[gmx-users] Free energy, frozen atoms and exclusions

[Ignacio Fernández Galván]

Hi all,

I'm trying to calculate solvation free energies and I've found
something "strange".

The thing is I'm interested in free energy of rigid molecules (I
consider other contributions separately), typically a small solute
molecule in water. I'm making a test with methanol. The setup is a
frozen methanol molecule in a box of of water molecules. Since I'm
considering rigid molecules, I create an .itp file with no bonded
terms, just atoms (with charges and LJ). The methanol molecule is
called MOH, and I have this in the .mdp:

energy_grps          = MOH
energygrp_excl       = MOH MOH
freezegrps           = MOH
freezedim            = y y y

which should freeze the methanol and disregard all nonbonded terms
between the methanol atoms (and there are no bonded terms).

Well, after the usual equilibration and such, I get this output for the
initial configuration with lambda=1:

LJ (SR): 5.70341e+03
Coulomb (SR): -3.65377e+04
...
dVpot/dlambda: 2.81161e+02

It seems right, but after simulation for several lambda values I get a
free energy of around 10 times what I would expect.

Now I repeat exactly the same simulation, but adding all pairs of
methanol atoms in the [exclusions] of the .itp file. For the same
initial configuration as before, I get this:

LJ (SR): 5.70341e+03
Coulomb (SR): -3.65377e+04
...
dVpot/dlambda: 2.14385e+00

(the only difference is in dVpot/dlambda). This (if I extrapolate)
would give a much more sensible value for the free energy.

As far as I could understand, having "energygrp_excl = MOH MOH" should
be the same as putting all atoms in the [exclusions] section, but it
obviously isn't, at least when it comes to calculating dVpot/dlambda.

Is there something I'm missing? A bug/feature in the code? A
limitation?

Thanks in advance

PS. As a side question, I'd say that, if I have a frozen molecule it
would be best *not* to remove the center of mass motion (not even of
the rest of the system). Am I right?


		
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