# Why does not the central force $-1/r^4$ violate Bertrand's theorem?

An orbit $$r(\theta)$$ is said to be bounded if $$0. Otherwise it is called unbounded.

Goldstein defines closed orbits as orbits in which the particle eventually retraces its own footsteps.

Bertrand's theorem says that the only forces whose all bounded orbits are closed are the inverse-square force and Hook's law.

My concern is the following: The attractive force $$-1/r^4$$ has only one bounded orbit, namely the circular orbit. This can be easily seen by the energy diagram bellow, where it is plotted the effective potential.

But the circular orbit is certainly closed which seems to violate Bertrand theorem. How to solve this inconsistency?

A possible solution to this paradox would be to consider an alternative definition of open/closed orbits: A bounded orbit is said to be open if, given a sufficiently large time, it completely fills the ring with internal and external radius $$r_{min}$$ and $$r_{max}$$. Otherwise it is called closed. Then a circular obit is open since $$r_{min}=r_{max}$$. It sounds a little weird to say a circular orbit is open though.

Any other explanation? Or am I missing something here?

• Bounded essentially means that under a small perturbation, the solution (orbit) remains bounded for all time, clearly your potential only admits a single unstable circular orbit. Oct 30 '18 at 18:20
• @Triatticus Wouldn't that be the definition of stable orbits instead of the definition of bounded orbits? Oct 30 '18 at 18:24

An English translation of Bertrand's original work can be found here. He opens by quoting the Binet equation, but this implicitly rules out circular orbits from everything that follows.

As a result, his proof holds only in the case of non-circular orbits. If your potential does not admit non-circular bounded orbits, then Bertrand's theorem does not apply. This is equivalent to the requirement that the potential admits stable bounded orbits.

Consider an object with energy $$E$$, as seen in the graph below:

The line of constant $$E$$ intersects $$V_\text{eff}(r)$$ at two radii, $$r_1$$ and $$r_2$$. These are the boundaries of two separate regions where motion is possible: $$r \leqslant r_1$$ and $$r \geqslant r_2$$ (solid red lines).

Let's only consider motion in the region $$r \leqslant r_1$$. At radius $$r_1$$, we have $$\dot{r} = 0$$ and $$\ddot{r} \sim -V'_\text{eff}(r_1) < 0$$, so it is a turning point of maximum radius. Since there is no corresponding turning point of minimum radius, this means that once $$\dot{r} < 0$$, the object will spiral towards the center, forever.

Therefore, the orbit of this object is bound by $$r_1$$, but it is not closed.

• I agree that the orbit with energy E and $r\leq r_1$ is not closed. But can I call this a limited orbit according to my first definition? Note that it requires two finite, non zero turning points. If I change that definition then the problem is solved. Oct 31 '18 at 11:13
• @Diracology A bounded orbit is an orbit for which $r_\max$ is finite. It's perfectly fine for $r_\min = 0$, in fact all radial orbits (i.e. orbits with zero angular momentum) in attracting central force fields have that property. Oct 31 '18 at 11:58

You have written it: Bertrand's theorem is concerned with central forces for which all bounded orbits are closed, i.e. either the initial conditions are such that the mass goes to infinity, or it they are such that the trajectory is a close orbit. Whether there exists some close orbits is irrelevant to Bertrand's theorem.

• But in my example there is only one bounded orbit and it is closed by Goldstein's definition. Hence all bounded orbits associated with that force is closed. Oct 30 '18 at 18:56
• ah, ok, I see your point. But from the top of my memory, there are orbits spiralling in with $1/r^4$. So I think your statement that there is only one bounded orbit is not true. Oct 30 '18 at 19:22

Bertrand Theorem is actually applicable for bounded orbits which are also stable. The potential $$-1/r^4$$ do allow a circular orbit, but that is unstable. For any force law of the form $$f=-kr^n$$, stable orbits are only possible for $$n>-3$$.

The force law (equation 3.48 of Goldstein) $$f(r)=-\frac{k}{r^{3-\beta ^2}}$$ corresponds to stable bounded orbits only for $$\beta ^2=1$$ and $$\beta ^2=4$$. Also follow the discussion after equation 3.48 of Goldstein where this is explained in details.