Regularization: What is so special about the Coulomb/Newtonian and harmonic potential?

Question

I wanted to know if the procedure for regularization of the Coulomb potential outlined in Celletti (2003): Basics of regularization theory could be generalized to arbitrary polynomial potentials. So instead of starting from an equation of motion with a force $F \sim x^{-2}$ I started with a generalized force $F \sim x^{n}$. So the equation of motion and the corresponding energy equation are: $$ \begin{align} \ddot{x} + K x^n = 0 && \frac{1}{2}\dot{x}^2 + \frac{K}{n+1}x^{n+1} = \text{const.} := - E \end{align} $$ Now following Celletti I looked at the generalized transformation $$ \begin{align} x = u^{-n} && \frac{dt}{ds} = x = u^{-n} \end{align} $$ Inserting these transformations into the equation of motion and the energy equation yields with $\frac{du}{ds}=u'$ $$ \begin{align} u'' - u \left[ \left(\frac{u'}{u}\right)^2 + \frac{K}{n}u^{n(n+1)}\right] = 0 && \frac{1}{2}\left(\frac{u'}{u}\right)^2 + \frac{K}{n^2(n+1)}u^{n(n+1)} = -E \end{align} $$ Now the whole exercise of regularization is to transform the equation of motion to the simpler form of an harmonic oscillator $u'' + 2Eu = 0$ (Bartsch(2003): The Kustaanheimo-Stiefel transformation in geometric algebra). But we see from the two equations above, that the terms in the square bracket are only equal to the twice the energy, if $$ n = \frac{n^2(n+1)}{2} $$ or after solving the corresponding quadratic equation $n^2+n-2=0$ $$ \begin{align} n_1 = 1 \ &\text{and}\ n_2 = -2 \end{align} $$ So it seems that one can only regularize Coulomb forces $F \sim x^{-2}$ and harmonic forces $F \sim x$. This result intrigued me a bit, because it singles out the same forces/potentials as Bertrand's theorem. But what does the the property that all bound orbits are also closed orbits (Bertrand's theorem) have to do with regularization? What is so special about the Coulomb and the harmonic potential that they get singled out in both cases? Do these potentials have some common deeper symmetry, so that regularization and Bertrand's theorem can only be fulfilled by them? Or is this just a spurious relationship between Bertrand's theorem and regularization theory?

Answer 1

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.

Answer 2

The Bertrand's theory

for a central force $~f(r)=\mu\,r^n~$, you obtain this equation

$$\ddot r-r\,\dot\varphi^2-\mu\,r^n=0$$

with $~\dot\varphi=\frac{h}{r^2}~$ (conservation of the angular momentum) $$\ddot r-\frac{h^2}{r^3}-\mu\,r^n=0\tag 1$$ and $ ~r=\frac{1}{u(\varphi)}~\quad\Rightarrow$

$$\dot r=- \left( {\frac {d}{d\varphi }}u \left( \varphi \right) \right) h\\ \ddot r=- \left( {\frac {d^{2}}{d{\varphi }^{2}}}u \left( \varphi \right) \right) {h}^{2} \left( u \left( \varphi \right) \right) ^{2} $$

thus equation (1)

$$ {\it u''}+u+\frac{\mu}{h^2}\,u^{-(n+2)} =0\tag 2$$

so only for $~n=-2~,f(r)=\frac{\mu}{r^2}$

$$u''+u+\frac{\mu}{h^2}=0$$