# 物理代写|光学代写Optics代考|ECOC2022

## 物理代写|光学代写Optics代考|Field-induced Director Axis Rotation in SmC Liquid Crystals

In practical implementations or switching devices, the logical thing to do is to involve only one or a small number of these distortions. If an external field is applied, the field-dependent terms (cf. Eqs. $4.5 \mathrm{a}$ and $4.5 \mathrm{~b}$ ) should be added to the total free-energy expression.

The process of field-induced director axis distortion in $\mathrm{SmC}$ is analogous to the nematic case. For example, the first three terms on the right-hand side of (4.79) correspond to the splay term in nematics:
$$\frac{1}{2} K_1\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)^2=\frac{1}{2} K_1\left(\frac{\partial \Omega_x}{\partial y}-\frac{\partial \Omega_y}{\partial x}\right)^2 .$$
Accordingly, if only such distortions (i.e. no layer displacement or coupling effects) are induced in an SmC cell by an applied field, Freedericksz transitions (discussed in the previous chapter for nematics) will occur.

Consider, for example, the effect caused by a magnetic field as depicted in Figure 4.19. The applied field has three components, $H_1, H_2$ and $H_3$, and the respective diamagnetic susceptibility components are $\chi_1^m, \chi_2^m$ and $\chi_3^m ; \chi_3^m$ corresponds to the director axis, usually denoted as the $C$ axis; $\chi_2^m$ is along $\mathrm{y}$, and $\chi_1^m$ is in a direction orthogonal to both the three and $z$ axes.

If the applied field is along the $C$ axis (i.e. $H_3$ ), a Freedericksz transition is possible for $\chi_2^m>\chi_3^m$. The director should rotate around the $z$-axis so that, in the strong field limit, its projection onto the smectic layer coincides with the $y$ axis. There is no change in the tilt angle $\theta$. The threshold field for this process is
$$H_{3 c}=\frac{\pi \sin \theta}{d}\left(\frac{B}{\chi_2^m-\chi_3^m}\right)^{1 / 2},$$

## 物理代写|光学代写Optics代考|Free Energy of Ferroelectric Liquid Crystals

The free energy of ferroelectric liquid crystals is more complicated; besides the elastic energy $F_E$, several others play an equally important role in determining the response of the ferroelectric-liquid crystal to an external field. These include the surface energy density $F_s$, the spontaneous polarization density $F_P$, and the dielectric interaction energy density $F_{\text {diel }}$ with the applied field. These interactions have been studied by various workers; here we summarized the main results.

The elastic part of the free energy is analogous to the chiral nematic phase. It has been derived by [45] and expressed as follows:
$$F_K=\frac{1}{2}\left{K_1(\nabla \cdot \hat{n})^2+K_2\left(\hat{n} \cdot(\nabla \times \hat{n})+Q_T\right)^2+K_3\left[\hat{n} \times(\nabla \times \hat{n})+Q_B\right]^2\right}$$
where $Q_T$ and $Q_B$ are the inherent twist and bend wave numbers and $b=\hat{n} \times \hat{k}$, where $\hat{k}$ is the layer normal unit vector.
Using the geometry shown in Figure 4.10, the director axis $\hat{n}$ becomes
$$\hat{n}=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$
Equation (4.59) becomes
$$F_E=F_0+\frac{1}{2} A\left(1+v \sin ^2 \phi\right)\left(\frac{\partial \phi}{\partial y}\right)^2+A(1+v) Q_0 \sin \phi\left(\frac{\partial \phi}{\partial y}\right)$$
where
$$\begin{gathered} F_0=\frac{1}{2}\left[K_2 Q_T^2+K_3 Q_B^2\right] \ A=K_1 \sin ^2 \theta \ B=\left(K_2 \cos ^2 \theta+K_3 \sin ^2 \theta\right) \sin ^2 \theta \ v=\frac{B-A}{A} \ Q_0=\frac{K_2 Q_T \sin \theta \cos \theta+K_3 Q_B \sin ^2 \theta}{A(1+v)} \end{gathered}$$

## 物理代写|光学代写Optics代考|Field-induced Director Axis Rotation in SmC Liquid Crystals

$$\frac{1}{2} K_1\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)^2=\frac{1}{2} K_1\left(\frac{\partial \Omega_x}{\partial y}-\frac{\partial \Omega_y}{\partial x}\right)^2 .$$

$$H_{3 c}=\frac{\pi \sin \theta}{d}\left(\frac{B}{\chi_2^m-\chi_3^m}\right)^{1 / 2}$$

## 物理代写|光学代写Optics代考|Free Energy of Ferroelectric Liquid Crystals

$$\hat{n}=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$

$$F_E=F_0+\frac{1}{2} A\left(1+v \sin ^2 \phi\right)\left(\frac{\partial \phi}{\partial y}\right)^2+A(1+v) Q_0 \sin \phi\left(\frac{\partial \phi}{\partial y}\right)$$

$$F_0=\frac{1}{2}\left[K_2 Q_T^2+K_3 Q_B^2\right] \quad A=K_1 \sin ^2 \theta B=\left(K_2 \cos ^2 \theta+K_3 \sin ^2 \theta\right) \sin ^2 \theta v=\frac{B-A}{A} Q_0=\frac{K_2 Q_T \operatorname{si}}{x}$$

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