Twin paradox problem from a perverse perspective

Question

Reformulation of the problem

I want to study twin paradox but considering that the reference frame of the twin that travels to Alpha Centauri has no velocity. Instead of considering that the twin who stays on the Earth is the second reference frame (you can forget about him now, he won't be mentioned again), I will consider that Alpha Centauri moves left until it reaches the twin and then moves right until its initial position.

Let us have two inertial frames $O$ (the twin that will remain static) and $O'$ (Alpha Centauri moving left). $O'$ is moving with a velocity $v'=-|v|$ with respect to $O$. Both inertial frames meet at $x=x'=0$ and $t=t'=0$. Let us consider an event A (Alpha Centauri starts its voyage to the twin) with coordinates $x_A=\gamma L, t_A=-\gamma L/|v|$ ($L$ is the distance from the Earth to Alpha Centauri in the standard reference frame). It can easily be proven that this event corresponds to $x'_A=0, t'_A=-L/|v|$. Event B is the event where both inertial frames meet, so $x_B=x'_B=0$ and $t_B=t'_B=0$. The time difference between both events for each inertial frame are $t_B-t_A=\gamma L/|v|$ and $t'_B-t'_A=L/|v|$.

Just as both inertial frames meet, another inertial frame $O''$, which moves at $v''=|v|$ with respect to $O$, also meets the other inertial frames at B, so $x''_B=0$ and $t''_B=0$. $O''$ keeps moving forward until it reaches event C (Alpha Centauri returns to its original position): $x_C=\gamma L$ and $t_C=\gamma L/|v|$. The coordinates for event $C$ for $O''$ are $x''_C=0, t''_C=L/|v|$. The time difference between events C and B for each inertial frames $O$ and $O''$ are $t_C-t_B=\gamma L/|v|$ and $t''_C-t''_B=L/|v|$.

If we compare the time elapsed for $O$ since event A to C with the time elapsed for $O'$ between A and B plus the time elapsed for $O''$ between B and C, we obtain that $t_C-t_A=2\gamma L/|v|$ and $(t''_C-t''_B)+(t'_B-t'_A)=2L/|v|<t_C-t_A$. This means that the time elapsed for a hypothetical traveller that moved velocity $|v|$ from A to B as $O'$ did and then from B to C as $O''$ did would measure a smaller time difference. This makes sense due to time dilatation.

Questions

Did Alpha Centauri age less than the twin between events A and C, while the twin aged less than Alpha Centauri (and the twin who remained on Earth) between the events of departure and arrival of the twin in the original formulation of the problem?

Is this a valid method to debunk the twin paradox?

Answer 1

In order to address your problem and questions, let me make some declarations first.

The Clock Effect is the statement that: between two timelike-related events, the elapsed proper-time along the inertial segment joining them is longer than the elapsed proper-time along any other timelike-worldline segment joining them. Geometrically speaking, this is the spacetime-analogue of the triangle-inequality.

The Twin Paradox is the attempt to dispute the Clock Effect by analyzing the problem from the viewpoint of the non-inertial worldline with the hope of obtaining the same conclusions as gotten from the viewpoint of the inertial worldline... thus nullifying the effect, resulting in no time dilation.

Let me apologize for the length of this answer.
The twin paradox has a minefield of gotchas.
(And, I think your setup has hit one of the gotchas.)
So, I'm looking to cover many of the bases.

With spacetime diagrams and calculations, I verify your first three paragraphs.
Then, I discuss the rest of your question.
You can, of course, skip down near the end for a discussion of your questions.

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For concreteness and arithmetic-simplicity, let us suppose that the "travelingTwin" has
$v_{outbound}=3/5$ and $v_{inbound}=-3/5$ as viewed in the inertial frame of the "homeTwin".
Further, let $L=3$ be the distance to AlphaCentauri, which is at rest in the hone-twin's frame.
We follow your convention that the turn-around event B on AlphaCentauri is the common origin $(x,t)=(0,0)$ of the travelingTwin and AlphaCentauri.

Let's work out some numbers for convenience:
With $v=\frac{3}{5}$ and $L=3$, we have $\gamma=(1-v^2)^{-1/2}=\left(\frac{5}{4}\right)=1.25$ ,
$\gamma L=\left(\frac{5}{4}\right)3=\left(\frac{15}{4}\right)=3.75$,
$\gamma (L/v) =\frac{\left(\frac{5}{4}\right)3}{\left(\frac{3}{5}\right)}=\frac{25}{4}=6.25$,
$L/v =\frac{3}{\left(\frac{3}{5}\right)}=5$,
$(L/v)/\gamma =\frac{3}{\left(\frac{3}{5}\right)\left(\frac{5}{4}\right)}=4$, and $(L/v)/\gamma^2 =\frac{3}{\left(\frac{3}{5}\right)\left(\frac{5}{4}\right)^2}=16/5=3.2$.

I have drawn a spacetime diagram on rotated graph paper so that we can more easily visualize the ticks along worldlines on the spacetime diagram.

So, starting at the separation event,
the ${\rm\color{red}{homeTwin\ ages\ 10\ ticks\ along\ the\ red\ worldline}}$ until the reunion event,
while the ${\rm\color{blue}{travelingTwin\ ages\ 4\ ticks\ along\ the\ blue\ worldline}}$ to the turn-around event B,
then ages ${\rm\color{green}{another \ 4\ ticks\ along\ the\ green\ worldline}}$ [for a total of 8 ticks] to the reunion event.

When the origin is at the separation event, this is the standard formulation of the problem.
Here, we have drawn it in the $\rm\color{magenta}{AlphaCentauri\ frame\ (what\ you\ call\ O'\ and\ O'')}$ with the origin at event B.
Note that, in this frame,
the event you will call A is simultaneous with the separation event, and
the event you will call C is simultaneous with the reunion event.

The AlphaCentauri-(x',t')-coordinates of the various events are shown.
In agreement with your second paragraph, $(x'_A,t'_A)=(0,-\frac{L}{v})=(0,-5)$ and $t'_B-t'_A=0-\left(-\frac{L}{v}\right)=\frac{L}{v}=5$.

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So, now we draw this is the frame of the $\rm\color{blue}{outbound\ frame\ (what\ you\ call\ O)}$. I have included the lines-of-simultaneity for this frame. Later, I will redraw the "standard diagram" with these lines.

The outboundTwin-(x,t)-coordinates of the various events are shown.
In agreement with your second paragraph, $(x_A,t_A)=(\gamma L ,-\gamma\frac{L}{v})=(3.75,-6.25)$ and $t_B-t_A=0-\left(-\gamma\frac{L}{v}\right)=\gamma\frac{L}{v}=6.25$.

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Up to event B, the travelingTwin (O-frame) and AlphaCentauri (O'-frame) are both inertial observers.
However, turning back and continuing onward along the travelingTwin's worldline,
the travelingTwin is no longer inertial because the travelingTwin's worldline now has a kink any inertial reference frame. (No Lorentz boost can straighten that kink.)
Physically, a ball initially at rest on a frictionless table on the travelingTwin's ship will move at event B if the travelingTwin turns back [by firing its rockets].
Since AlphaCentauri (O''-frame) maintains the same velocity (magnitude and direction) of the O'-frame, it continues to be an inertial observer [so, in fact, the O''-frame coincides with the O'-frame].

To be clear,
the $\rm\color{blue}{outbound\ leg\ of\ the\ travelingTwin}\ is\ inertial$,
the $\rm\color{green}{inbound\ leg\ of\ the\ travelingTwin}\ is\ inertial$,
but this piecewise-inertial combination of worldlines is non-inertial.

So, along this $\rm\color{green}{inbound\ leg}$, we draw another spacetime diagram (for convenience---one could learn to read everything from one spacetime diagram [on rotated graph paper]).
This leg and Its coordinate deserve its own set of "primes": O''' with $(x''',t''')$

Because the inbound speed is equal to the outbound speed, some of the coordinate-values are similar. However, if the inbound speed were different, the numbers would be different.. introducing some merely-numerical complication.

Note that the coordinates of an event in this inbound-leg's coordinates is different from that of the outbound-leg.
$(x_A''',t_A''')=(-3.75,-6.25)$ and $(x_{sep}''',t_{sep}''')=(0,-4)$ and $(x_{reu}''',t_{reu}''')=(-7.5,8.5)$
$(x_A,t_A)=(\phantom{-} 3.75,-6.25)$ and $(x_{sep},t_{sep})=(-7.5,-8.5)$ and $(x_{reu},t_{reu})=(0,4)$

Thus, for this non-inertial observer, One has to somehow splice the two spacetime diagrams together, effectively devising some rule to make coordinates in one-to-one correspondence with ALL events in spacetime. (No double counting of events and No omission of events.)
(For inertial observer, no special rule isn't needed... another indication that non-inertial observers are distinct from inertial ones.)

In agreement with your third paragraph (with $'''$), $(x'''_C,t'''_C)=(\gamma L ,\gamma\frac{L}{v})=(3.75,6.25)$ and $t'''_C-t'''_B=\left(\gamma\frac{L}{v}\right)-0=\gamma\frac{L}{v}=6.25$.

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Now for your comparisons in your fourth paragraph.

I will insert this hybrid diagram (blending the previous two diagrams) which has not yet had any rule devised to splice this into a valid spacetime diagram (as described above).
(If you naively follow
the $\rm\color{blue}{outbound\ leg}$ of the travelingTwin with its lines of simultaneity, then switch at B to
the $\rm\color{green}{inbound\ leg}$ of the travelingTwin with its lines of simultaneity,
then splice the lower half of the outbound diagram with the upper half of the inbound diagram ,
note that you have omitted the midpoint (tick 5) of the homeTwin's worldline on your spliced diagram.)

With the travelingTwin's noninertial hybrid $O-O'''$-frame,
the hybrid quantity $(t'''_C-t'''_B)-(t_B-t_A)=2\frac{\gamma L}{v}=6.25+6.25=12.5$.
Because of symmetry in the problem, $t'''_C=t_C$ and $t'''_B=t_B=0$,
so $t_C-t_A=2\frac{\gamma L}{v}=6.25+6.25=12.5$.

With the AlphaCentauri inertial $O'-O''$-frame,
the quantity $(t''_C-t''_B)+(t'_B-t'_A)=\frac{L}{v}+\frac{L}{v}=5+5=10$.

And, finally, to your questions.

Did Alpha Centauri age less than the twin between events A and C, while the twin aged less than Alpha Centauri (and the twin who remained on Earth) between the events of departure and arrival of the twin in the original formulation of the problem?

Here are various time-intervals.
I'll let you do the comparisons.

Along AlphaCentauri's inertial worldline ABC,
we have that AlphaCentauri aged (from the previous part),
$(t''_C-t''_B)+(t'_B-t'_A)=\frac{L}{v}+\frac{L}{v}=5+5=10$.

For travelingTwin's noninertial worldline, events A and C are not its worldline.
So, we use the travelingTwin's lines of simultaneity to assign times to events A and C.
We have (from the previous part),
$(t'''_C-t'''_B)-(t_B-t_A)=2\frac{\gamma L}{v}=6.25+6.25=12.5$.
(What would this mean?)

Along the homeTwin's inertial worldline sep-reunion,
we have that homeTwin aged
$(t^{home}_{reu}-t^{home}_{sep})=10$.

Along the travelingTwin's noninertial worldline sep-B-reunion,
we have that travelingTwin aged
$(t'''_{reu}-t'''_B)-(t_B-t_{sep})=2(L/v)/\gamma=4+4=8$.

For AlphaCentauri's inertial worldline, events sep and reunion are not its worldline.
So, we use AlphaCentauri's lines of simultaneity (which coincide with homeTwin's lines of simultaneity) to assign times to events sep and reunion.
We have,
$(t'_{reu}-t'_{sep})=\frac{L}{v}+\frac{L}{v}=5+5=10$.
(What would this mean?)

Is this a valid method to debunk the twin paradox?

NO.

As stated in the beginning of my answer,
the Clock Effect and Twin Paradox compare elapsed times along two worldlines with the same endpoint events... the spacetime-version of the triangle-inequality.

So, the comparison is between:
homeTwin's 10 ticks from sep-reunion
and
travelingTwin's 8 ticks from sep-B-reunion.
These describe the legs of a triangle.

Events A, B, and C do not form a triangle for the trip of the travelingTwin.
And the noninertial travelingTwin spacetime-diagrams and calculations of the time-intervals between events A and C it did not visit (to get 12.5) is suspect.

Answer 2

So what you have done is to translate the one twin in space, to the place which the other twin would have turned around, before starting the experiment. This certainly does makes the math a bit easier, as you have noticed that the relativity of simultaneity is no longer a factor in the mathematics. Unfortunately, the resolution to the twin paradox hinges crucially on the relativity of simultaneity, so I don't think your setup sheds light on why we cannot just "reverse the argument" so that Earth moves while Alpha Centauri stands still, and come to the reverse conclusion about which twin ages more slowly.

What is at stake is that whenever I accelerate in relativity, I see the clocks “ahead of me” tick faster and the ones “behind me” tick slower, in proportion to both my acceleration and the distance to those clocks. This is not just an important thing about special relativity, it is the only important thing in special relativity: it directly implies the relativity of simultaneity and it indirectly implies the Lorentz transform which implies length contraction and time dilation.

The twin who travels to Alpha Centauri has to accelerate to turn around and when they do, they see their twin’s clock speed up and tick incredibly fast, which is how they both agree that the twin who remained on Earth is older. That is the proper resolution of the paradox and it amounts to observing a relativity of simultaneity between $O'$ and $O''$. Anything which does not observe this crucial point cannot explain the twin paradox because the only reason that the twin paradox exists is that someone forgot to account for it.

With that said, one can introduce these effects in your case by having the traveling twin start and end at rest relative to the Earth twin. They immediately accelerate and see the Earth twin age, then the Earth twin appears to live in slow motion until the traveling twin makes the rendezvous with Earth and stops to have a drink with his/her brother/sister. One then does nothing need to consider the relativity of simultaneity at the deceleration because of that “in proportion to the distance ahead of me” aspect of how much faster the clocks tick. In such a case during the rendezvous the Earth twin appears significantly older while on the reverse trip the traveling twin legitimately sees the Earth twin age much less to make up the difference.