# How can I deduce the magnetic susceptibility tensor of a biaxial liquid crystal in its nematic phase?

I know from de Gennes' book  that the (magnetic) susceptibility tensor $\chi_{\alpha\beta}$ of an uniaxial liquid crystal in its nematic phase can be written as \begin{align} \chi_{\alpha\beta}&=\left(\frac{\chi_{xx}+\chi_{yy}+\chi_{zz}}{3}\right)\delta_{\alpha\beta}+\left(\chi_\|-\chi_\bot\right)\left(n_\alpha n_\beta-\frac{\delta_{\alpha\beta}}{3}\right)\\ &=A\delta_{\alpha\beta}+BQ_{\alpha\beta}, \end{align} where $Q_{\alpha\beta}=\left(n_\alpha n_\beta-\frac{\delta_{\alpha\beta}}{3}\right)$ is the tensorial order parameter, i.e., the anti-symmetric part of $\chi_{\alpha\beta}$ and $\chi_\|$ and $\chi_\bot$ are the molecular susceptibilities parallel and perpendicular to the director $\hat{n}$.

\begin{align}\end{align} Thus, what is the equivalent of this susceptibility for a biaxial nematic liquid crystal? As a trying, what is $B$ if I replace $Q_{\alpha\beta}$ by the ordering matrix  $S^{\alpha\beta}_{ij}=\frac{1}{2}\left(3i_\alpha j_\beta-\delta_{\alpha\beta}\delta_{ij}\right)$, regarding that $\vec{a}, \vec{b}, \vec{c}$ are the three orthogonal unit vectors linked to the molecule, with $i,j=a,b,c$ and $\alpha,\beta=x,y,z$? Can I expect that $B$ is a combination of the molecular susceptibilities $\chi_a, \chi_b, \chi_c$ on the $\vec{a}, \vec{b}, \vec{c}$ directions? If so, which combination is it?

UPDATE 2. I am more and more enjoying the possibility of using the ordering matrix $S^{\alpha\beta}_{ ij}$.

UPDATE 1. In  it is said that the diamagnetic anisotropy is written as $\Delta\chi=\chi_{zz}+\frac{1}{2}\left(\chi_{xx}+\chi_{yy}\right)$. If $\vec{a}, \vec{b}, \vec{c}$ are aligned to $\hat{x}, \hat{y}, \hat{z}$, can I regard $B\equiv\Delta\chi=\chi_{zz}+\frac{1}{2}\left(\chi_{xx}+\chi_{yy}\right)$ as the combination I'm looking for?