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 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.
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.
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).
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.
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.
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.
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.
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'= 4 line)!  / / !  / !  / / !  / ! t=5+     *   > (t = 5 line) !  * !  / * !  * ! t=3.2/ * !  * ! /  * !  * !  * ! * ! o> x !  !  ! 
t ^  t=10 + t'=8 *  *  *  *  * t=6.8\ *  *  \ *  * t=5+     *   > (t = 5  \ line)  \ \  \  \ \  \ (t'= 4  \ line)  \  t'  o>x   
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 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 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.
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?"
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.
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.
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.
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 ! e  < x > ! e < x > ! e < x > ! e < x > ! e <  x > ! e <  x> ! 3...3..........D ! e <  >x ! B2o2C>x ! e<  > x ! A..........1...1 ! <e  > x ! < e  > x ! < e > x ! < e > x ! < e > x ! < e >  x ! 
t e  < x > . e < x 4 e < x . > . e < .4 > . . e <  . x 3 . e < . .3>. . e < .  . .D . e 4 . . >x . B<.o.>C>x . e< . .  1 x . A .  . > x . . <2 . > x . .2 e .  > x . . < e. > x . < . 1 > x 1 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.
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.