Page:Eddington A. Space Time and Gravitation. 1920.djvu/74

$UOV$, $U^{\prime }OV^{\prime }$ in Fig. 3, whereas in three-dimensions light can traverse any straight line. This could be remedied by interposing some kind of dispersive medium, so that light of some wavelength could be found travelling with every velocity and following every track in space-time; then, looking at a solid which suddenly went out of existence, we should receive at the same moment light-impressions from every particle in its interior supposing them self-luminous). We actually should see the inside of it.

How our poor eyes are to disentangle this overwhelming experience is quite another question.

The interval is a quantity so fundamental for us that we may consider its measurement in some detail. Suppose we have a scale AB divided into kilometres, say, and at each division is placed a clock also registering kilometres. (It will be remembered that time can be measured in seconds or kilometres indifferently.)

When the clocks are correctly set and viewed from A the sum of the readings of any clock and the division beside it is the same for all, since the scale-reading gives the correction for the time taken by light, travelling with unit velocity, to reach $A$. This is shown in Fig. 9 where the clock-readings are given as though they were being viewed from $A$.

Now lay the scale in line with the two events; note the clock and scale-readings $t_{1}$, $x_{1}$, of the first event, and the corresponding readings $t_{2}$, $x_{2}$, of the second event. Then by the formula already given $s^{2}=(t_{2}-t_{1})^{2}-(x_{2}-x_{1})^{2}$. But suppose we took a different standard of rest, and set the scale moving uniformly in the direction $AB$. Then the divisions would have advanced to meet the second event, and $(x_{2}-x_{2})$ would be smaller. This is compensated, because $t_{2}-t_{1}$ also becomes altered. $A$ is now advancing to meet the light coming from any of the clocks along the rod; the light arrives too