« on: 17/09/2023 16:14:56 »
This is a slightly modified twin paradox to distinguish the effects of relative speeds and acceleration.In a different inertial reference frame (one in which both stars are moving at -0.4c), this is pretty much exactly the twin paradox, with twin A being stationary the entire time, and twin B going out and back, albeit at 0.4c only outbound, and faster on the return.
OK, so this is trivially computed in the Earth frame. The resolution for any exercise in special relativity is to simply choose a frame (we'll pick Earth since Earth time was specified), all you need to do is compute the dilation factor due to the speed relative to that frame. This is computed (in natural units) with
λ = 1/√(1-v2)
Twin A started a journey to Alpha Centauri 4 light years away in a space ship moving at 0.4c is expected to arrive 10 years later, according to earth observer.OK, so in Earth frame, λ=1.091 so 10 / 1.091 = 9 years 2 months elapsed time for twin A.
Twin B stayed home to improve the space ship, so he can go to Alpha Centauri 5 years later at 0.8 c.0.8c yields a λ of 1.667 so 5 years at home and 3 during the trip, so 8 total. Not so hard, right? You can do this yourself without asking each and every time.
Classical physics calculation predicts that they'll arrive at Alpha Centauri simultaneously.Classical in what sense? Classical physics refers to non-quantum physics. I think you mean Newtonian physics, not classical. Yes, every theory (including Newtonian) says they get there simultaneously. This is an objective fact, true in any reference frame.
How old are they when they meet up at Alpha Centauri?As computed above, 9y2m and 8y respectively.