Stereographic Projection of Circular Hodographs in Momentum 4-Space

[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:
\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 !

 

[eljose79]

Thank you for your interesting question. I am happy to help you with finding an elegant proof for the fact that the circular hodograph maps onto a great circle in momentum 4-space.

First, let's define the stereographic projection more precisely. The stereographic projection is a map from a sphere onto a plane, where the sphere is the 3-sphere in momentum 4-space and the plane is the equatorial hyperplane. This map is defined by projecting each point on the sphere onto the plane using a straight line from the north pole. In other words, every point on the sphere has a unique corresponding point on the plane, except for the north pole which does not map to any point on the plane.

Now, let's examine the hodograph equation given in the forum post. As you mentioned, the momentum vector traces out a circle in momentum 3-space. This circle can be thought of as a slice of the 3-sphere in momentum 4-space, where the radius of the circle is equal to the radius of the 3-sphere. This means that every point on the circle can be mapped to a unique point on the equatorial hyperplane using the stereographic projection.

Since the hodograph is a circle, it follows that the points on the circle are equidistant from the center of the circle. In other words, the distance from any point on the circle to the center is always the same. This is also true for the corresponding points on the equatorial hyperplane, as they are mapped from the same points on the circle.

Now, let's consider the inverse stereographic projection formula given in the forum post. Notice that the formulae for $p_1$, $p_2$, and $p_3$ are all proportional to the corresponding components of the momentum vector $\mathbf{p}'$. This means that the points on the equatorial hyperplane are also equidistant from the origin in momentum 4-space. In other words, the points on the equatorial hyperplane are all at the same distance from the north pole.

Since the points on the equatorial hyperplane are equidistant from the origin, it follows that they lie on a great circle in momentum 4-space. This is because a great circle on a 3-sphere is defined as a circle whose center is also the center of the 3-sphere.

In