Reaching a closed form solution for $S$ in the Hamilton-Jacobi equation

Question

I am trying to solve this problem

Discuss the use of parabolic coordinates to obtain separable Hamilton-Jacobi equations for the potential $V(\rho, \phi, z)=\frac{\alpha}{r} - Fz$ and give a closed expression for Hamilton's principal function $S$.

The relationship of parabolic coordinates and cylindrical coordinates is

$$ \rho = \sqrt{\xi\eta} $$

$$ z = \frac{1}{2}(\xi^2-\eta^2) $$

using that, and considering that $r = \sqrt{\rho^2+z^2}$, we reach the following Hamilton-Jacobi equation

$$ 2\xi(\frac{\partial S}{\partial \xi})^2+ \eta(\frac{\partial S}{\partial\eta})^2 + \frac{\xi+\eta}{\xi\eta}(\frac{\partial S}{\partial\phi})^2 + 2\alpha m - \frac{mF}{2}\xi^2(\xi+\eta) + \frac{mF}{2}\eta(\xi+\eta) = mE(\xi+\eta) $$

It is possible to separate the variable $\phi$, but we can see that there are crossed terms between $\xi$ and $\eta$ that can't be eliminated. I can add my calculations in an edit if needed.

My question is: when we stumble upon a Hamilton-Jacobi equation that isn't separable, is there another way to reach an expression for $S$? I know that a general method to solve differential equations doesn't exist, but I was thinking that maybe with a change of coordinates we could reach a separable equation?

Answer 1

The systems is already separable in parabolic coordinates, but you chose a strange convention. The usual coordinates are: $$ \begin{align} z &=\frac{\xi^2-\eta^2}{2} & \rho &= \xi\eta \end{align} $$ This is more natural as it is conformal. This is easiest to see by rewriting it as holomorphic map: $$ z+i\rho=\frac{(\xi+i\eta)^2}{2} $$ Btw, most coordinate changes can also be seen this way too. For example, polar coordinates are just the complex exponential (upon a change of radial function to make it conformal). If your interested in separating HJE’s then elliptic coordinates can be introduced in a similar fashion using hyperbolic functions.

The HJE is now (setting $m=F=\alpha=1$ without loss of generality up to the sign of $\alpha$): $$ \frac{1}{2(\xi^2+\eta^2)}\left(\frac{\partial S}{\partial \xi}\right)^2+ \frac{1}{2(\xi^2+\eta^2)}\left(\frac{\partial S}{\partial \eta}\right)^2 + \frac{1}{2\xi^2\eta^2}\left(\frac{\partial S}{\partial \phi}\right)^2+\frac{2}{\xi^2+\eta^2}-\frac{\xi^2-\eta^2}{2}=E $$ Separating: $$ S\to S(\xi,\eta)+L_z\phi $$ and multiplying by $\xi^2+\eta^2$, you can further separate: $$ \left[\frac{1}{2}\left(\frac{\partial S}{\partial \xi}\right)^2 +\frac{L_z^2}{2\xi^2} -\frac{\xi^4}{2} -E\xi^2 +1\right]+\left[\frac{1}{2}\left(\frac{\partial S}{\partial \eta}\right)^2 +\frac{L_z^2}{2\eta^2} + \frac{\eta^4}{2} -E\eta^2+1\right]=0 $$ The separation $2=1+1$ is purely cosmetic.

In general, it is hard to find coordinates that separate a given Hamiltonian. You usually focus on a coordinate system and deduce all the possible potentials that are separable. For parabolic coordinates, they are of the form: $$ V=\frac{f(\xi)+g(\eta)}{\xi^2+\eta^2} $$ or equivalently in terms of usual variables: $$ V=\frac{f(r+z)+g(r-z)}{r^2} $$ Check out the Mechanics by Landau-Lifschitz (tome 1) or Mathematical Methods for Classical Mechanics by Arnold for more.

Hope this helps.