# 馬克士威應力張量

## 導引

${\displaystyle \mathbf {f} =\rho \mathbf {E} +\mathbf {J} \times \mathbf {B} }$

${\displaystyle \mathbf {f} =\epsilon _{0}\left({\boldsymbol {\nabla }}\cdot \mathbf {E} \right)\mathbf {E} +{\frac {1}{\mu _{0}}}\left({\boldsymbol {\nabla }}\times \mathbf {B} \right)\times \mathbf {B} -\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\times \mathbf {B} }$

${\displaystyle {\frac {\partial }{\partial t}}(\mathbf {E} \times \mathbf {B} )={\frac {\partial \mathbf {E} }{\partial t}}\times \mathbf {B} +\mathbf {E} \times {\frac {\partial \mathbf {B} }{\partial t}}={\frac {\partial \mathbf {E} }{\partial t}}\times \mathbf {B} -\mathbf {E} \times ({\boldsymbol {\nabla }}\times \mathbf {E} )}$

{\displaystyle {\begin{aligned}\mathbf {f} &=\epsilon _{0}\left({\boldsymbol {\nabla }}\cdot \mathbf {E} \right)\mathbf {E} +{\frac {1}{\mu _{0}}}\left({\boldsymbol {\nabla }}\times \mathbf {B} \right)\times \mathbf {B} -\epsilon _{0}{\frac {\partial }{\partial t}}\left(\mathbf {E} \times \mathbf {B} \right)-\epsilon _{0}\mathbf {E} \times ({\boldsymbol {\nabla }}\times \mathbf {E} )\\&=\epsilon _{0}\left[({\boldsymbol {\nabla }}\cdot \mathbf {E} )\mathbf {E} -\mathbf {E} \times ({\boldsymbol {\nabla }}\times \mathbf {E} )\right]+{\frac {1}{\mu _{0}}}\left[-\mathbf {B} \times \left({\boldsymbol {\nabla }}\times \mathbf {B} \right)\right]-\epsilon _{0}{\frac {\partial }{\partial t}}\left(\mathbf {E} \times \mathbf {B} \right)\\\end{aligned}}}

${\displaystyle \mathbf {f} =\epsilon _{0}\left[({\boldsymbol {\nabla }}\cdot \mathbf {E} )\mathbf {E} -\mathbf {E} \times ({\boldsymbol {\nabla }}\times \mathbf {E} )\right]+{\frac {1}{\mu _{0}}}\left[({\boldsymbol {\nabla }}\cdot \mathbf {B} )\mathbf {B} -\mathbf {B} \times \left({\boldsymbol {\nabla }}\times \mathbf {B} \right)\right]-\epsilon _{0}{\frac {\partial }{\partial t}}\left(\mathbf {E} \times \mathbf {B} \right)}$

${\displaystyle \mathbf {A} \times ({\boldsymbol {\nabla }}\times \mathbf {A} )={\tfrac {1}{2}}{\boldsymbol {\nabla }}A^{2}-(\mathbf {A} \cdot {\boldsymbol {\nabla }})\mathbf {A} }$

${\displaystyle \mathbf {f} }$  的方程式內的旋度項目除去：

${\displaystyle \mathbf {f} =\epsilon _{0}\left[({\boldsymbol {\nabla }}\cdot \mathbf {E} )\mathbf {E} +(\mathbf {E} \cdot {\boldsymbol {\nabla }})\mathbf {E} \right]+{\frac {1}{\mu _{0}}}\left[({\boldsymbol {\nabla }}\cdot \mathbf {B} )\mathbf {B} +(\mathbf {B} \cdot {\boldsymbol {\nabla }})\mathbf {B} \right]-{\frac {1}{2}}{\boldsymbol {\nabla }}\left(\epsilon _{0}E^{2}+{\frac {1}{\mu _{0}}}B^{2}\right)-\epsilon _{0}{\frac {\partial }{\partial t}}\left(\mathbf {E} \times \mathbf {B} \right)}$

${\displaystyle \mathbf {S} ={\frac {1}{\mu _{0}}}\mathbf {E} \times \mathbf {B} }$

${\displaystyle T_{ij}\equiv \epsilon _{0}\left(E_{i}E_{j}-{\frac {1}{2}}\delta _{ij}E^{2}\right)+{\frac {1}{\mu _{0}}}\left(B_{i}B_{j}-{\frac {1}{2}}\delta _{ij}B^{2}\right)}$

${\displaystyle (\mathbf {A} \cdot {\stackrel {\longleftrightarrow }{\mathbf {T} }})_{j}=\textstyle {\sum _{i}}\ A_{i}T_{ij}}$

${\displaystyle \mathbf {f} =\nabla \cdot {\stackrel {\longleftrightarrow }{\mathbf {T} }}-\epsilon _{0}\mu _{0}{\frac {\partial \mathbf {S} }{\partial t}}}$

## 馬克士威應力張量的性質

${\displaystyle T_{ij}=\left({\begin{matrix}\epsilon _{0}(E_{x}^{2}-E^{2}/2)+{\cfrac {1}{\mu _{0}}}(B_{x}^{2}-B^{2}/2)&\epsilon _{0}E_{x}E_{y}+{\cfrac {1}{\mu _{0}}}(B_{x}B_{y})&\epsilon _{0}E_{x}E_{z}+{\cfrac {1}{\mu _{0}}}(B_{x}B_{z})\\\epsilon _{0}E_{x}E_{y}+{\cfrac {1}{\mu _{0}}}(B_{x}B_{y})&\epsilon _{0}(E_{y}^{2}-E^{2}/2)+{\cfrac {1}{\mu _{0}}}(B_{y}^{2}-B^{2}/2)&\epsilon _{0}E_{y}E_{z}+{\cfrac {1}{\mu _{0}}}(B_{y}B_{z})\\\epsilon _{0}E_{x}E_{z}+{\cfrac {1}{\mu _{0}}}(B_{x}B_{z})&\epsilon _{0}E_{y}E_{z}+{\cfrac {1}{\mu _{0}}}(B_{y}B_{z})&\epsilon _{0}(E_{z}^{2}-E^{2}/2)+{\cfrac {1}{\mu _{0}}}(B_{z}^{2}-B^{2}/2)\end{matrix}}\right)}$

## 動量守恆定律

${\displaystyle \mathbf {F} =\int _{\mathcal {V}}\ \mathbf {f} \mathrm {d} \tau =\int _{\mathcal {V}}\ \nabla \cdot {\stackrel {\longleftrightarrow }{\mathbf {T} }}\mathrm {d} \tau -\epsilon _{0}\mu _{0}{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{\mathcal {V}}\ \mathbf {S} \mathrm {d} \tau }$

${\displaystyle \mathbf {F} =\oint _{\mathcal {S}}\ {\stackrel {\longleftrightarrow }{\mathbf {T} }}\cdot \mathrm {d} \mathbf {a} -\epsilon _{0}\mu _{0}{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{\mathcal {V}}\ \mathbf {S} \mathrm {d} \tau }$

${\displaystyle \mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}}$

${\displaystyle {\frac {\mathrm {d} \mathbf {p} _{charge}}{\mathrm {d} t}}=\oint _{\mathcal {S}}\ {\stackrel {\longleftrightarrow }{\mathbf {T} }}\cdot \mathrm {d} \mathbf {a} -{\frac {\mathrm {d} \mathbf {p} _{em}}{\mathrm {d} t}}}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}(\mathbf {p} _{charge}+\mathbf {p} _{em})=\oint _{\mathcal {S}}\ {\stackrel {\longleftrightarrow }{\mathbf {T} }}\cdot \mathrm {d} \mathbf {a} }$

${\displaystyle {\frac {\partial }{\partial t}}({\mathfrak {p}}_{charge}+{\mathfrak {p}}_{em})=\nabla \cdot {\stackrel {\longleftrightarrow }{\mathbf {T} }}}$