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

Consider $\mathbb{R}^3$ in standard Cartesian co-ordinates, and the surface $S^2$ embedded within it defined by $(x^2+y^2+z^2)|_{S^2}=1$. A particular set of co-ords on $S^2$ are defined by

$\zeta = \frac{x}{z-1}$, $\eta = \frac{y}{z-1}$.

Express $1+\zeta^2+\eta^2$ in terms of $z$. By evaluating $d\zeta$ and $d\eta$, show that the line element on $S^2$ is given by

$ds^2|_{S^2}=(dx^2+dy^2+dz^2)|_{S^2} = \frac{d\zeta^2+d\eta^2}{f(\zeta,\eta)}$ (1),

where you should give the form of $f(\zeta,\eta)$

2. Relevant equations

$1+\zeta^2+\eta^2=\frac{x^2+y^2}{(z-1)^2}+1=\frac{-2}{z-1}$

$d\zeta=\frac{dx}{z-1}-\frac{xdz}{(z-1)^2}$ , $d\eta=\frac{dy}{z-1}-\frac{ydz}{(z-1)^2}$

$d\eta^2 + d\zeta^2= \frac{dx^2}{(z-1)^2} +\frac{x^2dz^2}{(z-1)^4}-\frac{2xdxdz}{(z-1)^3}+\frac{dy^2}{(z-1)^2} +\frac{y^2dz^2}{(z-1)^4}-\frac{2ydydz}{(z-1)^3}$ (2)

3. The attempt at a solution

So far I have been able to do the first two parts fine (the first two equations under 'Relevant equations'), the part I'm struggling with is trying to prove equation (1). So far I have tried computing $d\eta^2+d\zeta^2$, which is equation 2 above, as well as rearranging the differentials and trying $dx^2+dy^2$, but I feel like this is the wrong approach.

I'm just wondering if anyone has any tips or can see what to do? I feel like I'm over thinking and over complicating the problem, or there may be something I'm missing. Looking at equation (2), I can't see how to massage it into the desired form, so I feel I may either be attacking this in the wrong way. The usual approach is to compute and simplify $ds^2+dy^2+dz^2$, but I have not been able to get this to work.

This is my first post, hopefully I've done it right. And thanks in advance for any help!

Holty
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  • as $\zeta$ depends on $x,y,z=\sqrt{1-x^2-y^2}$, your $d\zeta$ should be calculated as $$d\zeta=\frac{\partial\zeta}{\partial x}dx+\frac{\partial\zeta}{\partial y}dy$$ and similar for $\eta$ – janmarqz Nov 14 '15 at 18:37

1 Answers1

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From $\displaystyle z=1-{2\over1+\zeta^2+\eta^2}$ you can find $dz$. Then from $x=\zeta(z-1)$ you can compute $dx=(z-1)d\zeta+\zeta dz$ and substitute there the value of $dz$ previously found. Do the same for $dy$, starting from $y=\eta(z-1)$, and a straightforward calculation will then yield $dx^2+dy^2+dz^2$.

EDIT

A possibly simpler way: you have $$ x=-{2\zeta\over1+\zeta^2+\eta^2}, \quad y=-{2\eta\over1+\zeta^2+\eta^2}, \quad z=1-{2\over1+\zeta^2+\eta^2}. $$ From that you can compute $dx$, $dy$ and $dz$.

Intelligenti pauca
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  • Thank you for the help. I had already tried this, but for $dz$ I get $dz=\frac{\partial z}{\partial\zeta}d\zeta + \frac{\partial z}{\partial\eta}d\eta = \frac{4\zeta d\zeta}{(1+\zeta^2+\eta^2)^2}+\frac{4\eta d\eta}{(1+\zeta^2+\eta^2)^2}$ , which just becomes unwieldy when substituted into $dx$, which then needs to be squared. Ultimately it seems to become too difficult to work with, and I feel there must be a simpler way. – Holty Nov 14 '15 at 20:04
  • There may be a simpler way, but the calculation is not so difficult after all. – Intelligenti pauca Nov 14 '15 at 22:30
  • @Holty See my edit, anyway. – Intelligenti pauca Nov 14 '15 at 22:41