# Noether's theorem for time dependent non-cyclic Lagrangian

I am asked to find the symmetries and conserved quantities for a system with the following Lagrangian:

$$\mathscr{L}=\frac{1}{2}m\dot{q}^2-af(t)q,$$

where $a$ is some constant and $f(t)$ is an arbitrary (but integrable) function of time.

I find this problem non trivial because the Lagrangian has no cyclic coordinates and it is a function of time, so neither the conjugate momentum $p$ or the energy $E$ are conserved quantities.

I proceed trying to find some symmetry such that $\delta\mathscr{L}=dg/dt$ (or maybe $=0$, the idea is that this condition is such that the Euler-Lagrange equations obtained via variational principle are left invariant). Then applying Noether's theorem, the conserved quantity would be:

$$C=\bigg(\frac{\partial\mathscr{L}}{\partial\dot{q}}\dot{q}-\mathscr{L}\bigg)\delta{t}-\frac{\partial\mathscr{L}}{\partial\dot{q}}\delta{q}-g,$$

where $g$ may, or may not be zero. So, for the Lagrangian in consideration:

\begin{align}\delta\mathscr{L}&=\frac{\partial\mathscr{L}}{\partial\dot{q}}\delta\dot{q}+\frac{\partial\mathscr{L}}{\partial q}\delta q+\frac{\partial\mathscr{L}}{\partial t}\delta t\\ &=(m\dot{q})\delta\dot{q}+(-af(t))\delta q+(-a\frac{\partial f}{\partial t}q)\delta t\end{align}

The problem here is that I can't think of any symmetry that can satisfy Noether's condition. Is there any other test that can give me the correct symmetries? Or maybe I can know the conserved quantities by looking at the form of the Lagrangian but I lack the intuition?

The conserved quantity that Frotaur got does not come from a symmetry.The way to get it is the following.We consider a transformation of the coordinates: $$q \rightarrow q + \epsilon$$ The tranformed Langrangian is : $$L'(q+\epsilon,\dot{q})=\dfrac{1}{2}m\dot{q}-af(t)q - af(t)\epsilon=L(q)-af(t)\epsilon$$ Taking the derivative with respect to $\epsilon$ at $\epsilon=0$ we have : $$\dfrac{dL'}{d\epsilon}\Bigr|_{\epsilon=0}=-af(t)$$ With a little work you can show from Taylors theorem and Euler-Lagrange equations that in the general family of transformations $$q\rightarrow q + \epsilon K(q,\dot{q})$$ you get: $$\dfrac{dL'}{d\epsilon}\Bigr|_{\epsilon=0}=\dfrac{d}{dt}\left( \dfrac{\partial L}{\partial \dot{q}}K(q,\dot{q})\right)$$ In our case $K(q,\dot{q})=1$ so we get the final result that Frotaur got: $$\dfrac{d}{dt}\left( \dfrac{\partial L}{\partial \dot{q}}\right)=-af(t)$$ $$m\dot{q}=-a\int f(t) +C$$

If you try naively without searching for symmetries, and instead writing the Euler-Lagrange equation you find:

$m\ddot{q}=-af(t)$

Integrating, you get a conserved quantity :

$m\dot{q}+a\int f(t)=C$

Not sure what symmetry it corresponds to yet, but I think it might be possible to reverse engineer it.

• Thank you. From the Hamiltonian formalism I can obtain the corresponding symmetries using the bracket formulation, $(\delta q=\epsilon,\delta p=0)$. But I don't see how to recover the corresponding conserved quantity when inserting the symmetries in $\delta\mathscr{L}$. – Saavestro Dec 23 '16 at 5:34