Twin's Paradox: Who is Older A or B?

[yykcw]

Assume A and B are at rest on Earth initially. Then B travel to to one light year far away from Earth with an extremely low speed (even slower than a car). So I can assume their ages are the same after B has arrived, right? Then, B returns to Earth with an extremely fast speed (~0.9c) and A and B met at one point. Assume B does not stop, at the moment they met, who is older?

 

[Orodruin]

So I can assume their ages are the same after B has arrived, right?

Not right. This depends on which inertial frame you use to describe things. Since they have spatial separation, their relative age is going to depend on the inertial frame. You will find inertial frames where A is younger and you will find inertial frames where B is younger. A big problem many people are facing when learning special relativity is the underlying assumption that there actually is a well defined way of comparing the times - there is not due to the relativity of simultaneity.

Regardless, in your setup, the traveling twin will be younger upon reuniting for the same reasons the traveling twin will be younger in the standard setup twin paradox.

 

[PWiz]

It is as Orodruin has pointed out. When two frames, initially co-located at some point (event, to be precise) A with their clocks synchronized meet again at point B, it is always the clock in the frame that accelerated that reads a lower time interval. This is the concept of maximum proper time (unaccelerated observers measure the greatest time interval).

 

[Nugatory]

B will be younger; this is still a twin-paradox scenario. However, the age difference will be small compared with the total elapsed time, so they may not notice.

It would be a good exercise to calculate about how long it takes B to travel one light-year at a speed that is less than that of a car.

If you haven't already seen the Twin Paradox FAQ take a look at it; the space-time diagram section may be the best way of understanding this particular variation on the problem.

 

[yykcw]

It is as Orodruin has pointed out. When two frames, initially co-located at some point (event, to be precise) A with their clocks synchronized meet again at point B, it is always the clock in the frame that accelerated that reads a lower time interval. This is the concept of maximum proper time (unaccelerated observers measure the greatest time interval).

What I don't understand is that, B should feel A is younger than him at any time during his return trip to earth, so if he does not stop, why at the point they met, to B, A suddenly becomes older?

 

[Orodruin]

What I don't understand is that, B should feel A is younger than him at any time during his back trip to earth, so if he does not stop, why at the point they met, to B, A suddenly becomes older?

No, you are failing to take the most important part of the resolution of the twin paradox into account: relativity of simultaneity.

In Bs returning frame, A is time dilated. However, the event A considers simultaneous with Bs turnaround is not simultaneous with Bs turnaround in Bs returning frame.

The only reason the twin paradox appears as a paradox at all is that people fail to take the relativity of simultaneity into account and think that the turnaround is simultaneous to the same event for A in all frames - this is simply not the case, see my first post in this thread.

 

[PWiz]

What I don't understand is that, B should feel A is younger than him at any time during his return trip to earth, so if he does not stop, why at the point they met, to B, A suddenly becomes older?

It's not as simple. In the first half of the journey (when person B is moving away from person A at constant velocity), both A and B see the other person as younger. This is the relativity of simultaneity, an inescapable consequence of the fact that time is not absolute. (No one is justified in saying who's older [or conversely, both of them are justified in saying that the other one is older] because we can't make a direct comparison until A and B are side by side again.) However, B cannot return to A without changing direction, and this means that B has to accelerate, so the situation is no longer symmetrical (acceleration breaks the symmetry because it means B is instantaneously changing inertial frames until B finally "settles" into a new inertial frame after changing the direction and speed to the required levels). Since B's rest frame is now a different frame compared to what B was initially going in during the first part of the journey, there is a different relation between this new frame and A's rest frame, and there is a disparity in the elapsed time for A and B.

 

[Smattering]

However, B cannot return to A without changing direction, and this means that B has to accelerate, so the situation is no longer symmetrical (acceleration breaks the symmetry because it means B is instantaneously changing inertial frames until B finally "settles" into a new inertial frame after changing the direction and speed to the required levels).

Exactly. The twin paradox is no longer paradox once the twins are reunited at the same location. Then, it becomes completely unambigious how old each of them will be.

 

[Herman Trivilino]

Assume A and B are at rest on Earth initially. Then B travel to to one light year far away from Earth with an extremely low speed (even slower than a car). So I can assume their ages are the same after B has arrived, right?

How about a speed of 3 m/s? Is that slow enough? My car usually has speeds greater than that. At that rate it would take 100 million years to get there, and each twin will measure the other's clock as having lost a small fraction of a second compared to his. So A will be 100 million years older, and B will be 100 million years older, give or take a couple tenths of a second, or so.

Then, B returns to Earth with an extremely fast speed (~0.9c) and A and B met at one point.

Imagine a spaceship moving towards Earth at a speed of 0.9c. B's plan is to climb aboard just as it passes him to catch a ride home. Ignoring the time it takes him to get up to speed and climb aboard, B's clock will have advanced about 0.48 years by the time he arrives to meet A. A's clock will advance about 1.11 years during that journey. So, B will be younger by the difference, which is less than 8 months out of the 100 million years.

It's easy to reckon this difference in aging from A's frame of reference. Because he's an inertial observer time dilation alone will do the job. It's a bit harder from B's frame of reference because there are two competing effects. One is time dilation during the journey home, the other is the relativity of simultaneity due to the shift in his frame of reference when he jumped aboard the spaceship to begin his journey home.

It really is worth the while, and kind of gratifying, to do both calculations and see that you get the same answer both ways. It seems Einstein got it right after all!