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Relativity and FTL Travel
by Jason W. Hinson (hinson@physicsguy.com)

Part II: More on Special Relativity
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Edition: 5.1
Last Modified: April 8, 2003
URL: http://www.physicsguy.com/ftl/
FTP (text version): ftp://ftp.cc.umanitoba.ca/startrek/relativity/
This is Part II of the "Relativity and FTL Travel" FAQ. It is an
"optional reading" part of the FAQ in that the FTL discussion in Part II
does not assume that the reader has read the information discussed below. If
your only interest in this FAQ is the consideration of FTL travel with
relativity in mind, then you may only want to read Part I: Special
Relativity and Part IV: Faster Than Light TravelConcepts and Their
"Problems".
In this part, we look more deeply into some points of special
relativity. By completing our discussion on the space time diagram as well
as explaining some of the paradoxes involved with SR, it should give the
reader a better understanding of the theory.
For more information about this FAQ (including copyright information
and a table of contents for all parts of the FAQ), see the Introduction to
the FAQ portion which should be distributed with this document.
Contents of Part II:
Chapter 3: Completing the SpaceTime Diagram Discussion
3.1 Comparing Time for O and O'
3.2 Comparing Space for O and O'
3.3 Once Again: The Light Cone
Chapter 4: Paradoxes and Solutions
4.1 The "Twin Paradox"
4.1.1 Viewing it with a SpaceTime Diagram
4.1.2 Explaining the "First Part"
4.1.3 Explaining the "Second Part"
4.1.4 Some Additional Notes
4.2 The "Car and Barn Paradox"
4.2.1 Viewing it with a SpaceTime Diagram
4.2.2 The explanation
Chapter 3: Completing the SpaceTime Diagram Discussion
Here we will complete the discussion of the spacetime diagrams which
we began in the previous chapter. We will do this by completely comparing
the coordinates our observers have for a particular event. To make that
comparison we will need to see how the lengths which represent one unit of
space and time in the reference frame of O compare with the lengths
representing the same units in O'. The easiest way for us to do this is to
use information we have already seenin particular, we use the fact that a
clock moving with respect to an observer seems to be running slowly to that
observer and a pole moving with respect to that observer seems to be shorter
to that observer by a factor of gamma. (Note: this was explained in Chapter
1. of this FAQ.) Understanding how to use this in the spacetime diagram in
order to completely construct the two observers' coordinate systems should
give some solid insight into time dilation and length contraction in special
relativity.
3.1 Comparing Time for O and O'
So, how do we show time dilation on our spacetime diagram. Well, the
key to this can be found by expressing time dilation in the following way:
In the O observer's frame of reference, let the tick t1 of his clock be
simultaneous with the tick t1' of the O' observer's clock. Also, let the
tick t2 of his (the O observer's) clock be simultaneous with the t2 tick of
the O' observer's clock. Then, we would find that
(Eq 3:1)
t2' t1'= (t2  t1)/gamma
where gamma (as defined in Section 1.4) would be calculated using the
relative velocity of O and O'. What Equation 3:1 says is that in the O
observer's frame of reference, the difference in the ticks of the O'
observer's clock is smaller than the difference in the O observer's own
ticks by a factor of gamma. Thus, we see that in the frame of O, the O'
observer's clock is running slowly.
As an example, from here on we will consider the case where the
relative velocity is 0.6 c such that gamma = 1.25. Using an example like
this will make the procedure easier to understand for the reader; however,
remember that we could redo this whole process with any speed (calculating a
new gamma factor, drawing a different speed for the observers, drawing
appropriate lines of simultaneity, etc.).
Now, what if we let the t1 tick be the "zero" tick. That means that at
the origin, when both of our observers are right next to one another, t1 =
t1' = 0. So, both of the observers agree (because there is no separation
between them in space at the origin) that t1 and t1' are simultaneous, and
happen at t = t'= 0. However, after some time, there will be a tick (t2) on
the O observer's clock. In the frame of reference of O, that tick is
simultaneous with the tick t2' of the O' observer's clock. Since t1 = t1' =
0, and we are using gamma = 1.25, we know (from Equation 3:1) that
(Eq 3:2)
t2' 0 = (t2  0)/1.25.
so
t2' = 0.8*t2
So, this says that in the frame of the O observer, the tick t2 of his
clock is simultaneous with the tick 0.8 t2 on the O' observer's clock. If we
draw a line of simultaneity in the O observer's frame of reference such that
it goes through the tick t2 of his clock, then it must also go through the
tick 0.8 t2 of the O' observer's clock. If we let t2 = 1 second, then we get
what is shown in Diagram 31. The distance from the origin, o, to the first
mark along t in that diagram is defined to be 1 second for our O observer.
Meanwhile, the distance from o to the "*" symbol along t' in that diagram is
0.8 second FOR THE O' OBSERVER. So, we begin to see that we can relate
distances in time along the axes of the different observers.
Diagram 31
t t'
^ /
 /
 /
 /
 /
 /
t = 1 +   *        line of simultaneity
 / t' = 0.8 for O at t = 1
 /
 /
 /
/
o> x


(Note: The line for t' only approximately represents an observer
moving at 0.6 c. It probably more closely represents 0.5 c, but
that's my ASCII for you. For our example, it _should_
represent an observer traveling at 0.6 c in the O observer's
frame of reference.)
This puts us on our way to understanding how, for example, different
lengths along t and t' relate to particular times on the clocks of the two
observers. Our next step to understanding this better will be to look at the
situation from the O' observer's frame of reference.
We have found what tick of the O' observer's clock is simultaneous with
the t = 1 tick of the O observer's clock
_in_the_O_observer's_frame_of_reference_. However, say we want to decide
what t' tick is simultaneous with the O observer's t = 1 tick
_in_the_O'_observer's_frame_of_reference_ (remember, the line of
simultaneity in Diagram 31 is only valid for the O observer's frame). To
figure this out, we need to draw a line of simultaneity in the O' observer's
frame of reference which passes through the event "the O observer's clock
ticks 1". When we do this, we want to note where that line passes the t'
axis, because that mark points out the tick on the O' observer's clock which
is simultaneous with O observer's t = 1 tick
_in_the_O'_observer's_frame_of_reference_. I have drawn this line in Diagram
32, but I have also left everything that was in Diagram 31.
Diagram 32
t t' line of simultaneity
^ / ' for O' at t' = 1.25
 / '
 / '
 / '
 % t' = 1.25
 ' /
t = 1 +   *       line of simultaneity
 / t' = 0.8 for O at t = 1
 /
 /
 /
/
o> x


(Note: The line of simultaneity for O' is a rough approximation)
Now, note that I marked the "%" symbol in that diagram (where the line
of simultaneity for O'which goes through t = 1crosses the t' axes) as
the event t' = 1.25. But how did I know that? Well, because in the frame of
reference of O', it is the O observer who is moving at 0.6 c, and thus it is
the O observer who's clocks are running slowly by a factor of 1.25. So, in
the frame of O', the event "t = 1 at the O observer's position" must be
simultaneous with the event "t'= 1.25 at the O' observer's position." That
way, in the O' observer's frame, it will be the O observer's clock which is
running slowly by a factor of 1.25. In addition, if the diagram were drawn
carefully I could use the length from the origin to "*" (which I know is 0.8
seconds for the O' observer) to figure out how much time passes between the
origin and the "%" symbol for O'. Either way, I find the same thing.
In Diagram 32, one can begin to see the power of using spacetime
diagrams to understand special relativity. Note that from one glance at that
diagram not only can we see that in the O observer's frame of reference the
O' observer's clock is running slow by a factor of 1.25 (i.e. the event "t =
1" is simultaneous with the event "t'= 0.8" in the O observer's frame) but
we also see that in the O' observer's frame it is the O observer's clock
which is running slow by a factor of 1.25 (i.e. the event "t = 1" is
simultaneous with the event "t'= 1.25" in the O' observer's frame). Thus, we
can see at once on this diagram that in each observer's own frame, the other
observer's clock is running slow. This happens to be one of the first, key
points to understanding the twin paradox (which will be discussed fully in
the next section).
3.2 Comparing Space for O and O'
So, we have found a correlation between the lengths which represent
certain times along the t axis for O and the lengths which represent certain
times along the t' axis for O'. We did this by using (1) the idea of time
dilation which was found earlier to be caused by the fact that light always
travels at c for all inertial observers and (2) the lines of simultaneity
for different observers which we learned how to draw by also using the fact
that light always travels at c for all inertial observers. Similarly, we can
find a correlation between lengths which represent certain distances along
the x axis for O and the lengths which represent certain distances along the
x' axis for O'. As an example, I have drawn a comparison of distances in
Diagram 33 which will be explained below.
Diagram 33
t t'
^ /
 /< line of constant position
 /  for O at x = 1
 / 
 /  / x'
 /  / '
t = 1 + /  / ' * = point where
 /  / ' x' = 0.8
 /  #
 / * /< line of constant position
 / '  / for O' at x' = 1.25
/ ' /
o+> x
 
 x = 1

(Note: The line for x' is a rough approximation)
Perhaps the best way to explain this diagram is as follows: Consider a
rod being held by the O observer such that one end of the rod follows the t
axis (and is thus always next to the O observer) while the other end follows
the vertical line drawn at x = 1. The rod then is obviously stationary in
the O observer's frame of reference. Second consider a rod being held by the
O' observer such that one end follows the t' axis and the other end follows
the line of constant position for O' which I have drawn.
Well, in the O observer's frame, his rod is obviously 1 lightsecond
long. But notice that in his frame the ends of the O' observer's rod are
next to the ends of the O observer's rod at t = 0. Thus, in the O observer's
frame, the O' observer's rod is also 1 lightsecond long. But length
contraction tells us that in the O observer's frame, the O' observer's rod
is shorter than its "rest length" by a factor of 1.25. Thus, in the O'
observer's frame (the frame in which his rod is at rest), his rod must
actually be 1.25 lightseconds long. That is how I know that the line of
constant position for O' I drew was for x'= 1.25.
Now, look at the distance along x' from the origin (o) to the point
marked "#". That distance represents the length of the O' observer's rod
from his own frame of reference (i.e. 1.25 lightseconds). Also, the
distance along x' from the origin to the point marked "*" represents the
length of the O observer's rod in the O' observer's frame of reference. That
distance must be 0.8 because in the O' frame, it is O and his rod which are
moving, and thus his rod seems length contracted by a factor of 1.25 from
its length in the frame of reference in which it is at rest (the O frame).
That number could have also been found by using the fact that the distance
from o to "#" was 1.25 lightseconds.
Finally, we again note the power of the spacetime diagram. At one
glance of Diagram 33 we are able to see that in the O' observer's frame,
his rod is 1.25 lightseconds long, while in the O observer's frame it is
only 1 lightsecond long. At the same time we are able to see that in the O
observer's frame, his rod is 1 lightsecond long, while in the O' observer's
frame, it is only 0.8 lightseconds long. Thus, each observer believes that
the other observer's rod is shorter than it is in the frame of reference in
which the rod is at rest. They each believe that the other is experiencing
length contraction, and with a spacetime diagram, we are able to see how
that is so.
3.3 Once Again: The Light Cone
Here I want to demonstrate how a light cone appears in the two
coordinate systems. In Section 2.8 I mentioned that the light cone is drawn
exactly the same for the two observers. Now that we understand how to draw
the two coordinate systems completely (i.e. we can now draw "tick" marks on
the x' and the t' axes as well as the x and t axes because of the discussion
above) we can make a diagram which clearly shows this. To start, in Diagram
34 I have shown the results of our discussion above in that I have
indicated approximately where the tick marks (+) would appear on the x' and
t' axes.
Diagram 34
t t'
 /
 +t'=2
 /
t=2+ /
 /
 / x'
 / '
 / '
 + t'=1 +'
t=1+ / ' x'=2
 / '
 / + '
 / ' x'=1
 / '
/ '
+o++> x
' / x=1 x=2
' / 
(Note: Again, x' and t' are rough approximation for v = 0.6 c)
Next, in Diagram 35 I have drawn the x and t axes along with lines of
simultaneity and lines of constant position (for O) at each tick mark. In
addition, the upper half of a light cone centered at the origin is shown
using # symbols. As you see (and as we would expect), it passes through the
points x = 1 lightsecond, t = 1 second; x = 2 lightseconds, t = 2 seconds;
etc.
Diagram 35
t

 # = Light
    #
+#
   # 
   # 
   # 
   # 
#    # 
#+#
 #  #  
 #  #  
 #  #  
 #  #  
 #  #  
+o++> x
   

Continuing with the diagrams, Diagram 36 shows the x' and t' axes
along with lines of simultaneity and lines of constant position (for O') at
each tick mark along those axes. Again, the upper half of a light cone
centered at the origin is also shown. As you see (and as we would again
expect), it passes through the points x' = 1 lightsecond, t' = 1 second;
etc. Note that the point x' = 1, t' = 1 is marked with an "@" symbol and the
tick marks on the x' and t' axes are marked with "+" marks to help make it
clear how the coordinate system works. Also notice that the light cone
itself is drawn exactly the same as it is in Diagram 35.
Diagram 36
t'
/
/ ' / / ' / /
/ ' / + ' / # / '
'/ / '/ / # '/
' / / ' / / # ' /
/ / ' / /# ' / x'
/ ' / @ / '
/ ' / / ' #/ / '
' / / ' # / / '
# ' / +' # / +' /
# / ' / # / ' / /
# / ' / # / ' / /
# / ' / # + ' / / '
'/# / # '/ / '/
' / # / # ' / / ' /
/ # /# ' / / ' /
/ o / ' /
/ ' / / ' / / '
/ ' / / ' / / '
Finally, I want to superimpose Diagram 35 and Diagram 36 to some
extent onto Diagram 37. It would be quite cluttered to put all the lines
included in the two diagrams, but I want to include the lines which make up
x = 1, t = 1, x'= 1, and t'= 1. These lines are thus drawn on Diagram 37,
but they terminate where they meet the light cone which is also shown. You
should begin to see the relationship between the two different frames of
reference and the fact that the light cone itself is exactly the same in
both coordinate systems. This is a direct result from the fact that every
step we took in producing these diagrams used the assumption that the speed
of light is the same in all inertial frames of reference.
Diagram 37
t t'
 /
 /
 + #
 / #
+ / #
 / # x'
 / # '
 / ' #/ '
 / ' # / +'
 +' # / '
+/# / '
 / #  / '
 / #  + '
 / # '
 / # ' 
/# ' 
+o+++>x
' /
' / 
Though this concludes our discussion of spacetime diagrams, we will
continue to see them in the next section, because they can be vital tools
for understanding paradoxes in special relativity.
Chapter 4: Paradoxes and Solutions
One misleading statement many people hear in connection with relativity
is something like this: "Time moves slower for you as your speed increases."
It is misleading because it implies some incorrect concepts. It implies that
there is an ABSOLUTE way to decide whether or not someone is truly at rest
or moving (at a constant, nonzero velocity) when in reality this depends on
your frame of reference. It implies that if you are moving at a constant
velocity, then your clock is moving slower than some sort of "correct" clock
which is truly not in motion. It also implies that you yourself might find
your clock ticking slower than usual.
However, as I have mentioned earlier, motion is relative. There is no
way to say that one object is truly at rest and another is truly moving at a
constant velocity. You can only say that one object is moving at a constant
velocity RELATIVE TO another object. You can say that in the frame of
reference of one observer (call him Joe) another observer (call her Jane) is
moving at a constant velocity. Then, in Joe's frame of reference, Jane's
clock is running slowly, and she is length contracted in the direction of
her motion. However, in Jane's frame of reference, JOE is the one who is
moving at a constant velocity relative to her. Because the laws of physics
are the same for all inertial frames, we must be able to apply the same laws
to Jane as we just applied to Joe. Thus, in Jane's frame, Joe's clock is the
one which is running slowly, and Joe is length contracted in the direction
of his motion.
This leads one to question whether or not relativity contradicts
itself. If all motion is relative, we have concluded that each observer
believes that the other observer's clock is running slowly, and each
believes that the other observer is length contracted in the direction of
motion. Isn't that a contradiction? For example, how can Jane's clock be
running slower than Joe's AND Joe's clock be running slower than Jane's?
Well, these questions lead to various solvable paradoxes in special
relativity.
As a note, the word "paradox" has a few different meanings, and when I
use it here, I will have this meaning in mine: "a paradox is a statement
that seems contradictory or absurd but that may in fact make sense." A
"solvable paradox" is then a paradox that does in fact make sense when
explained correctly, while an "unsolvable paradox" is a paradox for which
the statement "may in fact make sense" doesn't hold (i.e. an unsolvable
paradox is truly selfcontradictory).
The paradoxes in special relativity are solvable, and below I will
present two of these paradoxes along with their solutions.
4.1 The "Twin Paradox"
The twin paradox deals with the question of "who's clock is running
slower?" The story goes as followers: Two twins (say Sam and Ed) are both on
Earth when one of them (say Sam) decides to leave the Earth by very quickly
accelerating to a speed close to the speed of light. We then consider the
two frames of reference after Sam has reached a constant velocity. According
to special relativity, in Ed's frame of reference, Sam's clock is running
slowly, while in Sam's frame of reference, it is Ed's clock which is running
slowly.
Now, as long as the two are apart, it is not to hard to argue that the
question is strictly dependent on your point of view. By this I mean that we
can argue that there is no correct answer to the questionsthat who's clock
is running slower depends completely on what frame of reference you are in.
However, how would we continue this argument if we added the following to
the story:
At some point after Sam begins his trip away from the Earth, one of the
twins decides to go meet with the other twin. Either Ed decides to
accelerate away from the Earth and catch up to Sam, or Sam decides to
accelerate back towards the Earth to go back and meet with Ed. We then ask
this question: when the two twins are standing next to one another again,
which one is older?
With the above addition to the story, there must be a definite answer
to the final question. So, how can we continue to say that the answer
depends on your frame of reference? Well, as we will see, the final question
does have a definite answer, but the question of how this came about IS
dependent on who you ask.
4.1.1 Viewing it with a SpaceTime Diagram
So, now we will try to understand the twin paradox by using our old
friend, the spacetime 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 spacetime diagrams look
differently, the basic ideas we will discuss still hold.
Now we look at the spacetime diagrams. In Diagrams 41 and 42 below,
I have drawn the whole trip in two parts. In Diagram 41, you see Sam headed
away from Ed, and in Diagram 42, you see Sam after he has turned around and
is headed back to Ed.
Diagram 41 Diagram 42
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 spacetime 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 timewhich means that the distances
would be in lightyears). When Sam sees his own clock read 4, he turns
around with an instantaneous acceleration. At that point, we switch to
Diagram 42. 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 42 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 turnaround,
nothing significant happens when Sam turns around, and Sam's clock continues
to tick slowly after the turnaround (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 turnaround,
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
turnaround, 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.
4.1.4 Some Additional Notes
There are four specific things I want to make note of concerning the
twin paradox as I have explained it.
First, we should note that the outcome of the above thought experiment
(i.e. the fact that Sam ended up younger than Ed) is completely dependent on
the fact that Sam turned around and headed back to Ed. If instead Ed had
done the acceleration when he saw his own clock tick 4 years and had headed
over to meet Sam, then Ed would be the one who had aged a total of 8 years
while Sam had aged 10 years. Notice that the twin who undergoes the
acceleration must actually have a physical force applied to him to cause
that acceleration. During the acceleration he is no longer an inertial
observer (this is why his frame of reference shifts while the other twin's
frame does not shift). That differentiates his situation from the twin who
does not accelerate, and that breaks the symmetry between the two observers.
Unless one of them goes through an acceleration, their situations are
completely symmetric, and there is no absolute answer to the question "which
twin is younger?"
Second, I want to note something particular about the acceleration Sam
went through. Look back at the lines of simultaneity drawn for Sam's frame
before and after he accelerated. As we noted, the point where his "t' = 4"
line of simultaneity cross the t axis (Ed's position) shifts upward when Sam
turns around. Notice, however, that if Sam had taken a longer trip, then he
would have done the acceleration when he was further from Ed. Then that
"shift" would have been even larger, and after the acceleration, Sam's new
frame of reference would be one in which Ed's clock had "jumped" ahead an
even greater number of years. So, for Sam, the longer the trip he takes, the
bigger the change will be when he switches his frame of reference, and that
will make him an even greater number of years younger than Ed when they get
back together. Of course, for Ed, the longer the trip is for Sam, the longer
Sam's clock will be running slowly. So, Ed too agrees (with a different
explanation) that Sam will be more years younger than Ed in the end if the
trip is longer. As a final point on this, note that when Sam first
accelerates to start his trip, he is right next to Ed, so the acceleration
doesn't have much effect at all (as is true for his final acceleration at
the end of the trip). That is why we basically ignored those accelerations.
Third, I want to note something about Sam's explanation of the events.
Recall that when he changed frames of reference, his clock read 4 years
while (in his new frame) Ed's clock read 6.8 years. One may think that Sam
has thus changed to a frame where Ed's clock has been running faster;
however, we know that in Sam's new frame, Ed is still moving with respect to
Sam. Thus, in Sam's new frame Ed's clock has still been running slowly the
whole time. To understand how this can be, consider a third observer (Tim)
who has always been in the frame of reference which Sam has during the last
part of the trip. Let's say that Tim was traveling along (going to Earth)
when he saw Sam headed towards him, and to Tim's surprise, Sam turns around
and joins Tim in Tim's frame of reference as the two come together. Thus,
after Sam turns around, he and Tim are moving together, side by side. Now,
Tim notices that right after Sam turns around, Sam's clock reads 4 years.
Regardless of what Tim's clock reads, he can reset his clock to 4 years, and
we can backtrack 4 years along Tim's path to identify the origin of Tim's
frame (Sam's new frame). In Diagram 43 I have drawn (along with everything
in Diagram 42 Tim's path, the origin (o') of Sam's new frame of reference,
and a line of simultaneity for Tim's and Sam's frame at that origin.
Diagram 43
t
^

t=10 + t'=8
*
 *
 *
 *
 *
t=6.8\ *
 *
 \ *
 *
t=5+     *   > (t = 5
 \ line)
 \ \
t=3.6\ \
 \ \
 \ \ (t'= 4
 \ line)
 \ \
 \
 \ \
oo'>x
 \
 (Tim's path)
Notice that for Sam's new frame (the frame Tim has always been in) if
t'= 4 when Sam turns around, then the event at Ed's position which is
simultaneous with the origin in this frame (o') is the event "Ed's clock
reads 3.6 years". And there you have it. In Sam's new frame, while it is
true that Ed's clock is always been running slow, at the "beginning" for
this frame (i.e. at its origin) Ed's clock started at 3.6 years. In this new
frame of Sam's, Ed's clock had a "head start" (so to speak) when compared to
Tim's clock. That is why Ed's clock already reads 6.8 years while Sam's
clock reads only 4 years in Sam's new frame. In the end, we can describe the
events in whatever frame of reference we wish, and though they may each have
different explanation for what actually happens, they must all agree with
the final outcome when the two twins come back together.
The final note I want to make is, again, about Sam's "view" of the
events. When we say that before Sam's turnaround he is in a frame of
reference in which Ed's clock reads 3.2 years, and after the turnaround Sam
is in a frame of reference in which Ed's clock reads 6.8 years, one might be
tempted to say that as Sam accelerates, Ed's clock speeds up in Sam's frame
of reference. Of course, this doesn't change the way Ed sees his clock
running, but it is only the way things occur in Sam's changing frame of
reference. However, think about what would happen if Sam quickly changed his
mind after the turnaround and immediately turned BACK around to his
original heading. Then, in this new acceleration, Sam went from a frame
where Ed's clock read 6.8 years to a frame where Ed's clock reads 3.2 years
again. One would thus argue that Ed's clock went backwards in Sam's changing
frame of reference. Again, this doesn't have any real significance to the
way Ed is reading his own clock, but we have to come to terms with the fact
that Sam's new acceleration caused Ed's clock to go backwards in Sam's
changing frame. Perhaps the best way to think about this is simply to
realize that Sam is not in an inertial frame since he is accelerating.
Rather, Sam is simply changing into various inertial frames, and in each of
these inertial frames, moving clocks do tick slowly, time does goes forward
in all frames, etc. Either way you like to think about it, in the end, we
can explain the outcomes as needed.
4.2 The "Car and Barn Paradox"
The "Car and Barn" paradox deals with the question of "whose lengths
are shorter?" We have a barn whose front and back doors can be quickly open
and closed. There is also a car which is just long enough so that if you try
to fit it in the barn, and the barn doors close, they would close down on
the front and back bumpers of the car. Now, an observer in the car (say,
Carol) speeds the car towards the barn at a significant fraction of the
speed of light. One might then argue the following: from the point of view
of an observer sitting in the barn (say, Bob) the car will be length
contracted, and at some point it will be completely inside the barn. Bob
then reasons that he can close and open both barn doors while the car is
completely inside the barn. However, Carol will argue that it is the Barn
which moving with respect to here, and thus it the barn which is length
contracted. So, she argues, if Bob tries to close both doors at the same
time as the car goes through the barn, then the doors will smash into the
car.
We thus want to ask whether or not the barn doors do smash into the car
if Bob tries his idea, and how does each observer explain the outcome.
4.2.1 Viewing it with a SpaceTime Diagram
As we did with the twin paradox, here we will look at a spacetime
diagram of the car and barn experiment in order to explain the paradox. We
will draw our diagrams such the relative velocity of Carol and Bob is 0.6 c.
In Diagram 44 I have drawn the situation keeping Bob's frame of reference
in mind. To keep the diagram from getting too cluttered, a second diagram
(Diagram 45) of the same situation will be used to mark points according to
Carol's frame of reference.
Diagram 44 Diagram 45
t ! t
e  < x > ! e  < x > .
e < x > ! e < x 4
e < x > ! e < x . > .
e < x > ! e < .4 > . .
e <  x > ! e <  . x 3 .
e <  x> ! e < . .3>. .
3...3..........D ! e < .  . .D .
e <  >x ! e 4 . . >x .
B2o2C>x ! B<.o.>C>x
e<  > x ! . e< . .  1 x
A..........1...1 ! . A .  . > x
x ! . . <2 . > x
< e  > x ! . .2 e .  > x
< e > x ! . . < e. > x
< e > x ! . < . 1 > x
< e > x ! 1 e > x
< e >  x ! . < e >  x
In the diagrams we have the following: The ">" marks indicate the
positions of the front of the car at different points in time, while the "<"
marks show the back of the car. The e's mark the entrance to the barn, and
the x's mark the exit of the barn. Hopefully it is apparent to the reader
that the car travels from left to right (with respect to the barn) and
passes through the barn. Also note that at the point where the entrance and
exit of the barn cross the x axis (i.e. when the front and back of the barn
are both at t = 0 in Bob's frame), both the front and back doors quickly
close and open again. Those points are labeled B and C.
4.2.2 The explanation
We are interested in six different occurrences (though only 4 are shown
in the diagrams). The ones not shown in the diagrams are, first, the front
of the car enters the barn, and second, the back of the car exits the barn.
These would appear much lower and much higher (respectively) in the diagram
than is being shown here. The four events that we do note in the diagrams
are (A) the back of the car enters the barn, (B) the entrance door of the
barn closes and opens again, (C) the exit door of the barn closes and opens
again, and (D) the front of the car exits the barn. In the diagrams, I have
marked each of these events with the letters given and drawn lines of
simultaneity (marked with periods) for the observers.
In Diagram 44, we see that for Bob (whose lines of simultaneity are
drawn in that diagram), (A) is the first event which happens, and everything
that occurs simultaneous to (A) in Bob's frame of reference is marked with a
1. The next two events in Bob's frame are (B) and (C), which occur
simultaneously. Everything which occurs simultaneous to these events is
marked with a 2. Finally for Bob, (D) occurs, and everything which occurs
simultaneous to it is marked with a 3. Note that for Bob, as the back of the
car enters the barnevent (A)the front of the car has yet to exit the
barn. Also, when the doors close and openevents (B) and (C)simultaneous
in Bob's frame, the front and back of the car are inside the barn (they are
marked with 2's). Thus, in Bob's frame, the car is smaller than the barn,
and it is inside the barn when the doors close and open. Finally, after both
doors close and open, the front of the car exits the barnevent (D)in
Bob's frame.
However, in Diagram 45 we see simultaneous events marked from Carol's
frame of reference. Again, the lines of simultaneity at each event are
marked with periods (but here they are drawn from Carol's frame). Now, we
see that the "lowest" line of simultaneity on the diagram from Carol's frame
of reference passes through the event (C), the exit door of the barn closes
and opens. Thus, this event occurs first in Carol's frame. Everything
occurring simultaneous with it in Carol's frame is marked with a 1. Next in
Carol's frame, event (D) occurs, followed by event (A), while event (B)
occurs last. The events occurring simultaneous with these events are marked
2, 3, and 4, respectively. Thus, according to Carol's frame, things happen
as follows: First, while the front of the car is in the barn, but before the
back of the car enters the barn, the exit door of the barn closes and opens.
Next, the front of the car exits the barn. (Note that while the front of the
car is then outside the exit of the barn at this point, the back of the car
has yet to enter the barn in Carol's framelook along the x' axis, for
example. So for Carol, the barn is smaller than the car.) Next, the back of
the car enters the barn in Carol's frame. Finally, after the front of the
car has exited the barn and the back of the car has entered the barn, the
entrance door of the barn closes and opens.
And there you have it. In the end, each observer must agree that the
car gets through the barn without smashing into the doors. However, each
frame of reference offers a different explanation for how this comes to be,
because in each frame, different events are simultaneous with one another.
In Bob's frame, the car is in the barn all at once while the doors close and
open simultaneously. However, in Carol's frame, the doors do not close
simultaneously, and the car is never completely in the barn.
So, I hope you have seen the power of spacetime diagrams when it comes
to explaining things in special relativity. When we simply say that moving
clocks run slower and moving rulers length contract, we miss a real
understanding of special relativity. That understanding comes from realizing
that the actual coordinates in space and time for events are different for
different observers who are moving with respect to one another. This
relationship can be viewed with spacetime diagrams, and the answers to many
nagging questions in special relativity can be explained if one understands
these diagrams.