Page:Eddington A. Space Time and Gravitation. 1920.djvu/219

Note 2 (p. 47).

Suppose a particle moves from $(x_{1},y_{1},z_{1},t_{1})$ to $(x_{2},y_{2},z_{2},t_{2})$, its velocity $u$ is given by $u^{2}={\frac {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}{(t_{2}-t_{1})^{2}}}$ Hence from the formula for $s^{2}$$s^{2}=(t_{2}-t_{1}){\sqrt {(1-u^{2})}}$. (We omit a ${\sqrt {-1}}$, as the sign of $s^{2}$ is changed later in the chapter.)

If we take $t_{1}$ and $t_{2}$ to be the start and finish of the aviator's cigar (Chapter i), then as judged by a terrestrial observer, $t_{2}-t_{1}={\text{60 minutes,}}\quad {\sqrt {(1-u^{2})}}={\text{FitzGerald contraction}}={\tfrac {1}{2}}$.

As judged by the aviator, $t_{2}-t_{1}={\text{30 minutes,}}\quad {\sqrt {(1-u^{2})}}=1$.

Thus for both observers $s$ = 30 minutes, verifying that it is an absolute quantity independent of the observer.

Note 3 (p. 48).

The formulae of transformation to axes with a different orientation are $x=x^{\prime }\cos \theta -\tau ^{\prime }\sin \theta {\text{,}}\quad y=y^{\prime }{\text{,}}\quad z=z^{\prime }{\text{,}}\quad \tau =x^{\prime }\sin \theta +\tau ^{\prime }\cos \theta {\text{,}}$ where $\theta$ is the angle turned through in the plane $x\tau$.

Let $u=i\tan \theta$, so that $\cos \theta =(1-u^{2})^{-{\tfrac {1}{2}}}=\beta$, say. The formulae become $x=\beta \,(x^{\prime }-iu\tau ^{\prime }){\text{,}}\quad y=y^{\prime }{\text{,}}\quad z=z^{\prime }{\text{,}}\quad \tau =\beta \,(\tau ^{\prime }-iux^{\prime }){\text{,}}$ or, reverting to real time by setting $i\tau =t$, $x=\beta \,(x^{\prime }-ut^{\prime }){\text{,}}\quad y=y^{\prime }{\text{,}}\quad z=z^{\prime }{\text{,}}\quad t=\beta \,(t^{\prime }-ux^{\prime }){\text{,}}$ which gives the relation between the estimates of space and time by two different observers.

The factor $\beta$ gives in the first equation the FitzGerald contraction, and in the fourth equation the retardation of time. The terms $ut^{\prime }$ and $ux^{\prime }$ correspond to the changed conventions as to rest and simultaneity.

A point at rest, $x$ = const., for the first observer corresponds to a point moving with velocity $u$, $x^{\prime }-ut^{\prime }$ = const., for the second observer. Hence their relative velocity is $u$.