Problem using Noether's theorem in time-dependent lagrangian I have some problems calculating the conserved quantity for a lagrangian of the
form
$$
  L = \frac{1}{2}m\dot{q}^2 - f(t) a q,
$$
because I found the general problem too abstract, I tried at
first with $f(t) = t$, so, for the rest of the post I'll use
$$
  L = \frac{1}{2}m\dot{q}^2 - t a q.
$$
Using $L$ the resulting equation of motion is
$$
  m\ddot{q} = - at,
$$
so I think, this is probably the weakest point of my
reasoning, a possible transformation that left the equation
of motion invariant is
$$
  t \to T = t + \varepsilon, \quad
  q \to Q = q - a \varepsilon \frac{t^2}{2m}.
$$
Well, with this I compute the extended lagrangian*,
$L_{\hbox{ext}}$
$$
  L_{\hbox{ext}} = L(q, \frac{\bar{q}}{\bar{t}}, t) \bar{t},
$$
and using Noether's theorem, the conserved quantity is
$$
  I = 
  \frac{\partial L_{\hbox{ext}}}{\partial \bar{q}}
  \frac{\partial Q}{\partial \varepsilon}\vert_{\varepsilon = 0} + 
  \frac{\partial L_{\hbox{ext}}}{\partial \bar{t}}
  \frac{\partial T}{\partial \varepsilon}\vert_{\varepsilon
  = 0}, 
$$
computing the partials gives
$$
  I = - m \frac{\bar{q}}{\bar{t}}a \frac{t^2}{2} -
  \frac{1}{2} m \Bigl(
  \frac{\bar{q}}{\bar{t}}
  \Bigr)^2 - t a q.
$$
Unfortunately for my sanity, $\dot{I} \neq 0$. So, if you can help me to spot
any mistakes, I'll be very grateful since I have tried this procedure with other
time-dependent lagrangians and it worked. For example,
Example #1
Example #2
*NOTE if $\tau$ is the new time, then $\frac{\rm{d} t}{\rm{d} \tau} =
\bar{t}$, and $\bar{q} = \frac{\rm{d} q}{\rm{d} \tau}$, you can see
that $\dot{q} = \bar{q}/\bar{t}$.
FYI this form preserves the action.
 A: *

*$F(t)=-af(t)$ is an external force, which gives impulse $$J(t)~=~\int_{t_i}^t\! dt^{\prime} ~F(t^{\prime})\tag{A}$$ to the system, cf. my Phys.SE answer here. 

*The conserved quantity/constant of motion is the momentum minus the impulse $$Q~=~p-J.\tag{B}$$ (This can be identified with the initial momentum, cf. the above comment by user secavara.) 

*A constant of motion generates a quasisymmetry, cf. statement 3 in my Phys.SE answer here. 

*Therefore the sought-for quasisymmetry transformation is spatial translations
$$\delta q~=~\varepsilon \{q,Q\}~\stackrel{(B)}{=}~\varepsilon , \qquad 
\delta p~=~\varepsilon \{p,Q\}~\stackrel{(B)}{=}~0 , \qquad \delta t~=~0.\tag{C}$$

*The infinitesimal transformation of the Lagrangian
$$ \delta L ~\stackrel{(C)}{=}~\ldots~=~\varepsilon F~\stackrel{(A)}{=}~\varepsilon\frac{dJ}{dt} \tag{D} $$
is a total derivative.

*For spatial translations (C) the bare Noether charge is the conjugate momentum $p$. The full Noether charge is the quantity $Q$ in eq. (B).
The difference is because the transformation (C) is only a quasisymmetry (D) rather than a strict symmetry of the Lagrangian.
