4.1.1 Viewing it with a Space-Time Diagram (Twin Paradox)
Description
This article is from the Relativity and FTL Travel FAQ, by Jason W. Hinson jason@physicsguy.com with numerous contributions by others.
4.1.1 Viewing it with a Space-Time Diagram (Twin Paradox)
So, now we will try to understand the twin paradox by using our old
friend, the space-time diagram. To do this, we have to decide on some
specifics. First, we will say that the relative motion of Sam and Ed is 0.6
c. So, after Sam has accelerated to a constant speed, he will be traveling
at 0.6 c with respect to the Ed. (Of course, in Sam's frame, it is Ed who is
moving at a speed of 0.6 c away from Sam.) Next, we need to decide who will
be the one who eventually accelerates to go and meet with the other twin. In
our case, we will look at the situation where Sam turns around to go back
and meet with Ed. Finally, I should mention that the accelerations we will
be using will be "instantaneous" accelerations. This means that they take no
time to accomplish. In the real world, it would (of course) take time to
accelerate, and while this would make the space-time diagrams look
differently, the basic ideas we will discuss still hold.
Now we look at the space-time diagrams. In Diagrams 4-1 and 4-2 below,
I have drawn the whole trip in two parts. In Diagram 4-1, you see Sam headed
away from Ed, and in Diagram 4-2, you see Sam after he has turned around and
is headed back to Ed.
Diagram 4-1 Diagram 4-2
t ! t
^ ! ^
| ! |
| ! t=10 + t'=8
| ! |*
| ! | *
| ! | *
| t' ! | *
| / (t'= 4 line)! | *
| / / ! t=6.8\ *
| / ! | *
| / / ! | \ *
| / ! | *
t=5+ - - - - * - - -> (t = 5 line) ! t=5+ - - - - * - - -> (t = 5
| * ! | \ line)
| / * ! | \ \
| * ! | \
t=3.2/ * ! | \ \
| * ! | \ (t'= 4
/ | * ! | \ line)
| * ! | \
| * ! | t'
|* ! |
-----o------------------> x ! -----o------------------->x
| ! |
| ! |
(NOTE: Again, to make the ASCII diagram easy to draw, the positions drawn
for the "moving" observer more closely represents 0.5 c than 0.6 c for his
speed. I have had to try and approximate the lines of constant time for t'
accordingly in order to show how the paradox is solved.)
Now, to explain the diagrams: Ed (the twin on Earth) is represented by
the x and t axes while x' and t' denote the coordinate system for Sam. Sam's
motion through space-time is represented by the blue line marked t', as
usual. Now, at the origin, Sam instantaneously accelerates to the speed of
0.6 c. He then proceeds away from Ed until Sam sees that his own clock read
4 years (just to pick some unit of time--which means that the distances
would be in light-years). When Sam sees his own clock read 4, he turns
around with an instantaneous acceleration. At that point, we switch to
Diagram 4-2. In that diagram, Sam heads back to Ed.
4.1.2 Explaining the "First Part"
Now let's concentrate on the first of the two diagrams. Just before Sam
turns around, his clock reads 4 years. At that point I have drawn two lines
of constant time (or lines of simultaneity)--one for each observer. The line
parallel to the x axes is (of course) the line of simultaneity for Ed which
passes through the event "Sam's clock reads 4 years". Note that this line of
simultaneity for Ed also passes through the event "Ed's clock reads 5
years". Therefore, in Ed's frame of reference, the events "Sam's clock reads
4 years" and "Ed's clock reads 5 years" are simultaneous events. This
diagram thus explains how in Ed's frame of reference, Sam's clock is running
slower than Ed's by a factor of 0.8 (that's one over gamma when v = 0.6 c).
However, the line of simultaneity we were looking at is not a line of
simultaneity for Sam. Sam's line of simultaneity which passes through the
event "Sam's clock reads 4 years" is the one marked "t'= 4 line". This line
also passes through the event "Ed's clock reads 3.2 years". Therefore, in
Sam's frame of reference the events "Sam's clock reads 4 years" and "Ed's
clock reads 3.2 years" are the simultaneous events. This diagram thus
explains how in Sam's frame of reference, Ed's clock is running slower than
Sam's by a factor of 0.8.
So, the idea that they each believe the other person's clock is running
slowly can be explained. We see that it is, indeed, a question of which
frame of reference you are in, because different events are simultaneous in
different frames. It is interesting to note that this situation only seems
paradoxical in the first place because we are not use to the fact that
simultaneity isn't absolute. In everyday life, we get the idea that when two
events happen at the same time, then that is an absolute fact. However,
relativity shows us that this is not the case, and once we realize that, we
can understand how each observer can believe the other observer's clock is
running slowly.
With this "first part" of the paradox solved, we must now move to the
second part and ask this question: "how do we explain what happens when the
twins come back together?"
4.1.3 Explaining the "Second Part"
In Diagram 4-2 Sam has seen his own clock read 4 years, and he then
instantaneously accelerated to head back towards Ed. Right after the
acceleration, Sam's clock still basically reads 4 years. Note, however, that
Sam's frame of reference has changed. The inertial frame he was in before he
turned around is different from his inertial frame after he turned around. I
have thus drawn his new time line and a line of simultaneity (one which
passes through the event "Sam's clock reads 4 years") for his new frame of
reference.
Once again we will look at the simultaneous events in Ed's frame and in
Sam's (new) frame. Since Ed hasn't accelerated, he has remained an inertial
observer, and his frame of reference hasn't changed. Thus, in his frame the
events "Ed's clock reads 5 years" and "Sam's clock reads 4 years" are still
simultaneous. However, Sam is in a new frame of reference, and in this frame
the events "Ed's clock reads 6.8 years" and "Sam's clock reads 4 years" are
the simultaneous events.
So, each observer has his own explanation for the final outcome of the
situation. For Ed, Sam's clock is ticking slowly before the turn-around,
nothing significant happens when Sam turns around, and Sam's clock continues
to tick slowly after the turn-around (because he is still moving at 0.6 c
with respect to Ed). That is how Ed explains why he has aged 10 years and
Sam has only aged 8 years when they get back together at the end of Sam's
trip.
However, for Sam, the explanation is different. Before the turn-around,
Sam is in a frame of reference in which Ed's clock has been ticking slow,
and it has ticked 3.2 years while Sam's clock has ticked 4 years. After the
turn-around, Sam is in a frame in which Ed's clock (though it is still
ticking slowly) has already ticked 6.8 years while Sam's clock still reads
only 4 years have passed. Note that since Ed's clock is still running slowly
in Sam's new frame of reference, it will still only tick another 3.2 years
(in Sam's frame) during the last half of the trip, while Sam's clock ticks
another 4 years. However, since in Sam's new frame, Ed's clock has already
ticked 6.8 years, the additional 3.2 years will make a total of 10 years of
ticks for Ed's clock. Meanwhile, Sam has seen his own clock tick a total of
only 8 years.
And there you have it. Each observer agrees (as it must be) that when
the two are back together again, Ed will have aged a total of 10 years while
Sam has only aged a total of 8 years. They each have completely different
ways of explaining how this happened, but in the end, they each agree on the
final outcome.
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