Inverting parabolic and stereographic coordinates

AI Thread Summary
The discussion focuses on inverting parabolic coordinates defined by the equations μ = 2xy and λ = x² - y² to find x(μ, λ) and y(μ, λ). The initial attempt yields expressions for x and y, but the participant doubts their correctness due to discrepancies with lecture notes that suggest a relationship x² + y² = λ² + μ². The conversation also touches on the challenges of inverting nonlinear coordinate systems, particularly regarding stereographic projections. A method is proposed to express y in terms of x using the equation y = μ/(2x) and substituting it back to solve for x. The complexities of these transformations highlight the intricacies involved in nonlinear coordinate systems.
Epimetheus
Messages
2
Reaction score
0

Homework Statement



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

Homework Equations



None

The Attempt at a Solution



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

However, I don't think this is right ... I'm following a set of CM lecture notes, which immediately go on to claim x^2+y^2=\lambda^2+\mu^2 and
\dot x^2 + \dot y^2 = \frac14 \frac{\dot \lambda^2 + \dot \mu^2}{\sqrt{\lambda^2 + \mu^2}}
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
\frac{x}{\xi} = \frac{y}{\eta} = \frac1{1-\zeta}

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

It's hard to get rid of the "old" set of variables in this case.
 
Physics news on Phys.org
If you want to find x and y in terms of \lambda and \mu, then youre given \mu=2xy

y=\frac{\mu}{2x}.

Substitute this in the other equation and solve for x. Similarly for y.
 
Last edited:
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top