Zero Lagrangian

Question

If $L(q,dot{q},t)$ is a lagrangian of a system, then $L' = L + frac{dF(q,t)}{dt}$ is also a valid lagrangian and both lagrangians will lead to the same equation of motion.
But, what if I choose $F(q,t)$ such that $L'=0$?
begin{equation}
L + frac{dF}{dt} = 0 
F = - int{L dt} = -S 
end{equation}
where $S$ is the action.
Is this valid? If yes, can we not do this always? What does it mean?

J. Murray · Accepted Answer

If you can choose $F(q,t)$ such that $L'=0$, then $L$ can already be written in the form
$$Lbigg(q(t),dot q(t),tbigg) = frac{d}{dt}Gbigg(q(t),t)bigg)$$
for some $G$.  As a result, the action $$S[q] = int_{t_1}^{t_2} Lbigg( q(t),dot q(t), tbigg)dt = Gbigg(q(t_2),t_2bigg)-Gbigg(q(t_1),t_1bigg)$$
is independent of the path $q$ (since the endpoints $q(t_2)$ and $q(t_1)$ are fixed by the boundary conditions).  Since the action is the same for every path, there's no way to select a special one by demanding that $S$ be stationary with respect to small variations, so the action is not useful.

ACuriousMind · Answer

The action $S$ is a functional of paths $q(t)$, not a function of the variables $(q,dot{q},t)$, so the equality $F = -S$ makes no sense.

By claiming that you can choose $L = -frac{mathrm{d}F}{mathrm{d}t}$, you already have restricted the original Lagrangian to be a total time derivative (of $-F$). This is not the case for Lagrangians that usefully describe physical systems, since it would mean that the action
$$S[q(t)] = int_{t_i}^{t_f}L(q(t),dot{q}(t),t)mathrm{d}t =  - F(q(t_f),t_f) + F(q(t_i),t_i)$$
is only dependent on the start and end-points $(q(t_i),t_i)$ and $(q(t_f),t_f)$, so paths do not differ in their action and the principle of extremal action is useless.

Zero Lagrangian

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