the 1.57 constant for the cutoff distance in fix orient/fcc command

Hi Steve,

In the documentation for fix orient/fcc command, it’s written that there is a parameter used to compute “a cutoff distance of 1.57 * alat / sqrt(2) for finding the 12 nearest neighbors of each atom”.

I’m not sure if the following question should be addressed to this mailing list or somewhere else. Anyway, my question is where the number 1.57 comes from. I have checked the literature (Nature Materials, 5, 124-127 (2006)) cited, and didn’t find any mention of this number in this paper. I also checked the other literatures that reference this paper, but still didn’t find the answer to my question.

If the intention is to find only the 12 nearest neighbors (within the 1NN limit but not exceed 2NN), shouldn’t the cutoff distance be set as any number between alat/sqrt(2) and alat, i.e. between 1.0 to 1.414?

At 0K condition, setting cutoff distance of 1.57 will include the 12 neighbors that are within 1NN and the other 6 from 2NN.

While trying to make sense of this number, I came to a guess that 1.57 could somehow be derived from PI/2. But when I check the source code fix_orient_fcc.cpp, I found that the constant PI is defined using 4.0*atan(1.0) while the number 1.57 is defined separately. Therefore, I conclude that my initial guess is wrong (or is it?)

That concludes my question.

Thanks.

Tegar

That's not my bit of code in that part of fix_orient_fcc,cpp,
but I think it is just a fudge factor to insure a long enough
cutoff in an fcc lattice to insure there are at least 12 atoms
within it. So long as you find the 12 nearest, then I don't
think it matters how far away you look.

Does that sound right?

Steve

Thanks, Steve. I would think so, but, the follow-up paper (Acta Materialia 57 (2009) 3704–3713) containing a more detailed description of this method did mention that for this specific method (or LAMMPS command), i.e. synthetic GB motion of a fcc crystal, a coordination number of 12 is set (page3706).

Based on this paper, an order parameter Xi is computed by taking the magnitude of the relative distance vectors (i.e. the difference between the 12 ideal fcc 1NN position vectors to the actual 1NN position vectors). This paper explicitly said that it looks for the 12 nearest neighbors or all neighbors within a certain distance if there’s less than 12 neighbors.

Based on this description, I would conclude that that whichever number being used as that certain distance should always result in giving any atom a set of 12 nearest neighbors at most.

But this contradicts with the value of 1.57 that I asked earlier. The 1.57 number did give 12 nearest neighbors, but this is the minimum number of neighbors. At 0 K, the 1.57 cutoff distance will give at least 18 vectors of actual nearest neighbors.

This is the part where I see a problem with my understanding of this synthetic method, esp the 1.57 part. The method relies on how far each of the 12 1NN of an atom deviates from their ‘ideal grain’. If there are more than 12 1NN listed, e.g. 18, then the remaining 6 vectors would highly distort the calculation of Xi parameter. This distortion would create even more serious problems since the added potential and force are very much a function of Xi.

I hope I phrased my comments intelligibly.

Tegar

I'm forwarding your Q to Liz Holm and Stephen Foiles
who know more about the details of this grain-boundary
driving force method than I do. Perhaps they can
provide a justification for the 1.57 parameter.

Steve

Thank you, Steve.

Tegar