1
$\begingroup$

The tensorial form of Hooke's law for the strain-stress relationship in a crystal is (in the Voigt notation):

Hooke's law

where $\sigma$ is the strain, $\epsilon$ is the stress and C is the stiffness tensor:

stiffness tensor

For a crystalline system of the cubic symmetry class, the stiffness tensor reduces to:

enter image description here

The Born criterion for the stability of an unstrained crystal is that free energy must be represented by a positive defined quadratic form. In the case of a cubic crystal, it is known that this is equivalent to the following three conditions on the elastic constants:

$$C_{11} - C_{12} > 0$$ $$ C_{44} > 0$$ $$ C_{11} + 2 C_{12} > 0$$

But what about lower symmetry classes? What is the generic Born criterion for stability of a crystal? I have quite convinced myself that all the eigenvalues of $C$ must be positive, but I cannot find confirmation of that anywhere. Is it right? Is there a reference on that topic?

$\endgroup$

4 Answers 4

3
$\begingroup$

I've found a good analysis of the stability conditions for a crystal's elastic constants, both unstrained and under stress, in:

J. W. Morris Jr and C. R. Krenn, Philos. Mag. A 2000, 12, 2827–2840

To quote them:

In the linear elastic limit the conditions of internal stability reduce to the familiar condition that the 6 x 6 matrix $C_{ij}$ of elastic moduli have no negative eigenvalues.


So, almost two years later, after realizing a lot of people had trouble with this question (and there are mistakes made in some of the literature on the topic), we have published a short pedagogical reference on the issue: F. Mouhat and F.-X. Coudert, Phys. Rev. B, 90, 224104 (2014) (also on arXiv).

$\endgroup$
1
  • 1
    $\begingroup$ R. Cowley Phys Rev B 1976 is a great article as well for this subject. He covers many of the crystallographic point groups and shows their stability criterion. $\endgroup$
    – John M
    Oct 24, 2014 at 3:01
2
$\begingroup$

While I can't find a reference for this, I think your criterion is correct. Here's the argument I would use - if it's the same as yours, then maybe it's right! The elastic energy is (http://ciks.cbt.nist.gov/garbocz/manual/node8.html) $$E = \frac{1}{2}\int d^d r \epsilon_i C_{ij} \epsilon_j$$ where $C_{ij}$ is symmetric. Because $C_{ij}$ is symmetric and real, it can be diagonalized, and its eigenvectors are complete and orthogonal (https://en.wikipedia.org/wiki/Hermitian_matrix). Then you can expand $\epsilon_j$ in terms of the eigenvectors of $C_{ij}$, $\bf{v}^{(k)}$, i.e. $C_{ij} v_j^{(k)} = \lambda_k v_i^{(k)}$. $$\epsilon_j = \sum_k (\bf{v}^{(k)} \cdot \bf{\epsilon}) v^{(k)}_j$$ (This assumes that $\bf{v}^{(k)}$ is orthonormal, which we can always choose without loss of generality.) We then find that $$E = \frac{1}{2} \int d^d r \lambda_k (\bf{v}^{(k)} \cdot \bf{\epsilon})^2$$.
Stability means that there are no modes that will lead to an unbounded decrease in energy, and no marginal modes - i.e. that $\lambda_k > 0$ for all $k$ - your positive definite quadratic form.

$\endgroup$
1
  • $\begingroup$ Yes, that's how I figured it too… $\endgroup$
    – F'x
    Apr 2, 2013 at 7:43
1
$\begingroup$

This derivation is correct for cubic crystals without external pressure.

A fresh review can be found in Rev. Mod. Phys. 84, 945 (2012) Lattice instabilities in metallic elements http://rmp.aps.org/abstract/RMP/v84/i2/p945_1

$\endgroup$
-1
$\begingroup$

A detailed analysis may be found also in arXiv:1104.0173 [astro-ph.SR], D. A. Baiko "Shear modulus of neutron star crust", Eqs. (29)-(33). http://arxiv.org/abs/1104.0173

$\endgroup$
1
  • $\begingroup$ That paper is, at best, only very marginally related to my question :( $\endgroup$
    – F'x
    Apr 2, 2013 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.