B What is the effect of the twin paradoxon?

[Peter Strohmayer]

That's time dilation without any differential ageing.

A circular motion is complicated (where does the "time dialatation" come from, centripetal acceleration or relative motion?)

Let's stick with translational motion. Each of the twins is at the center of a long synchronized clock line. As the twins move past each other, they set the clocks directly in front of them - and thus their entire row of clocks - to "zero". The twin's clock continues to move past the other twin's row of clocks. Comparing the display of his clock with the display of the meeting clock, the twin notices that the display of his clock is lagging more and more behind the display of the just passing meeting clock. This is a symmetric process. Neither twin ages faster or slower as a result.

"It's time dilation without differential aging."

So what is the effect of the twin parodoxon? The acceleration of one twin!

 

[Ibix]

A circular motion is complicated (where does the "time dialatation" come from, centripetal acceleration or relative motion?)

You first need to define what you mean by time dilation in this context. I suspect @PeroK is using it to mean that "in the momentarily co-moving inertial frame of either observer the other one is always in motion and hence the other's clock is ticking slow". So in that case it obviously comes from relative motion.

If you mean something other than that by "time dilation" you need to define what you mean before the question can be answered.

This is a symmetric process. Neither twin ages faster or slower as a result.

More precisely, there is no way to answer which one is aging faster without picking a frame, and depending on which frame you pick the answer can be one, the other, or neither.

So what is the effect of the twin parodoxon? The acceleration of one twin!

All the acceleration does is invalidate a simple time-dilation based analysis, and allow the twins to meet up. There are at least two ways to do twin paradox type experiments without acceleration, and both still show differential aging effects - so the acceleration can't be that critical.

 

[Peter Strohmayer]

More precisely, there is no way to answer which one is aging faster without picking a frame, and depending on which frame you pick the answer can be one, the other, or neither.

In reference frame A, A would age slower than B, in reference frame B, B would age slower than A. This would be a contradiction, which is due to the fact that in the symmetrical situation described above, the lagging behind of clock displays compared to others is called "aging slower" or "... less time has passed". There is no justification whatsoever for making such speculations about the behavior of time.

The symmetry can only be broken by acceleration.

 

[Ibix]

In reference frame A, A would age slower than B, in reference frame B, B would age slower than A. This would be a contradiction,

No. This would be two different descriptions of the same thing. You can use the Lorentz transforms to switch between one and the other, so they are clearly not contradictory.

The symmetry can only be broken by acceleration.

As I said, there are at least two variations on the twin paradox that do not involve any acceleration but lead to differential aging.

 

[Dale]

A circular motion is complicated (where does the "time dialatation" come from, centripetal acceleration or relative motion?)

I disagree with this. It is circular motion that provides our most convincing experimental evidence for the clock hypothesis. This is precisely the hypothesis that there is no additional time dilation due to the acceleration that is not accounted for by the speed. In circular motion this hypothesis has been confirmed up to about $10^{18} \ g$

 

[Peter Strohmayer]

No. This would be two different descriptions of the same thing.

Of course, it is no contradiction if the clock of A lags behind one of the passing clocks of B, and if the clock of B lags behind one of the passing clocks of A.

But this is not what you claim.

The contradiction arises when the described facts are combined with speculative stories, e.g. that in unaccelerated motion A would age slower than B and B would age slower than A.

 

[Peter Strohmayer]

It is circular motion that provides our most convincing experimental evidence for the clock hypothesis.

I gladly believe you. But it irritates me that the twins move relative to each other and yet their watches always show the same time when they meet.

 

[Nugatory]

In reference frame A, A would age slower than B, in reference frame B, B would age slower than A. This would be a contradiction....

There is no contradiction here, just unclear thinking about what it means to say that one person is aging more slowly than another. The contradiction goes away when we define our terms carefully - including recognizing the relativity of simultaneity.

The symmetry can only be broken by acceleration.

Yet it is possible to construct a twin paradox situation that involves no acceleration at all, or when both twins experience identical accelerations.

 

[Peter Strohmayer]

just unclear thinking about what it means to say that one person is aging more slowly than another

I have no idea how I could have misunderstood the sentence, one ages slower than another, in view of the relativity of simultaneity.

 

[Ibix]

But this is not what you claim.

That is exactly what I claim. It was you who said it was a contradiction, in post #22. Either you have now changed your mind or you are discussing two different things and not making clear (at least to me) what you think is contradictory and what is not.

The contradiction arises when the described facts are combined with speculative stories,

Here's the problem: I don't know what you are calling "described facts" and what you are calling "speculative stories". You seem only to be describing two clocks in inertial motion and sometimes describing their behaviour as contradictory and sometimes not.

 

[Nugatory]

I have no idea how I could have misunderstood the sentence, one ages slower than another, in view of the relativity of simultaneity.

The assertion "A is aging more slowly than B" is equivalent to

At the same time that A's age is $A_0$ B's age is $B_0$. Later, at the same time that A's age is $A_0+\Delta T_A$ B's age is $B_0+\Delta B_T$, and $\Delta B_T\lt\Delta A_T$​

Clearly this will depend on the simultaneity convention used to define "at the same time"; the "A is aging more slowly than B" formulation serves only to obscure that dependency.

 

[Dale]

I gladly believe you. But it irritates me that the twins move relative to each other and yet their watches always show the same time when they meet.

Instead of being irritated, I look at this as an opportunity to demonstrate how these quantities can be calculated in an easy non-inertial frame to get consistent results.

In reference frame A, A would age slower than B, in reference frame B, B would age slower than A. This would be a contradiction

As others have stated, this is not a contradiction, but it is wrong. $$\frac{d\tau_A}{dt_A} > \frac{d\tau_B}{dt_A}$$ does not contradict $$\frac{d\tau_A}{dt_B} < \frac{d\tau_B}{dt_B}$$

 

[PeroK]

The symmetry can only be broken by acceleration.

Groan!

 

[Peter Strohmayer]

For example: consider the twins in the same circular orbit about a star or planet, but going in opposite directions. Each twin is moving relative to the other and each is continuously time dilated relative to the order. But, every time they pass each other their clocks show the same elapsed time.

Instead of being irritated, I look at this as an opportunity to demonstrate how these quantities can be calculated in an easy non-inertial frame to get consistent results.

Is the described counter-rotating circular motion perhaps equivalent to a mirror-image journey of two twins who start at the same time, turn the same distance and return at the same time (e.g. if their movements describe circles of the same size)?

 

[Peter Strohmayer]

You seem only to be describing two clocks in inertial motion and sometimes describing their behaviour as contradictory and sometimes not.

From the fact that the display of A's clock lags behind the display of one of B's passing clocks (which does not contradict a corresponding lag of B's clock), it cannot be concluded that A has aged more slowly than B (which, moreover, contradicts a corresponding slower aging of B). Even after a longer flyby, there is no change in the same age of A and B if they are treated equally in the sequence. When A and B meet again due to symmetric acceleration, they are still the same age.

 

[Peter Strohmayer]

Groan!

As described in #20, the reading of A's clock lags behind the readings of B's passing clocks, and the reading of B's clock lags behind the readings of A's passing clocks. This has nothing to do with differential aging. What would ever change in this result without acceleration?

There are at least two ways to do twin paradox type experiments without acceleration, and both still show differential aging effects - so the acceleration can't be that critical.

Will muons return?

 

[renormalize]

As I said, there are at least two variations on the twin paradox that do not involve any acceleration but lead to differential aging.

Can you expand on this statement by addressing a general question regarding travelers along two worldlines?

(Based on: https://sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters_May_30_2021/spacetime_tachyon/index.html)

Consider two observers $O_{1},O_{2}$ in flat spacetime, initially sharing the same worldline. At spacetime event $A$, their worldlines diverge and they each begin recording their proper-times and proper-accelerations. At event $B$, their worldlines re-converge and they compare readings. They find that their elapsed proper-times differ (i.e., they've experienced differential aging), so they rightly conclude that the individual worldlines they travelled between $A$ and $B$ must have different proper lengths. My question: are there any possible circumstances for which both $O_{1}$ and $O_{2}$ will have recorded no/zero proper accelerations between $A$ and $B$?

In other words: what possible mechanism in Minkowski spacetime, other than proper accelerations, can cause a difference in proper-lengths between two distinct worldline-segments that share common starting and ending points?

 

[jbriggs444]

In other words: what possible mechanism in Minkowski spacetime, other than proper accelerations, can cause a difference in proper-lengths between two distinct worldline-segments that share common starting and ending points?

Unless I am badly mistaken, in Minkowski spacetime two distinct geodesics can intersect only once.

You can make them intersect twice without a non-zero proper acceleration if you "bend" the rules with an event on a worldline where the proper acceleration is undefined. e.g. a clock handoff.

 

[PeterDonis]

Unless I am badly mistaken, in Minkowski spacetime two distinct geodesics can intersect only once.

You are not mistaken at all. This is correct; in fact it is at most once, since there are pairs of geodesics that do not intersect at all.

 

[renormalize]

Unless I am badly mistaken, in Minkowski spacetime two distinct geodesics can intersect only once.

Yes, that's really my point. Without accelerations, twins on different worldlines that started together can never meet again to compare ages. So I think we need to carefully distinguish between "differential aging where the twins start at event $A$ and end (or cross) at event $B$", which always requires acceleration (possibly infinite), and "differential aging where the twins never meet again". It's the later that I'd like see clarified to understand Ibix's statement that "...there are at least two variations on the twin paradox that do not involve any acceleration but lead to differential aging."

 

[PeterDonis]

Without accelerations, twins on different worldlines that started together can never meet again to compare ages.

This is true in flat Minkowski spacetime, as @jbriggs444 said, but it is not true in curved spacetime. In curved spacetime a pair of geodesics can cross more than once. Curved spacetime scenarios are probably what @Ibix was thinking of.

 

[Dale]

Is the described counter-rotating circular motion perhaps equivalent to a mirror-image journey of two twins who start at the same time, turn the same distance and return at the same time (e.g. if their movements describe circles of the same size)?

Yes. And as in that case you can use isotropy of the laws of physics to conclude that the readings will be the same.

what possible mechanism in Minkowski spacetime, other than proper accelerations, can cause a difference in proper-lengths between two distinct worldline-segments that share common starting and ending points?

The metric.

 

[PeroK]

Yes, that's really my point. Without accelerations, twins on different worldlines that started together can never meet again to compare ages. So I think we need to carefully distinguish between "differential aging where the twins start at event $A$ and end (or cross) at event $B$", which always requires acceleration (possibly infinite), and "differential aging where the twins never meet again". It's the later that I'd like see clarified to understand Ibix's statement that "...there are at least two variations on the twin paradox that do not involve any acceleration but lead to differential aging."

As has been explained in previous threads on this topic, acceleration is a physical constraint. The analogy is to claim that the hypotenuse of a right angle triangle is shorter than the sum of the other twin sides because of what happens to a pencil at the right angle corner. Rather than simply that the sum of the two lengths is greater..

Moreover, what is your explanation for more complicated scenarios where both twins accelerate? Do you have a complex system of measuring various accelerations and eventually determine which twin has aged more? Show me your equation for age based on proper acceleration! The alternative is simply to measure the length of each twin's worldline, which does not directly involve proper acceleration.

Finally, the same paradox can be described without acceleration, by the time measurement on the outbound journey being added to the time measurement on an inbound journey, with the turnaround being replaced by a passing of the baton, as it were. In this way, we see that the explanation is clearly and simply explained by the length of the different spacetime intervals. And, not necessarily by some mysterious force associated with proper acceleration.

 

[Peter Strohmayer]

You can make them intersect twice without a non-zero proper acceleration if you "bend" the rules with an event on a worldline where the proper acceleration is undefined. e.g. a clock handoff.

Please do not hand over any items from a moving car, including watches. The recipient could be injured. If you meant that a traveler shouts out the display of his watch to an oncoming traveler as he passes, that would be relativistic bookkeeping, but not physics.

 

[Peter Strohmayer]

It is circular motion that provides our most convincing experimental evidence for the clock hypothesis. This is precisely the hypothesis that there is no additional time dilation due to the acceleration that is not accounted for by the speed.

(The twins describe circles of the same size) Yes. And as in that case you can use isotropy of the laws of physics to conclude that the readings will be the same.

Then, in my opinion, the experimental evidence for the clock hypothesis is based on accelerated motion. I am not saying that the acceleration would cause an "additional time dilation", but that the detection of the time difference presupposes a return, and this return presupposes an acceleration.

 

[Ibix]

In other words: what possible mechanism in Minkowski spacetime, other than proper accelerations, can cause a difference in proper-lengths between two distinct worldline-segments that share common starting and ending points?

There are three scenarios that I was thinking of, although only one meets your criteria.

The first is to use curved spacetime, as others have noted. The traveller swings round a black hole and returns, having been inertial at all times. Yet their age is different.

The second is sometimes called the "triplet paradox", although that's a misnomer. You have three eternally inertial observers with speeds $0, \pm v$ arranged so that they don't all meet at the same time. Each observer records their proper time between meetings. You will find that the ratio between the sum of the shorter times and the longer time is the same $\gamma$ you would see in a twin paradox. Geometrically, this is because a triangle has the same properties whether you choose to extend the line segments making up its sides through the vertices or not. Physically this shows that the proper time of the travelling twin is the same as two inertial observers temporarily co-moving with him, so acceleration can't be the cause of the shorter time.

The third does meet your specifications. You use flat spacetime with a non-trivial topology - you give space two edges, then identify the edges. On a Minkowski diagram this is equivalent to printing it out and rolling it up into a cylinder with the time axis along the cylinder. Any pair of inertial observers with non-zero relative speed will meet repeatedly having experienced different elapsed times (except the special case where their velocities are equal and opposite in the frame whose timelike axis is along the cylinder), and again with no acceleration. The spacetime is still flat in the relevant sense.

 

[Ibix]

If you meant that a traveler shouts out the display of his watch to an oncoming traveler as he passes, that would be relativistic bookkeeping, but not physics.

You are saying that adding two quantities together and comparing them to a third is not physics? Is that a general rule for you?

 

[Peter Strohmayer]

I don't mean that apodictically. The transfer of the information about the display of the clock allows the calculation of a "slower aging", but to be really younger than his brother, the twin must return himself.

 

[Ibix]

but to be really younger than his brother, the twin must return himself.

Sure. But that only requires acceleration in flat spacetime with a trivial topology, and it doesn't mean acceleration causes differential aging.

 

[Peter Strohmayer]

Curved spacetime also leads to acceleration via gravity and equivalence. It is sufficient and simpler to analyze the twin paradox from the point of view of SRT.

 

[Ibix]

Curved spacetime also leads to acceleration via gravity and equivalence.

No it doesn't - gravity and acceleration are not the same thing and the equivalence principle does not say they are.

The twin paradox can be done in curved spacetime using only two inertial observers. The kind of acceleration you are referring to is coordinate acceleration, and if that is what you are referring to then you undermine your argument in flat spacetime because one can easily construct coordinate systems where the inertial twin undergoes larger coordinate accelerations than the non inertial one.

 

[renormalize]

Sure. But that only requires acceleration in flat spacetime with a trivial topology, and it doesn't mean acceleration causes differential aging.

Suppose I grant that acceleration doesn't cause differential aging. Nevertheless, under the conditions you list, differential aging always correlates with proper acceleration. That is, in a Minkowski spacetime with trivial topology, one cannot have the former without the latter. What physically distinguishes causation from correlation in this context?

 

[Ibix]

Nevertheless, under the conditions you list, differential aging always correlates with proper acceleration.

Not quite. Consider two twins who set out from Earth together at the same speed. One turns around and returns at the same speed, then stops at Earth and waits. The other continues on a bit more before returning at the same speed. Both undergo the same proper accelerations in the same sequence but end up having different ages - and the only difference is the timing and duration of the inertial phases. I can't see how you can interpret that one as "acceleration causes differential aging".

The other thing you can do is think of a Euclidean triangle. What makes the length along two sides greater than the length along one? "It has a corner" seems trite to me.

 

[Peter Strohmayer]

Not quite. Consider two twins who set out from Earth together at the same speed. One turns around and returns at the same speed, then stops at Earth and waits. The other continues on a bit more before returning at the same speed. Both undergo the same proper accelerations in the same sequence but end up having different ages - and the only difference is the timing and duration of the inertial phases. I can't see how you can interpret that one as "acceleration causes differential aging".

One can simplify this example of the different travel times of twins considerably and reduce it to the core of what is happening. I assume two clock lines (reference systems) moving towards each other.

After the meeting of A and B, the display of A's clock lags behind that of B's passing clocks. This time difference (which occurs symmetrically at B with respect to the passing clocks of A) increases proportionally to the length of the journey.

If A decides to jump over to one of the passing clocks of B (this is an acceleration in infinitesimal steps over the integral of all infinitesimal speed differences), then he takes the time of his clock A unchanged with him, but now rests in the frame of reference of B in the middle of the clocks of B, which now show the same time for A as for B. The display of A's clock (his lifetime) now remains constantly behind the displays of all clocks of B. This means that he has actually remained younger compared to B.

The difference in age increases the later A decides to jump to one of B's passing clocks (i.e. to accelerate and equalize the relative speed with respect to B). This does not change the fact that without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible.

 

[Dale]

Then, in my opinion, the experimental evidence for the clock hypothesis is based on accelerated motion.

Of course the experimental evidence for the clock hypothesis is based on accelerated motion. It is a hypothesis about accelerated motion. You could not experimentally test it without it.

I am not saying that the acceleration would cause an "additional time dilation", but that the detection of the time difference presupposes a return, and this return presupposes an acceleration

Let me ask you a closely related question. Given a triangle with sides $A$, $B$, and $C$ each side opposite angle $a$, $b$, and $c$ respectively. We know that $A<B+C$. Would you say that the angle $a$ caused the difference in length $B+C-A$?

This does not change the fact that without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible.

It certainly is possible to have differential aging without proper acceleration in the real universe, where spacetime is not flat.

 

[Ibix]

One can simplify this example of the different travel times of twins considerably and reduce it to the core of what is happening.

Unfortunately, your simpler example misses the point, which was that you can have identical accelerations and different aging. This strongly suggests that acceleration is not really relevant to differential aging. In fact, it's just one way to make the teins meet twice so that their clocks can be directly compared.

After the meeting of A and B, the display of A's clock lags behind that of B's passing clocks.

Maybe. That would depend on facts you haven't specfied about where A and B were when their clocks were zeroed.

without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible.

This statement is not correct. It is only true in flat spacetime with a trivial topology, and the fact that it's not true in general tells you that acceleration is not key to the differential aging. It's only the means you use to allow the twins to meet twice.

 

[Dale]

That is, in a Minkowski spacetime with trivial topology, one cannot have the former without the latter. What physically distinguishes causation from correlation in this context?

The real universe is not well described by Minkowski spacetime. In very ordinary curved spacetimes you can easily get different ages without any proper acceleration. You can also get proper acceleration without different ages. So proper acceleration is neither necessary nor sufficient for different ages.

 

[Sagittarius A-Star]

If you meant that a traveler shouts out the display of his watch to an oncoming traveler as he passes, that would be relativistic bookkeeping, but not physics.

A usual clock contains an oscillator, a counter and a display. What the counter does, is bookkeeping. There is no problem, to divide the work between two clocks.

 

[Dale]

A usual clock contains an oscillator, a counter and a display. What the counter does, is bookkeeping. There is no problem, to divide the work between two clocks.

Also, it is entirely physically possible for the two clocks to simply happen to read the same as they pass. Regardless how those clocks come to read the same, the fact remains that the time at the final comparison is smaller.

There is asymmetry, but no acceleration in this scenario. All the acceleration ever did was make the twins asymmetric. Other asymmetries suffice, not just acceleration.

 

[Peter Strohmayer]

Unfortunately, your simpler example misses the point, which was that you can have identical accelerations and different aging.

This is not correct: in #53 I described two identical accelerations (jumps to the passing encounter clock).

In fact, it's just one way to make the teins meet twice so that their clocks can be directly compared.

I have simplified the example: it is enough for one twin to switch to the other's frame of reference once to make him age less.

you haven't specfied about where A and B were when their clocks were zeroed

I find this objection too sophistical. In all these examples, the clocks are set to zero when the twins meet. I described this in #20.

It [i.e. "without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible"] is only true in flat spacetime with a trivial topology, and the fact that it's not true in general tells you that acceleration is not key to the differential aging. It's only the means you use to allow the twins to meet twice.

See above: It does not matter if the twins meet twice.

My statement applies to the acceleration due to recoil. I never claimed that this acceleration would play a role in slower aging in a gravitational field. It does not follow from a denial of this claim that recoil acceleration would not play a role even in a gravitationally-free field.

Rather, the question shifts to whether recoil acceleration and gravitational acceleration affect an observer's world line in a similar way. Can a heavy twin be equated with a recoil-accelerated twin? I do not feel called upon to answer this question. In any case, the answer does not change the dependence of the twin paradox on the recoil acceleration in a gravitationally free region.

 

[Dale]

I have simplified the example: it is enough for one twin to switch to the other's frame of reference once to make him age less

You are simply using the acceleration to have them start agreeing on simultaneity, which is always a matter of convention anyway. Even under that approach, it is during the inertial motion that the disagreement with the other’s clocks begins and that disagreement accumulates the longer the inertial motion continues.

However, it is just as easy for an observer to use a frame in which they are not at rest. Both observers could choose to use any inertial frame they agree on.

They could both use the frame of the other observer, there the one twin’s slow aging occurs on the first leg and the acceleration stops the one twin’s slow aging, with the one twin being younger.

They could both use the inertial frame where the one twin is initially at rest. In that frame the other twin is aging more slowly. The acceleration of the one twin makes the one twin age more slowly too. But in this frame the other twin is the younger.

They could also use an inertial frame where both twins are moving, with equal and opposite speed. Then both twins are at all times the same age, both before and after the acceleration.

It does not matter if the twins meet twice

It really does matter. As described above, if they do not meet twice then the final state is frame dependent. Any answer becomes possible.

 

[Dale]

The symmetry can only be broken by acceleration.

I think this may be your key lack of understanding. It is good to think in terms of symmetry, but it is incorrect that acceleration is the only thing that can break the symmetry.

In the three-observer scenario there is no acceleration. The symmetry is broken by the use of two inertial clocks instead of one. In curved spacetime you can have a twin scenario with no acceleration and the symmetry is broken by the curvature of the spacetime.

Furthermore, not only can you break the symmetry without acceleration, you can also accelerate without breaking the symmetry. This was described for the counter-rotating uniform circular motion. But it would apply for many other acceleration profiles too.

I think that there are two concepts you will need to correct before continuing:

1) the twins must meet at two events (at least) in order for their clock comparison to be frame invariant

2) acceleration breaks the symmetry in the usual scenario, but in general acceleration is neither necessary nor sufficient for broken symmetry

Please think on those ideas before continuing.

 

[Peter Strohmayer]

Let me ask you a closely related question. Given a triangle with sides A, B, and C each side opposite angle a, b, and c respectively. We know that A<B+C. Would you say that the angle a caused the difference in length B+C−A?

No, I wouldn't.

 

[Peter Strohmayer]

As described above, if they do not meet twice then the final state is frame dependent.

In order to have a basis for my thinking, I ask for the following clarifications:

We agree that in the example I gave, neither twin accelerates at the beginning of the trip (the twins are supposed to set their clocks to zero as they pass each other).

You use the terms "twins" and "observers" synonymously. (You speak of "two observers"). Then you speak of the two "observers" (=twins) agreeing on a frame of reference. So they obviously represent one and the same (imaginary) observer. - Could we then agree to call a twin an observer, if we take as a basis a reference frame in which the respective twin rests (at most up to the time of the first and at the same time only acceleration considered here)? It would be possible to choose an observer other than one of the twins (e.g. one moving relative to each of the twins), but we do not choose such an observer because it would obviously complicate the analysis.

Instead of speaking of "one twin" and "the other twin", I would suggest referring to the twins as twin A (which rests respectively moves to the left) and twin B (which moves to the right respectively rests), so that there are no misunderstandings (e.g. about which twin is accelerated when and to what speed, so that slower aging would be stopped; or about the claim that both twins would have aged the same since they set their clocks to zero
before and after "the acceleration", which would be impossible with a single acceleration of one of the twins I spoke of).

 

[Dale]

You use the terms "twins" and "observers" synonymously.

Yes

Then you speak of the two "observers" (=twins) agreeing on a frame of reference. So they obviously represent one and the same (imaginary) observer.

This is exactly the issue I am addressing. You have in your mind that “observer” = “frame”, such that if two observers agree to use the same reference frame they are the same observer. This is a result of the usual didactic approach where scenarios are populated by observers that each use their rest frame.

The principle of relativity states that all inertial frames are equivalent. So an observer may use any inertial frame. They may use one where they are at rest or they may use one where they are moving. I, as an observer, may use the frame where the ground is at rest and say “I went to the store”. I am not required to use the frame where I am at rest and say “the store came to me”.

Could we then agree to call a twin an observer, if we take as a basis a reference frame in which the respective twin rests (at most up to the time of the first and at the same time only acceleration considered here)?

This is precisely what I wish not to do. I am deliberately dissociating observers/twins from reference frames, as permitted by the principle of relativity.

Instead of speaking of "one twin" and "the other twin", I would suggest referring to the twins as twin A (which rests respectively moves to the left) and twin B

That is fine. The “one twin” and the “other twin” was your terminology which I simply continued.

 

[Dale]

@Peter Strohmayer I decided to go ahead and illustrate the issue so that maybe it is super-clear. This is the scenario you described where the blue line is the worldline of the inertial twin and the orange line is the worldline of the accelerating twin. On each worldline the first dot is $\tau=0 \ years$, the second dot is $\tau=4 \ years$, and the third dot is $\tau=8 \ years$ on each twin's clock. The speed is $v=3/5 \ c$.

They could both use the frame of the other observer, there the one twin’s slow aging occurs on the first leg and the acceleration stops the one twin’s slow aging, with the one twin being younger.

This frame is moving at $v=0$ with respect to the inertial twin.

They could both use the inertial frame where the one twin is initially at rest. In that frame the other twin is aging more slowly. The acceleration of the one twin makes the one twin age more slowly too. But in this frame the other twin is the younger.

This frame is moving at $v=3/5$ with respect to the inertial twin.

They could also use an inertial frame where both twins are moving, with equal and opposite speed. Then both twins are at all times the same age, both before and after the acceleration.

This frame is moving at $v=1/3$ with respect to the inertial twin.

By not having them co-located the relativity of simultaneity means that different frames disagree on which times are simultaneous. The result is that at the end you are free to choose an inertial frame where either twin is older or where they are the same age. It is entirely a matter of convention, and each is completely valid.

 

[Vanadium 50]

Tangent Thread

Huh. Usually in relativity you get hyperbolic tangents.

 

[Peter Strohmayer]

This is precisely what I wish not to do.

An "observer" is an ideal, massless, imaginary measuring station at the center of a chosen frame of reference.
It leads to misunderstandings, to call a mass point an "observer", which cannot observe anything, but only produce a world line.
Only if a twin fulfills the function of the mentioned measuring station in a chosen reference system, it should be called "observer".

That is fine. The “one twin” and the “other twin” was your terminology which I simply continued.

This is what I said:
-----
„#1: Each of the twins is at the center of a long synchronized clock line. As the twins move past each other, they set the clocks directly in front of them - and thus their entire row of clocks - to "zero". The twin's clock continues to move past the other twin's row of clocks. Comparing the display of his clock with the display of the meeting clock, the twin notices that the display of his clock is lagging more and more behind the display of the just passing meeting clock. This is a symmetric process. Neither twin ages faster or slower as a result."

#34: I assume two clock lines (reference systems) moving towards each other. After the meeting of A and B, the display of A's clock lags behind that of B's passing clocks. This time difference (which occurs symmetrically at B with respect to the passing clocks of A) increases proportionally to the length of the journey.

If A decides to jump over to one of the passing clocks of B (this is an acceleration in infinitesimal steps over the integral of all infinitesimal speed differences), then he takes the time of his clock A unchanged with him, but now rests in the frame of reference of B in the middle of the clocks of B, which now show the same time for A as for B. The display of A's clock (his lifetime) now remains constantly behind the displays of all clocks of B. This means that he has actually remained younger compared to B.

The difference in age increases the later A decides to jump to one of B's passing clocks (i.e. to accelerate and equalize the relative speed with respect to B). This does not change the fact that without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible.“

A = accelerating twin, B= inertial twin.
-----
To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place. The twins do not have to pull each other's beards. It is sufficient for them to rest in the same inertial frame, even if they are far apart in space.

I decided to go ahead and illustrate the issue so that maybe it is super-clear.

Your diagrams do not show what I have described.

The point is to determine which time the clock of the clock line of B just passing A shows when A decides to jump to this clock of B.

This cannot be read from the Minkowski diagrams you draw. You do not include the world line of the encounter clock. With proper analysis, it turns out that the age difference resulting does not depend on the choice of reference frame.

Here are the two graphs showing how a recoil acceleration leads to an age difference (strictly speaking, this schematic simplification is only valid for an infinitesimal acceleration step):
Red=Inertial twin B:

Green = accelerating twin:

different frames disagree on which times are simultaneous

Yes.

you are free to choose an inertial frame where either twin is older or where they are the same age

No.

 

[PeterDonis]

No.

You are incorrect here, and @Dale is correct. He showed you explicitly how. I strongly advise you to be careful, as you are getting close to a warning at this point.

 

[PeterDonis]

To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place.

If you want the comparison to be frame invariant, it is.

Again, I strongly advise you to be careful. You are getting close to a warning.