Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Length contraction and rod falling thorough a grate ?

  1. Aug 31, 2010 #1
    If i have a rod of length L at rest . And there is a hole up ahead that is L/2 and I move the rod at a velocity which will contract it to half its rest length. Will the rod fall into the grate?
    Well I think that from the rods point of view its still its rest length and it would appear that the hole is moving at it and the hole would appear length contracted and so the rod would not fall into the hole. Is this correct?
    But then I guess we could say that the hole is stationary and the rod is moving at it and the the rod would appear to be length contracted from the holes point of view.
     
    Last edited: Aug 31, 2010
  2. jcsd
  3. Aug 31, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    This is a variation on the barn-pole paradox, and your version is even mentioned on that Wikipedia page.

    If you don't want to read the solution right away, let me give you a hint: consider the events "front of the rod reaches the front of the gap" and "back of the rod reaches the back of the gap" and examine their simultaneity in both frames.
     
  4. Aug 31, 2010 #3
    Thanks for your answer, so in the one frame it will appear that the rod gets bent into the hole why does it appear to bend?
     
  5. Aug 31, 2010 #4

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    The whole clue to the paradox is, that events that are simultaneous for both ends in one frame, are not simultaneous in another frame.

    So for example, suppose that you assume that the rod falls vertically into the hole. Then physically: "the downward acceleration only starts once both ends are over the gap". The part in italics is the problem, because in one of the frames, both ends are never over the gap at the same time.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook