The
Twins Paradox

**Introduction**:
Terry and Stella cross each other traveling at a uniform speed of 99% the speed
of light w.r.t. each other, then one of them accelerates very rapidly to cruise
back toward the other at a uniform speed of 99% the speed of light.

**Question**: Does
relative aging happen during the way out (more precisely, will the entries in
each traveler’s diary on the day of the turnaround have a different date
depending on who is turning around), or does only time during travel after an
acceleration that can make the two reference frames not symmetric count toward
relative aging?

**Answer**: Relative
aging does happen during the way out.

**Question**: How
come the relative aging between the twins during the way out, while each of
their reference frames is equivalent and the situation is symmetrical, appears
to depend on what will happen **after**
the way out, namely on which of the two will boost his rockets to accelerate
toward the other one?

**Answer**: Have case
A be that Stella is the one who accelerates to return to Terry, and case B be
that in which Terry accelerates to return to Stella. Now take the ‘way out’,
i.e. the time before the rockets are turned on, to be the same for Terry in
both case A & B: 7 years time of Terry’s watch. The key to understand why
the aging of Stella is different in the two cases during those 7 years Terry’s
time is when the decision to start the rocket boosters is done, and the key to
that is who decides it. In case A (also the one in http://math.ucr.edu/home/baez/physics/twin_paradox.html
), Stella turns back after 1 yr *her time* since she left Terry --Terry
calculates her turnaround to be 7 yrs after they met, *his* time. In case B,
Terry decides to start his rocket 7 yrs his time after the first meeting; for
Stella, Terry's clocks go slower from her viewpoint and so his rocket started
at 49 yrs *her time* after the first meeting. So indeed we realize that in
order to have Terry’s aging at the time of the turnaround be the same in both
cases, we had to have the time elapsed for Stella be different depending on
which case, and that this is due to the fact that Stella sees Terry’s clock go
slower and Terry sees Stella’s clock go slower: in case A Stella does not use
Terry’s watch at all to determine when to turn around, but in case B Stella’s
time at the time of Terry’s turnaround depends on Terry’s time: it will not
happen until her perception of when Terry’s time marks 7 years since the first
crossing. This is so counterintuitive that it is worth restating it in yet a
different way: simultaneity
in special relativity is not a symmetrical function: Stella’s time when an event
happens in Terry’s worldline on year 7 according to Terry’s calendar is not the
same as Stella’s time when an event happens in Stella’s worldline such that
Terry’s calendar marks year 7 when it happens. So we see that indeed the
way out does count toward the relative aging, but there is no conflict with
causality. Even though in both cases the rocket is turned on simultaneously in
Terry's time, relativity of simultaneity dictates as shown above that in case
A, the rocket is turned on after 1 yr Stella's time, while in case B, the
rocket is turned on after 49 yrs Stella's time. That's why Stella's aging
during the way out is different depending on who turns the rocket out: because
the rocket is not turned on at the same time her time (or the accelerator's
time, either) in both cases.

**Question**: Given
that while the twins travel at uniform speed both of their reference frames are
equally valid and the situation is symmetrical so neither can be said to age
faster than the other until one accelerates: Does the relative aging depend on
how long the twins are traveling away from each other, or just on the portion
of the acceleration?

**Answer**:
Interestingly, the relative aging increases with how long the twins travel away
from each other; in fact if the acceleration happens fast enough (no limit on
how fast imposed by relativity), most of the relative aging happens during the
travel at constant speed during the way out and the way back. Sources: Modern
Physics; http://math.ucr.edu/home/baez/physics/twin_paradox.html
. As explained in the previous answer, the asymmetry lies in the fact that
given that the turnaround is located on the worldline of one traveler and not
the other, the time of the turnaround will be different for each traveler, and
so in essence what happens is that one traveler travels on that first ‘way-out’
leg of the trip longer than the other one does.

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