Action Principle (Classical)

In the theoretical mechanics course, it is taught that adding a total time derivative term \(f=f(q,t)\) to the Lagrangian, i.e., \(L→L+{df\over dt}\), does not change the EoM. This can be explained in two parts. The first is the simplest, from the variation of the endpoints: $$\delta S=\delta \int _a^b (L+{df\over dt})dt$$ $$=\delta \left[\int _a^b Ldt+\int _a^b {df\over dt} dt \right]$$ $$=\delta \left[\int _a^b Ldt+f(b)-f(a)\right]$$ $$=\delta \int _a^b Ldt+\delta f(b)-\delta f(a)$$ But since the endpoints are not varied, \(\delta f(b)=\delta f(a)=0\), naturally, $$\delta \int _a^b \left(L+{df\over dt}\right)dt=\delta \int _a^b Ldt=0$$ The second approach is to directly expand and observe whether \(L+\dot{f}\) satisfies \({d\over dt} {\partial \over\partial \dot{x}}\left( L+\dot{f} \right) -{\partial \over\partial x}\left( L+\dot{f} \right)=0\). $${d\over dt} {\partial \over\partial \dot{x}}\left( L+\dot{f} \right) -{\partial \over\partial x}\left( L+\dot{f} \right)={d\over dt} {\partial L\over\partial \dot{x}} -{\partial L\over\partial x}+\color{red}{{d\over dt} {\partial \dot{f} \over\partial \dot{x}} -{\partial \dot{f} \over\partial x}}$$ The red term allows us to observe two things, because \(f=f(x,t)\): $$df={\partial f\over\partial x} dx+{\partial f\over\partial \dot{x}} d\dot{x} +{\partial f\over\partial t} dt$$ $$\to \dot{f} ={\partial \over\partial x} \dot{x} +{\partial f\over\partial \dot{x}} \ddot{x} +{\partial f\over\partial t}$$ So, $${ \partial \dot{f} \over\partial \dot{x}} ={\partial f\over\partial x}\to dot\ cancellation$$
$${\partial \over\partial x} \dot{f}={\partial \over\partial x} \left({\partial \over\partial x} \dot{x} +{\partial f\over\partial \dot{x}} \ddot{x} +{\partial f\over\partial t}\right)$$ $$=\left({\partial \over\partial x} {\partial f\over\partial x}\right) \dot{x} +\left({\partial \over\partial x} {\partial f\over\partial \dot{x}} \right) \ddot{x} +\left({\partial \over\partial x} {\partial f\over\partial t}\right)$$ $$={d\over dt} \left({\partial f\over\partial x}\right)\to{d\over dt},{\partial \over\partial x} commute $$ We get the red term to be zero: $$\color{red}{{d\over dt} {\partial \dot{f} \over\partial \dot{x} }-{\partial \dot{f} \over\partial x}={d\over dt} {\partial f\over\partial x}-{d\over dt} {\partial f\over\partial x}=0}$$ Thus, $${d\over dt} {\partial \over\partial \dot{x}}\left( L+\dot{f} \right) -{\partial \over\partial x}\left( L+\dot{f} \right)={d\over dt} {\partial L\over\partial \dot{x}} -{\partial L\over\partial x}=0$$ Adding \(\dot{f}\) does not change the EoM.

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