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Relative aging without acceleration


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This posting presents a scenario with different relative aging absent acceleration (it does not include any change of magnitude or change of direction).


Three women are in a row along the axis of motion in the order C, A, B.


A and B are at rest with respect to each other on space stations 1LY apart. A and B agree on simultaneity and have synchronized their clocks. C is on a spaceship. C and AB are in inertial relative motion at 0.9c. All three women become pregnant (thus each child is inertial from conception).


A and B give birth simultaneously in the AB frame, and each sends a message to the other at the speed of light identifying the time of birth. By coincidence, C passes by A and gives birth at the same time that A does (both A and C have the same coordinate on the axis of motion and therefore agree on the simultaneity of the births).


C and her child continue toward B, in inertial relative motion. According to special relativity, B's child ages more slowly in C's frame than C's child. However, when C passes B's space station, B's child is older than C's (and A and B agree that A's child is also older than C's, because A and B's children were born simultaneously in the AB frame and have remained at rest with respect to each other). How can B's child age more slowly, yet end up older? The relativity of simultaneity.


A and B give birth simultaneously in their own frame. A and C give birth simultaneously in both frames. However, A and B do not give birth simultaneously in C's frame. In C's frame, her child is born at the same time that B's child is 0.9 years old. The rear of the approaching AB frame presents in C's frame older by the product of the proper distance in the AB frame (1LY) and the relative velocity (0.9c). As C and her child continue in relative motion along the length contracted distance from A to B (only 0.44LY), B's child ages more slowly than C's child. However, this slower marginal aging over the contracted distance is not enough to overcome the 0.9 years of additional age at inception, so B's child is older than C's child when they pass.


You can elaborate on this by considering a string of clocks between A and B, each at rest with respect to A and B, all synchronized in the AB frame. In the C frame, each successively distant clock is slightly more ahead (at the product of its distance from A in the AB frame and 0.9). Each individual clock runs slower in the C frame, as special relativity states. However, if C observes each clock in the series as she passes it, time in the AB frame runs faster by the inverse of the usual time dilation. Think of someone taking an express train across country and observing station clocks as he passes through (which is more practical than trying to look back at the clock of his departure station).


Several authors have pointed out the initial clock advancement in the C frame as described above. See for example Fowler at http://galileo.phys.virginia.edu/classes/252/lorentztrans.html, and Gollin at http://web.hep.uiuc.edu/home/g-gollin/relativity/p112_relativity_9.html.


McHugh has expressly explained the twin paradox using this analysis, at http://mathforum.org/kb/message.jspa?messageID=6688895.

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This is a fairly straightforward example of SR and the Lorentz transformations with inertial motion.


I think that using "age from birth" is useful for understanding a concrete example of the meaning of time in a real world situation. It also works the same if you use any clock and any initial clock settings (a baby ages one year per year from birth, but so does a 90-year-old, and an atomic clock measures 1s per sec and a container of yogurt grows 1 year worth of mold per year, etc). I think choosing the most absolute concrete example is useful, but then also considering the most abstract too can help prevent attaching too much meaning to absolute clock times.

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