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
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)
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, ..
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
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°
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.
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°.
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.
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.
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
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)
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)
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
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)
= 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
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
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
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.
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
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.
