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Relativistic Rod and Hole

  1. Oct 3, 2008 #1
    A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the rod. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

    What does this look like in the inertial frame of the rod?
     
  2. jcsd
  3. Oct 3, 2008 #2
    the hole would appear to be smaller than the rod but the rod wouldnt be parallel to the surface as it went through the hole due to loss of simultaneity.
     
  4. Oct 3, 2008 #3

    Am. J. Phys. Vol 74 (2006) pp998-1001
    A look at the paper quoted above brings you information and references about the subject you are interested in which is known as Rindler Effect. You can find the paper going on google and giving it to look for barn hole paradox
     
  5. Oct 4, 2008 #4
    The Rindler effect is about nonuniformly accellerating observers.
     
  6. Oct 4, 2008 #5

    Dale

    Staff: Mentor

    It sounds like you are making the rod perfectly rigid. Anyway, the answer is easy, and granpa already mentioned the essence. Simply Lorentz transform the worldline of the rod ends and the edges of the hole for the details.
     
  7. Oct 4, 2008 #6
    Phrak, the way you present the problem, you are setting yourself up for a long winded dialog about acceleration. If you don't want that, there is a strictly inertial version of the problem.

    Consider this reference frame: The rod is moving to the right, constant velocity. The surface (parallel to the rod) is moving up, constant velocity. The hole in the surface is on an intersection course with the rod. Now there is no need to include any acceleration. If the proper length of the rod is slightly larger than the proper length of the hole, and velocities are high enough, relativistic physics will predict that the rod will get swept into the hole as the surface rises.

    Now your question, describe the action from the ref frame of the rod is more straightforward.
     
  8. Oct 4, 2008 #7
    You're right. I'm guilty of posing an ill-posed question, but the acceleration of the rod is kinda central. I'll try again:


    A rod is traveling at a greatly obscene velocity parallel to a surface. The surface has a hole in it shorter than the length of the rod measured at rest. As the rod passes over the hole, you gently push it through, where at all times it remains parallel to the surface.

    What does this look like in the initial inertial frame of the rod?


    I'm curious to see if anyone can come up with the correct answer.
     
    Last edited: Oct 4, 2008
  9. Oct 4, 2008 #8

    Ich

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    Simply examine what this statement means in both reference frames. It still is an ill-posed question.
     
  10. Oct 4, 2008 #9

    Doc Al

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    Those gentle pushes may occur at the same time from the hole frame, but from the rod's frame the front of the rod gets pushed down first.
     
  11. Oct 4, 2008 #10
    What does the rod look like?
     
  12. Oct 4, 2008 #11

    Doc Al

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    Like a limp piece of spaghetti as it is bent into the hole. :smile:
     
  13. Oct 4, 2008 #12
    If you are saying the hole is shorter than the rod when they are together and no relative motion, what you are describing will never happen.
     
  14. Oct 4, 2008 #13

    Doc Al

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    Why is that? If the rod is moving fast enough, it will be shorter than the hole diameter in the hole frame.
     
  15. Oct 4, 2008 #14
    Phrak, I think I know what you are getting at.

    From an observer on the surface, the rod and surface remain parallel and the contracted rod passes thru the hole without incident. From an observer on the rod, the rod and surface are not parallel and this fact allows the uncontracted rod to again pass thru the hole without incident.

    If this is also your opinion, I don't think you will get much support for it in this forum.
     
  16. Oct 4, 2008 #15
    The observer is not on the rod. But could you explain what you mean?

    Support?? I'm not running for office, and I'm not selling anything, I swear. But if you're in the market I can get help you get some prime Wet Lands in Florida. But no---that wouldn't be right. Watch yourself, it's a concrete Rain Forest out there.
     
    Last edited: Oct 4, 2008
  17. Oct 4, 2008 #16
    What does the rod look like?
     
  18. Oct 4, 2008 #17

    Dale

    Staff: Mentor

    I thought granpa already described how it would look quite well. Perhaps you could be more specific on the form of the answer you are seeking. Otherwise, from what I can tell it has already been provided.
     
  19. Oct 4, 2008 #18
    "...wouldn't be parallel" can mean a lot of things.

    But Doc Al had something to say about spaghetti.
     
  20. Oct 5, 2008 #19

    Dale

    Staff: Mentor

    Then do you want the angle that it goes through the hole at? You haven't given enough information to specify that, so I think granpa's statement is the best you can do. If you want any more detail you will have to exactly specify everything in one frame and then simply transform it to any other frame you like.
     
  21. Oct 5, 2008 #20
    Dale, as far as I can tell granpa could, and probably was talking about a stright rod.

    In the inerial frame of the observer the rod remains parallel to the hole. Then, necessarily, the rod cannot be ridgid (Doc Al's oblique reference to spaghetti).

    In the inerital frame, in the initial trajectory of the rod, the rod is mechanically deformed as it is pushed through the hole. The leading and trailing ends remain parallel to the plane, above and below it.

    The rod has an lazy sigmiodal bend over the hole, as the force pushing it through the hole propagates from front to rear.
     
    Last edited: Oct 5, 2008
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