# 馬克士威應力張量

## 導引

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

$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$

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

\begin{align}\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{align}

$\mathbf{f}= \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B})\mathbf{B} - \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right) \right] - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)$

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

$\mathbf{f}$ 的方程式內的旋度項目除去：

$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)$

$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$

$T_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)$

$(\mathbf{A}\cdot \stackrel{\longleftrightarrow}{\mathbf{T}})_j=\textstyle{\sum_i}\ A_i T_{ij}$

$\mathbf{f} = \nabla \cdot \stackrel{\longleftrightarrow}{\mathbf{T}} - \epsilon_0 \mu_0 \frac{\partial \mathbf{S}}{\partial t}$

## 馬克士威應力張量的性質

$T_{ij} = \left( \begin{matrix} \epsilon_0(E_x^2 - E^2 /2)+\cfrac{1}{\mu_0}(B_x^2 - B^2 /2) & \epsilon_0 E_x E_y +\cfrac{1}{\mu_0}(B_x B_y) & \epsilon_0 E_x E_z+\cfrac{1}{\mu_0}(B_x B_z) \\ \epsilon_0 E_x E_y+\cfrac{1}{\mu_0}(B_x B_y) & \epsilon_0(E_y^2 - E^2 /2)+\cfrac{1}{\mu_0}(B_y^2 - B^2 /2) & \epsilon_0 E_y E_z +\cfrac{1}{\mu_0}(B_y B_z) \\ \epsilon_0 E_x E_z+\cfrac{1}{\mu_0}(B_x B_z) & \epsilon_0 E_y E_z+\cfrac{1}{\mu_0}(B_y B_z) & \epsilon_0(E_z^2 - E^2 /2)+\cfrac{1}{\mu_0}(B_z^2 - B^2 /2) \end{matrix} \right)$

## 動量守恆定律

$\mathbf{F}=\int_{\mathcal{V}}\ \mathbf{f}\mathrm{d}\tau=\int_{\mathcal{V}}\ \nabla \cdot \stackrel{\longleftrightarrow}{\mathbf{T}}\mathrm{d}\tau - \epsilon_0 \mu_0\frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathcal{V}}\ \mathbf{S}\mathrm{d}\tau$

$\mathbf{F}=\oint_{\mathcal{S}}\ \stackrel{\longleftrightarrow}{\mathbf{T}}\cdot\mathrm{d}\mathbf{a} - \epsilon_0 \mu_0\frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathcal{V}}\ \mathbf{S}\mathrm{d}\tau$

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

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

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

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