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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.

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    $\begingroup$ -1. Unclear. What is the "issue" you refer to in the last sentence? Why is the fact that x cannot be zero an "issue"? $\endgroup$ – sammy gerbil Jun 10 '17 at 13:04
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    $\begingroup$ A particle on the surface of an ellipsoid has no physical reason to avoid having the x coordinate zero. For example, I can simply place the particle somewhere on the ellipse formed by the intersection of the plane x = 0 with the surface of the ellipsoid, and then push it in any direction. So a coordinate system that does not permit the x coordinate to be zero cannot possibly describe all physically realizable trajectories of the system. $\endgroup$ – xdavidliu Jun 10 '17 at 15:13
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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.

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  • $\begingroup$ is it possible to remove this answer? There is evidently a misunderstanding of the originally posted question, and I don't feel that this answer adds anything. $\endgroup$ – xdavidliu Jun 10 '17 at 19:25
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    $\begingroup$ @xdavidliu - It most certainly does add something. Scale the cartesian coordinates of a unit sphere centered at the origin by your $\alpha$, $\beta$, and $\gamma$: $x=\alpha\cos\theta\cos\phi, y = \beta\cos\theta\sin\phi, z=\gamma\sin\theta$ and voila! there's your ellipsoid. $\endgroup$ – David Hammen Jun 11 '17 at 1:42
  • $\begingroup$ I considered doing this, but the issue seems to be that $\theta$ and $\phi$ then become non-orthogonal coordinates (i.e. the Jacobian matrix of this transformation no longer has determinant 1), so the resulting Lagrangian and equations of motion become quite messy. The advantage of the true ellipsoidal coordinates are that they are still orthonormal, so the Lagrangian and equations of motion are simple and elegant, and when we choose $\lambda = 0$, that results in $\mu$ and $\nu$ being "symmetric" with each other. $\endgroup$ – xdavidliu Jun 11 '17 at 16:31
  • $\begingroup$ i feel like ellipsoidal coordinates (which are an entirely different thing from simply rescaling spherical coordinates as you mentioned) were designed for this type of problem, so there should be a way to get it to work, especially since if we straightforwardly compute the Lagrange equations of motion using them, the equations are extremely elegant and simple. $\endgroup$ – xdavidliu Jun 11 '17 at 16:37
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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 $\leq-\mu\leq b^{2}$ and $b^{2}\leq-\nu\leq a^{2}$, which suggest the substitutions $$ \mu=-\left(\frac{b^{2}-c^{2}}{2}\right)\cos(2\theta)-\frac{b^{2}+c^{2}}{2} $$ $$ \nu=-\left(\frac{a^{2}-b^{2}}{2}\right)\cos(2\phi)-\frac{a^{2}+b^{2}}{2} $$ 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 $$ x^{2}=\frac{a^{2}(a^{2}+\mu)}{a^{2}-c^{2}}\sin^{2}\phi=a^{2}\left(1-\frac{b^{2}-c^{2}}{a^{2}-c^{2}}\cos^{2}\theta\right)\sin^{2}\phi $$ $$ y^{2}=b^{2}\sin^{2}\theta\cos^{2}\phi $$ $$ z^{2}=\frac{c^{2}(c^{2}+\nu)}{c^{2}-a^{2}}\cos^{2}\theta=c^{2}\left(1-\frac{a^{2}-b^{2}}{a^{2}-c^{2}}\sin^{2}\phi\right)\cos^{2}\theta $$ 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 $$ x=a\sqrt{1-\frac{b^{2}-c^{2}}{a^{2}-c^{2}}\cos^{2}\theta}\sin\phi $$ $$ y=b\sin\theta\cos\phi $$ $$ z=c\sqrt{1-\frac{a^{2}-b^{2}}{a^{2}-c^{2}}\sin^{2}\phi}\cos\theta $$

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.)

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