Lagrangian $L' = L + \frac{df}{dt}$ gives the same equations of motion

Question

It is well known that when a Lagrangian $L$ is incremented by the total time derivative of a function $f$ that does not depend on the time derivatives of the generalized coordinates, the same equations of motion are obtained. Usually the new Lagrangian is written as

$$L'(q,\dot q, t) = L(q, \dot q, t) + \frac{df}{dt}(q,t)$$

However, usually $L$ is considered as a function that depends on independent variables $q$ and $\dot q$, but that is obviously not the case with the term $\frac{df}{dt}(q,t)$. Wouldn't it be more correct (and more practical) to write

$$L'(q,\dot q, t) = L(q, \dot q, t) + \frac{\partial f}{\partial q}(q,t)\dot q + \frac{\partial f}{\partial t}(q,t)$$

so that in the action integral for a given trajectory it does have the form $\frac{df}{dt}(q(t),t)$, or is there something I misunderstood?

Answer 1

Rigorously speaking, yes, you are right if dealing with the Lagrangian function. Indeed E.-L. equations should be more properly written $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^k}\right) - \frac{\partial L}{\partial {q}^k}= 0\:, \quad \frac{d q^k}{dt} = \dot{q}^k\quad k=1,\ldots, n\:.$$ In other words $\dot{q}^k$ becomes $\frac{d q^k}{dt}$ just along the solutions of the equations but, otherwise $\dot{q}^k$ and ${q}^k$ are independent variables. This is because, geometrically speaking, $L$ is a map from the first jet bundle $j^1(S)$ where $T: S\to \mathbb R$ is the fiber bundle called spacetime of configurations, the basis $\mathbb R$ represents the axis of time whereas every fiber $T^{-1}(t)$ is the configuration space at time $t$. Natural local coordinates adapted to the fiber bundle structure are the standard coordinates $t, q^1,\ldots, q^n$. The jet bundle $J^1(S)$ adds kinematic coordinates $\dot{q}^1,\ldots, \dot{q}^n$.

In this picture the identity, in local natural coordinates, $$\frac{df(q(t),t)}{dt}=\sum_{k=1}^n\frac{\partial f}{\partial q}(q(t),t)\dot{q}^k(t) + \frac{\partial f}{\partial t}(q(t),t)$$ makes sense along the solutions of EL equations, but it does not without fixing a curve $q=q(t)$ (solution of EL equations or not) because the derivative in the left-hand side cannot be computed. Nevertheless the formalism is constructed just to encourage this intuitive and effective interpretation since, after all it is quite harmless. One may define something like $$\widetilde{\frac{df(q,t)}{dt}}=\sum_{k=1}^n\frac{\partial f}{\partial q}(q,t)\dot{q}^k + \frac{\partial f}{\partial t}(q,t)\:,$$ without fixing a section of $S$. As soon as a solution of EL is given, the notation becomes consistent with the standard one.

It is important to stress that if focusing on the action rather than the Lagrangian, in order to implement the variational principle, it is correct to always identify $\dot{q}^k$ with $\frac{dq^k}{dt}$, since the action is a functional over a space of curves and $\dot{q}^k=\frac{dq^k}{dt}$ is always assumed to be valid on each of theses curves no matter if they satisfy EL equations or not.

Answer 2

The OP has a point. If a dot denotes time-differentiation

$$\dot{q}~\equiv~ \frac{dq}{dt},$$

and if we add a total time derivative to the Lagrangian

$$\tilde{L}(q,\dot{q},t)-L(q, \dot{q}, t) ~=~ \frac{dF(q,t)}{dt}~\equiv~\frac{\partial F(q,t)}{\partial q}\dot{q} + \frac{\partial F(q,t)}{\partial t},$$

and if we want to view position $q$ and velocity $v$ as independent variables, cf. e.g. this Phys.SE post, then we should formally write

$$\tilde{L}(q,v,t)-L(q, v, t) ~=~ \frac{\partial F(q,t)}{\partial q}v + \frac{\partial F(q,t)}{\partial t}.$$