Is it possible to use two different metrics for one observer in distinct times (related to twin paradox)

Question

I am not going to rewrite another twin paradox, what i am interested now is that if we assume an observer who moves with acceleration first (during $0<t<T_0$), then he stops his acceleration and moves with constant velocity (during $T_0<t<\infty$), is it possible to use two different metrics (one for his acceleration (like Rindler) and another for his constant speed) to see what he observes in the universe? My first guess was that it should be ok, but actually it's not. And if it's not really ok, how are we going to describe his frame?

My problem: Assume two observers S and S' who are at the rest in $t=t'=0$ but at different places (they can synchronise their clock by exchanging signals, because they are the rest). Now observer S' accelerates with $g=2*10^8 m/s^2$ in its frame for one second, just after that he stops this acceleration and moves with constant speed toward S. We want to show that S will be older than S' according to S' when they meet each other at S's place, by using S' proper time which we can derive from its metric. However though, if our observer accelerate for a mere second we can see that S will become older only for 1.07 second,

https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

$t=c*sinh(gt'/c)/g=3*sinh(2/3)/2=1.075s$

Where $t$ indicate S elapsed time. This is expected and fine because our accelerated observer sees a faster clock for S. But this is where we encouter a big problem. If we assume that after acceleration (when S' becomes inertia) S' can use Minkowski metric, then due to the gamma factor for the remaining path, he will see a slower clock for S, and if we make their distance large enough, we will see that according to S', S will be younger in their meeting. (gamma factor should be around 1.34, if S' arrive at S location 1y later in S frame, according to S', S' will be 1.34y old while S is 1y old, and let's forget about that mere 0.07 second!). This is twin paradox because S will consider S' younger too. where did i go wrong? Do note that i just said this problem as an example, my question is in the title so don't try to solve paradox without answering the real question.

P.S: I saw around 6 pages of physics.stackexchange about twin paradox. Didn't find something like my question, in all of them S' was considered to be accelerated all of the time (when they wanted to calculate elapsed time directly in frame S') which solves the paradox clearly. However if my question is duplicated, please do enlighten me

Update: Note that we are going to write worldlines in a coordinate system that S' is always at the rest at the origin of coordinate system. We have no problem in the S coordinate system as it was done by @ChiralAnomaly

Answer 1

So S' and S are of equal age when they synchronize. Then there is 1 second of acceleration. At then end, S is now much older according to S', and the time when S synchronized his clock was a long time ago according to S', so it's no problem that S' watches him age slowly as they close in on each other.

Update: So the way this question is phrased isolates the heart of the Twin Paradox. It's not that the traveling twin ages less, it's that he ages less while seeing the at-home twin's clock run slower than his the whole time.

The paradox is resolved when you realize the at home twin's age jumps forward a lot during the turn around (acceleration). But: the acceleration is brief in BOTH reference frames...how can you account for years when each twin experienced only seconds?

That is the Andromeda Paradox: by changing your velocity, the definition of "now" at distance places changes. S' is far away and at rest w.r.t to S, so they can synchronize their clocks. Call those events $s_{\mu}'$ and $s_{\mu}$ respectively. They occur at the same time ($s_0 = s'_0$), with a large space like separation in the initial frames.

Once $S'$ is moving (fast) towards $S$, even if he accelerated only for a second in his frame and 1.07 s in $S$, the time coordinate of $s_0$ is way in his past, only because his definition of "now" at $s_i$ jumped into the future, relative to the initial conditions.

Worse: this change is reversible, just by turning his spaceship around.

To quote David Mermin:

"That no inherent meaning can be assigned to the simultaneity of distant events is the single most important lesson to be learned from relativity."

Answer 2

Set $c=1$ to save writing. The Minkowski metric can be expressed as $$ \dot\tau^2 = \dot t^2 - (\dot x^2+\dot y^2+\dot z^2), \tag{1} $$ where the worldline is described by functions $t(\lambda),x(\lambda),y(\lambda),z(\lambda)$, the dot denotes the derivative with respect to the parameter $\lambda$, and $\tau$ is the proper time along the worldline. Using equation (1), we can calculate the proper time along any finite segment of any given worldline, as long as the right-hand side is not negative anywhere (otherwise the proper time is undefined).

The OP describes two worldlines that intersect each other once. Both worldlines can be described using the same $t,x,y,z$ coordinate system, and the whole problem can be solved in that one coordinate system.

The worldline $S$ is $$ (t,x,y,z) = (\lambda_S,X,0,0) \tag{2} $$ where $X$ is a constant and $0\leq\lambda_S\leq\Lambda_S$. The value of $\Lambda_S$ will be determined below.

The worldline $S'$ has two segments: one with a non-zero acceleration, and one with constant velocity. I'll call these segments $A$ and $B$. To describe segment $A$, we can use $$ (t,x,y,z) = \big(a\sinh(\lambda_A),\, a\cosh(\lambda_A),\,0,\,0\big) \tag{3} $$ with $0\leq \lambda_A\leq \Lambda_A$, where $a$ is a constant that determines the acceleration and $\Lambda_A$ controls the duration of this segment. (Relating $a$ to $g$ is a separate exercise.) For segment $B$, we can use $$ (t,x,y,z) = \big(S+\lambda_B,\,C+v\lambda_B,\,0,\,0\big) \tag{4} $$ with $$ S \equiv a\sinh(\Lambda_A) \hskip2cm C \equiv a\cosh(\Lambda_A) \tag{5} $$ and $$ 0\leq\lambda_B\leq\Lambda_B \hskip2cm v = \tanh(\Lambda_A) \tag{6} $$ so that the segments and their first-derivatives both match at the interface. The condition that $S$ and $B$ intersect each other may be used to determine $\Lambda_S$ and $\Lambda_B$: $$ S+\Lambda_B = \Lambda_S \hskip2cm C+v\Lambda_B=X. \tag{7} $$ These can be used to express $\Lambda_S,\Lambda_B$ in terms of the inputs $a$, $X$, and $\Lambda_A$. To compute the elapsed proper times along each worldline:

• For worldline $S$, use (2) in (1) to get $\dot\tau$, and integrate over $0\leq \lambda_S\leq \Lambda_S$.

• For segment $A$, use (3) in (1) to get $\dot\tau$, and integrate over $0\leq\lambda_A\leq\Lambda_A$.

• For segment $B$, use (4) in (1) to get $\dot\tau$, and integrate over $0\leq\lambda_B\leq\Lambda_B$.

This shows how to formulate the problem. The rest is just rote calculation, so I'll stop here.

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By the way, by combining this scenario with its time-reflection through $t=0$, we get a scenario in which the worldlines intersect twice, turning it into a more traditional twin-paradox problem. No new calculations are required, because time-reflection symmetry says that we can just double all of the proper-time intervals that were calculated above.