[Halc]

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.

[hamdani yusuf (OP)]

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.

[Halc]

How very Newtonian

[hamdani yusuf (OP)]

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?

[hamdani yusuf (OP)]

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.

[hamdani yusuf (OP)]

How very Newtonian

The non-Newtonian 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.

[Halc]

The non-Newtonian 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.

[hamdani yusuf (OP)]

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.

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?

[Halc]

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.

[hamdani yusuf (OP)]

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.

[Halc]

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.

[hamdani yusuf (OP)]

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.

So to visualize the time jump, we need to draw tilted line of simultaneity, according to the object's speed.

[hamdani yusuf (OP)]

Here is the same scenario, but the twins start from earth simultaneously.

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.

[hamdani yusuf (OP)]

Here I found two common logical errors we can make in trying to explain asymmetric result in twin paradox.
https://hal.archives-ouvertes.fr/hal-02263337/document
Gerrit Coddens. What is the reason for the asymmetry between the twins in the twin paradox?. 2019.
hal-02263337v5

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 five-star 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 .

[hamdani yusuf (OP)]

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.

The result is the same as previous case, and the numbers are very similar.

[hamdani yusuf (OP)]

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?

[Halc]

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.

[Janus]

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. 

Space-time diagram from the Earth observer frame.
 
Image1.gif (6.81 kB . 360x566 - viewed 5054 times)

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:


Image2.gif (7.43 kB . 277x562 - viewed 5031 times)
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 Earth-planet 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.

[David Cooper]

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.

[hamdani yusuf (OP)]

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.