#57805 Twin Paradox

[vorpal blade]

I apologize, Vorpal. I'd not seen your previous post and was still under the assumption that we'd worked out the SR equations and you'd wanted to run it again with acceleration. Without crunching all the numbers, I'd guess where it seems the numbers fall out of sync is the assumption that t"=0, t'=0, and t=0 are simultaneous. Remember, different inertial reference frames will disagree on what is simultaneous.

We are free to set up the frames of reference and label the origins as we choose. Choosing a spatial origin on the earth for each reference frame, and agreeing on a clock setting of zero at one location and one time at that one location violates no principle of simultaneity. This is so basic to how the Lorentz transformations work that I am at a loss to explain it more clearly. I thought everyone acknowledged that the reference frame labels and coordinate markings were arbitrary, for mere convenience. I think you are going to have to dig a lot deeper to find fault with my analysis, if you can.

Different inertial reference frames disagree on simultaneity only when separated in space. Besides that, I'm basically using the same assumption on the origins as you are, just adding in the third reference frame.

Do I need more proof of my freedom to choose the 4-space origin?

[Tao]

Mmmkay, once again, hopefully with less screw-ups on my part: stationary observer regardless of local records time t, velocity v, and distance between space station and earth on x axis x. He records his observations of objects in motion as seeing time t’, velocity v, and distance x’.

Again t’=γ(t-vx/c²) and x’= γ(x-vt)

where γ = 1/√1-(v²/c²)

What the people on Earth see: at t₀, the outbound ship syncs with them, so t₀’=0 ship flies by at speed (v).8c, while the return ship is 2 ly away (x=2) approaching them at .8c. In 1.25 years (∆x/v=t₁) it passes the space station 1 ly out (x=1), noticing that the trip took .75 years for the ship. (t₁’=.75) In another 1.25 years (∆x/v=t₂) the return spaceship flys by and beams down their twin, having only taken had their passenger .75 years. They see their twin as being 2.5 yo, while the voyager remains 18 months.

What the outbound spaceship sees: at t₀=0 clocks on board and on earth sync as it passes them (x=0) at .8c. The space station is .6 ly away (x= 0.6) is approaching at the same velocity. However they notice that the time on the space station is t₀’=.8 As they reach the space station, their own clock reads that 0.75 years have passed (∆x/v=t₁), they show the space station’s reference frame as being dilated, leaving only 0.45 years gone by (t₁’)

What the return spaceship sees: at t₀=0 earth is -1.2 ly away at x=0 with a calendar reading t₀’=1.6 approaching at v. At t₁=.75 they pass a space station and another ship, beaming aboard an infant. They note that Earth's calendar reads t₁’= 2.05. At t₂= 1.5 they pass Earth, beaming down their bouncing baby with an earth time of t₂’=2.5

Thus, according to an earth observer, setting t₀=0=Jan 1 3001 their calendar and that of the space station reads t₀+ t₁=April 3002 (0+1.25=1.25) when the baby changes ships. But they observe the outgoing ship's calendar as reading t₀’+ t₁’ Oct 3001. (0 +.75=.75) Upon the arrival of the returning ship, their calendar reads July 1, 2003 (t₂=2.5) yet the traveling infant is 18 months. (.75 years outgoing, .75 returning)

The observers on either spaceship see that they their baby's time when transferred was, t₀+ t₁= Oct 3001 (0 + .75=.75), while they both note the space station's calendar reads t₀’+ t₁’ April 3002.{sorry about the earlier typo} (.8 + .45=1.25) {Figured following the baby's time would be easiest, correct me if I'm wrong}

The observers on the return spaceship saw their baby back at earth .75 years later (July 1, 3002 baby-time), noting that another .45 years have passed on earth, making the date of their arrival July 1, 3003 Earth-time. (t₀’+ t₁’+ t₂’= 1.6+ .45 + .45= 2.5 years)

Thus, each observer would agree, the infant that remained on earth has seen 2.5 years from Jan 1 3001, while the infant that went for the ride only aged 1.5 years from that time.

The reason for the addition of dates is for the calculation of what the calendar reads. t₁, t₁’, t₂, t₂’, etc, are already calculated as elapsed time, so adding them to original time (t₀’, t₀) gives us what the date would read for either the stationary or moving frame.

[Tao]

Different inertial reference frames disagree on simultaneity only when separated in space. Besides that, I'm basically using the same assumption on the origins as you are, just adding in the third reference frame.

Do I need more proof of my freedom to choose the 4-space origin?

Forgive my lack of clarity again. The setting of the origin is fine, you can put you origin where the Earth is at the beginning of the trip, as long as you express that it (the earth) is moving from x=0 to x=1.2 in the returning space ships reference frame. Which you did, your elapsed time figures look perfect. Where I'm wondering the confusion is is here:

II. Now let’s look at it from the reference frame of the incoming space ship.

A. At t=0 earth appears to be at x=0 and the return space ship at x=6/5. Observers in the return ship’s reference frame see the clock on earth to be

t0’’(x=0, t=0,v=4/5c) = (5/3)(0 – (4/5)(0) = 0

At this point, there are light years between the returning space ship and the earth. Simultaneity should crop up here. What I had meant by seeing t"=0, t'=0, and t=0 as simultaneous is that Jan 1 3001 does not occur at the same moment for both the Earth and the returning ship. Comparing calendars requires working in the fact that one frame is moving at 4/5c relative to the other and is either 1.2 or 2 ly away depending on which frame of reference you are considering.

By my calculations, this means the returning space ship see's earth's date as mid august 3002 when the outbound ship leaves. t’ = γ(t-vx/c²) t’ = (5/3)(0 - (4/5)(6/5)) = 8/5 years difference. Earth would see the ship's calendar read t’ = γ(t-vx/c²) t’ = (5/3)(0 - (4/5)(2)) = 8/3 years different then their Jan 1 3001 lift-off date.

[vorpal blade]

I'm not sure I have found all the places where I disagree with you, but I have found a few.

What the return spaceship sees: at t₀=0 earth is -1.2 ly away at x=0 with a calendar reading t₀’=1.6 approaching at v. At t₁=.75 they pass a space station and another ship, beaming aboard an infant. They note that Earth's calendar reads t₁’= 2.05. At t₂= 1.5 they pass Earth, beaming down their bouncing baby with an earth time of t₂’=2.5

I believe the correct calculations are (as I showed before) using your notation;
t₀’= t0’’(x=0, t=0,v=4/5c) = (5/3)(0 – (4/5)(0) = 0
t₁’= t1’’ (x=3/5, t=3/4, v=4/5c) = (5/3)(3/4 - (3/5)(4/5)) = 9/20 y.
t₂’= t2’(x=6/5, t=3/2, v=4/5c) = 5/3(3/2 – (4/5)(6/5)) = 9/10

The reason for the addition of dates is for the calculation of what the calendar reads. t₁, t₁’, t₂, t₂’, etc, are already calculated as elapsed time, so adding them to original time (t₀’, t₀) gives us what the date would read for either the stationary or moving frame.

This is not correct. The values of t are not calculated elapsed time, they are clock readings. To calculate elapsed time you must take the difference in clock readings. These values of t really depend solely on the way you set up the origin. I can give you arbitrarily large values of t. What is really important is the elapsed time differences. I was a little unclear on that before.

Where I'm wondering the confusion is is here:

II. Now let’s look at it from the reference frame of the incoming space ship.

A. At t=0 earth appears to be at x=0 and the return space ship at x=6/5. Observers in the return ship’s reference frame see the clock on earth to be

t0’’(x=0, t=0,v=4/5c) = (5/3)(0 – (4/5)(0) = 0

At this point, there are light years between the returning space ship and the earth. Simultaneity should crop up here. What I had meant by seeing t"=0, t'=0, and t=0 as simultaneous is that Jan 1 3001 does not occur at the same moment for both the Earth and the returning ship. Comparing calendars requires working in the fact that one frame is moving at 4/5c relative to the other and is either 1.2 or 2 ly away depending on which frame of reference you are considering.

By my calculations, this means the returning space ship see's earth's date as mid august 3002 when the outbound ship leaves. t’ = γ(t-vx/c²) t’ = (5/3)(0 - (4/5)(6/5)) = 8/5 years difference. Earth would see the ship's calendar read t’ = γ(t-vx/c²) t’ = (5/3)(0 - (4/5)(2)) = 8/3 years different then their Jan 1 3001 lift-off date.

This isn't correct. I think what you have to keep in mind is that in any given reference frame we may suppose that there are many observers located throughout that space in that reference frame. Clocks can be synchronized in a particular reference frame so all the clocks within that frame show t=0 at the same time. Einstein suggested a way to do this. What you can't do is talk about simultaneity in different locations in different reference frames. We previously discussed the synchronization of clocks on the space stations and on earth. It is possible to do this with supposed observers all along the way in the frame of reference of the returning ship, including one near the earth at time t=0.

[Tao]

Aye, you can have synchronicity among all clocks in one reference frame, but you cannot have it in clocks that are in two different reference frames and separated by space. Makes sense if you think of time as hash marks on grid paper: one sheet will always be consistent with itself, but won't always line up with another. Relativistic time dilation makes it like comparing 1 in grid paper to 1 cm paper: each is consistent within itself, and you line up origins to make any one point be in sync, but all other points will differ, at a measurable rate: t’=γ(t-vx/c²)

[vorpal blade]

Aye, you can have synchronicity among all clocks in one reference frame, but you cannot have it in clocks that are in two different reference frames and separated by space. Makes sense if you think of time as hash marks on grid paper: one sheet will always be consistent with itself, but won't always line up with another. Relativistic time dilation makes it like comparing 1 in grid paper to 1 cm paper: each is consistent within itself, and you line up origins to make any one point be in sync, but all other points will differ, at a measurable rate: t’=γ(t-vx/c²)

Yep, that is what I am saying. And my calculations have been consistent with that principle. We start with the earth as the zero point, and at a certain time (when the outgoing spaceship passes the earth) all of the clocks in each reference frame are synchronized within that frame, but corresponding clocks at points away from earth will not agree even though they may be located at the same point in space. I believe that I have been consistent in applying this rule, if you will read carefully my post.

Perhaps you did not notice that at time t=0 the incoming spaceship finds itself at coordinate x=1.2, in it's own inertial reference frame. Clearly at t=0, x=0 (the location of earth) the coordinate transform for t' for earth has to be zero, not 1.6 as you have given it. I don't see that you have a valid objection to my calculations.

[Tao]

That's fine then. All we would have to do then is be sure work in the time discrepancy on the end of the returning spaceship. In my mind I've always preferred to put my stationary point at the origin, so I tend to work that way, but it all works out either way.

Running the the Lorentz shift on the spaceships' end of things will still tack on that 1.6 years, answering the question:

How is it possible for the return ship to read the clock on earth to be 9/10 y, still the year 3001, while those on earth see the same clock, in the same location, at the same time to be 2.5 years, or mid-year 3003?

Note that 2.5 - 9/10 = 1.6 years.

[vorpal blade]

In the reference frame of the returning spaceship all points in that inertial reference frame are stationary.

We may assume that in the reference frame of the returning spaceship there is an observer opposite the earth at the time that the outgoing spaceship passes earth, or time t=0. The clocks in the inertial time frame of the incoming spaceships are all synchronized to t=0 in that reference frame. Therefore the clock on the incoming spaceship reads t=0 as does the clock of the observer near the earth in that time frame, as does the clock on the earth. There is no problem with simultaneity in this. I am not saying that those on the spaceship will read t=0 on any clocks in the inertial frame of earth, except the clock on earth. And every observer, no matter what position it finds itself in, will also read the clock on earth to read t=0 at the moment the outgoing spaceship passes earth.

There is no value of 1.6 years to tack on anywhere. The incoming spaceship will read the clock on earth to be zero when the outgoing spaceship passes the earth. That incoming spaceship will also read zero on it's own clock, because it's own clock will be synchronized with the observer near earth in the incoming spaceship's reference frame. More importantly, as time goes on the observers in the incoming spaceship will see the clock on the earth smoothly go from t=0 to t=9/10 years when it passes the earth.

That is just what the equations tell us.

[Tao]

I'm not sure we agree on relativistic simultaneity. How do you understand the principle of simultaneity to work?

The clocks in the inertial time frame of the incoming spaceships are all synchronized to t=0 in that reference frame. Therefore the clock on the incoming spaceship reads t=0 as does the clock of the observer near the earth in that time frame, as does the clock on the earth. There is no problem with simultaneity in this.

This is nearly a textbook example of simultaneity as I understand it. Events that are local in space are simultaneous in all local reference frames, so the clocks on the outbound spaceship and Earth will both sync up at their passing. The return spaceship is lightyears away. It will not agree as to the time at which the Earth's clocks state the outbound ship left.

I am not saying that those on the spaceship will read t=0 on any clocks in the inertial frame of earth, except the clock on earth.

In effect, you are, but in an obscure way, due to the choice of setting earth's original location as your origin. While we can arbitrarily set t=0 at any point we wish in any reference frame, that becomes the only place where the timelines mesh. Thus when you compare clocks when the earth reaches the returning spaceship (remember it is earth moving away from x=0, you are going to be lightyears away from that original meshpoint, even though it was on earth originally) you'll need to do a Lorentz shift on time. To do otherwise would be akin to arguing that that geometry is inconsistent because when lining up the first hash on different graph papers gives different results than lining up the last one.

In fact, your mention of an observer at rest with the returning ship may be the best way to show that simultanity needs to be taken into account. Where do you place such a ship?

Say we look at the set up and decide at t=0 earth and outbound pass and return is at 2 ly away, (1.2 in it's own frame) so if we put the new ship at -1.2 lightyears away from the return ship on the x axis, it would thus be at earth when the outbound spaceship passes, right? Nope. While it would be at earth, the outbound ship would be nowhere in sight. The people on earth would see only see .72 lightyears between this new ship and the return ship (length contraction).

So, lets put this new ship in the same moment and locale as the outbound ship's passing of Earth. In the Earth's reference frame, the time is t=0, the local ships are present, and the return ship is 2 light years away. In the new ship's reference frame, the outbound ship and earth is present, the time is t=0 and the return ship is 3 1/3 light years away. (2 ly in earth's {moving} frame of reference is 3 1/3 in the stationary {new/return spaceships} frame) How long until the Earth reaches the return ship? (3 1/3 ly)/(4/5c)= 4 1/6 years or 2.5 years measured from Earth (4 1/6)/(5/3) = 2.5 Which is what we had already anticipated, so our numbers check out.

But if the return ship is at 3 1/3 ly away when the new ship in its reference frame syncs with the outbound ship and earth, then t=0 can't occur when the return ship is 1.2 ly (or 6/5 ly) out. Lets do some number crunching on that extra space, shall we? 3 1/3 - 6/5 = 32/15 ly which at 4/5c would take 8/3 years to cross. This time in earth's reckoning, (8/3) / (5/3)= 24/15 or 1.6 years. Which checks with what we got by simply doing the Lorentz shift on time to account for simultaneity.

[Tao]

And sorry, NerdGirl, for effectively doubling your reading here.... but I would like to think that progress toward mutual understanding is being made.

[vorpal blade]

I'm not sure we agree on relativistic simultaneity. How do you understand the principle of simultaneity to work?

The clocks in the inertial time frame of the incoming spaceships are all synchronized to t=0 in that reference frame. Therefore the clock on the incoming spaceship reads t=0 as does the clock of the observer near the earth in that time frame, as does the clock on the earth. There is no problem with simultaneity in this.

This is nearly a textbook example of simultaneity as I understand it. Events that are local in space are simultaneous in all local reference frames, so the clocks on the outbound spaceship and Earth will both sync up at their passing. The return spaceship is lightyears away. It will not agree as to the time at which the Earth's clocks state the outbound ship left.

I am not saying that those on the spaceship will read t=0 on any clocks in the inertial frame of earth, except the clock on earth.

In effect, you are, but in an obscure way, due to the choice of setting earth's original location as your origin. While we can arbitrarily set t=0 at any point we wish in any reference frame, that becomes the only place where the timelines mesh. Thus when you compare clocks when the earth reaches the returning spaceship (remember it is earth moving away from x=0, you are going to be lightyears away from that original meshpoint, even though it was on earth originally) you'll need to do a Lorentz shift on time. To do otherwise would be akin to arguing that that geometry is inconsistent because when lining up the first hash on different graph papers gives different results than lining up the last one.

I think that the Wikipedia article you referenced in regard to the Relativity of simultaneity is correct. That is the way I look at it also.

When I say that the clocks in the inertial reference frame of the incoming spaceships are all synchronized I am just repeating what you said,

Aye, you can have synchronicity among all clocks in one reference frame, but you cannot have it in clocks that are in two different reference frames and separated by space.

I’m talking about the clocks in the one reference frame of the returning spaceship. We can make all those clocks synchronous with the clock on earth at one particular time, which is the time that the outgoing spaceship passes earth. I’m not trying to synchronize more than one clock in the reference frame of the earth with all of the clocks in the reference frame of the returning spaceship. The clock on the space station will not agree with the clock on the returning spaceship. But because all of the clocks in the reference frame of the returning spaceship are synchronized with each other, and those clocks are synchronized with the earth as the outgoing spaceship passes the earth, then the clock on the returning spaceship will agree with the clock on earth, at that one time only. Since the earth inertial reference frame is seen by the approaching spaceship to be traveling at a high speed the clocks in the earth inertial reference frame will appear to go more slowly than the clocks in the reference frame of the returning spaceship.

Sometimes I look at it this way: Imagine a very long spaceship – several light years long. This spaceship is traveling through our solar system, as it appears on earth. We can imagine that the earth is closest to the spaceship on January 1, and as the earth goes around the sun the earth gets further from the spaceship until about mid-year, then starts to approach the spaceship again. Meanwhile the spaceship is rushing through the solar system. Observers on the spaceship watch the earth move away and come back. We can imagine that on January 1, earth time an observer on the spaceship announces to all on the spaceship that the earth is now at its closest point. That observer will be long out of the solar system at the time when the earth is at its greatest distance from the spaceship, but another observer will be near earth’s orbit and will relay that information to the rest of the spaceship. The time will be noted and compared with previous observations and clock readings taken on the spaceship. When earth is again at its closest point to the spaceship one earth year will have passed away on earth. On the spaceship it appears the entire solar system is flying by the stationary spaceship.

We can assume that those on the spaceship have clocks identical to the clocks on earth. According to the Theory of Relativity then the clocks on earth move more slowly than the clocks in the spaceship. Therefore, as seen by the incoming spaceship during the time the earth makes one revolution around the sun, and comes close to the spaceship again, 5/3 or 1.6 years have passed on the spaceship.

In fact, your mention of an observer at rest with the returning ship may be the best way to show that simultanity needs to be taken into account. Where do you place such a ship?

Say we look at the set up and decide at t=0 earth and outbound pass and return is at 2 ly away, (1.2 in it's own frame) so if we put the new ship at -1.2 lightyears away from the return ship on the x axis, it would thus be at earth when the outbound spaceship passes, right? Nope. While it would be at earth, the outbound ship would be nowhere in sight. The people on earth would see only see .72 lightyears between this new ship and the return ship (length contraction).

I believe you are taking the length contraction multiple times. In the earth’s reference frame the distance between the earth and the second space station is 2 light years. To the people in the incoming spaceship that distance appears to be 1.2 light years. You cannot then take 3/5 of 1.2 light years to say that the distance is 0.72 light years as seen on earth. The people on earth still see the distance as 2 light years, as before. The distance between earth and the incoming spaceship is 2 light years in earth’s reference frame, and 1.2 light years in the spaceship’s reference frame.

So, lets put this new ship in the same moment and locale as the outbound ship's passing of Earth. In the Earth's reference frame, the time is t=0, the local ships are present, and the return ship is 2 light years away. In the new ship's reference frame, the outbound ship and earth is present, the time is t=0 and the return ship is 3 1/3 light years away. (2 ly in earth's {moving} frame of reference is 3 1/3 in the stationary {new/return spaceships} frame) How long until the Earth reaches the return ship? (3 1/3 ly)/(4/5c)= 4 1/6 years or 2.5 years measured from Earth (4 1/6)/(5/3) = 2.5 Which is what we had already anticipated, so our numbers check out.

No, the “new ship” is in the same reference frame as the incoming spaceship. We can consider it part of the incoming spaceship if we think of a really long spaceship. The return ship is at a distance of 2 light years away from the earth in the earth’s reference frame. This length is contracted to1.2 light years in the frame of reference of the incoming spaceship. Proper length contraction has already been accounted for, the “meter stick” between earth and the second space station has been contracted as it is seen moving in the stationary reference frame of the incoming spaceship, to be 3/5 reduced in length. It would be incorrect to assume a lengthening of the distance as seen in any reference frame.

The situation is similar to the outgoing spaceship. The distance between earth and the first space station is seen to be 1 light year in the earth reference frame. One does not then calculate that the distance in the frame of reference of the outgoing spaceship is 5/3 light years so that the moving earth reference frame distance is contracted to 1 light year. In a way we agree to set up an asymmetrical problem. The distance from earth to each space station is fixed, and we assume it to be contracted by the factor 3/5 as seen by stationary observers in either the incoming or outgoing space ship reference frames.

But if the return ship is at 3 1/3 ly away when the new ship in its reference frame syncs with the outbound ship and earth, then t=0 can't occur when the return ship is 1.2 ly (or 6/5 ly) out. Lets do some number crunching on that extra space, shall we? 3 1/3 - 6/5 = 32/15 ly which at 4/5c would take 8/3 years to cross. This time in earth's reckoning, (8/3) / (5/3)= 24/15 or 1.6 years. Which checks with what we got by simply doing the Lorentz shift on time to account for simultaneity.

This paragraph begins with the incorrect statement that the return ship is at 3 1/3 light year. It is at 2 light years in the earth reference frame, and 1.2 light years in its own reference frame.

If you think about it, I’m just being consistent with the theory. From the point of view of the incoming ship clocks move more slowly in the earth’s frame of reference. So, whatever time passes for the spaceship has to be greater than the time passing for earth. If 1.5 years seems to have passed on the spaceship, then only (3/5)(1.5) = 9/10 years must seem to have passed on earth as seen from the spaceship. You keep trying to make it a longer time as seen from the incoming spaceship, but that cannot be correct.

In summary, using the long spaceship thought experiment; those on the long spaceship watch the earth go around 9/10ths of a complete revolution while the twin jumps from outgoing spaceship to incoming spaceship and back to earth. Observers on the spaceship nearest the earth are supposed to watch the transition from earth to outgoing spaceship, and spread the word on the spaceship of the noted time. They note the time the twin jumps onboard their own spaceship, and when he jumps off at the earth, all the time they are observing the earth go around the sun. Meanwhile observers on earth see the earth make two and a half revolutions around the sun, twice coming close to the long spaceship before moving away again. The paradox is how those on the earth see themselves going 2.5 times around the sun while the twin is traveling, while those on the spaceship, near the earth, see the earth go around only 9/10 revolution while the twin is away from earth.

[Tao]

No, the “new ship” is in the same reference frame as the incoming spaceship. We can consider it part of the incoming spaceship if we think of a really long spaceship. The return ship is at a distance of 2 light years away from the earth in the earth’s reference frame. This length is contracted to1.2 light years in the frame of reference of the incoming spaceship. Proper length contraction has already been accounted for, the “meter stick” between earth and the second space station has been contracted as it is seen moving in the stationary reference frame of the incoming spaceship, to be 3/5 reduced in length. It would be incorrect to assume a lengthening of the distance as seen in any reference frame.

I'm afraid that's the way it works. Length contraction is only observed in the moving reference frame. Ignoring everything else, how long is your ship at rest? Whatever that length is then the passengers will see it as such for the duration of their trip. Speeds are irrelevant, as they are not moving, everything else is, so there is no relativistic influences upon their perception of themselves. If you say that the ship at rest is 1.2 ly long then passengers at the tail would say that they see the Earth at the nose when they see the earth 1.2 ly away. But from the earth's perspective, the ship is moving; thus under relativistic influences. Since at rest it is 1.2 ly long, an observer perceiving it moving at .8c (one on Earth, in our case) will see it contracted down to .72 ly. This isn't double dipping, it is taking something at rest and figuring its length contraction.

But if an Earth observer sees it as .72 ly long, how can the nose pass earth at the same time that they see the tail is 2 ly away? Seems to be a paradox. And it would be, if you fail to do a Lorentz shift on time. Why is such a shift needed? Because observers at different locals (nose and tail) will observe different synchronicity of events at relativistic speeds. They will agree that "it is now 12:00" but disagree that "we are now in sync with such a relativistic event"

If the ship is indefinitely long, (and moving at .8c) as I perceive this example of yours to be:

Sometimes I look at it this way: Imagine a very long spaceship – several light years long. This spaceship is traveling through our solar system, as it appears on earth. We can imagine that the earth is closest to the spaceship on January 1, and as the earth goes around the sun the earth gets further from the spaceship until about mid-year, then starts to approach the spaceship again. Meanwhile the spaceship is rushing through the solar system. Observers on the spaceship watch the earth move away and come back. We can imagine that on January 1, earth time an observer on the spaceship announces to all on the spaceship that the earth is now at its closest point.

I presume this observer is closest to earth? If so, then let's arbitrarily put a red band around the ship at his location.

That observer will be long out of the solar system at the time when the earth is at its greatest distance from the spaceship, but another observer will be near earth’s orbit and will relay that information to the rest of the spaceship.

Again, we'll presume that this observer is closest to earth's orbit at that moment, we'll put a green band around the spaceship where he is.

The time will be noted and compared with previous observations and clock readings taken on the spaceship. When earth is again at its closest point to the spaceship one earth year will have passed away on earth. On the spaceship it appears the entire solar system is flying by the stationary spaceship.

We'll say yet another observer was present on the ship for this and his seat is marked by a blue band around the spaceship. When the first observer makes his announcement, the entire ship has a clock running that ticks over from -0.000001 at to 0.000 and they call it January 1. All clocks all along the ship will read this and will agree with each other for the duration.

I'll try to run a general equation at .8c without filling in any numbers. Now how far apart are the bands? At rest, the green band is x ly away while the blue band is 2x. The earth's orbit is traveling the length of the ship at .8c so it takes them 2x/.8 years to go from the earth passing the red to the earth passing the blue. So when the second call is announced, it would show x/.8 years on all ship clocks, and when the third call was announced it would be 2x/.8 years.

Remember the ship is moving according to earth, so on Jan 1, the earth observers would see the green band as x/γ or .6x ly away and the blue band as .6(2x) ly away. Here's the crux though. Since the green bands and the blue bands aren't in the same locale nor reference frame, an earth observer won't see them at 0.000 as they do the clock at the red band. They see the green band's clock as reading... refer back to the wikipedia page.... t'=(t-vx/c²)/√(1-v²/c²). Now, if we can figure x, we can see where we go from there.

Crunch those numbers, and see what you come up with. If I haven't botched the equations, we should end up with the same results.

This paragraph begins with the incorrect statement that the return ship is at 3 1/3 light year. It is at 2 light years in the earth reference frame, and 1.2 light years in its own reference frame.

That is an attempt to show the various attempts that could be made to get around performing a Lorentz time shift. You have to decide what the distance between the ships is in their rest frame. If you want to make it one ship with it's nose at earth and it's tail where we'd calculated the return ship, fine, let me know how long you want that to be at rest. It will either be 1.2 ly long or 3 1/3ly long. You can pick. If we insist that it is 1.2 ly long, then that is what it is, at rest. And if we don't perform a length contraction and claim an Earth observer would perceive it as longer even though it is going .8c, we would not be being consistent with the math nor the theory, and it is no surprise if the conclusions show a paradox.

[FauxRaiden]

If anyone is going to come up with a working theory for time travel, it's gonna be Tao and Vorpal

[vorpal blade]

It does seem like time travel, FauxRaiden, when we talk about clocks that show the time in the future, or the past. But I don't think we can actually use this for time travel.

I have to think about how to respond to your latest post, Tao. I'm not trying to avoid it, just postpone it for a while. Meanwhile I'd like to get back to something you said earlier and see if you agreed or disagreed with what I said about it.

What the return spaceship sees: at t₀=0 earth is -1.2 ly away at x=0 with a calendar reading t₀’=1.6 approaching at v. At t₁=.75 they pass a space station and another ship, beaming aboard an infant. They note that Earth's calendar reads t₁’= 2.05. At t₂= 1.5 they pass Earth, beaming down their bouncing baby with an earth time of t₂’=2.5

I think what you are doing is putting the time coordinate origin on the return spaceship. Is that correct? If they observe the time coordinate on earth to be 1.6 years at the start of the twin's journey, and 2.5 years at the end, then the elapsed time on earth must be the difference between the two clock readings, or 2.5-1.6 = 0.9 years. The twin on earth, as observed by the return spaceship reference frame can only have aged 0.9 years.

You mentioned later that the clock readings of 2.05 and 2.5 are already calculated as elapsed time, but that can't be right, can it? These times depend entirely on where you set the origin, and we know that elapsed time cannot depend on where you set the origin. They have to be readings on the time coordinate, and the elapsed time is the difference between readings, not the addition. You wouldn't calculate the distance between points as the sum of coordinates on the axis, but the difference between coordinate values. The same is true of the time axis. Are we agreed on this?

Later you seem to acknowledge this when you state:

The observers on the return spaceship saw their baby back at earth .75 years later (July 1, 3002 baby-time), noting that another .45 years have passed on earth, making the date of their arrival July 1, 3003 Earth-time. (t₀’+ t₁’+ t₂’= 1.6+ .45 + .45= 2.5 years)

Another 0.45 years passed on earth means that 0.45 years has elapsed on earth. You have the elapsed times correct, and you can use that to calculate the calendar reading on earth, but the important point is that only 0.9 years has elapsed on earth during the time one of the twins was traveling. So, from the point of view of the traveling twin the twin on the earth cannot have aged more than 0.9 years. I don't see anything shaky about that logic.

[Tao]

That's the implications of dealing with dilated time in non accelerating frames. From the returning space ship's point of view Earth is moving at .8c and has a slower clock, stretching out the timeline. The timeline doesn't suddenly distend at the birth of a child, nor at the passing of a spaceship, the entire thing is dilated out evenly like a rubber band. Since we've defined t₀=0 was defined as the arbitrary point time when the returning spaceship was at 1.2 ly away from earth, we measure earth's elapsed time (the age of the earth twin) as 1.6 years at this point. As measured from earth, another .45 years pass from that point to the point where the outbound ship passes the return ship, and yet another .45 years elapse from that point to the point where the return ship passes earth. Yes the return ship would see .9 years elapse from the arbitrary point when earth is 1.2 ly away and they would agree that a 1.6 year old child would age .9 years from that point until his 2.5th birthday when he is reunited with his twin.

I think the key to this may be the fact that t’ is measuring earth time elapsed, the age of the stay at home twin. By putting t₀=0 based off earth's local at x=0, return ship at x=1.2, it makes the numbers seem pulled out of nowhere. These numbers mean nothing about the traveling baby, as t₀=0 is not when the return ship would see the outbound ship pass earth, it is simply when earth appears 1.2 ly away.

By putting the origin of the axis 1.2 ly away from the returning spaceship, it (as a stationary point) will always be at x=1.2. when the outbound spaceship passes earth and the twins are born and boarded, Earth and the outbound spaceship would be somewhere around x= -2.133 (runnin' on intuition here, didn't actually calculate it out) the 1.6 years earth time is the time it takes to get from that point to x=0, then the .45 years earth time is the time it takes to get to x=.6 when the outbound spaceship passes the stationary return one, and another .45 years for the earth to travel the remaining .6 ly to x=1.2 where it meets up with the stationary ship.

I'd never really worked with an origin translated away from the stationary reference point, and while it did cause me to stumble over the numbers a couple times, the fact that it still works out giving the exact same results comes as a support of the internal consistency of relativity, rather than a detraction.

[vorpal blade]

Since we've defined t₀=0 was defined as the arbitrary point time when the returning spaceship was at 1.2 ly away from earth, we measure earth's elapsed time (the age of the earth twin) as 1.6 years at this point.

I thought we agreed to put the origin of the earth coordinate system at earth. One could put the origin of the returning spaceship coordinate system at the spaceship, or at earth as I did. I elected to make the origins of both coordinate systems the same, at earth. This is not as arbitrary as you think, this is in fact the standard configuration. Consider this Wikipedia reference http://en.wikipedia.org/wiki/Lorentz_transformation :

Assume there are two observers O and Q, each using their own Cartesian coordinate system to measure space and time intervals. O uses (t,x,y,z) and Q uses (t',x',y',z'). Assume further that the coordinate systems are oriented so that the x-axis and the x' -axis are collinear, the y-axis is parallel to the y' -axis, as are the z-axis and the z' -axis. The relative velocity between the two observers is v along the common x-axis. Also assume that the origins of both coordinate systems are the same. If all these hold, then the coordinate systems are said to be in standard configuration.. [Emphasis in original]

The 1.6 years that we read on the clock on earth from the returning spaceship’s reference frame is not due to defining t=0 on earth, with the spaceship 1.2 ly away from earth. That comes from your analysis where you defined t=0 to be on the spaceship, in non-standard configuration. When we put the origins at earth then the clock reads t’=0 and not 1.6 years at t=0.

I also object to your use of the word “elapsed time.” These are time coordinate readings, not elapsed time readings.

As measured from earth, another .45 years pass from that point to the point where the outbound ship passes the return ship, and yet another .45 years elapse from that point to the point where the return ship passes earth. Yes the return ship would see .9 years elapse from the arbitrary point when earth is 1.2 ly away and they would agree that a 1.6 year old child would age .9 years from that point until his 2.5th birthday when he is reunited with his twin.

This is a surprising interpretation of what is happening. First of all, this is again assuming that we are using your non-standard configuration in order to get the number 1.6 on earth at time t=0. In the standard configuration which I used that number is zero.

Next point. Really, “we measure earth's elapsed time (the age of the earth twin) as 1.6 years at this point” and then we add 0.45 years to each leg of the trip? Let us suppose that we put the origin on the 10th space station, which is at a distance of 10 light years from earth in earth’s reference frame. In the reference frame of the returning spaceship that distance would be 6 light years from earth, so earth is at x = -6 ly. What does the Lorentz coordinate transform equation then say is the clock reading on earth at t=0? The value is t’ = -(5/3)(4/5)(-6) = 8. Let’s see then; if we now add 0.9 years to “earth's elapsed time (the age of the earth twin) as” 8 then the twin on earth now appears to be 8.9 years old to the traveling twin when they are reunited. This contradicts his own perception of himself being 2.5 years old, and is much older than the traveling twin who thinks he has aged 1.5 years. Hence, another paradox.

I think the key to this may be the fact that t’ is measuring earth time elapsed, the age of the stay at home twin. By putting t₀=0 based off earth's local at x=0, return ship at x=1.2, it makes the numbers seem pulled out of nowhere.

I have not found any source that says that the Lorentz transformations give elapsed times, but rather a relationship between coordinate points. The age of the stay-at-home-twin should not depend on the origin of the coordinate axes, but t’ definitely is linearly dependent on the distance between the earth and wherever you put the origin. And, as I have shown, putting spaceship t=0, x=0 at earth’s location, where the earth coordinate origin is located, is the standard configuration. The standard configuration avoids pulling numbers “out of nowhere.”

These numbers mean nothing about the traveling baby, as t₀=0 is not when the return ship would see the outbound ship pass earth, it is simply when earth appears 1.2 ly away.

By setting the origins at the same place, at the time when the twin is passed to the outbound ship, we do a lot more than pick a time when the earth appears to be 1.2 ly away from earth. The transfer occurs when t=0 in each reference frame. At t=0 each observer in every reference frame simultaneously sees the transfer of twin to outgoing spaceship. What they disagree about, and gives life to the relativity of simultaneity, is what different observers see on the clock on earth when the jump occurs.

I think a lot of our differences boil down to the interpretation of the meaning of the time readings on the clocks. You seem to think of them as elapsed time, giving them a physical significance. I think of them as somewhat abstract mathematical constructs, useful in calculating relative time differences, but without an intrinsic physical meaning. We can set up the problem in various ways, with the origins at various locations. The way we set it up gives different clock readings. I do not believe that people age according to the location of the coordinate origins.

The returning spaceship will see a clock reading of t’=1.6 on earth when he sees the newborn twin leave earth at t=0. That does not mean that the twin left 1.6 years after he was born. What it means is that the incoming spaceship must correlate 1.6 on the earth’s clock, as seen by him, with the date the newborn twin left earth on Jan 1, 3001. If it is in winter when the twin is born and leaves earth, the incoming ship will not see summer but winter. An observer at x=0 in the reference frame of the incoming spaceship will also see winter as he sees the newborn twin leave. His coordinate transform calculation shows the clock to be t’=0. However clock readings are really arbitrary; what are real are events and time differences (but time differences are reference frame dependent, according to Einstein, and not absolute.) Different spatially separated observers, who have synchronized clocks, will all agree when the event (such as the twin leaving the earth) occurred when comparing their own clock readings in their own reference frame. However, they will disagree with what the clock reading was in the moving reference frame.

Clock readings, as calculated by the Lorentz transformations, are also useful in determining relative simultaneity of events in a reference frame moving relative to an observer. They don’t give absolute times, but help to put events on a scaled timeline for that observer. Observers in other moving reference frames may order events differently.

I anticipate an objection to these ideas on the basis of the relativity of simultaneity. Let me explain it this way. Suppose at the moment that the outgoing spaceship passes the earth, and the newborn twin jumps onboard the spaceship, a flash of light it emitted. All observers will sooner or later see the flash, and knowing the distance traveled and relative velocities will be able to calculate the time this single event occurred. If the clocks were synchronized at t=0, and the flash occurred at t=0, then all observers will agree that the event occurred at t=0. If the incoming spaceship was close to space station 2 when the spaceship’s clock said t=0 then those on the spaceship will know that the twin left the earth as the spaceship passed space station 2.

The simultaneity that may not be agreed upon by observers in every reference frame, because of the relativity of simultaneity, are events occurring at separate locations, such as the incoming spaceship passing the second space station and the twin leaving the earth. I think this answers in a general way the questions you asked in the post I skipped over. I believe I was mistaken in the way I presented that before. But all of my conclusions remain the same. The twin on earth appears to the traveling twin to have aged 0.9 years while the traveling twin aged 1.5 years, and the twin on earth sees himself as aging 2.5 years. And that makes sense because the time dilation factor of 3/5 reduces 2.5 years to 1.5 years – the slowing down of the traveling twin relative to the earth – and the time dilation factor of 3/5 reduces 1.5 years to 0.9 years – the slowing down of time for the earth twin which is seen moving relative to the inertial stationary reference frames of the spaceships.

By the way, I’ve looked at many articles purporting to resolve the twin paradox. That is, that each twin sees the other as having aged more slowly because he was stationary while the other was moving. In most cases the articles conclude that you cannot resolve this paradox with Special Relativity alone. So, we haven’t really got to the controversial stuff yet; my point of view is that held by most scientists.

[Tao]

You've put the origin at earth, which is fine and dandy when working from Earth's perspective or from the perspective of the outbound spaceship, (which is analogous to the wilkipedia quote's standard configuration). I've no problem with that. Where I struggle is when dealing with the return ship's perspective. The return ship is stationary. Earth is moving towards it at .8c. I guess if you've been running your equations with a coordinate system that is moving relative to the stationary return ship, it would explain why our numbers continue to fail to mesh.

I had left the origin where you had desired it to be, at Earth's location at t=0 from Earth & the outbound spaceship's point of view. This is by definition 1.2 ly away from the return spaceship at t=0. Unfortunately for that particular coordinate system, Earth and the outbound spaceship aren't there at t=0, making it somewhat less intuitive to run the numbers.

The transfer occurs when t=0 in each reference frame. At t=0 each observer in every reference frame simultaneously sees the transfer of twin to outgoing spaceship.

Awesome, this might make things much easier. Though it does mean that t=0 will not be the same in all frames of reference, as I had thought you had desired. If t=0 is defined by the perception of baby transfer as the outbound passes Earth, the return ship will see it while much further away, thus making the 1.6 years seem much less contrived and tacked on.

... In most cases the articles conclude that you cannot resolve this paradox with Special Relativity alone.

I find this interesting. Looking at the first articles that pop up on a google search for twin paradox:

The first one:

How the seeming contradiction is resolved, and how the absolute effect (one twin really aging less) can result from a relative motion, can be explained within the standard framework of special relativity. The effect has been verified experimentally using precise measurements of clocks flown in airplanes[1] and satellites.

The second one:

The confusion arises not because there are two equally valid inertial rest frames, but (here's the tricky part) because there are three. A lot of explanations of the twin paradox have claimed that it is necessary to include a treatment of accelerations, or involve General Relativity. Not so. {emphasis mine}

The third one may not present the clearest example but still holds it up, as the entire page is based on special relativity and never mentions GR (but hey, it is a pbs site, aimed at a more general audience):

Einstein came up with an example to show the effects of time dilation that he called the "twin paradox." It's a lot like the Time Traveler game you just played. Let's try it out with a pair of pretend twins, Al and Bert, both of whom are 10 years old in their highly futuristic universe....

The fourth one:

If you think this whole affair is difficult to understand, then you're not alone. But it has been observed (in particle experiments), and can be explained (as, for example, in the above story), through Special Relativity.

It then goes on to set up a second example that avoids all acceleration, just as we did.

The fifth one is also less clear but does explain why SR is sufficient:

Observe that "time-dilation" is symmetrical between the two observers until the traveling twin turns around and becomes a non-inertial observer.
How do you resolve the Twin Paradox?
The "stay-at-home" twin is an inertial observer.

The outgoing "traveling" twin is an inertial observer.

Although the incoming "traveling" twin is also an inertial observer, it's not the same as the "outgoing observer". In other words, the complete trip (outgoing and incoming taken together) is not an inertial trip. ... Realize that no choice of boost Velocity (i.e. no Lorentz or Galilean transformation) will ever straighten out the kink on the traveling twin's worldline. Hence, the two observers are not equivalent.

And the sixth one: "It is important to point out, however, that appealing to General Relativity is not necessary to resolve the paradox, as demonstrated above."

I stopped there, but I'm sure if you desired, you could go further. That's six for six, and notice that most of these sites are university .edu addresses, lending credibility to their assertions for those who would otherwise doubt. If you adhere to all the steps of Special Relativity, not leaving things out, the twin paradox is solved, thus no paradox at all. Sure, GR brings things more into the realm of what we'd expect to see, but the math is less penetrable. If you don't mind dealing with integrals, there's a general solution using GR that proves that the traveling twin will age less for any given variable of acceleration, coasting speed and distance. So while not needed, it can be more mathematically satisfying. It wouldn't surprise me if something similar existed for SR, but it is usually not called for.

[Tao]

As a side note, I just noticed the number of page views this thing has accrued. I must say I'm surprised at the thought that there people other than Vorpal and myself slogging through all this. I know Nerdgirl and FauxRaiden have been trying to, are there others? Is there anything I can do to make it easier for any or all to understand?

[Marduk]

To be fair, I just click on this and peruse it to make sure you guys are playing nice. I don't actually care much for the content.

[vorpal blade]

That is interesting that the first six articles that came up on a google search of “twin paradox” all seem to you to support the idea that the twin paradox can be explained solely on the basis of special relativity. It was interesting to me to review the articles and compare our discussion with the discussion in those articles. I appreciate the opportunity to discuss this in this forum. I know I've learned a few things. Hopefully we will all have the patience to continue this dialog until we can more fully clarify the situation, and it would be wonderful if we could come to agree. I spend many hours studying and considering each of your posts, Tao.

You know, I studied relativity for many years, then put it aside 25 years ago. It is possible that the articles that I looked at back then put a different slant on things than the more recent articles. I apologize for not having fully explained what I meant by the assertion that you cannot fully resolve this paradox by special relativity alone. I hope when I have finished analyzing the articles you referenced it will be clear what I mean.

Before I start I would like to point out that the second article you referenced does support my claim that “A lot of explanations of the twin paradox have claimed that it is necessary to include a treatment of accelerations, or involve General Relativity.” The author of that particular article didn't think it was necessary, but he does agree with me that a lot of other scientific explanations do think it is necessary.

Now, in regard to the first article, found in Wikipedia. The next sentence after the one you quote states

Starting with Paul Langevin in 1911, there have been numerous explanations of this paradox, many based upon there being no contradiction because there is no symmetry—only one twin has undergone acceleration and deceleration, thus differentiating the two cases.

That is part of what I am talking about. Special relativity deals only with reference frames at rest which do not accelerate or decelerate. What you do about the times the spaceship is accelerating and decelerating (or switch between frames) falls outside of the theory of Special Relativity. Special Relativity cannot say any thing about what is happening at these times. Yet, if you look closely, most of the attempts to resolve the twin paradox depend completely on how you treat the times of acceleration and deceleration, or the time of transfer between frames, or simply allow the author to ignore looking at the problem from the point of view of the one who had to change reference frames, as “obviously” different from the stay at home twin. This used to be more explicitly noted that it is today. At least, from what I've seen.

Some of the explanations of the twin paradox deal solely with the observation that the stay at home twin will see himself as aging more slowly than the traveling twin sees himself age. They sometimes overlook the question, why doesn't the traveling twin see the stay at home twin as younger as well? The answer is often given, well, the fact that one experienced acceleration means we really know he is the traveling twin, and the problem is not symmetric. As reference one says:

The standard textbook approach treats the twin paradox as a straightforward application of special relativity. Here the Earth and the ship are not in a symmetrical relationship: the ship has a turnaround in which it undergoes non-inertial motion, while the Earth has no such turnaround. Since there is no symmetry, it is not paradoxical if one twin is younger than the other.

To me this means that Special Relativity alone cannot resolve the paradox; we must use the acceleration factor to indicate a preferred reference frame. And this places the resolution outside of the Special Theory.

The key to understanding this article's claim to resolving the paradox is found in the statement

It is during the acceleration at the U-turn that the traveling twin switches frames. That is when he must adjust his calculated age of the twin at rest. (Emphasis in original)

With the use of a handy graph the author concludes that

In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the resting twin. The traveling twin reckons that there has been a jump discontinuity in the age of the resting twin.

Now, I believe that this is convenient, and I understand the logic, but it falls outside of the Special Theory of Relativity. The theory makes no claims about what happens when you transfer reference frames. You can get this kind of answer using Lorentz transforms, but only if you are careful to select your reference plane origin in exactly the right place. Other origins are equally good, but ignored because they don't give the desired answer.

Furthermore, this trick of adjusting the age of the distant twin when the turn around occurs could with equal logic be applied to the traveling twin when viewing the stay at home twin. We could say that from the point of view of the traveling twin the stay at home twin jumps reference frames. The only reason this is not considered is because we know which twin has accelerated and which has not. Thus we apply a rule to the one traveling, and don't really consider the stay at home to be traveling at all. This defeats the relativity of the situation. As I say, I understand the justification used, but is not part of the theory of Special Relativity, and how do we know that this special treatment of the one not subject to acceleration is really justified? It requires a new ad hoc rule to apply for situations not covered by Special Relativity. I hope that is clear.

In my estimation the first article claims to resolve the paradox within the standard framework of Special Relativity, but fails to do so.

Moving on to the second article you referenced. The paradox is resolved in these words:

Now, since special relativity lets us use either rest frame, we assume Bob is the at-home twin. Ann speeds away at 3/5c. No problem so far. But after 4 years of waiting, Bob must change his inertial frame. If we allow Ann to return, we've only restated the problem with the names switched. In the first version, Ann stayed in an inertial frame, and she must stay in an inertial frame in this version. Bob zooms off after Ann at 15/17 light speed (now we know why it was important), and of course catches up. It takes him 4 years, and he has seen 8 years since Ann left. Ann has aged 10 years. Same result. No paradox.

As it stated here Ann “must stay in an inertial frame in this version.” In other words, we will apply the Special Theory preferentially because we really know who the traveler is and who it isn't. We are told that we are not allowed to consider Ann changing reference frames because of the acceleration thing, which is not covered by the Special Theory of Relativity.

So, I count the second article as having to rely on assumptions about what happens, or what we can and cannot consider in the calculations, based on things which are not part of the theory, and hence you cannot fully resolve this paradox by Special Relativity alone, as I said

The third article really doesn't address the issue of whether Special Relativity alone can answer the question. It says the twin paradox idea comes from Einstein, and that is about it. It isn't a scientific article, and I don't think it can be counted either way for determining what most scientists think.

The fourth article. I agree that the author believes he is using Special Relativity alone, but he is very careful to do his calculations in a way to support his narrative. Had he chosen different origins for his calculations he would have different readings on the clocks, and it would have spoiled his story. He assumes that he knows the effect of switching reference frames, that the traveler will suddenly see distant clocks to abruptly change time, but it isn't covered by the Special Theory. I believe I have shown that there are other legitimate ways to do the calculations and get different results. It is interesting that this author explains the differences of clock readings as “Doubleprime's explanation is that Unprime's clocks were running slowly and out of synchronization (again, just as in the above explanation).” Clock readings are interpreted to have real meanings when it fits the scenario, but if not they just look out of synchronization.

In the end the author justifies his use of two reference frames for the traveling twin, and only one reference frame for the stay at home twin on the basis that only one of them felt acceleration. Since that is not part of the Special Theory of Relativity he is in effect justifying his calculation procedure (and not considering the stay at home twin in two reference frames, as not “required”) by using something more than Special Relativity alone, even though “nobody experienced any accelerations.”

My biggest objection is that the author carefully orchestrates just the data that supports his claims.

I could not access the fifth article. But from what you have quoted it is similar to the others. The two observers are not equivalent because of acceleration. Acceleration is not part of the Special Theory of Relativity. To “resolve” the paradox you must make assumptions about the consequences of using non-equivalent observers, which is outside of the Special Theory of Relativity. The authors may not have realized it, but I count this as support of my claim.

The sixth article. Once again the author claims that General Relativity is not necessary. “But Special Relativity applies only to the relations between inertial frames of reference. In this regard, the situations of the twins are definitely not symmetrical.” My point is this, how can they justify using a theory which does not apply to the situation? It is by making assumptions about what happens when the theory does not apply, and by applying the theory in a manner that is calculated to give them the result they desire. “The naive interpretation--the reason why the situation is called a paradox--is to assume that the situation is competely[sic] symmetrical. If that were the case, Jane's diagram would simply be a mirror image of Joe's.” Exactly, so you have to invent one-sided arguments outside of the Special Theory to handle the differences.

So, in my observation most of these authors cannot claim to fully resolve the Twin Paradox by Special Relativity alone, whether they are aware of that or not.