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I have the set $$M=\{(x,y,z)\in\mathbb{R}^3:z^2-(a^2x^2+b^2y^2)=R^2,\ z>0\}$$ and I have to write the differential equation that describe the geodesic curve and draw it.

I used the parametrization $$ M=\left\{ \begin{pmatrix} x\\[1mm] y\\[1mm] \sqrt{R^2+(a^2x^2+b^2y^2)} \end{pmatrix}\in\mathbb{R}^3: x,y\in[0,\infty) \right\}, $$ and I found the Lagrangian $$ L(t,x,y,x',y')=\frac{\sqrt{(z(x,y)^2+a^4x^2)x'^2+(z(x,y)^2+b^4y^2)y'^2+2a^2b^2xyx'y'}} {z(x,y)} $$ where $z=\sqrt{R^2+(a^2x^2+b^2y^2)}$, and the first part of Euler-Lagrange equation is \begin{align*} \frac{\partial L}{\partial x}&= \frac{(2 x a^4 + 2 x a^2)x'^2 + 2 a^2 x y'^2 + 2 a^2 b^2 y x' y'}{2 z(x,y) \sqrt{(z(x,y)^2+a^4x^2)x'^2+(z(x,y)^2+b^4y^2)y'^2+2a^2b^2xyx'y'}} \\[3mm] &\phantom{=}- \frac{a^2 x \sqrt{(z(x,y)^2+a^4x^2)x'^2+(z(x,y)^2+b^4y^2)y'^2+2a^2b^2xyx'y'} }{z(x,y)^{2}} \end{align*} and this is awful.

Then, I used another parameterization: $$ M=\left\{ \begin{pmatrix} \frac{R}{a}\sinh u \cos v\\[1mm] \frac{R}{b}\sinh u \sin v\\[1mm] R\cosh u \end{pmatrix}\in\mathbb{R}^3:u>0,\ v\in[0,\pi) \right\}, $$ but, I can't simplify the expression: \begin{align*} ds^2&=dx^2+dy^2+dz^2\\[2mm] &=\frac{R^2}{a^2}(\cosh u \cos v\,du-\sinh u \sin v\,dv)^2 +\frac{R^2}{b^2}(\cosh u \sin v\,du+\sinh u \cos v\,dv)^2\\ &\phantom{={}}+(R\sinh u\,du)^2. \end{align*}

Are there a simpler way to find the equation?

MateAndres
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  • Most of the geodesics will have no pretty parametrization. Just the intersections of the surface with the yz plane and with the xz plane come out nice. However, you can still come to conclusions about the overall path of travel for the other geodesics.You should experiment first with $a=b,$ you get Clairaut's relation. Then think about $a \neq b.$ – Will Jagy Feb 16 '14 at 06:25
  • In order to solve these equations you need something called elliptic coordinates. This is a rich, beautiful and non-trivial subject, studied by some of the best mathematicians of the 19th century (like Jacobi). See http://en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid – Gil Bor Feb 16 '14 at 07:50
  • Arthur Cayley, Volume 8 Collected Works: On Geodesic Lines, In Particular Those of A Quadric Surface. Paper # 508. Agree with the above comment, check out what Jacobi did with the triaxial ellipsoid. – Alan Feb 16 '14 at 08:04

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enter image description here Here's a link to a demonstration project that does almost what you want. The code is available, you can modify it. Link: http://demonstrations.wolfram.com/HyperboloidGeodesics/

Alan
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