Twin paradox in SR

Question

I would like to explain the twin paradox in SR, but not considering that the paradox is due to the fact that the travelling twin is undergoing acceleration, which makes him a non-inertial observer, not explainable so with $SR$.

I mean: let the no-moving twin (S1) and the second moving twin (S2), that moves away from S1 of a constant velocity $v$ until a certain time in which he reverses its velocity from $v$ to $-v$, until he meets again his twin in D.

If we suppose $S1$ measure the interval of time associated to the first path as $T$, the total interval of time measured (by symmetry) is $2T$. Instead for the "time dilation phenomenon" the moving twin would measure an interval of time that is $\beta 2 T$, where from Lorentz transformations $\beta=(1-\frac{v^2}{c^2})^{-\frac{1}{2}}$. So since $\beta <1$ this causes that the second twin looks younger than the no-moving twin. But there is a mistake in using the rules of $SR$: it is referred to the turning point $P$.
In fact if for the no-moving twin we can adopt the symmetry property to consider as total amount of time $T+T$, it fails for the second twin.
In fact he does not adopt the same reference system in the first and second path, where the two paths are differentiated in according to the change of velocity (OP->PD). This means that instantly the straight line of instantaneous event with respect to $P$ rotates (passing from red to green) and so if before the reversing on velocity, the event $B$ was instantaneous to $P$, after it will be $C$.
So the length of the segment $BC$ gives us the missing time in $2T$ with respect to $2\beta T$. In P in fact we have to imagine an instantaneous growth by the rate given by $BC$ for the second twin.

$\textbf{Questions:}$

1)First of all do you think I have understood well the situation?
2) With my argument I have explained that we commit a mistake if we don't consider in total 3 different inertial observers...but what I can't understand is: now what is the conclusion? Is that the difference in time is given by $BC$? Or we can say that if we consider three different inertial observers is no more true that the second twin looks younger once he met again his twin?

Answer 1

This question, insofar as it is not a duplicate of previous questions, is founded on a couple of misunderstandings.

I would like to explain the twin paradox in SR, but not considering that the paradox is due by the fact that the travelling twin is undergoing acceleration, which makes him a non-inertial observer, not explainable so with

This assumes that acceleration is outside of SR, which is incorrect. SR is perfectly capable of handling acceleration. The statements of the two postulates refer to inertial frames, but that does not preclude the analysis of non-inertial objects from the perspective of an inertial frame. Indeed, that is usually how people learn about acceleration in the first place.

Furthermore, you can even use non-inertial frames in SR since the mathematics of transforming from an inertial frame to a non-inertial frame is not owned by GR. Similarly, you can use the mathematical framework of pseudo Riemannian geometry in SR. Indeed, Minkowski’s formulation of SR in terms of four-vectors is very close to that.

What distinguishes SR from GR is curved spacetime (I.e. tidal gravity). As long as there is not a non-uniform gravitational field then you are firmly in the realm of SR.

let the no-moving twin (S1) and the second moving twin (S2), that moves away from S1 of a constant velocity until a certain time in which he reverses its velocity from to −, until he meets again his twin in D.

The second misconception is that this is not acceleration. Acceleration is a change in velocity. Here the velocity changed from $v$ to $-v$. So that is an acceleration.

Having addressed those two misunderstandings the remainder of your question is simply the standard twins paradox and all of the usual answers apply. In particular, answers talking about acceleration apply since acceleration is indeed handled in SR.

Answer 2

One way the paradox is explained without acceleration is that at the point S2 would otherwise be turning around he encounters another traveler, S3, moving with speed v towards S1. S3 synchronizes his clock with S2's clock. S3 then completes the return to S1. Using the usual time dilation equation for S2's outward trip and S3's inward trip leads to the expected conclusion that the time for the trip out and back is less than the time measured by S1.

Answer 3

In 1918 Einstein informed the world that, as the traveling clock (twin) turns around and experiences acceleration, a HOMOGENEOUS GRAVITATIONAL FIELD emerges. This homogeneous gravitational field somehow affects the distant stay-at-home clock and makes it run very fast (which means that the stay-at-home twin suddenly gets very old) during the turning-around period:

Albert Einstein 1918: "A homogeneous gravitational field appears, that is directed towards the positive x-axis. Clock U1 is accelerated in the direction of the positive x-axis until it has reached the velocity v, then the gravitational field disappears again. An external force, acting upon U2 in the negative direction of the x-axis prevents U2 from being set in motion by the gravitational field. [...] According to the general theory of relativity, a clock will go faster the higher the gravitational potential of the location where it is located, and during partial process 3 U2 happens to be located at a higher gravitational potential than U1. The calculation shows that this speeding ahead constitutes exactly twice as much as the lagging behind during the partial processes 2 and 4." http://sciliterature.50webs.com/Dialog.htm

Modern versions of Einstein's 1918 argument:

David Morin, Introduction to Classical Mechanics With Problems and Solutions, Chapter 11, p. 14: "Twin A stays on the earth, while twin B flies quickly to a distant star and back. [...] For the entire outward and return parts of the trip, B does observe A's clock running slow, but enough strangeness occurs during the turning-around period to make A end up older." http://www.people.fas.harvard.edu/~djmorin/chap11.pdf

"When the twin in the spaceship turns around to make his journey home, the shift in his frame of reference causes his perception of his brother's age to change rapidly: he sees his brother GETTING SUDDENLY OLDER. This means that when the twins are finally reunited, the stay-at-home twin is the older of the two." http://topquark.hubpages.com/hub/Twin-Paradox

Richard Feynman: "So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger." http://www.feynmanlectures.caltech.edu/I_16.html

John Norton: "Moments after the turn-around, when the travelers clock reads just after 2 days, the traveler will judge the stay-at-home twin's clock to read just after 7 days. That is, the traveler will judge the stay-at-home twin's clock to have jumped suddenly from reading 1 day to reading 7 days. This huge jump puts the stay-at-home twin's clock so far ahead of the traveler's that it is now possible for the stay-at-home twin's clock to be ahead of the travelers when they reunite." http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html

Physics Girl (4:30): "One last question. What's happening to the clocks during the period of acceleration? We still get time dilation, but we have to use a different set of rules from the general relativity. General relativity states that clocks runs slower in accelerated reference frames. So while your twin is turning around, her clock runs slower, and she sees the same thing. She sees your clock running faster than hers, so you're aging quicker. It's during this period of acceleration that you become the older twin." https://www.youtube.com/watch?v=ERgwVm9qWKA