Parabolic Coordinates Radius

  • Thread starter bolbteppa
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  • #1
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Main Question or Discussion Point

Given Cartesian [itex](x,y,z)[/itex], Spherical [itex](r,\theta,\phi)[/itex] and parabolic [itex](\varepsilon , \eta , \phi )[/itex], where

[tex]\varepsilon = r + z = r(1 + \cos(\theta)) \\\eta = r - z = r(1 - \cos( \theta ) ) \\ \phi = \phi [/tex]

why is it obvious, looking at the pictures

WZ0CY.png


ccUqO.png


(Is my picture right or is it backwards/upside-down?)

that [itex]x[/itex] and [itex]y[/itex] contain a term of the form [itex]\sqrt{ \varepsilon \eta }[/itex] as the radius in

[tex]x = \sqrt{ \varepsilon \eta } \cos (\phi) \\ y = \sqrt{ \varepsilon \eta } \sin (\phi) \\ z = \frac{\varepsilon \ - \eta}{2}[/tex]

I know that [itex] \varepsilon \eta = r^2 - r^2 \cos^2(\phi) = r^2 \sin^2(\phi) = \rho^2[/itex] ([itex]\rho[/itex] the diagonal in the x-y plane) implies [itex] x = \rho \cos(\phi) = \sqrt{ \varepsilon \eta } \cos (\phi) [/itex] mathematically, but looking at the picture I have no physical or geometrical intuition as to why [itex] \rho = \sqrt{ \varepsilon \eta } [/itex].
 

Answers and Replies

  • #2
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##\varepsilon \eta = (r+z)(r-z)=r^2-z^2=h^2## and the height should occur at the position ##(x,y)##.
 

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