最小作用量原理（相對論性）

前面提到，當Lagrangian添加一個函數\(f=f(q,t)\)時間的全微分項df/dt，並不會改變EoM。在場論中也有對應的推廣，\(\mathcal{L}\to \bar{\mathcal{L}} =\mathcal{L}+\partial _\mu f^\mu\) ，一樣要注意的是\(f^\mu =f^\mu (\phi ,x^\mu )\)只能是\((\phi ,x^\mu )\)的函數。一樣可以從兩個部分來看。第一種是最簡單的從變分的邊界， $$\delta S=\delta \int (\mathcal{L}+\partial _\mu f^\mu ) d^4 x =\delta \left[\int \mathcal{L}d^4 x +\int \partial _\mu f^\mu d^4 x \right]$$ $$=\delta \int _a^b \mathcal{L}dt+\delta ∮ f^\mu d^3 S_\mu =\delta \int _a^b \mathcal{L}dt$$ 其中\(\int \partial _\mu f^\mu d^4 x =∮ f^\mu d^3 S_\mu \)利用高斯定理，\(\delta ∮ f^\mu d^3 S_\mu =0\)來自於變分邊界不變。 另外一種是直接展開，觀察\(\mathcal{L}+\partial _\mu f^\mu\) 是否滿足 $${\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial \phi} -\partial _\mu {\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }=0$$
$${\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial \phi} -\partial _\mu {\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }= {\partial\mathcal{L}\over\partial \phi} -\partial _\mu {\partial L\over\partial (\partial _\mu \phi ) }+{\partial (\partial _\mu f^\mu )\over\partial \phi }-\partial _\mu {\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }$$ $$={\partial (\partial _\mu f^\mu )\over\partial \phi }-\partial _\mu {\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }$$ 利用 $$df^\mu ={\partial f^\mu \over\partial \phi }\Big|_{x^\mu } d\phi +{\partial f^\mu \over \partial x^\nu }\Big|_\phi dx^\nu $$ $$\to \partial _\mu f^\mu ={\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \partial _\mu \phi +{\partial f^\mu \over\partial x^\mu } \Big|_\phi $$ 進一步偏微分 $${\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi )} ={\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \color{red}{ \text{dot cancelation analogy}}$$ 另外 $${\partial (\partial _\mu f^\mu )\over\partial \phi }\Big|_{x^\mu}={ \partial \over\partial \phi }\Big|_{x^\mu} \left({\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \partial _\mu \phi + {\partial f^\mu \over\partial x^\mu }\Big|_\phi\right) $$ $$={ \partial \over\partial \phi }\Big|_{x^\mu} \left({\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \right)\partial _\mu \phi +{\partial \over\partial \phi}\Big|_{x^\mu }\left( {\partial f^\mu \over\partial x^\mu }\Big|_\phi \right)$$ $$={ \partial \over\partial \phi }\Big|_{x^\mu } {\partial f^\mu \over\partial \phi }\Big|_{x^\mu } \partial _\mu \phi +{ \partial \over\partial x^\mu }\Big|_\phi { \partial f^\mu \over\partial \phi} \Big|_{x^\mu } $$ $$=\partial _\mu \left( {\partial f^\mu \over\partial \phi} \right)\to \partial _\mu\text{ and }{\partial\over\partial \phi}\text{ commute}$$ 所以 $${\partial (\partial _\mu f^\mu )\over\partial \phi} -\partial _\mu \left({\partial (\partial _\mu f^\mu )\over\partial (\partial _\mu \phi ) }\right)=\partial _\mu {\partial f^\mu \over\partial \phi} -\partial _\mu {\partial f^\mu \over\partial \phi} =0 $$ 故 $${\partial (\mathcal{L}+\partial _\mu f^\mu )\over\partial \phi }-\partial _\mu \left({ (\partial (L+\partial _\mu f^\mu )\over\partial (\partial _\mu \phi )}\right)={\partial \mathcal{L}\over\partial \phi} -\partial _\mu \left({\partial \mathcal{L}\over\partial (\partial _\mu \phi ) }\right)=0$$ 添加\(\partial _\mu f^\mu \)不會改變EoM。