free energy and phase equilibria - acmm · free energy and phase equilibria thermodynamic...
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Free Energy and Phase equilibria Thermodynamic integration 7.1
Chemical potentials 7.2 Overlapping distributions 7.2
Umbrella sampling 7.4 Application: Phase diagram of Carbon
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Why free energies? • Reaction equilibrium constants
• Examples: – Chemical reactions, catalysis, etc.... – Protein folding, binding: free energy gives binding constants
• Phase diagrams – Prediction of thermodynamic stability of phases, – Coexistence lines – Critical points – Triple points – First order/second order phase transitions
€
K =[B][A]
=pBpA
= exp −β(GB −GA )[ ]
€
A↔ B
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Critical point: no difference between liquid and vapor Triple point: liquid, vapor and solid in equilibrium. How do we compute these lines?
Along the liquid gas coexistence line increasing the pressure and temperature at constant volume the liquid density becomes lower and the vapor density higher.
Phase diagrams
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Phase equilibrium
TI = TII PI = PII µI = µII
If µI > µII : transport of particles from phase I to phase II.
Stable phase:
Lowest chemical potential (for single phase: lowest Gibbs free energy)
Criteria for equilibrium (for single component) Chemical potential
€
µ =∂F∂N#
$ %
&
' ( V ,T
=∂G∂N#
$ %
&
' ( P ,T
=Gm
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Relation thermodynamic potentials
Helmholtz free energy: F = U - TS
Gibbs free energy: G = F + PV
Suppose we have F(n,V,T)
Then we can find G from F from:
All thermodynamic quantities can be derived from F and its derivatives
TnVFVFG
,⎟⎠
⎞⎜⎝
⎛∂
∂−=
€
P = −∂F∂V$
% &
'
( ) n,T
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Phase equilibria from F(V,T) Common tangent construction
V
F
€
P = −∂F∂V$
% &
'
( ) n,T
Equal tangents liquid
gas Connecting line: equal G
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Common tangent construction
V
F TnVFVFG
,⎟⎠
⎞⎜⎝
⎛∂
∂−=liquid
gas
Helmholtz Free Energy Perspective
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Common tangent construction
V
G
€
G = F −V ∂F∂V$
% &
'
( ) n,T
liquid
gas
Only equilibrium when P,T is on coexistence line.
Both liquid and vapor G equal and minimal
Gibbs Free Energy Perspective
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We need F or µ • So equilibrium from F(V) alone or from P and µ
• So in fact for only 1 point of the equation of state the F is needed
• For liquid e.o.s even from ideal gas
€
F(V ) = F(V0) +∂F∂V#
$ %
&
' ( N,T
dVV0
V∫ = F(V0) − PdV∫
€
F(ρ) = F(ρ0) + N P( # ρ )ρ2
d # ρ ρ0
ρ
∫
€
βF(ρ) /N = βF id (ρ) /N +βP( $ ρ ) - $ ρ
ρ2d $ ρ
0
ρ
∫
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Equation of state
€
P = P ρ,T( )
€
∂F∂V#
$ %
&
' ( N ,T
= −P
€
F(ρ) = F(ρ0) + N P( # ρ )ρ2
d # ρ ρ0
ρ
∫
€
βF(ρ) /N = βF id (ρ) /N +βP( $ ρ ) - $ ρ
ρ2d $ ρ
0
ρ
∫
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Free Energies and Phase Equilibria
• Determine free energy of both phases relative to a reference state Free energy difference calculation General applicable: Gas, Liquid, Solid, Inhomogeneous systems, …
• Determine free energy difference between two phases Gibbs Ensemble Specific applicable: Gas, Liquid
General Strategies
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Statistical Thermodynamics
( )NVTQF ln−=β
€
A NVT =1
QNVT
1Λ3NN!
drN∫ A rN( )exp −βU rN( )[ ]
€
QNVT =1
Λ3NN!drN∫ exp −βU rN( )[ ]
Partition function
Ensemble average
Free energy
€
P rN( ) =1
QNVT
1Λ3NN!
dr'N∫ δ r'N −rN( )exp −βU r'N( )[ ]∝exp −βU rN( )[ ]Probability to find a particular configuration
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Ensemble average versus free energy
€
≈drN A rN( )∫ exp −βU rN( )[ ]drN exp −βU rN( )[ ]∫
= A NVT
Generate configuration using MC:
€
r1N ,r2
N ,r3N ,r4
N!,rMN{ }
€
A = 1M
Ai=1
M
∑ riN( )
€
βF = −lnQNVT = −ln 1Λ3NN!
drN∫ exp −βU rN( )[ ]
F is difficult, because requires accounting of phase space volume
Generate configuration using MD:
€
r1N ,r2
N ,r3N ,r4
N!,rMN{ }
€
A = 1M
Ai=1
M
∑ riN( ) ≈ 1T dtA(t)
0
T
∫ ≈ A∫ NVTergodicity
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I - Thermodynamic integration • Known reference state λ=0 • unkown target state λ=1
F(λ =1)−F(λ = 0) = dλ∂F λ( )∂λ
#
$%
&
'(N ,V ,Tλ=0
λ=1
∫
€
U λ( ) = 1− λ( )UI + λUII
Coupling parameter
€
QNVT λ( ) =1
Λ3NN!drN∫ exp −βU λ( )[ ]
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Thermodynamic integration
€
∂F λ( )∂λ
$
% &
'
( ) N ,T
= −1β∂∂λln Q( )
€
= −1β1Q∂Q∂λ
€
=drN ∂U λ( ) ∂λ( )∫ exp −βU λ( )[ ]
drN∫ exp −βU λ( )[ ]
( )λλ
λ∂
∂=
U
€
F λ =1( ) − F λ = 0( ) = d∫ λ∂U λ( )∂λ
λ
Free energy as ensemble average!
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Example • In general
• Specific example
€
∂U λ( )∂λ
λ
= UII −UI λ
€
U 0( ) =ULJ
U 1( ) =UStockm Stockmayer
Lennard-Jones
€
U λ( ) =ULJ + λU dipole-dipole
€
U λ( ) = (1− λ)UI + λUII
€
∂U λ( )∂λ
λ
= Udip−dipλ
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Free energy of solid More difficult. What is reference? Not the ideal gas. Instead it is the Einstein crystal: harmonic oscillators around r0
€
F = Fein + dλ∂U λ( )∂λλ= 0
λ=1
∫λ
U λ;rN( ) = (1−λ)U rN( )+λU r0N( )+λ α(ri − ri )
2
i=1
N
∑
€
F = Fein + dλλ= 0
λ=1
∫ −U rN( ) +U r0N( ) + α(ri − ri)
2
i=1
N
∑λ
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Hard sphere freezing
P
ρ
Solid free energy from Einstein crystal liquid free
energy from Ideal gas
Equal µ/P (and T)
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II - Thermodynamic perturbation Two systems:
System 1: N, V, T, U1 System 0: N, V, T, U0
ΔβF = βF1 −βF0 = −ln Q1 Q0( )
= −lndsN∫ exp −βU1$% &'dsN∫ exp −βU0( )
= −lndsN∫ exp −β U1 −U0( )$
%&'exp −βU0
$% &'
dsN∫ exp −βU0( )
€
Q0 =V N
Λ3NN!dsN∫ exp −βU0( )
€
Q1 =V N
Λ3NN!dsN∫ exp −βU1( )
( )1 0 0ln expF U Uβ βΔ = − − −$ %& '
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€
= −N ln 1Λ3ρ
%
& '
(
) * + N − ln dsN∫ exp −βU sN ;L( )[ ]( )
€
= −ln V N
Λ3NN!$
% &
'
( ) − ln dsN∫ exp −βU sN ;L( )[ ]( )
Chemical potential
€
µ ≡∂F∂N$
% &
'
( ) V ,T
€
QNVT =V N
Λ3NN!dsN∫ exp −βU sN ;L( )[ ]
€
βF = −ln QNVT( )
€
βF = βF IG + βF ex
} €
βµ = βµIG + βµex
TV
exex
NF
,!!"
#$$%
&
∂
∂≡
ββµ
TV
IGIG
NF
,!!"
#$$%
&
∂
∂≡
ββµ
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Widom test particle insertion
TVNF
,!"
#$%
&∂
∂≡
ββµ
€
βµ =βF N +1)( ) −βF N)( )
N +1− N
€
= −lnQ N +1( )Q N( )
€
= −ln
V N +1
Λ3N +3 N +1( )!V N
Λ3NN!
$
%
& & & &
'
(
) ) ) )
− lndsN +1∫ exp −βU sN +1;L( )[ ]dsN∫ exp −βU sN ;L( )[ ]
$
%
& &
'
(
) )
€
= −ln VΛ3 N +1( )
$
% &
'
( ) − ln
dsN +1∫ exp −βU sN +1;L( )[ ]dsN∫ exp −βU sN ;L( )[ ]
$
%
& &
'
(
) )
€
βµ = βµIG + βµex
€
βµex = −lndsN +1∫ exp −βU sN +1;L( )[ ]dsN∫ exp −βU sN ;L( )[ ]
%
&
' '
(
)
* *
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Widom test particle insertion
€
βµex = −lndsN +1∫ exp −βU sN +1;L( )[ ]dsN∫ exp −βU sN ;L( )[ ]
%
&
' '
(
)
* *
€
βµex = −lndsN∫ ds N+1∫ exp −β ΔU + +U sN ;L( )( )[ ]
dsN∫ exp −βU sN ;L( )[ ]
&
'
( (
)
*
+ +
€
U sN +1;L( ) = ΔU + +U sN ;L( )
€
= −lnds N+1∫ dsN∫ exp −βΔU +[ ]{ }exp −βU sN ;L( )[ ]
dsN exp −βU sN ;L( )[ ]∫
&
'
( (
)
*
+ +
€
= −ln ds N+1∫ exp −βΔU +[ ]NVT( ) Ghost particle!
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Hard spheres
€
βµex = −ln ds N+1∫ exp −βΔU +[ ]NVT( )
€
U r( ) =∞ r ≤σ0 r >σ
% & '
€
exp −βΔU +[ ] =0 if overlap1 no overlap% & '
probability to insert a test particle!
€
exp −βΔU +[ ]But, … may fail at high density
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III - Overlapping Distribution Method Two systems:
System 1: N, V,T, U1 System 0: N, V,T, U0
€
ΔβF = βF1 −βF0 = −ln Q1 Q0( ) = − lndsN∫ exp −βU1( )dsN∫ exp −βU0( )
#
$%%
&
'((= − ln
Q1Q0
#
$%
&
'(
Q0 =V N
Λ3NN!dsN∫ exp −βU0( )
€
Q1 =V N
Λ3NN!dsN∫ exp −βU1( )
€
p0 ΔU( ) =dsN∫ exp −βU0( )δ U1 −U0 −ΔU( )
dsN∫ exp −βU0( )
€
p1 ΔU( ) =dsN∫ exp −β U1 −U0( )[ ]exp −βU0[ ]δ U1 −U0 −ΔU( )
dsN∫ exp −βU1( )
p1 ΔU( ) = Q0
Q1exp −βΔU( ) p0 ΔU( )
=ΔU (δ function)
=1Q1
=Q0
Q11Q0
Q0
Q1= exp βΔF( )
€
p1 ΔU( ) =dsN∫ exp −βU1( )δ U1 −U0 −ΔU( )
dsN∫ exp −βU1( )
=Q0
Q1exp −βΔU( )
dsN∫ exp −βU0[ ]δ U1 −U0 −ΔU( )Q0
€
ln p1 ΔU( ) = β ΔF −ΔU( ) + ln p0 ΔU( )
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€
ln p1 ΔU( ) = β ΔF −ΔU( ) + ln p0 ΔU( )
€
βΔF = f1 ΔU( ) − f0 ΔU( )
Simulate system 0: compute f0 Simulate system 1: compute f1
€
f0 ΔU( ) ≡ ln p0 ΔU( ) − 0.5βΔU
€
f1 ΔU( ) ≡ ln p1 ΔU( ) + 0.5βΔU
Overlapping Distribution Method
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Chemical potential
System 1: N, V, T, U System 0: N-1, V, T, U + 1 ideal gas exFFF βµβββ ≡−=Δ 01
( ) ( )UfUfex Δ−Δ= 01βµ
System 0: test particle energy System 1: real particle energy 01 UUU −=Δ
Moderate density Hight density
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Umbrella sampling • Start with thermodynamic perturbation
€
ΔβF = −ln Q1 Q0( )
€
= −lndsN∫ exp −βU1( )dsN∫ exp −βU0( )
%
& ' '
(
) * *
€
exp −βΔF( ) =dsN∫ exp −βU0( )exp(−βΔU)
dsN∫ exp −βU0( )
&
' ( (
)
* + +
€
exp −βΔF( ) = exp −βΔU( ) 0
Can we use this for free energy difference between arbitrary systems?
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T1
T2
T5
T3
T4
E
P(E)
Overlap becomes very small
T0
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E
P(Δ
U)
Overlap becomes very small
System 1
System 0
ΔU
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Bridging function • Introduce function π(sN) altering distribution.
• This approach is called umbrella sampling
€
exp −βΔF( ) =dsN∫ π (sN )exp −βU1( ) /π (sN )dsN∫ π (sN )exp −βU0( ) /π (sN )
'
( ) )
*
+ , ,
exp −βΔF( ) =exp −βU1( ) /π
π
exp −βU0( ) /ππ
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E
P(Δ
U)
System 1
System 0
ΔU
umbrella
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Tracing coexistence curves • If we have a coexistence point on the phase diagram we can
integrate allong the line while maintaining coexistence.
dµα = dµβ
P
T
α phase
β phase dP
dT
P en T are equal along coexistence line
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Tracing coexistence curves
dµ = dG = - SdT + VdP
- SαdT + VαdP = - SβdT + VβdP
dPdT
=Sβ − SαVβ −Vα
dPdT
=ΔSΔV
=ΔHTΔV
Clapyeron equation
dPdT
=Δ(U +PV )TΔV dP = Δ(U +PV )
TΔVdT
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Example: Carbon Phase Diagram