# 位移電流

## 嚴格定義

$\mathbf{D}\ \stackrel{def}{=}\ \varepsilon_0 \mathbf{E} + \mathbf{P}$

$\mathbf{J}_D\ \stackrel{def}{=}\ \frac {\partial\mathbf{D}}{\partial t}$

$\mathbf{J}_ \mathbf{D} = \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} + \frac{\partial \mathbf{P}}{\partial t}$

## 原版安培定律的不足處

$\nabla \times \mathbf{B} = \mu_0 \mathbf{J }$

$\nabla\cdot(\nabla \times \mathbf{B}) = \mu_0 \nabla\cdot\mathbf{J }$

$\nabla\cdot(\nabla \times \mathbf{B}) =0$

$\nabla\cdot\mathbf{J }=0$

$\nabla\cdot\mathbf{J }+\frac{\partial \rho}{\partial t}=0$

$\oint_\mathbb{C} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} =\mu_0 I_{enc}$

## 馬克士威-安培方程式

$\nabla \times \mathbf{B} = \mu_0 \mathbf{J }+ \mu_0\varepsilon_0\frac{\partial \mathbf{E}}{\partial t}$ ;

$\nabla \times \mathbf{H} = \mathbf{J }_f+ \frac{\partial \mathbf{D}}{\partial t}$

## 從必歐-沙伐定律推導出位移電流

$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int_{\mathbb{V}'} \mathrm{d}^3r' \mathbf{J}(\mathbf{r}')\times \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3}$

$\nabla\times(\mathbf{A}_1\times\mathbf{A}_2) =(\mathbf{A}_2\cdot\nabla)\mathbf{A}_1 - (\mathbf{A}_1\cdot\nabla)\mathbf{A}_2 +\mathbf{A}_1(\nabla\cdot\mathbf{A}_2) - \mathbf{A}_2(\nabla\cdot\mathbf{A}_1)$

$\nabla\times\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi} \int_{\mathbb{V}'} \mathrm{d}^3r'\left\{ - [\mathbf{J}(\mathbf{r}')\cdot\nabla]\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} +\mathbf{J}(\mathbf{r}')\left[\nabla\cdot\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} -\mathbf{r}'|^3}\right] \right\}$

$\nabla\cdot\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}= 4\pi \delta(\mathbf{r}-\mathbf{r}')$

\begin{align}\nabla\times\mathbf{B}(\mathbf{r}) & =\mu_0\mathbf{J}(\mathbf{r})+\frac{\mu_0}{4\pi} \int_{\mathbb{V}'} \mathrm{d}^3r'\left\{ - [\mathbf{J}(\mathbf{r}')\cdot\nabla]\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} \right\} \\ & =\mu_0\mathbf{J}(\mathbf{r})+\frac{\mu_0}{4\pi} \int_{\mathbb{V}'} d^3r'\left\{[\mathbf{J}(\mathbf{r}')\cdot\nabla']\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} \right\} \\ \end{align}

$[\mathbf{J}(\mathbf{r}')\cdot\nabla']\frac{x - x'}{|\mathbf{r} - \mathbf{r}'|^3} =\nabla'\cdot\left[\mathbf{J}(\mathbf{r}')\frac{x - x'}{|\mathbf{r} - \mathbf{r}'|^3}\right] - \frac{x - x'}{|\mathbf{r} - \mathbf{r}'|^3} \nabla'\cdot\mathbf{J}(\mathbf{r}')$(1)

$\int_{\mathbb{V}'} \mathrm{d}^3r' \nabla'\cdot\left(\mathbf{J}(\mathbf{r}')\frac{x - x'}{|\mathbf{r} - \mathbf{r}'|^3}\right) = \oint_{\mathbb{A}'} \mathrm{d}\mathbf{a}' \cdot \mathbf{J}(\mathbf{r}')\frac{x - x'}{|\mathbf{r} - \mathbf{r}'|^3}$

$\nabla'\cdot\mathbf{J}(\mathbf{r}',\,t)+\frac{\partial \rho(\mathbf{r}',\,t)}{\partial t}=0$

$-\ \frac{x - x'}{|\mathbf{r} - \mathbf{r}'|^3} \frac{\partial \rho(\mathbf{r}',\,t)}{\partial t}$

$\nabla\times \mathbf{B} = \mu_0 \mathbf{J}+\frac{\mu_0}{4\pi} \int_{\mathbb{V}'} \mathrm{d}^3r'\left\{ \frac{\partial \rho(\mathbf{r}',\,t)}{\partial t}\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} \right\}$

$\mathbf{E} =\frac{1}{4\pi\epsilon_0}\int_{\mathbb{V}'} \mathrm{d}^3r' \rho(\mathbf{r}',\,t)\frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3}$

$\nabla\times \mathbf{B} = \mu_0 \mathbf{J}+\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$

## 電磁波的推導

$\nabla \cdot \mathbf{E} = 0$(2)
$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$(3)
$\nabla \cdot \mathbf{B} = 0$(4)
$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$(5)

$\mathbf{E}=\mathbf{B}=\mathbf{0}$

$\nabla \times \left(\nabla \times \mathbf{E} \right) = \nabla \times \left( - \frac{\partial \mathbf{B}}{\partial t} \right)$(6)

$\nabla \times \left(\nabla \times \mathbf{E} \right) = \nabla\left(\nabla \cdot \mathbf{E} \right) - \nabla^2 \mathbf{E} = - \nabla^2 \mathbf{E}$(7)

$\nabla \times \left( - \frac{\partial \mathbf{B}}{\partial t} \right) = - \frac{\partial}{\partial t} \left( \nabla \times \mathbf{B} \right) = - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$(8)

 $\nabla^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$ 。

 $\nabla^2 \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2}$ 。

$\Box \mathbf{E} = 0$
$\Box \mathbf{B} = 0$

$\mathbf{E} = \mathbf{E}_0 f\left( \mathbf{k} \cdot \mathbf{r} - \omega t \right)$

$\nabla^2 f\left(\mathbf{k} \cdot \mathbf{r} - \omega t \right) = \frac{1}{{c_0}^2} \frac{\partial^2}{\partial t^2} f\left(\mathbf{k}\cdot\mathbf{r} - \omega t \right)$

$\nabla \cdot \mathbf{E} =\mathbf{k}\cdot \mathbf{E}_0 f'\left(\mathbf{k} \cdot \mathbf{r} - \omega t \right) = 0$

$\mathbf{E} \cdot \mathbf{k} = 0$

$\nabla \times \mathbf{E} = \hat{\mathbf{k}} \times \mathbf{E}_0 f'\left(\mathbf{k} \cdot \mathbf{r} - \omega t \right) = - \frac{\partial \mathbf{B}}{\partial t}$

$\mathbf{B} = \frac{1}{\omega} \mathbf{k} \times \mathbf{E}$

## 參考文獻

1. ^ James C. Maxwell, On Physical Lines of Force, Philosophical Magazine and Journal of Science, 1961
2. ^ 2.0 2.1 2.2 2.3 Griffiths, David J. Introduction to Electrodynamics (3rd ed.). Prentice Hall. 1998: pp. 175–177, 323–325, 364–374. ISBN 0-13-805326-X.
3. ^ John D Jackson. Classical Electrodynamics 3rd Edition. Wiley. 1999: 238. ISBN 047130932X.
4. ^ 4.0 4.1 Daniel M. Siegel. Innovation in Maxwell's Electromagnetic Theory. Cambridge University Press. 2003. ISBN 0521533295.