Action Principle (Relativistic)

If we maintain gauge invariance, we can argue that the action will not include a term like \(A_\mu A^\mu\), as the variation of \(A^\mu\) (i.e., \(A^\mu \to A^\mu +\delta A^\mu\)) would lead to changes in the equations of motion: $$S=-{1 \over c^2} \int A_\mu A^\mu d^4 x $$ Adding a gauge \(\partial ^\mu G\): $$ \bar{A} ^\mu =A^\mu +\partial ^\mu G$$ Expanding the action: $$\bar{S}=-{1 \over c^2} \int \bar{A} _\mu \bar{A} ^\mu d^4 x $$ $$=-{1 \over c^2} \int (A_\mu +\partial _\mu G)(A^\mu +\partial ^\mu G) d^4 x $$ $$=-{1 \over c^2} \int A_\mu A^\mu d^4 x -{1 \over c^2} \int 2A^\mu \cdot \partial _\mu Gd^4 x -{1 \over c^2} \int \partial _\mu G\cdot \partial ^\mu Gd^4 x $$ Since the gauge \(\partial ^\mu G\) is independent of the 4-potential \(A^\mu\), the variation \(A^\mu \to A^\mu +\delta A^\mu\) does not affect \(\partial ^\mu G\). We compare the variations of \(S\) and \(\bar{S}\): **For \(S\):** $$\delta S=-{1 \over c^2} \delta \int A_\mu A^\mu d^4 x$$ $$ =-{2 \over c^2} \int A_\mu \delta A^\mu d^4 x$$ **For \(\bar{S}\):** $$\delta \bar{S}=-{1 \over c^2} \delta \int A_\mu A^\mu d^4 x -{1 \over c^2} \delta \int 2A_\mu \partial ^\mu Gd^4 x -\color{red}{0\,\,\left(\partial ^\mu G\text{ has no variation}\right)}$$ The first term is the original variation: $$\delta \bar{S}=-{2 \over c^2} \int A_\mu \delta A^\mu d^4 x -{2 \over c^2} \delta \int A^\mu \cdot \partial _\mu Gd^4 x $$ $$=-{2 \over c^2} \int (A_\mu +\partial _\mu G)\delta A^\mu d^4 x $$ Comparing, we find that they are different, which would lead to the breakdown of gauge invariance. Therefore, to maintain gauge invariance, the term \(A_\mu A^\mu\) would not appear. Of course, if one day experiments show that gauge invariance is incorrect, it might be possible to introduce an \(A_\mu A^\mu\) term.

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