# Why is $L_z$ not conserved despite rotational symmetry about the $z$-axis?

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?

• 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.
– Amit
Jan 29 at 18:32
• But I want to understand it from the "symmetry leading to conservation law" point of view. Jan 29 at 18:33
• It seems to me that the Lagrangian depends implicitly on $\phi$ in your statement because $\omega = \frac{\phi}{t}$
– Amit
Jan 29 at 18:35
• $\omega$ 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 $\phi$. Jan 29 at 18:37
• 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. Jan 30 at 0:50

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.
• @Solidification The exact same one used to derive the conservation of $E$ and $L_z$ in simpler problems. Jan 29 at 19:16