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Lorentz Transform Problem

  1. Jun 29, 2010 #1
    1. The problem statement, all variables and given/known data

    A space ship of proper length Lp = 400 m moves past a transmitting station at a speed of v = 0.76c. At the instant that the nose of the spaceship passes the transmitter, clocks at the transmitter and in the nose of the spaceship are synchronized to t = t' = 0. The instant that the tail of the spaceship passes the transmitter a signal is sent and subsequently detected by the receiver in the nose of spaceship.

    (a) When, according to the clock in the spaceship, is the signal sent?
    (b) When, according to the clock at the transmitter, is the signal received by the spaceship?
    (c) When, according to the clock in the spaceship, is the signal received?
    (d) Where, according to an observer at the transmitter, is the nose of the spaceship when the signal is received?

    2. Relevant equations

    Lorentz transform

    3. The attempt at a solution

    I let S be the reference frame in which the transmitter is stationary and S' be the reference frame in which the spaceship is stationary.

    Then the equation for the position of the nose of the spaceship is x' = 0 in frame S', and x = vt in frame S.

    Parts (a) and (c) are easy because they do not require the Lorentz transform.

    The answer to (a) is 400/(0.76c) = 1.7556 * 10^-6.

    The answer to (c) is 400/(0.76c) + 400/c = 3.0899 * 10^-6.

    For (b), I tried to use the Lorentz transform from what I got in (c).

    I get that (x', t') = (0, 3.0899 * 10^-6) maps to (x, t) = (1083.21, 4.7542 * 10^-6).

    But the correct answer is 6.32 * 10^-6, not 4.75 * 10^-6.

    I think the problem is that I am not interpreting the Lorentz transform correctly. What is the meaning of (x, t) = (1083.21, 4.7542 * 10^-6)? And also, how do I get to the correct answer?

    Thanks.
     
  2. jcsd
  3. Jun 29, 2010 #2

    vela

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    Everything you did looks right, and I'm getting the same answers you did.
     
  4. Jul 5, 2010 #3
    Thank you for the reply.

    I got the solutions manual from my teacher. In it, the answer to (b) is given by:

    [tex] \frac{1}{\sqrt{1-0.76^2}} \left[ 3.09 \times 10^{-6} \text{ s} - \frac{(-0.76c)(400 \text{ m})}{c^2} \right] = 6.32 \times 10^{-6} \text{ s} [/tex]

    Can someone explain that, or is it a mistake?
     
  5. Jul 5, 2010 #4

    vela

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    It's a mistake. The solution is incorrectly using x'=400 m, but the way the problem is set up, the nose of the spaceship is defined as x'=0 m.

    You could also do the calculations in the transmitter's frame. In this frame, the ship is length-contracted to 260 m, so the signal is sent when t=1.14 μs. If you calculate where the worldlines of the signal and the nose of the ship intersect, you'll get t=4.75 μs, as expected.
     
  6. Jul 9, 2010 #5
    Thank you! That made the problem a lot clearer!
     
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