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Runge-Lenz Vector

  1. Oct 28, 2011 #1
    I'm trying to prove that the time derivative of the Runge-Lenz vector is constant. Any ideas on how I would go about doing this?
  2. jcsd
  3. Oct 29, 2011 #2
    Any help on this?
  4. Oct 30, 2011 #3


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    Simply take the time derivative! The definition of the Lenz vector is

    [tex]\vec{A}=\vec{p} \times \vec{L}-m \alpha \frac{\vec{r}}{r}.[/tex]

    To show that this is conserved for the potential [itex]V(r)=-\alpha/r[/itex], we note that the angular momentum is conserved, and we have

    [tex]\vec{L}=m \vec{r} \times \dot{\vec{r}}=m r^2 \vec{\omega}=\text{const},[/tex]

    where [itex]\vec{\omega}[/itex] is the momentary angular velocity.

    Further we have

    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \frac{\vec{r}}{r}=\frac{\dot{\vec{r}}}{r}-\frac{\dot{r} \vec{r}}{r^2} = \vec{\omega} \times \frac{\vec{r}}{r}[/tex]


    [tex]\dot{\vec{p}}=-\vec{\nabla} V(r)=-\frac{\alpha}{r^3} \vec{r}[/tex]

    and thus

    [tex]\dot{\vec{A}}=\dot{\vec{p}} \times \vec{L}-m \alpha \vec{\omega} \times \frac{\vec{r}}{r}=-\frac{\alpha}{r^3} \vec{r} \times m r^2 \vec{\omega}-m \alpha \vec{\omega} \times \frac{\vec{r}}{r}=0.[/tex]
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