> Thanks for the link. As far as I understood, the vdw parameters r* and e
> are calculated in a particular way in AMBER force field such that the
> TIP3P oxygen is the 'reference' element/atom type in the vdw
> parameterization of cations. I am trying to convert some of the parameters
> from a paper (Langmuir 20, 3138, 2004) to AMBER format and would like to
> ask some question:
>
> The vdw energies in that paper is calculated using the following equation:
>
> E_vdw = D0 * [(R0/R)^12 - 2(R0/R)^6].
>
> And for sulphur (S), the vdw parameters are D0=0.25 kcal/mol and
> R0=3.98744 A. These parameters are from FFII force field, according to the
> paper.
It seems you can't have E_vdw for only 1 atom. Perhaps this is for S-S?
> The equation representing AMBER vdw energy is similar to the following
> equation (defined in http://ambermd.org/Questions/vdw.html)
>
> E_vdw_amber = e * [(r*_ij/r)^12 - 2 * (r*_ij/r)^6]
>
> but r*_ij is not what is defined in the AMBER .dat files, which is I think
> the crucial point here. For example, R0 for S is 3.98744 A in FFII, but if
> I want to use these parameters in the AMBER format, I have to set r* for S
> to be 3.98744/2, not 3.98744, is this right?
Amber has R*i for each atom type in the .dat file.
R*ij is for a pair of atom types.
Usually a given R*i should work for that atom i combined with
any other atom type j. However, for water params that involve
a single large vdw on O to enclose H's too, one needs to come
up with a R*i for the ion that allows for this. (Note that ion
params derived like this do not make sense for combining with
ordinary atoms, hence ions embedded in solute should ideally
have different vdw from the same ion in water.)
> R0 is the interatomic distance between two atoms
I think it's the _equilibrium_ distance between 2 atom types.
> while r* has a different meaning.
I'm suspecting R*ij = R0, maybe if type i == type j.
> If the atom type in interest is a cation, r* is [ (minimum energy
> distance) - 1.768 ] (which is the vdw radius of oxygen in TIP3P water).
That would be r*ion; r*ij would be equilibrium distance between
ion and O.
> If
> the atom in interest is not a cation, then it represent (minimum energy
> distance)/2. Thanks again.
The half-distance only applies when the atom types are the same.
Bill
>
> Best regards,
>
> PS: Jason, thanks for the .pdf file.
>
> Ilyas Yildirim, Ph.D.
> -----------------------------------------------------------
> = Department of Chemistry - 2145 Sheridan Road =
> = Northwestern University - Evanston, IL 60208 =
> = Ryan Hall #4035 (Nano Building) - Ph.: (847)467-4986 =
> = http://www.pas.rochester.edu/~yildirim/ =
> -----------------------------------------------------------
>
>
> On Mon, 4 Oct 2010, Bill Ross wrote:
>
> > Here is more on the L-J formulas:
> >
> > http://ambermd.org/Questions/vdw.html
> >
> > Bill
> >
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> >
>
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Received on Mon Oct 04 2010 - 18:00:03 PDT
