The twin Paradox revisited

[martillo]

15 minutes ago, Genady said:

No. In the last scenario, after B receives his signal, B ages by 10 while he observes C aging only by 5. The point is that when B receives his signal, C is already 5 in the B frame, because in this frame C received the signal before B.

But you say that at the end, when the twins encounter at the crossing point, both twins aged the same. Do you mean than after receiving the signal the other twin ages inversely, aging more and not less?

Edited February 12, 2023 by martillo

[Genady]

Just now, martillo said:

But you say that at the end, when the twins encounter at the crossing point, both twins aged the same. Do you mean than after receiving the signals the other twin ages inversely, aging more and not less?

No. When they meet, the age of B, since receiving his signal, is 10. The age of C, since receiving his signal, is 5+5=10. 10=10.

[swansont]

53 minutes ago, martillo said:

If the same answer is "the other age less" there's a problem. Both cannot age less than the other one...

Why, though? What prevents it? Time is relative. Saying this can’t be true stems from some assumption, and that assumption is flawed.

[martillo]

29 minutes ago, Genady said:

No. When they meet, the age of B, since receiving his signal, is 10. The age of C, since receiving his signal, is 5+5=10. 10=10.

But you said: " after B receives his signal, B ages by 10 while he observes C aging only by 5." This means that at some instant B having 10 observed C aged only 5 so how is that some time after they reach the same age? What happened for C first aging less and after recovering the same age than B? Everything as seen from B of course...

 

20 minutes ago, swansont said:

Why, though? What prevents it? Time is relative. Saying this can’t be true stems from some assumption, and that assumption is flawed.

As I mentioned, at the crossing point both twins can interchange photos to see how actually aged the other one. Only one can age less than the other one and for instance present a smaller beard.

Edited February 12, 2023 by martillo

[Genady]

4 minutes ago, martillo said:

But you said: " after B receives his signal, B ages by 10 while he observes C aging only by 5." This means that at some instant B having 10 observed C aged only 5 so how is that some time after they reach the same age? What happened for C first aging less and after recovering the same age than B?

Here it is step-by-step, in the B frame.

1. Signal goes out from the middle point in both directions.

2. One signal reaches C. C starts his clock.

3. Another signal reaches B. B starts his clock. At this moment in the B frame, B's clock shows 0 while C's clock already shows 5.

4. It takes 10 units from then for B to reach the crossing point. B observes that C's clock advances 5 units during this time.

5. At the crossing point, B's clock shows 10. C's clock also shows 10, because it was on 5 when B started his clock, and it advanced by 5 since then.

[swansont]

27 minutes ago, martillo said:

As I mentioned, at the crossing point both twins can interchange photos to see how actually aged the other one. Only one can age less than the other one and for instance present a smaller beard.

At the crossing point they are the same age

[martillo]

10 minutes ago, Genady said:

3. Another signal reaches B. B starts his clock. At this moment in the B frame, B's clock shows 0 while C's clock already shows 5.

But this would mean that at that instant C aged more than B by 5. So C first aging more than B and not less. This would not be the case if they are always travelling at some constant velocity relative to each other.

4 minutes ago, swansont said:

At the crossing point they are the same age

When the concept of the travelling twin aging less apply then?

[Lorentz Jr]

22 minutes ago, martillo said:

But this would mean that at that instant C aged more than B by 5. So C first aging more than B and not less. This would not be the case if they are always travelling at some constant velocity relative to each other.

3a. Another signal reaches B. B starts his clock. At this moment, neither B nor C is moving relative to the midpoint (because they've been waiting for the signal), so B's clock shows 0 and B calculates that C's clock also shows 0.

3b. B accelerates back toward the midpoint, and B also assumes that C has done the same. At this moment in the B frame, B's clock shows 0 and B calculates that C's clock already shows 5.

22 minutes ago, martillo said:

This would not be the case if they are always travelling at some constant velocity relative to each other.

Yes, but this is a problem with your example. B and C should really start at the midpoint (i.e. at the same point) if you want to make direct comparisons.

Edited February 12, 2023 by Lorentz Jr

[Genady]

7 minutes ago, martillo said:

But this would mean that at that instant C aged more than B by 5.

No. C did not age more than B. C started his clock before B. That's why by the time B starts his clock C has already 5 on his.

[Lorentz Jr]

4 minutes ago, Genady said:

C started his clock before B

... in B's reference frame.

(I know you already said that, but sometimes it helps for beginners to see it again.)

What were B and C doing before they received their signals?

Edited February 12, 2023 by Lorentz Jr

[Genady]

8 minutes ago, Lorentz Jr said:

What were B and C doing before they received their signals?

They were moving toward each other. The OP wanted to avoid acceleration and instead to "synchronize" their clocks by a signal from the middle point.

[martillo]

15 minutes ago, Lorentz Jr said:

3a. Another signal reaches B. B starts his clock. At this moment, neither B nor C is moving relative to the midpoint (because they've been waiting for the signal), so B's clock shows 0 and B calculates that C's clock also shows 0.

3b. B accelerates back toward the midpoint, and B also assumes that C has done the same. At this moment in the B frame, B's clock shows 0 and B calculates that C's clock already shows 5.

 

11 minutes ago, Genady said:

No. C did not age more than B. C started his clock before B. That's why by the time B starts his clock C has already 5 on his.

Well, that would mean twin C begins to age earlier than twin B does. I must think about all this much more...

14 minutes ago, Lorentz Jr said:

What were B and C doing before they received their signals?

Good question!

[Lorentz Jr]

5 minutes ago, martillo said:

C begins to age earlier than twin B does.

That's only true in B's reference frame. In C's reference frame, B begins to age earlier than twin C does. In the midpoint's reference frame, B and C begin to age at the same time.

You need to start and end B and C at the same point if you want to make fair comparisons, and that means they have to accelerate.

Edited February 12, 2023 by Lorentz Jr

[Genady]

17 minutes ago, martillo said:

C begins to age earlier than twin B does.

If B takes a picture of himself with a timestamp, say, 3, by his clock, and C takes a picture of himself with the same timestamp, 3, by his clock, then when they compare the pictures, they will look the same. The same at any timestamp.

[md65536]

1 hour ago, Lorentz Jr said:

In C's reference frame, B begins to age earlier than twin C does. In the midpoint's reference frame, B and C begin to age at the same time.

Careful, this can cause confusion. C has at least 2 reference frames. The frame you're calling "C's reference frame" is the frame where C is on the return journey to the meeting point. If that was clear to everyone in the discussion, that's fine.

Edit: Nevermind me, I missed where the discussion switched to a case with no acceleration. I guess that's fine because you're basically looking at only the last phase of a twin paradox experiment.

 

But in the full twin paradox experiment... in the C_outbound frame, B's clock is slower and B does everything that C does, later than C does. If they have a gradual acceleration phase, then when C is momentarily at rest with the "meeting point frame", B's clock has caught up and they've reached their maximum distance at exactly the same time in this frame. From then on, according to C, B's clock is ahead and anything that happens to C has happened earlier to B.

 

So, in the case of only the inbound B and C frames, and the midpoint frame where B and C are symmetrical and always the same age, it is relativity of simultaneity that makes them not the same age in other frames (except at the meeting event). In B's frame before meeting, C is older but ageing slower.

Edited February 12, 2023 by md65536

[Lorentz Jr]

3 minutes ago, md65536 said:

If that was clear to everyone in the discussion, that's fine.

I think it was. The only thing I don't see any discussion of is how the observers are supposed to know where the "midpoint" is when B and C receive their signals.

[Genady]

40 minutes ago, Lorentz Jr said:

how the observers are supposed to know where the "midpoint" is when B and C receive their signals.

It is the same point all the way, i.e., the starting point in the full process, the source point of the signals, and the meeting point at the end. IOW, the observer A. 

[martillo]

8 hours ago, Genady said:

Here it is step-by-step, in the B frame.

1. Signal goes out from the middle point in both directions.

2. One signal reaches C. C starts his clock.

3. Another signal reaches B. B starts his clock. At this moment in the B frame, B's clock shows 0 while C's clock already shows 5.

4. It takes 10 units from then for B to reach the crossing point. B observes that C's clock advances 5 units during this time.

5. At the crossing point, B's clock shows 10. C's clock also shows 10, because it was on 5 when B started his clock, and it advanced by 5 since then.

Seems to me now you have chosen a particular case of velocities for the twins. Seems to me that at other velocities the numbers don't match so well and it will be found the twins aging differently at the crossing point. Also seems to me you considered the twins' velocities as the same as the signals' velocities. If the signals are made of light this will bring the problem of travelling at light's velocity: time would not pass for them. Am I wrong in something?

Edited February 12, 2023 by martillo

[Genady]

18 minutes ago, martillo said:

Seems to me now you have chosen a particular case of velocities for the twins. Seems to me that at other velocities the numbers don't match so well and it will be found the twins aging differently at the crossing point. Also seems to me you considered the twins' velocities as the same as the signals' velocities. If the signals are made of light this will bring the problem of travelling at light's velocity: time would not pass for them. Am I wrong in something?

Yes, you are.

I picked some specific numbers for illustration. If you go through the math of Lorentz transformations, you'll find that it works out the same for any velocity.

Nowhere I considered the twins' velocities as the same as the signals' velocities. I assume that the signals are light. The twins' velocities are whatever they are.

[martillo]

15 hours ago, Lorentz Jr said:

You need to start and end B and C at the same point if you want to make fair comparisons, and that means they have to accelerate.

Fine, let us retake the case as formulated first:

 

19 hours ago, martillo said:

Considering the case of the twins A, B and C but forgetting twin A for a while. Just twins B and C. It is the case of two twins making a completely symmetrical travel and so both are affected the same way by any consideration in acceleration or time dilation. Each one will see the other one aging less and so there would be an inconsistency. The two different frames of reference observe contradictory results. How the inconsistency would be solved if possible?

Genady answered:

 

18 hours ago, Genady said:

Let's see it from the frame of B observing C.

By the time they reach the turning points, B aged 10 while C aged 5. After B turns, let's say almost momentarily, B will still be 10, but he will observe C being 15. By the time they reach the crossing point, B is 20, and C is 20.

The C will find all the same symmetrically.

I think now Genady considered here only the turn back of twin B aging less because of the acceleration in his own turning. Seems to me that the turning back of twin C would add the effect of less aging in C and they would not see the other one age the same at the end. B would observe C aged less at the end.

Edited February 12, 2023 by martillo

[Genady]

25 minutes ago, martillo said:

I think now Genady considered here only the turn back of twin B aging less because of the acceleration in his turning. Seems to me that the turning back of twin C would add the effect of less aging in C and they would not see the other one age the same at the end.

No, turning of C does not affect B. B just observes C moving with different velocities and calculates the total effect. However, turning of B affects B and its observations. 

The same in the classical scenario, with A staying on Earth and B going out and returning. Turning of B does not affect A. It does affect B and its observations.

We would have to consider the turning of C and its effect on the C observations, if we described the scenario in the C frame.

[martillo]

48 minutes ago, Genady said:

No, turning of C does not affect B.

I know, it just would affect his observations or calculations.

48 minutes ago, Genady said:

B just observes C moving with different velocities and calculates the total effect.

Do you mean that the turning acceleration of B compensates exactly the timing effects in C of both, the travel plus the acceleration in the turning, for the twins finally cross at the midpoint with the same age?

Edited February 12, 2023 by martillo

[Genady]

2 minutes ago, martillo said:

Do you mean that the turning acceleration of B compensates exactly the timing effects in C of both, the travel plus the acceleration in the turning?

I mean that in the B frame, the change in his observation of C caused by its turning compensates exactly the total time dilation that it observes in C.

[Genady]

1 hour ago, martillo said:

Do you mean that the turning acceleration of B compensates exactly the timing effects in C of both, the travel plus the acceleration in the turning, for the twins finally cross at the midpoint with the same age?

Maybe the most direct way to see what is going on there, is on a standard spacetime diagram. Here I've drawn it on the background of the A's frame, just because it is symmetrical and easy to draw. The analysis is still for the frame of twin B.

The vertical black line is A's time axis, and the horizontal black line is A's space axis. But I am not going to refer to them. I will focus on the colored lines.

The red line, IJK, is the worldline of B. The blue line, IMK, is the worldline of C. 

B starts at event I, turns at J, and returns at K. C starts at I, turns at M and returns at K.

When B is getting to the turning point J while outgoing on the section IJ, the green line JN is his simultaneity line, i.e., all events on this line are simultaneous with J, for B. So, coming to his turning point, J, B observes C at N, which is quite some time before the C's turning point, M. This is time dilation of C observed by B.

Then, B turns around, i.e., switches to the section JK. Now, the orange line JL is his simultaneity line, and he observes C at L, quite some time after C's turning point. While B makes the return trip JK, C makes only the part LK, which is again the time dilation of C as observed by B.

The total time that passed on B's clock is, IJ+JK. The total time that passed on C's clock as observed by B is IN+NML+LK. Obviously, because of the symmetry, these two lengths are equal. IOW, they aged by the same amount and cross at K with the same age.

[martillo]

Thanks for your clarifying explanations. I think the key point in the presented case is summarized in the statement:

2 hours ago, Genady said:

I mean that in the B frame, the change in his observation of C caused by its turning compensates exactly the total time dilation that it observes in C.

The timing variations in the clocks of the twins are strongly affected by their experienced accelerations at the turning back points. This case and the classic twins' paradox cannot be well explained without the considerations on the accelerations present. I remain wondering how they match perfectly with the time dilation effects but this would imply to enter in the area of General Relativity, something I do not pretend to do for now (I'm referring to the relation between clocks and accelerations).