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Question:
My question is the T/U paradox as explained in Taylor's spacetime physics (it is also one of the scenarios in the program rellab by physics academic software. the paradox is as follows a U made of steel has a detonator switch connected to the bottom of the uU and connected to an explosive. TA T made of the same steel just fits in t U when they are at trest wrt eraacth other. The T is moved far away from the U and accelerated to v=.9c. The T From the reference fromaeame of the U the T is lorentz contracted and the T is too short to reach the detonator From the reference frame od f the he T the U is arms are contracted and the T can strike the switch. Which occurs? I have been able to work my way through the other typical problems like the barn/pole paradox , and the runner train paradox. This one has me stumped.



Replies:
What a beeyootiful problem. The answer to the paradox is that the bomb goes off, since all observers agree after the collision that U and T are at rest and fit exactly into one another. Relativity does not alter the nature of events, just the order and interval between them. Two points are key to understanding what T and U would deduce from their observations: the first is that the speeding, shortened other object *lengthens* when it slows down during the collision. The second is that neither T nor U see as simultan- eous (1) the crossbar of the T hitting the ends of the U and (2) the point of the T hitting the bottom of the U. From T's point of view, 2 precedes 1, while U concludes 1 happens before 2. There is another frame of reference (moving with the center of mass) in which both events occur at the same time and in which T and U are the same length always. One always talks in relativity class about what observers *deduce* in their respective frames of reference. But it is fun to imagine what T and U would actually *see* according to our own habits of perception as the collision unfolds. To be concrete, let us imagine T and U at rest are each 100 million miles long (about the distance from Earth to the Sun), and they approach each other at a speed such that each would deduce with all the relativity math that the other is half its rest length. This speed turns out to be 87% of the speed of light c, or about 161,000 miles/sec. There are four interesting points of view: (A) sitting on the ends of the U, (B) sitting at the bottom of the U, (C) sitting on the point of the T, and (D) sitting on the crossbar of the T. Suppose everybody somehow synchronizes their watches so they all agree the point of the T passes the ends of the U at 5:00 pm exactly. Now as 5:00 pm comes on A sees the T approaching fast. Suppose A measures the length and speed of T naively, the way you usually would do it. To measure the length we suppose mile markers along the route T takes; A just notes which mile markers he sees next to the front and back of the T at a certain moment and subtracts. To measure the speed of T he starts his stopwatch when he sees the point pass a mile marker, then stops it when he sees the point pass the next, and divides. What will he see? The T will look 373 million miles long and will look like it is approaching at 6.5 times c, or about 1.2 million miles/sec. Whoa! How can this be? Because this is a *naive* measurement, A has forgotten to take into account the time it takes the image of the front and back of the T to get to him at c. This is the way we normally do things --- if you are standing at the finish line timing a footrace you start the watch when you see the runners start and stop it when they go past you. You do not add to the elapsed time the time it took for the image of the start to get to you. But you should, strictly speaking. If a runner ran *at* the speed of light, he would get to you *at the same time* as the image of him starting, and you would (wrongly) conclude by the naive measurement method that he covered the course in no time at all. Still, that is the way our habits of perception are set up. Of course if A takes the speed of light into account and does his relativity math correctly, he will conclude that, contrary to his naive impression, the T is 50 million miles long and approaching at 87% of the speed of light.

Now at 5:00 pm A sees the point of the T pass him. The crossbar is 373 million miles distant and still approaching at 6.5 times c, but the point is now receding from A, apparently at 0.5 times c or 86,000 miles/sec. Thus the T appears to shrink at the rate of 1.1 million miles/sec. At 5:05 pm the crossbar reaches A, hits the ends of the U, and stops. The point has now reached a point 27 million miles away from A, or about 1/3 the distance to the bottom of the U, so the T appears to be only 27 million miles long. The point continues to recede, so the T appears to A to be growing at 86,000 miles/sec. At 5:19 pm the T has grown to its rest length of 100 million miles long, and A sees it hit the bottom of the U and stop. Now everything is motionless and the U and T fit perfectly together.

Now let us consider it from the point of view of B (at the bottom of the U). B sees again a huge (373 million mile long) T approaching at 6.4 times c. B sees the point pass A at 5:09 pm (9 minutes after A sees it happen, because it takes 9 minutes for the image to go the 100 million miles between A and B). Barely a minute later, at 5:10 pm, the point of the T reaches B, hits the bottom of the U and stops. (Note that the point gets to B only a minute after B sees it leave A, even though light itself takes 9 minutes to get from A to B. The *naive* (wrong) conclusion is that the point is traveling a lot faster than light.) Now the crossbar is 373 million miles distant, and still approaching at 6.5 times c, so B sees the T shrinking at a rate of 1.2 million miles/sec. This goes on for 227 seconds until the T has shrunk to its rest length of 100 million miles. At this point, 5:14 pm, B sees the crossbar of the T hit the ends of the U and stop. Everything is motionless and the T and U fit perfectly. Everybody goes out for a beer and compares notes.



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