From: J.Dziedzic <J.D<ziedzic@so...>  20091215 13:59:01

Gernot Kieseritzky wrote: > Hi! > > On Mon, 20091214 at 10:35 +0000, J.Dziedzic wrote: >> Below I'm listing the \Delta{}G values produced by your script >> along with the value of the total electrostatic energy in sdie=1, >> pdie=\epsilon, obtained after \Delta{}G to the E1 value calculated above. > > Shouldn't it be sdie=\epsilon and pdie=1? Otherwise \Delta{}G would be > the solvation energy of set of point charges in a small \epsilon=80 > bubble surrounded by vacuum. Sorry, a typo on my part. I got it right in the input file, so the calculation was for pdie=1.0 and varying sdie, as it should be. >> \epsilon \Delta{}G E in dielectric of \epsilon >> 80 27119 8785 >> 50 26905 8571 >> 30 26526 8192 >> 10 24650 6316 >> 5 21865 3531 >> 4 20480 2146 >> 3 18181 153 >> 2 13606 4728 >> 1 0 18334 >> >> What is the reason for my getting negative energies? Could that >> be that the dielectric is polarizing sooo much? > > Well, it would suggest that the solvation energy in absolute values is > larger than the Coulomb repulsion of your system. Sounds kind of counter > intuitive with point charges of +6 and +1 at close distance. But I have > the feeling you mixed up sdie and pdie (see above). I mixed them up only when typing the email, the input file was fine. >> On a side note  I thought of a simple approach to remove the grid >> artefact, can you comment on the feasibility? >> >> 1) Output the charge distribution to an OpenDX file. >> 2) Locate all 3x3x3 pockets of isolated charge floating in a sea of >> zero charge. >> 3) Calculate the Coulomb sum for the interaction of these 27 charges. >> Since they will always be in a cavity, epsilon will be that of pdiel >> and constant. Repeat for all 3x3x3 charge pockets. >> 4) Subtract the obtained result from what APBS produces. >> >> This assumes that all point charges are separated by at least one >> grid point of zero charge and that chgm spl2 is used. >> >> Is this feasible? > > Maybe. I have tried this approach. It doesn't really work, because there is a shift in the potential by a constant, I believe. APBS uses the qphi integration and my approach uses a sum over pairs of q's. When phi is shifted, the results do not match. Is there any way to determine the constant term in the potential? > But in the end it's easier (and faster?) to simply subtract the > homogenious APBS result. I agree. I would have used the approach you have suggested, but I was worried about the negative results I got for the three point charges. So, do you think this is a valid result, just counterintuitive? I am also worried that with this technique the result will depend on the precise dimensions of the cavity. > If electrostatics is really timelimiting and > the grid artefact creates a serious problem in your application you > should consider using a Generalized Born approach. The currently fastest > implemention comes from the group of Prof. Caflisch: > > http://www3.interscience.wiley.com/journal/116327343/abstract?CRETRY=1&SRETRY=0 > > and is part of the latest CHARMM version 35. Thank you. The electrostatic calculation is not a time bottleneck in my case, far from it. It's just that I need to be able to get reliable values for electrostatic energy for a set of point charges embedded in an inhomogenous dielectric. I have another doubt as well. When a point charge is spilled onto the grid, the grid artefact occurs because all of these grid points are now in the potential generated by their own collection of charges. This artefact is meant to cancel out when calculating \Delta{}G, because a similar scenario occurs in the inhomogeneous dielectric case. However, as the boundary conditions are the same for both cases, the addition of the dielectric around the cavity now implies a subtle change in the potential in the cavity to accomodate the fact, that the same boundary conditions must be satisfied. Would this not change the grid artefact energy subtly, leading to the cancelingout being not 100% thanks,  Jacek Dziedzic 