[email protected] wrote:
Hello,
I wanted to know, if say my box is 30A (cubic) and if I set my
cut-off lengths i.e. rvdw-switch = 0.8 and rvdw =0.9 , does that mean,
the simulation will cut-0ff at 9A? Should I be changing that to 1.5nm
(to get half the box cut-off)? I am a little confused about that.
Setting rvdw = 0.9 means that van der Waals interactions beyond 0.9 nm are zero.
It has no effect on, for example, Coulombic interactions.
Most force fields have defined cutoffs that should be used in order to be
consistent with the original derivation of the parameter set, so it is generally
inadvisable to make ad hoc changes to the cutoffs in order to meet some
arbitrary criterion. The minimum image convention specifies that your longest
cutoff must always be less than half the smallest box vector. If you set up a
system with a cutoff equal to exactly one half of a box vector, if that box
vector decreases even slightly (i.e., under the influence of pressure coupling),
then you will be calculating spurious forces.
-Justin
Thanks.
Nisha Patel
--
========================================
Justin A. Lemkul
Ph.D. Candidate
ICTAS Doctoral Scholar
MILES-IGERT Trainee
Department of Biochemistry
Virginia Tech
Blacksburg, VA
jalemkul[at]vt.edu | (540) 231-9080
http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin
========================================
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