Theorem. In Euclidean space the ratio of projective and conformal representation is $\frac{r_p}{r_c} = \frac{2R^2}{r_c^2 + R^2}$.

Proof. From Fig. 2.17 every point on $S^+$ can be given as a vector $\textbf{p}^{SP}$ in cartesian coordinates, depending on $\varphi$ and $r_p$ with $SP$ being the origin: $$ \textbf{p}^{SP} = \left( \begin{array}{c} 0 \\ 0 \\ R \end{array} \right) + \left( \begin{array}{c} r_p \cos (\varphi) \\ r_p \sin (\varphi) \\ \sqrt{R^2 - r_p^2} \end{array} \right) $$ The vector $\textbf{v}^{SP} = \frac{\bf{p}^{SP}}{\lVert\bf{p}^{SP}\rVert}$ in direction of $\bf{p}^{SP}$ intersects with plane $z = R$, if its third coordinate equals $R$: $$ v^{SP}_z = s \frac{R + \sqrt{R^2 - r_p^2}}{\lVert\bf{p}^{SP}\rVert} = R, $$ where $s$ is a scalar parameter.

The square of the radius of the conformal representation $r_c$ in plane $z = R$ is the norm of the respective components of $\textbf{v}^{SP}$: $$ r_c^2 = \frac{s}{\lVert\bf{p}^{SP}\rVert} \Bigl\lVert \begin{array}{c} r_p \cos (\varphi) \\ r_p \sin (\varphi) \\ \end{array} \Bigr\rVert^2 = \frac{s}{\lVert\bf{p}^{SP}\rVert} r_p^2 = \frac{R}{R + \sqrt{R^2 - r_p^2}} r_p^2.$$ Solving for $r_p$ yields $$ r_p^2 \left( 1 + \left(\frac{R}{r_c}\right)^2 \right) + r_p \left( -\frac{2R}{r_c} \right) = 0.$$ Which holds for $$r_{p1} = 0, \qquad r_{p2} = r_c \frac{2R^2}{r_c^2 + R^2}.$$ $\Box$
2014 DEC 29 (v1.0)
Contact me: m.herrmann followed by an -at- followed by

This document created with the help of MathJax (Thank you guys!)