RE: [Apbs-users] factor of 1/2 in coulomb force calculations

[Nathan B. <ba...@bi...>]

I'm not sure if I understand your question.  Various factors of 1/2 appear
if you are double-counting the interactions:

1/2 \sum_i \sum_{j \neq i} u_{ij}

but are not present if you are counting interactions once

\sum_i \sum_{j > i} u_{ij}

There are other, different, factors of 1/2 that arise from the derivation of
the free energy functional.

Thanks,

Nathan

--
Nathan A. Baker, Assistant Professor
Washington University in St. Louis School of Medicine
Dept. of Biochemistry and Molecular Biophysics
Center for Computational Biology
700 S. Euclid Ave., Campus Box 8036, St. Louis, MO 63110
Phone:  (314) 362-2040, Fax:  (314) 362-0234
URL:  http://www.biochem.wustl.edu/~baker  

> -----Original Message-----
> From: apb...@ch... 
> [mailto:apb...@ch...] On Behalf Of 
> sergei grudinin
> Sent: Wednesday, December 08, 2004 10:11 AM
> To: apb...@ch...
> Subject: [Apbs-users] factor of 1/2 in coulomb force calculations
> 
> Dear Nathan,
> 
> I have a question concerning force calculation in coulomb 
> program. It seems to me that calculated forces have a 
> prefactor of 1/2, i.e. in case of two atoms the force acting 
> on each of them is calculated as 
> F=1/2*1/(4*pi*eps0)*q1*q2/r12. Does it have a certain sense?
> 
> Thanks in advance,
> Sergei Grudinin.
> 
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