general relativity - On the Twin Paradox Again

The search based on the term "Twin Paradox" gave (today) 538 results.

In all the answers, the answerers explained the phenomenon by referring to arguments sort of falling out of the framework of special relativity. I saw answers referring to acceleration and deceleration, changing coordinate systems, etc. Even Einstein referred to General Relativity when explaining the TP...

THE QUESTION

Is it not possible to explain the phenomenon purely within Special Relativity and without having to change the frame of reference?

EDIT

... and without referring to acceleration and deceleration, without having to stop and turn around or stop and start again one of the twins?

Answer

Is it not possible to explain the phenomenon purely within Special Relativity and without having to change the frame of reference?

It is possible to obtain the correct answer to the amount of time accumulated by either of the twins by using any single reference frame, without changing that frame at any point in the analysis. Whether or not such a calculation constitutes an “explanation” is a matter of opinion. I would tend to say “no” because the “paradox” is precisely about what happens when you incorrectly change reference frames.

To obtain the amount of time accumulated by any traveler we write their worldline as a parametric function of some parameter (using units where c=1), for example $r(\lambda)=(t(\lambda),x(\lambda),y(\lambda),z(\lambda))$ where $r$ is the worldline and $t$, $x$, $y$, and $z$ are the coordinates of the traveler in some reference frame whose metric is given by $d\tau$. Then, for any reference frame for any spacetime for any traveler, the amount of time is given by $\Delta\tau=\int_R d\tau$ where R is the total path of interest (i.e. all of the $r(\lambda)$ of interest). Because this is a completely general formula it applies for an inertial traveler or for a non inertial traveler, it also applies for an inertial reference frame or for a non inertial reference frame, it also applies in the presence of gravity or not.

For the specific case of an inertial frame we have $d\tau^2=dt^2-dx^2-dy^2-dz^2$ from which we can easily obtain $$\frac{d\tau}{dt}=\sqrt{1-\frac{dx^2}{dt^2}-\frac{dy^2}{dt^2}-\frac{dz^2}{dt^2}}=\sqrt{1-v^2}$$ So then $$\Delta\tau=\int_R d\tau=\int_R \frac{d\tau}{dt} dt = \int_R \sqrt{1-v^2} dt$$

Note, this last paragraph assumes an inertial frame (any inertial frame is the same). The usual mistake is to use the inertial frame expression in a non inertial frame. A similar procedure can be used in a non inertial frame, but you must use the appropriate expression for $d\tau$