vincent.chaptal at ibcp.fr
Wed Nov 21 08:50:13 PST 2012
I ran "phenix.explore_metric_symmetry" with the command:
Transforming point groups
From P 1 to C 1 2 1 (x-y,x+y,z) using :
I understand that going from P1 to C2, one needs to apply the
transformation matrix (x-y, x+y,z) on the P1 cell to form the C2 cell,
and (-k, -h, -l) on the reflections.
Naive question: why aren't the two matrices similar?
The reciprocal space is the fourier transform of the real space; i was
thinking that a reorientation matrix in the real space would be kept in
the reciprocal space. My maths are not that good, and in P1 it is more
complex than other space groups. Can someone tell me why the matrices
Also, in C2 there is a 2-fold axis parallel to b, so reflections (h,k,l)
are equivalent to (-h, k, -l).
In P1, they are not. Applying the above transformation matrix on the
reflections would give (hP1, kP1, lP1) transforms into (-kP1, -hP1,
-lP1), and these are equivalent to (kP1, -hP1, lP1)? Is this correct?
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