From: Gatti, Domenico <dgatti@me...>  20091211 15:54:10

Hi All, 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. Best, Domenico Domenico Gatti Biochemistry & Mol. Biology Wayne State University School of Medicine 540 E. Canfield Avenue Detroit, MI 48201 Tel: 3135770620 or 3139934238 Fax: 3135772765 dgatti@... 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. Thanks, Nathan On Dec 10, 2009, at 9:31 AM, Gernot Kieseritzky wrote: > Hi! > > On Wed, 20091209 at 17:49 +0000, J.Dziedzic wrote: >> Hi! >> >> 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 520, 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 selfinteraction 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 selfenergies: > > 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 
From: Nathan Baker <baker@bi...>  20091211 17:22:31
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Hello  Actually, all of the input files provided in the APBS examples/ directory demonstrate this in a variety of ways. Some of the examples remove selfenergies by calculating solvation energies while other examples remove selfenergies by using exactly the same grid positions for all atoms in each part of the calculation. Hope this helps, Nathan On Fri, Dec 11, 2009 at 9:23 AM, Gatti, Domenico <dgatti@...>wrote: > Hi All, > 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. > Best, > Domenico > > > Domenico Gatti > Biochemistry & Mol. Biology > Wayne State University School of Medicine > 540 E. Canfield Avenue > Detroit, MI 48201 > Tel: 3135770620 or 3139934238 > Fax: 3135772765 > dgatti@... > > > > > > > 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. > > Thanks, > > Nathan > > On Dec 10, 2009, at 9:31 AM, Gernot Kieseritzky wrote: > > > Hi! > > > > On Wed, 20091209 at 17:49 +0000, J.Dziedzic wrote: > >> Hi! > >> > >> 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 520, 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 selfinteraction 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 selfenergies: > > > > 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 > > > > >  > Return on Information: > Google Enterprise Search pays you back > Get the facts. > http://p.sf.net/sfu/googledev2dev > _______________________________________________ > apbsusers mailing list > apbsusers@... > https://lists.sourceforge.net/lists/listinfo/apbsusers >  Nathan Baker (http://bakergroup.wustl.edu) Associate Professor, Dept. of Biochemistry and Molecular Biophysics Director, Computational and Molecular Biophysics Graduate Program Center for Computational Biology, Washington University in St. Louis 
From: Gernot Kieseritzky <gernotf@ch...>  20091211 17:43:27

Hi! Based on Dziedzic's example: read mol pqr test.pqr end elec name test_inhom mgauto dime 65 65 65 nlev 4 cglen 50 50 50 fglen 12 12 12 cgcent mol 1 fgcent mol 1 mol 1 lpbe bcfl mdh pdie 1.0 sdie 80.0 chgm spl2 srfm smol swin 0.3 srad 1.4 sdens 10.0 temp 298.15 calcenergy total calcforce no end elec name test_hom mgauto dime 65 65 65 nlev 4 cglen 50 50 50 fglen 12 12 12 cgcent mol 1 fgcent mol 1 mol 1 lpbe bcfl mdh pdie 1.0 sdie 1.0 chgm spl2 srfm smol swin 0.3 srad 1.4 sdens 10.0 temp 298.15 calcenergy total calcforce no end print energy test_inhom  test_hom end quit where test.pqr is ATOM 1 H XXX A 1 0.000 0.000 0.000 1.000 1.000 ATOM 2 H XXX A 1 0.529 0.000 0.000 1.000 1.000 Result (APBS 1.1): test_inhom  test_hom = 2497.9 kJ/mol. For the total electrostatic energy we've got to add the Coulomb energy with sdie=pdie=1. In his example we have: Ec = 1 Hartree = 2625.5 kJ/mol. (You can also use the "coulomb.c" program in the APBS tools directory of the source tar ball). So we obtain for the total electrostatic energy with sdie=80 and pdie=1: E = 2625.5  2497.9 kJ/mol = +127.6 kJ/mol. When sdie approaches 1, E should tend to the Coulomb's law result 2625.5 +/ numerical error (in the order of 1E13), because test_inhom  test_hom vanishes in the limit. Best regards, Gernot On Fri, 20091211 at 10:23 0500, Gatti, Domenico wrote: > Hi All, > 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. > Best, > Domenico > > > Domenico Gatti > Biochemistry & Mol. Biology > Wayne State University School of Medicine > 540 E. Canfield Avenue > Detroit, MI 48201 > Tel: 3135770620 or 3139934238 > Fax: 3135772765 > dgatti@... > > > > > > > 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. > > Thanks, > > Nathan > > On Dec 10, 2009, at 9:31 AM, Gernot Kieseritzky wrote: > > > Hi! > > > > On Wed, 20091209 at 17:49 +0000, J.Dziedzic wrote: > >> Hi! > >> > >> 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 520, 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 selfinteraction 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 selfenergies: > > > > 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 > > > >  > Return on Information: > Google Enterprise Search pays you back > Get the facts. > http://p.sf.net/sfu/googledev2dev > _______________________________________________ > apbsusers mailing list > apbsusers@... > https://lists.sourceforge.net/lists/listinfo/apbsusers 