# 電磁張量

## 細節

数学注记：本文会使用到抽象的指标记号。


$F_{\alpha\beta} = \begin{bmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & -B_z & B_y \\ -E_y/c & B_z & 0 & -B_x \\ -E_z/c & -B_y & B_x & 0 \end{bmatrix}$

E電場
B磁場
c光速

### 性質

• 反對稱性$F^{\alpha\beta} \, = - F^{\beta\alpha}$（因此稱作雙向量（或稱雙矢、二重向量，bivector））。
• 零值的跡數或稱對角和
• 6個獨立分量——$E_x/c$$E_y/c$$E_z/c$$B_x$$B_y$$B_z$

$F_{\alpha\beta} F^{\alpha\beta} = \ 2 \left( B^2 - \frac{E^2}{c^2} \right) = \mathrm{invariant}$

$\epsilon_{\alpha\beta\gamma\delta}F^{\alpha\beta} F^{\gamma\delta} = - \frac{2}{c} \left( \vec B \cdot \vec E \right) = \mathrm{invariant} \,$

$\det \left( F \right) = \frac{1}{c^2} \left( \vec B \cdot \vec E \right) ^{2}$

$F_{ \alpha\beta } \ \stackrel{\mathrm{def}}{=}\ \frac{ \partial A_{\beta} }{ \partial x^{\alpha} } - \frac{ \partial A_{\alpha} }{ \partial x^{\beta} } \ \stackrel{\mathrm{def}}{=}\ \partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha}$

$A^{\alpha} = \left( \frac{\phi}{c} , \vec A \right)$，其協變（covariant）形式可以透過乘上閔可夫斯基度規$\eta \,$來得到：
$A_{\alpha} \, = \eta_{\alpha\beta} A^{\beta} = \left( \frac{\phi}{c}, -\vec A \right)$

$\eta = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

$\eta = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

$A_{\alpha} \, = \eta_{\alpha\beta} A^{\beta} = \left( -\frac{\phi}{c}, \vec A \right)$

## 導出電磁張量

$\partial_{\alpha} = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) = \left(\frac{1}{c} \frac{\partial}{\partial t}, \vec{\nabla} \right) \,$

$A_{\alpha} = \left(\frac{\phi}{c}, -A_x, -A_y, -A_z \right) \,$

$\vec A \,$向量勢，而$\left(A_x, A_y, A_z \right)$為其分量，
$\phi \,$純量勢
$c \,$光速

$\vec{E} = -\frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi \,$
$\vec{B} = \vec{\nabla} \times \vec{A} \,$

x分量為例：

$E_x = -\frac{\partial A_x}{\partial t} - \frac{\partial \phi}{\partial x} = c ( {1\over c} \frac{\partial}{\partial t}(-A_x) - \frac{\partial}{\partial x}({\phi\over c}) ) \,$
$B_x = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \,$

$E_x = c \left(\partial_0 A_1 - \partial_1 A_0 \right) \,$，或將c移動到等號左邊：$\frac{E_x}{c} = \partial_0 A_1 - \partial_1 A_0 \,$
$B_x = \partial_2 A_3 - \partial_3 A_2 \,$

$F_{\alpha\beta} = \partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha} \,$

## 與古典電磁學的關聯

$\mathcal{S} = \int \left( -\begin{matrix} \frac{1}{4 \mu_0} \end{matrix} F_{\mu\nu} F^{\mu\nu} \right) \mathrm{d}^4 x \,$

$\mathrm{d}^4 x \;$是對時間及空間的積分。

 $\mathcal{L} \,$ $= -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} F_{\mu\nu} F^{\mu\nu} \,$ $= -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu - \partial_\nu A_\mu \right) \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right) \,$ $= -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu - \partial_\mu A_\nu \partial^\nu A^\mu + \partial_\nu A_\mu \partial^\nu A^\mu \right) \,$

 $\mathcal{L} \,$ $= -\begin{matrix} \frac{1}{2\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu \right) \,$

$\partial_\nu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\nu A_\mu )} \right) - \frac{\partial \mathcal{L}}{\partial A_\mu} = 0 \,$

$\partial_\nu \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right) = 0 \,$

 $\partial_\nu F^{\mu \nu} = 0 \,$。

$~E^i /c \ \ = -F^{0 i} \,$
$\epsilon^{ijk} B^k = -F^{ij} \,$

## 場張量的重要性

$\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

$\vec{\nabla} \times \vec{B} - \frac{1}{c^2} \frac{ \partial \vec{E}}{\partial t} = \mu_0 \vec{J}$

$\partial_{\alpha} F^{\alpha\beta} = \mu_0 J^{\beta} \,$

$J^{\alpha} = ( c \, \rho , \vec{J} ) \,$四維電流密度

$\vec{\nabla} \cdot \vec{B} = 0$

$\frac{ \partial \vec{B}}{ \partial t } + \vec{\nabla} \times \vec{E} = 0$

$F_{ \alpha \beta , \gamma } + F_{ \beta \gamma , \alpha } + F_{ \gamma \alpha , \beta } = 0 \,$，或者利用反對稱化符號——方括號[]表示成
$F_{ [\alpha \beta , \gamma] } = 0 \,$

## 場張量與相對論

$F_{[\alpha\beta,\gamma]} \, = 0$
$F^{\alpha\beta}{}_{,\beta} \, = \mu_0 J^{\alpha}$

$J^\alpha{}_{,\alpha} \, = 0$

$F_{[\alpha\beta;\gamma]} \, = 0$
$F^{\alpha\beta}{}_{;\beta} \, = \mu_0 J^{\alpha}$

$J^\alpha{}_{;\alpha} \, = 0$

## 參考文獻

• Brau, Charles A. Modern Problems in Classical Electrodynamics. Oxford University Press. 2004. ISBN 0-19-514665-4.
• Peskin, Michael E.; Schroeder, Daniel V. An Introduction to Quantum Field Theory. Perseus Publishing. 1995. ISBN 0-201-50397-2.