- #176
zonde
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So we start with second twin waiting for one year while his home is moving away at 0.9c. After one year (from perspective of second twin) it turns around (from perspective of first twin) and meets first twin after another year (second twin's time).ghwellsjr said:Finally, since you want to have the traveling twin and the Earth twin jump frames at the moment of turn around, I beg you to provide us with the details of the calculations. Let's assume that the traveling twin turns around after one year on his clock and is traveling at 90%c. Can you do that? And can you also show the calculations for what each twin sees of the other twin's clock during the entire scenario, please?
And then, to address your comments to me, I'd like you to show us how you use the LT in this process, OK?
And we want to know what each twin sees of the other twin's clock.
We start with this diagram:
Both twins start at "A" and first (stay at home) twin is heading away at 0.9c.
Second twin traveling along AB will see first twin as traveling along AI. So that proper time along AI divided by proper time along AB will give what second twin sees of the first twin on foward trip.
Similarly we need AB/AJ, IC/BC and BC/JC.
Now I know that t coordinate of "B" is 1y(year).
First I will find coordinates of "I".
First twin will cover distance of 0.9ly(light years) in 1y and then signal at light speed will go back to second twin for another 0.9y. So we have that in 1.9y we would receive signal from 0.9ly distance. But as our time is only 1y then x coordinate of "I" is 0.9/1.9 ly and t coordinate is 1/1.9 y.
Now I want to find proper time along AI. So I will perform LT.
I(x=0.9/1.9ly,t=1/1.9y) transforms to I(x'=0, t'=0.1/sqrt(0.19)=0.2294y)
B(x=0,t=1y) transforms to B(x'=-0.9/sqrt(0.19)=-2.065ly,t'=1/sqrt(0.19)=2.294y)
So AI in first twin's rest frame is 0.2294y and AIp/ABp is 0.2294 (seconds of fist twin per second of second twin)
Now because of symmetry between top and bottom of the diagram we can find AC by taking twice t coordinate of B and it is 4.588y. And for the same reason BC in second twin's rest frame is 1y.
So we get that:
ABp/AJp=1/(4.588-0.2294)=0.2294 (seconds of second twin per second of first twin - what first twin sees of second twin before he turns around)
ICp/BCp=(4.588-0.2294)/1=4.359 (seconds of first twin per second of second twin - what second twin sees of first twin after he turns around)
BCp/JCp=1/0.2294=4.359 (seconds of second twin per second of first twin - what first twin sees of second twin after he turns around)