4
$\begingroup$

Suppose, that one wants to study a bulk structure which is doped by several impurities. More precisely, I want to calculate the elastic constants using a DFT code (VASP) by fitting stress to strain (Voigt Notation):

$\sigma_i = \sum_{i=1}^6 C_{ij} \varepsilon_j $

From the literature I know, that in a cubic crystal structure the number of independent elastic constant reduces to three. For my purpose, I construct a tetragonal supercell (large enough to cover a low concentration of impurities), e.g. 3x3x4 unit cells.
Following questions arise:
1)In the case without doping, there are still three elastic constants (due to the periodic boundary condition), because the symmetry of the unit cell transfers to the supercell ?
2) Considering the doped case, the relaxed supercell is probably not tetragonal anymore. The unit cell is also not cubic anymore, at least if impurities are present. Does this mean I have to regard all elastic constants ?

I ask this question, because with the loss of symmetry I have to regard more strained systems, resulting in more computations. My system is large (ca. 150 atoms) and the calculation IBRION=6 in VASP takes too long.

Best,
Luca

$\endgroup$
1
  • 2
    $\begingroup$ +1 Welcome to our forum! $\endgroup$
    – Camps
    Jan 23 at 15:50

1 Answer 1

1
$\begingroup$

In the undoped case, when considering a cubic crystal structure and using a tetragonal supercell not changes point symmetry, Hence number of independent elastic constants remains three. If you rotate the crystal to other orientation you may get different $C_{ij}$ components. However, Eigen values of $C_{ij}$ remains three.

When doping is introduced, the symmetry of the crystal can be altered, leading to changes in the elastic properties. Doping can disrupt the perfect arrangement of atoms in the crystal lattice, causing a deviation from the ideal symmetry. As a result, the relaxed supercell may no longer exhibit cubic symmetry especially in the vicinity of the dopant atoms.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .