[Origin]

Imagine that the CCTV on the turning point

Imagine you were able to stop talking about CCTVs.  One has to wonder how Einstein ever develop his theories without CCTVs!

[Janus]

The asymmetry was known once the itinerary was made. Everybody from anywhere can see it. I have no idea why you think a CCTV present at the star-clock would show anything not already visible to everybody else, including the ship which happens to actually be there.

Imagine that the CCTV on the turning point is also equipped with a powerful telescope which can observe the giant clocks as well as the clock on the space ship. What would it see during the experiment?
When the twin started the flight, the earth clock has already shown 4 y, while traveling clock is still 0, which will be seen by the distant CCTV 4 years later. When the signals arrive at the turning point, it's own clock would already show 8 y.
When the space ship arrives at the turning point, the giant clock there shows 4 + 4/0.999 = 8.004 y. The traveling clock shows 4/0.999/22.4 = 0.18 y

What any observer would visually see in a telescope is not the same thing as what they would determine the present reading is on the clock they are observing.  For example, a planet 4 ly from us, at this moment would be seeing events that occured on Earth in 2019.  But they would determine that, it is presently 2023 on Earth, because they know it took 4 years for the light they are presently seeing to reach them.
An observer traveling at 0.8 c relative to both Earth and the planet would, as he passed the planet see exactly the same thing, Earth in 2019. But he'd come to a different conclusion as to what year is was on Earth as that moment. Unlike the planet, which maintained a constant distance to Earth for the entire time, the distance between this observer and the Earth has been changing, and the distance between them was not 4 ly when the light left Earth.  And for the observer on the ship, it was the distance between Earth and himself when the light left Earth that determines how long it took the light to reach him at c.  If he had been traveling from Earth to the planet, he would have been less than 4 ly from Earth when the light left Earth, and it would have taken less than 4 years for the light to reach him.  Thus, when he sees 2019 on Earth as he passes the planet he cannot conclude that it is 2023 on Earth at that moment but an earlier year.
Adding CCTVs only means that you now have to take light propagation delays into account, on top of the Relativistic effects.

[Eternal Student]

Hi.

I'm a bit surprised this thread is still going - but it's an interesting topic and can be examined from many angles.
Some of the older posts have lost their special symbols and especially any equations but there's a diagram that survives.
   Post #45 had this diagram which is a great diagram:



   I would say a diagram like that explains most of what people want to know.   I'd just like to colour all the events that happen at Earth between the highest blue line and the lowest red line .   Here is that diagram:

Twins-green.jpg (41.62 kB . 511x421 - viewed 1075 times)

    Looking at the blue lines of simultaneity we can label those lines  (from bottom to top)   t'=1 unit,     t'=2   and t'=3 (where t' is the time elapsed for the travelling twin).   We see that the events in the green area are still in the future for the travelling twin on their outbound journey.    For example, assume the twin had NOT turned around at t'=3  (at spatial location = Alpha Centauri ) but instead had just continued on in uniform motion.   Here's the diagram for that:

Twins2.jpg (25.32 kB . 511x421 - viewed 1086 times)
   We see that events in the green area (which are spatial located at Earth) fall along a blue line of simultaneity with  t' > 3   (that is later than t'=3,  when the twin reached Alpha Centauri and should have turned around).

    However the travelling twin does turn around at t'=3.    Which immediately changes the slope of the lines of simultaneity,  we have the red lines of simultaneity.   We can start labelling those from bottom to top  as   t'=3,   t'=4 and t'=5.    Note the lowest red line is t'=3   exactly as for the latest blue line of simultaneity when the travelling twin reached Alpha Centauri -  we assume the turn-around took negligible time for the twin. 
    So on the return portion of the travelling twins journey all the green events are effectively in what they would consider as the past  -   e.g.  the green events would lie along the red lines of simultaneity for t' < 3.
Here's the diagram for that:

Twins4.jpg (25.48 kB . 511x421 - viewed 826 times)

    Anyway, that hopefully explains, with diagrams, what was discussed in earlier posts.   When the travelling twin changes their direction of motion at Alpha Centauri,  they change rest frames.   This has the consequence of making all the green events that were located at Earth, no longer be events that the travelling twin would consider as being in their future  (t' > 3).  When the travelling twin changed direction, the green events immediately change into events that the travelling twin would consider as being in their past  (t' <3).

Some events (located on earth) were in the travelling twins future (in the old rest frame) but they abruptly changed to being events that were in their past (in the new rest frame), they were never in their present or "now" , they were skipped over entirely.

   or, as Janus wrote:

An analogy would something along these lines:
Two men are back to back and then separate. According to each of them, the other is a given distance "behind" him. Man 1 then turns 180 degrees.  By his perspective, Man 2 goes from being behind him to being in front of him.

   
- - - - - - - - - - -

    Anyway, I quite like the diagram approach and just seeing where the "green events" will map to on the outward journey compared to the return journey.   It reduces the Twins paradox to a situation that is almost identical to "the Andromeda paradox"  (where an invasion fleet has or has not yet been launched from Andromeda depending on which direction you are moving in).    It is all seen to be a consequence of the idea that spacetime is like a loaf of bread and there is no preferred or correct way to cut it into time slices.  We can foliate it (slice it) at different angles to obtain very different "now" slices:  Boost to another frame of reference and we change the angle at which we slice through that bread,  so that events that were at some spatial distance from us can sometimes be moved into what we would consider as the past, the future or "now" just by changing your motion.
   So you (Hamdani and/or Dimensional who started the thread) can have that as another explanation of the "exact cause" of different aging of the twins if you want it:   It's a consequence of our ability to foliate spacetime in different ways just by changing the motion of one twin together with there being some physical distance between them when that change of motion occurred.  So you can recover all the criteria that has previously discussed:  For example, the distance between earth and Alpha Centauri is important because you're going to pivot around Alpha Centauri and hence sweep out a greater amount of what has been drawn as green events on Earth.  Similarly the speed of travel for the travelling twin is important because that determines the angle at which you will be foliating spacetime (if the travelling twin has high speed relative to the Earth twin the blue and red lines of simultaneity are severely angled from the horizontal on the diagram).   The acceleration of the travelling twin at Alpha Centauri is important but not numerically important:   The traveling twin HAD to change the direction of their motion at Alpha Centauri but exactly how that was done didn't matter much,  for convenience it was assumed to be almost instantaneous.   If the acceleration had been done over  0.1 s  instead of  1 s then the size of the acceleration would have been ten times larger but it doesn't matter and doesn't affect the age of either twin much.  All that matters is the total angle swept through when you foliate spacetime for the outbound journey compared to the return journey (the angle between the blue line of simultaneity vs. the red line at t'=3 when you pivot around spatial location= Alpha Centauri).
     
    Already too long.... I'll hide the next bit.   It may only add confusion anyway.   If you're happy with events being moved from the future to the past then that's fine, leave it there.
 

Spoiler: show

  Now, when you spend even more time considering Special Relativity you may start to realise that it is actually silly to assume an event is "definitely in my past" if it's spatially separated from me and has not yet entered my past light cone.  If an event has not yet entered my past light cone then I have almost complete control over whether it will get into my past.  I can accelerate away from the event and delay it getting into my past light cone and if I keep accelerating rapidly enough then it will never get into my past light cone.   Conversely, if I do what the travelling twin did,  I can turn to move towards the event instead of accelerating away from it.   Then those events that were spatially remote from my original starting position will rapidly start entering my past light cone:   The travelling twin can have much more than 1 earth second's of the green events entering their past light cone for every 1 second of their own elapsed time.   Of course they can change direction again and start accelerating away... only those events which have entered their past light cone are definitely "locked in":  They are most certainly in their past and cannot be prevented from having some influence on what they may experience locally.
   Anyway,  that's almost a completely different discussion about Special relativity.....  the causal structure of the universe is based on light cones and not on a time co-ordinate.  So some would say that we aren't even allowed to ask if the green events shown on the diagram above were in the travelling twins past or future when they were at Alpha Centauri -  they were outside the twins light cone and therefore not uniquely fixed or required to be in that twins past or future at that point in their journey.   The travelling twin had almost complete control over how quickly those green events would enter their past light cone,  they just change their motion to achieve that.

Best Wishes.

[hamdani yusuf (OP)]

I would say a diagram like that explains most of what people want to know.   I'd just like to colour all the events that happen at Earth between the highest blue line and the lowest red line .   Here is that diagram:

According to the observer at the turning point, the green events happen during the beginning and finishing of the trip. Although it is in the same frame of reference as the earth twin.

[hamdani yusuf (OP)]

Imagine that the CCTV on the turning point

Imagine you were able to stop talking about CCTVs.  One has to wonder how Einstein ever develop his theories without CCTVs!

You can replace the CCTV with other kind of observers.

[hamdani yusuf (OP)]

Adding CCTVs only means that you now have to take light propagation delays into account, on top of the Relativistic effects.

The simultaneity planes shown in the diagram are basically taking the propagation delays into account.

We can find many different explanations trying to solve the twin paradox, as shown by problem statement in wikipedia.

In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more. This result appears puzzling because each twin sees the other twin as moving, and so, as a consequence of an incorrect[1][2] and naive[3][4] application of time dilation and the principle of relativity, each should paradoxically find the other to have aged less. However, this scenario can be resolved within the standard framework of special relativity: the travelling twin's trajectory involves two different inertial frames, one for the outbound journey and one for the inbound journey.[5] Another way of looking at it is to realize the travelling twin is undergoing acceleration, which makes him a non-inertial observer. In both views there is no symmetry between the spacetime paths of the twins. Therefore, the twin paradox is not actually a paradox in the sense of a logical contradiction.

Starting with Paul Langevin in 1911, there have been various explanations of this paradox. These explanations "can be grouped into those that focus on the effect of different standards of simultaneity in different frames, and those that designate the acceleration [experienced by the travelling twin] as the main reason".[6] Max von Laue argued in 1913 that since the traveling twin must be in two separate inertial frames, one on the way out and another on the way back, this frame switch is the reason for the aging difference.[7] Explanations put forth by Albert Einstein and Max Born invoked gravitational time dilation to explain the aging as a direct effect of acceleration.[8] However, it has been proven that neither general relativity,[9][10][11][12][13] nor even acceleration, are necessary to explain the effect, as the effect still applies if two astronauts pass each other at the turnaround point and synchronize their clocks at that point. Such observer can be thought of as a pair of observers, one travelling away from the starting point and another travelling toward it, passing by each other where the turnaround point would be. At this moment, the clock reading in the first observer is transferred to the second one, both maintaining constant speed, with both trip times being added at the end of their journey.[14]
https://en.wikipedia.org/wiki/Twin_paradox

I want to see which explanation is more consistent with different observers in the same frame of reference as the earth observer. Is the asymmetry appears during the acceleration only, or it's when there's relative velocity instead.
Note that in this case, there are five observers instead of two like in standard twin paradox problem. They are two travelling twins, from earth and AC. The other three are stationary relative to each other: stay on earth twin, twin who stays on AC, and the CCTV at midway.
How to use transformation formula to convert observations from one observer to the others, given the relative motions between them?

[hamdani yusuf (OP)]

Since we are looking for the cause of asymmetry, we better start with a system that is perfectly symmetrical. After all the implications have been settled, we can change one parameter at a time, and see what happens, when the symmetry starts to break.

Supposed that there are also twins living on Alpha Centauri. One of them going through a journey just like the travelling twin from earth. To synchronize, the travelling twins were waiting for a signal transmitted by midway giant clock at t=-2 year. Thus travelling twin from earth and travelling twin from Alpha Centauri start simultaneously at t=0.
What the midway clock will see when the twins from earth and Alpha Centauri pass by at t=2 year?

By symmetry, it should see both travelling twins' clocks show the same value, whatever it is. But what the travelling twin from earth see of the clock carried by the travelling twin from Alpha CentauriCentauri,  and vice versa?

[hamdani yusuf (OP)]

In physics discussions, particularly when addressing concepts like time dilation and relativistic effects, precision in language is crucial to avoid confusion. If you believe that Don Lincoln's explanations have led to misunderstandings or misrepresented certain aspects of the topic, it's valid to provide additional context or clarification, especially if you have a strong background in the subject matter.

I'm open to anyone who wants to explain the problem I posted above. Which part of the thought experiment with symmetrical travelling twins isn't clear yet?

[Janus]

Since we are looking for the cause of asymmetry, we better start with a system that is perfectly symmetrical. After all the implications have been settled, we can change one parameter at a time, and see what happens, when the symmetry starts to break.

Supposed that there are also twins living on Alpha Centauri. One of them going through a journey just like the travelling twin from earth. To synchronize, the travelling twins were waiting for a signal transmitted by midway giant clock at t=-2 year. Thus travelling twin from earth and travelling twin from Alpha Centauri start simultaneously at t=0.
What the midway clock will see when the twins from earth and Alpha Centauri pass by at t=2 year?

By symmetry, it should see both travelling twins' clocks show the same value, whatever it is. But what the travelling twin from earth see of the clock carried by the travelling twin from Alpha CentauriCentauri,  and vice versa?

Twin leaving Earth: During his acceleration phase to AC, he will determine that the coordinate time for clocks at AC and the other traveling twin will have sped up compared to his own.  Once he ends his acceleration, Both the the other twin's clock will run slow. But, that other twin will also be further along their trip to the center point than he himself is, and will have ticked off more time. By the time they meet at the midpoint, the other twin's clock will read the same as his own. In other words, the other ship's clock runs fast, then slow, and ends up reading the same as his which ran at a constant rate the whole time when they meet.
It all goes back the relativity of simultaneity. As each twin transitions from being at rest in one inertial frame to another his notion of what events are simultaneous also changes.
The AC twin determines the same happening to the Earth traveling twin.

[hamdani yusuf (OP)]

Twin leaving Earth: During his acceleration phase to AC, he will determine that the coordinate time for clocks at AC and the other traveling twin will have sped up compared to his own.  Once he ends his acceleration, Both the the other twin's clock will run slow. But, that other twin will also be further along their trip to the center point than he himself is, and will have ticked off more time. By the time they meet at the midpoint, the other twin's clock will read the same as his own. In other words, the other ship's clock runs fast, then slow, and ends up reading the same as his which ran at a constant rate the whole time when they meet.
It all goes back the relativity of simultaneity. As each twin transitions from being at rest in one inertial frame to another his notion of what events are simultaneous also changes.
The AC twin determines the same happening to the Earth traveling twin.

Can you give quantitative values? Assume that the acceleration can be nearly instantaneous, or, if you think it's impossible, put the maximum acceleration you think is permitted by physical law.

[Janus]

Twin leaving Earth: During his acceleration phase to AC, he will determine that the coordinate time for clocks at AC and the other traveling twin will have sped up compared to his own.  Once he ends his acceleration, Both the the other twin's clock will run slow. But, that other twin will also be further along their trip to the center point than he himself is, and will have ticked off more time. By the time they meet at the midpoint, the other twin's clock will read the same as his own. In other words, the other ship's clock runs fast, then slow, and ends up reading the same as his which ran at a constant rate the whole time when they meet.
It all goes back the relativity of simultaneity. As each twin transitions from being at rest in one inertial frame to another his notion of what events are simultaneous also changes.
The AC twin determines the same happening to the Earth traveling twin.

Can you give quantitative values? Assume that the acceleration can be nearly instantaneous, or, if you think it's impossible, put the maximum acceleration you think is permitted by physical law.

Okay, simplest case of near instantaneous acceleration, with an assumption of 0.8c velocity for both ships.
E = Earth
ES = Earth ship
AC = Alpha Centauri
ACS = Alpha Centauri Ship

Events according to E, starting from the moment it reaches 0.8c relative to Earth
E time =0
AC time = 3.44 yr (relativity of simultaneity)
ACS time = ~1.26 yr
Distance of ACS to midpoint = ~0.283 ly
Relative velocity of ACS to ES = ~0.976c (addition of velocities)
Time dilation factor for ACS according to ES = ~0.22
closing speed between  ACS and midpoint = ~0.176c  (0.976c-0.8c)
Distance to midpoint for ES = ~1.29 ly (length contraction)
Time for ES to reach midpoint = ~1.61 yr
time for ACS to reach midpoint = ~1.61 yr (0.28 ly/.176c)*
Time accumulated for ACS while ES travels to midpoint = ~0.354 yr (1.61*0.22)
Time on ACS upon arrival to midpoint ~1.61 yr (1.26yr+0.354 yr)

Now what ES would visually see:
At moment of reaching 0.8c:
AC time = -4.3 yr (ES is basically seeing same light that E is, and E is seeing a 4.3 yr old image
ACS time = -4.3 (still waiting in port for departure)

When it first sees ACS leave AC:
Time on ES  = 1.43 yr
Doppler shift factor for AC = 3
Time ES sees on ACS was it leaves AC = 0 ( -4.3+1.43*3)
Doppler shift  factor for ACS once it is on its way = 9
Length of time ES see ACS Doppler shifted by above factor =~0.18 yr (1.61-1.43)
Amount of time accrued on ACs clock during this period ~1.61 yr (9*0.18 yr)

 If we change to the ACS frame we get the same results. Just swap E for AC and ES for ACS

* the values used in this example have been rounded off, so you are not going to get exact agreement between some answers.

[hamdani yusuf (OP)]

Let's start to analyze the situation from the simplest point of view, which is the observer staying on the midway point. After we get the observation values of each key events, we can try to transform them from other observers. They are the twins staying on earth, staying on AC, travelling earth twin, travelling AC twin.
The key events are :
1. Travelling twins start the journey. From midway perspective, they should be simultaneous due to symmetry.
2. Travelling twins arrive at midway. Again, from midway perspective, they should be simultaneous due to symmetry.
3. Travelling twins arrive at the destination and start the return journey.
4. Travelling twins arrive at midway in their return journey.
5. Travelling twins get back home.

In each events, the midway observer writes down the clocks that they observe:
1. The clock stays on earth.
2. The clock stays on midway.
3. The clock stays on Alpha Centauri.
4. The clock carried by travelling earth twin.
5. The clock carried by travelling AC twin.

The result would be a 5x5 matrix. This is before we transform to other observers.

[hamdani yusuf (OP)]

To be consistent, let's use the speed used in Don Lincoln's video, which is 0.999c.
Here are the observations of those 5 key events as recorded by midway observer.
1. The journey starts on earth at t=0 local time. The light signal takes 2 years to the midway observer, thus he will see the event when his clock shows t=2 years. Journey from Alpha Centauri is the same due to symmetry.
2. The travelling twins take 2 ly / 0.999c = 2.002... year to get to the midway. Thus the observer will see his clock showing t=2.002... years.
3. The journeys take 4 ly / 0.999c = 4.004... year to get to the turning point in on going leg. But it takes additional 2 years for light signal to the midway observer, thus he will see the event when his clock shows t=6.004... years.
4. The travelling twins take 6 ly / 0.999c = 6.006... years to get to the midway in return leg. Thus the observer will see his clock showing t=6.006... years.
5. The total journeys take 8 ly / 0.999c = 8.008... years to get home in return leg. But it takes additional 2 years for light signal to the midway observer, thus he will see the event when his clock shows t=10.008... years.

[Janus]

To be consistent, let's use the speed used in Don Lincoln's video, which is 0.999c.
Here are the observations of those 5 key events as recorded by midway observer.
1. The journey starts on earth at t=0 local time. The light signal takes 2 years to the midway observer, thus he will see the event when his clock shows t=2 years. Journey from Alpha Centauri is the same due to symmetry.
2. The travelling twins take 2 ly / 0.999c = 2.002... year to get to the midway. Thus the observer will see his clock showing t=2.002... years.

During which time the midpoint observer will see the traveling twins' clock Doppler shifted by a factor of 44.710, and accumulating 44.710 * .002 = 0.089 yr, thus having their clocks each reading 0.089 yr upon reaching the midpoint.

3. The journeys take 4 ly / 0.999c = 4.004... year to get to the turning point in on going leg. But it takes additional 2 years for light signal to the midway observer, thus he will see the event when his clock shows t=6.004... years.

After each twin passes the midpoint, and are receding from it, the Midpoint observer will seeing them Doppler shifted by a factor of 0.0224, with each accumulating 4.004 * .0224 = 0.89 yr on their clocks to read 0.178 yr upon arrival at their destinations. (while the clocks at the destinations will read 6.004-2= 4,004 yrs)

4. The travelling twins take 6 ly / 0.999c = 6.006... years to get to the midway in return leg. Thus the observer will see his clock showing t=6.006... years.
5. The total journeys take 8 ly / 0.999c = 8.008... years to get home in return leg. But it takes additional 2 years for light signal to the midway observer, thus he will see the event when his clock shows t=10.008... years.

The return legs are just mirrored versions of the outbound legs, and give the same results, with an additional 1.79 yrs added to each twin's clock and 4.004 yrs added to the planets' clocks as seen by the midpoint observer.

[hamdani yusuf (OP)]

During which time the midpoint observer will see the traveling twins' clock Doppler shifted by a factor of 44.710, and accumulating 44.710 * .002 = 0.089 yr, thus having their clocks each reading 0.089 yr upon reaching the midpoint.

Why the Doppler shift matters in clock reading? Isn't Lorentz' time dilation enough?

[hamdani yusuf (OP)]

During which time the midpoint observer will see the traveling twins' clock Doppler shifted by a factor of 44.710, and accumulating 44.710 * .002 = 0.089 yr, thus having their clocks each reading 0.089 yr upon reaching the midpoint.

Why the Doppler shift matters in clock reading? Isn't Lorentz' time dilation enough?

At starting event, all clocks are stationary according to midway observer. While at meeting event at midway, traveling clocks are right in front of the midway observer. They are neither approaching nor receding.

[Janus]

During which time the midpoint observer will see the traveling twins' clock Doppler shifted by a factor of 44.710, and accumulating 44.710 * .002 = 0.089 yr, thus having their clocks each reading 0.089 yr upon reaching the midpoint.

Why the Doppler shift matters in clock reading? Isn't Lorentz' time dilation enough?

The Lorentz time time dilation factor is what the observer determines is occurring at the other clock at a given moment.
Doppler shift is what he directly see and measures.
With a two clocks stationary with respect to each other and 1 light hour apart, if each sees the other clock as being 1 hr behind theirs, they still determine that the clocks are simultaneous once they account for the light propagation delay.
Likewise, Doppler shift is what you see, and time dilation is what remains after you account for the constantly changing light propagation delay. ( when the object is approaching you, the delay gets shorter and shorter, and when it is receding, it gets longer and longer.)
Of co

[Janus]

During which time the midpoint observer will see the traveling twins' clock Doppler shifted by a factor of 44.710, and accumulating 44.710 * .002 = 0.089 yr, thus having their clocks each reading 0.089 yr upon reaching the midpoint.

Why the Doppler shift matters in clock reading? Isn't Lorentz' time dilation enough?

At starting event, all clocks are stationary according to midway observer. While at meeting event at midway, traveling clocks are right in front of the midway observer. They are neither approaching nor receding.

But to get from starting point to midpoint they have to approach the midpoint, and to get to the end point from the midpoint, they have to recede from the midpoint.   It's during these periods that the "interesting stuff" takes place. It's also the bit you need to grasp to have any hope of understanding what is going on.

[hamdani yusuf (OP)]

when the object is approaching you, the delay gets shorter and shorter, and when it is receding, it gets longer and longer

When the object isn't approaching nor receding, the delay isn't changing.
Let's assume that the space ship doesn't hit the midway observer. He is 1 km away from the trajectory of the ship. The distance between the ship and observer can be plotted as such.
https://www.wolframalpha.com/input?i=plot+sqrt%281%2Bx%5E2%29+from+-10+to+10
This infinitesimaly short moment is what defines the second event.

[Janus]

when the object is approaching you, the delay gets shorter and shorter, and when it is receding, it gets longer and longer

When the object isn't approaching nor receding, the delay isn't changing.
Let's assume that the space ship doesn't hit the midway observer. He is 1 km away from the trajectory of the ship. The distance between the ship and observer can be plotted as such.
https://www.wolframalpha.com/input?i=plot+sqrt%281%2Bx%5E2%29+from+-10+to+10
This infinitesimaly short moment is what defines the second event.

The difference between the ship passing an infinitesimal distance from the observer and missing by a kilometer is negligible( it wouldn't even show up at the number of significant decimal places we are dealing with in this problem).
Why are you obsessed with one   "infinitesimaly short moment" when it takes an understanding of the whole of the problem to grasp what's occuring?
It's seems to me that you are trying as hard as possible to Not understand Relativity.