諾特定理

本節為筆者推論，還不確定是否正確
前面提到Lagrangian L的不唯一性，我們可以添加全微分項\(\dot{f}\) 而不改變EoM。以下來討論添加\(\dot{f}\) 對諾特定理的影響。我們只需先將\(L\to\bar{L} =L+\dot{f} \)改寫結論即可 $$\delta S=\int {d\over dt} \left[{\partial \bar{L} \over \partial \dot{q} } \Delta q-\left({\partial \bar{L} \over \partial \dot{q} } \dot{q} -\bar{L} \right)\delta t\right] dt$$ $$=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial \dot{f} \over \partial \dot{q} } \right)\Delta q-\left(\left({\partial L\over \partial \dot{q} } +{\partial \dot{f} \over \partial \dot{q} } \right) \dot{q} -L-\dot{f} \right)\delta t\right] dt$$ 注意\(f=f(q,t)\)，適用dot cancellation，\({\partial \dot{f} \over \partial \dot{q} } ={\partial f\over\partial q}\)，另外代入\( p={\partial L\over \partial \dot{q}}\)，所以 $$\delta S=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial f\over\partial q}\right)\Delta q-\left(\left(p +{\partial f\over\partial q}\right) \dot{q} -L-\dot{f} \right)\delta t\right] dt$$ $$=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial f\over\partial q}\right)\Delta q-\left(p\dot{q} +{\partial f\over\partial q}\dot{q} -L-\dot{f} \right)\delta t\right] dt$$ $$=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial f\over\partial q}\right)\Delta q-\left(H -{\partial f\over\partial t} \right)\delta t\right] dt$$ 如果當原本的Lagrangian \(L\)不具備對稱性，我們有機會透過\({\partial f \over \partial q}\)做修正，因為\(f=f(q,t)\)所以\({\partial f\over\partial q}={\partial \over\partial q}f(q,t)\)，只要\({\partial L\over \partial \dot{q} }\) 不是\(\dot{q}\) 的函數，我們可以用\({\partial f\over\partial q}=-{\partial L\over \partial \dot{q} } +const\)消除微分不為零的部分，使得\(\delta S=0\)，對應軌跡\(q\)變分不變的守恆量為 $${\partial L\over \partial \dot{q} } +{\partial f\over\partial q}=\bar{p}$$ ，我們可以稱作此Lagrangian \(\bar{L}= L+\dot{f}\) 對應的Action \(\bar{S}\)為Maximal Symmetry Action。