# Generic Born stability criteria

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

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

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

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?

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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

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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

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That paper is, at best, only very marginally related to my question :( –  F'x Apr 2 '13 at 7:43

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.

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Yes, that's how I figured it too… –  F'x Apr 2 '13 at 7:43
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.