free energies and phase transitions. condition for phase coexistence in a one-component system:
TRANSCRIPT
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Free energies and phase transitions
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Condition for phase coexistence in a one-component system:
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The Gibbs “Ensemble”
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NVT Ensemble
Fluid Fluid
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NVT Ensemble
G
LL
G
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Gibbs Ensemble
GL
Equilibrium!
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•Distribute n1 particles over two volumes•Change the volume V1
•Displace the particles
1 13
1 1 1
1 1 1 1
11 1 10 ( ) 0
1 1
G
2 2
( )
exp[ ( )] exp[ ( )]
( ) N
VN n N n
n V n N n
n n N n N n
dV V V V
d U d
Q N
U
V T
s s s s
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Distribute n1 particles over two volumes:
1 1 1
!
! !
N N
n n N n
31 1
1 1
1 1 1 1
11 1 10
1 1 2 2
1G 0 ( )
( )
exp
(
[ ( )] exp[ ( )]
) N
V n N n
n n N n N n
N
n V n N ndV V V V
d U d U
Q N V T
s s s s
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Integrate volume V1
31 1
1 1 1
1 1
1
1
1G 0 ( ) 1 1 1
1 2 2
0
1exp[ ( )] exp[ ( )]
(( ) )N
V n N
n n N n N
nN
n V n n
n
N
d U d U
dV V VQ N V T V
s s s s
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Displace the particles in box 1 and box2
1 13
1 1
1 1 1 1
1
1 1
1G 1 1 10 (
2 2
) 0
exp[ ( )] exp[
)
(
(
)]
( )N
VN n N n
n V n
n
n
n N n N n
N
d U d
Q N V T d
U
V V V V
s s s s
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1 13
1 1 1
1 1 1 1
1G 1 1 10 ( ) 0
1 1 2 2
( ) ( )
exp[ ( )] exp[ ( )]
N
VN n N n
n V n N n
n n N n N n
Q N V T dV V V V
d U d U
s s s s
Probability distribution
1 1
1 1 1 11 11 1 1 2 1 2
1 1
( )( ) exp [ ( ) ( )]
( )
n N nn N n n N nV V V
N n V U Un N n
s s s s
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Particle displacement
Volume change
Particle exchange
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Acceptance rules
1 1
1 1 1 11 11 1 1 2 1 2
1 1
( )( ) exp [ ( ) ( )]
( )
n N nn N n n N nV V V
N n V U Un N n
s s s s
Detailed Balance:
( ) ( )K o n K n o
( ) ( ) acc( ) ( ) ( ) acc( )N o o n o n N n n o n o
acc( ) ( ) ( )
acc( ) ( ) ( )
o n N n n o
n o N o o n
acc( ) ( )
acc( ) ( )
o n N n
n o N o
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1 1
1 1 1 11 11 1 1 2 1 2
1 1
( )( ) exp [ ( ) ( )]
( )
n N nn N n n N nV V V
N n V U Un N n
s s s s
1
1
1
1 1
1
1 11
1 1
1 11
2
12
1
( )( ) exp [ ( ) ( )]
( )
( )( ) exp [ ( ) ( )]
( )
n N n
n N
N n
N nn
V V VN U n U
n N n
V V VN o
n N
n
no U U
s
s
Displacement of a particle in box 1
1 1
1
1 1
1
1 11 2
1 1
1 11 2
1 1
( )exp [ ( ) ( )]
( )acc( )
( )acc( )exp [ ( ) ( )]
( )
n N nN n
n N nN n
V V VU n U
n N no n
V V Vn oU o U
n N n
s
s
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1 1
1 1 1 11 11 1 1 2 1 2
1 1
( )( ) exp [ ( ) ( )]
( )
n N nn N n n N nV V V
N n V U Un N n
s s s s
1
1
1
1 1
1
1 11
1 1
1 11
2
12
1
( )( ) exp [ ( ) ( )]
( )
( )( ) exp [ ( ) ( )]
( )
n N n
n N
N n
N nn
V V VN U n U
n N n
V V VN o
n N
n
no U U
s
s
Displacement of a particle in box 1
1
1
exp [ ( )]acc( )
acc( ) exp [ ( )]
U no n
n o U o
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Acceptance rules
1 1
1 1 1 11 11 1 1 2 1 2
1 1
( )( ) exp [ ( ) ( )]
( )
n N nn N n n N nV V V
N n V U Un N n
s s s s
Adding a particle to box 2
1
11
1
1 111
1 11 2
1 11 2
1 1
( )( ) exp [ ( ) ( )]
( )
(
acc(
)( ) exp [ ( ) (
1
)
)]( )
1
( )
acc( ) ( )
N
n N
nn
n
V V VN U U
N
V V VN o U o U o
n N n
o n N n
n o
n nn n
N o
n
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Moving a particle from box 1 to box 2
11
1 1
11
1 1
111
111 1
1 21 1
1 1
11 2
1 1
1 1
11 1
1
2
1
( )( ) exp [ ( ) ( )]
( 1) 1
( )( ) exp [ ( ) ( )
( )e
]( )
xp [ ( ) ( )]( 1) 1acc( )
( )acc( )exp [
( )
N nn
n N
N nn
n N n
n
V V VN n U
V V VU n U n
n N n
n U nn N n
V V VN o U o U o
n N n
o n
V V Vn oU
n N n
1 2( ) ( )]o U o
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11
1 1
11
1 1
111
111 1
1 21 1
1 1
11 2
1 1
1 1
11 1
1
2
1
( )( ) exp [ ( ) ( )]
( 1) 1
( )( ) exp [ ( ) ( )
( )e
]( )
xp [ ( ) ( )]( 1) 1acc( )
( )acc( )exp [
( )
N nn
n N
N nn
n N n
n
V V VN n U
V V VU n U n
n N n
n U nn N n
V V VN o U o U o
n N n
o n
V V Vn oU
n N n
1 2( ) ( )]o U o
Moving a particle from box 1 to box 2
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11
1 1
11
1 1
111
111 1
1 21 1
1 1
11 2
1 1
1 1
11 1
1
2
1
( )( ) exp [ ( ) ( )]
( 1) 1
( )( ) exp [ ( ) ( )
( )e
]( )
xp [ ( ) ( )]( 1) 1acc( )
( )acc( )exp [
( )
N nn
n N
N nn
n N n
n
V V VN n U
V V VU n U n
n N n
n U nn N n
V V VN o U o U o
n N n
o n
V V Vn oU
n N n
1 2( ) ( )]o U o
Moving a particle from box 1 to box 2
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11
1 1
111 1
2
21 2
1
1 21 1
1 11 2
1 1
1
1acc( )exp [ ]
acc
( )( ) exp [ ( ) ( )]
( 1) 1
( )( ) exp [ ( ) ( ]
( )
)( )
N nn
n N n
V V VN n U n U n
n N n
V V VN o U o U o
n
Vno n
U
n
n
Vn
N
Uo
Moving a particle from box 1 to box 2
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Analyzing the results (1)
Well below Tc Approaching Tc
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Analyzing the results (2)
Well below Tc Approaching Tc
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Analyzing the results (3)
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Condition for phase coexistence in a one-component system:
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With normal Monte Carlo simulations, we cannot compute “thermal” quantities, such as S, F and G, because they depend on the total volume of accessible phase space.
For example:
and
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F cannot be computed with importance sampling
NNNN
NVTNVT
UANQ
A rexprdr!
113
NNN PA rrdr
NN
NNN
P
PA
rdr
rrdr
NN
NNN
UC
UCA
rexpdr
rexprdr
NN
NNN
U
UA
rexpdr
rexprdr
Generate configuration using MC:
NMCNMC UCP rexpr
NM
NNNN r,r,r,r,r 4321 Ni
M
i
AM
A r1
1
with
NMCN
NMCNN
P
PA
rdr
rrdr
NMCN
NMCNN
UC
UCA
rexpdr
rexprdr
NN
NNN
U
UA
rexpdr
rexprdr
!
rexpr
3 NQ
UP
NNVT
NN
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Solutions:
1. “normal” thermodynamic integration
2. “artificial” thermodynamic integration
3. “particle-insertion” method
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How are free energies measured experimentally?
P
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Then take the limit V0 .
Not so convenient because of divergences. Better:
0, as V0
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This approach works if we can integrate from a known reference state - Ideal gas (“T=”), Harmonic crystal (“T=0”),
Otherwise: use “artificial” thermodynamic integration (Kirkwood)
Suppose we know F(N,V,T) for a system with a simple potential energy function U0: F0(N,V,T).
We wish to know F1(N,V,T) for a system with a potential energy function U1.
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Consider a system with a mixed potential energy function (1-)U0+ U1: F (N,V,T).
hence
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Or:
And therefore
Examples of application:
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F()
0 1
F(0)
F(1)
The second derivative is ALWAYS negative:
Therefore:
Good test of simulation results…
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1. Any atomic or molecular crystal.
Reference state: Einstein crystal
2. Nematic Liquid crystal:
Start from isotropic phase. Switch on “magnetic field” and integrate around the I-N critical point
isotropic nematic
density
field
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Chemical Potentials
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Particle insertion method to compute chemical potentials
But N is not a continuous variable. Therefore
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Does that help?
Yes: rewrite
s is a scaled coordinate: 0s<1
r = L s (is box size)
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Now write
then
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And therefore
but
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So, finally, we get:
Interpretation:
1. Evaluate U for a random insertion of a molecule in a system containing N molecule.
2. Compute
3. Repeat M times and compute the average “Boltzmann factor”
4. Then
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Lennard-Jones fluid
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NQ
NQ 1ln
NN
NGNG
1
))1
LUVPVVN
Q NNNNNPT ;sexpdsexpd
!
13
LUVPVV
LUVPVV
N
NNNN
NNN
N
N
;sexpdsexpd
;sexpdsexpd
!1
!11
ln111
3
33
Other ensembles: NPT
NPTQG lnTVN
F
,
TPN
G
,
NPT: Gibbs free energyNVT: Helmholtz free energy
LUVPVV
ULUVVPVV
N NNN
NNNN
;sexpdsexpd
expds;sexpdsexpd
1
1ln
1
3
LUVPVV
UVLUVPVV
N
PP
NNN
NNNN
;sexpdsexpd
expds;sexpdsexpd
1lnln
13
UN
PVP N expds
1lnln 1
3
NN
NGNG
1
))1 NVT:
NVT
UN expdslnln 1
The volume fluctuates!
UN
PVU
N
PVNN
expds1
expds1
11
NPT:
UN
PVP N expds
1lnln 1
3
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Hard spheres
NVT
ex UN expdsln 1
r
rrU
0
overlap no1
overlap if0exp U
Probability to insert a test particle!
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Particle insertion method to compute chemical potentials
But N is not a continuous variable. Therefore
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ACCEPTANCE OF RANDOM INSERTION DEPENDS ON SIZE
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ACCEPTANCE OF RANDOM INSERTION DEPENDS ON SIZE
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Particle insertion continued….
therefore
But also
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As before:
With s a scaled coordinate: 0s<1
r = L s (is box size)
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Now write
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And therefore
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Interpretation:
1. Evaluate U for a random REMOVAL of a molecule in a system containing N+1 molecule.
2. Compute
3. Repeat M times and compute the average “Boltzmann factor”
4. Then
DON’T EVEN THINK
OF IT!!!
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What is wrong?
is not bounded. The average that we compute can be dominated by INFINITE contributions from points that are NEVER sampled.
What to do?
Consider:
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And also consider the distribution
p0 and p1 are related:
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UpUFUp 01 lnln
UUpUf 5.0ln 00
UUpUf 5.0ln 11
UfUfF 01
3211 UcUbUaCUf
3200 UcUbUaCUf
01 CCF
Simulate system 0: compute f0
Simulate system 1: compute f1
Fit f0 and f1 to two polynomials that only differ by a constant.
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Chemical potentialSystem 1: N, V,T, USystem 0: N-1, V,T, U + 1 ideal gas
exFFF 01
UfUfex 01System 0: test particle energy System 1: real particle energy
01 UUU
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Does it work for hard spheres?
consider U=0
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Problems with Widom method:
Low insertion probability yields poor statistics.
For instance:
Trial insertions that consist of a sequence of intermediate steps.
Examples: changing polymer conformations, moving groups of atoms, …
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What is the problem with polymer simulations?
Illlustration:
Inserting particles in a dense liquid
Trial moves that lead to “hard-core” overlaps tend to be rejected.
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ANALOGY:
Finding a seat in a crowded restaurant.
Can you seat one person, please…
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Next: consider the random insertion of a chain molecule (polymer).
Waiter! Can you seat 100 persons… together
please!
Random insertions of polymers in dense liquids usually fail completely…
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(Partial) Solution: Biased insertion.
Berend Smit’s lecture tomorrow…
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Simulations of soft matter are time consuming:
1. Because of the large number of degrees of freedom (solution: coarse graining)
2. Because the dynamics is intrinsically slow. Examples:
1. Polymer dynamics
2. Hydrodynamics
3. Activated dynamics
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Slow dynamics implies slow equilibration. This is particularly serious for glassy systems.
6 hours 8 hours
Mountain hikes
..
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after a bit of global warming…
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after a bit of global warming…
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after a bit of global warming…
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after a bit of global warming…
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.. .. .. .. .. .. .. .. ..
20 minutes
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Sampling the valleys
..
Combine this… .. …with this
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Parallel Tempering
COMBINE Low-temperature and high-temperature runs in a SINGLE Parallel simulation
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In practice:
System 1 at temperature T1
System 2 at temperature T2
Boltzmann factor Boltzmann factor
Total Boltzmann factor
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SWAP move
System 1 at temperature T2
System 2 at temperature T1
Boltzmann factor Boltzmann factor
Total Boltzmann factor
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Ratio
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Systems may swap temperature if their combined Boltzmann factor allows it.
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number of MC cycles
Win
dow
0 10000 20000
300
200
100
NOTES:
1. One can run MANY systems in parallel
2. The control parameter need not be temperature
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Application: computation of a critical point INSIDE the glassy phase of “sticky spheres”:
GLASSY