From: Mark Abraham (Mark.Abraham_at_anu.edu.au)
Date: Wed Jul 07 2004 - 22:35:44 CDT
On Thu, 2004-07-08 at 12:34, Peter Jones wrote:
> >> Hi all,
> >> I wonder if anyone could explain how one defines hexagonal unit cell
> >> shapes when using periodic boundary conditions in NAMD, or direct me
> >> to
> >> a suitable reference,
See an inorganic chemistry text book for discussions on crystal packing
arrangements for some background here.
There are some small number of different ways you can define a repeating
volume defined by a given number of planar faces, depending on the
angles and distances between the planes. A cube is easy (three square
faces mutually at 90 degrees with the same edge length), a hexagonal
prism is almost as easy.
Once you have worked out the geometry you're packing, you need to
describe the three vectors that tell the computer the directions it
should move in order to replicate the periodic box. In
linear-algebra-speak these are the "basis vectors". With a cube, that's
easy - just move in the directions of the edges about any vertex. This
might gives basis vectors of (2,0,0), (0,2,0), (0,0,2) for a cubic
periodic cell of side length 2. You could also use (-2,0,0), (0,120) and
(0,0,-2) for reasons that are tedious to explain - try it and see. Think
about why (1,0,0), (0,1,0), (0,0,-1) wouldn't work.
With a hexagonal prism (i.e. a solid with parallel hexagonal faces
joined by six rectangular faces perpendicular to the hexagonal faces)
which I think you want to replicate, you need basis vectors that have
one 60 degree angle and two ninety degree angles between them. Try
writing out a tessellation of hexagons on paper to see why. Using the
two vectors at 60 degrees you can get from any hexagon center to any
other just using moves that correspond to those vectors (including moves
in the negative direction). The third vector (at 90 degrees to those
two) works like the cubic vectors...
a hexagon side length of h and,
a prism height of p, and
that has the hexagonal faces parallel to the xy-plane,
(h , 0 , 0)
(h/2, root(3)*h/2, 0)
(0 , 0 , p)
do the replication.
The sadistic might replace the second vector above by
(-h/2, root(3)*h/2, 0)
which should also work... :-)
Having said all that, I haven't ever done this in NAMD or anything else,
so I'm only guessing that that's the basis vectors it would want from
some orthorhombic examples they've given in the manual.
See page 61 for the periodic box syntax, of course.
P.S. Bonus question: Can you do a triangular prism periodic condition?
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