GeezerVison

In physics, $𝐐$-tensor is an orientational order parameter that describes uniaxial and biaxial nematic liquid crystals and vanishes in the isotropic liquid phase.^[1] The $𝐐$ tensor is a second-order, traceless, symmetric tensor and is defined by^[2][3][4]

$𝐐=S(𝐧\otimes 𝐧-\frac{1}{3}𝐈)+R(𝐦\otimes 𝐦-\frac{1}{3}𝐈)$

where $S=S\left(T\right)$ and $R=R\left(T\right)$ are scalar order parameters, $(𝐧,𝐦)$ are the two directors of the nematic phase and $T$ is the temperature; in uniaxial liquid crystals, $R=0$. The components of the tensor are

$Q_{ij}=S(n_i{n}_j-\frac{1}{3}{\delta }_{ij})+R(m_i{m}_j-\frac{1}{3}{\delta }_{ij})$

The states with directors $𝐧$ and $-𝐧$ are physically equivalent and similarly the states with directors $𝐦$ and $-𝐦$ are physically equivalent.

The $𝐐$-tensor can always be diagonalized,

$𝐐=\frac{1}{3}\left[\begin{array}{ccc}2S-R& 0& 0\\ 0& 2R-S& 0\\ 0& 0& -S-R\end{array}\right]$

The following are the two invariants of the $𝐐$ tensor,

$tr{𝐐}^2=Q_{ij}{Q}_{ji}=\frac{2}{3}(S^2-SR+R^2),tr{𝐐}^3=Q_{ij}{Q}_{jk}{Q}_{ki}=\frac{1}{9}\left[2\right(S^3+R^3)-3SR(S+R\left)\right];$

the first-order invariant $tr𝐐=Q_{ii}=0$ is trivial here. It can be shown that $(tr{𝐐}^2{)}^3\ge 6(tr{𝐐}^3{)}^2.$ The measure of biaxiality of the liquid crystal is commonly measured through the parameter

$\beta =1-6\frac{(tr{𝐐}^3{)}^2}{(tr{𝐐}^2{)}^3}=\frac{27S^2{R}^2(S-R{)}^2}{4(S^2-SR+R^2{)}^3}.$

In uniaxial nematic liquid crystals, $R=0$ and therefore the $𝐐$-tensor reduces to

$𝐐=S(𝐧𝐧-\frac{1}{3}𝐈).$

The scalar order parameter is defined as follows. If ${\theta }_{mol}$ represents the angle between the axis of a nematic molecular and the director axis $𝐧$, then^[2]

$S=⟨P_2(\cos {\theta }_{mol})⟩=\frac{1}{2}⟨3{\cos }^2{\theta }_{mol}-1⟩=\frac{1}{2}\int (3{\cos }^2{\theta }_{mol}-1)f\left({\theta }_{mol}\right)d\Omega$

where $⟨\cdot ⟩$ denotes the ensemble average of the orientational angles calculated with respect to the distribution function $f\left({\theta }_{mol}\right)$ and $d\Omega =\sin {\theta }_{mol}d{\theta }_{mol}d{\varphi }_{mol}$ is the solid angle. The distribution function must necessarily satisfy the condition $f({\theta }_{mol}+\pi )=f\left({\theta }_{mol}\right)$ since the directors $𝐧$ and $-𝐧$ are physically equivalent.

The range for $S$ is given by $-1/2\le S\le 1$, with $S=1$ representing the perfect alignment of all molecules along the director and $S=0$ representing the complete random alignment (isotropic) of all molecules with respect to the director; the $S=-1/2$ case indicates that all molecules are aligned perpendicular to the director axis although such nematics are rare or hard to synthesize.

• Landau–de Gennes theory