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Is the loss of closed Keplerian orbits in relativistic mechanics directly tied to the absence of the Runge-Lenz vector?

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3 Answers 3

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Yes, the two losses are directly related. As Goldstein explains, you can only expect a simple conserved quantity like the RL vector for degenerate cases like the Kepler problem, where $r$ is a single-valued function of $\theta$.

For the isotropic harmonic oscillator, motion is again periodic, and the corresponding conserved quantity is a tensor of the second rank, which also counts as "simple".

For other force laws (like GR), Goldstein comments that conserved quantities akin to the RL vector can be constructed, but "they are in general rather peculiar functions", since $r$ must be an infinite-valued function of $\theta$ for these non-closed orbits.

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What is the definition of "simple" in this context? What chapter/section and/or page number in Goldstein are you looking at? It's not obvious to me that it's valid to treat the motion of a test particle in the Schwarzschild spacetime as just another force law in classical mechanics, although I'd be willing to believe that it was a good enough approximation in some kind of semi-Newtonian limit. –  Ben Crowell Apr 8 '13 at 4:21
@BenCrowell: "Simple" appears to be something Professor Goldstein knows when he sees it: "Only where the orbits are closed ... can we expect the additional conserved quantity to be a simple algebraic function of r and p such as the Laplace-Runge-Lenz vector." Goldstein 2nd ed, Section 3-9, pp. 104-105 (and section 9-7 for the isotropic harmonic oscillator). –  Art Brown Apr 8 '13 at 4:26

In a classical context, LRL vector is conserved only for potentials behaving like $\frac{k}{r}$, indeed we can see the general construction of LRL vector :

\begin{eqnarray} \frac{d\vec{p}}{dt}\times \vec{L} &=& -\partial_r v(r) \frac{\vec{r}}{r} \times \vec{L}, \nonumber\\ \mu r^3 \frac{d\hat{r}}{dt} &=& -\vec{r} \times \vec{L}, \nonumber \\ \vec{A} &=& \vec{p} \times \vec{L} - \int dt \Big(\partial_r v(r)\mu r^2 \dot{\hat{r}}\Big), \nonumber \end{eqnarray} as we can see only a potential with the form $\frac{k}{r}$ give us vector $\vec{A}$ as a constant of motion...

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Some observations:

  1. For a fixed trajectory, you can change the Runge-Lenz vector simply by changing the speed of the particle as a function of time.

  2. When a planar orbit intersects itself, it looks to me from the form of the equation like you could obtain the same Runge-Lenz vector even if the motion is in a different direction at the revisited point.

  3. Motion in an $r^2$ potential is periodic, but doesn't conserve the Runge-Lenz vector.

For these reasons, I don't see any obvious connection between the conservation of the Runge-Lenz vector and the periodicity of the motion. If you have some specific reason for suspecting such a connection, please tell us.

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protected by Qmechanic May 19 at 0:15

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