# Action-angle variables for three-dimensional harmonic oscillator using cylindrical coordinates

I am solving problem 19 of ch 10 of Goldstein mechanics. The problem is:

A three-dimensional harmonic oscillator has the force constant k1 in the x- and y- directions and k3 in the z-direction. Using cylindrical coordinates (with the axis of the cylinder in the z direction), describe the motion in terms of the corresponding action-angle variables, showing how the frequencies can be obtained. Transform to the "proper" action-angle variables to eliminate degenerate frequencies.

This is my solution:

Representing Lagrangians in a cylindrical coordinate system:

\begin{align*}&T=\frac{1}{2}m\left(\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2\right),\;\,V=\frac{1}{2}k_1 r^2+\frac{1}{2}k_3 z^2\\&\rightarrow\mathcal{L}=T-V=\frac{1}{2}m\left(\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2\right)-\frac{1}{2}k_1 r^2-\frac{1}{2}k_3 z^2\end{align*}

Thus the generalized momenta are

$$p_r=\frac{\partial\mathcal{L}}{\partial\dot{r}}=m\dot{r},\;\,p_\theta=\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=mr^2\dot{\theta},\;\,p_z=\frac{\partial\mathcal{L}}{\partial\dot{z}}=m\dot{z}$$

And the Hamiltonian is

$$\mathcal{H}=p_r\dot{r}+p_\theta\dot{\theta}+p_z\dot{z}-\mathcal{L}=\frac{p_r^2}{2m}+\frac{p_\theta^2}{2mr^2}+\frac{p_z^2}{2m}+\frac{1}{2}k_1r^2+\frac{1}{2}k_3z^2=T+V.$$

Here $$\frac{\partial\mathcal{L}}{\partial \theta}=0$$ so $$\theta$$ is a cyclic coordinate and $$p_\theta$$ is an action variable itself.

Then I organize the Hamiltonian in terms of the conserved terms $$E_r$$, $$E_\theta$$, and $$E_z$$ as follows:

$$\mathcal{H}=\left(\frac{p_r^2}{2m}+\frac{1}{2}k_1r^2\right)+\frac{p_\theta^2}{2mr^2}+\left(\frac{p_z}{2m}+\frac{1}{2}k_3z^2\right)=E_r+E_\theta+E_z$$

Now find the action-angle variables for each coordinate. Expressing the expressions for $$E_r$$ and $$E_z$$ in the form $$\frac{1}{2}(p^2+\omega^2q^2)=E$$, and $$I=E/\omega$$, so we have

\begin{align*}&\omega_r=\sqrt{\frac{k_1}{m}}\rightarrow I_r=\frac{E_r}{\omega_r}=E_r\sqrt{\frac{m}{k_1}}\\&\omega_z=\sqrt{\frac{k_3}{m}}\rightarrow I_z=\frac{E_z}{\omega_z}=E_z\sqrt{\frac{m}{k_3}}\\&I_\theta=p_\theta\rightarrow\omega_\theta=\frac{E_\theta}{I_\theta}=\frac{p_\theta}{2mr^2}\end{align*}

Finally the Hamiltonian is represented as

$$\mathcal{H}=\omega_rI_r+\omega_\theta I_\theta+\omega_z I_z$$

Questions:

• Do I understand the meaning of "eliminate degenerate frequencies" correctly?
• I think $$\omega_\theta$$ may have the value of $$\dot{\theta}$$, but in my solution it is $$\dot{\theta}/2$$. Is it right?

There is also another issue, and that is the $$p_\theta/(2 m r^2)$$ term. You cannot just separate that one from $$E_r$$, since it depends on $$r$$! You will need to be a little bit more careful about that term.
Recall that the Hamilton-Jacobi equation is obtained by postulating a generating function of the second kind $$S(q^i, \alpha)$$, where $$\alpha$$ are separation constants. Now you substitute $$\partial S/\partial q^i$$ for canonical momenta $$p_i$$ into the Hamiltonian $$H(p_i,q^j)$$ and try to solve this equation $$H(\partial S/\partial q_i, q^j) = E$$ One trick that you can use is that if the equation is independent of any one coordinate $$q^j$$, the coordinate is "cyclical" and the principal function $$S$$ will only be a linear function of that coordinate. You will see that here the coordinate $$\theta$$ is cyclical. After you have separated such terms from the equation using some Ansatz for the principal function, you can try to guess that the rest of the function will be a sum of terms that only depend on one coordinate separately. Here such separability coordinates are $$r,z$$. As a result, you should get separate ordinary differential equations for the parts of the principal function and this should lead you to the transformation to AA coordinates very similarly to the case of the harmonic oscillator.
Finally, there is the issue of degenerate frequencies. There are two meanings of "degenerate" and both apply here, so you can approach this question from either side. You can understand "degenerate" as equivalent of indistinguishable. This meaning applies to the case of the $$\omega_x,\omega_y$$ frequencies in Cartesian coordinates, the frequencies are just the same. Sometimes "degenerate frequency" means simply a vanishing frequency. This would be the case of $$\omega_\theta$$, which you will to be equal to zero. Either way, Goldstein wants you to make a transform between the action-angle coordinates in cylindrical and Cartesian coordinates to show that these two facts are actually related.