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An apolar axis model with pair energy $-({\mathbf{s}}_i\cdot {\mathbf{s}}_j{)}^2$ is invariant when each molecular representative is independently reversed. The vector average changes under these replacements, but the nematic order parameter made from second moments does not. In an unbiased statistical ensemble the vector average vanishes; an individual finite representative need not have zero average. For independently chosen signs of a perfectly aligned axis, $⟨|N^{-1}{\sum }_i{\mathbf{s}}_i{|}^2⟩=1/N$. This distinguishes exact head-tail invariance of an observable from vanishing only after an average.^[ref]

The independent head-tail reversal symmetry means that the squared dot product interaction energy is unchanged when any individual ${\mathbf{s}}_i$ is replaced by $-{\mathbf{s}}_i$. All configurations ${\mathbf{s}}_i={\sigma }_i\mathbf{n}$, with ${\sigma }_i=\pm 1$ independently, attain the aligned ground-state pair energy. There is consequently no energetic preference for either end of a molecular axis. Averaging over these equally weighted signs gives $⟨{\mathbf{s}}_i⟩=0$ and $⟨\mathbf{m}⟩=0$ even in the nematically ordered phase. The surviving order is the unoriented axis, represented by a nematic director with $\mathbf{n}\sim -\mathbf{n}$. A second-moment tensor can detect this order while a polar vector cannot.^[ref]

The word “always” needs the statistical ensemble or apolar interpretation: it is not an identity for each finite configuration. For example, choosing all ${\sigma }_i=+1$ is also a ground-state representative and gives $\mathbf{m}=\mathbf{n}$. In an unbiased sign statistical ensemble the typical residual polar vector has size of order $N^{-1/2}$, because $⟨|\mathbf{m}{|}^2⟩=1/N$ for these independent ground-state signs. This does not make polar order an equilibrium nematic order parameter. The tensor below is unchanged by each sign flip and thus removes the arbitrary labeling of rod ends.

Choose a basis whose first axis is $\mathbf{n}$. The tensor with uniaxial nematic order then has eigenvalues $2Q,-Q,-Q$. Hence $\mathrm{Tr}{S}^2=6Q^2$, $\mathrm{Tr}{S}^3=6Q^3$ and $\mathrm{Tr}{S}^4=18Q^4$. Substitution givesFor $a,c>0$, $Q=0$ remains a local minimum with curvature $12a$. Its nonzero stationary points satisfy $2a+3bQ+12c{Q}^2=0$, soThey first appear at $b_s=-\sqrt{32ac/3}$ as a stationary inflection point; for more negative $b$, $Q_{-}$ is a local maximum and $Q_{+}$ a local minimum. The latter is initially metastable, becomes degenerate with $Q=0$ at $b_t=-\sqrt{12ac}$, and then becomes the global minimum as $b$ decreases further. This is a first-order transition in a cubic-quartic Landau potential: the order changes discontinuously. The saddle-node bifurcation creating a local minimum is not the coexistence condition. The isotropic minimum never loses local stability in this fixed positive-$a$ scan.

The illustrative plots set $a=c=1$ and show $f/6$. A cubic invariant is permitted because head-tail reversal leaves $S$ unchanged; it does not implement $S\mapsto -S$. The physical range of a uniaxial nematic order parameter imposes $-1/2\le Q\le 1$; the usual small-order-parameter Landau result assumes the displayed minimum lies in that interval. The unrestricted quartic graphs express the intended Landau model, with the effect of enforcing the bound stated in part (f).^[ref]

Author of Lectures on Landau Theory of Phase Transitions by Peter D. Olmsted, whose title page identifies the Department of Physics, Georgetown University, and the date 9 July 2015.