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.


Preliminary Lemma

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

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


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Consider the apsidial angle:

$\ds \Phi = \sqrt 2 \int_{r_\min}^{r_\max} \frac {M \rd r} {r^2 \sqrt {E - U_M} }$

where:

$E$ is the energy
$r_\min, r_\max$ are solutions to $\map {U_M} r = E$.

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


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


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$\Box$


Non-Perturbative Proof

In general $U_M$ is not monotonic on $\openint {r_\min} {r_\max}$.


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Therefore a unique inverse $\map r {U_M}$ does not exist.

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


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


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

Now write:

$\ds \Phi = \int_{U_0}^E \map F {U_M} \frac {\d U_{eff} } {\sqrt {E - U_M} }$

where:

$F = \sqrt 2 M \, \map {\dfrac {\d} {\d U_M} } {\dfrac 1 {r_1} - \dfrac 1 {r_2} }$
$U_0 \equiv \map {U_M} {r_0}$


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

$\ds \map F U = \dfrac 1 \pi \int_{U_0}^{U_M} \frac {\Phi \rd E} {\sqrt {U_M - E} }$

$\Phi$ may have energy dependence.

However, we require:

$\map \Phi E = 2 \pi \map q E$

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


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Finally, integrating gives:

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


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


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Write:

$\dfrac 1 r = \gamma \sqrt {U_M - U_0} + \map \Omega {U_0, U}$

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

$\map \Omega {U_0, U_0} = \dfrac 1 {r_0}$


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Explicitly:

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

It can be seen that the right hand side 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:

$\ds \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac {\d x} {\sqrt {E - \map W x}}$

where $\map W x \equiv \map U {\dfrac M x} + \dfrac 1 2 x^2$.

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:

$\omega = \sqrt {\dfrac {\map {V} {x_0} } m}$


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which means that:

$\Phi \approx 2 \pi \dfrac M {x_0^2} \sqrt {\map {W} {x_0} } = 2 \pi \sqrt {\dfrac {U'} {3 U' + x_0 U} }$

Suppose $\Phi$ were constant.

Then the solutions are:

$\map U r = k r^\alpha, \alpha \ge - 2, \alpha \ne 0$

and:

$\map U r = b \log r$

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 $\map U r \to \infty$ as $r \to \infty$.

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

$\ds \Phi = \sqrt 2 \int_{y_\min}^1 \frac {\d y} {\sqrt {\map W 1 - \map W y} }$

where:

$\map W y \equiv \dfrac {y^2} 2 + \dfrac 1 { {x_\max}^2} \map U {\dfrac M {y x_\max} }$

Therefore, when $E \to \infty$:

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

This means that:

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

Equating this with the previous result gives:

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

Therefore:

$\alpha=2$

Now let $\map U r = -k r^{-\beta}, 0 < \beta < 2$.

A similar approach gives:

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

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


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  • A non-perturbative proof of Bertrand's theorem - Santos, F C et al - arXiv:0704.0575
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