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Special Relativity: Twin Paradox Question

  1. Feb 12, 2015 #1
    I've already completed most of the question, it's an add on at the end that has stumped me.

    I've calculated using time dilation the difference in ages between the two twins, Joe traveled at v = 24/25 (c=1 units) to a planet for 7 years in HIS reference frame and returned at v = 12/25

    The question at the end is this:

    In Ed's frame (Ed stayed at home, whilst his twin Joe ventured through the galaxy) the distance between Earth and the distant planet is greater than 7 light years. So how does Joe explain that the outbound trip took less than seven years in his frame?

    I guessed this is something to do with length contraction, in Ed's frame the ship would contract. In Joe's frame he's standing still and space is wooshing past him, so would space contract? Doesn't seem to make sense to me, if we use the length contraction equation:

    L = L' / γ

    Where γ = (1 - v^2)^-1/2

    Applied to Joe's frame, I get 1.96 lyr (using L' = 7 lyr)
    But then traveling at v = 24/25 he would get the in 2-3 years, not 7?
     
  2. jcsd
  3. Feb 12, 2015 #2

    TSny

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    You are correct that the distance between the earth and the distant planet is contracted in Joe's frame. But you made a mistake in assuming L' is 7 Lyr.
     
  4. Feb 12, 2015 #3
    Ok, if thats the case then can I assume that L is greater than 7 so:

    7.1 = L'/γ
    7.1 * γ = L'
    L' = 25.36 lyr

    Surely that's not right either... is it because I'm using c = 1 units that I'm not getting the answer I expected?

    When i stick in γ = (1-v^2/c^2)^-0.5 I get out 7.1 again :S
     
  5. Feb 12, 2015 #4

    TSny

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    How did you get L = 7.1 for Ed?

    You should be able to deduce L' by using the information given for Joe. You can imagine a large stick stretching from the earth to the distant planet. From Joe's perspective that stick is of length L' and the stick moves past him in a time of 7 years. Knowing how fast the stick is moving past him, Joe can deduce L'
     
  6. Feb 12, 2015 #5
    Ok,

    Joe travels at v = 24/25 for 7 years so,
    L = 6.72 lyrs

    From Joe's reference frame the 'stick' travels at v = 24/25

    So Joe sees the stick at:

    L = L'γ
    L' = 6.72/(1-(24/25)^2)^-1/2
    L' = 1.88 lyrs

    Too small? Or is that right?
    Also if he has a speed of v = 24/25 then how does it take 7 years to travel that distance? Brain hurts.
     
  7. Feb 12, 2015 #6

    TSny

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    Which length are you trying to get here? The 7 years is the time as measured by Joe. In Joe's frame, he sees himself at rest (v = 0) and the stick is moving (v = 24/25 c). If the stick moves for 7 years (according to Joe) at a speed of v = 24/25 c (according to Joe), how far does the stick travel according to Joe? Would this give you L or L'?
     
    Last edited: Feb 12, 2015
  8. Feb 13, 2015 #7
    I think I'm getting it (hopefully),

    From Joe's frame,
    The stick is flying past at v = 24/25 for 7 years. Thus Joe measures it at 6.72lyr
    So, L = 6.72lyr in Joe's frame

    L = L'/γ
    Lγ = L' = 24 lyrs

    So at rest the stick is 24 lyrs long. This would be the measurement Ed takes also.

    But how can Joe get there faster than a pulse of light that would take 24 years?
     
  9. Feb 13, 2015 #8

    TSny

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    Yes, for Joe the distance between the earth and the planet is 6.72 lyrs.

    Good. For Ed, the distance of the trip is 24 lyrs. How long does the trip take according to Ed?
     
    Last edited: Feb 13, 2015
  10. Feb 13, 2015 #9
    According to Ed it takes 25 years!
    Ok, it's coming together now. Not as intuitively as I'd hoped, thanks a lot for your help though!
     
  11. Feb 13, 2015 #10

    TSny

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    OK, good work.
     
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