## [Apbs-users] trivial calculation in vacuum

 [Apbs-users] trivial calculation in vacuum From: J.Dziedzic - 2009-12-09 17:49:48 ``` 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. When presented with the same system: REMARK 1 Two hydrogen atoms one Bohr apart ATOM 1 H XXX 1 0.000 0.000 0.000 1.00 1.00 ATOM 2 H XXX 1 0.529 0.000 0.000 1.00 1.00 and a simple input file: read mol pqr test.pqr end elec name test mg-auto dime 193 193 193 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 elecEnergy test end quit ... 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? Thank you in advance, - J. Dziedzic ```

 [Apbs-users] trivial calculation in vacuum From: J.Dziedzic - 2009-12-09 17:49:48 ``` 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. When presented with the same system: REMARK 1 Two hydrogen atoms one Bohr apart ATOM 1 H XXX 1 0.000 0.000 0.000 1.00 1.00 ATOM 2 H XXX 1 0.529 0.000 0.000 1.00 1.00 and a simple input file: read mol pqr test.pqr end elec name test mg-auto dime 193 193 193 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 elecEnergy test end quit ... 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? Thank you in advance, - J. Dziedzic ```
 Re: [Apbs-users] trivial calculation in vacuum From: Gernot Kieseritzky - 2009-12-10 15:59:26 ```Hi! On Wed, 2009-12-09 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 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 ```
 Re: [Apbs-users] trivial calculation in vacuum From: Nathan Baker - 2009-12-10 21:35:10 ```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, 2009-12-09 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 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 > > > ------------------------------------------------------------------------------ > Return on Information: > Google Enterprise Search pays you back > Get the facts. > http://p.sf.net/sfu/google-dev2dev > _______________________________________________ > apbs-users mailing list > apbs-users@... > https://lists.sourceforge.net/lists/listinfo/apbs-users -- 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 ```