Special Relativity: Kinematics: Time Dilation and Length Contraction

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 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 =

âÉá

γ²

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 =

γ²âá’

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.