In principle, faulted interstitial Frank loops can unfault by dislocation shear reactions. This should occur at a critical stage in interstitial loop growth, when the energy of the faulted dislocation loop, with a relatively small Burgers vector, becomes equal to an equivalently sized, unfaulted dislocation loop, with a larger Burgers vector. (In the absence of a stacking fault, the energy of a dislocation scales as b2, where b is the magnitude of the Burgers vector.) From this critical point on, the energy cost to incre­mentally grow the size of a dislocation loop favors the unfaulted loop, since there is no cost in energy due to a stacking fault within the loop perimeter. We examine first the unfaulting of 1/3 [0001] (0001) loops in alumina.

To unfault a 1/3 [0001] (0001) dislocation loop in alumina, we must propagate a 1/3[1010] partial shear dislocation across the loop plane.6 This is described by the following dislocation reaction:

3[0001] + |[10T0] ! 1[10І1] (basal) [7

faulted loop partial shear unfaulted loop

1/3 [1010]

This reaction is shown graphically in Figure 4. Note that the magnitude of 1 / 3 [1010] is approximately the Al—Al (and O—O) first nearest-neighbor spacing in Al2O3. When we pass a 1/3[1010] shear through a 1/3 [0001] (0001) dislocation loop, the cation planes beneath the loop assume new registries such that in eqn [2], ab a2, and a3 commute as follows: a1 ! a3 ! a2 ! a1. The anion layers beneath the loop are left unchanged (B ! B, C! C). Taking the faulted (0001) stacking sequence in eqn [2] and assuming that the planes to the right are above the ones on the left, we perform the 1/3[1010] partial shear operation as follows:

a1 B аг C аз B а1 C аг B аз C a1 B аг |C a1 B аг C аз B a1 C аг B аз C (faulted)

C аз B a1 C аг B аз C a1 B аг C

a1 B аг C аз B a1 C аг B аз C a1 B аг C аз B a1 C аг B аз C a1 B аг C (unfaulted)

[8]

After propagating the partial shear through the loop, we are left with an unfaulted layer stacking sequence. The Burgers vector of the resultant dislocation loop, 1 / 3 [1011], is a perfect lattice vector; therefore, the newly formed dislocation is a perfect dislocation. The resultant 1/3 [1011] (0001) dislocation is a mixed dis­location, in the sense that the Burgers vector is canted (not normal) relative to the plane of the loop.