Statistical Physics. Solutions Sheet 11.

Transcription

1 Statistical Physics. Solutions Sheet. Exercise. HS 0 Prof. Manfred Sigrist Condensation and crystallization in the lattice gas model. The lattice gas model is obtained by dividing the volume V into microscopic cells which are assumed to be small enough such that they contain at most one gas molecule. In two and three dimensions, the result is a square and a cubic lattice, respectively. We neglect the kinetic energy of a molecule and assume nearest neighbornteractions. The total energy is given by ni nj () H = λ hi,ji where the sum runs over nearest-neighbor pairs and λ is the nearest-neighbor coupling. There is at most one particle in each cell (ni = 0 or ). This model is a simplification of hard-core potentials, like the Lennard-Jones potential, characterized by an attractive interaction and a very short-range repulsive interaction that prevents particles from overlapping. In order to study the case of a repulsive interaction, λ < 0, we divide the lattice into two alternating sublattices A and B. For square or cubic lattices, we find that all lattice sites A only have pointn B as their nearest neighbors. Figure : Schematic view of the lattice gas model. (a) Show the equivalence of the grand canonical ensemble of the lattice gas model with the canonical ensemble of an Ising model in a magnetic field. We consider the grand canonical Hamiltonian H µn = λ ni nj µ ni. (S.) i hi,ji By introducing Ising spins si through the relation ni = ( + si ), si = ±, (S.) we arrive at an Ising model H µn = J si sj h i hi,ji γ γ si h J NL = HI h J NL (S.3) with λ λ µ, h= γ+. (S.) Here, γ denotes the coordination number (number of nearest neighbors) and NL is the total number of lattice sites. The grand partition function Z = Tr [exp[ (H µn )]] of the lattice gas thus related to the canonical partition function ZI = Tr [exp( HI )] of the Ising model through J= λ µ ZG = ZI e ( 8 γ+ )NL with the relations (S.) for the exchange coupling J and the magnetic field h. (S.5)

2 (b) Introduce two mean-field parameters m A and m B, corresponding to the two sublattices A and B, and adapt the mean-field solution of the Ising model discussed in Sec. 5. of the lecture notes for these two parameters. What are the self-consistency conditions for m A and m B? The Hamiltonian of the Ising model is H I = J i,j s j h i. (S.6) We introduce the mean-field parameters m A and m B, which are defined as m A = i A, m B = s j j B. (S.7) Now we can write = m A,B + δ i := m A,B + ( m A,B), where we assume δ i to be small. Now we can expand the Hamiltonian as (S.8) H I = J i,j (m A + δ i)(m B + δ j) h i = J i,j (m Am B + m Bδ i + m Aδ j + δ iδ j) h i J i,j [m Am B + m B( m A) + m A(s j m B)] h i (S.9) = γn JmAmB γj i A m B γj j B m As j h i = γn JmAmB i A (γjm B + h) j B(γJm A + h)s j, where we used that nearest neighbors always belong to different sublattices and neglected the product δ iδ j. We find that the two sublattices A and B behave as paramagnetn the effective fields h A eff = γjm B + h, h B eff = γjm A + h. (S.0) The partition function of a paramagnet was already discussed previously, so the partition function of this mean-field Hamiltonian is Z I = exp [ ] [ ( N/ [ ( N/ γnjmamb cosh heff)] A cosh heff)] B. (S.) Thimmediately leads to the Helmholtz free energy F I(, h, N) = N ( Jγm Am B { [ ] [ log cosh(h A eff) + log cosh(heff)]} ) B. (S.) The self-consistent solutions are given by the local minima of the free energy. The conditions are therefore where h A eff and h A eff are given by (S.0). F [ ] I = 0 m B = tanh h A eff m A F I m B = 0 m A = tanh (S.3a) [ ] h B eff, (S.3b) (c) Use your results from parts (a) and (b) to calculate the grand potential for the lattice gas and determine the self-consistency relations for the two mean-field parameters ρ A = n i i A and ρ B = n i i B.

3 We use the mean-field approximation (S.) derived in part (b) and the relations (S.) in order to write the grand potential Ω(, µ, N L) = ( λ log ZG = FI(, h, NL) 8 γ + µ ) N L [ ( ) = NL λγ + µ + λγ (ρa )(ρb ) (S.) { [ ( )] [ ( )]}] log cosh (λγρa + µ) + log cosh (λγρb + µ), where we defined ρ = ( + m). Here, the effective magnetic fields (S.0) are replaced by h A,B eff (λγρb,a + µ). (S.5) We can now reformulate the self-consistency equations (S.3) for the lattice gas by inserting the relations (S.5). Using artanh x = log[( + x)/( x)] for x [, ], we obtain the two relations which can also be written in the form µ = log ρ A ρ A λγρ B = log ρ B ρ B λγρ A, (S.6) ρ A = + e, (λγρ B+µ) (S.7a) ρ B = + e. (λγρ A+µ) (S.7b) By inserting Eq. (S.7b) into Eq. (S.7a), we can in principle obtain the single condition [ ( [ ])] γλ ρ A = + exp + exp ( (γλρ A + µ)) + µ. (S.8) In the following we will use the mean-field solution of the lattice gas model in order to discuss the liquid-gas transition for an attractive interaction λ > 0. (d) Argue, why in this case the mean-field results can be simplified as the two densities must be equal, ρ A = ρ B = ρ. Use your knowledge of the Ising model to define a critical temperature T c, below which there are multiple solutions to the self-consistency equations, and discuss the solutions of ρ for temperatures above or below T c. Define also the critical chemical potential µ 0 corresponding to h = 0 in the Ising model and use this for a distinction of cases. The two self-consistency equations (S.7) are of the mathematical form a = φ(b) b = φ(a), (S.9) where the function is given by φ(x) =. (S.0) + e (λγx+µ) It is easy to see that for λ > 0 this function is monotonically increasing, while it is decreasing for λ < 0. Now if we assume b > a, thimplies f(b) > f(a). Thimmediately leads to a contradiction, as a = f(b) f(a) = b > a. The same contradiction follows for b < a. Therefore, for λ > 0 there are only symmetric solutions ρ A = ρ B for the self-consistency equations and we can simplify the whole treatment by just omitting the second mean-field parameter altogether. From Eq. (S.) we see that h = 0 corresponds to µ = λγ/ =: µ 0. For this case we can use the knowledge about the magnetic transition in the zero-field Ising model. In particular, there is a critical temperature k BT c = γλ/ = µ 0/ below which there exist two degenerate solutions. 3

4 liquid ρ ρ l (T ) ρ g (T ) µ = µ 0 µ > µ 0 µ < µ 0 gaseous T/T c.5 Figure : The density ρ as a function of temperature T for different values of the chemical potential µ. In the lattice gas, these solutions correspond to the liquid and to the gaseous phase and we will denote them by ρ l (T ) and ρ g(t ), respectively (see Fig. ). The third solution of Eq. (S.7) for µ = µ 0, namely ρ = /, is only stable above T c. In the general case, there is a unique solution of Eq. (S.7) for T > T c while for T T c there are three solutionn the neighborhood of µ = µ 0 = k BT c but only one minimizes Ω (see Figs. and 3). The solution with dρ/dµ > 0 is stable or metastable while the solution with dρ/dµ < 0 is unstable and corresponds to a local maximum of the grand potential Ω. Thus, for T < T c, the density ρ(t, µ) jumps at µ 0 reflecting the first-order liquid-gas transition (see Fig. 3). (e) Find the equation of state p = p(t, ρ) or p = p(t, v) and discuss the liquid-gas transition in the p v diagram. Thereby, v = /ρ is the specific volume. Compare with the van der Waals equation of state: ( p + ã ) ( b) v v = k B T. What is different in our model? Hint. For the lattice gas, we have b =. The pressure is given by p(, µ) = Ω(, µ, N L) N L = µ ( λγ (ρ ρ) log { [ ( λγ cosh ρ + µ )]}), (S.) where we used Eq. (S.). For ρ(, µ) ρ g() and ρ(, µ) ρ l () we can simply insert Eq. (S.6) into the above equation and obtain p(t, ρ) = λγ ρ log( ρ) (S.) or in terms of the specific volume v = /ρ p(t, v) = λγ v kbt log( v ). (S.3) But for ρ g() ρ(, µ) ρ l () there is coexistence of the liquid and the gas. We have to set µ = µ 0 and ρ = ρ g,l (T ) in Eq. (S.) (this corresponds to the Maxwell construction) leading to a constant pressure! This shown in the p v diagram Fig..

5 ρ 0. T = T c T = T c T = 0.5T c instable 0. stable metastable 0!3!.5!!.5! µ [k B T c ] Figure 3: The density ρ as function of the chemical potential µ for different temperatures. For T < T c there is a jump in ρ at µ = µ 0 = k BT c. We can rewrite the van der Waals equation of state as follows: p(t, v) = ã v + kbt v b. The elementary volume of the gas (hard core volume) b equals in our model, since the volume is given by the total number of lattice sites, N L. Comparing this with Eq. (S.3), we see that the first term identical if we set ã = λγ/, whereas the second term diverges either linearly (van der Walls) or logarithmically (our model) with v. This different behavior is present in the limiting case of high density and can be attributed to the short-range difference of the potential for the discrete lattice gas model and the continuous van der Waals gas. (f) Find the phase diagram (T p diagram). Determine the phase boundary (T, p c (T )) and, in particular, compute the critical point (T c, p c (T c )). The critical pressure is given by Eq. (S.) for µ = µ 0 = k BT c and ρ = ρ g,l (T ) p c(t ) = k BT cρ g,l(t ) k BT log( ρ g,l (T )), (S.) as shown in Fig. 5. In particular, for T = T c we have ρ g,l (T c) = / and p c(t c) = kbtc (log ). (S.5) Instead of the liquid-gas transition, which we have observed for an attractive interaction λ > 0, a crystallization transition (sublimation) can be observed for nearest-neighbor repulsion, λ < 0. In this case, we will find that the two mean-field parameters are different, ρ A ρ B, below some critical temperature T c. (g) Discuss the solutions below the critical temperature for λ < 0. Plot the densities ρ A and ρ B, as well as the average, (ρ A + ρ B )/ for both attractive and repulsive nearest-neighbor interaction at low temperature, T < T c. 5

6 T > T c p(t, v) [kbtc] T < T c T = T c critical point Coexistence v Figure : The isotherms p(t, v). The shaded region denotes the region of liquid-gas phase coexistence. p k B T C Liquid Gas Figure 5: p-t phase diagram of the lattice gas model. The two phases coexist when µ = µ 0 and T < T c (equilibrium line). Above T c there is only one phase (a single density for a given pressure). Below the same critical temperature k BT c = γ λ / as for an attractive interaction and in a certain range around µ 0 = γλ/, we find graphically that three different solutions for the self-consistency relations (S.7) exist, corresponding to the crossings of φ(ρ A) and φ(ρ B) shown in Fig. 7. There are two degenerate asymmetric solutions ρ A ρ B, which are related by ρ A ρ B, and one symmetric solution with ρ A = ρ B. The exact range µ [µ 0 µ, µ 0 + µ] for which three solutions exist is given by the condition φ (ρ) φ(ρ)=ρ <. (S.6) This can be understood by looking again at Fig. 7 and by noting that since φ(ρ) > 0, there must exist two asymmetric solutions whenever φ (ρ) < at the symmetric solution (the middle crossing). By inserting the definition of φ into Eq. (S.6) and solving for µ, one obtains with µ 0 µ < µ < µ 0 + µ, (S.7) µ = γλ θ + ( ) + θ log, θ = + θ γλ. (S.8) The asymmetric solutions, which are generally lower in energy, correspond to a crystal structure, where (at T = 0) one of the sublattices occupied while the other one is empty. The densities for attraction and 6

7 repulsion are shown in Fig. 6. While for a nearest-neighbor attraction the densities of the sublattices are identical, there is a symmetry-broken phase for nearest-neighbor repulsion. Ρ Ρ A Λ 0 Ρ B Λ 0 Ρ Λ 0 Ρ Λ Figure 6: Densities on the two sublattices for attractive (λ > 0) and repulsive (λ < 0) nearest-neighbor interaction at T = 0.5 T c. The thick lines show the average densities, the dashed and dotted lines the densities of the two sublattices..0 ΡB A Figure 7: Plot of φ(ρ A) and φ(ρ B), defined in Eq. (S.0), for µ [µ 0 µ, µ 0 + µ]. 7