[Eternal Student]

Hi.

What do you think would happen if they are both accelerated equally in magnitude, but in opposite direction?

   I don't know.  Where is the inertial frame?   You need to provide more information.

I have to assume some things.   I'll assume there is an inertial frame, F, with respect to which both people can be considered to have equal and opposite accelerations at all times, t, using the co-ordinates of frame F.    The situation is then perfectly symmetric for the two people, there is no dependance on which direction the people went in.   So the proper time elapsed for each person will be identical and if their clocks were synchronised at the start then they remain synchronised when they finally meet up again.

   This is different to the usual Twin paradox situation where only one of the twins would would have experienced an acceleration.   The usual twin paradox was clearly asymmetric  (not symmetric).

   The good news: There will always be a frame of reference where you can consider the people to have equal and opposite motions.
   The bad news:    However, that frame of reference may not be an inertial frame.  If it isn't an inertial frame, then a different result will follow.

Example 1:    The Hafele-Keating experiment discussed earlier / maybe in some vaguely related thread.   The clocks on the aeroplane showed a positive/negative time difference compared to the clock that stayed on earth depending on whether the aeroplane flew eastward or westward.   The frame in which the earth clock was at rest was not an inertial frame, if it had been then eastward or westward travel would have made no difference.

Example 2:    In the conventional twins paradox situation, it is tempting to consider a frame of reference where the origin will always be half-way between the travelling and earth-bound twin.   If the earth-bound twin considers the travelling twin to have velocity v then this   "half-way frame" must have a velocity v/2   relative to the earth-bound twin.     On the outward part of the journey the half-way frame was a Lorentz boost of the earth-bound frame by a velocity off-set parameter  +v/2     but when the travelling twin turns around at Andromeda (or wherever the furthest point was),   that frame has a velocity off-set parameter of -v/2   (it moves the other way).     Assuming the earth-bound frame was an inertial frame (which is what we do in the usual Twin paradox) then the "half-way frame" cannot be.   All inertial frames are just a consistent Lorentz boost of any other inertial frame and the half-way frame wasn't one consistent Lorentz boost throughout the whole journey.   As illustrated in the Hafele-Keating experiment, we are not surprised that westward or easward travel with respect to a non-inertial frame can make a difference to the elapsed time recorded by the clocks.
    This is another way in which you could resolve the so-called "paradox" of the twin paradox.   The only reference frame in which the motion of both twins is genuinely symmetric  (as regards displacement, velocity, acceleration and any other derived motion parameters like dⁿ x(t) / dtⁿ ) is a non-inertial frame.   We know from the Hafele-Keating experiment that motion in different directions (eastward or westward) w.r.t. a non-inertial frame really can affect the elapsed times recorded by the clocks.

Best Wishes.

[Halc]

I'll assume there is an inertial frame, F, with respect to which both people can be considered to have equal and opposite accelerations at all times, t, using the co-ordinates of frame F.    The situation is then perfectly symmetric for the two people

The situation is perfectly symmetric only if F is the inertial frame in which the two of them are initially stationary.

The good news: There will always be a frame of reference where you can consider the people to have equal and opposite motions.

Only if the people are initially stationary in that frame. If not, then they won't.

Example 1:    The Hafele-Keating experiment discussed earlier / maybe in some vaguely related thread.   The clocks on the aeroplane showed a positive/negative time difference compared to the clock that stayed on earth depending on whether the aeroplane flew eastward or westward.   The frame in which the earth clock was at rest was not an inertial frame, if it had been then eastward or westward travel would have made no difference.

But HK is not an example of identical proper acceleration. OK, it could have been identical coordinate acceleration relative to the rotating frame. If we idealize it (use trains on a uniform-altitude track to eliminate gravitational effects), and use trains that take a day to circumnavigate the globe so the start/stop point is identical for the frames we're using, the Eastbound train will experience less proper acceleration all the way than does the westbound one. They will have identical coordinate acceleration (0 all the way except at beginning/end) relative to the rotating frame, but not identical coordinate acceleration relative to the inertial frame of Earth.

[pzkpfw]

I think it's worth pointing out that identical proper accelerations can still result in differential aging.
In this badly drawn spacetime diagram (x) stays home, (y) and (z) fly off in rockets.
A: (y) and (z) apply identical accelerations to leave Earth.
B: (y) applies acceleration to reverse course and go back to Earth
C: later, (z) applies acceleration identical to what (y) used, to reverse and go back to Earth
D: (y) applies acceleration to stop at Earth
E: later, (z) applies acceleration identical to what (y) used, to stop at Earth
F: (x), (y), and (z) have three different ages.

Code: [Select]

.
 F|xyz
 E|xyZ   
  |xy z
  |xy  z
  |xy   z
 D|xY    z
 C|x y    Z
  |x  y  z
 B|x   Yz
  |x  yz
  |x yz
 A|xYZ
--+------------
  |
At all times of inertial travel, the situation(s) are symmetrical.
All proper accelerations are identical (B=C, D=E).
... but the overall situation is not symmetrical, the three paths through spacetime are different.

[hamdani yusuf (OP)]

 I don't know.  Where is the inertial frame?   You need to provide more information.

Let's say the twins started from the earth.

[hamdani yusuf (OP)]

I have to assume some things.   I'll assume there is an inertial frame, F, with respect to which both people can be considered to have equal and opposite accelerations at all times, t, using the co-ordinates of frame F.    The situation is then perfectly symmetric for the two people, there is no dependance on which direction the people went in.   So the proper time elapsed for each person will be identical and if their clocks were synchronised at the start then they remain synchronised when they finally meet up again.

We can make any assertion we like to believe, but our confidence in its validity depends on the quality of its supporting explanation.
Halc and many science Youtubers have shown that traveling twin can also have valid observation, and eventually get the same conclusion as the staying twin. He only need to take relativity of simultaneity into account. We don't just say that his observation is invalid, whatever it is, only because he changes his frame of reference.

We can also change the speed of one twin to make the situation no longer perfectly symmetrical. They go to the opposite direction in the same amount of time, say 4 years, then go back home. If your previous explanation is valid, it should also give the correct results in the slightly different situations.

[Eternal Student]

Hi.

The situation is perfectly symmetric only if F is the inertial frame in which the two of them are initially stationary.

   Not sure I agree - but I'll go half-way.     

1.  Relax the statement you ( @ Halc ) made slightly:
      The two people didn't have to be stationary in the frame F initially.   If the two people had initial velocities (in Frame F) that were equal in magnitude but opposite in direction then that would be just fine.

       Work in the frame F unless told otherwise
       Person 1 has initial velocity +v_initial,   while person 2 has velocity -v_initial   (I'll keep this 1 dimensional for simplicity and not make velocity into vectors).   We also have that the accelerations of each person are equal in magnitude but opposite in direction,   
     a₁ (t)   =  - a₂ (t).
     From which we obtain  that  v₁ (t) = - v₂ (t).         [Eqn 0]
The velocities of both people are equal and opposite at all times (in the frame F).

    Now we could just jump straight to the usual time-dilation formula for SR.   Select two close events, A and B along the path of person 1.   We'll choose to have Person 1 use their rest frame centred around themselves to describe co-ordinates of these events A and B.  There'll be different co-ordinates for events A and B in frame F.
The small time difference,  δt (in Frame F time co-ordinates)  between event A and B is related the difference in time co-ordinates allocated by person 1,  δT, by the following:
     δT  =  γ δt     where  γ is the usual  gamma factor.              [Eqn 1]
   (We had to use close events with only a small amount of time difference because the acceleration was of an unspecified nature,  person 1 may be accelerating and we can only apply [Eqn 1] with the velocity appearing in the gamma factor equal to the instantaneous velocity).
    Since person 1 was using their rest frame, this quantity δT is actually just the proper time elapsed for them between events A and B.  Additionally, the instantaneous velocity v(t) that will appear in that gamma factor γ is precisely the velocity of person 1 as determined in Frame F.
    We can integrate the expression  [Eqn 1]   to obtain a total time elapsed for person 1,
ΔT    =  ∫dT  =    ∫ γ(t) dt           [Eqn 2]
where this is a definite integral performed between limits we'll call t_early and t_later  as measured in the time co-ordinate of frame F.   It doesn't have to be the start and end of the entire journey, just some times t (in Frame F time co-ordinates) corresponding to two events along the path taken through spacetime by person 1.
(We'll pause to take a minor note that the exact position in space that person 1 has at some time doesn't influence the total elapsed time they experience along any segment of their journey.   The RHS of [Eqn 2]  involves only the gamma factor and is therefore only infuenced by their velocity during that segment of the journey.  So, for example, we could shift their initial position and it doesn't matter provided their motion remains the same).

An expression for the total time elapsed for person 2 can be obtained similarly (exactly the same as [Eqn 1] and [Eqn 2] but γ uses the velocity of person 2 in Frame F].  The γ(t) that appears in the RHS of [Eqn 2] is only dependant on  v² and not on the sign of the velocity, v.  So, from [Eqn 0] we know that the RHS will be identical.   So the total elapsed times recorded by either person will be identical when we progress from  t=0  to   t=now   as measured in the Frame F.

   Now we're all done.  Just press the <Play> button in Frame F and watch the people take their paths through spacetime.   They started at time t=0 in Frame F and somehow synchronized their stop watches.   That's easy if they were in the same initial position but there's an agreed procedure for synchronizing even if they were spatially separated.   The usual procedure is to have a light flash released half-way between them and have each person set their stop-watch going when the light flash reaches them.   (we would have needed to set off the light flash just before t=0, so let's assume it was).  Then they travel for a while and at some later time t_later  they compare the elapsed times displayed by their stopwatches somehow.   That's also easy if they are in the same place at time t_later but there can be an agreed procedure even if they are spatially separated.   The time interval in Frame F was   t_later - initial time  =  t_later  - 0   (or just  t_later to keep it simple).  So we see from [Eqn 2] that both people have recorded the same elapsed time on their stopwatches.
    We don't even need to be precise about whose time co-ordinate we are using.   [Eqn 2]  links the elapsed time for person 1  to the elapsed time in Frame F,   one is a monotonic function of the other because γ(t) is always a positive number.   The time shown on the stopwatch is the same for person 1 and person 2 at every instant  (use any one of the time co-ordinates to declare an instant of time).
   If we now inlcude the "minor note" made a little earlier (about our ability to shift the initial positions) then we have the following.... although it may be easier to show it in diagrams....
    When we have an inertial Frame F, so that particle 1 takes a path like this:


path1.jpg (15.74 kB . 368x379 - viewed 420 times)

  ....   or whatever you want.

   Then provided particle 2 takes a path like this:


path2.jpg (16.6 kB . 368x379 - viewed 422 times)

   ... where the motion is equal in magnitude but opposite in direction.

Then the proper time elapsed for both paticles will be the same, at every instant of time through that motion.

We don't need the start and end positions of both particles to be the same, they can be wherever you want.  Also we don't need the particles to have been at rest in Frame F initially - just moving in opposite directions with the same magnitude.

We can generalise the result even further if we wish:  We only needed the gamma factors to be the same at all instants of time.   Particle 1 and particle 2 can move at arbitrary angles to each other.  If the motion of particle 1 is the same as particle 2 upto some rotation and/or reflection of the frame, then the result holds.

- - - - - - - - - - -
 2.    I'll stiffen my earlier statement to inlcude that requirement and re-word it is as follows:

I'll assume there is an inertial frame, F, with respect to which both people can be considered to have equal and opposite accelerations motion at all times, t, using the co-ordinates of frame F.    The situation is then perfectly symmetric for the two people, there is no dependance on which direction the people went in.....
 

     Where motion is understood appropriately:    Identical (but directionally opposed) velocities initially along with identical but opposed accelerations at subsequent times.
 - - - - - -

This post is too long,  I won't go through the minor details related to the rest of your post, it hardly matters and it isn't sensible to side-track the OP any further.

Best Wishes.

[hamdani yusuf (OP)]

traveling twin can also have valid observation, and eventually get the same conclusion as the staying twin. He only need to take relativity of simultaneity into account. We don't just say that his observation is invalid, whatever it is, only because he changes his frame of reference.

We only need to explain how much his changes of reference frame affect his calculated age of the other twin.
So far, I haven't produced my own explanation for the twin paradox. I only cited various sources which I thought interesting or plausible, and then explored their implications, and found out how far they can be extrapolated until they break down and give nonsensical or clearly wrong results.

[Origin]

We only need to explain how much his changes of reference frame affect his calculated age of the other twin.
So far, I haven't produced my own explanation for the twin paradox. I only cited various sources which I thought interesting or plausible, and then explored their implications, and found out how far they can be extrapolated until they break down and give nonsensical or clearly wrong results.

Still not able to figure out why the twin paradox isn't actually a paradox?

[hamdani yusuf (OP)]

We only need to explain how much his changes of reference frame affect his calculated age of the other twin.
So far, I haven't produced my own explanation for the twin paradox. I only cited various sources which I thought interesting or plausible, and then explored their implications, and found out how far they can be extrapolated until they break down and give nonsensical or clearly wrong results.

Still not able to figure out why the twin paradox isn't actually a paradox?

If you think you can, please let us know. Let's start with this.

We can also change the speed of one twin to make the situation no longer perfectly symmetrical. They go to the opposite direction in the same amount of time, say 4 years, then go back home. If your previous explanation is valid, it should also give the correct results in the slightly different situations.

Let's make the staying twin in previous scenario move at 0.5c in the opposite direction of Alpha Centauri. To be consistent, the twin traveling to Alpha Centauri and back moves at 0.999c. How do you analyze the age of one twin from the perspective of the other twin?

[Origin]

If you think you can, please let us know. Let's start with this.

It has already been discussed.  The traveling twin accelerates to an inertial frame that is different than the at home twin and after the trip he then decelerates back to the at home twins inertial frame.

[hamdani yusuf (OP)]

Let's start with this.

Let's make a simpler case. One twin stay on earth, while another one travels to Alpha Centauri 4 light years away and back at 0.999c, just like before. One more twin travels with the second one, but only up to  midway, and then returns to earth, and then turns around again to midway, and then finally returns to earth. According to earth observer, both travelling twins always move at speed 0.999c during their 8 years plus journeys, which makes them younger by gamma factor around 22.4.
But according to the first travelling twin, the second travelling twin moves relative to him for half of the journey, which should produce time dilation. How should their calculations be reconciled?

[alancalverd]

both travelling twins always move at speed 0.999c

Not true. The fundamental error of almost everyone who talks about a paradox is to forget that a change of velocity is an acceleration. Even if you could have instantaneous acceleration, "go there at c and come back at c" is two different velocities.

[Origin]

Let's make a simpler case.

That isn't a simpler case.  The most simple case is a one way trip to another star or point in space.
The scenario is not important anyway since you will never figure out what is going on, you never reach a conclusion, you just go round and round and get more confused as your thread goes on.

[Eternal Student]

Hi.

But according to the first travelling twin, the second travelling twin moves relative to him for half of the journey, which should produce time dilation.

   It will - but that's not the only thing that influences the total time Emma would allocate to Fred.   The number and location of Emma and Fred's changes in motion will also be important.

Exactly as you stated here....

...traveling twin(s) can also have valid observation, and eventually get the same conclusion as the staying twin. He only need to take relativity of simultaneity into account.

   And relativitity of simultaneity includes treating the changes that follow when a twin changes direction.
See post #42,   https://www.thenakedscientists.com/forum/index.php?topic=86675.msg707537#msg707537

One of your twins   (triplets? since there were 3 twins?),  makes 3 changes in direction and sweeps out some events.   Another makes only 1 change of direction.    The earth twin made no changes.  The locations of every other twin relative to a twin who was changing direction is a complicated mess - but you could spend the time to identify it all, if you want.

 Anyway, you would see that the total time that Emma allocates to  Fred's total journey depends on quite a few things,  including   the number of -,   and the location of -,  changes in direction of motion that Fred and Emma may have had.

according to the first travelling twin, the second travelling twin moves relative to him for half of the journey,

    Simply knowing that Fred had some motion relative to Emma for half the total journey is not enough information.   Emma's allocation of time for Fred depends on a bunch of things.
- - - - - - -
Overall, it's easier to solve the problem a different way (avoid working in Emma's co-ordinates as much as possible).   
(i) Draw a spacetime diagram that shows everyones journey through spacetime - use the co-ordinates of the earth frame for this because it's easier.   
(ii) Measure space-time intervals between important events.  We know that's an invariant in any other inertial reference frame.   Along appropriate segments and for the appropriate twin these space-time intervals become purely time intervals in the rest frame of that twin.  That's the proper time elapsed for that twin and recorded on their stopwatch.     Add up the elapsed proper time for each twin along all the segments involved in their own journey to get their total elapsed time on their stopwatch.  Compare the final results.

Best Wishes.

[hamdani yusuf (OP)]

Let's make a simpler case.

That isn't a simpler case.  The most simple case is a one way trip to another star or point in space.
The scenario is not important anyway since you will never figure out what is going on, you never reach a conclusion, you just go round and round and get more confused as your thread goes on.

It's numerically simpler, since there is only one speed of the journey relative to the earth observer.
I'll let competing ideas compete, and let the best idea win. They don't have to be perfect, but I'll stick to the best one for the time being.
My criteria for best idea is effectiveness and efficiency, which in this case can be seen as generality and simplicity, respectively. The idea should be still applicable for some variations of the original thought experiment without giving clearly wrong or inconsistent results. When there are some ideas passing this first test, then apply Occam's razor to select the simplest one, which means that it uses least assumptions.

[hamdani yusuf (OP)]

Bad idea to reference your common sense or intuitions when discussing relativity that you obviously don't understand.

Here's what understanding means.

Gregory Chaitin propounds a view that comprehension is a kind of data compression.[19] In his essay "The Limits of Reason", he argues that understanding something means being able to figure out a simple set of rules that explains it. For example, we understand why day and night exist because we have a simple model?the rotation of the earth?that explains a tremendous amount of data?changes in brightness, temperature, and atmospheric composition of the earth. We have compressed a large amount of information by using a simple model that predicts it. Similarly, we understand the number 0.33333... by thinking of it as one-third. The first way of representing the number requires five concepts ("0", "decimal point", "3", "infinity", "infinity of 3"); but the second way can produce all the data of the first representation, but uses only three concepts ("1", "division", "3"). Chaitin argues that comprehension is this ability to compress data. This perspective on comprehension forms the foundation of some models of intelligent agents, as in Nello Cristianini's book "The shortcut", where it is used to explain that machines can understand the world in fundamentally non-human ways.
https://en.wikipedia.org/wiki/Understanding#As_a_model

It may sound unintuitive, but if you can recite a section of textbook verbatim, you don't necessarily understand it.
You need to compress the information so you can generalize it and apply the same idea in some different situations.

[alancalverd]

Similarly, we understand the number 0.33333... by thinking of it as one-third.

So how do we "understand" 1.414213......, 3.141459..... 2.71828.....  or any other irrational number?

[hamdani yusuf (OP)]

both travelling twins always move at speed 0.999c

Not true. The fundamental error of almost everyone who talks about a paradox is to forget that a change of velocity is an acceleration. Even if you could have instantaneous acceleration, "go there at c and come back at c" is two different velocities.

I mentioned speed instead of velocity.

[hamdani yusuf (OP)]

Similarly, we understand the number 0.33333... by thinking of it as one-third.

So how do we "understand" 1.414213......, 3.141459..... 2.71828.....  or any other irrational number?

By memorizing mathematical operations like square root, exponentiation, integration, etc. Then apply them to some natural numbers to get those irrational number algorithmically.

[Origin]

I'll let competing ideas compete, and let the best idea win. They don't have to be perfect, but I'll stick to the best one for the time being.

Great.  What is the best one?