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Den 09.09.2024 20:54, skrev Richard Hachel:Same as for the previous post.
Example #2:
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Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y
When Stella's clock show 9y 12h, Terrence will be at the position
x = 12 ly in Stella's frame. Terrence clock will now show 15 y.
When Stella's clock shows 9y 24h, Terrence will now be at the position
x = 0.8c⋅(9y 24h) = (7.2 ly + 19.2 lh) in Stella's frame.
Terrence clock will now show (30y 40h) - (9y 24h)/γ = 24.6y 25.6h
When Stella's clock shows 18y 24h, Stella is back.
Terrence will now be at the position x = 0 ly.
Terrence clock will now show (24.6y 25.6h) + (9y 24h)/γ = 30y 40h
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Note that when Stella's clock show 9y, Terrence clock simultaneously
in Stella's frame show 5.4y, and when Stella's clock show 9y 12h, Terrence clock simultaneously in Stella's frame show 15y.
That means that Terrence clock runs (15-5.5)y/12h = 7012.8 times
faster than Stella's clock during the first part of the U-turn.
But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
(It may be typos)
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