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I have an ellipsoid given by $S = \{ (x,y,z): \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$, for some fixed $a,b,c \in \mathbb{R}^{+} \}$. I need to find the umbilic points of this ellipsoid. However, everytime I try a parametrization $X$, the expressions for the first and second fundamental forms quickly become rather messy and the Gaussian curvature $K$ and the mean curvature $H$ are nigh impossible to work with by hand. I ask for help in finding a parametrization $X$ that the quantities therefrom derived are nice and tidy enough to work with by the time that I get to calculating the umbilic points. Also, if you have any tricks in which way to determine the umbilic points might be easiest in this case (which method to choose), that would be very helpful too.

I have tried $X_{1}(u,v) = (a \cos{u} \sin{v}, b \sin{u} \sin{v}, c \cos{v})$ and $X_{2}(u,v) = (u, v, \pm c \sqrt{1 - \frac{u^{2}}{a^{2}} - \frac{v^{2}}{b^{2}}})$ to no avail, along with a few others that I swiftly gave up on.

kevin
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1 Answers1

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Confocal Quadric coordinates have been of real importance for the ellipsoid. A typical situation is given in the example, the umbilic points are as in the reference above.

$$ \frac{x^2}{4} + \frac{y^2}{2} + \frac{z^2}{1} = 1 $$

$$ \frac{x^2}{4-\lambda} + \frac{y^2}{2-\lambda} + \frac{z^2}{1-\lambda} = 1 $$

$$ \frac{x^2}{4-\mu} + \frac{y^2}{2-\mu} + \frac{z^2}{1-\mu} = 1 $$

Here $\lambda$ and $\mu$ are given by: $ 4 > \mu > 2 > \lambda > 1 $ ( In the picture $\lambda = 1.5$ and $ \mu = 3$ ).

See: http://mathworld.wolfram.com/ConfocalEllipsoidalCoordinates.html

Alan
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  • What do $K$ and $H$ look like in this parametrization (preferably with more general $a$, $b$, and $c$)? – kevin Apr 14 '14 at 04:38
  • The computations for K and H are here: http://mathworld.wolfram.com/Ellipsoid.html , the coordinate system I was working with works well with this: http://en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid – Alan Apr 14 '14 at 05:25
  • So, no matter what I do, I am going to get to those rather messy quantities? (I guess that that is what "invariant" means, eh?) So what do I do in order to get the umbilic points? I do not want to have to set $K = H^{2}$ and solve for the appropriate coordinates with those brutish expressions! – kevin Apr 14 '14 at 05:32
  • $$a\sqrt( (a^2 - b^2)/(a^2 - c^2) ) , 0 , c\sqrt( (b^2 - c^2) / (a^2 - c^2) )$$ total of four points. What book are you using? I use "Modern Differential Geometry of Curves and Surfaces with Mathematica" by Alfred Gray. There are Mathematica Notebooks that make some of this stuff less tedious – Alan Apr 14 '14 at 05:46
  • K and H also computed here: http://math.stackexchange.com/questions/540710/gaussian-mean-curvature – Alan Apr 14 '14 at 05:57
  • I am using pencil and paper, unfortunately (and against my will). :/ – kevin Apr 14 '14 at 06:04
  • I only see three points (I think) above. I am using no book - it is just a homework assignment. I found the expressions for $K$ and $H$ given there, but solving for umbilic points from those is... tedious. – kevin Apr 14 '14 at 06:07
  • $x=\pm a \sqrt\frac{a^2 -b^2}{a^2 - c^2} $ , $y=0$ , $x=\pm c \sqrt\frac{b^2 - c^2}{a^2 - c^2} $ – Alan Apr 14 '14 at 07:19