#### Speakpigeon

**Valued Senior Member**

At constant speed, i.e. ignoring acceleration, going away and coming back is by definition symmetrical. Thus, without acceleration, the twin who goes away would come back still the same age as his twin. You could make the trip essentially a trip at constant speed for most of the time, with only two short periods both of acceleration and deceleration. So, the difference in age between the twins is all in these four periods of acceleration.

We can formalise that using three clocks, C0, C1 and C2.

C0 is assumed at rest. The other two are at constant speed relative to C0. C1 is going one way, C2 is coming from the opposite direction. When C1 comes to coincide with C0, the two clocks are synchronised. C1 then keeps going and therefore moves away from C0. After several years for example, C1 meets C2 going the other way. When they meet, C2 is synchronised with C1. Then they move appart and C2 inevitably gets to where C0 is. So, observers at C0 can look at C2. Guess what?

EB