Is there a high level reason why the inverse square law of gravitation yields periodic orbits without precession?

Question

Given a spherically symmetric potential $V: {\bf R}^d \to {\bf R}$, smooth away from the origin, one can consider the Newtonian equations of motion $$ \frac{d^2}{dt^2} x = - (\nabla V)(x)$$ for a particle $x: {\bf R} \to {\bf R}^d$ in this potential well. Under the spherical symmetry assumption, one has conservation of angular momentum (as per Noether's theorem, or Kepler's second law), and using this one can perform a "symplectic reduction" and reduce the dynamics to an autonomous second order ODE of the radial variable $r = |x|$ as a function of an angular variable $\theta$; see for instance https://en.wikipedia.org/wiki/Kepler_problem#Solution_of_the_Kepler_problem. In general (assuming an attractive potential and energy not too large), the energy surfaces of this ODE are closed curves, and this leads to the radial variable $r$ depending in a periodic fashion on the angular variable $\theta$, provided that one lifts the angular variable from the unit circle ${\bf R}/2\pi {\bf Z}$ to the universal cover ${\bf R}$.

In the special case of the inverse square law $V(x) = -\frac{GM}{|x|}$, it turns out that the period of the map $\theta \mapsto r$ is always equal to $2\pi$, which means in this case that the orbits are closed curves in ${\bf R}^d$ (whereas for almost all other potentials, with the exception of the quadratic potentials $V(x) = c |x|^2$, the orbits exhibit precession). Indeed, as was famously worked out by Newton (by a slightly different method), the calculations eventually recover Kepler's first law that the orbits under the inverse square law are ellipses with one focus at the origin.

The calculations are not too difficult - basically by applying the transformation $u = 1/r$ one can convert the aforementioned ODE into a shifted version of the harmonic oscillator - but they seem rather "miraculous" to me. My (rather vague) question is whether there is a "high level" (e.g., symplectic geometry) explanation of this phenomenon of the inverse square law giving periodic orbits without precession. For instance, in the case of quadratic potentials, the phase space has the structure of a toric variety ${\bf C}^d$ (with the obvious action of $U(1)^d$), and the Hamiltonian $\frac{1}{2} |\dot x|^2 + \frac{c}{2} |x|^2$ is just one linear component of the moment map, so the periodicity of the orbits in this case can be viewed as a special case of the behaviour of general toric varieties. But I wasn't able to see a similar symplectic geometry explanation in the inverse square case, as I couldn't find an obvious symplectic torus action here (in large part because the period of the orbits varies with the orbit, as per Kepler's third law). Is the lack of precession just a "coincidence", or is there something more going on here? For instance, is there a canonical transformation that transforms the dynamics into a normal form that transparently reveals the periodicity (similar to how action-angle variables reveal the dynamics on toric varieties)?

Answer 1

The gravitational or Coulomb potential has a "hidden" symmetry (hidden in the sense that it does not follow from the rotational symmetry). The resulting integral of the motion (the Runge-Lenz vector) prevents space-filling orbits in classical mechanics (all orbits are closed), and introduces a degeneracy of the energy levels in quantum mechanics (energy levels do not depend on the azimuthal quantum number).

The hidden symmetry raises the rotational symmetry group from three to four dimensions, so from SO(3) to SO(4). A geometric interpretation in four-dimensional momentum space of the SO(4) symmetry is given on page 234 of Lie Groups, Physics, and Geometry by Robert Gilmore. Historically, this interpretation goes back to V. Fock in Zur Theorie des Wasserstoffatoms [Z. Phys. 98, 145-154 (1935)]. The elliptic motion of the coordinate corresponds to a circular motion of the momentum. The circle in $\mathbb{R}^3$ is promoted to a circle in $\mathbb{R}^4$ by a projective transformation. SO(4) transformations in $\mathbb{R}^4$ rotate circles into circles, which then project down to circular momentum trajectories in the physical space $\mathbb{R}^3$.

You probably know that planets go around the sun in elliptical orbits. But do you know why? In fact, they’re moving in circles in 4 dimensions. But when these circles are projected down to 3-dimensional space, they become ellipses!
_{[John Baez, animation by Greg Egan. ]}

Answer 2

There are:

• Bertrand’s theorem, which says that the isotropic oscillator and Kepler potentials are the only analytic radial ones all of whose nonrectilinear bounded orbits are closed. (Recommendation: Albouy - Lectures on the two-body problem. But he says: “The proof of this theorem provides very little “explanation” of the phenomenon.”)

• The Levi-Civita–Kustaanheimo–Stiefel transformation, which maps the Kepler problem into a higher-dimensional isotropic oscillator subject to a constraint. (Recommendation: Cordani - The Kepler problem.)

Edit: Ultimately, I think everyone is right to say that the key to closed orbits for $\smash{\dfrac{d\mathbf v}{dt}=-\dfrac{k\mathbf r}{r^3}}$ (where $r=\|\mathbf r\|$) is conservation of $\mathbf L=\mathbf r\times\mathbf v$ and the (“Lenz”) eccentricity vector $$ \mathbf e = \frac{\mathbf L\times\mathbf v}k+\frac{\mathbf r}r. \tag1 $$ Indeed it follows that $\mathbf r$ satisfies the system: $\langle\mathbf L,\mathbf r\rangle=0$ and $\langle\mathbf e,\mathbf r\rangle=r-L^2/k$ which is, as Albouy stresses, Gauss’s preferred form of the equation of a (branch of) conic in the plane $\smash{\mathbf L^\perp}$, with focus at the origin, eccentricity $e=\|\mathbf e\|$, and semi-latus rectum $L^2/k$. Hence, if bounded, closed (an ellipse).

Once told about $\mathbf e$, checking $d\mathbf e/dt=0$ is easy. But to have the idea that it should exist is another matter, and (since you asked in comments) I won’t resist reproducing the wonderful way Lagrange did it in 1779. First he computes two derivatives of $r$ to obtain^a (as was well-known) $$ \frac{dr}{dt}=\frac{\langle\mathbf r,\mathbf v\rangle}r, \qquad\qquad\frac{d^2r}{dt^2}=\frac{L^2}{r^3}-\frac k{r^2}. \tag2 $$ Then he “very subtly” sets $s=r-L^2/k$, regards $r$ as a known function of $t$, and observes that as a result of $(2)$, $s$ then satisfies the same second order linear equation $$ \frac{d^2s}{dt^2}=-\frac{ks}{r^3} \tag3 $$ that $x,y,z$ are also known to satisfy. Therefore it is a linear combination of them, i.e. there is a fixed vector $\mathbf e$ such that $s=\langle\mathbf e,\mathbf r\rangle$ always. From that one quickly^b gets $(1)$, which Lagrange wrote in 1781.

Remarks: 1. The standard story of $\mathbf e$ omits Lagrange.^c

2. Eccentricity vector is terminology of Hamilton (1846, p. 349), revived by e.g. Temple (1931, p. 114). (The letter $\textbf e$ can also stand for Ermanno, which is the name Jacob Hermann used in 1710.)

3. The second equation $(2)$ is in Leibniz (1689, p. 91); e.g. Guicciardini (1999, p. 152) calls it “the first published differential equation applied to planetary motions.”

4. About Bertrand’s theorem, Wintner (1941, p. 420) makes a cryptic remark maybe relevant to you: “the topological nature of the problem (cf. §215), or, equivalently, the connection of the problem with existence of an additional integral in the large (cf. §218 bis), is usually not realized”.

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^{a. $\dfrac{dr}{dt}=\dfrac1r\dfrac d{dt}\dfrac{\langle\mathbf r,\mathbf r\rangle}2=\dfrac{\langle\mathbf r,\mathbf v\rangle}r$ and then (using this twice and the equation of motion once) $$ \frac{d^2r}{dt^2} =\frac d{dt}\frac{\langle\mathbf r,\mathbf v\rangle}r =\frac{r\dfrac d{dt}\langle\mathbf r,\mathbf v\rangle-\langle\mathbf r,\mathbf v\rangle\dfrac{dr}{dt}}{r^2} =\frac{r\left\{\|\mathbf v\|^2+\left\langle\mathbf r,\dfrac{d\mathbf v}{dt}\right\rangle\right\}-\dfrac{\langle\mathbf r,\mathbf v\rangle^2}{r}}{r^2} =\frac{\|\mathbf r\|^2\|\mathbf v\|^2-\langle\mathbf r,\mathbf v\rangle^2}{r^3}-\frac k{r^2}. $$ b. Indeed, by definition $ s=r-\dfrac{\langle\mathbf L,\mathbf L\rangle}k =\left\langle\dfrac{\mathbf r}r,\mathbf r\right\rangle - \dfrac{\langle\mathbf L,\mathbf r\times\mathbf v\rangle}{k} =\left\langle\dfrac{\mathbf r}r+\dfrac{\mathbf L\times\mathbf v}k,\mathbf r\right\rangle $, so $(1)$ does the trick.}

^{c. In summary (corrections welcome):
Year (re)discovered: |Denoted: |First cited (post-Lenz):   1710 Hermann (p. 465) (c/b,0) 1976 by Volk (p. 368)   1710 Bernoulli (p. 523) (c,h) 1976 by Volk (p. 372)   1779 Lagrange (p. 83) (f,g) 2002 by Albouy (p. 99)    1781 Lagrange (p. 206) (N,M,L) 1998 by Gutzwiller (p. 613)  1799 Laplace (p. 160) (λ,γ) 1941 by Wintner (p. 422)   1799 Laplace (p. 163) (f,f',f'') 1959 by Stumpff (p. 94)   1842 Jacobi (p. 21) (β,γ) 1941 by Wintner (p. 422)   1845 Hamilton (eq. 13) ε 1948 by Milne (p. 237)   1901 Gibbs (p. 135) eI 1971 by McIntosh (§II)   1919 Runge (p. 70) 𝖆 1924 by Lenz (p. 198)   1924 Lenz (p. 198) 𝕬 1926 by Pauli (p. 345)}  

Answer 3

Here is an interpretation using symmetry reduction, but without explicitly using the Lenz-Runge vector (it's essentially an extended version of the example given in Cushman & Bates "Global aspects of classical integrable systems", p. 75).

Let $Q = \mathbb{R}^3$ be the configuration space (ignoring regularization issues at the origin) and let $P = T^* Q = \mathbb{R}^6$ be the phase space. As we assume that the potential only depends on the radius, the Hamiltonian is $O(3)$ invariant. The corresponding conserved quantity is the angular momentum $J(q, p) = q \times p$, where we identified $\mathbb{R}^3$ with the Lie algebra of $O(3)$. Let $S^2_L \subseteq \mathbb{R}^3$ be the sphere with radius $L$. As alluded above, the Hamiltonian is $O(3)$-invariant and thus descends to the orbit reduced space $$J^{-1}(S^2_L) / O(3).$$ To identify this reduced space, it is convenient to use another symmetry. The group $SL(2, \mathbb{R})$ acts on $P$ and has momentum map $$K(q, p) = (x, y, z) = \Big(-q \cdot p, \frac{q^2}{2} - \frac{p^2}{2}, \frac{q^2}{2} + \frac{p^2}{2}\Big) \in \mathbb{R}^3.$$ As the $O(3)$- and $SL(2, \mathbb{R})$-action form a dual pair, the coadjoint orbit correspondence implies that the reduced phase spaces for the $O(3)$-action correspond to coadjoint actions of $SL(2, \mathbb{R})$ (in the present case, this can be seen quite easily using the formulas for $J$ and $K$). The only coadjoint orbits in the image of $K$ are the elliptic orbit, the parabolic one and zero. The last two are not interesting for us (they correspond to vanishing angular momentum). The $SL(2, \mathbb{R})$-orbit through the point $(0, 0, L)$ is the upper hyperbola $z^2 - y^2 - x^2 = L^2$ and is identified with the reduced space $J^{-1}(S^2_L) / O(3)$. The Hamiltonian descends to a Hamiltonian on the reduced phase space, which is given by $$H(x, y, z) = \frac{z-y}{2} + V(\sqrt{z+y}).$$ Your question is then equivalent to asking for a potential $V$ such that the intersection of the level set $H^{-1}(E)$ with the upper hyperbola is a closed curve. I don't have a definite answer to this last question but I guess the only possible potentials are the Harmonic oscillator and the Kepler potential.

This approach also suggests that all such potentials have to have a symmetry group $G$ that is an $U(1)$-extension of $O(3)$. The above reduction correspond then to a reduction by stages where you first quotient out the $O(3)$-symmetry and are left with a $U(1)$-symmetry on the reduced phase space.

Answer 4

The action-angle variables of the two-body graviational problem ('Kepler problem') are widely used in celestial mechanics community. These are called 'Delaunay variables' and make the toric structure of the phase space evident. See for example: Chang and Marsden - Geometric derivation of the Delaunay variables and geometric phases (CiteSeer published MSN) for a modern treatment. Note that this paper provides a canonical transformation.

Answer 5

There are some amazing aspects of hidden symmetry - Bertrand’s theorem connection. The first surprise is that it seems Runge-Lenz vector has relativistic (!) origin: https://www.sciencedirect.com/science/article/abs/pii/037596016891339X (Physical interpretation of the Runge-Lenz vector, by J.P.Dahl). The second surprise is that Runge-Lenz-like vectors do exist and may be defined for all systems with rotational symmetry including, in particular, centrally symmetric ones with open orbits: https://arxiv.org/abs/1005.1817 (Laplace-Runge-Lenz symmetry in general rotationally symmetric systems, by Uri Ben-Ya'acov).

Maybe the Eisenhart lift (see, for example, https://arxiv.org/abs/1503.07802 ) can shed some additional light to Bertrand’s theorem: https://arxiv.org/abs/1701.05783 (Superintegrable systems on 3-dimensional curved spaces: Eisenhart formalism and separability, by J.F. Cariñena, F.J. Herranz and M.F. Rañada).