From: Jérôme Hénin (jerome.henin_at_uhp-nancy.fr)
Date: Sat Nov 15 2003 - 08:30:50 CST
> I am sorry about I have made a mistake. I mean abouste hydration of a
> molecule, which mean transfer a molecule from vacuum to aqueous solutions.
No problem, actually it was in the title of your message but I hadn't seen
> Can you give me some example files about it?
I'll send you an example setup for methanol in a separate message. If anybody
on the list is interested, just ask !
The idea behind the calculation is the following :
To compute solvation free energies, you'll need to use a thermodynamic cycle :
MeOH (vacuum) ----> MeOH (water)
MeOH* (vacuum) ----> MeOH* (water)
where MeOH* is a "vanished" methanol molecule, with all nonbonded interaction
terms switched off. Let us call the solvation free energy Asolv, and let A*
be the free energy of anihilation.
The thermodynamic cycle tells that
Asolv(MeOH) - Asolv(MeOH*) = A*(MeOH, vacuum) - A*(MeOH, water)
Since MeOH* doesn't interact with anything, it has zero free energy of
solvation, so Asolv(MeOH*) = 0. The free energy we are looking for is thus :
Asolv(MeOH) = A*(MeOH, vacuum) - A*(MeOH, water)
So we have to setup two alchemical transformations : "methanol to nothing" in
vacuum and the same in water.
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