Time dilation in accelerating reference frames

[PeterDonis]

Why is B: x= 0?

Because event B, where the two twins meet up again, must be at $x = 0$ since the stay at home twin is at rest at $x = 0$ in this frame.

 

[Maxila]

Because event B, where the two twins meet up again, must be at $x = 0$ since the stay at home twin is at rest at $x = 0$ in this frame.

Both twins have traveled in scenario discussed above (one from A to B, the other from A to B to C), unlike the "classic twin paradox" where one twin stays on the Earth?

 

[PeterDonis]

Both twins have traveled in scenario discussed above (one from A to B, the other from A to B to C), unlike the "classic twin paradox" where one twin stays on the Earth?

My scenario is the classic twin paradox.

Both twins have "traveled" through spacetime, because it's impossible to "stand still" in spacetime; you're always moving into the future, whatever else you do.

Whether either twin is traveling through space depends on what coordinates you adopt. In the coordinates I used, the stay at home twin does not travel through space; he stays at $x = 0$ always. So he travels through spacetime from point A (the event where the twins separate) to point B (the event where the twins come back together), but he doesn't travel through space in the coordinates I used. The traveling twin, OTOH, travels through spacetime from point A to point C (where he turns around) to point B. (Note carefully the order of the points: he can't travel through spacetime from A to B to C, because the leg from B to C would be going backwards in time. He must travel from A to C to B.) In doing so, in the coordinates I used, he also travels through space, from where the stay at home twin is, to the turnaround point, back to the stay at home twin.

 

[Maxila]

Thank you for being patient in explaining how you arrived at the conclusion. However it is exactly why I am glad the Foundation Questions Institute (FQXi) is sponsoring an essay contest on "Physics and Mathematics" http://fqxi.org/community/forum/category/31424 I am a firm believer that Mathematics is by far the best form of communication we can use to explain physical phenomena because it is the most precise form of communication we have. Yet too many have faith akin to faith in a bible that when a model can accurately predict empirical outcomes all aspects of it, no mater how absurd must true of nature. As a person of science I keep an open mind to such possibilities but only when all possible more rational explanations have been exhausted.

 

[PeterDonis]

too many have faith akin to faith in a bible that when a model can accurately predict empirical outcomes all aspects of it, no mater how absurd must true of nature

I'm not sure how this is relevant here, since the only thing we have been saying is "true of nature" is the different proper times for the two twins. We have plenty of empirical evidence that that actually happens.

 

[Maxila]

Yes, time dilation and length contraction are proven phenomena; however if I were at position B, at rest with the twins before they start the journey from A, and we synchronized our clocks, when they both arrive at B and we compare clocks we would conclude (same assumptions of identical speeds, acceleration, and a non degenerate triangle apply):

1. I measured more time for the twin that went from A to C to B
2. Both twins would also agree more time passed for the twin that took A to C to B

While they all would disagree exactly how much more time A to B to C takes relative from A to B (due to time dilation), they would all be in agreement A to C to B was more time (and length) than the twin that went directly from A to B. Only mathematically (under the scenario specified) could you conclude A to B to C takes less time than A to B, again I have no doubt in the existence of time dilation or that all three observers would disagree on exactly how much more time. I am quite certain (for the scenario outlined) there would be no way empirically, to conclude the A to B to C trip takes less time. (We both know if the twin traveling from A to B to C traveled at a different rate of some higher speed then they could possibly conclude their trip time, and distance traveled was less)

 

[PeterDonis]

if I were at position B

B is not a position in space; it's a point in spacetime. You can't be "at rest" at a point in spacetime. Same for A and C; they are points in spacetime, not space.

Also, the lengths of the sides of the triangle in spacetime (AB, AC, and CB) are invariants; they're the same for all observers. Two observers in relative motion cannot measure different lengths for those sides. (All three sides are timelike, so these "lengths" are actually the elapsed proper times for observers following those paths in spacetime.)

 

[Maxila]

B is not a position in space; it's a point in spacetime. You can't be "at rest" at a point in spacetime. Same for A and C; they are points in spacetime, not space.

Yet empirically I can be at rest relative to the two twins.

Two observers in relative motion cannot measure different lengths for those sides.

As outlined they are rest together (not in relative motion) at B when they compare the clocks and distance traveled.

This is a good point for me to bow out of this thread as the conversation is no longer constructive.

 

[PeterDonis]

empirically I can be at rest relative to the two twins.

Once they come back together and are at rest relative to each other, yes. But this is not being "at position B". Being "at position B" would be like "being at noon on Tuesday"--it's not possible, you can't just stop yourself at "noon on Tuesday" and stay there.

As outlined they are rest together (not in relative motion) at B when they compare the clocks and distance traveled.

And when they do that, they will all find that the traveling twin, the one that took the path from A to C to B through spacetime, had less elapsed time on his clock than the twin that took the path straight from A to B through spacetime. That is what my computation of the sides of the triangle, which showed that AB > AC + CB, is telling you. What's more, they will all find the same difference in clock readings between the two twins, because the side lengths AB, AC, and CB are geometric invariants.

 

[PeterDonis]

empirically I can be at rest relative to the two twins.

Just to clarify this some more: if you are at rest relative to the two twins once they have come back together, and you have not done anything to change your state of motion (no firing rockets, using a solar sail, etc.), then you will have been at rest relative to the stay-at-home twin during the whole experiment. In other words, you, just like the stay-at-home twin, will have taken the path from A directly to B through spacetime (whereas the traveling twin takes the path A to C to B). So your calculation of the side lengths AB, AC, CB of the triangle in spacetime will be exactly the one I previously posted, which shows AB > AC + CB. Which means that you will certainly not conclude this:

1. I measured more time for the twin that went from A to C to B

Furthermore, this claim is also false:

2. Both twins would also agree more time passed for the twin that took A to C to B

The stay-at-home twin will obviously not conclude that the A to C to B path through spacetime is longer, since he is at rest relative to you for the entire experiment, so his calculation, like yours, is exactly the one I previously posted. But the traveling twin will also not conclude that the A to C to B path through spacetime (i.e., his own path) is longer, because, as I previously said, the side lengths AB, AC, and CB are geometric invariants, so all observers must agree on them.

The difficulty in actually modeling the calculation for the traveling twin is that he is not at rest in a single inertial frame the whole time. If we want to use inertial frames, we can use either the one in which the traveling twin is at rest outbound, or the one in which he is at rest inbound. If we want to use coordinates in which the traveling twin is at rest for the entire trip, we have to construct a non-inertial coordinate chart with that property, and there is no unique way to do that, so there are multiple possible calculations we could do this way. I won't go into any of those in detail unless you want me to.