Free energies and phase transitions. Condition for phase coexistence in a one-component system:
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Transcript of Free energies and phase transitions. Condition for phase coexistence in a one-component system:
Free energies and phase transitions
Condition for phase coexistence in a one-component system:
The Gibbs “Ensemble”
NVT Ensemble
Fluid Fluid
NVT Ensemble
G
LL
G
Gibbs Ensemble
GL
Equilibrium!
•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
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
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
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
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
Particle displacement
Volume change
Particle exchange
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
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
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
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
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
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
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
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
Analyzing the results (1)
Well below Tc Approaching Tc
Analyzing the results (2)
Well below Tc Approaching Tc
Analyzing the results (3)
Condition for phase coexistence in a one-component system:
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
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
Solutions:
1. “normal” thermodynamic integration
2. “artificial” thermodynamic integration
3. “particle-insertion” method
How are free energies measured experimentally?
P
Then take the limit V0 .
Not so convenient because of divergences. Better:
0, as V0
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.
Consider a system with a mixed potential energy function (1-)U0+ U1: F (N,V,T).
hence
Or:
And therefore
Examples of application:
F()
0 1
F(0)
F(1)
The second derivative is ALWAYS negative:
Therefore:
Good test of simulation results…
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
Chemical Potentials
Particle insertion method to compute chemical potentials
But N is not a continuous variable. Therefore
Does that help?
Yes: rewrite
s is a scaled coordinate: 0s<1
r = L s (is box size)
Now write
then
And therefore
but
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
Lennard-Jones fluid
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
Hard spheres
NVT
ex UN expdsln 1
r
rrU
0
overlap no1
overlap if0exp U
Probability to insert a test particle!
Particle insertion method to compute chemical potentials
But N is not a continuous variable. Therefore
ACCEPTANCE OF RANDOM INSERTION DEPENDS ON SIZE
ACCEPTANCE OF RANDOM INSERTION DEPENDS ON SIZE
Particle insertion continued….
therefore
But also
As before:
With s a scaled coordinate: 0s<1
r = L s (is box size)
Now write
And therefore
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!!!
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:
And also consider the distribution
p0 and p1 are related:
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.
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
Does it work for hard spheres?
consider U=0
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, …
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.
ANALOGY:
Finding a seat in a crowded restaurant.
Can you seat one person, please…
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…
(Partial) Solution: Biased insertion.
Berend Smit’s lecture tomorrow…
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
Slow dynamics implies slow equilibration. This is particularly serious for glassy systems.
6 hours 8 hours
Mountain hikes
..
after a bit of global warming…
after a bit of global warming…
after a bit of global warming…
after a bit of global warming…
.. .. .. .. .. .. .. .. ..
20 minutes
Sampling the valleys
..
Combine this… .. …with this
Parallel Tempering
COMBINE Low-temperature and high-temperature runs in a SINGLE Parallel simulation
In practice:
System 1 at temperature T1
System 2 at temperature T2
Boltzmann factor Boltzmann factor
Total Boltzmann factor
SWAP move
System 1 at temperature T2
System 2 at temperature T1
Boltzmann factor Boltzmann factor
Total Boltzmann factor
Ratio
Systems may swap temperature if their combined Boltzmann factor allows it.
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
Application: computation of a critical point INSIDE the glassy phase of “sticky spheres”:
GLASSY