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Pendulum on a Relativistic Train

  1. Mar 21, 2016 #1
    1. The problem statement, all variables and given/known data

    In a thought experiment, a train is moving at a speed of 0.95c relative to the ground. A pendulum attached to the ceiling of the train is set into oscillation. An observer T on the train and an observer G on the ground measure the period of oscillation of the pendulum. State and explain whether the pendulum period is a proper time interval for observer T, observer G or both T and G.

    2. Relevant equations

    Lorentz transformations and special relativity equation?


    3. The attempt at a solution

    I believe that T experiences the proper time period for the pendulum because we know that for points A and B that are static in space, a person stationary to either A or B measures the proper length and the person travelling from A to B at a speed of a fraction of light measures the proper time. However, since in this situation A and B are not stationary to G, they are stationary to T, therefore T measures the proper time for the oscillation period of the pendulum.

    Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Mar 21, 2016 #2
    So, you're thinking that I, as an observer watching the pendulum as the train passes through, will see a pendulum that is not swinging properly.

    Let's make this pendulum a simple sphere swinging from a line in a vacuum - as viewed from the train. Clearly, I will not see a sphere - because it will be foreshortened by Lorentz contraction. But the shape of that sphere doesn't matter. Also, I will see a sphere of much higher mass. But the mass of the sphere will also not affect it's period. Finally, I will see a sphere that is swinging slower than the train passengers see. So, at first it would appear that both the T's and the G's should calculate the same pendulum rate - and at least one of them should end up wondering why the pendulum isn't behaving properly.

    So your question is this: What is the difference between T's and G's calculation of the pendulum rate of swing?

    I will give you this hint: You're thinking from the "G" point of view. You need to consider what the world looks like from the "T" point of view.
     
  4. Mar 21, 2016 #3
    According to T, the pendulum appears to be swinging like a normal pendulum. T is also stationary relative to the equilibrium of the pendulum's oscillation so it should measure proper time for the period. T also measures the proper length. To G, the pendulum would appear to swing faster from left to right (assuming the train travels from left to right) and slower from right to left.
     
  5. Mar 21, 2016 #4
    You need more hints.
    First, let's make this simple. Have the train moving from west to east - and the pendulum swinging north and south.

    Let's make the length of the pendulum 2.3 meters. Since that length is perpendicular to motion, it will be 2.3 meters for both T and G.
    Here is a pendulum calculator:
    https://www.easycalculation.com/physics/classical-physics/simple-pendulum.php

    Compute the period as seen by T and G based on that calculator.
    Then, compute the ratio of the pendulum frequencies (T/G) based on the relativistic velocity.
    If the results do not agree, explain why.
    If you get to that point and it still hasn't struck you what is going on, I'll give you one more hint.
     
  6. Mar 22, 2016 #5
    You're right. Only the T sees proper time - because proper time is from the view of the pendulum.

    I was over-working the problem. The force of gravity in the proper time reference frame is much higher than for G, so the pendulum will be swinging much faster for T than for G. But that has nothing to do with proper time.
     
  7. Mar 22, 2016 #6
    Okay, thanks for your help!
     
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