A:
Hi Colin- This is a form of the 'twin paradox', which isn't really
a paradox. In fact, it's a terrific way of seeing how one can start
with the idea that the speed of light is the same constantin all frames
and come up with truly surprising consequences, such as that twins with
different travel itineraries will meet up again having aged differently
when they get back together. We already have some answers to questions
about the twin paradox on this web site, and so this answer will be a
different way to approach the situation.
A consequence of the speed of light being the same in all reference
frames is that for every pair of events, (t1,x1) and (t2,x2) (I'll only
worry about one dimension, along the x-axis), the quantity
tau = sqrt( (t1-t2)^2 - ((x1-x2)/c)^2 )
is the same in all reference frames, where c is the speed of light.
If an object stays put, that is, x1=x2, then tau is just t1-t2, and
corresponds to the amount of time elapsed on a clock on the object
between the two events. So if a clock reads zero time at event 1, it
will read tau time at event 2. This is true in all reference frames,
because specifying a unique location and time at which to read the
clock determines what number it will be showing. Note that this tau is
for the clock which goes from event 1 to event 2 without accelerating
-- if a traveler does not accelerate in one inertial reference frame he
accelerates in no other inertial reference frame.
If a traveler is accelerating, then one has to break up the journey
into infinitely small pieces and add up the contribution to the total
tau sum by integration.
Now there are three important events in the twin paradox situation.
1: Twin B departs A's location.
2: Twin B turns around (assume this happens quickly)
3: Twin B is reunited with twin A.
For simplicity, I'll assume twin B travels at a constant velocity
except for very short blast-off and turnaround accelerating periods.
Event 2 takes place at a faraway place, say a distance X in twin
A's frame, and at a time T. In Twin A's frame, these events are:
1: (t=0,x=0)
2: (t=T,x=X)
3: (t=2T,x=0)
Twin B doesn't have just one frame (he accelerates while turning
around -- he spends most of his time in either the outgoing frame or
the incoming one). Velocity is relative but acceleration isn't. You can
measure acceleration with an experiment which doesn't depend on the
environment.
The amount of time accumulated on Twin A's clock from event 1 to
event 3 is 2T, and the amount of time accumulated on Twin B's clock
from event 1 to event 3 is 2*(T^2-(X/c)^2), which is less. Since both
twins are at events 1 and 3 in all reference frames, this difference
between their clock readings can be compared in all reference frames --
twin B will be younger.
In your question you asked "what if twin A and twin B enter the
same reference frame?" The answer to this depends on whether twin A and
twin B have met up again. If they are not in the same place at the same
time, then their relative ages depends on the reference frame in which
they are viewed, unfortunately. Either A or B could be younger if they
are still apart, and different outside observers would come to
different conclusions. One consequence of the idea that the speed of
light is the same in all reference frames is that the idea of events
being "simultaneous" no longer holds. Events that happen at the same
times in different places in one reference frame happen at different
times in other reference frames.
Tom J.
(published on 10/22/2007)
