My question is the following: if we had the trajectory of a particle eventually reaching a point of a rotation axis $ \vec{u} $ (take that as being the z-axis for convenience) by an angle $ s $, would Noethers Theorem still give a conserved quantity?
More specifically (let me go through the calculations and details first)
- Statement of Noether's Theorem
If a Lagrangian $ \mathcal{L}(\vec{q_i}, \dot{\vec{q_i}}, t) $ admits a one-parameter group of diffeomorphisms $ h^s : \mathcal{M} \rightarrow \mathcal{M} $ such that $ h^{(s=0)} (\vec{q_i})= \vec{q_i} $, then there is a conserved quantity locally given by $$I = \sum_{i} \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \left.\frac{d}{ds}(h^s(q_i))\right\vert_{s=0}$$
- Applying to Simple Lagrangian
Assume a potential-free Lagrangian $ \mathcal{L} = \frac{m}{2}( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 ) $. A suitable transformation can be given by $$ h^s (x,y,z)= \begin{pmatrix} \cos(s) & -\sin(s) & 0 \\ \sin(s) & \cos(s) & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
Working out the conserved quantity, we get that the z-component of angular momentum $ L_z = m \dot{y}(t) x(t) - m \dot{x}(t) y(t) $ is conserved for any path $ (x(t),y(t),z(t)) $.
- The problem: If this trajectory would include any point on the rotation axis z, $ h^s(q_i) $ would be 0 there and so by conservation, valued 0 all along the path. However, we know that angular momentum is conserved. So, in all rigor - is this inconsistency amendable or a sign of some bigger problem?