Action Principle (Relativistic)

As mentioned earlier, adding a total time derivative term \(f=f(q,t)\) to the Lagrangian does not change the EoM. There is a corresponding extension in field theory, where \(\mathcal{L}\to \bar{\mathcal{L}} =\mathcal{L}+\partial _\mu f^\mu\). It is important to note that \(f^\mu =f^\mu (\phi ,x^\mu )\) can only be a function of \((\phi ,x^\mu )\). This can be analyzed from two perspectives. The first is the simplest, considering the boundary of variation: $$\delta S=\delta \int (\mathcal{L}+\partial _\mu f^\mu ) d^4 x =\delta \left[\int \mathcal{L}d^4 x +\int \partial _\mu f^\mu d^4 x \right]$$ $$=\delta \int _a^b \mathcal{L}dt+\delta ∮ f^\mu d^3 S_\mu =\delta \int _a^b \mathcal{L}dt$$ Where \(\int \partial _\mu f^\mu d^4 x =∮ f^\mu d^3 S_\mu \) uses Gauss's theorem, and \(\delta ∮ f^\mu d^3 S_\mu =0\) comes from the invariance of boundary variation. Another method is to directly expand and observe whether \(\mathcal{L}+\partial _\mu f^\mu\) satisfies: $${\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial \phi} -\partial _\mu {\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }=0$$
$${\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial \phi} -\partial _\mu {\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }= {\partial\mathcal{L}\over\partial \phi} -\partial _\mu {\partial L\over\partial (\partial _\mu \phi ) }+{\partial (\partial _\mu f^\mu )\over\partial \phi }-\partial _\mu {\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }$$ $$={\partial (\partial _\mu f^\mu )\over\partial \phi }-\partial _\mu {\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }$$ Using $$df^\mu ={\partial f^\mu \over\partial \phi }\Big|_{x^\mu } d\phi +{\partial f^\mu \over \partial x^\nu }\Big|_\phi dx^\nu $$ $$\to \partial _\mu f^\mu ={\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \partial _\mu \phi +{\partial f^\mu \over\partial x^\mu } \Big|_\phi $$ Further partial differentiation: $${\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi )} ={\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \color{red}{ \text{dot cancelation analogy}}$$ Additionally, $${\partial (\partial _\mu f^\mu )\over\partial \phi }\Big|_{x^\mu}={ \partial \over\partial \phi }\Big|_{x^\mu} \left({\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \partial _\mu \phi + {\partial f^\mu \over\partial x^\mu }\Big|_\phi\right) $$ $$={ \partial \over\partial \phi }\Big|_{x^\mu} \left({\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \right)\partial _\mu \phi +{\partial \over\partial \phi}\Big|_{x^\mu }\left( {\partial f^\mu \over\partial x^\mu }\Big|_\phi \right)$$ $$={ \partial \over\partial \phi }\Big|_{x^\mu } {\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \partial _\mu \phi +{ \partial \over\partial x^\mu }\Big|_\phi { \partial f^\mu \over\partial \phi} \Big|_{x^\mu } $$ $$=\partial _\mu \left( {\partial f^\mu \over\partial \phi} \right)\to \partial _\mu\text{ and }{\partial\over\partial \phi}\text{ commute}$$ Thus, $${\partial (\partial _\mu f^\mu )\over\partial \phi} -\partial _\mu \left({\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }\right)=\partial _\mu {\partial f^\mu \over\partial \phi} -\partial _\mu {\partial f^\mu \over\partial \phi} =0 $$ Therefore, $${\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial \phi }-\partial _\mu \left({ (\partial (L+\partial _\mu f^\mu )\over\partial (\partial _\mu \phi )}\right)={\partial \mathcal{L}\over\partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi ) }\right)=0$$ Adding \(\partial _\mu f^\mu \) does not change the EoM.

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