Ch121a Atomic Level Simulations of Materials
and Molecules
Room BI 115
Lecture: Monday, Friday 2-3pm
Lab Session: Wednesday 2:30-3:30pm
Extra help: Monday 3-3:30pm and Wednesday 22:30pm
Lecture 3, April 8, 2013
FF1: valence, QEq, vdw
William A. Goddard III, wag@wag.caltech.edu
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
1
Ch121a Atomic Level Simulations of Materials
and Molecules
Room BI 115
Hours: Monday, Wednesday, Friday 2-3pm
Lecture 3, April 8, 2013
FF1: valence, QEq, vdw
William A. Goddard III, wag@wag.caltech.edu
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
TA’s Caitlin Scott and Andrea Kirkpatrick
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
2
Homework and Research Project
First 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 project
The final is a research report describing the calculations and
conclusions
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
3
The full Quantum Mechanics Hamiltonian:
Htotal(1..Nel,1..Mnuc)
M 2
M
N 2
2
2
H  
 
 
k 2M k
i 2mi
k
nuclei

l k
N
Z k Zl e2

Rkl
i

j i
e 2 M,N Z k e 2

rij
rkj
kj
electronnucleuselectronnucleus
nucleus
electron
Born-Oppenheimer approximation, fix nuclei (Rk=1..M), solve for Ψel(1..N)
Hel(1..Nel) Ψel(1..Nel) = Eel Ψel(1..Nel)
Zk
1 2
1
 
Hel(1..Nel) 
=    
rij
i 2
kj rkj
i
j i
Ψel(1..Nel) and Eel(1..Mnuc) are functions of the nuclear coordinates.
electrons
Next solve the nuclear QM problem
Hnuc(1..Mnuc) Ψnuc(1..Mnuc) = Enuc Ψnuc(1..Mnuc)
Z k ZZlkeZ2 l e 2
2 2 2 2 2 2 2 2
e2 e2
+E
 
 )


Hnuc(1..MnucH
) =H  
el






(1..M


nuc
2
M
2
m
R
r
k
i
k
i
i mi
kl Rkl
k 2kM k
i 2
kl  k l  k
i j i j i ij rij kj
FF is analytic
fit to
this
(1..MIII,nuc
© copyright
2012
WilliamE
A.PE
Goddard
all )rights reserved
Ch120a-Goddard-L03
4
The Bending potential surface for CH2
C

H
1B
1
H
1D
g
1A
1
9.3 kcal/mol
3B
1
3S g
linear
Note that E(p)  E(p)
Symmetric about   p180°
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
5
Force Field
2
2

2
2
 



EPEH(1..M
)
=
E
(1..M
)

nuc
el
nuc  
k 2M k
i 2mi
k

l k
Z k Zl e2

Rkl
i

j i
Z k e2
e2

rij kj rkj
The Force Field (FF) is an analytic fit to such expressions
but formulated to be transferable so can use the same
parameters for various molecules and solids
General form:
E  Evalence  Enonbond  (Euser  Econstra
E valence  E bond  E angle  E torsion  E inversion  EEbacross-terms
 E bb'  E aa' 
Involves covalent bonds and the coupling between them (2-body, 3body, 4-body, cross terms)

E nonbond = Evdw + Eelectrostatic
Involves action at a distance, long range
interactions
Ch120a-Goddard-L03
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6
Bond Stretch Terms: Harmonic oscillator form
Taylor series expansion about the equilibrium, Re, d = R-Re
E(d)=E(Re)+(dE/dR)e[d]+½(d2E/dR2)e[d]2+(1/6)(d3E/dR3)e[d]3 + O(d4)
ignore
ignore
zero
E(R) = (ke/2)(R-Re
) 2,
Kb
Simple Harmonic
ignore
Re = 0.9 – 2.2 Å
Ke = 700 (Kcal/mol)/Å 2
Units: multiply by 143.88 to convert
mdynes/ Å to (Kcal/mol)/Å 2
R
QM eigenstates, En = n + ½ , where n=0,1,2,…
Re
Ground state wavefunction (Gaussian)
0(d)  (mw/pЋ) exp[- (mw/2Ћ)d2]
Some force fields, CHARRM, Amber use k=2Ke to
avoid multiplying by 2
Ch120a-Goddard-L03
Problem: cannot break the bond, E  ∞
© copyright 2012 William A. Goddard III, all rights reserved
7
Bond Stretch Terms: Morse oscillator form
R
Want the E to go to a constant as break
bond, popular form is Morse function
Evdw (Rij) = De[X2-2X] where X = exp[-a(R-Re)] = exp[-(g/2)(R/Re -1)]
Note that E(Re) = 0, the usual convention for bond terms
De is the bond energy in kcal/mol 0.04
Re is the equilibrium bond distance 0.03
a is the Morse scaling parameter 0.02
calculate from Ke and De
0.01
a= sqrt(ke/2De),
where ke = d2E/dR2 at R=Re
(the force constant)
Re
0
-0.01
E=0
De
-0.02
-0.03
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard
III, all
2
3 rights reserved
4
5
8
6
Morse Potential – get analytic solution for
quantum vibrational levels
EMorse(Rij) = De[X2-2X] where X = exp[-a(R-Re)] = exp[-(g/2)(R/Re -1)]
a = sqrt(ke/2De), where ke = d2E/dR2 at R=Re (the force constant)
Ev = hv0(n + 1/2 ) – [hv0(n + ½)]2/4De
n0 = (a/2p)sqrt(2De/m)
De
D0 Level separations decrease linearly
with level
En+1 – En = hv0 – (n+1)(hv0)2/2De
Write
Ch120a-Goddard-L03
En/hc = we(n + ½ ) - xewe(n + ½ )2
© copyright 2012 William A. Goddard III, all rights reserved
9
Angle Bend Terms: cosine harmonic
Expand E() as Fourier series.
J
Note only cos(n) since must
be symmetric about =0 and p

I
K
E() = C0 + C1 cos() + C2 cos(2) + C3 cos(3) + ….
E() ~ ½ Ch [cos  - cos e]2
E=0 at  = e
Using cos(2) = 2 cos2 - 1
E’() = dE()/d = - Ch [cos  - cos e] sin  = 0 at  = e
E”() = d2E()/d2 = - Ch [cos  - cos e] cos  + Ch (sin )2
k = E”(e) = Ch (sin e)2
Ch120a-Goddard-L03
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10
The Bending potential surface for CH2
C

H
1B
1
H
1D
g
1A
1
9.3 kcal/mol
3B
1
3S g
linear
Note that E(p)  E(p)
Symmetric about   p180°
Ch120a-Goddard-L03
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11
Barriers for angle term
The energy is a maximum at  = p.
Thus energy barrier to “linearize” the molecule becomes
1 cos(o ) 2
2
E barrier 1/2C1 cos(o ) 1/2K (
)
sin( o )
By symmetry the angular energy satisfies
E(p   )  E(p   )
E( )  E( )
This is always satisfied for the cosine expansion but
A
second more popular form is the Harmonic theta expansion
E() = ½ K  [ - e]2

However
except for linear molecules this does NOT satisfy the
symmetry relations, leading to undefined energies and forces for
 = 180° and 0°. This is used by CHARMM, Amber, ..
12
Ch120a-Goddard-L03
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Angle Bend Terms for linear molecule,
If e = 180° then we write E() = K [1 + cos()]
since for  = 180 – d this becomes
E(d) = K [1 + (-1 + ½
d2)
~ 1/2 K
K
d2
J
I
Ch120a-Goddard-L03

© copyright 2012 William A. Goddard III, all rights reserved
13
Simple Periodic Angle Bend Terms
For an Octahedral complex
The angle function should have the form
E() = C [1 - cos4] which has minima for
 = 0, 90, 180, and 270°
With a barrier of C for
 = 45, 135, 225, and 315°
Here the force constant is K = 16C = CP2 where P=4
is the periodicity, so we can write C=K/P2
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
14
Simple Periodic Angle Bend Terms
For an Octahedral complex
The angle function should have the form
E() = C [1 - cos4] which has minima for
 = 0, 90, 180, and 270°
With a barrier of C for
 = 45, 135, 225, and 315°
Here the force constant is K = 16C = CP2 where P=4
is the periodicity, so we can write C=K/P2
However if we wanted the minima to be at
 = 45, 135, 225, and 315° with maxima at
 = 0, 90, 180, and 270°
then we want to use E() = ½ C [1 + cos4]
Thus the general form is
E() = (K/P2)[1 – (-1)Bcos4]
Where B=1 for the case with a minimum at 0° and
B=-1
for a maximum at
0° 2012 William A. Goddard III, all rights reserved
Ch120a-Goddard-L03
© copyright
15
Dihedral barriers
7.6 kcal/mol Cis barrier
HOOH
Part of such barriers can
be explained as due to
φe ~ 111°
vdw and electrostatic
interactions between the
H’s.
1.19 kcal/mol But part of it arises from
Trans barrier covalent terms (the pp
lone pairs on each S or
O)
This part has the form
HSSH: φe ~ 92°
E(φ) = ½ B [1+cos(2φ)]
5.02 kcal/mol trans barrier
Which is
E=0 for φ=90,270° and
7.54 kcal/mol cis barrier
16
E=B
φ=0,180°
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard
III, allfor
rights
reserved
dihedral or torsion terms
J
Given any two bonds IJ and KL attached
to a common bond JK, the dihedral angle
φ is the angle of the JKL plane from the
IJK plane, with cis being 0
E(φ) = C0 + C1 cos(φ) + C2 cos(2 φ) + C3
cos(3 φ) + ….which we write as
p
φ
L
K
I
L
I
J
K
E( )  1/2[kn (1 d cos(n )]
n1
Where each kn is in kcal/mol, n = 1, is the
periodicity of the potential and
d=+1 the cis conformation is a minimum
d=-1 the cis conformation is thea maximum
Input data is kn and dn for each n.
Ch120a-Goddard-L03
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17
Consider the central CC bond in Butane
There are nine possible dihedrals: 4 HCCH,
2 HCCC, 2 CCCH, and 1 CCCC.
I
Each of these 9 could be written as
E(φ) = ½ B [1 + cos(3 φ)],
For which E=0 for φ = 60, 180 and 300°
and E=B the rotation barrier for φ = 0, 120 and 240°.
Ch120a-Goddard-L03
L
J
© copyright 2012 William A. Goddard III, all rights reserved
K
18
Consider the central CC bond in Butane
There are nine possible dihedrals: 4 HCCH,
L
2 HCCC, 2 CCCH, and 1 CCCC.
I
Each of these 9 could be written as
J
K
E(φ) = ½ B [1 – cos(3 φ)],
For which E=0 for φ = 60, 180 and 300°
and E=B the rotation barrier for φ = 0, 120 and 240°.
For ethane get 9 HCCH terms.
The total barrier in ethane is 3 kcal/mol but standard vdw and
charges of +0.15 on each H will account for ~1 kcal/mol.
Thus only need explicit dihedral barrier of 2 kcal/mol.
Ch120a-Goddard-L03
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19
Consider the central CC bond in Butane
There are nine possible dihedrals: 4 HCCH,
L
2 HCCC, 2 CCCH, and 1 CCCC.
I
Each of these 9 could be written as
J
K
E(φ) = ½ B [1 + cos(3 φ)],
For which E=0 for φ = 60, 180 and 300°
and E=B the rotation barrier for φ = 0, 120 and 240°.
For ethane get 9 HCCH terms.
The total barrier in ethane is 3kcal/mol but standard vdw and
charges of +0.15 on each H will account for ~1 kcal/mol.
Thus only need explicit dihedral barrier of 2 kcal/mol.
Can do two ways.
Incremental: treat each HCCH as having a barrier of 2/9 and add
the terms from each of the 9 to get a total of 2 (Amber, CHARRM)
Or use a barrier of 2 kcal/mol for each of the 9, but normalize by
the total number of 9 to get a net of 2 (Dreiding)
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
20
Twisting potential surface for ethene
The ground state (N) of ethene prefers φ=0º to obtain the highest
overlap of the pp orbitals on each C
The rotational barrier is 65 kcal/mol.
We write this as E(φ) = ½ B [1-cos2 φ)]. Since there are 4 HCCH
terms, we calculate each using the full B=65, but divide by 4.
T
The triplet excited
state (T) prefers φ
=90º to obtain the
lowest overlap.
N
φ = -90°
Ch120a-Goddard-L03
φ = 0°
φ = 90°
© copyright 2012 William A. Goddard III, all rights reserved
21
Inversions
When an atom I has exactly 3 distinct bonds IJ, IK, and IL, it is
often necessary to include an exlicit term in the force field to
adjust the energy for “planarizing” the center atom I.
E( )  1/2C(cos(w )  cos(w o ))
L
2
Where the force constant in kcal/mol is
J
Kw  C sin 2 w o
I

K
L
J
w

I
K
Ch120a-Goddard-L03

Umbrella inversion:
ω is the angle between
The IL axis and the
JIK plane
© copyright 2012 William A. Goddard III, all rights reserved
22
AMBER Improper Torsion JILK
Amber describes inversion using an improper torsion.
E( )  1/2K cos[n(   o )]
n=2 for planar angles (0=180°) and n=3 for the tetrahedral
Angles (0 =120°).

J
L
I

Improper Torsion:  is the
angle between the JIL plane
and the KIL plane
K

Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
23
CHARMM Improper Torsion IJKL
In the CHARMM force field, inversion is defined as if it were a
torsion.
E( )  1/2K [   o ]
2
For a tetrahedral carbon atom with equal bonds this angles is
φ=35.264°.
Improper Torsion: φ is the angle
between the IJK plane and the
L

LJK plane
J
K
Ch120a-Goddard-L03

I
© copyright 2012 William A. Goddard III, all rights reserved
24
Summary: Valence Force Field Terms
example
Description
Typical Expressions
Kb
(R  Ro )2
E ( R) 
2
Bond Stretch
Harmonic Stretch
Angle bend
Harmonic-cosine E  [k / 2 sin( 0 )][cos( )  cos( 0 )]2
Torsion
dihedral
Inversion
E(φ) = ½ B [1 – cos(3 φ)],
Umbrella inversion
Ch120a-Goddard-L03
Ew  [1/ 2kw / sin(w0 )][cos(w )  cos(w0 )]2
© copyright 2012 William A. Goddard III, all rights reserved
25
Coulomb (Electrostatic) Interactions
Most force fields use fixed partial charges on each nucleus,
leading to electrostatic or Coulomb interactions between each
pair of charged particles. The electrostatic energy between
point charges Qi and Qk is described by Coulomb's Law as
EQ(Rik) = C0 Qi Qk /(e Rik)
where Qi and Qk are atomic partial charges in electron
units
TA check
C0 converts units:
numbers
if E is in eV and R in A, then C0 = 14.403
If E is in kcal/mol and R in A, then C0 = 332.0637
e = 1 in a vacuum (dielectric constant)
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
26
How estimate charges?
Even for a material as ionic as NaCl diatomic, the dipole
moment  a net charge of +0.8 e on the Na and -0.8 e on the
Cl not the idealized charges of +1 and -1
We need a method to estimate such charges in order to
calculate properties of materials.
A side note about units.
In QM calculations the unit of charge is the magnitude of the
charge on an electron and the unit of length is the bohr (a0)
Thus QM calculations of dipole moment are in units of ea0 which
we refer to as au. However the international standard for
quoting dipole moment is the Debye = 10-10 esu A
Where m(D) = 2.5418 m(au)
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
27
QEq Electrostatics energy
•Self-consistent Charge Equilibration (QEq)
•Describe charges as distributed (Gaussian)
•Thus charges on adjacent atoms shielded
(interactions  constant as R0) and include
interactions over ALL atoms, even if bonded (no
exclusions)
•Allow charge transfer (QEq method)
atomic
interactions
I
1 2

E ({qi })J ij (qi ,q j ,rij )  i qi  J i qi 
2

i j
i 
Jij
(
)
E int rij , Qik , Q lj 

Erf 

ij kl
ij kl
rij

rij 

Keeping:
q Q
i
i
Qik Qkl
Hardness (IP-EA)
1/rij
Electronegativity (IP+EA)/2
rij
r i0 + r j0
Three universal parameters for each element:  io , J io , Ric , Ris & qic
1991: used experimental IP, EA, Ri;
Ch120a-Goddard-L03
© copyright 2012
William
A. Goddard
III, all rights reserved
ReaxFF
get all parameters
from
fitting
QM
28
Charge dependence of the energy (eV) of an
atom
E=12.967
Harmonic fit
The 2nd order
expansion in Q is ok
E=0
E=-3.615
Cl+
Q=+1
Cl
Q=0
Ch120a-Goddard-L03
ClQ=-1
= 8.291
Get minimum at Q=-0.887
Emin = -3.676
= 9.352
© copyright 2012 William A. Goddard III, all rights reserved
29
QEq: the optimum charges lead to Equilibration
that is, equal chemical potential
Expand the energy of an atom in a power series of the net
charge on the atom, E(Q)
Including terms through 2nd order leads to
Charge Equilibration for Molecular
Dynamics Simulations;
A. K. Rappé and W. A. Goddard III;
J. Phys. Chem. 95, 3358 (1991)
(2)
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
(3)
30
QEq parameters
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
31
Interpretation of J, the hardness
Define an atomic radius as
RA0
Re(A2) Bond distance of
homonuclear
H
0.84 0.74
diatomic
C
1.42 1.23
N
1.22 1.10
O
1.08 1.21
Si
2.20 2.35
S
1.60 1.63
Li
3.01 3.08
Thus J is related to the coulomb energy of a charge the size of the
32
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
atom
The total energy of a molecular complex
Consider now a distribution of charges over the
atoms of a complex: QA, QB, etc
Letting JAB(R) = the Coulomb potential of unit
charges on the atoms, we can write
Taking the derivative with respect to charge leads to the
chemical potential, which is a function of the charges
or
The definition of equilibrium is for all chemical potentials to be
equal.
This leads to © copyright 2012 William A. Goddard III, all rights reserved
Ch120a-Goddard-L03
33
The QEq equations
Adding to the N-1 conditions
The condition that the total charged is fixed (say at 0)
leads to the condition
Leads to a set of N linear equations for the N variables QA.
AQ=X, where the NxN matrix A and the N dimensional vector A
are known. This is solved for the N unknowns, Q.
We place some conditions on this. The harmonic fit of charge to
the energy of an atom is assumed to be valid only for filling the
valence shell.
Thus we restrict Q(Cl) to lie between +7 and -1 and
Q(C) to be between +4 and -4
Similarly Q(H) is between +1 and -1
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
34
The QEq Coulomb potential law
We need now to choose a form for JAB(R)
A plausible form is JAB(R) = 14.4/R, which is valid when the
charge distributions for atom A and B do not overlap
Clearly this form as the problem that JAB(R)  ∞ as R 0
In fact the overlap of the orbitals leads to shielding
shielded
unshielded
The plot shows the
shielding for C atoms using
various Slater orbitals
And l = 0.5
Ch120a-Goddard-L03
Using RC=0.759a0
© copyright 2012 William A. Goddard III, all rights reserved
35
QEq results for alkali halides
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
36
QEq for Ala-His-Ala
Amber
charges in
parentheses
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
37
QEq for deoxy adenosine
Amber
charges in
parentheses
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
38
QEq for polymers
Nylon 66
PEEK
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
39
Polarizable QEq
Allow each atom to have two charges:
A fixed core charge (+4 for Si) with a Gaussian shape
A variable shell charge with a Gaussian shape but subject to
displacement and charge transfer
Electrostatic interactions between all charges, including the core and
shell on same atom
Allow Shell to move with respect to core, to describe atomic
polarizability
Self-consistent charge equilibration (QEq)

 c 2
ic 3 2 c
c
 (r )  ( p ) Qi exp( i  | r  ri | )
 s 2
is 3 2
s 
s
s
i (r )  ( p ) Qi exp( i  | r  ri | )
c
i
Four universal parameters for each element:
Get from QM
 io , J io , Ric , Ris & qic
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
40
Noble gas dimers
s
Ar2
Re
De
All nonbonded atoms and
molecules exhibit a very repulsive
interaction at short distances due
to overlap of electron pairs (Pauli
Repulsion) and a weak attractive
interaction scaling like 1/R6 at long
R (London Dispersion. Together
these are called van der Waals
(vdW) interactions
Most popular form: Lennard-Jones 12-6
E=A/R12 –B/R6
= De[12 – 26] where = R/Re
Note that E is a min at Re, note the 2
A 2nd form for LJ 12-6 is
 4 De[t12 – t6] where tR/s
Note that E=0 at R = s, note the 1
Here s = Re(1/2)1/6 =0.89 Re
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van der Waals interaction
What do vdW interactions account for?
 Short range: Pauli Repulsion of overlapping electron pairs
Pauli repulsion: Erep 
e
 CRij
(where C  0)
i j
Born repulsion: or Erep 
1
(where N  9)

N
i  j Rij
 Long range attraction  London dispersion due to
Instantaneous fluctuations in the QM charges
Edisp   (
i j
C6,ij
6
ij
R

C8,ij
8
ij
R

C10,ij
10
ij
R
...)
Dipole-dipole Dipole-quadrupoleQuadruole-quadrupole
We usually neglect higher order (>R6) terms.
Ch120a-Goddard-L03
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42
Popular vdW Nonbond Terms: LJ12-6
Lennard-Jones 12-6
E(R)=A/R12 – B/R6
NonBond
=Dvdw{1/12 – 2/6} where  = R/Rvdw
R
Here Dvdw=well depth,
nd Rvdw = Equilibrium distance for vdw dimer
Alternative form
E(R) = 4Dvdw{[svdw/R)12]- [svdw/R)6]}
Where s = point at which E=0 (inner wall, s ~ 0.89 Rvdw)
Dimensionless force constant k = {[d2E/d2]=1}/Dvdw = 72
The choice of 1/R6 is due to London Dispersion (vdw Attraction)
However there is no special reason for the 1/R12 short range form
(other that it saved computation time for 1950’s computers)
LJ 9-6
would lead to a© more
accurate
wall.
Ch120a-Goddard-L03
copyright 2012
William A. inner
Goddard III,
all rights reserved
43
Popular vdW Nonbond Terms- Exponential-6
NonBond
Buckingham or exponential-6
E(R)= A e-CR - B/R6 which we write as
R
where R0 = Equilibrium bond distance;
 = R/R0 is the scaled distance
Do=well depth
z = dimensionless parameter related to force constant at R0
We define a dimensionless force constant as k = {(d2E/d2)=1}/D0
k = 72 for LJ12-6
z = 12 leads to –D0/6 at long R, just as for LJ12-6
z = 13.772 leads to k = 72 just as for LJ12-6
A problem with exp-6 is that E- ∞ as R 0. To avoid this BioGraf, LinGraf, CeriusII calculate the
inner maximum and reflect E about this point for smaller R so that E and E’ are continuous. When
z>10 this point is well up the inner wall and not important
44
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
Converting between X-6 and LJ12-6
Usual convention: use the same R0 and D0 and set z = 12.
I recall that this leads to small systematic errors.
Second choice: require that the long range form of exp-6 be
the same as for LJ12-6 (ie -2D0/6) and require that the inner
crossing point be the same. This leads to
This was published in the
Dreiding paper but I do not know
if it has ever been used
Ch120a-Goddard-L03
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45
Popular vdW Nonbond Terms: Morse
NonBond
E(R) = D0 {exp[-b(R-R0)] - 2 exp[(-(b/2)(R-R0)]}
At R=R0, E(R0) = -D0
At R=∞, E(R0) = 0
D0 is the bond energy in Kcal/mol
R0 is the equilibrium bond distance
b is the Morse scaling parameter
I prefer to write this as
E(R) = D0 [X2 - 2X] where X = exp[(a/2)(1-)]
At R=R0,  = 1 X = 1  E(R0) = -D0
At R=∞, X = 0  E(R0) = 0
k = {(d2E/d2)=1}/D0 = a2/2
Thus a = 12  k = 72 just as for LJ12-6
R
0.04
0.03
0.02
0.01
0
-0.01
D0
R0
-0.02
-0.03
2
3
4
5
Few theorists believe that Morse makes sense for vdw
parameters since it does not behave as 1/R6 as R∞
However, the vdw curve matches 1/R6 only for R>6A, and for our
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vdW combination rules
Generally the vdw parameters are provided for HH, CC,
NN etc (diagonal cases) and the off-diagonal terms are
obtain using combination rules
D0IJ = Sqrt(D0II D0JJ)
R0IJ = Sqrt(R0II R0JJ)
zIJ = (zIJ + zIJ)/2
Sometimes an arithmetic combination rule is used for vdw
radii
R0IJ = (R0II + R0JJ)/2 but this complicates vdw calculations
(the amber paper claims to do this but the code uses
geometric combinations of the radii)
Ch120a-Goddard-L03
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47
1-2 and 1-3 Nonbond exclusions
For valence force fields, it is assumed that the bond and angle
quantities include already the Pauli Repulsion and electrostatic
contributions so that we do not want to include them again in the
nonbond list.
Thus normally we exclude from the vdw and coulomb sums
the contributions from bonds (1-2) and
next nearest neighbor (1-3) interactions
Ch120a-Goddard-L03
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48
1-4 Nonbond exclusions
There is disagreement about 1-4 interactions. In fact the origin of
the rotational barrier in ethane is probably all due to Pauli
repulsion orthogonality effects.
Thus a proper description of the vdW interactions between 1-4
atoms should account for the barrier.
In fact it accounts only for 1/3 of the barrier.
I believe that this is because the CH bond pairs are centered at
the bond midpoint not on the H atoms as assumed in the vdw.
Thus using atom centered vdw accounts for only part of the
barrier necessitating an explicit dihedral term.
Thus I believe that the1-4 vdw and electrostatic terms should be
included. However some FF, such as Amber, CHARRM reduce
the 1-4 interactions by a factor of 2. I do not know of a
justification for this.
Ch120a-Goddard-L03
© copyright 2012 William A. Goddard III, all rights reserved
49