Does the Twin Paradox Break Symmetry in the Zig-Zag Scenario?

[ash64449]

Thank you all! I think i understand now!

I think i haven't sufficiently understood PeterDonis's #35 post where he explains what frame of reference is. It must be because there are several terms which i haven't been exposed previously.( for example: meaning of timelike and spacelike vectors and they way vectors have physical meaning-contracting the frame field vectors at that point with other vectors)

I completely understand PeterDonis's #19 post, that B is only changing the state of motion w.r.t to A and not changing frames.

Actually changing frames do not have any meaning. All objects are in all frame of reference.( the post that i made do not have any meaning at all!)

While calculating,it is important that you work all the calculation in only one frame. We should not change reference frame for calculation mid-way.

You can describe the entire scenario using just one frame

An example of this was given by ghwellsjr in #5 post of this thread: https://www.physicsforums.com/threa...pace-time-diagram-analysis-resolution.814805/

I would like to point specifically to the part of the post where he used a space-time diagram in which he used rest frame of Stella during the first part of the scenario of Twin paradox being discussed there.

That post helped me to understand that frames are not used in general to understand one's point of view( since Stella is rest(at the origin) in frame being used for the first part). But one can find that during returning part of the journey,in the frame concerned, Stella changed position in that frame.

Reference frame are actually used for calculation purpose only. It may describe one's view if and only if object don't change the state of motion w.r.t it.
Therefore i understood that

co-ordinate system having the clocks to measure time at which events happen(I think they are called reference frame) are attached to a particular object located in the origin of the reference frame

is wrong since by this statement i was referring that reference frame is used for understanding one's point of view. I thought that this reference frame helped the object describe him when event's happen to him. This is totally wrong since this is possible so long as object don't change the reference frame.

As a result of this understanding i obtained the answer for my post:

So conclusion is if object tries to move,reference frame(which object uses to perform calculations)remains there?

 

[ash64449]

If i consider that a reference frame which helps an observer describe 'time' at which event happen(by putting clocks all over the coordinate system),coordinates at which event happen to be his frame, Can't i consider an observer an the origin of the reference frame at rest to be his frame?

If an object moves away from origin with a certain velocity, now observer has changed frames according to my consideration,right? Since now this reference frame doesn't help him describe 'time' at which event happen and coordinates at which an event happen?

 

[PeterDonis]

there are several terms which i haven't been exposed previously.( for example: meaning of timelike and spacelike vectors and they way vectors have physical meaning-contracting the frame field vectors at that point with other vectors)

These ideas are very important; if you don't understand them, it's going to be very difficult for you to make sense of any discussion of frames of reference. I suggest working through Sean Carroll's online lecture notes on relativity:

http://arxiv.org/abs/gr-qc/9712019

Chapter 1, on SR and flat spacetime, introduces the basic concepts.

Can't i consider an observer an the origin of the reference frame at rest to be his frame?

No, because, as you go on to show in your very next sentence, the observer can change his state of motion, but the frame can't. So the observer and the frame can't be identical.

If an object moves away from origin with a certain velocity, now observer has changed frames according to my consideration,right?

No, he's changed his state of motion. He was at rest relative to one frame before; now he's at rest relative to another, different frame.

this reference frame doesn't help him describe 'time' at which event happen and coordinates at which an event happen?

"Time" and "coordinates" are not absolutes; they depend on the coordinate chart you choose. Here you are implicitly choosing a coordinate chart corresponding to the first frame (the chart whose coordinate basis vectors are the frame vectors). The coordinates in this chart do give "times" and "positions" that would be measured by the first frame.

But you could just as easily choose a coordinate chart corresponding to the second frame (the one the observer is at rest relative to after he changes his state of motion). This chart would give "times" and "positions" that would be measured by the second frame, and after the observer changes his state of motion, he is at rest relative to the second frame, so its measurements of time and position would correspond to his.

 

[CycoFin]

To all these SR twin paradoxes I can give two basic rules.

1. Moving clock seems to run slower.
2. If you are looking moving clocks that are moving past you, that time seems to go faster.

At first these two rules can be seen to be in contradiction with each other but it is capturing the main idea of SR.

 

[Stephanus]

To all these SR twin paradoxes I can give two basic rules.

1. Moving clock seems to run slower.
2. If you are looking moving clocks that are moving past you, that time seems to go faster.

At first these two rules can be seen to be in contradiction with each other but it is capturing the main idea of SR.

But, who moves?
The clock or us?

 

[CycoFin]

But, who moves?
The clock or us?

does not matter, result is same

 

[Stephanus]

does not matter, result is same

Okay...,
But if one of the twin comes back, why he's younger than the one who stay? Should the result be same?

 

[CycoFin]

Okay...,
But if one of the twin comes back, why he's younger than the one who stay? Should the result be same?

Because he made the turn back.

Why he knows that he made turn back and not his twin? Because all the clocks from his point of view lost sync on turn.

 

[PeterDonis]

To all these SR twin paradoxes I can give two basic rules.

1. Moving clock seems to run slower.
2. If you are looking moving clocks that are moving past you, that time seems to go faster.

The second rule doesn't make sense, and neither of them really seem helpful for understanding the twin paradox.

all the clocks from his point of view lost sync on turn.

What clocks? If the traveling twin carries a clock with him, it doesn't do anything unusual when he turns around.

If you're talking about clocks that are spatially separated from the traveling twin, then what, if anything, "happens" to them when the traveling twin turns around depends on what simultaneity convention you adopt.

 

[PeterDonis]

if one of the twin comes back, why he's younger than the one who stay?

Because his path through spacetime is shorter. The correct general rule for deciding who has less elapsed time is not "who is moving". The correct general rule is to look at the different paths through spacetime and compare their lengths.

In the case of the standard twin paradox, the comparison is simple because one of the twins, the stay-at-home twin, is in free fall the whole time; he never feels any acceleration. It's easy to show that, between any two events in flat spacetime (i.e., in the absence of gravity), the path between them with the most elapsed time is the free-fall path. One way of seeing this is to observe that the traveling twin's path through spacetime, if we idealize his turning around to take place instantaneously, forms two sides of a triangle, and the stay-at-home twin's path through spacetime is the third side of the triangle. There is a triangle inequality in spacetime similar to the one in ordinary Euclidean geometry; the only difference is that, because of the minus sign in the metric of spacetime, the third side of the triangle (the stay-at-home twin's path) is longer than the sum of the other two (the traveling twin's path).

In other words, the key thing is not "motion", it's spacetime geometry.

 

[CycoFin]

The second rule doesn't make sense, and neither of them really seem helpful for understanding the twin paradox.

If you look clocks that are moving by, each of those clocks is running slower than yours. But if you look time by those clocks that pass you, that time is running faster than yours. It seems that those moving clocks are running slower but they are out sync so that the time they show (at your point) is running faster.

What clocks? If the traveling twin carries a clock with him, it doesn't do anything unusual when he turns around.

If you're talking about clocks that are spatially separated from the traveling twin, then what, if anything, "happens" to them when the traveling twin turns around depends on what simultaneity convention you adopt.

Yes, I mean two clocks spatially separeted with traveling twin (let say front and back of the spaceship). They are not in sync after turn. Also even staying twin didn't have his clocks on sync on first place (see above, from point of view of the turning twin), they are now even different way out of sync after turn (from point of view of the turning twin).

 

[PeterDonis]

if you look time by those clocks that pass you, that time is running faster than yours.

You're going to have to give a specific scenario to illustrate this. It doesn't look right to me.

 

[Ibix]

He's imagining a stream of clocks passing in front of him at velocity v. If he always looks at the clock infront of him and ignores the fact that it isn't the same clock, it appears to be running fast since the Lorentz transform simplifies to $t'=\gamma t$ if x=0. The usual time dilation formula, of course, comes from holding x'=0, rather than x.

What relevance this has to the twin paradox I am not sure.

 

[CycoFin]

You're going to have to give a specific scenario to illustrate this. It doesn't look right to me.

Maybe it is my english that it is causing troubles, this should be very basic thing to person who knows SR.

Let say we have stationary clocks between Earth and pluto. They are on sync from our point of view.

If one person rides with spaceship from Earth to pluto with relativistic speeds, he will notice that clock on Earth and pluto and every clock between them is running slower than his own spaceship clock. But if he watch clocks from his spaceship window, the time the clocks are showing is running faster than his own spaceship clock. This is because from his point of view our clocks are NOT in sync. For him, clocks on near Pluto are advanced, not showing same time as on earth. "Why your clocks are not in sync and are running slow?" he complains to us. "Why your clocks are not in sync and are running slow", we complain to him.

I need to add two more rules

3. moving clocks are out of sync
4. stationary clocks remain on sync unless you change speed or direction

 

[PeterDonis]

If one person rides with spaceship from Earth to pluto with relativistic speeds, he will notice that clock on Earth and pluto and every clock between them is running slower than his own spaceship clock. But if he watch clocks from his spaceship window, the time the clocks are showing is running faster than his own spaceship clock.

No, that is not correct. The person on the ship will observe the clocks on Pluto running faster than his own, but he will observe the clocks on Earth running slower than his own. When he corrects both of these observations for light travel time, he computes that both sets of clocks (on Earth and Pluto) are running slower than his own.

This is because from his point of view our clocks are NOT in sync.

That is true, but it has nothing to do with the different rates of the clocks. Clock synchronization and clock rate are two different things.

 

[ash64449]

Because his path through spacetime is shorter

Clocks of both the twins measure proper time.
Clock with the traveling twin traverses lesser distance in space-time,therefore lesser proper time is for the clock with the traveling twin.

But the home twin traverses bigger distance in the space-time and therefore has more proper time than traveling twin.

Is that the reason why you consider that traveling twin ages less than home-twin?

 

[ash64449]

No, because, as you go on to show in your very next sentence, the observer can change his state of motion, but the frame can't. So the observer and the frame can't be identical.

No, he's changed his state of motion. He was at rest relative to one frame before; now he's at rest relative to another, different frame.

The only reason why you did not agree with me is that you have ignored this consideration or assumption i made:

If i consider that a reference frame which helps an observer describe 'time' at which event happen(by putting clocks all over the coordinate system),coordinates at which event happen to be his frame

You said:

he is at rest relative to the second frame, so its measurements of time and position would correspond to his.

I assumed that when such a correspondence is produced i would "call" it observer's frame.

Since you have ignored it, i as well will ignore it. I will talk in such a manner that 'changes frame' and 'his frame' would not come next time.

"Time" and "coordinates" are not absolutes; they depend on the coordinate chart you choose. Here you are implicitly choosing a coordinate chart corresponding to the first frame (the chart whose coordinate basis vectors are the frame vectors). The coordinates in this chart do give "times" and "positions" that would be measured by the first frame.

But you could just as easily choose a coordinate chart corresponding to the second frame (the one the observer is at rest relative to after he changes his state of motion). This chart would give "times" and "positions" that would be measured by the second frame, and after the observer changes his state of motion, he is at rest relative to the second frame, so its measurements of time and position would correspond to his.

I completely agree with you.

 

[PeterDonis]

I assumed that when such a correspondence is produced i would "call" it observer's frame.

And I was telling you that this way of speaking leads to confusion, because it doesn't keep distinct observers and frames. It looks like you are not going to use that way of speaking any longer, which was my objective in pointing all this out.

 

[PeterDonis]

Clock with the traveling twin traverses lesser distance in space-time,therefore lesser proper time is for the clock with the traveling twin.

But the home twin traverses bigger distance in the space-time and therefore has more proper time than traveling twin.

Is that the reason why you consider that traveling twin ages less than home-twin?

Yes.

 

[ash64449]

It is right that twin paradox can be solved by using any frame of reference-since in any frame of reference concerned traveling twin always ages less than the home twin.

But the original reason why twin paradox arises is due to one's point of view-because each twin considers other's clock to run slower-which correspond's to one's point of view.

But it is to be understood that any frame of reference in general doesn't express one's point of view.

For example, we can describe home-twin's point of view by using a frame of reference in which the home-twin is at rest. Since home-twin is always at rest and consider him to be at the origin,frame's measurement of 'time' and position would correspond to home-twin's.

But not so for traveling twin.

For example,if we use the frame of reference in which the traveling twin is rest in the first part of the scenario, traveling twin moves in this frame of reference in the second part of the scenario.
Similarly,we cannot use a frame of reference in which traveling twin is rest in the second part of the scenario.(i.e either frame unable to express traveling twin's point of view)

To express what i am saying, i am going to state what 'paradox' does twin paradox presents.

To keep the calculations simple,

Let the distance between the Earth and planet be 8 light years.(rest length between them)
Let speed of the traveling twin(TT) be 0.8c in both part's of the scenario.

Now the factor 'gamma' is 1/0.6

Now let the Home Twin (HT) calculate the time TT take to come back according to his point of view.Since the frame he uses has distance between the planet and Earth as 8 light years and speed is 0.8c,
Time taken by the TT to reach the planet is 10 years according to HT.
Consequently, Total time to reach the Earth is 20 years. So according to HT, TT takes 20 years to reach earth.
Now the HT makes calculations how much time TT would take for the same.

In the frame HT uses, TT is moving and hence time runs slower for TT. So HT predicts TT would measure 20*0.6=12 years to reach earth. And hence concludes that in between the events 'TT leaves the earth' and 'TT reaches the earth', proper time elapsed for TT is 12 years and proper time elapsed for HT is 20 years and hence TT is 8 years younger than HT.

Now let us go for the calculation of TT

In TT's point of view, Earth and the planet are moving and hence represent moving length. the length is 8*0.6=4.8 light years.

Speed of both planet and Earth is 0.8c, Therefore time taken for TT to reach planet is 4.8/0.8=6 years and therefore 12 years to reach the Earth for TT-same as predicted by HT.

Now TT calculates how much time HT measures for the same from his point of view.

According to TT, HT is moving. So time is running slower for HT. In all parts of the journey, it is the HT that moves, therefore time measured by HT should be 12*0.6=7.2 years for HT. So TT calculates elapsed proper time for him is 12 years but for HT is 7.2 years Therefore HT is 4.8 years younger that TT.This is the manner how both get contradicting result. Each predicts the other one should be younger.So the solution to the paradox should be given which includes one's point of view. Since the calculation error comes when one predicts the elapsed proper time of the other from his point.

Error is in the analysis on TT's and not HT. In fact HT's point of view is same as that of calculations made in a frame of reference where HT is rest.

But not for TT. TT's point of view can't be explained by any single frame since he is always moving in one part of the scenario or the other in whatever frame of reference one takes.

So original question of twin paradox should be: 'What is the error in the analysis of the age of the HT done by TT'?

 

[PeterDonis]

if we use the frame of reference in which the traveling twin is rest in the first part of the scenario, traveling twin moves in this frame of reference in the second part of the scenario.

This is only true if we use inertial frames. But that is not required. We can use a non-inertial frame (more precisely, a non-inertial coordinate chart) in which the traveling twin is at rest the whole time. But the metric in this chart will be different than it would be for an inertial chart, so when we use this non-inertial chart to calculate the elapsed proper time for the traveling twin and the home twin, we find that the home twin still experiences more elapsed time.

This is the manner how both get contradicting result. Each predicts the other one should be younger.

The HT's calculation is correct, because he is calculating the length of the entire path through spacetime followed by the TT. But the TT's calculation is not correct, because he is not calculating the length of the entire path through spacetime of the HT. There is a section of the HT's path that is not covered by the TT's calculation because of the change in simultaneity convention when the TT turns around.

I highly recommend reading the Usenet Physics FAQ article on the twin paradox:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Pay particular attention to the Spacetime Diagram Analysis; it clearly shows how everything fits together, and what is missing in the TT's analysis that you have described.

 

[ash64449]

This is only true if we use inertial frames. But that is not required. We can use a non-inertial frame (more precisely, a non-inertial coordinate chart) in which the traveling twin is at rest the whole time. But the metric in this chart will be different than it would be for an inertial chart, so when we use this non-inertial chart to calculate the elapsed proper time for the traveling twin and the home twin, we find that the home twin still experiences more elapsed time.

This method of treatment is new to me.

No, they don't. The HT's calculation is correct

Yes. I did say HT's calculation is correct. As i said here :

Error is in the analysis on TT's and not HT

There is a section of the HT's path that is not covered by the TT's calculation because of the change in simultaneity convention when the TT turns around.

I actually know the answer! i was trying to say that This paradox's error should be pointed out from one's point of view.

But definitely the answer i thought was also due to change in simultaneity convention but method is not in terms of space-time diagram analysis.

 

[PeterDonis]

This paradox's error should be pointed out from one's point of view.

I'm not sure exactly what "one's point of view" means, but it seems to me that pointing out to the TT that his calculation leaves out a portion of the HT's worldline should count as pointing out his error "from his point of view".

method is not in terms of space-time diagram analysis.

You can use any method you like, as long as you get the right answer. If the TT does an analysis that leaves out a portion of the HT's worldline, he is not going to get the right answer, so whatever method he is using is wrong.

 

[ash64449]

non-inertial frame (more precisely, a non-inertial coordinate chart) in which the traveling twin is at rest the whole time.

I don't think this is absolutely necessary. I mean what is the reason why we want TT rest the whole time?

Because we want to describe his point of view. The technique you think is not familiar to me. But i do know that switching frames helps to describe one's point of view.( here to describe TT's point of view, for the first part of the scenario,frame of reference in which Earth and the planet are moving towards the left. Second part of the scenario: frame of reference in which Earth and the planet move towards the right)

This helps us to understand the change in simultaneity convention satisfactorily to solve twin paradox.

 

[PeterDonis]

switching frames helps to describe one's point of view

If you're willing to accept that the TT's "point of view" can switch simultaneity conventions instantly, sure.

You also have to ignore the portions of the TT's trip where he is accelerating, not inertial--for any real scenario, the TT's turnaround will take a finite time, during which he is accelerating and so is not at rest in any inertial frame. During these times, the only way to describe the TT as being at rest is to use non-inertial coordinates. And once you've admitted that you need to do that, why not just construct a non-inertial coordinate chart in which the TT is at rest the whole time?

 

[ash64449]

I'm not sure exactly what "one's point of view" means

Well, for example each one sees other one's clock to run slower.( "We" can see that happen when we analyse same thing in a different frame,but that is not object's point of view)

his calculation leaves out a portion of the HT's worldline should count as pointing out his error "from his point of view".

yes. agreed.

EDIT: I don't understand how that attached file came from. I had problems while quoting your messages.

 

[ash64449]

If you're willing to accept that the TT's "point of view" can switch simultaneity conventions instantly, sure

Well, that's the most important part of my method of solution. His simultaneity convention changes at the turn around But definitely not the method your talking about i used.(the non-inertial chart you are talking about)

You also have to ignore the portions of the TT's trip where he is accelerating, not inertial--for any real scenario,

Yes. in fact i have ignored that. we can introduce situation in twin paradox in which acceleration is ignored but rest remain the same.

the only way to describe the TT as being at rest is to use non-inertial coordinates.

OK. It is a method. It is the only way if we there is a condition that i have to stick to the same co-ordinates the whole time.

But there is another method to make TT at rest. This involves switching between two inertial frames that helps TT to be at rest.

Note: the only reason why switching frame is needed is to describe one's point of view. But if that's not the case, Then there are chances you get error because we must stick to the same frame of reference for any calculation,except those calculation which describe one's point of view(for example, we can take a frame in which Earth and planet are moving towards the left and TT at rest initially and solve twin paradox but it does not solve the problem addressed by me " what is the error in the analysis of TT's point of view"?)

 

[PAllen]

Emboldened by Peter's essay on frames versus coordinates, I thought of producing a little essay on the relationship between my post #31 and Peter's #35 (along the way making terminology consistent between these posts). However, I think the a more important concept to address is the pervasive (and false) idea that it is reasonable for a non-inertial observer to assume distant positions and times correspond to those of an inertial observer momentarily co-located and stationary with respect to the non-inertial observer. This gets at some of the same issues as my prior plan, but with a more physical slant.

The first thing to ask is how does an inertial observer get from a local frame as described in #35 to a global, natural, coordinate system (which, in SR, is often called - sloppily - a global inertial frame, rather than an inertial coordinate system)? There are actually several ways, which all happen to produce the same result for the inertial observer. Simultaneity can be established with Einstein's simultaneity procedure. Distances can either be measured by two way light speed or rulers matched to a standard ruler, or by parallax. For inertial observers, it doesn't matter what (reasonable) procedure you choose, the result will be the same. One way to conceptualize the resulting coordinates is a grid of clocks connected by stiff springs that are never under tension or compression (in a moment it will be clear why specify this rather than rulers). The clocks, once synchronized, keep mutually consistent time (comparing them any time later shows they are in synch, accounting for light delays). Obviously, you don't actually construct such a thing, but you certainly would think reasonable coordinates would be equivalent to such a thing. A fundamental point is that the equivalence of different procedures produces a unique natural choice.

Then, you should ask, what are reasonable (global) coordinates for a non-inertial observer? The surprising answer (expanded on below), is that there aren't any! A little more precisely, there are many possible choices, but none of them satisfy the expectations you might have from the inertial case. Further, each fails in different ways, so there is none to prefer. As a result, there is simply no fixed meaning to be attached to a statement like "how far away is that distant object" for a non-inertial observe. The best you can do is say: "choosing to use procedure x (rather than y), the distance is blah". Even more surprisingly, while available procedures include radar ranging and parallax, one that is not available for all non-inertial observers is rulers!

Ok, so let's try by build physically motivated coordinates for a non-inertial observer. One might assume we can posit a grid of clocks connected by springs such that the springs don't stretch or compress. We allow that each clock is independently propelled as needed, to achieve this. We don't even worry, for now, about synchronizing the clocks. The question is, can we set up such a grid for a rocket that starts at rest and accelerates [edit: and turns], then coasts. The surprising answer is no, this is impossible, in principle in SR! The relevant theorem is Herglotz-Noether. Thus, the most basic concept of what a coordinate grid means is unachievable for a [edit: some ] non-inertial observer. There are different ways of understanding this, but one way is that trying to apply the constraint that the springs don't stretch or compress requires that some clocks move super-luminally (or worse, backwards in time). This same theorem implies that the concept of a long rigid ruler for a [edit: general] non-inertial observer does not exist, even conceptually.

On recognizing that it is impossible to build rigid grid coordinates for a non-inertial observer, what can you do? The two most common choices are radar coodinates or Fermi-Normal coordinates (developed for GR but useful for non-inertial observers in SR). At laboratory scales (or within a rocket), they are identical to plausible experimental precision. Radar coordinates are operationally defined and can be extended globally as long as the defining observer's world line is inertial before some event, and inertial after some event. The beginning and end inertial motions need not be the same. Fermi-Normal coordinates try to glue together momentarily co-moving inertial coordinates, to the extent that it is possible. As implied by Herglotz-Noether, this has a few unexpected consequences:

1) A world line of constant position, varying coordinate time, will not generally represent a timelike world line (this is a consequence of the impossibility of a rigid ruler per Herglotz-Noether).
2) To produce a valid coordinate chart (e.g. to avoid multiply labeling events), the coverage of the coordinates may be very limited. For a zigzag trajectory, they will be limited to a world tube around the origin.

 

[ash64449]

why not just construct a non-inertial coordinate chart in which the TT is at rest the whole time?

How to construct such non-inertial coordinate chart? is it the same as space-time diagram analysis?

 

[PeterDonis]

each one sees other one's clock to run slower.

It depends on what you mean by "see".

If you mean what each one actually sees, literally, with their eyes or a telescope (or a measuring device that detects Doppler shifts), then they will each see the other's clock running slower if they are moving away from each other, but they will each see the other's clock running faster if they are moving towards each other.

If you mean what each one calculates after correcting the above observations for light travel time, then the result depends on what kind of coordinates they are using. If each one is using inertial coordinates in which he is at rest, then yes, each one will "see" (calculate) the other's clock to be running slower. But if they decide to use some other coordinates, that might not be the case. There is no absolute fact of the matter about relative clock rates for spatially separated observers; the only absolute facts of the matter are about the elapsed times each one experiences on their own clocks when following particular paths through spacetime.

that's the most important part of my method of solution. His simultaneity convention changes at the turn around

But this "most important part" of your method is also the part that makes the TT's calculation of the HT's elapsed time wrong. The change in simultaneity convention at the turnaround is what causes the TT to leave out a portion of the HT's worldline in his calculations.

 

[ash64449]

But this "most important part" of your method is also the part that makes the TT's calculation of the HT's elapsed time wrong. The change in simultaneity convention at the turnaround is what causes the TT to leave out a portion of the HT's worldline in his calculations.

This is an image in the Wikipedia article "twin paradox".

You see that there is a portion left out(in between the red lines and blue lines), I think you are talking about this.

I think i will describe my solution to Twin paradox and that will help to find out what you and I was actually trying to say.

Now error in the analysis of TT's point of view in the calculation of HT's age is this- He ignored Relativity Of simultaneity that leads to non-synchronization of the clocks of Earth and the planet.

Now When TT started the journey, Both the TT's clock and HT's clocks read zero. But when TT left, From his point of view(use an inertial frame where both Earth and planet are moving to describe TT's point of view), Clocks of Earth and the planet are non-synchronous. This time Clock of the planet is at the back side of motion, Therefore planet clock is ahead than Earth's clock by 6.4 years(by calculation), Now when TT started the journey, Earth's clock was reading zero. Therefore Planet's clock should read 6.4 years.

Now HT's clock is running slower. TT takes 6 years to reach planet. By this time both the clocks( Earth clock and planet clock) will go ahead by 3.6 years.(6*0.6=3.6)

So Earth's clock will be reading 3.6 years and planet's clock will be reading 10 years when TT reaches planet.

Now for the return part of the journey, it is the Earth's clock that is at the rear and the planet's clock is at the back and hence their reading should differ by 6.4 years with Earth clock's time ahead. But during the start of the return journey,planet's clock is reading 10 years, therefore Earth's clock is reading 16.4 years( see the change in simultaneity convention, To describe TT's point of view we had to switch frames,(why? when describing his point of view,TT should be rest) which changes simultaneity convention)

Now TT takes another 6 years by which both clocks-earth's and the planet's clock go ahead by 3.6 years.

So by the time TT reaches earth, Earth's clock is reading 20 years of elapsed proper time and TT's clock 12 years of elapsed proper time.

Change is simultaneity convention at the turn around is key here.

 

[ash64449]

I see,,The portion in between 3.6 and 16.4 years is left out at the turn around from TT's point of view.
But that should not be worried. Why?

Because at the end from both's point of view calculation show that when they meet, HT shows 20 years and TT shows 12 years.

Since clocks show proper time, then both agree that HT has 20 years of elapsed proper time and TT 12 years of elapsed proper time and hence TT ages less than HT.

But really my method of analysis is about point of view analysis because original question asked for it- Point of view's are something which do not have absolute significance or no existence.

It is the space-time interval that has absolute significance since it doesn't change when we change frames. So that method(Peter's analysis) of analysis is better.

 

[PeterDonis]

The portion in between 3.6 and 16.4 years is left out at the turn around from TT's point of view.

Because at the end from both's point of view calculation show that when they meet, HT shows 20 years and TT shows 12 years.

These two statements do not seem consistent. If the first statement is true, then the calculation from the TT's point of view should be off by 16.4 - 3.6 = 12.8 years.

 

[PeterDonis]

Point of view's are something which do not have absolute significance or no existence.

Not only that, there is no unique definition of what an observer's "point of view" is; there are multiple ways of constructing one. Your analysis uses one particular way: adopt the simultaneity convention of the inertial frame in which the observer is at rest, and when the observer changes his state of motion, change simultaneity conventions instantly. But that is not the only possible way of doing it. As I said before, there is no absolute fact of the matter about relative clock rates for spatially separated observers; there is also no absolute fact of the matter about which spacelike separated events are simultaneous. It's a matter of convention, and there are multiple possible conventions.

 

[ash64449]

These two statements do not seem consistent. If the first statement is true, then the calculation from the TT's point of view should be off by 16.4 - 3.6 = 12.8 years.

OK. the first statement is false. Is there any false issue in my solution of the paradox?