One Way Twin Paradox

  • Thread starter GRDixon
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In the following, "G" stands for "gamma". Clocks O and O' coincide when they mutually read zero. POV is that of K'.

Prove: When O' reads T'>0, the coincident K clock reads more than T'.

Proof: When O' reads T', O reads T'/G and O' coincides with K clock at x=GvT'. That clock reads xv/(cc) more than O:

T'/G + GvT'v/(cc) = GT' > T'.
 

JesseM

Science Advisor
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Your proof is correct--another way to see it is just to take the event of O' reading T' as having coordinates x'=0, t'=T' in the K' frame and then plug those into the reverse version of the Lorentz transformation, t=G*(t' + vx'/c^2) which gives t=GT' in the K frame. But why do you consider this a paradox? In the K' frame, this clock was still ticking slower than O' the whole time, it's just that it started out reading more than 0 at t'=0. In fact, at t'=0 it must have read GT' - T'/G, so that it did tick forward by T'/G between t'=0 and t'=T'. You can check this using the Lorentz transformation--if we plug in x'=vT' (the position of this clock at t'=0), t'=0 into the reverse version of the Lorentz transformation we get t=GT'*(v^2/c^2), and you can see that GT' - T'/G = GT'*(1 - 1/G^2) = GT'*(1 - (1 - v^2/c^2)) = GT'*(v^2/c^2).
 
Last edited:
249
0
Your proof is correct--another way to see it is just to take the event of O' reading T' as having coordinates x'=0, t'=T' in the K' frame and then plug those into the reverse version of the Lorentz transformation, t=G*(t' + vx'/c^2) which gives t=GT' in the K frame. But why do you consider this a paradox? In the K' frame, this clock was still ticking slower than O' the whole time, it's just that it started out reading more than 0 at t'=0. In fact, at t'=0 it must have read GT' - T'/G, so that it did tick forward by T'/G between t'=0 and t'=T'. You can check this using the Lorentz transformation--if we plug in x'=vT' (the position of this clock at t'=0), t'=0 into the reverse version of the Lorentz transformation we get t=GT'*(v^2/c^2), and you can see that GT' - T'/G = GT'*(1 - 1/G^2) = GT'*(1 - (1 - v^2/c^2)) = GT'*(v^2/c^2).
A very nice analysis. And of course the Lorentz Xform is always the preferred way to go. It was just that I read about the K clock at x=vT reading more than the K origin clock (from the K' POV) in Griffiths "Intro to ElecDyn." And I wanted to "exercise the idea," so to speak. I would think that, so long as a K' observer remains at rest in K', then the idea of the twin "paradox" applies. If I were to take such a trip, however, I'd probably always switch back to K for my rest frame, at trip's conclusion, and realize that (a) the K origin clock really WAS synchronized with the adjacent K clock at x=vT, and (b) my twin back at the origin really WAS older than I. Bottom line: Same Old Same Old. If I had a dollar for every time this so-called paradox has been discussed in the literature ... Ya Hoo! Thanks for the alternate POV.
 

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