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Homework Help: Again with the Kepler problem

  1. Feb 3, 2007 #1

    quasar987

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    There must be something I'm totally missing here.

    The situation is the following.

    I am asked to show that given the lagrangian for the Kepler problem,

    [tex]L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}[/tex]

    the k-th component of the Runge-Lenz vector,

    [tex]A_k=\mathbf{\dot{q}}^2q_k-\mathbf{q}\cdot\mathbf{\dot{q}}
    \dot{q}_k-q_k/q[/tex]

    is the conserved quantity associated (in the sense of Noether's thm) with the infinitesimal coordinate transformation [itex]\mathbf{q}\rightarrow\mathbf{q}+\delta \mathbf{q}[/itex], where [itex]\delta q_i = \epsilon(\dot{q}_iq_k-\frac{1}{2}q_i\dot{q}_k-\frac{1}{2}\mathbf{q}\cdot \mathbf{\dot{q}}\delta_{ik})[/itex], epsilon being the infinitesimal parameter.

    Following Noether's theorem, I know that if [tex]\delta L=L(\mathbf{q}+\delta \mathbf{q}, \mathbf{\dot{q}}+\delta \mathbf{\dot{q}},t)-L(\mathbf{q},\mathbf{\dot{q}},t)[/itex] can be written as

    [tex]\delta L=\epsilon \frac{d}{dt}\Lambda(\mathbf{q},\mathbf{\dot{q}},t)+\mathcal{O}(\epsilon^2)[/tex]

    then the quantity

    [tex]F_k:=\sum_{i=1}^3\frac{\partial L}{\partial \dot{q}_i}(\dot{q}_iq_k-\frac{1}{2}q_i\dot{q}_k-\frac{1}{2}\mathbf{q}\cdot \mathbf{\dot{q}}\delta_{ik}) - \Lambda[/tex]

    is conserved. By direct comparison of F_k with A_k I find that Lambda must be

    [tex]\Lambda = \frac{q_k}{q}[/tex]

    (also, this is confirmed by the wiki article on the Runge-Lenz vector: http://en.wikipedia.org/wiki/Runge-Lenz#Noether.27s_theorem )

    So what remains to be done is to show by direct calculation that indeed,

    [tex]\delta L=\epsilon \frac{d}{dt}(\frac{q_k}{q})+\mathcal{O}(\epsilon^2)[/tex]

    So I expand [itex]\delta L[/itex]:

    [tex]\delta L= \frac{1}{2}(\mathbf{\dot{q}}^2+2\mathbf{\dot{q}}\cdot \delta\mathbf{\dot{q}}+(\delta\mathbf{\dot{q}})^2)+(\mathbf{q}^2+2\mathbf{q}\cdot \delta\mathbf{q}+(\delta\mathbf{q})^2)^{-\frac{1}{2}}-\frac{1}{2}\mathbf{\dot{q}}^2-(\mathbf{q}^2)^{-\frac{1}{2}}[/tex]

    And here I find it impossible to put this in a form [itex]\delta L=\epsilon A+\mathcal{O}(\epsilon^2)[/itex] because of all these guys in the numerator and shielded by a square root. I have also tried "cheating", i.e. say "since epsilon is arbitrarily small, I can neglect this and this term" but nothing even comes close to the form I want.

    So I concluded that there must be something fundamentally flawed about the reasoning laid above. Anyone sees?

    Thanks for reading!


    P.S. I would appreciate feedbacks, so that if I get a few feedbacks that the above is right, I will post more of my work so we can find where I'm going wrong.
     
    Last edited: Feb 4, 2007
  2. jcsd
  3. Feb 5, 2007 #2

    quasar987

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    Solved, thanks. Turns out I had the right answer but in an hostile form that made it difficult to recognize.
     
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