Tuesday, 1 January 2019

Exercise 2.07 Prolate spheroidal coordinates (Kepler problem)

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?

Answer

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.
Read the full story (3 pages) at Ex 2.07 Kepler problem.pdf.

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