On Thu, Aug 11, 2011 at 4:48 AM, Jan-Philip Gehrcke <jgehrcke.googlemail.com
> wrote:
>
> > Delta_G = 1/2 Integral( rho(X) * phi(X) dX)
> > [[ This is the 'magical equation' you were looking for. It comes from
> basic
> > electrostatics, which is probably why it isn't mentioned often -- Energy
> =
> > potential * charge summed over a continuous distribution, at its most
> basic.
> > It's certainly easy to get lost here when all you're doing is looking
> > through literature that assumes this is common knowledge and the buck
> stops
> > with calculating the reaction field ;) ]]
> >
> > Here, rho(X) is your charge distribution (just point charges here) and
> > phi(X) is this reaction field potential I described above, dependent on
> > atomic positions and integrated over all space. (The 1/2 is to account
> for
> > double-counting in the double sum/integral). Since we know the charge
> > distribution, the problem now becomes that we need to solve for the
> reaction
> > field potential, where we solve for the electrostatic potential in both
> > environments. The PBE is the equation that we use to solve for this
> > potential.
>
>
> Thanks, let me summarize:
>
> 1) Calculate phi_vac(x) by solving PBE with rho(x) and epsilon_vac
>
> 2) Calculate phi_sol(x) by solving PBE with rho(x) and epsilon_sol
>
> 3) Take the difference of phi_vac(x) and phi_sol(x) to get phi_reaction(x).
>
> 4) Calculate the energy E of rho(x) in the reaction field potential.
>
> E is then considered to be the electrostatic contribution to solvation
> free energy.
>
> Now, I still have to think/read about what this reaction field potential
> means and why E is the energy we are looking for. It's still a bit
> paradox for me that rho(x) is placed into a potential that was
> calculated by using rho(x).
>
The reaction field potential is how much the electrostatic potential changes
going from vacuum/gas phase into solution. Since your electrostatic energy
is just the integral of the potential over the charge distribution, you find
the energy difference via integrating over the vacuum potential and
subtracting the energy found by integrating over the solution potential.
You can, in this case, factor out the charge distribution and just subtract
the 2 potentials when calculating the difference, arriving at the reaction
field potential (and thus only needing to integrate over the "difference"
potential).
As for your latter comment -- the very presence of charges creates an
electric potential, so it would in fact be very worrying if rho(x) was *not*
used to determine your electrostatic potential. Think of the first form of
electrostatic energy you ever saw in physics: E = k*q1*q2/r. You can think
of calculating this energy as a product of the electric potential created by
the first charge at the second charge. That is, phi = k*q1/r. Now you
integrate/sum over the interaction with other point charges (namely, q2).
Thus, you arrive at E = phi * q2. If you run over all particles with your
double-sum, then you wind up doing the same for q2 and q1 as you did for q1
and q2 (which is why you needed that 1/2 term I mentioned earlier). In this
case, the electrostatic potential is determined by the point charges from
your charge distribution.
For real-world cases, you need an equation to determine your potential from
your charge distribution -- that's the PBE.
>
> As you've pointed out, the electrostatic potential itself
> > depends on the charge distribution (which doesn't change in our MM
> framework
> > since everything's a fixed point charge; ignoring polarizable force
> fields,
> > but will affect the electron density in a QM framework if taken into
> account
> > properly). Therein lies the difficulty (solving the PBE). The
> generalized
> > Born approximation is an approximation based on a "trivial" solution to
> the
> > PBE that attempts to provide an efficient, analytic way of calculating
> the
> > solvation free energy. Good references I think are Chris Cramer's book
> and
> > Andrew Leach's book, as this is only a very, very broad overview.
> >
> > The problem with just adding up pairwise potentials, oversimplified, is
> that
> > it neglects any screening effects caused by a dielectric solvent.
> Because
> > solvent partial charges can react to solute partial charges,
> electrostatic
> > effects are screened compared to what they'd be in vacuum (this is the
> > effect of a dielectric).
>
> While adding up pairwise coulomb energies, screening effects could be
> included by using a (maybe distance dependent) dielectric constant,
> right? So that does not convince me :-) Any other important reason, why
> the result of that sum is not what we are looking for?
>
A distance-dependent dielectric *does* add screening effects, which is why
it is sometimes used in computer simulations to mimic solvation effects.
However, the formula I gave above for solvation free energy is the real
formula (integrating a charge distribution over its potential) which you
will find in any textbook that talks about this subject. Moreover,
electrostatic theory states that the electrostatic potential, which you need
in the previously mentioned formula, obeys Poisson's equation (or the
Poisson-Boltzmann equation in the presence of mobile ions). Thus, if the
distance-dependent dielectric *doesn't* provide the same solution as the
PBE, then it's not correct :).
Like I said, this topic is difficult to get your head around completely
(undoubtedly less difficult for Physicists that have made their way through
graduate level E&M successfully), so it won't necessarily come easily.
HTH,
Jason
--
Jason M. Swails
Quantum Theory Project,
University of Florida
Ph.D. Candidate
352-392-4032
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Received on Thu Aug 11 2011 - 07:00:04 PDT
