5
$\begingroup$

I'm having trouble wrapping my mind around the Laplace-Runge-Lenz vector.

Conservation of momentum can be visualized as an object moving in a straight line with constant speed. One can even visualize the conservation of energy thinking that if it wasn't conserved in some system it would eventually dissipate and come to rest, etc. But how can I visualize the conversation of the Laplace-Runge-Lenz vector?

Or, put another way, what physical consequences would there be if the Laplace-Runge-Lenz vector was not conserved?

$\endgroup$
5
+50
$\begingroup$

The first hint is in the well-known fact that the Laplace-Runge-Lenz (LRL) vector is always along the perihelion, that is to say that it is the direction of the major axis of the ellipse. The second hint is that with a potential

$$V(r) \propto \frac{1}{r^{1+\epsilon}},$$

however small is $\epsilon > 0$, the non-circular orbits are not closed: the perihelion precesses, which can be shown to correspond to a slow rotation of the LRL vector. Thus it can be said that the conservation of the LRL is intimately linked with the fact that the orbit closes.

However the Keplerian potential is not the only one with closed orbits: the 3D harmonic potential

$$V(r) = k r,$$

also has that property (those two potentials are the only ones with that property that all orbits are closed: this is Bertrand's theorem). However the extra conserved quantity in this case is not the LRL vector but an unrelated tensor, specifically,

$$T_{ij} = \frac{p_ip_j}{2m}+\frac{kr_ir_j}{2}.$$

This symmetric tensor brings two orthogonal axes of symmetry to the orbits instead of just one in the Keplerian case. Thus my statement above can be refined by stating that the LRL vector is associated with closed orbits with only one axis of symmetry.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice answer. But... I could not follow the last paragraph $\hat{A} = \hat{p}\times\hat{L} - mk\hat{r}$ how does that form a tensor? $\endgroup$ – Abhijeet Melkani Oct 28 '17 at 18:10
  • $\begingroup$ That tensor is not related to the LRL tensor. This is the point: being a symmetric tensor it naturally has two axes of symmetry, thus making the harmonic case completely separate from the Keplerian one. I made a few edits to clarify. $\endgroup$ – user154997 Oct 28 '17 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.