Twin Paradox: Who Is Right, A or B?

[Sagittarius A-Star]

#1 - Einstein's objection to the criticism quoted was that non-inertial frames invalidate the analysis.

Non-inertial reference frames can be used in SR, but it is more complicated to calculate in non-inertial frames than in inertial frames.

#2 - All the descriptions of the twin paradox have the traveler hopping onto a moving frame at the start, decelerating and accelerating to turn around, and hopping off a moving frame at the end. Hence these descriptions are not realizable (because of the hopping) and also not valid (because of the non-inertial frames).

In 1918, Einstein described the twin paradox in both frames, without hopping:
https://en.wikisource.org/wiki/Translation:Dialog_about_Objections_against_the_Theory_of_Relativity
(Remark: When he speaks in flat spacetime about "gravitational field", this is named by today's physicist mostly "pseudo-gravitational field".)

#3 - What experiments confirm time dilation?

For example experiments with myons in storage rings:

Bailey et al. (1977) measured the lifetime of positive and negative muons sent around a loop in the CERN Muon storage ring. This experiment confirmed both time dilation and the twin paradox, i.e. the hypothesis that clocks sent away and coming back to their initial position are slowed with respect to a resting clock.

Source:
https://en.wikipedia.org/wiki/Experimental_testing_of_time_dilation#Twin_paradox_and_moving_clocks

Is there a realizable experiment confirming time dilation that doesn't involve non-inertial frames?

Almost without acceleration are myons in the upper atmosphere (moving inertially with almost c):
http://www.atmosp.physics.utoronto.ca/people/strong/phy140/lecture32_01.pdf

 

[Will Flannery]

It is easy to formulate it using only inertial frames: simply do the analysis in the home twin’s frame.

Not quite. The problem is that in the original post A stopped at B. He didn't note B's age as he passed by. I regarded having the traveler stop as part of the scenario. It's a question of physical realizability.

"Thus when A arrives at B's location A expects to see a younger B."

However, it seems like you specifically want experiments that confirm time dilation for inertial objects.

Inertial object? What is an inertial object?

I assume that you mean you are trying to reformulate it using only inertial objects. That can be done in terms of what is sometimes called a “triplets scenario”. In that case you have three inertially moving clocks, say P, Q, R. In P’s frame Q is moving to the right at v and R is moving to the left at v. As Q passes P they reset their clocks to 0. Then as R passes Q we have R set their clock to match Q. Then we compare the clocks of R and P when they meet.

The difference is the same as the twin paradox, but all clocks are inertial.

This is an improvement to the Lincoln explanation in my view as it doesn't have the hopping. But it still lacks the realizable angle. In the twin paradox the twin (as it is usually formulated) the twin has to arrive home, not flash by at fraction of the speed of light.

So, the twin paradox scenario as stated, with the twin arriving back and stopping, has not been analyzed at all as far as I know. To do so would require analyzing the effect of acceleration/deceleration on the traveling twin's clock. What if he arrived and his clock matched that of his twin ?

 

[Dale]

Inertial object? What is an inertial object?

It is an object which is moving inertially. I.e. an object where an attached accelerometer would read 0.

The point is that inertial objects are physical. You can physically take an object and attach an accelerometer and determine if it reads 0. In contrast an inertial frame is a mathematical construct that requires an arbitrary synchronization convention and which has no impact on any physical measurement.

But it still lacks the realizable angle.

Why not? It is perfectly realizable. Of course it is not identical to the twin scenario, but that is obvious: since you wanted to eliminate acceleration then you must make some changes.

 

[Will Flannery]

Why not? It is perfectly realizable. Of course it is not identical to the twin scenario, but that is obvious: since you wanted to eliminate acceleration then you must make some changes.

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact. I harbor the suspicion the the twins' clocks would read the same :).

 

[Dale]

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact

Huh? Most analyses of the twin paradox do not eliminate acceleration. That is, in fact, the standard approach. You are making no sense.

I harbor the suspicion the the twins' clocks would read the same :).

Your suspicion is unfounded. Accelerations have been tested up to about 10^18 g. No additional acceleration-based time dilation effect was found.

 

[Ibix]

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact. I harbor the suspicion the the twins' clocks would read the same :).

We've been assuming acceleration was instantaneous because that way you don't need calculus. Finite acceleration is fairly straightforward to include if you want to do it, and it will yield the same result that the traveller is younger (assuming that the twins' ages are compared by Einstein synchronisation before the experiment starts).

 

[Will Flannery]

Huh? Most analyses of the twin paradox do not eliminate acceleration. That is, in fact, the standard approach. You are making no sense.

? Analysis of acceleration/deceleration is entirely absent from the standard approach.

Your suspicion is unfounded. Accelerations have been tested up to about 10^18 g. No additional acceleration-based time dilation effect was found.

I base it on this (but it's too late in the day for me to check) - Need to rethink - I'll go over this tomorrow (or not).

 

[Nugatory]

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact. I harbor the suspicion the the twins' clocks would read the same :).

There are several acceleration-free ways of stating the problem. Here’s one, described using a frame in which the Earth is at rest:
We start with spaceship A somewhere to the left of the earth, moving right at speed v relative to earth, and spaceship B somewhere farther to the right of earth, moving left at speed v relative to earth. A will pass by Earth and then eventually meet B moving in the opposite direction; and then B will continue on past the Earth while A sails off into the distance.

As A passed Earth we zero a clock on Earth and on ship A; this works because A and Earth are at the same place at the same time. Later A passes B and and as they pass the clock on B is set to $T_1$, the time on A’s clock as they pass one another; again this works because A and B are colocated. B’s clock continues ticking as B approaches earth. The time on B’s clock as B passes Earth will be the total time along the outbound leg and the inbound leg; it’s the time that would be registered by a hypothetical acceleration-tolerant clock that was transferred from A to B as they passed one another.

When we compare B’s clock and the Earth clock as they pass, we will find that more time passed on Earth than on the two legs of the round trip.

 

[Ibix]

An object undergoing constant proper acceleration $\alpha$ has coordinate acceleration $dv/dt=\alpha/\gamma^3$. You can integrate this to obtain its coordinate velocity:$$v=\frac{\alpha c t}{\sqrt{c^2+\alpha^2t^2}}$$That allows you to write the Lorentz $\gamma$ factor:$$\gamma=\sqrt{1+\frac{\alpha^2}{c^2}t^2}$$Noting that $d\tau/dt=1/\gamma$, you can substitute the expression for $\gamma$ and integrate to determine the proper time elapsed on the ship while coordinate time $t$ elapses:$$\tau=\frac c\alpha\sinh^{-1}\left(\frac\alpha ct\right)$$If a ship, initially at rest, accelerates at some constant proper acceleration $\alpha$ for some coordinate time $T$ then decelerates at the same proper acceleration for the same amount of time, we can calculate the elapsed time for the ship to be$$\begin{eqnarray*}
\tau&=&\frac c\alpha\sinh^{-1}\left(\frac\alpha cT\right)+\frac c{-\alpha}\sinh^{-1}\left(\frac{-\alpha}cT\right)\\
&=&2\frac c\alpha\sinh^{-1}\left(\frac\alpha cT\right)
\end{eqnarray*}$$You can confirm for yourself that $\sinh^{-1}(x)\neq x$ and hence $\tau\neq 2T$. In fact $|\sinh^{-1}(x)|<|x|$, and hence $\tau<2T$ and the traveling twin will be younger on arrival.

Note that this flight profile has no inertial phase at all - it is under acceleration all the way, and the result is qualitatively the same as the purely inertial case. This is, in fact, obvious from examining a spacetime diagram in the rest frame of the planets the ship flew between. The ship's worldline is slanted (albeit not a constant slant); its elapsed time is, therefore, lower.

 

[Dale]

? Analysis of acceleration/deceleration is entirely absent from the standard approach.

What are you talking about. The standard scenario is that the traveling twin accelerates at the beginning, middle, and the end! And the most common variant is that the traveling twin accelerates in the middle. How can you say it is absent from the standard approach?

I base it on this (but it's too late in the day for me to check) - Need to rethink - I'll go over this tomorrow (or not).

Bailey et al., “Measurements of relativistic time dilation for positive and negative muons in a circular orbit,” Nature 268 (July 28, 1977) pg 301. They confirmed the clock hypothesis (no additional acceleration effect for time dilation) to 10^18 g. The acceleration effect is 0. The same experiment is a direct test of the twin paradox also.

Your suspicion has no basis, neither in theory nor in experiment.

 

[Sagittarius A-Star]

Thus when A arrives at B's location A expects to see a younger B.

That's wrong. This can be easily understood at the example of myons in a storage ring. Myons can be used as clocks, because they have a defined half-life period.

Assume, myon A ist rotating in the storage ring and myon B is at rest in the center of the ring. A theoretical model of this is the rotating disk, which Einstein described in his popular book from 1917 (Appendix III c):

“From this it follows that
ν = ν₀ (1+ Δϕ/c²)
In the first place, we see from this expression that two clocks of identical construction will go at different rates when situated at different distances from the centre of the disc. This result is also valid from the standpoint of an observer who is rotating with the disc.
Now, as judged from the disc, the latter is in a gravitational field of potential ϕ “

Source:
https://en.wikisource.org/wiki/Rela...isplacement_of_Spectral_Lines_Towards_the_Red

Einstein showed there, why A and B do not disagree.

 

[Ibix]

What are you talking about. The standard scenario is that the traveling twin accelerates at the beginning, middle, and the end!

I suspect that @Will Flannery has observed that if you can write a standard twin paradox with no accelerations (as you can by replacing the traveller with two inertial travellers moving in opposite directions) then something may have been left out by using infinite acceleration. That's why I wrote down the maths for a constant acceleration case in #44, to show that there are no qualitative differences with finite acceleration.

More generally, @Will Flannery, the elapsed time for any worldline between (inertial) coordinate times $t=T_1$ and $t=T_2$ is given by$$\tau=\int_{T_1}^{T_2}\sqrt{1-\frac{v^2(t)}{c^2}}dt$$which follows from the definition of proper time. This is clearly maximised by $v=0$, and always smaller for any $v(t)$ that is non-zero at any time. Thus the traveling twin is younger, whatever their velocity profile, and we haven't had to consider any non-inertial frames.

Edit: furthermore, as Dale noted, this theoretical argument has been tested to some pretty high accelerations.

 

[SiennaTheGr8]

OP might find it useful to read the "Trip to Canopus" chapter of Taylor & Wheeler's Spacetime Physics. In particular, this page may help develop some intuition for how the relativity of simultaneity factors in.

 

[Will Flannery]

I suspect that @Will Flannery has observed that if you can write a standard twin paradox with no accelerations (as you can by replacing the traveller with two inertial travellers moving in opposite directions) then something may have been left out by using infinite acceleration.

No, I said that the standard scenario didn't consider acceleration/deceleration not because the traveling twin didn't accelerate/decelerate, but because the analysis of the traveling twin never analyzed the acceleration/deceleration. E.g. in Lincoln's vid the traveler hops on and off inertial frames traveling at different velocities, which seemed to me to be avoiding acceleration/deceleration.

But, now I see that it didn't need to be mentioned because it doesn't make much difference, there are no extraordinary effects from acceleration/deceleration.

But then, I abandoned my original version of the paradox because of the unanalyzed acceleration/deceleration, which I (now see) correctly thought didn't make much difference in terms of time dilation.

However, the acceleration/deceleration does make a big difference in which frame is considered inertial, so that was the problem in my original version, as I figured you should be able to analyze the motion in both frames ignoring the acceleration/deceleration.

So, that's what I'm puzzling over now. And I'm wondering ... was Lincoln wrong when he said, quoting "All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them'? Because if he is right then any observer can consider himself at rest in an inertial frame (?), and if that's true then I think my original version of the paradox is OK !

 

[Ibix]

I figured you should be able to analyze the motion in both frames ignoring the acceleration/deceleration.

You can analyse the trip in any frame you like. However, naively patching together the inertial frames associated with inertial segments of a worldline does not produce a valid frame - the result has gaps and overlaps. The overlaps aren't relevant here, but the gaps are, because the worldline of the Earth crosses one such gap. The time elapsed in that gap is not accounted for by such a naive approach. Failing to realize this is how your analysis of the traveller concluding that the Earth twin is younger goes wrong. It's not that you can't do the analysis in a non-inertial frame - it's that you aren't doing the analysis in a frame at all. You are doing it in a kludge that doesn't account for all of the relevant portion of the Earth twin's history.

Lincoln is correct that you can always regard yourself as at rest. You cannot always regard yourself as inertial, and if you are not inertial you cannot construct a rest frame by patching together parts of inertial frames.

 

[PeroK]

So, that's what I'm puzzling over now. And I'm wondering ... was Lincoln wrong when he said, quoting "All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them'? Because if he is right, then I think my original version of the paradox is OK !

On the one hand what Lincoln is saying is trivial: every observer, regardless of their state of motion, has a valid rest frame. The question is what laws of physics apply in an arbitrary frame of reference?

In SR, we have the Lorentz transformation between inertial reference frames (this unifies TD, LC and RoS into one neat mathematical package). In the case of the twin paradox, A changes from the initial rest frame into a new rest frame. It's valid to do this. But, he/she must apply the Lorentz transformation. Any data that A had from the initial rest frame (e.g. the location and time at planet B), must be transformed to the new "space shuttle" frame. This transformation changes the time "now" at planet B into some future time. Thereafter, time dilation applies to clocks on planet B, but they were already well ahead when A joined the space shuttle. And, when you do the maths, A gets the same answer that $B$ got using the initial rest frame: A's clock shows less time.

 

[Janus]

No, I said that the standard scenario didn't consider acceleration/deceleration not because the traveling twin didn't accelerate/decelerate, but because the analysis of the traveling twin never analyzed the acceleration/deceleration. E.g. in Lincoln's vid the traveler hops on and off inertial frames traveling at different velocities, which seemed to me to be avoiding acceleration/deceleration.

It's not really avoiding acceleration, it is invoking an "undefined" acceleration. Acceleration is a cahnge of velocity, the magnitude of which is determined by the magnitude of velocity change and the time over which the change is made; A = dV/t . If you "hop" between inertial frames, there is a change of velocity, but it occurs over zero time; thus A =dV/0. and division by zero is undefined.

But, now I see that it didn't need to be mentioned because it doesn't make much difference, there are no extraordinary effects from acceleration/deceleration.

The "effects" of acceleration are only measured by the accelerated observer.
Even in the "hopping" between frames you can't avoid this.
I start at rest with respect to A and B, while next to A and the clocks are both are synchronized.
If I suddenly "hop" to a frame where I have a significant velocity with respect to both in the direction of B, The clock a A doesn't change, but for me, the clock at B jumps forward in time.
On the other hand, if I were to accelerate up to that same velocity over some extremely short but non-zero time, B's clock would run very fast according to me during that period, and would again end up being quite a bit ahead of the Clock at A once I reached that relative velocity.
All "hopping" between frames did was to make the advance in B's clock discontinuous, instead of gradual.

There is one other difference between the 'instantaneous' change of velocity approach and the finite acceleration approach. You can only use the first one if you are dealing with single "point" observers which make the velocity change. If you try to apply such an instantaneous velocity change to an extended system, you'll create contradictions.

 

[Nugatory]

. was Lincoln wrong when he said, quoting "All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them'?

Not wrong, and...

Because if he is right then any observer can consider himself at rest in an inertial frame (?)

No. An observer who is subject to proper acceleration (as opposed to coordinate acceleration) may consider themself to be at rest, but the frame in which they are at rest is not inertial.[/quote][/QUOTE]

 

[Will Flannery]

Not wrong, and...No. An observer who is subject to proper acceleration (as opposed to coordinate acceleration) may consider themself to be at rest, but the frame in which they are at rest is not inertial.

OK, I'll buy that ... but ... at 4:20 Lincoln says there must be something that breaks the symmetry of the experience of Don (homebound) and Ron (traveler) ... and then goes on to say that the fact that one possible distinction is that Ron accelerates and Don doesn't (and that the time dilation is due to acceleration) but that is wrong wrong wrong. Of course it is wrong that the time dilation occurs because of acceleration/deceleration, but the distinction is that Don gets to be in an inertial frame and Ron doesn't, and that is because Ron accelerates in order to acquire velocity and to turn around and to stop. So the fact that Ron accelerates/decelerates is what breaks the symmetry of their experiences ... right ?

 

[PeroK]

OK, I'll buy that ... but ... at 4:20 Lincoln says there must be something that breaks the symmetry of the experience of Don (homebound) and Ron (traveler) ... and then goes on to say that the fact that one possible distinction is that Ron accelerates and Don doesn't (and that the time dilation is due to acceleration) but that is wrong wrong wrong. Of course it is wrong that the time dilation occurs because of acceleration/deceleration, but the distinction is that Don gets to be in an inertial frame and Ron doesn't, and that is because Ron accelerates in order to acquire velocity and to turn around and to stop. So the fact that Ron accelerates/decelerates is what breaks the symmetry of their experiences ... right ?

Acceleration is one way to break the symmetry. And, given that physical objects cannot change inertial reference frames without accelerating, that's a natural explanation.

But, what acceleration is doing is changing reference frames. So, you can equally explain the scenario without acceleration: either by hypothesising an instantaneous hop onto a space shuttle; or, by measuring time by clock readings on various spacetime paths. In other words, instead of hopping on the shuttle, A simply has a ship clock on the shuttle synchronized with his as it goes past (this is essentially a local process).

Either way, it amounts to the same thing.

 

[Ibix]

So the fact that Ron accelerates/decelerates is what breaks the symmetry of their experiences ... right ?

In this case, yes. Your twins have symmetric experiences of each other while they are on the two planets. But one of them chooses to get into a rocket and fly to the other - this is an asymmetry, and their experiences must be asymmetric if they do different things.

 

[Dale]

Lincoln says there must be something that breaks the symmetry of the experience of Don (homebound) and Ron (traveler) ... and then goes on to say that the fact that one possible distinction is that Ron accelerates and Don doesn't (and that the time dilation is due to acceleration) but that is wrong wrong wrong.
...
So the fact that Ron accelerates/decelerates is what breaks the symmetry of their experiences ... right ?

Time dilation does not depend on acceleration, but the acceleration does break the symmetry.

 

[Nugatory]

but the distinction is that Don gets to be in an inertial frame and Ron doesn't, and that is because Ron accelerates in order to acquire velocity and to turn around and to stop

The acceleration is a bit of a red herring here. The two are following different paths through spacetime; the two paths begin and end at the same points (the separation event and the reunion event) but have different lengths; the length of a path through spacetime is measured by the time elapsed on a clock following that path.

Acceleration only comes into the picture because there’s no way of setting them on different paths through spacetime without accelerating at least one of them... without that they’d both follow the same path, side by side. (The acceleration-free variation in post #43 in this thread is actually a technique for measuring the length the accelerated path without accelerating anything).

 

[Ibix]

You can analyse the trip in any frame you like. However, naively patching together the inertial frames associated with inertial segments of a worldline does not produce a valid frame - the result has gaps and overlaps.

Here's a graphical explanation of that. First, here's a spacetime diagram drawn in the frame of the planets. The red lines are the two planets (the left hand one is the distant planet, the right hand one is Earth), and the green line shows the twin traveling from one to the other.

I've shaded in the regions of spacetime that this frame regards as "before" and "after" the journey. Below, I've drawn the same diagram, but this time I have shaded the region of spacetime that the traveling twin's inertial frame regards as "during" the journey. This region is slanted on this diagram, because it is using a different frame's simultaneity criterion.

So these are the three regions of spacetime you are trying to patch together to make a single frame:

As you can see, the result doesn't cover all of spacetime, and does double-cover some regions (darker shaded). So the problem you have with this approach is that you've failed to count the time elapsed on Earth for the bit of the right hand red line that doesn't lie in a shaded region. That's why you are predicting that the Earth twin will be younger - you are only counting a part of their actual lifetime.

It is, of course, possible to construct non-inertial frames in a non-naive manner so that they do cover all the relevant regions. My favourite is Dolby and Gull, but plenty of others are possible. Note that the maths for using such non-inertial frames is significantly more complicated - for a start, the expression for the interval will not be so trivial.

 

[Will Flannery]

Here's a graphical explanation of that. First, here's a spacetime diagram drawn in the frame of the planets. The red lines are the two planets (the left hand one is the distant planet, the right hand one is Earth), and the green line shows the twin traveling from one to the other.
View attachment 264699
I've shaded in the regions of spacetime that this frame regards as "before" and "after" the journey. Below, I've drawn the same diagram, but this time I have shaded the region of spacetime that the traveling twin's inertial frame regards as "during" the journey. This region is slanted on this diagram, because it is using a different frame's simultaneity criterion.
View attachment 264700
So these are the three regions of spacetime you are trying to patch together to make a single frame:
View attachment 264701
As you can see, the result doesn't cover all of spacetime, and does double-cover some regions (darker shaded). So the problem you have with this approach is that you've failed to count the time elapsed on Earth for the bit of the right hand red line that doesn't lie in a shaded region. That's why you are predicting that the Earth twin will be younger - you are only counting a part of their actual lifetime.

It is, of course, possible to construct non-inertial frames in a non-naive manner so that they do cover all the relevant regions. My favourite is Dolby and Gull, but plenty of others are possible. Note that the maths for using such non-inertial frames is significantly more complicated - for a start, the expression for the interval will not be so trivial.

#1 - and I suppose that the frame for an object as it progresses through time has to cover all of spacetime, that is its now line has to sweep out all of spacetime? (that makes sense, but nothing is obvious to me in this game).

#2 - I do believe that the Lincoln vid was while not technically wrong it was misleading, on his insistence that acceleration did not account for the distinction in the twins perspectives. Yes? No? Or, maybe it is technically wrong.

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly on the inbound flight. Note that both these vids are relatively recent and have 1+ million views, no mean achievement for a physics vid I think. Yes? No?

 

[Ibix]

#1 - and I suppose that the frame for an object as it progresses through time has to cover all of spacetime, that is its now line has to sweep out all of spacetime? (that makes sense, but noth

It's possible to have coordinate systems that don't cover all of spacetime. But they must cover the region of interest once and only once for them to be useful for a given problem.

#2 - I do believe that the Lincoln vid was while not technically wrong it was misleading, on his insistence that acceleration did not account for the distinction in the twins perspect

Not sure without re-watching. It certainly seems to have mislead you.

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly on the inbound flight.

I haven't watched the video, but it is possible that they are referring to what you literally see. If you travel towards a clock, the Doppler effect makes it appear to tick faster and if you travel away it makes it appear to tick slower. If you subtract out the changing lightspeed delay, you will be left with time dilation, which is the same in both directions.

Note that this analysis shows another asymmetry. The traveling twin will see the apparent rate of the stay-at-home's clock switch from slow to fast half way through the journey when they turn around. The stay-at-home will see the traveller's clock tick slowly until they see the turnaround, which will be a long time aftef half way through the trip.

 

[Janus]

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly on the inbound flight. Note that both these vids are relatively recent and have 1+ million views, no mean achievement for a physics vid I think. Yes? No?

I'm pretty sure what is being considered here is what they would see due to relativistic Doppler shift. The "speed up" seen during the return leg is just due to the decreasing travel distance for the light. The traveling twin would still conclude that time was passing more slowly during the return leg.
I

 

[Mister T]

However, the acceleration/deceleration does make a big difference in which frame is considered inertial, so that was the problem in my original version, as I figured you should be able to analyze the motion in both frames ignoring the acceleration/deceleration.

Did you not read Post #35 where I showed that both of your "twins" are inertial? At the end, T whizzes past B and compares his clock to B's clock.

 

[Halc]

It is, of course, possible to construct non-inertial frames in a non-naive manner so that they do cover all the relevant regions. My favourite is Dolby and Gull, but plenty of others are possible.

As for choice of systems (hard to call them frames) that do cover all regions, and only once, my favorite is the one the popular media often uses: What you see is what's happening now.

So for instance, a headline of 'Betelgeuse is expected to go supernova next week' is more digestible by the public than 'Betelgeuse is expected to have gone supernova 640 years ago'.

So A and B (not twins) look at each other from 20 light year distance, and each looks 20 years younger than himself, so each considers the other to be 20 years younger.
When A is 40, he embarks on a trip to B at .8c, when B appears to be 20. During the 15 years of the trip (ship clock), B's age accelerates to 3x normal, so B is 45 years older, or 65 (20 + 45) at trip end, and A is 55 (40 + 15).
From B's perspective, A looks 20 years younger until B's apparent departure at A's age 60 when A appears 40. B then appears to age 3x normal for the 5 year duration of the trip. At arrival 5 years later, A has aged 15 years during the 5 apparent traveling years, and B has aged 5 more, so they're 55 and 65 respectively.

All nice and consistent, symmetrical even, and no white or double grey areas. No discontinuities of anybody's age just because they're not present at some acceleration event. The rule is: for any observer, all of local spacetime is foliated by his past light cone. It lacks commutativity: If A is 40, B is 20 to A, but A is not 40 to 20 year old B, but that jives with reality. B cannot communicate with A anyway without the 40 year round trip lag, just like talking to the Mars rover.

#1 - and I suppose that the frame for an object as it progresses through time has to cover all of spacetime

Actually, there's no such rule, but sometimes it's nice to keep it that way locally. For instance, the inertial frame of the solar system hardly covers all of spacetime. Only the local stuff, not things say beyond the radius of the observable universe.

 

[Will Flannery]

Did you not read Post #35 where I showed that both of your "twins" are inertial? At the end, T whizzes past B and compares his clock to B's clock.

There was no 'T' in my scenario. I do understand that simultaneity is frame dependent.

 

[kuruman]

In the twin paradox there is a time in which A and B are at the same age at the same location (Earth). That's the starting point. This scenario has B already on the distant planet before he is joined by A. Obviously, if the two traveled together to the planet, there would be no difference in their ages throughout the trip and thereafter.

Here is my simple thought: If the two travel separately, assuming that the two trips are identical in terms of accelerations and velocities, when the twins meet on the distant planet, they will have zero age difference relative to each other. The relative aging process should be time-invariant and cannot depend on departure delays.

@Will Flannery's mistake is in A's reasoning. Suppose we start with a triplet instead of twins. Sibling C stays on the Earth, B leaves first and is followed by A. When A and C receive word that B has reached his destination safely, they know that B is younger than both of them. When A reaches his destination safely, he knows that he has aged relative to C exactly the same way that B has. Therefore, he expects to see B as aged as he is but younger than C. Similarly, when he lands on the planet, B knows that he is younger than A and C and that he will age at the same rate as them thereafter. Thus, when B is joined by A on the planet, he expects him to be as aged as he is, with both of them younger than C.

 

[Will Flannery]

On the one hand what Lincoln is saying is trivial: every observer, regardless of their state of motion, has a valid rest frame. The question is what laws of physics apply in an arbitrary frame of reference?

In SR, we have the Lorentz transformation between inertial reference frames (this unifies TD, LC and RoS into one neat mathematical package). In the case of the twin paradox, A changes from the initial rest frame into a new rest frame. It's valid to do this. But, he/she must apply the Lorentz transformation. Any data that A had from the initial rest frame (e.g. the location and time at planet B), must be transformed to the new "space shuttle" frame. This transformation changes the time "now" at planet B into some future time. Thereafter, time dilation applies to clocks on planet B, but they were already well ahead when A joined the space shuttle. And, when you do the maths, A gets the same answer that $B$ got using the initial rest frame: A's clock shows less time.

I missed this when you posted it, but now I see that that's it exactly, and I think it applies to Lincoln's version of the twin paradox where the traveling twin begins his return trip by hopping from his inertial frame heading outward to an inertial frame heading back to the home planet. When he hops he changes the basis for the spacetime vector space, and therefore he needs to apply the appropriate coordinate transformation to his current measurements, and if he does that he will 'get the right answer' when he returns home and will not be surprised by the age difference. So it's not the acceleration/deceleration per se that creates the problem, the problem is that the acceleration/deceleration produces a change in velocity and hence a basis change for the traveler's measurements, and he must apply the appropriate coordinate transformation before proceeding. Sweet!

 

[Mister T]

There was no 'T' in my scenario.

That has no effect on the issue. I re-stated your one-way twin paradox using only inertial observers and participants.

I do understand that simultaneity is frame dependent.

So you understand why the time on T's clock doesn't match the time on B's clock when T passes B. That should resolve the issue you raised in your original post.

 

[Will Flannery]

In the London vid he says
"All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them?"

And this seems to me to be clearly wrong. If I can say that I am not moving then it naturally follows that I am not accelerating, as if I were accelerating I would have to be moving. However, whether or not an object is accelerating is not 'mathematical' property only, it is a physical property that is frame independent. If I can say that I am not accelerating then I can say that I'm the center of an inertial frame. Thus, if London is right every observer can say he is at the origin of an inertial frame. It doesn't get any more wrong than that. ... :).

 

[PeroK]

In the London vid he says
"All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them?"

And this seems to me to be clearly wrong. If I can say that I am not moving then it naturally follows that I am not accelerating, as if I were accelerating I would have to be moving.

Absolute (proper) acceleration is not the same as absolute motion. For example, those of us on the surface of the Earth are accelerating upwards from the force of the surface of the Earth. But, we can still consider ourselves legitimately at rest in our frame of reference - which is, therefore, not an inertial frame. Meanwhile, objects in freefall towards the Earth have no forces on them, hence no proper acceleration.

Proper acceleration implies you do not have an inertial rest frame, but it does not imply some sort of absolute motion. You can measure the absolute value of the acceleration, but you cannot measure any absolute value of velocity at any time.