# Connection between Kepler Problem and Harmonic Oscillator

Background. Take the Kepler Lagrangian as

$L^K = \frac{1}{2}\dot{q}_i\dot{q}_i + \frac{k}{q}$,

and the Lagrangian for the isotropic harmonic oscillator as

$L^H = \frac{1}{2}\dot{q}_i\dot{q}_i - \frac{1}{2}kq_i q_i$.

Consider coordinate transformations of the form $q_i \rightarrow q_i + \lambda_{(jk)} g_{i(jk)}$. For the Kepler problem, use

$g^K_{ij} = q_i\dot{q}_j - (1+\beta)\dot{q}_i q_j + \beta \delta_{ij}\mathbf{q}.\mathbf{\dot{q}}$

(i.e. $\lambda_j$ is a vector). For the harmonic oscillator use

$g^H_{ijk} = \delta_{ij}\dot{q}_k + \beta\delta_{ik}\dot{q}_j$

(i.e. $\lambda_{jk}$ is a second rank tensor). Then we find from Noether's theorem the conserved charges

$Q^K = -(1+\beta)\left(\mathbf{q}.\mathbf{\dot{q}}\dot{q}_j - \dot{q}^2 q_j + k q_j/q \right)$

and

$Q^H = (1+\beta)(\dot{q}_j\dot{q}_k + k q_j q_k)$,

which are the Laplace-Runge-Lenz vector and the second rank energy tensor respectively.

Now the following interesting observation can be made: In the degenerate case, where $\beta = -1$ and both Noether charges vanish, there exist simple relationships between the transformation tensors:

$q_k g^H_{ijk} = -g^K_{ij},\quad [g^K_{ij},\dot{q}_k] = g^H_{ijk}$.

Question. Intuitively, I would say that these relationships between coordinate transformations belonging to two different Lagrangians are due to a common symmetry (i.e. isotropy, the angular momentum tensor is conserved for both Lagrangians), but the precise origin of this connection is not clear to me.