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Special Relativity and Ship Problem

  1. Oct 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A ship is traveling at .8c, and sends a message to earth to saying he's fine. 15 minutes later he realizes he made a typo and sends another message. How long does it take for the second to message to arrive at earth after the first one arrives?




    2. Relevant equations



    3. The attempt at a solution
    So my class is currently in a debate whether the correct answer is 45 minutes or 20 minutes. The teacher says it's 20 minutes. I feel like 20 minutes is the answer it takes for the light to travel from the ships final position to the final position. I feel like the 20 minute answer is failing to take into account the 15 minutes the spaceship travels in it's reference frame BEFORE it sends the signal. Isn't 20 minutes only the correct answer if the signals are sent from point A and point B simultaneously???

    Any grad student out there want to crack this problem for me real quick? Is my teacher correct? We have to add 25 minutes (15*γ) to the 20 minutes it takes from the signal to cover the distance the spaceship has moved in the earths frame.
     
  2. jcsd
  3. Oct 18, 2013 #2

    UltrafastPED

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    The 15 minutes is in the ship reference frame - so from the author's point of view the two messages are sent and received 15 minutes apart.

    But the author knows that clocks are running slow on earth, so their reaction is delayed.

    So multiply the 15 minutes by the clock rate that goes with .8c. What do you get then?
     
  4. Oct 18, 2013 #3
    That would turn it into 25 minutes. But then we also have to add to that to the time it takes for light to travel the distance the spaceship covers in that time.
     
  5. Oct 18, 2013 #4

    UltrafastPED

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    Yes ... at .8c for 15 minutes what additional time do you get? This would be added to the original 15 minute delay; then apply the Lorentz factor.
     
  6. Oct 18, 2013 #5
    Let's use the Lorentz Transformation to look at the following 3 events:

    Event 1: Message 1 sent at x' = 0, t' = 0, x = 0, t = 0 (Ship just passing earth)
    Event 2: Message 2 sent at x' = 0, t' = 15 min
    Event 3: Message 2 arrives at earth x = 0.

    What are the S frame coordinates of Event 2?

    x = γ(x' + vt')=γvt'
    t = γ(t'+vx'/c^2)=γt'

    Now for event 3: Event 3 takes place in S frame.
    Distance in S frame that signal travels = γvt'
    Time signal arrives = t+γvt'/c=γt'+γvt'/c = γt'(1+v/c)

    Substituting the numbers: 25(1+0.8)=45min

    So it looks like the teacher is wrong. Of course, this assumes that the 15 minute time delay between signals is measured on the ship's clock.
     
  7. Oct 18, 2013 #6

    UltrafastPED

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    Ship is traveling at 0.8c wrt earth, so Lorentz factor = 10/6.

    Ship sends message 1 (at light speed), travels for 15 minutes at 0.8c (12 light minutes) and sends message 2.
    So by ship time the messages are 15 minutes apart, but the second message takes 12 minutes longer to arrive -
    so by ship time they will be seen to arrive at earth 15+12=27 minutes apart.

    By earth time this will be 27 * Lorentz factor = 27*10/6 = 45 minutes between message 1 and message 2.

    So I agree w/Chestermiller calculation - and the teacher's answer needs review and a careful explanation.
     
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