## Question

Prolate spheroidal coordinates can be used to simplify the Kepler problem in celestial mechanics. They are related to the usual Cartesian coordinates $(x,y,z)$ of Euclidean three-space by $$x={\mathrm{sinh} \chi \ }{\mathrm{sin} \theta \ }{\mathrm{cos} \phi \ }$$ $$y={\mathrm{sinh} \chi \ }{\mathrm{sin} \theta \ }{\mathrm{sin} \phi \ }$$ $$z={\mathrm{cosh} \chi \ }{\mathrm{cos} \theta \ }$$ Restrict your attention to the plane $y=0$ and answer the following questions.

(a) What is the coordinate transformation matrix ${\partial x^{\mu }}/{\partial x^{\nu'}}$ relating $(x,z)$ to $(\chi ,\theta )$?

(b) What does the line element ${ds}^2$ look like in prolate spheroidal coordinates?

 why ${{\mathrm{sinh}}^{\mathrm{2}} \chi \ }{{\mathrm{cos}}^{\mathrm{2}} \theta \ }+{{\mathrm{cosh}}^{\mathrm{2}} \chi \ }{{\mathrm{sin}}^{\mathrm{2}} \theta \ }$ = ${{\mathrm{sinh}}^{\mathrm{2}} \chi \ }+{{\mathrm{sin}}^{\mathrm{2}} \theta \ }$
The answer to the first part was tricksy. Should you find  $\chi,\theta$  in terms of $x,z$? That is very hard and differentiating the expression would also be. After circling a bit I decided that the prime in ${\partial x^{\mu }}/{\partial x^{\nu'}}$ was for the $\chi,\theta$ system and the 'coordinate transformation matrix' would be found from the general tensor coordinate transformation law. I was justified when I could use that in part (b). ${ds}^2$ is the metric and the 'coordinate transformation matrix' can be reused to calculate it in the $\chi,\theta$.
I also found a sample answer at the University of Utah.  They have it in assignments 2 and 3. Maybe none of the students could do it before Feb 7. Their solutions agreed with mine (give or take a $\pm$) but did not explicitly use the great general tensor coordinate transformation law. They use more informal methods. It also contains a few typos. The solution to part (b) is $${ds}^2=\left({{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}}^{\mathrm{2}} \chi \ }{{\mathrm{cos}}^{\mathrm{2}} \theta \ }+{{\mathrm{cosh}}^{\mathrm{2}} \chi \ }{{\mathrm{sin}}^{\mathrm{2}} \theta \ }\right)\left({\mathrm{d}\chi }^2+{\mathrm{d}\theta }^2\right)$$ which is their second last line. Their last line is $$\Rightarrow {ds}^2=\left({{\mathrm{sinh}}^{\mathrm{2}} \chi \ }+{{\mathrm{sin}}^{\mathrm{2}} \theta \ }\right)\left({\mathrm{d}\chi }^2+{\mathrm{d}\theta }^2\right)$$It remained a mystery [until Feb 2019] to me how to prove they are the same, but I could show it numerically with Desmos or in the graphic above. Comparison of Utah and my solutions.