Why are there no edge dislocations in a face-centered cubic (FCC) lattice?

Question

In crystallography, we know that the Burgers vector is parallel to the dislocation direction in a screw dislocation, and perpendicular in a edge dislocation.

Furthermore, I have in my lecture notes "FCC cannot have edge dislocations" but there is no explanation of why, aside from "in FCC, the Burger's vector belongs to the family of the dislocation direction vectors"

I don't understand what this means.

Answer 1

I have to stretch my memory to remember how this goes (and read a bit from here).

First, how do you define burgers vector? You take the end point of the dislocation and make a circle around it as it were a perfect lattice, the extra (or missing step) from the full loop is the burgers vector.

Now, is the burgers vector also a lattice vector? If yes, then it is (edge)dislocation. If no, then it is only a partial (edge)dislocation (partial dislocations have their own name).

Now, how does this work on a fcc-lattice case. On quick read for link I pasted, it seems that fcc-lattice cannot have edge dislocations, because the burgers vector is not a lattice vector.

You may want to go this through by yourself and check that I didn't make any mistakes.