Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: hamdani yusuf on 03/03/2018 21:44:11

This is a modified thought experiment about twin paradox.
Twin A goes to planet X which is 10 light years away from earth at speed 0.5c, hence the journey takes 20 years. A then wait for B who goes to X afterwards at speed 0.2c, hence the journey takes 50 years.
When A and B meet again on planet X, who is the older one?

If we assume that B starts after A after A reaches his destination and come to a halt, then
A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 years
B will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

If we assume that B starts after A after A reaches his destination and come to a halt, then
A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 years
B will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.
so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.

If we assume that B starts after A after A reaches his destination and come to a halt, then
A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 years
B will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.
so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.
There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.

In the original paradox, the twin A reverses its velocity as soon he arrives planet X. When back to earth he aged 34,64, while the earth twin aged 40 years.
But during all his trip, earth was moving from him at 0.5c. So, from the reference frame of the traveller twin, at first glance, his brother at earth should have aged only 34,64 x 0,866 = 30 years.
But when reversing the engines at planet X, the traveller twin observed a sudden change of the clocks at earth. Just before reversing velocity, t' = 20  0,5 x 10 = 15 years, what should be expected, because for him, the earth was moving, and his brother at earth was aging slower (15 = 17,32 x 0,866).
But just after reversing velocity, the clock at earth shows t' = 20 + 0,5 x 10 = 25 years.
So, in the returning trip, more 17,32 passes for him, and more 15 years at earth. During all the time, the brother at earth aged slowly for the travelling brother, but the change of syncronization when reversing of the engines, made all the difference.
A consequence is that for an observer at earth, the sudden reversing of velocity of the ship would have taken 10 years. During all that time the ship (and any clock inside it) would be seen as stopped at planet X.

A consequence is that for an observer at earth, the sudden reversing of velocity of the ship would have taken 10 years. During all that time the ship (and any clock inside it) would be seen as stopped at planet X.
After I posted it, I realized it was wrong. The "news" of the arriving ship would come to earth at t = 20 + 10 = 30 years. And during the next 10 years the ship would be observed as coming back.

Saspinski, Have you read this?
http://home.earthlink.net/~owl232/twinparadox.pdf
It might help.

A consequence is that for an observer at earth, the sudden reversing of velocity of the ship would have taken 10 years.
There is no turnaround in the scenrio posted by the OP

Saspinski, Have you read this?
http://home.earthlink.net/~owl232/twinparadox.pdf
It might help.
Yes, it is a comprehensive explanation, using the same reference frame for all the journey.
But it is also possible to change the reference frame together with the traveller twin, since you don't forget the sudden change in the earth time from his point of view at the turning point.

This is a modified thought experiment about twin paradox.
Twin A goes to planet X which is 10 light years away from earth at speed 0.5c, hence the journey takes 20 years. A then wait for B who goes to X afterwards at speed 0.2c, hence the journey takes 50 years.
When A and B meet again on planet X, who is the older one?
Who's still alive?
Clearly the math says a faster rate of travel incurs less age, yet who's aging more and through what time period? You'll find an answer there based on the math.

Who's still alive?
Well, 69 and 67 seem reasonable life expectancy, unless they have succumbed to disease or forgotten to pack food.
Clearly the math says a faster rate of travel incurs less age, yet who's aging more and through what time period? You'll find an answer there based on the math.
Here to be exact:
https://www.thenakedscientists.com/forum/index.php?topic=72502.msg535232#msg535232

If we assume that B starts after A after A reaches his destination and come to a halt, then
A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 years
B will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.
so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.
There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.
There is no paradox from A's perspective, PROVIDED he realizes that B (she) instantaneously ages during A's instantaneous stop at his destination. If he DOESN'T realize that, then he DOES perceive a paradox when she arrives, because he would expect her to be younger at arrival. That's the SAME issue that occurs in the standard twin "paradox" scenario.

Instantaneous aging? Is this because the mathematics are too hard?

Instantaneous aging? Is this because the mathematics are too hard?
"Near" instantaneous would be better. There is nothing that prevents us with dealing with low accelerations over extended periods, it's just that for the purposes of illustration of the principles involved, it jut not worth the trouble. For example, let's consider A during that part of their outbound trip where they come to a rest with respect to B some ten light years distant from B.
If he were to do this by maintaining a constant 1g acceleration( as measured by himself), then he would have to start his deceleration when he is 0.15 ly (as measured by B) from where he wants to stop. If you want to calculate how much time passes on B's clock according to A during this period, you have to factor in the fact that his relative velocity with respect to A is constantly changing, and thus the time dilation due to that. You also have to figure in the effects caused by his being in an accelerated frame. Clocks in the direction of the Acceleration (towards B) will run fast, by a factor determined by the magnitude of his acceleration and the distance to those clocks. In the case of B's clock this distance is increasing as A goes travels between the point where they first start the deceleration and where they come to a stop with respect to B.
Not an impossible task, but an extra complication that doesn't really aid in the example. It is much simpler to assume that the acceleration is very high and over a very short distance.
So, for example, I gave the respective ages for A and B as being 67.32 and ~69 years. But these answers are a result of rounding. As long as we assume that the acceleration phases for either A and B occurred over a very short distance any difference in the answer made by accounting for them are too small to effect the rounding.
The problem with using truly Instantaneous changes in velocity is that this equates to an infinite acceleration over zero time, and anytime you bring infinities (especially coupled with a division by zero) in, you are inviting headaches.

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.
There is no paradox from A's perspective, PROVIDED he realizes that B (she) instantaneously ages during A's instantaneous stop at his destination. If he DOESN'T realize that, then he DOES perceive a paradox when she arrives, because he would expect her to be younger at arrival. That's the SAME issue that occurs in the standard twin "paradox" scenario.
I should also add that B (she) also doesn't perceive a paradox, PROVIDED she realizes that when she instantaneously changes velocity to leave home, A (he) will instantaneously age. But if she DOESN'T realize that, then she DOES perceive a paradox when she arrives at the destination, because she finds A older than she expects him to be. Again, that's the SAME issue that occurs in the standard twin "paradox" scenario.

Instantaneous aging? Is this because the mathematics are too hard?
The mathematics isn't too hard (at least for piecewiseconstant accelerations), but it's not as easy as for the instantaneous velocitychange case. Instantaneous velocity changes are of course not actually possible, but it is a mistake to think that the results they provide are so unlike the realistic cases as to be of no use. There are realistic scenarios, involving acceleration segments of only 1g, that are qualitatively very similar to the instantaneous velocitychange case. On my webpage
https://sites.google.com/site/cadoequation/cadoreferenceframe (https://sites.google.com/site/cadoequation/cadoreferenceframe)
I give an example where the traveler (he) is about 40 ly away from his home twin (according to her), traveling away from her at about .77 ly/y, and then accelerates for two years (of his life) at 1g in a direction TOWARDS her. During that acceleration, he concludes that she gets about 64 years older. So during the acceleration, she ages more than 30 times faster than he does.
If he then decides to accelerate for another two years (of his life), but this time in the direction AWAY from her, he will conclude that she gets about 59 years YOUNGER during the acceleration.

This may do nothing to stem the flow of questions/speculations about the terrible twins, but it is interesting.
https://www.laboratoryequipment.com/news/2018/03/nasastudykellyastronauttwinsshowsspacegenesdifferences?et_cid=6288235&et_rid=517749120&type=cta&et_cid=6288235&et_rid=517749120&linkid=https%3a%2f%2fwww.laboratoryequipment.com%2fnews%2f2018%2f03%2fnasastudykellyastronauttwinsshowsspacegenesdifferences%3fet_cid%3d6288235%26et_rid%3d%%subscriberid%%%26type%3dcta

We looked at the twin paradox in our programme on the science of Star Wars  "Star Wars: The Science Strikes Back" (https://www.thenakedscientists.com/podcasts/nakedscientistspodcast/starwarssciencestrikesback)  late last year, which might prove interesting to readers of this thread...

Searched for twin paradox in youtube, here are some of the top results.
Which one do you like best? Why?
Based on the comment of the videos, it seems that they still leave many people in confusion.

If we assume that B starts after A after A reaches his destination and come to a halt, then
A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 years
B will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.
so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.
There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.
What would happen if the calculation is done in A's reference frame?
According to A, he never moves. Instead he will see B moves away at 0.5c for 20 years and then moves back at 0.2c for 50 years. Accounting for time dilation, after 70 years according to A, B would only age 17.32+48.99=66.31 years.
On the other hand, according to B, he never moves. Instead he will see A moves away at 0.5c for 20 years and then moves back at 0.2c for 50 years. Accounting for time dilation, after 70 years according to B, A would only age 17.32+48.99=66.31 years.

What would happen if the calculation is done in A's reference frame?
According to A, he never moves. Instead he will see B moves away at 0.5c for 20 years and then moves back at 0.2c for 50 years. Accounting for time dilation, after 70 years according to A, B would only age 66.31 years.
The ages of the respective twins upon there meeting is computed to be the same regardless of which frame is used for the computations. That meeting is an objective event. Hence, A will be ~67.3, B will be ~69.
Neither person is inertial for the duration of the exercise, so there is no one frame for either of them.

Here is the space time diagram for the case.
The left is in earth reference frame, the middle is in A reference frame, while the right is in B reference frame.
The question is how to get the same result using Lorentz transform in A and B's reference frame? (Probably using a similar method as in Minute Physics' or The Rest of Us' video)
To make it clear, the height of lower lines are 20 unit, while the upper lines are 50 units to represent duration of A's and B's journeys, respectively.

How very Newtonian

Here is an alternative diagram showing time dilation for the moving observer.
The bold red line shows when A is moving, hence his time is dilated to 17.32 years instead of 20 years for his journey.
The bold blue line shows when B is moving, hence his time is dilated to 48.99 years instead of 50 years for his journey.
This calculating method is often used to explain experiment of muon under relativistic effect.
The diagram is scaled so that objects moving at light speed will be represented by 45 degree lines. The height of lower lines in the middle diagram are 17.3 units, while the upper lines of right diagram are 49 units to represent dilated time.
The remaining question is, how can each twin calculate the correct age of his brother by accounting their difference in frame of reference?

Here is another alternative accounting for relative simultaneity. According to A, B didn't start the journey yet as he arrive on X.
According to B, A hasn't arrived on X yet as he start going.
Here the height of bold red line in the middle diagram is 17.3 units while its blue counterpart is kept at 20 units.
The bold blue line in the right diagram is 49 units while its red counterpart is kept at 50 units.

How very Newtonian
The nonNewtonian relativistic methods are presented in my posts above. The theory of special relativity has provided a tool to calculate distance and period of time in a frame of reference when observed from another frame of reference. The question here is how to utilize the tool appropriately to get consistent answers among different frames of reference.

The nonNewtonian relativistic methods are presented in my posts above. The theory of special relativity has provided a tool to calculate distance
Yes, it provides tools for computing the distance, but no distance is computed in your pictures. It isn't hard, but the pictures have no labels or numbers. Try computing distance. In A's outgoing frame, just before X gets to A, how far away is B? Your picture just doesn't show that, and if you actually compute it, you'll get a better picture.

Yes, it provides tools for computing the distance, but no distance is computed in your pictures. It isn't hard, but the pictures have no labels or numbers. Try computing distance. In A's outgoing frame, just before X gets to A, how far away is B? Your picture just doesn't show that, and if you actually compute it, you'll get a better picture.
Here is the picture showing length contraction as well as time dilation experienced by moving observer.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=72502.0;attach=30405;image)
FYI, the picture is made using shapes in excel. Their width and height are set in format pane.
The unit of space axis is light years, while the time axis is in years.
In A frame, there is sudden jump in his distance to B when A is stopping on planet X. When A was moving, B was 8.7 light years behind, but when A already stopped, B's position becomes 10 lightyear behind.
In B frame, there is sudden jump in his distance to A when B is starting to move from earth. Just before B was moving, A was 10 light years ahead, but when B already moved, A's position becomes 9.8 lightyear ahead.
Is this the corect explanation for the case here?

In A frame, there is sudden jump in his distance to B when A is stopping on planet X. When A was moving, B was 8.7 light years behind, but when A already stopped, B's position becomes 10 lightyear behind.
Excellent. That's what I was looking for. The funny jump wasn't there before.
In B frame, there is sudden jump in his distance to A when B is starting to move from earth. Just before B was moving, A was 10 light years ahead, but when B already moved, A's position becomes 9.8 lightyear ahead.
Yes. Hard to see, but putting the numbers there helps.
All correct now. The explanation is that the distance to observers changes (dilates) given the inertial frame change of the observer. The top of the two graphs are also correctly different times, each reflecting the proper time of the 'primary' observer, the one with the straight line.

So in the last picture, we get the speed of the moving observer as measured by the stationary one is the same as the speed of stationary observer measured by the moving one. In the example above, they are 0.5c and 0.2c. The effect of length contraction and time dilation cancels out in the calculation of speed. But consequently, we get jump of distance as measured by moving observers when they are changing their speed. it's interesting that this jump isn''t explicitly mentioned anywhere in the videos explaining twin paradox.
In B's frame, during quick deceleration on planet X, A (and earth) appear to change position very rapidly from 8.7 to 10 light years. If the deceleration happens in 1 second, then A appears to move much higher than speed of light. This can be observed by looking at the gradient of the line during the jump.
Even a mundane acceleration/deceleration of 1 g would make a very far away object to move/change position very quickly, even exceeding the speed of light.

So in the last picture, we get the speed of the moving observer as measured by the stationary one is the same as the speed of stationary observer measured by the moving one. In the example above, they are 0.5c and 0.2c.
Not necessarily. There is a time jump as well and if the distant twin accelerated during that time, the symmetry is lost. So what you say is true only for inertial observers. It works for this example because of them both accelerating somewhat near the same time, and the acceleration is instant, not gradual.
The effect of length contraction and time dilation cancels out in the calculation of speed. But consequently, we get jump of distance as measured by moving observers when they are changing their speed. it's interesting that this jump isn''t explicitly mentioned anywhere in the videos explaining twin paradox.
In the standard scenario, one twin never accelerates, so no jump from the home perspective. In the standard scenario, the outbound speed equals the return speed, but if they differ, you get that jump from his perspective.
In B's frame, during quick deceleration on planet X, A (and earth) appear to change position very rapidly from 8.7 to 10 light years. If the deceleration happens in 1 second, then A appears to move much higher than speed of light.
That's nothing special. From my perspective here in my back yard, every star in the sky except our sun is moving at far greater than light speed. No biggie. It is only forbidden in inertial frames.
Even a mundane acceleration/deceleration of 1 g would make a very far away object to move/change position very quickly, even exceeding the speed of light.
or would make time go backwards for a sufficiently distant object. The Andromeda 'paradox' is all about that.

Not necessarily. There is a time jump as well and if the distant twin accelerated during that time, the symmetry is lost. So what you say is true only for inertial observers. It works for this example because of them both accelerating somewhat near the same time, and the acceleration is instant, not gradual.
(https://upload.wikimedia.org/wikipedia/commons/thumb/c/ce/Twin_Paradox_Minkowski_Diagram.svg/500pxTwin_Paradox_Minkowski_Diagram.svg.png)
So to visualize the time jump, we need to draw tilted line of simultaneity, according to the object's speed.

Here is the same scenario, but the twins start from earth simultaneously.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=72502.0;attach=30415;image)
In earth frame, A arrive on planet X at 20 years, while B arrive at 50 years. To help us track the events, let's put a checkpoint at 4 light years milestone where B is located when A just arrived on X.
In A frame, the distance from earth to planet X contracts to 8.66 lightyear and the travel last for only 17.32 years.
B moves 0.2c relative to earth, thus with relativistic velocity addition formula, It becomes 0.3333c relative to A. So when A arrives on planet X, B's position is 5.77 light year in A's frame.
Then A stops on planet X, which makes B apparently changes his velocity to +0.2c. This change in velocity makes the position jumps from 5.77 to 6 light years. The next part of B's journey is observed as the same events from earth as well as from A. B finishes the remaining 6 light year travel in 30 years.
A and B meet on planet X when A has increased his age by 47.32 years, counted from start of the journey.
In B frame, A moves at 0.3333c. B arrives at the checkpoint at 3.92 light years journey in 19.6 years. At the same time, A arrives on planet X located at 6.53 light years away from B.
Then A stops on planet X, which makes him changes his velocity to 0.2c as observed from B. Since B doesn't change speed, he doesn't observe position jump. He sees A move at 0.2c from 6.53 light years to 0 in 29.39 years.
A and B meet on planet X when B has increased his age by 48.99 years, counted from start of the journey.
It looks like our numbers are still consistent with previous scenario, although the calculation is more complicated.

Here I found two common logical errors we can make in trying to explain asymmetric result in twin paradox.
https://hal.archivesouvertes.fr/hal02263337/document
Gerrit Coddens. What is the reason for the asymmetry between the twins in the twin paradox?. 2019.
hal02263337v5
Appendix: Two possible logical errors we can make in trying to explain P2
6.1 Boobytrap 1: Sarah does not forcedly feel her accelerations
In this subsection we want to show that an argument often used in the discussion is wrong. Many approaches are based on the narrative which has it that Sarah must accelerate her spacecraft at the points P and Q, that she can bring about the accelerations needed by firing her rockets, that she will then feel these accelerations and that this in turn will clearly show that the accelerations could be responsible for the asymmetry. And when we fail to imagine any other paradigm to solve the paradox, we readily make the leap that they should be responsible for it. Well, they are not! The whole narrative is a leading argument and not in the least compelling. Firing the rockets of a spacecraft is not the only way we can accelerate it. To illustrate this we can refer to the way NASA reduces the cost of bringing satellites to distant planets, by putting to profit accelerations exerted by other planets along a carefully planned trajectory. These accelerations are gravitational and as such not necessarily felt.
A spacecraft in circular orbit around the Earth is in an accelerated motion and the people inside do not feel anything of this motion: They are “weightless”. This is so symmetrical that in the analogous problem of the motion of the Earth around the Sun we have believed for millenia that the Earth was standing still with the rest of the Universe revolving around us. If we replaced the Sun by a black hole and we kept far enough away from it in order not to feel tidal effects at the length scale of our bodies, we could circle around that black hole at relativistic velocities and think we are standing still because we do not feel any acceleration: In the language of general relativity we would be travelling on a geodesic. The argument contains thus a pars pro toto. One cannot always notify to Sarah that she should admit that she has felt accelerations in order to convince her that she would not be entitled to turn the tables by claiming that from her viewpoint it was Théo who made the travel. Of course, felt or otherwise, all accelerations will introduce additional effects of time dilatation in addition to those of special relativity.
Our considerations about their contributions must intervene in the analysis of the paradox P2 ∈ T2, but it is absolutely not our intention to address the solution of P2 ∈ T2. We suspect that the accelerations may displace the problem rather than solving it, because under certain conditions accelerated motion can also be relative, as illustrated by the examples given.
Boobytrap 2: The time dilatation during accelerated motion has no bearing on the time dilatation during uniform motion
The following argument shows how we can render the relative contribution of the accelerations to the age difference negligible in a physical experiment with real twins. Let us consider a straight line segment PQ as the trajectory of the journey. We can make the accelerations take place over short distances PP1, Q1Q, QQ1 and P1P. The motions over P1Q1 and Q1P1 are then uniform. Whatever the effect of the accelerations may be, we can choose the distance P1Q1 and make it so long that the effect of the accelerations becomes negligible with respect to the age difference that builds up during the uniform motions over P1Q1 and Q1P1. This is is because the age difference builds up linearly along the distance P1Q1. Due to the homogeneity of space, increasing the distance P1Q1 will not change the effect of the accelerations over PP1, Q1Q, QQ1 and P1P. We can thus make the age difference as large as we like by increasing the distance P1Q1, a fact that must condemn any quantitative attempt to account for the assymetry between the two twins on the sole basis of the accelerations. The time dilatations due to the accelerations and those due to the uniform motion are completely independent. Each part of the journey must be considered as making a contribution in its own right, if we want to end up with a viable account of the global time dilatation.
This shows that attributing the asymmetry uniquely to the accelerations is a fivestar glaring error and raises the question of the true origin of the asymmetry when the motion is strictly uniform, as is the case for time dilatation within the pure context of special relativity .

Here is the same scenario, but the twins start from earth simultaneously.
And here is the same scenario, but the twins finish on planet X simultaneously.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=72502.0;attach=30417)
The result is the same as previous case, and the numbers are very similar.

Quote from: hamdani yusuf on Yesterday at 08:47:43
Here is the same scenario, but the twins start from earth simultaneously.
And here is the same scenario, but the twins finish on planet X simultaneously.
We can see that in B's frame, B doesn't change velocity since beginning of the trip until the end, although the diagrams don't show B's position before the journey start and after it ends.
Would it make a difference if B never changes velocity at all?
Let's say that the twin were born in a space ship moving at 0.2c on a line connecting earth and Planet X. When the ship is passing earth, twin A stops by on earth for 30 years in earth time, and then catching up B with faster space ship at 0.5c relative to earth, which makes them meets again after 20 years journey from earth to planet X (measured in earth time). The question is, what's the symmetry breaker between B and earth observer, giving that they never really change their frame of reference?

Would it make a difference if B never changes velocity at all?
B is inertial from the beginning to the end, so what he does before or after that makes no difference to the facts depicted.
The question is, what's the symmetry breaker between B and earth observer, giving that they never really change their frame of reference?
Same as the symmetry breaker before: A is not inertial for the duration of the exercise and B is.

Quote from: hamdani yusuf on Yesterday at 08:47:43
Here is the same scenario, but the twins start from earth simultaneously.
And here is the same scenario, but the twins finish on planet X simultaneously.
We can see that in B's frame, B doesn't change velocity since beginning of the trip until the end, although the diagrams don't show B's position before the journey start and after it ends.
Would it make a difference if B never changes velocity at all?
Let's say that the twin were born in a space ship moving at 0.2c on a line connecting earth and Planet X. When the ship is passing earth, twin A stops by on earth for 30 years in earth time, and then catching up B with faster space ship at 0.5c relative to earth, which makes them meets again after 20 years journey from earth to planet X (measured in earth time). The question is, what's the symmetry breaker between B and earth observer, giving that they never really change their frame of reference?
If B never changes velocity, then the symmetry between B and the Earth observer is never broken.
Spacetime diagram from the Earth observer frame.
[ Invalid Attachment ]
Dark blue line  Earth
Green line  distant planet
blue line  B
Red line  A (if he leaves Earth at the same time as B)
magenta line A ( if he waits at Earth and then leaves)
Times for each clock shown. It is assumed that in the Earth rest frame clocks at the distant planet are synchronized to the Earth clock.
The two times separated by the / represents Distant planet time/ A's time
If you shift to B's frame of reference with no changes in velocity, you get this:
[ Invalid Attachment ]
When B passes Earth the Earth clock reads zero, B sets his clock to zero also (or B is born, if you like.) At this moment the clock at the distant planet already reads 2 years past 0.
B takes ~ 49 years by his clock to reach the distant planet, during which time, Both the Earth clock and distant planet clock advances ~48 years. Thus the planet's clock reads 2+48 = 50 years upon his arrival, while according to him the Earth clock reads 0+48 = 48 years at that same moment.
The cyan line represents the "checkpoint" that you mentioned earlier and where B would be at when A(red line ) reaches the planet (as measured in the Earthplanet frame.) In B's frame he doesn't reach this point until after A has reached the distant planet.
The point is that if B never changes velocity with respect to the Earth, B and and the Earth will disagree as to who aged more between B passing Earth and passing the planet.

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.
There is always a paradox with STR: the paradox is found in the issue of which clock was ticking faster during either one of the legs of the trip. STR has both clocks ticking faster than each other at the same time, and that's a mathematical impossibility.

Same as the symmetry breaker before: A is not inertial for the duration of the exercise and B is.
My question is, why the separation period of the twins is measured lower in B's frame compared to earth's, while both of them are equally inertial. The only difference I can find is that planet X is stationary in earth frame while moving in B frame.

My question is, why the separation period of the twins is measured lower in B's frame compared to earth's, while both of them are equally inertial. The only difference I can find is that planet X is stationary in earth frame while moving in B frame.
The end of the separation period is defined by event R (reunion) which is on B's worldline (and A's as well), but not Earth's. In any frame, B is 48.99 years old at that event. In Earth frame, Earth is age 50 at an event simultaneous with R. In B's frame, 48 years have gone by on Earth when the reunion takes place.
Is that what you're asking? The separation period is lowest from A's POV actually. Earth's frame doesn't really count since Earth is not present at event R.
None of this has to do with the frame of Planet X, since X's frame does not effect the reunion. If both A and B assume frame X when they arrive together, then we can do it in X's frame, but I'm talking about the frame of B enroute when referencing B's frame above.

Same as the symmetry breaker before: A is not inertial for the duration of the exercise and B is.
My question is, why the separation period of the twins is measured lower in B's frame compared to earth's, while both of them are equally inertial. The only difference I can find is that planet X is stationary in earth frame while moving in B frame.
If you are asking why B measure's 48.99 yrs between passing Earth and meeting up with A at planet X, it is because in B's frame the spatial separation between Earth and planet X is ~9. 798 light years. So, according to B, the Earth passes by at 0.2c, with planet x trailing 9.798 light years behind at the same velocity. Thus it takes 9.798 ly/ 0.2c = 48.99 yrs between Earth passing and Planet X passing as measured by B.

Let's make the situation more symmetrical by adding planet Y, which is stationary in B frame and located 10 light years behind, measured in B frame. How far is the distance between B and planet Y when measured by earth observer?
Let's say the earth observer also has a twin, called D. When B is passing by and A jump to earth, D jump into B's space ship. After 30 years in the ship, D return to earth at 0.5c measured in B frame. Will he arrive on earth at the same time as planet Y passing by earth?

Let's make the situation more symmetrical by adding planet Y, which is stationary in B frame and located 10 light years behind, measured in B frame. How far is the distance between B and planet Y when measured by earth observer?
Still 9.8 LY. They're moving in Earth frame, so the distance is contracted.
Let's say the earth observer also has a twin, called D.
No C? What happened to poor Charlie? Run over by a bus at birth. Tragic...
When B is passing by and A jump to earth, D jump into B's space ship. After 30 years in the ship, D return to earth at 0.5c measured in B frame. Will he arrive on earth at the same time as planet Y passing by earth?
All in Bob's frame: Dewey is stationary for 30 years, then travels 10 LY to planet Y at 0.5c, so it takes 20 Bob years to get there. 50 years total. Earth (with Alice) is moving steadily to Y at 0.2c, so it also takes 50 years to do it. So yes, they get there at the same time, and I didn't need to invoke relativity to figure that out. Relativity is only needed to figure out their ages when Dewey is reunited with Alice and they can grieve over Charlie's grave together.

No C? What happened to poor Charlie? Run over by a bus at birth. Tragic...
The letter c is already occupied frequently for speed of light. D was chosen because it's right before E, which stands for Earth observer in previous scenario.All in Bob's frame: Dewey is stationary for 30 years, then travels 10 LY to planet Y at 0.5c, so it takes 20 Bob years to get there. 50 years total. Earth (with Alice) is moving steadily to Y at 0.2c, so it also takes 50 years to do it. So yes, they get there at the same time, and I didn't need to invoke relativity to figure that out. Relativity is only needed to figure out their ages when Dewey is reunited with Alice and they can grieve over Charlie's grave together.
In the end of the journey, D will be reuniting with his twin, E when planet Y is passing Earth. Meanwhile, A will be reuniting with B when they are passing planet X.

Let's make the situation more symmetrical by adding planet Y, which is stationary in B frame and located 10 light years behind, measured in B frame. How far is the distance between B and planet Y when measured by earth observer?
Let's say the earth observer also has a twin, called D. When B is passing by and A jump to earth, D jump into B's space ship. After 30 years in the ship, D return to earth at 0.5c measured in B frame. Will he arrive on earth at the same time as planet Y passing by earth?
From the Earth frame:
Y is 9.798 ly away as B passes, and is approaching at 0.2c. It will take 48.99 yrs to reach Earth.
D has jumped on to B's ship, an stays for 30 yrs (B's time)., This takes 30.62 years by Earth's clock. This means that at that point B and D will be 0.2c x 30.62 = 6.124 ly from Earth.
D heads back towards Earth at 0.5c relative to B as measured by B, this equates to (0.2c0.5c)/(1(0.2c)(0.5c)) = 0.333...c
(minus sign represents direction back towards Earth.)
6.124 ly/ 0.333c = 18.372 y
Added to the 30.62 yrs equals 48.99 yrs. The same amount of time that passed between B passing and Planet Y arriving.
In B's frame:
Y is 10 light years away when he passes Earth and D jumps aboard. After 30 yrs of flight, B and D are 6 ly from Earth. And Y is 4 ly from the Earth. With D moving at 0.5c towards the Earth and the Earth receding at 0.2c, the closing speed between D and the Earth, according to B is 0.3c, at which rate, it will take 20 years for D to reach the Earth. It also will take 20 yrs for planet Y to cover 4 ly at 0.2c. So according to B, it takes 50 yrs between his passing Earth and both D and Y to meet up at the Earth. Meanwhile, according to B, the clock on Earth has been time dilated and accumulated 48.99 yrs. The exact same amount of time as determined by the Earth.

The letter c is already occupied frequently for speed of light. D was chosen because it's right before E, which stands for Earth
I'm pretty sure that people wouldn't have been confused by using "c" for the speed of light and "C" for the extra participant. And even if you messed up here and there and used "c" when you meant "C", the context in which it was used would have made it clear at to what you meant.

With D moving at 0.5c towards the Earth and the Earth receding at 0.2c, the closing speed between D and the Earth, according to B is 0.3c, at which rate, it will take 20 years for D to reach the Earth.
That doesn't seem like relativistic velocity addition.
To make the scenario perfectly symmetrical using those numbers, in B frame D should aim to planet Y which is 10 lightyears away. At speed of 0.5c, it will take 20 years to reach planet Y.

With D moving at 0.5c towards the Earth and the Earth receding at 0.2c, the closing speed between D and the Earth, according to B is 0.3c, at which rate, it will take 20 years for D to reach the Earth.
That doesn't seem like relativistic velocity addition.
It's not. Relativistic velocity addition is used in situations where you add Y's velocity relative to X, and Z's velocity relative to Y. In other words the velocities are measured in different frames. Here, all velocities are relative to B's frame, so regular arithmetic is used to add/subtract them. Here, Janus was expressing the difference in velocities as measured in B's frame. Notice that he said 'according to B'. To compute that same relative velocity relative to either Earth or D would require the relativistic addition.
To make the scenario perfectly symmetrical using those numbers, in B frame D should aim to planet Y which is 10 lightyears away. At speed of 0.5c, it will take 20 years to reach planet Y.
Right, and if you look at how I did the math at the bottom of post 43, you see I computed it just like that: how long Dewey takes to get to Y.

With D moving at 0.5c towards the Earth and the Earth receding at 0.2c, the closing speed between D and the Earth, according to B is 0.3c, at which rate, it will take 20 years for D to reach the Earth.
That doesn't seem like relativistic velocity addition.
To make the scenario perfectly symmetrical using those numbers, in B frame D should aim to planet Y which is 10 lightyears away. At speed of 0.5c, it will take 20 years to reach planet Y.
As point out by Halc, it isn't needed here.
Example, You have Alice, Bob, and Charlie.
Alice "stays home", while as measured by her, Bob and Charlie both head off in opposite directions at 0.5c
According to Alice, With Bob going in one direction and Charlie in the other, the distance between them is increasing at a rate of 0.5c + 0.5c = 1c . After 1 hr they will be 1 light hr apart from each other ( Bob will be 0.5 light hr from Alice in one direction, and Charlie 0.5 light hr from Alice in the the Other.) etc.
For Bob to get the relative speed between himself and Charlie, he adds the 0.5c he measures relative to Alice, to the 0.5c Alice measures as the relative velocity between herself and Charlie, using relativistic velocity addition.
(.05c+..5c)/1(1+0.5c(0.5c)/c^2) = 0.8c
The difference between Charlie's velocity and Alice's velocity as measured by Bob is 0.8c0.5c = 0.3c
After 1 hr ( by Bob's clock) Bob measures Alice to be 0.5 light hr away, and Charlie as being 0.8 light hr way. A difference of 0.3 light hr.

One thing about this socalled paradox that is worth a little thought touches on the very basics of how we think about time travel.
Dan and Emily are twins, and Dan makes the obligatory journey, after which he will be 20 yrs younger than his sister. Fast forward to the point of return in (say) the year 2150. Dan has aged 20 yrs less than Emily; he is at the stage she would have been at 20 yrs earlier. It is easy to think of this as his having travelled 20 yrs into his sister’s past, but is that the wrong way of looking at it? Both are at the same point in time – 2150. In what sense has Dan travelled into the past?
Consider another scenario. This time there is no journey into space, but in 2150 Dan uses a time machine to travel back 20 yrs. He meets Emily. She is as she was in 2130. He has travelled 20 yrs into her past and she is younger, not older, as is the case in the former scenario.
Dan comes back from space younger than Emily. Pop Sci books and online discussions provide the "hitchhiker" with abundant explanations for that. However, if, instead of the highspeed journey, Dan had been placed in some sort of “stasis chamber” in which his development and ageing had been halted for 20 yrs, the result in 2150 would be the same as in the first scenario. He would appear to be 20 yrs younger than Emily. So, would we claim that he had timetravelled?

One thing about this socalled paradox that is worth a little thought touches on the very basics of how we think about time travel.
Dan and Emily are twins, and Dan makes the obligatory journey, after which he will be 20 yrs younger than his sister. Fast forward to the point of return in (say) the year 2150. Dan has aged 20 yrs less than Emily; he is at the stage she would have been at 20 yrs earlier. It is easy to think of this as his having travelled 20 yrs into his sister’s past, but is that the wrong way of looking at it? Both are at the same point in time – 2150. In what sense has Dan travelled into the past?
In no sense has he traveled into Emily's past. He just experienced 20 yrs less time than she did between their separation and rejoining. Relativity never claims that Dan travels into the past, only that the time interval he measured between the two events was shorter. Time passed differently for the two. If Dan left Emily when both their calendars read 2100, then when they met up again, Emily's calendar would read 2150, while Dan's would read 2130.
When They meet up again, it is 50 yrs into Emily's "future" and 30 yrs into Dan's future.
When you state that they are are the same point in time, 2150. you are implying some type of absolute universal nature to time, and that 2150 is the "real" time. It is only 2150 by the Earth clocks and this no more the "real time" than the 2130 according Dan's clocks. It is just that when they meet up again, Dan agrees that by the Earth clock it is 2150 (twenty more years passed for Earth than did for him.)
There is the time as measured by Emily, and there is the time as measured by Dan. and there is no more meaning to time than that in this scenario.
Consider another scenario. This time there is no journey into space, but in 2150 Dan uses a time machine to travel back 20 yrs. He meets Emily. She is as she was in 2130. He has travelled 20 yrs into her past and she is younger, not older, as is the case in the former scenario.
Again, the SR twin scenario does not claim any travel "into the past".
Dan comes back from space younger than Emily. Pop Sci books and online discussions provide the "hitchhiker" with abundant explanations for that. However, if, instead of the highspeed journey, Dan had been placed in some sort of “stasis chamber” in which his development and ageing had been halted for 20 yrs, the result in 2150 would be the same as in the first scenario. He would appear to be 20 yrs younger than Emily. So, would we claim that he had timetravelled?
Once again. SR does not imply "time travel" other than the time travel we all experience everyday from this moment to the next. What SR does say is the there is no universal meaning to "the passage of time". Every inertial reference frame measures time by its own independent standard, and that is the only meaning to "time" there is.
It's like the notions of left and right. Everyone's "left" and "right" is unique to them. If we are standing next to each and not facing the same direction, my "left" will not be the same as your "left'. And there is no "universal" concept of "leftness".
Putting Dan in a stasis chamber, while he stays at rest with respect to Emily might, for Dan, "seem" the same personally, but only If Dan were not allowed to measure what was happening outside his chamber. It might "mimic" the end result of Relativistic effects, but it wouldn't be the same.

The Special Relativity is reciprocal.
The time dilation is reciprocal and therefore the twin paradox is a logical fallacy.
It is easy to prove, because Einstein introduced light propagation frame dragging with the postulate about the light propagation.
Jano

Here is a quick explanation of the problem, train car moving at 0.866c, there are train (red lines) and platform (blue lines) observers:
(https://i.imgur.com/1pavPu2.png)
The red arrow up is crossing ydirection at speed c in the train frame
… and because the light beam is frame dragged then the ycomponent speed in the platform frame is c/2.
Copy and paste, replacing red and blue, changing frames.
The blue arrow up is crossing ydirection at speed c in the platform frame
… and because the light beam is frame dragged then the ycomponent speed in the train frame is c/2.
s  platform second, s'  train second... for better readability
There is no length contraction issue. The red arrow moves along x’=0cs’, it leads straight to time dilation definition.
The same goes for the blue arrow up, it moves along x=0cs and it leads straight to time dilation definition in the reciprocal way.
(https://i.imgur.com/mSng7qh.png)
The events:
AB – red arrow up on the train > 1cs'.
AE – blue arrow to the right > 2cs – AB time dilation.
AC – blue arrow up on the platform > 1cs.
AD – red arrow to the left > 2cs' – AC time dilation.
AE – blue arrow up and down on the platform > 2cs – the full round trip on the platform.
AF – 4cs' – time dilation of the full platform round trip AE.
The two observers see each others time to tick slower. They do not agree on time,
Jano

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.
There is always a paradox with STR: the paradox is found in the issue of which clock was ticking faster during either one of the legs of the trip. STR has both clocks ticking faster than each other at the same time, and that's a mathematical impossibility.
The Special Relativity is reciprocal.
The time dilation is reciprocal and therefore the twin paradox is a logical fallacy.
It is easy to prove, because Einstein introduced light propagation frame dragging with the postulate about the light propagation.
Jano
Not true. Time dilation is reciprocal only with inertial motion. In every twin type scenario at least one of them has to accelerate which will desync their watches (unless they both accelerate at the same rate).
Doppler shift also cause two watches in inertial motion to each run slower from the other's perspective, nothing to do with relativity. This is direction dependant though, they'll each run faster from the other's perspective if they're moving towards each other.
Doppler shift is usually ignored because it will always cancel itself out once the twins meet back in the same frame as each other again but any question as to what an observer would actually experience in that situation needs to take it into account.

Thanks Janus. I think that helps to confirm my suspicion that when PS authors use this sort of thing as “evidence” for pastdirected time travel, they are probably promoting their own ideas, rather than established scientific thought.
Interestingly, Wolfson, when talking of future directed TT, says: “If you don’t like the Earth you find 100 years in the future, there’s no going back. You’re either stuck where (or, rather, when) you are, or you can take your chances on a jump further into the future. But you can’t go back.”

Not true. Time dilation is reciprocal only with inertial motion. In every twin type scenario at least one of them has to accelerate which will desync their watches (unless they both accelerate at the same rate).
...
Please, support this statement: "Time dilation is reciprocal only with inertial motion." by quoting some textbooks.
Good luck!
The clock desynchronization is part of the Special Relativity,
Jano
Edit: Check this video, linked on the first page of this thread.
The acceleration has nothing to with the time dilation.

Not true. Time dilation is reciprocal only with inertial motion. In every twin type scenario at least one of them has to accelerate which will desync their watches (unless they both accelerate at the same rate).
...
Please, support this statement: "Time dilation is reciprocal only with inertial motion." by quoting some textbooks.
Good luck!
The clock desynchronization is part of the Special Relativity,
Jano
Clock desynchronization happens with Doppler shift too, that causes no contradiction just as the time dilation caused by relative motion causes none.
In every case two observers that are in the same frame will always agree on the amount of time that has passed on each clock.
In instances where like the twin scenario the twin that accelerated will be the one who experiences less time passing so their watch will be behind the other's once they meet back up.
There's no need to quote any textbook, this is all well known in sr.

Well, let us analyze the video with some additional actions:
(https://i.imgur.com/VUcemI3.png)
The observer A sends a light beam for a roundtrip in the ydirection to his mirror at 1cs distance in the ydirection at EVENT I.
The roundtrip is 2s for the A observer.
The B observer is watching this roundtrip.
Question 1: How much time elapsed for the observer B?
Jano

Well, let us analyze the video with some additional actions:
I'm not watching any videos.
If you have a scenario where one observer makes a journey and then returns back to the same frame as the other observer (which looks to be the case with observer A making the 'journey', in other words A is the one that changes frames) then less time has passed on observer A's watch once they return.
Then both observers agree on the amount of time that's passed on A's watch and both agree on the amount of time that's passed on B's watch. They agree that it's less time on A's watch. B's watch was time dilated from A's frame while they were in inertial motion relative to each other but B's watch sped up from A's perspective every time they (A) changed frames, enough that A's own watch is behind B's once they meet back up despite B's watch being time dilated from A's frame while they were in inertial motion relative to each other.

Well, let us analyze the video with some additional actions:
I'm not watching any videos.
If you have a scenario where one observer makes a journey and then returns back to the same frame as the other observer (which looks to be the case with observer A making the 'journey', in other words A is the one that changes frames) then less time has passed on observer A's watch once they return.
Then both observers agree on the amount of time that's passed on A's watch and both agree on the amount of time that's passed on B's watch. They agree that it's less time on A's watch. B's watch was time dilated from A's frame while they were in inertial motion relative to each other but B's watch sped up from A's perspective every time they (A) changed frames, enough that A's own watch is behind B's once they meet back up despite B's watch being time dilated from A's frame while they were in inertial motion relative to each other.
It is difficult to have a discussion if we do not have a basis for it.
If you checked the video you would see that the observer A is not moving, not making a journey.
The question is about state of the affairs before any frame changes.
Your statements do not apply to the question I posted,
Jano

It is difficult to have a discussion if we do not have a basis for it.
If you checked the video you would see that the observer A is not moving, not making a journey.
The question is about state of the affairs before any frame changes.
Your statements do not apply to the question I posted,
Jano
Okay. Well the description I gave applies to any situation where you're comparing two observers who start off in the same frame, one of them changes frames and then changes back to the original frame.
I don't really get why you'd be asking about before either changes frames. At that point their watches will obviously be running at the same rate as each other and the experiment hasn't started yet.
What normally confuses people is that they can't see how from the perspective of the observer who accelerates, the observer who remains inertial ends up older than them when they see the inertial observer moving slower through time while they're in intertial motion relative to each other. The fact that the inertial observer's watch speeds up from the other observer's perspective whenever they accelerate and this always adds more time to the inertial watch than was lost while it was running slowly is usually left out of people's explanations for some reason. Without that it doesn't make sense. How could one watch end up ahead of the other if they're only ever running slowly from each other's frame?

It is difficult to have a discussion if we do not have a basis for it.
If you checked the video you would see that the observer A is not moving, not making a journey.
The question is about state of the affairs before any frame changes.
Your statements do not apply to the question I posted,
Jano
Okay. Well the description I gave applies to any situation where you're comparing two observers who start off in the same frame, one of them changes frames and then changes back to the original frame.
I don't really get why you'd be asking about before either changes frames. At that point their watches will obviously be running at the same rate as each other and the experiment hasn't started yet.
What normally confuses people is that they can't see how from the perspective of the observer who accelerates, the observer who remains inertial ends up older than them when they see the inertial observer moving slower through time while they're in intertial motion relative to each other. The fact that the inertial observer's watch speeds up from the other observer's perspective whenever they accelerate and this always adds more time to the inertial watch than was lost while it was running slowly is usually left out of people's explanations for some reason. Without that it doesn't make sense. How could one watch end up ahead of the other if they're only ever running slowly from each other's frame?
Well, you really should watch the video.
The time dilation has nothing to do with acceleration. It is a claim made in the video.
Why would the author made such a claim?
Because there are experiments that prove that.
I am not sure why you are talking about the acceleration?
Jano

Well, you really should watch the video.
The time dilation has nothing to do with acceleration. It is a claim made in the video.
Why would the author made such a claim?
Because there are experiments that prove that.
I am not sure why you are talking about the acceleration?
Jano
You've either misunderstood the video or the video is wrong. Acceleration is how you change frames. The one that changes frames is always the younger once they're back in the same frame again. Without that distinction how could a different amount of time possibly pass for each? It's probably just poorly explained in the video, it usually is.

Well, let us analyze the video with some additional actions:
(https://i.imgur.com/VUcemI3.png)
The observer A sends a light beam for a roundtrip in the ydirection to his mirror at 1cs distance in the ydirection at EVENT I.
The roundtrip is 2s for the A observer.
The B observer is watching this roundtrip.
Question 1: How much time elapsed for the observer B?
Jano
I am going to answer the question I posted above.
The video analysis is wrong.
The author or any relativity experts are welcome to defend the video claim here.
(https://i.imgur.com/Z0BktrL.png)
The frame dragging:
(https://i.imgur.com/1pavPu2.png)
It is clear that A's light roundtrip beam crosses ydirection at c but the B observer's ydirection speed is c/gamma = c/22.4 for the same light beam.
The A's 2s are equal 2*22.4=44.8s' on the rocket for the B observer.
This is the opposite of what the video claims,
Jano

Well, you really should watch the video.
The time dilation has nothing to do with acceleration. It is a claim made in the video.
Why would the author made such a claim?
Because there are experiments that prove that.
I am not sure why you are talking about the acceleration?
Jano
You've either misunderstood the video or the video is wrong. Acceleration is how you change frames. The one that changes frames is always the younger once they're back in the same frame again. Without that distinction how could a different amount of time possibly pass for each? It's probably just poorly explained in the video, it usually is.
From here: http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html
The clock postulate generalises this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, this ratio depends only on v, and does not depend on any derivatives of v, such as acceleration. So this says that an accelerating clock will count out its time in such a way that at any one moment, its timing has slowed by a factor (γ) that depends only on its current speed; its acceleration has no effect at all.
Although the clock postulate is just that, a postulate, it has been verified experimentally up to extraordinarily high accelerations, as much as 1018 g in fact (see the FAQ What is the experimental basis of Special Relativity?).
This is the basis for the video claim,
Jano

Well, you really should watch the video.
The time dilation has nothing to do with acceleration. It is a claim made in the video.
Why would the author made such a claim?
Because there are experiments that prove that.
There are not experiments that 'prove' that. Fact of the matter is, barring gravity scenarios, two clocks initially in sync in each other's presence cannot later be out of sync in each other's presence without at least one of them having accelerated at some point. That means that it very much at least has something to do with acceleration, even if acceleration alone is not enough.
There are several explanations about 'the cause' of the twins not being the same age, and most of them revolve around relativity of simultaneity in one form or another, but there is no one correct way to explain it.
You've either misunderstood the video or the video is wrong. Acceleration is how you change frames.
Technically, a change of frames is just an abstract operation and requires no acceleration. What you mean to say is that acceleration is how one changes the frame in which one is stationary. I'm kind of anal about the difference between the two wordings, because I'm 'in' all frames at a given time, but stationary in only one of them.
The clock postulate generalises this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, this ratio depends only on v, and does not depend on any derivatives of v, such as acceleration.
This comment (admittedly out of context) is seriously misleading, if not outright wrong. When a clock is accelerating, it is in an accelerating reference frame and ARFs do not have the properties of an inertial reference frame (IRF). The time of some remote clock in the momentary interial frame of the accelerating clock may be stopped, running fast, slow, or backwards. It is a function of v, but not only of v, as this statement seems to assert.
Although the clock postulate is just that, a postulate, it has been verified experimentally up to extraordinarily high accelerations, as much as 1018 g in fact (see the FAQ What is the experimental basis of Special Relativity?).
Actually, I'm not sure what this 'clock postulate' is that they're talking about, nor do I have the context of the referenced 'above quantity'. I'm just commenting on the incomplete quote you posted. Maybe in full context this might make some sense, but then you shouldn't have quoted it out of context like that.
As for the 1018g thing, SR does not suggest that acceleration has any direct effect on dilation. I can have two clocks on Earth, one sitting in a building at the equator and the other in a centrifuge at thousands of g's at the north pole and they'll stay perfectly in sync. Similarly, I can have two twins that, for the duration of an experiment, accelerate continuously at 0.1g and 3g respectively, and the one at the low acceleration turns out younger than the highG one when reunited at the end. But if neither ever accelerates, they're not going to be different ages at the end of the experiment, so any statement that acceleration is unrelated is also wrong.

Halc,
the clock postulate states that the time dilation is a function of velocity and not a function of acceleration.
There could be two clocks undergoing a constant acceleration of the same magnitude but they will have a time dilation between them. That's all.
Having said that, Dr. Lincoln video shows a setup of a twin paradox experiment purely based on the inertial frames, no acceleration, that's doable.
He claims that there is SR time dilation between the inertial observers.
He is wrong, the time dilation is reciprocal due to light propagation frame dragging.
That's the reason the observers do not agree on each others time,
Jano

You've either misunderstood the video or the video is wrong. Acceleration is how you change frames.
Technically, a change of frames is just an abstract operation and requires no acceleration. What you mean to say is that acceleration is how one changes the frame in which one is stationary. I'm kind of anal about the difference between the two wordings, because I'm 'in' all frames at a given time, but stationary in only one of them.
Yes. But it's common practice to refer to an observer's frame as the one in which they're stationary.
Halc,
the clock postulate states that the time dilation is a function of velocity and not a function of acceleration.
There could be two clocks undergoing a constant acceleration of the same magnitude but they will have a time dilation between them. That's all.
Having said that, Dr. Lincoln video shows a setup of a twin paradox experiment purely based on the inertial frames, no acceleration, that's doable.
He claims that there is SR time dilation between the inertial observers.
He is wrong, the time dilation is reciprocal due to light propagation frame dragging.
That's the reason the observers do not agree on each others time,
Jano
The difference in elapsed time on watches of the twins is not directly caused by acceleration. You could set it up again with the same accelerations but a greater difference in elapsed time if the twin that leaves Earth stays in relative inertial motion for longer than the first time, or less difference in elapsed time if the twin that leaves Earth is in relative inertial motion for less time before coming back.
Hopefully those examples help anyone who's trying to grasp it and using Doppler shift simplifies rather than complicated the situation.
Ignoring the effects of relative time dilation for a moment, if observers are moving away from each other each will see the other's watch running slowly due to Doppler shift, if they accelerate when they reach one light year away from each other so they're now in the same frame then each will see that an extra year has passed on their own watch compared to the other observer's watch. If they then move towards each other each will see the other's watch running fast and it will catch up with their own watch once they meet up again, as long as they mirrored each other's acceleration so that relative effects are the same for both of them.
With the twin scenario it's usually one twin that accelerates away and then comes back. In this situation they are still both time dilated from the frame of other twin but the observers aren't mirrored, when they're back in the same frame again at the end they are in a frame in which one twin was time dilated but the one that stayed on Earth wasn't so more time will have passed on the watch of the twin that stayed on Earth.
Now if we combine the two scenarios using the mirrored acceleration from the first scenario, the twins both accelerate away from Earth and then stop accelerating so they're in inertial motion relative to each other. Each will see the other's watch running slowly due to Doppler shift and running slower still due to relative time dilation. If they then accelerate once they reach one light year away from each other so they're now in the same frame then each will see the watch of the other observer speeding up while they're accelerating and one year behind their own watch once they're in the same frame and their watches are ticking at the same rate again.
Now they accelerate back towards Earth and each sees the other's watch ticking faster than their own during the acceleration and a combination of running slower due to relative time dilation while they're inertially travelling towards each other but faster due to Doppler shift, so you could set it up so that each twin sees the other's watch ticking at the same rate as (but still behind) their own watch during the return trips. Once they reach Earth they accelerate to Earth's frame and each sees the other's watch ticking fast during the acceleration so they're the same age as each other back at Earth, because their accelerations were mirrored the whole time.
The key take away is that the difference in elapsed time due to Doppler shift always cancels itself out when two observers are back together because it's always mirrored but time dilation caused by relative inertial motion doesn't (unless their accelerations are mirrored) so it will depend on which watches were time dilated and for how long in the frame they're in when they meet back up to compare watches.

...
The key take away is that the difference in elapsed time due to Doppler shift always cancels itself out when two observers are back together because it's always mirrored but time dilation caused by relative inertial motion doesn't (unless their accelerations are mirrored) so it will depend on which watches were time dilated and for how long in the frame they're in when they meet back up to compare watches.
Awal,
please, define time dilation caused by the relative inertial motion with some drawing/diagram.
... or feel free to point out what is wrong with my post https://www.thenakedscientists.com/forum/index.php?topic=72502.msg597904#msg597904 (https://www.thenakedscientists.com/forum/index.php?topic=72502.msg597904#msg597904).
I am confused how is the Doppler shift related to time dilation in your description,
Jano

I am confused how is the Doppler shift related to time dilation in your description
Doppler shift is caused by the fact that light takes time to travel from one place to another. If you're 1LY away from the source it will take the light one year to reach you, so if you're travelling away from the source you'll see a watch at the source ticking slowly as the light waves get stretched, think of the peaks of each wave and how they'll get further apart. If you're moving towards the watch the light waves get squashed and the watch ticks fast.
Doppler shift always cancels itself out whenever two observers separate and then meet up again because it's caused by increasing the delay in the time it takes for the light to reach you while you're moving away from each other and decreasing the delay in the time it takes for the light to reach you while you're moving towards each other.
The point is that the difference in tick rates of moving clocks caused by Doppler shift is always mirrored with any two observers so each will see the other's watch as moving slower than their own or faster than their own depending on whether they're moving towards or away from each other, but it always cancels itself out if they meet up again in the same frame.
Relative time dilation on the other hand is different. If one observer changes frames then while they're in relative inertial motion to each other each sees the other's watch as moving slower than their own but unless they're accelerations are mirrored, different amounts of time will have passed on the watches when they meet up again.
Which watch is behind depends on which watch was time dilated and who long for in the frame that they're in at the end when comparing their watches.

...
Relative time dilation on the other hand is different. If one observer changes frames then while they're in relative inertial motion to each other each sees the other's watch as moving slower than their own but unless they're accelerations are mirrored, different amounts of time will have passed on the watches when they meet up again.
...
I am still confused. Where is a change of frames for an observer needed for time dilation?
Please, support your claims by quoting textbooks,
Jano

Question to all, what it is that one would not understand the frame dragging?
(https://i.imgur.com/1pavPu2.png)
The red arrow up is crossing ydirection at speed c in the train frame
… and because the light beam is frame dragged then the ycomponent speed in the platform frame is c/2.
Copy and paste, replacing red and blue, changing frames.
The blue arrow up is crossing ydirection at speed c in the platform frame
… and because the light beam is frame dragged then the ycomponent speed in the train frame is c/2.
It is obvious to me, please, help me out.
Is there anything confusing?
Jano

...
Relative time dilation on the other hand is different. If one observer changes frames then while they're in relative inertial motion to each other each sees the other's watch as moving slower than their own but unless they're accelerations are mirrored, different amounts of time will have passed on the watches when they meet up again.
...
I am still confused. Where is a change of frames for an observer needed for time dilation?
Please, support your claims by quoting textbooks,
Jano
If they start off in the same frame then one (or both) of them has to change frames in order for them to be in motion relative to each other. If one changes frames and then comes back so they're in the same frame as each other again the it's the watch of the observer that changed frames that's behind the other observer's watch because in this frame they are the ones that were time dilated while the two were in inertial motion relative to each other.
Both watches will be time dilated from the frame of the other while they're in inertial motion relative to each other but when the observer who changes frames accelerates they will see the other watch speed up so that once they're in the same frame again when they meet back up the watch that didn't change frames is ahead of their own watch despite that other watch running slow while they were in inertial motion relative to it.
Clear? :)
Question to all, what it is that one would not understand the frame dragging?
(https://i.imgur.com/1pavPu2.png)
The red arrow up is crossing ydirection at speed c in the train frame
… and because the light beam is frame dragged then the ycomponent speed in the platform frame is c/2.
Copy and paste, replacing red and blue, changing frames.
The blue arrow up is crossing ydirection at speed c in the platform frame
… and because the light beam is frame dragged then the ycomponent speed in the train frame is c/2.
It is obvious to me, please, help me out.
Is there anything confusing?
Yes, that's not what frame dragging means. Wrong term.

Awal,
not clear.
I'll leave you with a textbook: https://www1.maths.leeds.ac.uk/~serguei/teaching/sr.pdf
Jano

Awal,
not clear.
I'll leave you with a textbook: https://www1.maths.leeds.ac.uk/~serguei/teaching/sr.pdf
Jano
Thanks for that but I'm fine, nothing I've said is in disagreement with SR.
You're clearly struggling however. That's the last time I try to help you understand something that's obviously beyond your ability to grasp.

But it's common practice to refer to an observer's frame as the one in which they're stationary.
Agree, but in these funny threads where the discussion is in disagreement (especially with a denier, which I don't think is the case here), it pays to be precise about one's language. I know what you meant, and I don't disagree with you.
The clock postulate states that the time dilation is a function of velocity and not a function of acceleration.
That seems to be true only relative to inertial frames. Relative to the ARF of the traveler, it is the twin at home that has all the nonzero velocity, and yet that twin is older, not younger, upon reunion. The age of the twin back home is not just a function of his velocity.
There could be two clocks undergoing a constant acceleration of the same magnitude but they will have a time dilation between them.
This is also demonstrably false. Two clocks strapped to each other will follow the same worldline, and hence will have no dilation between them, despite 'a constant acceleration of the same magnitude'. You can do whatever you want to them and they won't get out of sync. They're measuring the same thing.
Having said that, Dr. Lincoln video shows a setup of a twin paradox experiment purely based on the inertial frames, no acceleration, that's doable.
The tagteam scenario is well known, and there is no objective demonstration of one person aging more than another in this example. Two measurements are needed to demonstrate that, but no two pair pf twins/clocks are ever in each other's presence more than once.
He is wrong, the time dilation is reciprocal due to light propagation frame dragging.
I actually don't know what you mean by 'reciprocal'. Frame dragging is more of a gravitational term, not an SR term. You're not using the term correctly.
Now they accelerate back towards Earth and each sees the other's watch ticking faster than their own during the acceleration and a combination of running slower due to relative time dilation while they're inertially travelling towards each other but faster due to Doppler shift, so you could set it up so that each twin sees the other's watch ticking at the same rate as (but still behind) their own watch during the return trips.
As for looking at it from Doppler effect, say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.

Halc,
I did not say the accelerated clocks would be next to each other.
Clocks moving in a circle, the first at radius r=1m with omega=1rad/s and the second at radius r=100m with omega=0.1rad/s ... the acceleration magnitude is the same but the clocks will be time dilated.
Hence the time dilation is a function of velocity and not acceleration.
Please, let us postpone this topic and let us focus on the frame dragging.
I understand your point about the frame dragging being a gravitational term.
Still, let us make an observation.
At t=0s = t'=0s' and x=0cs = x'=0cs' when and where the origins are aligned the train observer sends a light beam across the train car. The red arrow up.
At t=2s the moving origin O' is at x=1.732cs.
The red arrow stays at x'=0cs' all the time in the moving reference frame, it is just moving along the y axis.
The platform observer sees the red arrow frame dragged along the x axis.
I call it a frame dragging because the red arrow is at 90 degrees to x axis at each point where the moving frame origin O' is located along the x axis.
Would you have a better name?
Does this help?
Jano

Now they accelerate back towards Earth and each sees the other's watch ticking faster than their own during the acceleration and a combination of running slower due to relative time dilation while they're inertially travelling towards each other but faster due to Doppler shift, so you could set it up so that each twin sees the other's watch ticking at the same rate as (but still behind) their own watch during the return trips.
As for looking at it from Doppler effect, say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.
But without mentioning that from the perspective of the twin that leaves Earth, the Earth watch has to tick faster than their own when they accelerate it looks like they'll be they same age as each other when they meet back up. They can never be different ages without one of them accelerating more than the other even though acceleration isn't the actual cause.
I really just brought Doppler shift in to show a simplified example of two watches each running slowly from the other's frame but showing the same elapsed time when they meet back up and trying to show that with time dilation the only way to do this is if their acceleration mirror each other but I think it might have just muddied the water, hopefully some readers got some benefit from it.

... say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.
How did you get these numbers?
Thanks,
Jano

As for looking at it from Doppler effect, say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.
But without mentioning that from the perspective of the twin that leaves Earth, the Earth watch has to tick faster than their own when they accelerate it looks like they'll be they same age as each other when they meet back up.
First, I'm not talking about how fast another clock IS ticking, I'm talking about how fast it appears to tick as you just look at it, or more realistically, as you get reports of the current clock reading sent as signals to each other. And from that perspective, both see just what I say above: About 0.268x slow when you see a receding clock, and 3.732 when you see an approaching clock (at 0.866c in both cases). Do you disagree with that? It very much results in one twin aging twice as much as the other during the exercise.
They can never be different ages without one of them accelerating more than the other even though acceleration isn't the actual cause.
Well sure. Without acceleration somewhere, they'll never meet again. I'm just explaining the age difference without explicitly working the acceleration into the computation. I'm doing it the Doppler way instead of 12 other valid ways to explain it. Doppler has to do with what is observed, and doesn't involve so much what is going on in a given frame.
That's what's wrong with the one video: It claims it has the one correct way to explain it, but there are many different ways, all correct. Many more incorrect ways too, but we don't count those.

As for looking at it from Doppler effect, say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.
But without mentioning that from the perspective of the twin that leaves Earth, the Earth watch has to tick faster than their own when they accelerate it looks like they'll be they same age as each other when they meet back up.
First, I'm not talking about how fast another clock IS ticking, I'm talking about how fast it appears to tick as you just look at it, or more realistically, as you get reports of the current clock reading sent as signals to each other. And from that perspective, both see just what I say above: About 0.268x slow when you see a receding clock, and 3.732 when you see an approaching clock (at 0.866c in both cases). Do you disagree with that? It very much results in one twin aging twice as much as the other during the exercise.
No I don't disagree, it's just missing what the accelerating observer sees the other clock doing. If you're using instant acceleration as a simplified example then the twin that accelerates sees the the other clock jump forward in time at the moment they instantly accelerate at the half way point to move back towards Earth at .886c for example.
Here's an interesting exercise, at what velocity do you have to be moving towards an observer to see their watch ticking at the same rate as yours after taking both time dilation and Doppler shift into account?

I understand your point about the frame dragging being a gravitational term.
Still, let us make an observation.
At t=0s = t'=0s' and x=0cs = x'=0cs' when and where the origins are aligned the train observer sends a light beam across the train car. The red arrow up.
At t=2s the moving origin O' is at x=1.732cs.
The red arrow stays at x'=0cs' all the time in the moving reference frame, it is just moving along the y axis.
The platform observer sees the red arrow frame dragged along the x axis.
While not disputing the numbers, that's not an example of frame dragging. The platform observer can compute (not observe) that the light travels 2 seconds as it moves 2 light seconds from the one event (origin) to the other, and yes, at a diagonal so the target event is at coordinates 1.732, 1. That's not frame dragging, it's just light taking 2 seconds to travel 2 light seconds distance. The train frame doesn't seem to come into play at all in that computation.
I call it a frame dragging because the red arrow is at 90 degrees to x axis at each point where the moving frame origin O' is located along the x axis.
Would you have a better name?
No, but if you really feel the need to name the concept, you should not use a term that means a different thing in relativity theory. 'Frame moving' maybe.
How did you get these numbers?
Well, consider a trip out and back to a point 1.732 lighthours away at 0.866c, so it takes two hours to do that in the Earth frame, and two hours to get back. At 16.08 minutes (not 16:08), the Earth clock sends a signal saying it's that time. It gets to the turnaround point the same time as the ship. According to the ship clock (running at half speed), it takes an hour to get there, and so if he's watching the Earth clock, it has advanced 16.08 minutes per hour, which is 0.268 the normal rate. The Earth guy sees the same thing of the receding ship clock. It is entirely symmetrical, just like the train/platform symmetry.
Likewise, during the 2nd hour, the returning guy sees the Earth clock advance steadily from 16.08 minutes to 240 minutes, which is 3.732 times the rate of his own clock. Again, the Earth observer must see the same rate on the ship as it returns.

No I don't disagree, it's just missing what the accelerating observer sees the other clock doing. If you're using instant acceleration as a simplified example then the twin that accelerates sees the the other clock jump forward in time at the moment they instantly accelerate at the half way point to move back towards Earth at .886c for example.
He sees no such thing. In my example just above, the Earth clock appears to read 16.08 minutes both just before and just after hitting the trampoline out there. Sure, the Earth person ages 3.5 hours due to the frame change, but that's a computation, not what he actually sees.
You'd be able to watch the same supernova a dozon times if acceleration changed what you see.
Here's an interesting exercise, at what velocity do you have to be moving towards an observer to see their watch ticking at the same rate as yours after taking both time dilation and Doppler shift into account?
Easy. Both observers would have to be stationary relative to each other.
Mind you, that's not the only solution. Two observers could circle each other at some rate, and thus have a relative velocity. You can get really creative and plot a funny spiral path in so the Earth guy sees a 11 rate between them, but then the incoming guy would still see a different rate.

Halc,
This is not how the time dilation is determined.
Three clocks are needed to determine a time dilation.
Two synchronized clocks in a grid of inertial observers  the first inertial reference frame.
One moving clock in the second inertial reference frame.
Stationary clock 1  SC1
Stationary clock 2  SC2
Moving clock  MC
The first event  when SC1 is aligned with MC  the time on the clocks are recorded.
The second event  when SC2 is aligned with MC  the time on the clocks are recorded.
(t2t1) = gamma * (t'2t'1)
where
t'1  is the time on the moving clock at the first event
t'2  is the time on the moving clock at the second event
t1  time on the SC1 at the first event
t2  time on the SC2 at the second event
Does this help?
Jano

I hope this is going to help. From the textbook I mentioned earlier:
(https://i.imgur.com/1sxGOeK.png)
See where the first and the second events are, the right bottom corner on the figure,
Jano

This is not how the time dilation is determined.
There are multiple ways to determine it. Not all of them require as many as (or as few as) three clocks. When it was first demonstrated by experiment (HK), only two clocks were used.
See where the first and the second events are, the right bottom corner on the figure,
Only the third event is shown in the bottom right figure. The other points of the triangle shown are not the other two events, but rather points in space where those events occurred in the past, which is a frame dependent relation. An event is not a point in space, but a point in spacetime, and events are frame invariant.

HafeleKeating experiment is a mystery, very unreliable,
Jano
(https://i.imgur.com/bruLA1s.png)

...
See where the first and the second events are, the right bottom corner on the figure,
Only the third event is shown in the bottom right figure. The other points of the triangle shown are not the other two events, but rather points in space where those events occurred in the past, which is a frame dependent relation. An event is not a point in space, but a point in spacetime, and events are frame invariant.
Right, and I showed spacetime diagrams in my previous posts.
How did Einstein derive the time dilation?
He did it in a similar way, except his events happened along the x axis.
Einstein did not do any observer roundtrip as you are suggesting.
What is your point?
Please, show me a textbook where the time dilation is defined by an observer roundtrip,
Jano

What is your point?
I have no particular point. You seem to be making all these assorted assertions, and I'm giving counterexamples when I don't feel the asserted thing is true, which is often.

Halc,
you said:
As for looking at it from Doppler effect, say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.
Your claim is in bold.
I got curious how can you say that and your reply was:
...
Well, consider a trip out and back to a point 1.732 lighthours away at 0.866c, so it takes two hours to do that in the Earth frame, and two hours to get back. At 16.08 minutes (not 16:08), the Earth clock sends a signal saying it's that time. It gets to the turnaround point the same time as the ship. According to the ship clock (running at half speed), it takes an hour to get there, and so if he's watching the Earth clock, it has advanced 16.08 minutes per hour, which is 0.268 the normal rate. The Earth guy sees the same thing of the receding ship clock. It is entirely symmetrical, just like the train/platform symmetry.
Likewise, during the 2nd hour, the returning guy sees the Earth clock advance steadily from 16.08 minutes to 240 minutes, which is 3.732 times the rate of his own clock. Again, the Earth observer must see the same rate on the ship as it returns.
I am just pointing out that you have no time dilation definition.
I showed that the time dilation definition done in the textbooks is reciprocal and therefore it is a logical contradiction.
Please, show me what assertions are not true,
Jano

I'm doing it the Doppler way instead of 12 other valid ways to explain it. Doppler has to do with what is observed, and doesn't involve so much what is going on in a given frame.
Among those 12 ways, which one do you think is the most understandable for lay people? Why so?

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Relativistic_longitudinal_Doppler_effect
HafeleKeating experiment is a mystery, very unreliable,
Jano
(https://i.imgur.com/bruLA1s.png)
Can you provide a link to the source?

Here's an interesting exercise, at what velocity do you have to be moving towards an observer to see their watch ticking at the same rate as yours after taking both time dilation and Doppler shift into account?
Any non zero longitudinal speed will generate different magnitudes of Doppler blue shift and time dilation which don't cancel each other.
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=72502.0;attach=30454;image)

I showed that the time dilation definition done in the textbooks is reciprocal and therefore it is a logical contradiction.
Right, I forget you're coming into this as a denier. Your misrepresentations of the theory are deliberate, not just done out of ignorance of the theory.
Again, I don't know what you mean by 'reciprocal'. I use the word 'symmetric', so maybe you mean something like that. If it wasn't symmetric, you'd have an objective test for which person is the one moving, which violates the principle of relativity, first postulate of SR.

I'm doing it the Doppler way instead of 12 other valid ways to explain it. Doppler has to do with what is observed, and doesn't involve so much what is going on in a given frame.
Among those 12 ways, which one do you think is the most understandable for lay people? Why so?
The number 12 is arbitrary. I meant to say there are quite a number of ways to explain it, and not one explanation is the one correct one.
I think the best explanation is by relativity of simultaneity: that an event on Earth that is simultaneous with the turnaround event in the outbound frame is not simultaneous with that same event in the inbound frame.
Any non zero longitudinal speed will generate different magnitudes of Doppler blue shift and time dilation which don't cancel each other.
Yes, which is why my nonstationary solutions posted were not examples of longitudinal motion. The web site mentions the mutual circling case, but I didn't see my spiral case. The inbound traveler can come home on an inefficient path that makes his clock appear to run at normal rate the whole way in.

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Relativistic_longitudinal_Doppler_effectHafeleKeating experiment is a mystery, very unreliable,
Jano
...
Can you provide a link to the source?
Yusuf,
here it is:
(https://i.imgur.com/qaAhj0C.png)
Link at the bottom of that page: http://handle.dtic.mil/100.2/ADA489971 is currently broken.
I have a copy of the pdf though. If you want one, just send me a pm and we can coordinate,
Jano
(https://i.imgur.com/W0OWx2v.png)

I showed that the time dilation definition done in the textbooks is reciprocal and therefore it is a logical contradiction.
Right, I forget you're coming into this as a denier. Your misrepresentations of the theory are deliberate, not just done out of ignorance of the theory.
Again, I don't know what you mean by 'reciprocal'. I use the word 'symmetric', so maybe you mean something like that. If it wasn't symmetric, you'd have an objective test for which person is the one moving, which violates the principle of relativity, first postulate of SR.
Halc,
I hope you understand the three clocks scenario as a definition of the time dilation?
... as explained here:
Halc,
This is not how the time dilation is determined.
Three clocks are needed to determine a time dilation.
Two synchronized clocks in a grid of inertial observers  the first inertial reference frame.
One moving clock in the second inertial reference frame.
Stationary clock 1  SC1
Stationary clock 2  SC2
Moving clock  MC
The first event  when SC1 is aligned with MC  the time on the clocks are recorded.
The second event  when SC2 is aligned with MC  the time on the clocks are recorded.
(t2t1) = gamma * (t'2t'1)
where
t'1  is the time on the moving clock at the first event
t'2  is the time on the moving clock at the second event
t1  time on the SC1 at the first event
t2  time on the SC2 at the second event
Does this help?
Jano
I'll map the events to the spacetime diagram below.
(https://i.imgur.com/B3u77eH.png)
The first event  when SC1 is aligned with MC  is the event A on the spacetime diagram.
The second event  when SC2 is aligned with MC  is the event D on the spacetime diagram.
AD  2s' on the train car.
AD  4s on the platform.
4s = gamma * 2s'
The simultaneity line between G and D is due to the fact that the platform observers form a grid of inertial observers with synchronized clocks.
So what happens when we turn the tables around?
We are going to have the grid of inertial observers on the train and just one clock on the platform?
(https://i.imgur.com/Y1GnBQG.png)
The first event  when SC1 (now on the train) is aligned with MC (now on the platform)  is the event A on the spacetime diagram.
The second event  when SC2 (now on the train) is aligned with MC (now on the platform)  is the event E on the spacetime diagram.
AE  2s on the platform.
AE  4s' on the train car.
4s' = gamma * 2s
The simultaneity line between E and F is due to the fact that the train car observers form a grid of inertial observers with synchronized clocks.
Does this help in explaining why the time dilation is reciprocal?
The reciprocity is a contradiction.
The time on the train moves faster than on the platform. Don't you agree?
The postulate about the constancy of the light propagation in an inertial reference frame can be rephrased:
Any light beam roundtrip in an inertial reference frame will appear time dilated for any other moving inertial reference frame.
Jano

No I don't disagree, it's just missing what the accelerating observer sees the other clock doing. If you're using instant acceleration as a simplified example then the twin that accelerates sees the the other clock jump forward in time at the moment they instantly accelerate at the half way point to move back towards Earth at .886c for example.
He sees no such thing. In my example just above, the Earth clock appears to read 16.08 minutes both just before and just after hitting the trampoline out there. Sure, the Earth person ages 3.5 hours due to the frame change, but that's a computation, not what he actually sees.
Oh I think I see what you mean, the difference in elapsed time on their watches at the end is because of the difference in proper time that the journey took from each frame.As for looking at it from Doppler effect, say a twin goes out and back at .866c, when the two are moving away from each other, each will see the other's clock running at about 0.268 normal rate, but when each sees the other approaching, he will see the other's clock run at about 3.732 times the normal rate. All very symmetrical, and it still nicely works out to one twin aging twice as much as the other when reunited, and thus is one valid way to explain the discrepancy in their ages without mentioning acceleration.
3.732 times the normal rate on the return journey, are you sure? That's seems too fast.
You'd be able to watch the same supernova a dozon times if acceleration changed what you see.
No amount of slowing or speeding up clocks in other frames could allow you to see the same event twice. You'd need to overtake light for that so the same light reached you fora second time.
Here's an interesting exercise, at what velocity do you have to be moving towards an observer to see their watch ticking at the same rate as yours after taking both time dilation and Doppler shift into account?
Any non zero longitudinal speed will generate different magnitudes of Doppler blue shift and time dilation which don't cancel each other.
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=72502.0;attach=30454;image)
There must be a longitudinal speed that will generate magnitudes of Doppler blue shift and time dilation that do cancel each other out. According to that graph it's less than 0.1c.
The reciprocity is a contradiction.
The time on the train moves faster than on the platform.
Of course it's not a contradiction. Time on the train moves faster than on the platform for someone on the train but for someone on the platform time moves faster on the platform.
Do you understand that Doppler shift will cause two objects moving away from each other to each seethe other's watch ticking slowly? Because it takes time for light to travel so each sees the other's watch ticking slowly because of the increasing delay for light to cover the greater distance. Do you think that's a contradiction.
Now have two twins accelerating away from Earth and then coming back with mirrored accelerations to compare watches and both will show the same time. Now have one accelerate away and come back to the Earth in which they themselves were time dilated during the journey and Earth wasn't so their watch will be behind a watch that stayed on Earth.

...
The reciprocity is a contradiction.
The time on the train moves faster than on the platform.
Of course it's not a contradiction. Time on the train moves faster than on the platform for someone on the train but for someone on the platform time moves faster on the platform.
Do you understand that Doppler shift will cause two objects moving away from each other to each seethe other's watch ticking slowly? Because it takes time for light to travel so each sees the other's watch ticking slowly because of the increasing delay for light to cover the greater distance. Do you think that's a contradiction.
Now have two twins accelerating away from Earth and then coming back with mirrored accelerations to compare watches and both will show the same time. Now have one accelerate away and come back to the Earth in which they themselves were time dilated during the journey and Earth wasn't so their watch will be behind a watch that stayed on Earth.
Awal,
please, show me where the Doppler effect comes into play when two stationary clocks are aligned with the moving clock.
The bold part is not true. The SC1 observer is not watching the MC going away. The SC1 recorded the local time when MC was aligned. When the SC2 observer sees the MC locally then SC2 records the local time.
The Doppler effect is not a cause for the time dilation. Please, read the textbooks.
I hope this helps. If it does then, please, don't be shy to 'Say Thanks' at the action button,
Jano

...
The reciprocity is a contradiction.
The time on the train moves faster than on the platform.
Of course it's not a contradiction. Time on the train moves faster than on the platform for someone on the train but for someone on the platform time moves faster on the platform.
Do you understand that Doppler shift will cause two objects moving away from each other to each seethe other's watch ticking slowly? Because it takes time for light to travel so each sees the other's watch ticking slowly because of the increasing delay for light to cover the greater distance. Do you think that's a contradiction.
Now have two twins accelerating away from Earth and then coming back with mirrored accelerations to compare watches and both will show the same time. Now have one accelerate away and come back to the Earth in which they themselves were time dilated during the journey and Earth wasn't so their watch will be behind a watch that stayed on Earth.
Awal,
please, show me where the Doppler effect comes into play when two stationary clocks are aligned with the moving clock.
The bold part is not true. The SC1 observer is not watching the MC going away. The SC1 recorded the local time when MC was aligned. When the SC2 observer sees the MC locally then SC2 records the local time.
The Doppler effect is not a cause for the time dilation. Please, read the textbooks.
I hope this helps. If it does then, please, don't be shy to 'Say Thanks' at the action button,
Jano
Of course Doppler shift isn't the cause of relative time dilation. You really need to read more carefully.
The point I'm making is that you need to wrap your head around Doppler shift first before you try to tackle time dilation because you keep saying that reciprocal time dilation is a contradiction when it clearly isn't.
The part in bold...Because it takes time for light to travel so each sees the other's watch ticking slowly because of the increasing delay for light to cover the greater distance.
... has nothing to do with relative time dilation. When two objects are apart it takes time for the light to get from one to the other, so obviously while they're moving away from each other each will see the other's watch ticking slower than their own.
Do you understand why two objects moving away from each other will each see the other's watch ticking slower than their own due to Doppler shift alone? And you you think that it's in any way logically inconsistent?
How can you seriously think that you understand sr and that's it's wrong if you can't even understand the difference between time dilation and Doppler shift?

Awal,
You can check Einstein's 1905 paper.
It has two parts:
I. KINEMATICAL PART
II. ELECTRODYNAMICAL PART
The time dilation is derived in the PART I.
Guess how many times Einstein talks about the Doppler effect in PART I? Zero!
Please, stop mixing the Doppler effect with the time dilation,
Jano

Awal,
You can check Einstein's 1905 paper.
It has two parts:
I. KINEMATICAL PART
II. ELECTRODYNAMICAL PART
The time dilation is derived in the PART I.
Guess how many times Einstein talks about the Doppler effect in PART I? Zero!
Please, stop mixing the Doppler effect with the time dilation,
Jano
All I'm trying to do is show you that reciprocal clock retardation happens with Doppler shift as well as relative time dilation and is in no way logically inconsistent, in either case.

3.732 times the normal rate on the return journey, are you sure? That's seems too fast.
I ran the numbers in post 82. The traveler takes an hour to get back home, and in that time sees the Earth clock go from 16.08 minutes to 240 minutes, a total time of about 224 minutes, which is 3.732 hours.
The Earth observer sees the same numbers on the traveler clock: about a quarter speed on the way out, and 3.732x on the way back. If they saw different rates from each other it would violate the first postulate of SR.
Halc,
I hope you understand the three clocks scenario as a definition of the time dilation?
Never seen that definition. It isn't nearly as objective as the one you get if you just google 'time dilation':
Time dilation is a difference in the elapsed time measured by two clocks, either due to them having a velocity relative to each other, or by there being a gravitational potential difference between their locations.

Halc,
now you know. You can go ahead and fix the wiki.
Time dilation is a difference in the elapsed time measured by two clocks, either due to them having a velocity relative to each other...
It is impossible to compare time of two inertially moving clocks in a straight line.
They meet together only once. The second clock from the inertial observers grid is required to do the second reading from the moving clocks.
I hope this helps,
Jano

It is natural that the experts are not convinced.
I am going to make counter argument to my claims based on this textbook:
(https://i.imgur.com/XuZht5U.png)
(https://i.imgur.com/GbrhHNf.png)
(https://i.imgur.com/pAikI3n.png)
(https://i.imgur.com/vGki517.png)
It is official, I am a 'newcomer' to SR :)
Jano

In short, I do not understand simultaneity, the length contraction, the clock desynchronization and ...
The fact that different observers disagree on clock rates or simultaneity just means that such concepts are not invariant: they are coordinate dependent.
... and specifically this one...
All the socalled paradoxes of relativity involve, not the inconsistency of a single observer’s deductions, but the inconsistency of assuming that certain concepts are independent of the observer when they are in fact very observer dependent.
The simultaneity is the obvious example...
Observers define 'now' as described in § 1.5 for observer O' (O bar) , and this is the most reasonable way to do it. The problem is that two different observers each define 'now' in the most reasonable way, but they don’t agree. This is an inescapable consequence of the universality of the speed of light.
The meaning of the last quote is that if there are two inertial observers moving relatively to each other then they do not agree on 'now' moment, they do not agree on the simultaneity.
Essentially, the simultaneity means one time slice through space. The x axis represents that for the platform reference frame and x’ represents one slice of time for the train car reference frame. The shortest time interval based on the current physics calculations is the Planck time interval as per wiki. The simultaneity time slice interval means there is no dt. The clocks are stuck, frozen. There is no Planck time tick. The space is stuck in a moment.
We will use our train thought experiment. L'=3.4641cs' long train car moves at constant speed v=0.866c through a train station. The Lorentz factor gamma=2 between the train car reference frame S' and the station platform reference frame S. A train conductor at the front of the train car travels along the train station platform that is L=1.732cs long in the platform reference frame and reaches the end of the station platform. The platform observer at the end of the platform is aligned with the front of the train car and the front conductor. This is an event A where x=0cs=x'=0cs' and when t=0s=t'=0s'. The second train conductor at the end of the train car has synchronized clocks with the front conductor.
The first question  what is the x position and what is the time t on the ground that the second conductor at the end of the train car sees outside at train car time t'=0s'? The Lorentz Transformation (LT) gives us the answer x=6.9282cs and the time is t=6s. This is an event B.
The second question  what is the x position and what is the time t' when the second conductor at the end of the train car sees time t=0s on the platform? The LT gives us the answer again x=1.732cs and the time on the train is t'=3s'. This is an event C at the beginning of the station platform aligned with the end of the train car.
(https://i.imgur.com/39TO8kV.png)
The disagreement on the simultaneity is a logical fallacy!!!
Why??? How come?
It follows from the Lorentz Transformation that any tiny dt has a corresponding dt’. This works both ways. If there is not dt then there is no dt’!!!
The B, C events are bilocated in space and time and the event A cannot be true for both of them. When the second train conductor at the end of the train car has time t’=0s’ then the conductor sees t=6s in the platform frame. The platform frame grid of inertial observers sees time t’=3s’ on the front conductor clock when t=6s. This invalidates the initial condition t=0s=t'=0s'. This is a catch 22, the logical contradiction,
Jano

This is the 'Special Relativity Magic'!!!
The front of the train car is stuck in a moment and the back shrinks in 3s'.
Jano
(https://i.imgur.com/46oY7RU.png)

This is the 'Special Relativity Magic'!!!
Near as I can tell, they're trying unsuccessfully to depict a fast moving train car.
Event A is not clearly identified, but presumably it is the front of the car at location 0 in both frames.
They have length contracted way too much. The gamma is only two, but the diagram seems to suggest that it is 4x, which is wrong.
In the x frame, the rear of the car at time 0 (event C) is at 1.7cs, which I assume to mean light seconds.
In the x' frame, the rear, at time 0, is at 3.46cs (event B), but the picture depicts it way out at 6.9cs, which is the first frame's location of the rear of the car 6 seconds before, so the car was never that long in any frame.
Not sure where you got that or what they're trying to demonstrate. Clearly some denial site which is in the business of misrepresenting what the actual theory says.
The front of the train car is stuck in a moment and the back shrinks in 3s'.
The picture doesn't in any way depict the situation at a different time except for that event B thing in the x frame where it doesn't say where the front of the car is at that time. Nothing 'shrinks' because nothing accelerates in the scenario depicted.

Halc,
The train car length is in the frame.
The first question – what is the position and what is the time on the ground that the second conductor at the end of the train car sees outside at train car time ?
This is the event .
The second question  what is the position and what is the time when the second conductor at the end of the train car sees time on the platform?
The moving train car is contracted due to the Lorentz contraction.
The solution of the equation gives us
for .
Now we can calculate time .
This is the event .
I hope this helps,
Jano

Halc,
The train car length is in the frame.
The first question – what is the position and what is the time on the ground that the second conductor at the end of the train car sees outside at train car time ?
I'm not disputing any of the numbers in the picture. They're correct. I'm disputing them depicting a train car of length 6.9cs, which occurs at no time in any frame.
I'm also disputing your comment that anything is shrinking in so many seconds.
The second question  what is the position and what is the time when the second conductor at the end of the train car sees time on the platform?
The moving train car is contracted due to the Lorentz contraction.
No it isn't. You're asking questions from the conductors' PoV, in whose frame there is no contraction at all. It is the platform that is contracted in that frame, which is why 6.92cs of platform is the same length as a 3.46cs car.

gravitational potential difference between their locations.
This means that a clock on the front side of accelerating space ship ticks at different rate than a clock on the rear side of the ship.

gravitational potential difference between their locations.
This means that a clock on the front side of accelerating space ship ticks at different rate than a clock on the rear side of the ship.
Absolutely. That's a classic application of equivalence principle.

Halc,
I'm not disputing any of the numbers in the picture. They're correct. I'm disputing them depicting a train car of length 6.9cs, which occurs at no time in any frame.
This is the Lorentz Transformation ... at the event where and when .
The grid of inertial observers that has the clocks synchronized will see the moving length contracted.
Therefore the observer at the end of the train car of the length sees the plaform length contracted so
This is very important to settle!
Do you agree that this is correct conclusion?
This is the only correct end result based on the Lorentz Transformation, as you said, the numbers are good.
Why do you doubt then?
Jano

Halc,
This is the Lorentz Transformation ... at the event where and when .
The grid of inertial observers that has the clocks synchronized will see the moving length contracted.
Therefore the observer at the end of the train car of the length sees the plaform length contracted so
You don't seem to have read my response at all:
I'm not disputing any of the numbers in the picture. They're correct. I'm disputing them depicting a train car of length 6.9cs, which occurs at no time in any frame.
So here you are defending the numbers, which I said were correct.
Didn't do so well in reading comprehension in 3rd grade?
This is very important to settle!
Do you agree that this is correct conclusion?
Well, the picture says x' for event C is .346, but you've now changed its sign to .346, which is wrong. The rear conductor is always at x' = .346 so that location does not define an event for him. He's at that location at all times.
Again, read my reply that you quoted. I said the numbers in the picture were correct. I said what wasn't correct, and I don't see you defending any of that.

Halc,
I'm disputing them depicting a train car of length 6.9cs, which occurs at no time in any frame.
... then I have no idea what this means.
What do you mean at no time in any frame?
Can you describe what you mean in the spacetime diagram?
Do you mean the full black horizontal line at t=6s?
(https://i.imgur.com/39TO8kV.png)
I had to repeat 3rd grade 3 times, how did you know? ;) :D
Jano

What do you mean at no time in any frame?
Can you describe what you mean in the spacetime diagram?
Do you mean the full black horizontal line at t=6s?
The diagram show the train car in the track frame, and it has constant length of 1.73cs at each of four different times, and that is depicted in the picture. That part seems OK, but in your comment in post 107 said "the back shrinks in 3s'", whatever that means. It doesn't, since it is an inertial object. It remains 1.73cs at all times in that frame.
The picture shows a train car (the upper one) having length of 6.92cs. That's wrong. Regardless of frame, it cannot be longer than its proper length of 3.46cs.
The picture also calls (presumably) the three events an 'impossibility triangle', but doesn't expand on what exactly is impossible about it. But such language identifies the image as being from some sort of denial site.

What do you mean at no time in any frame?
Can you describe what you mean in the spacetime diagram?
Do you mean the full black horizontal line at t=6s?
The diagram show the train car in the track frame, and it has constant length of 1.73cs at each of four different times, and that is depicted in the picture. That part seems OK, but in your comment in post 107 said "the back shrinks in 3s'", whatever that means. It doesn't, since it is an inertial object. It remains 1.73cs at all times in that frame.
The picture shows a train car (the upper one) having length of 6.92cs. That's wrong. Regardless of frame, it cannot be longer than its proper length of 3.46cs.
The picture also calls (presumably) the three events an 'impossibility triangle', but doesn't expand on what exactly is impossible about it. But such language identifies the image as being from some sort of denial site.
Halc,
there are many things to address in your post.
I'd like to do it one by one.
First, I'd like to get an agreement on this statement:
It follows from the Lorentz Transformation that any tiny dt has a corresponding dt’. This works both ways. If there is not dt then there is no dt’!!!
Do you agree with it?
Jano

...
The picture shows a train car (the upper one) having length of 6.92cs. That's wrong. Regardless of frame, it cannot be longer than its proper length of 3.46cs.
...
Halc,
this is not wrong.
I am not saying that the train car is longer, I am saying .
...but when the train conductor at the end of the train car looks outside at then he is going to see .
This is based on the LT.
Does this help?
Jano

this is not wrong.
I am not saying that the train car is longer, I am saying .
...but when the train conductor at the end of the train car looks outside at then he is going to see .
This is based on the LT.
Does this help?
4th time: I never said that part was wrong.
Read what I actually said.

this is not wrong.
I am not saying that the train car is longer, I am saying .
...but when the train conductor at the end of the train car looks outside at then he is going to see .
This is based on the LT.
Does this help?
4th time: I never said that part was wrong.
Read what I actually said.
The picture shows a train car (the upper one) having length of 6.92cs. That's wrong. Regardless of frame, it cannot be longer than its proper length of 3.46cs.
I should go back to 3rd grade then.
Where does it show?
The picture shows that when the clocks are synchronized on the train then the train observers see
length contracted to
Jano

Halc,
The inertial observers do not agree on the x/x' units of measure.
The length contraction reciprocity is the other side of the SR inconsistency coin.
The first side of the coin being the time dilation reciprocity,
Jano

Halc,
Look:
(https://i.imgur.com/46oY7RU.png)
The platform x units of measure are constant in the upper and the lower part of the figure.
The x' is prolonged in the upper and contracted in the lower,
Jano

Where does it show?
The picture shows the upper train car being 4 times the length of the lower one, when λ is only 2.

Where does it show?
The picture shows the upper train car being 4 times the length of the lower one, when λ is only 2.
Halc,
I explained it in the post right above yours,
Jano

Halc,
the answer to my post #117 is very important.
It follows from the Lorentz Transformation that any tiny dt has a corresponding dt’. This works both ways. If there is not dt then there is no dt’!!!
The answer has to be yes, it is a true statement.
In that case the B, C events require dt, dt' change because they are bilocated in space and time.
The event A must not change  no dt, dt'!!!
This is a contradiction.
The back of the train cannot undergo dt, dt' change when the front of the train is stuck in a moment of the event A.
Therefore the events A, B, C form an 'impossibility triangle',
Jano

Apparently, this would be a better 'relativity of simultaneity' description:
(https://i.imgur.com/nCoP8Hy.png)
The reasoning is that the top two train cars represent the train point of view only.
The bottom contracted train car represents one stuck in a moment time slice, no dt allowed, of the platform point of view only.
Nevertheless, the question from my above post does not go away.
Is the bold part a true statement?
Jano

All,
Please, I'd like to get some feedback.
Which of my SR assertions are false?
Speak now or forever hold your peace. ;)
There is more to come, more SR 'gems'.
Have you ever heard about of the moving reference frame travelling through the stationary reference frame?
It is at the core of the clock desynchronization, very tightly linked to the simultaneity problem.
The explanation is coming next (soon),
Jano

gravitational potential difference between their locations.
This means that a clock on the front side of accelerating space ship ticks at different rate than a clock on the rear side of the ship.
Absolutely. That's a classic application of equivalence principle.
How much is the difference?
Can the effect be more obvious by increasing the length of the spaceship, say 1 light year?

How much is the difference?
Can the effect be more obvious by increasing the length of the spaceship, say 1 light year?
One can make the effect more obvious by increasing acceleration or yes, by lengthening the ship. A lightyear ship can only accelerate at somewhere around 1G at the front maximum.
I did a thread about it.
https://www.thenakedscientists.com/forum/index.php?topic=75140.msg556948#msg556948
The topic was more about acceleration and less about time dilation, but the ratio of the two is direct.
If the front accelerates at half the G's of the rear, then the clock at the rear must be dilated at half the rate of the front clock, in the accelerating frame of the ship. Note that a ship accelerating at different levels from point to point still has one proper ship frame.
In general, this cannot be true if the thrust is applied all at one point in the ship. The ship must physically lose proper length if it is pushed from the rear (as is typically the case for a rocket), or gain proper length if it is pulled from the front. The thread I linked assumes each point on the ship has its own propulsion to eliminate this complication.

Here is a figure:
(https://i.imgur.com/S3eqChe.png)
I hope the numbers are OK, please, double check them.
The position of the train follows the spacetime diagrams.
(https://i.imgur.com/39TO8kV.png)
Now, we can see how moves from the back of the train car to the front.
Let us ask some 'silly' questions.
For example, what happens to when it reaches the front of the train car?
I'll keep asking this question again.
represents the simultaneity line. moves and but there is no ?
Is the the line, the x' axis where really that much simultaneous?
Jano

There must be a longitudinal speed that will generate magnitudes of Doppler blue shift and time dilation that do cancel each other out. According to that graph it's less than 0.1c.
It's exactly 0.

What is the meaning of dotted black line connecting AB and continues up at different angle?
(https://i.imgur.com/39TO8kV.png)

What is the meaning of dotted black line connecting AB and continues up at different angle?
Hi Yusuf,
I see only solid blue line connecting events AB, it is x' axis, simultaneity line and A, B is the full length of the train car on the x' axis, it is L'=3.4641cs'.
The events A, B happened at the same time on the train but they are still events.
The A, B interval is 3cs' interval and it is called spacelike interval.
The dotted line B, C is the back of the train time line. That is how the back of the train moves from 0s' to 3s'.
The dotted line from C to 3's on the ct' time line is the train position at t'=3s'. It is the simultaneity line on the train, parallel to the x' axis.
I hope this helps,
Jano

There must be a longitudinal speed that will generate magnitudes of Doppler blue shift and time dilation that do cancel each other out. According to that graph it's less than 0.1c.
It's exactly 0.
Yes. I was thinking that blue shift starts off greater than time dilation but time dilation would have to overpower it before reaching a relative velocity of c.
So it's never possible to see a clock moving directly towards you ticking fast due to time dilation because blue shift will always be greater.
I still need to get my mind in gear, it's been a while.
Here's an interesting exercise, at what velocity do you have to be moving towards an observer to see their watch ticking at the same rate as yours after taking both time dilation and Doppler shift into account?
Easy. Both observers would have to be stationary relative to each other.
Mind you, that's not the only solution. Two observers could circle each other at some rate, and thus have a relative velocity. You can get really creative and plot a funny spiral path in so the Earth guy sees a 11 rate between them, but then the incoming guy would still see a different rate.
I'm not quite understanding this.
If two observers are circling each other without the distance between them changing then of course they would each see the other's watch ticking at the same rate as their own watch but how could following a path that spirals inwards cause a different result than a direct path?
Imagine they are the only two objects in the universe. Any spiral path is now meaningless because there's no point of reference to create a spiral, all they can do is move directly towards or away from each other. Don't you need to include other objects to create the spiral?

At what velocity do you have to be moving towards an observer to see their watch ticking at the same rate as yours after taking both time dilation and Doppler shift into account?
Both observers would have to be stationary relative to each other.
Mind you, that's not the only solution. Two observers could circle each other at some rate, and thus have a relative velocity. You can get really creative and plot a funny spiral path in so the Earth guy sees a 11 rate between them, but then the incoming guy would still see a different rate.
I'm not quite understanding this.
First of all, my reply concerned the point of view of the observer, who we're trying to get to see an incoming clock tick at normal rate. You kind of worded it the other way around, but it can work that way as well.
If two observers are circling each other without the distance between them changing then of course they would each see the other's watch ticking at the same rate as their own watch but how could following a path that spirals inwards cause a different result than a direct path?
Easiest case: From PoV of Earth, the remote clock is getting closer (blueshift) but is moving (redshift). If you adjust the angle just right, the two cancel. The relativistic effect is a function of speed, but the Doppler effect is a function of the rate of reduction of separation. A circular path would be red shifted (all dilation), but a direct path is dominated by Doppler. Somewhere between is a balance. The math isn't too hard to work out.
Imagine they are the only two objects in the universe. Any spiral path is now meaningless because there's no point of reference to create a spiral
Earth is the reference. Remember, rotation is still absolute, not relative.
So no need for a 3rd object for it to be a spiral.

What is the meaning of dotted black line connecting AB and continues up at different angle?
Hi Yusuf,
I see only solid blue line connecting events AB, it is x' axis, simultaneity line and A, B is the full length of the train car on the x' axis, it is L'=3.4641cs'.
The events A, B happened at the same time on the train but they are still events.
The A, B interval is 3cs' interval and it is called spacelike interval.
The dotted line B, C is the back of the train time line. That is how the back of the train moves from 0s' to 3s'.
The dotted line from C to 3's on the ct' time line is the train position at t'=3s'. It is the simultaneity line on the train, parallel to the x' axis.
I hope this helps,
Jano
I'm sorry, I meant the line BC. The way it suddenly change gradient at C makes me wonder.

(https://upload.wikimedia.org/wikipedia/commons/thumb/c/ce/Twin_Paradox_Minkowski_Diagram.svg/500pxTwin_Paradox_Minkowski_Diagram.svg.png)
So to visualize the time jump, we need to draw tilted line of simultaneity, according to the object's speed.
Yusuf,
can we talk about this time jump?
What is that? Do you know?
Jano

It's what happens when they accelerate to change frames.
It's only a jump if you use an instant acceleration approximation.
You should have used the red and blue the other way round for red and blue shift.

(https://upload.wikimedia.org/wikipedia/commons/thumb/c/ce/Twin_Paradox_Minkowski_Diagram.svg/500pxTwin_Paradox_Minkowski_Diagram.svg.png)
So to visualize the time jump, we need to draw tilted line of simultaneity, according to the object's speed.
Yusuf,
can we talk about this time jump?
What is that? Do you know?
Jano
You can find the explanation in the video.
Searched for twin paradox in youtube, here are some of the top results.
Which one do you like best? Why?
Based on the comment of the videos, it seems that they still leave many people in confusion.

Thank you Yusuf,
how about this deceleration and acceleration after 3s' on the train car?
The line can be 'smoother' if more steps where done in the deceleration and acceleration stage.
(https://i.imgur.com/Ir59wNR.png)
Is there anything wrong with the spacetime diagram?
What is the time dilation for the roundtrip?
Is it 8s' to 16s ratio? Yes, no?
Jano

Is there anything wrong with the spacetime diagram?
Yes.
1, you have the left side of the object (train?) moving back and forth at turnaround time.
2. You have the object accelerating far faster than it possibly can
3 You have the object begin acceleration at different proper times, which will destroy the object.

Is there anything wrong with the spacetime diagram?
Yes.
1, you have the left side of the object (train?) moving back and forth at turnaround time.
2. You have the object accelerating far faster than it possibly can
3 You have the object begin acceleration at different proper times, which will destroy the object.
Halc,
that's cool :)
1. It is a long train car going away, stopping, accelerating in the opposite (returning) direct and coming back. No problem there.
2. Where, can you tell me the sequence of events?
3. I do not know what you mean, please, tell me the events on the spacetime diagram. I can mark them for better discussion.
Thanks,
Jano

1, you have the left side of the object (train?) moving back and forth at turnaround time.
2. You have the object accelerating far faster than it possibly can
3 You have the object begin acceleration at different proper times, which will destroy the object.
1. It is a long train car going away, stopping, accelerating in the opposite (returning) direct and coming back. No problem there.
You depict otherwise. The left end turns around, goes left, goes back to the right again, then finally left at full speed. It shouldn't do that.
2. Where, can you tell me the sequence of events?
3. I do not know what you mean, please, tell me the events on the spacetime diagram. I can mark them for better discussion.
You seem to have a million km train trying to reverse as quickly as possible centered on 4 seconds as measured by a train clock on the right. Your gamma seems to be 2, based on 4 seconds (train) lining up with 8 seconds on the vertical time coordinate (Earth?), but its speed is a tad too fast for that since it gets out to 7 light seconds on the picture, not 6.93. Close enough.
Sequence of events: As measured in the Earth coordinate system depicted, after 8 seconds, the right end of the train turns around immediately. No curve. Hard angle. Just the right end does this.
Simultaneously (train frame), the entire train begins to accelerate left at uniform proper acceleration (uniform meaning that any given point in the train will experience unvarying G force, but each point a different G force than other points).
The left of the train is at Earth coordinate x=.33, t=2.31 when it begins its acceleration to the left. It finishes at x=.33, t=13.69 and thereafter is inertial.
At the 8 second mark in Earth frame, this curve will take the left end to its rightmost point at x=3.6 (and the right at 6.93). This should be a fair picture of a sort of maximum G turnaround for the train without ever altering its proper length. It takes around 11.4 seconds in Earth time to complete the turnaround. It cannot be done faster than that without compressing the proper length of the train. It can be done slower if you want the right end to curve like you depict in your diagram.
All this assumes that whatever force is needed to do this is applied everywhere, not just at one end or the other like a real train.

1, you have the left side of the object (train?) moving back and forth at turnaround time.
2. You have the object accelerating far faster than it possibly can
3 You have the object begin acceleration at different proper times, which will destroy the object.
1. It is a long train car going away, stopping, accelerating in the opposite (returning) direct and coming back. No problem there.
You depict otherwise. The left end turns around, goes left, goes back to the right again, then finally left at full speed. It shouldn't do that.
...
Halc,
I'll respond to this now, the rest in the morning.
The horizontal line at platform 8s is stopped train car. It is not moving. The length L' is equal to L = 3.4641cs.
When the train is slowing down it is stretching out to normal length in the platform frame,
Jano

1, you have the left side of the object (train?) moving back and forth at turnaround time.
2. You have the object accelerating far faster than it possibly can
3 You have the object begin acceleration at different proper times, which will destroy the object.
1. It is a long train car going away, stopping, accelerating in the opposite (returning) direct and coming back. No problem there.
You depict otherwise. The left end turns around, goes left, goes back to the right again, then finally left at full speed. It shouldn't do that.
...
Halc,
I'll respond to this now, the rest in the morning.
The horizontal line at platform 8s is stopped train car. It is not moving. The length L' is equal to L = 3.4641cs.
When the train is slowing down it is stretching out to normal length in the platform frame,
Jano
I know all that, except the 3.4642 part. I was guessing a million km, which is about 3.3cs. All my comments still stand. The left side of the car should not move left and right like that. Draw it like I describe it and it will be full proper length at turnaround without all this back and forth nonsense.

1, you have the left side of the object (train?) moving back and forth at turnaround time.
2. You have the object accelerating far faster than it possibly can
3 You have the object begin acceleration at different proper times, which will destroy the object.
1. It is a long train car going away, stopping, accelerating in the opposite (returning) direct and coming back. No problem there.
You depict otherwise. The left end turns around, goes left, goes back to the right again, then finally left at full speed. It shouldn't do that.
...
Halc,
I'll respond to this now, the rest in the morning.
The horizontal line at platform 8s is stopped train car. It is not moving. The length L' is equal to L = 3.4641cs.
When the train is slowing down it is stretching out to normal length in the platform frame,
Jano
I know all that, except the 3.4642 part. I was guessing a million km, which is about 3.3cs. All my comments still stand. The left side of the car should not move left and right like that. Draw it like I describe it and it will be full proper length at turnaround without all this back and forth nonsense.
Halc,
my scenario is real. There is no need to complicated the analysis with turning around.
If we work with x and t dimensions then there is no turn around, only stop and accelerate back.
A spaceship does not have to turn. It can stop and go straight back.
It is impossible to flip the train car around without involving y, z dimensions.
If you trust the Special Relativity then the back of the train car will go back in the platform frame.
Now we are getting onto another SR 'gem' and I see you are getting confused.
I am just a messenger here, please, don't shoot me, :)
Jano

my scenario is real.
Your scenario can be real, yes, but you're depicting smashing the train car into much to short of a proper length, and then yanking it to pieces a moment later. There will be nothing left but shrapnel.
You asked for help in doing it without ever putting strain on the long object, and I showed you what must be done to do that. You can ignore that at your choice, as you've done with prior posts.
There is no need to complicated the analysis with turning around.
I didn't turn it around. I stopped it and then went back the way it came, still facing the same way.
If we work with x and t dimensions then there is no turn around
I expressed everything as x and t (or x' and t') dimensions. At no point did I put it sideways into y territory in attempt to swap ends. That would be a complicated operation indeed.
I am just a messenger here, please, don't shoot me, :)
If you're getting this incorrect diagrams from elsewhere, then your source is wrong. If they're of your own creation, then you're not just a messenger

Does this make more sense?
(https://i.imgur.com/p9ytDos.png)
No, not really.
The deceleration and acceleration in the spacetime diagram is another 'gem' of the Relativity.
The post #140 shows backward acceleration for the back of the train car.
This is the instantaneous turn around at the front but the back of the train car is 'screwed' again.
The Relativity is a 'beauty' ;)
... to be continued.
Jano

Does this make more sense?
No. Train still gets destroyed, since (in the proper frame of the train) you accelerate the right end leftward while the left end is still stationary, which crushes the train.
Then suddenly you yank the left end twice, which rips the poor thing apart.
Draw it with the left end as a curve like i said, starting and ending at the events I identified. That will keep the whole object the same proper length at all times in the process, so it comes out undamaged.

Does this make more sense?
(https://i.imgur.com/p9ytDos.png)
No, not really.
The deceleration and acceleration in the spacetime diagram is another 'gem' of the Relativity.
The post #140 shows backward acceleration for the back of the train car.
This is the instantaneous turn around at the front but the back of the train car is 'screwed' again.
The Relativity is a 'beauty' ;)
... to be continued.
Jano
Where is the rear end of the train at t=7.9s? Shouldn't it still moving to the right at the same speed as the front end?
AFAIK, the train should lie in the simultaneity planes (depicted in blue and red lines in the picture below)
(https://upload.wikimedia.org/wikipedia/commons/thumb/c/ce/Twin_Paradox_Minkowski_Diagram.svg/500pxTwin_Paradox_Minkowski_Diagram.svg.png)

No. Train still gets destroyed, since (in the proper frame of the train) you accelerate the right end leftward while the left end is still stationary, which crushes the train.
Then suddenly you yank the left end twice, which rips the poor thing apart.
What if the middle of the train is removed, leaving only the front and rear ends. Will this make the situation more manageable?

Where is the rear end of the train at t=7.9s?
Assuming the left is the rear, pretty close to where you have it. The worldline for the right side is fine in your picture, but the left side needs to start uniform proper acceleration to the left at x=.33, t=2.31 as I said in post 143. Make it a nice clean curve that results in a full proper length when it is stationary in that frame at t=8 seconds.
Shouldn't it still moving to the right at the same speed as the front end?
Only in its own frame, not in some other frame like the frame of x/t like you show.
If both ends of an accelerating object moved at the same speed in the other frames, it could not contract.
AFAIK, the train should lie in the simultaneity planes (depicted in blue and red lines in the picture below)
Your 2nd picture does not show lines of simultaneity of any of the frames in question. They need to be angled more. The turnaround point is simultaneous with x=0 t=2 in the outbound frame, and t=14 in the inbound frame. You have it at t=4 and 12 respectively which is wrong.

The acceleration/deceleration are problems.
(https://i.imgur.com/lLfcjmB.png)
It comes back to the simultaneity.
The platform observers see the deceleration from the event A to the event C.
The train observers see the deceleration from the event B to the event C. The yellow line would be a nice hyperbolic line if done over the tiny delta t's.
Who's deceleration is correct?
Is the back observer going to measure any backward acceleration.
It cannot based on the train point of view.
It has to based on the platform point of view.
Who is right?
What is uniform acceleration/deceleration?
Jano

The platform observers see the deceleration from the event A to the event C.
The train observers see the deceleration from the event B to the event C.
Events are objective, so every observer sees a given worldline pass through the same events, so what you're saying (different worldlines taken (black or yellow) from B to C) cannot be correct.
Who's deceleration is correct?
Neither, but the hyperbola idea is closer.
Is the back observer going to measure any backward acceleration.
The object is accelerating towards the back, so of course everybody on the train observes backward acceleration.
Try this:
turnaround.jpg (60.55 kB . 310x458  viewed 1706 times)
The yellow lines are lines of simultaneity from the train PoV. The entire train is always moving at some uniform speed at each one of those lines of simultaneity. The thick black lines are the contracted train in the platform frame, the same way you use them.

The platform observers see the deceleration from the event A to the event C.
The train observers see the deceleration from the event B to the event C.
Events are objective, so every observer sees a given worldline pass through the same events, so what you're saying (different worldlines taken (black or yellow) from B to C) cannot be correct.
Who's deceleration is correct?
Neither, but the hyperbola idea is closer.
Is the back observer going to measure any backward acceleration.
The object is accelerating towards the back, so of course everybody on the train observes backward acceleration.
Try this:
turnaround.jpg (60.55 kB . 310x458  viewed 1706 times)
The yellow lines are lines of simultaneity from the train PoV. The entire train is always moving at some uniform speed at each one of those lines of simultaneity. The thick black lines are the contracted train in the platform frame, the same way you use them.
Halc,
Right, that's how the yellow hyperbolic line looked like, your back grey line.
Here is the problem:
(https://i.imgur.com/p9ytDos.png)
The front still has instantaneous rotation.
The horizontal line on your plot at platform 6s is too long from the platform point of view.
The deceleration starts at 3s'. That is 6s on the platform. If the line is longer we broke the platform simultaneity!
That is the reason the Relativity is logically inconsistent.
Two observers do not agree on the length of the train because they do not agree on the simultaneity!!!!
This is exactly the point where the disagreement on the simultaneity bites back.
Imagine a big display on the back of the train car showing the deceleration value.
It can show only one true local deceleration value!
No disagreement on deceleration like on the simultaneity BS.
The question is at what point the real deceleration starts?
Is it the platform event position or train event position?
It cannot be both.
This is the same events impossibility triangle as shown before!!!!
Jano

Here is the problem:
The front still has instantaneous rotation.
I did it in a short enough time that it was beyond the resolution of the picture. You can take longer to do it, but then it takes longer to turn around, and you've not said how long you want the front to take during the process. Maybe starting at t=3, platform time, in which case the rear of the train would need to start accelerating left before it event gets to x=0.
The horizontal line on your plot at platform 6s is too long from the platform point of view.
It is the correct length as best as I could draw it, a little over 2ls.
The deceleration starts at 3s'.
OK, you want a slower curve. Curve the right end (no hard angles) starting at the point you indicate. The left end then starts where the lowermost yellow line is drawn. Now both sides are curves.
Two observers do not agree on the length of the train because they do not agree on the simultaneity!!!!
This is exactly the point where the disagreement on the simultaneity bites back.
Imagine a big display on the back of the train car showing the deceleration value.
It can show only one true local deceleration value!
Define that. You mean G force? There is no such thing as 'deceleration value' in physics. G force is a measure of local acceleration, which would be negative in this instance since we've assigned the right direction as positive. Yes, you can have a big display at the rear showing the acceleration there. It will real a different value than a similar display at the front, which reads a much higher magnitude.
The question is at what point the real deceleration starts?
At the bottom yellow line I drew if you want the front to start at t=3. That bottom yellow line intersects that event.
A line is not a 'point', so there is no one point for acceleration to commence for an object that is not a point mass.
Is it the platform event position or train event position?
It is that bottom yellow line, which is a line of simultaneity for the train. That means the entire train begins accelerating leftward at the same time in its frame. It is not deceleration since the train, while inertial before then, was always stationary in its frame.

...
It is that bottom yellow line, which is a line of simultaneity for the train. That means the entire train begins accelerating leftward at the same time in its frame. It is not deceleration since the train, while inertial before then, was always stationary in its frame.
Halc,
this is another Relativity 'gem'.
I am going to explain it on the Triplets Paradox.
Imagine a double train car, two tracks next to each other.
The original Twin paradox, the two train cars accelerate to the right.
One train car decelerates, accelerates back at 4s' (instantaneously), like in the Twin paradox, but the other train car continues ahead, to the right.
(https://i.imgur.com/OcApne3.png)
We can see that velocity between x', ct' frame and x'', ct'' frame is v''=0.9897433186c.
The gamma between x', ct' frame and x'', ct'' frame is 7.
When two triplets meet again they conclude that the third one is very very very old compare to them.
This seals the deal!
The time on the moving train is ticking faster than on the platform.
Right!!??
Jano

Imagine a double train car, two tracks next to each other.
The original Twin paradox, the two train cars accelerate to the right.
One train car decelerates, accelerates back at 4s' (instantaneously), like in the Twin paradox, but the other train car continues ahead, to the right.
We can see that velocity between x', ct' frame and x'', ct'' frame is v''=0.9897433186c.
The gamma between x', ct' frame and x'', ct'' frame is 7.
OK so far. It assumes the train cars have insignificant length, which wasn't the case before.
When two triplets meet again they conclude that the third one is very very very old compare to them.
Depends on the frame used. In the c/x platform frame, the distant twin has aged 8 seconds, same as the one that turns around. In the returning twin's ct'' frame, he's 4.57 seconds old. Neither answer sounds like 'very very very' old.
Your conclusion indicates that you seem bent on 'sealing the deal' on some kind of contradiction when there is none.
Have fun with that. You seem uninterested in my replies.

Imagine a double train car, two tracks next to each other.
The original Twin paradox, the two train cars accelerate to the right.
One train car decelerates, accelerates back at 4s' (instantaneously), like in the Twin paradox, but the other train car continues ahead, to the right.
We can see that velocity between x', ct' frame and x'', ct'' frame is v''=0.9897433186c.
The gamma between x', ct' frame and x'', ct'' frame is 7.
OK so far. It assumes the train cars have insignificant length, which wasn't the case before.
When two triplets meet again they conclude that the third one is very very very old compare to them.
Depends on the frame used. In the c/x platform frame, the distant twin has aged 8 seconds, same as the one that turns around. In the returning twin's ct'' frame, he's 4.57 seconds old. Neither answer sounds like 'very very very' old.
Your conclusion indicates that you seem bent on 'sealing the deal' on some kind of contradiction when there is none.
Have fun with that. You seem uninterested in my replies.
Halc,
where did you get that number???
The 4s'' in the ct'' frame are 28s' in the ct' frame.
Gamma is 7!
The distant triplet is two times older than the platform frame triplet.
The 16s in the platform frame are 32s' in the ct' frame.
Gamma is 2!
Just check the spacetime diagram,
Jano

Try this:
turnaround.jpg (60.55 kB . 310x458  viewed 1706 times)
Halc,
Do you claim this is a correct deceleration/acceleration?
The back of the train car deceleration/acceleration does not match what platform observers see due to the simultaneity.
The front of the train car has instantaneous change of frames.
What is this spacetime diagram?
Jano

The Triplets Challenge
Here is a very smart book:
(https://i.imgur.com/QOjHogH.png)
(https://i.imgur.com/GhVhU9K.png)
Now the Triplets Paradox.
(https://i.imgur.com/ERk774M.png)
The left triplet travels left, the right triplet travels right, the third one does not travel.
I challenge every relativist out there to show us a solution what time will be displayed on the left triplet clock and on the right triplet clock when all triplets meet again!
Jano

In the returning twin's ct'' frame, he's 4.57 seconds old.
where did you get that number???
The 4s'' in the ct'' frame are 28s' in the ct' frame.
Yes, but nobody at the reunion has reason to use the ct' frame. I said what frames I used, which are the respective frames of each of the two twins comparing notes at the reunion.
Do you claim this is a correct deceleration/acceleration?
It is a correct illustration of a very rapid turnaround with negligible time to reverse the front. It is not instant on the right, but higher G than the resolution of the picture.
The back of the train car deceleration/acceleration does not match what platform observers see due to the simultaneity.
Of course the observations don't match. The people on the platform don't see it until the light from the maneuver gets to them. They're not present with the observers in the train. They measure the duration as being longer than the guy in the back of the train.
The left triplet travels left, the right triplet travels right, the third one does not travel.
I challenge every relativist out there to show us a solution what time will be displayed on the left triplet clock and on the right triplet clock when all triplets meet again!
Obviously the same time, due to symmetry.
Assuming the middle clock reads 1 (year? The bits you quote don't indicate units) at the end, it appears that both traveling clocks will read about 0.875 at reunion. What's supposed to be even remotely different about that situation compared to the first one with only one traveler?

Yes, but nobody at the reunion has reason to use the ct' frame. I said what frames I used, which are the respective frames of each of the two twins comparing notes at the reunion.
It is a correct illustration of a very rapid turnaround with negligible time to reverse the front. It is not instant on the right, but higher G than the resolution of the picture.
The back of the train car deceleration/acceleration does not match what platform observers see due to the simultaneity.
Of course the observations don't match. The people on the platform don't see it until the light from the maneuver gets to them. They're not present with the observers in the train. They measure the duration as being longer than the guy in the back of the train.
There is a grid of inertial observers on the platform.
Their observation does not match your spacetime diagram.
The left triplet travels left, the right triplet travels right, the third one does not travel.
I challenge every relativist out there to show us a solution what time will be displayed on the left triplet clock and on the right triplet clock when all triplets meet again!
Obviously the same time, due to symmetry.
Assuming the middle clock reads 1 (year? The bits you quote don't indicate units) at the end, it appears that both traveling clocks will read about 0.875 at reunion. What's supposed to be even remotely different about that situation compared to the first one with only one traveler?
The Relativity says: "When two observers move in relation to each other then they cannot have the same proper time on their clocks."
That's it, that's the LT.
Does the left triplet move in relation to the right triplet?
If yes then they cannot have the same time on their clocks!!!
This is the Relativity!!!
If they end up with the same time on their clocks as per the middle triplet then the middle triplet reference frame is preferred one!
Jano

There is a grid of inertial observers on the platform.
Their observation does not match your spacetime diagram.
The spacetime diagram is in the frame of the platform, so by definition, the 'grid' observers on said platform must witness the events that occur at their respective locations. Translation: your statement above is self contradictory.
If you say my placement of events/worldlines is wrong, that's a different assertion, but you didn't say that this time. You seem to say something different each time, but consistent in being wrong at least.
I drew a highacceleration reverse of a million km rigid object moving at .866c, centered at t=4 platform time. In that regard, the diagram is exactly correct. Your diagrams depict a nonrigid turnaround since the proper length of the object is often too long or too short. That doesn't make it wrong, it just means your train needs to be rubber in order to survive what you're doing to it.
The Relativity says: "When two observers move in relation to each other then they cannot have the same proper time on their clocks."
Relativity theory says no such thing (assuming "the relativity" means the theory).

There is a grid of inertial observers on the platform.
Their observation does not match your spacetime diagram.
The spacetime diagram is in the frame of the platform, so by definition, the 'grid' observers on said platform must witness the events that occur at their respective locations. Translation: your statement above is self contradictory.
If you say my placement of events/worldlines is wrong, that's a different assertion, but you didn't say that this time. You seem to say something different each time, but consistent in being wrong at least.
I drew a highacceleration reverse of a million km rigid object moving at .866c, centered at t=4 platform time. In that regard, the diagram is exactly correct. Your diagrams depict a nonrigid turnaround since the proper length of the object is often too long or too short. That doesn't make it wrong, it just means your train needs to be rubber in order to survive what you're doing to it.
Halc,
The deceleration starts at 6s for the platform frame.
This is not the case in your spacetime diagram.
The back observer starts to decelerate at 0s in your diagram.
The Relativity says: "When two observers move in relation to each other then they cannot have the same proper time on their clocks."
Relativity theory says no such thing (assuming "the relativity" means the theory).
What is the Twin paradox then?
How can two observers have the same time?
Please, explain how Twin paradox is wrong,
Jano

The deceleration starts at 6s for the platform frame.
Yours does, which is why your object gets crushed at first.
This is not the case in your spacetime diagram.
The back observer starts to decelerate at 0s in your diagram.
At 2.3 seconds actually. It's inertial up to that point. That event is simultaneous with the front of the train in the train frame, when the entire train begins to decelerate all at the same time.
Please, explain how Twin paradox is wrong,
I think it's wrong to call it a paradox. I didn't assert that the twins scenario is wrong. I said your assertion above is wrong.

The deceleration starts at 6s for the platform frame.
Yours does, which is why your object gets crushed at first.
This is not the case in your spacetime diagram.
The back observer starts to decelerate at 0s in your diagram.
At 2.3 seconds actually. It's inertial up to that point. That event is simultaneous with the front of the train in the train frame, when the entire train begins to decelerate all at the same time.
Please, explain how Twin paradox is wrong,
I think it's wrong to call it a paradox. I didn't assert that the twins scenario is wrong. I said your assertion above is wrong.
Halc,
what is the position of the front of the train car for x=0.2598cs and t=2.3s in the train frame coordinates?
The Triplets challenge  is the right triplet going to say that the left triplet proper time is the same as his/her (the right triplet)?
If the proper time is not the same then the middle triplet reference frame analysis is useless, essentially, we do not have physics.
If the left triplet proper time is observed/analyzed as to be same as the proper time of the right triplet from the right triplet frame then we need to see those claims/analysis because according to MCF (momentarily comoving inertial reference frame) the left frame is moving and the Lorentz Transformation does not allow for the proper time to be the same.
MCF analysis is an approximation for the accelerated frames time dilation analysis.
The claim is:
 if the left time is not the same from the right frame point of view ... then we do not have physics because it contradicts the middle triplet's frame.
 if the left time is the same as the right frame time ... then we do not have the Lorentz Transformation, we do not have the Relativity
I am just talking about the thought experiment analysis but if this would be a real life experiment then either result contradicts the Relativity,
Jano

what is the position of the front of the train car for x=0.2598cs and t=2.3s in the train frame coordinates?
I didn't do any calculations to 4 digits of precision. The yellow lines show lines of simultaneity along the changing frame of the train as it turns around. You can see the line from that event being simultaneous with the short duration at t'=4 x'=0 measured at the front of the train.
The Triplets challenge  is the right triplet going to say that the left triplet proper time is the same as his/her (the right triplet)?
Proper time is frame independent, so all observers will agree that the proper times of the left and right worldlines between events A and B are identical.
If the proper time is not the same then the middle triplet reference frame analysis is useless, essentially, we do not have physics.
If the left triplet proper time is observed/analyzed as to be same as the proper time of the right triplet from the right triplet frame then we need to see those claims/analysis because according to MCF (momentarily comoving inertial reference frame) the left frame is moving and the Lorentz Transformation does not allow for the proper time to be the same.
Maybe you should not only read, but actually comprehend that book you linked.
The claim is:
 if the left time is not the same from the right frame point of view ... then we do not have physics because it contradicts the middle triplet's frame.
 if the left time is the same as the right frame time ... then we do not have the Lorentz Transformation, we do not have the Relativity
Have fun justifying these claims.

what is the position of the front of the train car for x=0.2598cs and t=2.3s in the train frame coordinates?
I didn't do any calculations to 4 digits of precision. The yellow lines show lines of simultaneity along the changing frame of the train as it turns around. You can see the line from that event being simultaneous with the short duration at t'=4 x'=0 measured at the front of the train.
The Triplets challenge  is the right triplet going to say that the left triplet proper time is the same as his/her (the right triplet)?
Proper time is frame independent, so all observers will agree that the proper times of the left and right worldlines between events A and B are identical.
If the proper time is not the same then the middle triplet reference frame analysis is useless, essentially, we do not have physics.
If the left triplet proper time is observed/analyzed as to be same as the proper time of the right triplet from the right triplet frame then we need to see those claims/analysis because according to MCF (momentarily comoving inertial reference frame) the left frame is moving and the Lorentz Transformation does not allow for the proper time to be the same.
Maybe you should not only read, but actually comprehend that book you linked.
The claim is:
 if the left time is not the same from the right frame point of view ... then we do not have physics because it contradicts the middle triplet's frame.
 if the left time is the same as the right frame time ... then we do not have the Lorentz Transformation, we do not have the Relativity
Have fun justifying these claims.
Halc,
using this: http://www.trell.org/div/minkowski.html
(https://i.imgur.com/9un6KKK.png)
The time on the train car is 4.15s'.
The train starts to decelerate at 2.3s according to your calculations.
The front of the train car is already past the turning point for the platform 2.3s.
What are you trying to show us with your spacetime diagram?
Jano

using this: http://www.trell.org/div/minkowski.html
Cool tool.
The time on the train car is 4.15s'.
The train starts to decelerate at 2.3s according to your calculations.
So much for me just eyeballing it.
It appears that my acceleration begins more at t=2.13, not 2.3, but nowhere near the 6 second mark where you put it.
That means it also ends around t=13.87
I said the object was a million km long, so that should make d' always equal to 3.3356 or so.

using this: http://www.trell.org/div/minkowski.html
Cool tool.
The time on the train car is 4.15s'.
The train starts to decelerate at 2.3s according to your calculations.
So much for me just eyeballing it.
It appears that my acceleration begins more at t=2.13, not 2.3, but nowhere near the 6 second mark where you put it.
That means it also ends around t=13.87
I said the object was a million km long, so that should make d' always equal to 3.3356 or so.
Halc,
Your t=2.13s will be more than t'=4s' because anything above 2s is above 4s'. This is the Lorentz transformation.
The deceleration is arbitrary it can last longer or shorter.
I said from 3s' to 4s'. There is nothing wrong with that.
But there is the simultaneity problem. It is 0s for the back and 6s for the front,
... or it is 6s from the platform point of view.
Let us assume that the train would be a spaceship with engines at the front.
The back of the spaceship starts to decelerate at 0s and the engine is ignited at 6s.
Where is causality in this?
This does not make any sense,
Jano

Your t=2.13s will be more than t'=4s' because anything above 2s is above 4s'. This is the Lorentz transformation.
You have no idea what you're doing. The front is well above the 2s mark, and is not above 4s', so your statement is obviously false. It all depends on how far down x' you want to go, and 3.3356 isn't enough to push it below the 2s mark at that speed.
But there is the simultaneity problem. It is 0s for the back and 6s for the front,
... or it is 6s from the platform point of view.
Your problem is that you're doing things to the train in the platform frame. Do it in the train frame like any train controller would. Assuming the clocks in the train are initially in sync while inertial, all points in the train begin to accelerate left at t'=4. Those clocks will run at different speeds relative to the train while accelerating, so they'll not be in sync anymore one the inertial phase is over. This is consistent with SR and also with the equivalence principle.
Let us assume that the train would be a spaceship with engines at the front.
The engines need to be everywhere, else any changes in thrust cannot be felt elsewhere in the ship an sooner than the speed of sound will allow, and there can be no material that transmits pressure waves faster than light. In other words, it is physically impossible to accelerate a ship from thrust at one end without introducing strain (deformation) to the ship. So since I am doing this as a bornrigid object, the thrust must be applied everywhere to maintain that constant proper length.
The back of the spaceship starts to decelerate at 0s and the engine is ignited at 6s.
Where is causality in this?
Exactly so. This is why you can't thrust from one point. Causality says the ship must bend if you do it that way.

Your t=2.13s will be more than t'=4s' because anything above 2s is above 4s'. This is the Lorentz transformation.
You have no idea what you're doing. The front is well above the 2s mark, and is not above 4s', so your statement is obviously false. It all depends on how far down x' you want to go, and 3.3356 isn't enough to push it below the 2s mark at that speed.
...
Halc,
let us just talk about the first part of your post:
(https://i.imgur.com/nnu41T3.png)
As you can see 2s in the platform frame is 4s' in the train frame.
Where did you get the 3.3356cs' when the train car length is L'=3.4641cs'?
Jano

You have no idea what you're doing. The front is well above the 2s mark, and is not above 4s', so your statement is obviously false. It all depends on how far down x' you want to go, and 3.3356 isn't enough to push it below the 2s mark at that speed.
...
Halc,
let us just talk about the first part of your post
OK. The first part of my post was saying you have no idea what you're doing. To illustrate:
As you can see 2s in the platform frame is 4s' in the train frame.
A time in one frame cannot correspond to a specific time in another. An event does, but the time 2s on the platform intersects every time in the train frame, depending on location. This is a good example of you showing that you don't know what you're doing.
Where did you get the 3.3356cs' when the train car length is L'=3.4641cs'?
I was guessing a million km, something I've said in multiple posts. Your pictures never said a different figure. Yes, if you extend the train to 3.464, it will need to begin acceleration left immediately as it fully passes the x=0 mark on the platform in two seconds.

Halc,
just check my posts #106 and #107. I never changed the initial point of the discussion.
I know what I am talking about.
If the back of the train car is at platform x=0cs and t=2s then L' ()  the proper length to the right, the simultaneity line of the train car, will end up at 4s'.
That's what I am saying.
Your deceleration after 2s does not make sense for the original setup.
The length units of measure were cs and not km at the beginning.
Even if we consider the 'bornrigid object' and the deceleration everywhere there is a problem.
The train car will start to stretch from the back side of the train car for the platform observer.
Remember the post where I showed ct, ct' and ct'' frames?
Here is what is happening.
The train car is decelerating in ct frame and stretching out from the back, the left side.
The train car is accelerating in ct' frame and shrinking from the front, the right side.
A total chaos and disagreement between the frames on what is going on.
The observers disagree on the acceleration because of different simultaneity.
Jano

Halc,
just check my posts #106 and #107. I never changed the initial point of the discussion.
That did not involve the current turnaround scenario, so I didn't know the length came from there.
The picture there pictures a train car contracted by a factor of 4 despite a gamma of only 2. All very deceptive.
I know what I am talking about.
I suspect you do, but that makes you trolling: deliberate misrepresentation. Your posts in the energy/momentum conservation thread bear this out.
Even if we consider the 'bornrigid object' and the deceleration everywhere there is a problem.
The train car will start to stretch from the back side of the train car for the platform observer.
Exactly as it needs to. It length is going to change from 1.73 to twice that when it matches the platform speed.

Halc,
just check my posts #106 and #107. I never changed the initial point of the discussion.
That did not involve the current turnaround scenario, so I didn't know the length came from there.
The picture there pictures a train car contracted by a factor of 4 despite a gamma of only 2. All very deceptive.
I know what I am talking about.
I suspect you do, but that makes you trolling: deliberate misrepresentation. Your posts in the energy/momentum conservation thread bear this out.
Even if we consider the 'bornrigid object' and the deceleration everywhere there is a problem.
The train car will start to stretch from the back side of the train car for the platform observer.
Exactly as it needs to. It length is going to change from 1.73 to twice that when it matches the platform speed.
Halc,
This is a disagreement on physics!!!
The platform observer sees the deceleration at the back but the front does not have any deceleration.
The train car is not a rigid body. It is stretching out.
Are you saying that the train car is a preferred frame?
Jano

...
I suspect you do, but that makes you trolling: deliberate misrepresentation. Your posts in the energy/momentum conservation thread bear this out.
...
Halc,
I am not trolling!
I started my argument with pure, clean SR logic.
The end result is a logical fallacy.
Pointing out the truth is NOT trolling!
Jano