# Question about a particular statement on plastic deformation of single crystals

"...the wider the dislocation, the lower is the elastic energy of the crystal because the atomic spacing in the slip direction is closer to its equilibrium spacing."

What has elastic energy to do with the atomic spacing in the slip direction?

(The quote is apparently from Dieter's Mechanical Metallurgy.)

Consider a simple edge dislocation model, corresponding to an extra half plane of atoms within a material:

Plastic deformation can occur by rightward movement of this dislocation, thus carrying slip through the material.

Now consider the following schematic delineating the region where the atomic positions are disturbed from their normal spacing (Fig. 4-16b of Dieter, 1988 SI metric edition):

Dieter is saying that if the width $$W$$ of this interfacial region (which is a reasonable surrogate for the dislocation size) were larger, then the atomic spacing would be less disturbed. In other words, the compressive and tensile effects from adding the extra half plane would be spread out among a larger number of atoms.

Thus:

"...the wider the dislocation [meaning the wider the interfacial region], the lower is the elastic energy of the crystal [corresponding to stretching and squeezing of bond lengths] because then the atomic spacing in the slip direction is closer to its equilibrium spacing [because the disturbance could be accommodated over more atoms]."

Does this make sense? If it's still confusing, try drawing an interfacial region that includes two or more additional atoms on both sides [crude adaptation of Dieter's figure follows]:

The stretching and squeezing of bond lengths is clearly less severe with this wider dislocation.

• How are we differentiating a slipped region from an unslipped region? – user586228 Feb 22 at 2:17
• In the slipped region, atoms are no longer next to all of their original neighbors. – Chemomechanics Feb 22 at 2:30