Hi Ignacio,
> associated with that process. By that, I mean the energy necessary to
> deconstruct, block by block, the original system (of course, there is
> something strange here that I am ignoring by the moment, because I am
> assuming the system is in an ensemble with constant temperature, but
> what's
I guess whats puzzling me here is in what thermodynamic cycle you would
include this free energy of transformation to get something that's
comparable to an experimental value.
> In terms of the TI integral equation, what I want is the integral from 0
> (whole system) to 1 (only a couple of atoms) of <dv/dl>, where lambda (in
> some form) is applied to every interaction of the original system, so that
> the system is effectively deconstructed.
>
> So, the question is: if I set up V0 as the whole system and V1 as a couple
> of atoms (from the original ones) and use ifsc=1, will sander give the
> dv/dl
> corresponding to what's above?
If you set most of your system to be unique to V0 and keep a few atoms as
common in both V0 and V1 then you can run such a calculation. Since that
is a very big change, you will probably have some convergence issues.
However, this is still not equal to removing the complete system, as the
disappearing part will keep its internal interactions intact. Otherwise,
e.g. bonds would get softer and softer with increasing lambda and in the
end your system consists of just non-interacting particles, in which case
your convergence problems would be even worse. So, in a sense, you can
never really disappear something in TI, you can only remove interactions
between some parts of your system.
Kind Regards,
Dr. Thomas Steinbrecher
BioMaps Institute
Rutgers University
610 Taylor Rd.
Piscataway, NJ 08854
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Received on Wed May 20 2009 - 12:32:33 PDT
