Action Principle (Relativistic)

This section is a conjecture by the author and is not yet confirmed to be correct
As mentioned earlier, adding a total time derivative term \(f=f(q,t)\) to the Lagrangian does not change the EoM. There is a corresponding extension in field theory, where \(\mathcal{L}\to \bar{\mathcal{L}} =\mathcal{L}+\partial _\mu f^\mu\). It is important to note that \(f^\mu =f^\mu (\phi ,x^\mu )\) can only be a function of \((\phi ,x^\mu )\). Here, we discuss the conditions under which \(f=f(q,\dot{q} ,t)\) and \(f^\mu =f^\mu (\phi ,\partial _\mu \phi ,x^\mu )\) might not change the EoM. In principle, we require: $$ {\partial \dot{f} \over\partial q}-{d\over dt} {\partial \dot{f} \over \partial \dot{q} } =0$$ $${\partial (\partial _\alpha f^\alpha )\over\partial \phi} -\partial _\mu \left({\partial (\partial _\alpha f^\alpha )\over\partial (\partial _\mu \phi )}\right )=0)$$ Extended dot cancellation for \(f=f(q,\dot{q} ,t)\) type: $${\partial \dot{f} \over \partial \dot{q} } ={d\over dt} {\partial f\over\partial \dot{q} }+{\partial f\over\partial q}$$ $$ {\partial \dot{f} \over\partial q}-{d\over dt} {\partial \dot{f} \over \partial \dot{q} } =0$$ $$\to {d\over dt} {\partial f\over\partial q}-{d\over dt} \left({d\over dt} {\partial f\over\partial \dot{q} }+{\partial f\over\partial q}\right)=0$$ $$\to {d^2\over dt^2} {\partial f\over\partial \dot{q} }=0$$ $$\to f=(at+b) \dot{q} +g(q,t)$$


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

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