Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need a little help please

  1. Sep 20, 2007 #1
    im at a youtube discussion and if it would be possible i would like anyone with understanding of Einsteins special theory of relativity to give me some information on a certain issue: it is about the famous train experiment, the one about simultaneity. if you use a rod to mesure a distance from the front of the train to the middle, it would mesure a shorter distance than a slower reference frame because light takes less time to arrive at the middle from the front, that is, the front tip "mesures first". what about from the back to the middle? i know from high school that distances shorten in the direction of the movement, but, if it is the fact that light arrives first that determines lenght and time, what about the light rays that take longer from the back to reach the front?

    another problem i have is with this, posted by a youtube user:
    "Very good loupax! You're right! This video is incorrect in it's depiction of the paradox. The paradox is that BOTH observers see the light hitting them AT THE SAME TIME and BOTH observers see the SAME light hitting the other person at DIFFERENT TIMES. Watch "The Mechanical Universe and Beyond", episode 42 of 52, the Lorentz Transformation for a full explanation."
    is this true?

    thanks in advance
     
  2. jcsd
  3. Sep 20, 2007 #2

    JesseM

    User Avatar
    Science Advisor

    The other youtube user is wrong, if some events coincide at a single time and place in one frame, like two light beams hitting a person at the same moment, then those events must coincide in all frames--different frames cannot disagree about purely local facts like this.

    The thing to understand about simultaneity is that if the observer on the train has clocks at either end which are synchronized in his frame, then the same two clocks will be out-of-sync in the frame of the observer on the tracks. This is because in relativity, each observer defines what it means for clocks to be "synchronized" using the assumption that light travels at the same speed in both directions in his frame. So, the train observer might synchronize clocks at the front and back of the train by setting off a flash of light at the midpoint between them, then setting both clocks to read the same time at the moment the light from the flash reaches them. But if he uses this method to synchronize his clocks, naturally the observer on the tracks will judge the clocks to be out-of-sync, since in his frame the clock at the back of the train is moving towards the point the flash was set off, while the clock at the front is moving away from that point, so if he assumes light travels at the same speed in both directions in his own frame he must conclude the light reached the back clock before it reached the front clock. If you work things out it turns out that any time two clocks are synchronized and a distance x apart in their own rest frame, then in a frame where they are moving at speed v along the line they both lie on, the clock in front must be behind the clock in back by a time of vx/c^2.

    So, in the reverse situation where two flashes happen at the front and back of the train and travel inwards, you have to keep in mind that the two frames disagree about whether the flashes happened "at the same time" or "at different times". If the train is 10 light-seconds long in its own frame, and it's moving at 0.6c relative to the track observer, then the time of the clock in back must be ahead of the time of the clock in front by 6 seconds in the track-observer's frame. So if both flashes happen at t'=50 seconds according to the clocks on the train, in the track-observer's frame the first flash happens next to the clock in back when it reads 50 seconds, while the front clock only reads 44 seconds at that moment...and the clock in front is ticking at 0.8 the normal rate due to time dilation, so it takes an additional 7.5 seconds according to the track-observer's clock for the clock in front to tick 6 more of its own seconds and reach a time of 50 seconds, at which point the lightning hits next to the front of the train.

    Let's say in the track-observer's frame, the first strike at the back happens at t=0 seconds according to his own time-coordinate, and it happens at a position of x=-10 light-seconds according to his space-coordinate. If the train is 10 light-seconds long in its own rest frame, in the track-observer's frame it's shrunk to 8 light-seconds due to length contraction. So, summing up what's happening when the first lightning bolt strikes:

    t=0 seconds in track-observer's frame
    back of train at x=-10 l.s.
    middle of train at x=-6 l.s.
    front of train at x=-2 l.s.
    clock at back of train reads t'=50 seconds, clock at front reads t'=44 seconds (clock at middle would be halfway between, showing t'=47 seconds)

    Then the lightning hits the front of the train 7.5 seconds later in the track-observer's frame, at which point the train has moved forward a distance of (7.5 s)*(0.6c) = 4.5 light-seconds. So, summing up what's happening when the second lightning bolt strikes:

    t=7.5 seconds in track-observer's frame.
    back of train at x=-5.5 l.s.
    middle of train at x=-1.5 l.s.
    front of train at x=2.5 l.s.
    clock at back of train reads t'=56 seconds, clock at front reads t'=50 seconds (clock at middle shows t'=53 seconds)

    And finally, at t=10 seconds in the track-observer's frame, we have:
    back of train at x=-4 l.s.
    middle of train at x=0 l.s.
    front of train at x=4 l.s.
    clock at back of train reads t'=58 seconds, clock at front reads t'=52 seconds (clock at middle shows t'=55 seconds)

    and this is when the light from both lightning flashes is now meeting at the origin at x=0 l.s. too. Remember, the first flash happened 10 seconds earlier at a distance of 10 l.s., so if it moves at c in the track observer's frame, this is where it should be now; and the second flash happened 2.5 seconds earlier at a distance of 2.5 l.s. from the origin, so if it moves at c in the track observer's frame, this is where it should be now too. So although the two flashes happened at different distances from the origin where the track-observer is standing, they also happened at different times in his frame, in such a way that the light from each one reaches him at the same time.

    Meanwhile, in the train-observer's frame both flashes happened simultaneously at t'=50 seconds according to synchronized clocks next to each flash when it happened, and the light from each one reached the center of the train when the clock there read t'=55 seconds, which makes sense since in the train-observer's frame the distance from the middle of the train to either end is 5 l.s.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Need a little help please
  1. A little math help? (Replies: 22)

  2. A little geodesic help (Replies: 5)

Loading...