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I've got a question concerning the special theory of relativity. Once again twins paradox, but somewhat more complicated. That is one-dimensional problem and gravity is neglected. Acceleration/deceleration are present but not always mentioned.

Suppose, before take off of a spaceship from Earth, it was loaded with three containers X, Y, Z of freshly made isotope. Concentration of decayed atoms are x0 = y0 = z0 = 0. When the spaceship reaches constant velocity V relative to Earth, concentrations of decayed atoms in containers are equal, x1 = y1= z1. Now, the spaceship is an inertial system of coordinates, and we use it as a reference point.

Container X stays on the spaceship, while containers Y and Z are immediately loaded into small identical rockets. Container Y is sent forward with velocity V relative to the spaceship. Relative to Earth it reaches constant velocity V < Uy < 2V. It moves in this state for a time Ty in its own reference frame. This time is arbitrary larger than the acceleration time.

Container Z is sent backward with velocity V relative to a spaceship. Relative to Earth it reaches constant velocity Uz = 0 and stays like this for a time Tz = Ty in its own reference frame.

Then, containers Y and Z are returned on the spaceship. Because relative to spaceship both containers performed equivalent journey, concentrations of decayed atoms should be now y2 = z2 < x2. This relation should hold from this moment on.

But when spaceship returns to Earth, this relation is hard to to explain. Because relative to Earth containers X, Y and Z performed partially different journeys. Barring common parts and accelerations and using now Earth time, the round trip time Tr is the same for all three containers. But container Z spends most of this time with zero velocity Uz, container X spends most of this time with velocity V, while container Y spends most of the time with velocity Uy > V. Accordingly, upon return to Earth concentrations of decayed atoms should be y3 < x3 < z3.

What am I missing?

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  • $\begingroup$ How do you propose setting $T_z = T_y$? Frame z doesn't measure $T_y$ from its own frame. $\endgroup$
    – Triatticus
    Commented May 19, 2021 at 14:34
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    $\begingroup$ Have you drawn a spacetime diagram in the reference frame of X? Once you have that, transform it into a spacetime diagram in Earth’s reference frame, and you’ll see where the cancellation happens. $\endgroup$
    – rob
    Commented May 19, 2021 at 14:45
  • $\begingroup$ While you are right about $X$ container, your conclusion about $Y$ and $Z$ are dubious. In order for them to return to the spaceship $Y$ would have to travel slower and $Z$ faster than $X$ $\endgroup$
    – nwolijin
    Commented May 19, 2021 at 15:02
  • $\begingroup$ All of the "twin paradox" type "paradoxes" are the result of failure to apply the Lorentz transformation correctly (or failure to even apply the Lorentz transformation, using some pop science explanation.) It is impossible in SR for an event to happen in one inertial frame but a contradictory event happen in another inertial frame. It just can't happen. The different frames merely provide different coordinates in space and time for when the same event happens. People need to give up on these "twin paradoxes" because all they show is a lack of understanding of relativity. $\endgroup$ Commented May 19, 2021 at 15:48
  • $\begingroup$ @Triatticus Yes, you are right. But both rockets have clocks. They start their journey back after prearranged time. It is set as Ty for one rocket and Tz = Ty for the other $\endgroup$
    – user36225
    Commented May 19, 2021 at 16:25

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The first point I should make is that the effects in question will be the same whether you use decaying isotopes, clocks, or any other mechanism or phenomenon that happens with a certain time-cycle. What you are asking, is whether time will have passed at a different rate for X, Y and Z.

You are right in pointing out that when Z sets off backwards from X it is stationary relative to the Earth, while Y spends that period moving away at twice the speed of X, ie 2V.

What you overlook, however, is that for Z to return to X it must reverse its motion and go for a time at twice the speed of X, ie 2V. Y on the other hand, will reverse its motion to return to X, during which period it too will be stationary relative to the Earth. So you see that, contrary to your initial impression, relative to the Earth, Z and Y each spend exactly the same amount of time stationary and moving at 2V, they just do so on opposite legs of their journeys.

As a result, relative to Earth time passes at the same overall rate for Z and Y, which is what X observes too.

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  • $\begingroup$ You are completely right. I definitely overlooked return paths to the spaceship. I figured it out some time ago, but, unfortunately, after posting a question. Thank you $\endgroup$
    – user36225
    Commented May 20, 2021 at 0:39
  • $\begingroup$ No problem. It took me a while to figure it out too! $\endgroup$ Commented May 20, 2021 at 6:31

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