Free Energy Calculation
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- Βάλιος Γεννάδιος
- 5 χρόνια πριν
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Transcript
1 Free Energy Calculation 1
2 Outline Free energies: classical thermodynamics Free energies: statistical thermodynamics Monte Carlo simulations: what went wrong? Experiments: how to make a chemical potentialmeter? Free energy techniques: Thermodynamic integration Krikwood coupling parameter Widom test particle insertion Overlapping distribution Histogram method Umbrella sampling 2 2
3 Chemical potential Until now we have assumed that most of our systems are closed: NVE Let us now assume that our system can exchange matter First law du = TdS pdv + i Chemical potential µ i U n i U n i S,V,n j S,V,n j dn i Change in U if we add one mole of component i at constant S, V, n j 3
4 How is the chemical potential is defined in term of A,G,H A U TS da = du TdS SdT Using du = TdS pdv + U n i i U S,V,n j dn i da = SdT pdv + i n i A dn i S,V,n j da = SdT pdv + i Chemical potential A µ i n i n i T,V,n j = dn i T,V,n j U n i S,V,n j 4
5 Energy du = TdS pdv + i µ i dn i Enthalpy dh = TdS + Vdp + i µ i dn i Helmholtz free energy da = SdT pdv + i µ i dn i Gibbs free energy dg = SdT + Vdp + i µ i dn i Chemical potential µ i U n i S,V,n j = H n i S,p,n j = A n i T,V,n j = G n i T,p,n j 5
6 G is extensive How does the Gibbs free energy change if the system is increased by a factor k? G = kg G =(k 1)G dg = SdT + Vdp + Integration gives i µ i dn i G = i µ i n i G = i µ i (k 1)n i Which gives: G = i µ i n i 6
7 α β Equilibrium Let us consider a system with a constant number of particles, volume, and energy. This system consists of two phases α and β Question: When are these two systems in equilibrium? ds = ds α + ds β = 0 ds α = duα T α + pα T α dvα µ α i T α dnα i 1 i ds α + ds β = T α 1 p T β du α α + T α pβ T β dv α µ α i T α µβ i T β dn α i i Equilibrium: T α = T β p α = p β µ α i = µ β i 7
8 Vapor-liquid equilibria Equilibrium: T α = T β p α = p β µ α = µ β Van der Waals equation of P = k BT v b a v 2 We need to compute the chemical potential 8
9 Chemical potential In equilibrium we have: µ α (T, V α )=µ β (T, V β ) Equilibrium if: How to relate this to the equation of state? µ αβ = µ α (T, V α ) µ β (T, V β ) α µ µ αβ = dv = 0 V Recall that we derived the relation between changes in the chemical potential and temperature and pressure (Gibbs-Duhem) G = µ i n i or, for a pure component G = nµ i n i dµ i = SdT + Vdp or ndµ = SdT + Vdp dµ = sdt + vdp i β T,n 9
10 µ αβ = α β dv µ v We need to find µ v µ v T T Which gives: = µ T T,n p p = 0 + v v µ αβ = = 0 dµ = sdt + vdp µ µ = µ(t, P) v T T µ p + v p v α = T dvv β α β T T =0 =v p v T (p p coex )dv Equation of state gives us the chemical potential T = vp β α α β pdv 10
11 Vapor-liquid equilibria Equilibrium: T α = T β p α = p β µ α = µ β Equal chemical potential if α β (p p coex )dv = 0 Equilibrium if these areas are equal 11
12 NVT ensemble Define de Broglie wave length: Partition function: Free energy: F = 1 lnq( N,V,T ) β 12 12
13 Example: ideal gas Free energy: 13
14 Thermo recall (3) Helmholtz Free energy: Pressure Energy: 14
15 Free energy: Pressure: βf = N ln Λ 3 + N ln N V Energy: Chemical potential: βµ IG = ln Λ 3 + ln ρ +1 µ i = F N i T,V,N j βµ IG = βµ 0 + ln ρ 15 15
16 Monte Carlo simulation What is the difference between <A> and F? 16 16
17 The price of importance sampling A NVT = 1 1 dr N Q NVT Λ 3N N! A( r N )exp βu ( r N ) = dr N A( r N )P( r N ) = dr N A( r N )P( r N ) dr N P( r N ) = dr N A( r N )C exp βu ( r N ) = dr N A( r N ) exp βu r N dr N C exp βu ( r N ) dr N exp βu ( r N ) Generate configuration using MC: with r N 1,r N 2,r N 3,r N N { 4,r M } A = 1 M M A P MC ( r N ) = C MC exp βu r N i=1 N ( r i ) = dr N A( r N )P MC r N dr N P MC r N = dr N A( r N )C MC exp βu r N dr N C MC exp βu ( r N ) = dr N A( r N )exp βu ( r N ) dr N exp βu ( r N ) 17 17
18 Experimental (2) F V N,T = P F( P) F P 0 = V V o F V N,T dv? Works always for vapor-liquid Requires a large number of simulations to fit the equation of state Does not work for solid-liquid 18 18
19 Thermodynamic integration Coupling parameter U ( λ) = ( 1 λ)u I + λu II Lennard-Jones Stockmayer Q NVT ( λ) = 1 Λ 3N N! dr N exp βu λ 19 19
20 Q NVT F λ ( λ) = λ N,T 1 Λ 3N N! = 1 β dr N λ ln Q exp βu λ ( λ) = dr N U λ exp βu λ dr N exp βu λ F( λ = 1) F( λ = 0) = d Free energy as ensemble average! λ U λ λ λ U ( λ) = ( 1 λ)u I + λu II 20 20
21 Molecular association B + A G 1 Experimentally we can measure the free energies of association. B A B G 2 + A A B We would like to have the molecule which has the optimal free energy of association 21 21
22 How to compute the free energy of association? B + A G 1 B A Solution: compute the potential of mean force = the reversible work required to bring B to the host 22 22
23 A+B G 1 AB G solv G bind A+B G 2 AB We do not need to know the absolute value, but to see whether B or B is better we need to know only: ΔΔG = ΔG 1 ΔG 2 U ( λ) = U ( B) + λ U ( B' ) U ( B) In a simulation it is easier to compute: ΔΔG = ΔG bind ΔG solv 23 23
24 B A G 1bond B A U ( λ) = U ( B) + λ U ( B' ) U ( B) F( λ = 1) F( λ = 0) = d λ U λ λ λ When will this computation be accurate? 24 24
25 Q NVT = 1 Λ 3N N! Chemical potential dr N exp βu r N Scaled coordinates: s=r/l Q NVT = V N Λ 3N N! = ln = N ln ds N βf = ln( Q ) NVT V N Λ 3N N! ln exp βu s N ; L 1 Λ 3 ρ ln ( dsn exp βu ( s N ; L) ) ( dsn exp βu ( s N ; L) ) 25 25
26 Chemical potential: Widom test particle method βf = ln( Q ) NVT = N ln 1 Λ 3 ρ ln ( dsn exp βu ( s N ; L) ) βf = βf IG + βf ex µ F N V,T } 26 26
27 βf ( N) ) βf N + 1) βµ = N + 1 N = ln Q( N + 1) Q N = ln V N +1 Λ 3N + 3 ( N + 1)! V N Λ 3N N! V = ln Λ 3 N + 1 βµ = βµ IG + βµ ex βµ ex = ln ds N +1 ln ds N +1 exp βu s N +1 ; L ds N exp βu ( s N ; L) ln ds N +1 exp βu s N +1 ; L ds N exp βu ( s N ; L) exp βu s N +1 ; L ds N exp βu ( s N ; L) 27 27
28 βµ ex = ln ds N +1 ds N exp βu s N +1 ; L exp βu ( s N ; L) U ( s N +1 ; L) = ΔU + + U ( s N ; L) βµ ex = ln = ln ds N ds N +1 ds N +1 ( ) exp β ΔU + + U s N ; L ds N exp βu ( s N ; L) ds N { exp βδu + }exp βu s N ; L ds N exp βu ( s N ; L) = ln ds N +1 exp βδu + NVT Ghost particle! 28 28
29 29 29
30 Hard spheres = r σ U r βµ ex = ln ds N +1 exp βδu + NVT 0 r > σ exp βδu + = 0 if overlap 1 no overlap Probability to insert a test particle! 30 30
31 Lennard-Jones fluid 31 31
32 βf( N ) βµ = βf N + 1 N + 1 N = ln Q( N + 1) Q N = ln βµ = βµ IG + βµ ex V N +1 Λ 3N + 3 ( N + 1)! V N Λ 3N N! V = ln Λ 3 N + 1 βµ ex = ln ds N +1 Real-particle method ln ds N +1 exp βu s N +1 ; L ds N exp βu ( s N ; L) ln ds N +1 exp βu s N +1 ; L ds N exp βu ( s N ; L) exp βu s N +1 ; L ds N exp βu ( s N ; L) 32 32
33 βµ ex = ln ds N +1 exp βu s N +1 ; L ds N exp βu ( s N ; L) U ( s N +1 ; L) = ΔU + + U ( s N ; L) U ( s N ; L) = U ( s N +1 ; L) ΔU + βµ ex = ln = ln ds N +1 exp βu s N +1 ; L exp +βδu + βu s N +1 ; L ds N +1 = + ln exp +βδu + N +1VT ds N +1 exp βu s N +1 ; L ds N +1 exp +βδu + exp βu s N +1 ; L real particle! 33 33
34 Real particle energy 34 34
35 Other ensembles: NPT NVT: Helmholtz free energy µ F N V,T NPT: Gibbs free energy µ G N P,T βg = ln( Q NPT ) Q NPT = 1 Λ 3N N! ds N dvv N exp βvp exp βu s N ; L βµ = βg ( N) ) βg N + 1) N + 1 N βµ = ln Q N + 1 Q N 35 35
36 βµ = ln Q N + 1 Q N βµ = ln 1 Λ 3N + 3 ( N + 1)! 1 Λ 3N N! dvv N +1 exp βvp ds N +1 ds N dvv N exp βvp exp βu s N +1 ; L exp βu ( s N ; L) βµ = ln 1 Λ 3 N + 1 dvv N exp( βvp)v ds N exp βu s N ; L ds N +1 exp βδu + dvv N exp( βvp) ds N exp βu ( s N ; L) 36 36
37 βµ = ln 1 Λ 3 N + 1 dvv N exp( βvp)v ds N exp βu s N ; L ds N +1 exp βδu + dvv N exp( βvp) ds N exp βu ( s N ; L) ln βµ = ln Λ 3 βp βp N + 1 ds N dvv N exp βvp exp βu s N ; L ds N +1V exp βδu + dvv N exp( βvp) ds N exp βu ( s N ; L) The volume fluctuates! βpv N + 1 ds N +1 exp βδu + βpv N + 1 ds N +1 exp βδu + βµ = ln( Λ 3 βp) ln βpv N + 1 ds N +1 exp( βδu + ) 37 37
38 NVT versus NPT NVT: βµ = β ln( ρ) ln ds N +1 exp βδu + NVT NPT: βµ = ln( Λ 3 βp) ln βpv N + 1 ds N +1 exp( βδu + ) 38 38
39 Overlapping Distribution Method Two systems: System 0: N, V,T, U 0 System 1: N, V,T, U 1 Q 0 = V N Λ 3N N! ds N exp( βu 0 ) Q 1 = V N ds N exp( βu Λ 3N 1 ) N! ΔβF = βf 1 βf 0 = ln( Q 1 Q 0 ) = ln ds N exp βu 1 ds N exp βu 0 = ln q 1 q 0 ds N exp( βu 1 ) = q 1 q 0 ds N exp( βu 0 ) 39
40 System 0: N, V,T, U 0 System 1: N, V,T, U 1 Let us define p 1 ( ΔU ) ds N exp( βu 1 )δ U 1 U 0 ΔU and ds N exp βu 1 p 0 ( ΔU ) ds N exp( βu 0 )δ U 1 U 0 ΔU ds N exp( βu 0 ) 40 40
41 ΔβF = ln p 1 ds N exp βu 1 ds N exp βu 0 = ln q 1 q 0 ( ΔU ) = ds N exp( βu 1 )δ U 1 U 0 ΔU ds N exp( βu 1 ) p 0 ( ΔU ) = ds N exp( βu 0 )δ U 1 U 0 ΔU ds N exp( βu 0 ) p 1 ( ΔU ) = ds N exp β ( U 1 U 0 ) exp[ βu 0 ]δ U 1 U 0 ΔU ds N exp( βu 1 ) exp[ βδu ] ds N exp[ βu 0 ]δ ( U 1 U 0 ΔU ) = ds N exp( βu 1 ) =ΔU: Because of the δ function it can be taken outside the integration ds N exp( βu 1 ) = q 1 q 0 ds N exp( βu 0 ) 41 41
42 Overlapping Distribution Method p 1 ( ΔU ) = exp[ βδu ] ds N exp[ βu 0 ]δ U 1 U 0 ΔU ds N exp( βu 1 ) = q 0 exp( βδu ) q 1 ds N exp[ βu 0 ]δ U 1 U 0 ΔU ds N exp( βu 0 ) q 0 = exp( βδf) q 1 ds N exp( βu 1 ) = q 1 q 0 ds N exp( βu 0 ) 42 42
43 Let us define two new functions: f 0 f 1 ( ΔU ) ln p ( 0 ΔU ) 0.5βΔU ( ΔU ) ln p ( 1 ΔU ) + 0.5βΔU βδf = f 1 ( ΔU ) f 0 ( ΔU ) Fit f 0 and f 1 to two polynomials that only differ by the offset. f 1 f 0 ( ΔU ) C 1 + aδu + bδu 2 + cδu 3 βδf = C ( ΔU ) 1 C 0 C 0 + aδu + bδu 2 + cδu 3 Simulate system 0: compute f 0 Simulate system 1: compute f
44 System 0: N-1, V,T, U + 1 ideal gas System 1: N, V,T, U ΔU = U 1 U 0 System 0: test particle energy System 1: real particle energy 44 44
45 System 0: N-1, V,T, U + 1 ideal gas System 1: N, V,T, U f 0 ( ΔU ) ln p 0 ( ΔU ) 0.5βΔU f 1 ( ΔU ) ln p 1 ( ΔU ) + 0.5βΔU p 0 ( ΔU ) Accurate sampling only if there is overlap between p 1 ( ΔU ) the two distributions How to create this overlap? 45 45
46 Intermezzo: order parameters and Landau Free energies (1) Landau free energy density: β f ( q) = ln 1 Λ 3N N! δ ( q q ( r N ))exp βu r N Probability to find the system with order parameter q Free energy: P( q)dq exp β f q dq dr N F = f ( q)dq 46
47 Probability to find the system with order parameter q P( q)dq exp β f q dq P( q) q 47 47
48 Histogram method (1) P( q) Define a set of windows: q U j ( q) = min if q < q j 0 if q min max j < q < q j max if q > q j For each window determine: P j q 48 48
49 P j ( q) For each window separately Connect the windows using some overlap between Windows 49 49
50 Umbrella sampling (1) exp β f q = 1 Λ 3N N! δ ( q q ( r N ))exp βu r N exp βδf q Δf q = f ( q) f IG ( q) = dr N δ ( q q ( r N ))exp βu r N ( ) δ q q r N dr N dr N exp βδf q exp βδf q = π ( r N )π 1 ( r N )δ ( q q( r N ))exp βu r N π ( r N )π 1 ( r N )δ q q( r N ) dr N dr N = π 1 ( r N )δ ( q q( r N ))exp βu ( r N ) π ( r N )dr N π 1 ( r N )δ q q( r N ) π ( r N )dr N 50 50
51 exp βδf q exp βδf q = π 1 ( r N )δ ( q q( r N ))exp βu ( r N ) π ( r N )dr N π 1 ( r N )δ q q( r N ) π ( r N )dr N δ ( q q( r N ))exp βu ( r N ) = π 1 δ ( q q( r N ))π 1 r N π ( r N ) π This is NOT a Boltzmann distribution How to choose π? P( q) Ideal sampling q 51 51
52 P( q) q βf( q) = exp( βf est q ) π r N q 52 52
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