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Relativity Problem

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data

    An inertial observer O bounces a radar signal off an arbitrary event P. If the signal is emitted and received by O at times τ1 and τ2 respectively, as indicated by O’s clock, prove that the squared interval ∆s 2 between O’s origin event (i.e., its spatial origin at time τ = 0) and P is c 2 τ1τ2.
    2. Relevant equations
    ∆s^2=ct^2-|r|^2

    3. The attempt at a solution
    u'=γ(u1-βu4)?? I am not even sure if I need to do a Lorentz transformation. Please give me some direction!
     
  2. jcsd
  3. Feb 10, 2015 #2
    Hi. I assume your solution is supposed to read: c2τ1τ2...
    So:
    What is the distance between O and P?
    At what time on the clock does the signal reach P?
     
  4. Feb 10, 2015 #3
    Sorry, I am not the smartest. We have to use the Lorentz transformation to find that, right?
     
  5. Feb 10, 2015 #4

    phinds

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    Well then, you are not living up to your log-on name now are you? :D
     
  6. Feb 10, 2015 #5
    No: you're looking at everything from O's perspective, and O is not moving so try to see this without thinking about relativity.
    - You start the clock at 0;
    - At τ1 you send a radar signal (what does that tell you about the speed of your signal?);
    - At τ2 the signal is back to you after having bounced from P.
    So now: at what time did the signal get to P? How far did it travel from you?
     
  7. Feb 10, 2015 #6
    Is it (τ1+τ2)/2?
     
  8. Feb 10, 2015 #7
    If that's the answer to the first question: yes. Do you see why? (The trip to P takes (τ2–τ1)/2 but the clock started at 0 so you have to add τ1.)
    Now what's the speed of your signal? How far is P, then?
     
  9. Feb 10, 2015 #8
    r=c(τ2-τ1)/2
     
  10. Feb 10, 2015 #9
    Yep. Now the first answer was your Δτ, the second your Δr. So what's the invariant interval?
     
  11. Feb 10, 2015 #10
    ∆s^2=c∆t^2-|∆r|^2 (this is the invariant interval right?)
     
  12. Feb 10, 2015 #11
    Yes. Don't forget to square c as well...
     
  13. Feb 10, 2015 #12
    Okay, well I think I have it from here. Thank you!
     
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