Help with Cartesian to Ellipsoidal Coordinates

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Homework Help Overview

The discussion revolves around transforming Cartesian coordinates to ellipsoidal coordinates, specifically isolating expressions for ellipsoidal coordinates given certain equations involving parameters a, b, c, λ, μ, and ν. The original poster is attempting to derive new equations from a set of three equations related to the coordinates.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve a system of equations but finds the process increasingly complicated. They express uncertainty about the geometric implications that might simplify the transformation. Participants suggest manipulating the equations by multiplying and rearranging terms to find common denominators and simplify the expressions.

Discussion Status

Participants are actively engaging with the problem, offering hints and suggestions for manipulation of the equations. There is a sense of progress as the original poster acknowledges the algebraic practice involved, indicating that they are moving toward a resolution, though no explicit consensus or final solution has been reached.

Contextual Notes

The original poster notes that there is more to the problem, including finding scale factors and calculating divergence and curl in ellipsoidal coordinates, but their current focus is on deriving the equations for each coordinate surface.

jtleafs33
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Homework Statement


I need to isolate the expressions for ellipsoidal coordinates (see below)...

I'm given:

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

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

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

For -\lambda<c^2<-\mu<b^2<-\nu<a^2

And, I need to transform this to:

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

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

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

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 \lambda,\mu,\nu in terms of only x,y,z,a,b,c, 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!
 
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hi jtleafs33! :smile:

hint: start by multiplying both top and bottom of the first fraction by (b2 - c2) :wink:
 
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 (b^2-c^2)
Multiply the second equation by (a^2-c^2)
Multiply the third equation by (a^2-b^2)

Rearranging, I let:

k=(a^2-b^2)(a^2-c^2)(b^2-c^2)

So now I have:

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

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...
 
jtleafs33 said:
kx^2=(a^2+\lambda)(a^2+\mu)(a^2+\nu)(b^2-c^2)
ky^2=(b^2+\lambda)(b^2+\mu)(b^2+\nu)(a^2-c^2)
kz^2=(c^2+\lambda)(c^2+\mu)(c^2+\nu)(a^2-b^2)

ok, now divide both sides of the first one by a2 + λ (and so on), and then expand what's left :smile:
 
tiny-tim said:
ok, now divide both sides of the first one by a2 + λ (and so on), and then expand what's left :smile:

Ok, so:

\frac{kx^2}{a^2+\lambda}=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

by "and so on" what do you mean?
If I divide the second one by (b^2+\lambda) and the third by (c^2+\lambda) i'll just have 3 rearranged expressions for \lambda, 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!
 
jtleafs33 said:
by "and so on" what do you mean?
If I divide the second one by (b^2+\lambda) and the third by (c^2+\lambda) i'll just have 3 rearranged expressions for \lambda …

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

and when you add them all, you should find that nearly everything cancels! :smile:
 
tiny-tim said:
no, you'll have 3 rearranged expressions for μ and \nu :wink:

and when you add them all, you should find that nearly everything cancels! :smile:

Ahh, well at least it's been a nice algebra practice. I got it now, Thanks! :biggrin:
 

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