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This is a slightly modified twin paradox to distinguish the effects of relative speeds and acceleration.

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.

Twin B stayed home to improve the space ship, so he can go to Alpha Centauri 5 years later at 0.8 c.

Classical physics calculation predicts that they'll arrive at Alpha Centauri simultaneously.

How old are they when they meet up at Alpha Centauri?

As computed above, 9y2m and 8y respectively.

Quote from: Halc on 17/09/2023 16:14:56As computed above, 9y2m and 8y respectively.Thanks for the answers.Do we get the same answers if the calculations are made from the perspectives of the travelling twins?

None of these phenomena are paradoxes, they are just somewhat counterintuitive when given a cursory examination. Once one does the maths, all becomes clear.

What's the formula necessary to get the same answers as calculated from earth perspective?

Let's describe the same case from twin A's perspective. He stays in his own reference frame

while Alpha Centauri moves closer at 0.4c. Length contraction reduces the duration of the journey from 10y to 9y2m.

It's what his clock will show when Alpha Centauri arrive at his location.

Meanwhile, twin B moves away to the other direction at the same speed.

Then at half way, he returns to A's position.

Somehow, B's clock will age less than A's when they meet, ie 8y.

Half way to what?

QuoteSomehow, B's clock will age less than A's when they meet, ie 8y.We know that took 5.455 to get to turnaround (in frame A), so that leaves 9.167 = 5.455 = 3.712 years to get back. Clearly he'll be going faster on the return than on the way out.2.182 ly / 3.712 years is 0.588c which is his return speed in frame A. It's that easy. λ = 1/√(1-v2), so at that speed, λ is 1.236. To verify, 3.712 / λ = 3, so he ages 3 years during the 2nd leg, which is what we got in the first example.

Now, how would it look like from B's perspective?

B is not inertial.

I think it would be best to work your way backwards, computing the second leg first, and then the prior leg of the journey. You know most of the relevant numbers from the posts above, such as how fast is twin A is moving relative to this frame, and how long (his clock) it takes him to do it. Start with that. All you need is the one formula (which involves a square root, scary...), and the rest can be done with a 4-function calculator.

Quote from: Halc on 20/09/2023 00:14:46We know that took 5.455 to get to turnaround (in frame A), so that leaves 9.167 = 5.455 = 3.712 years to get back. 2.182 ly / 3.712 years is 0.588c which is his return speed in frame A. It's that easy. λ = 1/√(1-v^{2}), so at that speed, λ is 1.236. To verify, 3.712 / λ = 3, so he ages 3 years during the 2nd leg, which is what we got in the first example.How do you get 8y as B's age when calculated from A's perspective?

We know that took 5.455 to get to turnaround (in frame A), so that leaves 9.167 = 5.455 = 3.712 years to get back. 2.182 ly / 3.712 years is 0.588c which is his return speed in frame A. It's that easy. λ = 1/√(1-v^{2}), so at that speed, λ is 1.236. To verify, 3.712 / λ = 3, so he ages 3 years during the 2nd leg, which is what we got in the first example.

Is A inertial?

Should we calculate the Doppler effect to get the correct answer?

Would it make a difference whether those twins stop at the end of the journey, or continue going at their travelling velocity, or turn around to return to the earth, as measured by earth observer?

Nothing you do at a given moment in time can effect what your own clock says at that moment in time.

All that matters is that you pick a frame and figure out the speed of each clock in that frame, and for how long it travels at that speed. The one formula that is needed (quoted several times above) takes speed as an input, and makes no reference to acceleration.

No. The acceleration is irrelevant other than providing the SPEED at which relativistic effects occur.