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Homework Help: Relativistic signal passing

  1. Sep 28, 2010 #1
    I was solving this problem, and I was comparing my answers with a classmate, and we got very different answers. I don't think I made a mistake here, but I may have missed something.

    1. The problem statement, all variables and given/known data
    Tipler - Modern Physics 5e 1-57.
    "Two observers agree to test time dilation. They use identical clocks and one observer in frame S' moves with speed v = 0.6c relative to the other observer in frame S. When their origins coincide, they start their clocks. They agree to send a signal when their clocks read 60 min and to send a confirmation signal when each receives the other's signal.

    (a) When does the observer in S receive the first signal from the observer in S'

    (b) When does he receive the confirmation signal?

    (c)Make a table showing the times in S when the observer sent the first signal, received the first signal, and received the confirmation signal. How does this table compare to the one constructed by S?"

    [tex]\beta = 0.6[/tex]

    [tex]\gamma=\frac{1}{\sqrt{1 - \beta^{2}}}[/tex]


    3. The attempt at a solution

    a) The observer in S' sends their first signal at d' = 0,ct'. To find where this is in S we do an inverse 1D Lorentz transform.

    [tex]\gamma\left[\begin{array}{cc}1&\beta\\\beta&1\end{array}\right]\left[\begin{array}{cc}0\\ct'\end{array}\right]=\left[\begin{array}{cc}\gamma\beta ct'\\\gamma ct'\end{array}\right][/tex]

    This is the point in S when the observer in S' sends the signal. This signal travels at the speed of light in the -d direction until it hits the ct axis.(d = 0)


    [tex]ct_{light} = -d_{light} = \gamma\beta ct'[/tex]

    The total ct is

    [tex]ct\ =\ ct_{light} + ct_{hour}\ =\ \gamma\beta ct' + \gamma ct'\ =\ \gamma(1 + \beta)ct'[/tex]

    [tex]\gamma\ =\ \frac{1}{\sqrt{1 - 0.6^{2}}}\ =\ 1.25,\ \beta = 0.6,\ t' = 1\ hr.[/tex]

    we get t = 2 hrs.

    b)First, we need to find our when S' received the first message. Here we want to transform S --> S', so we use the forward Lorentz transform:

    [tex]\gamma\left[\begin{array}{cc}1&-\beta\\-\beta&1\end{array}\right]\left[\begin{array}{cc}0\\ct\end{array}\right]=\left[\begin{array}{cc}-\gamma\beta ct\\\gamma ct\end{array}\right][/tex]

    This is the point in S' when the observer in S sends the signal. This signal travels at the speed of light in the d' direction until it hits the ct' axis.(d' = 0)


    [tex]ct'_{light} = d'_{light} = -(-\gamma\beta ct) = \gamma\beta ct[/tex]

    The total ct' is

    [tex]ct'\ =\ ct'_{light} + ct'_{hour}\ =\ \gamma\beta ct + \gamma ct\ =\ \gamma(1 + \beta)ct[/tex]

    This comes out to t' = 2 hr.

    Now we can use the general formula in S for the confirmation pulse:

    [tex]ct\ =\ ct_{light} + ct_{hour}\ =\ \gamma\beta ct' + \gamma ct'\ =\ \gamma(1 + \beta)ct'[/tex]

    with t' = 2 hrs. we get t = 4 hrs. as the arrival time of the confirmation pulse.

    Time first signal sent: +1 hour.
    Time first signal received: +2 hours.
    Time confirmation received: +4 hours.

    The table should be the same with observer S'.

    Attached is a Minkowski diagram of the scenario from the perspective of S which seems to confirm my calculations.

    My classmate got a bit over 4000 seconds for a) and just over 2 hours for b).

    Attached Files:

    Last edited: Sep 28, 2010
  2. jcsd
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