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Please, my question is at the end of this formulated problem.

In the case of the twin paradox, the travel can be made without never accelerating higher than $g$. So the one who remains on the earth will always be subject to $g$, while the traveler can be under $g$ only for some part of the trip for example:

  • First he leaves earth, after overcoming $g$, and keeps accelerating at $g$ for 180 days, where he reaches around $0.508 c$;

  • Then the acceleration is removed and he travels at $0.508 c$ away from earth for 5 years;

  • After this, G is applied in the opposite direction for 180 days so that its speed in relation to earth slows down to 0 m/s;

  • When the relative speed to earth is null, the traveler spends 1 day slowly rotating 180 degrees, so that he points back to earth;

  • Then he starts accelerating at G for 180 day again;

  • After that G is removed and he travels at constant speed of 0.508 C for 5 years towards the earth;

  • At the end of this 5 years a contrary G force is applied for 180 days, so that he can slow down and return to earth;

  • Now he traveled at constant speed for 10 year, and during this period of inertial trip, his time suffered dilation of 1.16 as supposed by the observer on earth. But he also supposes that the time on earth was extended.

  • During this 10 years + 720 days + 1 day he was subject to G for 720 day (Never more than G);

  • Meanwhile his brother on earth was subject to G during the whole time;

Who really suffered time dilation and why? What causes time dilatation, speed or acceleration?

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    $\begingroup$ You mean $g$, not $G$. They’re different. $\endgroup$
    – G. Smith
    Oct 28 '20 at 4:08
  • $\begingroup$ What does "suffered time dilation" mean? $\endgroup$
    – WillO
    Oct 28 '20 at 4:33
  • $\begingroup$ Yes, g and not G. $\endgroup$ Oct 28 '20 at 23:33
  • $\begingroup$ in "suffered time dilation" I mean the one whose clock run slowly. $\endgroup$ Oct 29 '20 at 1:20
  • $\begingroup$ Also $c$, not $C$. $\endgroup$
    – Qmechanic
    Oct 29 '20 at 6:31
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You need to use the correct relativistic formulae to calculate these things. Please see The Relativistic Rocket.

It looks like your value of 0.508c after 180 days (ship time) comes from the Newtonian formula $v=at$. The relativistic value is 0.468954c. And 180 days ship time corresponds to almost 187 days, 20 hours, 46 minutes Earth time. If you then cruise at a constant speed of 0.468954c you can use the Lorentz factor $\gamma\approx1.132217$ to calculate the time dilation.

Who really suffered time dilatation and why?

As per usual in the twin paradox, when the traveller gets back to Earth, the Earthbound twin will be older.

What causes time dilatation, speed or acceleration?

In Special Relativity, speed causes time dilation, but with constant speed the situation is symmetrical. If observers A & B have a constant relative velocity then A measures B's clock to be running slow by a factor of $\gamma$, and B measures A's clock to be running slow by a factor of $\gamma$.

To break the symmetry, (at least) one of the observers needs to make one or more changes of reference frame. It's not so much that the acceleration causes time dilation, it's merely the mechanism whereby the reference frame is changed.

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  • $\begingroup$ Thanks PM 2Ring. Yes, I employed Newtonian formula hoping it was not very high difference. I will use the correct model in the future. $\endgroup$ Oct 28 '20 at 23:35
  • $\begingroup$ Would you explain how the change of reference frame would really determine time dilatation? Is there any mathematical model that shows this fact? $\endgroup$ Oct 28 '20 at 23:41
  • $\begingroup$ @TheCircuitCracker When you do the time calculations for the simplest version if the twin paradox (with instantaneous velocity changes) using the Lorentz transformations you use 1 equation for the Earthbound twin, but you need to use 2 equations for the traveller because their worldline (trajectory through spacetime) isn't a straight line, it has a kink in it (and that kink exists no matter which inertial frame you use to describe their trip). So you have to use 1 equation for the outbound half of the journey, and another for the inbound half. $\endgroup$
    – PM 2Ring
    Oct 29 '20 at 7:16
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    $\begingroup$ @TheCircuitCracker There are already a lot of answers about this on the site. Quite a few of them are here: physics.stackexchange.com/q/242043/123208 In particular, see robphy's, which has several nice diagrams: physics.stackexchange.com/a/507592/123208 And Marco Occram's: physics.stackexchange.com/a/507416/123208 John Rennie's answer (the top-voted one) is certainly good, but please see the criticisms written by Elio Fabri. $\endgroup$
    – PM 2Ring
    Oct 29 '20 at 7:36
  • $\begingroup$ FWIW, when I was learning this material, what clicked for me was the variation using 3 observers and no acceleration, only frame changes. You can read about that in the last section here: www1.phys.vt.edu/~jhs/faq/twins.html I'm sure there's a version of that on this site, but I'm not able to find it. $\endgroup$
    – PM 2Ring
    Oct 29 '20 at 7:37
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Who really suffered time dilation and why? What causes time dilatation, speed or acceleration?

The twin at earth will be older. The reason can be understood by the theorem mentioned in this question.

The twin at earth followed a straight line in spacetime, that is the t-axis between the 2 events: departing and returning of the other twin.

On the other hand, if the path of the traveller twin is plotted in the time x space graph, it consist of several steps between the same initial and end points.

It is only possible to deviate from a straight line with some acceleration. So we could say that it is a necessary condition for the mismatch of clocks at the end point.

But we can not use the equivalence principle in this case, and compare $g$ acting on the twin at earth with $g$ acting inside the rocket on the traveller twin. The EP is valid only locally, and that includes a short $\Delta t$. Ex: both can throw an object in the air and observe the same accelerated path for some seconds until falling on the ground.

But if the earth twin throws on object with 12 km/s it will not return (supposing no air drag). But it will return for the twin in the rocket in less than one hour. So the artificial gravity there is not comparable with the earth's one.

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