Noether's Theorem

This section is a conjecture by the author and is not yet confirmed to be correct
As mentioned earlier, the Lagrangian \(L\) is not unique, and we can add a total derivative term \(\dot{f}\) without changing the EoM. Here, we discuss the impact of adding \(\dot{f}\) on Noether's theorem. We simply need to modify the conclusion by substituting \(L\to\bar{L} =L+\dot{f}\): $$\delta S=\int {d\over dt} \left[{\partial \bar{L} \over \partial \dot{q} } \Delta q-\left({\partial \bar{L} \over \partial \dot{q} } \dot{q} -\bar{L} \right)\delta t\right] dt$$ $$=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial \dot{f} \over \partial \dot{q} } \right)\Delta q-\left(\left({\partial L\over \partial \dot{q} } +{\partial \dot{f} \over \partial \dot{q} } \right) \dot{q} -L-\dot{f} \right)\delta t\right] dt$$ Note that \(f=f(q,t)\), so the dot cancellation applies, \({\partial \dot{f} \over \partial \dot{q} } ={\partial f\over\partial q}\). Additionally, substituting \( p={\partial L\over \partial \dot{q}}\), we have: $$\delta S=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial f\over\partial q}\right)\Delta q-\left(\left(p +{\partial f\over\partial q}\right) \dot{q} -L-\dot{f} \right)\delta t\right] dt$$ $$=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial f\over\partial q}\right)\Delta q-\left(p\dot{q} +{\partial f\over\partial q}\dot{q} -L-\dot{f} \right)\delta t\right] dt$$ $$=\int {d\over dt} \left[\left({\partial L\over \partial \dot{q} } +{\partial f\over\partial q}\right)\Delta q-\left(H -{\partial f\over\partial t} \right)\delta t\right] dt$$ If the original Lagrangian \(L\) does not possess symmetry, we may have the opportunity to modify it through \({\partial f \over \partial q}\) since \(f=f(q,t)\) so that \({\partial f\over\partial q}={\partial \over\partial q}f(q,t)\). As long as \({\partial L\over \partial \dot{q} }\) is not a function of \(\dot{q}\), we can use \({\partial f\over\partial q}=-{\partial L\over \partial \dot{q} } +const\) to eliminate the non-zero differential part, making \(\delta S=0\). The conserved quantity corresponding to the invariance under the variation of trajectory \(q\) is: $${\partial L\over \partial \dot{q} } +{\partial f\over\partial q}=\bar{p}$$ We can call the Lagrangian \(\bar{L}= L+\dot{f}\) the Maximal Symmetry Action corresponding to the Action \(\bar{S}\).

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