13.4: *Kepler Orbit Action-Angle Variables

We have not yet covered Kepler orbits, so skip this section for now: it's here to refer back to later. It's from Landau, p 167. For motion confined to a plane, we can take the central potential analysis with \(\theta=\pi / 2, p_{\theta}=0 \text { and } p_{\phi}=m v_{\phi} r\), the angular momentum, so the Hamiltonian is

\begin{equation}H=\frac{1}{2 m}\left(p_{r}^{2}+\frac{p_{\phi}^{2}}{r^{2}}\right)+V(r)\end{equation}

The Hamilton-Jacobi equation is therefore

\begin{equation}
\frac{1}{2 m}\left(\frac{\partial S_{0}}{\partial r}\right)^{2}+V(r)+\frac{1}{2 m r^{2}}\left(\frac{\partial S_{0}}{\partial \phi}\right)^{2}=E
\end{equation}

So, following the previous analysis of separation of variables for motion in a central potential, here

\begin{equation}
S_{0}(r, \phi)=S_{r}(r)+S_{\phi}(\phi)=S_{r}(r)+p_{\phi} \phi
\end{equation}

The action variable for the angular motion is just the angular momentum itself,

\begin{equation}
I_{\phi}=\frac{1}{2 \pi} \int_{0}^{2 \pi} p_{\phi} d \phi=L
\end{equation}

And the radial action variable, with potential \(V(r)=-k / r, \text { is }\)

\begin{equation}
I_{r}=2 \frac{1}{2 \pi} \int_{r_{\text {min }}}^{r_{\max }} \sqrt{\left[2 m\left(E+\frac{k}{r}\right)-\frac{L^{2}}{r^{2}}\right]} d r=-L+k \sqrt{\frac{m}{2|E|}}
\end{equation}

(Details on doing the integral are given in the Appendix, Mathematica can do it too.)

So the energy is

\begin{equation}
E=-\frac{m k^{2}}{2\left(I_{r}+I_{\phi}\right)^{2}}
\end{equation}

The motion is degenerate: the two fundamental frequencies coincide, \(\partial I_{\phi} / \partial E=\partial I_{r} / \partial E\)

This has major consequences in quantum mechanics: the actions are all quantized in units of Planck's constant, for the hydrogen atom, from the formula above, the energy depends only on the sum of the quantum numbers: above the ground state, energy levels are degenerate, which is why the energy spectrum has the deceptively simple form so successfully explained by the Bohr model.

The orbital parameters, semi-latus rectum and eccentricity, from \(|E|=k / 2 a \text { and } L^{2}=k m a\left(1-e^{2}\right)\), are

\begin{equation}
\mathrm{l}=\frac{I_{\phi}^{2}}{m k}, \quad e^{2}=1-\left(\frac{I_{\phi}}{I_{\phi}+I_{r}}\right)^{2}
\end{equation}

Recall the semi-major axis is given by \(|E|=k / 2 a\) and from the above expression

\begin{equation}
\frac{b}{a}=\frac{I_{\phi}}{I_{\phi}+I_{r}}=\frac{|m|}{n}
\end{equation}

in the hydrogen atom quantum number notation.

Appendix: Doing the Integral for The Radial Action Ir

The integral can be put in the form

\begin{equation}
I=\frac{C}{2 \pi} \int_{\alpha}^{\beta} \sqrt{(x-\alpha)(\beta-x)} \frac{d x}{x}
\end{equation}

which can be integrated by taking a contour encircling the cut from \(\alpha\) to \(\beta\). The integral will have a contribution from the pole at the origin equal to \(\begin{equation}C \sqrt{-\alpha \beta}\end{equation}\) and another from the circle at infinity, which is

\begin{equation}
I_{\infty}=\frac{C}{2 \pi} \oint\left(1-\frac{\alpha}{2 z}\right)\left(1-\frac{\beta}{2 z}\right) i d z=-\frac{C(\alpha+\beta)}{2}
\end{equation}

Equating coefficients (multiplying the term inside the square root by \(r^{2}\))

\begin{equation}C^{2}=2 m|E|, \quad C^{2}(\alpha+\beta)=2 m k, \quad C^{2} \alpha \beta=L^{2}\end{equation}

So the contribution from the origin gives the \(\begin{equation}
-L, \text { the circle at infinity } m k / \sqrt{2 m}|E|=k \sqrt{m / 2}|E|
\end{equation}\).