[gmx-users] Tabulated potentials make newbies crazy
gmx3 at hotmail.com
Tue Dec 1 09:39:33 CET 2009
> Date: Tue, 1 Dec 2009 07:44:51 +1100
> From: Mark.Abraham at anu.edu.au
> To: gmx-users at gromacs.org
> Subject: Re: [gmx-users] Tabulated potentials make newbies crazy
> ms wrote:
> > Mark Abraham ha scritto:
> > > Sorry, I was a bit incomplete last night. Charge groups are the
> >> fundamental unit for neighbour-searching (3.4.2) to construct lists of
> >> charge groups for nonbonded interactions, which determine lists of
> >> atom-atom interactions. However, the nonbonded interactions are
> >> evaluated as nested sums, first over energy groups. So for energy groups
> >> of Protein and SOL, the neighbour search finds all pairs of charge
> >> groups that are both Protein and inside the cutoffs, and lists them.
> >> Then all Protein-SOL similarly, then all SOL-SOL. This requires that the
> >> energy groups be a union of only complete charge groups (and I am not
> >> aware that this is spelled out anywhere in the manual!). So for energy
> >> groups of Calpha, Rest_of_Protein and SOL, it would be necessary to use
> >> an individual charge group for each Calpha. This would usually mean it
> >> has a net non-integral charge that is equal in magnitude of the charge
> >> of the group from which it is taken. It is well known that small charge
> >> groups of non-integral charge can then wander back and forth across
> >> cut-off boundaries and generate artefacts.
> > Ok, thanks for the clarification. This doesn't suggest a trivial
> > solution to the problem, quite the opposite: I understood correctly that
> > charge groups must be neutral, and this is impossible to do if we put
> > each C-alpha as a charge group.
> > I can coarse the thing further -that's quite the plan, actually- and
> > eliminate electrostatics, but I hoped to have a look at what happens
> > with the new potential and getting it right, before going so far.
> > So, even if the following:
> >>> The problem is that since I have a single molecule now, and the single
> >>> molecule must be neutral, so it must be all a single charge group
> >>> ("Therefore we have to keep groups of atoms with total charge 0
> >>> together. These groups are called charge groups.", 4.6.2).
> > is not entirely correct, it is indeed correct that charge groups
> > *should* be neutral. Isn't it?
> Indeed. See 3.4.2 and ref therein.
Charge groups do not have to be neutral at all.
If you use plain cut-off's or a reaction field without buffer, neutral charge
groups will help a lot to reduce cut-off artifacts.
However with PME or reaction-field-zero (with a buffer) the charge of
a charge group does not matter at all.
We should add a few lines to the manual explaining this.
> >> If you're running in single precision, that precision cannot represent
> >> values as large as 10^41. Since in any simulation (but particularly
> >> coarse-grained one) non-bonded atoms aren't going to get this close, the
> >> values are next to irrelevant. Just choose 10^38 for anything larger
> >> than that.
> > Right, it is probably precision problem. Thanks.
> >>>>> Now, my questions are:
> >>>>> - What is the accepted range of values in tables?
> >>>> I don't think this is the problem.
> >>> It is the least problem probably, given my confusion on energy-charge
> >>> groups, but it seems it is too...
> >>>>> - How do I define a steep repulsion potential correctly?
> >>>> It's terse, but manual 6.7 seems to have the necessary information.
> >>> 6.7 is one of the references I am obviously using, but it gives only
> >>> general (even if essential!!) information, nothing speficic on "good" or
> >>> "bad" potential shapes/values. But probably that's the least problem :)
> >> Knowing a sensible shape is your problem, if you're choosing to
> >> unbalance the force field by changing one of the contributing potentials...
> > I meant "sensible" in the meaning of "can be interpolated more or less
> > faithfully / will be calculated with more or less artefacts" -of course
> > if it makes sense in the model is my problem...
> The cubic spline interpolation will do a mighty fine job of any function
> that is suitably continuous provided that the density of points is
> sufficiently fine... interpolating a sinusoid with a density comparable
> with the period would obviously be a disaster!
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