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 blaetterundsterne.org.
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