To unfault a 1/4 [110] (110) dislocation loop in spinel, we must propagate a 1 /4[112] partial shear dislocation across the loop plane.12 This is described by the following dislocation reaction:

4[110] + i[H2] ! 1[101] [11]

faulted loop partial shear unfaulted loop

This reaction is shown graphically in Figure 5. When we pass a 1 /4[112] shear through a 1/4 [110] (110) dislocation loop, the atomic planes beneath the loop assume new registries, such that in eqn [6], ajpj and a2p2 commute as follows: a1p1 ! a2p2 ! a1p1. The anion layers beneath the loop are left unchanged (B! B, C! C). Also the Al p’ layers are left unchanged (p’ ! p’). Taking the faulted (110) stack­ing sequence in eqn [6] and assuming that the planes to the right are above the ones on the left, we perform the 1/4[112] partial shear operation as follows:

(P’S) (a, P,C) (P’S) KP2C) (P’S (a, b,C) |(P’S)| (a, P,C) (P’S) ^C) (faulted)

P’S a2P2C P’S a, P,C

(P’S) (a, P,q (P’S) (a2P2C) (P’S (a, P,C) (P’S) (a2P2C) (P’S) (a, P,C) (unfaulted)

[12]

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, 1/2[101], is a perfect lattice vector; therefore, the newly formed dislocation is a per­fect dislocation (equal to the Mg-Mg first nearest — neighbor spacing). The resultant 1/2[101](110) dislocation is a mixed dislocation, in the sense that the Burgers vector is canted (not normal) relative to the plane of the loop.