Computing Potential of Mean Force --SMD-cv

From: gamini_at_ncbs.res.in
Date: Tue Jul 11 2006 - 15:36:45 CDT

Dear Namd users,
Hi !!

I would like to understand on how to computationally evaluate the
potential of mean force for a given non-equlibrium process using SMD-cv
approach.
To start with, I have considered the process of unfolding of ubiquitin.
Ubiquitin unfolding was simulated using Steered Molecular Dynamics
-constant velocity pulling, as illustrated in the tutorial.

I have the following data(output) in log file:
The positions of the SMD atom, x y z
>From this, I have computed, the extension as center-center distance
euclidean)between the fixed(x,y,z) and SMD(x,y,z)atom.
Therefore, I have the timeseries for extension.

Also given in the log file are the three components of forces( fx,fy,fz).
I have calculated force as (nx*fx+ny*fy+nz*fz), where nx,ny,nz are
normalized vetors.
Now, I have the timeseries for force.

I have the following questions regarding the same.

a)I would like to understand, given the available output from such
simulation (SMD-cv), how to reconstruct the potential of mean force
computationally?

b)From what limited understanding I have from the literature
(Reference: Jensen et al;PNAS (2002), vol 99,6731-6736 ),
the potential of mean force can be computed from Jarzynski's identity by
cumulant expansion(to second order for accuracy)as :

<W(t)> - 1/2KbT{[<W(t)^2>] -[<W(t)>^2] }

Is it correct that I evaluate the above expression for computing potential
of mean force ?

c)And, Is W(t) evaluated as:
W(t)= {/ 0 to t(v*f(t)*dt)} - k/2[R(t)- R0 -v*t]^2
where,
/ 0 to t =integral 0 to t (sorry!! I had to limit my use of symbols)

R(t)-R0 = Extension
v= constant velocity
k= spring constant
f(t) = force

Kindly, help me understand if I am going right.

I also have a brief query on the use of dielectric constant( in the
configuration file).
Is it wise to use a dielectric of 80 for water instead of performing the
simulation in water box ?

Will I have to compromise on a significant contribution to energy in doing
so?
Please comment.

Thanks in advance.

Sincerely,
Ramya Gamini

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