In the book of Classical Mechanics by Goldstein, at page 88, it is given that:

$$ \frac{d^{2} u}{d t^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right) . $$ The preceding equation is such that the resulting orbit is symmetric about two adjacent turning points. To prove this statement, note that i= the orbit is symmetrical.

enter image description here

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?

  • $\begingroup$ If the orbit has has multiple turning points, then the orbit is symmetric about any two adjacent $\endgroup$
    – caverac
    Dec 2, 2018 at 8:20
  • $\begingroup$ @caverac the the author states that the equation is always symmetric about two axes; there is no extra assumption $\endgroup$
    – Our
    Dec 2, 2018 at 8:30
  • 3
    $\begingroup$ Goldstein just forgets to inform the reader that in this portion of the text he is considering bounded orbits only. $\endgroup$
    – Qmechanic
    Dec 11, 2018 at 2:17

1 Answer 1


The author isn't using the elliptic orbit of the 1/r potential (which would have changed the statement of symmetry to be "around the semiminor and semimajor axes the paths are symmetric"), because he is still talking about a generalized central force, hence turning points are not necessarily axes. Fig 3.12 shows two examples combined: once for the periapsis (which would look more like a semimajor axis, if it were corresponding to a 1/r potential) and again on the adjacent turning point (would be semiminor axis), so that $u$ is the same in going to the two turning points on either side of each of the $\theta=0$ points. The adjacent turning points are on either side of one turning point, an unbounded path is still symmetric as Goldstein later analyzes scattering and states between eq. 3.93 and 3.94: "[T]he orbit must be symmetric about the direction of the periapsis," even though he is talking about scattering the symmetry is the same. The turning points in fig 3.12 can be considered each as $\theta=0$, by coordinate freedom, and so each path has a symmetry under $\theta=-\theta$ swap.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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