Gauge Symmetry of the Lagrangian

Question

My teacher told the following statement to me during office hours. Is it correct and if so, how could one go about proving it?

Given a material system subject to holonomic and smooth constraints whose Lagrangian is $L$ (we are assuming that it can be derived from a generalized potential rather than just a potential), then for a given function $g = g(\underline{q}, \underline{\dot q}, t)$: \begin{equation*} \frac{d}{dt} \left( \frac{\partial g}{\partial \dot q_k} \right) - \frac{\partial g}{\partial q_k} = 0 \hspace{5mm} \iff \hspace{5mm} \exists f=f(\underline{q}, t): g = \frac{d}{dt} f \tag{1}\end{equation*}

By straightforward computation one can prove the second implication ($\impliedby$) but I am having trouble with the other one. So far, by using the fact that $$ \frac{\partial}{\partial \dot q_k} \left( \frac{d}{dt}f \right) = \frac{d}{dt} \left( \frac{\partial}{\partial \dot q_k} f \right) + \frac{\partial}{\partial q_k}f $$ and integrating with respect to time I have arrived to: $$\frac{\partial}{\partial \dot q_k}f = Ct + D$$ where $C,D$ are constants of integration. However it does not seem right and it doesn't help me in showing that a primitive of $g$ with respect to time can not depend on lagrangian velocities.

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Any comment or answer is much appreciated!

Answer 1

Your teacher undoubtedly implicitly meant to say that eq. (1) for $g$ should hold for all (possible virtual) paths $t\mapsto q(t)$. Then the trivial implication $\Longleftarrow$ is still true, while the non-trivial implication now $\Longrightarrow$ holds (as long as there are no topological obstructions) due to an algebraic Poincare lemma of the so-called bi-variational complex, see Ref. 1. For an elementary proof, see Ref. 2.

See also this related Phys.SE post.

References:

1. G. Barnich, F. Brandt and M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.

2. J.V. Jose and E.J. Saletan, Classical Dynamics: A Contemporary Approach, 1998; Section 2.2.2, p. 67.

Answer 2

Well, the non-trivial implication is in fact false.

Take $L = \dot{q}^2$ and $g(t,q, \dot{q}) := \dot{q}^3$ so that the E-L equations are $\frac{d\dot{q}}{dt}=0$.

We have

(1) $\frac{d\dot{q}}{dt}=0$ form the E-L equations.

(2) $\frac{d}{dt}\frac{\partial g}{\partial \dot{q}} - \frac{\partial g}{\partial q}= \frac{d\dot{q}}{dt} \frac{\partial^2g}{\partial \dot{q}^2} - 0=0$.

So there should exist $f=f(t,q)$ with $\frac{d}{dt}f(t,q)= \dot{q}^3$. In other words $$\frac{\partial f}{\partial t}+ \frac{\partial f}{\partial q} \dot{q} = \dot{q}^3\:.$$ Since this must be valid for every $\dot{q}$ we conclude that $\frac{\partial f}{\partial t}=0$ and thus $$\frac{\partial f}{\partial q} \dot{q} = \dot{q}^3\:.$$ But $\frac{\partial f}{\partial q}$ does not depend on $\dot q$ and thus, taking two derivatives in $\dot{q}$ on both sides we have $$0= \dot{q}\:.$$ This is even false along to motions as the E-L equations permit $\dot{q}= constant \neq 0$.

Presumably, in your statement it is assumed that $g$ is linear (non-homogeneous possibly) in $\dot{q}^k$. In that case both implicatios are locally true.