User:Chhe/Sandbox

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:
 * $$U=\frac{1}{2}K_2\left(\frac{d\theta}{dz}\right)^2-\frac{1}{2}\epsilon_0\Delta\chi_eE^2\sin^2{\theta}$$

The free energy per unit area is then:
 * $$F_A=\int_0^d\frac{1}{2}K_2\left(\frac{d\theta}{dz}\right)^2-\frac{1}{2}\epsilon_0\Delta\chi_eE^2\sin^2{\theta}\,dz \,$$

Minimizing this using the calculus of variations gives:
 * $$\left(\frac{\partial U}{\partial \theta}\right)-\frac{d}{dz}\left(\frac{\partial U}{\partial\left(\frac{d\theta}{dz}\right)}\right)=0$$
 * $$K_2\left(\frac{d^2\theta}{dz^2}\right)+\epsilon_0\Delta\chi_eE^2\sin{\theta}\cos{\theta}=0$$

Rewriting this in terms of $$\zeta=\frac{z}{d}$$ and $$\xi_d=d^{-1}\sqrt{\frac{K_2}{\epsilon_0\Delta\chi_eE^2}}$$ 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 $$E_t$$.
 * $$E_t=\frac{\pi}{d}\sqrt{\frac{K_2}{\epsilon_0\Delta\chi_e}}$$