AMBER Archive (2009)Subject: Re: [AMBER] Solvation potential and free energies
From: Carlos Simmerling (carlos.simmerling_at_gmail.com)
Date: Thu Feb 05 2009  06:33:08 CST
this is pretty much beyond what can be explained in email, I suggest
that you look at some review articles on solvation energies for
details. however, it might help to note that the solvation in implicit
models like GB is a free energy because we have integrated over all
solvent degrees of freedom, thus it becomes a potential of mean force.
there really isn't a "solvation potential energy" with implicit
solvent models. With explicit solvent, each snapshot with specific
water coordinates provides a potential energy with zero solvent
entropy, and in order to get solvation free energy one needs to do the
appropriate averaging over all possible water configurations. this is
one reason why implicit models are popular for calculation of free
energies of solvation they do it in 1 step.
regarding your equations, I find it difficult to read them in that
format, perhaps others can comment.
On Thu, Feb 5, 2009 at 7:27 AM, <guardiani_at_fi.infn.it> wrote:
> Dear Amber experts,
>
> I am currently performing REMD simulations
> using the Generalized Born solvent model
> (igb=7). Everything is working very well,
> and I have just a little theoretical question.
>
> In the Amber9 manual I read that, in order to
> attain the GB/surfacearea force field, it is
> sufficient to add to the vacuum force field
> the term (Eq 5.3 of Amber9 Manual):
>
> \sum_{ij}\frac{q_{i}q_{j}}{f^{gb}(R_{ij})} + \sigma*A
>
> This expression, however, is different from the
> one that can be derived from Equations 2 and 3 of
> your paper J. Phys. Chem. B, Vol 104, No 15, 2000.
>
> As I understand, Equation 2 represents the Coulomb
> contribution in a proteinlike environment, and this
> term corresponds to the Coulomb contribution in the
> Amber vacuum forcefield. In order to account for
> the solvation effects, it is then necessary to add
> Equation 3:
>
> \frac{1}{2}(\frac{1}{\epsilon_{p}}  \frac{1}{\epsilon_{w}})
> \sum_{ij}\frac{q_{i}q_{j}}{f^{gb}(R_{ij})}
>
> This expression, however, is not the same as
> equation Eq 5.3 of Amber9 Manual due to the
> presence of the factor
> \frac{1}{2}(\frac{1}{\epsilon_{p}}  \frac{1}{\epsilon_{w}})
>
> Could you please explain the reason for this difference ?
>
> In general, I noticed that in all papers about continuum
> solvent models the goal is to compute the electrostatic
> component of the solvation free energy. No mention is made
> of potential energy. Could you briefly explain (or refer me
> to the appropriate literature) the relation between solvation
> free energy and solvation potential energy ?
>
> Lots of thanks for your help.
>
> Best regards,
>
> Carlo Guardiani
>
>
>
>
> 
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>
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