Hi Yan,
1.) The spring constant you need depends on whether or not you use the
nmropt module or the ncsu_pmd module. What is actually needed in WHAM (and
similar variants like MBAR) is the bias potential energy in all
configurations and simulations; this is easily recomputed if the spring
constants and bias anchor points are known and you have the bias coordinate
timeseries saved.
2.) This question is interesting. I believe the answer is yes, but with
some interesting qualifiers. Why don't you just try it? Playing with the
WHAM equations would be useful too.
3.) If you can sample without any restraints, then you have no need for
WHAM since the sampling weights are all unity in that case (although I'm
not 100% a program will converge to this result). All you need to do in
that case is generate an estimate of the probability density (probably a
histogram) and take its negative logarithm (you could also integrate the
mean force in that case too).
Regards,
Brian
On Wed, Aug 8, 2012 at 1:07 PM, Yan Li <liyantiger.yahoo.com> wrote:
> Dear Amber users,
>
> I am trying to calculate PMF with WHAM, and have a question about the
> spring constant. In Amber manual, it says PMF can be computed using
> umbrella sampling. And in WHAM input file, the spring constant from
> umbrella sampling is needed. It looks the spring constant is necessary, but
> I wonder how the spring constant is used in WHAM after reading WHAM manual.
> The first question is that, in many MD trajectories with umbrella
> sampling, if all time series data have the same spring constant, does the
> absolute value of the spring constant affect WHAM results? For example, if
> the spring constant is set to 2 or 4 in WHAM input file no matter what it
> is in MD, will it generate different energy surfaces?
> Another question is that, suppose that I have a very long MD trajectory
> (long enough to sample) without any constraint, is it possible to calculate
> PMF from this trajectory? If possible, what should the spring constant be?
> It looks like a stupid question. I would very appreciate that you could
> correct my misunderstanding. Thank you for your time.
>
> best wishes,
> Yan
>
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>
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Received on Wed Aug 08 2012 - 13:30:03 PDT
