- #36
JesseM
Science Advisor
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S wouldn't subtract 4 yrs. In S's frame, the distance between Earth and AC is not 4 light years, it's shorter by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex] due to Lorentz contraction. Also, in S's frame AC is moving towards him, so its distance when the light was emitted is different from its distance when the light actually reaches him.phyti said:S sees the AC clock at 1996 as he leaves earth.
S sees the AC clock at 2004.1 when he arrives at AC.
S sees the elaplsed time on his (S) clock as .88 yrs.
S sees nothing wrong with his clock.
It is running slower only to the other observers but not to him.
Even if S subtracts 4 yrs for the light to reach earth
To make the calculations a little easier, let's say S is traveling at 0.8c relative to Earth and Alpha Centauri, rather than 0.975c as in my previous calculations. This means that in S's frame, the distance between E and AC is not 4 light years but [tex]4 * \sqrt{1 - 0.8^2}[/tex] = 4*0.6 = 2.4 light years. So how far away would AC have had to have been when it emitted the light, in order for it to be 2.4 light years away at the moment the light reaches S? If the distance is x, and the time the light took to reach S is t, then we have:
x - ct = 0
and in this time AC moved closer a distance of 0.8c * t:
x - 0.8ct = 2.4
Solving for t:
0.8ct + 2.4 = ct
2.4 = 0.2ct
t = 12 years
Which means x = 12 light years. So in S's frame, if he sees Alpha Centauri's clock read 1996 as he passes Earth, he concludes this event actually happened 12 years earlier in his frame. And S reaches AC 2.4 ly/0.8c = 3 years after passing Earth in his frame, at which point AC's clock reads 2005. So, S concludes that it took AC's clock 12+3 = 15 years to advance forward 9 years from 1996 to 2005. This means that in S's frame, AC's clock is slowed down by a factor of 9/15 = 0.6, which is exactly the time dilation factor [tex]\sqrt{1 - v^2/c^2}[/tex]
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