page 1 of 2
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.
|tB = =|
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 v0âáγ1 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.
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 .
|tB = + = âÉáγ2|
|γtA = γ = tB = γ2âá = γâálB =|
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.
Take a Study Break!