How to generate a fcc crystal matrice?

Question

Chenyue Xu 2017-7-10

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编辑： Walter Roberson 2024-2-5

I would like to generate a fcc crystal lattice with its atomic positions in a matrice of 3 columns. I tried to expand the crystal by translating the unit cell, but I struggle with the translation vectors and loops. Can someone suggest me a fast method? Thanks!

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John D'Errico 2017-7-10

Then my answer provides the desired coordinates.

Chenyue Xu 2017-7-10

Yes, it's perfect if you can help me on the coordinates! Thanks

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Answer 1

John D'Errico 2017-7-10

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https://ww2.mathworks.cn/matlabcentral/answers/348118-how-to-generate-a-fcc-crystal-matrice#answer_273620

编辑：John D'Errico 2017-7-10

在 MATLAB Online 中打开

Easy enough if all you need are the FCC nodes in an unstructured sequence.

fccbase = [dec2bin(0:7) - '0';.5 .5 0;.5 .5 1;.5 0 .5;.5 1 .5;0 .5 .5;1 .5 .5];
[latx,laty,latz] = ndgrid(0:2,0:2,0:2);
latxyz = [latx(:),laty(:),latz(:)];
latxyz = repmat([latx(:),laty(:),latz(:)],14,1);
latxyz = latxyz + reshape(permute(repmat(fccbase,[1,1,prod(size(latx))]),[3 1 2]),[],3);
latxyz = unique(latxyz,'rows');
plot3(latxyz(:,1),latxyz(:,2),latxyz(:,3),'o')

The basic idea should be clear. I've taken advantage of one nice fact, that unique will be exactly able to locate numbers that are integers, or integers plus 0.5. All of those numbers are exactly representable in double precision. So there is no need even to use a tolerance.

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Chenyue Xu 2017-7-10

Thanks a lot!

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Answer 2

Jose Martinez 2019-5-26

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https://ww2.mathworks.cn/matlabcentral/answers/348118-how-to-generate-a-fcc-crystal-matrice#answer_376637

在 MATLAB Online 中打开

function FCCLattice = FCCMake(a,V)
% "a" is the Lattice Parameter:
a = a;
%Set the Volume of the cube in variable V:
V = V; %This means your cube will be of dimensions VxVxV.
%%%----Code-----%%%
%Create an array that stores all your coordinates:
%First you will need to establish the number of atoms
%that the array will have, we will do this by establishing
%the known number of atoms, in this case FCC 4xVxVxV.
N = V*V*V*4;
%This array will gives is a matrix of Nx3, that will store data.
FCCLattice = zeros(N,3);
%Create the vectors for the atoms position in the FCC Lattice:
FCCatoms = [0 0 0;(a/2) (a/2) 0;(a/2) 0 (a/2);0 (a/2) (a/2)];
%Set a variable to change rows in the array storing the coordinates:
 n = 0;
%Create 3 For loops to mix the positions between xyz
for x = 0:V-1
    for y = 0:V-1
        for z = 0:V-1
            for i = 1:4
                %Create a vector that translate your location in the
                %coordinate system into the position in the crystal:
                coordinatestranslation = a*[x y z];
                %Add 1 to move one row down in the array:
                n = n+1;
                FCCLattice(n,:) = coordinatestranslation +FCCatoms(i,:);
            end
        end
    end
end
end