Noether's Theorem

Classical Noether's Theorem Noether's theorem, as an important theorem in classical mechanics, is also the key reason why Lagrangian and Hamiltonian mechanics transcend Newtonian mechanics. The statement of Noether's theorem is that when a system satisfies the EoM or the principle of least action, if we apply a variation \(\delta \alpha\) to the Action S and still have \(\delta S=0\), Noether's theorem indicates that there is a corresponding conserved quantity. Here we discuss simultaneous variations in time \(t\) and the physical trajectory \(q\): $$q \to \bar{q} = q+\delta q$$ $$t \to \bar{t} =t+\delta t$$ However, it is worth noting that the trajectory \(q\), as a function of time \(t\), will be affected not only by its own variation but also by the change in time. Define \(\Delta q\) to represent the total effect: $$\Delta q\equiv \bar{q} (\bar{t} )-q(t)$$ $$=\bar{q} (\bar{t} )-q(\bar{t} )+q(\bar{t} )-q(t)$$ $$=\delta q+\dot{q} \delta t$$ Consider the difference before and after the variation: $$\delta S=\delta \int L dt=\int \delta L dt+\int L d\delta t$$ The first term can be expanded as: $$\delta L=L \left(\bar{q} (\bar{t} ),\dot{\bar{q}}(\bar{t} ),\bar{t} \right)-L\left(q(t),\dot{q} (t),t\right)$$ $$=L \left(\bar{q} (\bar{t} ),\dot{\bar{q}}(\bar{t} ),\bar{t} \right)-L\left(q(\bar{t} ),\dot{q} (\bar{t} ),\bar{t} \right)+L\left(q(\bar{t} ),\dot{q} (\bar{t} ),\bar{t} \right)-L\left(q(t),\dot{q} (t),t\right)$$ $$={\partial L\over\partial q} \delta q+{\partial L\over \partial \dot{q} } \delta \dot{q} +{dL\over dt} \delta t$$ $$={\partial L\over\partial q} \delta q-\left({d\over dt} {\partial L\over \partial \dot{q} }\right) \delta q+{d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t$$ $$=\left[{\partial L\over\partial q} -{d\over dt} {\partial L\over \partial \dot{q} }\right]\delta q+{d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t$$ Since the Lagrangian satisfies the EoM: $${\partial L\over\partial q} -{d\over dt} {\partial L\over \partial \dot{q} } =0 $$ The first term simplifies to: $$\delta L={d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t$$ The second term can be rewritten as: $$\int L d\delta t=\int L {d\delta t\over dt} dt$$ Combining the two terms: $$\delta S=\int \left[{d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t\right] dt+\int {L d\delta t\over dt} dt$$ $$=\int {d\over dt} \left[{\partial L\over \partial \dot{q} } \delta q+L\delta t\right] dt$$ However, since \(\delta q\) is just the variation of the trajectory itself, the total variation \(\Delta q\) must be considered. Using $$\delta q=\Delta q-\dot{q} \delta t$$ Substitute into the equation: $$\delta S=\int {d\over dt} \left[{\partial L\over \partial \dot{q} } \left(\Delta q-\delta t \right)+L\delta t\right] dt$$ $$=\int {d\over dt} \left[{\partial L\over \partial \dot{q} } \Delta q-\left({\partial L\over \partial \dot{q} } \dot{q} -L\right)\delta t\right] dt$$ If the variation leaves the action unchanged, i.e., \(\delta S=0\), then: $$ {d\over dt} \left[{\partial L\over \partial \dot{q} } \Delta q\right]=0 \to {\partial L\over \partial \dot{q} } \Delta q =const.$$ $$ {d\over dt} \left[ \left({\partial L\over \partial \dot{q} } \dot{q} -L\right)\delta t\right]=0 \to \left({\partial L\over \partial \dot{q} } \dot{q} -L\right)\delta t = const.$$ The conserved quantity corresponding to the invariance under trajectory \(q\) variation is: $${\partial L\over \partial \dot{q} } =p$$ This is momentum conservation.
The conserved quantity corresponding to the invariance under time \(t\) variation is: $$H={\partial L\over \partial \dot{q} } \dot{q} -L$$ This is energy conservation.

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