Could we post an EXAMPLE SCRIPT of how the corrections suggested by
Gernot and Nathan for Dziedzic's simple calculation of two point charges
should be implemented? Does this mean that the grid artifact subtraction
should be carried out always in all the APBS calculations? Perhaps, this
occurs already by default, and I did not realize it.
Biochemistry & Mol. Biology
Wayne State University School of Medicine
540 E. Canfield Avenue
Detroit, MI 48201
Tel: 313-577-0620 or 313-993-4238
Hi All --
Gernot is absolutely correct. I would also add that, after correcting the
issues Gernot raised below, you should also examine the sensitivity of your
results on "chgm spl0" vs. "chgm spl2" since charge discretization can
affect these types of calculations as well.
On Dec 10, 2009, at 9:31 AM, Gernot Kieseritzky wrote:
> On Wed, 2009-12-09 at 17:49 +0000, J.Dziedzic wrote:
>> I am confused about the result of a very simple calculation when done
>> with APBS. Consider a trivial system of two point charges with unit
>> charges at a separation of 1 Bohr length, in vacuum. The Coulombic
>> energy of this system is exactly 1 Hartree, that is 2625.5 kJ/mole.
>> ... APBS yields 31740 kJ/mole, which off by a factor of 12. Making the
>> grid finer only makes things worse, the results being:
>> dime Energy (kJ/mole)
>> 65 1.209E04
>> 129 2.222E04
>> 193 3.174E04
>> 257 4.058E04
>> 289 4.539E04
>> which are all way off, by a factor of 5-20, from the correct value of
>> 2625.5 kJ/mole.
>> I understand that normally one is interested in energy differences
>> between a system in vacuo and a solvated system, and any discretization
>> errors introduced are canceled if the grid is the same in both
>> calculations. Yet, with a system so trivial, without any dielectric
>> at all and, thus, without the arbitrariness of the cavity, what is the
>> underlying reason for the calculation being so off from the mark?
> Two words: grid artefact! Basically, the deviation is not due to
> numerical problems, rather the high energy values you observe are the
> result of the self-interaction of the grid points. That's why the
> deviation is increasing with higher resolution as the grid points are
> getting closer. What you have to do to get the total electrostatic
> energy of your system without self-energies:
> 1) Compute the Coulomb energy in the homogeneous continuum.
> 2) Compute the solvation energy of the same charge distribution using
> APBS. The grid artefact cancels as you calculate an energy difference
> here. This, of course, requires that you use the same grid setup in both
> APBS runs.
> 3) Add the values together.
> Best regards,
> Gernot Kieseritzky
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