Correct realizations of hexagonal lattice using MPB? #2484
Replies: 2 comments 6 replies
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Technically, this is a honeycomb structure in a hexagonal lattice. The MPB primitive unit cell is a rhombus (not a hexagon — that is a Wigner–Seitz cell, whereas in MPB you need to specify the primitive lattice vectors, which give a parallelogram/parallelopiped), and has two holes per primitive cell. (You also need to understand MPB's coordinate system: the geometry centers are not Cartesian coordinates, they are in the basis of the lattice vectors.*) You should start with the triangular lattice (i.e. hexagonal lattice) tutorial and add a second hole to the unit cell. Actually, I just remembered that there is an MPB example file for a honeycomb lattice, albeit in Scheme — should be easy for you to copy from. |
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PS. Even if you use a hexagonal Wigner–Seitz cell, there are still only two cylinders per primitive cell. In your diagram, you drew your hexagons too big — the primitive Wigner–Seitz cell is a hexagon whose corners are the centers of the cylinders. So, the cell contains only 1/3 of each cylinder, and 6/3 = 2 whole cylinders per Wigner–Seitz cell. |
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I want to realize the real hexagonal lattice using MPB, there will be six rods in the unit cell,
However, the generated material distribution is not correct (I have used
rectify=True
to correct the distributions ), How can I resolve this?Beta Was this translation helpful? Give feedback.
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