# 電磁波方程式

${\displaystyle \left(\nabla ^{2}-{\frac {1}{{c}^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\right)\mathbf {E} \ =\ 0\,\!}$
${\displaystyle \left(\nabla ^{2}-{\frac {1}{{c}^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\right)\mathbf {B} \ =\ 0\,\!}$

— 詹姆斯·麦克斯韦

## 理論推導

${\displaystyle \nabla \cdot \mathbf {E} =0\,\!}$ (1)
${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}\,\!}$ (2)
${\displaystyle \nabla \cdot \mathbf {B} =0\,\!}$ (3)
${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\,\!}$ (4)

${\displaystyle \nabla \times (\nabla \times \mathbf {E} )=-{\frac {\partial }{\partial t}}(\nabla \times \mathbf {B} )=-\mu _{0}\varepsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}\,\!}$
${\displaystyle \nabla \times (\nabla \times \mathbf {B} )=\mu _{0}\varepsilon _{0}{\frac {\partial }{\partial t}}(\nabla \times \mathbf {E} )=-\mu _{o}\varepsilon _{o}{\frac {\partial ^{2}\mathbf {B} }{\partial t^{2}}}\,\!}$

${\displaystyle \nabla \times \left(\nabla \times \mathbf {V} \right)=\nabla \left(\nabla \cdot \mathbf {V} \right)-\nabla ^{2}\mathbf {V} \,\!}$

${\displaystyle \left(\nabla ^{2}-{\frac {1}{{c}^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\right)\mathbf {E} \ =\ 0\,\!}$ (5)
${\displaystyle \left(\nabla ^{2}-{\frac {1}{{c}^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\right)\mathbf {B} \ =\ 0\,\!}$ (6)

## 齊次的波動方程式的協變形式

${\displaystyle A^{\mu }\ {\stackrel {def}{=}}\ (\phi /c,\,\mathbf {A} )\,\!}$

${\displaystyle {\frac {\partial A^{\mu }}{\partial x^{\mu }}}=0\,\!}$

${\displaystyle \ \Box A^{\mu }=0\,\!}$

## 彎曲時空中的齊次的波動方程式

${\displaystyle {A^{\mu }}_{;\mu }\ {\stackrel {def}{=}}\ {\frac {\partial A^{\mu }}{\partial x^{\mu }}}=0\,\!}$

${\displaystyle -{A^{\alpha ;\beta }}_{\beta }+{R^{\alpha }}_{\beta }A^{\beta }=0\,\!}$

## 波動方程式的解

${\displaystyle {\frac {1}{{c}^{2}}}{\partial ^{2}f \over \partial t^{2}}-\nabla ^{2}f\ =\ 0\,\!}$ (7)

${\displaystyle f(\mathbf {r} ,t)=g(\mathbf {k} \cdot \mathbf {r} -\omega t)\,\!}$

${\displaystyle \omega ^{2}=c^{2}k^{2}\,\!}$

${\displaystyle \omega =c|k|\,\!}$

### 正弦波

${\displaystyle f=f_{0}\cos(\mathbf {k} \cdot \mathbf {r} -\omega t+\phi _{0})\,\!}$

${\displaystyle e^{i\theta }=\cos \theta +i\sin \theta \,\!}$

${\displaystyle f=\operatorname {Re} \{f_{0}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t+\phi _{0})}\}\,\!}$

${\displaystyle {\tilde {f}}=f_{0}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t+\phi _{0})}={\tilde {f}}_{0}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\,\!}$

${\displaystyle f=\operatorname {Re} \{{\tilde {f}}\}\,\!}$

### 線性疊加

${\displaystyle f(\mathbf {r} ,t)=\int _{0}^{\infty }{\tilde {f}}_{0}(\mathbf {r} ,\omega )e^{-i\omega t}\ d\omega \,\!}$

## 齊次的電磁波方程式的解

### 單色正弦平面波的解

${\displaystyle {\tilde {\mathbf {E} }}(\mathbf {r} ,t)={\tilde {\mathbf {E} }}_{0}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\,\!}$
${\displaystyle {\tilde {\mathbf {B} }}(\mathbf {r} ,t)={\tilde {\mathbf {B} }}_{0}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\,\!}$

${\displaystyle \mathbf {k} \cdot {\tilde {\mathbf {E} }}(\mathbf {r} ,t)=\mathbf {k} \cdot {\tilde {\mathbf {E} }}_{0}=0\,\!}$
${\displaystyle \mathbf {k} \cdot {\tilde {\mathbf {E} }}(\mathbf {r} ,t)=\mathbf {k} \cdot {\tilde {\mathbf {B} }}_{0}=0\,\!}$

${\displaystyle \nabla \times {\tilde {\mathbf {E} }}=\left(\nabla e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right)\times {\tilde {\mathbf {E} }}_{0}=i\mathbf {k} \times {\tilde {\mathbf {E} }}=-{\frac {\partial {\tilde {\mathbf {B} }}}{\partial t}}=i\omega {\tilde {\mathbf {B} }}\,\!}$

${\displaystyle {\tilde {\mathbf {B} }}={\frac {\mathbf {k} }{\omega }}\times {\tilde {\mathbf {E} }}={\frac {1}{c}}{\hat {\mathbf {k} }}\times {\tilde {\mathbf {E} }}\,\!}$

### 電磁波譜分解

${\displaystyle \mathbf {E} (\mathbf {r} ,t)=\mathbf {E} _{0}\cos(\omega t-\mathbf {k} \cdot \mathbf {r} +\phi _{0})\,\!}$
${\displaystyle \mathbf {B} (\mathbf {r} ,t)=\mathbf {B} _{0}\cos(\omega t-\mathbf {k} \cdot \mathbf {r} +\phi _{0})\,\!}$

${\displaystyle k=|\mathbf {k} |={\omega \over c}={2\pi \over \lambda }\,\!}$

### 圓柱對稱性解

${\displaystyle {\mathbf {E} }(\mathbf {r} ,t)={\frac {{\mathbf {E} }_{0}}{s}}\cos(kz-\omega t){\hat {s}}\,\!}$
${\displaystyle {\mathbf {B} }(\mathbf {r} ,t)={\frac {\mathbf {E} _{0}}{cs}}\cos(kz-\omega t){\hat {\phi }}\,\!}$

### 球對稱性解

${\displaystyle {\mathbf {E} }(\mathbf {r} ,t)={\frac {\mathbf {E} _{0}\sin \theta }{r}}\left[\cos(kr-\omega t)-{\frac {1}{kr}}[\sin(kr-\omega t)\right]{\hat {\phi }}\,\!}$
${\displaystyle {\mathbf {B} }(\mathbf {r} ,t)=-{\frac {\mathbf {E} _{0}\sin \theta }{cr}}\left[\cos(kr-\omega t)-{\frac {1}{kr}}[\sin(kr-\omega t)\right]{\hat {\theta }}\,\!}$

## 參考文獻

1. ^ 馬克士威, 詹姆斯, A Dynamical Theory of the Electromagnetic Field (PDF): pp. 499, 1864 [2009-12-15], （原始内容存档 (PDF)于2011-07-28）
2. Griffiths, David J. Introduction to Electrodynamics (3rd ed.). Prentice Hall. 1998: pp. 411–412, 451–453. ISBN 0-13-805326-X.