AMBER Archive (2009)
Subject: Re: [AMBER] RE: VDW and TI for free energy
From: Qicun Shi (qshi01_at_gmail.com)
Date: Sat Oct 10 2009 - 23:05:20 CDT
Thanks for analyzing in detail. I am getting to the point of such a
cancelling effect between the two half cycles.
On Sat, Oct 10, 2009 at 4:44 AM, <steinbrt_at_rci.rutgers.edu> wrote:
> Hi Qicun,
> > *(1) For OH all FF9X VDW radius is zero, including FF02 or latest
> > right?*
> the hydrogen type HO in ff99 (and later ones I believe) has a vdw radius
> of zero. If you want to make sure, take a look at the parmXX.dat you load
> when building your system, it has the vdW-parameters at the end (the files
> are in $AMBERHOME/dat/leap/parm/).
> > *(2) Generally, is it right, using R-XH as an example, that*
> > * disappearing charge R-XH -> R-X^-(p0) is different from *
> > * disappearing atom (charge + mass) R-XH -> R-X(^-)*
> > * (p0 is proton mass with zero charge and **X^- an anion of X)?*
> In the limit of enough sampling, having the chargeless, interactionsless
> mass p0 attached to an atom should not change your system behaviour. You
> would however get different free energy results from the two processes you
> describe, because in one of them, you also remove the bond potential of
> the H-X bond from the system. This is a contribution that should cancel
> out if you do the other leg of your free energy cycle the same way though.
> So in principle the two ways to remove a hydrogen should give the same
> final answer, but possibly different intermediate results.
> Dr. Thomas Steinbrecher
> BioMaps Institute
> Rutgers University
> 610 Taylor Rd.
> Piscataway, NJ 08854
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