諾特定理

PDF Reference:
\( ^a \) Peir-Ru Wang(王培儒), On the Natural Equivalence Between Canonical and Hilbert Energy Momentum Tensors via Noether's Theorem, arXiv:2507.05780
\( ^b \) Peir-Ru Wang(王培儒), Yen-Cheng Chang(張晏誠), Canonical Energy-Momentum Tensor of Abelian Fields, arXiv:2503.15031

諾特定理作為古典力學重要的定理，也是Lagrangian、Hamiltonian超越牛頓力學的重要原因。諾特定理的表述為，當系統滿足EoM、或最小作用量原理之下，我們若給予Action S一個變化\(\delta \alpha\)卻可以保持\(\delta S=0\)，諾特定理表明會對應到一個守恆量。在這邊我們討論同時對時間\(t\)跟物理軌跡\(q\)做變分 $$q \to \bar{q} = q+\delta q$$ $$t \to \bar{t} =t+\delta t$$ 但值得注意的是，軌跡\(q\)作為時間\(t\)的函數，會受到自身變分的影響之外，也會受到時間改變的有影響，定義\(\Delta q\)表示完整影響 $$\Delta q\equiv \bar{q} (\bar{t} )-q(t)$$ $$=\bar{q} (\bar{t} )-q(\bar{t} )+-q(\bar{t} )-q(t)$$ $$=\delta q+\dot{q} \delta t$$ 考慮變分前後的差別 $$\delta S=\delta \int L dt=\int \delta L dt+\int L d\delta t$$ 第一項仔細寫下為 $$\delta L=L \left(\bar{q} (\bar{t} ),\dot{\bar{q}}(\bar{t} ),\bar{t} \right)-L\left(q(t),\dot{q} (t),t\right)$$ $$=L \left(\bar{q} (\bar{t} ),\dot{\bar{q}}(\bar{t} ),\bar{t} \right)-L\left(q(\bar{t} ),\dot{q} (\bar{t} ),\bar{t} \right)+L\left(q(\bar{t} ),\dot{q} (\bar{t} ),\bar{t} \right)-L\left(q(t),\dot{q} (t),t\right)$$ $$={\partial L\over\partial q} \delta q+{\partial L\over \partial \dot{q} } \delta \dot{q} +{dL\over dt} \delta t$$ $$={\partial L\over\partial q} \delta q-\left({d\over dt} {\partial L\over \partial \dot{q} }\right) \delta q+{d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t$$ $$=\left[{\partial L\over\partial q} -{d\over dt} {\partial L\over \partial \dot{q} } \right]\delta q+{d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t$$ 因為Lagrangian滿足EoM $${\partial L\over\partial q} -{d\over dt} {\partial L\over \partial \dot{q} } =0 $$ 所以第一項只剩下 $$\delta L={d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t$$ 第二項單純改寫 $$\int L d\delta t=\int L {d\delta t\over dt} dt$$ 將兩項合併 $$\delta S=\int \left[{d\over dt} \left({\partial L\over \partial \dot{q} } \delta q\right)+{dL\over dt} \delta t\right] dt+\int {L d\delta t\over dt} dt$$ $$=\int {d\over dt} \left[{\partial L\over \partial \dot{q} } \delta q+L\delta t\right] dt$$ 但因為\(\delta q\)只是軌跡自身的變分，必須考慮到完整的變化\(\Delta q\)，利用 $$\delta q=\Delta q-\dot{q} \delta t$$ 代入 $$\delta S=\int {d\over dt} \left[{\partial L\over \partial \dot{q} } \left(\Delta q-\delta t \right)+L\delta t\right] dt$$ $$=\int {d\over dt} \left[{\partial L\over \partial \dot{q} } \Delta q-\left({\partial L\over \partial \dot{q} } \dot{q} -L\right)\delta t\right] dt$$ 如果經過變分後不變，即\(\delta S=0\)，代表 $$ {d\over dt} \left[{\partial L\over \partial \dot{q} } \Delta q\right]=0 \to {\partial L\over \partial \dot{q} } \Delta q =const.$$ $$ {d\over dt} \left[ \left({\partial L\over \partial \dot{q} } \dot{q} -L\right)\delta t\right]=0 \to \left({\partial L\over \partial \dot{q} } \dot{q} -L\right)\delta t = const.$$ 對應軌跡q變分不變的守恆量為 $${\partial L\over \partial \dot{q} } =p$$ ，為動量守恆。
對應時間t變分不變的守恆量為 $$H={\partial L\over \partial \dot{q} } \dot{q} -L$$ 為能量守恆。