Action Principle (Relativistic)

In this section, we discuss how adding a gauge does not change the variation of the action, meaning that the equations of motion are unaffected by gauge changes, representing gauge invariance. When adding a gauge: $$ \bar{A} ^\mu =A^\mu +\partial ^\mu G$$ We observe which terms in the action \(S=S_P+S_{PF}+S_F\) are affected: $$S_P=-mc^2 \int _a^b d\tau =\int -\rho_m cd^4 x $$ $$S_{PF}=-{e\over c} \int _a^b A_\mu dx^\mu =-{1 \over c^2} \int A_\mu J^\mu d^4 x $$ $$S_F=\int -{1 \over 16 \pi c} F_{\mu \nu} F^{\mu \nu} d^4 x $$ Since we have already proven that \(F^{\mu \nu}\) is not affected by gauge transformations and \(S_P\) does not involve \(A^\mu\), the only affected term is \(S_{PF}\): $$\bar{S}_{PF}=-{e\over c} \int _a^b \bar{A} _\mu dx^\mu =-{1 \over c^2} \int \bar{A} _\mu J^\mu d^4 x $$ Now, let's vary \(\bar{S}_{PF}\): $$\delta \bar{S}_{PF}=-{1 \over c^2} \delta \int \bar{A} _\mu J^\mu d^4 x =-{1 \over c^2} \delta \int A_\mu J^\mu d^4 x \color{red}{-{1 \over c^2} \delta \int \partial _\mu G \cdot J ^\mu d^4 x }$$ Applying the chain rule: $$\partial _\mu G \cdot J ^\mu =\partial _\mu (GJ^\mu )-G\partial _\mu J^\mu $$ This yields: $$\delta \bar{S}_{PF}=-{1 \over c^2} \delta \int A_\mu J^\mu d^4 x \color{red}{-{1 \over c^2} { \delta \int \partial _\mu \left(GJ^\mu \right) d^4 x + {1 \over c^2} \delta \int G\partial _\mu J^\mu d^4 x} } $$ The middle term, using the Divergence Theorem: $$\delta \bar{S}_{PF}=-{1 \over c^2} \delta \int A_\mu J^\mu d^4 x \color{red}{-{1 \over c^2} {\delta \left[ ∮ GJ^\mu dS_\mu \right]}{ +{1 \over c^2} \delta \int G \left[\partial _\mu J^\mu \right] d^4 x }}$$ The middle term is zero since the variation on the boundary is zero: $$\color{red}{\delta \left[∮ GJ^\mu dS_\mu \right]=0}$$ Finally, recall the continuity equation: $${\partial \rho_e \over \partial t}+ \nabla \cdot \overrightarrow{J}=0$$ In relativistic form: $$\partial _\mu J^\mu =0$$ Thus, the last term is also zero: $$\color{red}{\delta \int G \left[\partial _\mu J^\mu \right] d^4 x =0}$$ Therefore: $$\delta \bar{S}_{PF}=-{1 \over c^2} \delta \int A_\mu J^\mu d^4 x =\delta S_{PF}$$ This shows that adding a gauge does not change the variation of the action, indicating gauge invariance.

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