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When the birefringence separation of optical propagation constants $\Delta k=2\pi \Delta n/\lambda$ is large compared with the spatial rate of rotation of the optical axes, light with linear polarization in a local eigenpolarization follows the twisting nematic director. Thickness alone is insufficient if birefringence vanishes. Strong field alignment along the propagation direction suppresses the transverse optical anisotropy and hence the quarter-turn polarization rotation.

The dielectric response differs along and perpendicular to the nematic director. In Gaussian units the orientational electric contribution at prescribed field is $-\Delta ϵ(\mathbf{E}\cdot \mathbf{n}{)}^2/(8\pi )$; positive anisotropy favors alignment with the electric field. The electric term does not receive the additional one-half convention used in writing the Frank elastic terms.

For a unit nematic director, the bulk elastic density is $[K_1(\nabla \cdot \mathbf{n}{)}^2+K_2(\mathbf{n}\cdot \nabla \times \mathbf{n}{)}^2+K_3|\mathbf{n}\times \nabla \times \mathbf{n}{|}^2]/2$. The three positive moduli penalize nematic splay, nematic twist, and nematic bend. Boundary elastic terms, chirality, or varying nematic order require additional contributions.

The leading scalar coupling of an electric field to the nematic order parameter is quadratic in the field and linear in the tensor. For uniaxial nematic order, it is $-3\chi \lambda E^2({\cos }^2\vartheta -1/3)/2$. When $\chi \lambda >0$, minimization aligns the nematic director with the field axis and adds $-\chi E^2\lambda$ to the scalar Landau free energy.

Nematic bend curves the integral lines of the nematic director. For a unit field, $(\mathbf{n}\cdot \nabla )\mathbf{n}=-\mathbf{n}\times (\nabla \times \mathbf{n})$, so the bend energy measures curvature along those lines.

Nematic splay describes local spreading or convergence of the nematic director field. Its leading elastic cost is quadratic in the divergence and invariant under the head-tail identification $\mathbf{n}\sim -\mathbf{n}$.

Nematic twist rotates the nematic director around an axis perpendicular to it. The helix $\mathbf{n}=(\cos qz,\sin qz,0)$ has zero nematic splay and nematic bend, and twist density $K_2{q}^2/2$.

For a laterally uniform nematic director $(\cos \theta \cos \varphi ,\cos \theta \sin \varphi ,\sin \theta )$, its divergence is $\cos \theta {\theta }^{\prime }$ and its twist invariant is $-{\cos }^2\theta {\varphi }^{\prime }$. Thus $f=K_1{\cos }^2\theta +K_3{\sin }^2\theta$ and $g=K_2{\cos }^4\theta +K_3{\sin }^2\theta {\cos }^2\theta$. A normal electric field contributes $-c{\sin }^2\theta$. The Euler-Lagrange equations for two fields are $(g{\varphi }^{\prime }{)}^{\prime }=0$ and $f{\theta }^{\prime \prime }+f_{\theta }{\theta }^{\prime 2}/2-g_{\theta }{\varphi }^{\prime 2}/2+2c\sin \theta \cos \theta =0$.

The three contributions to the Frank elastic energy penalize distinct slow distortions of the nematic director. Nematic splay measures local spreading or convergence and is governed by $K_1(\nabla \cdot \mathbf{n}{)}^2/2$. Nematic twist measures rotation around an axis perpendicular to the director, with cost $K_2(\mathbf{n}\cdot \nabla \times \mathbf{n}{)}^2/2$. Nematic bend measures curvature of director integral lines; for a unit vector, $(\mathbf{n}\cdot \nabla )\mathbf{n}=-\mathbf{n}\times (\nabla \times \mathbf{n})$, giving cost $K_3|\mathbf{n}\times \nabla \times \mathbf{n}{|}^2/2$. These terms respect the head-tail symmetry of a nematic liquid crystal. Positive elastic constants oppose these distortions.

For the one-dimensional reduction assume lateral uniformity, so $\theta$ and $\varphi$ depend only on $z$. A prime below means $d/dz$. Directly differentiating the nematic director givesThe squared curl is ${\sin }^2\theta {\theta }^{\prime 2}+{\cos }^2\theta {\varphi }^{\prime 2}$. Subtracting the squared twist invariant givesConsequently the one-dimensional twisted nematic energy has elastic coefficientsA normal electric field has $\mathbf{E}\cdot \mathbf{n}=E_0\sin \theta$. There is a genuine factor-of-two inconsistency in the source: its initial functional places the electric term inside the overall factor $1/2$, whereas its subsequent density and threshold use the full electric term. For the requested density and threshold, use the usual orientational coupling $-ϵE_0^2{\sin }^2\theta /(8\pi )$. Here $ϵ$ is the dielectric anisotropy of a nematic, in the printed Gaussian-unit convention. If the first functional is instead read literally, this coupling is halved and the resulting threshold voltage is multiplied by $\sqrt{2}$.

Put $c=ϵE_0^2/(8\pi )$ and $\mathcal{L}=f{\theta }^{\prime 2}/2+g{\varphi }^{\prime 2}/2-c{\sin }^2\theta$. Apply the Euler-Lagrange equations for two fields. Variation in $\varphi$, which does not appear explicitly, givesVariation in $\theta$ gives $(f{\theta }^{\prime }{)}^{\prime }-f_{\theta }{\theta }^{\prime 2}/2-g_{\theta }{\varphi }^{\prime 2}/2+2c\sin \theta \cos \theta =0$, orAt $\theta =0$, $f=K_1$, $g=K_2$, and $f_{\theta }=g_{\theta }=0$. With zero field, ${\theta }^{\prime }=0$ and ${\varphi }^{\prime }=\pi /(2L)$ satisfy both equations and the imposed azimuthal endpoints. Thus the quarter-turn planar texture is a stationary solution. Stationarity alone is not a claim of stability for arbitrary elastic constants.

To find the Fréedericksz transition, impose the additional strong planar anchoring $\theta (0)=\theta (L)=0$. Write $\alpha =\pi /(2L)$. The small-angle expansions areThe conserved $g{\varphi }^{\prime }$ and fixed total twist imply ${\varphi }^{\prime }=\alpha +O({\theta }^2)$ for a pure tilt perturbation. Linearizing the tilt Euler-Lagrange equation therefore givesSubstitute $\theta ={\theta }_0\sin (\pi z/L)$. A nonzero neutral amplitude requiresSince $V_c=E_{c}L$, the quarter-turn nematic instability threshold isThis is the first zero-endpoint mode: higher sine modes replace $K_1$ by $n^2{K}_1$ and cost more elastic energy. The quadratic energy change is $\frac{1}{2}\int [K_1{\theta }^{\prime 2}+\{(K_3-2K_2){\alpha }^2-ϵE_0^2/(4\pi )\}{\theta }^2]dz$, so the first mode has negative stiffness above this voltage. The result assumes $ϵ>0$ and $K_1+(K_3-2K_2)/4>0$. If the latter is negative the planar state is already unstable at zero field; for negative dielectric anisotropy of a nematic, a normal field opposes this tilt. The literal initial electric prefactor would give $V_{c,\mathrm{literal}}=\sqrt{2}{V}_c$.

For reflective operation of the twisted nematic field effect, orient each polarizer's transmission axis along the adjacent planar nematic director. Light entering through the top polarizer initially has $y$ linear polarization. At low voltage, adiabatic optical following in a twisted nematic rotates it through a quarter turn to $x$ as it travels downward, so it passes the bottom polarizer and reaches the mirror. On returning, it follows the reverse twist back to $y$ and exits the top polarizer: the pixel appears bright. At sufficiently high voltage, positive dielectric anisotropy of a nematic aligns most of the director with the normal field. The bulk no longer produces the quarter-turn optical rotation, so the crossed bottom polarizer blocks the incident light and the pixel appears dark. Elastic relaxation restores the bright twisted state after voltage is removed. The adiabatic requirement is more precisely $2\pi |\Delta n|/\lambda \gg |{\varphi }^{\prime }|$, involving birefringence, not thickness alone. Voltage switches director orientation and hence the reflected brightness; it does not require the liquid crystal to emit light.

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.

For a strictly planar nematic director, the order-parameter manifold at fixed nonzero amplitude is ${\mathbb{RP}}^1$, homeomorphic to a circle. Its fundamental group is $\mathbb{Z}$. The topological charge of a two-dimensional nematic disclination is $q=\Delta \chi /(2\pi )$ for director angle $\chi$, defined modulo $\pi$; the integer winding of the physical doubled angle $2\chi$ is $m=2q$. Thus a $q=+1$ texture has $m=2\ne 0$ and cannot be continuously replaced by a uniform nonvanishing planar texture while the surrounding loop is held fixed:It can nevertheless split into two $q=+1/2$ defects, because their charges add to one. Such splitting preserves the total winding and is not topological unwinding. Allowing escape into the third dimension changes the manifold to ${\mathbb{RP}}^2$, with fundamental group $\mathbb{Z}/2\mathbb{Z}$; the doubled winding of this integer defect is then trivial. That escape is excluded by the stated two-dimensional director model.

Energetically, the defect is not the preferred isolated charge-one configuration at large separation of scales. At constant amplitude, the planar nematic divergence-elasticity identity gives $F_{\mathrm{el}}=(K{\lambda }_0^2/2)\int |\nabla \chi {|}^2$, so a charge $q$ has self-energy $\pi K{\lambda }_0^2{q}^2\log (L/r_0)$. Separating two $+1/2$ defects by $d$, with $r_0\ll d\ll L$, makes the far field still charge one but reduces the near-field logarithm:The logarithmic gain eventually dominates the finite core cost. HenceThis is energetic fission of an integer nematic disclination. It proves failure of global energetic stability, not the absence of every microscopic kinetic barrier or a full linear stability spectrum of the radial core. Core structure, boundaries or pinning can make a charge-one texture metastable over a finite time or length scale.

The total charge of the widely separated $+1,-1$ pair is zero, so a uniform planar nematic director is compatible with the far boundary if it imposes no additional winding. Each integer defect is susceptible to energetic fission of an integer nematic disclination: the positive defect can split into two $+1/2$ defects and the negative defect into two $-1/2$ defects. Like-sign elementary defects repel, whereas opposite signs attract through their elastic distortion. This commonly produces subsequent encounters and annihilations of opposite half-integer defects.

The attraction can also be seen before fission. In the fixed-amplitude outer theory, a neutral pair of charges $q,-q$ separated by $R$ hasIts energy therefore falls as the separation decreases. In nematic defect-antidefect annihilation, opposite defects meet, the amplitude can vanish in their merging cores, and the local winding disappears; the remaining smooth director distortion relaxes towards the uniform energy minimum.

Direct approach and annihilation of the unsplit integer pair is also topologically allowed. Whether fission happens first, the paths of the four elementary defects, and the annihilation time require a specified dynamical law and core mobilities; the free energy alone does not fix them. Pinning, imposed boundary textures, or sustained thermal production of defect pairs can prevent a strictly defect-free final state. The stated final state presumes dissipative relaxation with no such obstruction.

At $\xi =0$, the polar order parameter obeysFor $a<0$, $b>0$, the magnitude relaxes to $\sqrt{-a/b}$ while the direction rotates continuously. Thus the likely ordered behaviour is tumbling rather than a fixed alignment.

More generally, for $|\xi |<1$ and $g\ne 0$, $\xi \cos 2\theta -1$ never vanishes, so the tumbling of a polar order parameter persists. The polar order parameter completes a full $2\pi$ rotation inobtained by integrating $dt=2|d\theta |/(|g|(1-\xi \cos 2\theta ))$ over a full rotation. An unoriented nematic director repeats after half this period, whereas a polar order parameter distinguishes opposite directions. For $\xi \ne 0$ its magnitude can also oscillate. If the polar order parameter decays to zero, its orientation is no longer a physical observable; the tumbling conclusion concerns a persistent ordered state.

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.

A scalar coupling linear in the nematic order parameter requires a second-rank anisotropy. A single electric field supplies no such tensor, while $E_i{E}_j$ does. Its isotropic part gives $E^2\mathrm{Tr}Q=0$. Hence the lowest joint order in $E$ and $Q$ of the electric-field coupling to nematic order isTerms depending only on $E$ do not affect minimization over $Q$. With $\vartheta$ the angle between the nematic director and the electric field,For $\chi \lambda >0$, the lowest free energy occurs at ${\cos }^2\vartheta =1$, so the nematic director aligns along either $+\mathbf{E}$ or $-\mathbf{E}$. Then $E_i{Q}_{ij}{E}_j=\lambda E^2$, givingUnlike the linear composition term in a conserved binary fluid mixture, this term changes the equilibrium nematic order parameter: its spatial integral is not fixed.

A real symmetric traceless $2$ by $2$ matrix has eigenvalues $\lambda$ and $-\lambda$ and can be writtenfor a unit nematic director $\mathbf{n}\sim -\mathbf{n}$. Since $Q_{ij}{Q}_{ji}=2{\lambda }^2$, the uniform Landau free energy isFor $a<0<b$, its nonzero minima satisfyand henceThe signs $\pm {\lambda }_0$ amount to exchanging the two orthogonal eigenvectors, so the chosen convention takes ${\lambda }_0\ge 0$.