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Paul B. Andersen :
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some problems of contradiction, or even absurdity.
We will return to the eternal problem, which has scared away all the contributors I have dealt with for 40 years, although the opposite is always said, and that I am the one who would be a wimp.
So we return to the problem, which is nevertheless very simple.
Stella goes into the stars for a journey of 24 light years.
The speed is 0.8c on the way there (12 ly), and 0.8c on the way back (12 ly).You will find the answer here:
We quickly obtain the following data by the Doppler effect.
Stella and Terrence agree to beep every month.
It is clear that, for Stella, on the way there, as on the way back, she will beepOK. Stella has aged Tₛ = 2L/vγ = 18 y
108 times (216 times in total). >
SO, Terrence will receive 216 beeps.
For his part, Terrence will beep for 30 years. He will therefore sendOK
360 beeps.
Except that if Terrence receives the same number of beeps on the way there as on the way back (108 each time), Stella receives many more beeps on the way back than on the way there.Doppler shift when Stella is moving out Do = √((c−v)/(c+v)) = 1/3
It is then interesting to know how many beeps Stella receives on the way there, and how many she receives on the way back.
Once this is done, we can still progress in understanding things.Terrence will age C = 2L/v = 30y
We just have to not be afraid, and have a playful spirit.
R.H.--
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