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What is wrong with my thinking?

  1. Dec 16, 2005 #1
    Einstein stated that the laws of physics are the same in all inertial reference frames.

    Lets do the old thought experiment

    There are a pair of twins each with identical watches. One remains on earth while the other travels out into space near the speed of light and back again.

    The twin on earth will see the effects of relativity on the travelling twin because they are not in the same inertial reference frames.

    To the twin inside the spaceship, his watch ticks at the same rate as his brother because the traveller and his watch are always at rest with respect to each other. The twin on earth is also at rest with respect to his watch. Hence both twins keep their own time at the same rate (although during the trip, they will see each other differently due to relativity) .

    When the travelling twin completes the trip and comes back to earth thereby reuniting with the reference frame of his brother, the amount of time ticked off by the travelling twin should be identical to the amount ticked off by his brother who was on earth all that time. This is because they have been keeing time in an inertial frame that is idential with each other. Hence the two twins have aged the same when they finall meet.

    The aging difference should only occur when they do not meet in other words, when the are at different inertial references frames with respect to each other.

    But experts have stated that I am wrong and the travelling twin will only have aged a small proportion compared to how much it has aged on earth.
  2. jcsd
  3. Dec 16, 2005 #2


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    Each twin's watch ticks at the normal rate in their own reference frame. However, in each twin's reference frame, the other twin's watch is genuinely ticking slower--it's not just that they see each other differently, even if the twins correct for the light delay each will conclude the other twin's clock is ticking slower than their own.
    No, for the other twin to turn around and return to earth, he has to accelerate during the turnaround phase and switch into a different inertial reference frame. "Inertial reference frame" means the frame of an observer who always travels at constant velocity and never accelerates. An inertial observer travelling alongside the twin during the outbound leg would see him at rest, but then when he turns around and heads back towards earth, this inertial observer (who continues to move away from the earth at the same constant velocity) sees him acquire a large velocity in the direction of the earth during the inbound leg of the trip.
  4. Dec 16, 2005 #3
    If you're not afraid of a little math there's an excellent treatment of this problem that I just read up on arXiv: http://arxiv.org/abs/gr-qc/0104077 . I was actually looking for a good k-calculus link to give you when I stumbled across this, it is the best treatment of the problem that I can recall seeing.
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