最小作用量原理（古典）

理論力學的課程中，都有學過當\(Lagrangian\)添加一個函數\(f=f(q,t)\)時間的全微分項\({df\over dt}\)，\(L→L+{df\over dt}\)並不會改變EoM。這可以從兩個部分來看。第一種是最簡單的從變分的端點， $$\delta S=\delta \int _a^b (L+{df\over dt})dt$$ $$=\delta \left[\int _a^b Ldt+\int _a^b {df\over dt} dt \right]$$ $$=\delta \left[\int _a^b Ldt+f(b)-f(a)\right]$$ $$=\delta \int _a^b Ldt+\delta f(b)-\delta f(a)$$ 但因為端點不做變分，所以\(\delta f(b)=\delta f(a)=0\)，自然的 $$\delta \int _a^b \left(L+{df\over dt}\right)dt=\delta \int _a^b Ldt=0$$ 另外一種是直接展開，觀察\(L+\dot{f}\) 是否滿足 \({d\over dt} {\partial \over\partial \dot{x}}\left( L+\dot{f} \right) -{\partial \over\partial x}\left( L+\dot{f} \right)=0\) $${d\over dt} {\partial \over\partial \dot{x}}\left( L+\dot{f} \right) -{\partial \over\partial x}\left( L+\dot{f} \right)={d\over dt} {\partial L\over\partial \dot{x}} -{\partial L\over\partial x}+\color{red}{{d\over dt} {\partial \dot{f} \over\partial \dot{x}} -{\partial \dot{f} \over\partial x}}$$ 紅色項我們可以觀察兩件事，因為\(f=f(x,t)\) $$df={\partial f\over\partial x} dx+{\partial f\over\partial \dot{x}} d\dot{x} +{\partial f\over\partial t} dt$$ $$\to \dot{f} ={\partial \over\partial x} \dot{x} +{\partial f\over\partial \dot{x}} \ddot{x} +{\partial f\over\partial t}$$ 所以 $${ \partial \dot{f} \over\partial \dot{x}} ={\partial f\over\partial x}\to dot\ cancellation$$
$${\partial \over\partial x} \dot{f}={\partial \over\partial x} \left({\partial \over\partial x} \dot{x} +{\partial f\over\partial \dot{x}} \ddot{x} +{\partial f\over\partial t}\right)$$ $$=\left({\partial \over\partial x} {\partial f\over\partial x}\right) \dot{x} +\left({\partial \over\partial x} {\partial f\over\partial \dot{x}} \right) \ddot{x} +\left({\partial \over\partial x} {\partial f\over\partial t}\right)$$ $$={d\over dt} \left({\partial f\over\partial x}\right)\to{d\over dt},{\partial \over\partial x} commute $$ 會得到紅色項為零 $$\color{red}{{d\over dt} {\partial \dot{f} \over\partial \dot{x} }-{\partial \dot{f} \over\partial x}={d\over dt} {\partial f\over\partial x}-{d\over dt} {\partial f\over\partial x}=0}$$ 故 $${d\over dt} {\partial \over\partial \dot{x}}\left( L+\dot{f} \right) -{\partial \over\partial x}\left( L+\dot{f} \right)={d\over dt} {\partial L\over\partial \dot{x}} -{\partial L\over\partial x}=0$$ 添加\(\dot{f}\) 不會改變EoM。