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Stereographic Projection of Circular Hodographs in Momentum 4-Space

  1. Jun 22, 2012 #1
    Dear all,

    I want to prove that a circular hodograph (planetary orbit in momentum 3-space) stereographically projects onto a great circle of a 3-sphere in momentum 4-space.

    The equation for the hodograph is given by:
    \begin{equation}
    \left( \frac{mk}{L} \right)^2 = p_1'^2 + \left( p_2' - \frac{M}{L} \right)^2,
    \end{equation}
    with ## m ## the mass of the planet, ## L ## the angular momentum, ## M ## the Laplace-Runge-Lenz vector, and ## p_1' ## and ## p_2' ## the components of the momentum vector ## \mathbf{p}' = \left( p_1', p_2', 0 \right) ##. The momentum vector ## \mathbf{p}' ## is seen to trace out a circle in momentum 3-space, centered at ## \left( 0, \frac{M}{L} \right) ## on the ## p_2' ##-axis and with radius ## \frac{mk}{L} ##. The hodograph intersects the ## p_1' ##-axis in two points at a distance denoted ## p_0 ##. The magnitude of ## p_0 ## can be determined by setting ## p_2' = 0 ## in the above equation, yielding:
    \begin{equation}
    p_0 = \sqrt{- 2mE},
    \end{equation}
    where I used the relation ## \mathbf{M}^2 = 2m \mathscr{H} \mathbf{L}^2 + m^2 k^2 ##.

    I now want to stereographically project this curve on a 3-sphere ## \mathbb{S}^3 ## of radius ## p_0 ## in momentum 4-space ## \mathbb{R}^4 ##. That is:

    Let ## \mathbb{S}^3 = \left\{ \mathbf{p} = \left( p_1, p_2, p_3, p_4 \right) \; | p_1^2 + p_2^2 + p_3^2 + p_4^2 = p_0^2 \right\} ## represent a hypersphere in momentum 4-space of radius ## p_0 ##, and let ## \hat{\mathbf{n}} = \left( 0, 0, 0, p_0 \right) \in \mathbb{R}^4 ## denote the north pole. Construct the 3-dimensional hyperplane ## \mathbb{R}^3 = \left\{ \left( p_1, p_2, p_3, 0 \right) \in \mathbb{R}^4 \right\} ##, which cuts the 3-sphere in half. The intersection ## \mathbb{R}^3 \cap \mathbb{S}^3 ## defines the equator of ## \mathbb{S}^3 ##. Given any point ## \mathbf{p} = \left( p_1, p_2, p_3, p_4 \right) \in \mathbb{S}^3 ##, other than the north pole ## \hat{\mathbf{n}} ##, there is a unique line connecting ## \mathbf{p} ## with ## \hat{\mathbf{n}} ## which intersects the equatorial hyperplane ## \mathbb{R}^3 ## at a point ## \mathbf{p}' = \left( p_1', p_2', p_3', 0 \right) ##. The inverse stereographic projection of ## \mathbb{S}^3 ## is then defined as the map:
    \begin{equation}
    \xi^{-1}: \mathbb{R}^3 \rightarrow \mathbb{S}^3 \backslash \{ \hat{\mathbf{n}} \}: \mathbf{p}' \mapsto \mathbf{p},
    \end{equation}
    with ## \mathbf{p} ## the inverse projection of ## \mathbf{p}' ##, given by the formulae:
    \begin{equation}
    p_1 = p_0 \frac{2 p_0 p_1'}{p_0^2 + p\,'^{\,2}} \quad ; \quad p_2 = p_0 \frac{2 p_0 p_2'}{p_0^2 + p\,'^{\,2}} \quad ; \quad p_3 = p_0 \frac{2 p_0 p_3'}{p_0^2 + p\,'^{\,2}} \quad ; \quad p_4 = p_0 \frac{p\,'^{\,2} - p_0^2}{p\,'^{\,2} + p_0^2}.
    \end{equation}
    Could someone help me with finding an elegant proof for the fact that the hodograph in the first equation maps onto a great circle ?

    Thank you so much in advance !
     
  2. jcsd
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