[hamdani yusuf (OP)]

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. He 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. Does special theory of relativity predict the same?
How old are they when they meet up at Alpha Centauri?

[Halc]

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.

[hamdani yusuf (OP)]

As 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?

[Bored chemist]

As 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?

Yes.
Because the alternative would be a contradiction.

[paul cotter]

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

[hamdani yusuf (OP)]

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

Have you done the calculations?
Do you use length contraction for the distance?
Do you use relativistic velocity addition?
Do you use Doppler effect?
What's the formula necessary to get the same answers as calculated from earth perspective?

[Origin]

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

Using the equations from special relativity, you will find that there are no actual paradoxes as paul cotter said.

[paul cotter]

1/√(1-vsq/csq) is all you need. I did not calculate this scenario, Halc had already done so and thus there was no need. With all due respect, and I mean this sincerely, I believe you, Hamdani, look at too many videos and get confused messages from these. A reliable physics textbook would clear up any doubts you may have.

[hamdani yusuf (OP)]

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.

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

[Halc]

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

Just for some terminology, 'perspective' is a point of view.  I think you mean his frame, or more specifically, the frame in which twin A is stationary, so we can call it 'frame A'.  Frame E was the one where Earth is stationary.  Everybody is in all frames, so there is no way to exit it, so 'stays in' should read 'is stationary in'.
As for perspective, two people stationary relative to each other still have different perspectives since they're not in the  same place and thus see different things.  The twins scenario isn't about perspectives, or even about twins for that matter. The twins is just a funny way to emphasize that people can act as clocks.

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

Length contraction is about length, not time. Earth and AC are 9.166 light years apart in that frame. That's length contraction.  AC needs to travel that distance to where 'twin A' is, so yes, you are correct in that the journey taken by AC is of length 9 1/6 light years.

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

Yes, exactly.

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

Riding 'spaceship Earth' away from twin A at the speed of 0.4c.

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

Half way to what?  No, he simply rides Earth for 5 of his years.  λ=1.091 for 0.4c, remember?  So that's going to take 5λ years in the frame we're using, which is 5.455 years. 5.455 years at 0.4c takes him 2.182 ly away from where he started, which isn't particularly halfway to anything.

Somehow, 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.

[hamdani yusuf (OP)]

Half way to what? 

Halfway of A's journey, calculated from earth's perspective.

[hamdani yusuf (OP)]

Somehow, 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.

How do you get 8y as B's age when calculated from A's perspective?

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

[hamdani yusuf (OP)]

B is not inertial.

Is A inertial?

[hamdani yusuf (OP)]

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.

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

[hamdani yusuf (OP)]

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?

[Halc]

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-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.

How do you get 8y as B's age when calculated from A's perspective?

By adding 3 to the 5 subjective years he took on the outbound trip.
5 years was given in the description. 3 was computed as per the bit that you quoted, 3.712 years / λ  where λ is computed to be 1.236 for the return leg.

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

B is not inertial. Yes, he has a perspective, but he doesn't define any one inertial frame, so you need to be more specific.  You probably want the frame in which his 2nd leg has him stationary since the 1st leg frame is the same one as Earth/AC, and we already did that one.

Go through steps similar to above and you can do it yourself.  In that frame, A moves towards the destination at constant speed, and B and Earth move toward it at some higher speed, both getting there simultaneously and before twin A.  At that point, twin B stops, waiting for both A and AC to get to him simultaneously, but at different speeds..
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.

Is A inertial?

A is moving at constant velocity for the duration of the exercise, so yes, A is inertial. Sure, there is acceleration at one end or both, but no duration is spent at those alternate velocities, so those zero-duration periods are irrelevant to what the respective clocks (ages) say when compared at the reunion event.

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

Doppler effect only concerns a perspective, a point of view, but not a frame since a point of view is frame independent. What a particular observer sees cannot be calculated until you know what is, and if you already know what is, the correct answer is already in your hands.
So no.  Computing Doppler (or worrying about observers at all) is useless for this exercise. Use the techniques shown above. Use the values you already know and it is simple to figure the missing ones. The only formula needed is the one given. Ditch the fact that there are twins at all and just assume clocks are taking the trips indicated, on ships with no windows while we're at it.

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. In fact, I said above that time dilation is all about speed relative to a specific frame, so accelerations don't matter anywhere. 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.

So no, any acceleration done at the final event cannot effect the comparison of ages(times on the clocks) in each other's presence at the reuniting event. One cannot make a clock jump in time except by just setting it to a different value, such as I need to do on my microwave every time I get a power glitch.

Similarly, one's future plans (say a trip back to Earth or not) cannot effect the result of a comparison at a given event. If it did, that would be backwards causality.

[hamdani yusuf (OP)]

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

Can it change what a relatively moving observer see on our accelerating clock?

[hamdani yusuf (OP)]

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.

Change of reference frame requires acceleration. It's the cause of asymmetric time dilation in twins paradox.
Without it, no twin can conclude that the other twin is older than himself.

[paul cotter]

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

[hamdani yusuf (OP)]

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

You're saying no, except yes.