From: David Hardy (dhardy_at_ks.uiuc.edu)
Date: Thu Sep 20 2007 - 13:39:45 CDT
Jason,
The switched potential has the form: U(r) s(r),
where U(r) is your unmodified Lennard-Jones potential
and s(r) is a smooth, piecewise defined polynomial:
s(r) = 1 for r <= switchdist
s(r) = 0 for r >= cutoff
otherwise, s(r) is a decreasing function
So there is no danger of the function being entirely positive, although
your choice of switchdist will probably give you a much shallower well
depth than you want.
Regards,
Dave
On Sep 20, 2007, at 1:00 PM, Jason O'Young wrote:
> Hi All,
>
> I set my switchdist to 0.01 and my cutoff to 10.
>
> I am not too sure how the switching function would look like, but is
> there a danger of the function being entirely positive?
>
> For an example of what I mean, see this picture from the users guide:
> http://www.ks.uiuc.edu/Research/namd/2.6/ug/img14.png
>
> If the switching was to begin at such a low value, would the function
> just proceed from infinitely positive to 0? Or would the general form
> of the potential function be "squeezed"?
>
> Thanks in advance!
> Jason
>
----------------------------------------------------------
David J. Hardy, PhD
Theoretical and Computational Biophysics
Beckman Institute, University of Illinois
dhardy_at_ks.uiuc.edu
http://www.ks.uiuc.edu/~dhardy/
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