Slip Systems

How to analyze slip transmission at grain boundaries

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Burgers vector and normal directions

Consider hexagonal symmetry of alpha-Titanium

`cs = crystalSymmetry('622',[3,3,4.7],'x||a','mineral','Titanium (Alpha)')`
```
cs = crystalSymmetry

mineral        : Titanium (Alpha)
symmetry       : 622
a, b, c        : 3, 3, 4.7
reference frame: X||a, Y||b*, Z||c

```

Then basal slip is defined by the Burgers vector (or slip direction)

`b = Miller(2,-1,-1,0,cs,'UVTW')`
```
b = Miller
size: 1 x 1
mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
U  2
V -1
T -1
W  0
```

and the slip plane normal

`n = Miller(0,1,-1,0,cs,'HKIL')`
```
n = Miller
size: 1 x 1
mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
h  0
k  1
i -1
l  0
```

Accordingly we can define a slip system in MTEX by

`sSBasal = slipSystem(b,n)`
```
sSBasal = slipSystem
mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
size: 1 x 1
U    V    T    W  | H    K    I    L CRSS
2   -1   -1    0    0    1   -1    0    1
```

The most important slip systems for cubic, hexagonal and trigonal crystal lattices are already implemented into MTEX. Those can be accessed by

`sSBasal = slipSystem.basal(cs)`
```
sSBasal = slipSystem
mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
size: 1 x 1
U    V    T    W  | H    K    I    L CRSS
1    1   -2    0    0    0    0    1    1
```

Obviously, this is not the only basal slip system in hexagonal lattices. There are also symmetrically equivalent ones, which can be computed by

`sSBasalSym = sSBasal.symmetrise('antipodal')`
```
sSBasalSym = slipSystem
mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
size: 3 x 1
U    V    T    W  | H    K    I    L CRSS
1    1   -2    0    0    0    0    1    1
1   -2    1    0    0    0    0    1    1
-2    1    1    0    0    0    0    1    1
```

The length of the burgers vector, i.e., the amount of displacment is

`sSBasalSym.b.norm`
```ans =
3.0000
3.0000
3.0000
```