# 電磁張量

（重定向自电磁张量

## 細節

数学注记：本文会使用到抽象的指标记号。


${\displaystyle F_{\alpha \beta }={\begin{bmatrix}0&E_{x}/c&E_{y}/c&E_{z}/c\\-E_{x}/c&0&-B_{z}&B_{y}\\-E_{y}/c&B_{z}&0&-B_{x}\\-E_{z}/c&-B_{y}&B_{x}&0\end{bmatrix}}}$

E電場
B磁場
c光速

### 性質

• 反對稱性${\displaystyle F^{\alpha \beta }\,=-F^{\beta \alpha }}$ （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——${\displaystyle E_{x}/c}$ ${\displaystyle E_{y}/c}$ ${\displaystyle E_{z}/c}$ ${\displaystyle B_{x}}$ ${\displaystyle B_{y}}$ ${\displaystyle B_{z}}$

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

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

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

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

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

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

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

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

## 導出電磁張量

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

${\displaystyle A_{\alpha }=\left({\frac {\phi }{c}},-A_{x},-A_{y},-A_{z}\right)\,}$

${\displaystyle {\vec {A}}\,}$ 向量勢，而${\displaystyle \left(A_{x},A_{y},A_{z}\right)}$ 為其分量，
${\displaystyle \phi \,}$ 純量勢
${\displaystyle c\,}$ 光速

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

x分量為例：

${\displaystyle E_{x}=-{\frac {\partial A_{x}}{\partial t}}-{\frac {\partial \phi }{\partial x}}=c({1 \over c}{\frac {\partial }{\partial t}}(-A_{x})-{\frac {\partial }{\partial x}}({\phi \over c}))\,}$
${\displaystyle B_{x}={\frac {\partial A_{z}}{\partial y}}-{\frac {\partial A_{y}}{\partial z}}\,}$

${\displaystyle E_{x}=c\left(\partial _{0}A_{1}-\partial _{1}A_{0}\right)\,}$ ，或將c移動到等號左邊：${\displaystyle {\frac {E_{x}}{c}}=\partial _{0}A_{1}-\partial _{1}A_{0}\,}$
${\displaystyle B_{x}=\partial _{2}A_{3}-\partial _{3}A_{2}\,}$

${\displaystyle F_{\alpha \beta }=\partial _{\alpha }A_{\beta }-\partial _{\beta }A_{\alpha }\,}$

## 與古典電磁學的關聯

${\displaystyle {\mathcal {S}}=\int \left(-{\begin{matrix}{\frac {1}{4\mu _{0}}}\end{matrix}}F_{\mu \nu }F^{\mu \nu }\right)\mathrm {d} ^{4}x\,}$

${\displaystyle \mathrm {d} ^{4}x\;}$ 是對時間及空間的積分。

 ${\displaystyle {\mathcal {L}}\,}$ ${\displaystyle =-{\begin{matrix}{\frac {1}{4\mu _{0}}}\end{matrix}}F_{\mu \nu }F^{\mu \nu }\,}$ ${\displaystyle =-{\begin{matrix}{\frac {1}{4\mu _{0}}}\end{matrix}}\left(\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)\left(\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }\right)\,}$ ${\displaystyle =-{\begin{matrix}{\frac {1}{4\mu _{0}}}\end{matrix}}\left(\partial _{\mu }A_{\nu }\partial ^{\mu }A^{\nu }-\partial _{\nu }A_{\mu }\partial ^{\mu }A^{\nu }-\partial _{\mu }A_{\nu }\partial ^{\nu }A^{\mu }+\partial _{\nu }A_{\mu }\partial ^{\nu }A^{\mu }\right)\,}$

 ${\displaystyle {\mathcal {L}}\,}$ ${\displaystyle =-{\begin{matrix}{\frac {1}{2\mu _{0}}}\end{matrix}}\left(\partial _{\mu }A_{\nu }\partial ^{\mu }A^{\nu }-\partial _{\nu }A_{\mu }\partial ^{\mu }A^{\nu }\right)\,}$

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

${\displaystyle \partial _{\nu }\left(\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }\right)=0\,}$

 ${\displaystyle \partial _{\nu }F^{\mu \nu }=0\,}$ 。

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

## 場張量的重要性

${\displaystyle {\vec {\nabla }}\cdot {\vec {E}}={\frac {\rho }{\epsilon _{0}}}}$

${\displaystyle {\vec {\nabla }}\times {\vec {B}}-{\frac {1}{c^{2}}}{\frac {\partial {\vec {E}}}{\partial t}}=\mu _{0}{\vec {J}}}$

${\displaystyle \partial _{\alpha }F^{\alpha \beta }=\mu _{0}J^{\beta }\,}$

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

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

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

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

## 場張量與相對論

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

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

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

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

## 參考文獻

• Brau, Charles A. Modern Problems in Classical Electrodynamics. Oxford University Press. 2004. ISBN 978-0-19-514665-3.
• Peskin, Michael E.; Schroeder, Daniel V. An Introduction to Quantum Field Theory. Perseus Publishing. 1995. ISBN 978-0-201-50397-5.