a partial dislocation
A perfect dislocation slips the crystal by a full lattice vector, so it leaves a perfect crystal behind. But it often turns out to be energetically cheaper for the slip to happen in two smaller HALF-steps rather than one full step. Each of those smaller steps is carried by a partial dislocation — a dislocation whose Burgers vector is NOT a full lattice vector, but only a fraction of one. Because a partial does not restore the perfect stacking, it cannot travel alone without leaving a flaw in its wake; two partials work as a team.
The classic case is in face-centred-cubic metals, where the close-packed layers stack ABCABC. A perfect glide dislocation with b = a/2[10-1] would slide one B-layer atom directly over the A-layer atom below — but that path runs the atom uphill over the atom in its way. It is easier for the atom to zig-zag: first slide into the nearby hollow (a partial of type a/6[2-1-1]), then slide on to the proper site (a second partial a/6[11-2]). Check the sum: a/6[2-1-1] + a/6[11-2] = a/6[3 0 -3] = a/2[10-1], the original. These two a/6<112> partials are Shockley partials, and each has a shorter Burgers vector than the perfect one.
Why split? Frank's energy rule again: line energy scales as b^2, and here b_perfect^2 = a^2/2 = 0.5 a^2, while the two partials sum to 2 x (a^2/6) = 0.33 a^2 — less, so the split lowers energy and happens spontaneously. There is a catch that ties the partials together: between them the stacking is wrong (an ABCACABC slip creates a thin slab of hexagonal-like stacking), a stacking fault, which costs energy. So the two partials repel each other elastically but are held apart by the fault they bound, settling at an equilibrium separation. Partials, and the faulted ribbon between them, control cross-slip, work hardening, and whether a metal deforms by slip or by twinning.
FCC dissociation: a/2[10-1] -> a/6[2-1-1] + a/6[11-2] + stacking fault. Energy check by Frank's rule: 0.5 a^2 (before) versus 0.33 a^2 (after) means the reaction lowers elastic line energy, so it proceeds. The two Shockley partials then sit a few nanometres apart, held by the balance of their mutual repulsion against the stacking-fault energy that wants to pull them back together.
A perfect dislocation splits into two partials (shorter b, lower b^2) bounding a strip of stacking fault.
A partial's Burgers vector is not a lattice translation, so a lone partial ALWAYS trails a stacking fault; partials come in bound pairs (or bound to a fault) and never stand alone in the way a perfect dislocation can.