Conservation of ang. momentum for paths reaching a rotation axis

1. Nov 25, 2013

quantum52

Hey everyone.
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)

1. 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 $$I = \sum_{i} \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \left.\frac{d}{ds}(h^s(q_i))\right\vert_{s=0}$$

2. 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 $(x(t),y(t),z(t))$.

3. 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 rigour - is this inconsistency amendable or a sign of some bigger problem?