The following paradox is sometimes found puzzling. Take coordinates for an observer~$A$ at rest at the origin, and let $B$~be at rest at the time~$t$ at a considerable distance from the origin. The vibrations of an atom at~$B$ are slower (as measured in the time~$t$) than those of an atom at~$A$, and since the coordinate-system is static this difference will be detected experimentally by \emph{any} observer who measures the frequency of the light he receives. Accordingly $B$~must detect the difference, and conclude that the light from~$A$ is displaced towards the violet relatively to his standard atom. This is absurd since, if we choose $B$ as origin, the light from~$A$ should be displaced towards the red. The fallacy lies in ignoring what has happened during the long time of propagation from~$A$ to~$B$ or $B$ to~$A$; during this time the two observers have ceased to be in relative rest, so that compensating Doppler effects are superposed. To obtain a clearer geometrical idea of de~Sitter's world, we consider only one dimension of space, neglecting the coordinates$\theta$ and~$\phi$. Then by~\Eq{(67.31)} \begin{align*} -ds^{2} &= R^{2} (d\omega^{2} + \sin^{2}\omega\, d\zeta^{2}) = R^{2} (d\chi^{2} - \cos^{2}\chi\, dt^{2}) \\ &= dx^{2} + dy^{2} + dz^{2}, \end{align*} \PageSep{165} where \begin{alignat*}{2} x &= R\sin\omega \cos\zeta &&= R\cos\chi \sin it, \\ y &= R\sin\omega \sin\zeta &&= R\sin\chi, \\ z &= R\cos\omega &&= R\cos\chi \cos it, \end{alignat*} and \[ x^{2} + y^{2} + z^{2} = R^{2}. \] It will be seen that real values of $\chi$ and~$t$ correspond to imaginary values of~$\omega$ and~$\zeta$ and accordingly for real events $x$~is imaginary and $y$~and $z$ are real. Introducing a real coordinate $\xi = -ix$, real space-time will be represented by the hyperboloid of one sheet with its axis along the axis of~$\xi$, \[ y^{2} + z^{2} - \xi^{2} = R^{2}, \] the geometry being of the Galilean type \[ ds^{2} = d\xi^{2} - dy^{2} - dz^{2}. \] We have \begin{gather*} r = R\sin\chi = y, \\ \tanh t = -i\tan it = -ix/z = \xi/z, \end{gather*} so that the space-partitions are made by planes perpendicular to the axis of~$y$, and the time-partitions by planes through the axis of~$y$ cutting the hyperboloid into lunes. The light-tracks, $ds = 0$, are the generators of the hyperboloid. The tracks of undisturbed particles are (non-Euclidean) geodesics on the hyperboloid; and, except for $y = 0$, the space-partitions will not be geodesics, so that particles do not remain at rest.