### Time Dilation

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 *v*0âá’*γ*1 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.