How do I extrac the anisotropic part of a tensor? Given the elements $\chi_{ij}$ of a tensor in cartesian coordinates, with
\begin{pmatrix}
 \chi_\bot& 0 &0 \\ 
0 & \chi_\| &0\\ 
 0&0 & \chi_\| 
\end{pmatrix},
where the blank entries are zero, in de Gennes' book "The physics of liquid crystals" is said that the anisotropic part $Q_{ij}$ of the $\chi_{ij}$ is $Q_{ij}=Q_0\left( \chi_{ij}-\frac{1}{3}\delta_{ij}\sum_\gamma{\chi_{\gamma\gamma}}\right)$, where $Q_0\equiv Q_{zz}$. 
Why this is the anisotropic part? I don't know any basic classical book that could orient me. Please, forgive for the elementary question.
 A: The isotropic part is the one that you can represent with a scalar. The condition is that the rest is traceless. Isotropic in physics means independent on the direction, which is exactly the condition that it acts on every vector equally. So,
$$A_{ij}v_j=av_i$$
for every $v_j$ simply means $A_{ij}=a\delta_{ij}$.
You actually already wrote down the solution how to split this. You split the tensor into the part that acts on every vector as an equal stretch, and the rest. By convention, you take "as much" as you can into the isotropic part (tracelessness usually has a special meaning in physics) and you can also say that taking the trace, you get the average stretch as your isotropic contribution. $\chi_{ij}$ is the full tensor, and you subtract the isotropic tensor that has the same trace as $\chi_{ij}$ (the $1/3 \delta_{ij} \operatorname{Tr} \chi$ part). The result is that $Q_{ij}$ has a zero trace.
I don't think you need a book about it. It's sort of a projection - to make something traceless, you subtract an isotropic tensor that has the exact same trace. It's sort of a 3D version of this very stupid example:
$$\begin{bmatrix}5 & 0\\0 & 9\end{bmatrix}=
\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}+
\begin{bmatrix}-2 & 0\\0 & 2\end{bmatrix}$$
This is a special case of the more general decomposition of an asymmetric tensor: there, you get in addition to the isotropic (scalar) part and the anisotropic symmetric part (shear), also the antisymmetric part.
Btw this is probably a bit off-topic for this SE, physics would be a better location for it.
