The most important and famous results in Special Relativity are that of time dilation and length
contraction. Here we will proceed by deriving time dilation and then deducing length contraction from it. It is important to note that we could do it the other way: that is, by beginning with length contraction.

Consider the situations shown in the diagram. In i) we have the first observer O_{A} at rest with respect to
a moving train, which has velocity v to the right with respect to the ground. The carriage has a height h
and has a mirror on the roof. O_{A} designs a clock that measures the passage of time by firing a laser placed on the floor at
the roof of the carriage and registering the time taken for it to hit the floor of the carriage again (after
bouncing off the mirror on the roof). In O_{A}'s frame the time taken for the laser light to reach is roof is
just h/c and the roundtrip time is:

t_{A} =

In the frame of an observer on the ground, call her O_{B}, the train is moving with speed v (see ii) in
). The light then follows a diagonal path as shown, but still with speed c. Let
us calculate the length of the upward path: we can construct a right-triangle of velocity vectors since we
know the horizontal speed as v and the diagonal speed as c. Using the Pythagorean Theorem we can
conclude that the vertical component of the velocity is as shown on the diagram. Thus
the ratio the diagonal (hypotenuse) to the vertical is . But we know that the
vertical of the right-triangle of lengths is h, so the hypotenuse, must have length . This is the length of the upward path. Thus the overall length of the path taken by the light in
O_{B}'s frame is . It traverses this path at speed c, so the time taken is:

t_{B} = =

Clearly the times measured are different for the two observers. The ratio of the two times is
defined as γ, which is a quantity that will become ubiquitous in Special Relativity.

= γâÉá

All this might seem innocuous enough. So, you might say, take the laser away and what is the problem?
But time dilation runs deeper than this. Imagine O_{A} waves to O_{B} every time the laser
completes a cycle (up and down). Thus according to O_{A}'s clock, he waves every t_{A} seconds. But
this is not what O_{B} sees. He too must see O_{A} waving just as the laser completes a cycle, however
he has measured a longer time for the cycle, so he sees O_{A} waving at him every t_{B} seconds. The
only possible explanation is that time runs slowly for O_{A}; all his actions will appear to O_{B} to be in
slow motion. Even if we take the laser away, this does not affect the physics of the situation, and the result
must still hold. O_{A}'s time appears dilated to O_{B}. This will only be true if O_{A} is stationary next
to the laser (that is, with respect to the train); if he is not we run into problems with simultaneity and it
would not be true that O_{B} would see the waves coincide with the completion of a cycle.

Unfortunately, the most confusing part is yet to come. What happens if we analyze the situation from
O_{A}'s point of view: he sees O_{B} flying past at v in the backwards direction (say O_{B} has a laser
on the ground reflecting from a mirror suspended above the ground at height h). The relativity
principle tells us that the same reasoning must apply and thus that O_{A} observes O_{B}'s clock
running slowly (note that γ does not depend on the sign of v). How could this possibly be right?
How can O_{A}'s clock be running slower than O_{B}'s, but O_{B}'s be running slower than O_{A}'s?
This at least makes sense from the point of view of the relativity principle: we would expect from the
equivalence of all frames that they should see each other in identical ways. The solution to this mini-paradox
lies in the caveat we put on the above description; namely, that for t_{B} = γt_{A} to hold, O_{A}
must be at rest in her frame. Thus the opposite, t_{A} = γt_{B}, must only hold when O_{B} is at
rest in her frame. This means that t_{B} = γt_{A} holds when events occur in the same place in
O_{A} frame, and t_{A} = γt_{B} holds when events occur in the same place in O_{B}'s frame.
When v0âá’γ1 this can never be true in both frames at once, hence only one
of the relations holds true. In the last example described (O_{B} flying backward in O_{A}'s frame), the
events (laser fired, laser returns) do not occur at the same place in O_{A}'s frame so the first relation we
derived (t_{B} = γt_{A}) fails; t_{A} = γt_{B} is true, however.

Length Contraction

We will now proceed to derive length contraction given what we know about time dilation. Once again
observer O_{A} is on a train that is moving with velocity v to the right (with respect to the ground). O_{A}
has measured her carriage to have length l_{A} in her reference frame. There is a laser light on the back
wall of the carriage and a mirror on the front wall, as shown in .

O_{A} observes how long the laser light takes to make a roundtrip up-and-back through the carriage, bouncing back
from the mirror. In O_{A}'s frame this is simple:

t_{A} =

Since the light traverses the length of the carriage twice at velocity c. We want to compare the length as
observed by O_{A} to the length measured by an observer at rest on the ground (O_{B}). Let us call the length
O_{B} measures for the carriage to be l_{B} (as far as we know so far l_{B} could equal l_{A}, but we will soon
see that it does not). In O_{B}'s frame as the light is moving towards the mirror the relative speed of the light
and the train is c - v; after the light has been reflected and is moving back towards O_{A}, the relative speed
is c + v. Thus we can calculate the total time taken for the light to go up and back as:

t_{B} = + = âÉáγ^{2}

But from our analysis of time dilation above, we saw that
when O_{A} is moving past O_{B} in this manner, O_{A}'s time is dilated, that is: t_{B} = γt_{B}. Thus we can write:

γt_{A} = γ = t_{B} = γ^{2}âá’ = γâá’l_{B} =

Note that γ is always greater than one; thus O_{B} measures the train to be shorter than O_{A} does. We say that the train is length contracted for an observer on the ground.

Once again the problem seems to be that is we turn the analysis around and view it from O_{A}'s point of view: she
sees O_{B} flying past to the left with speed v. We can put O_{B} in an identical (but motionless) train and apply
the same reasoning (just as we did with time dilation) and conclude that O_{A} measures O_{B}'s identical carriage to
be short by a factor γ. Thus each observer measures their own train to be longer than the other's. Who is right? To
resolve this mini-paradox we need to be very specific about what we call 'length.' There is only one meaningful definition of
length: we take object we want to measure and write down the coordinates of its ends simultaneously and take the
difference. What length contraction really means then, is that if O_{A} compares the simultaneous coordinates of his own
train to the simultaneous coordinates of O_{B}'s train, the difference between the former is greater than the difference between
the latter. Similarly, if O_{B} writes down the simultaneous coordinates of his own train and O_{A}'s, he will find the
difference between his own to be greater. Recall from Section
1 that
observers in
different frames have different notions of simultaneous. Now the 'paradox' doesn't seem so surprising at all; the times
at which O_{A} and O_{B} are writing down their coordinates are completely different. A simultaneous measurement for
O_{A} is not a simultaneous measurement for O_{B}, and so we would expect a disagreement as to the observers concept of
length. When the ends are measured simultaneously in O_{B}'s frame l_{B} = , and when events are measured
simultaneously in O_{A}'s frame l_{A} = . No contradiction can arise because the criterion of simultaneity
cannot be met in both frames at once.