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Help with Cartesian to Ellipsoidal Coordinates

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to isolate the expressions for ellipsoidal coordinates (see below)...

    I'm given:

    x2=[itex]\frac{(a^2+\lambda)(a^2+\mu)(a^2+\nu)}{(a^2-b^2)(a^2-c^2)}[/itex]

    y2=[itex]\frac{(b^2+\lambda)(b^2+\mu)(b^2+\nu)}{(b^2-a^2)(b^2-c^2)}[/itex]

    z2=[itex]\frac{(c^2+\lambda)(c^2+\mu)(c^2+\nu)}{(c^2-b^2)(c^2-a^2)}[/itex]

    For [itex]-\lambda<c^2<-\mu<b^2<-\nu<a^2[/itex]

    And, I need to transform this to:

    [itex]\frac{x^2}{a^2+\lambda}[/itex]+[itex]\frac{y^2}{b^2+\lambda}[/itex]+[itex]\frac{z^2}{c^2+\lambda}[/itex]=1

    [itex]\frac{x^2}{a^2+\mu}[/itex]+[itex]\frac{y^2}{b^2+\mu}[/itex]+[itex]\frac{z^2}{c^2+\mu}[/itex]=1

    [itex]\frac{x^2}{a^2+\nu}[/itex]+[itex]\frac{y^2}{b^2+\nu}[/itex]+[itex]\frac{z^2}{c^2+\nu}[/itex]=1

    3. The attempt at a solution
    I've tried to solve the 3 equations in the first part as a system of equations to end up with 3 new equations, one each for [itex]\lambda,\mu,\nu[/itex] in terms of only [itex]x,y,z,a,b,c[/itex], but this just keeps getting more and more complicated. Solving in this 'traditional' way gives me expressions that even Maple or Mathematica refuse to isolate for one variable. I think there maybe is something I'm supposed to note about the geometry or something that helps me develop the second 3 equations?. Any help would be greatly appreciated!!
     
  2. jcsd
  3. Oct 1, 2012 #2

    tiny-tim

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    hi jtleafs33! :smile:

    hint: start by multiplying both top and bottom of the first fraction by (b2 - c2) :wink:
     
  4. Oct 1, 2012 #3
    Right. I didn't think of that, but I see you're trying to get me a common denominator here to simplify things.

    Multiply the first equation by [itex](b^2-c^2)[/itex]
    Multiply the second equation by [itex](a^2-c^2)[/itex]
    Multiply the third equation by [itex](a^2-b^2)[/itex]

    Rearranging, I let:

    k=[itex](a^2-b^2)(a^2-c^2)(b^2-c^2)[/itex]

    So now I have:

    [itex]kx^2[/itex]=[itex](a^2+\lambda)(a^2+\mu)(a^2+\nu)(b^2-c^2)[/itex]
    [itex]ky^2[/itex]=[itex](b^2+\lambda)(b^2+\mu)(b^2+\nu)(a^2-c^2)[/itex]
    [itex]kz^2[/itex]=[itex](c^2+\lambda)(c^2+\mu)(c^2+\nu)(a^2-b^2)[/itex]

    This is still stumping me on where to go from here, I can feel this thing getting more complicated as I try to solve this system of eqns by simplifying and substituting in...
     
  5. Oct 1, 2012 #4

    tiny-tim

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    ok, now divide both sides of the first one by a2 + λ (and so on), and then expand what's left :smile:
     
  6. Oct 1, 2012 #5
    Ok, so:

    [itex]\frac{kx^2}{a^2+\lambda}[/itex]=[itex]a^4b^2+a^2b^2\mu+a^2b^2\nu+b^2\mu\nu-a^4c^2-a^2c^2\mu-a^2c^2\nu-c^2\mu\nu[/itex]

    by "and so on" what do you mean?
    If I divide the second one by [itex](b^2+\lambda)[/itex] and the third by [itex](c^2+\lambda)[/itex] i'll just have 3 rearranged expressions for [itex]\lambda[/itex], so you must mean something different?

    I really appreciate your help. There is a lot more to this problem (finding scale factors, calculating divergence and curl in these coordinates) but I know how to do these things. The hiccup I'm having is finding the equations for each coordinate surface, which is what you're helping me with. It's frustrating when you already know what you're trying to show, but you can't figure out how to show it!
     
  7. Oct 1, 2012 #6

    tiny-tim

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    no, you'll have 3 rearranged expressions for μ and [itex]\nu[/itex] :wink:

    and when you add them all, you should find that nearly everything cancels! :smile:
     
  8. Oct 1, 2012 #7
    Ahh, well at least it's been a nice algebra practice. I got it now, Thanks!!! :biggrin:
     
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