# Bertrand's Theorem/Non-Perturbative Proof

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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

### 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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$\Box$

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:

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

$\blacksquare$

## Sources

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