电动力学汇总

⚪ 介质中麦克斯韦方程：

\begin{matrix} \nabla \cdot D = ρ_{f} \\ \nabla \times E = - \frac{\partial B}{\partial t} \\ \nabla \cdot B = 0 \\ \nabla \times H = J_{f} + \frac{\partial D}{\partial t} \\ w h e r e D \equiv ϵ_{0} E + P, H \equiv \frac{B}{μ_{0}} - M \end{matrix}

$\rho_p=-\div \bold P,~~\bold J_m=\curl\bold M$ 。

$\Delta W=\int \Delta\bold D\cdot \bold E\dd \tau$ $W=\int \bold D\cdot \bold E\dd \tau$ $W={1\over 2\mu_0}\int B^2\dd \tau$ 。

⚪ 真空麦克斯韦方程组：

\begin{matrix} \nabla \cdot E = \frac{1}{ϵ_{0}} ρ \\ \nabla \times E = - \frac{\partial B}{\partial t} \\ \nabla \cdot B = 0 \\ \nabla \times B = μ_{0} J + ϵ_{0} μ_{0} \frac{\partial E}{\partial t} \end{matrix}

$\rho={1\over 2}\epsilon_0 E^2+{1\over 2\mu_0}B^2$ 。能量守恒：

\frac{\partial ρ}{\partial t} + \nabla \cdot (\frac{1}{μ_{0}} E \times B) + J \cdot E = 0

${1\over \mu_0}\bold E \times \bold B=\bold P$ $\bold J\cdot \bold E$ 为单位体积的电力做功。

$\bold f=\rho\bold E+\bold J\times\bold B$ 。可将其写为：

\begin{matrix} f_{i} = \partial_{j} T_{i j} - ϵ_{0} \frac{\partial}{\partial t} (E \times B)_{i} \\ w h e r e T_{i j} = ϵ_{0} E_{i} E_{j} + \frac{1}{μ_{0}} B_{i} B_{j} - \frac{ϵ_{0}}{2} δ_{i j} E_{k} E_{k} - \frac{1}{2 μ_{0}} δ_{i j} B_{k} B_{k} \end{matrix}

$T_{ij}$ $\bold g=\epsilon_0\bold E\times \bold B=\mu_0\epsilon_0\bold S$ $-\bold T$ $\bold l=\bold r\times \bold g$ 。

⚪ 电磁波：单色平面波解：

\begin{matrix} E = E_{0} e^{i (k \cdot r - ω t)}, B = B_{0} e^{i (k \cdot r - ω t)} \\ k \times E = ω B B = \frac{1}{c} \hat{k} \times E \end{matrix}

$c={1\over \sqrt{\epsilon_0\mu_0}}$ 。光的能量密度、能流密度、动量密度：

\begin{matrix} u = \frac{1}{2} ϵ_{0} E^{2} + \frac{1}{2 μ_{0}} B^{2} = ϵ_{0} E^{2} \\ S = c u \hat{k} = c ϵ_{0} E^{2} \hat{k} \\ g = \frac{1}{c^{2}} S = \frac{1}{c} u \hat{k} \end{matrix}

$v={1\over \sqrt{\epsilon\mu}}$ 。

⚪ 单色平面波于界面的行为：

$\bold E$ 平行于入射反射面：

\begin{matrix} E_{R} = \frac{α - β}{α + β} E_{I}, E_{T} = \frac{2}{α + β} E_{I} \\ α = \frac{\cos θ_{T}}{\cos θ_{I}} = \frac{\sqrt{1 - [(n_{1} / n_{2}) \sin θ_{I}]^{2}}}{\cos θ_{I}} \\ β = \frac{μ_{1} v_{1}}{μ_{2} v_{2}} \end{matrix}

$\alpha=\beta$ 。折射反射率：

R = {(\frac{α - β}{α + β})}^{2}, T = α β {(\frac{2}{α + β})}^{2}

$\bold E$ 垂直于入射反射面：

\begin{matrix} E_{R} = \frac{1 - α β}{1 + α β} E_{I}, E_{T} = \frac{2}{1 + α β} E_{I} \\ R = {(\frac{1 - α β}{1 + α β})}^{2}, T = α β {(\frac{2}{1 + α β})}^{2} \end{matrix}

⚪ 导体对波的吸收与反射

导体中的麦克斯韦方程组：

\begin{matrix} \nabla \cdot E = \frac{ρ_{f}}{ϵ}, \nabla \times E = - \frac{\partial}{\partial t} B \\ \nabla \cdot B = 0, \nabla \times B = μ σ E + μ ϵ \frac{\partial E}{\partial t} \end{matrix}

$\rho_f(t)=\rho(0)e^{-\sigma t/\epsilon}\rho_f(0)$ 。

电磁场的传播方程：

\begin{matrix} E (z, t) = E_{0} e^{i (\tilde{k} z - ω t)} = E_{0} e^{- κ z} e^{i (k z - ω t)} \\ B (z, t) = \frac{\tilde{k}}{ω} E_{0} e^{- κ z} e^{i (k z - ω t)} = \frac{K}{ω} E_{0} e^{- κ z} e^{i (k z - ω t + ϕ)} \\ w h e r e \tilde{k} = k + i κ = K e^{i ϕ} \\ k = ω \sqrt{\frac{ϵ μ}{2}} {(\sqrt{1 + {(\frac{σ}{ϵ ω})}^{2}} + 1)}^{2}, κ = ω \sqrt{\frac{ϵ μ}{2}} {(\sqrt{1 + {(\frac{σ}{ϵ ω})}^{2}} - 1)}^{2} \\ K = ω \sqrt{ϵ μ \sqrt{1 + {(\frac{σ}{ϵ ω})}^{2}}}, ϕ = \arctan \frac{κ}{k} \end{matrix}

⚪ 波导：

沿 z 方向的真空直理想导体管中的波形式设为：

\begin{matrix} E = (E_{x} \hat{x} + E_{y} \hat{y} + E_{z} \hat{z}) \exp i (k z - ω t) \\ B = (B_{x} \hat{x} + B_{y} \hat{y} + B_{z} \hat{z}) \exp i (k z - ω t) \end{matrix}

$E_i,B_i$ $x,y$ 的函数。边界条件：

E_{/ /} |_{b o u n d a r y} = 0, B_{⊥} |_{b o u n d a r y} = 0

$\omega\neq kc$ ，有：

\begin{matrix} E_{x} = \frac{i}{(ω / c)^{2} - k^{2}} (k \frac{\partial E_{z}}{\partial x} + ω \frac{\partial B_{z}}{\partial y}) \\ E_{y} = \frac{i}{(ω / c)^{2} - k^{2}} (k \frac{\partial E_{z}}{\partial y} - ω \frac{\partial B_{z}}{\partial x}) \\ B_{x} = \frac{i}{(ω / c)^{2} - k^{2}} (k \frac{\partial B_{z}}{\partial x} - \frac{ω}{c^{2}} \frac{\partial E_{z}}{\partial y}) \\ B_{y} = \frac{i}{(ω / c)^{2} - k^{2}} (k \frac{\partial B_{z}}{\partial y} + \frac{ω}{c^{2}} \frac{\partial E_{z}}{\partial x}) \end{matrix}

$E_z,B_z$ 为两独立方程的解，且满足边界条件。

\begin{matrix} [\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + {(\frac{ω}{c})}^{2} - k^{2}] E_{z} = 0 \\ [\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + {(\frac{ω}{c})}^{2} - k^{2}] B_{z} = 0 \end{matrix}

⚪ 真空中麦克斯韦方程的通解

定义标势、矢势：

\begin{matrix} E = - \nabla V - \frac{\partial A}{\partial t} \\ B = \nabla \times A \end{matrix}

可改写麦克斯韦方程组为：

\begin{matrix} \frac{ρ}{ϵ_{0}} = - \nabla^{2} V + \frac{\partial}{\partial t} \nabla \cdot A \\ μ_{0} J = \nabla (\nabla \cdot A + ϵ_{0} μ_{0} \frac{\partial V}{\partial t}) + ϵ_{0} μ_{0} \frac{\partial^{2} A}{\partial t^{2}} - \nabla^{2} A \end{matrix}

$\div\bold A+\epsilon_0\mu_0{\part V\over \part t}=0$ 。上式化简为：

\begin{matrix} \frac{ρ}{ϵ_{0}} = ϵ_{0} μ_{0} \frac{\partial^{2}}{\partial t^{2}} V - \nabla^{2} V \\ μ_{0} J = ϵ_{0} μ_{0} \frac{\partial^{2}}{\partial t^{2}} A - \nabla^{2} A \end{matrix}

其通解为：

\begin{matrix} V (r, t) = \frac{1}{4 π ϵ_{0}} \int d^{3} r^{'} \cdot \frac{ρ (r^{'}, t - ⲝ / c)}{ⲝ} \\ A (r, t) = \frac{μ_{0}}{4 π} \int d^{3} r^{'} \cdot \frac{J (r^{'}, t - ⲝ / c)}{ⲝ} \\ w h e r e ⲝ = r - r^{'} \end{matrix}

利用它可得Jefimenko公式：

\begin{matrix} E (r, t) = \frac{1}{4 π ϵ_{0}} \int d^{3} r^{'} \cdot (\frac{ρ (r^{'}, t_{r})}{ⲝ^{2}} \hat{ⲝ} + \frac{\dot{ρ} (r^{'}, t_{r})}{c ⲝ} \hat{ⲝ} - \frac{\dot{J} (r^{'}, t_{r})}{c^{2} ⲝ}) \\ B (r, t) = \frac{μ_{0}}{4 π} \int d^{3} r^{'} (\frac{J (r^{'}, t_{r})}{ⲝ^{2}} + \frac{\dot{J} (r^{'}, t_{r})}{c ⲝ}) \times \hat{ⲝ} \end{matrix}

⚪ 动点电荷的场：

Lienard-Wiechert势：

\begin{matrix} V (r, t) = \frac{1}{4 π ϵ_{0}} \frac{q c}{ⲝ (c - v \cdot \hat{ⲝ})} \\ A (r, t) = \frac{μ_{0}}{4 π} \frac{q c v}{ⲝ (c - v \cdot \hat{ⲝ})} \end{matrix}

$\bold v=\bold v(t_r)$ 。由此可得：

\begin{matrix} E (r, t) = \frac{q}{4 π ϵ_{0}} \frac{ⲝ}{(ⲝ \cdot u)^{3}} [(c^{2} - v^{2}) u + ⲝ \times (u \times a)] \\ B (r, t) = \frac{q}{4 π ϵ_{0} c} \frac{ⲝ}{(ⲝ \cdot u)^{3}} \times [(c^{2} - v^{2}) u + ⲝ \times (u \times a)] \\ w h e r e u = c \hat{ⲝ} - v \end{matrix}

$\bold a=0$ ，可将上式简化为：

\begin{matrix} E (r, t) = \frac{q}{4 π ϵ_{0}} \frac{1 - v^{2} / c^{2}}{(1 - v^{2} \sin^{2} θ / c^{2})^{3 / 2}} \frac{\hat{R}}{R^{2}} \\ B (r, t) = \frac{1}{c} \hat{ⲝ} \times E = \frac{1}{c^{2}} v \times E \\ w h e r e R = r - r^{'} (t) = r - v t, θ = ⟨ R, v ⟩ \end{matrix}

$\bold p=\bold p_0\cos \omega t$ $d\ll \lambda\ll r$ $(r,\theta,\phi)$ 处的势和场和辐射为：

\begin{matrix} V (r, t) = - \frac{p_{0} ω}{4 π ϵ_{0} c} \frac{\cos θ}{r} \sin [ω (t - r / c)] \\ A (r, t) = - \frac{μ_{0} p_{0} ω}{4 π} \sin [ω (t - r / c)] \\ E (r, t) = - \frac{μ_{0} p_{0} ω^{2}}{4 π} \frac{\sin θ}{r} \cos [ω (t - r / c)] \hat{θ} \\ B (r, t) = - \frac{μ_{0} p_{0} ω^{2}}{4 π c} \frac{\sin θ}{r} \cos [ω (t - r / c)] \hat{ϕ} \\ ⟨ S (r, t) ⟩ = \frac{μ_{0} p_{0}^{2} ω^{4}}{32 π^{2} c} \frac{\sin^{2} θ}{r^{2}} \hat{r} \\ ⟨ P ⟩ = \frac{μ_{0} p_{0}^{2} ω^{4}}{12 π c} \end{matrix}

$\bold m=\bold m_0\cos \omega t$ $d\ll \lambda\ll r$ $(r,\theta,\phi)$ 处的和场和辐射为：

\begin{matrix} V (r, t) = 0 \\ A (r, t) = - \frac{μ_{0} m_{0} ω}{4 π c} \frac{\sin θ}{r} \sin [ω (t - r / c)] \hat{ϕ} \\ E (r, t) = \frac{μ_{0} m_{0} ω^{2}}{4 π c} \frac{\sin θ}{r} \cos [ω (t - r / c)] \hat{ϕ} \\ B (r, t) = - \frac{μ_{0} m_{0} ω^{2}}{4 π c^{2}} \frac{\sin θ}{r} \cos [ω (t - r / c)] \hat{θ} \\ ⟨ S (r, t) ⟩ = \frac{μ_{0} p_{0}^{2} ω^{4}}{32 π^{2} c^{3}} \frac{\sin^{2} θ}{r^{2}} \hat{r} \\ ⟨ P ⟩ = \frac{μ_{0} p_{0}^{2} ω^{4}}{12 π c^{3}} \end{matrix}

$r'\ll {c\over |\rho^{(n)}/\dot\rho|^{1/n}} \ll r$ 下。

\begin{matrix} V (r, t) \approx \frac{1}{4 π ϵ_{0}} (\frac{Q}{r} + \frac{r \cdot p (t - r / c)}{r^{3}} + \frac{r \cdot \dot{p} (t - r / c)}{r^{2} c}) \\ A (r, t) \approx \frac{μ_{0}}{4 π} \frac{\dot{p} (t - r / c)}{r} 利 用 \int_{V} J d^{3} r = \dot{p} \end{matrix}

$\propto r^{-1}$ ）的项，得：

\begin{matrix} E (r, t) \sim \frac{μ_{0}}{4 π r} [\hat{r} \times (\hat{r} \times \ddot{p} (t - r / c))] \\ B (r, t) \sim - \frac{μ_{0}}{4 π r c} [\hat{r} \times \ddot{p} (t - r / c)] \end{matrix}

$\ddot{\bold p}// \hat{\bold z}$ ，则有：

\begin{matrix} E (r, t) \sim \frac{μ_{0} \ddot{p} (t - r / c)}{4 π} \frac{\sin θ}{r} \hat{θ} \\ B (r, t) \sim \frac{μ_{0} \ddot{p} (t - r / c)}{4 π c} \frac{\sin θ}{r} \hat{ϕ} \\ S (r, t) = \frac{μ_{0}}{16 π^{2} c} [\ddot{p} (t - r / c)]^{2} \frac{\sin^{2} θ}{r^{2}} \hat{r} \\ P (t) = \frac{μ_{0}}{6 π c} [\ddot{p} (t - r / c)]^{2} \end{matrix}

$\bold p=q\bold r'$ ，则有拉莫尔公式：

P = \frac{μ_{0} q^{2} a^{2}}{6 π c}

⚪ 点电荷的辐射：接收处能流密度（坡印廷矢量）仅由加速度场决定：

\begin{matrix} E_{r a d} (r, t) = \frac{q}{4 π ϵ_{0}} \frac{ⲝ}{(ⲝ \cdot u)^{3}} ⲝ \times (u \times a) \\ S (r, t) = \frac{\hat{ⲝ}}{c μ_{0}} E_{r a d}^{2} \end{matrix}

$v\ll c$ 可得拉莫尔公式：

\begin{matrix} S (r, t) = \frac{a^{2} \sin^{2} θ}{16 π ϵ_{0} c^{3} ⲝ^{2}} \hat{ⲝ} \\ P (r, t) = \frac{μ_{0} q^{2} a^{2}}{6 π c} \end{matrix}

从接收处反推辐射处：

\begin{aligned} \frac{d P}{d Ω} |_{e m i t} & = (1 - \frac{\hat{ⲝ} \cdot v}{c}) S ⲝ^{2} = \frac{ⲝ \cdot u}{c ⲝ} S ⲝ^{2} \\ = \frac{q^{2}}{16 π^{2} ϵ_{0}} \frac{| \hat{ⲝ} \times (u \times a) |^{2}}{(\hat{ⲝ} \cdot u)^{5}} \\ P & = \int d Ω \frac{d P}{d Ω} = \frac{μ_{0} q^{2} γ^{6}}{6 π c} (a^{2} - {| \frac{v \times a}{c} |}^{2}) \end{aligned}

$\bold v,\bold a$ $\bold v$ 方向的受力最简可能表达式（Abraham-Lorentz）：

F_{r a d} = \frac{μ_{0} q^{2}}{6 π c} \dot{a}

⚪ 相对论时空观：

洛伦兹变换矩阵：

\begin{matrix} Λ^{μ}_{ν} = (\begin{matrix} γ & - γ β \\ - γ β & γ \\ 1 \\ 1 \end{matrix}), μ, ν = 0, 1, 2, 3 \end{matrix}

坐标（位移对偶矢量）变换：

\begin{matrix} x^{μ} = (x^{0}, x^{1}, x^{2}, x^{3}) = (c t, x, y, z) \\ {\bar{x}}^{ν} = Λ^{ν}_{μ} x^{μ} \end{matrix}

四速度（世界线切矢量）变换：

\begin{matrix} η^{μ} = (\frac{c}{\sqrt{1 - u^{2} / c^{2}}}, \frac{u}{\sqrt{1 - u^{2} / c^{2}}}) \\ {\bar{η}}^{ν} = Λ^{ν}_{μ} η^{μ} \end{matrix}

对应速度变换：

\begin{matrix} {\bar{u}}_{x} = \frac{u_{x} - v}{(1 - v u_{x} / c^{2})} \\ {\bar{u}}_{y} = \frac{u_{y}}{γ (1 - v u_{x} / c^{2})} \\ {\bar{u}}_{z} = \frac{u_{z}}{γ (1 - v u_{x} / c^{2})} \end{matrix}

⚪ 相对论动力学：

四动量变换：

\begin{matrix} p^{μ} = m η^{μ} = (\frac{m c}{\sqrt{1 - u^{2} / c^{2}}}, \frac{m u}{\sqrt{1 - u^{2} / c^{2}}}) \\ {\bar{p}}^{ν} = Λ^{ν}_{μ} p^{μ} \end{matrix}

四力：

\begin{matrix} K^{μ} = \frac{d p^{μ}}{d τ} \\ K^{μ} = (\frac{d E}{c d τ}, \frac{1}{\sqrt{1 - u^{2} / c^{2}}} F) \end{matrix}

⚪ 相对论电动力学：

电磁场变换：

\begin{matrix} E_{x}^{'} = E_{x}, E_{y}^{'} = γ (E_{y} - v B_{z}), E_{z}^{'} = γ (E_{z} + v B_{y}) \\ B_{x}^{'} = B_{x}, B_{y}^{'} = γ (B_{y} + v E_{z} / c^{2}), B_{z}^{'} = γ (B_{z} - v E_{y} / c^{2}) \end{matrix}

电磁场张量：

\begin{matrix} F^{μ ν} = (\begin{matrix} 0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\ - \frac{E_{x}}{c} & 0 & B_{z} & - B_{y} \\ - \frac{E_{y}}{c} & - B_{z} & 0 & B_{x} \\ - \frac{E_{z}}{c} & B_{y} & - B_{x} & 0 \end{matrix}) \\ G^{μ ν} = (\begin{matrix} 0 & B_{x} & B_{y} & B_{z} \\ - B_{x} & 0 & - \frac{E_{z}}{c} & \frac{E_{y}}{c} \\ - B_{y} & \frac{E_{z}}{c} & 0 & - \frac{E_{x}}{c} \\ - B_{z} & - \frac{E_{y}}{c} & \frac{E_{x}}{c} & 0 \end{matrix}) \\ {\bar{F}}^{δ σ} = Λ^{δ}_{μ} Λ^{σ}_{ν} F^{μ ν}, {\bar{G}}^{δ σ} = Λ^{δ}_{μ} Λ^{σ}_{ν} G^{μ ν} \end{matrix}

$\rho_0$ 为固有密度）：

J^{μ} = \sum ρ_{0} (η^{μ}) η^{μ} = (c ρ, J)

电荷守恒：

\partial_{μ} J^{μ} = \frac{\partial J^{μ}}{\partial x^{μ}} = 0

麦克斯韦方程组可写作：

\partial_{ν} F^{μ ν} = μ_{0} J^{μ}, \partial_{ν} G^{μ ν} = 0

势矢量：

\begin{matrix} A^{μ} = (V / c, A) \\ F^{μ ν} = \partial^{μ} A^{ν} - \partial^{ν} A^{μ} \end{matrix}

Lorentz规范与对应的d'Alembert方程：

\begin{matrix} \partial_{ν} A^{ν} = 0 \\ \partial_{ν} \partial^{ν} A^{μ} = - μ_{0} J^{μ} \end{matrix}