- #1
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...
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...
Last edited by a moderator: