# RE: About the correction of charging two ion water box by using FEP

From: JC Gumbart (gumbart_at_ks.uiuc.edu)
Date: Sat Jun 02 2012 - 07:43:15 CDT

Hmm…if you instead assume that you are going from total charge 2 to charge
0, the formula works perfectly (2^2-0^2 = 4). So I think you can’t neglect
ion1-ion2 periodic interactions after all. Read the derivation in Hummer et
al. (1996) very carefully and see if it makes sense to you.

From: owner-namd-l_at_ks.uiuc.edu [mailto:owner-namd-l_at_ks.uiuc.edu] On Behalf
Of Win Liu
Sent: Saturday, June 02, 2012 3:34 AM
To: JC Gumbart
Cc: namd-l_at_ks.uiuc.edu
Subject: Re: namd-l: About the correction of charging two ion water box by
using FEP

Hi JC,

I see your point, but the data from the FEP didn't show this. I do the FEP
for two knids of box sizes, 70A and 30A. The two sodiums are fixed with 8A
apart in center of the box. The FEP value for them are 201.1 and 164.3
kcal/mol, the corrections for each box size are 13.4 and 32.4(according to
your formula which accounts for the self energy). Then the free energy for
charging such two ions are 201.1+13.4=214.5 which is quite different with
164.3 +32.4=196.7, So the corrections should not be the formula like this(in
FEP module of NAMD)? Am I right? Thanks.

The best wishes.

Wenhao

2012/5/29 JC Gumbart <gumbart_at_ks.uiuc.edu>

Thinking about it again, it probably doesn't matter. The self-energy is a
sum over individual charges, so I think the correction for two separate
charges should be 2*(1/2*\xi_EW (1^2-0^2) = \xi_EW. Of course, you can
check this yourself.

I would use alchDecouple on, in which case I think the two ions will
interact with each other, but these interactions will not change during the
simulation so they won't contribute to the free energy. But really, this
seems to be a bit of an odd simulation, so I recommend thinking very
carefully about what you should expect to happen.

On May 28, 2012, at 11:59 AM, Win Liu wrote:

Hi JC,

Thanks for the reply.
2012/5/28 JC Gumbart <gumbart_at_ks.uiuc.edu>

Read the paper of Hunenberger and McCammon carefully: they say that the
inter-ion pmf is not affected dramatically by artifacts, which happen to
cancel.

I see your point.

My intuition is that you will have self-energy corrections for ion1 with
ion1, ion2 with ion2, and ion1 with ion2.

Also the ion interaction between ion1 and ion2? Do you mean the interaction
between ion1 and the ion2s from other boxes? In addition, in my case, what
the parameter alchDecouple should be, on or off? If I set the alchDecouple
on, the FEP will not include the Coulomb interaction between ion1 and ion2
in one box, but will include the self energy. Is my understanding correct?

Thanks a lot.

Wenhao

From: owner-namd-l_at_ks.uiuc.edu [mailto:owner-namd-l_at_ks.uiuc.edu] On Behalf
Of Win Liu
Sent: Monday, May 28, 2012 2:06 AM
To: namd-l_at_ks.uiuc.edu
Subject: namd-l: About the correction of charging two ion water box by using
FEP

Hi NAMD people,

I tried to charge two sodium atoms into two ions in water box by using FEP
module. For one ion system, as written in FEP tutorial, there is a formula
to estimate the artifatcs caused by the finite box size(PME artifacts) which
was derived and explained in Gerhard Hummer's paper. So how about the
artifacts in such two ions system? Actually, I read a paper written by
Philippe H. Hünenberger and J. Andrew McCammon, J. Chem. Phys. 110, 1856
(1999), in it, they concluded that in two identical ions system, artifacts
caused by PME and PBC will be surprisingly almost 0. So is this means that
the box size will not affect the value of free energy change in such
process? But I tried the 30A and 70A size, the values are quite different.
So can anyone point out my fault? Thanks a lot.

The best wishes.

Wenhao

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