Twin Paradox Revisited?

Question

Once again, look at Alexander's question and the follow up question:

http://answers.yahoo.com/question/index;_ylt=ArVMRKyqf5weyNEklQHMoXDsy6IX?qid=20070515120146AA2qnnG

Now, we have twins A and B, both with wristwatches, set at noon at the time twin B departs. Twin A remains stationary. Twin B travels at speed (1/2)c for 1 hour, and returns at speed (1/2)c for 1 hour, per twin A's watch, so that A's watch now shows 2 pm. Both A and B have telescopes trained on each other's wristwatches. What is the function of times B(Ta) and A(Tb), where B is B's time as seen by A in A's time Ta, and A is A's time as seen by B in B's time Tb?

I wish the answer to this would be explained more often to those totally puzzled by the Twin Paradox. Are functions B(Ta) and A(Tb) symmetrical?

How wierd, Y!A has deleted my previous question because of "content"! Does anyone have a clue what's going on?

U98, sorry about that, but Y!A has decided that my previous question, even after couple of highly intelligent answers were given, was a menace to the public, so they deleted it! I will have to repost the question.

Bekki, sorry about that, I had liked both yours and Alexander's answers, and I was using it to warm up to this one.

To make this problem numerically easier, you can use v = (3/4)c, so that the gamma factor is (5/4).

Bekki, that's right, the functions are "symmetric" only that the clock rates observed by A and B are same both in red and blue shift, but while for B the "turnaround time" is at his own halftime, for A it comes much later. Moreover, while B will come to a point where he sees A's clock being in agreement with B's own, A never sees such at thing.

Answer 1

Boooo! I tried to refer to your previous question, in which I at least partly figured this out, and they erased it. Yahoo answers is glitchy at the moment. I'll answer in a sec if I get a chance.

Up to turnaround, it goes like this. At time t, A looks at B's watch. But he sees the watch at a time T in the past (t-T) at a position v(t-T). The time of light travel, T, is just the position over c.

T = v (t-T)/c, so T = vt / (c + v), so the observed time is given by
(t-T) gamma = gamma t / (1 + v/c).

So the time is slowed by relativistic time dilation (gamma) and a Doppler redshift factor.

This is how it looks until (t-T) gamma = 1 hour for A and until t = 1 hour for B. Then B turns around. Now the time dilation is the same, but the redshift turns into a blueshift, so at highly relativistic speeds, the other guys clock seems to be moving faster if you don't correct for light travel time.

So on the way there, the functions are symmetric. On the way back, they are somewhat symmetric (ie, run at the same rate). The difference between A&B is that the travelling twin thinks he turned around after an hour. The stationary twin thinks he spent longer travelling before turn around because of time dilation and the redshift. So ultimately, more time passes for the stationary twin.

Answer 2

your link is done.