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.
|t A =|
|t B = =|
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 0âáγ 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.
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 .
|t A =|
|t B = + = âÉá γ 2|
|γt A = γ = t B = γ 2âá = γâál B =|
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.
Take a Study Break!