What is the proof for Bertrand's theorem in celestial mechanics?

[cristianbahena]

Homework Statement

I don't know how I can find the equation (2.78) and the next equations. It is from Classical Dynamics: A Contemporary Approach: Jorge V.

Relevant Equations

integral equation of energy (central problem)

 

[etotheipi]

Usually you will start from $\mathcal{L} = \frac{1}{2}\mu v^2 - V$ and then write\begin{align*}
\frac{1}{2}\mu v^2 + V &= E \\

\frac{1}{2} \mu (\dot{r}^2 + r^2\dot{\varphi}^2) + V(r) &= E
\end{align*}Also $\varphi$ doesn't appear in the Lagrangian so $\partial \mathcal{L} / \partial \dot{\varphi} = \mu r^2 \dot{\varphi} := L$ is conserved, so you can replace $\tilde{V}(r) := \frac{1}{2} \mu r^2 \dot{\varphi}^2 + V(r) = \dfrac{L^2}{2\mu r^2} + V(r)$. Further, write $\dot{r} = \dfrac{dr}{d\varphi} \dot{\varphi} = \dfrac{dr}{d\varphi} \dfrac{L}{\mu r^2}$ so that\begin{align*}
\frac{L^2}{2\mu^2 r^4} \left( \dfrac{dr}{d\varphi} \right)^2 + \tilde{V}(r) = E
\end{align*}which is simple to invert to the integral\begin{align*}

\varphi(r) = \sqrt{\frac{L^2}{2\mu}} \int^r \frac{dr}{r^2\sqrt{E - \tilde{V}(r)}}

\end{align*}The classic substitution is now $u=1/r$, which transforms this into\begin{align*}

\tilde{\varphi}(u) = -\sqrt{\frac{L^2}{2\mu}} \int^{u} \frac{du}{\sqrt{E - \tilde{W}(u)}}

\end{align*}Of course we can also take the integral between the minimum and maximum $u$ of the orbit, and using the scaling $u = yu_{\mathrm{max}}$ given in the image the integral becomes\begin{align*}

\Theta := \varphi(r_{\mathrm{max}}) - \varphi(r_{\mathrm{min}}) = \sqrt{\frac{L^2}{2\mu}} \int_{v_{\mathrm{min}}}^1 \frac{u_{\mathrm{max}} dy}{\sqrt{E - \tilde{U}(y)}}

\end{align*}You should notice that since $\tilde{V}(r) = \dfrac{L^2}{2\mu r^2} + V(r)$, that $\tilde{W}(u) = \dfrac{L^2 u^2}{2\mu } + V\left(\dfrac{1}{u} \right)$. If we let $V(r) = \alpha r^k$ then $\tilde{W}(u) = \dfrac{L^2 u^2}{2\mu } + \alpha u^{-k}$. Finally when we substitute for $y$ we see that\begin{align*}
\tilde{U}(y) = \dfrac{L^2 y^2 u_{\mathrm{max}}^2}{2\mu } + \alpha u_{\mathrm{max}}^{-k} y^{-k} = y^2 u_{\mathrm{max}}^2 \left( \dfrac{L^2 }{2\mu } + \alpha (yu_{\mathrm{max}})^{-(k+2)}\right) = u_{\mathrm{max}}^2 \mathcal{Y}(y)
\end{align*}Furthermore $u_{\mathrm{max}}^2 \mathcal{Y}(1) = \dfrac{L^2 u_{\mathrm{max}}^2}{2\mu} + \alpha (u_{\mathrm{max}})^{-k} = E$ is just the energy evaluated at the minimum radius, hence\begin{align*}
\Theta = \sqrt{\frac{L^2}{2\mu}} \int_{v_{\mathrm{min}}}^1 \frac{dy}{\sqrt{\mathcal{Y}(1) - \mathcal{Y}(y)}}
\end{align*}Although I can't comment on the line at the very end of the proof without seeing "equation (2.77)", hope it's somewhat helpful.