FCC stacking faults and PO

rowlesmr

I've got some printed steels (FCC, austenite) which look like there are stacking faults involved; the 200 peak is too low compared to the 111.

But I've also got some PO involved, which I can (only) model with 6th order SH (two parameters; c41, c61).

Looking at the stacking faults tutorial, Cu (FCC) is done in hexagonal coordinates in order to make the stacking faults align with the c axis. I can do the same with the steel (Fm-3m --> R-3m:H), and I've already done the calculations of the correct unit cell prms and atom positions.

Questions:
How do I constrain the 6 x HCP SH prms (6th order; c40, c43-, c60, c63-, c66+) to have two unique prms as per the FCC unit cell?

OR

How do I make stacking faults work when not parallel to a unit cell axis?

Thanks

Matthew

rowlesmr

Answer for 1:

By Inspection of the refined coefficients:
c63m = -c66p;
c43m = 2 c66p;

c60 = -0.5 c40;
c20 = -4 c40;

How can I check the theory for this?

johnsoevans

Dear Matthew: I never have good answers to your questions. You only ask difficult ones!

Answer for (2) is that I don't think you can. You have to either be parallel to a cell axis, or shift to a related cell like your R3m:h example. As in the separate thread, the diffax manual is a good source of ideas. The pdf is normally available online (though not this afternoon).

My only thought on (1( would be whether Alan would share that part of the code to compare equations.

rowlesmr

Thanks John

Reading up on how stacking faults are done, that's what I thought. Just need to wrap my head around the nomenclature for describing faulting, and how I can combine stacking faults with twinning.

I'll contact Alan re the SH.

rowlesmr

Now I'm getting into the dim dark recesses of crystallography.

Converting BCC (Im-3m) to orthorhombic

The twinning plane in BCC is (112), so following [1] (cited by [2]), (and fiddling with the order of axes to maintain right handedness), the orthorhombic basis vectors are defined, in terms of the BCC vectors (2.880 Å), as:

A = a - b (4.073 Å)
B = (1/2) (a + b - c) (2.494 Å)
C = a + b + 2c (7.054 Å)

this gives the c-axis in the faulting direction. This is also triples the volume, so I expect 6 atoms in the unit cell.

I've then manually picked out the atom positions in the new unit cell as
0 0 0
1/2 0 1/2
0 1/3 1/3
1/2 1/3 5/6
0 2/3 2/3
1/2 2/3 1/6

(and it is 6)

By observation, I think it is a B base-centred orthorhombic cell, but I can't get a space group. If I plug it into Topas in P1, the calculated pattern matches BCC, so I'm good there. Platon wants to bring it back to P-3m1. FINDSYM goes back to Im-3m.*

Looking through International Tables, I can't find a SG with the right symmetry. What I think I need is:

B base-centred orthorhombic cell
(0,0,0)+ and (1/2,0,1/2)+
Multiplicity Coords
2 0,0,0 (or equivalent, eg 0,0,z or x,y,z)
2 0,y,y (or equivalent, eg 0,u,z or x,y,z)

Is there such a beast?

*Is there any way to limit the space groups it searches?

[1] Hirsch and Otte (1957) Acta Cryst. 10: 447
[2] Warren "X-Ray Diffraction", p. 305

mathiasmorch

Instead of using FINDSYM maybe you could use mode decomposition with ISODISTORT?
Plugging in your sites and lattice in P1 and running it vs the parent gives 107 structures.

At a glance I don't think any of the structures matched your given lattice parameters, centering and atomic coordinates.
However there a few orthorombic space groups proposed, so maybe one can work for you.
I really don't know much about modelling stacking faults, so maybe it's not a suitable way to find a candidate for a given faulting direction.

I attached the distortion file in case you want to have a look yourself.

(In case you're not familar with ISODISTORT, the radio button "subgroup tree" + toggling Generate .cif/.STR further down on the page generates the tree + zip files with .cif and .STR for TOPAS.)

rowlesmr

There wasn't a file attached?

I put in the Im-3m structure as the parent, and the P1 str as the distorted one, and got 107 subgroups, so I assume i did the same as you.

I chose blah blag [1,1,2] as the transformation matrix.

.

Looking through the subgroups, the only ones which have a (1,1,2) basis for an axis are monoclinic or triclinic.

I'll make up bigger versions of my P1 cell and see what happens.

rowlesmr

Yep. 2x2x2 P1 cell gave a few more than 107 structures.

No structures more symmetric than monoclinic had an axis parallel to the cubic [112].

It looks like C2/m is the symmetry of choice, just with all angles == 90.