If one were to read old texts on mathematical physics, like Maxwell, Morse & Feshbach, Hilbert and Courant, Jacobi, etc... they'd find ellipsoidal coordinates popping up, but the authors derive the coordinates using un-illuminating algebra.

Is there a quick, geometric, way of writing down the Cartesian coordinates

$x^2 = (a^2+\xi)(a^2+\eta)(a^2+\zeta)/(b^2-a^2)(c^2-a^2)$

$y^2 = (b^2+\xi)(b^2+\eta)(b^2+\zeta)/(a^2-b^2)(c^2-b^2)$

$z^2 = (c^2+\xi)(c^2+\eta)(c^2+\zeta)/(a^2-c^2)(b^2-c^2).$

of the ellipsoidal coordinate system by drawing a picture, analogous to the way one derives the Cartesian components of the spherical system by drawing a sphere and projecting:

enter image description here

(It can be done quickly algebraically, but there's no geometry there)?

There seems to be an unintelligible, un-intuitive trigonometric method for the special case of oblate-spheroidal coordinates

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(Pages 139 - 145 here give another un-intuitive description with a bad picture)

but this isn't the general case, thus I'm hoping for a picture that that literally copies the spherical picture above, including the details of what the $x$, $y$ & $z$ are by projections are, which these oblate pictures don't do.

If you have similar quick intuition as to why the scale factors and Laplacian take the forms they do, that would be a fantastically helpful bonus, thanks very much.


  1. Morse, Feshbach - "Methods of Mathematical Physics", P. 512
  2. Featherstone, Claessens - "Closed-Form Transformation Between Geodetic and Ellipsoidal Coordinates".
  3. Mathworld Confocal Ellipsoidal Coordinates Article
  4. Landau, - "Electrodynamics of Continuous Media", Section 4
  5. Tang, - "Mathematical Methods for Engineers and Scientists 2", P. 139 - 145
  • $\begingroup$ I actually think the link you gave to Tang's book is good geometrical one. $\endgroup$
    – Ali
    Jun 30, 2014 at 17:25
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    $\begingroup$ Yeah Tang's is the best thing I can find, but I still think it's still very convoluted, unnatural and takes far too much work to write down $x$, $y$ and $z$, there ought to be a 'proof without words' for something like this! $\endgroup$
    – bolbteppa
    Jun 30, 2014 at 18:57
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    $\begingroup$ P. Moon and D. Spencer's "Field Theory Handbook" has a diagram for this coordinate system, but I'm not sure it's terribly intuitive either. $\endgroup$
    – paisanco
    Jul 3, 2014 at 21:10
  • $\begingroup$ As great as Moon and Spencer's books are, the diagram's are no help unfortunately. Similar diagrams/exposition in Willatzen's 'Separable Boundary Value Problems in Physics' book. $\endgroup$
    – bolbteppa
    Jul 4, 2014 at 9:01

1 Answer 1


$\def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\th{\theta} \def\f{\varphi}$The oblate spheroidal coordinates can be easily motivated geometrically. Begin with spherical coordinates. From the figure \begin{align*} x &= r\sin\th\cos\f \\ y &= r\sin\th\sin\f \\ z &= r\cos\th. \end{align*} We have $$x^2+y^2+z^2=r^2.$$ Scale the axes so that \begin{align*} X &= \frac{\a}{r}x = \a\sin\th\cos\f \\ Y &= \frac{\b}{r}y = \b\sin\th\sin\f \\ Z &= \frac{\g}{r}z = \g\cos\th. \end{align*} We have $$\left(\frac{X}{\a}\right)^2+\left(\frac{Y}{\b}\right)^2+\left(\frac{Z}{\g}\right)^2=1,$$ so for a given $\a,\b,\g$ these are ellipses. We require the ellipses to be confocal. Recall that the foci of the ellipse $(x/a)^2+(y/b)^2=1$ are located at $x=\pm c = \pm\sqrt{a^2-b^2}$, where we assume $a\geq b$. We require $\a\geq\b\geq\g$ and $\a^2-\g^2=\b^2-\g^2=c^2$, where $c$ is the linear eccentricity of the ellipses in the $xz$- and $yz$-planes, a constant. Thus, $\a = \b = \sqrt{\g^2+c^2}$ and so \begin{align*} X &= \sqrt{\g^2+c^2}\sin\th\cos\f \\ Y &= \sqrt{\g^2+c^2}\sin\th\sin\f \\ Z &= \g\cos\th. \end{align*} The coordinate system is given by $(\g,\th,\f)$.


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