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special relativity Watch

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    Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of u = 250 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4 hours. By how much will the reading of the two clocks differ, and which clock will show the shorter elapsed time? (Hint : Since u << c, you can simplify sqrt(1 ? u2/c2) by a binomial expansion).

    ok so i think i have this right, but im confused as to which should be the rest frame and which should be the moving frame, i.e which should be t and whcih should be t'.

    I used gamma ~ 1-(u2/c2)/2

    then said that t=4 hours and t' =?

    then used t'=t/gamma

    so i ended up with t'-t=-5x10-9s

    is this the right way around, i.e the moving clock should have a longer time? i dont think that is right....

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    If you want to read up on this topic some more try googling "Twin Paradox". It's the same concept.

    The clock that takes the trip to NYC will be the one reading the smaller time. The moving frame experiences time dilation (ie the clock ticks slower) and so less time is recorded. Look again at your approx. of gamma, it should be 1+((v^2)/(2(c^2)).

    You may well ask: what about equivalence of inertial frames? Can't the one going to NYC say that the one at home was moving? The answer is NO. Equivalence only holds for inertial frames, whereas the act of turning back to return home must have involved an acceleration.
 
 
 
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