The NI transition is a first-order phase transition, albeit it is very weak. The order parameter is the \mathbf{Q} tensor, which is symmetric, traceless, second-order tensor and vanishes in the isotropic liquid phase. We shall consider a uniaxial \mathbf Q tensor, which is defined by :\mathbf Q = S(\mathbf n\otimes\mathbf n - \tfrac{1}{3}\mathbf I) where S=S(T) is the scalar order parameter and \mathbf n is the director. The \mathbf Q tensor is zero in the isotropic liquid phase since the scalar order parameter S is zero, but becomes non-zero in the nematic phase. Near the NI transition, the (
Helmholtz or
Gibbs) free energy density \mathcal{F} is expanded about as :\mathcal{F} = \mathcal{F}_0 + \frac{A}{2} Q_{ij}Q_{ji} - \frac{B}{3} Q_{ij}Q_{jk}Q_{ki} + \frac{C}{4} (Q_{ij}Q_{ij})^2 or more compactly :\mathcal{F} = \mathcal{F}_0 + \frac{A}{2}\mathrm{tr} \,\mathbf{Q}^2 - \frac{B}{3}\mathrm{tr} \,\mathbf{Q}^3 + \frac{C}{4}(\mathrm{tr} \,\mathbf{Q}^2)^2 where (A,B,C) are functions of temperature. Near the phase transition, we can expand A(T)=a (T-T_*)+\cdots, B(T) = b + \cdots and C(T)=c + \cdots with (a,b,c) being three positive constants. Now substituting the \mathbf Q tensor results in :\mathcal{F} - \mathcal{F}_0 = \frac{a}{3}(T-T_*)S^2 - \frac{2b}{27} S^3 + \frac{c}{9}S^4. This is minimized when :3a(T-T_*) S - b S^2 + 2c S^3=0. The two required solutions of this equation are :\begin{align}\text{Isotropic:} & \,\,S_I = 0,\\ \text{Nematic:} & \,\,S_N = \frac{b}{4c} \left[1+\sqrt{1-\frac{24ac}{b^2}(T-T_*)}\,\right]>0. \end{align} The NI transition temperature T_{NI} is not simply equal to T_* (which would be the case in second-order phase transition), but is given by :T_{NI} = T_* + \frac{b^2}{27ac}, \quad S_{NI} = \frac{b}{3c} S_{NI} is the scalar order parameter at the transition. ==References==