Last Modified 23 February 2001

The ``Static'' Tight-Binding Program: Example XV

The basic theory used to find the shear elastic constant C₄₄ can be found in Example 5. Although this formally covers only the bcc structure, it is easy to adapt to the fcc and simple cubic structures. Adaption to diamond is not as easy as for C₁₁-C₁₂, because of internal parameter relaxation. We will handle diamond separately at a later date.

In the following we use option 3 in the strain table to generate an orthorhombic strain along the <110>, <1(-1)0>, and <001> directions:

Then the energy behaves as

Notes:

1. We determine C₄₄ at the equilibrium lattice constant found in the equilibrium/bulk modulus calculation.
2. Because of the distortion, the symmetry of these structures is lower than in the cubic case. This is reflected in the space group, and also in the choice of the k-point mesh. Note that the k-points used here are equivalent to the k-points used in calculating the bulk modulus. Thus, if you use the ``regular.08'' k-point mesh and e₁ = 0 here you should get exactly the same energy using as the corresponding cubic lattice constant in the calculations above.
3. Note that the space group is determined by applying the distortion and then rotating the system by 45 degrees around the z axis. The space group files listed here then reverse the rotation so that we are back in the original coordinate system.

Note that C₄₄ for simple cubic Ruthenium is negative. This means the simple cubic Ruthenium is elastically unstable and so cannot form unless it is stablized by entropic effects or anharmonic phonons. The non-existence of s.c. Ru isn't all that surprising, of course, since the ground state of Ru is hcp.

Find the Bulk Modulus

Find C₁₁-C₁₂

Cubic Elastic Constant Home Page

Look at other examples.

Get other parameters from the Tight-binding periodic table.

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