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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.

Qmechanic
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David
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    Hint: Notice that the initial velocity, $\dot{q}{(0)} , = , \dot{q}{(t)} , + , \frac{a}{m} , \int^t_0 , f_{(t')} , \mathrm{d}t' $, or any function of it, is time independent. – secavara Dec 24 '18 at 19:37
  • If the lagrangian is time dependent then there is exchange of energy with an external degree of freedom. The Noether theorem will tell you that the systems energy is not conserved as a term $\partial L /\partial t$ will appear on the right hand side of the conservation law. – my2cts Dec 24 '18 at 22:51

1 Answers1

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  1. $F(t)=-af(t)$ 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.

  2. 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.)

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

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

  5. 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.

  6. 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.

Qmechanic
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  • Thanks @qmechanic. I see and corroborated what you kindly wrote. But I still don't see why my transformation, that leaves the equation of motion invariant, doesn't work. Using Noether's theorem with your transformation, results in $m\dot{q}$. In that case, is valid to conclude that for other lagrangians I should add $-J$to the resulting $I$ calculated using Noether's theorem? – David Dec 25 '18 at 19:52