Q tensor

In physics, $$\mathbf Q$$ tensor is an orientational order parameter that describes uniaxial and biaxial nematic liquid crystals and vanishes in the isotropic liquid phase. The $$\mathbf Q$$ tensor is a second-order, traceless, symmetric tensor and is defined by


 * $$\mathbf{Q} = S\left(\mathbf n\mathbf n - \frac{1}{3}\mathbf I\right) + P\left(\mathbf m\mathbf m - \frac{1}{3}\mathbf I\right) $$

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


 * $$Q_{ij} = S\left(n_in_j - \frac{1}{3}\delta_{ij}\right) + P\left(m_im_j - \frac{1}{3}\delta_{ij}\right)$$

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

The $$\mathbf Q$$ tensor can always be diagonalized,



\mathbf Q= -\frac{1}{2}\begin{pmatrix} S+P & 0 &0 \\ 0 &S-P & 0 \\ 0 & 0& -2S\\ \end{pmatrix} $$

The following are the invariants of the $$\mathbf Q$$ tensor


 * $$\delta = Q_{ij}Q_{ij} = \frac{1}{2}(3S^2+P^2), \quad \Delta = Q_{ij}Q_{jk}Q_{ki} = \frac{3}{4}S(S^2-P^2);$$

the first-order invariant $$Q_{ii}=0$$ is trivial here. It can be shown that $$\delta^3\geq 6\Delta^2.$$

Uniaxial nematics
In uniaxial nematic liquid crystals, $$P=0$$ and therefore the $$\mathbf Q$$ tensor reduces to


 * $$\mathbf{Q} = S\left(\mathbf n\mathbf n - \frac{1}{3}\mathbf I\right).$$

The scalar order parameter is defined as follows. If $$\theta_{\mathrm{mol}}$$ represents the angle between the axis of a nematic molecular and the director axis $$\mathbf n$$, then


 * $$S = \langle P_2(\cos \theta_{\mathrm{mol}})\rangle = \frac{1}{2}\langle 3 \cos^2 \theta_{\mathrm{mol}}-1 \rangle = \frac{1}{2}\int (3 \cos^2 \theta_{\mathrm{mol}}-1)f(\theta_{\mathrm{mol}}) d\Omega$$

where $$\langle\cdot\rangle$$ denotes the ensemble average of the orientational angles calculated with respect to the distribution function $$f(\theta_{\mathrm{mol}})$$ and $$d\Omega = \sin \theta_{\mathrm{mol}}d\theta_{\mathrm{mol}}d\phi_{\mathrm{mol}}$$ is the solid angle. The distribution function must necessarily satisfy the condition $$f(\theta_{\mathrm{mol}}+\pi) = f(\theta_{\mathrm{mol}})$$ since the directors $$\mathbf n$$ and $$-\mathbf n$$ are physically equivalent.

The range for $$S$$ is given by $$-1/2\leq S\leq 1$$, 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.