Action Principle (Relativistic)

In electromagnetism, we define the relationship between the potentials \(\phi\), \(\overrightarrow{A}\), and the electric field \(\overrightarrow{E}\), and the magnetic field \(\overrightarrow{B}\): $$ \overrightarrow{E}=- \nabla \phi -{1 \over c} {\partial \overrightarrow{A}\over \partial t}$$ $$\overrightarrow{B}= \nabla ×\overrightarrow{A} $$ However, the potentials \(\phi\) and \(\overrightarrow{A}\) are not unique, and we can introduce a gauge function \(G(ct,\overrightarrow{x})\): $$\bar{\phi} =\phi +{1 \over c} {\partial G \over \partial t}$$ $$\overrightarrow{\bar{A}}=\overrightarrow{A}- \nabla G$$ The electric field \(\overrightarrow{E}\) and the magnetic field \(\overrightarrow{B}\) remain unchanged: $$\overrightarrow{\bar{E}}=- \nabla \bar{\phi}-{1 \over c} {\partial \overrightarrow{\bar{A}}\over \partial t}$$ $$=- \nabla \phi - \color{red}{\nabla {1 \over c} {\partial G \over \partial t}}-{1 \over c} {\partial \overrightarrow{A}\over \partial t}+\color{red}{{1 \over c} {\partial ( \nabla G)\over\partial t}}$$ $$=- \nabla \phi -{1 \over c} {\partial \overrightarrow{A}\over \partial t}=\overrightarrow{E}$$ $$\overrightarrow{\bar{B}}= \nabla ×\overrightarrow{\bar{A}}= \nabla ×\overrightarrow{A}- \nabla × \nabla G= \nabla ×\overrightarrow{A}=\overrightarrow{B} $$

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

In relativity, the above operation can be written in the form of 4-potentials: $$(\bar{\phi},\overrightarrow{\bar{A}} )=(\phi ,\overrightarrow{A} )+(\partial _{ct} G,- \nabla G)=(\phi ,\overrightarrow{A} )+(\partial _{ct},- \nabla )G$$ Recall: $$\partial ^\mu =(\partial _{ct},- \nabla )$$ Thus: $$\bar{A} ^\mu =A^\mu +\partial ^\mu G$$

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

Lorenz gauge condition: Relativistic, satisfies Lorentz transformation $$ \nabla \cdot \overrightarrow{A}+{1 \over c} {\partial \phi \over \partial t}=0$$ Written in scalar form: $$\partial _\mu A^\mu =0$$ Thus, it satisfies the relativistic transformation.

As previously shown, the electric field \(\overrightarrow{E}\) and the magnetic field \(\overrightarrow{B}\) remain unchanged under a gauge transformation: $$\overrightarrow{\bar{E}}=\overrightarrow{E}$$ $$\overrightarrow{\bar{B}}=\overrightarrow{B}$$ This indicates that \(F^{\mu \nu}\) also remains unchanged under a gauge transformation: $$ \bar{F} ^{\mu \nu} =F^{\mu \nu} $$ We can also prove this: $$\bar{F} ^{\mu \nu} =\partial ^\mu \bar{A} ^\nu -\partial ^\nu \bar{A} ^\mu$$ $$=\partial ^\mu (A^\nu +\partial ^\nu G)-\partial ^\nu (A^\mu +\partial ^\mu G)$$ $$=\partial ^\mu A^\nu +\color{red}{\partial ^\mu \partial ^\nu G}-\partial ^\nu A^\mu -\color{red}{\partial ^\nu \partial ^\mu G}$$ $$=\partial ^\mu A^\nu -\partial ^\nu A^\mu =F^{\mu \nu} $$

Originally written in Chinese by the author, these articles are translated into English to invite cross-language resonance.