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For a scalar Landau-de Gennes free energy with $\bar{b}>0$, $\bar{c}<0$, and coupling $-\chi E^2\lambda$ with $\chi >0$, the critical endpoint of a quartic Landau free energy is reached at the displayed field strength and ${\bar{a}}_c=3{\bar{c}}^2/(8\bar{b})$, ${\lambda }_c=-\bar{c}/(4\bar{b})$. It terminates the field-biased first-order phase transition when both field and quadratic coefficient can be tuned. This is a quartic small-field mean-field approximation; higher terms may alter its location.

A nematic director is an unoriented axis: $\mathbf{n}$ and $-\mathbf{n}$ describe the same molecular alignment. A vector average cannot represent this head-tail symmetry. The orientational second moment $⟨n_i{n}_j⟩$ is a symmetric second-rank tensor, and removing its isotropic part gives the traceless second-rank tensor $Q_{ij}\propto ⟨n_i{n}_j-{\delta }_{ij}/2⟩$. It vanishes for an isotropic angular distribution and transforms correctly under rotation matrices.

Every nonzero real symmetric traceless $2\times 2$ matrix has eigenvalues $\lambda /2,-\lambda /2$ and can be written $Q=\lambda (\mathbf{n}{\mathbf{n}}^{\mathsf{T}}-I/2)$ by choosing the eigenvector of the positive eigenvalue as director and $\lambda \ge 0$. Consequently $\mathrm{Tr}{Q}^2={\lambda }^2/2$. The bulk Landau-de Gennes free energy becomesFor $a<0$, its minimum occurs at ${\lambda }^2=-a/b$. The elastic density $K|\nabla \cdot Q{|}^2/2$ is nonnegative and vanishes for a uniform field. Thus, with no boundary condition imposing a texture, the pointwise lower bound is attained byThe negative-amplitude representation is equivalent in two dimensions to rotating the director by $\pi /2$, so the positive choice loses no tensor minima. This is minimization of the specified coarse-grained functional, not a claim of true thermodynamic long-range nematic order at every temperature in an infinite two-dimensional system. The PDF explicitly uses the divergence elastic term; the TeX's $|\nabla Q{|}^2$ loses the dot and would give different elastic coefficients.

The nematic order parameter is a symmetric second-rank tensor and a traceless second-rank tensor describing orientational anisotropy. For molecular unit axes $\mathbf{s}$, $Q_{ij}$ is proportional to $⟨s_i{s}_j-{\delta }_{ij}/3⟩$. It vanishes in an isotropic phase and is unchanged by head-tail reversal $\mathbf{s}\mapsto -\mathbf{s}$, unlike a polar order parameter. For uniaxial nematic order, the nematic director is thus an unoriented axis.

One convention for the Landau-de Gennes free energy through fourth order, respecting rotational symmetry, isFor a traceless second-rank tensor in three dimensions, the Cayley-Hamilton theorem gives $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$, so there is only one independent quartic rotational invariant of a symmetric traceless tensor. The only possible linear scalar is $\mathrm{Tr}Q=0$; without an external anisotropy no linear term survives.

The cubic rotational invariant of a symmetric traceless tensor distinguishes prolate and oblate forms of uniaxial nematic order. Head-tail reversal leaves $Q$ unchanged and does not impose $Q\mapsto -Q$. Thus a cubic term is allowed in three dimensions and generically produces a first-order phase transition, rather than a symmetry-enforced continuous onset. In two dimensions a symmetric traceless second-rank tensor has eigenvalues $s,-s$, hence $\mathrm{Tr}{Q}^3=0$: the cubic invariant vanishes identically in two dimensions. This is a statement about the Landau free energy; fluctuations in two dimensions require a separate treatment.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.

Under rotation by a rotation matrix, a symmetric second-rank tensor transforms by conjugation, so its traces of powers are invariant. For a three-dimensional traceless second-rank tensor, the Cayley-Hamilton theorem implies $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$. In two dimensions its eigenvalues are $s,-s$, so $\mathrm{Tr}{Q}^3=0$. These identities organize the Landau-de Gennes free energy.