Thermodynamic integration

Transcription

1 Thermodynamic integration Localizing liquid-solid phase transitions Christoph Tavan Freie Universität Berlin / Technische Universität Berlin December 7, 2009

2 Overview Problem Theoretical basics Thermodynamic integration Localization of phase equilibria

3 60 Problem 50 P* random init. conf. fcc init. conf. bcc init. conf S. Grandner and S. H. L. Klapp: Freezing of charged colloids in slit pores. J. Chem. Phys., 129, (2008). ρ b * Simulations predict solid and fluid phase Metastable density range

4 60 Problem 50 P* ρ b * Where is the coexistence line?

5 60 Problem 50 P* ρ b * Where is the coexistence line?

6 60 Problem 50 P* ρ b * Where is the coexistence line?

7 Thermodynamics of phase equilibria Condition for equilibrium between solid (s) and liquid (l) phase: µ s (P, T) = µ l (P, T) With chemical potential µ = g = G N and Gibbs free energy G = F PV and Helmholtz free energy F.

8 Thermodynamics of phase equilibria So calculate Gibbs free energy for both phases: g = G N = F N PV N

9 Thermodynamics of phase equilibria So calculate Gibbs free energy for both phases: g = G N = F N PV N Easy: Pressure P as an ensemble average

10 Thermodynamics of phase equilibria So calculate Gibbs free energy for both phases: g = G N = F N PV N Easy: Pressure P as an ensemble average Hard: Helmholtz free energy F: F(N,V,T) = 1 ln Z(N,V,T) β

11 Thermodynamics of phase equilibria So calculate Gibbs free energy for both phases: g = G N = F N PV N Easy: Pressure P as an ensemble average Hard: Helmholtz free energy F: F(N,V,T) = 1 ln Z(N,V,T) β 1 Z(N,V,T) = Λ 3N exp [ βu(r) ] d 3N r N! with Λ = h β/2πm and β = 1/k B T.

12 Thermodynamics of phase equilibria So calculate Gibbs free energy for both phases: g = G N = F N PV N Easy: Pressure P as an ensemble average Hard: Helmholtz free energy F: F(N,V,T) = 1 ln Z(N,V,T) β 1 Z(N,V,T) = Λ 3N exp [ βu(r) ] d 3N r N! with Λ = h β/2πm and β = 1/k B T. In general impossible to evaluate directly

13 Observation: Derivatives of F Derivatives of the thermodynamic potentials are easy to clalculate: ( ) F = P(N,V,T) V NT

14 Observation: Derivatives of F Derivatives of the thermodynamic potentials are easy to clalculate: ( ) F = P(N,V,T) V NT Intensive quantities: ( ) f (ρ,t) = P(ρ,T) ρ ρ 2 T

15 Observation: Derivatives of F Derivatives of the thermodynamic potentials are easy to clalculate: ( ) F = P(N,V,T) V NT Intensive quantities: ( ) f (ρ,t) = P(ρ,T) ρ ρ 2 Calculate f = F/N by integration: T f (ρ) f (0) = ρ 0 P( ρ,t) ρ 2 d ρ

16 Observation: Derivatives of F Derivatives of the thermodynamic potentials are easy to clalculate: ( ) F = P(N,V,T) V NT Intensive quantities: ( ) f (ρ,t) = P(ρ,T) ρ ρ 2 Calculate f = F/N by integration: T f (ρ) f (0) = ρ 0 P( ρ,t) ρ 2 d ρ

17 Good for fluid phase: P* Thermodynamic integration fest (fcc) flüssig f (ρ) = ρ 0 P( ρ,t) ρ 2 d ρ ρ*

18 Good for fluid phase: Thermodynamic integration f (ρ) = ρ 0 P( ρ,t) ρ 2 d ρ For solid phase: Phase transitions break the reversibility of the integration path

19 Good for fluid phase: Thermodynamic integration f (ρ) = ρ 0 P( ρ,t) ρ 2 d ρ For solid phase: Phase transitions break the reversibility of the integration path But thermodynamic integration is still possible

20 Thermodynamic integration: solid phase Define effective interaction potential: Ũ(λ) = (1 λ)u + λu Ref λ [0,1]

21 Thermodynamic integration: solid phase Define effective interaction potential: Ũ(λ) = (1 λ)u + λu Ref λ [0,1] Calculate derivative of the free energy: ( ) F(λ) = 1 ln Z(N,V,T,λ) λ β λ NVT = 1 Z βz λ

22 Thermodynamic integration: solid phase Define effective interaction potential: Ũ(λ) = (1 λ)u + λu Ref λ [0,1] Calculate derivative of the free energy: ( ) F(λ) = 1 ln Z(N,V,T,λ) λ β λ NVT = 1 Z βz λ Ũ(λ) = λ λ

23 Thermodynamic integration: solid phase Define effective interaction potential: Ũ(λ) = (1 λ)u + λu Ref λ [0,1] Thermodynamic integration: 1 F = F Ref 0 U Ref U λ dλ Choose reference system with known free energy F Ref.

24 Einstein crystal Potential: U Ein (r) = N i=1 α ( ) ri r 2 i,0 2 Free energy: F Ein = 3 ( ) N 2π 2 β ln αβλ 2

25 Einstein crystal: choice of α Choose α such that fluctuations in the integrand become minimal: r 2 λ=1 = r 2 λ=0

26 Einstein crystal: choice of α Choose α such that fluctuations in the integrand become minimal: r 2 λ=1 = r 2 λ=0 Mean square displacement for the einstein crystal is known: r 2 λ=1 = 3 2 k BT 2 α

27 Einstein crystal: choice of α Choose α such that fluctuations in the integrand become minimal: r 2 λ=1 = r 2 λ=0 Mean square displacement for the einstein crystal is known: r 2 λ=1 = 3 2 k BT 2 α Condition: α = 3k BT r 2 λ=0

28 Constrained center of mass For λ 0 the particles are no longer bound to their lattice sites. The integrand in 1 F = F Ein becomes sharply peaked. 0 U Ein U λ dλ } {{ } F

29 Constrained center of mass For λ 0 the particles are no longer bound to their lattice sites. The integrand in 1 F = F Ein becomes sharply peaked. 0 U Ein U λ dλ } {{ } Constrain the center of mass movement. Free energy changes: F F CM = F CM Ein FCM

30 Constrained center of mass Free energy changes: F CM = F CM Ein FCM Change due to constraint for the einstein crystal: β ( ( ) F Ein F CM 3 β 2 ) Ein = 2 ln α 4π 2 m and for arbitrary interacting crystals: β ( F F CM) = ln ( ρ ) ln ( β 2πNm )

31 Constrained center of mass Free energy changes: F CM = F CM Ein FCM Change due to constraint for the einstein crystal: β ( ( ) F Ein F CM 3 β 2 ) Ein = 2 ln α 4π 2 m and for arbitrary interacting crystals: β ( F F CM) = ln ( ρ ) ln ( β 2πNm )

32 Constrained center of mass Excess free energy F ex F F id of the unconstrained crystal: βf ex N 3 2 ln 2 ln N N ( ) 2π β FCM αβ N + 3 ( ) 2π 2N ln αβ + ln ρ N ln ρ + 1 ln(2π) 2N.

33 Constrained center of mass Excess free energy F ex F F id of the unconstrained crystal: βf ex N 3 2 ln 2 ln N N ( ) 2π β FCM αβ N + 3 ( ) 2π 2N ln αβ + ln ρ N ln ρ + 1 ln(2π) 2N.

34 Constrained center of mass Excess free energy F ex F F id of the unconstrained crystal: βf ex N 3 2 ln 2 ln N N ( ) 2π β FCM αβ N + 3 ( ) 2π 2N ln αβ + ln ρ N ln(2π) ln ρ + 1 2N. System size dependence: β F CM N = const. ln N N + O ( N 1)

35 Constrained center of mass Excess free energy F ex F F id of the unconstrained crystal: βf ex N + ln N ( ) 2π N 3 2 ln const. O ( N 1) + 3 ( ) 2π αβ 2N ln αβ + ln ρ N System size dependence: β F CM N ln(2π) ln ρ + 1 2N. ln N = const. N + O ( N 1) So βf ex /N + ln N/N depends linear on 1/N.

36 System size dependence Soft sphere fcc-crystal at ρ = , using α = 132 and N = 216, 392, 810, 1728, 3600 (β F ex* )/N + ln(n)/n Diese Arbeit (3) - 5.5(2)/N Polson et al (7) (7)/N /N

37 Summary Thermodynamic integration for fluid phase: βf ex N ρ = β 0 P( ρ,t) ρ/β ρ 2 d ρ

38 Summary Thermodynamic integration for fluid phase: βf ex N ρ = β 0 P( ρ,t) ρ/β ρ 2 d ρ Thermodynamic integration for solid phase: βf ex N = 3 2 ln 2 ln N N ( ) 2π β 1 U Ein U αβ N λ dλ + 3 2N ln 0 + ln ρ N ln ρ + 1 ln(2π) 2N ( ) 2π αβ

39 Can we now find phase-equilibria? Theoretically expected form of the chemical potential vs. pressure: fest flüssig µ P f P c P s P

40 Can we now find phase-equilibria? Computer simulation results for U(r) = ε(r ) 12 soft spheres at T = 1: fest (fcc) flüssig µ* P*

41 Can we now find phase-equilibria? Difference between the previously shown curves: Polynom 2. Grades Polynom 3. Grades Polynom 4. Grades µ* s -µ* f P* Phase coexistence at P = 22.6(5)

42 Equation of state for soft spheres Coexistence at P = 22.6(5), ρ f = 1.152(7), ρ s = 1.194(7) P* P* flüssig fest ρ* ρ* fest (fcc) flüssig Phasen koexistenz

43 Summary Thermodynamic integration for liquid and solid phase. Allows calculation of Helmholtz free energy in NVT-Ensemble. Gibbs free energy follows directly. Phase coexistence can be found. Equation of state can be drawn.

44 What about the charged colloids? 60 50??? P* ρ* b Coexistence pressure still to be found... Effects due to walls?