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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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1 Answers1

3

  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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