Action variables for the Kepler problem

[etotheipi]

Homework Statement

Express the Hamiltonian of the orbit in terms of the action variables $I_{\phi}$ and $I_r$ and hence show that the orbit is closed

Relevant Equations

N/A

Let's consider the Hamiltonian $$H = \frac{1}{2m} p_r^2 + \frac{1}{2mr^2} p_{\phi}^2 - \frac{k}{r}$$where the generalised momenta are here $p_r = m\dot{r}$ and $p_{\phi} = mr^2 \dot{\phi}$ conjugate to the coordinates $r$ and $\phi$. Since $p_{\phi}$ does not depend on $\phi$ it can easily be integrated to obtain the action variable $I_{\phi} = p_{\phi}$ (which is also by construction a constant of motion), however the action variable corresponding to $r$ proves more difficult. Since $\partial H / \partial t = 0$, we have $H \equiv E = \text{constant}$ because $dH/dt = -\dot{p}_i \dot{q}_i + \dot{p}_i \dot{q}_i + \partial H / \partial t = 0$. The integral for $I_r$ becomes$$I_r = \frac{1}{2\pi} \oint p_r dr = 2\times \frac{1}{2\pi} \int_{r_{\text{min}}}^{r_{\text{max}}} \sqrt{2mE + \frac{2mk}{r} - \frac{I_{\phi}^2}{r^2}} dr$$We must show this equals$$I_r = k\sqrt{\frac{m}{2|E|}} - I_{\phi}$$Two problems, I can't see how to do that integral, and we'd also need to somehow eliminate $r_{\text{min}}$ and $r_{\text{max}}$. Anyway, the integral is probably more important right now... let $A \equiv 2mE$, $B \equiv 2mk$ and $C \equiv -I_{\phi}^2$, and then$$I_r = \frac{1}{\pi} \int_{r_{\text{min}}}^{r_{\text{max}}} \frac{\sqrt{Ar^2 + Br + C}}{r} dr$$I wonder if there's supposed to be a nice way of solving it, because all I tried was this$$I_r = \frac{1}{\pi} \int_{r_{\text{min}}}^{r_{\text{max}}} \frac{1}{r} \sqrt{A \left(r+ \frac{B}{2A} \right)^2 + \left(\frac{C}{A} - \frac{B^2}{4A^2} \right)} dr = \frac{1}{\pi} \int_{u(r_{\text{min}})}^{u(r_{\text{max}})} \frac{\sqrt{Au^2 + D}}{u - \frac{B}{2A}} du$$with $D \equiv \left(\frac{C}{A} - \frac{B^2}{4A^2} \right)$, which does not seem to help. Maybe I could let $u \equiv \sqrt{D/A} \sinh{\theta}$, in which case$$I_r = \frac{1}{\pi} \int_{\theta(r_{\text{min}})}^{\theta(r_{\text{max}})} \frac{\sqrt{Au^2 + D}}{u - \frac{B}{2A}} du = \frac{D}{\pi \sqrt{A}} \int_{\theta(r_{\text{min}})}^{\theta(r_{\text{max}})} \frac{\cosh^2{\theta}}{\sqrt{\frac{A}{D}} \sinh{\theta} - \frac{B}{2A}} d\theta$$but this just gets messier and messier. Is there a trick or something that would make the integral easier? Thank you in advance

EDIT: By the way, I forgot to write that also we're told that a useful result is$$\int_{r_{\text{min}}}^{r_{\text{min}}} \sqrt{ \left(1-\frac{r_{\text{min}}}{r} \right) \left(\frac{r_{\text{max}}}{r} - 1 \right) } dr = \frac{\pi}{2}(r_{\text{min}} + r_{\text{max}}) - \pi \sqrt{r_{\text{min}}r_{\text{max}}}$$although it's not obvious to me at all how that helps...

 

[TSny]

$$I_r = \frac{1}{\pi} \int_{r_{\text{min}}}^{r_{\text{max}}} \frac{\sqrt{Ar^2 + Br + C}}{r} dr$$

$r_\text{min}$ and $r_\text{max}$ correspond to the roots of the argument of the square root in the integrand. Thus, the argument may factored to be proportional to $(r - r_{\text{max}})(r - r_{\text{min}})$. Be careful with signs. For example, $A$ is a negative number.

 

[etotheipi]

Ah perfect, that's the key insight! I suppose that would be because the roots of $p_{r} = m\dot{r} = 0$ correspond to $r = r_{\text{max}}$ and $r = r_{\text{min}}$ because these two are the local maximum and minimum with respect to $t$ in the coordinate $r = r(t)$. So, using a funny letter for the proportionality constant to keep things interesting,$$2mEr^2 + 2mkr - I_{\phi}^2 = \Xi(r-r_{\text{min}})(r-r_{\text{max}}) = \Xi r^2 - \Xi(r_{\text{min}} + r_{\text{max}}) r+ \Xi r_{\text{min}} r_{\text{max}}$$Since $r^2$, $r$ and $1$ are linearly independent functions and this holds for all $r \in [r_{\text{min}}, r_{\text{max}}]$ we have $\Xi = 2mE = (-2mk)/(r_{\text{min}} + r_{\text{max}}) = (-I_{\phi}^2)/(r_{\text{min}} r_{\text{max}})$, which we keep for later. Back to the integral, since $|r- r_{\text{max}}| = r_{\text{max}} -r \geq 0$,$$I_r = \frac{\sqrt{|\Xi|}}{\pi} \int_{r_{\text{min}}}^{r_{\text{max}}} \frac{1}{r} \sqrt{|(r-r_{\text{min}})(r-r_{\text{max}})|} dr = \frac{\sqrt{|\Xi|}}{\pi} \int_{r_{\text{min}}}^{r_{\text{max}}} \sqrt{\left(1-\frac{r_{\text{min}}}{r}\right) \left(\frac{r_{\text{max}}}{r} -1 \right)} dr$$And using the magic hint$$I_r = \frac{\sqrt{|\Xi|}}{\pi} \left(\frac{\pi}{2}(r_{\text{min}} + r_{\text{max}}) - \pi \sqrt{r_{\text{min}}r_{\text{max}}} \right) = \frac{\sqrt{|\Xi|}}{2}\left( r_{\text{min}} + r_{\text{max}}\right) - \sqrt{|\Xi| r_{\text{min}}r_{\text{max}} }$$and after a little more re-writing, and remembering that $\Xi = - |\Xi| < 0$, as well as that $E = -|E| < 0$ and $I_{\phi} = |I_{\phi}| > 0$,$$I_r = \frac{\sqrt{-2mE}}{2} \frac{-2mk}{2mE} - \sqrt{I_{\phi}^2} = k\sqrt{\frac{m}{2|E|}} - I_{\phi}$$Hence the Hamiltonian takes the form$$H = -\frac{mk^2}{2(I_r + I_{\phi})^2}$$which is symmetric in the two action variables, and thus the rate of change of both of the corresponding angle variables are equal, i.e. $$\dot{\theta}_r = \frac{\partial H}{\partial I_r} \equiv \frac{\partial H}{\partial I_{\phi}} = \dot{\theta}_{\phi}$$and since the frequencies of both of these variables in the phase space are the same, the orbit is closed.

Thanks you!