The volume is one degree of freedom of a unit cell. But apart from the very high symmetry crystals, the volume is not sufficient to determine the shape of the unit cell. Even if one does not consider the freedom due to the positions of the atoms inside the unit cell, and even one sticks to a given symmetry, there are many ways to construct a unit cell with a given volume. This video shows how to find the shape of a unit cell (at a given volume or at a given pressure), both in a manual way and in an automatic way.
For your report of this week about the Fe-Al crystal, make a few variations of the b/a-ratio of the orthorhombic cell in the range 1.58-1.70, keeping the c/a-ratio constant to √3, and keeping the volume constant to the optimal volume you found before (this can be done at medium precision, expect 2-3 minutes per calculation). Inspect how the total energy behaves. Keep in mind that this b/a-ratio corresponds to the c/a-ratio of the underlying hcp cell. Plot total energy as a function of b/a-ratio, and determine the b/a ratio that has the lowest energy. How far away is this from the c/a ratio for the ideal hcp stacking? Compare the amount of energy variation when you change the shape of the unit cell with the amount of energy variation when you changed the volume of the unit cell a few pages ago.
The shape of the orthorhombic cell is determined not only by b/a, but also by c/a. Hence, in principle one should now vary c/a (keeping b/a constant) to find its optimal value. But at this optimal value for c/a, the previous value for b/a is not necessarily optimal any longer: b/a should be re-optimized (and will vary probably less than before). And then again for c/a, etc., etc., until nothing changes any longer. Alternatively, one could immediately scan a surface of b/a and c/a values, and search the energy minimum in two dimensions. Moreover, at these optimal b/a and c/a, the previous optimal volume should be checked again. This is rapidly becoming tedious… don’t do it.
For these reasons, shapes are usually not optimized in this way. Using the stress tensor instead is the easier and more general option. This will be discussed later in this chapter. If your DFT code does not have a stress tensor, then the above manual procedure is what you’ll have to use anyway.
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Started by: Ruben Quiroz
expected time: 20m
report time spent (page code AW05C)