[物理学与PDEs]第1章习题15 媒介中电磁场的电磁动量密度向量与电磁动量流密度张量

对媒质中的电磁场, 推导其电磁动量密度向量及电磁动量流密度张量的表达式 (7. 47) 及 (7. 48).

解答: 由 $$\beex \bea \cfrac{\rd}{\rd t}\int_\Omega \cfrac{1}{c^2}{\bf S}\rd V &=\cfrac{\rd }{\rd t}\int_\Omega\cfrac{1}{c^2}{\bf E}\times{\bf H}\rd V\\&=\cfrac{1}{c^2}\int_\Omega \sex{\cfrac{\p{\bf E}}{\p t}\times{\bf H}+{\bf E}\times\cfrac{\p{\bf H}}{\p t}}\rd V\\ &=\cfrac{1}{c^2}\int_\Omega \sez{ \cfrac{1}{\ve} (\rot{\bf H}-{\bf j})\times {\bf H} +{\bf E}\times\cfrac{1}{\mu}(-\rot{\bf E}) }\rd V\\ &=\cfrac{1}{c^2}\int_\Omega \sez{ \cfrac{1}{\ve}\rot{\bf H}\times{\bf H} +\cfrac{1}{\mu}\rot{\bf E}\times{\bf E} -\cfrac{1}{\ve}{\bf j}\times {\bf H} }\rd V\\ &=\cfrac{1}{c^2\ve}\int_\Omega \sez{\Div({\bf H}\otimes {\bf H}) -(\Div{\bf H}){\bf H}-\cfrac{1}{2}\Div(H^2{\bf I})}\rd V\\ &\quad+\cfrac{1}{c^2\mu}\int_\Omega \sez{ \Div({\bf E}\otimes {\bf E})-(\Div{\bf E}){\bf E}-\cfrac{1}{2}\Div(E^2{\bf I}) }\rd V\\&\quad-\cfrac{1}{2}\int_\Omega{\bf j}\times{\bf H}\rd V\\ &=\int_{\p\Omega} \sez{ \mu{\bf H}\otimes{\bf H}+\ve{\bf E}\otimes {\bf E}-\cfrac{1}{2}(\mu H^2+\ve E^2){\bf I} }\cdot {\bf n}\rd S\\ &\quad-\int_\Omega \sez{\ve(\Div{\bf E}){\bf E} +\mu(\Div{\bf H}){\bf H}}\rd V-\cfrac{1}{\ve}\int_\Omega {\bf j}\times{\bf H}\rd V. \eea \eeex$$ 即知结论 (注意物理意义).