Ch121a Atomic Level Simulations of Materials and Molecules

© copyright 2013-William A. Goddard III, all rights reservedCh120a-Goddard-L06 1

Ch121a Atomic Level Simulations of Materials and Molecules

William A. Goddard III, [email protected] and Mary Ferkel Professor of Chemistry,

Materials Science, and Applied Physics, California Institute of Technology

Lecture 6 and 7, April 16 and 21, 2013MD3: vibrations

Room BI 115Lecture: Monday, Wednesday Friday 2-3pm

TA’s Jason Crowley and Jialiu Wang

Lecture 6 Presented by Jason Crowley

© copyright 2013-William A. Goddard III, all rights reservedCh120a-Goddard-L06 2

Homework and Research ProjectFirst 5 weeks: The homework each week uses generally available computer software implementing the basic methods on applications aimed at exposing the students to understanding how to use atomistic simulations to solve problems. Each calculation requires making decisions on the specific approaches and parameters relevant and how to analyze the results. Midterm: each student submits proposal for a project using the methods of Ch121a to solve a research problem that can be completed in the final 5 weeks.The homework for the last 5 weeks is to turn in a one page report on progress with the projectThe final is a research report describing the calculations and conclusions

© copyright 2012 William A. Goddard III, all rights reservedCh121a-Goddard-L07 3

Outline of today’s lecture

Vibration of moleculesClassical and quantum harmonic oscillatorsInternal vibrations and normal modesRotations and selection rulesExperimentally probing the vibrationsDipoles and polarizabilitiesIR and Raman spectraSelection rulesThermodynamics of moleculesDefinition of functionsRelationship to normal modesDeviations from ideal classical behavior


Simple vibrations

Starting with an atom inside a molecule at equilibrium, we can expand its potential energy as a power series. The second order term gives the local spring constantWe conceptualize molecular vibrations as coupled quantum mechanical harmonic oscillators (which have constant differences between energy levels)Including Anharmonicity in the interactions, the energy levels become closer with higher energySome (but not all) of the vibrational modes of molecules interact with or emit photons This provides a spectroscopic fingerprint to characterize the molecule


Vibration in one dimension – Harmonic Oscillator

Consider a one dimensional spring with equilibrium length xe which is fixed at one end with a mass M at the other. If we extend the spring to some new distance x and let go, it will oscillate with some frequency, w, which is related to the M and spring constant k.To determine the relation we solve Newton’s equation M (d2x/dt2) = F = -k (x-xe)Assume x-x0=d = A cos(wt) then –Mw2 Acos(wt) = -k A cos(wt) Hence –Mw2 = -k or w = Sqrt(k/M). Stiffer force constant k higher w and higher M lower w

No friction

E= ½ k d2


Reduced Mass

M1M2

Put M1 at R1 and M2 at R2

CM = Center of mass Fix Rcm = (M1R1 + M2R2)/(M1+ M2) = 0Relative coordinate R=(R2-R1)Then Pcm = (M1+ M2)*Vcm = 0 And P2 = - P1

Thus KE = ½ P12/M1 + ½ P2

2/M2 = ½ P12/m

Where 1/m = (1/M1 + 1/M2) or m = M1M2/(M1+ M2) Is the reduced mass.Get w = sqrt(k/m). Thus we can treat the diatomic molecule as a simple mass on a spring but with a reduced mass, m


mw

ww

2

)2/1()2/1(2

2

a

vnE

e

en

=

= (n + ½)2

For molecules the energy is harmonic near equilibrium but for large distortions the bond can break.

The simplest case is the Morse Potential:

2/1

2

)2

(

)1()(

e

axe

hcDka

ehcDxV

=

=

Exact solution

Real potentials are more complex; in general: ....)2/1()2/1()2/1( 2 = een nvnE wdww (n + ½)2 ....)2/1()2/1()2/1( 2 = een nvnE wdww (n + ½)3

Successive vibrational levels are closer by

mw

ww

2

)2/1()2/1(2

2

a

vnE

e

en

=

=

(Philip Morse a professor at MIT, did not manufacture cigarettes)


Vibration for a molecule with N particles

Fk = -(∂E(Rnew)/∂Rk) = -(∂E/∂Rk)0 - Sm (∂2E/∂Rk∂Rm) (dR)m

Where we have neglected terms of order d2. Writing the 2nd derivatives as a matrix (the Hessian) Hkm = (∂2E/∂Rk∂Rm) and setting (∂E/∂Rk)0 = 0, we get

Fk = - Sm Hkm (dR)m = Mk (∂2Rk/∂t2) To find the normal modes we write (dR)m = Am cos wt leading to

Mk(∂2Rk/∂t2) = Mk w2 (Ak cos wt) = Sm Hkm (Amcos wt)

Here the coefficient of cos wt must be {Mk w2 Ak - Sm Hkm Am}=0

There are 3N degrees of freedom (dof) which we collect together into the 3N vector, Rk where k=1,2..3NThe interactions then lead to 3N net forces, Fk = -(∂E(Rnew)/∂Rk) all of which are zero at equilibrium, R0

Now consider that every particle is moved a small amount leading to a 3N distortion vector, (dR)m = Rnew – R0

Expanding the force in a Taylor’s series leads to

Newton’s equation


Solving for the Vibrational modesThe normal modes satisfy {Mk w2 Ak - Sm Hkm Am}=0To solve this we mass weight the coordinates as Bk = sqrt(Mk)Ak

leading toSqrt(Mk) w2 Bk - Sm Hkm [1/sqrt(Mm)]Bm}=0 leading to

Sm Gkm Bm = wk2 Bk where Gkm = Hkm/sqrt(MkMm)

G is referred to as the reduced HessianFor M degrees of freedom this has M eigenstatesSm Gkm Bmp = dkp Bk (w2)p

where the M eigenvalues (w2)p are the squares of the vibrational energies.If the Hessian includes the 6 translation and rotation modes then there will be 6 zero frequency modes, wp = 0


Saddle points

If the point of interest were a saddle point rather than a minimum, G would have one negative eigenvalue, (w2)p = - A2 where A is a positive number This leads to an imaginary frequency, wp = iA ,

Saddle point


For practical simulationsWe can obtain reasonably accurate vibrational modes from just the classical harmonic oscillators, usually within a few %N atoms => 3N degrees of freedomHowever, there are 3 degrees for translation, n = 0 3 degrees for rotation for non-linear molecules, n = 0 2 degrees if linearThe rest are vibrational modes


Normal Modes of Vibration H2O

1595 cm-1

3657 cm-1

3756 cm-1

H2O D2O

1178 cm-1

2671 cm-1

2788 cm-1

Sym. stretch

Antisym. stretch

Bend

Isotope effect: n ~ sqrt(k/M):Simple nD/nH ~ 1/sqrt(2) = 0.707:

Ratio: 0.730

Ratio: 0.735

Ratio: 0.742

More accurately, reduced massesmOH = MHMO/(MH+MO)mOD = MDMO/(MD+MO)Ratio = sqrt[MD(MH+MO)/MH(MD+MO)] ~ sqrt(2*17/1*18) = 0.728

Most accuratelyMH=1.007825MD=2.0141MO=15.99492Ratio = 0.728


•EM energy absorbed by interatomic bonds in organic compounds•frequencies between 4000 and 400 cm-1 (wavenumbers) •Useful for resolving molecular vibrations

http://webbook.nist.gov/chemistry/http://wwwchem.csustan.edu/Tutorials/INFRARED.HTM

The Infrared (IR) SpectrumCharacteristic vibrational modes


Normal Modes of Vibration CH4

2917 cm-1 3019 cm-1 1534 cm-1CH4

CD4 1178 cm-1 2259 cm-1 1092 cm-1

Sym. stretch

1

Anti. stretch

3

Sym. bend

2

Sym. bend

3

A1 T2 E T2

1306 cm-1

996 cm-1

4 independent CH bonds 4 CH stretch modes, by symmetry one is triply degenerate6 possible angle terms HCH 5 HCH modes, one doubly degenate, on triply deg.Reason only 5 linearly independent HCH


Fitting force fields to Vibrational frequencies and force constants

Hessian-Biased Force Fields from Combining Theory and Experiment; S. Dasgupta and W. A. Goddard III; J. Chem. Phys. 90, 7207 (1989)

H2CO

MC: Morse bond stretch and cosine angle bendMCX: include 1 center cross terms

4 atoms 12-6=6 vibrations

CH sym str CO stretch CH2 scis

CH asym strCH2 rock CH2 wag


The Schrödinger equation H = e

for harmonic oscillator22

22

21

2kx

xm

=

,2,1,0 )21( == nnn we

xym

eyy

ey

ye

e

y

y

y

y

w

==

=

=

=

=

)32(3

1)(

)12(2

1)(

2)(

)(

2/34/13

2/24/12

2/4/11

2/4/10

2

2

2

2

The QM Harmonic Oscillator

energy ,2,1,0 )21( == nnn we

wavefunctions

reference http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c1

xym

eyy

ey

ye

e

y

y

y

y

w

==

=

=

=

=

)32(3

1)(

)12(2

1)(

2)(

)(

2/34/13

2/24/12

2/4/11

2/4/10

2

2

2

2

xym

eyy

ey

ye

e

y

y

y

y

w

==

=

=

=

=

)32(3

1)(

)12(2

1)(

2)(

)(

2/34/13

2/24/12

2/4/11

2/4/10

2

2

2

2

H

Gaussian


Raman and InfraRed spectroscopy

IRVibrations at same frequency as radiationTo be observable, there must be a finite dipole derivativeThus homonuclear diatomic molecule (O2 , N2 ,etc.) does not lead to IR absorption or emission. Raman spectroscopy is complementary to IR spectroscopy.radiation at some frequency, n, is scattered by the molecule to frequency, n’, shifted observed frequency shifts are related to vibrational modes in the molecule IR and Raman have symmetry based selection rules that specify active or inactive modes


IR and Raman selection rules for vibrations

The electrical dipole moment is responsible for IR

= rrr 3),()( dtt m

The polarizability is responsible for Raman

)()()( ttt em = For both, we consider transition matrix elements of

the form

=

iini Q

t

)(

|)(|'

,

m

The intensity is proportional to dm/dR averaged over the vibrational state

where e is the external electric field at frequency n


IR selection rules, continued

For IR, we expand dipole moment

....)( 00

= ii i

QQmmm

We see that the transition elements are

The dipole changes during the vibration

Can show that n can only change 1 level at a time

iiii

nQnQ

||')( 0m


Raman selection rules

For Raman, we expand polarizability

....)( 00

= ii i

QQ

substitute the dipole expression for the induced dipole

Same rules except now it’s the polarizability that has to change

For both Raman and IR, our expansion of the dipole and alpha shows higher order effects possible

iiii

nQntQ

||')()( 0 e

iiii

nQntQ

||')()( 0 e

=


•center of mass translation Dx= Dx Dy=0 Dz=0

Dx=0 Dy=Dy Dz=0

Dx=0 Dy=0 Dz=Dz •center of mass rotation (nonlinear molecules) Dx=0 Dy=-cDqx Dz=bDqx Dx= cDqy Dy=0 Dz=-aDqy Dx= -bDqz Dy=aDqx Dz=0•linear molecules have only 2 rotational degrees of freedom•The translational and rotational degrees of freedom can be removed beforehand by using internal coordinates or by transforming to a new coordinate system in which these 6 modes are separated out

Translation and Rotation Modes

E is a constant dE/dx = 0 d2E/dx2 = 0Thus the eigenmode l=0

E is a constant dE/dx = 0 d2E/dx2 = 0Thus the eigenmode l=0


Classical Rotations

The moment of inertia about an axis q is defined as

)(2 qxmIk

kkqq = xk(q) is the perpendicular distance to the axis q

Can also define a moment of inertia tensor where (just replace the mass density with point masses and the integral with a summation. Diagonalization of this matrix gives the principle moments of inertia!

the rotational energy has the form

qqqq

q qq

q

qqqqrot

IJ

IJ

qIE

w

w

=

== 2)(

21 2

2

=

kk

kkk

m

mrR0


Quantum RotationsThe rotational Hamiltonian has no associated potential energy

zz

z

yy

y

xx

x

IJ

IJ

IJH

222

222

=

JJJMM

JJJKJ

KIII

JJMKJE

J

J

JJrot

=

==

=

,...,1,

,...,1,,...2,1,0

)21

21(

2)1(),,( 22

2

zJIII

JH )21

21(

2

2

=

For symmetric rotors, two of the moments of inertia are equivalent, combine:

Eigenfunctions are spherical harmonic functions YJ,K or Zlm with eigenvalues


Transition rules for rotations

For rotationsWavefunctions are spherical harmonicsProject the dipole and polarizability due to rotationIt can be shown that for IRDelta J changes by +/- 1Delta MJ changes by 0 or +/-1Delta K does not changeFor RamanDelta J could be 1 or 2Delta K = 0But for K=0, delta J cannot be +/- 1


Raman scattering

• Phonons are the normal modes of lattice vibrations (thermal + zero point energy)

• When a photon absorbs/emits a single phonon, momentum and energy conservation the photon gains/loses the energy and the crystal momentum of the phonon. – q ~ q` => K = 0– The process is called anti-Stokes for absorption and

Stokes for emission.– Alternatively, one could look at the process as a

Doppler shift in the incident photon caused by a first order Bragg reflection off the phonon with group velocity v = (ω/ k)*k


Raman selection rules

For Raman, we expand polarizability

....)( 00

= ii i

QQ

substitute the dipole expression for the induced dipole

Same rules except now it’s the polarizability that has to change

For both Raman and IR, our expansion of the dipole and alpha shows higher order effects possible

iiii

nQntQ

||')()( 0 e

iiii

nQntQ

||')()( 0 e

=


Another simple way of looking at Raman

)cos()cos(21)cos(2)(

)cos()cos(212)(

)cos()(2)()()(

intint00

0int0

0

tttttt

ttt

ttttt

wwwwewem

wewm

weem

D=

D=

==

Take our earlier expression for the time dependent dipole and expose it to an ideal monochromatic light (electric field)

We get the Stokes lines when we add the frequency and the anti-Stokes when we substractThe peak of the incident light is called the Rayleigh line


•The external EM field is monochromatic•Dipole moment of the system•Interaction between the field and the molecules•Probability for a transition from the state i to the state f (the Golden Rule)

•Rate of energy loss from the radiation to the system

•The flux of the incident radiation

)ωcos(ε)( 0 tEtE =

==

==N

j

ij

ij

n

ii qr

1i

1

μ DipoleMolecular μμ Dipole Total

)(μ)( tEt=

)]ωω(δ)ωω(δ[|με|2

π)ω(2

20 = fififi ifEP

fffi ωωω =

)ω(ωρ)ω( fifii f

irad PE =

20π8

EcnS =c: speed of lightn: index of refraction of the medium

Skip The Sorption lineshape - 1

==

==N

j

ij

ij

n

ii qr

1i

1

μ DipoleMolecular μμ Dipole Total


•Absorption cross section (w)

•Define absorption linshape I(w) as

•It is more convenient to express I(w) in the time domain

SErad )ω()ω(α

=

)ωω(δ|με|ρ3)1(ω4π

)ω(α3)ω(2

ωβ2 =

= fii f

i ife

cnI

dtet

dteiffiI

dte

ti

tEE

i

i fi

ti

if

ω

ω]-[

ω

)(μ)0(μ2π1

|με||με|ρ2π3)ω(

2π1ω)(δ

=

=

=

=

I(w) is just the Fourier transform of the autocorrelation function of the dipole moment

ensemble average

Beer-Lambert law Log(P/P0)=bc

Skip The Sorption lineshape - II


Non idealities and surprising behavior

Anharmonicity – bonds do eventually dissociateCoriolis forces

Interaction between vibration and rotationInversion doublingIdentical atoms on rotation – need to obey the Pauli PrincipleTotal wavefunction symmetric for Boson and antisymmetric for Fermion ),()1(),( ,, qq

JJ MJJ

MJ

NVEN

YY =

=


Figure taken from Streitwiser & Heathcock,

Introduction to Organic Chemistry, Chapter

14, 1976Electromagnetic Spectrum

How does a Molecule response to an oscillating external electric field (of frequency w)? Absorption of radiation via exciting to a higher energy state ħw ~ (Ef - Ei)


Using the vibrational modes: thermodynamicsIn QM and MM the Energy at minima = motionless state at 0KBUT, experiments are made at finite T, hence corrections are required to allow for rotational, translational and vibrational motion.The internal energy of the system: U(T)=Urot(T)+Utran(T)+Uvib(T)+Uvib(0)

32-AJB

Uvib T )= hn i

2 hn i

exp hn i KBT 1

i=1

Nmod

The vibrational frequencies ni) of the normal modes (Nmod) calculated from the eigenvalues λi) of the force-constant equivalent of Hessian matrix of second derivatives

Vibrational frequencies can be used to calculate entropies and free energies, or to compare with results of spectroscopic experiments

n i =λi

2

From equipartition theorem: Urot(T) = (3/2)KBT , Utran(T) = (3/2)KBT per molecule (except Urot(T)=KBT for linear molecules)BUT, vibrational energy levels are often only partially excited at room T, thus Uvib(T) requires knowledge of vibrational frequenciesUvib(T) = vibrational enthalpy @ T - vibrational enthalpy @ 0K


Thermodynamics

Describe a system in terms of Hamiltonian H(p,q) where p is generalized momentum and q is generalized coordinate

Here Q, the Partition function, is a normalization constant

For a system in equilibrium, probability of a state with energy H(p,q) is

P(p,q) = exp[-H(p,q)/kBT]/Q

which is referred to as a Boltzmann distribution,

Q = S exp[-H(p,q)/kBT] summed over all states of the system


Thermodynamic functions can all be derived from Q

QkTTQkTQkS

TQkTE

VN

VN

lnAEnergy Free Helmholtz

)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

VNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,

)ln()ln(2 Capacity Heat

)ln( Pressure

)ln( Potential Chemical

=

=

=m


,2,1 8 2

22

== xx

n nmLnh

xe

VhmkTdneeeeTVq mL

nβh

nnn yyx

2/32

3

08

1

βε

1

βε

1

βεtrans )π2()(),( 2

22

ynynxn ===

=

=

=

The partition function for translation

Assume a cubic periodic box of side L

The QM Hamiltonian is

The QM eigenfunctions are just periodic functions for x, y, and z directions, sin(nxxp/L) etc

Leading to

2

22

2 xm

=H

Thus the partition function for translation becomes


Thermodynamic functions for translation

VNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,


)ln( Pressure


=

=

=m

VhMkT 2/3

2 )π2(Q =

= (3/2) kT

2/5

2/3

2

π2ln eNV

hMkT

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=QkT

TQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTT

QkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

e

NV

hMkT 2/3

2

π2ln= -kT

= (3/2) k

VNkT

= k

Ideal gas

equipartition


= 2

2

2

2

sin1

θθsin

θθsin1

2 IH

,2,1,0 )12(

2)1( 2

J

2

==

= JJI

JJJ we

2/12

22/1

2

22/1

2

21/2

02

)1(2

rot )π8()π8()π8(σ

π)12(σ1)(

2

hkTI

hkTI

hkTIdJeJTq CBAI

JβJ

==

khv

k

kIh

v

AA

==

=

ω re temperatulvibrationa

π8 re temperaturotational 2

2

The partition function for rotation

This leads to energy levels of

Thus the partition function becomes

This is 2 the Laplacian

I = moment of inertia


Thermodynamic functions for rotation (non linear)

VNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,


)ln( Pressure


=

=

=m

Q =

= (3/2) kT

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=QkT

TQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTT

QkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

= 0

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

= -kT

= (3/2) k

= k

2/131/2

)(σ

π

CBA

T

2/132/31/2

)(σ

πlnCBA

Te

2/131/2

)(σ

πlnCBA

T

equipartition

equipartition


,2,1,0 )21( == nnn we

=

=

=

=

=

e

1ωβ

ω/2β

1 0

β

1jj nvib e

eeq n

khv

k

kIh

v

AA

==

=

ω re temperatulvibrationa

π8 re temperaturotational 2

2

Summing over all normal modes leads to

The partition function for vibrations

An isolated harmonic oscillator with vibrational frequency ω

Has a spectrum of energies

Substituting into the Boltzmann expression leads to

=

=

=

=

=

e

1ωβ

ω/2β

1 0

β

1jj nvib e

eeq n

q =


Thermodynamic functions for vibration (harmonic oscillator)

VNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,


)ln( Pressure


=

=

=m

Q =

= (3/2) kTQkT

TQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

= 0

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

= -kT

= k

= k

=

63

1/

/2

1

n

jT

T

jv

jv

e

e

=

63

1/

1

/

2

n

jT

vv

jv

jj

e

T

T

=

63

1

/

/ )1ln(1

/n

j

T

Tv jv

jv

j ee

T

=

63

1

/)1ln(

2

n

j

Tv jvj eT

2/

/263

1 )1(

= T

Tn

j

v

jv

jv

j

e

eTVNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,


)ln( Pressure


=

=

=m

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

© copyright 2012 William A. Goddard III, all rights reservedCh121a-Goddard-L08 41VNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,


)ln( Pressure


=

=

=m

Q =

= - kT DeQkT

TQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

= 0

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

= -De - kT

= 0

= k

VNVNV

TN

TV

TQkT

TQkTC

VQkTp

NQkT

,2

22

,

,

,


)ln( Pressure


=

=

=m

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

QkTTQkTQkS

TQkTE

VN

VN


)ln(ln Entropy

)ln( Energy

,

,2

=

=

=

kTDe

ee /1w

1ln ew

1ln ee

kTD

w

1nuclear

β1

β1electronic

==

qeeq ee D

eD

e ww

Assuming the reference state has free atoms

Thermodynamic functions for electronic states

we will assume qelect=1


)/2βexp(-1)/2βexp()βexp()( n

eh

hqn

QHO

==

Harmonic oscillator Partition function

)(ln)(ln0

HOqSdQ

=Write partition function of the systemAs a continuous superposition of oscillators

)()(βlnβ0

10

,

10 E

VN

WSdVTQTVE

=

=

)()(lnβln0,

1 SBVN

WSdkTQQkS

=

=

)()(βlnβ0

10

10 AWSdVQVA

==

Thermodynamic properties

1)βexp(β

2β)(

=

hhhW Q

E

)]βexp(1ln[1)βexp(

β)( h

hhW Q

S

=

)/2βexp()βexp(1ln)(

hhW Q

A

=

Zero point energyReference energy

)()(0,

vCB

VNv WSdk

TEC

=

=2

2

)]βexp(1[)βexp()β()(

hhhW Q

Cv =

e hnn )21( =

Weighting functions

Thermodynamic Properties for a Crystal

where


Where do we get the vibrational density of states DoS(n)?

Experimentally from Inelastic neutron scattering

Compare to phonon dispersion curves. Peak is for phonons with little dispersion

“Phonon Densities of States and Related ThermodynamicProperties of High Temperature Ceramics” C.-K Loong, J.European Ceramic Society, 1998

Can use to calculate thermodynamic properties


Jason stopped on April 16, 2014

TOD A. PASCAL MSC, CALTECH

Validation of 2PT method for entropy and free energy for amino acid side chains

Tod A PascalRavi AbrolWilliam A Goddard IIIMSC 2013 Research Conference


General OutlineOverview of the 2PT method for

calculating accurate entropies and free energies

Application to the solvation thermodynamics of Amino Acid sidechains


Entropy

“Any method involving the notion of entropy, the very existence of which depends on the second law of thermodynamics, will doubtless seem to many far-fetched, and may repel beginners as obscure and difficult of comprehension.”

--Willard Gibbs, Graphical Methods in the Thermodynamics of Fluids (1873)

Rudolf Clausius - originator of the concept of "entropy".

ttt SG DD=D

http://en.wikipedia.org/wiki/Image:Clausius.jpg

http://en.wikipedia.org/wiki/Image:Gibbs_Josiah_Willard.jpg


Entropy: How to calculate it?• Test Particle Methods (insertion or deletion)

• Good for low density systems • Perturbation Methods • Thermodynamic integration, Thermodynamic perturbation• Applicable to most problems • Require long simulations to maintain “reversibility”

• Nonequilibrium Methods (Jarzinski’s equality)• Obtaining differential equilibrium properties from irreversible processes • Require multiple samplings to ensure good statistics

• Normal mode Methods• Good for gas and solids• Fast• Not applicable for liquids

References: Frenkel, D.; Smit, B. Understanding Molecular Simulation from Algorithms to Applications. Academic press: Ed., New York, 2002. McQuarrie, A. A. Statistical Mechanics. Harper & Row: Ed., New York, 1976. Jarzynski, C. Nonequilibrium Equality for Free Energy Differences. Phys. Rev. Lett. 1997, 78, 2690.

49

Remaining issue: experimental energies are free energies, need to calculate entropy

Free Energy, F = U – TS = − kBT ln Q(N,V,T)Kirkwood thermodynamic integration

J. G. Kirkwood. Statistical mechanics of fluid mixtures, J. Chem. Phys., 3:300-313,1935

Enormous computational cost required for complete sampling of the thermally relevant configurations of the system often makes this impractical for realistic systems. Additional complexities, choice of the appropriate approximation formalism or somewhat ad-hoc parameterization of the “reaction coordinate”

General approach to predict Entropy, S, and Free Energy

50

Solvation free energies amino acid side chainsPande and Shirts (JCP 122 134508 (2005) Thermodynamic

integration leads to accurate differential free energies

51

Solvation free energies amino acid side chainsPande and Shirts (JCP 122 134508 (2005) Thermodynamic

integration leads to accurate differential free energies

But costs 8.4 CPU-years on 2.8 GHz processor

52

Starting point: Get Density of states from MD (Velocity autocorrelation) Partition Function entropy

DoS(n) is the vibrational density of States

Problem: as n 0 get DoS ∞ unless DoS(0) = 0

zero

zero zero

DoS(n)

Calculate entropy from DoS(n)

Velocity autocorrelation function

53

S ( )Finite density of states at n =0Proportional to diffusion coefficient

Also strong anharmonicity at low frequencies

Problem with Liquids: S(0)≠0

DoS(n)

n

where D is the diffusion coefficientN=number of particlesM = mass

zero

zero zero

54

•Decompose liquid DoS(n) to a gas and a solid contribution•DoS(n) total = DoS(n) gas + DoS(n) solid •S(0) attributed to gas phase diffusion, fit to hard sphere theory•Gas component contains small n anharmonic effects•Solid component contains quantum effects

S ( )

Gasexponentialdecay

S ( )

SolidDebye crystalS(v) ~v2

S ( )

solid-likegas-like

Two-Phase Thermodynamics Model (2PT)

Total

=

)()()()(00

gP

gHOP

s WSdWSdP

=Property =The two-phase model for calculating thermodynamic properties of liquids from molecular dynamics: Validation for the phase diagram of Lennard-Jones fluids; Lin, Blanco, Goddard; JCP, 119:11792(2003) wag536


The 2PT Method The basic idea

The DoS Thermodynamics

The gas component VAC for hard sphere gas

DoS for hard sphere gas

)()()( solidgas SSS =

)()()()(00

HOP

sgP

g WSdWSdP

=

S ( )

) exp(3)( tmkTtcHS = tcoefficienfriction :),,( HSHST

2

0

0222

61

412)(

=

=

fNs

sNSgas

HS

gasHS

gas

NSs

fNN

12)0(0 ==

=

Two unknowns (a and Ngas) or (s0 and f)

TOD A. PASCAL MSC, CALTECHDetermining so and f from MD Simulation so (DoS of the gas component at

v=0) completely remove S(0) of the fluid

f (gas component fraction) T -> ∞ or ρ -> 0: f -> 1 (all gas) ρ -> ∞: f-> 0 (all solid)

0)0( ),0(0 == solidSSs

); ,(

),(

0HSHS TD

TDf

=mN

kTsTD12

)0(),( =

Enskog)-(Chapman )(183);,( 2/1

20 mkTTD

HS

HSHS

=

one unknown σHS


Determining σHS

σHS (hard sphere radius for describing the gas molecules)

gas component diffusivity should agree with statistical mechanical predictions at the same T and ρ

gas component diffusivity from MD simulation

HS diffusivity from the Enskog theory

mfNkTsfTDHS

12),( 0=

1)(4);,(),( 0

=fyz

fyfTDfTD HSHSHS

3

6HSy

=3

32

)1(1)(

yyyyyz

=Hard sphere packing fraction

Compressibility


And finally….A universal equation for f

Graphical representation

022662 2/52/32/72/3532/152/9 =DDDD ffyff

3/23/12/100 )6()(

92),,,( :ydiffusivit normalized

mkT

NssmT =D

0.0

0.2

0.4

0.6

0.8

1.0

1.E-05 1.E-03 1.E-01 1.E+01 1.E+03D (normalized diffusivity)

f or f

y

ffy


Computational SchemeRun a MD simulation(trajectory information saved)

Calculate VAC

Calculate DoS (FFT of VAC)Apply HO

approximationTo S()

1PT thermodynamic

predictions

Calculate S(0) and D

Solve for f

Determine Sgas(), Ssolid()

Apply HO statisticsTo Ssolid()

Apply HS statistics to Sgas()

2PT thermodynamic

predictions


Test case 1: Lennard Jones Fluid

●stable

●metastable

●unstable

T - diagram for Lennard Jones Fluid

0.6

1.0

1.4

1.8

0.0 0.4 0.8 1.2*

T*

Solid

LiquidGas

Supercritical Fluid

• Intermolecular potential

• Phase diagram

= 612 )()(4)(

rrrV e

Lennard-Jones Potential

r = V(r)

r

0-e

(T*=kT/e *=3)

– critical point

– triple point006.0304.0

006.0316.1*

*

=

=

c

cT

69.0* tpT


LJ VAC and DoS

0

10

20

30

40

50

60

70

80

90

100

0 20 40 60 80 100

frequency v(cm-1)

DoS

S(c

m)

=====

Density of StatesVelocity Autocorrelation

-800

-400

0

400

800

1200

1600

0 0.5 1 1.5 2 2.5 3time (ps)

C (t)

=====

gasliquidsolid

gasliquidsolid

)()0()( tvvtC = dtetCkT

S ti

=

πυ2)(lim 2)υ(


2PT DoS Decomposition

0

200

400

600

800

1000

1200

0 5 10 [cm-1]

S(

) [cm

]

0

5

10

15

20

25

30

0 50 100 150 [cm-1]

S(

) [cm

]

0

5

10

15

20

25

30

35

0 50 100 150 [cm-1]

S(

) [cm

]

• Examples

LJ gas liquid FCC solid

solid-likegas-likegas-like

solid-like

solid-likegas-like


Pressure and Energy for LJ GasPressure

-2

0

2

4

6

8

10

12

14

16

18

0 0.2 0.4 0.6 0.8 1 1.2 *

P *

T*=1.8T*=1.4T*=1.1T*=0.9MD

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

0 0.2 0.4 0.6 0.8 1 1.2 *

E *

T*=1.8T*=1.4T*=1.1T*=0.9MD2PT(Q)

Pressures and MD Energies agree with EOS valuesQuantum Effect (ZPE) most significant for crystals (~2%)

Total Energy


Entropy

4

6

8

10

12

14

16

18

20

0 0.2 0.4 0.6 0.8 1 1.2 *

S *

T*=1.8

T*=1.4

T*=1.1

T*=0.9

1PT(Q)

4

6

8

10

12

14

16

18

20

0 0.2 0.4 0.6 0.8 1 1.2 *

S *

T*=1.8T*=1.4T*=1.1T*=0.92PT(Q)2PT(C)

1PT 2PT model

•Overestimate entropy for low density gases

•Underestimate entropy for liquids•Accurate for crystals

•Accurate for gas, liquid, and crystal•Accurate in metastable regime•Quantum Effects most important for crystals (~1.5%)

gas

liquid

crystal


Gibbs Free Energy

-30

-25

-20

-15

-10

-5

0

5

0 0.2 0.4 0.6 0.8 1 1.2 *

G *

T*=1.8T*=1.4T*=1.1T*=0.91PT(Q)

1PT 2PT model

•Underestimate free energy for low density gases

•overestimate entropy for liquids•Accurate for crystals

•Accurate for gas, liquid, and crystal•Accurate in metastable regime

gas

liquid

crystal

-30

-25

-20

-15

-10

-5

0

5

0 0.2 0.4 0.6 0.8 1 1.2 *

G *

T*=1.8T*=1.4T*=1.1T*=0.92PT(Q)2PT(C)


Why does 2PT work?

0

1

2

3

4

5

0 50 100 150 [cm-1]

WS(

)

0

5

10

15

20

25

30

0 50 100 150 [cm-1]

S(

) [cm

]

0

200

400

600

800

1000

1200

0 2 4 6 8 10 [cm-1]

S(

) [cm

]

HS fy = 0.036

QHO

HS fy = 0.309

CHO

)()()()(00

gP

gHOP

s WSdWSdP

=

gas

liquid • 1PT overestimates Wsgas for gas for

modes < 5 cm-1

• 1PT underestimates Wsgas for liquid for

modes between 5 and 100 cm-1

• 2PT properly corrects these errors


Convergence of 2PT

6.5

7.5

8.5

9.5

10.5

11.5

12.5

13.5

14.5

15.5

100 1000 10000 100000 1000000

MD steps

S*

2PT(Q)2PT(C) MBWR EOS

gas (*=0.05 T*=1.8)

liquid (*=0.85 T*=0.9)

• For gas, the entropy

converges to within 0.2% with

2500 MD steps (20 ps)

• For liquid, the entropy

converges to within 1.5% with

2500 MD steps (20 ps).


Melting and Solidification

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0.00 0.40 0.80 1.20

T*

Simulation conditions

solid

supercritical fluid

3

4

5

6

7

8

0.80 1.20 1.60 2.00T*

S*liquid (EOS)solid (EOS)heatingcoolingclassical

Ent

ropy

solidmetatstable

unstablesupercritical

fluid

starting withfcc crystal

starting with amorphous liquid

• Initial amorphous structure is used in the cooling process

•The fluid remains amorphous in simulation even down to T*=0.8 (supercooled)

•The predicted entropy for the fluid and supercooled fluid agree well with EOS for LJ fluids

• Initial fcc crystal is used in the heating process

•The crystal appears stable in simulation even up to T*=1.8 (superheated)

•The predicted entropies for the crystal and superheated crystal agree well with EOS for LJ solids


0

2

4

6

8

10

12

14

16

0 500 1000 1500 2000 2500 3000

S_hs(v)[cm]S_s(v)[cm]Stot(v)[cm]

Water Power Spectrum (DoS) (25 ps, 1fs steps)

w (cm-1)

w)

Power spectrum for water at 300 K. The power spectrum is decomposed into a gas (diffusive) and a solid (fixed) spectra and their contributions added to yield the free energy of the liquid state.

0

2

4

6

8

10

12

14

16

1 10 100 1000 10000




Entropy of water box

Theory: 69.6 +/- 0.2 J/K*molExperimental Entropy: 69.9 J/K*mol

(NIST) Statistics collected over 20ps of dynamics , no additional computation

cost


Thermodynamics of water on POPCPOPC Bulk Shell 1 Shell 2 Shell 3 Exp

temp (K) 303.40 298.72 298.39 294.27 298.28ZPE_(kJ/mol/SimBox)= 3101.97 63.28 65.21 61.52 63.32

Vo__(kJ/mol/SimBox)=

-19542.5

1 -436.24 -436.71 -436.07 -436.29

Eq__(kJ/mol/SimBox)=

-16270.7

6 -365.83 -364.49 -367.32 -365.85 -378.23

Ec__(kJ/mol/SimBox)=

-18529.6

6 -414.84 -414.84 -414.84 -414.84Sq_(J/mol_K/SimBox)= 1211.99 68.08 60.07 74.77 69.14 69.90Sc_(J/mol_K/SimBox)= -1275.28 23.47 14.07 31.00 24.42

Aq__(kJ/mol/SimBox)=

-16638.4

8 -386.17 -382.42 -389.33 -386.47

Ac__(kJ/mol/SimBox)=

-18142.7

4 -421.85 -419.04 -423.97 -422.13Cvq(J/mol_K/SimBox)= 1019.57 37.84 38.16 38.81 37.94 37.27Cvc(J/mol_K/SimBox)= 3338.36 71.62 73.29 72.12 71.89S(0)(cm/mol/SimBox)= 0.05 0.02 0.00 0.01 0.01fluidicity___factor= 0.19 0.26 0.12 0.22 0.23constant__________D= 0.10 0.19 0.04 0.13 0.16 0.13

TOD A. PASCAL MSC, CALTECHAccurate entropies of liquid systems Fourier Transform VAC

to get Density of Vibrational StatesGet Enthalpy and Free Energy using quantum partition function

Do 25 picosec MD, Extract Velocity Autocorrelation (VAC) Function

0

10

20

30

40

50

60

70

80

90

100

0 20 40 60 80 100

frequency v(cm-1)

DoS

S(c

m)

=====

gasliquidsolid

-800

-400

0

400

800

1200

1600

0 0.5 1 1.5 2 2.5 3time (ps)

C (t)

=====

gasliquidsolid

Works well for solids


0

2

4

6

8

10

12

14

16

0 500 1000 1500 2000 2500 3000


Water Power Spectrum (DoS) (25 ps, 1fs steps)

w (cm-1)

w)


0

2

4

6

8

10

12

14

16

1 10 100 1000 10000



74

Entropy of water box

Theory: 69.6 +/- 0.2 J/K*molExperimental Entropy: 69.9 J/K*mol (NIST)

Statistics collected over 20ps of MD , no additional cost

75

Absolute Entropies of common solventssolvent Expa AMBER 03 Dreiding GAFF OPLS AA/L acetic acid 37.76 43.49 ± 1.04 32.58 ± 0.24 acetone 47.90 40.92 ± 0.18 40.99 ± 0.20 41.01 ± 0.29 43.58 ± 0.44 acetonitrile 35.76 33.78 ± 0.24 30.20 ± 0.35 30.63 ± 0.52 31.17 ± 0.20 benzene 41.41 38.20 ± 0.43 36.61 ± 0.48 36.63 ± 0.89 38.65 ± 0.39 1,4 dioxane 46.99 36.91 ± 0.36 39.32 ± 0.12 35.92 ± 0.43 40.15 ± 0.32 DMSO 45.12 36.94 ± 0.28 36.40 ± 0.20 35.23 ± 0.33 37.11 ± 0.17 ethanol 38.21 30.90 ± 0.47 33.29 ± 0.44 27.90 ± 0.20 30.90 ± 0.22 ethylene glycol 39.89 28.53 ± 0.09 29.44 ± 0.22 26.66 ± 0.30 31.43 ± 0.23 furan 42.22 35.45 ± 0.10 35.65 ± 0.56 34.60 ± 0.43 36.96 ± 0.25 methanol 30.40 25.21 ± 0.39 26.38 ± 0.31 23.57 ± 0.34 25.57 ± 0.12 THF 48.71 38.61 ± 0.36 35.76 ± 0.24 34.96 ± 0.22 43.53 ± 0.45 toluene 52.81 45.56 ± 0.35 42.44 ± 0.25 41.85 ± 0.27 45.40 ± 0.34 M.A.D.b 2.09 2.27 2.62 1.47 Avg. Error -7.13 -6.43 -9.13 -5.85 R.M.S. Error 7.63 7.84 9.59 6.08 R2 0.82 0.85 0.81 0.92

Best estimate* chloroform 48.28 45.49 ± 0.63 44.90 ± 0.56 51.52 ± 0.23 42.43 ± 0.56 nma 46.14 39.64 ± 0.32 37.85 ± 0.20 37.94 ± 0.18 40.29 ± 0.11 TFE 49.32 45.06 ± 0.29 43.18 ± 0.10 41.35 ± 0.31 43.48 ± 0.29

Accuracy predicted entropy only limited by accuracy of force field

Thermodynamics of liquids: standard molar entropies and heat capacities of common solvents from 2PT molecular dynamics; Pascal; Lin; Goddard; Phys. Chem. Chem. Phys., 13: 169-181 (2011) wag897


How accurate is it? water model F3C SPC SPC/E TIP3P TIP4P-

EwStrn(2PT)

50.590.25

53.050.14

49.870.14

55.590.15

49.790.07

Srot(2PT)

11.540.06

12.030.03

10.410.04

12.900.04

9.530.07

Svib(2PT)a

0.040.00 - - - -

ftrn(2PT)b

0.250.01

0.290.01

0.230.01

0.340.00

0.240.01

frot(2PT)b

0.060.00

0.070.00

0.050.00

0.080.00

0.050.00

S(2PT) 62.180.30

65.090.13

60.280.16

68.490.14

59.320.12

S(1PT) 53.820.13

56.240.13

52.280.18

59.370.17

51.390.09

S(CT)c - 70.10 66.60 72.70 66.30S(NN)d - 73.51 66.91 80.19 65.46S(FD)d - 65.10 64.48 70.86 -S(FEP)d - 68.20 63.36 72.58 63.62

S(expt)e

69.95

Within 3% of FEP method, while being 3 orders of magnitude more efficient

Lin, Maiti and Goddard, JPC-B, 2011


Amino Acid Side Chains

TOD A. PASCAL MSC, CALTECHAA side-chain analogs

ssAmino-acid abbreviation

analogue pKaa

Alanine ala Methane Arginine arn N-

Propylguanidine

13.65

Asparagine ash Acetamide Aspartate asn Acetic Acid 3.86Cysteine cys Methanethi

ol

Glutamine gln Propionamide

Glutamine glh Propionic Acid

4.24

Histidine hie 4-methylimidazole

6.00

Histidine hid 4-methylimidazole

6.00

Isoleucine ile 1-butane Leucine leu isobutane Lysine lyn N-

butylamine10.79

Methionine met Methyl Ethyl Sulfide

Phenylalanine

phe Toluene

Serine ser Methanol Threonine thr Ethanol Tryptophan trp 3-

methylindole

Tryosine tyr 4-cresol Valine val Propane


Calculating the solvation thermodynamics of AA analogs with 2PT

ss


Thermodynamics of water around small solutes

TOD A. PASCAL MSC, CALTECHConvergence of the solvation energies

Convergence from ~ 2ns of MD: time required for system equilibration


Performance of the 2PT method

94% correlation with experiment Solvation entropy and enthalpy underestimated

TOD A. PASCAL MSC, CALTECHComparison of 2Pt with other methods

94% agreement with Free Energy Perturbation 3 orders of magnitude more efficient!

84

2PT has 98% correlation with results of Shirts and Pande2PT has 88% correlation with experiments – measure of accuracy

of forcefieldBut Total Simulation time: 36.8 CPU hrs doing each 10 timesFactor of 2000 improvement


ConclusionsAfter equilibration, can obtain

converged thermodynamics of solvation from as little as 10ps of MD dynamics

Both solvation enthalpy and entropy of common forcefields are significantly underestimated

Water structure around “hydrophobic”/hydrophilic small molecules have dramatically different enthalpic signatures


Thermodynamics and temperature dependence of interfacial water molecules

Tod A PascalWilliam A Goddard IIIMSC 2013 Research Conference


What to expect from this talkSurface tension of water:

What is it Why is it important

Methods of Calculating Surface tension Kirkwood-Buff Theory Evaluation of Surface Free Energy Other schemes

Comparison of KB theory and 2PT methodTemperature dependence and energy

profilesMolecular orientations at the air-water

interface


Surface tension

Liquid Temperature °C

Surface tension, γ

Acetic acid 20 27.60Acetic acid (40.1%) + Water

30 40.68

Acetic acid (10.0%) + Water

30 54.56

Acetone 20 23.70Diethyl ether

20 17.00Ethanol 20 22.27Ethanol (40%) + Water

25 29.63

Ethanol (11.1%) + Water

25 46.03

Glycerol 20 63.00n-Hexane 20 18.40Hydrochloric acid 17.7M aqueous solution

20 65.95

Isopropanol

20 21.70Liquid helium II

-273 [19]0.37Liquid nitrogen

-196 8.85Mercury 15 487.00Methanol 20 22.60n-Octane 20 21.80Sodium chloride 6.0M aqueous solution

20 82.55

Sucrose (55%) + water

20 76.45

Water 0 75.64Water 25 71.97Water 50 67.91Water 100 58.85Toluene 25 27.73

Hydrophobicity Terrestrial Life (Insects)

“contractive tendency of the surface of a liquid that allows it to resist an external force”

http://en.wikipedia.org/wiki/Acetic_acid_(data_page)

http://en.wikipedia.org/wiki/Acetone_(data_page)

http://en.wikipedia.org/wiki/Diethyl_ether_(data_page)

http://en.wikipedia.org/wiki/Diethyl_ether_(data_page)

http://en.wikipedia.org/wiki/Ethanol_(data_page)

http://en.wikipedia.org/wiki/Glycerol_(data_page)

http://en.wikipedia.org/wiki/Hexane_(data_page)

http://en.wikipedia.org/wiki/Hydrochloric_acid

http://en.wikipedia.org/wiki/Hydrochloric_acid

http://en.wikipedia.org/wiki/Molar_solution

http://en.wikipedia.org/wiki/Isopropanol_(data_page)

http://en.wikipedia.org/wiki/Isopropanol_(data_page)

http://en.wikipedia.org/wiki/Helium

http://en.wikipedia.org/wiki/Helium

http://en.wikipedia.org/wiki/Surface_tension#cite_note-20

http://en.wikipedia.org/wiki/Nitrogen

http://en.wikipedia.org/wiki/Nitrogen

http://en.wikipedia.org/wiki/Mercury_(element)

http://en.wikipedia.org/wiki/Methanol_(data_page)

http://en.wikipedia.org/wiki/Octane

http://en.wikipedia.org/wiki/Sodium_chloride

http://en.wikipedia.org/wiki/Sodium_chloride

http://en.wikipedia.org/wiki/Molar_solution

http://en.wikipedia.org/wiki/Sucrose

http://en.wikipedia.org/wiki/Water_(data_page)

http://en.wikipedia.org/wiki/Toluene

TOD A. PASCAL MSC, CALTECHCalculating Interfacial surface Tension

Kirkwood-Buff Theory (or is it Tolman Theory)?

sVzn

z)(

)( =

ij

ij

ji ij

ij

sBN r

rdurz

VTkzzP

)(1)()(),(

2

=

ij

ij

ji ij

ijij

sBT r

rdur

yxV

TkzzP)(

21)()(

),(

22

=

• Density profile

• Stress profile

• Interfacial tension = )()( zPzPdz TN

zLLV yxs D=

x

y

z

Richard C. Tolman

John Gamble Kirkwood


F3C water box IFT 75x75x100

cell 6900 F3C

water molecules (20Å thick layer)

500ps NVT equilibration dynamics with LAMMPS

Stress per atom dumped every 10fs for 100ps

59.9 dynes/cm3 obtained (Experimental: 70 dynes/cm3)

TOD A. PASCAL MSC, CALTECHCalculating Interfacial Surface TensionDirect evaluation of surface free energy

Requires evaluation of the surface entropy Requires extensive simulation time for

convergence Can be approximated from potential of mean

force calculations – large uncertainties

𝛾=( 𝜕𝐺𝜕 𝐴 )𝑁 ,𝑇 ,𝑃

=(𝐺𝑠𝑢𝑟𝑓𝑎𝑐𝑒−𝐺𝑏𝑢𝑙𝑘

𝜕 𝐴 )𝑁 ,𝑇 ,𝑃

𝐺=𝐻− 𝑃𝑉¿𝑈 −𝑇𝑆−𝑃𝑉

surface

bulk


The Recipe: the 2PT Method

92

S ( )

Finite density of states at n =0Proportional to diffusion coefficient

Harmonic Approximation Entropy= ∞•Also strong anharmonicity at low frequencies

Liquid

S ( )

Gasexponentialdecay

S ( )

SolidDebye crystalS(v) ~v2New Model

2 phase theory (2PT )Liquid Solid + Gas •Two-Phase Thermodynamics

Model (2PT)•Decompose liquid S(v) to a gas and a solid contribution

•S(0) attributed to gas phase diffusion•Gas component contains anharmonic effects

•Solid component contains quantum effects

S ( )

solid-likegas-like

The two-phase model for calculating thermodynamic properties of liquids from molecular dynamics: Validation for the phase diagram of Lennard-Jones fluids; Lin, Blanco, Goddard; JCP, 119:11792(2003)

)/2βexp(-1)/2βexp()βexp()( n

e

hhq

n

QHO

==


Comparison of Surface Tensions calculated with both methods

Excellent agreement Systematic underestimation of experimental value: Deficiency of forcefield (SPC-Ew) used


The thermodynamic factors that determine the IFT

The decrease in surface tension with increasing temperature is an entropically driven process


Where does the increased entropy with temperature originate?

Both rotational and translational entropy increases with increasing temperature


What happens to the enthalpy with increasing temperature?

Hydrogen bonding in supercooled water is different from ambient water Subsurface water molecules are enthalpically stabilized Effect is reduced with increasing temperature


The thermodynamics stability of interfacial water

Surface tension effects propagate into subsurface Implications for propensity of ions at the interface


ConclusionsCan evaluate the surface tension of

liquids from direct evaluation of the surface energy

Reduction in surface tension with temperature is entropically driven

Sub-surface water molecules are preferentially stabilized enthalpically, especially for super-cooled water


Hydrogen bonding and entropy of confined water molecules under high pressure

Collaborators:Yousung Jung (KAIST – Korea)William A. Goddard III (Caltech)

TOD A. PASCAL MSC, CALTECHThe Hydrophobic effect: Importance/Consequences

Driving force in formation of Membranes Micelles Globular proteins

Hydrophobic DNA bases stack so as to exclude water molecules

"The antipathy of the paraffin chain for water is, however, frequently misunderstood. There is no question of actual repulsion between individual water molecules and paraffin chains, nor is there any very strong attraction of paraffin chains for one another. There is, however, a very strong attraction of water molecules for one another in comparison with which the paraffin-paraffin or paraffin-water attractions are slight." - G. S. Hartley 1936

Nonpolar groups (alkane chains) are hydrophobic

Polar groups (alcohols) are hydrophilic

What are the microscopic thermodynamic forces involved? Enthalpy and/or Entropy?

How does structure of water molecules change as you move away from the surface?

How do these microscopic differences impact the macroscopic bulk properties?

ttt SG DD=D

http://en.wikipedia.org/wiki/Image:Water_drop_on_a_leaf.jpg

http://images.google.com/imgres?imgurl=http://www.steve.gb.com/images/science/hydrophobic_effect.png&imgrefurl=http://www.steve.gb.com/science/lipids_and_membranes.html&h=415&w=382&sz=35&hl=en&start=1&sig2=ScGrX3NYIysUciE672_SXg&um=1&tbnid=1NnvkwsiBDw1VM:&tbnh=125&tbnw=115&ei=jOs_R8qNLaKuggOsi-ngCA&prev=/images?q=hydrophobic+effect&svnum=10&um=1&hl=en


Using CNTs for fast water conduction (Theory!)

Synthetic analogue of biological aquaporins


And then the experiments…

TOD A. PASCAL MSC, CALTECHMore confirmation from experiments…

sub-2nm vertically aligned CNTs, microfabricated into membranes Flux estimated: 10-40 water/nm^2/ns (1000-10,000 times faster) Slip length 1.4 micro-m, breakdown of continuum Hagen-Poiseuille

theory

Holt, Park et al, Science (2006)

TOD A. PASCAL MSC, CALTECHWhy does water conduct through CNTs?

Atomic smoothness? Depletion layer (hydrophobicity) & dangling OH bonds near the

interface?

Aluru, Nano Lett (2008)

TOD A. PASCAL MSC, CALTECHYes… but why would water want to do this??

Enthalpically, water-water H-bonds are broken upon creating a surface (unfavorable)

Entropically, going into a confined space reduces entropy (unfavorable)

Nonetheless, experimentally

water spontaneously wets the internal

pores of CNTWhat is the physical origin?

Spontaneous filling of CNT with water appears to be against textbook concept!


MD simulation setup SPC-E water

model QMFF-Cx

forcefield for graphite

Carbon – water interactions obtained from QM

50ns MD simulation

LAMMPS simulation engine

12 612 6 4LJE

r r e

=

εO-C = 0.65 kJ/mol εH-C = 0.29σO-C = 3.166 ÅσC-H = 3.390


Structure of water inside CNT

Single file waters insiide (6,6) CNT located in center of CNT Ice – like waters inside (8,8) and (9,9) CNTs absorbed on walls


Enthalpy of waters inside CNT

Decreased enthalpy is observed due to confinement and shows excellent correlation to the average number of HB per water molecule

Enthalpy of ice-like (8,8) and (9,9) waters more favorable than bulk Enthalpy of waters in all other CNTs less favorable than bulk; worse in case of

single-file (6,6)


So what about the entropy…

Entropic trends exactly opposite to enthalpy Free rotations contributes 60% (6,6) - 20%

(11,11 and onwards) to entropic gain Translational entropy (due to reduced density

near hydrophobic interface) responsible for rest


Finally a complete picture of free energy!

water molecules inside the CNTs have lower free energies than bulk water

entropy dominates for tube diameters less than 1.0 nm (gas phase), the enthalpy dominates for tubes between 1.1 and 1.2 nm (ice phase), and both energies compensate for tubes larger than 1.4 nm (liquid phase)


The origin of the trends in rotational entropy

Water inside (6,6) and (7,7) resemble as gas – increase rotational entropy Water inside (8,8) and (9,9) resemble ice/water – decrease rotational entropy Water inside larger CNTs resemble bulk water – same rotational entropy

TOD A. PASCAL MSC, CALTECHThere is something special about water…

LJ liquid with same interactions with CNT as water (M3B) has unfavorable free energies

Thermodynamics recovered by including 3-body H-bond (mW)

SPC-E

qH: +0.4238 e-

M3B

mWStillinger-Weber 3-body

TOD A. PASCAL MSC, CALTECHSo waters flow into CNTs because…

Favorable local chemical potential inside CNT Lower free energy due to lower

enthalpy for ice-like CNTs but higher entropy for all others

Loss of hydrogen bonding inside tube overcome by increased entropy due to confinement

Ch121a Atomic Level Simulations of Materials and Molecules

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