# Stereographic Projection of Circular Hodographs in Momentum 4-Space

1. Jun 22, 2012

### Mendeleeff

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:

\left( \frac{mk}{L} \right)^2 = p_1'^2 + \left( p_2' - \frac{M}{L} \right)^2,

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:

p_0 = \sqrt{- 2mE},

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:

\xi^{-1}: \mathbb{R}^3 \rightarrow \mathbb{S}^3 \backslash \{ \hat{\mathbf{n}} \}: \mathbf{p}' \mapsto \mathbf{p},

with $\mathbf{p}$ the inverse projection of $\mathbf{p}'$, given by the formulae:

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}.

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 !