AMBER Archive (2009)Subject: Re: [AMBER] Solvation potential and free energies
From: Lorenzo Gontrani (l.gontrani_at_caspur.it)
Date: Thu Feb 05 2009  06:34:47 CST
Dear Carlo, I suggest you read some papers/reviews from Prof. J. Tomasi
group (PCMDepartement of Chemistry Pisa University); if I am not mistaken,
some papers dealt in particular with the question free energy/potential
energy.
I hope this helps
Lorenzo
guardiani_at_fi.infn.it scrive:
> 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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