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Consider a bead on parabolic wire satisfying $z=\alpha \rho^2$ and rotating about the $z$-axis with a uniform angular velocity $\omega$ so that $\phi=\omega t$. The Lagrangian is given by $$L=\frac{1}{2}m[(1+4\alpha^2\rho^2)\dot{\rho}^2+\rho^2\omega^2]-mg\alpha\rho^2.$$ See here. In this problem, a rotation about the $z$-axis does not change $\rho=\sqrt{x^2+y^2}$ and $\dot{\rho}$. So there is a symmetry of the Lagrangian about the z-axis. But despite that why is the $z$-component of the angular momentum, $L_z=m(x\dot{y}-y\dot{x})=m\rho^2\omega$ not conserved? Why? What do I misunderstand here conceptually?

Could it be that since the coordinate $\phi$ disappeared from $L$, it does not make sense to talk about rotational symmetry under a rotation about the $z$-axis through angle $\psi$ i.e., $\phi\to\phi+\psi$? Is there rotational symmetry in this problem or it does not? I am not sure. If not why?

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  • $\begingroup$ At the most basic level the angular momentum as a vector is $\vec{L} = \vec{r} \times \vec{p}$ , now it is clear that both $\vec{r}$ and $\vec{p}$ of the bead have components in the $xy$ plane here, which vary with time according to the details of the problem. Hence by the property of the cross product, there will be a varying component of $\vec{L}$ in the $z$ direction. $\endgroup$
    – Amit
    Jan 29 at 18:32
  • $\begingroup$ But I want to understand it from the "symmetry leading to conservation law" point of view. $\endgroup$
    – Solidification
    Jan 29 at 18:33
  • $\begingroup$ It seems to me that the Lagrangian depends implicitly on $\phi$ in your statement because $\omega = \frac{\phi}{t}$ $\endgroup$
    – Amit
    Jan 29 at 18:35
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    $\begingroup$ $\omega$ is a constant (say $\omega=10s^{-1}$). It is a rate at which the wire is being moved by an external agent and $\phi=\omega t$ is thus a time-dependent constraint that allows us to eliminate $\phi$. $\endgroup$
    – Solidification
    Jan 29 at 18:37
  • $\begingroup$ There are forces of constraint that keep the particle on the parabola $z=\alpha \rho^2$. You can investigate those via Lagrange multipliers to understand the situation better. $\endgroup$
    – ZeroTheHero
    Jan 30 at 0:50

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Once you've eliminated $\phi$, you have a Lagrangian in the single variable $\rho$, so you can't even talk about $\phi$ translations anymore and you shouldn't expect a conserved quantity.

On the other hand, if you work in terms of Cartesian coordinates, then you can talk about a translation in $\phi$, but this isn't a symmetry because the wire picks out a specific angle at each moment. However, since the wire's rotation speed is uniform, there is a symmetry under the simultaneous rotation and time translation $\delta \phi = \omega \alpha$, $\delta t = \alpha$ which by Noether's theorem implies that $E + \omega L_z$ is conserved.

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  • $\begingroup$ I have a few questions. What form of Noether's theorem are you using which led you to this conserved quantity? $\endgroup$
    – Solidification
    Jan 29 at 19:16
  • $\begingroup$ @Solidification The exact same one used to derive the conservation of $E$ and $L_z$ in simpler problems. $\endgroup$
    – knzhou
    Jan 29 at 19:16

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