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Length contraction falling down a hole.

  1. Sep 4, 2009 #1
    Ok, so imagine observer A is moving along with a measuring rod of length L that is being slid along the ground by a rod that holds it steady by pushing it against the ground as it slides (no tilting whatsoever, the base of the rod remains firm against the ground), at a velocity comparable to c. Let's say that the velocity is such that observer B who is standing on the ground and not moving with the measuring rod will measure the length of the rod to be L/2. At some point, the measuring rod passes over a hole of length L/2. If the observer B on the ground is correct in his measurement of the length, the connecting rod will push it down into the hole when it passes over, if observer A that moves with the measuring rod is correct, the connecting rod will not be able to push it down into the hole and the rod will continue along its path.

    I got this question from http://www.youtube.com/watch?v=202fU9qIVK4", and the professor resolves the paradox by saying that one observer must conclude that the measuring rod was able to tilt. This is the reason for my forbidding the tilting of the measuring rod in the experiment. So what happens in this scenario? How are the two observations resolved?
     
    Last edited by a moderator: Apr 24, 2017
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  3. Sep 4, 2009 #2

    PeterDonis

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    You don't get to stipulate that the rod can't tilt. You can only stipulate that the rod is held *at one end* as it is being pushed along (if it were being held at both ends, it wouldn't fall through the hole, even if you worked the problem in the frame where the hole looked large enough to let it). But the rod is moving fast enough that the force exerted at the end where it's being pushed, which is the back end, can't "catch up" to the front end of the rod in time to prevent it from tilting and falling through the hole.
     
    Last edited by a moderator: Apr 24, 2017
  4. Sep 5, 2009 #3

    Dale

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    You are forgetting that there is no such thing as a perfectly rigid material (although you can have perfectly rigid motion). Any change in strain will propagate through that material at the speed of sound in the material which is always much less than c. Because of this supporting structure will actively push the rod into the hole because the front end will act like a coiled spring.
     
  5. Sep 5, 2009 #4
    I think I have somewhat of a more intuitive understanding of the tilting now, please correct me if this is the wrong way to think about it. So let's say that observer A is in a train of length L moving at a speed v comparable to c, and observer B is in a train, also of length L, which is at rest. Before the relativistic correction, when observer A passes observer B, the front ends of the train should meet up, and the back ends of the train should also meet up. One should be able to draw a line perpendicular from the side of the train at both the front and back end to show this. With the relativistic correction, however, each observer sees the other observer's train as having been shortened by a certain amount, and either the line at the front end, back end, or both must be skewed by a certain amount to connect the front ends and back ends of the trains, hence the apparent tilting.
     
  6. Sep 5, 2009 #5

    PeterDonis

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    This seems OK as far as it goes, but you're leaving out the time element, which is never a good idea with relativity problems. It's a necessary part of the picture that the events where the front and back ends of the trains meet up happen at different times in the two different frames.
     
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