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1© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Ch121a Atomic Level Simulations of Materials and Molecules
William A. Goddard III, wag@wag.caltech.eduCharles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics, California Institute of Technology
BI 115Hours: 2:30-3:30 Monday and Wednesday
Lecture or Lab: Friday 2-3pm (+3-4pm)
Teaching Assistants Wei-Guang Liu, Fan Lu, Jose Mendozq, Andrea Kirkpatrick
Lecture 3, April 6, 2011Force Fields – 1 valence, QEq, vdw
2© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
kj kj
k
ijijikl
lk
klki ik k r
eZ
r
e
R
eZZ
mMH
2222
22
2
22
Born-Oppenheimer approximation, fix nuclei (Rk=1..M), solve for Ψel(1..N)
The full Quantum Mechanics Hamiltonian: Htotal(1..Nel,1..Mnuc)
nuclei electrons nucleus-nucleus
electron-nucleus
electron-electron
Hel(1..Nel) =
NM M M,NN
Hel(1..Nel) Ψel(1..Nel) = Eel Ψel(1..Nel)
Ψel(1..Nel) and Eel(1..Mnuc) are functions of the nuclear coordinates.
Next solve the nuclear QM problem
+Eel(1..Mnuc)Hnuc(1..Mnuc) =
Hnuc(1..Mnuc) Ψnuc(1..Mnuc) = Enuc Ψnuc(1..Mnuc)
EPE(1..Mnuc)FF is analytic fit to this
3© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Force Field
Involves covalent bonds and the coupling between them (2-body, 3-body, 4-body, cross terms)
= Eel(1..Mnuc)EPE(1..Mnuc)
The Force Field (FF) is an analytic fit to such expressions but formulated to be transferable so can use the same terms for various molecules and solids
General form:
Ecross-terms
E nonbond = Evdw + Eelectrostatic
Involves action at a distance, long range interactions
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Re = 0.9 – 2.2 ÅKe = 700 (Kcal/mol)/Å 2
Units: multiply by 143.88 to convert mdynes/ Å to (Kcal/mol)/Å 2
Simple Harmonic
Bond Stretch Terms: Harmonic oscillator form
R
ignore
Taylor series expansion about the equilibrium, Re, = R-Re
E()=E(Re)+(dE/dR)e[]+½(d2E/dR2)e[]2+(1/6)(d3E/dR3)e[]3 + O(ignore zero ignore
Kb
Some force fields, CHARRM, Amber use k=2Ke to avoid multiplying by 2
Re
Problem: cannot break the bond, E ∞
Ground state wavefunction (Gaussian)mЋ) exp[- (m2Ћ)2]
eigenstates, E = + ½ , where =0,1,2,…
E(R) = (ke/2)(R-Re)2,
5© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Note that E(Re) = 0, the usual convention for bond termsDe is the bond energy in kcal/molRe is the equilibrium bond distance is the Morse scaling parametercalculate from Ke and De
Bond Stretch Terms: Morse oscillator form
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
2 3 4 5 6
D0
R0
RWant the E to go to a constant as break bond, popular form is Morse function
E = 0
where X = exp[-(R-Re)] = exp[-(/2)(R/Re -1)]
Evdw (Rij) = De[X2-2X]
= sqrt(ke/2De), where ke = d2E/dR2 at R=Re (the force constant)
6© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Morse Potential – get analytic solution for quantum vibrational levels
where X = exp[-(R-Re)] = exp[-(/2)(R/Re -1)]
Evdw (Rij) = De[X2-2X]
= sqrt(ke/2De), where ke = d2E/dR2 at R=Re (the force constant)
Ev = hv0( + 1/2 ) – [hv0( + ½)]2/4De
0 = (/2)sqrt(2De/m)
Level separations decrease linearly with level
E+1 – E = hv0 – (+1)(hv0)2/2De
Write
En/hc = e( + ½ ) - xee( + ½ )2
7© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
The Bending potential surface for CH2
3B1
1A1
1B1
3g-
1g
9.3 kcal/mol
8© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Angle Bend Terms: cosine harmonic
I K
J
E() = C0 + C1 cos() + C2 cos(2) + C3 cos(3) + ….
~ ½ Ch [cos - cos e]2 = 0 at = e
E”() = dE()/d = - Ch [cos - cos e] sin = 0 at = e
E”() = d2E()/d2 = - Ch [cos - cos e] cos + Ch (sin )2
= Ch (sin e)2 at = e
Thus k = Ch (sin e)2
9© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
The energy barrier to “linearize” the molecule becomes
Ebarrier 1/2C 1 cos(o) 2 1/2K (1 cos(o)
sin(o))2
E( ) E( )By symmetry the angular energy satisfy
E( ) E()This is always satisfied for the cosine expansion but
Barriers for angle term
A second more popular form is the Harmonic theta expansionE() = 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, ..
10© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Angle Bend Terms for linear molecule,
I
K
J
If e = 180° then we write E() = K [1 + cos()]
since for = 180 – this becomes
E() = K [1 + (-1 + ½ K
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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
Simple Periodic Angle Bend Terms
12© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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°
Simple Periodic Angle Bend Terms
13© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Rotational barriers
HOOH
1.19 kcal/mol Trans barrier
7.6 kcal/mol Cis barrier
HSSH:
5.02 kcal/mol trans barrier
7.54 kcal/mol cis barrier
Part of these barriers can be explained as due to vdw and electrostatic interactions between the H’s.
But part of it arises from covalent terms (the p lone pairs on each S or O
This part has the form E(φ) = ½ B [1+cos(2φ)]
Which is
E=0 for φ=90,270° and E=B for φ=0,180°
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dihedral or torsion terms
Where each kn is in kcal/mol, n = 1, is the periodicity of the potential and
d=+1 the cis conformation is a minimumd=-1 the cis conformation is thea maximumInput data is K and d for each n.
E() 1/2[kn (1 d cos(n)]n1
p
IJ K
LE(φ) = C0 + C1 cos(φ) + C2 cos(2 φ) + C3 cos(3 φ) + ….which we write as
I K
Jφ LGiven 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
15© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Consider the central CC bond in Butane
There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC.Each of these 9 could be written asE(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°.
IJ K
L
16© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Consider the central CC bond in Butane
IJ K
L
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.
There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC.Each of these 9 could be written asE(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°.
17© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Consider the central CC bond in Butane
IJ K
L
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)
There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC.Each of these 9 could be written asE(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.
18© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Twisting potential surface for etheneThe ground state (N) of ethene prefers φ=0º to obtain the highest overlap with a rotational barrier of 65 kcal/mol for prefers φ =90º to obtain the lowest overlap.
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.
φ = 0° φ = 90°φ = -90°
T
N
19© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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() cos(o))2
I
K
LJ
K C sin2o
Where the force constant in kcal/mol is
LJ
I
K
Umbrella inversion:ω is the angle betweenThe IL axis and the JIK plane
20© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Amber describes inversion using an improper torsion.
E() 1/2K cos[n( o)]
N=2 for planar angles (=180°) and n=3 for the tetrahedralAngles ( =120°).
Improper Torsion: is the angle between the JIL plane and the KIL planeL
J
I
K
AMBER Improper Torsion JILK
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CHARMM Improper Torsion IJKL
In the CHARMM force field, inversion is defined as if it were a torsion.
E() 1/2K [ o]2
Improper Torsion: φ is the angle between the IJK plane and the LJK plane
L
J
IK
For a tetrahedral carbon atom with equal bonds this angles is φ=35.264°.
22© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Bond Stretch
Angle bend
Torsion
Inversion
Typical ExpressionsDescription Illustration
2
00)]cos())][cos(sin(2/[ kEHarmonic-cosine
Dreiding dihedral
2
00)]cos())][cos(sin(/2/1[ kE
Umbrella inversion
2
2)( o
b RRK
RE Harmonic Stretch
E() 1/2[kn (1 dcos(n)]n1
p
Summary: Valence Force Field Terms
23© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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 /(Rik)
where Qi and Qk are atomic partial charges in electron units
C0 converts units: if E is in eV and R in A, then C0 = 14.403If E is in kcal/mol and R in A, then C0 = 332.0637= 1 in a vacuum (dielectric constant)
TA check numbers
24© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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.
We need a method to estimate such charges in order to calculate properties of materials.
First a bit more 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 (D) = 2.5418 (au)
25GODDARD - MSC/CaltechGODDARD - MSC/Caltech Berlin 3/22/04Berlin 3/22/04
ReaxFF Electrostatics energy – Critical element
Three universal parameters for each element:1991: used experimental IP, EA, Ri;
ReaxFF get all parameters from fitting QM
ci
si
ci
oi
oi qRRJ &,,,
Keeping: i
i Qq
•Self-consistent Charge Equilibration (QEq)•Describe charges as distributed (Gaussian)•Thus charges on adjacent atoms shielded (interactions constant as R0) and include interactions over ALL atoms, even if bonded (no exclusions)•Allow charge transfer (QEq method)
ji iiiiiijjiiji qJqrqqJqE
2
1),,(}{
Electronegativity (IP+EA)/2
Hardness (IP-EA)
interactions atomic
lk
kir
rErflj
kiij QQQQrE
ij
ijklij
klij
,,int
Jij
rij
1/rij
ri0
+ rj0
I
2
26© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Charge Equilibration
Charge Equilibration for Molecular Dynamics Simulations;
A. K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991)
First consider how the energy of an atom depends on the net charge on the atom, E(Q)
Including terms through 2nd order leads to
(2) (3)
27© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Charge dependence of the energy (eV) of an atom
E=0
E=-3.615
E=12.967
Cl Cl-Cl+
Q=0 Q=-1Q=+1
Harmonic fit
= 8.291 = 9.352
Get minimum at Q=-0.887Emin = -3.676
28© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
QEq parameters
29© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Interpretation of J, the hardness
Define an atomic radius as
H 0.84 0.74C 1.42 1.23N 1.22 1.10O 1.08 1.21Si 2.20 2.35S 1.60 1.63Li 3.01 3.08
RA0 Re(A2) Bond distance of
homonuclear diatomic
Thus J is related to the coulomb energy of a charge the size of the atom
30© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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
or
Taking the derivative with respect to charge leads to the chemical potential, which is a function of the charges
The definition of equilibrium is for all chemical potentials to be equal. This leads to
31© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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
32© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
The QEq Coulomb potential lawWe 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 overlapClearly this form as the problem that JAB(R) ∞ as R 0In fact the overlap of the orbitals leads to shielding The plot shows the shielding for C atoms using various Slater orbitals
And = 0.5 Using RC=0.759a0
33© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
QEq results for alkali halides
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QEq for Ala-His-Ala
Amber charges in
parentheses
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QEq for deoxy adenosine
Amber charges in
parentheses
36© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
QEq for polymers
Nylon 66
PEEK
37GODDARD - MSC/CaltechGODDARD - MSC/Caltech Berlin 3/22/04Berlin 3/22/04
Four universal parameters for each element:Get from QM
Polarizable QEq
)||exp()()(
)||exp()()(
2
2
23
23
si
si
si
si
ci
ci
ci
ci
rrQr
rrQrsi
ci
Allow each atom to have two charges:A fixed core charge (+4 for Si) with a Gaussian shapeA variable shell charge with a Gaussian shape but subject to displacement and charge transferElectrostatic interactions between all charges, including the core and shell on same atomAllow Shell to move with respect to core, to describe atomic polarizabilitySelf-consistent charge equilibration (QEq)
ci
si
ci
oi
oi qRRJ &,,,
38© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Noble gas dimers
Ar2
Re
De
Most popular form: LJ 12-6E=A/R12 –B/R6
= De[12 – 26] where = R/ReDe[12 – 6] where R/Here = Re(1/2)1/6 =0.89 Re
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
39© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
van der Waals interaction
6, 8, 10,
6 8 10( ...)ij ij ij
dispi j ij ij ij
C C CE
R R R
What vdW interactions account for?
Short range repulsion Overlap of orbitals and the Pauli Principle
Dipole-dipole Dipole-quadrupole Quadruole-quadrupole
We often neglect higher order (>R6) terms.
(where 0)
1or (where 9)
ijCR
repi j
rep Ni j ij
E e C
E NR
Long range attraction Instantaneous multipolar interaction (London dispersion)
Pauli repulsion:
Born repulsion:
40© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Popular vdW Nonbond Terms: LJ12-6NonBond
RLennard-Jones 12-6
E(R)=A/R12 – B/R6
=Dvdw{1/12 – 2/6} where = R/Rvdw
= 4Dvdw{[vdw/R)12]- [vdw/R)6]}
Dvdw=well depth,
Rvdw = Equilibrium distance for vdw dimer
= point at which E=0 (inner wall, ~ 0.89 Rvdw)
Dimensionless force constant = {[d2E/d2]=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 inner wall.
41© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Popular vdW Nonbond Terms- Exponential-6
Buckingham or exponential-6E(R)= A e-CR - B/R6 which we write as
NonBond
R
whereR0 = Equilibrium bond distance; = R/R0 is the scaled distanceDo=well depth = dimensionless parameter related to force constant at R0
We define a dimensionless force constant as = {(d2E/d2)=1}/D0 = 72 for LJ12-6= 12 leads to –D0/6 at long R, just as for LJ12-6= 13.772 leads to = 72 just as for LJ12-6A 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 >10 this point is well up the inner wall and not important
42© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Converting between X-6 and LJ12-6
Usual convention: use the same R0 and D0 and set = 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
43© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
Popular vdW Nonbond Terms: MorseNonBond
RE(R) = D0 {exp[-R-R0)] - 2 exp[(-(R-R0)]}At R=R0, E(R0) = -D0
At R=∞, E(R0) = 0D0 is the bond energy in Kcal/molR0 is the equilibrium bond distance is the Morse scaling parameter
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
2 3 4 5 6
D0R0
I prefer to write this asE(R) = D0 [X2 - 2X] where X = exp[(/2)(1-)]At R=R0, = 1 X = 1 E(R0) = -D0
At R=∞, X = 0 E(R0) = 0 = {(d2E/d2)=1}/D0 = 2/2Thus = 12 = 72 just as for LJ12-6Few 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 systems there will be other atoms in between. So Morse is ok.
44© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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)
IJ = (IJ + IJ)/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)
45© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
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 included in the nonbond interactions.
Thus normally we exclude from the vdw and coulomb sums the contributions from bonds (1-2) and next nearest neighbor (1-3) interactions
46© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03
1-4 Nonbond exclusions
There is disagreement about 1-4 interactions. In fact the original 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.
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.