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Time-dialation and light clock

  1. Jun 1, 2013 #1
    Can someone explain exactly(or to the best of your abilities) the changes that the light clock goes through from being in a stationary position, to reaching the speed of light? How does the light clock work, and how does the distance between "ticks" change to result in the slowing down of time when traveling at the speed of light? Thanks.
  2. jcsd
  3. Jun 2, 2013 #2
    The lightclock doesn't travel at the speed of light.

    The most fundamental thing (I think) to understanding relativity is to know the experimental observation that the speed of light is constant for everyone. Always. Regardless of motion.
    Think about why that is strange. If you are running toward me at 10m/s and I throw a ball at you at 10m/s you will see it moving to you at 20m/s right?
    Now if this is a photon and you are running toward me at speed v, and I shine light at you it will travel at speed c. You will NOT observe it coming toward you at speed c+v, you will just see it coming at speed c. That's like if the ball I threw at you were coming toward you at 10m/s instead of 20m/s.

    So now that you understand that, the rest comes directly from it. Think of a lightclock as simply 2 mirrors one on top of the other with a photon in between them bouncing back and forth (being reflected back and forth between the mirrors).
    Now imagine a situation in which a lightclock is on a spaceship traveling near the Earth but going very fast, and there is observer A standing in the spaceship (not moving relative to the lightclock also on the ship) and there is observer B standing on Earth (observer B sees the lightclock moving very fast because it's on the ship)

    Now, in 1 tick of the lightclock (photon bounces down and up) how far does observer A say the light has travelled? The answer is simply twice the distance that separates the 2 mirrors.

    But what about observer B? How far did the photon travel in 1 tick? Check out this animation:
    The moving lightclock is what observer B sees and the stationary one (on the left) is what observer A sees.

    See how observe B sees that the lightclock travels a larger distance?

    Now, simply combine this fact with the observation that both observers must still see the light traveling at the same speed and you see that the time it takes for each observer to witness 1 tick of the lightclock is:
    t = d/c
    c is constant, but observer B sees a larger 'd', so then observer B will calculate a larger 't' for 1 tick of the lightclock than observer A does.

    So that means that observer B witnesses events in observer A's reference frame (the ships frame) to take more time than a person on board does.
  4. Jun 2, 2013 #3


    Staff: Mentor

    Basically, the speed of light is invariant, which means that the time between ticks is equal to the distance the light travels (in units where c=1). That distance is greater in frames where the clock is moving, so the time is similarly longer (dilated).
  5. Jun 2, 2013 #4
    Thank you all so much for taking the time to explain this very interesting experiment/topic.
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