[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: cad in R32/ccp4 asymmetric units
*** For details on how to be removed from this list visit the ***
*** CCP4 home page http://www.dl.ac.uk/CCP/CCP4/main.html ***
Erica Ollmann Saphire wrote:
> I am running cad on a data set in the space group R32 and am
> wondering why cad does what it does.
>For example, input.mtz has these two reflections first:
>4 -2 6
>4 -1 -7 with a certain F and sigF.
> output.mtz (output from cad) has these two reflections first, with
> the same F and sigF as the first dataset, so they ought to be
>2 2 6
>3 1 -7 with the same F and sigF
> So, the question is, how are these reflections equivalent?
CAD is correct, and those pairs of reflexions are indeed equivalent. It's
always easier to see what's going on in trigonal & hexagonal if you use
four-index Miller-Bravais symbols (h k (-h-k) l):
(4 -2 6) becomes (4 -2 -2 6)
(4 -1 -7) becomes (4 -1 -3 -7)
Next, study the general symmetry operations for 3bar m 1, the Laue class of
R32. These show that:
Rotating h, k, i in any reflexion gives an equivalent one.
(h k i l) = (k i h l) = (i h k l)
Swapping any two of h, k, i and then changing all three of their signs in any
reflexion gives an equivalent one. e.g.
(h k i l) = (-k -h -i l)
Applying these two rules to your reflexions gives:
(4 -2 -2 6) = (-2 -2 4 6) = (2 2 -4 6)
(in the last step, the two -2's have been swapped!)
(4 -1 -3 -7) = (-3 4 -1 -7) = (3 1 -4 -7)
Finally, dropping the i index gives CAD's output.
Ian Clifton Phone: +44 1865 275631
Dyson Perrins Laboratory Fax: +44 1865 275674
Oxford University OX1 3QY UK firstname.lastname@example.org