Almost all metals found in nature are polycrystalline so that there must be grain boundaries. My understanding is that individual grains are tiny defectless crystals and different grains are rotated w.r.t each other at the grain boundaries where the periodicity is interrupted. Are the grain boundaries same as line defects or in particular, same as dislocations? Even if they are not same, does the polycrystalline nature of metals i.e., the existence of grain boundaries imply the existence of line defects? The answer here tends to suggest that.

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    $\begingroup$ No, dislocations are line defects, grain boundaries are 2D defects. Grain boundaries can be constructed out of some arrangement of screw and edge dislocations, most easily seen for high-symmetry boundaries. $\endgroup$ – Jon Custer Apr 30 '18 at 14:18
  • $\begingroup$ @JonCuster Can I say that the polycrystalline nature of metal causes it to have line defects? The answer here physics.stackexchange.com/questions/368262/… impresses upon me that metals being polycrystalline must have line defects. Is that wrong? $\endgroup$ – mithusengupta123 Apr 30 '18 at 14:26
  • $\begingroup$ Take a bicrystal, an intensively studied system, which is two perfect crystals joined with a twist boundary between them. The twist boundary is an array of screw dislocations. No line defects in the two crystals on either side are required. So, you need line defects to construct the grain boundary, but they are not required in the interior of the otherwise perfect crystals. $\endgroup$ – Jon Custer Apr 30 '18 at 14:53
  • $\begingroup$ You can have either a polycrystal or a single crystal with either a large or small dislocation density within the crystalline region. The two aspects are not intrinsically related. $\endgroup$ – Chemomechanics Apr 30 '18 at 15:24

Grain boundaries are 2-dimensional crystallographic defects. Dislocations are 1-dimensional crystallographic defects. Small angle grain boundaries (<15°) can be considered to be composed of an array of step dislocations. Large angle grain boundaries are more complicated.

  • $\begingroup$ I didn't get the answer to my question yet. My question was whether the existence of grain boundaries imply the existence of dislocations. Thanks. $\endgroup$ – mithusengupta123 Apr 30 '18 at 16:48
  • $\begingroup$ @mithusengupta123 - Small-angle grain boundaries consist of an array of dislocations. Therefore, the existence of such a grain boundary does imply the existence of dislocations! See C. Kittel, Introduction to Solid State Physics, 8th ed., 2005, Chapter 21, Figure 11. $\endgroup$ – freecharly Apr 30 '18 at 18:43

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