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Stereographic Projection

  1. Mar 25, 2007 #1
    Hi there,

    Look at the topic from my textbook "Stereographic Projection". Please inform me if I have understood it correctly :)

    Lets take specific example from my textbook [tex]S' = \{(x,y,z)|x^2 + y^2 + (z-1)^2 = 1\}[/tex] is a sphere where [tex]N = (0,0,2)[/tex] and [tex]P = (x,y,z)[/tex] can be viewed as steographic projection [tex]g: \sim\{N \} \rightarrow \mathbb{R}^2[/tex] which carries a point of the sphere S' minus the North Pole N onto the intersection of line from N to p with the xy-plane. Then [tex]g(x,y,z) = (u,v)[/tex] and [tex](x,y,z) \in S' \sim \{N\}[/tex] and finally [tex](u,v) \in \mathbb{R}^2 [/tex].

    (1) show that [tex]g^{-1}: \mathbb{R}^2 \rightarrow S' [/tex] is [tex]\left( \begin{array}{ccc}x\\y\\z\end{array} \right) = \left( \begin{array}{ccc}\frac{2(u^2 + v^2)}{u^2 + v^2 +4} \\\\ \frac{2u}{u^2 + v^2 +4} \\\\ \frac{2v}{u^2 + v^2 +4} \end{array} \right)[/tex]


    Solution (1):

    If I write S' in terms of u and v I get

    [tex]u^2 + v^2 = \frac{z}{2-z}[/tex]

    then [tex]z = \frac{2(u^2 + v^2)}{u^2 + v^2 +1}[/tex]

    where its possible to see [tex]2-z = \frac{2}{u^2 + v^2 +1}[/tex]

    Therefore [tex]x = \frac{2u}{u^2 + v^2 +1}[/tex]

    and

    [tex]y = \frac{2v}{u^2 + v^2 +1}[/tex]

    p.s. I treat [tex]g(x,y,z) = \left\{\begin{array}{ccc}\frac{x + iy}{2-z} \ \ \mathrm{if \ z \neq 2} \\ \infty \ \ \mathrm{if \ z=2} \end{array}[/tex]

    and that [tex]g: S' \rightarrow \mathbb{C}^*[/tex] is a unique plane which containes (0,0,1) and (0,0,2) and (x,y,z). Futhermore [tex]g(x,y,z) = u + iv[/tex]

    This is true is [tex]g^{-1}(u+iv) = (\frac{2u}{u^2 + v^2 +1})^2 + (\frac{2v}{u^2 + v^2 +1})^2 + ((\frac{2(u^2 + v^2)}{u^2 + v^2 +1})-1)^2=1[/tex]

    How can this solution be true, since the textbook claims that

    [tex](x,y,z) = g^{-1} = \begin{array}{ccc}\(\frac{2u}{u^2 + v^2 +4}, \frac{2v}{u^2 + v^2 +4},\frac{2(u^2 + v^2)}{u^2 + v^2 +4})\end{array} \)[/tex] ??

    [tex]\mathbb{C}^* = \mathbb{C} \cup \{\infty\}[/tex] is a Riemann Sphere. Then the stereographic projection from S' to [tex]\mathbb{C}^{*}[/tex] is defined by setting g(x,y,z) equal to the point where the line from (x,y,z) to (0,0,2) intersects with the plan z=0. This is defined for all points on S' except (0,0,2).

    2) I am then told to show that sphere S' can be covered by two coordinant neighborhoods. I have an idear on howto show this: If their exist a Coordinant neighborhood around the Northpole and Southpole of S', then if their respective domains intersect, then the sphere is covered by these two coordinant neighborhoods? Aren't they?

    Best Regards

    Fred

    p.s. Could somebody here please inform me what I am doing wrong here in 1?
     
    Last edited: Mar 25, 2007
  2. jcsd
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