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Old 18-July-2004, 07:22 PM
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Default Re: (dt')=(1+ax/c^2)(dt)/sqrt[1-(v/c)^2]

Quote:
Originally Posted by Lorentz
I stand by my argument. A ground-based observer sees clocks flying west run slow compared to a stationary clock on the ground and sees eastbound clocks run slower still. An observer on the westbound plane sees the clock on the ground run slow, but not as slow as an observer on the eastbound plane would.
[Jim]

Sorry, I have to correct you there. (snip)
Not my quote. I was quoting Taibak.
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