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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.