## C6    Our Sample Problem as Epstein Diagram

We want to present the sample problem, which we solved at the end of B6 as an Epstein diagram. To keep everything as simple and clear as possible we will draw two diagrams: One from the point of view of black with a pipe at rest and the other from ‘the red’ point of view of the particle, which sees the pipe racing by.

First to the angle between the two time-axes: The sine value of v/c = 0.8 need not be converted into an angle. We need only count the little squares on our graph paper and select 20 squares for the circle radius, e.g.: 16/20 = 8/10 provides the correct angle (note the green auxiliary lines).

While the particle moves in space-time from O to B, the pipe ages according to the same segment, that is, OA = CD. The clocks at the two tube ends are synchronized for black: O and C (later A and D) project on the same point of the black time-axis. Reaching point B the particle leaves the pipe. The units are selected as follows: 3 squares correspond spatially to a distance of 3 m and to an interval of 10 ns on the time-axis! Importantly: One is free to select only one of the two scales, the other then being fixed by the speed of light. We read: The transit lasts for black OB = OA = 50 ns, while only OE = 30 ns applies for red.

That is the view of black. Now we consider the point of view of red.

For red the pipe OC moves through space-time to AD. This requires a time of OA = CD = OB = 30 ns. Red measures the length of the pipe to be OE = FO = OC • cos(φ) = OC • 0.6 = 7.2 m. Red knows that the clock of black indicates 50 ns at point D. Red says nevertheless that the whole transit required the time OG = FA = OA • cos(φ) = OB • cos(φ) = 30 ns • 0.6 = 18 ns for black. Black measures 50 ns, because its rear clock (the one that moves on segment CD) exhibits a constant advance of EC = GD = 32 ns over the other one moving on OA. This is the desynchronization of black's clocks as stated by red.

Carefully compare this presentation with B6 and convince yourself that we can read all results correctly from the diagrams without the need to do any calculations!