# All single crystalline anisotropic?

1. Oct 10, 2011

### xxh418

Hi all:
I have a question about the anisotropy properties of SINGLE CRYSTAL. The definition of the isotropy in WiKi is that the properties of the materials are the identical in ALL directions. If so, none of the single crystal is isotropic even though you can find such XYZ planes that the properties are the same(such as BCC FCC). But if you rotate like 1 degree, the properties in the X and X' directions will be different. Am I right? Or I have make mistakes. I know that the macro-properties of the properties are ofter isotropic just due to the polycrystalline structure.

2. Oct 10, 2011

### Mapes

What property are you asking about? Second-rank tensors (diffusivity, thermal conductivity) of cubic single crystals are isotropic. Fourth-rank tensors (stiffness, compliance) aren't. See Nye's book for discussion.

3. Oct 10, 2011

### xxh418

Mapes, Thanks for your reply. I am talking about the diffusivity. But I can not understand how the diffusivity can be all the same in all directions. For example, for Si crystal, If I pick {110},{1 1 -1}{1 1 2} directions, the arrangement of the atoms in every directions are very different. Thus, I suppose the diffusivisty should be different.

4. Oct 10, 2011

### Mapes

The arrangement of atoms looks different in that direction, but the diffusive flux adds up to the same value (for cubic crystals). Try calculating it.

5. Oct 17, 2011

### xxh418

Mapes:
I still can not get it. I drew the crystal lattice and think it over and over. Just assume we have simple cubic structure. The diffusion tension is [D 0 0; 0 D 0; 0 0 D]. If we rotate an small angle thita. Then the new diffusion coefficient in the new direction would be D multiplied by a function of thia. Then given the same temperature gradients for two directions, the diffusion coefficients are different (one has thita, one not). The flux should be different. I know there must be something wrong of my logic. But I do not know what is the problem.

Regards
Xu

6. Oct 17, 2011

### Mapes

It's easier to keep the structure stationary and imagine the driving force (the temperature gradient) changing direction. Decompose the gradient vector $\textbf{G}$ into the principal axes of the structure: $\textbf{G}=G\textbf{i}\cos \theta+G\textbf{j}\sin \theta$. Then the flux is $\textbf{F}=\textbf{D}\textbf{G}=DG\textbf{i}\cos \theta+DG\textbf{j}\sin \theta$. What is the magnitude of vector $\textbf{F}$?

7. Oct 18, 2011