Twin Paradox Explained: No Math Required

[Mark1980]

Hello I have been trying to understand the twin paradox (without math) but I’m still trying to grasp the idea. I have seen and read enough tutorials to know that acceleration is not needed for the twin paradox to be solved. For anyone who doesn’t know the twin paradox without acceleration thought experiment it goes like this.

An outbound traveller from beyond Earth is moving at a constant velocity 0.5c relative to Earth towards a distant star which is also on the same inertial frame as the earth. As they pass the earth, the Earth observer and the outbound traveller start a clock at zero. At the same time there is an inbound traveller beyond the distant star traveling at the same constant velocity as the outbound traveller but in the opposite direction. Both travellers will pass the distant star at the same time at which point the inbound traveller will synchronise a clock with the time of the outbound traveller. Later on when the inbound traveller passes by the Earth and compares their clock with the earth, their clock will read less time than the Earth's. This is explained by the fact that the travellers clocks had been on two inertial frames.

I am struggling with this idea a little so let me propose a similar thought experiment below to further illustrate my thinking.

Suppose you have three planets lined up an equal distance apart on the same inertial frame, let’s call them Planet A, B & C. Now let’s say there are two spaceships, let’s call them spaceship X and spaceship Y. X is traveling at a constant velocity v (0.5c relative to the planets) from beyond A towards C via B. Y is traveling at the same constant velocity but in the opposite direction. X will pass by A at the same time as Y passes by C.

Now suppose as X and Y pass A and C respectively they will both start a clock to time their journeys. An observer on planet A will also start a clock at this time. When the spaceships reach B (and pass each other by) Y will synchronise a second clock with the clock from X.

Now we have three clocks, one on planet A and two on Spaceship Y; his own original clock started at intersection of planet C and the one he synchronised with X’s clock at the intersection of planet B.

As Y passes by A they will compare all the clocks times. According to the thought experiments I have read on the twin paradox, the clock that was synchronised from Spaceship X should read less than the clock on Planet A (for arguments sake let’s say 20 years). Planet A should read more (for arguments sake let’s say 30 years) than this and therefore the traveling clock has aged less.

However my confusion is coming from the fact that Y’s original clock would also read 20 years would it not? 10 years from planet C to B and another 10 years from B to A. But this clock is only on one inertial frame so why would their original clock have read 20 years whilst the planet A’s clock reads 30 as both can be considered at rest?

Apologies if this thought experiment is too confusing I tried to make it as simple as possible. I’m certain my thinking is wrong somewhere here but if someone can point out to me where I am going wrong it would be appreciated.

 

[PeterDonis]

is moving at a constant velocity

Relative to what? Velocity is relative. If you don't say what the traveler's velocity is relative to, you have not specified a well-defined scenario. I assume you mean that the Earth and the distant star are at rest relative to each other, and that velocity is relative to the Earth and the star, but you should not leave it to me to assume; you should specify it explicitly. This will not just help me; it will help you, by forcing you to think explicitly about every assumption you are making and what you have to specify to make sure you have a well-defined scenario.

This is explained by the fact that the travellers clocks had been on two inertial frames.

No, it's not, it's explained by the fact that the path through spacetime followed by the travelers (outbound until meeting, then inbound) is shorter than the path through spacetime followed by the earth. It's the Minkowski spacetime version of the triangle inequality; because of the minus sign in the metric, the two sides of the triangle in spacetime that are followed by the travelers are together shorter than the third side, followed by the earth.

if the inbound traveller had of started their own clock beyond the distant star at double the distance from the Earth to the star then the same result would have occurred no?

The zero point of the clocks doesn't matter. All that matters is the difference in the times recorded on the clocks between meetings. The inbound traveler has to subtract the reading on his clock when he met the outbound traveler, from the reading on his clock when he reaches the earth, to get his elapsed time between those two events. That will be independent of when he set his clock to zero (as long as he didn't reset it while he was traveling between the two meetings).

I suggest that you rethink all your scenarios with the above in mind.

 

[Dragon27]

But this clock is only on one inertial frame so why would their original clock have read 20 years whilst the planet A’s clock reads 30 as both can be considered at rest?

They're certainly not at rest with each other, otherwise they wouldn't meet in one place having started in different places. If I read you assumptions right, we have one clock at the planet A, and another clock at the planet C. The planets are stationary in our main inertial frame. In this frame both clocks start at zero at the same time, and then the clock from planet C moves toward the stationary clock at planet A. As it reaches planet A obviously it will show less elapsed time than the clock from planet A. Now what's the problem with that?

 

[anuttarasammyak]

However my confusion is coming from the fact that Y’s original clock would also read 20 years would it not? 10 years from planet C to B and another 10 years from B to A. But this clock is only on one inertial frame so why would their original clock have read 20 years whilst the planet A’s clock reads 30 as both can be considered at rest?

Say the clocks of planets A and C are synchronized in their common rest IFR. In this IFR
Departure of Y, clock A: 0 clock C: 0 clock original Y:0
Arrival of Y, clock A: 30 clock C: 30 clock original Y:20

Y regards this correspondence is meaningless because clocks A and C are not only moving so under time dilation which is easily treated but also not synchronized in his IFR.

 

[Mark1980]

Ok Thanks for the replies. I have had a think about my scenarios and read some more and I realize there may have been some incorrect assumptions. Let me point them out and see if someone can confirm them for me. Note: I am removing spaceship X in the below examples.

First off the distance between the planets is something the observers on different inertial frames will disagree about. Let’s say the planets observe the distance between each other at 7.5 light years away. At 0.5c relative velocity they will observe spaceship Y taking 15 years to travel between planets. From the perspective of A, it therefore takes 30 years in total for spaceship Y to go from planet C to A and during this journey A will observe Y’s clock going slower and indeed Y will say it only took 20 years so A will feel vindicated in his observations.

However from the perspective of spaceship Y, they will see the three planets travel towards them at 0.5c relative velocity but due to length contraction Y will observe the distance between them being less than 7.5 light years and therefore when they start their clock at the passing of C, they will record A passing by them in only 20 years as according to them the relative velocity of the traveller (planet A) was the same but the distance traveled was shorter. Is this correct?
Assuming this is correct the only part of this that is still causes me problems is the fact that Y will still see the clock of A going slower during the journey so how does he account for the fact that A’s clock has actually aged more. Am I correct in saying this is due to the relativity of simultaneity?

Let me explain my thinking on this. The three planets who are on same inertial frame will agree that all events occur simultaneously right? That is if they start a clock when Y passes by C they will agree they started their clocks simultaneously. However spaceship Y will not agree on this. According to Y’s perspective prior to planet C even passing him by, 15 light years away he would observe planet A start their clock (planet A now 15 light years away traveling at 0.5c relative velocity and therefore taking 30 years to reach him). 10 years later C finally passes by him at which point Y starts his clock which then reads 20 years when A finally reaches him. The point being here that Y did not agree that the starting of A’s clock was simultaneous with him passing by C which is when he starts his clock?

Have I got everything correct here or am I still wrong in my assumptions?

 

[PeterDonis]

the distance between the planets is something the observers on different inertial frames will disagree about

Yes.

Let’s say the planets observe the distance between each other at 7.5 light years away. At 0.5c relative velocity they will observe spaceship Y taking 15 years to travel between planets. From the perspective of A, it therefore takes 30 years in total for spaceship Y to go from planet C to A

Yes.

during this journey A will observe Y’s clock going slower and indeed Y will say it only took 20

No. The time dilation factor for a speed of 0.5c is only $\sqrt{1 - 0.5^2} = \sqrt{0.75} = 0.866$, so $25.98$ years will elapse on Y's clock between leaving planet C and arriving at planet A.

Another way to get this result is to compute the spacetime interval between the events of Y leaving planet C and Y arriving at planet A, in the common rest frame of the planets. We have a distance of 30 light-years traveled in a total of 15 years, so the interval is $\sqrt{30^2 - 15^2} = \sqrt{675} = 25.98$ years.

A general comment: it really, really, really helps to think in terms of events. Pick a convenient inertial frame and give all the events coordinates in that frame; for example, we could say that the events of Y leaving planet C, call that event C, and Y arriving at planet A, call that event A, have coordinates in the planets' common rest frame of:

C: $(x, t) = (15, 0)$

A: $(x, t) = (0, 30)$.

(This assumes that planet A is at the spatial origin of the common rest frame.)

from the perspective of spaceship Y, they will see the three planets travel towards them at 0.5c relative velocity but due to length contraction Y will observe the distance between them being less than 7.5 light years

Yes. See below.

therefore when they start their clock at the passing of C, they will record A passing by them in only 20 years

No, 25.98 years. See above.

according to them the relative velocity of the traveller (planet A) was the same but the distance traveled was shorter

Yes. You can easily calculate how much shorter from the above: 25.98 years at 0.5c gives 12.99 light-years. So planet A travels 12.99 light-years, according to Y, and the distance between each planet is half that, or 6.495 light-years, in Y's rest frame.

Y will still see the clock of A going slower during the journey so how does he account for the fact that A’s clock has actually aged more.

A's clock hasn't aged more according to Y. But that's because of relativity of simultaneity: the event on A's worldline that happens "at the same time" as Y leaving planet C, according to Y, is not the same as the event on A's worldline that happens "at the same time" as Y leaving planet C, according to A. See below.

Y did not agree that the starting of A’s clock was simultaneous with him passing by C which is when he starts his clock?

Yes. The key is to ask the question: what event on A's worldline is simultaneous with event C above? This question has two different answers, one for A and one for Y.

For A, the answer is simple: it's the event, call it event O, with coordinates $(x, y) = (0, 0)$ in A's frame. That's because A is always at $x = 0$ in this frame (the planets are all at rest in this frame), and $t = 0$ is the time coordinate of event C (when Y leaves planet C).

For Y, the answer takes a little bit of work to find, but not much. Rather than deal with Lorentz transforms, we can simply ask: how much time elapses on A's clock, according to Y, during A's traversal of the 12.99 light-years, in Y's frame prior to meeting Y at event A? This will just be the 25.98 years that elapse on Y's clock, times the same time dilation factor we saw above (since the relative velocity is the same), which is 0.866, so we have $25.98 \times 0.866 = 22.499$ years. So according to Y, the event on A's worldline that is simultaneous with event C--i.e., the event that Y thinks of as "where A is when I leave planet C"--will happen $30 - 22.499 = 7.501$ years after event O (since the time elapsed on planet A between event O and event A is 30 years). So that event, call it event AY, will have coordinates $(x, t) = (0, 7.501)$ in the planets' common rest frame.

 

[Mark1980]

Great. I think I have a pretty good understanding of it now. Apologies for not including the math. I knew all along my numbers weren't strictly correct but I was just simplifying the numbers in order to conceptualise the whole thing. I have been reading and watching videos on the twin paradox for a while trying to wrap my head around this concept. There seems to be so much misleading or incomplete information about this. People going on about acceleration and jumps from one inertial frame to another causing the "time jumps". Once you understand the relativity of simultaneity the whole concept of time dilation and length contraction make much more sense. Thanks for the help.

 

[Ibix]

I was just simplifying the numbers in order to conceptualise the whole thing.

Use velocities of 0.6c or 0.8c. They give rational $\gamma$ factors (5/4 and 5/3 respectively), which makes everything cleaner.

Once you understand the relativity of simultaneity the whole concept of time dilation and length contraction make much more sense.

The relativity of simultaneity is the key to everything, yes.

 

[PeroK]

Great. I think I have a pretty good understanding of it now. Apologies for not including the math. I knew all along my numbers weren't strictly correct but I was just simplifying the numbers in order to conceptualise the whole thing. I have been reading and watching videos on the twin paradox for a while trying to wrap my head around this concept. There seems to be so much misleading or incomplete information about this. People going on about acceleration and jumps from one inertial frame to another causing the "time jumps". Once you understand the relativity of simultaneity the whole concept of time dilation and length contraction make much more sense. Thanks for the help.

You could try learning SR properly. Try here, for example:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

 

[Mark1980]

Thanks for the shared paper. That gives a pretty good explanation of SR and the twin paradox. I wish somebody had sent that to me before as it would have saved me a lot of trouble. I have a couple of queries I want to follow up on though since I've thought about this a bit more.

My examples above did not include any acceleration as I was trying to remove as many variables as possible and therefore only deals with observations from inertial frames but I'm interested in exploring observations made during accelerations. So let’s say there are two planets with individual observers on the same inertial frame as each other (A & B). Then a traveller (X) sets out from A to B. X accelerates for the first half of the trip and decelerates for the second half.

If we break the observations down to each group of observers then A & B see the same thing correct? They see X’s clock start at the same rate then as he departs and accelerates they see his clock slow down for the whole journey although at different rates depending on the speed they observe at any given moment? The rate they observe X’s clock slow down will increase more and more for the first half and then start to decrease more and more for the second half until he lands at B where it will be the same rate again.

X is a bit more complicated. Am I correct in assuming that he will see the clock on B move faster than his own and the clock on A move slower than his own during the whole trip but again at different rates depending on the relativistic speeds at any given moment during the trip?

 

[PeroK]

Thanks for the shared paper. That gives a pretty good explanation of SR and the twin paradox. I wish somebody had sent that to me before as it would have saved me a lot of trouble. I have a couple of queries I want to follow up on though since I've thought about this a bit more.

My examples above did not include any acceleration as I was trying to remove as many variables as possible and therefore only deals with observations from inertial frames but I'm interested in exploring observations made during accelerations. So let’s say there are two planets with individual observers on the same inertial frame as each other (A & B). Then a traveller (X) sets out from A to B. X accelerates for the first half of the trip and decelerates for the second half.

If we break the observations down to each group of observers then A & B see the same thing correct? They see X’s clock start at the same rate then as he departs and accelerates they see his clock slow down for the whole journey although at different rates depending on the speed they observe at any given moment? The rate they observe X’s clock slow down will increase more and more for the first half and then start to decrease more and more for the second half until he lands at B where it will be the same rate again.

X is a bit more complicated. Am I correct in assuming that he will see the clock on B move faster than his own and the clock on A move slower than his own during the whole trip but again at different rates depending on the relativistic speeds at any given moment during the trip?

If by "see" you mean calculated observations, then A and B share a reference frame so, by definition, they make the same observations. SR is a theory of spacetime dealing with the relationship between different inertial reference frames. In this case, a change of origin is not significant.

X has an accelerating rest frame, so saying what X "sees" is doubly ambiguous. You could say that at any instant X's frame of reference coincides with an inertial frame, with a given velocity relative to the A-B frame. But, as this set of inertial frames is continuously changing, piecing together what X observes over time is complicated and problematic.

 

[Mark1980]

Problematic mathematically or conceptually? Or both? I mean if you break X’s trip into a series of moments as you say so that at any given instant he is on an inertial frame then during the entire trip he is continually changing inertial frames. The first half he is changing from one inertial frame to another which has a faster speed relative to the planets and then for the second half that speed relative to the planets continually starts to decrease. For the first half as his speed increases would he not calculate that B’s clock is going at a faster and faster rate compared to A? That is to say if he calculated the clock times for both A and B between each change of inertial frames he would calculate that the clock for B compared to A is getting further and further ahead and this increasing disparity is due to the relativity of simultaneity?

When he decelerates he begins to see the rate of time decrease between A and B although X would still calculate that B’s clock is going at a faster rate at any given moment during the deceleration. He would only see the disparity in the rate of time disappear when he lands at B.

Is acceleration just something that SR is not equipped to deal with and an understanding of GR is needed?

 

[PeterDonis]

Is acceleration just something that SR is not equipped to deal with and an understanding of GR is needed?

Definitely not. But constructing non-inertial frames in SR involves issues that do not arise if you only use inertial frames. So in many cases the best way to deal with acceleration in SR is to describe accelerated motion using inertial frames.

The biggest issue with objects that are not moving inertially is that there is no such thing as "the" rest frame of such an object. There are multiple different ways of constructing frames in which an accelerated object is at rest, and all of them have limitations. Also, many intuitions that people have about inertial frames in SR are simply invalid for non-inertial frames.

 

[PeroK]

Is acceleration just something that SR is not equipped to deal with and an understanding of GR is needed?

SR deals with flat spacetime; GR deals with curved spacetime (gravity). But, dealing with accelerated reference frames in flat spacetime requires more sophisticated mathematics. See, for example, Rindler coordinates:

https://en.wikipedia.org/wiki/Rindler_coordinates

 

[Ibix]

I mean if you break X’s trip into a series of moments as you say so that at any given instant he is on an inertial frame then during the entire trip he is continually changing inertial frames.

The problem is that each frame covers all of spacetime, so using them all is redundant. You actually want slices of each one covering the slice of spacetime "during" the instant X is at rest in that frame. Unfortunately, the relativity of simultaneity bites you here - two frames don't agree on the boundary between "during" one instant and the next, so that naive construction falls over because each slice overlaps the next in some places and gaps open between them in others.

There are solutions to this - essentially, you define an arbitrary cutoff line between X's instantaneous rest frames. But this doesn't really have the characteristics of a rest frame for X. Better to come up with asingle coherent strategy.

 

[Mark1980]

Yes I didn’t like the idea of breaking the problem down into a series of inertial frames because that seems to defeat the purpose of the query which was to understand the difference between accelerating frames and inertial ones but it was alluded to before so I went along with it.

The problem is that each frame covers all of spacetime, so using them all is redundant. You actually want slices of each one covering the slice of spacetime "during" the instant X is at rest in that frame. Unfortunately, the relativity of simultaneity bites you here - two frames don't agree on the boundary between "during" one instant and the next, so that naive construction falls over because each slice overlaps the next in some places and gaps open between them in others.

Yes because the fact X is accelerating and reaching higher speeds (at least for the first half of the journey) he would observe a larger and larger disparity between Planets B’s clock to Planets A clock due to the relativity of simultaneity meaning he would constantly need to observe Planets B clock going faster and faster than A’s. Am I correct here? On the second half of the journey my thinking is this disparity begins to reduce although I'm a little confused in the difference in X’s observations between the acceleration and deceleration phases. I will get to this later. For now let me restart as I have had a rethink on this.

In the version of the twin paradox where the traveller reaches very fast speeds relative to the planets in a negligible amount of time before cruising at a constant speed, he will become non-inertial for a very brief period during which he sees the (for lack of a better term) “behind clock” (planet B’s clock) accelerate much faster than his own. For no part of the journey to B will X see his home planet A’s clock move faster than his own unless and until he turns around at planet B whereby he momentarily becomes non-inertial again and Planet A now becomes the “behind clock”.

For the purpose of understanding observations made in accelerating frames we can forget the return journey to planet A and focus on X’s journey to B. Now in the example above in the brief moment the traveller becomes non-inertial Planet B’s clock moves rapidly compared to his own. But if the traveller is instead non-inertial for the entire journey would he not see Planet B’s clock move faster than his own for the entire journey albeit not as rapidly? His proper acceleration will dictate the rate at which he sees B’s clock move; the faster his proper acceleration the more rapidly he will see the Planet B’s clock move compared to his own. Am I correct here? As I touched on before this also means X will see B’s clock move faster than A’s clock for the duration of the journey.

Now let's say X maintains his proper acceleration at 1g for the first half and then decelerates at 1g for the second half will he observe B’s clock move faster at the same rate compared to his own for the entire journey or will he have different observations for the acceleration and deceleration phases?

 

[PeroK]

Yes because the fact X is accelerating and reaching higher speeds (at least for the first half of the journey) he would observe a larger and larger disparity between Planets B’s clock to Planets A clock due to the relativity of simultaneity meaning he would constantly need to observe Planets B clock going faster and faster than A’s. Am I correct here?

If you think of how SR was developed: we started with the postulates and developed the Lorentz Transformation for Inertial Reference Frames.

If you want to study accelerated frames, you must apply the postulates to develop a transformation for accelerated frames. You can't simply apply ad hoc elements of the Lorentz Transformation.

 

[vanhees71]

You can calculate physical quantities in whatever frame you like, because they are defined by covariant objects. In SR it is most convenient to work in a (global) inertial frame using a pseudo-orthonormal basis, defining the usual Minkowski coordinates $x^{\mu}$.

Now take two observers starting at the same space-time point $P_1$ and meeting again at $P_2$. Let $x^{\mu}(\lambda)$ be the parametrization of the time-like worldline of the first observer and $y^{\mu}(\lambda')$ that of the second observer (in terms of the usual Minkowski coordinates). Then the twins' clocks comoving with each of them along their world lines read
$$c \tau_1 = \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda_1 \sqrt{\dot{x}^{\mu} \dot{x}^{\nu} \eta_{\mu \nu}}, \quad c \tau_2 = \int_{\lambda_1'}^{\lambda_2'} \mathrm{d} \lambda_1' \sqrt{\dot{x}^{\mu} \dot{x}^{\nu} \eta_{\mu \nu}}.$$
The only constraint on the world lines is that they begin and end at the same spacetime points, i.e.,
$$x^{\mu}(\lambda_1)=y^{\mu}(\lambda_1'), \quad x^{\mu}(\lambda_2)=y^{\mu}(\lambda_2')$$
and that they are timelike.

It's not very surprising that in general $\tau_1 \neq \tau_2$ from this geometrical point of view of spacetime: We defined the "duration" for each twins' trip as a physical scalar quantity, which is entirely independent on the coordinates and reference frames we use to do the calculation. Here we used a global inertial frame of SR and Minkowski coordiantes, because it's most convenient choice. The use of manifest covariant quantities to calculate the proper times for the duration of the respective trip of both twins ensures without further calculation that their "aging" during the trip is completely independent of the choice of the reference frame(s) you calculate it, because the proper times are scalar quantities.

That's all what's behind the "twin paradox". It's only a paradox, because we are used to Galilean thinking about space and time. From the point of view of special relativity, it's not so surprising anymore since from the geometrical point of view it's not more surpring than the fact that in a Euclidean plane connecting two points with different curves leads to different arc-lengths of this curve.

Another interesting question is, which time-like worldline we have to choose for an observer to age most. That's a typical problem for the calculus of variations. The proper time for an arbitrary time-like worldline connecting to fixed points is (again written in terms of the usual Minkowski coordinates) is
$$c \tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \underbrace{\sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}}_{L}.$$
The variational principle is exactly what physicists are used to in using Hamilton's principle, i.e., you get the Euler-Lagrange equations,
$$\dot{p}_{\mu}=\frac{\partial L}{\partial x^{\mu}}=0.$$
So all we need is
$$p_{\mu} = \frac{\partial L}{\partial \dot{x}^{\mu}} = \eta_{\mu \nu} \frac{\dot{x}^{\nu}}{\sqrt{\eta_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}}}=\text{const}.$$
Now we can use proper time itself to parametrize the curve, which leads to
$$ \eta_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=c=\text{const}$$
and we get
$$p_{\mu}=\dot{x}_{\mu}=\text{const},$$
i.e., an observer with constant four-velocity, i.e., an inertial observer ages most.

 

[PeterDonis]

dealing with accelerated reference frames in flat spacetime

Just to be clear, there is no need to deal with accelerated reference frames in SR in order to deal with accelerated motion in SR. Accelerated motion can be dealt with just fine using inertial frames. The only reason you would need an accelerated reference frame is if you insisted on having a "rest frame" for an accelerated object; but the real lesson of trying to deal with accelerated frames in SR is that insisting on a "rest frame" for an accelerated object is a lot less useful and meaningful than your original intuitions think it ought to be.

 

[Mark1980]

I'm still not getting it so let me break this down again but his time I will introduce some mathematics into this (I used online calculators here so correct me if any of them are wrong). I have tried to make this as short as possible but it may drag on.

We have our 2 planets A & B which are in the same inertial frame 10 light years (ly) apart. There is 1 observer on planet B and 3 on Planet A. The observer on Planet B and 1 observer on planet A will stay put for the duration of the experiment. The 2nd observer from planet A will warp to 0.6c instantaneously and travel to planet B remaining at 0.6c for the duration of the journey before stopping instantaneously again upon arrival. The final and 3rd observer starting on Planet A will accelerate for the entire journey to planet B (with no deceleration phase) until arrival where he will stop instantaneously to land at B regardless of what speed he is at at that point.

I will first focus on observations from the planetary observers and the first traveller who travels at a constant velocity. The planetary observers observe the 1st traveller go at 0.6c for 10ly. This journey takes 16.67 years for them but they view the travellers clock go at 80% the speed of their own clocks meaning they observe his clock go only 13.33 years. Simple enough.

The 1st traveller upon instantly reaching 0.6c sees the distance to Planet B contract to only 8ly but his relativistic speed remains at 0.6c therefore he calculates his journey at 13.33yrs. During this journey he observes both planets clocks go slower by the same factor that they observed his clock go slower which is 80% the speed of his own. Ergo he sees the planets clocks only move 10.67 years. The exception to this is during the brief moments when he becomes non-inertial. In the instant he sets off he views planet B’s clock move forward 6 years (6 + 10.67 = 16.67) and in the instant he stops at Planet B he sees the same effect on Planet A whereby Planet A’s clock leaps 6 years ahead.

The 2nd traveller who set off at the same time as traveller 1 accelerates at a comfortable 0.1089g and maintains this acceleration for the entire journey until he reaches planet B where he stops instantly. He reaches a cool 0.88c velocity by the time he reaches planet B and clocks his journey at a brisk 12.33yrs. The planetary observers will see his clock go slower and slower as he reaches faster and faster speeds and they will agree overall that his clock only passed 12.33 yrs.

Now in my head I'm thinking that because the first traveller observed planet B’s clock go faster for the brief moment he was non-inertial, that the 2nd traveller can also see Planet B’s clock go faster for the moment he is non-inertial which for him happens to be the entire journey. So instead of seeing a rapid jump in time like the 1st traveller he sees that spread out for his entire journey which for him translates to viewing a faster clock on Planet B. If my calculations are correct he would see Planet B’s clock go faster by ~35% compared to his own clock for the entire journey. Therefore where he sees his journey take 12.33 years he sees Planet B’s clock go 16.67 years.

At this point I do have some other questions that popped up in my head during the formulation of this thought experiment but I want to throw it to the forum first to see if my calculations and assumptions are correct at this point. If I am wrong please let me know at which point or which sections were wrong.

 

[Ibix]

The exception to this is during the brief moments when he becomes non-inertial. In the instant he sets off he views planet B’s clock move forward 6 years (6 + 10.67 = 16.67) and in the instant he stops at Planet B he sees the same effect on Planet A whereby Planet A’s clock leaps 6 years ahead.

It's important to realize that he sees none of this. He sets off at time zero in the rest frame of the planets, but the light showing him his destination planet's clocks reading zero is 10ly away. He can't see it yet. What he sees is the clocks reading -10 years, and that does not change when he accelerates.

What changes is his interpretation of that information. When he's at rest with respect to the planets he says "those clocks are 10ly away and stationary, so I need to add 10 years to account for the light travel time so they now show zero". After the acceleration he still sees the clock reading -10, but the planet is now doing 0.6c towards him, and the distance away it was when that light was emitted was 20ly. It is now 20 years later, but the planet's clocks are ticking slow, so he only adds 4/5 of 20 years to deduce that the planet's clocks now show +6 years. Also in 20 years the planet has moved 12ly, so is now 8ly distant as you note.

So, to repeat, what changes is not what the observer sees. Rather, his interpretation of the things he sees changes. In the constant acceleration case, trying to apply this process of reinterpretation continually won't tell you anything helpful. You can never deduce an arrival time because you are always just about to reinterpret everything again. You need a more sophisticated mechanism. Or you could just adopt a single inertial reference frame and work in that one.

 

[anuttarasammyak]

The 2nd observer from planet A will warp to 0.6c instantaneously and travel to planet B remaining at 0.6c for the duration of the journey before stopping instantaneously again upon arrival.

Warp is Sci-Fi word so "jump on board the spaceship when it passes by" and "jump it off on the planet" would be a more appropriate way to explain your case. He/she should be tough enough against the shock and needs infinite small or zero time to get on/off board ideally. On board he/she notices that planet A clock and planet B clock is badly synchronized with 6 years advance of B as you showed, perhaps by radio reading of B clock broadcast.

Here jumps or shocks or accelerations or forces have nothing to do with it. Such cases as
- no person but a clock is thrown in,
- an avatar robot on board is switched on when the ship is passing by
- a clock in the ship is adjusted with a clock on the planet when the ship is passing by
show the same result.