I Does the "space twin" benefit from length contraction?

[Ibix]

The crux is that the traveling twin has to experience acceleration in order to make a return. So if he makes a linear trip it is during the interval when he decelerates from +0.6c to -0.6c where all of the asymmetry in aging must take place! So while he is turning around some weird stuff happens...

Add a third party, one who travels at 0.6c in the opposite direction. So we have a stay-at-home, an outbound traveller, and an inbound traveller. The outgoing traveller never stops, simply reports their clock reading to the inbound traveller when they pass each other. When the inbound traveller passes the stay-at-home they report the outbound's outbound time plus their own time since they passed the outbound. This must give the same answer as the twin paradox, but there is no acceleration whatsoever

The acceleration is necessary for two twins to meet up again. But it can't be the cause of the age difference. (I believe PeroK has already pointed this out).

 

[hutchphd]

The acceleration is both necessary and sufficient for again meeting. It is the only such asymmetry in the problem. It therefore must be the "cause" of the age difference. QED

 

[Ibix]

The acceleration is both necessary and sufficient for again meeting. It is the only such asymmetry in the problem. It therefore must be the "cause" of the age difference. QED

I just described a variant of the experiment with no acceleration and the same result. So "necessary" is false.

It's also possible to construct experiments where twins undergo the same acceleration phases in the same order, just with different duration inertial segments. They end up having different ages. So it is very difficult to conclude that the acceleration is causing the age difference.

 

[hutchphd]

What number comment describes that variant?...I missed it...sorry

 

[Ibix]

What number comment describes that variant?...I missed it...sorry

#51. See also PeroK's #45 for a longer discussion.

 

[hutchphd]

This must give the same answer as the twin paradox

Oh I found it. In fact this is exactly my point...the asymmetry comes from the change in direction. All the time the rover twin is traveling uniformly, he will see the clocks on the Earth ticking more slowly than his (and vice versa). Imagine his surprise when his Earth twin is older! That extra time on Earth must have transpired during his deceleration! When he gets home he will be younger yet except for the noninertial parts of the trip each twin will agree (correctly!) the other's biological clocks are running slower. There are pieces of general relativity that deal with acceleration but I am not able to reproduce them out of memory and too lazy to work it out right now...

 

[Nugatory]

There are pieces of general relativity that deal with acceleration...

This is a very common misconception. Special relativity handles problems involving acceleration just fine as long as the spacetime is flat, as it is in this discussion. General relativity is needed only if the spacetime is not flat because there are gravitating bodies involved, and then it's needed even if no acceleration is involved.

For an example of special relativity properly handling acceleration, google for "Rindler coordinates".

More generally, acceleration just means that the worldline of the accelerated object is curved instead of straight, so you may have to do a line integral along the curve instead of just calculating $-\Delta{t}^2+\Delta{x}^2+\Delta{y}^2+\Delta{z}^2$ from the endpoints of a line segment. This additional mathematical complexity is why many introductory SR texts don't discuss the acceleration cases, inadvertently feeding the misconception that SR cannot handle them.

 

[Nugatory]

That extra time on Earth must have transpired during his deceleration!

If you are not already familiar with the Twin Paradox FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html, give it a try, and pay particular attention to the part about the "Time gap objection".

 

[Ibix]

In fact this is exactly my point...the asymmetry comes from the change in direction

You said it came from acceleration in #50 & 52, which is a different point.

All the time the rover twin is traveling uniformly, he will see the clocks on the Earth ticking more slowly than his (and vice versa).

This is not true if you mean "see" literally. Due to the Doppler effect, the traveling twin will see the stay-at-home's clock ticking slowly on the outbound leg and quickly on the inbound leg. The stay-at-home will see the traveller's clock ticking slowly for the 80% of the trip time the traveller can be seen to be heading outwards, and quickly for the 20% of the time the traveller is seen to be inbound. (Edit: the 80/20 figure assumes a speed of 0.6c and that my mental arithmetic is reliable.)

It's only once the twins subtract out the effects of the lightspeed delay that they calculate that the other's clock ticks slowly. Time dilation is not directly observable.

Imagine his surprise when his Earth twin is older!

He won't be surprised at all if he's actually been watching the other's clock.

That extra time on Earth must have transpired during his deceleration!

This is true in a limited sense. As I noted above, the slow running of the other's clock is not directly observable. It's something that each twin calculates based on their direct observations plus some assumptions about how one should synchronise separated clocks. If the traveller does that process naively before and after the turn around then they find that what they were calling "now, on Earth" just before the turn around is not the same as what they now call "now, on Earth" just after the turn around. The difference in "now" will account for the extra time.

But this is not an effect of acceleration. It's not even an effect of changing direction. It's an effect of changing which clocks the traveling twin chose to regard as synchronised with theirs. In other words, it boils down to using one definition of "during" for the outbound trip, a different definition for the inbound trip, and ascribing the resulting inconsistency to something that happened "during" the turnaround.

For one possible non-naive resolution to this, see https://arxiv.org/abs/gr-qc/0104077

There are pieces of general relativity that deal with acceleration

As Nugatory noted, general relativity is not needed to analyse this. You can do an analysis (for the finite acceleration case, not the instantaneous turn around) using gravitational time dilation, but that has always seemed over-complicated to me.

 

[PeroK]

I used to teach relativity to freshman engineering students for fun, so here is just one more thought to slightly mess with your head. The traveling twin could start his "life" already at speed 0.6c passing Earth (hence no acceleration) and the final comparison upon his return could by made by taking a snapshot as he zoomed past Earth on the return (again no acceleration). The crux is that the traveling twin has to experience acceleration in order to make a return. So if he makes a linear trip it is during the interval when he decelerates from +0.6c to -0.6c where all of the asymmetry in aging must take place! So while he is turning around some weird stuff happens...

The fundamental flaw in your logic is this.

Imagine a) the scenario where the traveller has only one acceleration phase, at the turn around.

b) The scenario where the traveller has two acceleration phases: one at the start and one at turn around.

c) The scenario where the traveller has a third acceleration phase at the end.

All of these scenarios result in approx the same differential ageing.

If "weird stuff" happens when you accelerate that caused differential ageing, then more "weird stuff" would happen in case c) than in b) etc.

Moreover, if the acceleration caused the differential ageing, then it wouldn't matter how far the traveller cruised though space. The ageing would depend only on the acceleration profile, not on the overall journey.

In fact, the opposite is the case: the differential ageing depends simply on the velocity profile, but not on the acceleration profile.

 

[Herman Trivilino]

All the time the rover twin is traveling uniformly, he will see the clocks on the Earth ticking more slowly than his (and vice versa).

There is no such notion as "all the time" as you are using it here. If one twin says that during the time I was getting married you were reading page 12, the other twin will say that during the time you were getting married I was reading a different page.

That extra time on Earth must have transpired during his deceleration! When he gets home he will be younger yet except for the noninertial parts of the trip each twin will agree (correctly!) the other's biological clocks are running slower.

To your use of "when" the same comment as above applies. The notion of "when" upon which the twins will agree occurs at the departure and reunion. Therefore it's possible to say that they agree when they share the same location.

 

[hutchphd]

Just looked at Parok's post...sorry but I don' t understand what he is doing. The description from every reference frame will be consistent within. the special theory.

This is not true if you mean "see" literally. Due to the Doppler effect, the traveling twin will see the stay-at-home's clock ticking slowly on the outbound leg and quickly on the inbound leg. The stay-at-home will see the traveller's clock ticking slowly for the 80% of the trip time the traveller can be seen to be heading outwards, and quickly for the 20% of the time the traveller is seen to be inbound. (Edit: the 80/20 figure assumes a speed of 0.6c and that my mental arithmetic is reliable.)

I don't understand the argument. If we use this definition of "see" then each twin will agree instead that an equal time has passed for the other (is that what you are saying?) and yet Rover is still younger than Homer when they meet ! Do they agree that they were not actually born twins? What are you saying here?

 

[PeroK]

Just looked at Parok's post...sorry but I don' t understand what he is doing. The description from every reference frame will be consistent within. the special theory.

I don't understand the argument. If we use this definition of "see" then each twin will agree instead that an equal time has passed for the other (is that what you are saying?) and yet Rover is still younger than Homer when they meet ! Do they agree that they were not actually born twins? What are you saying here?

There is the "Doppler" explanation of the paradox, which analyzes what each twin observes directly. Rather than what they measure in their respective reference frames. The Doppler explanation, in particular, does not need all the "weirdness" to happen during one or more acceleration phases.

 

[hutchphd]

I thought you were giving the Doppler explanation. Where is the asymmetry in this explanation?t

 

[PeroK]

I thought you were giving the Doppler explanation. Where is the asymmetry in this explanation?t

It's the change of direction.

 

[hutchphd]

So what does Rover observe about Homer's clock during the "change in direction" ?

 

[PeroK]

So what does Rover observe about Homer's clock during the "change in direction" ?

Nothing. The change of direction can be instantaneous - or near instantaneous.

More precisely it's the timing of the change of direction. For example, if the relative velocity is $\frac35 c$ and the distance is $2.4$ light years (Earth frame), then at the turn around point, Rover's clock reads $4$ years and the light from Homer's second birthday has just reached him.

Whereas at the turn around point Homer's clock reads $5$ years etc.

 

[PeroK]

In fact, to put it more simply. In Homer's frame, Rover travels for 5 years and a greater distance before he turns round.

Whereas, in Rover's frame Homer will travel for only 4 years and a shorter distance before turning round.

So, the asymmetry is inherent in the problem from the outset - owing to length contraction, at least.

 

[hutchphd]

For clarity let us assume that each of the Travelers is emitting a 1 Hz signal and counting this signal is how they characterize each others time (correctly or not...that is how they do it) . The only time where they will disagree about the rate of tick tock of each others clock is during the deceleration where they must measure very different things. What does each measure from that interval ??.

 

[PeroK]

For clarity let us assume that each of the Travelers is emitting a 1 Hz signal and counting this signal is how they characterize each others time (correctly or not...that is how they do it) . The only time where they will disagree about the rate of tick tock of each others clock is during the deceleration where they must measure very different things. What does each measure from that interval ??.

There is no deceleration. We are hypothesising an instantaneous change of direction.

 

[hutchphd]

In fact, to put it more simply. In Homer's frame, Rover travels for 5 years and a greater distance before he turns round.

Whereas, in Rover's frame Homer will travel for only 4 years and a shorter distance before turning round.

So, the asymmetry is inherent in the problem from the outset - owing to length contraction, at least.

Does this not imply that Rover will expect Homer to be younger?

 

[PeroK]

Does this not imply that Rover will expect Homer to be younger?

No.

 

[hutchphd]

There is no deceleration. We are hypothesising an instantaneous change of direction.

You mean you are hypothesizing an infinite deceleration...not a zero. Not a good idea to introduce infinities. I would prefer to call it large but finite and re-ask the same question.

 

[jbriggs444]

[

For clarity let us assume that each of the Travelers is emitting a 1 Hz signal and counting this signal is how they characterize each others time (correctly or not...that is how they do it) . The only time where they will disagree about the rate of tick tock of each others clock is during the deceleration where they must measure very different things. What does each measure from that interval ??.

The word "during" for two separated observers in relative motion is a non-starter.

Edit: further, there is no time at which either observer will either "see" or "observe" the other's tick rate to be 1 Hz -- with the possible exception of an instant at turnaround where the traveller might be momentarily at rest in the stay-at-home twin's frame and "see" 1 Hz. Or a pair of instants at turnaround where the traveler "observes" a tick rate momentarily at 1 Hz. The stay at home twin would "see" a 1 Hz tick rate momentarily significantly after the halfway time and would "observe" a 1 Hz tick rate momentarily exactly at the halfway time.

 

[PeroK]

You mean you are hypothesizing an infinite deceleration...not a zero. Not a good idea to introduce infinities. I would prefer to call it large but finite and re-ask the same question.

Nothing special happens to what they see at the turn around. See the link in post #46 for a Doppler analysis.

In any case, the underlying implication of your questions that only acceleration can cause the differential ageing is flawed.

Many analyses can happily assume an instantaneous ( or nearly if you prefer) turn around.

 

[hutchphd]

OK I will rephrase. Each twin carries a machine that plots the received signal on a graph. When they get back together they can look at these graphs and the number of ticks each sees for the other must correspond to their relative ages (events do not disappear). We have already agreed about symmetry during the inertial parts of the journey. So there will be a section of the graphs that do not agree. What does that look like??

 

[jbriggs444]

OK I will rephrase. Each twin carries a machine that plots the received signal on a graph. When they get back together they can look at these graphs and the number of ticks each sees for the other must correspond to their relative ages (events do not disappear). We have already agreed about symmetry during the inertial parts of the journey. So there will be a section of the graphs that do not agree. What does that look like??

Let's get more specific. This machine has a pen that moves up and down to reflect the signal state and a strip of paper moving right to left under the pen. I.e. a chart recorder.

We agree that the two strips of chart paper will have different numbers of peaks. But until we specify how the paper moves under the two pens, we cannot know where the two strips will differ. [Yes, there is a natural assumption that might be made -- but it's still an assumption]

 

[PeroK]

We have already agreed about symmetry during the inertial parts of the journey.

No, we haven't. See post #68, for example. The experiment is fundamentally asymmetric.

Rover, as we have already explained, will record a shorter outbound journey than Homer.

 

[hutchphd]

It is a normal strip recorder bought by for each twin at birth from acme strip recorder company and each works perfectly.

 

[Nugatory]

OK I will rephrase. Each twin carries a machine that plots the received signal on a graph. When they get back together they can look at these graphs and the number of ticks each sees for the other must correspond to their relative ages (events do not disappear). We have already agreed about symmetry during the inertial parts of the journey. So there will be a section of the graphs that do not agree. What does that look like??

This is covered in some detail in the "Doppler Shift Analysis" section of the FAQ that I linked to above.

The earthbound twin receives ticks from the traveller at a rate less than his own tick rate for more than half the time between separation and reunion. Shortly before the reunion, the rate at which the earthbound twin receives ticks increases to be greater than his own. This transition from receives-slow to receives-fast does not happen at the same time as the deceleration for turn around, it happens much later.

The space twin receives ticks at the slow rate until the turnaround, and after the turnaround receives ticks at the faster rate. So the two graphs disagree only in where the transition happens: after the midpoint for earthbound, and at the midpoint for the traveller.

 

[hutchphd]

Got it. That's a very satisfying explanation. I greatly appreciate your patience and hope it has been edifying for others as well!

 

[Herman Trivilino]

Got it. That's a very satisfying explanation. I greatly appreciate your patience and hope it has been edifying for others as well!

If you haven't done so yet, you might consider going to YouTube and doing a search for "Hewitt Twin Trip". He does the analysis in a concrete way that makes it clear that no matter which twin is the receiver of signals sent from the other twin, they will both agree based on their own counts how much more proper time passed for the staying twin.

By the way, the relative speed in that video is $\frac{3}{5} c$ making $\gamma$ equal to $\frac{5}{4}$.

Edit: Oh, and that makes the Doppler shift factor $\sqrt{\frac{1+\beta}{1-\beta}}$ equal to $2$.

 

[hutchphd]

Ah yes the old 3,4,5 triangle. I fear I am about to be addicted to this site. Great stuff thanks

 

[hutchphd]

While the "who is older" analysis does not strictly require it, I am still interested in what Romer will experience during a real life (finite non instantaneous) turn around. Let me imagine his space filled with a field of clocks which he synchronizes at birth (while sitting next to Homer). On his travels the clocks will accompany him.
If I understand the Rindler coordinate transform, as soon as Romer feels any acceleration his field of clocks will:
1) no longer be synchronized...they will run at different rates during the acceleration depending upon their position along the acceleration axis.
2) In fact I guess each clock will experience a slightly different acceleration?
v3) When returned to Homers inertial frame the deviations from acceleration will sum to zero and the Romer clocks will again be synchronized (but still disagree with Homer's clocks!)
Again I appreciate your indulgence...Is this correct?

 

[PeroK]

While the "who is older" analysis does not strictly require it, I am still interested in what Romer will experience during a real life (finite non instantaneous) turn around. Let me imagine his space filled with a field of clocks which he synchronizes at birth (while sitting next to Homer). On his travels the clocks will accompany him.
If I understand the Rindler coordinate transform, as soon as Romer feels any acceleration his field of clocks will:
1) no longer be synchronized...they will run at different rates during the acceleration depending upon their position along the acceleration axis.
2) In fact I guess each clock will experience a slightly different acceleration?
v3) When returned to Homers inertial frame the deviations from acceleration will sum to zero and the Romer clocks will again be synchronized (but still disagree with Homer's clocks!)
Again I appreciate your indulgence...Is this correct?

If you consider an acceleration phase at the beginning, then a clock at the front of Rover's ship will run faster than one at the rear. During the journey, therefore, the ship clocks may be out of sync. Then, if there is a deceleration phase at the turnaround, the clocks will end up back in sync again.

The clocks don't experience a different acceleration, but their different positions in the accelerating frame leads to time dilation between them. There is a heuristic argument for gravitational time dilation based on this and the equivalence principle.

 

[hutchphd]

The clocks don't experience a different acceleration, but their different positions in the accelerating frame leads to time dilation between them. There is a heuristic argument for gravitational time dilation based on this and the equivalence principle.

If I am riding with Romer some distance down the axis and feel (measure) the same acceleration but my clock reads a different time than must I not be traveling at different speed ??

 

[PeroK]

If I am riding with Romer some distance down the axis and feel (measure) the same acceleration but my clock reads a different time than must I not be traveling at different speed ??

Not necessarily. You get internal time dilation in an accelerating reference frame.

The factor is $1 + \frac{ah}{c^2}$ where $h$ is the distance in the direction of acceleration $a$.

The clock to the rear runs slow by this amount, in the accelerating frame.

Note that if the ship is accelerating homogeneously in the initial IRF. I.e the front and rear always have the same velocity in that frame, then the clocks will stay synchronised in that frame. And, if at the end of the acceleration phase, they are in sync in the initial IRF, then they must be out of sync - by $hv/c^2$ in the ship frame.

This ties in with the "leading clocks lag" rule, if you think about it.

 

[Nugatory]

If I am riding with Romer some distance down the axis and feel (measure) the same acceleration but my clock reads a different time than must I not be traveling at different speed ??

It takes some care to properly describe an extended object (such as your collection of clocks moving along with Romer) undergoing acceleration. The basic problem is that when we say that all parts of the object are experiencing the same acceleration at the same time... we have to assign some meaning to "at the same time" and "the distance between the clocks at any given moment", and of course both of these notions are frame-dependent.

This has become a good time to mention two other google searches: "Born rigid motion" and "Bell spaceship paradox".

 

[PeroK]

It takes some care to properly describe an extended object (such as your collection of clocks moving along with Romer) undergoing acceleration. The basic problem is that when we say that all parts of the object are experiencing the same acceleration at the same time... we have to assign some meaning to "at the same time" and "the distance between the clocks at any given moment", and of course both of these notions are frame-dependent.

This has become a good time to mention two other google searches: "Born rigid motion" and "Bell spaceship paradox".

Yes. Note that the heuristic argument in post #87 ignores the length contraction of the ship. But, if acceleration is modest, then the effect of a few metres contraction over, say, one light year of travel results in a tiny difference in velocity across the ship in the IRF.

 

[hutchphd]

I don't disagree with any of these details, but my simple question should have a simple answer I believe. Me and Romer synchronize our clocks to zero before the acceleration and agree after the acceleration that we are still attached to the same spacecraft but disagree as to the time for acceleration. Our delta V is manisfestly the same yet our times for it differ...does that not require the acceleration we measure (eg with our buttocks) to be different?

 

[PeroK]

I don't disagree with any of these details, but my simple question should have a simple answer I believe. Me and Romer synchronize our clocks to zero before the acceleration and agree after the acceleration that we are still attached to the same spacecraft but disagree as to the time for acceleration. Our delta V is manisfestly the same yet our times for it differ...does that not require the acceleration we measure (eg with our buttocks) to be different?

First, for an acceleration phase of months, say, the difference in time is only $h/c$ even for a final velocity close to $c$.

This is the same margin for error as ignoring length contraction in the IRF.

To resolve this, you would need to analyse the problem exactly. For example, in the IRF view, the ship could stay the same length - i.e. its proper length would increase and then you wouldn't have exactly a rigidly accelerating frame.

 

[PeroK]

PS it would be important to analyse coming out of the acceleration phase. If they both stop accelerating when their respective clocks reach a certain time, then what does that look like in the IRF?

 

[PAllen]

I don't disagree with any of these details, but my simple question should have a simple answer I believe. Me and Romer synchronize our clocks to zero before the acceleration and agree after the acceleration that we are still attached to the same spacecraft but disagree as to the time for acceleration. Our delta V is manisfestly the same yet our times for it differ...does that not require the acceleration we measure (eg with our buttocks) to be different?

That’s where the notion of rigidity comes into play. If the ship is taken to be rigid, then an accelerometer at the front will measure lower acceleration than the rear. The amount is extremely small, a lot less than the change in acceleration from front to back if the ship were sitting on a planet with the same acceleration at the bottom.

 

[PAllen]

Here are two scenarios:

1) In an initial inertial frame, front and back begin accelerating at the same time by the same amount. Then, per the same frame, deceleration begins at the same time, by the same amount until front and back are both at rest again. These will be simultaneous for this frame. Assume acceleration for a month. Then, the back will see the front move slowly away, to many times the initial separation. Then, before the back starts decelerating, they will see the front moving towards them. When the back finally stops decelerating, the original length will be restored. Both rocket clocks will show the exact same time, much, much less than clocks in the original frame.

2) In an inertial frame, back starts accelerating, accelerates for a month per the initial frame, then decelerates to stop. The motion of the front is governed by the rule that it always looks the same distance away, per the back. To achieve this, the front accelerates and decelerates less. This is a frame independent statement. Per the initial frame, the distance between them will shrink to a tiny fraction of the initial distance, then grow back. The front clock will have accumulated more time than the back, but this difference will be very small compared to the how much either is behind the initial frame clocks.

[edit: A couple of corrections to case 2, above. During deceleration, for born rigid motion, what was the front will have decelerate faster than the back. And, without doing the math, I don’t know how the clocks will end up, except they will be extremely close, because the front will be ticking slower than back, per the back, during deceleration. Also, from the original frame, both front and back will have covered the exact same distance, but with different velocity profile, so I see no obvious argument as to exactly how they end up comparing to each other.]

 

[PeroK]

That’s where the notion of rigidity comes into play. If the ship is taken to be rigid, then an accelerometer at the front will measure lower acceleration than the rear. The amount is extremely small, a lot less than the change in acceleration from front to back if the ship were sitting on a planet with the same acceleration at the bottom.

... looking at my notes, for a rigidly accelerating ship,we have a proper acceleration of:

$\frac{a}{1 + ah/c^2}$

And, we have a proper time of:

$\tau = (1 + ah/c^2)\tau_0$

Hence, as you say, the proper acceleration varies inversely with proper time along the ship.

 

[hutchphd]

Thank you all is now copacetic in my relativity brains...I appreciate your help.

 

[nomadreid]

This thread has lots of very helpful analyses, but despite having read the thread and Baez’s explanation, watched Hewitt’s videos, followed the calculations and so forth, there is one point which intrudes upon my getting a good intuition of the asymmetry. I will start with the geometric viewpoint, that between two points in spacetime there is only one straight line, so that an asymmetry is inevitable. On one side one says that acceleration is not important, on the other side one emphasizes that it is the turn-around, the switching IRF’s, that is important. But isn’t the turn-around an acceleration? So the traveler must assume that she is the one turning around, not the earth, which she can only do by noticing a force. No? So isn't the classic explanation that it is the acceleration which makes the difference still valid, even if one adds the cautionary note that one doesn't need GR, only SR, to handle the difference that the acceleration makes?

Putting the question another way: the only way that the traveler knows that she is turning around is either to experience a force or acceleration (the popular explanation that the thread quickly discards) or to assume that the homebody’s IRF is the one with the straight line: but if the Traveler does not assume this and ignores the force (or acceleration), then what is to stop her from assuming that hers is the straight line and the homebody’s the one with the crook?

To make the question in the spirit of the diagrams in Baez’s post (rotated 90 degrees,), one could compare, say, the graphs
{ Trav: y=|x|; Home: the line from y=(-5,5) to(5,5) }
and
{Trav: the line from (-5,0) to (5,0); home: y=5-|x|. }
The only way to find out (before one gets back to Earth to compare clocks) which one of the two twins in on the straight line is by noticing the acceleration, why is it said that the acceleration is not the important factor?

 

[Nugatory]

then what is to stop her from assuming that hers is the straight line and the homebody’s the one with the crook?

The presence or absence of a crook is a physical and mathematical fact about a path through spacetime.

It is true that you cannot force an object to follow a path with a crook in it without subjecting the object to acceleration, but we can measure the proper time along a path without forcing an object to follow that path (transfer the reading on the outbound clock to an inbound clock as they pass one another, similar to the way that we avoid acceleration at the beginning and end of the round trip), and these measurements will confirm the presence of the crook.

 

[Herman Trivilino]

So isn't the classic explanation that it is the acceleration which makes the difference still valid, even if one adds the cautionary note that one doesn't need GR, only SR, to handle the difference that the acceleration makes?

Perhaps parts of what you're saying are correct, but the part of it that I quoted is not correct. But the thing is, it's the straight-line paths through spacetime that are the cause of the difference in twin ages. If all the traveling twin did after departure is slow down, turn around, and return to the reunion, the difference is twin ages would be smaller than if the traveling twin moved away at a steady speed before executing the very same turn-around to make it back to the reunion.

In other words, you can have the very same amount of acceleration in each case with very different outcomes. So it cannot be that the acceleration accounts for the difference in ages.

 

[PeroK]

In other words, you can have the very same amount of acceleration in each case with very different outcomes. So it cannot be that the acceleration accounts for the difference in ages.

And, you can have very much the same differential ageing with different acceleration profiles. If we insist on acceleration, then we get the same result regardless of whether there are two, three or four acceleration phases. Which is the same result as when we simply switch reference frames by transferring clock readings.

The small differences in these cases can be explained by the different velocity profiles during the various acceleration phases.

So, it's really the word "cause" that is the issue. The acceleration cannot be seen as a direct cause of differential ageing.