# 位移电流

## 严格定义

${\displaystyle \mathbf {D} \ {\stackrel {def}{=}}\ \varepsilon _{0}\mathbf {E} +\mathbf {P} }$

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

${\displaystyle \mathbf {J} _{\mathbf {D} }=\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}+{\frac {\partial \mathbf {P} }{\partial t}}}$

## 原版安培定律的不足处

${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} }$

${\displaystyle \nabla \cdot (\nabla \times \mathbf {B} )=\mu _{0}\nabla \cdot \mathbf {J} }$

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

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

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

${\displaystyle \oint _{\mathbb {C} }\mathbf {B} \cdot \mathrm {d} {\boldsymbol {\ell }}=\mu _{0}I_{enc}}$

## 麦克斯韦-安培方程

${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} +\mu _{0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}}$ ;

${\displaystyle \nabla \times \mathbf {H} =\mathbf {J} _{f}+{\frac {\partial \mathbf {D} }{\partial t}}}$

## 从毕奥-萨伐尔定律推导出位移电流

${\displaystyle \mathbf {B} (\mathbf {r} )={\frac {\mu _{0}}{4\pi }}\int _{\mathbb {V} '}\mathrm {d} ^{3}r'\mathbf {J} (\mathbf {r} ')\times {\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}}$

${\displaystyle \nabla \times (\mathbf {A} _{1}\times \mathbf {A} _{2})=(\mathbf {A} _{2}\cdot \nabla )\mathbf {A} _{1}-(\mathbf {A} _{1}\cdot \nabla )\mathbf {A} _{2}+\mathbf {A} _{1}(\nabla \cdot \mathbf {A} _{2})-\mathbf {A} _{2}(\nabla \cdot \mathbf {A} _{1})}$

${\displaystyle \nabla \times \mathbf {B} (\mathbf {r} )={\frac {\mu _{0}}{4\pi }}\int _{\mathbb {V} '}\mathrm {d} ^{3}r'\left\{-[\mathbf {J} (\mathbf {r} ')\cdot \nabla ]{\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}+\mathbf {J} (\mathbf {r} ')\left[\nabla \cdot {\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}\right]\right\}}$

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

{\displaystyle {\begin{aligned}\nabla \times \mathbf {B} (\mathbf {r} )&=\mu _{0}\mathbf {J} (\mathbf {r} )+{\frac {\mu _{0}}{4\pi }}\int _{\mathbb {V} '}\mathrm {d} ^{3}r'\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^{3}r'\left\{[\mathbf {J} (\mathbf {r} ')\cdot \nabla ']{\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}\right\}\\\end{aligned}}}

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

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

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

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

${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} +{\frac {\mu _{0}}{4\pi }}\int _{\mathbb {V} '}\mathrm {d} ^{3}r'\left\{{\frac {\partial \rho (\mathbf {r} ',\,t)}{\partial t}}{\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}\right\}}$

${\displaystyle \mathbf {E} ={\frac {1}{4\pi \epsilon _{0}}}\int _{\mathbb {V} '}\mathrm {d} ^{3}r'\rho (\mathbf {r} ',\,t){\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}}$

${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} +\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}}$

## 电磁波的推导

${\displaystyle \nabla \cdot \mathbf {E} =0}$(2)
${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}}$(3)
${\displaystyle \nabla \cdot \mathbf {B} =0}$(4)
${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}}$(5)

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

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

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

${\displaystyle \nabla \times \left(-{\frac {\partial \mathbf {B} }{\partial t}}\right)=-{\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {B} \right)=-\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}}$(8)

 ${\displaystyle \nabla ^{2}\mathbf {E} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}}$。

 ${\displaystyle \nabla ^{2}\mathbf {B} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {B} }{\partial t^{2}}}}$。

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

${\displaystyle \mathbf {E} =\mathbf {E} _{0}f\left(\mathbf {k} \cdot \mathbf {r} -\omega t\right)}$

${\displaystyle \nabla ^{2}f\left(\mathbf {k} \cdot \mathbf {r} -\omega t\right)={\frac {1}{{c_{0}}^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}f\left(\mathbf {k} \cdot \mathbf {r} -\omega t\right)}$

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

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

${\displaystyle \nabla \times \mathbf {E} ={\hat {\mathbf {k} }}\times \mathbf {E} _{0}f'\left(\mathbf {k} \cdot \mathbf {r} -\omega t\right)=-{\frac {\partial \mathbf {B} }{\partial t}}}$

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