# Zero Lagrangian

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?

• Oct 5, 2020 at 19:13

If you can choose $$F(q,t)$$ such that $$L'=0$$, then $$L$$ can already be written in the form $$L\bigg(q(t),\dot q(t),t\bigg) = \frac{d}{dt}G\bigg(q(t),t)\bigg)$$ for some $$G$$. As a result, the action $$S[q] = \int_{t_1}^{t_2} L\bigg( q(t),\dot q(t), t\bigg)dt = G\bigg(q(t_2),t_2\bigg)-G\bigg(q(t_1),t_1\bigg)$$ 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.
1. 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.
2. 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.