the_emi_guy said:
These folks are better served by showing them how simple application of time dilation to all three inertial frames (see post #18) shows this in a trivial and intuitive manner (and I do mean "inertial frames" here, with each observer at rest in his frame).
Whether this is "trivial and intuitive" is debatable (see below), but it seems to me that these folks would be served better still by giving them the most general method of all: spacetime geometry and lengths of curves. What we have here is a triangle in spacetime: one side is the stay-at-home twin's worldline, and the other two sides are the traveling twin's outbound and inbound worldlines. Then all we need is the Minkowski spacetime analogue to the Pythagorean theorem--of which the time dilation equation you are implicitly using in your method is just a special case.
the_emi_guy said:
As a follow on we could consider "what if there were only two observers?"
That's not a follow-on; it's how the original scenario was posed. When you formulate it in terms of three observers, you are, on the face of it, talking about a
different scenario. You then have to explain how the result you derive with three observers gives the right answer for the case of only two, which was the actual scenario posed. This is certainly doable, but I question whether it is "trivial and intuitive" to a person who is struggling with time dilation and differential aging.
the_emi_guy said:
what did the second observer do? Did he turn around instantaneously experiencing infinite acceleration (a non-physical physics problem)? What if he accelerates and decelerates gradually over time? What if he slingshots himself around another planet experiencing weightlessness the whole way? It may very well turn out that all of these produce the same result because spacetime does not get curved and SR still applies (or maybe not).
The first case is the one that is equivalent to your analysis--although I would prefer to phrase it in terms of taking the limit as the turnaround time (the time elapsed on the traveling twin's clock during the period when he is accelerating to turn around) goes to zero. The key point is that, if the answer we are interested in is elapsed proper time for the two twins, we don't really care about the details of how the traveling twin turns around, as long as the time elapsed on his clock while doing so is negligible compared to the time elapsed on his clock during the inbound and outbound legs.
The second case, accelerating and decelerating gradually over time, is
not equivalent to your analysis. And the simplest way to state the reason why is by using spacetime geometry and worldlines: the worldline of the traveling twin is
different in this case than it is in the idealized case you are analyzing (which is equivalent to the idealized case of instantaneous turnaround). So you would not expect the traveling twin's elapsed time to be the same. It will still be shorter than the stay-at-home twin's elapsed time when they meet up again, but by a different amount (which will depend on the details of how the twin accelerates and decelerates gradually). The spacetime geometry method adapts easily to this case; it just amounts to evaluating arc length along a curve instead of a pair of straight lines.
The two cases above can both be formulated in flat spacetime. The third, by contrast, requires curved spacetime, because in flat spacetime there is no way for the traveling twin to turn around without feeling acceleration, whereas in this third "slingshot" case, the traveling twin is weightless, feeling no acceleration, the whole way. The spacetime geometry method carries over to this case with no problem at all, since, as I noted, it is completely general.