[gmx-users] changing ewald_rtol

[Johnny Lu]

Thank you very much for the answer.

On Tue, Nov 18, 2014 at 2:03 PM, Mark Abraham <mark.j.abraham at gmail.com>
wrote:

> On Tue, Nov 18, 2014 at 7:38 PM, Johnny Lu <johnny.lu128 at gmail.com> wrote:
>
> > Hi.
> >
> > from online manual:
> >
> > "*ewald-rtol (1e-5)* The relative strength of the Ewald-shifted direct
> > potential at *rcoulomb* is given by *ewald-rtol*. Decreasing this will
> give
> > a more accurate direct sum, but then you need more wave vectors for the
> > reciprocal sum."
> >
> > Is it ok to use ewald-rtol=1e-9 for Amber99SB-ILDN forcefield?
> >
>
> That's not a question that has a useful single answer.
> 1) It wasn't parameterized with any Ewald method, so any subsquent use of
> that method is valid only upon empirical evidence.
> 2) Different observables in different systems have different dependencies
> on the quality of the electrostatics model, and this one parameter is only
> part of any story.
> 3) A tiny magnitude of forces at the cutoff matters little if they're being
> added to other forces at shorter distances - and if there's more than about
> 5 orders of magnitude difference, then the small ones get lost in the
> accumulation in floating-point representation
> 4) Making one part of the electrostatics approximation super accurate while
> ignoring the other doesn't do any good when you add them together, because
> the largest error dominates.
> 5) Some other error might be more dominant still (e.g. incomplete sampling,
> interior residues have the same charges as exterior residues, no
> polarizability)
>
> Will the sum in reciprocal space become very inaccurate?
> >
>
> Yes, like the docs say, because ewald-rtol and rcoulomb are used to compute
> the Ewald splitting parameter beta, which is then used in reciprocal space
> also.
>
> Or this will just make the simulation run slower?
> >
>
> By itself, it's just multiplication by a different scaling factor. Fixing
> the issues in reciprocal space (increasing number of grid points and/or
> spline interpolation order) will be expensive.
>
> Mark
>
>
> > Thanks again.
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