来源：‣（矢量分析与电磁场 → 麦克斯韦）
迁移说明：从正式 Notes 对应小节迁入，正文与数学表达式依原笔记保留；原图片和附件仍请查看来源。内容尚待校对。
$$ \begin{align*} c|\vec B|=&|\vec E| \Rightarrow v|\vec B|=|\vec E| \tag{11} \\ c =& \frac{1}{\sqrt {\mu_0\epsilon_0}} \Rightarrow v=\frac{1}{\sqrt {\mu\epsilon}}=\frac c {n}\Rightarrow n=\sqrt{\mu\epsilon}\tag{12}\\ \eta =& \sqrt{\frac{\mu}{\epsilon}}\Rightarrow \vec H = \frac{\vec{e_k}\times\vec E}{\eta}\tag{13}\\ \vec{S} = \vec{E}\times\vec{H}=&\frac{|\vec E|^2}{\eta}\vec{e_k}=vw\vec{e_k},w=\epsilon E^2 = \frac 1 \mu B^2=\frac 1 2\left(\epsilon E^2 + \frac 1 \mu B^2\right) \tag{14}\\ k =& \frac{\omega}{v}=\omega\sqrt {\mu\epsilon}\Rightarrow \frac{k_1}{k_2}=\frac{n_1}{n_2}=\frac{v_2}{v_1} \tag{15} \end{align*} $$
$$ \begin{align*} \vec{e_n}\times(E+E')=\vec{e_n}\times \vec{E''}\\ E+E'=E''\\
\Gamma_\perp = \frac{\eta_2\cos\theta_i-\eta_1\cos\theta_t}{\eta_2\cos\theta_i+\eta_1\cos\theta_t}=\frac{\sqrt\epsilon_1\cos\theta_i-\sqrt\epsilon_2\cos\theta_t}{\sqrt\epsilon_1\cos\theta_i+\sqrt\epsilon_1\cos\theta_t}\\
\tau_\perp=\frac{2\eta_2\cos\theta_i}{\eta_2\cos\theta_i+\eta_1\cos\theta_t}=\frac{2\sqrt\epsilon_1\cos\theta_i}{\sqrt\epsilon_1\cos\theta_i+\sqrt\epsilon_2\cos\theta_t}\\
\Gamma_{//} = \frac{\eta_1\cos\theta_i-\eta_2\cos\theta_t}{\eta_1\cos\theta_i+\eta_2\cos\theta_t}=\frac{\sqrt\epsilon_2\cos\theta_i-\sqrt\epsilon_1\cos\theta_t}{\sqrt\epsilon_2\cos\theta_i+\sqrt\epsilon_1\cos\theta_t}\\
\tau_{//} = \frac{2\eta_2\cos\theta_i}{\eta_1\cos\theta_i+\eta_2\cos\theta_t}=\frac{2\sqrt\epsilon_2\cos\theta_i}{\sqrt\epsilon_2\cos\theta_i+\sqrt\epsilon_1\cos\theta_t}\\ 平行入射无反射（布儒斯特）：\theta_{Bt}=\arctan\sqrt{\frac{\epsilon_2}{\epsilon_1}}\\ 全反射：\theta_{Cs}=\arcsin\sqrt{\frac{\epsilon_2}{\epsilon_1}}\\
\end{align*} $$
$$ \vec E(\vec x,t)=\vec E_0 e^{-\alpha z}e^{i(\beta z - \omega t)}, \alpha\approx\beta\approx\sqrt{\frac{\omega\mu\sigma}{2}}=\frac{1}{\delta}\\ \vec H = \frac{\vec k \times \vec E} {\omega \mu} $$
$$ \begin{align*} E_x =& A_1 \cos(k_xx)\sin(k_yy)e^{ik_zz}\\ E_y =& A_2 \cos(k_yy)\sin(k_xx)e^{ik_zz}\\ E_x =& A_3 \sin(k_xx)\sin(k_yy)e^{ik_zz}\\ \vec H =& - \frac{i}{\omega \mu}\nabla \times \vec E\\ k_xA_1&+k_yA_2-ikA_3=0 \end{align*} $$