Homework 8 Solutions Problem 1: Kittel 10-4 (a) The partition function of a single oscillator that can move in three dimensions is given by:

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1 Homework 8 Solutions Problem : Kittel 0-4 a The partition function of a single oscillator that can move in three dimensions is given by: Z s e ɛ nx+ny+nz hω/τ e ɛτ e n hω/τ e ɛ/τ e hω/τ n x,n y,n z n where ɛ is the ground state energy of an atom in the solid lattice. Here, the zero of the energy is taken to be the energy of a free atom at rest which is the ground state energy of an atom in the gas. We may estimate the chemical potential of the atoms in the solid: µ s G s F s + pv s F s λ s e µs/τ e Fs/τ e log Zs Z s e ɛ/τ e hω/τ G s, F s, and v s are the Gibbs energy, Helmholtz energy, and volume per particle for the atoms in the solid. p is the vapor pressure of the gas-solid system. We have neglected the term pv s, which is presumably very small for a solid. We know the chemical potential for the gas phase, which we treat as an ideal gas: µ g τ log n n Q λ g e µg/τ n n Q p τn Q p τ 2π h 2 Mτ /2 Because the system is in diffusive equilibrium, we have that λ g λ s. From this, we may obtain the vapor pressure: p τ 2π h 2 /2 e ɛ/τ e hω/τ p Mτ M 2π /2 τ 5/2 h e hω/τ e ɛ/τ M /2 τ 5/2 h 2π ω h τ e ɛ/τ M /2 ω 2π e ɛ/τ τ/2 where we have taken a high temperature limit τ >> hω b We start with the Clausius-Clapeyron equation K&K equation 0-5: dp dτ L τv g v s L Lp τv g τ 2 where the last step involved the ideal gas law. Next, we differentiate our expression for the vapor pressure: dp M /2ω e ɛ/τ dτ 2π 2τ + ɛ /2 τ 5/2

2 M /2 ω 2π τ p ɛ τ 2 τ 2 /2 e ɛ/τ τ 2 ɛ τ 2 Problem 2 Kittel 0-5 a The total free energy of the system is: F F s + F g U s + U g + pv s + V g U s + U g + pv U s + F g N s ɛ o + N g τ[logn g /V n Q ] In this derivation, we have assumed V g V and V s 0. b If the total particle number N N s + N g is constant, then we may write: FN g N g Nɛ o + N g τ[logn g /V n Q ] Minimizing this with respect to N g gives the condition: df dn g 0 ɛ o + τ[logn g /V n Q ] + τ N g n Q V exp ɛ o /τ c The equilibrium vapor pressure follows from the ideal gas law: p N gτ V n Qτ exp ɛ o /τ Problem - Kittel 0-6 a On Kittel page 256, we have an expression for the difference in free energy between the normal and superconducting phases of a material: [F N τ F S τ]/v B 2 cτ/2µ o Using the relation: σ F/ T V, we see immediately that: [σ S τ σ N τ]/v B 2 τ 2µ cτ B c db c o µ o dτ As τ 0, both entropies go to zero by the third law. This implies that at τ 0, db c /dτ 0. We are also given that B c decreases with increasing temperature. The figure shows a curve consistent with these facts not the only possibility. b Since B c τ c 0, we have that F N τ c F S τ c. Because the entropies are also equal, we also know that F N τ c F S τ c. Therefore, the curves are tangent at τ τ c. 2 By defintion,f U τσ. Because at τ τ c, F N F S and σ N σ S,which together imply that U N U S. The latent heat per volume of a transition between states N and S is defined as L τσ N σ S Bc db c µ 0 dτ. At τ τ c, B c 0 so therefore L 0. c C C S C N τ σ S/V τ τ σ N/V τ 2 τ d 2 Bc 2 2µ 0 dτ 2

3 Bc T Figure : Cartoon of B c vs τ. For τ << τ c, we are told that C is dominated by C N. Also, we are told that C N γτ + Oτ 2 and C 2 Oτ 2. Therefore, C γτ + Oτ 2 τ d [ db ] c B c µ 0 dτ dτ γ Oτ + [ dbc 2 d 2 B ] c + Bc µ 0 dτ dτ 2 τ [ dbc 2 d 2 B ] c + Bc µ 0 dτ dτ 2 Since the above holds for any low temperature, we know that in particular, it should hold for τ 0. Using the fact that db c /dτ 0 at τ 0, we have: γ µ 0 B c d 2 B c dτ 2 τ0 Problem 4 - Kittel 0-8 a We seek the free energy of a gas of phonons in the Debye approximation. A fast way to obtain this is to note that for a photon gas, the free energy density is F photon /V π 2 τ 4 /45 h c. The Debye theory follows the same derivation where the sound speed v is used instead of c and the number of polarizations is instead of 2. Following the photon derivation see Homework, problem 5, we see that we need to multiply the photon expression by /2: F phonon /V 2 F photon/v π2 τ 4 0 h v b Whether the α or β configuration is favored depends on which state has the lower free energy. When τ τ c, the two free energies are equal. The free energy of state α per volume is the sum of the free energy per volume of the ground state crystal, U α τ 0, plus the free energy density of the phonons, which are excited when the crystal experiences a finite temperature. Likewise

4 for β. π2 τ 4 c F α F β U α 0 0 h vα U β 0 π2 τ 4 c 0 h v β τ 4 c 0 h /π 2 [U β 0 U α 0]/v β v α c The latent heat per volume is: dfβ L τ c σ α τ c σ β τ c τ c dτ df α dτ τ c 4 π2 τc 0 h 4 π2 τ c π 2 τc 4 vβ 0 h 4 vα 0 h vβ 4[U β 0 U α 0] ττ c π2 τ 4 c 0 h v α Problem 5 The equation for the isotherm is given on page 2-2 of the notes: pv 8τ V V 2 where p is given in units of p c, τ in units of τ c, and V in units of V c. For the case of τ 0.9, the isotherm is shown in the accompanying figure. Also drawn are the lines p p max and p p min which define the pressure range over which the volume is triple-valued. We find the pressure range using the fsolve function of Maple. The numbers turn out to be p min and p max The corresponding volumes are v min,l 0.79, v min,g 4.6, v max,l 0.59, and v max,g.5. In order to determine the pressure where we get liquid-gas coexistence, we use the procedure outlined in the hint. V2 V dp V dp dv dv V V2 2.4ln V fv, V 2 + V V V V 2 Here V and V 2 are the liquid and gas volumes respectively corresponding to a particular value of the pressure. Coexistence occurs at the pressure which causes fv, V 2 0, for reasons discussed more fully in the lecture notes. By numerical iteration, we find this happens for a pressure p Problem 6 On page 02 of Kittel, the model of ferromagnetism is discussed. The energy per unit volume of this system is given by U/V λm 2 /2 nτ c m 2 /2, where the variable are defined in the problem set. We also know that m tanhm/t where t τ/τ c. Armed with this information, we may calculate the heat capacity per volume. First, we need to find dm/dτ: dm dt sech 2 m/t m t 2 + dm t dt dm dτ msech 2 m/t τ c tsech 2 m/t t 2 4

5 V Figure 2: Plot of P/P c vs V/V c where p p max and p p min are also shown. Then: C du/v dτ nτ c m dm dτ nm2 sech 2 m/t t 2 tsech 2 m/t The following figure is a plot of C vs t for t < as requested. For t >, m 0 is the only solution to the self-consistent equation m tanhm/t. This implies that C 0 above the critical temperature. The jump discontinuity is finite. Problem 7 a Suppose a system can be in a number of states and the probability of it being in state i is given by p i. We want to calculate the average value of some property A of the system, < A > i A ip i. Formally, this involves summing an infinite number of terms. Monte-Carlo simulation is a way to get an approximate answer by summing a finite number of terms. The basic idea is to randomly choose some states {i} and calculate the average of the {A i }. The Metropolis algorithm is a way to ensure that, if we take enough states, then the probability that a given state i will appear in our random sequence is p i. Therefore, our Monte-Carlo average will well-approximate the actual average. Here is the idea of Metropolis: Suppose our system begins in some state i. We randomly choose another state j. If p i < p j, then we accept the choice and our system is in state j. If p i > p j, then we accept the choice with probability p j /p i. Depending on the outcome of the random number generator, our system will stay in state i or move to state j. The important point is that either way we update our running average for A. b Suppose we have two states 0 and which occur with probabilities p and p. Then if our random number generator gives a number x < p, then 5

6 .6 h8_p6.txt Figure : Plot of C vs t τ/τ c. we say our system is in state 0. Otherwise, it is in state. c If our system is in state 0 after n steps, then after the n + st step, we know that it is either in state 0 with probability t 00 or in state with probability t 0. Thus, t 00 + t 0. Likewise, t 0 + t. d Suppose we impose the condition that if our inputs are canonical, then the outputs must be canonical. This means: which means that: e E/τ t 00 t t 00 t t 00 + t e E/τ t 00 e E/τ t e E/τ e The transition matrix for Metropolis is given by: T Metro e E/τ e E/τ 0 To see this is Metropolis, we note that if we start in state, then it must transition to the lower state 0, which implies that t 0 and t On the other hand, if the system is in state 0, then it will transition to state with a transition probability which is the ratio of the probabilities p /p 0 e E/τ. 6

7 f If we want canonical outputs regardless of the inputs, then we need the following matrix. You may explicitly verify that it behaves as required: T Canon + e E/τ e E/τ e E/τ 7

Phase Transitions. Phys112 (S2012) 8 Phase Transitions 1

Phase Transitions. Phys112 (S2012) 8 Phase Transitions 1 Phase Transitions cf. Kittel and Krömer chap 10 Landau Free Energy/Enthalpy Second order phase transition Ferromagnetism First order phase transition Van der Waals Clausius Clapeyron coexistence curve