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Barn Pole Paradox with Silo Variation

  1. Jan 14, 2013 #1
    This is the barn pole paradox using a slightly curved pole and a silo instead of a perfectly straight pole and doors on a barn. Let's say we have a pole of proper length of 1 light-year that is traveling around a circle that has a curvature of 1 meter over 1 light-year of length. Let that pole rotate around the circumference of a circle with each point of the pole having an angular velocity of V = 0.866c.

    From the point of view of the rest frame of the circle, this rotating pole's length when it is virtually parallel to the x-axis is 0.5 light-years in length (due to length contraction), with a curvature of slightly less than 1 meter over this 0.5 light-year length. We then build two cylinders with centers at the center of this circle, one with a diameter slightly larger than the circle the rotating pole is traveling around and one with a diameter slightly smaller than the circle the rotating pole is traveling around. These two cylinders now encase the rotating pole as it travels along the circumference of a very large circle. Since these cylinders are in the rest frame of the circle, we build the larger one with a diameter that has a curvature slightly larger than 1 meter over a 0.5 light-year length (just slightly larger than the pole's curvature) and the inner cylinder with a slightly smaller diameter. The rotating curved pole just fits in the space between these two cylinders as it rotates around the circle.

    Now we look at another inertial reference frame that is moving with V = 0.866c relative to the rest frame of the circle and cylinders. When the pole is moving in that same direction as this moving reference frame they have virtually zero relative velocity. This moving frame measures the length of the pole to be just slightly less than 1 light-year in length with a curvature of slightly less than 1 meter over that 1 light-year of length. This same moving frame measures the curvature of the two cylinders to have about 1 meter of curvature over 0.25 light-years of length (due to length contraction) over the arc length where the moving pole is at this point in time.

    How does the moving frame explain how this slightly curved pole fits into the space between the two cylinders that have four times the curvature?

    Thanks,
    David Seppala
    Bastrop TX
     
    Last edited: Jan 14, 2013
  2. jcsd
  3. Jan 14, 2013 #2

    PeterDonis

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    This scenario actually combines elements of the barn and pole paradox and another paradox which I have not seen a "standard" name for, but which I'll call the "rod and hole" paradox, and which I'll describe briefly below. As such, the solution is going to be considerably more complicated than the solution of either one of the above taken by itself. All I'm going to do here is briefly sketch the key concepts that are involved.

    (1) In the barn and pole paradox, the key concept is relativity of simultaneity, or more generally the fact that the time ordering of spacelike separated events is frame dependent.

    (2) In the "rod and hole" paradox, the key concept is that there is no such thing as a completely rigid body in relativity; internal forces inside objects can't be transmitted any faster than the speed of light, so if one end of an object starts responding to a force, the other end won't start responding immediately, but only after (at least) the light travel time delay from one end of the object to the other.

    The "rod and hole" paradox goes like this: suppose a rod is sliding along a frictionless surface with a hole in it. The rod is 1.01 meter long in its rest frame, and the hole is 1 meter long in its rest frame (and wider than the rod). So if the rod is at rest relative to the hole, it won't fall through it.

    Now suppose the rod is sliding over the hole at a large fraction of the speed of light. The rod is then length contracted, so it falls through the hole. But in the rod's rest frame, the hole is length contracted, so the rod should not fall through the hole. Which is correct?

    The answer is that the rod does fall through the hole. But what does the process look like in the rod's rest frame? The answer is that, when the front end of the rod starts to slide over the hole, the rod bends; the front end starts to fall while the back end is still being supported by the frictionless surface. The reason this happens is that, when the front end is no longer supported by the surface, the internal forces exerted by the back end have to change to keep the front end from falling: but the front end is moving too fast for the change in the internal forces exerted by the back end to "catch up" in time to prevent the front end from falling.

    I believe the resolution of the "paradox" you propose will basically be a combination of these two key concepts. I haven't worked through it in detail to verify that; however, I would still recommend that, before tackling the scenario you propose, you first make sure you fully understand the resolutions of the two "paradoxes" given above.
     
  4. Jan 14, 2013 #3

    Bill_K

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    "Thin Man and the Grid", I believe.
     
  5. Jan 14, 2013 #4

    PeterDonis

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  6. Jan 18, 2013 #5
    This is incorrect. Since each length element of the rotating pole is moving in a slightly different direction relative to the rest frame from the next, it is contracted in a different direction. The overall contraction is such as to maintain the curvature of 1 metre per light-year.

    That invalidates the rest of your analysis.

    Sylvia.
     
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