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"? $\endgroup$ Commented Jun 10, 2017 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
    Commented Jun 10, 2017 at 15:13

2 Answers 2


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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    $\begingroup$ @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. $\endgroup$ Commented Jun 11, 2017 at 1:42

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 $c^2 \leq -\mu\leq b^{2}$ 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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