B Understanding twin paradox without math

[PeterDonis]

where/when does the thought experiment start and end in his thought experiment?

The same places as in yours. As several posters have already pointed out, the curves in spacetime are exactly the same.

To compare World lines you need a start and ending event.

Same as in your version.

He says in the beginning of his thought experiment everyone sets their clock to zero

You don't actually care how anyone's clocks are set, since you are measuring differences in clock readings, not absolute clock readings. All you need are the differences in clock readings between the start and end of the three line segments in spacetime: the "stay at home" line segment, the "traveling outbound" line segment, and the "traveling inbound" line segment. Then you just compare the sum of the second and third with the first; the latter will be larger.

 

[PeroK]

In any event to have different velocities they would have to accelerate he just kind of starts in the middle of the thought experiment after the acceleration and then claimed acceleration is not needed kind of disingenuous in my opinion.

But, you do get the same numerical answer, which ought to make you stop and think.

In other words, if we measure the elapsed time on clocks on the outbound and inbound journeys we get the equivalent result as the idealised physical experiment with a single traveling clock where the acceleration is extremely large.

It also doesn't matter whether we have 0, 2, or 4 acceleration phases. That should also make you think. Whether the experiment starts and ends with the traveling clock ar rest or not is essentially irrelevant to the result.

The only role acceleration plays is that it is necessary to enable the physical experiment. But, it is not itself necessary for the difference in elapsed times on the two clocks.

 

[Herman Trivilino]

They would have to jump for one spaceship to another which is acceleration.

All they need do is synchronize clocks as they pass each other.

 

[obtronix]

You don't actually care how anyone's clocks are set, since you are measuring differences in clock readings, not absolute clock readings.

Well that's not true he's saying you start the clocks simultaneously and measure time from that start. (e.g. spaceship C to event 2).

 

[PeterDonis]

that's not true he's saying you start the clocks simultaneously and measure time from that start

He's probably expecting that you will exercise some intelligence and not take everything he is saying exactly literally.

In any case, the fact that how each leg's clocks are handled was not specified correctly (if that is the case), does not mean that what is being described "is not the twin paradox". What makes the scenario "the twin paradox" is, as has already been pointed out multiple times, the geometry of the curves in spacetime. There are multiple ways to realize those curves in an actual scenario.

 

[Ibix]

Well that's not true he's saying you start the clocks simultaneously and measure time from that start. (e.g. spaceship C to event 2).

No, he has A (stay at home) and B (outbound) zero their watches as they pass, and C (returner) zero her watch as she passes B, and make a note of B's clock reading. B and C's times get reported to A, and their sum is the same as a single out-and-back traveller's total time. (At the 5.15 mark in the "no maths" video.)

 

[obtronix]

No, he has A (stay at home) and B (outbound) zero their watches as they pass, and C (returner) zero her watch as she passes B, and make a note of B's clock reading. B and C's times get reported to A, and their sum is the same as a single out-and-back traveller's total time. (At the 5.15 mark in the "no maths" video.)

Okay he doesn't say that but maybe he misspoke.

But I could assume C at rest, perform similar SpaceTime interval calculations, and have them record the longest time.

 

[Ibix]

But I could assume C at rest, perform similar SpaceTime interval calculations, and have them record the longest time.

No you can't - all those measures are invariant. Your own diagrams show you that!

 

[obtronix]

No you can't - all those measures are invariant. Your own diagrams show you that!

My diagrams have acceleration in them! He's creating a "ghost being" that is jumping from one ship to the other without accelerating. So he's kind of adding apples with oranges intervals.

If you think that's untrue what makes A so different than B and C? Relativity says motion is relative. I certainly can make a situation where events start and stop on the C World line. He just arbitrarily picked A.

 

[Motore]

My diagrams have acceleration in them!

No, they have instantaneous acceleration, which in your words would be a "ghost acceleration" as it's physically not possible.

 

[PeterDonis]

He's creating a "ghost being" that is jumping from one ship to the other without accelerating.

You are quibbling. There is no need for anyone to jump from the outbound to the inbound ship at the turnaround event. All that needs to happen is that the inbound ship records the outbound ship's clock reading as they pass each other. And, as has already been pointed out, that's what the video specifies.

what makes A so different than B and C?

Spacetime geometry. As has already been pointed out, there is a triangle in spacetime with three sides, A, B, and C. The fact that A > B + C is just the spacetime version of the triangle inequality; we have > instead of < because of the minus sign in the spacetime metric.

Relativity says motion is relative.

Yes, but that doesn't imply the other (wrong) claims you are making.

I certainly can make a situation where events start and stop on the C World line.

You could define a different triangle in spacetime, sure. That would be a different scenario from the one we are discussing in this thread. And it would have a different relationship between the lengths of its sides. But that relationship would still be invariant, independent of any choice of frame.

What you can't do is have the same triangle in spacetime but then claim that "events start and stop on the C worldline". Because with that triangle in spacetime, they don't.

He just arbitrarily picked A.

If he did, then so did you when you posed the scenario in your OP.

I think you are not thinking clearly about what you are saying.

 

[Ibix]

My diagrams have acceleration in them! He's creating a "ghost being" that is jumping from one ship to the other without accelerating.

No, he's just copying information from one to the other.

If you think that's untrue what makes A so different than B and C?

The experimental setup. For example, A is the only one who sees the others as moving with equal and opposite velocities.

 

[Ibix]

I certainly can make a situation where events start and stop on the C World line. He just arbitrarily picked A.

Of course. But that would be a different scenario.

 

[obtronix]

Of course. But that would be a different scenario.

Not a different scenario a different point of view, everybody will be traveling exactly in the same way, same speed same distance, just the handoffs of times will be different.With the original twin paradox thought experiment you get the same answer from everyone's perspective.

 

[Ibix]

Not a different scenario a different point of view, everybody will be traveling exactly in the same way, same speed same distance, just the handoffs of times will be different.

But not the proper times, because they are invariant. And that's what we're measuring.

With the original twin paradox thought experiment you get the same answer from everyone's perspective.

And this one too. The triangle in the Minkowski diagram is identical to yours!

 

[PeterDonis]

everybody will be traveling exactly in the same way, same speed same distance, just the handoffs of times will be different

In which case whatever it is that you are calling "the handoffs of times" have no physical meaning and are irrelevant to the scenario.

 

[Sagittarius A-Star]

If we draw the lines of simultaneous events for the "left spaceship" mirror, at the turnaround event there's a velocity discontinuity, and so we can draw two lines of simultaneous events, one line connecting with T=2, and the other connecting with T=3 (so to speak, obviously).

According to the OP, $\gamma = \frac{5}{4}$. Than means, the turnaround event of the "left spaceship" mirror ($\tau = 2$) has in the "left spaceship's" outbound rest-frame a simultaneity-line to the $T= 2 \cdot \frac{4}{5} = 2- \frac{2}{5}$ event of the (time-dilated) "left earth" mirror and (for symmetry reasons) in the "left spaceship's" inbound rest-frame a simultaneity-line to the $T= 3+\frac{2}{5}$ event of the "left earth" mirror.

 

[FactChecker]

IMO, a lot of the "solutions" to the paradox are dodging the real paradox by changing the scenario so that the paradox is removed. They do not really explain the original paradox.
IMO, the real paradox is due to confusing the second derivative of position with acceleration.
The second derivative of position is just as "relative" as the first derivative, velocity, is. Looking only at the first and second derivatives, the problem is completely symmetric wrt the twins. One might think that a traveling-twin-centered calculation should be just as valid as the Earth-centered calculation. Yet the two calculations give conflicting answers. The real "solution" to the original paradox is to point out that the Earth-centered calculation uses an intertial reference frame whereas the traveling-twin-centered calculation is in an accelerating frame. Acceleration is the key in distinguishing between an IRF and a non-IRF. To explain the original paradox, acceleration can not be ignored. The original paradox can be bypassed (but not explained) by changing the calculations to use only IRFs.
PS. I have seen a calculation that used the traveling-twin-centered reference frame from beginning to end, took an acceleration profile into account, and came up with the same answer as the Earth-centered IRF calculation. That was intellectually satisfying.

 

[PeterDonis]

Looking only at the first and second derivatives, the problem is completely symmetric wrt the twins. One might think that a traveling-twin-centered calculation should be just as valid as the Earth-centered calculation. Yet the two calculations give conflicting answers.

I'm not sure what "traveling-twin centered calculation" you are referring to here. Do you have a reference?

I have seen a calculation that used the traveling-twin-centered reference frame from beginning to end, took an acceleration profile into account, and came up with the same answer as the Earth-centered IRF calculation. That was intellectually satisfying.

Again, can you give a reference?

 

[PeterDonis]

IMO, a lot of the "solutions" to the paradox are dodging the real paradox by changing the scenario so that the paradox is removed.

The real "solution" to the original paradox is to point out that the Earth-centered calculation uses an intertial reference frame whereas the traveling-twin-centered calculation is in an accelerating frame. Acceleration is the key in distinguishing between an IRF and a non-IRF. To explain the original paradox, acceleration can not be ignored.

I think you are misidentifying the "paradox". The "paradox" is that the two twins have aged differently when they meet up again. But that "paradox" is resolved by spacetime geometry: the two twins follow different worldlines between the same starting and ending events, and those different worldlines have different lengths.

Proper acceleration in the original "paradox" scenario is best viewed as the means by which the traveling twin's worldline is enabled to meet up a second time with the stay-at-home twin's (since in flat spacetime it is impossible for two inertial worldlines to meet more than once).

Note, also, that proper acceleration has nothing to do with "an accelerating frame". You can perfectly well describe proper acceleration using an inertial frame, or inertial motion using a non-inertial frame. The best way to avoid confusion in cases like this is to not even mention frames at all, but to focus on the invariants: the spacetime geometry, the worldlines of the twins, and whatever invariant properties (such as the proper acceleration of the traveling twin in the standard scenario) are required to enable the two worldlines to meet a second time.

 

[member 728827]

enable the two worldlines to meet a second time.

But is the same paradox if they meet but the moving one doesn't stop, but continues moving away at relativistic speed, or for the paradox, they both need to meet at rest? Well, really, they don't meet if they do not do it at rest.

 

[FactChecker]

I'm not sure what "traveling-twin centered calculation" you are referring to here. Do you have a reference?

If we truly (erroneously) ignore acceleration, the distinction between an IRF and others disappears. We have no "acceleration", only the second derivative of position. Then the situation of the twins is completely symmetric. If we swap the twin in whose reference frame we do any calculation, all relative positions, first derivatives, and second derivatives are swapped. The resulting ages are swapped accordingly.
IMO, this is the true misconception of the original paradox. I think that if you look closely at what the confused people are saying, this is at the heart of it. They are thinking of the second derivative of position as acceleration.

CLARIFICATION: I am not saying that the above is valid, I am just saying that, IMO, this is what confuses many people about the paradox. And to say that we do not need to consider acceleration to get the correct answer is not the same as saying that acceleration is not important in understanding the root cause of their confusion.

Again, can you give a reference?

Sorry, the only thing I can find in notes is this: https://en.wikipedia.org/wiki/Twin_...psed_times:_how_to_calculate_it_from_the_ship
I don't know if this was it. I can not say that I follow the calculations or looked up the references.

 

[PAllen]

I'm not sure what "traveling-twin centered calculation" you are referring to here. Do you have a reference?Again, can you give a reference?

Hmm. Any valid treatment in non-inertial coordinates in which the rocket maintains constant coordinate position would have these characteristics. Radar and Fermi-Normal are the most common, but uncountably infinite other choices are possible.

 

[PeterDonis]

But is the same paradox if they meet but the moving one doesn't stop, but continues moving away at relativistic speed, or for the paradox, they both need to meet at rest?

If you want to quibble over minor details, the worldlines will not be exactly the same if the traveling twin starts and ends at rest relative to the stay-at-home twin, as compared to the case described in this thread, where the traveling twin passes by the stay-at-home twin at relativistic speed at the starting and ending events. However, the differences are minor and don't affect the main point of the analysis, which is that different paths through spacetime between the same two events can have different elapsed times.

Well, really, they don't meet if they do not do it at rest.

Nonsense. If they pass each other at relativistic speed, they still meet in any sense that matters for this discussion.

 

[Nugatory]

The real "solution" to the original paradox is to point out that the Earth-centered calculation uses an intertial reference frame

That’s not sufficient to eliminate the paradox. Throughout the entire journey the traveling twin can correctly apply the time dilation formula and correctly conclude that the earth clock is running slower than their own - yet somehow the earth clock is ahead when the twins reunite.

 

[PeroK]

That’s not sufficient to eliminate the paradox. Throughout the entire journey the traveling twin can correctly apply the time dilation formula and correctly conclude that the earth clock is running slower than their own - yet somehow the earth clock is ahead when the twins reunite.

Another vital point about the paradox without acceleration is that it shows that differential ageing is fundamentally different from time dilation.

The idea that "because he/she accelerated the traveling twin's time is really dilated and not the Earthbound twin's" is plain wrong.

In fact, this treatment explodes several of the myths around the twin paradox and should lead the student to the conclusion that it is fundamentally about spacetime geometry and nothing else.

 

[PeterDonis]

Any valid treatment in non-inertial coordinates in which the rocket maintains constant coordinate position would have these characteristics.

Would have what characteristics? Not the ones that FactChecker described here:

One might think that a traveling-twin-centered calculation should be just as valid as the Earth-centered calculation. Yet the two calculations give conflicting answers.

That's why I asked him for a reference.

 

[vanhees71]

Since the proper times of each twin are scalars, they cannot depend on the choice of coordinates nor reference frames. That's what resolves the paradox: The aging of each twin is independent of the choice of the arbitrary coordinates and reference frames.

 

[robphy]

It seems to me that there are two issues, which seem to get combined/mixed-up.

• Clock Effect: between two timelike-related events, the elapsed time along a worldline depends on the worldline [unlike what happens in Galilean physics].
  (A similar thing happens for two points on a plane: between points in the plane, the distance along a path joining the points depends on the path.)
• Twin Paradox [in attempt to disprove the Clock Effect by contradiction]:
  the mistaken assertion that:
  since the observer along any worldline can regard him "at rest",
  then that observer can regard also themselves as "inertial" [which is false].

 

[Kairos]

differential ageing is fundamentally different from time dilation.

The idea that "because he/she accelerated the traveling twin's time is really dilated and not the Earthbound twin's" is plain wrong.

It's a pleasure to read that!
One of the reasons for the intensive debates about the twin paradox may be that its official solution looks like an asymmetric application of time dilation of special relativity... which is symmetric by nature.
Acceleration is indeed asymmetric but cannot be invoked as an intrinsic cause of differential aging as perfect atomic clocks are unaffected by acceleration.

This is why I find the diagrams using the light clock at the beginning of this post very illuminating: accelerations play there only a simple role of transition between inertial phases and the same result is predicted seen from the three point of views (earth, outward traveler and return traveler). This is basically the same solution as that using the simultaneity lines, but in an even clearer way

 

[vanhees71]

There is nothing asymmetric. The aging of each twin is an invariant (scalar). Both twins must start and meet again at the same events to compare their clocks. For any timelike world line $C$ connecting two fixed events in any coordinates you have
$$\tau[C]=\int_C \sqrt{g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\mu}}.$$
That's manifestly invariant under coordinate transformations since the $g_{\mu \nu}$ transform as 2nd-rank covariant tensorcomponents, and it's a functional (sic!) of the time-like world line, $C$. That's all behind the twin paradox, which of course is not a paradox. It's simply strange for us as we usually don't move with speeds close to the speed of light, and the gravitational fields we experience are negligible, so that have the impression that our clocks all show the same time when synchronized once.

That's of course not the case for high-precision clocks, where even the small changes of the gravitational field of the Earth as well as the effect from moving clocks relative to each other can be measured.

 

[PAllen]

Would have what characteristics? Not the ones that FactChecker described here:

That's why I asked him for a reference.

I was focusing on this statement in the same post:

"PS. I have seen a calculation that used the traveling-twin-centered reference frame from beginning to end, took an acceleration profile into account, and came up with the same answer as the Earth-centered IRF calculation. That was intellectually satisfying."

rather than what you were questioning. I hadn't noticed the conflicting statements in one post.

 

[FactChecker]

Would have what characteristics? Not the ones that FactChecker described here:

That's why I asked him for a reference.

My post was trying to illuminate the misconceptions that people have that lead to the paradox. I was not trying to say that those misconceptions were correct. Without referring to acceleration, the distinction between an IRF and other reference frames is removed. A person who erroneously mistakes acceleration for the second derivative of relative position would think that the situation of the twins was completely symmetric.

 

[Orodruin]

Always worth a re-post: https://www.physicsforums.com/insights/geometrical-view-time-dilation-twin-paradox/

Not understanding the resolution to the twin paradox is exactly analogous to not understanding the triangle inequality in Euclidean space.

Which path is longer - red or green?

 

[Ibix]

Which path is longer - red or green?

...and (relevant to this thread and Lincoln's video) is the green path different if I extend its straight line segments beyond the intersections?

 

[Dale]

Not a different scenario a different point of view

This is not correct. If it were a different (inertial) point of view then there would exist a Lorentz transformation from one scenario to the other. There is not.

The issue is that in all frames the time-order of the three meeting events is invariant. A-meets-B is first, then B-meets-C is second, and A-meets-C is last in all frames. So it is an invariant fact that A is at the first and last meeting. In all frames, the one at the first and last meeting will have a larger time measure between meetings than the sum of the other two.

 

[Dale]

My post was trying to illuminate the misconceptions that people have that lead to the paradox. I was not trying to say that those misconceptions were correct. Without referring to acceleration, the distinction between an IRF and other reference frames is removed. A person who erroneously mistakes acceleration for the second derivative of relative position would think that the situation of the twins was completely symmetric.

I agree with this. Mathematically and theoretically, of course, the three-clock-no-acceleration scenario is indeed equivalent to the two-clock-yes-acceleration scenario. But the issue is that mathematically and theoretically there is no paradox; the twin's paradox is a pedagogical paradox not a genuine mathematical paradox. So a replacement scenario that matches theoretically but not pedagogically is not a good replacement scenario.

The pedagogical issue is misapplied symmetry. Students get the idea that all inertial observers are equivalent pounded into their heads. They learn that there is symmetry between an observer on a train and an observer on an embankment. So students that fail to recognize that the equivalence is only between inertial observers think that there is symmetry between the two observers in the standard scenario. I don't know of any students that think there is symmetry between one observer and two observers. So the three-observer scenario is not pedagogically equivalent, which is the whole paradox.

 

[robphy]

FactChecker,

Here's the spacecraft version, since the spacecraft is not in a single inertial frame you need two graphs, one from the outgoing spacecraft 's point of view and one from the incoming spacecraft 's point of view, you get the same answer in both.

Some pedagogical suggestions for your diagram (which are nice and it's good that you have created it via a calculation [in Excel]).
(To those who know about my diagram,... yes... these suggestions are features of my diagram.)

• It seems that your observer-astronauts separate at the origin event.
  As others have suggested, their implied worldlines should get the labels "Earth" and "Space Ship". Their mirror-worldlines should have "mirror" in their labels.
• At the separation event, you have their clocks zeroed then-and-there. However, you use the "light-signal meets the left(rear) mirror-worldline" to mark these zeroes for these clocks. As drawn, these spacelike-related events are simultaneous only this Earth frame. (The calculation and diagram work out in the end because of symmetry... but I fear a more general situation will lead to problems.)
• Instead, it would be better to mark-off ticks by "light-signal meets the astronaut-worldline".
  In this way, you are using the light-cone of their common separation event
  (rather than the light-cones of their "signals at their rear-mirror")
• In addition, rather than using only the forward-directed signal,
  you can also use the rearward-directed light-signal, which together trace out "causal diamonds".
• With these changes, the diagram for the turn-around events (which are a result of using "nice numbers" to use fractions) will look more continuous. [Nice numbers are gotten with velocities whose Doppler factors are rational, leading to triangles with Pythagorean triples.]
• Furthermore, you may realize a symmetry that is not easily recognized or appreciated in your diagram: the "causal diamonds" traced out by the light-signals of the earth-clock has the same area as those diamonds of the astronaut-clock. Thie equality encodes the idea that each observer is using the same "standard issue clock".
• This "equality of light-clock-diamond areas" implies length-contraction of the mirror-worldtubes. In fact, you might then appreciate that the mirror-worldlines are implied... and you don't need to draw them. Indeed, the physical results are unchanged by the choice of "standard time unit".
• Keep exploring with your diagram (and your calculation to create them)! Good luck!

Spoiler: https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
Spoiler: https://physics.stackexchange.com/a/507592/148184

 

[PeterDonis]

My post was trying to illuminate the misconceptions that people have that lead to the paradox.

That's fine, but if you are going to describe a specific misconception, which you did, you need to give a reference. Just saying "some people claim this" is not sufficient.

Without referring to acceleration, the distinction between an IRF and other reference frames is removed.

This is not correct. An inertial frame and a non-inertial frame are different coordinate charts with different forms for the metric. You can describe all that without mentioning acceleration at all. It is true that there will, in general, be a parameter in the metric in the non-inertial frame that has units of proper acceleration, but that in itself does not mean that parameter has to have the physical meaning of the proper acceleration of something.

A person who erroneously mistakes acceleration for the second derivative of relative position would think that the situation of the twins was completely symmetric.

Not if they look at the spacetime geometry and the worldlines of each twin.

 

[FactChecker]

That's fine, but if you are going to describe a specific misconception, which you did, you need to give a reference. Just saying "some people claim this" is not sufficient.

I am not going to focus on one, or even ten examples when there are dozens in this forum. The misconception is a natural consequence of how relative motion is presented to millions of beginning physics students.

This is not correct. An inertial frame and a non-inertial frame are different coordinate charts with different forms for the metric. You can describe all that without mentioning acceleration at all. It is true that there will, in general, be a parameter in the metric in the non-inertial frame that has units of proper acceleration, but that in itself does not mean that parameter has to have the physical meaning of the proper acceleration of something.

Is that usually presented in answers to the Twin Paradox? If so, I have missed it.

Not if they look at the spacetime geometry and the worldlines of each twin.

IMO, the root cause of the misconception of the Twin Paradox is usually ignored in favor of proposing other IRFs to get the same result as the Earthbound calculation. I would prefer that the real misconception be addressed head-on by explaining why the solution in an accelerating reference frame must be more complicated. The situation of the twins is not symmetric.

 

[PeterDonis]

I am not going to focus on one, or even ten examples when there are dozens in this forum.

I have never seen any post in this forum that meets this description of yours:

One might think that a traveling-twin-centered calculation should be just as valid as the Earth-centered calculation. Yet the two calculations give conflicting answers.

I have seen plenty of incorrect claims by posters that the traveling twin would calculate a conflicting answer, but I have never seen any poster who makes such a claim give an actual calculation to back up such a claim. But in the above quote you are claiming that there is such a calculation. That's why I asked for a reference to one.

 

[FactChecker]

I have seen plenty of incorrect claims by posters that the traveling twin would calculate a conflicting answer, but I have never seen any poster who makes such a claim give an actual calculation to back up such a claim. But in the above quote you are claiming that there is such a calculation. That's why I asked for a reference to one.

The assumption that the situation of the Twins is completely symmetric leads to the Earth-centered calculations being swapped to the traveling-twin and wrongly concluding, from that perspective, that the Earth-bound twin has aged less. There is no need for a new calculation. You must have seen this a hundred times.
(see naive[3][4] application of time dilation and the principle of relativity, each should paradoxically find the other to have aged less )

 

[PeterDonis]

The assumption that the situation of the Twins is completely symmetric leads to the Earth-centered calculations being swapped to the traveling-twin

In a sort of hand-waving way, perhaps. But again, I have never seen anyone who makes this claim actually try to do the math.

There is no need for a new calculation.

Yes, there is: in order to do the math, you need to actually define the traveling twin's rest frame, since it is not the same as the Earth twin's rest frame. If that frame were inertial, as the Earth twin's is, you would be able to find a single inertial worldline that was at rest in the frame for the whole trip. But of course you can't. And any actual attempt to do the math would quickly run up against that insurmountable difficulty. The only reason people are able to make such a claim is that they have not actually tried to do the math.

(see naive[3][4] application of time dilation and the principle of relativity, each should paradoxically find the other to have aged less )

As far as I can tell, all of the references given in that passage of the Wikipedia article start right out by saying explicitly that the viewpoint being described is incorrect, and explain why, along the same lines that I gave above. So anyone who reads those references should already be on notice that this viewpoint is wrong and that any claims made based on it are also wrong.

 

[vanhees71]

This is not correct. An inertial frame and a non-inertial frame are different coordinate charts with different forms for the metric. You can describe all that without mentioning acceleration at all. It is true that there will, in general, be a parameter in the metric in the non-inertial frame that has units of proper acceleration, but that in itself does not mean that parameter has to have the physical meaning of the proper acceleration of something.

What's physical is the spacetime itself, not the coordinate charts used to describe them. In special relativity you have Minkowski space as the spacetime manifold, and this is not changed by choosing another coordinate system, defining a non-inertial frame of reference. In GR the specific spacetime (a pseudo-Riemannian manifold) is determined by the physical situation, but it's also completely independent of the choice of coordinates.

The aging of the twins is also independent of the choice of coordinates, it's given by the proper times of the twins along their world lines (measured from the common start to the common end).

 

[Sagittarius A-Star]

What's physical is the spacetime itself, not the coordinate charts used to describe them.
...
The aging of the twins is also independent of the choice of coordinates, it's given by the proper times of the twins along their world lines (measured from the common start to the common end).

That's true. The twin's world lines are usually mathematically described using coordinates of a reference frame, often the rest-frame of the non-inertial twin.

There is sometimes confusion like "according to special relativity, from twin A's perspective twin B will of aged less but from twin B's perspective twin A will of aged less".

Source (see OP):
https://www.physicsforums.com/threads/symmetrical-twin-paradox.663787/

I think, the error made in such a case is, that the commonly known time-dilation factor $\sqrt{1-v^2/c^2}$ is applied to the rest frames of both twins in the calculations of the elapsed proper time of the respective other twin, although one of the rest-frames is non-inertial. It is not understood, that with reference to a non-inertial frame, an additional "gravitational" time-dilation exists. For example, the time-dilation factor in Kottler–Møller coordinates for a Rindler-frame (with constant acceleration $\alpha$) is
$\sqrt {(1+ \frac{\alpha\,x} {c^2})^2 - \frac{v^2}{c^2}}$.

 

[SiennaTheGr8]

There's a chapter in Taylor/Wheeler's Spacetime Physics that I recall being very helpful for understanding the twin paradox ("Trip to Canopus" I think?). It does emphasize geometry, but it also shows how the relativity of simultaneity comes into play when you analyze the problem in terms of inertial frames. In particular, it discusses what the Earth-twin's wristwatch reads "at the same time as" the turnaround event for both the outbound-frame and the inbound-frame.

 

[vanhees71]

That's true. The twin's world lines are usually mathematically described using coordinates of a reference frame, often the rest-frame of the non-inertial twin.

Sure, how else do you want to describe the worldline? This is no contradiction to the fact that the aging of each twin is a property of the spacetime and not of any choice of coordinates or frames. In SR it's of course most convenient to use a global inertial frame and pseudo-Cartesian coordinates.

There is sometimes confusion like "according to special relativity, from twin A's perspective twin B will of aged less but from twin B's perspective twin A will of aged less".

This cannot be, because the aging of the twins are scalar quantities and thus independent of any choice of coordinates and frames.

Source (see OP):
https://www.physicsforums.com/threads/symmetrical-twin-paradox.663787/

I think, the error made in such a case is, that the commonly known time-dilation factor $\sqrt{1-v^2/c^2}$ is applied to the rest frames of both twins in the calculations of the elapsed proper time of the respective other twin, although one of the rest-frames is non-inertial. It is not understood, that with reference to a non-inertial frame, an additional "gravitational" time-dilation exists. For example, the time-dilation factor in Kottler–Møller coordinates for a Rindler-frame (with constant acceleration $\alpha$) is
$\sqrt {(1+ \frac{\alpha\,x} {c^2})^2 - \frac{v^2}{c^2}}$.

Of course you have to use the metric components according to each choice of coordinates/frames. Otherwise you simply produce errors.

 

[vanhees71]

There's a chapter in Taylor/Wheeler's Spacetime Physics that I recall being very helpful for understanding the twin paradox ("Trip to Canopus" I think?). It does emphasize geometry, but it also shows how the relativity of simultaneity comes into play when you analyze the problem in terms of inertial frames. In particular, it discusses what the Earth-twin's wristwatch reads "at the same time as" the turnaround event for both the outbound-frame and the inbound-frame.

Of course, comparing clock readings of clocks at different places depends on the observers due to the relativity of simultaneity. That's why the clocks have to be synchronized at the beginning of the journey, when the twins start at the same place and then compared after the trip, when both twins again meet at the same place.

 

[FactChecker]

This cannot be, because the aging of the twins are scalar quantities and thus independent of any choice of coordinates and frames.

He says it is a misconception. That is exactly why it is a paradox. The answer to the problem is NOT the same as the explanation of the paradox. There are many ways to get the correct answer to the problem that do nothing to illuminate why the misconception of the paradox is wrong. IMO, most of the "answers" in this forum to the Twin Paradox are like that.

 

[Ibix]

He says it is a misconception. That is exactly why it is a paradox. The answer to the problem is NOT the same as the explanation of the paradox. There are many ways to get the correct answer to the problem that do nothing to illuminate why the misconception of the paradox is wrong. IMO, most of the "answers" in this forum to the Twin Paradox are like that.

You are drawing a distinction between explaining why the right answer is what it is and explaining what's left out of the naive (wrong) analysis. I agree that the paradox arises when one doesn't realise that there's a change in simultaneity when one changes reference frames, and the change in simultaneity exactly accounts for the difference between the inertial twin's elapsed time $T$ and the "paradoxical" $T/\gamma^2$. However, my experience of twin paradox questions at PF is that they don't usually ask how to resolve the paradox, but usually ask some variant on "what mechanism makes the moving clock run slow", to which some variant on "clocks measure interval and interval is path dependent just like distance" is a reasonable answer. The OP's diagrams (with the needed corrections) are another approach. I think that's why a lot of answers don't invoke the simultaneity change explanation you're using.