# Degree of anisotropy of crystal tensors

Does there exist a scalar that can describe how anisotropic the elasticity of a crystal is? What about other tensors such as the permittivity or susceptibility? I found a Wikipedia article that was particularly illuminating:

Fractional anisotropy is a scalar value between zero and one that describes the degree of anisotropy of a diffusion process. A value of zero means that diffusion is isotropic, i.e. it is unrestricted (or equally restricted) in all directions. A value of one means that diffusion occurs only along one axis and is fully restricted along all other directions._

Could this be extended to $C_{ijkl}$? If so, how do I construct this parameter that is between 0 and 1? I'm assuming it starts by somehow contracting the elastic tensor. This can be very useful if you have a bimaterial system in which a particular physical phenomena emerges from the mismatch of this anisotropic parameter.

• How about a ratio of the magnitudes of the smallest and biggest wife vectors of the tensor? Nov 7, 2014 at 22:09
• "wife" vector? I'm not familiar with the concept. Nov 7, 2014 at 22:19
• Yeah - stupid phone. Eigen vector. Sigh... Nov 8, 2014 at 1:16
• The PRL you pointed out is intriguing (and I will need to ponder it more), but does seem to point out one direction. However, as it is an 'ensemble averaged' method, I still think some measure of the anisotropy of the elastic tensor itself would be useful as well (hence the eigen vector approach). But the way to go there might be to split the elastic tensor into two - an isotropic part and an anisotropic part. Hmmmm... Nov 10, 2014 at 15:33
• So for reference here we are talking about PRL 101, 055504 (2008). My issue with that work is that the authors aren't really clear how they construct $A^{U}$ for the noncubic crystal classes. However, after rereading it Eq (9) is the important one. They are looking at the expansion of the $\textit{spherical}$ AND $\textit{deviatoric}$ parts of the unit fourth-order tensor. A previous answer here was only looking at the deviatoric part. Perhaps that is what I was missing. Nov 10, 2014 at 16:18

I'm going to follow a paper Phys. Rev. Lett. 101, 055504 that seems to answer this question very concisely. Usual Voight notation: $C_{ijkl} \to C_{mn}$ here, we define Voight and Reuss estimators as defined in Proc. Phys. Soc. A 65 349. For example : $$K^V= \frac{1}{9}\left(C_{11} + C_{22} + C_{33} + 2 (C_{12} + C_{23} + C_{31})\right)$$ and so on for $K^R, G^V, \mathrm{and} \,G^R$. I will further define these in this answer later, if needed. The authors of the PRL then go to define $$A^U = 5 \frac{G^R}{G^V} + \frac{K^R}{K^V} - 6 \geq 0$$ as the universal elastic anisotropic index. Their claims, that $A^U = 0$ for isotropic crystals where $C_{11} = C_{22} = C_{33}$, $C_{44} = C_{55} = C_{66} = \frac{C_{11} - C_{12}}{2}$, and $C_{12} = C_{13} = C_{23}$ are exact, as well as the claims about cubic classes. This quantity covers a wide range of crystal classes(all of them) and does not have the conflicting issues that the Zener anisotropic index has, as I understand.

The idea of defining a spherical part and an anisotropic part is the right approach, I think. Here's my attempt.

The fractional anistropy parameter for the diffusion process can be rewritten as $$\text{FA}^2 = \frac{3}{2} \frac{ \mathbf{\tilde{D}}^2}{\textbf{D}^2},$$ where $\mathbf{D}$ is the diffusion tensor and $\tilde{\mathbf{D}} = \mathbf{D} - \frac{1}{3} \mathbf{I} \, \text{tr} (\mathbf{D})$ is the traceless version of the same. In this form, it's not too hard to see what we're doing: we're taking the tensor, subtracting out its spherically symmetric part, squaring the remainder, and then normalizing it relative to the square of the original tensor. In particular, if the tensor is already spherically symmetric, then $\text{FA} = 0$ automatically.

To that end, let's define $$\tilde{C}_{ijkl} = C_{ijkl} - K \delta_{ij} \delta_{kl} - \mu \left( 2 \delta_{k(i} \delta_{j)l} - \frac{2}{3} \delta_{ij} \delta_{kl} \right)$$ where we have defined $K$ and $\mu$ to be $$K = \frac{1}{9} \delta^{ij} \delta^{kl} C_{ijkl}$$ and $$\mu = \frac{1}{20} \left( 2 \delta^{k(i} \delta^{j)l} - \frac{2}{3} \delta^{ij} \delta^{kl} \right) C_{ijkl}.$$ By construction, we have $$\delta^{ij} \delta^{kl} \tilde{C}_{ijkl} = \left( 2 \delta^{k(i} \delta^{j)l} - \frac{2}{3} \delta^{ij} \delta^{kl} \right) \tilde{C}_{ijkl} = 0.$$ To see this, contract each of these tensors with the right-hand side of the definition of $\tilde{C}_{ijkl}$ above, and note that $$\left(2 \delta^{k(i} \delta^{j)l} - \frac{2}{3} \delta^{ij} \delta^{kl} \right) \delta_{ij} \delta_{kl} = 0.$$

Effectively, this process projects the tensor $C_{ijkl}$ onto the space of spherically symmetric tensors with the appropriate index structure (now a two-dimensional vector space rather than a one-dimensional vector space as in the case of the diffusion tensor.) In particular, if $C_{ijkl}$ already corresponds to the form of a stiffness tensor for an isotropic medium (for some $K$ and $\mu$), then we will get $\tilde{C}_{ijkl} = 0$. An analog of the fractional isotropy parameter would then be $$\text{FA}^2 = \alpha \frac{\tilde{C}_{ijkl}\tilde{C}^{ijkl}}{C_{ijkl} C^{ijkl}},$$ where $\alpha$ is a normalization parameter that I unfortunately don't have time to calculate right now. I'll return to this answer later when I have time to figure out how to calculate it.

Edit: $\alpha = 5/4$, I think. If we plug in a $C_{ijkl}$ with $C_{1111} = 1$ and the rest of the components zero, we get $\text{FA}^2 = \frac{4}{5}\alpha$. Requiring $\text{FA} = 1$ in this case then implies that $\alpha = 5/4$.

Note that we're constrained by the fact that the stiffness tensor can't be an arbitrary tensor, but must be "positive semi-definite" in the sense that the gradient energy density $C_{ijkl} \partial_i u_j \partial_k u_l$ must always be non-negative. (Otherwise, the state with $\partial_i u_j = 0$ everywhere wouldn't be a stable equilibrium state for the system.) This constraint is what prevents us from picking a tensor with $K = \mu = 0$ and $\tilde{C}_{ijkl} = C_{ijkl}$.