9780030839931, Chapter 4, Problem 4P

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Chapter 4, Problem 4P Problem (a) Prove that the Wigner-Seitz cell for any two-dimensional Bravais lattice is either a hexagon or a rectangle. (b) Show that the ratio of the lengths of the diagonals of each parallelogram face of the Wigner-Seitz cell for the face-centered cubic lattice (Figure 4.16) is (c) Show that every edge of the polyhedron bounding the Wigner-Seitz cell of the body- centered cubic lattice (Figure 4.15) is times the length of the conventional cubic cell. (d) Prove that the hexagonal faces of the bcc Wigner-Seitz cell are all regular hexagons. (Note that the axis perpendicular to a hexagonal face passing through its center has only threefold symmetry, so this symmetry alone is not enough.) Step-by-step solution Step of 4 a) We know that by construction, for every vector in the Bravais lattice, the inverse vector is also in the Bravais From the construction of the Wigner-Seitz cell using bisecting lines, we find that for every side of the Wigner-Seitz cell, there is another that is the inverse of The number of sides is four(rectangle), six (hexagon) or more. Let's take one Bravais lattice point as the origin O. The nearest Bravais Lattice point is let's say A. Then there is a second point as well at -A. Let's take the closest point to O not on the line OA as the point B. Now let's construct the line normal to OA through O and assuming that B is on the left of this line, we find that there is a Bravais Lattice point C on the line through B parallel to OA and on the right of this line with BC=OA. Any other Bravais Lattice point on BC will not contribute a side to the Wigner-Seitz cell, since it has to be further out than A, B, C and -A. Similarly, on the other side of OA we find the points -B, -C. The Wigner-Seitz cell is a hexagon, unless B is on the normal line constructed before, in which case the cell would be a rectangle. Comments (1) Anonymous Not graphical Step of 4 (b) One diagonal connects the lattice points are So the length of the diagonal is The other diagonal connects the lattice points are So the length of the other diagonal is = a 2 Therefore, the ratio of the lengths of the diagonals of each parallelogram face of the Weigner-Seitz for the face centered cubic lattice is Comments (1) Anonymous How do you know it is (1,1,3)? Step 3 of 4 (c) The corners of the square on top are given by The length of this side is Points on the square on the right face are given by The distance between a point on the top and side square is the distance between This is Therefore and b 4 and the length is Step of 4 (d) From the figure given below, take the central point as the origin. It looks as if the hexagon faces could have sides of two different lengths, but opposite sides of the hexagon must have the same length since they are in contact when we add a second Wigner-Seitz cell centered on a corner point.