最小作用量原理（相對論性）

在電磁學中，我們定義Potential \(\phi\)、\(\overrightarrow{A}\)和電場\(\overrightarrow{E}\)、磁場\(\overrightarrow{B}\)的關係： $$ \overrightarrow{E}=- \nabla \phi -{1 \over c} {\partial \overrightarrow{A}\over \partial t}$$ $$\overrightarrow{B}= \nabla ×\overrightarrow{A} $$ 但Potential \(\phi\) 、\(\overrightarrow{A}\)不唯一，可以引入一個Gauge function \(G(ct,\overrightarrow{x} )\) $$\bar{\phi} =\phi +{1 \over c} {\partial G \over \partial t}$$ $$\overrightarrow{\bar{A}}=\overrightarrow{A}- \nabla G$$ 一樣保持電場\(\overrightarrow{E}\)、磁場\(\overrightarrow{B}\)不變 $$\overrightarrow{\bar{E}}=- \nabla \bar{\phi}-{1 \over c} {\partial \overrightarrow{\bar{A}}\over \partial t}$$ $$=- \nabla \phi - \color{red}{\nabla {1 \over c} {\partial G \over \partial t}}-{1 \over c} {\partial \overrightarrow{A}\over \partial t}+\color{red}{{1 \over c} {\partial ( \nabla G)\over\partial t}}$$ $$=- \nabla \phi -{1 \over c} {\partial \overrightarrow{A}\over \partial t}=\overrightarrow{E}$$
$$\overrightarrow{\bar{B}}= \nabla ×\overrightarrow{\bar{A}}= \nabla ×\overrightarrow{A}- \nabla × \nabla G= \nabla ×\overrightarrow{A}=\overrightarrow{B} $$

Note \(\nabla × \nabla G=0\) $$( \nabla × \nabla G)_i=\varepsilon _{ijk} \partial _j \partial _k G$$ $$={1 \over 2} \varepsilon _{ijk} \partial _j \partial _k G+{1 \over 2} \varepsilon _{ijk} \partial _j \partial _k G$$ $$={1 \over 2} \varepsilon _{ijk} \partial _j \partial _k G+{1 \over 2} \varepsilon _{i\color{red}{kj}} \partial _{\color{red}{k}} \partial _{\color{red}{j}} G$$ $$={1 \over 2} \varepsilon _{ijk} \partial _j \partial _k G \color{red}{-}{1 \over 2} \varepsilon _{i\color{red}{jk}} \partial _j \partial _k G=0$$

在相對論中，上述的操作可以寫為4-Potential的形式： $$(\bar{\phi},\overrightarrow{\bar{A}} )=(\phi ,\overrightarrow{A} )+(\partial _{ct} G,- \nabla G)=(\phi ,\overrightarrow{A} )+(\partial _{ct},- \nabla )G$$ 回憶 $$\partial ^\mu =(\partial _{ct},- \nabla )$$ 故寫成 $$\bar{A} ^\mu =A^\mu +\partial ^\mu G$$

Coulomb gauge condition：非相對論性Non-relativistic $$ \nabla \cdot \overrightarrow{A}=0$$

Lorenz gauge condition：相對論性，滿足Lorentz transformation $$ \nabla \cdot \overrightarrow{A}+{1 \over c} {\partial \phi \over \partial t}=0$$ 寫成Scalar form $$\partial _\mu A^\mu =0$$ 所以滿足相對論轉換。

剛剛已經寫下電場\(\overrightarrow{E}\)、磁場\(\overrightarrow{B}\)在加入Gauge之下保持不變 $$\overrightarrow{\bar{E}}=\overrightarrow{E}$$ $$\overrightarrow{\bar{B}}=\overrightarrow{B}$$ 這其實說明，\(F^{\mu \nu}\) 也在加入Gauge之下保持不變： $$ \bar{F} ^{\mu \nu} =F^{\mu \nu} $$ 當然我們也可以寫下證明 $$\bar{F} ^{\mu \nu} =\partial ^\mu \bar{A} ^\nu -\partial ^\nu \bar{A} ^\mu$$ $$=\partial ^\mu (A^\nu +\partial ^\nu G)-\partial ^\nu (A^\mu +\partial ^\mu G)$$ $$=\partial ^\mu A^\nu +\color{red}{\partial ^\mu \partial ^\nu G}-\partial ^\nu A^\mu -\color{red}{\partial ^\nu \partial ^\mu G}$$ $$=\partial ^\mu A^\nu -\partial ^\nu A^\mu =F^{\mu \nu} $$