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J.G.
J.G.
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I think there are three components to this.

The first is that the basic physical implications of a radial force law - in particular, something as binary as whether we get a closed stable orbit - don't change when we take non-classical effects into account, though these do make some small quantitative changes people wouldn't detect in the nineteenth century. ThisThus each Bertrand option works.

The second is that by definition if $u$ oscillates then $x$ also repeats, as expected of a closed stable orbit.

The third is that it just so happens the closed-stable-orbit $n$, of which there are only finitely many because we end up solving a polynomial as you found, all lead in particular to the oscillation of $u$ being harmonic, so there aren't non-Bertrand options. We could call this a coincidence, but it's hard to write down a closed-stable-solutions second-order ODE whose solutions aren't harmonic (after a suitable transformation; after all, orbital radii aren't sinusoidal), especially since the force term will likely be linearizable under most reasonable transformations.

I think there are three components to this.

The first is that the basic physical implications of a radial force law - in particular, something as binary as whether we get a closed stable orbit - don't change when we take non-classical effects into account, though these do make some small quantitative changes people wouldn't detect in the nineteenth century. This each Bertrand option works.

The second is that by definition if $u$ oscillates then $x$ also repeats, as expected of a closed stable orbit.

The third is that it just so happens the closed-stable-orbit $n$, of which there are only finitely many because we end up solving a polynomial as you found, all lead in particular to the oscillation of $u$ being harmonic, so there aren't non-Bertrand options. We could call this a coincidence, but it's hard to write down a closed-stable-solutions second-order ODE whose solutions aren't harmonic (after a suitable transformation; after all, orbital radii aren't sinusoidal), especially since the force term will likely be linearizable under most reasonable transformations.

I think there are three components to this.

The first is that the basic physical implications of a radial force law - in particular, something as binary as whether we get a closed stable orbit - don't change when we take non-classical effects into account, though these do make some small quantitative changes people wouldn't detect in the nineteenth century. Thus each Bertrand option works.

The second is that by definition if $u$ oscillates then $x$ also repeats, as expected of a closed stable orbit.

The third is that it just so happens the closed-stable-orbit $n$, of which there are only finitely many because we end up solving a polynomial as you found, all lead in particular to the oscillation of $u$ being harmonic, so there aren't non-Bertrand options. We could call this a coincidence, but it's hard to write down a closed-stable-solutions second-order ODE whose solutions aren't harmonic (after a suitable transformation; after all, orbital radii aren't sinusoidal), especially since the force term will likely be linearizable under most reasonable transformations.

Source Link
J.G.
J.G.
  • 25.4k
  • 2
  • 40
  • 70

I think there are three components to this.

The first is that the basic physical implications of a radial force law - in particular, something as binary as whether we get a closed stable orbit - don't change when we take non-classical effects into account, though these do make some small quantitative changes people wouldn't detect in the nineteenth century. This each Bertrand option works.

The second is that by definition if $u$ oscillates then $x$ also repeats, as expected of a closed stable orbit.

The third is that it just so happens the closed-stable-orbit $n$, of which there are only finitely many because we end up solving a polynomial as you found, all lead in particular to the oscillation of $u$ being harmonic, so there aren't non-Bertrand options. We could call this a coincidence, but it's hard to write down a closed-stable-solutions second-order ODE whose solutions aren't harmonic (after a suitable transformation; after all, orbital radii aren't sinusoidal), especially since the force term will likely be linearizable under most reasonable transformations.