*时域Maxwell方程

∇×E⃗=−∂B⃗∂t∇×H⃗=μ0J⃗+μ0ε0∂E⃗∂t∇⋅D⃗=rhoε0∇⋅B⃗=0\begin{aligned} &\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \\ &\nabla \times \vec{H} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t} \\ &\nabla \cdot \vec{D} = \frac{rho}{\varepsilon_0} \\ &\nabla \cdot \vec{B} = 0 \end{aligned}

*对于传播问题，ρ=0,J⃗=0\rho = 0,\vec{J} = 0

Maxwell方程组等价为

∇2E⃗−1c2∂2E⃗∂t2=0∇2B⃗−1c2∂2B⃗∂t2=0∇⋅E⃗=0∇⋅B⃗=0\begin{aligned} &\nabla^2 \vec{E} - \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} = 0 \\ &\nabla^2 \vec{B} - \frac{1}{c^2} \frac{\partial^2 \vec{B}}{\partial t^2} = 0\\ &\nabla \cdot \vec{E} = 0 \\ &\nabla \cdot \vec{B} = 0 \end{aligned}

其中c=1μ0ε0c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}为光速

*对于辐射问题，ρ≠0,J⃗≠0\rho \neq 0,\vec{J} \neq 0

考虑

E⃗=−∇φ−∂A⃗∂tB⃗=∇×A⃗∇⋅A⃗+1c2∂φ∂t=0\begin{aligned} &\vec{E} = -\nabla \varphi - \frac{\partial \vec{A}}{\partial t} \\ &\vec{B} = \nabla \times \vec{A} \\ &\nabla \cdot \vec{A} + \frac{1}{c^2} \frac{\partial \varphi}{\partial t} = 0 \end{aligned}

得到另一个等价形式

∇2φ−1c2∂2φ∂t2=−ρε0∇2A⃗−1c2∂2A⃗∂t2=−μ0J⃗\begin{aligned} &\nabla^2 \varphi - \frac{1}{c^2} \frac{\partial^2 \varphi}{\partial t^2} = -\frac{\rho}{\varepsilon_0} \\ &\nabla^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = -\mu_0 \vec{J} \\ \end{aligned}

*频域Maxwell方程

∇×E⃗=−iωB⃗∇×H⃗=J⃗−iωD⃗∇⋅D⃗=ρ∇⋅B⃗=0\begin{aligned} &\nabla \times \vec{E} = -i \omega \vec{B} \\ &\nabla \times \vec{H} = \vec{J} - i \omega \vec{D} \\ &\nabla \cdot \vec{D} = \rho \\ &\nabla \cdot \vec{B} = 0 \end{aligned}

频域本构关系

D⃗(ω)=ε(ω)E⃗(ω)B⃗(ω)=μ(ω)H⃗(ω)\begin{aligned} &\vec{D}(\omega) = \varepsilon(\omega) \vec{E}(\omega) \\ &\vec{B}(\omega) = \mu(\omega) \vec{H}(\omega) \end{aligned}

频域边值关系

n^×(E⃗2−E⃗1)=0n^×(H⃗2−H⃗1)=J⃗Sn^⋅(D⃗2−D⃗1)=ρSn^⋅(B⃗2−B⃗1)=0\begin{aligned} &\hat{n} \times (\vec{E}_2 - \vec{E}_1) = 0 \\ &\hat{n} \times (\vec{H}_2 - \vec{H}_1) = \vec{J}_S \\ &\hat{n} \cdot (\vec{D}_2 - \vec{D}_1) = \rho_S \\ &\hat{n} \cdot (\vec{B}_2 - \vec{B}_1) = 0 \end{aligned}

*时谐场

复指数形式：E⃗(x⃗,t)=E⃗(x⃗)e−iωt\vec{E}(\vec{x},t)=\vec{E}(\vec{x})e^{-i\omega t}
瞬时值形式：E⃗(x⃗,t)=Re{E⃗(x⃗)e−iωt}\vec{E}(\vec{x},t)=Re\{\vec{E}(\vec{x})e^{-i\omega t}\}

*频域传播问题

∇×E⃗=iωB⃗∇×H⃗=−iωD⃗\begin{aligned} &\nabla \times \vec{E} = i \omega \vec{B} \\ &\nabla \times \vec{H} = -i \omega \vec{D} \\ \end{aligned}

等价于

均匀平面波

E⃗(x⃗)=E0⃗eik⃗⋅x⃗\vec{E}(\vec{x}) = \vec{E_0} e^{i \vec{k} \cdot \vec{x}}

H⃗(x⃗)=H0⃗eik⃗⋅x⃗\vec{H}(\vec{x}) = \vec{H_0} e^{i \vec{k} \cdot \vec{x}}

H⃗0=k^×E⃗0η\vec{H}_0 = \hat{k} \times \frac{\vec{E}_0}{\eta}

其中η=με\eta = \sqrt{\frac{\mu}{\varepsilon}}为波阻抗

*在无损介质面的反射与折射

边值关系：

n^×(E⃗2−E⃗1)=0n^×(H⃗2−H⃗1)=0\begin{aligned} \hat{n} \times (\vec{E}_2 - \vec{E}_1) &= 0 \\ \hat{n} \times (\vec{H}_2 - \vec{H}_1) &= 0 \\ \end{aligned}

  1. 方向：Snell定律
    θr=θi\theta_r = \theta_i
    sin⁡θisin⁡θt=n2n1\frac{\sin \theta_i}{\sin \theta_t} = \frac{n_2}{n_1}
    振幅：Frenel公式
    N波，P波
  2. 关于反射波：
    (1)n1<n2n_1 < n_2，N波有半波损失（光疏入射光密）
    (2)n1>n2n_1 > n_2，θi>θc=arcsin⁡n21\theta_i > \theta_c = \arcsin n_{21}
    全反射
    (3)Brewster角：θB=arctan⁡n2n1\theta_B = \arctan \frac{n_2}{n_1}，P波无反射，反射波中只有N波
  3. 良导体σωε>>1\frac{\sigma}{\omega \varepsilon} >> 1

∇×E⃗=−iωB⃗∇×H⃗=−iωε~E⃗\begin{aligned} &\nabla \times \vec{E} = -i \omega \vec{B} \\ &\nabla \times \vec{H} = -i \omega \tilde{\varepsilon} \vec{E} \\ \end{aligned}

其中ε~=ε0+iσω\tilde{\varepsilon} = \varepsilon_0 + i\frac{\sigma}{\omega}
穿透深度：δ=2ωμσ\delta = \sqrt{\frac{2}{\omega \mu \sigma}}

E⃗(x⃗)=E0⃗e−α⃗⋅x⃗eiβ⃗⋅x⃗\vec{E}(\vec{x}) = \vec{E_0} e^{-\vec{\alpha}\cdot \vec{x}} e^{i\vec{\beta}\cdot\vec{x}}

表面电阻：RS=1σδR_S = \frac{1}{\sigma\delta}
表面电流：JS⃗≈n^×H⃗∣S\vec{J_S} \approx \hat{n}\times \vec{H}|_S
单位面积焦耳热：Pd=12∣JS⃗∣RSP_d = \frac{1}{2}|\vec{J_S}|R_S
4. 理想导体：σ→∞\sigma \rightarrow \infty
边值关系：

n^×E⃗=0本质约束条件n^×H⃗=J⃗sn^⋅D⃗=ρsn^⋅B⃗=0\begin{aligned} &\hat{n} \times \vec{E} = 0 \quad \text{本质约束条件} \\ &\hat{n} \times \vec{H} = \vec{J}_s \\ &\hat{n} \cdot \vec{D} = \rho_s \\ &\hat{n} \cdot \vec{B} = 0 \end{aligned}

a. 波导
b. 谐振腔

*辐射问题

∇⋅J⃗−iωρ=0∇⋅A⃗−iωc2φ=0\begin{aligned} &\nabla \cdot \vec{J} - i\omega \rho = 0\\ &\nabla \cdot \vec{A} - \frac{i\omega}{c^2} \varphi = 0\\ \end{aligned}

A⃗(x⃗)=μ04π∫J⃗(x⃗′)eikrrdV′\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x}')e^{ikr}}{r}dV'

分别考虑近场和远场

A⃗≈μ04πeikrR∫J⃗(x⃗′)dV′=−iωμ04πeikRRP⃗0\vec{A} \approx \frac{\mu_0}{4\pi} \frac{e^{ikr}}{R} \int \vec{J}(\vec{x}')dV' = -\frac{i\omega\mu_0}{4\pi} \frac{e^{ikR}}{R} \vec{P}_0

以下内容没抄上

*能流

时域：瞬时值能流 S⃗=E⃗×H⃗\vec{S} = \vec{E}\times\vec{H}
瞬时值电场能量密度：we=E⃗⋅D⃗2w_e = \frac{\vec{E}\cdot \vec{D}}{2}
磁场能量密度：wm=B⃗⋅H⃗2w_m = \frac{\vec{B}\cdot \vec{H}}{2}
Poynting定理：

−∇⋅S⃗=∂∂t(we+wm)+E⃗⋅J⃗-\nabla \cdot \vec{S} = \frac{\partial}{\partial t}(w_e + w_m) + \vec{E} \cdot \vec{J}

复数能流：也没抄上，开摆！