My question is in reference to Landau's Vol. 1 Classical Mechanics. On Page 6, the starting paragraph of Article no. 4, these lines are given:
If an inertial frame $К$ is moving with an infinitesimal velocity $\mathbf\epsilon$ relative to another inertial frame $K'$, then $\mathbf v' = \mathbf v+\mathbf \epsilon$. Since the equations of motion must have the same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of co-ordinates and time.
1) Doesn't this hold for same frame? Why is Landau changing the Lagrangian of frame $K$, $L$, to $L'$ with the change satisfying this condition? So, how can he assume that the action would be minimum for the same path in $K'$ as there was in $K$? In two frames the points $q_1$ and $q_2$ aren't same which are at $t_1$ and $t_2$.
2) How did he assume that this is the one and only way to change the Lagrangian without changing the path of least action? Can we prove this?
With respect to first question, I feel that there is something fundamentally amiss from my argument as the Lagrangian is dependent only on magnitude of velocity, so $q_1$ and $q_2$ won't matter. I have made an explanation myself that since the velocity is changed infinitesimally, it should essentially be the same path governed by the previous Lagrangian, the path it took with constant velocity $v$. But, still I am not convinced. The argument isn't concrete in my head. Please build upon this argument or please provide some alternative argument.
I know the question (1) and argument above are very poorly framed but I am reading Landau alone without any instructor and so have problems forming concrete ideas.