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Another length contraction paradox

  1. Apr 20, 2008 #1
    This is similar to the pole and barn paradox, but different enough that I'm not quite confident in my answer:

    A train and tunnel have the same length when at rest relative to each other. The train approaches the tunnel with constant velocity. An engineer on the train places rockets on the front and rear, equipped with timing devices such that the rockets will be launched simultaneously, in a vertical direction, when the midpoint of the train passes the midpoint of the tunnel. Do the rockets cause any damage to the tunnel?

    My thought here is that "timing devices" is key. It seems to mean that the engineer somehow calculates at what time he will pass the midpoint, as opposed to sending a signal to the rockets when that happens.

    In any case, the obvious answer is that since the tunnel is contracted in the train's frame, the rockets don't hit the tunnel. That also must mean that the rear rocket it launched first in the tunnel frame--before the midpoint is reached. Is that right?

    It doesn't seem right to me if I consider a timing device on the train. Say it's set to go off at t=10 (calculated by the engineer to correspond with the midpoint passing) and it's started when the front of the train passes the front of the tunnel. In the tunnel frame, the timer runs slow, so if the rocket goes off at some time before the rear enters the tunnel, it has gone off before the timer reaches 10. What am I missing here?
     
  2. jcsd
  3. Apr 20, 2008 #2
    I think can actually understand this if the engineer inside the train calculated when the train will be in the middle from the train's reference frame and setup the rockets be launched simultaneously when this happens.

    In the train's reference frame, the tunnel is contracted, the rockets are launched simultaneously when the train is in the middle and don't hit the tunnel. In the tunnel's frame, the back rocket is launched first before the back of the train enters the tunnel, then the train reaches the middle, then the front rocket is fired just as the train's front exits the tunnel.

    I think your problem lies in the fact that you're imagining the timer as an absolute time inside of the entire train, which I think is not possible. Is there a timer on each side of the train next to the rockets? Or is there a timer in the middle of the train that sends a signal to the rockets in each side when t=10?
     
    Last edited: Apr 20, 2008
  4. Apr 20, 2008 #3

    JesseM

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    Are you familiar with the relativity of simultaneity as a separate phenomenon from time dilation? In SR, two events that are simultaneous in one frame can happen at different times in another frame, and two clocks which are synchronized in their own rest frame, with a distance of L between them in that frame, will be out-of-sync by vL/c^2 in a frame where they're moving at speed v along the axis between them (the clock in the back will show a more advanced time in this frame).
     
  5. Apr 20, 2008 #4
    That makes a bit more sense now. I'm trying to keep it in my head that if two events at different locations are simultaneous in one frame, they can't be simultaneous in another (correct?) Which means that if two clocks at the front and back of the train were synchronized in the train's frame, they wouldn't be synchronized in the tunnel's frame. My problem, I think, was assuming that they would be synchronized in both frames.

    I guess a similar way to look at it would be that if there was one clock on the train, calculated to send out a pulse of light to the rockets so that the light would reach them right when the midpoint passed, then obviously the rear rocket would receive the signal first in the tunnel's frame (and still both would receive it at the same time in the train's frame)
     
  6. Apr 21, 2008 #5

    JesseM

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    Well, if you have more than one space dimension, then if two clocks are synchronized in their rest frame, they'll still be synchronized in a frame which is moving perpendicular to the axis between the clocks. But in any frame which has a nonzero velocity along that axis, they'll be out-of-sync.
    Right, that's exactly why different frames disagree about simultaneity--each frame defines simultaneity using the assumption that light travels at the same speed in all directions in that frame, so two clocks are synchronized in their own rest frame if you can set off a pulse of light at the midpoint between them and they both show the same time when the light reaches them. This necessarily means they'll be out-of-sync in other frames, for the same kind of reason you describe.
     
  7. Apr 22, 2008 #6
    I started thinking a bit about the timer in the middle and I think there's no way one can avoid hitting the back of the tunnel using a signal from the middle of the train.

    To simplify things, let's imagine a sensor in the middle of the train that reacts to a signal in the middle of the tunnel, when the sensor goes over the signal (meaning the middle of the train is directly above the middle of the tunnel) a signal is sent to both sides of the train to trigger the rockets (let's assume this signal has a speed of c in both directions.).

    If we think about the tunnel's frame of reference, it is impossible for the back rocket of the contracted train to miss the tunnel, since the only way this could happen would be if the back rocket was fired before the signal had been sent. This is impossible, if an event A causes an event B in any frame of reference, B cannot precede A in any frame of reference unless the signal from A to B moves faster than c.

    In the train's reference, this means there is no way the signal from the middle of the train can reach the back of the train before it enters the tunnel, no matter how contracted the tunnel is in the train's frame of reference (the more contracted it is, the faster it approaches).

    Edit: of course if you use a timer inside the train like JaWiB suggested and calculate it so the signal will reach both ends simultaneously exactly when the train is in the middle of the tunnel you can avoid blasting it, the signal would of course need to be sent way before the train is in the middle of the tunnel. In the tunnel's frame of reference, the signal would simply reach the back of the train before it enters the tunnel, and reach the front until after it has exited the tunnel.
     
    Last edited: Apr 22, 2008
  8. Apr 22, 2008 #7

    JesseM

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    That wasn't the original thought-experiment as I understood it; I understood it to mean that the rockets at the front and back of the train have timers which are programmed to fire them simultaneously with the event of the middle of the train passing the middle of the tunnel, with simultaneity defined in terms of the train's frame of reference (no actual physical signal is sent from the middle of the train to the rockets, the engineer just calculates when the middle of the train will reach that point based on the train's velocity, and sets the timers accordingly). In your version, where an actual signal has to go from the middle of the train to the rockets at either end, the firing of the rockets won't be simultaneous with the middle of the train passing the middle of the tunnel, in any frame (although in the train's frame the firing of the two rockets will be simultaneous with each other).
     
  9. Apr 22, 2008 #8
    wow, you are fast, I think you actually wrote this before I finished my edit (though it depends on the frame of reference, I guess :P ).
     
  10. Apr 22, 2008 #9

    JesseM

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    If you use a timer, you don't need a signal at all; you can just have clocks at the ends and middle which are synchronized in the train's frame, calculate what time the middle clock will show as it passes the middle of the tunnel, and then set the rockets to go off when their own clocks show the same time.
     
  11. Apr 22, 2008 #10
    Yes, of course, but I was refering to
    I don't need to avoid hitting a contracted tunnel with rockets launched from a train moving at a relativistic speed either :P
     
    Last edited: Apr 22, 2008
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