电动力学 第7周作业

Chasse_neige

1. 理想绝缘介质中的波动方程及平面电磁波解

(a) 书4.4

若 \(\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}\) 仍按 \(e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)}\) 变化，但 \(\mathbf{D}\) 不再与 \(\mathbf{E}\) 平行（即 \(\mathbf{D} = \epsilon \mathbf{E}\) 不成立）。

(1) 证明： $$ \mathbf{k} \cdot \mathbf{B} = \mathbf{k} \cdot \mathbf{D} = \mathbf{B} \cdot \mathbf{D} = \mathbf{B} \cdot \mathbf{E} = 0 $$ 但一般 \(\mathbf{k} \cdot \mathbf{E} \neq 0\)。

证明： $$ \nabla \cdot \mathbf{B} = 0 $$ 所以 $$ \nabla \cdot \mathbf{B}_{0} e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} = i \mathbf{k} \cdot \mathbf{B} = 0 $$ 非线性极化介质中，由于\(\mathbf{D}\) 不再与 \(\mathbf{E}\) 平行，所以 \(\mathbf{E}\) 不一定在与 \(\mathbf{k}\) 垂直的平面内，所以一般 \(\mathbf{k} \cdot \mathbf{E} \neq 0\)。

同理，在没有自由电荷的空间中 $$ \nabla \cdot \mathbf{D} = 0 $$

\[ \nabla \cdot \mathbf{D}\_{0} e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} = i \mathbf{k} \cdot \mathbf{D} = 0 \]

由于 $$ \nabla \times \mathbf{B} = \mu \frac{\partial}{\partial t} \mathbf{D} $$ 所以 $$ \nabla \times \mathbf{B}_{0} e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} = \mu \frac{\partial}{\partial t} \mathbf{D}_{0} e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} $$

\[ \mathbf{k} \times \mathbf{B} = - \mu \omega \mathbf{D} \]

\[ \mathbf{B} \cdot \mathbf{D} = 0 \]

同理 $$ \nabla \times \mathbf{E} = - \frac{\partial}{\partial t} \mathbf{B} $$ 所以 $$ \nabla \times \mathbf{E}_{0} e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} = - \frac{\partial}{\partial t} \mathbf{B}_{0} e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} $$

\[ \mathbf{k} \times \mathbf{E} = \omega \mathbf{B} \]

\[ \mathbf{E} \cdot \mathbf{B} = 0 \]

(2) 证明： $$ \mathbf{D} = \frac{1}{\omega^2 \mu} \left[ k^2 \mathbf{E} - (\mathbf{k} \cdot \mathbf{E}) \mathbf{k} \right] $$

证明： $$ \mathbf{D} = - \frac{1}{\mu \omega} \mathbf{k} \times \mathbf{B} = - \frac{1}{\mu \omega} \mathbf{k} \times \left( \frac{1}{\omega} \mathbf{k} \times \mathbf{E} \right) = \frac{1}{\omega^2 \mu} \left[ k^2 \mathbf{E} - (\mathbf{k} \cdot \mathbf{E}) \mathbf{k} \right] $$

(3) 证明：

能流 \(\mathbf{S}\) 与波矢 \(\mathbf{k}\) 一般不在同一方向上。 证明： $$ \therefore \,\, \mathbf{k} \times \mathbf{S} = \mathbf{k} \times (\mathbf{E} \times \mathbf{H}) = - (\mathbf{k} \cdot \mathbf{E}) \mathbf{H} \neq 0 $$

所以能流 \(\mathbf{S}\) 与波矢 \(\mathbf{k}\) 一般不在同一方向上。

(b) 试证：对 \(\rho = 0, \vec{J} = 0\) 的麦克斯韦方程组，四个方程中只有两个是独立的。

证明：在无自由电荷和电流的情况下，麦克斯韦方程组为：

\[ \begin{aligned} \nabla \cdot \mathbf{E} &= 0 \\\\ \nabla \cdot \mathbf{B} &= 0 \\\\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\\\ \nabla \times \mathbf{B} &= \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \end{aligned} \]

可以通过两个旋度方程导出散度方程如下：

\[ \nabla \cdot (\nabla \times \mathbf{E}) = - \frac{\partial}{\partial t} (\nabla \cdot \mathbf{B}) = 0 \]

\[ \nabla \cdot (\nabla \times \mathbf{B}) = \mu_{0} \epsilon_{0} \frac{\partial}{\partial t} (\nabla \cdot \mathbf{E}) = 0 \]

所以 $$ \begin{aligned} \frac{\partial}{\partial t} \nabla \cdot \mathbf{E} &= 0 \\ \frac{\partial}{\partial t} \nabla \cdot \mathbf{B} &= 0 \end{aligned} $$ 所以对 \(\rho = 0, \vec{J} = 0\) 的麦克斯韦方程组，四个方程中是独立的，我认为无法从两个方程推出其余的两个。

2. 定态波动方程及平面波解

(a) 书4.1

考虑两列振幅相同、偏振方向相同、频率分别为 \(\omega + d\omega\) 和 \(\omega - d\omega\) 的线偏振平面波，它们都沿 z 轴方向传播。

(1) 求合成波，证明波的振幅不是一个常数，而是一个波。 合成波：\((k = \frac{\omega}{c})\) $$ A \cos ((k + dk)z - (\omega + d \omega) t) + A \cos ((k - dk)z - (\omega - d \omega) t) = 2 A \cos (dk z - d \omega t) \cos(kz - \omega t) $$ 所以合成波的振幅为 \(2 A \cos (d k z - d \omega t)\)，是一个波。 (2) 求合成波的相位传播速度和振幅传播速度。 相位传播速度： $$ v_{p} = \frac{\omega}{k} $$ 振幅传播速度： $$ v_{g} = \frac{d \omega}{d k} $$ (b) 试证,用复数方法得到的麦克斯韦方程组平面波解(包括 \(\gamma \neq 0\)),在取实部后仍是麦克斯韦方程组的解.

证明：

使用复数法解麦克斯韦方程组： $$ \begin{aligned} \nabla \cdot \mathbf{E} &= 0 \\ \nabla \times \mathbf{E} &= - \frac{\partial}{\partial t} \mathbf{B} \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{B} &= \mu (\gamma \mathbf{E} + \epsilon \frac{\partial}{\partial t} \mathbf{E}) \end{aligned} $$

带入波动假设： $$ \begin{aligned} \mathbf{k} \cdot \mathbf{E} &= 0 \\ \mathbf{k} \times \mathbf{E} &= \omega \mathbf{B} \\ \mathbf{k} \cdot \mathbf{B} &= 0 \\ \mathbf{k} \times \mathbf{B} &= - \mu (i \gamma + \epsilon \omega) \mathbf{E} \end{aligned} $$ 所以 $$ \mathbf{k} \times (\mathbf{k} \times \mathbf{B}) = - k^{2} \mathbf{B} = - \mu (i \gamma + \epsilon \omega) \omega \mathbf{B} $$ 即 $$ \mathbf{k}^{2} = \mu \epsilon \omega^{2} + i \gamma \mu \omega $$ 假设 \(\mathbf{k} = \mathbf{k_{R}} + i \mathbf{k_{I}}\)，则 $$ \begin{aligned} \mathbf{k_{R}^{2}} - \mathbf{k_{I}^{2}} &= \mu \epsilon \omega^{2} \\ 2 \mathbf{k_{R}} \cdot \mathbf{k_{I}} &= \gamma \mu \omega \end{aligned} $$ 所以对应的平面波解为： $$ \begin{aligned} \mathbf{E} &= (\mathbf{E}_{0R} + i \mathbf{E}_{0I}) e^{i(\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{k}_{I} \cdot \mathbf{r}} \\ \mathbf{B} &= (\mathbf{B}_{0R} + i \mathbf{B}_{0I}) e^{i(\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{k}_{I} \cdot \mathbf{r}} \end{aligned} $$ 并且有 $$ \begin{aligned} \mathbf{E}_{0R} \cdot \mathbf{k}_{R} - \mathbf{E}_{0I} \cdot \mathbf{k}_{I} = 0 \\ \mathbf{E}_{0I} \cdot \mathbf{k}_{R} + \mathbf{E}_{0R} \cdot \mathbf{k}_{I} = 0 \\ \mathbf{B}_{0R} = \frac{\mathbf{k}_{R}}{\omega} \times \mathbf{E}_{0R} - \frac{\mathbf{k}_{I}}{\omega} \times \mathbf{E}_{0I} \\ \mathbf{B}_{0I} = \frac{\mathbf{k}_{I}}{\omega} \times \mathbf{E}_{0R} + \frac{\mathbf{k}_{R}}{\omega} \times \mathbf{E}_{0I} \\ \mathbf{B}_{0R} \cdot \mathbf{k}_{R} - \mathbf{B}_{0I} \cdot \mathbf{k}_{I} = 0 \\ \mathbf{B}_{0I} \cdot \mathbf{k}_{R} + \mathbf{B}_{0R} \cdot \mathbf{k}_{I} = 0 \end{aligned} $$

取实部： $$ \begin{aligned} \Re [\mathbf{E}] &= (\mathbf{E}_{0R} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{E}_{0I} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} \\ \Re [\mathbf{B}] &= (\mathbf{B}_{0R} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{B}_{0I} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} \end{aligned} $$

带入麦克斯韦方程组：

$$ \begin{aligned} &\nabla \cdot (\mathbf{E}_{0R} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{E}_{0I} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} \\ &= \mathbf{E}_{0Ri} \partial_{i} (\cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}}) - \mathbf{E}_{0Ii} \partial_{i} (\sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}}) \\ &= \mathbf{E}_{0Ri} (- \mathbf{k}_{Ri} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)- \mathbf{k}_{Ii} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} - \\ &\mathbf{E}_{0Ii} (\mathbf{k}_{Ri} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{k}_{Ii} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) e ^{- \mathbf{k}_{I} \cdot \mathbf{r}} \\ &= e^{- \mathbf{k}_{I} \cdot \mathbf{r}} (- (\mathbf{E}_{0R} \cdot \mathbf{k}_{I} + \mathbf{E}_{0I} \cdot \mathbf{k}_{R}) \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) + (\mathbf{E}_{0I} \cdot \mathbf{k}_{I} - \mathbf{E}_{0R} \cdot \mathbf{k}_{R}) \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) \\ &= 0 \end{aligned} $$ 磁场同理。

考虑旋度方程： $$ \begin{aligned} &\nabla \times (\mathbf{E}_{0R} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) - \mathbf{E}_{0I} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t)) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} \\ &= \nabla (\cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}}) \times \mathbf{E}_{0R} - \nabla (\sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}}) \times \mathbf{E}_{0I} \\ &= (- \mathbf{k}_{R} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} - \mathbf{k}_{I} \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}}) \times \mathbf{E}_{0R} - \\ &( \mathbf{k}_{R} \cos(\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} - \mathbf{k}_{I} \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}}) \times \mathbf{E}_{0I} \\ &= - \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} \mathbf{B}_{0I} \omega - \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} \mathbf{B}_{0R} \omega = - \frac{\partial}{\partial t} \mathbf{B} \end{aligned} $$

\[ \begin{aligned} \nabla \times \mathbf{B} &= \nabla \times (\mathbf{B}\_{0R} \cos (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) - \mathbf{B}\_{0I} \sin (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t)) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}} \\\\ &= \nabla (\cos (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}}) \times \mathbf{B}\_{0R} - \nabla (\sin (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}}) \times \mathbf{B}\_{0I} \\\\ &= (- \mathbf{k}\_{R} \sin (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}} - \mathbf{k}\_{I} \cos (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}}) \times \mathbf{B}\_{0R} - \\\\ & ( \mathbf{k}\_{R} \cos(\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}} - \mathbf{k}\_{I} \sin (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}}) \times \mathbf{B}\_{0I} \\\\ &= - \cos (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}} (\mathbf{k}\_{I} \times \mathbf{B}\_{0R} + \mathbf{k}\_{R} \times \mathbf{B}\_{0I}) - \sin (\mathbf{k}\_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}\_{I} \cdot \mathbf{r}} (\mathbf{k}\_{R} \times \mathbf{B}\_{0R} - \mathbf{k}\_{I} \times \mathbf{B}\_{0I}) \end{aligned} \]

由于 $$ \begin{aligned} &\mathbf{k}_{I} \times \mathbf{B}_{0R} + \mathbf{k}_{R} \times \mathbf{B}_{0I} \\ &= \mathbf{k}_{I} \times (\frac{\mathbf{k}_{R}}{\omega} \times \mathbf{E}_{0R} - \frac{\mathbf{k}_{I}}{\omega} \times \mathbf{E}_{0I}) + \mathbf{k}_{R} \times (\frac{\mathbf{k}_{I}}{\omega} \times \mathbf{E}_{0R} + \frac{\mathbf{k}_{R}}{\omega} \times \mathbf{E}_{0I}) \\ &= \frac{1}{\omega} ((\mathbf{k}_{R} \cdot \mathbf{E}_{0R}) \mathbf{k}_{I} + (\mathbf{k}_{I} \cdot \mathbf{E}_{0R}) \mathbf{k}_{R} - 2 (\mathbf{k}_{I} \cdot \mathbf{k}_{R}) \mathbf{E}_{0R} + (\mathbf{k}_{R} \cdot \mathbf{E}_{0I}) \mathbf{k}_{R} - (\mathbf{k}_{I} \cdot \mathbf{E}_{0I}) \mathbf{k}_{I} + (\mathbf{k}_{I}^{2} - \mathbf{k}_{R}^{2}) \mathbf{E}_{0I}) \\ &= \frac{1}{\omega} (- \mu \epsilon \omega^{2} \mathbf{E}_{0I} - \gamma \mu \omega \mathbf{E}_{0R}) = - \mu \epsilon \omega \mathbf{E}_{0I} - \gamma \mu \mathbf{E}_{0R} \end{aligned} $$

\[ \begin{aligned} &\mathbf{k}\_{R} \times \mathbf{B}\_{0R} - \mathbf{k}\_{I} \times \mathbf{B}\_{0I} \\\\ &= \mathbf{k}\_{R} \times (\frac{\mathbf{k}\_{R}}{\omega} \times \mathbf{E}\_{0R} - \frac{\mathbf{k}\_{I}}{\omega} \times \mathbf{E}\_{0I}) - \mathbf{k}\_{I} \times (\frac{\mathbf{k}\_{I}}{\omega} \times \mathbf{E}\_{0R} + \frac{\mathbf{k}\_{R}}{\omega} \times \mathbf{E}\_{0I}) \\\\ &= \frac{1}{\omega} (- (\mathbf{k}\_{R} \cdot \mathbf{E}\_{0I}) \mathbf{k}\_{I} - (\mathbf{k}\_{I} \cdot \mathbf{E}\_{0I}) \mathbf{k}\_{R} + 2 (\mathbf{k}\_{I} \cdot \mathbf{k}\_{R}) \mathbf{E}\_{0I} + (\mathbf{k}\_{R} \cdot \mathbf{E}\_{0R}) \mathbf{k}\_{R} - (\mathbf{k}\_{I} \cdot \mathbf{E}\_{0R}) \mathbf{k}\_{I} + (\mathbf{k}\_{I}^{2} - \mathbf{k}\_{R}^{2}) \mathbf{E}\_{0R}) \\\\ &= \frac{1}{\omega} (- \mu \epsilon \omega^{2} \mathbf{E}\_{0R} + \gamma \mu \omega \mathbf{E}\_{0I}) = - \mu \epsilon \omega \mathbf{E}\_{0R} + \gamma \mu \mathbf{E}\_{0I} \end{aligned} \]

所以 $$ \begin{aligned} &- \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} (\mathbf{k}_{I} \times \mathbf{B}_{0R} + \mathbf{k}_{R} \times \mathbf{B}_{0I}) - \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} (\mathbf{k}_{R} \times \mathbf{B}_{0R} - \mathbf{k}_{I} \times \mathbf{B}_{0I}) \\ &= - \cos (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} (- \mu \epsilon \omega \mathbf{E}_{0I} - \gamma \mu \mathbf{E}_{0R}) - \sin (\mathbf{k}_{R} \cdot \mathbf{r} - \omega t) e^{- \mathbf{k}_{I} \cdot \mathbf{r}} (- \mu \epsilon \omega \mathbf{E}_{0R} + \gamma \mu \mathbf{E}_{0I}) \\ &= \mu (\gamma \mathbf{E} + \epsilon \frac{\partial}{\partial t} \mathbf{E}) \end{aligned} $$ 所以该实部满足麦克斯韦方程组。

(c) 试求穿透深度的严格表达式，并从它出发证明在 \(\gamma\) 很大时 \(\delta \to \sqrt{\frac{2}{\omega \mu \gamma}}\)

穿透深度： $$ \begin{aligned} k_{R}^{2} - k_{I}^{2} &= \mu \epsilon \omega^{2} \\ 2 k_{R} k_{I} \cos \theta &= \gamma \mu \omega \end{aligned} $$ 所以 $$ \begin{aligned} \left( \frac{\gamma \mu \omega}{2 k_{I} \cos \theta} \right )^{2} - k_{I}^{2} &= \mu \epsilon \omega^{2} \\ k_{I}^{4} + \mu \epsilon \omega^{2} k_{I}^{2} - \left( \frac{\gamma \mu \omega}{2 \cos \theta} \right )^{2} &= 0 \\ k_{I}^{2} &= \frac{- \mu \epsilon \omega^{2} + \sqrt{(\mu \epsilon \omega^{2})^{2} + 4 \left( \frac{\gamma \mu \omega}{2 \cos \theta} \right )^{2}}}{2} \end{aligned} $$ 所以穿透深度的严格表达式为 $$ \delta = \frac{1}{k_{I} \cos \theta} = \frac{1}{\cos \theta \sqrt{\frac{- \mu \epsilon \omega^{2} + \sqrt{(\mu \epsilon \omega^{2})^{2} + 4 \left( \frac{\gamma \mu \omega}{2 \cos \theta} \right )^{2}}}{2}}} $$ 当\(\gamma \gg 1\) 时， $$ \frac{- \mu \epsilon \omega^{2} + \sqrt{(\mu \epsilon \omega^{2})^{2} + 4 \left( \frac{\gamma \mu \omega}{2 \cos \theta} \right )^{2}}}{2} \approx \frac{- \mu \epsilon \omega^{2} + 2 \left( \frac{\gamma \mu \omega}{2 \cos \theta} \right )}{2} \approx \frac{\gamma \mu \omega}{2 \cos \theta} $$ 所以 $$ \delta \to \sqrt{\frac{2}{\gamma \mu \omega \cos \theta}} $$ 当相位传播方向与衰减方向相同时 \(\delta \to \sqrt{\frac{2}{\omega \mu \gamma}}\)

(d) 试讨论在什么条件下,椭圆偏振是左旋或右旋的?

当椭圆偏振光对应的电矢量在两正交坐标轴上投影分量相等，且\(x\) 分量相位领先 \(y\) 分量 \(\frac{\pi}{2}\) 时为左旋光；\(x\) 分量相位落后 \(y\) 分量 \(\frac{\pi}{2}\) 时为右旋光。

3. 电磁波在界面上的反射和折射

(a) 书4.2, 4.6

4.2 一平面电磁波以 \(\theta = 45^\circ\) 从真空入射到 \(\varepsilon_r = 2\) 的介质，电场强度垂直于入射面。求反射系数和折射系数。

电场强度垂直于入射面，所以是 \(s\) 光，入射的介质折射率为 \(\sqrt{2}\)。

入射角\(\theta = 45^\circ\) ，折射角 \(\sqrt{2} \sin \theta' = \sin \theta\)，所以 \(\theta' = \frac{\pi}{6}\)

利用 \(s\) 光的菲涅耳公式： $$ r_{s} = \frac{n_{1} \cos \theta - n_{2} \cos \theta'}{n_{1} \cos \theta + n_{2} \cos \theta'} = \frac{\frac{\sqrt{2}}{2} - \sqrt{2} \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2} + \sqrt{2} \frac{\sqrt{3}}{2}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} $$

所以 $$ R = r_{s}^{2} = \frac{2 - \sqrt{3}}{2 + \sqrt{3}} $$

\[ T = 1 - R = \frac{2\sqrt{3}}{2 + \sqrt{3}} \]

4.6 平⾯电磁波垂直射到⾦属表⾯上，试证明透入⾦属内部的电磁波能量全部变为焦⽿热。

证明：假设入射波矢为 \(\mathbf{k_{1}}\)，透射为 \(\mathbf{k_{2}}\) ，反射为\(\mathbf{k_{3}}\)

当平⾯电磁波垂直射到⾦属表⾯上时（金属内部\(\gamma \gg 1\)）：

我们假设振幅透射率为 \(t\)

透射部分的衰减：

由于此时相位传播方向和衰减方向同向，所以 $$ k_{I}^{2} = \frac{- \mu \epsilon \omega^{2} + \sqrt{(\mu \epsilon \omega^{2})^{2} + \left( \gamma \mu \omega \right )^{2}}}{2} $$ 进入部分的电磁波能流： $$ \mathbf{S} = \Re[\mathbf{E} \times \mathbf{H}] = \Re[\mathbf{E} \times (\frac{\mathbf{k}}{\mu \omega} \times \mathbf{E})] = t^{2} \mathbf{E}_{0}^{2} \frac{\mathbf{k_{R}}}{\mu \omega} $$ 所以 $$ <\mathbf{S}> = \frac{1}{2} t^{2} \mathbf{E}_{0}^{2} \frac{\mathbf{k}_{R}}{\omega} $$ 焦耳热： $$ \mathbf{j} = \gamma \mathbf{E} $$ 发热功率密度 $$ w = \frac{1}{\gamma} j^{2} = \gamma \mathbf{E}^{2} $$ 带入上一题推导出的衰减结果 $$ \mathbf{E} = t \mathbf{E}_{0} e^{i(k_{R} z - \omega t) - k_{I} z} $$

\[ <w> = \frac{1}{2} \gamma t^{2} \mathbf{E}_{0}^{2} e^{- 2 k_{I} z} \]

所以

\[ \int_{0}^{\infty} \langle w \rangle d z = \frac{\gamma t^{2} \mathbf{E}_{0}^{2}}{4 k_{I}} \]

由于\(2 k_{R} k_{I} = \gamma \mu \omega\)，所以 $$ \frac{\gamma t^{2} \mathbf{E}_{0}^{2}}{4 k_{I}} = \frac{1}{2} t^{2} \mathbf{E}_{0}^{2} \frac{k_{R}}{\mu \omega} $$ 即入射能量全部转化为焦耳热。

(b) 试证：当入射波电场垂直入射面时，反射和透射振幅满足：

\[ \vec{E}_{20\perp} = \frac{2 E_{10\perp} \vec{e}_y}{1 + \frac{\mu_1 k_{2z}}{\mu_2 k_1 \cos \theta_1}} \]

\[ \vec{E}_{30\perp} = \frac{1 - \frac{\mu_1 k_{2z}}{\mu_2 k_1 \cos \theta_1}}{1 + \frac{\mu_1 k_{2z}}{\mu_2 k_1 \cos \theta_1}} E_{10\perp} \vec{e}_y \]

进一步当 \(\mu_1 = \mu_2, \gamma_2 = 0\) 时，它们化为：

\[ \vec{E}_{20\perp} = \frac{2 \cos \theta_1 \sin \theta_2}{\sin (\theta_1 + \theta_2)} E_{10\perp} \vec{e}_y \]

\[ \vec{E}_{30\perp} = \frac{\sin (\theta_2 - \theta_1)}{\sin (\theta_2 + \theta_1)} E_{10\perp} \vec{e}_y \]

证明：当入射波电场垂直入射时，电磁波在介质界面处满足以下边界条件：（下述所有电磁场和波矢均为复数）

电场的切向分量连续： $$ E_{1y} + E_{3y} = E_{2y} $$ 磁场的切向分量连续： $$ H_{1x} + H_{3x} = H_{2x} $$ 磁感应强度法向连续： $$ \mu_{1}(H_{1z} + H_{3z}) = \mu_{2} H_{2z} $$ 入射波的磁场： $$ \mathbf{H}_{1} = \frac{\mathbf{k}_{1} \times \mathbf{E}_{1}}{\omega \mu_1} $$ 反射波的磁场： $$ \mathbf{H}_{3} = \frac{\mathbf{k}_{1} \times \mathbf{E}_{3}}{\omega \mu_1} $$ 透射波的磁场： $$ \mathbf{H}_{2} = \frac{\mathbf{k}_{2} \times \mathbf{E}_{2}}{\omega \mu_2} $$ 将磁场分量代入边界条件：

磁场的 \(x\)分量连续： $$ \frac{k_{1z}}{\mu_1} (E_{10\perp} - E_{30\perp}) = \frac{k_{2z}}{\mu_2} E_{20\perp} $$ 磁感应强度的z分量连续： $$ k_{1x} (E_{10\perp} + E_{30\perp}) = k_{2x} E_{20\perp} $$ 从磁场的 \(z\)分量连续方程： $$ \frac{k_{1x}}{\mu_1} (E_{10\perp} - E_{30\perp}) = \frac{k_{1x}}{\mu_2} E_{20\perp} $$ $$ E_{10\perp} + E_{30\perp} = E_{20\perp} $$ 将此结果代入磁场强度的 \(x\)分量连续方程： $$ \frac{k_{1z}}{\mu_1} (E_{10\perp} - E_{30\perp}) = \frac{k_{2z}}{\mu_2} (E_{10\perp} +E_{30\perp}) $$ 解得： $$ E_{30\perp} = \frac{- \frac{k_{2z}}{\mu_{2}} + \frac{k_{1z}}{\mu_{1}}}{\frac{k_{1z}}{\mu_{1}} + \frac{k_{2z}}{\mu_{2}}} E_{10\perp} $$

带入 $k_{1z} = k_{1} \cos \theta_{1} $，进一步整理为： $$ E_{30\perp} = \frac{1 - \frac{\mu_1 k_{2z}}{\mu_2 k_1 \cos \theta_1}}{1 + \frac{\mu_1 k_{2z}}{\mu_2 k_1 \cos \theta_1}} E_{10\perp} $$

透射振幅由 \(E_{10\perp} + E_{30\perp} = E_{20\perp}\)给出： $$ E_{20\perp} = \frac{2 E_{10\perp}}{1 + \frac{\mu_1 k_{2z}}{\mu_2 k_1 \cos \theta_1}} $$

当 \(\mu_1 = \mu_2\) 且 \(\gamma_2 = 0\) 时， \(k_{2z} = k_{2} \cos \theta_{2}\)，并且 \(k_{1} \sin \theta_{1} = k_{2} \sin \theta_{2}\)： $$ E_{20\perp} = \frac{2 \cos \theta_1 \sin \theta_2}{\sin (\theta_1 + \theta_2)} E_{10\perp} $$ $$ E_{30\perp} = \frac{\sin (\theta_2 - \theta_1)}{\sin (\theta_2 + \theta_1)} E_{10\perp} $$

(c) 试证明：

\[ D_{\parallel} = \frac{4 \mu_1 \cos \theta_1 \text{Re} \left( \frac{k_2 k_{2z}^*}{k_{2}^{*}} \right)}{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2} \]

\[ R_{\parallel} = \left| \frac{\frac{k_{2z}}{k_2} - \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1}{\frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1} \right|^{2} \]

\[ D_{\parallel} + R_{\parallel} = 1 \]

证明：

对于平行时的能流透射与反射：用和（b）问相似的方法，容易得到

\[ E_{30 \parallel} = \frac{- \frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}}{\frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}} E_{10 \parallel} \]

\[ E_{20 \parallel} = \frac{2 \cos \theta_{1}}{\frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}} E_{10 \parallel} \]

所以能流的透射：

\[ D_{\parallel} = \frac{\mathbf{S}\_{2 \parallel z}}{\mathbf{S}\_{1 \parallel z}} = \frac{\frac{1}{2 \mu_{2} \omega} (k_{2Rz} \mathbf{E}\_{2 \parallel}^{2} - \Re [E_{2 \parallel z} (\mathbf{k}\_{2} \cdot \mathbf{E}\_{2 \parallel}^{*})])}{\frac{1}{2 \mu_{1} \omega} (k_{1Rz} \mathbf{E}\_{1 \parallel}^{2})} \]

带入\(\frac{\mathbf{E}\_{2 \parallel}^{2}}{\mathbf{E}\_{1 \parallel}^{2}} = \left|\frac{2 \cos \theta_{1}}{\frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}}\right|^{2}\)以及 \(k_{1Rz} = k_{1} \cos \theta_{1}\)

由于 \(\mathbf{k}\_{2} \cdot \mathbf{E}\_{2 \parallel} = 0\) ，所以

\[ \begin{aligned} \Re [E_{2 \parallel z} (\mathbf{k}\_{2} \cdot \mathbf{E}\_{2 \parallel}^{*})] &= - \Re [E_{2 \parallel z} (\mathbf{k}\_{2} \cdot 2 \Im [\mathbf{E}\_{2 \parallel}])] \\\\ &= - \Re [|\mathbf{E_{2 \parallel}}| \frac{k_{2z}}{ |\mathbf{k_{2}}|} (\mathbf{k}\_{2} \cdot 2 \Im [\mathbf{E}\_{2 \parallel}])] = \mathbf{E}\_{2}^{2} \left( - \Re[k_{2} \frac{k_{2z}}{k_{2}}^{*}] + \Re[k_{2z}] \right) \end{aligned} \]

得到：

\[ \begin{aligned} \frac{\frac{1}{2 \mu_{2} \omega} (k_{2Rz} \mathbf{E}\_{2 \parallel}^{2} - \Re [E_{2 \parallel z} (\mathbf{k}\_{2} \cdot \mathbf{E}\_{2 \parallel}^{*})])}{\frac{1}{2 \mu_{1} \omega} (k_{1Rz} \mathbf{E}\_{1 \parallel}^{2})} &= \frac{\mu_{1}}{\mu_{2} k_{1} \cos \theta_{1}} \left| \frac{2 \cos \theta_{1}}{\frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}} \right|^{2} \Re [k_{2} \frac{k_{2z}^{*}}{k_{2}^{*}}] \\\\ &= \frac{4 \mu_1 \cos \theta_1 \text{Re} \left( \frac{k_2 k_{2z}^*}{k_{2}^{*}} \right)}{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2} \end{aligned} \]

\[ R_{\parallel} = \frac{\mathbf{S}\_{3 \parallel z}}{\mathbf{S}\_{1 \parallel z}} = \frac{\frac{1}{2 \mu_{1} \omega} (k_{2Rz} \mathbf{E}\_{3 \parallel}^{2})}{\frac{1}{2 \mu_{1} \omega} (k_{1Rz} \mathbf{E}\_{1 \parallel}^{2})} \]

带入\(\frac{\mathbf{E}\_{3 \parallel}^{2}}{\mathbf{E}\_{1 \parallel}^{2}} = \left| \frac{- \frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}}{\frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}} \right|^{2}\)

以及 \(k_{1Rz} = k_{1} \cos \theta_{1}\) \(k_{3Rz} = k_{3} \cos \theta_{1}\)

\[ R_{\parallel} = \left| \frac{- \frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}}{\frac{k_{2z}}{k_{2}} + \frac{\mu_{1} k_{2} \cos \theta_{1}}{\mu_{2} k_{1}}} \right|^{2} \]

\[ \begin{aligned} D_{\parallel} + R_{\parallel} &= \frac{4 \mu_1 \cos \theta_1 \text{Re} \left( \frac{k_2 k_{2z}^*}{k_{2}^{*}} \right)}{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2} + \left| \frac{\frac{k_{2z}}{k_2} - \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1}{\frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1} \right|^{2} \\\\ &= \frac{4 \mu_1 \cos \theta_1 \text{Re} \left( \frac{k_2 k_{2z}^*}{k_{2}^{*}} \right) + \mu_{2} k_{1} \left| \frac{k_{2z}}{k_2} - \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^{2}}{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2} \\\\ &= \frac{\mu_{2} k_{1} \left(4 \frac{\mu_1}{\mu_{2} k_{1}} \cos \theta_1 \text{Re} \left( \frac{k_2 k_{2z}^*}{k_{2}^{*}} \right) + \left| \frac{k_{2z}}{k_2} - \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^{2}\right)}{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2} \\\\ &= \frac{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2}{\mu_2 k_1 \left| \frac{k_{2z}}{k_2} + \frac{k_2 \mu_1}{k_1 \mu_2} \cos \theta_1 \right|^2} = 1 \end{aligned} \]

其中利用了

\[ (z_{1} + z_{2}) (z_{1}^{*} + z_{2}^{*}) = (z_{1} - z_{2}) (z_{1}^{*} - z_{2}^{*}) + 4 \Re [z_{1} z_{2}^{*}] \]