Relativist twins... 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? Please Register or Log in to view the hidden image! EB

How do you propose that you "go away and come back" without acceleration? You might as well say you can "go away and come back" without velocity. So basically you are saying if this impossible thing happens then this would happen. That is called science fiction. Good we are going to use acceleration again. OK, both are at constant velocity. Lets label them to make it easy to keep track. C1 is going east and C2 is going west. OK Which one of the travelers changed directions (accelerated)? Or did they both change directions? I can think of only 1 specific case where the clocks would still be synchronized. In most cases they would not be synchronized. What? I don't know what I am supposed to be guessing at. I know that the Co will see that C2 has aged less than Co. What is the point you are trying to make?

In your example with the three clocks, when C0 reads C2 as they pass, it will read that C2 reads less time than C0 does. Example: C1 leaves C0 at 0.6c. C1 meets up with C2 6 ly from Earth as measured by the Earth. This takes 10 y according to C0. According to C1, due to length contraction, this distance is only 4.8 ly from C0, and it only took 8 y between separating from C0 and meeting C2, so it passes a reading of 8 y to C2. It take another 10 y for C2 to get back to C0 according to C0, so it reads 20 y when they meet. Again, for C2, the distance between where it met C1 and C0 is only 4.8 ly, which means it ticks off 8 y between these two events and reads 16 y upon meeting up with C0. According to C0, both C1 and C2 ticked 0.8 as fast as itself , each for 10 years, and it also expects C2 to read 16 yrs when they meet up.

Clear and to the point. Thanks! It's a bit late here according to my natural clock and so for now I'll have to take your word for it but I'm not suspecting any mistake or misunderstanding. So, congratulation... EB