Physical equivalence of Lagrangian under addition of dF/dt

genericusrnme

You can do it without going all that dirty work by looking at the action

[itex]S=\int L dt = \int (L' + \frac{dF}{dt})dt [/itex]

That method seems much neater to me

Mod note: removed remaining steps.

genericusrnme	You can do it without going all that dirty work by looking at the action [itex]S=\int L dt = \int (L' + \frac{dF}{dt})dt [/itex] That method seems much neater to me Mod note: removed remaining steps.