AMBER Archive (2005)

Subject: Re: AMBER: Comparing EELEC from amber and that calculated in the trivial fashion.

From: Carlos Simmerling (carlos_at_ilion.bio.sunysb.edu)
Date: Mon Nov 28 2005 - 06:41:30 CST


don't forget to add in all of the interactions with the periodic images

Pavan G wrote:

>Hello all,
>
>I am trying to reproduce the value of EELEC for the first time step
>recorded in a md run:
>" NSTEP = 100 TIME(PS) = 1626.100 TEMP(K) = 298.76 PRESS = -18.6
> Etot = -76468.3208 EKtot = 35432.1212 EPtot = -111900.4421
> BOND = 22540.2259 ANGLE = 2334.1316 DIHED = 2373.1196
> 1-4 NB = 749.3074 1-4 EEL = 24187.2168 VDWAALS = 20399.8387
> EELEC = -184484.2821 EHBOND = 0.0000 RESTRAINT = 0.0000
> EKCMT = 10710.9469 VIRIAL = 10864.5298 VOLUME = 383436.0683
> Density = 1.0553
> Ewald error estimate: 0.3583E-05"
>
>This is a run containing ~40,000 atoms. I got the charges of the atoms
>from prmtop file and distances using the coordinates of the mdcrd
>file. Now, the crudest way I could calculate the Electrostatic
>interaction is to pick 2 atoms from the 40,000 and calculate energy
>using
>E = charge.1 * charge.2 / distance
>REF: http://amber.scripps.edu/Questions/units.html
>But this obviously includes atoms which are covalently bonded and have
>hydrogen bonds.
>This gives me a number which is ~12 times the number by amber(EELEC =
>-184484.2821).
>
>Now, where is the mistake. What set of atoms should I consider/not consider ?
>
>Thanks a lot
>Pavan
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