## C5 Desynchronization in the Epstein Diagram

Clocks, which are synchronized for red, project onto the same point of the red time-axis. All red synchronized clocks lie on lines parallel to the red space-axis. For black, however, which reads time always from its black time-axis, such clocks exhibit a time difference:

Black ascertains: ∆t' • c = ∆x' • sin(φ) = ∆x' • v/c, which is, except for the sign, precisely the formula of **B6**! In addition, we can determine the sign: From the point of view of black, the leading clock of red at D runs behind the spatially trailing clock at C. Therefore it must mean ∆t' • c = -∆x' • sin(φ) = -∆x' • v/c in perfect agreement with **B6**. Both red clocks are moving at the same speed, but the one in front (spatially) is permanently behind (temporally) the other by the same amount.

It is worth considering (and also non-trivial) how the synchronization of distant passing clocks of **B6** would look in an Epstein diagram. One must note the fact that light in each reference frame always moves with c and thus the quotient v/c is 1 and the angle φ is always 90º! Light moves in each reference frame only spatially, parallel to the space-axis. Thus: Photons do not age!