B Relativity: Twin Paradox - Is Age Determinable?

[PeroK]

The thing this doesn't do is provide an explanation for why naive application of the time dilation formula leads to a paradox. It is indeed a complete and very general solution to the scenario and any other similar one. But it doesn't explain the mistake, which is essentially the same one from Round the World in Eighty Days - forgetting to change your calendar when you change timing convention.

To summarise. The flat spacetime metric tells you everything you need to know about any path or paths through spacetime. The simplest explanation for the twin paradox, therefore, involves analysis in any IRF. It doesn't need to be the Earth's rest frame. Any IRF will do and the metric does the rest.

This, however, does not explain things in terms of a physical explanation for an accelerating observer. When we try to find an accelerating coordinate system to cover the entire experiment we immediately find ambiguities. The same event may be mapped to different coordinates, for example. This undermines the attemept to give a clear and unambiguous record of what happened where and when in an accelerating reference frame.

This takes us into the background material presented in this thread: attempts to give a consistent coordinate system for the accelerating observer etc. In particular, if a simultaneity convention is seen as something non-physical, then a physical explanation for the accelerating observer is elusive.

 

[PAllen]

I assume this has been stated earlier, so apologize for the repetition. But there is a simple invariant statement that can be made (whose description will differ by coordinate choice, but not the result):

If you Einstein synchronize the Earth and Mars clock, and synchronize an Earth rocket clock with the Earth clock, then have the rocket travel to mars, then the rocket clock will be behind the Mars clock. This is true no matter what travel path or acceleration is used, but these latter choices determine whether the amount is miniscule or enormous (e.g. a looping path near c per earth, that takes 1k years in earth-mars frame to reach mars).

Obvious simplifications: gravity ignored, Earth and Mars considered mutually stationary.

 

[vanhees71]

I think one saves oneself a lot of trouble, if one simply uses covariant quantities. If this is not possible, something usually is at least problematic, if not simply ill-defined or wrong, with the way a problem is stated or thought about.

In the case of the "one-way twin paradox" one can simply refer to the proper time of each twin. That's a local concept too, because it just is the time each twin reads off from his or her wristwatch. Using these proper times for the "aging" of the twins thus you get an unanimous answer to who aged more or less compared to the other. Then there's no frame dependence nor is there any paradox to be thought about left!

 

[morrobay]

Isn't the proper time equal to the space-time interval in this situation? So obviously less proper time for traveler.

 

[PAllen]

I think one saves oneself a lot of trouble, if one simply uses covariant quantities. If this is not possible, something usually is at least problematic, if not simply ill-defined or wrong, with the way a problem is stated or thought about.

In the case of the "one-way twin paradox" one can simply refer to the proper time of each twin. That's a local concept too, because it just is the time each twin reads off from his or her wristwatch. Using these proper times for the "aging" of the twins thus you get an unanimous answer to who aged more or less compared to the other. Then there's no frame dependence nor is there any paradox to be thought about left!

But for the one way example here, there is no way to avoid a synchronization assumption, because that is the sole determinant of what the start event is for the Mars clock. There is only one incident of colocation. The interval beginnings are determined solely by a synchronization decision, which can be a physical procedure, thus invariant, but it is still a choice, and effectively defines a frame.

 

[PeterDonis]

"The essence of Einstein's first insight into General Relativity was this: (a) you can derive time dilation for uniform pseudo-gravitational fields, and (b) the Principle of Equivalence then implies time dilation for gravitational fields. A stunning achievement, but irrelevant to the twin paradox. "

The use of the EP to predict time dilation in real gravitational fields is irrelevant to the twin paradox, yes. But the interpretation that a pseudo-gravitational field (which has time dilation) is present while Stella fires her thrusters, and that this pseudo-gravitational field explains how Terence ages much faster than Stella during the turnaround, is certainly not irrelevant.

 

[PeroK]

The use of the EP to predict time dilation in real gravitational fields is irrelevant to the twin paradox, yes. But the interpretation that a pseudo-gravitational field (which has time dilation) is present while Stella fires her thrusters, and that this pseudo-gravitational field explains how Terence ages much faster than Stella during the turnaround, is certainly not irrelevant.

I was just thinking though ... what happens if Stella executes the turnaround more than once? Terence must age faster every time. Which means that he must get physically younger during the first phase of the second turnaround. Which seems very unphysical.

To go through the steps:

Stella is heading away, Terence has age $T$ years in Stella's frame.
Stella executes a turnaround of a few hours, say. Terence has age $T + 1$ years, say, in Stella's frame.
Stella brakes and heads away again. Terence is back to age $T$ years (approx) in Stella's frame.
Stella executes the turnaround again and Terence is back to $T + 1$ years.

Or, alternatively, if Stella orbits a distant star at relativistic speeds, then Terence's age is going backwards and forwards during each orbit.

 

[PeterDonis]

he must get physically younger during the first phase of the second turnaround

No, he just ages slower, because he's deeper in the gravity well when Stella is accelerating away from him.

Stella brakes and heads away again. Terence is back to age $T$ years (approx) in Stella's frame.

I don't see why. Terence would simply be at age $T + 1$ years plus much, much less than a few hours (the time Stella takes to brake and head away again by her own clock).

Stella executes the turnaround again and Terence is back to $T + 1$ years.

No, he is at $T + 2$ years (plus the small increment of time he aged during the brake and head away again phase).

 

[PeterDonis]

if Stella orbits a distant star at relativistic speeds, then Terence's age is going backwards and forwards during each orbit

Stella has zero proper acceleration in this case (she's in a free fall orbit), so there is no pseudo-gravitational field, so the EP analysis does not apply.

If Terence is indeed far enough away for his "age" to be fluctuating this way if Stella adopts a "naive" extrapolation of her local comoving inertial frame, that just means that this method of extrapolating her local comoving inertial frame does not produce a valid coordinate chart that far away, because the mapping of coordinate time to proper time along Terence's timelike worldline is not one-to-one.

 

[PeroK]

No, he is at $T + 2$ years (plus the small increment of time he aged during the brake and head away again phase).

That can't be right. The differential ageing relative to Terence can't depend on the number of changes of direction.

 

[PeroK]

Stella has zero proper acceleration in this case (she's in a free fall orbit), so there is no pseudo-gravitational field, so the EP analysis does not apply.

Yes, forget the star orbit. I meant a powered orbit in any case: fly round, take a look at the solar system and head home!

 

[DaveC426913]

Or, alternatively, if Stella orbits a distant star at relativistic speeds, then Terence's age is going backwards and forwards during each orbit.

No. Terence is always getting older. The rate will alternate between aging slowly and aging rapidly, but it will always be positive.

That can't be right. The differential ageing relative to Terence can't depend on the number of changes of direction.

When Stella is closing the distance he will age rapidly, when Stella is receding he will age slowly.

 

[PeroK]

No. Terence is always getting older. The rate will alternate between aging slowly and aging rapidly, but it will always be positive.When Stella is closing the distance he will age rapidly, when Stella is receding he will age slowly.

The problem is that ageing rapidly and ageing slowly don't cancel out. Let's assume Stella's turnaround is a day, say. Half a day to slow down and half a day to speed back up (or a day to go in a loop).

If Terence ages a year in half a day (Stella's time) and next to nothing in the next half day, then that's still a year older, give or take.

If, therefore, Stella keeps repeating the turnaround, then the years pile up for poor Terence! Unless, of course, there is a part of the repeated turnaround in which Terence actually gets younger.

Fundamentally, the age differential when Stella returns is determined by Stella's velocity profile over the duration. The turnaround, physically, does not do anything special to the age difference (in a short period of Stella's time). It only changes the simultaneity convention.

This is the fundamental problem with ascribing the age differential to something physical during the acceleration phase.

 

[DaveC426913]

The problem is that ageing rapidly and ageing slowly don't cancel out.

Agree. Though not a problem, as you put it..

If, therefore, Stella keeps repeating the turnaround, then the years pile up for poor Terence!

They sure do. If Stella spends a great deal of time moving at relativistic velocities - in any direction - she's going to come back to a very old Terence.

Unless, of course, there is a part of the repeated turnaround in which Terence actually gets younger.

Nope.

Fundamentally, the age differential when Stella returns is determined by Stella's velocity profile over the duration. The turnaround, physically, does not do anything special to the age difference (in a short period of Stella's time). It only changes the simultaneity convention.

Agree! Post 79:

Yes. Velocity change (specifically, sign from + to -). Not acceleration change.

Although, note that Terence's very slow aging immediately begins to speed up (to normal) as soon as Stella begins her deceleration (negative acceleration) on approach to the turn around, while she still has positive velocity - and not when her velocity actually reverses.

Though only when her velocity actually flips from positive to negative does Terence start to age faster than her.

So positive acceleration decreases Terence's rate of aging,
while negative acceleration increases his rate of aging
(independent of what that rate was - slower or faster than Stella's - at the time).

 

[jbriggs444]

No. Terence is always getting older. The rate will alternate between aging slowly and aging rapidly, but it will always be positive.

It will alternate between forward and backward. It will not always be positive. This is a feature of the naive construction of an accelerated frame. For any given [powered] orbital acceleration, the hyperplane of simultaneity in the tangent inertial frame of the traveler will swing back and forth on the worldline of a sufficiently distant stay at home twin.

Analogously, draw a straight line and a curved line side by side vertically up a piece of paper. Draw lines of "simultaneity" at regular intervals perpendicular to the curved line. As one moves up the curved line, the "simultaneous" point on the vertical line will move both up and down.

 

[PeterDonis]

The differential ageing relative to Terence can't depend on the number of changes of direction.

Why not? You're changing Stella's path through spacetime. That changes the differential aging.

 

[DaveC426913]

It will alternate between forward and backward. It will not always be positive

We may be talking past each other here.
Terence will always be getting older (his aging will always be positive) according to Stella.
His rate of aging will decrease and increase - it may be negative or positive compared to Stella's - but it will always be > 0.

 

[PeterDonis]

It will alternate between forward and backward. It will not always be positive. This is a feature of the naive construction of an accelerated frame.

If you have an invalid frame, any statements about relative aging made using that frame are also invalid. So I wouldn't say "it will alternate between forward and backward", since that would be treating statements made using an invalid frame as valid.

 

[PeterDonis]

Terence will always be getting older (his aging will always be positive).

This will be true if Stella is using a valid frame (more precisely, coordinate chart) that covers both her and Terence during the entire trip.

However, the "frame" that is used to make claims about Terence's aging going backwards is not valid.

 

[Herman Trivilino]

During inertial flight, the clocks and people in other, relatively moving IRFs always appear to have slow clocks and be aging slower.

Two additional clocks and a synchronization convention are required to reach this conclusion.

That is, each twin needs two clocks, separated along the line of motion, and synchronized.

 

[DaveC426913]

This will be true if Stella is using a valid frame (more precisely, coordinate chart) that covers both her and Terence during the entire trip.

?
Are the real, physical scenarios where Stella could ever conclude that Terence is not aging positively?

 

[PeterDonis]

Are the real, physical scenarios where Stella could ever conclude that Terence is not aging positively?

As I said, if Stella is using a valid coordinate chart that covers both her and Terence during the entire trip, she will conclude that Terence ages positively during the entire trip.

If Stella is not using such a chart, she can't make any valid statement whatever about how Terence ages during her trip.

And since there are an infinite number of possible valid coordinate charts she could use, which will give different answers for how Terence ages during her trip, no single such statement can claim to be "the" correct statement of how Terence ages during her trip. As far as physical invariants are concerned, there is simply no fact of the matter about how someone spatially separated from you ages while you are apart. The only invariant is the comparison of your elapsed times when you meet again.

 

[jbriggs444]

?
Are the real, physical scenarios where Stella could ever conclude that Terence is not aging positively?

The Andromeda paradox is a classic one.

Walking forward or backward on the street can affect one's assessment of "time now" in Andromeda. However, as @PeterDonis may well point out, this amounts to using an invalid accelerated frame of reference to make statements of no physical relevance.

 

[DaveC426913]

As I said, if Stella is using a valid coordinate chart that covers both her and Terence during the entire trip, she will conclude that Terence ages positively during the entire trip.

If Stella is not using such a chart, she can't make any valid statement whatever about how Terence ages during her trip.

And since there are an infinite number of possible valid coordinate charts she could use, which will give different answers for how Terence ages during her trip, no single such statement can claim to be "the" correct statement of how Terence ages during her trip. As far as physical invariants are concerned, there is simply no fact of the matter about how someone spatially separated from you ages while you are apart. The only invariant is the comparison of your elapsed times when you meet again.

You are suggesting that in an Einsteinian universe, I cannot assume that someone is aging positively, even if I make no judgement about any other aspect of their aging?

I am pretty sure I'm safe to say their aging will always be positive, since for it to be negative would literally require going back in time.

But that being, said, we have been taking about Stella being able to observe Terence.

 

[jbriggs444]

I am pretty sure I'm safe to say their aging will always be positive, since for it to be negative would literally require going back in time.

I believe that the reasoning is a bit different than that. If one requires valid coordinate charts to be suitably continuous, one to one and onto over a particular connected region then it'll turn out that any transformations will be monotone in t.

This admits the possibility of coordinate charts that have t and t' advancing in opposite directions, but not the possibility that the direction will switch in mid world-line.

 

[PeterDonis]

as @PeterDonis may well point out, this amounts to using an invalid accelerated frame of reference to make statements of no physical relevance.

Indeed.

 

[PAllen]

In general, if a world line B has multiple turnarounds (towards and away from some distant inertial world line A), there will exist no single valid Fermi-Normal chart based B that covers all of A. Fermi-Normal charts are what you get if glue MCIFs together. They are also the type of chart where pesudogravity takes its simplest form. The charts you build for different parts of B will overlap in their coverage of A.

However, there is a standard GR technique to deal with this. Instead of trying to pretend you have one invalid chart, you simply recognize you are describing spacetime with a set of overlapping charts , based on different sections of B. As A traverses an overlap between charts, you can integrate along this section of A using either chart, but you better not use count it twice. With this in mind, you can say that while B accelerates towards A, A’s clock is running fast by a precise computable amount based on distance per this chart, and acceleration. While B accelerates away from A, the potential gradient is opposite, and A runs slower. Charts corresponding to when B is not accelerating are obvious (Fermi-Normal = Minkowski in this case). HOWEVER, when you add up such independent computations, you must account for overlaps, making sure you use only one result for the portion of A that traverses an overlap. It does not matter what choices you make, as long as you count any part of A only once. Done consistently you will always get the same result. These overlaps have nothing to do with time reversal - they just reflect multiple description of some part of spacetime by inclusion in more than one chart.

With this major complication, you can claim to have a pseudogravity model of any possible trajectory. Note, it is not unique, in that there are multiple choices for each overlap. In fact, infinite consistent choices, because there is no need to make your changes in use of one chart to another coincide with overlap bounaries. You can make the switch anywhere in the overlap.

 

[PeterDonis]

You are suggesting that in an Einsteinian universe, I cannot assume that someone is aging positively, even if I make no judgement about any other aspect of their aging?

You can always assume that someone is aging positively along their own worldline, since that's how proper time is defined.

But that is not the same as saying they are aging positively "in your reference frame". The latter statement requires you to be using a frame (more precisely, coordinate chart) that validly covers both you and the other person. And as this discussion has illustrated, finding such a chart can be a non-trivial exercise and the "obvious" ways of doing it don't work.

 

[DaveC426913]

But that is not the same as saying they are aging positively "in your reference frame". The latter statement requires you to be using a frame (more precisely, coordinate chart) that validly covers both you and the other person.

But that's tantamount to allowing the possibility that someone I can't observe might be going backward in time - or might be traveling > c when I'm not looking.

I don't need to look for a valid reference frame that covers us both to know that that can't happen - any more than I need to know that a spaceship heading away from me, both of us doing .99c cannot be exceeding c. I can conclude these things from first principles, without needing to see them.

*barring Cosmological Expansion of course

 

[PeterDonis]

that's tantamount to allowing the possibility that someone I can't observe might be going backward in time

No, it's just saying that "nobody goes backward in time" is not a statement about their time "relative to you" if you are spatially separated from them. It's only a statement about how their time progresses along their own worldline.

 

[DaveC426913]

No, it's just saying that "nobody goes backward in time" is not a statement about their time "relative to you" if you are spatially separated from them. It's only a statement about how their time progresses along their own worldline.

I'm not sure that cancels what I'm saying: that I can be confident someone's aging is always positive (whether relative to me or not), regardless of a common FoR. It is an inevitable consequence of an Einsteinian universe, since it does not allow time to go backward - even at any distance (again barring CosEx).

 

[PeterDonis]

It is an inevitable consequence of an Einsteinian universe, since it does not allow time to go backward - even at any distance

The only physical invariant involved with "aging" or "time flow" is proper time along timelike worldlines. In a time orientable spacetime all such aging will be in the same direction--more precisely, if we pick any single timelike worldline, and put an arrow on it in the "future" direction, which distinguishes the halves of the light cones all along that worldline, and then follow the continuous progression of the light cones everywhere else in spacetime, we will find that the arrows all point into the same halves of the light cones everywhere; they will never "flip". If that's a valid translation of what you're saying here, then OK.

 

[DaveC426913]

we will find that the arrows all point into the same halves of the light cones everywhere; they will never "flip". If that's a valid translation of what you're saying here, then OK.

Yes. Which is why we don't need the specification of a valid reference frame that includes them both - it seems to me.

In any scenario that doesn't violate the laws of physics, Terence's aging will always be positive (right?) Oh, and non-zero to-boot.

*again, CosEx aside

 

[PeterDonis]

we don't need the specification of a valid reference frame that includes them both

Not for what I described, no. You don't need a reference frame at all; worldlines and light cones are invariants.

In any scenario that doesn't violate the laws of physics, Terence's aging will always be positive (right?)

In the sense I described, yes, since Terence's worldline will be timelike.

 

[PeroK]

I'm not sure that cancels what I'm saying: that I can be confident someone's aging is always positive (whether relative to me or not), regardless of a common FoR. It is an inevitable consequence of an Einsteinian universe, since it does not allow time to go backward - even at any distance (again barring CosEx).

What I was attempting to analyse was the "acceleration causes ageing" interpretation of the twin paradox. I was trying to highlight an issue with this interpretation. Consider the following:

First, in the Earth's approximately inertial rest frame:

A stays on Earth and B makes a journey of 4 light years at relativistic velocity of $0.8c$ with a gamma factor of $5/3$. As previously discussed, we can neglect the initial and final acceleration phases that occur close to Earth (or, in fact, remove from the experiment altogether). The critical acceleration phase is the turnaround. We assume this phase is short - let's assume a day.

First, we analyse this in the Earth frame. B makes a journey of 10 years (Earth frame) with only 6 years proper time. A is 4 years older than B upon B's return. Give or take the extra day for the turnaround.

Second, we have the "acceleration causes ageing" analysis from B's perspective. We have two inertial phases of 3 years, where A "ages" less than B. In fact, A "ages" only a total of 3.6 years during the inertial phases of the journey. The conclusion is that A must age by 6.4 years during the turnaround.

In this analysis ,therefore, A ages 1.8 years on B's outward journey, 6.4 years during the turnaround and 1.8 years during the return journey.

The problem with this analysis that I have been trying to highlight is what happens if, instead of a simple turnaround, B makes a full orbit and a half at the turnaround? I'm assuming this turnaround takes 3 days, where B changes direction three times.

It seems logical that if the first turnaround caused A to age by 6.4 years, then so must the third change of direction. This would lead to A ageing by 12.8 years during the orbit and a half and being 10.4 years older than B upon B's return.

But, in this scenario, A should still be only 4 years older than B (upon B's return), give or take a day or two for the extra orbit.

The only logical explanation, therefore, is that A must get younger during the second turnaround (the one where B turns back away from Earth again). And, of course, A must get younger by 6.4 years during this middle turnaround.

This is what, in my view, makes the "acceleration causes ageing" an unphysical explanation. The above rapid ageing and getting younger phenomena are artefacts of a simultaneity convention; and not direct physical effects.

Another example along these lines is to look at the distance back to Earth during a (powered) orbit of a distant star. When the ship is moving in the direction to or from Earth we have length contraction and the distance, in the above example, is only 2.4 light years. But, when the ship is moving perpendicular to this direction, the Earth is the full 4 light years away.

This again seems unphysical to me. This alternating distance is not something of any physical relevance to the distant spaceship. And nor is the measurement that the Earth is getting 6.4. years older and 6.4 years younger during every orbit of the distant star.

 

[A.T.]

instead of a simple turnaround, B makes a full orbit and a half at the turnaround?

Sorry, I'm not sure if I understand. Is this a 2D problem now?

This is what, in my view, makes the "acceleration causes ageing" an unphysical explanation.

It's about answering the question: "How does the whole process look like in the rest frame of the traveling twin?", and that analysis must take the acceleration of that frame into account.

 

[PeroK]

Sorry, I'm not sure if I understand. Is this a 2D problem now?

It should make no difference whether B makes a turning circle or a linear deceleration/acceleration. Three linear changes of direction might be simpler, but an orbit and a half seemed more natural visually.

It's about answering the question: "How does the whole process look like in the rest frame of the traveling twin?", and that analysis must take the acceleration of that frame into account.

How a process "looks" can be ambiguous in an accelerating reference frame. If you use the "light signals" analysis, there is no rapid ageing of A during the turnaround.

 

[A.T.]

It should make no difference whether B makes a turning circle or a linear deceleration/acceleration. Three linear changes of direction might be simpler, but an orbit and a half seemed more natural visually.

Sorry I still cannot picture it. Can you draw a diagram? Or can you make your point using a 1D scenario and constant acceleration that merely flips direction?

 

[PeroK]

Sorry I still cannot picture it. Can you draw a diagram? Or can you make your point using a 1D scenario and constant acceleration that merely flips direction?

Linear motion:

Assume the star is in the +ve x direction. B's outbound has $v = +0.8c$

Simple turnaround:

Accelerates from $v = +0.8c$ to $v = -0.8c$ in 1 day (Earth time) and returns home.

Extended turnaround:

1) Accelerates from $v = +0.8c$ to $v = -0.8c$ in 1 day (Earth time)
2) Accelerates from $v = -0.8c$ to $v = +0.8c$ in 1 day (Earth time) (moving away from Earth again)
3) Repeat step 1) and return home.

Physically, 1) and 3) are identical. Whatever "ageing" happens as a result of 1) must also happen as a result of 3). Therefore, the first "ageing" must be reversed by 2).

 

[A.T.]

Linear motion:

Assume the star is in the +ve x direction. B's outbound has $v = +0.8c$

Simple turnaround:

Accelerates from $v = +0.8c$ to $v = -0.8c$ in 1 day (Earth time) and returns home.

Extended turnaround:

1) Accelerates from $v = +0.8c$ to $v = -0.8c$ in 1 day (Earth time)
2) Accelerates from $v = -0.8c$ to $v = +0.8c$ in 1 day (Earth time) (moving away from Earth again)
3) Repeat step 1) and return home.

Physically, 1) and 3) are identical. Whatever "ageing" happens as a result of 1) must also happen as a result of 3). Therefore, the first "ageing" must be reversed by 2).

Thanks. Just to clarify: Are you assuming that the "Extended turnaround" will result in the same age difference on return as the "Simple turnaround"? Or why do you think that the effects of 1) and 2) must cancel?

 

[PeroK]

Thanks. Just to clarify: Are you assuming that the "Extended turnaround" will result in the same age difference on return as the "Simple turnaround"? Or why do you think that the effects of 1) and 2) must cancel?

From analysis in an IRF, all these scenarios depend only on B's spacetime interval - not on short-term acceleration profiles - isn't that the whole issue?

To within a day or two either way, the spacetime interval has a fixed length for all these scenarios. As long as B is traveling at 0.8c for almost all of the journey (in the Earth frame), then the differential ageing is determined (to within a day or two) solely by that.

This is the fundamental problem with the acceleration-based analysis. You can include all sorts of additional accelerations, but it doesn't make any significant difference - as long as the accelerations themselves are short lived.

 

[A.T.]

To within a day or two either way, the spacetime interval has a fixed length for all these scenarios.

You can set it up to have the same spacetime intervals for both scenarios. But then 1) and 3) of the extended turnaround are not identical to the acceleration in the the simple scenario, because they happen closer to A.

 

[jbriggs444]

You can set it up to have the same spacetime intervals for both scenarios. But then 1) and 3) of the extended turnaround are not identical to the acceleration in the the simple scenario, because they happen closer to A.

Presumably one would arrange for the all three proper accelerations at the extended turnaround to be equal in magnitude and proper duration. A might then disagree on their equality, but we are not particularly concerned with A's view of the turn-around. We are trying to wrap our heads around the difficulties with B's perspective.

Nonetheless, A should see that the velocity changes achieved by all three accelerations are equal in magnitude.

 

[A.T.]

Presumably one would arrange for the all three proper accelerations to be equal in magnitude and proper duration.

I was talking about the difference of acceleration 1 (or 3) to the single acceleration the simple scenario. Not a difference between 1 and 3.

I see no reason for the claim that the effects on differential aging of acceleration 1 and 2 have to cancel. It seems to be based on the wrong assumption that acceleration 3 is identical to the single acceleration in the simple scenario. In accelerated frames the clock rates are position dependent, so the distance plays a role.

 

[jbriggs444]

effects on differential aging of acceleration

Differential aging has no role to play in proper acceleration. It seems clear by symmetry that the same proper acceleration profile over proper time will achieve the same reversal of coordinate velocities in either direction.

Accelerations 1, 2 and 3 (and on up to 2n+1) at the extended turnaround are identical except in direction. Though they need not be for the scenario to achieve its pedagogical goal.

 

[A.T.]

Differential aging has no role to play in proper acceleration.

I'm referring to this :

Whatever "ageing" happens as a result of 1) must also happen as a result of 3). Therefore, the first "ageing" must be reversed by 2).

I don't see why 2) must "reverse the ageing" during 1).

 

[PeroK]

I was talking about the difference of acceleration 1 (or 3) to the single acceleration the simple scenario. Not a difference between 1 and 3.

I see no reason for the claim that the effects on differential aging of acceleration 1 and 2 have to cancel. It seems to be based on the wrong assumption that acceleration 3 is identical to the single acceleration in the simple scenario. In accelerated frames the clock rates are position dependent, so the distance plays a role.

The accelerations all take place at the same position in B's frame, by definition!

You can easily arrange for 1) and 3) to take place at the same position relative to A.

Fundamentally, though, look at what you are saying:

All accelerations for B must be physically unique in some way?? There's no such thing as periodic motion??

What if we have B execute SHM in A's frame? Each cycle of the motion is physically different? Or, the first is unique in some way?

 

[PeroK]

I'm referring to this :

I don't see why 2) must "reverse the ageing" during 1).

Because if it doesn't you get cumulative ageing of A (assuming multiple changes of direction) which is not supported by analysis in an IRF.

 

[A.T.]

You can easily arrange for 1) and 3) to take place at the same position relative to A.

But 1) and 3) do not take place at the same position relative to A as the acceleration in the simple case, assuming the same total space time intervals for both scenarios.

To clarify: Are you assuming that A will age the same amount in B's frame during these two accelerations?:
S1: Single acceleration in the simple scenario
E3: 3rd acceleration in the extended turnaround scenario

 

[jbriggs444]

Here is a drawing depicting extended turnaround in Euclidean geometry. The traveling twin B is on the right. The stay at home twin A on the left. Lines of (B-relative) simultaneity are drawn in the middle.

In order to shift from Minkowsky to Euclid, it was necessary to change the direction of the accelerations.

If you imagine yourself as a bug named B crawling up the right hand line, the progression of "simultaneous" positions sweeps both forward and backward on bug A's world line.

Of course, the multi-mapping of the point of intersection makes the naive B-relative coordinate chart invalid -- when extended as far as A's world line. The naive B-relative coordinate chart remains valid when applied over a sufficiently small world-tube surrounding B.