Notes on Relativity and Cosmology - Physics Department, UCSB

94 CHAPTER 4. MINKOWSKIAN GEOMETRY
x = -12/5 Lyr.
G
x = 0
G
D
t = 6 yr.
G
C
B
t = 3 yr.
G
A
t = 0
G
There are a couple of weird things here. For example, what happened to event
E? In fact, what happened to all of the events between B and C? By the way,
how old is Alphonse at event B? In Gaston’s frame of reference (which is inertial
before t G = 3, so we can safely calculate things that are confined to this region of
time), Alphonse has traveled (12/5)Lyr. in three years. So, Alphonse must experience
a proper time of √ 3 2 − (12/5) 2 = √ 9 − 144/25 = √ 81/25 = (9/5)years.
Similarly, Alphonse experiences (9/5)years between events C and D. This means
that there are 10 − 18/5 = (32/5)years of Alphonse’s life missing from the diagram.
(Oooops!)
It turns out that part of our problem is the sharp corner in Gaston’s worldline.
The corner means that Gaston’s acceleration is infinite there, since he changes
velocity in zero time. Let’s smooth it out a little and see what happens.
Suppose that Gaston still turns around quickly, but not so quickly that we
cannot see this process on the diagram. If the turn-around is short, this should
not change any of our proper times very much (proper time is a continuous
function of the curve!!!), so Gaston will still experience roughly 6 years over the
whole trip, and roughly 3 years over half. Let’s say that he begins to slow down
(and therefore ceases to be inertial) after 2.9 years so that after 3 years he is
momentarily at rest with respect to Alphonse. Then, his acceleration begins to
send him back home. A tenth of a year later (3.1 years into the trip) he reaches
.8 c, his rockets shut off, and he coasts home as an inertial observer.
We have already worked out what is going on during the periods where Gaston
is inertial. But, what about during the acceleration? Note that, at each instant,
Gaston is in fact at rest in some inertial frame – it is just that he keeps changing
from one inertial frame to another. One way to draw a spacetime diagram for
Gaston is try to use, at each time, the inertial frame with respect to which he
is at rest. This means that we would use the inertial frames to draw in more
of Gaston’s lines of simultaneity on Alphonse’s diagram, at which point we can
again copy things to Gaston’s diagram.
A line that is particularly easy to draw is Gaston’s t G = 3year line. This is
because, at t G = 3years, Gaston is momentarily at rest relative to Alphonse.
This means that Gaston and Alphonse share a line of simultaneity! Of course,
they label it differently. For Alphonse, it it t A = 5years. For Gaston, it is