[gmx-users] more free energy questions

[Ilya Chorny]

The balls are frozen so the configuration does not change. Only the
water configuration could change. 50 kj/mol seems crazy because if each
ball cost 50 kj/mol the total would be 250 not 99 as was obtained from
the first simulation. 

On Tue, 2003-11-11 at 13:16, David wrote:
> On Tue, 2003-11-11 at 21:09, Ilya Chorny wrote:
> > Hello,
> > 
> > 	I ran a simulation to calculate the solvation free energy between soft
> > spheres where the potential energy is (v=kt(3/r)^12 and six of the same
> > spheres fused together at a distance of three angstroms. I ran the
> > simulation at constant NVT. My box contains about 4000 H2O molecules and
> > is 5 X 5 X 5 nm^3. The used ten windows used lambda values
> > (0,.1,.2,.3,.4,.5,.6,.7,.8,.9,10) starting with one ball at lmabda = 0
> > and ending up with 6 balls at lambda equals 1. I get a reasonable result
> > of about 97 kj/mole. Previous work has shown the solvation free energy
> > of one ball is about 20 kj/mol and I would expect six balls fused
> > together to be less. I then take the six ball and mutate them to to five
> > balls  using just one lambda value of .5. Doing this I get crazy results
> > (50-60 kj/mol). Any ideas what is going on? Any help would be greatly
> > appreciated. Statistical convergence tests were used for all the
> > simulations
> Is it really tha crazy? It could be due to a particular conformation of
> the balls.
> 
> > 
> > P.S. I noticed in my first simulation that the value I got at lambda =
> > .5 was about the same as my overall result.
> could be coincidence.
> 
> > 
> > 
> > Ilya