The twin Paradox revisited

[Lorentz Jr]

2 minutes ago, martillo said:

The initial situation is when A at the midpoint emits the synchronizing signals to B and C. The clocks are synchronized to 0 after when receiving the signals.

And what do B and C do while the signals are traveling toward them? How do we know where they will be when they receive their signals?

[martillo]

8 minutes ago, Lorentz Jr said:

And what do B and C do while the signals are traveling toward them?

They are just travelling.

9 minutes ago, Lorentz Jr said:

How do we know where they will be when they receive their signals?

That is what I did in part of the calculations. You must read them.

On 2/14/2023 at 2:44 PM, martillo said:

Well, here my calculations. I hope the notation would not complicate to follow the reasoning. The aim was to facilitate the understanding. May be I'm wrong in some thing(s) I cannot see. I would appreciate to be aware of them.

RELATIVISTIC SYMMETRIC TRAVELERS

 

INITIAL CONSIDERATIONS:

B and C travelling in opposite directions to cross at midpoint A. They all reunite at the same position.

It is considered a stationary frame with B. The case of frame in C is totally symmetric.

A emit light synchronization signals to B and C : vel__B(signal) = c = 3x10⁸ m/s

Velocity of C in frame B: vel__B(C) = (3/5)c

Relativistic factor γ = 1/(1-v²/c²)^1/2 = 3/2

Relativistic relation of elapsed times: time__C = (1/γ)time__B

Distances observed by frame at B: dist__B(B,A) = dist__B(A,C) = 3x10⁸ mts

D : point where clock of C is synchronized: clock__C = 0

E: point where clock of B is synchronized: clock__B = 0

 

CALCULATIONS:

vel = dist/time --> time = dist/vel

Times of signals travelling as seen by B and C :

time__B(signal(A,B)) = dist__B(A,B)/c = 3x10⁸/3x10⁸ = 1 sec

time__C(signal(A,B)) = time__C(C,E) = (1/ γ).1 = 1/ γ = 2/3 sec

 

Distance from A to D travelled by signal of synchronization of C :

X = dist__B(A,D) which verifies:

time__B(C,D) = (dist__B(A,C) - X )/vel__C = (3x10⁸ – X)/(3/5)3x10⁸ =

                     = time__B(signal(A,D)) = X/(3x10⁸)

Substituting values:

(3x10⁸ – X)/(3/5)(3x10⁸) = X/(3x10⁸)

(5/3)3x10⁸ = (1 + 5/3)X

X = (5/8)3x10⁸ mts

Times of travelling of C from D to B as seen by B and C :

time__B(B,D) = (3x10⁸ + X)/(3/5)(3x10⁸) = (1 +5/8)/(3/5) = (13/8)/(3/5) = (13x5)/(3x8)  

                     = 65/24 sec

time__C(B,D) = (1/ γ)time__B(B,D) = (2/3).(65/24) = 65/36 sec

Time of clock of C at the end of the travel:

clock__C(B=C) = time__C(B,D) =  65/36 sec

(65/36 = 1.80555555…)

Time of clock of B at the end of the travel:

(time from the initial situation to the end – time of synchronization)

clock__B(B=C) = time__B(B,C) – time__B(signal(A,B)) = (dist__B(B,C)/vel__C) – 1 =

                        = (2(3x10⁸))/((3/5)(3x10⁸)) – 1 = 10/3 – 1 = 7/3 sec

(7/3 = 2.33333333…)

 

CONCLUSION:

Clock__C (B=C) < clock__B(B=C)

These calculations conclude in a final lecture of the clock of C with a smaller value than the value of the clock of B what would mean less time passed for C than the time passed for B.

 

FINAL NOTE:

Just to mention here, the calculation of the lecture of clock of C at the synchronization of B (point E) gives the value ¼ sec and it was verified that adding this value to the time elapsed in the travel of C from E to B (time_C(B,E)) matches the value of the total travel of C in 65/36 = 1.80555555… sec.

 

 

 

Each clocks start when the owner traveler receives the synchronizing signal from A.

I wrote the calculations in Word and copied and pasted to the window of the post. Some emoticons appeared! For instance the C : without the space is replaced by a smile. I just edited that with a space.

 

[Lorentz Jr]

9 hours ago, martillo said:

That is what I did in part of the calculations. You must read them.

Okay, let's start with an observer at rest with respect to A and at B's location x = -L when the signals arrive, and let's set c and L equal to 1 for simplicity, so we don't have to keep writing them in every step. Then we have the following:

• B receives a signal at t = x = 0.
• C receives a signal at  t = 0 and x = 2.

Now let's switch to B's reference frame, which moves at speed +v:

• B receives a signal at x' = t' = 0.
• C received a signal at x_i' = γ(2 - 0v) = 2γ  and  t_i' = γ(0 - v(2)) = -2γv.

Now we need the speed of C relative to B. We get that from the velocity-addition formula:

u' = (u-v) / (1 - uv) = ((-v) - v) / (1 - (-v)v) = -2v/(1 + v²)

So the (positive) relative speed between B and C is v' = 2v/(1 + v²), and C's motion in B's reference frame is

x' = x_i' - v'(t' - t_i') = 2γ - v'(t' - (-2γv)) = 2γ - v'(t' + 2γv))

In B's reference frame, B and C will meet when C gets to x' = 0:

0 = 2γ - v'(t_f' + 2γv)

v'(t_f' + 2γv) = 2γ

t_f' + 2γv = 2γ/v'

t_f' = 2γ/v' - 2γv = 2γ((1 + v²)/2v - v) = (γ/v) ((1 + v²) - 2v²) = 1/γv

L/γv is just the time-dilated time for B to return to the midpoint at A.

Now, what happens in C's frame, as calculated from B's frame?
We need the time when C received a signal (at  x_i' = 2γ  and  t_i' = -2γv  in B's frame).
Defining Γ to be the gamma factor for v' (i.e. between B and C), we have

t_i'' = Γ(t' + v'x') = Γ(- 2γv + 2γv') = 2Γγ(v' - v)

(1 + v²) t_i'' = 2Γγ( 2v - v(1 + v²) = 2Γγv(1 - v²) = 2Γv/γ

Of course, C's clock was set to zero, but that just means it's not synchronized correctly in B's frame. The important thing is that it passed through the desired point in spacetime.

And the time when C meets B is t_f'' = Γ(t' + v'x') = Γ(1/γv + 0v') = Γ/γv.

So the elapsed time in C's frame is Δ = t_f'' - t_i'' = Γ/γv + 2Γγ(v - v'),

and now the math gets a little grungy, so we'll define w = v² and w' = v'².

γvΔ/Γ = 1 + 2γ²v(v - v')

(1+w)γvΔ/Γ = 1 + w + 2γ²(w(1+w) - 2w) = 1 + w + 2γ²(w² - w)

(1-w)(1+w)γvΔ/Γ = (1-w)(1+w) + 2w(w-1)

(1+w)γvΔ/Γ = (1+w) - 2w = 1-w

(1+w)² wΔ² (1-w') = (1-w)³ = wΔ² ((1+w)² - 4w) = wΔ²(1-w)²

(1-w) = wΔ²

Δ = 1/γv

So the same amount of time elapses for C as does for B.

Edited February 16, 2023 by Lorentz Jr

[martillo]

1 hour ago, Lorentz Jr said:

Okay, let's start with an observer at rest with respect to A and at B's location x = -L when the signals arrive, and let's set c and L equal to 1 for simplicity, so we don't have to keep writing them in every step.

This are not the same assumptions I made for the initial state. This is a similar but different problem. In my problem the initial instant is that of the emission of the signals from A to B and C and in that instant B is located at -L and C at +L. B and C receive the signals at some time after when they set their clocks to zero. I consider just one stationary frame in B and a moving one in C. The reciprocal problem is just symmetric with symmetric results. It must be noted (as Genady pointed out) that in B stationary frame the traveler C receives the synchronizing signal before B because C is moving towards the initial place of A where the signal was emitted while B is not. This means that at the instant when B receives his signal (setting his clock to zero), C has already received its own before and so at that B instant the clock of B the time is greater than zero.

I have expected from you something about a possible error in my problem as I formulated it. You come with another problem. I will try to understand this your problem and point out something if I could see some issue in your resolution but unfortunately the followers of the thread have now two different problems to analyze making the things much hard I think. That is not something good I think but I leave the other ones to decide which problem to stay with...

NOTE: In my problem no velocity-addition formula is required to be applied. Is a much simpler problem I think. I don't switch frames...

Edited February 16, 2023 by martillo

[Lorentz Jr]

49 minutes ago, martillo said:

This are not the same assumptions I made for the initial state. This is a similar but different problem. In my problem the initial instant is that of the emission of the signals from A to B and C and in that instant B is located at -L and C at +L. B and C receive the signals at some time after when they set their clocks to zero.

So x = -ct for B's signal and x = vt - L for B. The signal arrives when -ct = vt - L.
(c+v)t = L
t = L/(c+v)
x = -L/(1 + v/c) for B and x = +L/(1 + v/c) for C.

It's exactly the same problem, martillo. The only difference is B and C receive their signals at a distance of L/(1 + v/c) from A instead of L, and the signals arrive at t = L/(c+v) instead of zero. Same problem with different numbers.

49 minutes ago, martillo said:

I consider just one stationary frame in B and a moving one in C.

That doesn't work. You need to start in A's frame, because that's the frame in which B and C's positions are symmetric and they receive their signals simultaneously.

49 minutes ago, martillo said:

The reciprocal problem is just symmetric with symmetric results. It must be noted (as Genady pointed out) that in B stationary frame the traveler C receives the synchronizing signal before B because C is moving towards the initial place of A where the signal was emitted while B is not. This means that at the instant when B receives his signal (setting his clock to zero), C has already received its own before and so at that B instant the clock of B the time is greater than zero.

Right. That's -2γv.

49 minutes ago, martillo said:

I have expected from you something about a possible error in my problem as I formulated it. You come with another problem.

I come with the correct solution to your problem, martillo. A slight variation with different numerical values, but the same math and same result.

49 minutes ago, martillo said:

NOTE: In my problem no velocity-addition formula is required to be applied. Is a much simpler problem I think. I don't switch frames...

Any solution without the velocity-addition formula will be incorrect, unless you solve the problem in A's frame. Otherwise, you need to know the speed of C relative to B in order to calculate the correct gamma factor between them.

Edited February 16, 2023 by Lorentz Jr

[martillo]

5 minutes ago, Lorentz Jr said:

Any solution without the velocity-addition formula will be incorrect. You need to know the speed of C relative to B in order to calculate the correct gamma factor between them.

No, in my problem there's just one stationary frame in B and a moving one in C. The only velocity considered is that of C in relation to B and I set it to be (3/5) of light velocity for a γ to be 3/2. No other velocity is needed to be considered. A is just the place where the synchronizing signals are emitted. At the end B and C reunite in that initial point A so to calculate when C reaches A is the same to calculate when C reaches B (independent of A) and this is what I do. 

[md65536]

12 hours ago, martillo said:

In frame A at the initial state of t = 0 we have:

A at position x = 0, traveler B at some x = -L and traveler C at some position x = +L.

Now in frame B by substituting values in the first equation of Lorentz transform we have:

A at position x = 0, traveler B at x' = -γL and traveler C at x' = +γL

We can see that A is preserved as a midpoint in the transform and my assumption would be correct then.

I'm not saying all my calculations are correct, just that the initial assumption dist__B(B,A) = dist__B(A,C) is correct.

I'm looking for other possible errors now but I cannot see anyone...

I still think you should start with what you understand and build up from there, but...

One thing I see is that you have A at the origin, and are looking only at time t=0 where the origins of A's frame and B's coincide. Yes, distances from the origin should scale by gamma. However, that stops working for t != 0, when the 2 frames' origins no longer coincide. Another way to look at it is that in B's frame, the velocity of C is not 2 times the velocity of A, so A does not maintain its place in the middle of B and C.

Also, to get things to align at this one time, you've set up everything really unconventionally. You have B not at the origin of its own frame. I think you're transforming "the starting location of B, in A's frame" into a coordinate in B's frame... but B is not actually at that coordinate at time t'=0 in B's frame! Since you have B far from the origin in its own frame, the time at that location will be transformed. So basically you're saying "In B's frame, A is in the middle of where C will be at some time in the past or future, and where B will be some time in the past or future," which I think is correct. This is too complex and confusing for me.

Edited February 16, 2023 by md65536

[Lorentz Jr]

On 2/14/2023 at 11:44 AM, martillo said:

time__C(signal(A,B)) = time__C(C,E) = (1/ γ).1 = 1/ γ = 2/3 sec

This is incorrect, martillo. You can't just use gamma for your calculations. You have to use the Lorentz transformations.

[martillo]

46 minutes ago, md65536 said:

One thing I see is that you have A at the origin, and are looking only at time t=0 where the origins of A's frame and B's coincide. Yes, distances from the origin should scale by gamma. However, that stops working for t != 0, when the 2 frames' origins no longer coincide. Another way to look at it is that in B's frame, the velocity of C is not 2 times the velocity of A, so A does not maintain its place in the middle of B and C.

The movement of A is not considered in the problem. Is no necessary. Just the movement of C in relation to B matters. The only things that matters in A is its initial location where the emission of the synchronizing signals takes place. At the end C crosses with B at that location but as the three coincide the calculation is made for when C crosses with B what is what really matters.

46 minutes ago, md65536 said:

Also, to get things to align at this one time, you've set up everything really unconventionally.

It could be unconventionally but precisely that is why it is simple to solve. 

46 minutes ago, md65536 said:

You have B not at the origin of its own frame. You say that A B and C will meet at one point, but then you set it up so they won't. I think this is way too complex and confusing.

I think you are mixing the way I solved the problem with the way I demonstrated the assumption dist__B(B,A) = dist__B(A,C) at the initial state to be correct. For this demonstration I used a frame centered at A for the Lorentz's transform to be applied in an easier way to show that an initial midpoint is preserved as a midpoint in the change of frames. In the problem of the travelers the frame of B is centered at the initial position of B. I think the same assumption is valid in both cases. Am I wrong in this?

 

23 minutes ago, Lorentz Jr said:

This is incorrect, martillo. You can't just use gamma for your calculations. You have to use the Lorentz transformations.

I have seen many times in the literature that for elapsed times, I mean for variations in time, it is valid:

Δt = γΔt'

All times considered in the problem are elapsed times so I think is right to be applied. Am I wrong in this? Good point to review...

Edited February 16, 2023 by martillo

[md65536]

18 minutes ago, martillo said:

I think you are mixing the way I solved the problem with the way I demonstrated the assumption dist__B(B,A) = dist__B(A,C) at the initial state to be correct. For this demonstration I used a frame centered at A for the Lorentz's transform to be applied in an easier way to show that an initial midpoint is preserved as a midpoint in the change of frames. In the problem of the travelers the frame of B is centered at the initial position of B. I think the same assumption is valid in both cases. Am I wrong in this?

Yes, I made a mistake but revised it. You didn't show that the distance from A "to  B" is the same as between A and C, in B's frame. You showed it for the distance from A "to the location of B at time t=0 as measured in A's frame", which is not the location of B in B's frame at time t'=0, because you have B not at the origin of B's frame.

I really think you've overcomplicated it.

Edited February 16, 2023 by md65536

[martillo]

1 minute ago, md65536 said:

I really think you've overcomplicated it.

Just separate the two things and focus just in the problem now. With the right assumptions is an easy problem to solve...

[Lorentz Jr]

25 minutes ago, martillo said:

I have seen many times in the literature that for elapsed times, I mean for variations in time, it is valid:

Δt = γΔt'

All times considered in the problem are elapsed times so I think is right to be applied. Am I wrong in this? Good point to review...

That applies when the initial and final events are the same for both intervals and the two events occur at the same location in the primed frame. In the current problem, B and C receive two different signals, so the simple gamma formula doesn't work. More time passes for B if you start counting from when C receives a signal, but the elapsed time for B starts when B receives a signal, not at the earlier time when C does.

[md65536]

20 minutes ago, martillo said:

Just separate the two things and focus just in the problem now. With the right assumptions is an easy problem to solve...

Easy for you, maybe... with your own assumptions.

But it's a good demonstration of the point of the thread's topic. You can apply SR and use its equations, then slip one little detail from one frame into to another inappropriately, and you come up with a different answer.

[martillo]

45 minutes ago, Lorentz Jr said:

That applies when the initial and final events are the same for both intervals and the two events occur at the same location in the primed frame. In the current problem, B and C receive two different signals, so the simple gamma formula doesn't work. More time passes for B if you start counting from when C receives a signal, but the elapsed time for B starts when B receives a signal, not at the earlier time when C does.

Well, I agree, I must review this in deep...

I realized about other two problems in my calculations:

1) If v = (3/5)c then γ = 5/4 and not 3/2 as I used in the calculations. That's a bad problem of mine. I use to make silly mistakes in numerical calculations. I think it does not alter the conclusion of less time passing for traveler C but is an error...

2) I used: Δt = γΔt' but what appear everywhere is the inverse: Δt' = γΔt . I don't understand. If for example γ = 5/4 and the stationary time is 10 using Δt' = γΔt will give Δt' = 50/4 what would mean the moving clock running faster because showing a bigger time. I used the inverse relation in the problem. What do you think about?

40 minutes ago, md65536 said:

Easy for you, maybe... with your own assumptions.

But it's a good demonstration of the point of the thread's topic. You can apply SR and use its equations, then slip one little detail from one frame into to another inappropriately, and you come up with a different answer.

Do you mean the assumption dist__B(B,A) = dist__B(A,C) at the initial state of the travelers' problem is not correct for you?

Edited February 16, 2023 by martillo

[martillo]

May be the second comment does not apply...

I apologize. I got confused now. I think I need some rest now...

Edited February 16, 2023 by martillo

[md65536]

5 hours ago, martillo said:

Do you mean the assumption dist__B(B,A) = dist__B(A,C) at the initial state of the travelers' problem is not correct for you?

Yes, it's not correct for anybody.

What I wrote earlier makes no sense, I'll try again...

You have frames A, B, and C coincident at time t=0. In A's frame, B and C are equidistant from A. Let's call events B_A and C_A the events on B's and C's respective world lines that are at time t=0 in A's frame. Then you transformed B_A and C_A into B's frame and found that they're also equidistant from the event "A at proper time 0" in B's frame because of the way you put B's origin at that event. So far so good.

In B's frame, at time t'=0 where you've set it up so that A is at the origin at that time, neither events B_A nor C_A have coordinate time t'=0. You've set it up so that they are away from the origin (x' != 0), so if you transform their time t into t' you should find that those events are not simultaneous in this frame. Therefore it's not the distances between the objects A, B, and C that you're comparing, but distances to events at different times.

I guess the basic idea is that the Lorentz transformation you used applies to events, not the objects.

(Note that the location of B is fixed in B's frame, so the distance to the origin is always the same and t' doesn't really matter for measuring the distance to B---I messed that up in an earlier reply---but the distance to C does depend on what time t' you measure it.)

Edited February 17, 2023 by md65536

[martillo]

10 hours ago, Lorentz Jr said:

This is incorrect, martillo. You can't just use gamma for your calculations. You have to use the Lorentz transformations.

You are right Lorentz Jr, this is the main error in my calculations. There´s a gamma factor of proportionality in the transform between frames but there's also the translation of the moving frame. That was my big error.

3 hours ago, md65536 said:

Yes, it's not correct for anybody.

And everybody is right. I apologize for my error.

On 2/14/2023 at 4:11 PM, Genady said:

Here is a problem:

If B and C are symmetrical relative to the signal in A, i.e, dist__A(B,A) = dist__A(A,C), then dist__B(B,A) ≠ dist__B(A,C).

If, on the other hand, dist__B(B,A) = dist__B(A,C), then they are not symmetrical in A and thus their clocks don't need to be the same when they meet.

I apologize Genady for not agreeing with you before. I couldn't see it. My error, my fault. I could see it applying the Lorentz's transform at the initial instant. It's all about relativity of simultaneity, something I didn't consider in the problem...

Edited February 17, 2023 by martillo

[Lorentz Jr]

1 hour ago, martillo said:

I apologize for my error.

Thank you, martillo. I appreciate your candor.

1 hour ago, martillo said:

It's all about relativity of simultaneity, something I didn't consider in the problem...

Right! Except guess what, martillo: Using your gamma formula is a simpler ending for the solution!

I was thinking about what you said, that the simple gamma formula works for events on the primed world line, and it turns out that finishing the solution in B's frame is simpler than that huge mess at the end of my first solution (in C's frame).

We need to start with the parameters in B's frame:

• t_i' = -2γv
• v' = 2v/(1 + v²)
• t_f' = 1/γv

Then calculate the elapsed time from C's signal, still in B's frame:

Δ' = t_f' - t_i' = 1/γv + 2γv

vΔ'/γ = 1 - v² + 2v² = 1 + v²

And finally, just divide that by Γ:

Δ = Δ'/Γ

(vΔ/γ)² = (1 - v'²)(1 + v²)² = (1 + v²)² - 4v² = (1 - v²)² = 1/γ⁴

Δ = 1/γv

MUCH easier! 😋

Edited February 17, 2023 by Lorentz Jr

[martillo]

47 minutes ago, Lorentz Jr said:

Thank you, martillo. I appreciate your candor.

Right! Except guess what, martillo: Using your gamma formula is a simpler ending for the solution!

I was thinking about what you said, that the simple gamma formula works for events on the primed world line, and it turns out that finishing the solution in B's frame is simpler than that huge mess at the end of my first solution.

We need to start with the parameters in B's frame:

• t_i' = -2γv
• v' = 2v/(1 + v²)
• t_f' = 1/γv

Then calculate the elapsed time from C's signal, still in B's frame:

Δ' = t_f' + 2γv

vΔ'/γ = 1 - v² + 2v² = 1 + v²

And finally, just divide that by Γ:

Δ = Δ'/Γ

(vΔ/γ)² = (1 - v'²)(1 + v²)² = (1 + v²)² - 4v² = (1 - v²)² = 1/γ⁴

Δ = 1/γv

Much easier! 😋

Glad to see that all the discussion left something positive not only for me but for you too. Thanks!

Edited February 17, 2023 by martillo

[Genady]

5 hours ago, martillo said:

I apologize Genady for not agreeing with you before. I couldn't see it. My error, my fault. I could see it applying the Lorentz's transform at the initial instant. It's all about relativity of simultaneity, something I didn't consider in the problem...

Sure, accepted. I am really glad that you have found it and thank you, @Lorentz Jr and @md65536.

[md65536]

23 hours ago, martillo said:

It's all about relativity of simultaneity, something I didn't consider in the problem...

Yes, it seems that was the only thing missing? Don't worry about errors, that's the point of the thread, where something appears paradoxical when presented in a way that each part of it makes sense, but some missing thing is hidden.

You were right---at least if we're talking about events only---that if (event) A is in the middle of (events) B and C in its own frame, it's in the middle in every frame. I was wrong, it doesn't matter where where the origin of the frame is. The Lorentz transformation is a linear transformation, so it preserves the 4D linearity of those 3 events (preserving the midpoint), but not distance between them or simultaneity.

[martillo]

14 hours ago, md65536 said:

Yes, it seems that was the only thing missing? Don't worry about errors, that's the point of the thread, where something appears paradoxical when presented in a way that each part of it makes sense, but some missing thing is hidden.

You were right---at least if we're talking about events only---that if (event) A is in the middle of (events) B and C in its own frame, it's in the middle in every frame. I was wrong, it doesn't matter where where the origin of the frame is. The Lorentz transformation is a linear transformation, so it preserves the 4D linearity of those 3 events (preserving the midpoint), but not distance between them or simultaneity.

I'm still confused with some things...

My reasoning without math:

If someone has a clock running slower he will measure less time between two events than other one with a clock running faster. Well, consider two ones travelling to each other in uniform motion (without any acceleration). If the first event is two travelers at some distance and the second event is when they cross each other then one clock measures less and the other more. I don't understand how that cannot be true.

But now imagine the two travelers shave their beards at the first event. When they cross each other they would present different beards because different time passed for each other. That is not real, they must present the same beard.

Where am I wrong in this?

I just tried to figure out the problem with synchronizing signals from the middle point to make the math with clocks but I only got to be more confused now. I would need time to rethink now. May be in some future in some new thread I could discuss this subject better, I don't know...

 

 

Edited February 18, 2023 by martillo

[Genady]

41 minutes ago, martillo said:

first event is two travelers at some distance

If they are at some distance, then it is not an event. It is two events.

[martillo]

54 minutes ago, Genady said:

If they are at some distance, then it is not an event. It is two events.

Well, this is what I must think in then...

As designing time-space diagrams seems to me is easy for you, can you post a space-time diagram for the perspective of the traveler at B? I mean for the "stationary" frame attached to B. That is my approach. I think it would be interesting for everyone.

Edited February 18, 2023 by martillo

[Genady]

18 minutes ago, martillo said:

Well, this is what I must think in then...

As designing time-space diagrams seems to me is easy for you, can you post a space-time diagram for the perspective of the traveler at B? I mean for the "stationary" frame attached to B. That is my approach. I think it would be interesting for everyone.

Yes. 5-10 minutes...

@martillo: