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Conservation of ang. momentum for paths reaching a rotation axis

  1. Nov 25, 2013 #1
    Hey everyone.
    My question is the following: if we had the trajectory of a particle eventually reaching a point of a rotation axis [itex] \vec{u} [/itex] ((take that as being the z-axis for convenience) by an angle [itex] s [/itex] , 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 [itex] \mathcal{L}(\vec{q_i}, \dot{\vec{q_i}}, t) [/itex] admits a one-parameter group of diffeomorphisms [itex] h^s : \mathcal{M} \rightarrow \mathcal{M} [/itex] such that [itex] h^{(s=0)} (\vec{q_i})= \vec{q_i} [/itex], then there is a conserved quantity locally given by [tex] I = \sum_{i} \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \left.\frac{d}{ds}(h^s(q_i))\right\vert_{s=0} [/tex]

    2. Applying to Simple Lagrangian

    Assume a potential-free Lagrangian [itex] \mathcal{L} = \frac{m}{2}( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 ) [/itex].
    A suitable transformation can be given by [tex] h^s (x,y,z)= \begin{pmatrix}
    cos(s) & -sin(s) & 0 \\
    sin(s) & cos(s) & 0 \\
    0 & 0 & 1
    \end{pmatrix} [/tex]

    Working out the conserved quantity, we get that the z-component of angular momentum [itex] L_z = m \dot{y}(t) x(t) - m \dot{x}(t) y(t) [/itex] is conserved for any path [itex] (x(t),y(t),z(t)) [/itex].

    3. The problem:
    If this trajectory would include any point on the rotation axis z, [itex] h^s(q_i) [/itex] 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?

    Any comments are very welcome.
    David
     
  2. jcsd
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