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Modificaition in pole and barn paradox

  1. Mar 27, 2009 #1
    In the pole and barn paradox, if the observer (person A) closes the both doors simultaneusly (with respect to himself), when he sees the runner's stick (person B) is just inside and keeps it closed forever after that, how will things happen, will the runner be traped or the door will hit the stick?
    For those who don't know it here is the paradox.
    Person A is on the ground and person B is running with a long horizontal stick at relativistic speed. Then the length of the stick will appear to contract for A. So, he develops a cage equal in length of the observed length of the stick. He then asks the runner to run into the cage, and at the instant the runner appears to be completely in he closes both the door, for an instant. So, for that instant the stick should have to be contracted for Person B as well as it is fitting in a cage smaller than its proper length for B.
    (it is explained by saying that the closing of the door doesn't occurs simultaneusly so, even if A sees the stick completely inside for a instant its not the case with B). But what for my above modification?
     
    Last edited: Mar 27, 2009
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  3. Mar 28, 2009 #2

    Fredrik

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    From the runner's point of view, the door on the far side of the barn closes first. So the pole and the runner crashes through it when the back end of the pole is still outside. A short time after that, the back end of the pole is inside the barn, and the other door is closed.
     
  4. Mar 28, 2009 #3

    HallsofIvy

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    What does the runner do once he is inside the barn? If he keeps running then, of course, the pole hits the back of the barn. If he stops running,then, from the point of view of an observer stationary with respect to the barn, his pole regains its full length and- hits the back of the barn.
     
  5. Mar 28, 2009 #4
    Ok, fine Fredrik. But it may be interesting to see what the still observer sees. As he sees, both doors closing simultaneusly will he see the first door break even befor it is closed or what?
     
  6. Mar 28, 2009 #5

    Doc Al

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    The ground observers will "see" both doors close, followed by the front end of the pole smashing into the forward door.
     
  7. Mar 28, 2009 #6

    Fredrik

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    Nothing magical happens. The pole is shorter than the barn, so it doesn't touch either of the two doors at the precise moment when they're both slammed shut.

    What's important to note here is that the order of the following two events isn't the same in the two frames:

    Event 1. The pole hits the second door.
    Event 2. The first door is closed.

    If the runner could somehow grab hold of the front of the pole and stop instantaneously at an event that the outside observer considers to be after the first door is closed, then the force he applies to the pole would compress it. (More like converting the whole thing into a quark-gluon plasma and blowing up both the barn and the city it's in actually). If the pole is made of "unobtainium" so that it doesn't break, then it would expand back like a spring that's been compressed, and punch a hole in the first door.

    Of course there's no such thing as unobtainium, but then there aren't really any pole vaulters running at 0.8c either.
     
  8. Mar 28, 2009 #7
    For whom is the pole shorter than the Barn???? Fredrik. ----I think you mistyped it. Anyway--- I am thinking like this
    Ok, the runner goes inside, just when front end of the pole is about to escape out of the barn, front door slams in (and if he could breakthrough it, just when the back end goes in, back door would slam in) since I said, he can't breakthrough, he had to stop immediatly so the for him the barn will attain its full length now, and his pole will be traped inside, when then the back door slams in. Whats wrong here?
     
  9. Mar 28, 2009 #8
    The information that the front of the rod has stopped dead travels at the speed of sound through the rod. Even though the front of the rod is stopped by the door, the back of the rod keeps going. He can't "stop immediately." He doesn't even know immediately that the front of the rod is stopped.
     
  10. Mar 28, 2009 #9

    sylas

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    The original problem statement said that the pole is trapped in the barn.

    Hence. From the perspective of the the barn, the pole enters the barn. Both doors are slammed shut while the pole is entirely inside the barn. The pole slams into a door and comes to a screeching halt. It takes some amount of time for the compression wave from the front of the pole to propagate through to the back of the pole. Hence the back of the pole stops sometimes after the front of the pole. The pole is compressed as a result of the collision.

    Let's put some numbers on this. For convenience, I'll use units in which the speed of light is equal to 1. I'll measure time in microseconds. I'll measure distance in light-microseconds, which would be about 300 meters. I'll make the pole as rigid as I possibly can, to minimize compression.

    (A) From the point of view of the barn

    There's a pole of length 4, velocity 0.6, moving into a barn of length 5. At some point while the pole is entirely within the barn, both doors are shut.

    The pole hits the rear door at a time I'll call zero. From the collision, a compression wave propagate back up the pole. By making the pole as rigid as possible, I'll let this be the speed of light. At the same time, the rear of the pole is moving in the other direction at 0.6. The two meet at a distance of 2.5 from the collision. With maximum possible rigidity, the pole compresses to this size, simply by virtue of being squashed in the collision. The compression wave moves back up the pole at speed 1, and the rear of the pole comes to rest at 2.5 micoseconds after the collision. The length of the compressed pole is now 2.5, half the size of the barn, and the pole is at rest inside the barn.

    (B) From the point of view in which the pole is initially at rest

    Now what does someone running with the pole see? Assume that they remain inertial, by virtue of watching from outside.

    The gamma factor is 1.25. Hence they see a barn approaching the pole, with velocity 0.6C, and with a length of 4. The pole, at rest, has a length of 5.

    The barn approaches the pole at high speed. Sometime after the front of the pole is swallowed by the moving barn, the rear door of the barn slams shut, and there's a collision between the pole and the door. The barn does not slow down in the slightest, and the front of the pole is pushed at high speed (0.6C) towards the back of the pole, which is still at rest. The pole, being infinitely rigid, has a pressure wave propagating back along the pole at velocity c. It takes 5 microseconds to reach the end of the pole, because that is the length of the pole. After this, the entire pole is accelerated up to the velocity of the barn. It is also compressed. In those 5 microseconds, the front of the pole has been moving towards the back at 0.6, and so the front of the pole is pushed up a distance 3 towards the back. The pole is now of length 2; half the length of the barn. It's inside the barn, moving at the same speed of 0.6.

    In the meantime, the rear of the pole has been approaching the front of the barn. At the instant of collision, the rear of the pole is still distance 1 outside the barn, and it takes times 1/0.6, or about 1.67 microseconds, for the back of the pole to pass inside the barn; still well before the compression is complete.

    The spacetime distance between the closing of the two doors is a space-like distance of 5. This is an invariant for all inertial observers. From the the frame in which the pole is initially at rest, the front door closes some time t after the back door has closed. Hence the distance between the two door closing events is 4 (length of the barn) + 0.6t. We have (4+0.6t)2 - t2 = 25, and so 9 - 4.8t + 0.64t2 = 0, for which t = 3.75. This is well after the whole compressing pole has passed into the barn (at t = 1.67), but before the pole has finished compressing (at t = 5).

    That is, 3.75 microseconds after the rear door of the moving barn shuts, the front door closes. This is a bit before the pole is fully compressed, but it is significantly after the back of the pole has entered the barn.

    Cute, heh?

    Cheers -- Sylas
     
  11. Mar 29, 2009 #10

    Fredrik

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    I didn't. For the observer at rest relative to the barn, the pole is shorter than the barn...at least in the version of this problem that I'm used to seeing, but I see now that you specified that the two lengths are exactly the same in this frame. That makes the problem really weird. The doors can't close with the pole inside unless we assume that they move at infinite speed, and in addition to that, we must also specify e.g. that the barn and pole occupies an open interval (a,b) of the x axis while the doors are at a and b. (Alternatively, that the barn and pole occupies [a,b] and the doors [a-d/2,a) and (b,b+d/2] where d is the thickness of the door).

    You can avoid these weird complications completely (and they really are completely uninteresting) just by choosing different numbers, e.g. that the rest length of the pole is L, the rest length of the barn is 3/4 L and v=0.8, so that the length of the pole in the barn's rest frame is 0.6L.

    OK, let's disregard the fact that the doors would have to move at infinite speed. You have specified two events: A) The front of the pole reaches the second door and B) the rear of the pole gets to the point where the first door will be when it's closed. Those events are simultaneous in the barn's rest frame, but not in the runner's rest frame. To the runner, A happens first. You have already specified that B will happen too, so that rules out the possibility that we're making all parts of the pole stop at the same time in the runner's frame. (We could, in principle, have a machine that grabs every part of the pole simultaneously in the runner's rest frame, but this would guarantee that B never happens. This machine clearly stops the rear end of the pole when it's outside the barn).

    If the only external force on the pole in the runner's rest frame (and I mean the inertial frame that's co-moving with him before the pole hits the door) is the force from the door, then I have already explained what will happen. See my previous post, and also ZikZak's comment about the speed of sound. If instead, we imagine a machine that grabs every part of the pole simultaneously in the runner's rest frame, then it will be "trapped" inside the barn, as you said. The forces from the machine have squeezed it into a shorter length. The pole has been forcefully compressed by exactly the right amount to compensate for the Lorentz contraction that's no longer present.

    Note that the machine that grabs every part of the pole in the runner's rest frame is forcefully compressing it, and the machine that grabs every part of the pole in the pole's rest frame is forcefully stretching it. This is a good way to see that there can't be any absolutely rigid bodies in SR. "Rigid" is supposed to mean that every part moves the same way at all times, but if they move the same way at all times in one frame, they don't in another.
     
  12. Mar 29, 2009 #11
    If you use yet another variation of the paradox, it makes things a lot simpler while giving a good feel for what's going on:

    Imagine the barn contains a set of brakes that can brake a part of the pole infinitely quickly. So as soon as one of the brakes slams shut, that part of the pole remains where it is. The bakes are quite small, but there are a lot of them along the length of the barn.

    From the point of view of an observer who is stationary relative to the barn, all the brakes act simultaneously as soon as the pole is in the barn, so it is held in position inside. It will be under a lot of stress as it has been prevented from expanding while being brought to a standstill but, since the brakes are very strong, it will remain contracted.

    From the point of the view of the runner, the brake on the forward part of the pole slams shut first, while the back end continues to move forward, contracting the pole. Then the next brake is activated, then the next, etc... Finally the last brake slams shut when the pole is inside. The pole will be under a lot of stress because of the way different parts of it were decellerated, but since the brakes are very strong, it will remain contracted.

    Of course the situation becomes a bit more complex when you don't use brakes but just let the pole slam into the door. You just have to figure out what happens in one reference frame (taking into account things like the fact that information about hitting doors can never travel faster than light speed), and you'll find a perfectly equivalent observation from the other point of view.


    By the way, I'd love to see the guys from MythBusters try this one out some day ;)
     
  13. Mar 29, 2009 #12
    The pole-barn experiment has been banned in most rural areas because too many high speed poles have missed the barn and impaled the farmers cows.

    Even if allowed, the ground observer would see an expanded pole, and would not expect to capture it within the barn.

    Using the x-transformation, plug in -v, and see what results.
     
  14. Mar 30, 2009 #13
    ROTFL :rofl:
     
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