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Inverting parabolic and stereographic coordinates

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Given the parabolic co-ordinate system defined, given Cartesian coordinates x and y, as
    [tex]$\mu=2xy$[/tex]
    [tex]$\lambda = x^2-y^2,$[/tex]
    find the inverse transformation [tex]x(\mu, \lambda)[/tex] and [tex]y(\mu,\lambda)[/tex].

    2. Relevant equations

    None

    3. The attempt at a solution

    We compute [tex]\lambda/\mu^2[/tex] in order to obtain
    [tex]x=\pm \sqrt{\frac{\lambda\pm\sqrt{\lambda^2+1}}2}[/tex]
    and
    [tex]y=\frac{2\mu}{\pm \sqrt{\frac{\lambda\pm\sqrt{\lambda^2+1}}2}}[/tex]

    However, I don't think this is right ... I'm following a set of CM lecture notes, which immediately go on to claim [tex]x^2+y^2=\lambda^2+\mu^2[/tex] and
    [tex]\dot x^2 + \dot y^2 = \frac14 \frac{\dot \lambda^2 + \dot \mu^2}{\sqrt{\lambda^2 + \mu^2}} [/tex]
    which looks tantalizingly similar to spherical polars, but doesn't seem to follow from what I have ...

    Any thoughts on inverting general "nonlinear" co-ordinate systems (other than cylindrical and spherical)? What about finding the inverse transformation for the stereographic projection defined by
    [tex]\frac{x}{\xi} = \frac{y}{\eta} = \frac1{1-\zeta}[/tex]

    so that [tex]x=\frac{\xi}{1-\zeta}; y=\frac{\eta}{1-\zeta} [/tex]?

    It's hard to get rid of the "old" set of variables in this case.
     
  2. jcsd
  3. Mar 1, 2008 #2
    If you want to find x and y in terms of [tex]\lambda[/tex] and [tex]\mu[/tex], then youre given [tex]\mu=2xy[/tex]

    [tex]y=\frac{\mu}{2x}[/tex].

    Substitute this in the other equation and solve for x. Similarly for y.
     
    Last edited: Mar 1, 2008
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