Wednesday, April 22, 2015

Lenz-Runge-Laplace vector

the Lenz-Runge-Laplace vector is stationary for the Newton-potential $\Phi\propto 1/r$. is that potential the only one with that property?

2 comments:

  1. the LRL-vector is defined as $\vec{A}=\vec{p}\times\vec{L}-m\alpha\vec{e}_r$, whose terms have the time evolution $\mathrm{d}_t(\vec{p}\times\vec{L}) = mr^2\partial_r V\vec{\omega}\times\vec{e}_r$ and $\mathrm{d}_t m\alpha\vec{e}_r = m\alpha\vec{\omega}\times\vec{e}_r$. the derivative of $\vec{A}$ only vanishes if $\partial_r V=\alpha/r$, therefore the Newtonian gravitational potential is the only one with a constant LRL-vector.

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  2. CQW would like to add that the conservation of the LRL-vector generates the degeneracy of of atomic states with the quantum number $\ell$, while the degeneracy with $m$ is due to spherical symmetry.

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