最小作用量原理（古典）

Lagrangian equation不只可以在\((q,\dot{q} ,t)\)中描述，也可以在動量空間\((p,\dot{p} ,t)\)中描述，轉換如下： $$ p={\partial L \over \partial \dot{q}} $$ $$\dot{p} ={d\over dt} {\partial L\over\partial \dot{q}}={\partial L\over\partial q}$$ \(p\)為廣義動量。利用\(L=L(q,\dot{q} ,t)\)， $$dL=\dot{p} dq+pd\dot{q} +{\partial L\over\partial t} dt$$ $$=d(\dot{p} q)-qd\dot{p} +d(p\dot{q} )-\dot{q} dp+{\partial L\over\partial t} dt$$ $$=\color{red}{d(\dot{p} q+p\dot{q} )}-qd\dot{p} -\dot{q} dp+{\partial L\over\partial t} dt$$ 紅色項移項 $$d \left(L\color{red}{-\dot{p} q-p\dot{q}} \right)=-qd\dot{p} -\dot{q} dp+{\partial L\over\partial t} dt$$ 定義新的Lagrangian \(\bar{L}\) $$\bar{L} \equiv L-\dot{p} q-p\dot{q} =L-{d\over dt} (pq)$$ 得到\(d\bar{L}\) $$ d\bar{L}=-qd\dot{p} -\dot{q} dp+{\partial L\over\partial t} dt$$ 比較左右兩邊可以得到 $${\partial \bar{L}\over\partial \dot{p} }=-q$$ $${\partial \bar{L}\over\partial p}=-\dot{q}$$ $$\to {d\over dt} {\partial \bar{L}\over\partial \dot{p}}={\partial \bar{L}\over\partial p}$$ 為動量空間的EoM。可以注意到形式不變(Form invariant)。