For simplicity we set $m = 1$, so that the effective potential becomes: :$U_M = U + \dfrac {M^2} {2 r^2}$ {{explain|What is $m$ in this context?}} Consider the apsidial angle: :$\displaystyle \Phi = \sqrt 2 \int_{r_\min}^{r_\max} \frac {M \rd r} {r^2 \sqrt {E - U_M} }$ 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). {{explain|Apocenters or pericenters? Or either? This could be clarified.}} Recall that if $\Phi$ is commensurable with $\pi$, then an orbit is closed. {{explain|Link to a proof demonstrating this.}} === Non-Perturbative Proof === In general $U_M$ is not monotonic on $\openint {r_\min} {r_\max}$. {{explain|Link to a proof of that statement.}} 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 $\openint {r_\min} {r_\max}$. {{explain|What does $r_{1, 2}$ mean in this context?}} Note that the orbit corresponding to $r_0$ is the stable circular orbit. {{explain|Link to a proof of the above.}} Therefore this will be possible for orbits sufficiently close to it.  Now write: :$\displaystyle \Phi = \int_{U_0}^E F \left({U_M}\right) \frac {\d U_{eff} } {\sqrt {E - U_M} }$ where: :$F = \sqrt 2 M \, \map {\dfrac {\d} {\d U_M} } {\dfrac 1 {r_1} - \dfrac 1 {r_2} }$ :$U_0 \equiv \map {U_M} {r_0}$ {{explain|What does $U_{eff}$ mean?}} This is Abel's Integral Equation which can be solved for $F \left({U_M}\right)$, giving: :$\displaystyle F \left({U}\right) = \dfrac 1 \pi \int_{U_0}^{U_M} \frac {\Phi \rd E} {\sqrt{U_M - E} }$ $\Phi$ 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. {{Explain|Link to a result explaining why a rational continuous function must be constant.}} Finally, integrating gives: :$\dfrac 1 {r_1} - \dfrac 1 {r_2} = \gamma \sqrt{U_M - U_0}, \gamma = \dfrac {\sqrt 2 \Phi} {\pi M}$ {{explain|Provide a link demonstrating the solution of that integral.}} Now consider defining the {{LHS}} to be a single function $\dfrac 1 {\map r U}$, meromorphic with a simple pole at $r = 0$. {{explain|The definitions for meromorphic and simple pole as given on {{ProofWiki}} are specifically for complex functions. Either these need to be expanded to define them in the specific context of real functions, or the above definition is actually for a complex function. In any case, the domain of all the functions defined in this proof need to be clarified.}} 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: :$\Omega \left({U_0, U_0}\right) = \dfrac 1 {r_0}$ {{explain|What are $\left({U_0, U}\right)$ and $\left({U_0, U_0}\right)$ in this context?}} 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: : $U \propto r^2, -\dfrac 1 {r^2}$ {{qed}} === Proof based on Asymptotic Behaviour === The substitution $x = M / r$ gives: :$\displaystyle \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac{\d x} {\sqrt{E - W \left({x}\right)}}$ 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: :$\omega = \sqrt{\dfrac {V'' \left({x_0}\right)} m}$ {{explain|Provide a link to a proof of the above, and explain what $m$ is.}} 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: :$\displaystyle \Phi = \sqrt 2 \int_{y_\min}^1 \frac {\d y} {\sqrt {W \left({1}\right) - W \left({y}\right)} }$ 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. {{qed}} {{Namedfor|Joseph Louis François Bertrand|cat = Bertrand}}