# 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

- 1974: V.I. Arnold:
*Mathematical Methods of Classical Mechanics*: $\S 2.\text D$, problems $1$. to $6$.

This page may be the result of a refactoring operation.As such, the following source works, along with any process flow, will need to be reviewed. When this has been completed, the citation of that source work (if it is appropriate that it stay on this page) is to be placed above this message, into the usual chronological ordering.In particular: Each of these is to be replaced by $\mathsf{Pr} \infty \mathsf{fWiki}$ house style citation links.If you have access to any of these works, then you are invited to review this list, and make any necessary corrections.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{SourceReview}}` from the code. |

- 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