Bertrand's Theorem

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Theorem

Let $U: \R_{>0} \to \R$ be analytic for $r > 0$.

Let $M > 0$ be a nonvanishing angular momentum such that a stable circular orbit exists.

Suppose that every orbit sufficiently close to the circular orbit is closed.


Then $U$ is either $k r^2$ or $-\dfrac k r$ (for $k > 0$) up to an additive constant.


Proof

For simplicity we set $m = 1$, so that the effective potential becomes:

$U_M = U + \dfrac {M^2} {2 r^2}$


Consider the apsidial angle:

$\displaystyle \Phi = \sqrt 2 \int_{r_\min}^{r_\max} \frac {M \ \mathrm d r} {r^2 \sqrt {E - U_M} }$

where:

$E$ is the energy
$r_\min, r_\max$ are solutions to $U_M \left({r}\right) = E$.

By definition, this is the angle between adjacent apocenters (pericenters).



Recall that if $\Phi$ is commensurable with $\pi$, then an orbit is closed.


Non-Perturbative Proof

In general $U_M$ is not monotonic on $\left({r_\min \,.\,.\, r_\max}\right)$.


Therefore a unique inverse $r \left({U_M}\right)$ does not exist.

However, suppose it is possible to construct separate inverse functions $r_{1, 2}$ for the intervals $\left({r_0 \,.\,.\, r_\min}\right)$ and $\left({r_0 \,.\,.\, r_\max}\right)$, where $r_0$ is the minimum of $U_M$ on $\left({r_\min \,.\,.\, r_\max}\right)$.


Note that the orbit corresponding to $r_0$ is the stable circular orbit.



Therefore this will be possible for orbits sufficiently close to it.

Now write:

$\displaystyle \Phi = \int_{U_0}^E F \left({U_M}\right) \frac {\mathrm d U_{eff} } {\sqrt {E - U_M} }$

where:

$F = \sqrt 2 M \dfrac {\mathrm d} {\mathrm d U_M} \left({\dfrac 1 {r_1} - \dfrac 1 {r_2} }\right)$
$U_0 \equiv U_M (r_0)$



This is Abel's Integral Equation which can be solved for $F \left({U_M}\right)$, giving:

$\displaystyle F \left({U}\right) = \dfrac 1 \pi \int_{U_0}^{U_M} \frac {\Phi \ \mathrm d E} {\sqrt{U_M - E} }$

$\Phi$ may have energy dependence.

However, we require:

$\Phi \left({E}\right) = 2 \pi q \left({E}\right)$

such that $q: \R \to \Q$ is continuous, and a rational continuous function must be constant.



Finally, integrating gives:

$\dfrac 1 {r_1} - \dfrac 1 {r_2} = \gamma \sqrt{U_M - U_0}, \gamma = \dfrac {\sqrt 2 \Phi} {\pi M}$



Now consider defining the left hand side to be a single function $\dfrac 1 {r \left({U}\right)}$, meromorphic with a simple pole at $r = 0$.



Write:

$\dfrac 1 r = \gamma \sqrt {U_M - U_0} + \Omega \left({U_0, U}\right)$

where $\Omega$ is an analytic function in $U$ in an open neighborhood of $U_0$ such that:

$\Omega \left({U_0, U_0}\right) = \dfrac 1 {r_0}$


Explicitly:

$\sqrt {U_M - U_0} = \sqrt {U + \dfrac {M^2} {2 r^2} - U_0} = \dfrac 1 r \sqrt{\dfrac {M^2} 2 + \left({U - U_0}\right) r^2}$

It can be seen that the RHS also has a simple pole at $r=0$.

However, it also has a branching point at $r = r_0$.

To avoid this, the radicand must be the square of an analytic function with a zero at $r_0$.

The only possibilities are:

$U \propto r^2, -\dfrac 1 {r^2}$

$\blacksquare$


Proof based on Asymptotic Behaviour

The substitution $x = M / r$ gives:

$\displaystyle \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac{\mathrm d x} {\sqrt{E - W \left({x}\right)}}$

where $W \left({x}\right) \equiv U \left({\dfrac M x}\right) + \dfrac 1 2 x^2$.

In general, the frequency of oscillations around a stable equilibrium at $x = x_0$ for a particle of mass $m$ in a potential $V$ is given by:

$\omega = \sqrt{\dfrac {V'' \left({x_0}\right)} m}$


which means that:

$\Phi \approx 2 \pi \dfrac M {x_0^2} \sqrt {W'' \left({x_0}\right)} = 2 \pi \sqrt{\dfrac {U'} {3 U' + x_0 U''}}$

Suppose $\Phi$ were constant.

Then the solutions are:

$U \left({r}\right) = k r^\alpha, \alpha \ge - 2, \alpha \ne 0$

and:

$U \left({r}\right) = b \log r$

Therefore:

$\Phi = \dfrac {2 \pi} {\sqrt{\alpha + 2}}$

with $\alpha = 0$ corresponding to the logarithmic solution.

This is discarded, since in that case $\Phi$ is not commensurable with $\pi$.


Now consider $\alpha > 0$, so that $U \left({r}\right) \to \infty$ as $r \to \infty$.

The substitution $x = y x_\max$ reduces $\Phi$ to:

$\displaystyle \Phi = \sqrt 2 \int_{y_\min}^1 \frac {\mathrm d y} {\sqrt {W \left({1}\right) - W \left({y}\right)} }$

where:

$W \left({y}\right) \equiv \dfrac {y^2} 2 + \dfrac 1 { {x_\max}^2} U \left({\dfrac M {y x_\max}}\right)$

Therefore, when $E \to \infty$:

$x_\max \to \infty$ and $y_\min \to 0$

This means that:

$\displaystyle \lim_{E \mathop \to \infty} \Phi = \pi$

Equating this with the previous result gives:

$\dfrac {2 \pi} {\sqrt {\alpha + 2}} = \pi$

Therefore:

$\alpha=2$

Now let $U \left({r}\right) = -k r^{-\beta}, 0 < \beta < 2$.

A similar approach gives:

$\displaystyle \lim_{E \mathop \to -0} \Phi = \frac {2 \pi} {2 - \beta}$

so that:

$\dfrac {2 \pi} {2 + \alpha} = \dfrac {2 \pi} {\sqrt {2 + \alpha} }$

It follows that:

$\alpha = -1$

and the result follows.

$\blacksquare$


Source of Name

This entry was named for Joseph Louis François Bertrand.


Sources



  • A non-perturbative proof of Bertrand's theorem - Santos, F C et al - arXiv:0704.0575
  • Yoel Ticochinsky, Am. J. Phys 56 (12) December 1988