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.

Figure %: Time dilation on a moving train.
Consider the situations shown in the diagram. In i) we have the first observer OA 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. OA 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 OA's frame the time taken for the laser light to reach is roof is just h/c and the roundtrip time is:

tA =    

In the frame of an observer on the ground, call her OB, 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 OB's frame is . It traverses this path at speed c, so the time taken is:

tB = =    

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 OA waves to OB every time the laser completes a cycle (up and down). Thus according to OA's clock, he waves every tA seconds. But this is not what OB sees. He too must see OA waving just as the laser completes a cycle, however he has measured a longer time for the cycle, so he sees OA waving at him every tB seconds. The only possible explanation is that time runs slowly for OA; all his actions will appear to OB 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. OA's time appears dilated to OB. This will only be true if OA 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 OB 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 OA's point of view: he sees OB flying past at v in the backwards direction (say OB 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 OA observes OB's clock running slowly (note that γ does not depend on the sign of v). How could this possibly be right? How can OA's clock be running slower than OB's, but OB's be running slower than OA'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 tB = γtA to hold, OA must be at rest in her frame. Thus the opposite, tA = γtB, must only hold when OB is at rest in her frame. This means that tB = γtA holds when events occur in the same place in OA frame, and tA = γtB holds when events occur in the same place in OB'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 (OB flying backward in OA's frame), the events (laser fired, laser returns) do not occur at the same place in OA's frame so the first relation we derived (tB = γtA) fails; tA = γtB is true, however.

Length Contraction

We will now proceed to derive length contraction given what we know about time dilation. Once again observer OA is on a train that is moving with velocity v to the right (with respect to the ground). OA has measured her carriage to have length lA 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 .

Figure %: Length contraction in a moving train.
OA observes how long the laser light takes to make a roundtrip up-and-back through the carriage, bouncing back from the mirror. In OA's frame this is simple:

tA =    

Since the light traverses the length of the carriage twice at velocity c. We want to compare the length as observed by OA to the length measured by an observer at rest on the ground (OB). Let us call the length OB measures for the carriage to be lB (as far as we know so far lB could equal lA, but we will soon see that it does not). In OB'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 OA, the relative speed is c + v. Thus we can calculate the total time taken for the light to go up and back as:

tB = + = âÉáγ2    

But from our analysis of time dilation above, we saw that when OA is moving past OB in this manner, OA's time is dilated, that is: tB = γtB. Thus we can write:

γtA = γ = tB = γ2âá’ = γâá’lB =    

Note that γ is always greater than one; thus OB measures the train to be shorter than OA 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 OA's point of view: she sees OB flying past to the left with speed v. We can put OB in an identical (but motionless) train and apply the same reasoning (just as we did with time dilation) and conclude that OA measures OB'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 OA compares the simultaneous coordinates of his own train to the simultaneous coordinates of OB's train, the difference between the former is greater than the difference between the latter. Similarly, if OB writes down the simultaneous coordinates of his own train and OA'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 OA and OB are writing down their coordinates are completely different. A simultaneous measurement for OA is not a simultaneous measurement for OB, and so we would expect a disagreement as to the observers concept of length. When the ends are measured simultaneously in OB's frame lB = , and when events are measured simultaneously in OA's frame lA = . No contradiction can arise because the criterion of simultaneity cannot be met in both frames at once.