@Warp: A more in-depth explanation:
Minkowski diagrams are very similar to standard x-y graphs: you read off points as you would read in any such plots -- to get the t coordinate in any given reference frame, you trace a line parallel to that frame's simultaneity line until it reaches the time axis for that reference frame and read the value at the time axis (and vice versa to get the position). The only difference is the distance function -- in a x-y graph, you use Pythagoras theorem to get the distance along slanted curves, in a Minkowski diagram, you measure the space-time interval given by the Minkowski metric. I would link to the Wikipedia page on this, but it is highly technical; so the short version:
In space-time, you can define a "distance" between any two events by the following formula:
S = (c * delta-t)^2 - (delta-x)^2 - (delta-y)^2 - (delta-z)^2
If S > 0, you have what is called a
timelike interval, and sqrt(S)/c is the proper time interval that elapses on an inertial reference frame that considers these two events to have the same spatial coordinates -- and there will be such frames (this would be an absolute "temporal distance" between the two events). Moreover, you will have a fixed, definite, temporal ordering of these two events -- there will not be any reference frame (inertial or otherwise) which considers these two events to be simultaneous or which reverses the order in which these two events happen (this would require either traveling back in time or superluminal speeds). Conversely, you cannot define any absolute spatial distance between these events.
If S < 0, you have a
spacelike interval, and sqrt(-S) is the proper distance between the two events in any inertial reference frame that considers them to have the same time coordinates -- i.e., simultaneous (this will be an absolute spacial distance between the two events). You cannot define any absolute temporal order between these two events, and there are reference frames (inertial or otherwise) in which either event may happen before the other; moreover, you cannot find any reference frame (inertial or otherwise) in which these two events are at the same spatial location (this would require superluminal speeds).
If S = 0, you have a
null or
lightlike interval -- the interval is zero, and this means that the two events can be connected by a beam of light.
The Pythagorean distance is invariant with regard to rotations of the coordinate systems (as well as translations); likewise, the Minkowski interval is invariant with regards to purely spatial rotations, as well as rotations in a plane involving the time axis -- these are called Lorentz transformations. If you add translations of the coordinate system to the Lorentz transformations, you reach the Poincaré transformations, which also preserve the Minkowski interval.
All of this is based on the prevailing modern view of Relativity as being a study of the geometry of space-time, by the way; and while, historically, the Minkowski interval was derived from the Lorentz transformations, it is now known today that it is the interval -- the space-time metric -- which is the important thing.
In particular, according to this distance function, you have one important consequence (which I am not going to prove): return to the first 3 diagrams I posted and look at either the left or right diagrams. In these diagrams, the more vertical gray lines (the ones which have arrows in the last diagram I posted) are orthogonal to the more horizontal gray lines. That is a Lorentz transformation, or a rotation in space-time for you :-)
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So, with all of that in mind, let us return to the last diagram I posted. If you notice, some of the gray lines have arrows in them, as does one of the black lines. These are the temporal axis of the moving ships. While I drew two for each ship, one for the bow and one for the stern of each ship, you only need one -- any one, including any other you wish to draw which is parallel to these. Selecting one means selecting what position each ship considers to be position x = 0 in its frame. Lets pick the time axis on the bow of each ship.
The gray lines without arrows, the simultaneity lines, work similarly. You can pick one to indicate what you consider the time t = 0 according to that frame (including one parallel to the ones I drew). Lets pick the lower gray lines to be the time t = 0. Combined with the above, all 3 reference frames have the same event as the origin of their coordinate systems.
Hopefully, now things are starting to become more clear. So I will now start some analysis of the diagram to illustrate how it goes.
I will assume that the ships have proper lengths equal to L from now on.
First, the point where the two black lines meet -- the origin. This point has the coordinates (t, x) = (0, 0) in the 3 coordinate systems.
Now, the yellow line has two endpoints: the origin and the point where the stern of B coincides with the bow of A. Now, the bow of A is always at position x = 0 according to A, and in the problem, we assumed that this happens when the clock at A marks 1 s. Thus, the other endpoint is at (t, x) = (1 s, 0) according to ship A. The Minkowski interval tells us that this is a proper time interval, and the time that elapses in A is 1 s -- which we already knew. Moving this to other reference frames will be tricky yet, so I will leave it for later.
Take now the orange lines: they have as one endpoint the bow of ship A and ad the other endpoint the stern of ship A. At this point, we can obtain the coordinates of the endpoints of all of the orange lines in the reference frame of A: the lowest orange line goes from (0, -L) to (0, 0), while the middle orange line goes from (1 s, -L) to (1 s, 0). Both of these come from the description of the problem. Now to find the endpoints of the topmost orange line: the defining characteristic is that the stern of ship B has moved enough (from the point of view of ship A) that it now coincides with the stern of ship A. By hypothesis, ship B has then moved (according to ship A) 3 times its length (in A's point of view) -- since it took B 1 s to move its own length (again, according to A), then the topmost orange line
has coordinates going from (3 s, -L) to (3 s, 0) according to A.
And so on. The trickiest thing to calculate are the speeds of A and B according to the external reference frame, as well as the gamma factor for this speed; this is because you need to calculate the hyperbolic angle and use inverse hyperbolic tangents and so on (see
this for something about hyperbolic functions).
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Now, there is what HHS said. You can (hopefully) verify it from the diagram now.
This picture is the combination of the 3 images above, plus 3 more lines. The 3 new lines show, in the 3 different frames, the time that the bow of ship B takes to move from bow to stern of ship A. The yellow line and the pink line show the times as measured at the bow of A and B, respectively, and are on the same scale (as the ships have the same speed in this diagram); the cyan line is in the external reference frame, and is in another scale. If the yellow line is 1 second, then the cyan line is sqrt(3/2) (as HHS stated) and the pink line is 2 seconds (it is roughly twice as long in the image, and doing the math will show this to be the case). If the time interval in question is measured at another point in the ships, you can just slide the respective line keeping the endpoints in the same simultaneity lines (for the cyan line, this means "horizontally").
As you can see, these time intervals measure quite different things. And since they only share one event (the origin), it doesn't really make sense to compare them.
Now we go to the clocks. If we are going to assume that the folks at ship A are smart enough to check the clock of B at both relevant events and measure the delta, this question can be answered objectively. Otherwise, it will depend on where the clock is located in ship B.
Assuming smart folks, we can place the clock somewhere convenient -- say, at the stern of ship B. Then, we have a Minkowski triangle with two known sides (the yellow line and the lowest light blue line) and one unknown side (the gray line between them). This unknown side is the time interval elapsed in the bow of ship B between the two relevant events as seen by A.
Now, the side can be calculated if we have the rest length of the ships. I will represent this length by L (this is the length of the orange and purple lines). We know that whatever this rest length is, the light blue lines (and the green lines too) will be half of this: L/2. The yellow line has "length" equal to c * 1 s. Putting in the proper time formula (which is a special case of the space-time version of Pythagoras theorem), we have that the desired time interval dt is
dt = sqrt[(1 s)^2 - (L/2c)^2]
This is how much time will elapse in a clock on ship B as seen by ship A in the problem at hand.
I hope this is clear enough; this kind of thing is much easier to explain in person, having instantaneous feedback from the subject if something is unclear...
In all pictures, time is on the vertical. The pictures are roughly to the scale of the problem (A views B as having about half its length). There are Minkowski diagrams, by the way.
On the center is the situation according to the external reference frame in which ships A and B have the same length and move at the same speed; ship A is red, ship B is blue.
On the left is the situation as viewed by ship A plotted on the external reference frame; ship A is orange, ship B is light blue. In gray are the simultaneity lines of A (the more horizontal gray lines) and the lines of constant position of A (the more vertical gray lines). In English, this means 'the points which A considers to be simultaneous' and 'the points which A considers to be stationary'.
On the right is the situation as viewed by ship B plotted on the external reference frame; ship A is purple, ship B is green. In gray are the simultaneity lines of B (the more horizontal gray lines) and the lines of constant position of B (the more vertical gray lines). In English, this means 'the points which B considers to be simultaneous' and 'the points which B considers to be stationary'.
Note that there is some overlap between A and B in the 3 images in the middle lines; these are where the stern of ship B coincides with the bow of ship A according to ship A (left), the external observer (center) and B (right).
In the left and right pictures, you have to explicitly include the Lorentz factor on any distances and times measured along the gray lines (due to Minkowski geometry).
As you can see, the situation is quite asymmetrical as viewed from within the ships (not so much for the external observer).
If anything is unclear, please say so and I will attempt to explain it better.