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Relative motion of two reference frames

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    Two clocks located at the origins of the [itex]K[/itex] and [itex]K'[/itex] systems (which have a relative speed [itex]v[/itex]) are
    synchronized when the origins coincide. After a time [itex]t[/itex], an observer at the origin of the [itex]K[/itex]
    system observes the [itex]K'[/itex] clock by means of a telescope. What time does the [itex]K' [/itex]clock read?


    2. Relevant equations

    [itex] t' = \gamma t[/itex]

    3. The attempt at a solution

    If [itex] v [/itex] is a significant fraction of the speed of light [itex] c [/itex]:

    [itex] t' = \gamma t [/itex] So the observer in the [itex] K [/itex] frame would see a time equivalent to [itex] t\gamma [/itex] has passed for the moving clock

    However, in the Newtonian limit [itex] \gamma \rightarrow 1 [/itex] so that [itex] t'=t [/itex]

    In which case the observer in the [itex] K [/itex] frame would see that his/her clock would agree with the moving clock.

    Sometimes I get turned around a bit with relativity so I'm just posting this to make sure I'm not getting turned around again, thanks for checking my work everyone.
     
  2. jcsd
  3. May 3, 2014 #2
    You're forgetting about the time delay. when you look at things with a telescope you see them in the past because it takes light some time to reach you.
     
  4. May 3, 2014 #3
    Yeah I know I'm not really sure how to account for that. What I got was what the observers in the [itex] K [/itex] frame would see, but the question is asking for what the [itex] K' [/itex] clock currently reads when the the observer in the [itex] K [/itex] frame looks at the clock. The observer won't see the same time that is actually on the clock, right?
     
  5. May 3, 2014 #4
    Let t's represent the time that the light signal left the clock in the K' frame of reference. The location of that clock is x' = 0. From the Lorentz Transformation, at what time and location ts and xs in the K frame of reference was the signal sent? How long did it take the light signal to travel from x = xs to x = 0? In terms of t's, at what time t did the light signal arrive at x = 0?

    Chet
     
  6. May 3, 2014 #5
    At what time and location ts and xs in the K frame of reference was the signal sent?

    [itex] t_s=\gamma (t_s'+\frac{\beta}{c} x') \rightarrow t_s=\gamma(t_s') [/itex]

    [itex] x_s = \gamma(x'+\beta ct_s') \rightarrow x_s = \gamma\beta(ct_s')[/itex]

    How long did it take the light signal to travel from x = xs to x = 0?

    Would it be [itex] \frac{x_s}{c} [/itex]?

    In terms of t's, at what time t did the light signal arrive at x = 0?

    [itex]\gamma\beta(t_s')[/itex]

    Would finding the distance in space-time between the events [itex](0,t)[/itex] and [itex](\gamma\beta(ct_s'),\gamma(t_s'))[/itex] also be useful in solving this problem?
     
    Last edited: May 3, 2014
  7. May 3, 2014 #6
    Yes. [itex] \frac{x_s}{c}= \gamma \beta t_{s'}[/itex]
    No. You have to add this travel time to the time that the signal was sent in the K frame of reference. It would be [itex]t_s+\gamma\beta(t_s')=\gamma(t_s')+\gamma\beta(t_s')=\gamma(t_s')(1+
    \beta)[/itex]

    So, [itex]t=\gamma(t_s')(1+\beta)[/itex]

    So, now solve for ts', which is the reading on the clock in K' that is seen by the observer at x = 0 at time t in K. (You may be able to cancel some terms between γ and (1+β)

    Chet
     
  8. May 3, 2014 #7
    [itex] t = \gamma(t_s')(1+\beta) \rightarrow t = \frac{t'_s(1+\beta)(1-\beta)}{(1-\beta^2)^{\frac{1}{2}}(1-\beta)}[/itex]

    [itex] t = \frac{t'_s(1-\beta^2)^{\frac{1}{2}}}{(1-\beta)} \rightarrow t = t'_s[\frac{(1-\beta^2)}{(1-\beta)^2}]^{\frac{1}{2}}[/itex]

    [itex] t = t'_s[\frac{1+\beta}{1-\beta}]^{\frac{1}{2}} \rightarrow t'_s = t[\frac{1+\beta}{1-\beta}]^{\frac{-1}{2}}[/itex]
     
  9. May 3, 2014 #8
    Nice job.

    Chet
     
  10. May 3, 2014 #9
    Thank you, by the way: are there any helpful hints for relativity you can offer? I seem to get turned around very easily by these problems. Thanks for the advice and help too.
     
  11. May 3, 2014 #10
    I'm not sure what works for me also works for other people. When I get bogged down, I tend to gravitate toward the Lorentz Transformation, and not toward formulas for length contraction and time dilation.

    Chet
     
  12. May 3, 2014 #11
    Okay I'll try to focus on the Lorentz transformations instead of the length contraction and time dialation formulas and see if that helps iron out some of my problems with relativity, thanks again for your help.
     
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