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I've the impression that any attempt to get the simple statement "more pedagogical" rather raises the confusion about a pretty simple "no-brainer".
Telling a student that the thing that confuses them is a “pretty simple no brainer” may not be as helpful to the student as you think. Usually they require a teacher to address their actual misunderstanding, not insult them.

Orodruin, vanhees71, weirdoguy and 3 others
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This shows immediately that there is no paradox. I have no clue, which more many ways you have in mind to answer the problem. I've the impression that any attempt to get the simple statement "more pedagogical" rather raises the confusion about a pretty simple "no-brainer".
The paradox is that they believe there are two equally valid calculations that give conflicting answers. It is not enough to say that one calculation is a "no-brainer". It is necessary to convince them that the other calculation is invalid in its own right.

Sagittarius A-Star, vanhees71 and robphy
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"no brainer" likely refers to the person who already accepts the result,
but not so much (as others have said)
for the student who is struggling to understand why it's true
and why alternative attempts against the result aren't true or valid.

SiennaTheGr8
I think typically somebody struggling with the twin paradox might have a little background in SR from an inertial-frames perspective, and might understand something about time dilation, but presumably doesn't understand the full implications of the Lorentz transformation (read: the relativity of simultaneity) and almost certainly isn't well-acquainted with the invariant-based geometric picture.

So whatever explanation you give them, they're going to try to reconcile it with what they already know (as they should). And if their background knowledge is all about inertial frames and expressing everything in terms of Cartesian coordinates and coordinate time, then the geometric explanation—while "simplest" (and clearly best)—is probably not going to be the easiest for them to digest.

For me, the relativity of simultaneity was the key to understanding the twin paradox.

From a coordinate-based inertial-frames perspective, what distinguishes the rocket-twin's situation from the Earth-twin's? Answer: the Earth-twin remains in one inertial frame, while the rocket-twin switches frames at the turnaround event. [EDITED FOR MORE CAREFUL WORDING] Answer: the Earth-twin's inertial rest frame never changes, while the rocket-twin's does (at the turnaround event).

Why does that matter if both twins are always moving inertially? Don't they both always say that the other's wristwatch is running slow? Answer: yes, but at the turnaround event, the rocket-twin's answer to the question "what time does the Earth-twin's wristwatch display right now?" changes dramatically—the Earth "suddenly skips ahead in time" for the rocket-twin, by decades in the usual setup (though it would of course have to happen "smoothly" in real life, since acceleration is required). By contrast, the Earth-twin's answer to the question "what time does the rocket-twin's wristwatch display right now?" never changes dramatically at any point.

vanhees71
Fair enough, but at a quick glance those threads do cover the issues with simultaneity, or use the Doppler analysis (preferred by @ghwellsjr).

Sagittarius A-Star
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Telling a student that the thing that confuses them is a “pretty simple no brainer” may not be as helpful to the student as you think. Usually they require a teacher to address their actual misunderstanding, not insult them.
This was not against the students but the teachers. In my experience the explanation that the different aging is expressed in terms of scalars, i.e., the proper times of the twins, and thus independent of the coordinates used to calculate them is well undrstood.

I have no clue, which more many ways you have in mind to answer the problem.
You can explain where the naive analysis that gives the paradoxical ##T/\gamma^2## result for the stay-at-home's elapsed time goes wrong (i.e. they forgot that they changed simultaneity convention, and thus failed to account for some of the stay-at-home's worldline). I agree that there are better and easier ways of getting to the right answer, but it's also good to explain why the wrong method doesn't work.

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i.e. they forgot that they changed simultaneity convention, and thus failed to account for some of the stay-at-home's worldline
Or you could introduce radar coordinates for the non inertial twin and show that the trip is not symmetrical from his perspective. But I suspect that students that are confused with this “paradox” will probably be totally lost with that.

robphy and SiennaTheGr8
Lluis Olle
An (enough) good explanation for me is:

- The Alice (Earth) observer has a lattice of synchronized clocks that fill the whole space. Specifically there's an "Earth's clock" at the star position, that reports back its readings to Earth (at each time, what happened there).

- In Bob's (traveler) perspective, during the outbound path and just when he observes the star reach his position, his clock reads 2.0 and as he knows that clocks run slower in Earth, he predicts Earth's clock to be ##2 \cdot \frac{1}{\gamma}=1.6##.

- Leading clocks lag, which means that for two synchronized clocks separated by a distance of D in the same frame, the leading clock (respect the velocity) will lag ##\frac{D \cdot v}{c^2}## respect the rear clock, as observed by a rest frame. So the "Earth's clock" at the star position is ahead its correspondent sibling at Earth by 0.9, and it records 2.5 (1.6+0.9) at the event of the star reaching Bob at his time of 2. When the Earth's observer collect the reported data, all fits nicely.

- The velocity is then suddenly inverted and the inbound path started. As soon as this happens, and Bob is put abruptly into another frame, that "Earth's clock" at the star is now the leading clock of the play, so it lags by 0.9 respect its correspondent sibling clock at Earth. Obviously, the reading of that clock in that "after" instant is the same as "before", but now lags by 0.9 respect the correspondent sibling at Earth. So that Earth's clock must be 3.4. In no time for Alice, Earth clock has gone from 1.6 to 3.4

- Then, it takes another 1.6 Earth-years for Earth to reach Bob's position. And that accounts for the 5 years elapsed in Earth, but only 4 for the traveler twin.

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Or you could introduce radar coordinates for the non inertial twin and show that the trip is not symmetrical from his perspective. But I suspect that students that are confused with this “paradox” will probably be totally lost with that.
For me, it was radar methods that made things click. This provided a measurement procedure and operational definitions for assigning coordinates. (This was more understandable than first trying to imagine shrunken rulers and slow clocks…. That is, I use the radar methods first to then interpret all of the various traditional but likely confusing ways of thinking about things.)

vanhees71 and Dale
obtronix
I think typically somebody struggling with the twin paradox might have a little background in SR from an inertial-frames perspective, and might understand something about time dilation, but presumably doesn't understand the full implications of the Lorentz transformation (read: the relativity of simultaneity) and almost certainly isn't well-acquainted with the invariant-based geometric picture.

So whatever explanation you give them, they're going to try to reconcile it with what they already know (as they should). And if their background knowledge is all about inertial frames and expressing everything in terms of Cartesian coordinates and coordinate time, then the geometric explanation—while "simplest" (and clearly best)—is probably not going to be the easiest for them to digest.

For me, the relativity of simultaneity was the key to understanding the twin paradox.

From a coordinate-based inertial-frames perspective, what distinguishes the rocket-twin's situation from the Earth-twin's? Answer: the Earth-twin remains in one inertial frame, while the rocket-twin switches frames at the turnaround event. [EDITED FOR MORE CAREFUL WORDING] Answer: the Earth-twin's inertial rest frame never changes, while the rocket-twin's does (at the turnaround event).

Why does that matter if both twins are always moving inertially? Don't they both always say that the other's wristwatch is running slow? Answer: yes, but at the turnaround event, the rocket-twin's answer to the question "what time does the Earth-twin's wristwatch display right now?" changes dramatically—the Earth "suddenly skips ahead in time" for the rocket-twin, by decades in the usual setup (though it would of course have to happen "smoothly" in real life, since acceleration is required). By contrast, the Earth-twin's answer to the question "what time does the rocket-twin's wristwatch display right now?" never changes dramatically at any point.

is almost never accepted as a reasonable answer by most people, I suppose you need to introduce the block universe and angled slices idk

"Earth-twin's answer to the question "what time does the rocket-twin's wristwatch display right now?" never changes dramatically at any point."

Exactly, this is Brian Greene's explanation of the twin paradox the only measurements you can trust are from an observer in an inertial frame. No time skips. Which is why I drew those diagrams, from three inertial frames.

Dale
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This seems like a good time to reference the Usenet Physics FAQ article on the twin paradox:

is almost never accepted as a reasonable answer by most people
See the "time gap" page in the above article.

"Earth-twin's answer to the question "what time does the rocket-twin's wristwatch display right now?" never changes dramatically at any point."
If you use the Earth twin's inertial rest frame to define "now", then of course not. But the best answer is that "now" has no physical meaning at all in relativity.

this is Brian Greene's explanation of the twin paradox the only measurements you can trust are from an observer in an inertial frame.
And this is wrong. It is perfectly possible to construct non-inertial frames that do not have "time skips" and work just as well as inertial frames for analyzing physics. You just need to use the correct equations for the non-inertial frame.

PeroK
Exactly, this is Brian Greene's explanation of the twin paradox the only measurements you can trust are from an observer in an inertial frame. No time skips. Which is why I drew those diagrams, from three inertial frames
I'd be interested in a reference to where Greene says this. Either there's some context you're missing or Greene is saying wrong things in his popularisations (again). There are plenty of non-inertial frames one can work with where there are no time skips (and Greene knows this - the study of non-inertial frames was a step on the road to the discovery of general relativity). They are just mathematically harder to use than inertial frames.

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vanhees71
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Exactly, this is Brian Greene's explanation of the twin paradox the only measurements you can trust are from an observer in an inertial frame. No time skips. Which is why I drew those diagrams, from three inertial frames.
It seems to me you are going in the wrong direction here. In GR there are no global inertial frames, so the idea of relying on global inertial frames is no longer possible. If you now try to learn GR, then the whole ground has been pulled out from under your feet.

Also, in GR you can more easily have a version of the paradox where there is no proper acceleration. I.e. by executing half an orbit of a black hole as a turnaround.

There is also the opposite paradox of two clocks in the same circular orbit around the Earth or a star in opposite directions. There is continuous time dilation (same gravitational potential and non-zero relative velocity), yet complete symmetry. Despite the continuous mutually symmetric time dilation, the clocks show the same time when they cross paths twice per orbit.

This is where you need a broader (not narrower) understanding of these concepts. In particular, it is an understanding of spacetime geometry that allows you not only to understand the twin paradox fully, but to move on to GR.

robphy, Dale, vanhees71 and 1 other person
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In particular, it is an understanding of spacetime geometry that allows you not only to understand the twin paradox fully, but to move on to GR.

robphy and vanhees71
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Exactly, this is Brian Greene's explanation of the twin paradox the only measurements you can trust are from an observer in an inertial frame. No time skips. Which is why I drew those diagrams, from three inertial frames.
I'd be interested in a reference to where Greene says this. Either there's some context you're missing or Greene is saying wrong things in his popularisations (again). There are plenty of non-inertial frames one can work with where there are no time skips (and Greene knows this - the study of non-inertial frames was a step on the road to the discovery of general relativity). They are just mathematically harder to use than inertial frames.

WSU: Special Relativity with Brian Greene (The Twin Paradox, Explanation #1 &t=09h26m30s )

Here is how I would paraphrase his statements:
You must be an "inertial observer for the whole trip"
in order to claim that you (tautologically) are an "inertial observer for the whole trip".
Otherwise, the full analysis from a non-inertial observer (Gracie)
is not valid as if Gracie were an "inertial observer".
I think Greene's presentation of this point is fine (**in this format [of a recorded non-interactive lecture]**)...
although sometimes soundbites with missing context can lead to misinterpretations of the full message.
(More succinctly, as I often say, "𝐁𝐞𝐢𝐧𝐠 𝐚𝐛𝐥𝐞-𝐭𝐨-𝐛𝐞-𝐚𝐭-𝐫𝐞𝐬𝐭"≠"𝐁𝐞𝐢𝐧𝐠 𝐢𝐧𝐞𝐫𝐭𝐢𝐚𝐥". )
**I think that if subtleties were raised by a live questioner, Greene would clarify his statements.**

In my interpretation, it's too strong to draw a conclusion that any analysis (any measurement) from Gracie is invalid.
As @Ibix suggests, one needs more care to handle non-inertial Gracie's measurements compared to inertial George's measurements.

Ibix and PeroK