[gmx-users] workaround for PME coulomb energy (fwd)
David L. Bostick
dbostick at physics.unc.edu
Tue Nov 26 17:40:37 CET 2002
Hello David,
Thanks for your help with the workaround. I have an additional question
pertaining to your last reply below:
> E(lr) = E(lr) (ion-ion) + E(lr) (ion-rest) + E(lr) (rest-rest)
> Setting the ion charge to zero will give you E(lr) (rest-rest)
> setting the rest to zero will give you E(lr) (ion-ion)
> but if you have only one ion that term will be zero. Do check the charge
> correction to PME, as that may be different when the total charge
> changes.
How do I check the charge correction to PME?
Thanks,
David
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David Bostick Office: 262 Venable Hall
Dept. of Physics and Astronomy Phone: (919)962-0165
Program in Molecular and Cellular Biophysics
UNC-Chapel Hill
CB #3255 Phillips Hall dbostick at physics.unc.edu
Chapel Hill, NC 27599 http://www.unc.edu/~dbostick
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On 23 Oct 2002, David wrote:
> On Wed, 2002-10-23 at 17:44, David L. Bostick wrote:
> >
> > Hello all,
> >
> > I am interested in using David van der Spoel's method for calculating the
> > total coulomb energy of a single ion with the rest of the system. I just
> > want to ask a few questions so I can feel comfortable with what I'm doing.
> >
> > 1)If I take the original .edr file from the simulation, is the total
> > coulomb energy of the system Coulomb (SR) + Coulomb (LR) as extracted by
> > g_energy? That is.. Coulomb 1-4 doesn't really factor in does it?
> In this case it is constant, yes.
>
> >
> > 2)Then if I set the ion charge to zero and rerun to generate a new .edr
> > file and calculate the total coulomb energy again as Coulomb (SR) + Coulomb
> > (LR), this will be the total PME energy for System_&_!ion-System_&_!ion?
> Coulomb (SR) you can do in groups anyway, so you can get that more
> accurate from the original run.
>
> > 3)Then I can subtract the result of #2 from the result of #1 and obtain the
> > total coulomb PME energy of System_&_!ion-ion? Since there is only one
> > ion, the coulomb energy of ion-ion is zero so I don't have to worry about
> > it.
> E(lr) = E(lr) (ion-ion) + E(lr) (ion-rest) + E(lr) (rest-rest)
> Setting the ion charge to zero will give you E(lr) (rest-rest)
> setting the rest to zero will give you E(lr) (ion-ion)
> but if you have only one ion that term will be zero. Do check the charge
> correction to PME, as that may be different when the total charge
> changes.
>
> > 4)If I write the total ewald sum in the traditional way,
> >
> > E = E(real) + E(reciprocal) + E(self) + E(dipole_correction)
> >
> > is the coulomb energy as extracted by g_energy as follows:
> >
> > Coulomb (SR) = E(real)
> >
> > and
> >
> > Coulomb (LR) = E(reciprocal) + E(self) + E(dipole_correction) ?
> >
> > How exactly are these terms divided among the two quantities Coulomb(SR)
> > and Coulomb(LR)?
> Coulomb(LR) = E(reciprocal) + E(self) + E(dipole_correction)
> Coulomb(SR) = E(real)
>
>
> Groeten, David.
> ________________________________________________________________________
> Dr. David van der Spoel, Biomedical center, Dept. of Biochemistry
> Husargatan 3, Box 576, 75123 Uppsala, Sweden
> phone: 46 18 471 4205 fax: 46 18 511 755
> spoel at xray.bmc.uu.se spoel at gromacs.org http://zorn.bmc.uu.se/~spoel
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