User:Chhe/Sandbox

Derivation

Twist Geometry

If a 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 the an electric field along the y-axis the director will be given by:

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

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

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

Under this arrangement the distortion free energy density becomes:

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

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

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

The free energy per unit area is then:

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

Minimizing this using the calculus of variations gives:

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

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

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

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

This equation simplifies further to:

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

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

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

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

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