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Special relativity of space traveler and reference frames

  1. Jul 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A space traveler takes off from Earth and moves at speed 0.99c toward star Vega, which is 26.00 ly distant. How much time will have elapsed by earth clocks when the traveler reaches Vega?


    2. The attempt at a solution

    I looked at the problem solution, which demonstrated this in a much simpler way -- but I want to make sure that I can understand the process of doing it the longer way.

    I thought to use the Lorentz transformation equation:

    ∆t = [∆t' + (v∆x')/c^2)] * 1/[sqrt(1-(v/c)2]

    where ∆t is the time as experienced by earth and ∆t' is the time experienced by the traveler.

    In order to find ∆t' I first found the new distance between Earth and Vega as experienced by the traveler according to length contraction, and then divided by the speed of the traveler.
    I'll use D° for the proper distance (26 ly) and D for the adjusted distance:

    D = D°/gamma = 2.459 x 10^17 meters * sqrt(1-(v/c)2)
    D = 3.4699 x 1016 meters.

    So now I can find my ∆t' based on this distance and the given velocity:

    ∆t' = D/v = 3.4699x1016/(.99c) = 1.168 x 108 seconds, which is 3.70 years.

    Now, because I have ∆x' (the contracted distance) and ∆t', I should be able to solve for ∆t based on the equation from above:

    ∆t = [∆t' + (v∆x')/c^2)] * 1/[sqrt(1-(v/c)2]

    ∆t = (1.168 x 108 + [(0.99c)(3.46989x1016)]/c2)*gamma

    For the second term, I calculated 1.145 x 108 seconds, added that to my ∆t' and then multipled by the Lorentz factor 1/sqrt(1-(v/c)2) and ended up getting 51.9 years -- the answer should be 26.26 years.

    I'm not totally sure where I went wrong -- maybe it was my algebra/calculator but I assume it's something else. It seems that my ∆t' is correct because I got the second part of this question right (not included). Any thoughts?
     
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  3. Jul 29, 2012 #2

    TSny

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    Hi 09jlk.

    As you noted, you're getting it the long way around as a check on the shorter method. Good. That's a great way to test your understanding.

    You're dealing with the time between two events (traveler leaving earth and traveler arriving at Vega). What is the value of x' for each of these two events?
     
  4. Jul 29, 2012 #3
    So for x value -- I start at x=0 (traveler leaving Earth)...I know that if the traveler is going at 0.99c along this path of 26 light years, then her perceived length of path should be x' = x/γ = 26 ly * sqrt(1-(v/c)2). Converting to meters from light years and plugging in 0.99 for v/c, I got 2.459 x 10^17 meters * 7.09 (Lorentz factor) to get a ∆x' or 3.469 x 10^16 meters.
     
  5. Jul 29, 2012 #4

    TSny

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    You have to think of the traveler as carrying along her own coordinate system (the x' coordinate system). Suppose the traveler is located at the origin of this coordinate system. What is the value of x' at the event where she leaves the earth? What is the value of x' at the event of arriving at Vega?
     
  6. Jul 29, 2012 #5
    Hmm, I'm not sure. This is where I'm getting most confused -- the stationary (earth) coordinate system versus her own moving coordinate system... So if she is carrying the coordinate system with her, then does that mean I should find a negative change in x?

    Why wouldn't I be able to use the number I calculated for the lorentz contraction path? (x/γ)?
     
  7. Jul 29, 2012 #6

    TSny

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    Always remember that the Lorentz transformation equations t = γ*(t'+x'v/c2), etc. apply to the space and time coordinates of a single event.

    So, if we let subscript 1 refer to the event of the traveler leaving the earth, we would have
    t1 = γ*(t'1+x'1*v/c2)

    Likewise you get an equation with subscript 2 representing the event of arriving at Vega.

    You may then subtract the two equations to get the equation you are using. Then Δx' is just standing for x'2 - x'1. But, Δx' does not represent the distance between Vega and the earth as measured in the x' frame.

    [Edited to correct a sign error]
     
  8. Jul 29, 2012 #7

    TSny

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    You need to ask yourself, where does event 1 occur along the x' coordinate axis and where does event 2 occur along the x' coordinate axis.
     
  9. Jul 29, 2012 #8
    Hm, that seems counterintuitive, since I am able to use the new distance (the one I was using before) and divide by the traveler's velocity in order to get ∆t' (at least, doing that got me the correct answer for part two.) -- but I still have to calculate ∆x' by saying ∆x' = γ(∆x - v∆t), substituting in 26 ly for ∆x, 0.99c for v, and ∆t for 26 ly/0.99c. Is that correct?
     
  10. Jul 29, 2012 #9

    TSny

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    No. Δx' is actually very easy to calculate. For the moment, let's just consider event 1 where the traveler leaves the earth. We agree to choose the origin of the x' axis at the location of the traveler. (So, the origin of the x' axis is always coincident with the traveler - she "carries it along"). So, what is the value of x' for event 1? Hint: Where does this event occur in relation to the traveler?
     
  11. Jul 29, 2012 #10
    Event one -- she leaves and so it happens at zero. Event two, she arrives, and she's still with the origin if she's carrying it... so... are you saying that event two is happening again at zero? (I'm confused! I'm not sure if I am reading this correctly.)
     
  12. Jul 29, 2012 #11

    TSny

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    You're correct! The two events occur at the "same place" in the x' frame (that is, they both have the same value of x' = 0). So, what is Δx'?
     
  13. Jul 29, 2012 #12
    Oh wow, so ∆t = (∆t')gamma -- you can just use the time dilation formula! So much easier than I was making it! Thank you -- I appreciate you leading me through it.
     
  14. Jul 29, 2012 #13

    TSny

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    Good work. You've shown how the time dilation formula is derivable from the Lorentz transformation equations!
     
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