I Twin Paradox: Who Is Right, A or B?

[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.
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[Herman Trivilino]

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!

 

[Herman Trivilino]

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.

 

[Nugatory]

If I can say that I am not accelerating then I can say that I'm the center of an inertial frame.

That is not correct. For a counterexample, consider me sitting at rest in my chair right now: It is very convenient to describe myself as at rest, choosing a frame in which my coordinate velocity and acceleration is zero. Yet that frame is not inertial; the proper acceleration is non-zero.

 

[Sagittarius A-Star]

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.

No. The following "hopping" is only choosen as "twin paradox" scenario to avoid calculus:

where the traveling twin begins his return trip by hopping from his inertial frame heading outward to an inertial frame heading back

A smooth turn-around of the traveling twin A with a finite acceleration towards the "stay at home"-twin B is also allowed. While the turn-around, twin B is aging in A's frame faster than twin A. Twin A can regard himself as beeing at rest.

The following animation of a Minkowski diagram demonstrates this:

https://www.geogebra.org/m/M3YVE6eT

The green dashed line, that is a tanget line to the world line of A, is A's ct'-axis.
The other green dashed line is A's x'-axis. It is also a "simultaneity-line" in A's frame.

Please move the red "Time" control button on the upper right corner from left to right, with constant speed on your screen.

Then you can observe in parallel, how fast the intersection between A's green "simultaneity-line" and B's black t-axsis is moving up. You will see, that B ist aging fastest in A's frame while A's smooth turn-around.

A's acceleration can be regarded as a sequence of infinitesimal small "hoppings" from one "co-moving" inertial frame to the next "co-moving" inertial frame.

 

[Sagittarius A-Star]

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?"

Here is another video. It shows the traveling twin at rest in his non-inertial frame:



(Remark: The title "Twin Paradox in General Relativity" is misleading, because the video shows a twin paradox in flat spacetime, but with "pseudo-gravitational time-dilation" in an accelerated frame. That is a topic of SR.)

 

[Ibix]

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.

This does not follow. The point is that there is no absolute meaning to "not moving". I have to say "not moving with respect to something", and if the something I pick is not inertial then I also need to accelerate.

There is an absolute meaning to "inertial" and "not inertial". But that isn't the same thing.

 

[Will Flannery]

The point is that there is no absolute meaning to "not moving".

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

 

[Ibix]

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

See the sentence after the one you quoted. Or @Nugatory's response.

 

[PeroK]

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

Have you studied uniform circular motion? The acceleration is centripetal (towards) the centre. But, there is no "motion" towards the centre. The acceleration is at right angles to the direction of motion .

The analogy isn't perfect, as you are moving in a circle, but never in the direction of the acceleration.

You are accelerating; you are changing your inertial reference frame. But, at any instant you will be at rest relative to some inertial observer. The key point is that you cannot give an absolute value to your state of motion. Perhaps forget about imprecise statements like "claim to be moving". The precise statement is that you may attribute no absolute velocity at any time. You may, however, attribute an absolute proper acceleration.

 

[kuruman]

Folks, tell me I am wrong if I am, but isn't it true that the twins at some point were together on Earth and were aged identically until B left? Isn't it also true that whatever affected the aging of B during his trip relative to A is time-invariant? In other words, B's aging process relative to A while traveling does not depend on whether B leaves for his trip today or a month later. The same applies to A. When A decides to leave, if his trip is identical in all respects to B's, he will age identically with B while traveling.

Obviously, the twins would have the same relative age if they traveled together. Unless delaying the departure of A somehow affects A's aging process differently from B's, when the twins are reunited on the planet they will have the same age.

I posted as much in #66, but there were no replies. If I am naively wrong, I would like to know. Thanks.

 

[PeroK]

Folks, tell me I am wrong if I am, but isn't it true that the twins at some point were together on Earth and were aged identically until B left? Isn't it also true that whatever affected the aging of B during his trip relative to A is time-invariant? In other words, B's aging process relative to A while traveling does not depend on whether B leaves for his trip today or a month later. The same applies to A. When A decides to leave, if his trip is identical in all respects to B's, he will age identically with B while traveling.

Obviously, the twins would have the same relative age if they traveled together. Unless delaying the departure of A somehow affects A's aging process differently from B's, when the twins are reunited on the planet they will have the same age.

I posted as much in #66, but there were no replies. If I am naively wrong, I would like to know. Thanks.

That's a good answer, but I think we are assuming that the twins have ended up on different planets after perhaps several space journeys and, by a process of Einstein clock synchronisation, established that, lo and behold, they are still approximately the same biological age in their common rest frame.

PS I have in the past tentatively suggested that using twins is a singularly unscientific way to measure time!

 

[Nugatory]

PS I have in the past tentatively suggested that using twins is a singularly unscientific way to measure time!

It is indeed - identically constructed clocks would be much better. But when constructing example for people unfamiliar with the concepts, biological aging feels more "real", maps better to what somene's notion of proper time ought to be.

 

[PeroK]

It is indeed - identically constructed clocks would be much better. But when constructing example for people unfamiliar with the concepts, biological aging feels more "real", maps better to what somene's notion of proper time ought to be.

I know my views are not shared, but when I was learning SR I decided to clear all the extraneous material away: twins, pole vaulters and barns, lightning strikes. Everything I considered "garbage". Not only did I consider these things extraneous, but that they were designed to deceive, cloud the mind and obscure the physics.

We have events, particles, measurements of position and time, reference frames, clocks and metre sticks. What austerity!

 

[kuruman]

That's a good answer, but I think we are assuming that the twins have ended up on different planets after perhaps several space journeys and, by a process of Einstein clock synchronisation, established that, lo and behold, they are still approximately the same biological age in their common rest frame.

PS I have in the past tentatively suggested that using twins is a singularly unscientific way to measure time!

Thank you for your response. The discussion may have digressed to different planets and multiple journeys, but the original post states that "A gets on a rocket and travels to B's planet."

I agree with your sentiments about the twin paradox. IMO it is as scientific as Schrodinger's cat.

 

[PeterDonis]

surely if I'm accelerating I cannot claim to be not moving

As you sit at your computer typing your posts, you are accelerating, because you are on the surface of the Earth and you feel weight. Are you moving?

 

[Shirish]

I'll take a stab at explaining this through spacetime diagrams. Other members can feel free to correct if there are some things I got wrong.

The thick axes correspond to $A$'s planet's frame, and the orthogonal axes on the right are those of $B$'s planet. The above spacetime diagram is from $B$'s perspective. The origin clocks on both planets are synchronized. But suddenly $A$ jumps into the rocket, and you can see that $t=0$ for the rocket coincides with $t=t_0$ for the planet $B$ origin clock in the rocket frame. You see that the time elapsed in the journey in $B$'s frame is $T$.

The demarcations on the time axes represent "ticks" on the clocks in the respective frames. $B$ sees that the rocket's time axis demarcations are larger, i.e. the clock ticks in the rocket take longer, i.e. $B$ will say that fewer ticks were counted by the rocket frame clock.

When $A$ stops on planet $B$, even though $A$'s clock switches its ticking rate to what $B$'s clock's ticking rate is, that still doesn't change the fact that fewer ticks were counted by $A$'s clock so far from $B$'s planet's perspective, which is where they're finally comparing times.

Why can't we reverse this argument and switch roles? Let's look at another ST diagram from the rocket's perspective:

Remember we'd concluded that $t=0$ for the rocket coincides with $t=t_0$ for the planet $B$ origin clock? This is what we're trying to account for above and that's why planet $B$'s spacetime diagram is shifted below. So now $t_0$ of $B$ planet's origin clock coincides with $0$ on the rocket clock, which is what we want as this ST diagram is from the rocket frame perspective.

In terms of aging, $B$ has a headstart of $t_{\text{extra}}$ on $A$. So even though $B$'s clock ticks slower from $A$'s perspective, some time has already elapsed on $B$'s clock which makes up for it.

Hope that helps!

 

[Dale]

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

Sure you can. You just have to use a non inertial frame.

 

[Will Flannery]

I'm working on a complete and definitive deconstruction of the twin paradox, accompanying it with space time diagrams and the Lorentz transforms for each of the frames involved. And in this process I've hit a snag ...
# I start with frame A stationary, frame B moving with velocity V = .6, giving a gamma of 1.25. The arrival at the distant planet is at (10, 6) in the A frame and (0, 8) in the B frame.
#When the moving twin reaches the distant planet we go from frame B stationary to frame A' moving with velocity V=-.6. The departure from the near planet is at (-10,-6) in the A' frame. All is well so far.
#Now we go from frame A' stationary to frame B' moving at velocity V =-.6 for the return trip. However, applying the Lorentz transform, I get the location of the departure from the near planet at (-

 

[Dale]

Looks like you are missing some text. Are your coordinates (t,x) or (x,t)? Do your frames all share an origin? I am a little confused about the primed frames.

 

[Will Flannery]

Looks like you are missing some text. Are your coordinates (t,x) or (x,t)? Do your frames all share an origin? I am a little confused about the primed frames.

I didn't mean to post the above, hit the wrong button. I'm still working on it, there is a conceptual error lurking somewhere in there and I haven't given up on it yet ... :)

 

[PeroK]

I didn't mean to post the above, hit the wrong button. I'm still working on it, there is a conceptual error lurking somewhere in there and I haven't given up on it yet ... :)

Let me give you a hand. Consider a journey from $A$ to $B$ in two frames. The first frame is the rest frame of $A, B$ and the second is a frame moving at speed $v$ from $A$ to $B$.

We'll take the common spacetime origin to be the departure event at $A$, event $E_0$; and the second event to be the arrival at $B$: $E_1$:

In the first frame we have, with the time coordinate first:
$$E_0 = (0, 0), \ \ E_1 = (t, x) = (\frac d v, d)$$
And, in the second (primed) frame we have:
$$E'_0 = (0, 0), \ \ E'_1 = (t', x') = (\gamma(t - \frac v {c^2} x), \gamma(x - vt)) = (\gamma(\frac d v - \frac{vd}{c^2}), 0)$$
That looks good: the second event also takes place at the $x' = 0$ and we have:
$$t' = \frac{\gamma d}{v}(1 - \frac{v^2}{c^2}) = \frac{d}{\gamma v} = \frac t {\gamma}$$

 

[Will Flannery]

I'm working on a complete and definitive deconstruction of the twin paradox, accompanying it with space time diagrams and the Lorentz transforms for each of the frames involved. Down to business -

# I start with frame A stationary, frame B moving with velocity V = .6, giving a gamma of 1.25. The arrival at the distant planet is at (10, 6) in the A frame and (0, 8) in the B frame.

The relevant Lorentz transforms are from the Stationary A to Moving B frames :
SMT(10,6,V) = 8 % A to B time of arrival event
SMX(10,6,V) = 0 % A to B location of arrival event

#When the moving twin reaches the distant planet we go from the moving frame B to a stationary to frame A' moving with velocity -V. The A' frame is the same as the A frame with the origin at (10,6). The departure from the near planet is at (-10,-6) in the A' frame. All is well so far.

#Now we go from frame A' stationary to frame B' moving at velocity -V for the return trip. However, applying the Lorentz transform, I get the location of the departure from the near planet at (-17,-15) in the B' frame ! Of course we don't want to go back to that event, but still this isn't right.

Well, it is right after all. The space time diagram for the return flight, showing the A' frame and the B' frame, is

The Lorentz transforms are:
SMT(-10, -6, -V) = -17
SMX(-10, -6, -V) = -15 % this was a big surprise to me, and inexplicable without the diagram
The original departure event is at (-17, -15) in the B' frame, and B' traveler timeline intersects the world line for the near planet at (8,0) in the B' frame.

So we have the travelers analysis of the trip, with a total trip time of 16 years, while the twin on the near planet ages 20 years.

All that is missing is a discussion of the asymmetry of the twins journeys which has been discussed. I've just added the spacetime diagrams, but I think that's an important addition towards making everything explicit.
Notes:
SMT = @(t,x,V) gamma*(t - x*V); % Stationary to moving Lorentz transform
SMX = @(t,x,V) gamma*(x - t*V);

 

[Ibix]

This seems overly complex. Why not just write down the coordinates of the events in the rest frame of the planets (departure is $t=0,x=0$, arrival is $t=10,x=6$, return is $t=20,x=0$) and then use the Lorentz transforms to get the coordinates in other frames? Why are you moving the origin about? That just confuses things and doesn't add anything.

Also note that the usual convention is to draw Minkowski diagrams with the time axis vertical.

 

[robphy]

I'm working on a complete and definitive deconstruction of the twin paradox, accompanying it with space time diagrams and the Lorentz transforms for each of the frames involved. Down to business -

[snip]

So we have the travelers analysis of the trip, with a total trip time of 16 years, while the twin on the near planet ages 20 years.

Sorry, I haven't been following closely
but...
Is this still the "one-way" twin paradox posed in the OP and in the title?
Or, is this now the standard twin paradox? with the one-way variation now abandoned?

 

[Will Flannery]

This seems overly complex. Why not just write down the coordinates of the events in the rest frame of the planets (departure is $t=0,x=0$, arrival is $t=10,x=6$, return is $t=20,x=0$) and then use the Lorentz transforms to get the coordinates in other frames? Why are you moving the origin about? That just confuses things and doesn't add anything.

Also note that the usual convention is to draw Minkowski diagrams with the time axis vertical.

I think my version is as simple as it can be and still cover the necessary considerations, and it does have the advantage that all the steps are detailed. But, maybe I'm wrong. If you 'filled in the details' of your approach I could get a better idea. And, generally, I've found that I think I understand something, and then set out to do the spacetime diagrams, and I find out my understanding is not as good as I thought. I think space time diagrams are am essential heuristic aid. I know I got the convention wrong, but I started wrong, tried to change, but couldn't.

 

[Will Flannery]

Sorry, I haven't been following closely
but...
Is this still the "one-way" twin paradox posed in the OP and in the title?
Or, is this now the standard twin paradox? with the one-way variation now abandoned?

Yes, I abandoned the one-way paradox long ago, after about my 3rd post.

 

[Ibix]

I think my version is as simple as it can be and still cover the necessary considerations,

I don't understand why you moved origins. That's unnecessary and confusing. If you want your origin to be the turnaround event, why not just have the turnaround occur at (0,0) in all frames?

If you 'filled in the details' of your approach I could get a better idea.

What is there to fill in? Once you've specified the coordinates of the key events, you just need to transform them into whatever frames you wish to work in. It's just far easier to specify the coordinates of all three events in the rest frame of the Earth - you can do it pretty much by inspection. If we change to using the turnaround as the origin then by simple arithmetic the coordinates are:

• Departure: $t=-10,x=6$
• Turnaround: $t=0, x=0$
• Return: $t=10, x=6$

Then you can use the Lorentz transforms to get them in the outbound rest frame:

• Departure: $t'=-8, x'=0$
• Turnaround: $t'=0, x'=0$
• Return: $t'=17, x'=15$

Likewise you can get the coordinates in the inbound frame - given the above, the answers are obvious from symmetry. The large values for the return coordinates in the outbound frame are easy to explain. In this frame the Earth is moving at 0.6c in the +x direction and has a sizeable headstart over the return leg, which is only done at 0.88c, not a huge velocity difference.

I think space time diagrams are am essential heuristic aid.

I agree. I don't understand your diagram, though, since the three black lines, which I take to be the worldlines of the traveling and Earthbound twins don't meet. They should form a triangle, isosceles in the Earth frame.

I know I got the convention wrong, but I started wrong, tried to change, but couldn't.

I'd note that using a non-standard convention will put you at a significant disadvantage in trying to understand other people's diagrams. If you truly find switching axes challenging, I strongly suggest you put in the effort to overcome it or you will struggle with interpreting almost every spacetime diagram you encounter.

 

[Will Flannery]

I don't understand why you moved origins. That's unnecessary and confusing. If you want your origin to be the turnaround event, why not just have the turnaround occur at (0,0) in all frames?

What is there to fill in? Once you've specified the coordinates of the key events, you just need to transform them into whatever frames you wish to work in. It's just far easier to specify the coordinates of all three events in the rest frame of the Earth - you can do it pretty much by inspection. If we change to using the turnaround as the origin then by simple arithmetic the coordinates are:

• Departure: $t=-10,x=6$
• Turnaround: $t=0, x=0$
• Return: $t=10, x=6$

Then you can use the Lorentz transforms to get them in the outbound rest frame:

• Departure: $t'=-8, x'=0$
• Turnaround: $t'=0, x'=0$
• Return: $t'=17, x'=15$

Likewise you can get the coordinates in the inbound frame - given the above, the answers are obvious from symmetry. The large values for the return coordinates in the outbound frame are easy to explain. In this frame the Earth is moving at 0.6c in the +x direction and has a sizeable headstart over the return leg, which is only done at 0.88c, not a huge velocity difference.

I agree. I don't understand your diagram, though, since the three black lines, which I take to be the worldlines of the traveling and Earthbound twins don't meet. They should form a triangle, isosceles in the Earth frame.

I'd note that using a non-standard convention will put you at a significant disadvantage in trying to understand other people's diagrams. If you truly find switching axes challenging, I strongly suggest you put in the effort to overcome it or you will struggle with interpreting almost every spacetime diagram you encounter.

# The origins followed the traveling twin, which seems like the natural thing for me. He was changing frames, he was at the origin of his frame, so the new frame should have the same origin. The origins of all the frame changes is (0, 0) in the traveler's frame, and the new frame has the same origin. I assumed that was the norm.

# "What is there to fill in?" Believe it or not, I cannot follow your method at all. Maybe if I studied it a lot, but, I'm happy with my method. :) ... All this stuff seems very confusing to a newbie, I think, and I spent 10 years doing inertial nav for rockets !

# I didn't provide sufficient explanation. I'll explain the three black lines in more detail. The spacetime diagram is when the traveler changes frames when he starts the trip home. I thought there was an error when I saw that the coordinates of the original departure from the near planet were (-17, -15) in the new frame. I thought it had to be a mistake and after much gnashing of teeth I decided to tweak my spacetime diagram program to handle negative velocities so I could draw the diagram. And after a while I was able to draw the diagram and saw that the (-17,-15) was right ! So, that's the only purpose of the lines labeled -15 and -17 in the diagram. The line labeled 8 is the traveler's world line on the return trip and it intersects the near planet's worldline after 8 years in the traveler's frame.

 

[Ibix]

The origins followed the traveling twin, which seems like the natural thing for me.

Note that the frame origins are events - they can't follow anyone. In his rest frame(s) the traveller's spatial coordinates are constant and you can choose his location to be the spatial origin, yes. My point was thhat if you pick the origin event of all frames to be the turnaround event then you get the traveller at the spatial origin on both legs of the journey for free, without having to introduce a fourth frame (the one you called A') that's the same as another frame (the one you called A) except for a translation.

"What is there to fill in?" Believe it or not, I cannot follow your method at all.

I did the same thing you did, but with a better choice of origin to achieve the "traveller at rest at the spatial origin" effect you wanted. I used $x=x_0+vt$ to write down the coordinates of the three important events (departure, turnaround and return) for a ten year journey at -0.6c and another ten year journey at +0.6c given an instantaneous turnaround at the origin. Then I used the Lorentz transforms to get the coordinates in the frame where the traveller is at rest on the outbound leg. I didn't do the calculation for the other leg - you can do so explicitly or, as I suggested, just cheat because this problem is highly symmetric.

On closer inspection, I see that this is in fact a description of the second spacetime diagram in your post #90. So all I'm saying is that your first diagram is unnecessary and confusing. You also haven't drawn the journey in any other frame, only the Earth twin's frame. You may find it enlightening to draw those frames' versions of events.

The spacetime diagram is when the traveler changes frames when he starts the trip home.

As written, this doesn't make sense. A spacetime diagram covers - in principle - all of spacetime, so "when" is meaningless applied to the diagram. I think you mean that the line marked 8 is (approximately, since it doesn't connect the dots marking the events) the traveller's worldline on the return leg. The other two lines don't make sense in that context. The 15 and 17 are coordinate differences in the traveller's frame, so lines referring to them ought to be parallel to the traveller's frame's coordinate grid, which they manifestly are not.

I thought there was an error when I saw that the coordinates of the original departure from the near planet were (-17, -15) in the new frame

As I attempted to explain in my last post this is because, in the traveller's rest frame for the second part of the journey, the Earth is moving towards him at 0.6c and the first leg of the journey was at speed 0.88c, not that much faster. So the departure event has to be a long time in the past for the first leg to have built up enough of a lead over the Earth that it takes eight years sitting still for the Earth to catch up.

I note that I actually explained it last time with reference to the second leg as seen from the first leg rest frame, so the explanation was the other way around.

 

[Nugatory]

He was changing frames, he was at the origin of his frame, so the new frame should have the same origin. The origins of all the frame changes is (0, 0) in the traveler's frame, and the new frame has the same origin. I assumed that was the norm.

i do not believe it is possible to write a clear and complete description while using phrases like “He was changing frames”, and similar. The traveller isn’t switching frames, you are in your description of the problem.

There is an inertial frame in which the Earth is at rest. There is an inertial frame in which the Earth is moving to the left with a constant speed $v$ and the traveller is at rest until the turnaround. There is an inertial frame in which the Earth is moving to the right with constant speed $v$ and the traveller is at rest after the turnaround.

We can use any of these frames to describe the situation and calculate the elapsed time between separation event and reunion event for both twins, and we’ll get the same answer. We can use any of these frames along with the Doppler effect to describe what the twins see when they watch each other’s clocks through a telescope and we’ll get an intuitive description of their experience throughout.

That’s the clear way. There’s also an unclear way, which is what you’re trying to do: Use one frame for the first half of the analysis and a second frame for the second half. If so we are the ones changing frames, not the traveller - the traveller is “in” all frames throughout. This approach is complicated and error-prone and if we are not careful with the relativity of simultaneity will lead to results that appear paradoxical.

It is worth noting that when Einstein introduced differential aging in the 1905 paper, he did so without using the language of changing frames... and was able to dismiss the subject in a few sentences without any paradox to explain.

 

[PeroK]

Just to re-emphaise that the one-way journey was analysed, for an arbitrary speed, with a few lines of algebra in post #89. It was only one application of the Lorentz Transformation.

 

[Will Flannery]

Note that the frame origins are events - they can't follow anyone. In his rest frame(s) the traveller's spatial coordinates are constant and you can choose his location to be the spatial origin, yes. My point was thhat if you pick the origin event of all frames to be the turnaround event then you get the traveller at the spatial origin on both legs of the journey for free, without having to introduce a fourth frame (the one you called A') that's the same as another frame (the one you called A) except for a translation.

I did the same thing you did, but with a better choice of origin to achieve the "traveller at rest at the spatial origin" effect you wanted. I used $x=x_0+vt$ to write down the coordinates of the three important events (departure, turnaround and return) for a ten year journey at -0.6c and another ten year journey at +0.6c given an instantaneous turnaround at the origin. Then I used the Lorentz transforms to get the coordinates in the frame where the traveller is at rest on the outbound leg. I didn't do the calculation for the other leg - you can do so explicitly or, as I suggested, just cheat because this problem is highly symmetric.

On closer inspection, I see that this is in fact a description of the second spacetime diagram in your post #90. So all I'm saying is that your first diagram is unnecessary and confusing. You also haven't drawn the journey in any other frame, only the Earth twin's frame. You may find it enlightening to draw those frames' versions of events.

As written, this doesn't make sense. A spacetime diagram covers - in principle - all of spacetime, so "when" is meaningless applied to the diagram. I think you mean that the line marked 8 is (approximately, since it doesn't connect the dots marking the events) the traveller's worldline on the return leg. The other two lines don't make sense in that context. The 15 and 17 are coordinate differences in the traveller's frame, so lines referring to them ought to be parallel to the traveller's frame's coordinate grid, which they manifestly are not.

As I attempted to explain in my last post this is because, in the traveller's rest frame for the second part of the journey, the Earth is moving towards him at 0.6c and the first leg of the journey was at speed 0.88c, not that much faster. So the departure event has to be a long time in the past for the first leg to have built up enough of a lead over the Earth that it takes eight years sitting still for the Earth to catch up.

I note that I actually explained it last time with reference to the second leg as seen from the first leg rest frame, so the explanation was the other way around.

# "if you pick the origin event of all frames to be the turnaround event"
-That is exactly what I did !

#I didn't do the calculation for the other leg - you can do so explicitly or, as I suggested, just cheat because this problem is highly symmetric.
-The purpose of my explanation was to analyze the trip from the traveler's perspective, and in particular the turnaround at the distant planet where he has to reverse direction, in London's vid he jumps from one inertial frame to another. That was the focus, but if you try to do the Lorentz transform for that jump, the velocity is - v + (-v) = -2v which is greater than c and I think the Lorentz transform fails or doesn't make sense for that case, so I had the traveler jump to a frame at traveling at -v to put him back in the original A frame but with a different spatial origin, and then hop to a frame traveling at -v to get him heading back to the original planet. I see all this as necessary ... :)

#The 15 and 17 are coordinate differences in the traveller's frame, so lines referring to them ought to be parallel to the traveller's frame's coordinate grid, which they manifestly are not.
-You're right. I've fixed it ... sort of ... and it does clarify what's going on ...

# You also haven't drawn the journey in any other frame, only the Earth twin's frame. You may find it enlightening to draw those frames' versions of events.
- Each spacetime diagram shows two frames, in the first the Earth frame and the frame for the traveler's outward journey, the 2 for the translated Earth frame and the traveler's home journey. In the traveler's frame the spatial coordinate of the near and distant planets is always 0, so drawn separately ... would that be interesting ?