February 27, 2021, 02:07:37 PM
Forum Rules: Read This Before Posting


Topic: X-Ray Crystallography: SPACE GROUP problem :(  (Read 8353 times)

0 Members and 1 Guest are viewing this topic.

Offline roonie1988

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
X-Ray Crystallography: SPACE GROUP problem :(
« on: April 20, 2010, 05:03:25 AM »
hi, does anyone knows how to draw the diagram for space group "P 21 m a" and showing the distribution of objects and distribution of symmetry elements??

someone told me it's looks similar to the space group "P m c 21,  No. 26" (mm2 group), but I am really confuse and I thought it actually looks like "P n m a, No. 62" or "P m m a, No. 51" which both belong to (mmm group)

could anyone explain what's "P 21 m a" looks like? (the space group multiplicity, the coordinates of equivalent positions, and special positions...etc)

thanks~~~ ???

Offline cth

  • Full Member
  • ****
  • Posts: 223
  • Mole Snacks: +36/-8
Re: X-Ray Crystallography: SPACE GROUP problem :(
« Reply #1 on: April 20, 2010, 02:31:54 PM »
someone told me it's looks similar to the space group "P m c 21,  No. 26" (mm2 group),
That someone is right. ;) You go from Pmc21 to P21ma by simply changing the reference axes:
- the x axis becomes y,
- the y becomes z
- and the z becomes x.
It is just a swap, with nothing else changed. To give you an example, consider a box: it has a length, a width and a height. Now, you flip that box on one side:
- the height has now become length,
- the length becomes width,
- the width is now height.
It works exactly the same. And notice that the box still remains perfectly unchanged.

but I am really confuse and I thought it actually looks like "P n m a, No. 62" or "P m m a, No. 51" which both belong to (mmm group)
P21ma and Pnma can't be equivalent because they do not contain the same symmetry elements. For example, Pnma has no 21 screw axis while P21ma does. The fact that those two space groups have two different numbers in the international tables for crystallography illustrates that they are different.

could anyone explain what's "P 21 m a" looks like? (the space group multiplicity, the coordinates of equivalent positions, and special positions...etc)
Can you get access to the book "International Tables for Crystallography" volume A? It would be much easier.

Offline roonie1988

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
Re: X-Ray Crystallography: SPACE GROUP problem :(
« Reply #2 on: April 20, 2010, 11:11:05 PM »
hmm...I don know if it is correct??
For P21ma, a combination of a mirror-plane perpendicular to the b-axis and a a-glide perpendicular to the c-axis which creates a inversion centers in the lattice.
Do those coordinates and operations exactly same as the Pmc21??

thanks for your *delete me*

Offline cth

  • Full Member
  • ****
  • Posts: 223
  • Mole Snacks: +36/-8
Re: X-Ray Crystallography: SPACE GROUP problem :(
« Reply #3 on: April 21, 2010, 06:03:25 PM »
Nice picture, very well done :) :o. I guess you drew it yourself, am I right? Given the symmetry elements present, it is missing some coordinates. If you apply one of the 21 screw axis to the position x,y,z, it creates a new position at x+1/2,-y,-z which is not accounted for in the drawing. The same for the inversion centres, the position -x,-y,-z is missing.
Unfortunately, this is not P21ma because it has too many symmetry elements.

a combination of a mirror-plane perpendicular to the b-axis and a a-glide perpendicular to the c-axis which creates a inversion centers in the lattice.
Why do you think that? It is not an inversion centre they will create, but rather a 21 screw axis along a, which is indeed present in P21ma. When you have two mirrors perpendicular to one another, they create a 2-fold rotation axis. If one of the mirror is a glide-plane instead, then they form a 21 screw axis.

In your drawing, you should remove the inversion centres and the central mirror plane perpendicular to the a-axis. Then, you have the symmetry elements for P21ma. The coordinates need to be recalculated.

Do those coordinates and operations exactly same as the Pmc21??
You just need to rearrange the a, b and c-axis in the right order when moving from Pmc21 to P21ma (a-axis becomes b, b-axis becomes c and c-axis becomes a. It is a simple permutation.). But except that, nothing else changes.

Offline roonie1988

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
Re: X-Ray Crystallography: SPACE GROUP problem :(
« Reply #4 on: April 21, 2010, 08:10:25 PM »
yeah I drew the diagram by "paint" lol sorry it is not nice....
here is another diagram i just made:
your help will be greatly appreciated!!


Offline cth

  • Full Member
  • ****
  • Posts: 223
  • Mole Snacks: +36/-8
Re: X-Ray Crystallography: SPACE GROUP problem :(
« Reply #5 on: April 22, 2010, 01:00:53 PM »
Those pictures are nicely done. You have more patience with Paint than I do!  ::)

Your last diagram is nearly perfect, with only one problematic detail.
First, what is right in the picture:
- All the symmetry elements are there with the 21 screw axis parallel to the a-axis. The mirror perpendicular to the b-axis. The a-glide plane perpendicular to the c-axis. No problem, you have the space group P21ma.
- All the symmetry-generated positions are correct. For example if you start from the position x,y,z and apply the mirror perpendicular to b-axis, you end up at the position x,-y,z. No problem.
- The stereographic projection is correct.

- The numbers on the right showing (general positions, symmetry operations and reflection conditions) look great but, and here is the problem, they correspond to the space group Pmc21.
For example, if you look at the mirror in the symmetry operations list, you see: m[0,y,z]. The mirror is perpendicular to the a-axis.
If you look at the screw axis, you can read: 21(0,0,z)[0,0,1/2]. (0,0,z) means that 21 is parallel to the c-axis.
What you have here corresponds to Pmc21, not to P21ma.
To summarise, the picture corresponds to P21ma but the table corresponds to Pmc21. I guess you copied the picture from the international tables and added the right axis permutation, but you forgot to permute the coordinates as well.  :)

This is easily corrected by a cyclic permutation:
  in Pmc21             becomes      in P21ma
  1. x,y,z                                1. x,y,z
  2. -x,y,z                               2. x,-y,z
  3. x,-y,1/2+z                         3. 1/2+x,y,-z
  4. -x,-y,1/2+z                        4. 1/2+x,-y,-z

Are you OK with it? If yes, permuting the rest of the Pmc21 table in a way that it belongs to P21ma should be easy game. It works exactly the same as above. After that, it's all done.

Offline roonie1988

  • Regular Member
  • ***
  • Posts: 9
  • Mole Snacks: +0/-0
Re: X-Ray Crystallography: SPACE GROUP problem :(
« Reply #6 on: April 24, 2010, 08:17:07 AM »
great!!
now i think i can determines those coordinates

for P21ma,coordinates and its symmetry operations should be:
1. x,y,z                    1
2. x,-y,z                   m(x,0,z)
3. 1/2+x,y,-z         a(x,y,0) (1/2,0,0)
4. 1/2+x,-y,-z       21(1/2,0,0) (x,0,0)
Reflection Conditions:
h00: h=2n  (21 screw axis along the a-axis)
hk0: h=2n  (a-glide perpendicular to c-axis)

thank you "cth"  :)

Sponsored Links