C3 Time Dilation in the Epstein Diagram
Thus, for black, B moves with velocity v in the direction of the positive x-axis. With the meeting at O both A and B set their clocks to 0000. The meeting took place at x=0 and x' =0. Then A and B moved on in space-time (with the speed of light) and somewhat later we have the following situation:
According to C2, OA = OB = c • ∆t and B has moved spatially for black from O to X. Thus
OX = OB • sin(φ)= (c • ∆t) • (v / c) = v • ∆t
What is the meaning of the segment OY?
OY = OB • cos(φ) = OA • cos(φ)
If we no longer interpret OA in space-time, but rather assume the point of view of black and consider OA as “the time ∆t that has passed for me since our meeting at O”, then OY represents ∆t • cos(φ), and that is nothing else but ∆t' (according to the statement at the end of C2). That is, it is the time, from the point of view of black (and also from that of red!) that has elapsed for red since the meeting in O! Since cos(φ) corresponds directly to the radical term which we use in the time dilatation calculation, black can directly read from its time-axis, how much time elapsed for everyone since the meeting in O.
To repeat: Black projects the space-time position of everyone on its time-axis and therefore knows how much time elapsed for everyone since the meeting in O. Since the clocks were set to zero at the meeting, we have
OA = ∆t = t and OY = ∆t’ = t’
From its time-axis black reads t and t', obtaining the elapsed time of all clocks involved!
We can avoid the conversion using factor c, if we compare only times. The whole representation would be simpler and more elegant, if we would measure lengths in light seconds. c would then simply have the value 1 or 1 light-second per second, that is '1 light'. However there are good reasons to stick to the standard system of units.