Twist geometry

A diagram showing the twist geometry, where ${\displaystyle E_{t}}$ is the threshold electric field.

If a nematic liquid crystal that is confined between two parallel plates that induce a planar anchoring is placed in a sufficiently high constant electric field then the director will be distorted. If under zero field the director aligns along the x-axis then upon application of an electric field along the y-axis the director will be given by:

${\displaystyle \mathbf {\hat {n}} =n_{x}\mathbf {\hat {x}} +n_{y}\mathbf {\hat {y}} }$

${\displaystyle n_{x}=\cos {\theta (z)}}$

${\displaystyle n_{y}=\sin {\theta (z)}}$

Under this arrangement the distortion free energy density becomes:

${\displaystyle {\mathcal {F}}_{d}={\frac {1}{2}}K_{2}\left({\frac {d\theta }{dz}}\right)^{2}}$

The total energy per unit volume stored in the distortion and the electric field is given by:

${\displaystyle U={\frac {1}{2}}K_{2}\left({\frac {d\theta }{dz}}\right)^{2}-{\frac {1}{2}}\epsilon _{0}\Delta \chi _{e}E^{2}\sin ^{2}{\theta }}$

The free energy per unit area is then:

${\displaystyle F_{A}=\int _{0}^{d}{\frac {1}{2}}K_{2}\left({\frac {d\theta }{dz}}\right)^{2}-{\frac {1}{2}}\epsilon _{0}\Delta \chi _{e}E^{2}\sin ^{2}{\theta }\,dz\,}$

Minimizing this using calculus of variations gives:

${\displaystyle \left({\frac {\partial U}{\partial \theta }}\right)-{\frac {d}{dz}}\left({\frac {\partial U}{\partial \left({\frac {d\theta }{dz}}\right)}}\right)=0}$

${\displaystyle K_{2}\left({\frac {d^{2}\theta }{dz^{2}}}\right)+\epsilon _{0}\Delta \chi _{e}E^{2}\sin {\theta }\cos {\theta }=0}$

Rewriting this in terms of ${\displaystyle \zeta ={\frac {z}{d}}}$ and ${\displaystyle \xi _{d}=d^{-1}{\sqrt {\frac {K_{2}}{\epsilon _{0}\Delta \chi _{e}E^{2}}}}}$ where ${\displaystyle d}$ is the separation distance between the two plates results in the equation simplifying to:

${\displaystyle \xi _{d}^{2}\left({\frac {d^{2}\theta }{d\zeta ^{2}}}\right)+\sin {\theta }\cos {\theta }=0}$

By multiplying both sides of the differential equation by ${\displaystyle {\frac {d\theta }{d\zeta }}}$ this equation can be simplified further as follows:

${\displaystyle {\frac {d\theta }{d\zeta }}\xi _{d}^{2}\left({\frac {d^{2}\theta }{d\zeta ^{2}}}\right)+{\frac {d\theta }{d\zeta }}\sin {\theta }\cos {\theta }={\frac {1}{2}}\xi _{d}^{2}{\frac {d}{d\zeta }}\left(\left({\frac {d\theta }{d\zeta }}\right)^{2}\right)+{\frac {1}{2}}{\frac {d}{d\zeta }}\left(\sin ^{2}{\theta }\right)=0}$

${\displaystyle \int {\frac {1}{2}}\xi _{d}^{2}{\frac {d}{d\zeta }}\left(\left({\frac {d\theta }{d\zeta }}\right)^{2}\right)+{\frac {1}{2}}{\frac {d}{d\zeta }}\left(\sin ^{2}{\theta }\right)\,d\zeta \,=0}$

${\displaystyle {\frac {d\theta }{d\zeta }}={\frac {1}{\xi _{d}}}{\sqrt {\sin ^{2}{\theta _{m}}-\sin ^{2}{\theta }}}}$

The value ${\displaystyle \theta _{m}}$ is the value of ${\displaystyle \theta }$ when ${\displaystyle \zeta =1/2}$. Substituting ${\displaystyle k=\sin {\theta _{m}}}$ and ${\displaystyle t={\frac {\sin {\theta }}{\sin {\theta _{m}}}}}$ into the equation above and integrating with respect to ${\displaystyle t}$ from 0 to 1 gives:

${\displaystyle \int _{0}^{1}{\frac {1}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}\,dt\,\equiv K(k)={\frac {1}{2\xi _{d}}}}$

The value K(k) is the complete elliptic integral of the first kind. By noting that ${\displaystyle K(0)={\frac {\pi }{2}}}$ one finally obtains the threshold electric field ${\displaystyle E_{t}}$.

${\displaystyle E_{t}={\frac {\pi }{d}}{\sqrt {\frac {K_{2}}{\epsilon _{0}\Delta \chi _{e}}}}}$

As a result, by measuring the threshold electric field one can effectively measure the twist Frank constant so long as the anisotropy in the electric susceptibility and plate separation is known.