The two clocks really haven't been through equivalent histories.
Neither of them is stationary in an inertial reference frame,
unfortunately, as the Earth turns on its axis once a day. We've
answered questions on the twin paradox already, explaining how
different travel itineraries for clocks can result in their reading
different amounts of elapsed time when they are brought together.
Special relativity tells us that comparing the time on two clocks which
are not brought together at the same place produces ambiguous results
-- the difference in time read by two clocks which are separated by
some distance depends on the speed of the observer measuring the
difference. But you framed the problem correctly, asking about clocks
which are compared at the same location after different journeys.
The amount of time accumulated on a clock on a journey from one
location in spacetime to another is called the "proper time" for the
journey, and it is a "Lorentz invariant". That is, when a clock travels
from point A to point B (and by "point" I mean also that the time is
specified) along a particular path, then the amount of time accumulated
on the clock during the journey is independent of the speed of an
observer looking at it. The clock can be equipped with a paper stamper
and write the time on arrival -- say, 3:00 PM, and this piece of paper
has to read the same number in all frames of reference.
It turns out the Lorentz-invariant proper time in Special
Relativity (which is all that's needed here) is dtau^2 =
dt^2-dx^2/c^2-dy^2/c^2-dz^2/c^2, where c is the speed of light and dx,
dy, dz are the displacements in three-dimensional space corresponding
to a bit of the path the clock takes on its trip during a time interval
dt. Take the square root of both sides and integrate this quantity up
over the path, and you've got the amount of time accumulated on the
clock along that path (gravity messes this up, but we're all at the
same gravitational potential, more or less, and the special
relativistic effect is the biggie here). You have to do this separately
for both clocks, and the miraculous thing is that the answer you get
doesn't depend on the frame of reference you use, as long as it's
inertial (Earth's spinning frame of reference isn't going to work), and
you use the proper Lorentz transformations between frames.
Because the proper time added up is less for something that's
moving, the clock that's moving (and accelerating, because you cannot
move and come back to the same place without turning around somehow)
more will read less time when reunited. So if your aeroplane takes off
and flies eastward around the globe, it will be moving more than the
stay-at-home clock, which is also going in an eastward circle, but at a
slower speed. An eastbound clock will read less time than the
stay-at-home clock. A westbound clock, however, travels less than the
stay-at-home clock (unless it's going so fast as to go around the globe
twice in one day, in which case it is going faster), and will read more
time when reunited.
People doing the experiment for real should probably fly planes
around both ways, with several clocks on each and on the ground, to
control the unavoidable uncertainties in the measurement.
Incidentally, satellites, which are far above the surface of the
ground, have general-relativistic effects to be concerned about too.
For a Global Positioning System (GPS) satellite, its motion around the
Earth slows down its clock, but the fact it's up higher in Earth's
gravitational field speeds it up. The second factor, the speedup part,
is actually bigger than the special relativistic time dilation factor.
GPS receivers account for this in their built-in electronics and
software. The general-relativistic effect should be smaller for an
airplane flying in the atmosphere, but airplanes do in fact fly about
five miles up in the sky. Maybe if the airplane is really really really
slowly (say, do this in a balloon) then the general relativistic effect
can be bigger than the special relativistic effect.
Tom
(published on 10/22/2007)
