Equal area parameterization of a torus?

Question

I am trying to parameterise a surface of revolution such that each infinitesimal area element is uniform across the surface. The cross-sections of the surface are shown in the picture below. The title of the post refers to a torus as I thought this might be an easier place to start. However, I really want to know how to do this for a surface with a general cross-section like the one below.

Here is what I have tried so far...

Let $l_\theta$ denote the length and coordinate along the contour shown below in the $x$-$z$ plane. Where I am using $\theta$ here as a reference to the poloidal direction in "Toroidal and poloidal coordinates".

Let $R(l_\theta)$ denote the distance to the corresponding point from the $z$-axis.

Let $$\phi=\mathrm{atan2}(y, x),$$ denote the toroial angle.

Let $$s_\theta = \frac{A_{tot}}{2\pi R(l_\theta)}\frac{l_\theta}{l_{tot}},$$ where $A_{tot}$ gives the total area of the surface and $l_{tot}$ give the total length of the contour.

Note that $$\begin{aligned} ds_{\theta} &= dl_\theta\frac{ds_\theta}{dl_\theta} \\ &= dl_\theta\frac{A_{tot}}{2\pi l_{tot}}\left(\frac{1}{R(l_\theta)}-\frac{l_\theta}{R(l_\theta)^2}\frac{dR}{dl_\theta}\right) \end{aligned}$$

Using $\phi$ and $s_\theta$ as my coordinates nearly gives what I want but not quite. The infinitesimal area elements are given by $$\begin{aligned} R(l_\theta)\,d\phi\,ds_\theta &=\frac{A_{tot}}{2\pi l_{tot}}d\phi dl_\theta\left(1-\frac{l_\theta}{R(l_\theta)}\frac{dR}{dl_\theta}\right) \\ &\approx \frac{A_{tot}}{2\pi l_{tot}}d\phi dl_\theta \end{aligned}$$ for $$\frac{l_\theta}{R(l_\theta)}\frac{dR}{dl_\theta}\ll1.$$ Do you know if it's possible to get a better parameterization where the infinitesimal areas are completely uniform across the surface? If not, do you know how I can improve on the parameterization above?

Answer 1

$\newcommand{\Bar}[1]{\overline{#1}}$Briefly, yes it's possible to reparametrize a surface of rotation so the area element is exactly the product of the coordinate differentials.

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Abstractly, in the terminology of the linked wikipedia page, a surface of rotation with poloidal angle $\Bar{\theta}$ and toroidal angle $\phi$ has metric $$ g = \Bar{E}(\Bar{\theta})\, d\Bar{\theta}^{2} + \Bar{G}(\Bar{\theta})\, d\phi^{2}. $$

Lemma: If $f$ is an arbitrary positive, continuous function, there exists a reparametrization $\theta = \theta(\Bar{\theta})$ so that the metric becomes $$ f(G(\theta))\, d\theta^{2} + G(\theta)\, d\phi^{2}. $$

The desired coordinate system satisfies $E = f(G) = 1/G$, so the area element in the new coordinates is $$ dA = \sqrt{EG}\, d\theta\, d\phi = d\theta\, d\phi. $$

Proof of the lemma: Let $$ \theta = \int^{\Bar{\theta}(\theta)} \sqrt{\frac{\Bar{E}(u)}{f(\Bar{G}(u))}}\, du. $$ (The lower limit of integration is geometrically immaterial, changing $\theta$ by an additive constant.) This function $\theta$ satisfies $$ 1 = \sqrt{\frac{\Bar{E}(\bar{\theta})}{f(\Bar{G}(\Bar{\theta}))}} \frac{d\Bar{\theta}}{d\theta}, $$ or $$ \Bar{E}(\bar{\theta})\, d\Bar{\theta}^{2} = f(\Bar{G}(\Bar{\theta}))\, d\theta^{2}. $$ Substituting, and writing $\Bar{G}(\Bar{\theta}) = G(\theta)$, gives $$ g = \Bar{E}(\Bar{\theta})\, d\Bar{\theta}^{2} + \Bar{G}(\Bar{\theta})\, d\phi^{2} = f(G(\theta))\, d\theta^{2} + G(\theta)\, d\phi^{2}. $$

(You didn't ask, but it's a neat fact that the Gaussian curvature of the metric in these "symplectic" or "equareal" coordinates is the formally linear expression $K = -\frac{1}{2}G''(\theta)$.)

If your surface is generated by revolving a parametric plane path $(x(\Bar{\theta}), z(\Bar{\theta}))$, the parametrization may be taken as $$ \mathbf{x}(\Bar{\theta}, \phi) = (x(\Bar{\theta})\cos\phi, x(\Bar{\theta})\sin\phi, z(\Bar{\theta})), $$ and the "initial" metric components are $$ E(\Bar{\theta}) = x'(\Bar{\theta})^{2} + z'(\Bar{\theta})^{2},\qquad G(\Bar{\theta}) = x(\Bar{\theta})^{2}. $$ Sadly, but as might be expected, the change-of-coordinates integral to equareal coordinates is usually non-elementary.

Additional details of possible interest may be found in L'Enseignement Math. 49 (2003), pp. 157–172.