1. 设真空中运动的带电体的总机械动量为 ${\bf G}_m$, 则 $$\beex \bea \cfrac{\rd {\bf G}_m}{\rd t} &=\int_\Omega {\bf f}\rd V\\ &=\int_\Omega (\rho {\bf E}+\rho {\bf v}\times {\bf B})\rd V\\ &=\int_\Omega \sex{ \ve_0\Div {\bf E}\cdot {\bf E}+\cfrac{1}{\mu_0}\rot{\bf B}-\ve_0{\bf E}_\times {\bf B} }\rd V\\ &\quad\sex{\rho=\ve_0\Div {\bf E},\ \rho {\bf v}={\bf j}=\cfrac{1}{\mu_0}\rot{\bf B}-\ve_0{\bf E}_t}\\ &=\int_\Omega \sez{ \ve_0\Div {\bf E}\cdot {\bf E} +\cfrac{1}{\mu_0}\rot{\bf B}\times{\bf B} -\ve_0({\bf E}\times{\bf B})_t+\ve_0{\bf E}\times{\bf B}_t }\rd V\\ &=\int_\Omega \sez{ \ve_0\Div{\bf E}\cdot {\bf E} -\ve_0{\bf E}\times\rot{\bf E} +\cfrac{1}{\mu_0}\rot{\bf B}\times{\bf B} -\ve_0({\bf E}\times {\bf B})_t }\rd V\\ &\quad\sex{

{\bf B}_t=-\rot{\bf E}}\\ &=-\ve_0\cfrac{\rd}{\rd t}\int_\Omega {\bf E}\times {\bf B}\rd V\\ &\quad\sex{\Div {\bf E}\cdot {\bf E}+\rot{\bf E}\times {\bf E}=\Div({\bf E}\otimes {\bf E})-\cfrac{1}{2}\Div(E^2{\bf I}),\ \Div{\bf B}=0}\\ &=-\ve_0\mu_0\cfrac{\rd}{\rd t}\int_\Omega {\bf S}\rd V\quad\sex{

{\bf S}=\cfrac{1}{\mu_0}{\bf E}\times {\bf B}}\\ &=-\cfrac{\rd}{\rd t}\int_\Omega \cfrac{1}{c^2}{\bf S}\rd V\\ &\quad\sex{c=\cfrac{1}{\sqrt{\ve_0\mu_0}}=2. 997925\times 10^8\ m/s^2:\mbox{ 真空中的光速}}. \eea \eeex$$

 

 

2. 上式即代表电磁动量守恒与转化定律.