麦克斯韦方程组

库仑定律 叠加原理 电场 高斯定理 静电场旋度 电势

$$\boldsymbol{F}_{12}=\frac{q_1q_2\boldsymbol{R}_{12}}{4\pi\varepsilon_0 R^3}=-\boldsymbol{F}_{21}\\
\boldsymbol{F}_q=\frac{1}{4\pi\varepsilon_0}\int\frac{q\rho\boldsymbol{R}}{R^3}\dd V\\
\boldsymbol{E}\overset{定义}{=}\frac{\boldsymbol{F}_q}{q}=\frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho\boldsymbol{R}}{R^3}\dd V\\
\int_V\nabla\cdot\boldsymbol{E}\dd V=\oint_S\boldsymbol{E}\cdot\dd\boldsymbol{S}=\frac{1}{4\pi\varepsilon_0}\int_V\oint_S\frac{\rho\boldsymbol{R}}{R^3}\cdot\dd\boldsymbol{S}\;\dd V=\frac{1}{\varepsilon_0}\int_V\rho\dd V\\
\nabla\cdot\boldsymbol{E}=\frac{\rho}{\varepsilon_0}\\
\boldsymbol{E}(\boldsymbol{r})=\frac{1}{4\pi\varepsilon_0}\int_{V’}\frac{\rho(\boldsymbol{r}’)(\boldsymbol{\boldsymbol{r}-\boldsymbol{r}’})}{|\boldsymbol{r}-\boldsymbol{r}’|^3}\dd V’\\=-\nabla\left[\frac{1}{4\pi\varepsilon_0}\int_{V’}\frac{\rho(\boldsymbol{r}’)}{|\boldsymbol{r}-\boldsymbol{r}’|}\dd V’\right]
\overset{定义}{=}-\nabla\varphi\\
\nabla\times \boldsymbol{E}=0$$

连续性方程 欧姆定律 安培定律 磁感应强度 静磁场散度 静磁场旋度 磁矢势

$$\int_V\nabla\cdot\boldsymbol{j}\dd V=\oint_S\boldsymbol{j}\cdot\dd \boldsymbol{S}=-\frac{\dd}{\dd t}\int_V\rho\dd V\\
\nabla\cdot\boldsymbol{j}+\frac{\partial \rho}{\partial t}=0\\
\boldsymbol{j}=\sigma_c\boldsymbol{E}\\
\dd \boldsymbol{F}_{12}=\frac{\mu_0}{4\pi}\frac{\boldsymbol{j}_1\dd V_1\times(\boldsymbol{j}_2\dd V_2\times\boldsymbol{R}_{12})}{R^3}\neq -\dd \boldsymbol{F}_{21}\\
\boldsymbol{F}_{12}=\frac{\mu_0 I_1I_2}{4\pi}\oint_{l_1}\oint_{l_2}\frac{\dd\boldsymbol{l}_1\times(\dd\boldsymbol{l}_2\times \boldsymbol{R}_{12})}{R^3}\\
=\frac{\mu_0 I_1I_2}{4\pi}\oint_{l_1}\oint_{l_2}\frac{\dd\boldsymbol{l}_1\cdot \boldsymbol{R}_{12}\dd\boldsymbol{l}_2-\dd\boldsymbol{l}_1\cdot\dd\boldsymbol{l}_2\boldsymbol{R}_{12}}{R^3}\\
=\frac{\mu_0 I_1I_2}{4\pi}\oint_{l_1}\oint_{l_2}\left[\dd\boldsymbol{l}_1\cdot\nabla_{l_2}\frac{1}{R}\dd\boldsymbol{l}_2-\frac{\dd\boldsymbol{l}_1\cdot\dd\boldsymbol{l}_2\boldsymbol{R}_{12}}{R^3}\right]\\
=-\frac{\mu_0 I_1I_2}{4\pi}\oint_{l_1}\oint_{l_2}\frac{\dd\boldsymbol{l}_1\cdot\dd\boldsymbol{l}_2\boldsymbol{R}_{12}}{R^3}=-\boldsymbol{F}_{21}\\
\dd\boldsymbol{F}_{\boldsymbol{j}_1\dd V_1}=\frac{\mu_0}{4\pi}\int_{V_2}\frac{\boldsymbol{j}_1\dd V_1\times(\boldsymbol{j}_2\dd V_2\times\boldsymbol{R}_{12})}{R^3}\\
\boldsymbol{F}_{\boldsymbol{j}_1\dd V_1}\overset{定义}{=}\boldsymbol{j}_1\dd V_1\times\boldsymbol{B}\\
\boldsymbol{B}=\frac{\mu_0}{4\pi}\int_{V}\frac{\boldsymbol{j}\times\boldsymbol{R}}{R^3}\dd V\\
\boldsymbol{B}(\boldsymbol{r})=\frac{\mu_0}{4\pi}\int_{V’}\frac{\boldsymbol{j}(\boldsymbol{r}’)\times(\boldsymbol{r}-\boldsymbol{r}’)}{|\boldsymbol{r}-\boldsymbol{r}’|^3}\dd V’\\=\nabla\times\left[\frac{\mu_0}{4\pi}\int_{V’}\frac{\boldsymbol{j}(\boldsymbol{r}’)}{|\boldsymbol{r}-\boldsymbol{r}’|}\dd V’\right]\overset{定义}{=}\nabla\times\boldsymbol{A}\\
\nabla\cdot\boldsymbol{B}=0\\
\nabla\times\boldsymbol{B}=\nabla(\nabla\cdot\boldsymbol{A})-\nabla^2\boldsymbol{A}\\
=\nabla\left[\frac{\mu_0}{4\pi}\int_{V’}\boldsymbol{j}(\boldsymbol{r}’)\nabla\cdot\frac{1}{|\boldsymbol{r}-\boldsymbol{r}’|}\dd V’\right]+\mu_0\boldsymbol{j}\\
=-\nabla\left[\frac{\mu_0}{4\pi}\int_{V’}\boldsymbol{j}(\boldsymbol{r}’)\nabla’\cdot\frac{1}{|\boldsymbol{r}-\boldsymbol{r}’|}\dd V’\right]+\mu_0\boldsymbol{j}\\
=-\nabla\left[\frac{\mu_0}{4\pi}\int_{V’}\left(\nabla’\cdot\frac{\boldsymbol{j}(\boldsymbol{r}’)}{|\boldsymbol{r}-\boldsymbol{r}’|}-\frac{1}{|\boldsymbol{r}-\boldsymbol{r}’|}\nabla’\cdot\boldsymbol{j}(\boldsymbol{r}’)\right) \dd V’\right]+\mu_0\boldsymbol{j}\\
\overset{静磁场}{=}-\nabla\left[\frac{\mu_0}{4\pi}\oint_{S’}\frac{\boldsymbol{j}(\boldsymbol{r}’)}{|\boldsymbol{r}-\boldsymbol{r}’|}\cdot \dd \boldsymbol{S}’\right]+\mu_0\boldsymbol{j}\overset{局域性}{=}\mu_0\boldsymbol{j}$$

洛伦兹公式 法拉第定律 位移电流 真空麦克斯韦方程组

$$\boldsymbol{f}=\rho\boldsymbol{E}+\boldsymbol{j}\times\boldsymbol{B}\\
\mathscr{E}_{感应}=\oint_l\boldsymbol{E}_{感应}\cdot\dd\boldsymbol{l}=-K\frac{\dd}{\dd t}\int_S\boldsymbol{B}\cdot\dd\boldsymbol{S}\\
\nabla\times \boldsymbol{E}\overset{静止}{=}-K\frac{\partial \boldsymbol{B}}{\partial t}\\
\nabla\times (\boldsymbol{E’}-K\boldsymbol{v}\times\boldsymbol{B})\overset{v}{=}-K\frac{\partial \boldsymbol{B}}{\partial t}\\
\boldsymbol{E’}\overset{点电荷}{=}\boldsymbol{E}+\boldsymbol{v}\times\boldsymbol{B}\qquad K\overset{类比}{=}1\\
\nabla\times \boldsymbol{E}=-\frac{\partial \boldsymbol{B}}{\partial t}\\
0=\nabla\cdot(\nabla\times\boldsymbol{B})=\nabla\cdot(\mu_0\boldsymbol{j})\overset{矛盾}{=}-\mu_0\frac{\partial\rho}{\partial t}=\\
\mu_0\frac{\partial\rho}{\partial t}=\nabla\cdot\left[\mu_0\varepsilon_0\frac{\partial\boldsymbol{E}}{\partial t}\right]\\
\nabla\times\boldsymbol{B}\overset{修正}{=}\mu_0\boldsymbol{j}+\mu_0\varepsilon_0\frac{\partial\boldsymbol{E}}{\partial t}\\
\left\lbrace\begin{array}{l}\displaystyle\nabla\cdot \boldsymbol{E}=\frac{\rho}{\varepsilon_0}\\\displaystyle\nabla\times\boldsymbol{E}=-\frac{\partial\boldsymbol{B}}{\partial t}\\\nabla\cdot\boldsymbol{B}=0\\\displaystyle\nabla\times\boldsymbol{B}=\mu_0\boldsymbol{j}+\mu_0\varepsilon_0\frac{\partial\boldsymbol{E}}{\partial t}\end{array}\right.$$

极化强度 极化电荷 极化电流 磁化强度 磁化电流

$$\boldsymbol{P}\overset{定义}{=}\frac{\sum\boldsymbol{p}}{\Delta V}\\
\int_V\rho_{P}\dd V=-\oint_S\boldsymbol{P}\cdot\dd\boldsymbol{S}=-\int_V\nabla\cdot\boldsymbol{P}\dd V\\
\rho_{P}=-\nabla\cdot\boldsymbol{P}\\
\nabla\cdot\boldsymbol{j}_p=-\frac{\partial\rho_P}{\partial t}=\nabla\cdot\frac{\partial\boldsymbol{P}}{\partial t}\\
\boldsymbol{j}_P=\frac{\partial\boldsymbol{P}}{\partial t}\\
\boldsymbol{M}\overset{定义}{=}\frac{\sum\boldsymbol{m}}{\Delta V}\\
\int_S\boldsymbol{j}_m\cdot\dd\boldsymbol{S}=\oint_l\boldsymbol{M}\cdot\dd\boldsymbol{l}=\int_S\nabla\times\boldsymbol{M}\cdot\dd\boldsymbol{S}\\
\boldsymbol{j}_m=\nabla\times\boldsymbol{M}\qquad \nabla\cdot\boldsymbol{j}_m=0\\
\rho_t=\rho+\rho_P=\rho-\nabla\cdot\boldsymbol{P}\\
\boldsymbol{j}_t=\boldsymbol{j}+\boldsymbol{j}_P+\boldsymbol{j}_m=\boldsymbol{j}+\frac{\partial\boldsymbol{P}}{\partial t}+\nabla\times\boldsymbol{M}
$$

电位移矢量 磁场强度 介质麦克斯韦方程组 弱场各向同性线性响应

$$\left\lbrace\begin{array}{l}\displaystyle\nabla\cdot \boldsymbol{E}=\frac{\rho-\nabla\cdot\boldsymbol{P}}{\varepsilon_0}\\\displaystyle\nabla\times\boldsymbol{E}=-\frac{\partial\boldsymbol{B}}{\partial t}\\\nabla\cdot\boldsymbol{B}=0\\\displaystyle\nabla\times\boldsymbol{B}=\mu_0\left(\boldsymbol{j}+\frac{\partial\boldsymbol{P}}{\partial t}+\nabla\times\boldsymbol{M}\right)+\mu_0\varepsilon_0\frac{\partial\boldsymbol{E}}{\partial t}\end{array}\right.\\
\boldsymbol{D}\overset{定义}{=}\varepsilon_0\boldsymbol{E}+\boldsymbol{P}\\
\boldsymbol{H}\overset{定义}{=}\frac{1}{\mu_0}\boldsymbol{B}-\boldsymbol{M}\\
\left\lbrace\begin{array}{l}\displaystyle\nabla\cdot \boldsymbol{D}=\rho\\\displaystyle\nabla\times\boldsymbol{E}=-\frac{\partial\boldsymbol{B}}{\partial t}\\\nabla\cdot\boldsymbol{B}=0\\\displaystyle\nabla\times\boldsymbol{H}=\boldsymbol{j}+\frac{\partial\boldsymbol{D}}{\partial t}\end{array}\right.\\
\boldsymbol{P}=\chi\varepsilon\boldsymbol{E}\\
\boldsymbol{D}=\varepsilon\boldsymbol{E}=\varepsilon_r\varepsilon_0\boldsymbol{E}=(\chi+1)\varepsilon_0\boldsymbol{E}\\
\boldsymbol{M}=\kappa\boldsymbol{H}\\
\boldsymbol{B}=\mu\boldsymbol{H}=\mu_r\mu_0\boldsymbol{H}=(\kappa+1)\mu_0\boldsymbol{H}
$$

边界条件 真空麦克斯韦方程组完备性

$$\hat{\boldsymbol{n}}_{12}\cdot(\boldsymbol{D}_1-\boldsymbol{D}_2)=\sigma\\
\hat{\boldsymbol{n}}_{12}\cdot(\boldsymbol{B}_1-\boldsymbol{B}_2)=0\\
\hat{\boldsymbol{n}}_{12}\times(\boldsymbol{E}_1-\boldsymbol{E}_2)=0\\
\hat{\boldsymbol{n}}_{12}\times(\boldsymbol{H}_1-\boldsymbol{H}_2)=\boldsymbol{\alpha}\\
\hat{\boldsymbol{n}}_{12}\cdot(\boldsymbol{j}_1-\boldsymbol{j}_2)=\frac{\partial\sigma}{\partial t}\\
\hat{\boldsymbol{n}}_{12}\cdot(\boldsymbol{P}_1-\boldsymbol{P}_2)=-\sigma_P\\
\hat{\boldsymbol{n}}_{12}\times(\boldsymbol{M}_1-\boldsymbol{M}_2)=\boldsymbol{\alpha}_m\\
\boldsymbol{E}\overset{定义}{=}\boldsymbol{E}’-\boldsymbol{E}’’\qquad\boldsymbol{B}\overset{定义}{=}\boldsymbol{B}’-\boldsymbol{B}’’\\
\boldsymbol{E}(\boldsymbol{r},0)=0\qquad\boldsymbol{B}(\boldsymbol{r},0)=0\\
\boldsymbol{E}|_S=0\qquad\boldsymbol{B}|_S=0\\
I\overset{定义}{=}\frac{\dd}{\dd t}\int_V(\varepsilon_0E^2+\frac{1}{\mu_0}B^2)\dd V\\
=\int_V(\frac{2}{\mu_0}\boldsymbol{E}\cdot(\nabla\times\boldsymbol{B})-\frac{2}{\mu_0}\boldsymbol{B}\cdot(\nabla\times\boldsymbol{E}))\dd V\\
=-\frac{2}{\mu_0}\int_V\nabla\cdot(\boldsymbol{E}\times\boldsymbol{B})\dd V\\
=-\frac{2}{\mu_0}\int_S(\boldsymbol{E}\times\boldsymbol{B})\cdot\dd \boldsymbol{S}=0\\
\boldsymbol{E}’\equiv\boldsymbol{E}’’\qquad\boldsymbol{B}’\equiv\boldsymbol{B}’’
$$

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