Variation on twin paradox involving a common timekeeper

Question

To try to understand the twin paradox better I thought of a variant involving a third party time keeper that both twins can agree on, and am trying to understand what each party would actually observe while the astronaut is moving as well as when he decelerates back to earth speed and when both return home.

Specifically, let's say Alice is on earth and Bob is an astronaut zooming away from Alice at a velocity (v_ab) of 0.97c which gives a time dilation factor of roughly 4:1.

Suppose that Chrono is going to act as a third-party time keeper, and takes off right behind Bob in the same direction but at half the speed, such that v_ac and v_bc are both 0.48c. This yields a time dilation factor of 1.1:1. Let's also pretend that everyone can accelerate and decelerate effectively instantaneously so that the amount of time they're all travelling at anything other than their target speeds is negligible.

Chrono, at a regular interval from his perspective, sends a signal to both Alice and Bob. When they get Chrono's signal, they each snap a selfie and send it to Chrono, who then immediately relays it to the other twin. A couple initial questions:

• Will the twins appear to be aging at the same rate when they compare selfies? (Put differently, do they agree on how much time has passed between the receipt of consecutive signals from Chrono?)
• Do the two twins each perceive the same amount of time between their receipt of consecutive selfies relayed by Chrono?
• Assuming yes to both, can the time dilation between the twins be physically detected in ANY way prior to one of them decelerating, or is it only happening "on paper" at this point?

Now for the part that really stumps me. Suppose after 10 years from his perspective, Bob decelerates to the same frame as Chrono, so v_ab = v_ac = 0.48c, while v_bc = 0. Because the time dilation between Bob and Chrono was relatively small (1.1:1), while Bob aged 10 years, Chrono aged 11 in that same time.

But now let's say Chrono and Bob both immediately decelerate back down to Alice's frame, so v_ab = v_ac = v_bc = 0. Because of the time dilation caused while v_ac = 0.48c, Alice should have aged 12 years while Chrono aged 11. But because of the time dilation caused while v_ab = 0.97c, Alice should have aged 40 years compared to Bob's 10. In other words, when Bob and Chrono decelerate down to Alice's frame, it seems they should disagree on how much Alice has aged, which is of course absurd.

How does this apparent paradox get resolved? I'm guessing there's an error in my assumptions about what everyone's relative velocities and time dilation factors are, and I should be using the addition of velocities formula somewhere, but I'm not sure where exactly the error lies.

PS - I know this might feel like a homework problem but I assure everyone it's not. This is just my attempt to wrap my head around time dilation involving 3 parties rather than two.

Answer 1

Here's a graphical method (a spacetime diagram on "rotated graph paper") that you could use to analyze various situations you might want to dream up. This should hopefully give some intuition that you might not get from just applying formulas, especially if you want to consider piecewise-inertial trips.

I've drawn a different situation in timekeeper-Chrono's frame (but I'm sure that you can change it to your own scenario) where Alice and Bob move in opposite directions with the same outgoing and incoming speeds. Chrono is the only inertial observer that is at the separation event O and reunion event Z.

I have also chosen nicer numbers so that many of the calculations are rational numbers (with $v_{BC}=3/5$ and $v_{AC}=-3/5$ so that $v_{BA}=\frac{v_{BC}-v_{AC}}{1-v_{BC}v_{AC}}=\frac{15}{17}\approx 0.882$).

(Nicer calculations occur when the doppler factor $k=\sqrt{\frac{1+v}{1-v}}$ is rational.
For $v=3/5$, the doppler factor $k=2$.
For $v=-3/5$, the doppler factor $k=1/2$.
For $v=4/5$, the doppler factor $k=3$.
You can also work backwards using $v=\frac{k^2-1}{k^2+1}$. Check it.
The time-dilation factor $\gamma=\frac{1}{\sqrt{1-v^2}}=\frac{k^2+1}{2k}=\frac{k+k^{-1}}{2}$.)

A ticking longitudinal light-clock traces out what I call a light-clock diamond.
When a longitudinal light-clock is moving with respect to the frame of the diagram, the diamonds of its clock are stretched in the frontward-future direction by the doppler factor $k$ and shrunk in the rearward-future direction by the same factor, thus preserving the area of the diamond [as expected since the boost has determinant one]).

Here's the initial situation with the two travelers Alice and Bob in the "center of momentum frame" Chrono:

Note $v_{BC}=\tanh\theta=\frac{OPP}{ADJ}=\frac{PQ}{OP}=\frac{6}{10}=3/5$, $v_{AC}=\frac{PR}{OP}=\frac{-6}{10}=-3/5$ and $v_{AB}=\frac{BdtR}{OBdt}=\frac{-15}{17}$.

Note that (blue) Bob's ticks are reshaped, stretched by 2 and shrunk by 2.
When $k$ is a rational number, this is easy to draw on rotated graph paper.

Note that $\gamma_{BC}=\cosh\theta=\frac{ADJ}{HYP}=\frac{OP}{OQ}=\frac{10}{8}=5/4$, as expected.

The dash-dot line QA is parallel to the spacelike-diagonals of (blue) Bob's diamonds, and is a set of events simultaneous according to Bob. (In fact, BdtR is parallel to QA. BdtR is the instantaneous displacement to Alice (at -15) according to Bob when Bob's clock reads 17.) The other dotted line is simultaneous according to Bob during his return trip.

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If you want to broadcast and receive light-signals, then follow lines parallel to the rotated grid.

Two ticks after separation Chrono broadcasts his wristwatch image (reading 2 ticks) and is received by Alice when her wristwatch reads 4 ticks. That ratio 4/2 is equal to the Doppler factor and says that Alice has speed $\frac{(4/2)^2-1}{(4/2)^2+1}=3/5$ with respect to Chrono, and similarly for Bob.

Two ticks after separation Alice broadcasts her wristwatch image (reading 2 ticks) and is received by Chrono when his wristwatch reads 4 ticks (as expected by the relativity principle). Furthermore, Bob receives her broadcast when Bob's clock reads 8, which means that Bob's speed is $\frac{(8/2)^2-1}{(8/2)^2+1}=15/17$ with respect to Alice.

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• So, now you have the tools to analyze what happens with regard to comparing selfies that each traveler broadcasts to the others.
• You can also alter the scenario. For example, 4 ticks after separation, Bob decides to assume the same speed as Chrono.

For more on this approach, consult my earlier answers to

• Triplets paradox

• What if two twins flew off in opposite directions and were reunited in a perfectly symmetric way, would they have aged same?

• Question about Special Relativity similar to twin clock experiment

• Equivalence of two definitions of proper time in special relativity displays different trips between separation and reunion

• What is the proper way to explain the twin paradox? (My answer has a reference to my article on AJP.)

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If you really want to use this to approximate your original proposed speed of $v=0.97$, use $v_{BC}=4/5$ (so, $k_{BC}=3$) and $v_{AC}=-4/5$ so that $v_{BA}=40/41\approx 0.97561$... and use several sheets of graph paper that you'll have to tape together.

Answer 2

If you don't mind, it is possible to simplify putting C in a ship close to Earth, and in the same frame of it. A and B travels each one at 0.485c in opposite directions. The relative velocity between A and B is: $$v_{AB} = \frac{0.485+0.485}{1+0.485*0.485} = 0,785c$$

That way A and B have the same magnitude of velocity with respect to C, that is the idea (I suppose). Of course all selfies send to C from A and B will show the same age of A and B, if C receives them at the same time.

But to compare the time dilation of each one in relation to the other, it is better to suppose 'trains of ships' instead of just one per frame of reference. This approach avoids the complications of the delays of exchanging signals.

So, C has a clone each light-day in the both directions, and all of them in its own frame. The same for A and for B for their frames. The relative distance is kept constant between the clones of A, B and C in theirs respective frames of reference.

C observes the A and B clones passing by younger and younger than him. The same for A and B with respect to the C clones, and all of them at the same rate ($\approx 1:1.14)$.

But A observes B clones passing by younger and younger at a higher rate ($\approx 1:1.61$) because of their bigger relative speed. The same for B observing A clones.

When B changes its velocity to stay in the same frame of C, he now ages at the same speed of the nearby C clone, and also all of C clones including the original. If they exchange selfies, they will show a younger B compared to C.

When B and C change their velocities to stay in A frame, A will see a nearby C clone as younger with respect to her as C with respect to B in the previous case. But A will see a nearby B clone more younger than C. Of course, as they are now all in the same frame, if the originals exchange selfies they will confirm the age relative difference.

Note that in all the cases, even before the acceleration, the time dilation is transparent for each (original) ship with respect to the clones of the other frames. The effect of a sudden acceleration, moving to other frame, in only to freeze the time difference accumulated until that moment, always visible in the clone nearby.

Answer 3

I'm going to put down an answer which does not bother about multiple observers but simply does the calculation. Here it is.

To solve the twin paradox draw the worldlines of the two twins on a diagram, and then work out the proper time passing for each of them (exactly if you can, approximately otherwise). For a section of worldline crossing distance $\delta x$ in elapsed time $\delta t$ (both as observed in some inertial frame which should be left fixed throughout), the elapsed proper time is $\delta t / \gamma$ where $\gamma = 1/\sqrt(1-v^2/c^2)$ with $v = \delta x/\delta t$.

That's really all there is to it.

If one of the worldlines you choose to draw is a straight from start to finish, then for convenience you may as well do the calculation in the inertial frame where that particle (or observer or whatever) is not moving. For that line you get $v=0$ always and so $\gamma = 1$ and so total proper time $\tau = (t_f - t_i)$ where $t_i$ is the initial time in that frame and $t_f$ is the final time in that frame.

For any other line (i.e. a non-straight one) you get some sections with $v \ne 0$ and for those sections $\gamma > 1$, but there are no sections with $\gamma < 1$ because that can't happen. So once you have added up all the sections you must find the total $\tau$ for the non-straight line has to be less than $(t_f - t_i)$. To repeat, this is because the $\gamma$ values on the non-straight line go above 1 for some sections, but never go below 1.

The general conclusion can now be stated: taking a wiggly path in spacetime is guaranteed to reduce your overall elapsed proper time between two given events.

I want to emphasize that the above is the whole twin paradox in complete generality, and brings out the essential point. Once you have understood this then you are well on the way to understanding whatever other scenarios you may wish to bring in.