最小作用量原理（相對論性）

本節為筆者推論，還不確定是否正確
前面提到，當Lagrangian添加一個函數\(f=f(q,t)\)時間的全微分項\({df\over dt}\)，並不會改變EoM。在場論中也有對應的推廣，\(\mathcal{L}\to \bar{\mathcal{L}} =\mathcal{L}+\partial _\mu f^\mu\) ，一樣要注意的是\(f^\mu =f^\mu (\phi ,x^\mu )\)只能是\((\phi ,x^\mu )\)的函數。在此我們討論看看怎麼樣的\(f=f(q,\dot{q} ,t)\)和\(f^\mu =f^\mu (\phi ,\partial _\mu \phi ,x^\mu )\)有可能不改變EoM。原則上我們要求 $$ {\partial \dot{f} \over\partial q}-{d\over dt} {\partial \dot{f} \over \partial \dot{q} } =0$$ $${\partial (\partial _\alpha f^\alpha )\over\partial \phi} -\partial _\mu \left({\partial (\partial _\alpha f^\alpha )\over\partial (\partial _\mu \phi )}\right )=0)$$ Extended dot cancellation for \(f=f(q,\dot{q} ,t)\) type $${\partial \dot{f} \over \partial \dot{q} } ={d\over dt} {\partial f\over\partial \dot{q} }+{\partial f\over\partial q}$$ $$ {\partial \dot{f} \over\partial q}-{d\over dt} {\partial \dot{f} \over \partial \dot{q} } =0$$ $$\to {d\over dt} {\partial f\over\partial q}-{d\over dt} \left({d\over dt} {\partial f\over\partial \dot{q} }+{\partial f\over\partial q}\right)=0$$ $$\to {d^2\over dt^2} {\partial f\over\partial \dot{q} }=0$$ $$\to f=(at+b) \dot{q} +g(q,t)$$


$${\partial (\partial _\alpha f^\alpha ) \over \partial (\partial _\mu \phi )} =\partial _\alpha {\partial f^\alpha\over\partial (\partial _\mu \phi )} +{\partial f^\mu \over\partial \phi }$$ $${\partial (\partial _\alpha f^\alpha )\over\partial \phi} -\partial _\mu \left({\partial (\partial _\alpha f^\alpha )\over\partial (\partial _\mu \phi )}\right )=0$$ $$\partial _\alpha {\partial f^\alpha\over\partial \phi} -\partial _\mu \left(\partial _\alpha {\partial f^\alpha\over\partial (\partial _\mu \phi )} +{\partial f^\mu \over\partial \phi }\right)=0$$ $$\partial _\mu \partial _\alpha {\partial f^\alpha\over\partial (\partial _\mu \phi )} =0$$ $$\partial _\mu \partial _\alpha {\partial f^\alpha\over\partial (\partial _\mu A^\gamma )} =0$$