The travelling twin possesses two synchronized clocks - one (A) at the front end and the other (B) at the back end of a very long spaceship moving with constant speed towards the sedentary twin. As A passes the sedentary twin, the latter sets his clock to read the same as A. According to special relativity, when later B passes the sedentary twin, B shows more time elapsed than the sedentary twin's clock. That is, special relativity predicts that, as the travelling twin performs the forward part of the trip, he measures the sedentary twin's clock to run SLOWER than his own.

In this scenario things are so planned in advance by the travelling twin that, at the moment B reaches the sedentary twin's clock, the spaceship stops and both A and B stop ticking (simultaneously as judged in the spaceship). The sedentary twin's clock is also stopped by adjacent B. In this pause between the forward and the backward part of the trip, A and B read the same and the sedentary twin's clock reads less. This is what special relativity predicts if measurements are done in the travelling twin's (inertial) system.

Einsteinians,

Do you agree that, in the pause between the forward and the backward part of the trip as defined above, both twins see less time elapsed on the sedentary twin's clock than on A and B? Can you suggest a scenario for the remaining backward part of the trip (with B at the front end and A at the back end of the spaceship) such that, when A and the sedentary twin's clock meet at the end, the latter shows more time elapsed than A, Divine Einstein, yes we all believe in relativity, relativity, relativity?

Pentcho Valev