Periodic Orbits: The Role of the Coulomb Potential in Producing Periodic Motion

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The discussion centers on the motion of a test particle in a central field, specifically questioning whether the Coulomb potential (1/r) is the only potential that produces periodic motion. It is noted that while many potentials can yield periodic orbits, not all will result in closed periodic orbits. The concept of an "effective potential" is introduced to simplify the analysis, with periodicity being defined by the time interval between apocenter and pericenter, even if the orbit does not cover a full 2π radians. Reference is made to Landau-Lifchitz, which identifies only two potentials—1/r and r²—that lead to closed trajectories, prompting further inquiry into the underlying reasons for this phenomenon. The discussion highlights a desire for a deeper understanding beyond the mathematical derivation of these properties.
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Let's consider the motion of a test particle in a central field.
Is the Coulomb potential, 1/r, the only one that produces a periodic motion?
If no, is there a condition for periodicity to occur?

Thanks,

Michel
 
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Most potentials will have periodic oribts, but many of them won't be closed periodic orbits. You can use the idea of an "effective potential" to make the problem one-dimensional, the period of the orbit would then be the time interval between apocenteron or pericentron. But if the angle covered wasn't 2*pi radians, the orbit will be periodic, but not closed.

There's a list in Goldstein "Classical Mechanics" of the force-laws that give closed orbits, IIRC.
 
Thanks a lot for your useful comment.

You gave me the idea to check in Landau-Lifchitz. (I don't have Goldstein unfortunately).
He states that there are only two potentials that result in closed trajectories: 1/r and r² . That's already good to know. However, I don't see where this magic comes from. The algebra is simple and clear, but it does not indicate some more "fundamental" reason.

Michel

PS: In other words, can a property of an integral be understood in another way?
 
I've never had any deeper insight than the mathematical demonstration you've already seen, sorry.
 
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