Particle on surface of ellipsoid

Question

I found this problem in a book called Structure and Interpretation of Classical Mechanics. The problem asks to write down the Lagrange equations of motion for a particle confined to the surface

$$
frac{x^{2}}{alpha^{2}}+frac{y^{2}}{beta^{2}}+frac{z^{2}}{gamma^{2}}=1,
$$

using a "suitable" coordinate system.

Naturally, I thought to use ellipsoidal coordinates. From that page, we have

$$
x^{2}=frac{(a^{2}+lambda)(a^{2}+mu)(a^{2}+nu)}{(a^{2}-b^{2})(a^{2}-c^{2})}
$$

and similar equations for $y^2$ and $z^2$. The quantities $a, b, c$ are parameters of the transformation that can be suitably chosen. $lambda, mu, nu$ are the ellipsoidal coordinates themselves. The inequality

$$
-lambda<c^{2}<-mu<b^{2}<-nu<a^{2}
$$

must be satisfied, according to the definition.

To solve the problem, I chose $lambda = 0$ (this collapses the 3 degrees of freedom of ellipsoidal coordinates to the two-dimensional manifold of the ellipsoid surface), $alpha, beta, gamma = a, b, c$, then directly computed $dot{x}^2$, etc., computed the Lagrangian, and straightforwardly differentiated to obtain the equations of motion. The final result is quite elegant due to many nice cancellations.

However, upon reflection, I realized that ellipsoidal coordinates is completely inappropriate here. A very simple way to see this: from the definition, it is easy to see that $x^2$ can never reach zero, but the particle has no reason to avoid having $x = 0$. Equivalently, the ellipsoidal coordinate system does not seem to be a simple one-to-one transformation, as it is the squares of $x, y, z$ that are written in the formula, not $x, y, z$ themselves.

Can anyone resolve this issue?

Edit: In response to comments and answers, the scheme presented here is a 2d projection of the 3d ellipsoidal coordinate system onto the surface given by $lambda = 0$, which is an ellipsoid. Additionally, the reason that $x = 0$ being inaccessible becomes a problem is that there are clearly physical trajectories of the particle that must, at some time, have the $x$ coordinate zero (not to mention both $y$ and $z$ also), so any coordinate system that does not permit $x = 0$ must necessarily not be sufficient to describe all possible trajectories of the particle, and hence the coordinate system cannot possibly be used to obtain the general Lagrange equations of motion.

WAH · Answer

The main problem I see is that the system you propose to use has 3 coordinates, but the manifold on which you wish to do your physics is 2D.

Let's take the simple example of a sphere of radius 1, in which case $alpha=beta=gamma=1$. Then the coordinates you want to use are $theta$ and $phi$.  Notice that there are only two coordinates because you're starting in 3D space, but restricting your motion with one equation of constraint.

What you really want to do is find a similar system of angular coordinates for your more general sphere, where $alpha,,beta,$ and $gamma$ can be any real number.

xdavidliu · Answer

First, to address the apparent "not being able to reach zero" issue in the OP, we relax the inequalities from strict, e.g. $<$, to monotonic, e.g. $leq$. With this relaxation, $x^2$ can just hit zero.
Now as an additional step, we want to convert the square ellipsoidal coordinates to non-squared ones that actually account for the signs of $x, y, z$, we consider the inequalities
$c^2 leq -muleq b^{2}$ and $b^{2}leq-nuleq a^{2}$, which suggest the substitutions
$$
mu=-left(frac{b^{2}-c^{2}}{2}right)cos(2theta)-frac{b^{2}+c^{2}}{2}
$$
$$
nu=-left(frac{a^{2}-b^{2}}{2}right)cos(2phi)-frac{a^{2}+b^{2}}{2}
$$
We then have the identities
$$
a^{2}+nu=(a^{2}-b^{2})sin^{2}phi
$$
$$
b^{2}+mu=(b^{2}-c^{2})sin^{2}theta
$$
$$
b^{2}+nu=-(a^{2}-b^{2})cos^{2}phi
$$
$$
c^{2}+mu=-(b^{2}-c^{2})cos^{2}theta
$$
With these, the original coordinate equations become
$$
x^{2}=frac{a^{2}(a^{2}+mu)}{a^{2}-c^{2}}sin^{2}phi=a^{2}left(1-frac{b^{2}-c^{2}}{a^{2}-c^{2}}cos^{2}thetaright)sin^{2}phi
$$
$$
y^{2}=b^{2}sin^{2}thetacos^{2}phi
$$
$$
z^{2}=frac{c^{2}(c^{2}+nu)}{c^{2}-a^{2}}cos^{2}theta=c^{2}left(1-frac{a^{2}-b^{2}}{a^{2}-c^{2}}sin^{2}phiright)cos^{2}theta
$$
Note that $a^2+mu$ and $c^2+nu$ are certainly nonzero and positive, from the inequality constraints. Hence, taking square roots and inferring signs, we have
$$
x=asqrt{1-frac{b^{2}-c^{2}}{a^{2}-c^{2}}cos^{2}theta}sinphi
$$
$$
y=bsinthetacosphi
$$
$$
z=csqrt{1-frac{a^{2}-b^{2}}{a^{2}-c^{2}}sin^{2}phi}costheta
$$
The above is a better version of the ellipsoidal coordinates, since it contains the sign information.
(As an aside: I worked these equations out about a year ago, shortly after asking the question in the OP, but didn't get around to posting my answer until now. I wasn't able to find this calculation in any of the textbooks or the wikipedia article, so as far as I know this may be original research.)

Particle on surface of ellipsoid

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