[email protected])
----------------------------------------------------------------------
PART II: More on Special 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
IV 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
and PART IV.
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
"Relativity and FTL Travel--Introduction to the FAQ" portion which
should be distributed with this document.
Contents of PART II:
3. Completing the Space-Time 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
4. Paradoxes and Solutions
4.1 The "Twin Paradox"
4.1.1 Viewing it with a Space-Time 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 Space-Time Diagram
4.2.2 The explanation
3. Completing the Space-Time Diagram Discussion
Here we will complete the discussion of the space-time diagrams
which we began in the previous section. 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
seen--in 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. Understanding how to use this in the
space-time 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 space-time diagram. Well,
the key to this can be found by explaining time dilation with the
following: 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
t2'-t1' = (t2-t1)/gamma,
where gamma would be defined from the relative velocity of O and O'.
That is, 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. As an example, 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 answers 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) 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. And since t1 = t1' = 0, we know that (from the
equation above)
t2'- 0 = (t2 - 0)/1.25. (Remember, we are considering
the case where gamma = 1.25)
so
t2' = 0.8*t2
So, we find that if we draw a line of simultaneity in the O
observers frame of reference such that it goes through the tick t2 of
the O observer's 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 3-1. The distance from the origin, o, to the "+"
symbol 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 3-1
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.
Let's say we want to decide what tick of the O' observer's clock
is simultaneous with the t = 1 tick of the O observers clock in the O'
observer's frame of reference (remember, the line of simultaneity in
Diagram 3-1 is only valid for the O observer's frame). To find 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 _in_the_O'_observers_frame_of_reference_
with tick t = 1 on the O observer's clock. I have drawn this line in
Diagram 3-2, but I have also left everything that was in Diagram 3-1.
Diagram 3-2
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, how did I know that the "%" (where the line of simultaneity
for O'--which goes through t = 1--crosses the t' axes) was the event
t' = 1.25? Well, because for O', the O observer's clock is the one
which is running slowly by a factor of 1.25. So, in his frame of
reference, the event t = 1 at the O observer's position must be
simultaneous with the event t' = 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 "%" sign for O'. Either
way, I find the same thing.
In Diagram 3-2, one can begin to see the power of using space-
time 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 3-3 which will be explained below.
Diagram 3-3
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 axes (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' axes 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 light-
second 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
light-second long. But length contraction tells us that in the O
observer's frame, the O' observer's rod is short 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 light-seconds long. That is how
I know that the line of constant position for O' I drew was for x' =
1.25.
Now, if you look at the distance along x' from the origin (o) to
the point marked #. That represents the length of the O' observer's
rod from his own frame of reference (i.e. 1.25 light-seconds). 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 light-seconds
(as we showed above)
Finally, we again note the power of the space-time diagram. At
one glance of Diagram 3-3 we are able to see that in the O' observer's
frame, his rod is 1.25 light-seconds long, while in the O observer's
frame it is only 1 light-second long. At the same time we are able to
see that in the O observer's frame, his rod is 1 light-second long,
while in the O' observer's frame, it is only 0.8 light-seconds 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 space-time 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 2.6 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 3-4 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 3-4
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 3-5 I have drawn the x and t axes along with
lines of simultaneity and lines of constant position 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 light-second, t = 1 second; x = 2
light-seconds, t = 2 seconds; etc.
Diagram 3-5
t
|
| # = Light
| | | | #
-------------+-----------------------#-
| | | # |
| | | # |
| | | # |
| | | # |
# | | | # |
-#-----------+-----------#------------
| # | # | |
| # | # | |
| # | # | |
| # | # | |
| # | # | |
--+-----------o-----------+-----------+----> x
| | | |
|
Continuing with the diagrams, Diagram 3-6 shows the x' and t'
axes along with lines of simultaneity and lines of constant position
at each tick mark along those axes. Again, the upper half of a light
cone centered at the origin is shown. As you see (and as we would
again expect), it passes through the points x' = 1 light-second, 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 3-5.
Diagram 3-6
t'
/
/ ' / / ' / /
/ ' / + ' / # / '
'/ / '/ / # '/
' / / ' / / # ' /
/ / ' / /# ' / x'
/ ' / @ / '
/ ' / / ' #/ / '
' / / ' # / / '
# ' / +' # / +' /
# / ' / # / ' / /
# / ' / # / ' / /
# / ' / # + ' / / '
'/# / # '/ / '/
' / # / # ' / / ' /
/ # /# ' / / ' /
/ o / ' /
/ ' / / ' / / '
/ ' / / ' / / '
Finally, I want to superimpose Diagrams 3-5 and 3-6 to some
extent onto Diagram 3-7. It would be quite cluttered to put all the
lines included in the two diagrams, but I want to include the lines
which make x = 1, t = 1, x' = 1, and t' = 1. These lines are thus
drawn on the Diagram 3-7, 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 at the speed of light is
the same in all inertial frames of reference.
Diagram 3-7
t t'
| /
| /
| + #
| / #
+ / #
| / # x'
| / # '
| / ' #/ '
| / ' # / +'
| +' # / '
+-----/-----# / '
| / # | / '
| / # | + '
| / # '
| / # ' |
|/# ' |
--+-----------o-----------+-----------+-----------+--->x
' /|
' / |
Though this concludes our discussion of space-time diagrams, we
will continue to see them in the next section, because they can be
vital tools for understanding paradoxes in special relativity.
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 a particular way to decide whether
or not someone is moving at a constant velocity. 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 self-contradictory).
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
questions--that 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 space-time diagram
So, now we will try to understand the twin paradox by using or
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 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 coordinates
of Sam. Sam's positions at different times are marked with * symbols.
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 (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, though
Ed's clock still ticks slowly, it 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 frame of reference, it
will still only tick another 3.2 years (in Sam's frame) during the
rest 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, while Sam sees 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. Unless one of them goes through an acceleration, their
situations are completely symmetric, and there is no real answer to
the question "which twin is younger?"
Second, I want to note something particular about the
acceleration Sam went through. Note that as Sam gets further from Ed,
the shift in the space axis which occurs as Sam turns around will make
a bigger effect on which of Ed's ticks is simultaneous (in Sam's
frame) with the point where Sam turns around. If Sam had accelerated
when he was closer to Ed, then the point where Sam's space axes
crosses Ed's time line before the turn-around would have been much
closer to the point where Sam's new space axes crosses Ed's time line
after the turn-around. (Look at the diagrams and see if this doesn't
make sense.) So, for Sam, the longer the trip he takes, the bigger
the change will be when he switches his frame of referse, 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. And as a final point on this topic, 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)--which 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 watch
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 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 clock reads, he can reset his watch to 4 years,
and we can backtrack 4 years along Tim's path to identify the origin
of Sam's new frame. In Diagram 4-3 I have drawn (along with
everything in Diagram 4-2) 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 4-3
t
|
t=10+ t'=8
|*
| *
| *
| *
| *
t=6.8\ *
| *
| \ *
| *
t=5+ - - - - * - - -> (t = 5
| \ line)
| \ \
t=3.6|\ \
| \ \
| \ \ (t'= 4
| \ line)
| \ \
| \
| \ \
-----o-------------------o'----->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 it's 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 turn-around he is in a
frame of reference in which Ed's clock reads 3.2 years, and after the
turn-around 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 Ed quickly changed his mind after the turn-
around 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 the barn doors while the car is completely inside
the barn. However, Carol will argue that it is the Barn which is
length contracted, and that if Bob tries to close both doors at the
same time as the car goes through the barn, then the doors will be
smashed.
We thus want to ask whether or not the barn doors do get smashed
if Bob tries his idea, and how does each observer explain the outcome.
4.2.1 Viewing it with a Space-Time Diagram
As we did with the twin paradox, here we will look at a space-
time 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 4-4 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 4-5) of the same
situation will be used to mark points according to Carol's frame of
reference.
Diagram 4-4 ! Diagram 4-5
!
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 .
-------B-2-----o-----2-C----->x ! ---------B-<---.-o.---->-C------>x
e< | > x ! . e< . . | 1 x
A.......|...1...1 ! . A . | . > x
<e | > 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 front and back of the barn cross the x axis, both
the front and back door's quickly close and open again.
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 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 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 by periods) for the observers.
In Diagram 4-4, we see that for Bob (whose lines of simultaneity
are drawn in at 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 barn--event (A)--the
front of the car has yet to exit the barn. Also, when the doors close
and open--events (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 barn--event (D)--
in Bob's frame.
However, in Diagram 4-5 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
frame. 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 space-time 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 space-
time diagrams, and the answers to many nagging questions in special
relativity can be explained if one understands these diagrams.
======================================================================
Relativity and FTL Travel
by Jason W. Hinson (