Action Principle (Classical)

The Lagrangian equation can be described not only in \((q,\dot{q} ,t)\) space but also in momentum space \((p,\dot{p} ,t)\). The transformation is as follows: $$ p={\partial L \over \partial \dot{q}} $$ $$\dot{p} ={d\over dt} {\partial L\over\partial \dot{q}}={\partial L\over\partial q}$$ \(p\) is the generalized momentum. Using \(L=L(q,\dot{q} ,t)\), $$dL=\dot{p} dq+pd\dot{q} +{\partial L\over\partial t} dt$$ $$=d(\dot{p} q)-qd\dot{p} +d(p\dot{q} )-\dot{q} dp+{\partial L\over\partial t} dt$$ $$=\color{red}{d(\dot{p} q+p\dot{q} )}-qd\dot{p} -\dot{q} dp+{\partial L\over\partial t} dt$$ Moving the red term to the other side: $$d \left(L\color{red}{-\dot{p} q-p\dot{q}} \right)=-qd\dot{p} -\dot{q} dp+{\partial L\over\partial t} dt$$ Define a new Lagrangian \(\bar{L}\): $$\bar{L} \equiv L-\dot{p} q-p\dot{q} =L-{d\over dt} (pq)$$ We get \(d\bar{L}\): $$ d\bar{L}=-qd\dot{p} -\dot{q} dp+{\partial L\over\partial t} dt$$ Comparing the left and right sides, we get: $${\partial \bar{L}\over\partial \dot{p} }=-q$$ $${\partial \bar{L}\over\partial p}=-\dot{q}$$ $$\to {d\over dt} {\partial \bar{L}\over\partial \dot{p}}={\partial \bar{L}\over\partial p}$$ This is the EoM in momentum space. It is noteworthy that the form is invariant.

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