Re: twin clock problem - SR experts help!

dseppala@xxxxxxxxxxxxx wrote:
> Explanations of problems similar to this one have been posted, but
> when I change the parameters to the problem, I realize that those
> explanations are wrong. So I'm looking for an expert who can explain
> the following SR problem.
>
> Let there be two clocks, Clock A and Clock B, moving along the x-axis.
> They have a relative velocity of 3 meters / second. At time t0, they
> are at the same position in space. At this point in space and time,
> both clocks are set to zero. They now continue to move away from each
> other still traveling at 3 meters / second. According to Einstein's
> theory, observers in Frame A (Clock A's rest frame) measure that Clock
> B is running at a slower rate than identical synchronized clocks that
> are at rest in Frame A. After 10**16 seconds, observers in Frame A
> measure Clock B to be one second slower than Clock A. (The clocks are
> about 10**8 light-seconds apart).
> When Clock A reads 10**16 seconds, let Clock A undergo an
> acceleration over a 0.1 second interval in the direction toward Clock
> B. Likewise, when Clock B reads 10**16 seconds, let Clock B undergo an
> acceleration over a 0.1 second interval towards Clock A. When the two
> accelerations have ended, Clock A and Clock B are now approaching each
> other with a relative velocity of 3 meters / second. Now to clarify
> who's frame the 0.1 second interval was measured over, all frames
> used in this problem measure 0.1 seconds as the acceleration interval
> plus or minus some extremely small delta since the relative velocity
> of 3 meters / second is so slow relative to c. So any variation in the
> measurement of 0.1 second versus the 10**16 seconds has negligible
> effect on the answer to this problem.
> Observers in this new Clock A frame observe that as Clock B is
> approaching Clock A, Clock B is running slower than Clock A
> (Einstein's theory). When Clock A meets Clock B, both clocks
> indicate the same time. This is true because of the symmetry of the
> problem.
> The question is, as measured by Clock A, if Clock B was running
> slower than Clock A for the 10**16 seconds of inertial motion when the
> clocks are moving apart, and Clock B is running slower than Clock A
> for the 10**16 seconds of inertial motion when the clocks are moving
> toward each other, and both clocks were set to zero when they first
> meet, how do they end up showing the same time when they meet again?

During the turnaround A measures B's clock to quickly gain a lot of
seconds. This compensates for the time dilation periods. Keep in mind
that this "measures" means "reading the A-synchronised clocks at the
relevant events along B's trajectory" - it doesn't mean B's clocks
suddenly go bananas because A turned his head.

This is fully analogous to the Euclidean situation with two cars going
from P to Q, one car along PXQ, the other along PYQ, like so:

Q
/ \
/ \
/ \
/ \
/ \
X Y
\ /
\ /
\ /
\ /
\ /
P

Both cars start at the same time and drive at the same speed. Each
driver looks *perpendicularly* out the window in the direction of the
other driver and concludes "he is driving slower than me" (that
corresponds to "his clock is slower") since every foot (say) of one
driver's path corresponds to more than a foot of the other driver's
path when viewed in the perpendicular direction. This holds for both
the outgoing and the incoming parts of their journey.

One may then ask your type of question: how is it possible then that
they meet at the same time? Well, during the turnaround the
perpendicularly stretched out finger will trace a very large and quick
*backward* sweep against the other driver's path. The amount of
distance "lost" that way compensates for the distance gained during the
straight ("inertial") portions of the journey.

Someone might say: "but this makes the notion of simultaneity into a
mere convention". And it's true - simultaneity in relativity is a
convention. It allows one to set up reference frames with particularly
nice properties plus for v << c they correspond to Newtonian frames.
Simultaneity is not really needed for SR (just like algebra of
coordinates is not needed for Euclidean geometry), it's "merely" mighty
convenient because it allows us to quantify things.

> Now some may think something happens to the clocks when they are
> accelerated to to 3 meters / second. However, as is easily
> demonstrated experimentally, when an electronic clock or most
> mechanical clocks change their velocity by 3 meters / second over a
> 0.1 second interval, they do not gain or lose anywhere close to a
> second or two relative to nearby clocks that haven't accelerated.
> Observers in the original rest frame of Clock A near the position of
> Clock A do not measure a substantial change in Clock A as it changes
> its velocity by 3 meters / second. Likewise observers in the original
> rest frame of Clock B in the vicinity of Clock A do not observe any
> substantial change in Clock A as it changes its velocity by 3 meters /
> second. And Observers moving with Clock A through the acceleration
> likewise do not measure any substantial change in the Clock A time
> relative to nearby clocks in the original Frame A and Frame B that
> haven't accelerated.

I think the Euclidean analogy above answers some of your questions.

> Also we can demonstrate logically that Clock A and Clock B do not
> gain or lose sufficient time during the acceleration to make any
> difference in the indicated time when they meet.

It's not that the clocks A and B suddenly "do" something during the
acceleration. What physically happens is that you have two
symmetrically moving clocks so they display the same elapsed times upon
reunion. The whole "time dilation" during the inertial portion of the
trip and the sudden burst of "time contraction" during the accelerated
portion of the trip is an artifact of the observation method called
"synchronised clocks" and "using instantaneously co-moving inertial
frames for periods of acceleration" - ponder that Euclidean analogy
again. Of course different elapsed times for non-symmetrically moving
cloks is a real phenomenon, it's only that the method of keeping track
of the clocks that uses the above coordinate method features those
coordinate-related artifacts.

--
Jan Bielawski

.

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