engine eﬃciency η in terms of the temperatures τA, τB , τC , τD and γ (= Cp/CV ).

29 The eﬃciency is η = Qh−Ql . So the heat comes and leaves the system in only two places, because AB and CD are Qh isentropic. The heat in is Qh = Cp (τC − τB ). The heat out is Ql = CV (τD − τA). So the eﬃciency is

Q − Q C (τ − τ ) − C (τ − τ ) η = h l = p C B V D A Qh Cp (τC − τB )

τD − τA η = 1 − X γ (τC − τB )

(b) Find η/ηC , where ηC is the eﬃciency of a Carnot engine operating at the highest and lowest temperatures of the cycle.

First find the highest and lowest temperatures. From the graphs and defined process: VC τB = VB τC and VC > VB → γ−1 γ−1 τC > τB ; pDτA = pAτD and pD > pA → τD > τA; and τAVA = τB VB and VA > VB → τB > τA. The Carnot γ−1 γ τA VB VC γ−1 VB τC efficiency is ηC = 1 − τ . Also from the ideal gas law, τB = τC . τD = V τC . VD = V τ . Combining C VC D C A γ VC the statements τD = τA The efficiency is then VB

γ VC τA V − 1 η = 1 − B γτC (1 − VB /VC )

Then the ratio is

γ VC τA V −1 η 1 − B = γτC (1−VB /VC ) τA X ηC 1 − τC

From previous knowledge, it should be η/ηC < 1. Now the check

γ VC γ τA −1 V VB τ C − 1 1 − A V τ γτC (1−VB /VC ) < 1 ⇒ B > A τA 1 − γτC (1 − VB /VC ) τC τC γ (VC /VB ) − 1 > γ (1 − VB /VC )

This statement is true as long as γ > 1, which is true by the fact that γ = Cp/CV = (CV + 1)/CV > 1. X

9.3 Problem 3: Vibrational Modes of a Molecule

Suppose that in a certain polyatomic ideal gas, the vibrational motions of each molecule can be described in terms of three independent simple harmonic oscillators having the same frequency ω/2π. The vibrational energy of each 3 molecule is then given by εn = na + nb + nc + 2 ~ω, where n = na + nb + nc with all the n’s being 0, 1, 2, ... (a) What is the degeneracy of the vibrational states of each molecule as a function of n?

(n + 2)! The degeneracy for S independent harmonic oscillators is (S +n−1)!/(S −1)!(n)!. Here S = 3 then g(n) = n!2 X

(b) Write down an expression to represent the vibrational part of the molecular partition function (call it Zvib).

∞ X 1 2 −3 ω/2τ −n ω/τ The partition function is the sum of the Boltzmann factor. So Z = n + 3n + 2 e ~ e ~ vib 2 X n=0

~ω (c) Calculate Zvib at high temperature, i.e. assume τ 1.

30 The partition function at high temperature

−3~ω/2τ Z ∞ e 2 −n~ω/τ Zvib = n + 3n + 2 e dn 2 0 e−3~ω/2τ Γ(3) 3Γ(2) 2Γ(1) = + + 2 (~ω/τ)3 (~ω/τ)2 ~ω/τ

Now applying ~ω τ

3 2 3 ~ω 3τ 3τ τ Zvib = 1 − + + 2 τ (~ω)3 (~ω)2 ~ω

(d) Evaluate the heat capacity and energy per molecule at high temperature.

2 ∂ log Zvib The energy is Uvib = τ ∂τ , which is

3 2 2 ∂ 3 ~ω 3τ 3τ τ Uvib = τ log 1 − + + ∂τ 2 τ (~ω)3 (~ω)2 ~ω 18τ 2 − 6τ − z 18τ 2 = τ 2 ~ω⇒τ U = τ 2 (2τ − 3) (3τ 2 + 3τ + 1) vib 6τ 3

9 The total energy per molecule is U = U + U ⇒ U = 3 τ + 3τ → U = τ . The heat capacity is tot ideal vib tot 2 tot 2 9 C = ∂U ⇒ C = ∂ 9 τ ⇒ C = V ∂τ V ∂τ 2 V 2 X

9.4 Problem 4: Relativistic Massless Bosons

Consider an ideal gas of massless bosons in thermal equilibrium. The number of such bosons in the system is constant.

(a) Derive an expression for the number of thermally excited bosons above the ground state if the energy of the bosons happens to exhibit the dispersion relation ε = ~ck. ∞ First k is the wavenumber or k = nπ/L. The partition function would be Z = P e−n~π/Lτ . Then n=0

−1 X −n π/Lτ 1 hni = Z ne ~ ⇒ hni = e~π/Lτ − 1

Then

X 1 Z ∞ 4πn2dn V τ 3 Z ∞ x2dx Ne = hni = n π/Lτ = 2 3 3 x 8 0 e ~ − 1 2π ~ c 0 e − 1 V τ 3 V τ 3 Ne = 2ζ(3) ⇒ Ne ≈ 1.202 X π2~3c3 π2~3c3

(b) Find the critical temperature if the number density of 1020 cm−3 below which Bose-Einstein condensation (BEC) occur.

For a BEC, N = Ne. Then

3 r 20 2 20 V τ 3 10 π 10 V = 1.202 → τ = ~c ⇒ T = 21458 K X π2~3c3 1.202

31 9.5 Problem 5: Kittel & Kroemer 7.9 and more

(a) Calculate the integral for Ne(τ) for a one-dimensional gas of noninteracting bosons, and show that the integral does not form in one dimension. Take λ = 1 for the calculation.

L 2m 1/2 From Equation 7.86, D() = 2 . Then π ~ Z Z ∞ L 2m 1/2 1 Ne = dD()f(, τ) = d 2 /τ 0 π ~ e − 1 1/2 Z ∞ −1/2 L 2m 1/2 x dx = 2 τ x π ~ 0 e − 1

Taking the ﬁrst order term on the exponent R ∞ x−1/2dx/(1 + x − 1) = −1/2x−1/2 ∞, which does not converge. Any 0 0 higher order terms will diverge also. X (b) What functional form does the density of state (DOS) need to have for an abrupt transition to occur in a 1-D and 2-D Boson gases? What is the range of α in 1-D? In 2-D?

32 Chapter 10

Problem Set 10

10.1 10.1: Collisions with a wall for a Fermi Gas

Calculate the average rate of collision per unit area with a wall in a Fermi gas at T = 0, which will be called Φ.

Let’s think of this problem in momentum space. With T = 0, f = 1 of the fermi-dirac distribution. For an isotropic 2 3 sphere in momentum space: sin θdθdφp dp/(2π~) . Then the velocity is v · nˆ = v cos θ. Remembering that v = p/m. So

pF π/2 2π Z Z Z (2s + 1) p3 Φ = cos θ sin θdφdθdp (2π~)3 m 0 0 0 pF π(2s + 1) Z (2s + 1)π p4 = p3dp = F m(2π~)3 4(2π~)3 m 0

2 2 2/3 ~ 6π N Now, pF = 2mF , and F = 2m (2s+1)V

π(2s + 1) 4m2 2 2 6π2N 4/3 Φ = ~ 4(2π~)3 m 2m (2s + 1)V 3(6π2)1/3 N 4/3 Φ = ~ 16(2s + 1)1/3 m V X

10.2 10.2: Free energy and pressure of a Boson gas

Determine the free energy F and pressure P on a Boson gas at τ τE . Include the degeneracy factor (2s + 1) in your answers.

1.306V 2M 3/2 5/2 2 From the class notes U = NeτI, where Ne = 2 τ (2s + 1) and I = .7702. Now U = −τ ∂(F/τ)/∂τ. 4 π~ Then Z 1 F/τ = − Udτ τ 2 τ .7702(1.306)V 2M 3/2 Z F/τ = − (2S + 1) τ 1/2dτ 4 π~2 0 .7702(1.306)V 2M 3/2 2 F = − (2s + 1)τ 5/2 4 π~2 3 3/2 M 5/2 2 F = −.4742V (2s + 1) τ = − U X π~2 3

33 Now for the pressure 3/2 ∂F 2(.7702)(1.306) 2M 5/2 p = − ⇒ p = 2 (2s + 1)τ ∂V τ (3)4 π~ M 5/2 2 U p = .4742(2s + 1) τ = X π~2 3 V

10.3 10.3: Discontinuity in the slope of the heat capacity of a Bose gas

Provide the missing steps on the derivation of the heat capacity of a Bose gas near τE . That is, use ∆U expression ∂CV −3.66N to show that ∆CV = 0 and ∆ = at τ = τE ∂τ τE

From the lecture notes 3π2 6 N − N 2 U − U = ∆U = − ~ N µ=0 ex o m3 µ=0 V τ

3/2 τ Now Nµ=0 = N . So at τ = τE → Nµ=0 = N. It is obvious that the squared term is 0 from Nµ − N = 0, τE ∂U ∂(Uo+∆U) ∂Uo ∂∆U then ∆U = 0. Now CV = ∂τ , and ∆CV = ∂τ − ∂τ = ∂τ = ∆CV . Which is 3π2 6 ∂N N − N 2 ∆C = − ~ µ=0 µ=0 V m3V 2 ∂τ τ N − N 1 ∂N N − N + 2N µ=0 µ=0 − µ=0 µ=0 τ τ ∂τ τ 2

∂CV Every term is multiplied by Nµ=0 − N, which is 0 at τ = τE . Now for ∆ ∂τ , which is ∂C ∂2(U + ∆U) ∂2U ∂∆C ∆ V = o − o = V ∂τ ∂τ 2 ∂τ 2 ∂τ The only nonzero term is the second term by the same argument as before. 2 6 " 2 2 # ∂∆CV 6π ~ ∂Nµ=0 Nµ=0 − N ∂ Nµ=0 Nµ=0 − N 1 ∂Nµ=0 ∂Nµ=0 Nµ=0 − N = − + Nµ=0 + Nµ=0 − 2Nµ=0 ∂τ m3V 2 ∂τ τ 2 ∂τ 2 τ 2 τ 2 ∂τ ∂τ τ 3

Again, by Nµ=0 − N = 0, only one term remains

∂∆C 6π2 6 1 ∂N 2 V = − ~ N µ=0 ∂τ m3V 2 µ=0 τ ∂τ

3/2 Now using Nµ=0 = N(τ/τE ) . Then

2 2 6 3/2 ! 2 6 3 ∂∆CV 6π ~ τ 3 N ∂∆CV 6π ~ 9 N = − 3 2 N 3/2 ⇒ = − 3 2 3 ∂τ m V τE 2 1/2 ∂τ m V 4 ττ τ τE E τ=τE 2 6 3 2 ∂∆CV 6π ~ 9 2 m 2.612V N N = 3 2 N 2 = −3.66 X ∂τ m V 4 2π~ N τE τE

10.4 10.4: Maximum work extracted from an ideal gas

Determine the maximum work that can be extracted from an ideal gas system held at constant volume by cooling it from temperature τ1 to a temperature τo of the medium.

The amount of heat that the system gives oﬀ can be turned into work is

τ Z 1 ∆Qgain = NCV dτ ⇒ ∆Q = NCV (τ1 − τo)

τo

34 But there is some energy lost to the change in entropy

τ Z 1 NCV τ1 ∆σ = dτ ⇒ ∆σ = NCV log τ τo τo τ1 The heat is ∆Qlost = τ∆σ = NCV τo log . The diﬀerence of the two is the maximum work that can be done by τo the ideal gas

τ1 W = NCV τ1 − τo − τo log X τo

35 36 Chapter 11

Problem Set 11

11.1 11.1:A review problem

The entropy of a monatomic ideal gas can be written as σ = (3N/2) ln τ+ terms that involve N, V , M and some constants. By including the internal degrees of freedom in the entropy of a polyatomic ideal gas, prove that the factor N 3N/2 must be replaced by the factor γ−1 .

From problem 11.3, ∂σ CV CV = τ ⇒ dσ = dτ ⇒ σ = CV ln τ ∂τ V τ Now ∂σ ∂σ ∂p ∂V ∂p CV = τ = Cp + τ = Cp − τ ∂τ V ∂p τ ∂τ V ∂τ p ∂τ V Now applying the ideal gas law pV = Nτ ∂V ∂p N N Cp − τ = Cp − τ = Cp − N ∂τ p ∂τ V p V

Then CV = Cp − N.

N CV N CV N N = = = CV → = CV γ − 1 Cp − CV N γ − 1 Then from the ﬁrst equation

N σ = C ln τ ⇒ σ = ln τ V γ − 1 X

11.2 11.2: Dissociation of water

A certain number of moles of H2O are introduced into a container of volume V . At some high temperature T dissociation takes place according to 2H2O → 2H2 + O2. Let x denote the fraction of water vapor molecules which are dissociated at T corresponding to a total pressure P . (a) Derive an expression that relates x to p and a constant that depends only on temperature.

The partial pressure is pi = niτ, so the total pressure is X p = niτ i Now for each water molecule that dissociates, there are 3/2 molecules. Then 3 x p = (1 − x)nτ + xnτ = 1 + nτ 2 2

37 Then from class N 2 H2O = K (τ) N 2 H N H2 O2 Factoring the concentration n = N/V

(1 − x)2n2 2(1 − x)2 2 x 3 = Kn(τ) ⇒ 3 = n (x ) 2 n x Kn(τ) Placing that into the pressure

2(1 + x/2)(1 − x)2 τ p = 3 X x Kn(τ)

(b) Write down an equation that relates the chemical potentials of H2O, H2 and O2 in equilibrium. In equilibrium, the Gibb’s free energy is 0.

G = µiNi = 0

Now, rewriting the chemical equation as

−2H2O + 2H2 + O2 = 0 ⇒ −2NH2O + 2NH2 + NO2

Combining the two equations

µiNi = −2µH2ONH2O + 2µH2 NH2 + µO2 NO2 = 0

Setting the number equal as part of the equilibrium conditions

− 2µH2O NH2O + 2µH2 NH2 + µO2 NO2 = 0

1 −2µH O + 2µH + µO N = 0 ⇒ µH + µO = µH O 2 2 2 2 2 2 2 X

11.3 11.3: Practice with the Jacobian

Since any thermodynamical quantity of a system with a ﬁxed number of particles can be written as a function of two variables that are not necessarily the same as those used in class, thus you may select the appropriate functional forms to derive the following expressions:

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

Express the Helmholtz free energy F as a function of T and V to prove that ∂σ = ∂p so that one of the above ∂V τ ∂τ V expressions can be also be written as ∂U = −p + τ ∂p ∂V τ ∂τ V

Writing the energy in terms of entropy and volume

U(σ, V ) ⇒ dU = τdσ − pdV ∂σ ∂σ dU = τ dτ + τ − p dV ∂τ V ∂V τ Now for the relationship between the energy and temperature holding the volume constant

∂U ∂σ = τ X ∂τ V ∂τ V

38 Then the relationship between the energy and volume holding the temperature constant

∂U ∂σ = −p + τ X ∂V τ ∂V τ Writing the free energy in terms of temperature and volume

dU = d(τσ) − σdτ − pdV ⇒ dF = −σdτ + pdV

Then ∂F ∂F dF = dτ + dV ∂τ V ∂V τ ∂F ∂F = −σ; = −p ∂τ V ∂V τ ∂2F ∂2F ∂σ ∂p = ⇒ = X ∂τ∂V ∂V ∂τ ∂V τ ∂τ V

11.4 11.4: More practice

2 ∂CV ∂ p Prove that ∂V = τ ∂τ 2 τ V From problem 11.3

∂C ∂ ∂σ ∂2σ ∂ ∂σ V = τ = τ = τ ∂V τ ∂V ∂τ τ ∂V ∂τ ∂τ V ∂V τ Again from problem 11.3

∂ ∂σ ∂ ∂p ∂2p τ = τ = τ 2 X ∂τ V ∂V τ ∂τ V ∂τ V ∂τ V

11.5 11.5: van der Waals Gas

Use some of the results from problems 11.3 and 11.4 above to obtain a formula for the energy of a van der Waals 2 gas which obeys the equation of state p + a/V (V − b) = (R/kB ) τ, where R is the gas constant. Note that your answer will also involve the speciﬁc heat function CV .

The pressure is R τ a p = − 2 kB V − b V From problem 11.3 ∂p 0 00 ∂p dU = CV dτ + τ − p dV ⇒ U + dU = CV τ + τ − p dV ∂τ V ∂τ V Then with the pressure from above

∂p R τ R τ a a τ − p = − + 2 = 2 ∂τ V kB V − b kB V − b V V Z a a dV = − V 2 V Placing this in the second equation

a U = C τ − + constant V V X

39 40 Chapter 12

Problem Set 12

12.1 12.1: Maxwell Relations

∂σ ∂p The formula Cp − CV = −τ ∂p ∂τ was derived in class τ V ∂p 2 1 ∂V (a) Show that Cp − CV = τV Kτ ∂τ , where Kτ = − V ∂p is isothermal compressibility. V τ ∂σ ∂V ∂p ∂p ∂σ So what needs to be shown here is that ∂p = ∂p ∂τ . Now using the Jacobian and that ∂τ = ∂V τ τ V V τ ∂σ ∂V ∂σ ∂(V, τ) ∂(σ, τ) ∂(σ, τ) ∂σ = = = = ∂p τ ∂p τ ∂V V ∂(p, τ) ∂(V, τ) ∂(p, τ) ∂p τ From this

∂σ ∂p ∂V ∂σ ∂p V ∂V ∂p ∂p ∂p 2 Cp − CV = −τ = −τ = −τ Cp − CV = τV Kτ X ∂p τ ∂τ V ∂p τ ∂V V ∂τ V V ∂p τ ∂τ V ∂τ V ∂τ V

n (b) If the entropy σ(p, T ) = a(p)τ at the low temperature, where a(p) is a function of pressure, show that Cp −CV = 2 1 da 2n+1 τ . This result implies that there is no distinction between Cp and CV at very low temperatures. VKτ dp 1 ∂p ∂p ∂σ ∂p From the ﬁrst part, K = −V ∂V . Matching the terms, it needs to be shown that ∂V ∂p = ∂τ . Then τ τ τ τ V ∂p ∂p ∂σ ∂(p, τ) ∂(σ, τ) ∂(σ, τ) ∂σ = = = = ∂τ V ∂V τ ∂p τ ∂(V, τ) ∂(p, τ) ∂(V, τ) ∂V τ

∂σ ∂p ∂p ∂σ ∂p From the ﬁrst part ∂V = ∂τ . So ∂V ∂p = ∂τ , Placing this into the original equation with the τ V τ τ V deﬁnition of Kτ . ∂σ ∂p 1 ∂σ 2 Cp − CV = −τ = τ ∂p τ ∂τ V VKτ ∂p Now σ(p, T ) = a(p)τ n, so

n 2 2 1 ∂(a(p)τ ) 1 da 2n+1 τ ⇒ Cp − CV = τ X VKτ ∂p VKτ dp

12.2 12.2: Equilibrium conditions

A substance is in equilibrium in the presence of externally applied forces. A small number of particles or a small amount of energy, or both, are allowed to transfer between any two inﬁnitesimal volume elements of this substance. Prove that the temperature and chemical potential must each have a constant value.

From the ﬁrst law of thermodynamics, ∆W = ∆U − ∆Q. The system is at equilibrium so ∆W = ∆U = 0. Now let’s consider two subsystems 1 and 2, then ∆U1 = ∆U2 = 0, ∴ ∆Q1 = ∆Q2. Now from the deﬁnition of heat, δ(σ1τ1) = δ(σ2τ2). From equilibrium terms σtot is constant, so are any smaller sections. Then σ1δτ1 = σ2δτ2 → δτ1 = δτ2. The only way this can be true at equilibrium is if δτ1 = 0 = δτ2. Now the total number of particles is constant. This means δ(µ1N1) = δ(µ2N2) → µ1δN1 = µ2δN2 ∴ µ1 = µ2 or that the chemical potential is constant. X

41 12.3 12.3: Fluctuation in number of a Fermi gas

Determine (∆N)2 for an electron gas at temperatures much lower than the Fermi temperature.

From lecture notes ∂N (∆N)2 = τ ∂µ τ,V At temperatures below the Fermi temperature, the number is

V 2mµ 3/2 N = F 3π2 ~2 Then

3/2! 3/2 3/2 2 !1/2 2 !1/3 ∂N ∂ V 2mµF V 2m V 2m 3π N (∆N)2 = τ = τ = τ µ1/2 = ~ 2 2 2 2 F 2 2 ∂µ τ,V ∂µ 3π ~ 2π ~ 2π ~ 2m V

31/3mV τ N 1/3 (∆N)2 = 4/3 2 X π ~ V

12.4 12.4: Fluctuation in volume, pressure, entropy and temperature

Find (a) ∆V ∆p From class

! ! ! ! CV 2 ∂σ ∂p ∂p 2 CV 2 ∂p 2 ∆σ∆τ − ∆V ∆p = ∆τ + ∆V ∆τ − ∆τ∆V − ∆V = ∆τ − ∆V τ ∂V τ ∂V V ∂V τ τ ∂V τ

By the fact that we are looking for function of p and V means we need h∆V ∆pi = ∂p h∆V 2i. From the class ∂V τ 2 ∂V notes h∆V i = −τ ∂p . Substituting this into equation above τ ∂p ∂V h∆V ∆pi = −τ = −τ X ∂V τ ∂p τ (b) ∆σ∆τ 2 From the previous problem, h∆σ∆τi = CV h∆τ 2i. From class, h∆τ 2i = τ . Combining these two τ CV

2 CV 2 CV τ h∆σ∆τi = h∆τ i = = τ X τ τ CV

12.5 12.5: Kittel 10.5: Gas-solid equilibrium

Consider the gas-solid equilibrium under the extreme assumption that the entropy of the solid may be neglected over the temperature range of interest. Let −εo be the cohesive energy of the solid, per atom. Treat the gas as ideal and monatomic. Make the approximation that the volume accessible to the gas is the volume V of the container, independent of the much smaller volume occupied by the solid.

(a) Show that the total Helmholtz free energy of the system with the total number of atoms N = Ns +Ng is constant.

F = Fs + Fg = −Nsεo + Ngτ [log (Ng/V nQ) − 1] (12.1)

The entropy of the solid can be neglected, because it has no eﬀect in the free energy. Fs = Us = −Nsεo The free energy of a gas is given in Kittel and Kroemer as Fg = Ngτ [log (n/nQ) − 1] here n = Ng/V . The total free energy is then

F = −Nsεo + Ngτ [log (Ng/V nQ) − 1]

42 (b) Find the minimum of the free energy with respect to Ng; show that in the equilibrium condition

−εo/τ Ng = nQV e (12.2)

∂F At equilibrium ∂N = 0. Then g N,τ,V

∂ h i h i N − Ng εo + Ng τ log Ng /V nQ − 1 = εo + τ log Ng /V nQ − 1 + τ = εo + τ log Ng /V nQ = 0 ∂Ng Finally

−εo/tau εo + τ log (Ng/V nQ) = 0 ⇒ Ng = nQV e X

3/2 Ngτ −εo/τ 5/2 m −εo/τ p = = τnQe ⇒ p = τ e X V 2π~2

43 44 Chapter 13

Problem Set 13

13.1 13.1: Superconduction and Heat Capacity

A certain metal in zero magnetic ﬁeld and at atmospheric pressure has a heat capacity Cn = ατ in the normal state, 3 and a heat capacity Cs = γτ in the superconducting state. Here α and γ are constants.

(a) Express these constants in terms of each other and the critical temperature τc. The heat capacity is given by C = τ ∂σ . Then i ∂τ i ∂σ ∂σ C = τ n ⇒ ατ = τ n ⇒ σ = ατ n ∂τ ∂τ n ∂σ ∂σ 1 C = τ s ⇒ γτ 3 = τ s ⇒ σ = γτ 3 s ∂τ ∂τ s 3

At the critical temperature σn = σs. Then

1 γ ατ = γτ 3 ⇒ α = τ 2 c 3 c 3 c X

(b) What is the diﬀerence between the internal energy of the metal in the normal and superconducting states τ = 0? Express the answer in terms of γ and τc. R τc 2 R τc 4 The internal energy diﬀerence comes from Un = 0 Cndτ = 1/2ατc and Us = 0 Csdτ = 1/4γτc . Then 1 1 1 1 γ ∆U = ατ 2 − γτ 4 = γτ 4 − γτ 4 ⇒ ∆U = − τ 4 2 c 4 c 6 c 4 c 12 c X

13.2 13.2: Kittel 10.8: First order crystal transformation

Consider a crystal that can exist in either of two structures, denoted by α and β. We suppose that the α structure is the stable low temperature form and the β structure is the stable high temperature form of the substance. If the zero of the energy scale is taken as the state of separated atoms at inﬁnity, then the energy density U(0) at τ = 0 will be negative. The phase stable at τ = 0 will have the lower value of U(0); thus Uα(0) < Uβ (0). If the velocity of sound vβ in the β phase is lower than vα in the α phase, corresponding to lower values of the elastic moduli for β, then the thermal excitations in the β phase will have larger amplitudes than in the α phase. The larger the thermal excitation, the large the entropy and the lower the free energy. Soft systems ten to be stable at high temperatures, hard systems at low. (a) Show from Chapter 4 that the free energy density contributed by the phonons in a solid at a temperature much 2 4 3 3 less than the Debye temperature is given by −π τ /30v ~ , in the Debye approximation with v taken as the velocity of all phonos. The energy density of the phonons at τ θ is

3π4Nτ 4 π2τ 4 U = 3 = 3 5 (kB θ) 10 (~v)

45 Now

U Z π2τ 2 ∂(F/τ) = − 2 ∂τ ⇒ F/τ = − 3 dτ τ 10 (~v) π2τ 3 π2τ 4 F/τ = U(0) − 3 ⇒ F = U(0) − 3 X 30 (~v) 30 (~v)

(b) Show that at the transformation temperature

4 3 2 −3 −3 τc = 30~ /π [Uβ (0) − Uα(0)] / vβ − vα (13.1)

There will be a ﬁnite real solution if vβ < vα. This example is simpliﬁed model of a class of actual phase transformations in solids.

At the phase transformation the Gibb’s free energy are equal. If there are no volume changes Fα(τc) = Fβ (τc). From above

2 4 2 4 π τc π τc Fα(τc) = Fβ (τc) ⇒ Uα(0) − 3 = Uβ (0) − 3 30 (~vα) 30 (~vβ )

! 3 3 4 1 1 30~ 4 30~ Uβ (0) − Uα(0) τ − = (Uβ (0) − Uα(0)) ⇒ τ = c v3 v3 π2 c π2 −3 −3 X β α vβ − vα

(c) The latent heat of transformation is deﬁned as the thermal energy that must be supplied to carry the system through the transformation. Show that the latent heat of this model is

L = 4 [Uβ (0) − Uα(0)] (13.2)

The latent heat is the change in enthalpy. If the volume change is small, the change is just the change of internal energy.

L = Hβ − Hα = Uβ (τc) − Uα(τc) 2 4 2 4 π τc π τc = Uβ (0) − Uα(0) − 3 − 3 10 (~vα) 10 (~vβ ) = Uβ (0) − Uα(0) − (−Uβ (0) + Uα(0)) = 4 (Uβ (0) − Uα(0)) X

13.3 13.3: Kittel 11.2: Mixing energy in 3He −4 He and P b − Sn mixtures

The phase diagram of liquid 3He−4 He mixtures in Figure 11.8 shows that the solubility of 3He in 4He remains finite (about 6 percent) as τ → 0. Similarly, the P b − Sn phase diagram of Figure 11.14 shows a finite residual solubility of PB in solid Sn with decreasing τ. What do such finite residual solubilities imply about the form of the function u(x)?

The ﬁnite residual solubilities imply that at straight line can be drawn from x = 0 that is tangential to the minimum of u(x), for Helium is x ≈ .94 and for P b − Sn x ≈ .97. X

13.4 13.4: Kittel 11.4: Solidiﬁcation range of a binary alloy

Consider the solidification of a binary alloy with the phase diagram of Figure 11.10. Show that, regardless of the initial composition, the melt will always become fully depleted in component B by the time the last remnant of the melt solidifies. That is, the solidification will not be complete until the temperature has dropped to TA.

Starting as some concentration x, as it is cooled some will solidify, xS1. This amount is removed from x. From the diagram, it will consist of B, so the remaining liquid, xL1, will have a lower concentration of B. Again, cooling the sample more some more will solidify, xS2. Now the liquid, xL2, will have more of B removed, so xL2 < xL1. This can be done continuously until all of B has been removed.X

46 13.5 13.5: Kittel 11.5: Alloying of gold into silicon

(a) Suppose a 1000 A˚ layer of Au is evaporated onto a Si crystal, and subsequently heated to 400o C. From the Au − Si phase diagram, Figure 11.11, estimate how deep the gold will penetrate into the silicon crystal. The densities of Au and Si are 19.3 and 2.23 g cm−3.

The total number of moles per unite area is n, and nAu, nSi, which are the number of moles per unite area of gold and silicon, respectively. At a given temperature, the number of gold moles in a mixture is nAu = xAun. The same can be said for silicon. The number of moles of gold per unit area is nAu = tAuρAu/MAu. Then same can be side for o silicon. From ﬁgure 11.11, at 400 C, xSi = .32 ∴ xAu = .62. Then ˚ −3 −1 nAu nSi tAuρAu tSiρSi tAuρAuMSixSi 1000A19.3g cm 28g mol .32 ˚ = ⇒ = ⇒ tSi = = −1 ⇒ tSi = 554A X xAu xSi MAuxAu MSixSi ρSiMSixAu 2.33g cm−3197g mol .68

(b) Redo the estimate for 800o C. o For 800 C, xSi = .45 then xAu = .55. Then ˚ −3 −1 tAuρAuMSixSi 1000A19.3g cm 28g mol .45 ˚ tSi = = −1 ⇒ tSi = 963A X ρSiMSixAu 2.33g cm−3197g mol .55