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Special relativity argument

  1. Nov 2, 2009 #1
    My friend and I disagree on this problem:

    A spaceship of proper length 200m moves with respect to
    us at 0.6c. There are two clocks on the ship, at the bow and stern, that
    have been synchronized with each other in their rest frame. We, also
    have a number of clocks synchronized in our frame. Just as the bow
    of the ship reaches us, both our clocks and the clock at the bow read
    t = 0. At this time t = 0 (to us), what does the clock in the stern of the
    ship read?

    My friend claim that the time at the stern must be zero since clocks are synchronized. But I said no, I want to apply lorentz transformation and found a positive time. Which of us is right? I used this argument:

    [tex]t=\gamma (t'+\frac{ux'}{c^2})[/tex]
    In which the prime denotes quantity in the spaceships's frame. I then set t=0 and x'=-200m and solved for t'.

    I really think that I am right but don't know why he is wrong. Please help us settle this.
     
  2. jcsd
  3. Nov 2, 2009 #2

    Doc Al

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    Staff: Mentor

    You are correct. Clocks synchronized in one frame (that of the ship, for instance) will not be synchronized when observed from a frame in which they are moving. Another way of saying this is that simultaneity is frame dependent.
     
  4. Nov 2, 2009 #3

    Doc Al

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    Staff: Mentor

    Let me add that according to the ship when its bow clock passes your clock all of the ship clocks will read zero. But to them your clocks are not synchronized. (Clock desynchronization works both ways.)
     
  5. Nov 2, 2009 #4
    Think of it like this. Say a sychronising flash at the centre of the spaceship triggers the clocks fore and aft to start ticking. The rear going start signal gets to the back first (from the point of view of the ground observers) because the rear of the ship and the signal are heading towards each other, so the rear clock starts ticking first. In the meantime the forward going signal is catching up with the front of the spaceship which is going away from it. By the time the forwards going signal reaches the front and starts the front clock the rear clock will already have:

    [tex] T = \frac{L_o v}{c^2} [/tex] seconds elapsed on it, where [itex]L_o[/itex] is the proper length of the spaceship. In your example the rear clock will show 200*0.6/1^2 = 120 seconds in the ground frame, when the front clock is reading zero.

    This fairly simple formula takes length contraction and time dilation into account and can be obtained from rearranging the transformation you quoted and using a displacement of -200m for x. This negative displacement comes about because the origin of the moving frame is the nose of the ship and the displacement of the rear of the ship from the origin is in the opposite direction to the velocity vector which is taken as positive. I hope that makes some sort of sense :P
     
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