5
$\begingroup$

Is there a way to see that Bertrand's theorem is true intuitively. I mean without getting into too much mathematics ?

$\endgroup$

1 Answer 1

3
$\begingroup$

Not really. This is the type of question that a high school student can ask, but it takes differential equations to solve. Never the less, you can see pieces of it from simple explanations.

Bertrand's Theorem states that only two types of central force potentials produce stable, closed orbits.

If we use radial coordinates, an orbit is completed when $\theta$ goes from $0$ to $2\pi$. $r$ can vary during the orbit. So can the velocity of the orbiting particle. Velocity has two components, $\dot{r}$ and $\dot{\theta}$.

For the orbit to be closed, the particle must return to the same position with the same speed and direction. This means that $r$, $\dot{r}$ and $\dot{\theta}$ start of with some values. They all return to the same values at the same time, just when $\theta$ reaches $2\pi$.

Problems with this type of constraint often have multiple solutions, where each solution is discrete.

To see this, consider a simpler problem of this type. You are required to run around a circular track that is 60 m long, arriving at the start just when the minute hand of your watch returns to its original position. You can do this by running at $1 m/s$ or $2 m/s$ or $3 m/s$. But not at $1.5 m/s$ or $1.001 m/s$.

Note that all the solutions are completely different. At $1 m/s$, when $\theta = 2\pi$, the second hand has returned to the start. At 2 m/s, when $\theta = 2\pi$, the second hand is nowhere near the start. The second hand reaches the start when $\theta = 4\pi$.


For Bertrand's theorem, we want to find what kind of potential, or equivalently what kind of force, generates closed orbits. We start with circular orbits. Under any central force law, these will be constant velocity closed orbits.

Next consider small perturbations of a circular orbit. For some force laws, the particle is sucked into the center or flies away to infinity. We won't consider those because they are not stable.

For the rest, we can simulate the force law by laying a rain gutter along a horizontal circular track, and rolling a marble around it. The rain gutter has a smoothly curved cross section. The marble may roll up and down the sides a little as it orbits, but it never gets very far from the circle. Given a force law, you could calculate what the cross section of the rain gutter would have to be to reproduce the same orbit as the force law.

For large perturbations, the shape of the rain gutter would affect how quickly to marble rolls back and forth. But for small perturbations, all rain gutters, and all central force laws, are very much alike. At the very bottom of the rain gutter, the cross section will closely match some circle. The only thing that controls how quickly the marble rolls back and forth is the radius of the circle.

Suppose the marble to rolls back and forth $m$ times as it completes $n$ orbits. For the orbit to be closed, $m$ and $n$ must both be integers. Another way of saying this is that $m/n$ must be a rational number.

It is possible to relate $m/n$ to the radius of the rain gutter cross section, and to relate that to the force law. In this way, a great many force laws can be shown to produce non-closed almost-circular orbits. The only force laws that produce closed almost-circular orbits are like this

$\displaystyle F = -\frac{k}{r^{3-(m/n)^2}}$

where k is some positive number.

For gravity, $m = n = 1$, and $k = GM_1M_2$

$\displaystyle F = -\frac{GM_1M_2}{r^2}$


By considering the shape of the gutter, it is possible to eliminate more force laws. The end result is that only two types of force law produce closed stable orbits.

$\endgroup$
4
  • 1
    $\begingroup$ This is wrong! Bertrand's theorem is that the only central force potentials for which every bounded orbit is closed are $kr^2$ and $-k/r$, $k>0$. $\endgroup$
    – auxsvr
    Commented Apr 27, 2014 at 7:33
  • $\begingroup$ @auxsvr - Sorry to be misleading. This is incomplete. The Wikipedia article in the question derives Bertrand's Theorem with a lot of math. I was trying to make some steps in the article intuitive. The answer was getting long. I didn't do all the steps. If you follow the article, you will find more constraints on $m/n$ (which they call $\beta$). The end result is that $\beta^2$ must be 0, 1, or 4. This leads to the the result you mentioned. $\endgroup$
    – mmesser314
    Commented Apr 27, 2014 at 13:14
  • $\begingroup$ The statement "Bertrand's Theorem states that only two types of central force potentials produce stable, closed orbits" is wrong. This is not Bertrand's theorem. $\endgroup$
    – auxsvr
    Commented Apr 27, 2014 at 13:30
  • $\begingroup$ @auxsvr - I just quoted the top line of the Wikipedia article. I think the difference may be that small perturbations from a circular orbit are closed with a $1/r^3$ force law, but larger perturbations are not. Bertrand's Theorem identifies force laws for which all orbits are closed. If this is wrong, please let me know. I will be happy to correct a mistake. $\endgroup$
    – mmesser314
    Commented Apr 27, 2014 at 13:33

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.