Translate and Rotate Rigid Body

mfisch

Hi all

According to the Tech Ref, translate ta 0.5 tb 0.5 tc 0.5 should translate the origin of a rigid body along a, b, c within the cell (in fract. coordinates).

rotate 0 qa 1 qb 0 qc 0
rotate 0 qa 0 qb 1 qc 0
rotate 0 qa 0 qb 0 qc 1

should then rotate the rigid body through its origin along the axes of the unit cell. This works fine for an orthogonal cell. How can I rotate around the axes in a e.g. triclinic cell?

Below is an example that can be run in Topas. Check what happens when xou change the first two "rotate 0 ..." lines to "rotate 90 ...". then change it back and do the same after changing the cell angles to 85°.

'--------------

iters 0
yobs_eqn = 100;
min 1
max 30
del 0.001
bkg 100
x_calculation_step 0.001
LP_Factor(90)
lam ymin_on_ymax 0.0001 la 1
lo !lambda 0.5
lh !l_h 0.001

str
a 5
b 6
c 7
al 90 '85
be 90 '85
ga 90 '85
space_group "P1"

view_structure

site A x 0.50000 y 0.50000 z 0.50000 occ O 1
site B x 0.50000 y 0.50000 z 0.78571 occ B 1
site C x 0.10000 y 0.50000 z 0.50000 occ H 1

rigid
load z_matrix
{ A
B A 2
C A 2 B 90 }

' rotate 90 qx 1 qy 0 qz 0
' rotate 90 qx 0 qy 1 qz 0
' rotate 0 qx 0 qy 0 qz 1

rotate 0 qa 1 qb 0 qc 0 '90
rotate 0 qa 0 qb 1 qc 0 '90
rotate 0 qa 0 qb 0 qc 1

translate
ta 0.5
tb 0.5
tc 0.5

' translate
' tx = 5 / 2;
' ty = 6 / 2;
' tz = 7 / 2;

alancoelho

Hi Martin

I used the rigid body editor with the INP segment below and the display of both structures seem fine as seen in the Screen shot.

str
a 5
b 6
c 7
al 85
be 85
ga 85
space_group "P1"

site A x 0.50000 y 0.50000 z 0.50000 occ O 1
site B x 0.50000 y 0.50000 z 0.78571 occ B 1
site C x 0.10000 y 0.50000 z 0.50000 occ H 1

rigid
load z_matrix
{ A
B A 2
C A 2 B 90 }

rotate 90 qx 1 qy 0 qz 0
rotate 90 qx 0 qy 1 qz 0
rotate 0 qx 0 qy 0 qz 1

str
a 5
b 6
c 7
al 90
be 90
ga 90
space_group "P1"

site A x 0.50000 y 0.50000 z 0.50000 occ O 1
site B x 0.50000 y 0.50000 z 0.78571 occ B 1
site C x 0.10000 y 0.50000 z 0.50000 occ H 1

rigid
load z_matrix
{ A
B A 2
C A 2 B 90 }

rotate 90 qx 1 qy 0 qz 0
rotate 90 qx 0 qy 1 qz 0
rotate 0 qx 0 qy 0 qz 1

mfisch

Hi Alan

Maybe, I was a bit unclear:

What I understand:

Orthogonal cell:
- translate tx, ty, tz in carthesian system
- translate ta, tb, tc in crystal system
-> ta 0.5 tb 0.5 tc 0.5 equal to tx = 5/2; ty = 6/2; tz = 7/2; in my case.

rotate 0 qx 1 qy 0 qz 0
rotate 0 qx 0 qy 1 qz 0
rotate 0 qx 0 qy 0 qz 1

does the same as

rotate 0 qa 1 qb 0 qc 0
rotate 0 qa 0 qb 1 qc 0
rotate 0 qa 0 qb 0 qc 1

but what is their difference? Through which points do the rotation axes go?

Triclinic system:
- translate tx, ty, tz in carthesian system
- translate ta, tb, tc in crystal system
-> ta 0.5 tb 0.5 tc 0.5 NOT equal to tx = 5/2; ty = 6/2; tz = 7/2; in my case.

As mentioned in the tech ref, I would expect

rotate 0 qx 1 qy 0 qz 0
rotate 0 qx 0 qy 1 qz 0
rotate 0 qx 0 qy 0 qz 1

to rotate around the orthogonal system axes and

rotate 0 qa 1 qb 0 qc 0
rotate 0 qa 0 qb 1 qc 0
rotate 0 qa 0 qb 0 qc 1

to rotate around the crystal axes.

If I start with my rigid body in the triclinic case, the distance A-C is parallel to a, A and C a re on the same y-height.

rotate 0 qa 1 qb 0 qc 0
rotate 90 qa 0 qb 1 qc 0
rotate 0 qa 0 qb 0 qc 1

I assume, the above would now be a pure rotation around b. A-C is rotated 90° around b, but the y-height is no longer the same, so other rotations are involved, too.

How exactly, can I rotate a rigid body around the crystal axes in a non-orthogonal system?

Cheers
Martin

----------------------
EDIT: I think the problem (my problem :-) is to see what happens when one rotates something orthogonal in a non-rothogonal system. I will play around a bit with the rigid body visualizer...
Is it somehow possible to display a*, b* and c* in the rigid body editor?

metty

How exactly, can I rotate a rigid body around the crystal axes in a non-orthogonal system?

If you use qa, qb, qc with unity vectors from your examples you can rotate around the crystallographic axes...
For the orthorhombic case the rotation axes are congruent. In the triclinic case the rotation around qx, qy, qc is around a cartesian coordinate system whereas the rotation around qa, qb, qc is around the triclinic axes.

mfisch

- orthogonal system:

use ta, tb, tc to translate and qa, qb, qc to rotate around a, b, c (which are parallel a*, b*, c* and x, y, z of the system of the rigid body)

- non-orthogonal system:

use ta, tb, tc to translate within the cell and qx, qy, qz to rotate around the orthogonal system of the rigid body or qa, qb, qc to rotatae the orthogonal system of the rigid body around the triclinic cell.
But what is the relation between the orthogonal system of the rigid body and a, b, c, a*, b*, c*? Can someone please make a drawing that could clarify this?

Thanks,
Martin

alancoelho

>But what is the relation between the orthogonal system of the rigid body
>and a, b, c, a*, b*, c*?

The direction of a* is b x c. Is there another question in your question that I am missing.

mfisch

Hi Alan

I managed to understand with a triclinic cell, a cubic rigid body and chapter 7.24.7. Should I ever have time, I'll draw a picture...

Cheers
Martin