12
$\begingroup$

I've just run across a nomenclature for coincidence lattices at grain boundaries which uses a capital greek letter sigma followed by an integer. For example A.D. Rollett's 2016 Grain Boundary Engineering & Coincident Site Lattice (CSL) Theory (also archived) includes references to $\Sigma 3$, $\Sigma 5$, $\Sigma 7$ and $\Sigma 13$.

Question(s):

  1. Is it possible to explain what this means and how to know which sigma applies a given grain boundary?
  2. Does this apply to 1D grain boundaries between 2D grains as well? (e.g. adjacent islands of graphene side-by-side on a supporting surface)
Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ +1. Interesting question. I've forwarded it to two people I know that have answered questions with the grain-boundaries tag in the past. $\endgroup$
    – Nike Dattani - No Free Time
    Commented Aug 30, 2020 at 21:06
  • 1
    $\begingroup$ The linked resource has some explanation on the meaning of the $\Sigma$ values on Slide 37. $\endgroup$
    – Mythreyi
    Commented Aug 31, 2020 at 14:45
  • 1
    $\begingroup$ Mythreyi was one of the two people I sent this to. She told me she knows the answer to the first question but not the second (as can be seen in his answer below). Perhaps you want to ask another question about whether or not this notation can apply to 1D boundaries between 2D grains, but let's see what the community thinks. $\endgroup$
    – Nike Dattani - No Free Time
    Commented Aug 31, 2020 at 17:50
  • $\begingroup$ slightly related: What tools are used for the preparation of grain boundary models? $\endgroup$
    – uhoh
    Commented May 23, 2022 at 23:33

1 Answer 1

Reset to default
11
$\begingroup$

The $\Sigma$ values represent the volume of the Coincident Site Lattice (CSL) of the grain boundary in terms of the volume of the unit cell of the crystal. In general, grain boundaries with higher symmetry have lower $\Sigma$ values.

Note that CSL boundaries are special grain boundaries. So, they do not represent all grain boundaries comprehensively. But, for CSL boundaries, one way to calculate the $\Sigma$ value by first constructing the CSL, measuring its volume, and dividing the result by the volume of the unit cell of crystal.

Unfortunately, it can be difficult to understand CSLs without visuals, but the explanation in the link in the question (to Prof. Rollett's notes, also archived) is a good place to start. The notes on crystal defects by Prof. Dr. Föll also has a helpful section on CSL. Apart from these resources, most texts on grain boundaries will have discussions on CSL boundaries and $\Sigma$ values.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ To answer the second question: since it's a geometric construction, it should be possible to get similar values for boundaries in 2D materials. However, I leave the question of whether it will be useful to do so to others who know the field of 2D materials better. $\endgroup$
    – Mythreyi
    Commented Aug 31, 2020 at 14:41
  • 1
    $\begingroup$ Thanks for your post. I think you mean to refer to the discussions on slides 39 and 40 and the expression $$\frac{1}{2}(m^2+n^2) \ \text{if} \ m^2+n^2 \text{even,}$$ $$(m^2+n^2) \ \text{if} \ m^2+n^2 \text{odd}$$ "Sigma denotes the ratio of the volume of coincidence site lattice to the regular lattice." 1) If so do you think the expression should appear in your answer so that if the breaks the answer will still have value? 2) 2) Any idea why the $1/2$ is dropped when the sum is even? Why would $\Sigma=5$ for $m,n=3,1$ but $\Sigma=17$ and not $8.5$ for $m,n=4, 1$? $\endgroup$
    – uhoh
    Commented Sep 1, 2020 at 4:56
  • $\begingroup$ @uhoh: These expressions seem to be a way to get the volumes. I don't think they are necessary to understand what the $\Sigma$ values mean. I do agree that giving a way to "measure the volume" will make the answer more comprehensive, but for that, one must first talk about how to construct the CSL and proceed to further discussions on the calculation of the volumes (which is what the slides do). $\endgroup$
    – Mythreyi
    Commented Sep 1, 2020 at 11:57
  • $\begingroup$ Okay, I'll write an additional answer myself then. Thanks! $\endgroup$
    – uhoh
    Commented Sep 1, 2020 at 12:09
  • 1
    $\begingroup$ @uhoh I think the difference by a factor of $\frac{1}{2}$ appears because there will be one more CSL lattice point in the CSL lattice unit cell. So, the volume is perhaps normalised. You can see the additional CSL lattice point in the figure on Slide 33. $\endgroup$
    – Mythreyi
    Commented Sep 1, 2020 at 12:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .