Notes on Relativity and Cosmology - Physics Department, UCSB

4.2. THE TWIN PARADOX 93
But, you might say, Gaston is almost always in an inertial reference frame. He
is in one inertial frame on the trip out, and he is in another inertial frame on the
trip back. What happens if we just put these two frames of reference together?
Let’s do this, but we must do it carefully since we are now treading new ground.
First, we should draw in Gaston’s lines of simultaneity on Alphonse’s spacetime
diagram above. His lines of simultaneity will match 2 simultaneity in one inertial
frame during the trip out, but they will match those of a different frame during
the trip back. Then, I will use those lines of simultaneity to help me draw a
diagram in Gaston’s not-quite-inertial frame of reference, much as we have done
in the past in going from one inertial frame to another.
Since Gaston is in a different inertial reference frame on the way out than on the
way back, I will have to draw two sets of lines of simultaneity and each set will
have a different slope. Now, two lines with different slopes must intersect......
t = 6 yrs.
G
G
a
s
t
o
n
Alphonse
+
t = 3 yrs.
G
t A =5
D
C
E
t = 3 −
G yrs. t = 0
G
t =0 A
B
A
x =0
A
Alpha
Centauri
x A
=4
Here, I have marked several interesting events on the diagram, and I have also
labeled the lines of simultaneity with Gaston’s proper time at the events where
he crosses those lines. Note that there are two lines of simultaneity marked
t G = 3years!. I have marked one of these 3 − (which is “just before” Gaston
turns around) and I have marked one 3 + (which is “just after” Gaston turns
around).
If I simply knit together Gaston’s lines of simultaneity and copy the events from
the diagram above, I get the following diagram in Gaston’s frame of reference.
Note that it is safe to use the standard length contraction result to find that
in the inertial frame of Gaston on his trip out and in the inertial frame of
Gaston on his way back the distance between Alphonse and Alpha Centauri
is 4Lyr √ 1 − (4/5) 2 = (12/5)Lyr.
2 Strictly speaking, we have defined lines of simultaneity only for observers who remain
inertial for all time. However, for an observer following a segment of an inertial worldline,
it is natural to introduce lines of simultaneity which match the lines of simultaneity in the
corresponding fully inertial frame.