Code for drawing nanotubes

It is a Matlab script nanotube.m. Matlab 6.5 was used.

Nanotube pictures

Click thumbnails below to see the nano tubes full size

n =31, m = 12

n =0, m = 10 n =1, m = 9 n =2, m = 8 n =3, m = 7 n =4, m = 6 n =5, m = 5
n =10,m = 0 n =9, m = 1 n =8, m = 1 n =7, m = 3 n =6, m = 4 n =5, m = 5



Remarks on (n,m) and chirality

There are symmetries that will yield equivalent nanotubes (up to a geometric transformation of three space). I believe you folks call this a point group of graphite (at a carbon) and are probably familiar with the reduction. The symmetries when realized on the (n,m) turn out to be:

(n,m) -> (n,m)
(n,m) -> (-m,n+m)
(n,m) -> (-n-m,n)
(n,m) -> (-n,-m)
(n,m) -> (m,-n-m)
(n,m) -> (n+m,-n)

(n,m) -> (m,n)
(n,m) -> (-n,n+m)
(n,m) -> (-n-m,m)
(n,m) -> (-m,-n)
(n,m) -> (n,-n-m)
(n,m) -> (n+m,-m)

Applying the transforms allows one reduce consideration to 0<=n<=m

Following a luncheon conversation with D. Jelski it appears that chiral means something like the nanotube is not similar to its mirror image. In the above discussion, the first six transformations are orientation preserving and the last six are not. I have not checked it completely but two nanotubes constructed from vectors equivalent by transformations in the first list should be rotationally equivalent and vectors equivalent by transformations in the second list should be mirror images of each other. I think it can be proven that vector leads to a non-chiral nanotube if an only if it is fixed by one of the last six transformations This leads to the following equations.

n = m
n = 0
2n = -m
n = -m
2m = -n
m = 0

Thus the nanaotubes satisfying 0 <= n <= m are chiral unless n = m or m = o, the armchair and zigzag nanotubes.