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8. (Bonus) Laplace-Runge-Lenz vector via Noether's theorem: The Lagrangian for the reduced one-body problem for the central potential $V(r) = -\frac{\kappa}{r}$ is given by: $\mathcal{L} = \frac{1}{2}m\dot{r}^2 + \frac{\kappa}{r}$ Consider the following variation is coordinates: $\delta r = a \times (\dot{r} \times r) + \dot{r} \times (r \times a)$, $= 2(a \cdot r)\dot{r} - (a \cdot \dot{r})r - (r \cdot \dot{r})a$. where $a$ is arbitrary vector of infinitesimal magnitude. Show that the vari- ation in $\mathcal{L}$ becomes: $\delta \mathcal{L} = \frac{dF}{dt} = \frac{d}{dt} \left\{ m \left[ (a \cdot r)\dot{r}^2 - (a \cdot \dot{r})(r \cdot \dot{r}) \right] + (a \cdot r)\frac{\kappa}{r} \right\}$ Show that the associated constant of motion is $G = m^{-1}a \cdot [p \times L - \kappa m e_r]$. Since $a$ is arbitrary, the Laplace-Runge-Lenz vector is the quantity that is conserved. It turns out that the conservation of $L$ and $A$ in the Kepler problem in addition to the energy is due to the problem being invariant under $SO(4)$ rotations (rotations in 4 dimensions). The rotations are in phase space and hence the components of $L$ and the components of $A$ are the generators of $SO(4)$ rotations in phase space (which is six-dimensional for the reduced problem).

          8. (Bonus) Laplace-Runge-Lenz vector via Noether's theorem: The Lagrangian
for the reduced one-body problem for the central potential $V(r) = -\frac{\kappa}{r}$ is
given by:
$\mathcal{L} = \frac{1}{2}m\dot{r}^2 + \frac{\kappa}{r}$
Consider the following variation is coordinates:
$\delta r = a \times (\dot{r} \times r) + \dot{r} \times (r \times a)$,
$= 2(a \cdot r)\dot{r} - (a \cdot \dot{r})r - (r \cdot \dot{r})a$.
where $a$ is arbitrary vector of infinitesimal magnitude. Show that the vari-
ation in $\mathcal{L}$ becomes:
$\delta \mathcal{L} = \frac{dF}{dt} = \frac{d}{dt} \left\{ m \left[ (a \cdot r)\dot{r}^2 - (a \cdot \dot{r})(r \cdot \dot{r}) \right] + (a \cdot r)\frac{\kappa}{r} \right\}$ 
Show that the associated constant of motion is
$G = m^{-1}a \cdot [p \times L - \kappa m e_r]$.
Since $a$ is arbitrary, the Laplace-Runge-Lenz vector is the quantity that is
conserved. It turns out that the conservation of $L$ and $A$ in the Kepler
problem in addition to the energy is due to the problem being invariant
under $SO(4)$ rotations (rotations in 4 dimensions). The rotations are in
phase space and hence the components of $L$ and the components of $A$ are
the generators of $SO(4)$ rotations in phase space (which is six-dimensional
for the reduced problem).

$8. (Bonus) Laplace-Runge-Lenz vector via Noether's theorem: The Lagrangian for the reduced one-body problem for the central potential V(r) = -(κ)/(r) is given by: ℒ = (1)/(2)mṙ^2 + (κ)/(r) Consider the following variation is coordinates: δ r = a × (ṙ× r) + ṙ× (r × a), = 2(a · r)ṙ - (a ·ṙ)r - (r ·ṙ)a. where a is arbitrary vector of infinitesimal magnitude. Show that the vari- ation in ℒ becomes: δℒ = (dF)/(dt) = (d)/(dt){ m [ (a · r)ṙ^2 - (a ·ṙ)(r ·ṙ) ] + (a · r)(κ)/(r)} Show that the associated constant of motion is G = m^-1a · [p × L - κ m er]. Since a is arbitrary, the Laplace-Runge-Lenz vector is the quantity that is conserved. It turns out that the conservation of L and A in the Kepler problem in addition to the energy is due to the problem being invariant under SO(4) rotations (rotations in 4 dimensions). The rotations are in phase space and hence the components of L and the components of A are the generators of SO(4) rotations in phase space (which is six-dimensional for the reduced problem).$

Added by Stacey A.

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