The twin Paradox revisited

[Genady]

1 hour ago, martillo said:

this would imply to enter in the area of General Relativity

Reference to the gravitational time dilation is helpful for a qualitative understanding of what is going on there, but to calculate the effect of the acceleration you don't need GR. It all is computed using SR, although the math is somewhat involved. Here you can find the calculations, and the author also applies them in the general case of two twins with various spacetime trajectories, of which two scenarios considered above - the twins A and B, and the twins B and C - are special cases:

acceleration.pdf (uconn.edu) (The Relation Between Acceleration and Time Dilation in Special Relativity, James G. Bridgeman, December 8, 2018)

Edited February 12, 2023 by Genady

[martillo]

I'm working now in the solution to the case of the twins travelling in opposite directions without accelerations involved but with just a signal emitted from the middle point for clocks synchronization. I'm trying to make the calculations following the approach given by Genady (no Minkowsky diagrams) with some particular values for the velocity of the twins and the distance between them but I'm not getting the expected result. The ages does not match at the end. Is there any detailed calculation for this case available on the web that could help? I made my searches but couldn't find it. 

On 2/11/2023 at 8:57 PM, Genady said:

If a synchronizing signal is sent from a point in the middle in the two directions, then, in the B frame, it will reach C before it reaches B, because C moves toward B in twice the speed with which the middle point moves toward B. Thus, by the time the signal reaches B, the C clock will be already on 5. Then by the time they meet, the B clock advances 10, while the C clock advances 5, which makes them equal, 10.

 

On 2/11/2023 at 10:04 PM, 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.

 

Edited February 14, 2023 by martillo

[Genady]

3 minutes ago, martillo said:

twins travelling in opposite directions without accelerations involved but with just a signal emitted from the middle point for clocks synchronization

I don't think there is a way to consistently synchronize these frames when they are in relative motion and at a distance from each other. You can see it on my spacetime diagram above. Except the starting and the crossing points, there are no points on the B's and C's worldlines, which are simultaneous in B frame and in which the twins are the same age.

Specifically, when A sends the signals, the twins are of the same age in A frame, but not of the same age in B frame.

If you want, you can show your calculations, and I'll try to point where there is a wrong assumption.

[martillo]

3 minutes ago, Genady said:

If you want, you can show your calculations, and I'll try to point where there is a wrong assumption.

Good. I'm currently reviewing the calculations and I'm preparing something to post but I could late some time.

[Genady]

1 minute ago, martillo said:

Good. I'm currently reviewing the calculations and I'm preparing something to post but I could late some time.

Take your time. I take mine, too.  

What is your time zone?

[martillo]

28 minutes ago, Genady said:

Specifically, when A sends the signals, the twins are of the same age in A frame, but not of the same age in B frame.

I'm not considering their "absolute" age but how they age after they receive the synchronizing signals from the midpoint. For instance I'm considering the twins could shave at the time of receiving the signals and compare their beards at the final crossing midpoint.

20 minutes ago, Genady said:

What is your time zone?

Uruguay, UTC+3 If I'm not wrong. 21:42 hs right now. Night coming I know...

Edited February 14, 2023 by martillo

[Genady]

1 minute ago, martillo said:

For instance I'm considering the twins could shave at the time of receiving the signals and compare their beards at the final crossing midpoint.

I don't see why their beards would be of equal 'age' at the crossing point. I expect them to be different.

4 minutes ago, martillo said:

Uruguay, UTF+3 If I'm not wrong. 21:42 hs right now. Night coming I know...

One hour ahead of me.

[martillo]

1 minute ago, Genady said:

I don't see why their beards would be of equal 'age' at the crossing point. I expect them to be different.

But isn't expected in the relativistic prediction that the same time must pass (clocks' timing) in both twins after the receiving signal and reaching the midpoint? I mean, after they receive the signals they should age the same and have same beard. You said they present the same timestamp...

[Genady]

11 minutes ago, martillo said:

But isn't expected in the relativistic prediction that the same time must pass (clocks' timing) in both twins after the receiving signal and reaching the midpoint? I mean, after they receive the signals they should age the same and have same beard. You said they present the same timestamp...

I see. I've replied too fast, sorry. Send your calculations whenever they are ready.

[Lorentz Jr]

56 minutes ago, martillo said:

Uruguay

GMT-3

[Genady]

2 hours ago, martillo said:

I'm considering the twins could shave at the time of receiving the signals and compare their beards at the final crossing midpoint.

I've added these two events to the diagram, see below. Yellow lines are two signals. C receives the signal at event P. B receives his signal at event Q. You might refer to them in your calculations.

[martillo]

10 hours ago, Lorentz Jr said:

GMT-3

Right. I'm retired now and so anytime could e good for Physics now . I just follow my intuition. Sometimes I prefer the late advanced night when the city is very quiet and I can concentrate better...

9 hours ago, Genady said:

I've added these two events to the diagram, see below. Yellow lines are two signals. C receives the signal at event P. B receives his signal at event Q. You might refer to them in your calculations.

Thanks for the diagram. In my calculations I consider just one frame, the frame of B as the stationary and C travelling while comparing the clocks of B and C as observed by B. If the final lecture on both clocks were the same then there would be no problem, isn't it? The same time would have passed for both travelers since the synchronization. But as for now is not the case. I don't get what makes me reach different results with C aging less. I do not consider frame in A, I assume A just an emitter of the synchronization signals. The case of a stationary frame in C is just totally symmetric to the case in B. I worked hard in the night, making some final verifications now to type the calculations later.

Edited February 14, 2023 by martillo

[Genady]

2 hours ago, martillo said:

I do not consider frame in A, I assume A just an emitter of the synchronization signals.

That is right. But the situation is described in A frame, specifically, B and C each move with speed v toward A, in the frame A. When you go to the frame B, what is the speed of C toward B? It is not 2v.

[martillo]

2 hours ago, Genady said:

That is right. But the situation is described in A frame, specifically, B and C each move with speed v toward A, in the frame A. When you go to the frame B, what is the speed of C toward B? It is not 2v.

I just consider C travelling at some velocity towards A as A travels to B as seen by a stationary B and calculate what happens. I calculate the lectures in the clocks of B and C considering time = distance/velocity. Nothing else. A frame in C is a totally symmetric case. Frame in A is not considered at all. Just an emitter of synchronizing signals at A and the end point. The point of interest is how the travelers observe the things in the relativistic prediction. How the stationary one observe the things compared with how the relativistic prediction dictates the travelling one would observe. The stationary and the travelling clocks are compared. Just that. You are right I must take care about the velocities. I have found an error in that. I will late some time in the correction...

Edited February 14, 2023 by martillo

[Genady]

3 minutes ago, martillo said:

I just consider C travelling at some velocity towards a stationary B and calculate what happens. I calculate the lectures in the clocks of B and C. Nothing else. A frame in C is a totally symmetric case. Frame in A is not considered at all. Just an emitter of synchronizing signals at A. The point of interest is how the travelers observe the things in the relativistic prediction. How the stationary one observe the things compared with how the relativistic prediction dictates the travelling one would observe. The stationary and the travelling clocks are compared. Just that.

Now I don't understand what you expect in the result. If A does not have any special properties, then the signal might go from B as well, i.e. A might be B itself. In this case, the signal goes off from B and B starts his clock, the signal gets received by C and C starts his clock, and sometime later they meet and compare their clocks? What is your prediction?

Maybe I am confused because of the word "lectures" above (bold)? What do you mean?

[martillo]

16 minutes ago, Genady said:

Now I don't understand what you expect in the result. If A does not have any special properties, then the signal might go from B as well, i.e. A might be B itself. In this case, the signal goes off from B and B starts his clock, the signal gets received by C and C starts his clock, and sometime later they meet and compare their clocks? What is your prediction?

Maybe I am confused because of the word "lectures" above (bold)? What do you mean?

I was editing while you posting. The moving A is considered as the emitter of the synchronizing signals and the final crossing point of the travelers where the three travelers A, B and C coincide in position. to say C travels to A or to B is the same since they will all coincide at the end. B does not emit the signals. The intention is to consider exactly the same posted case but just as observed by B only.

I meant by lectures in the clocks as the time (timestamp?) the clocks show.

 

Edited February 14, 2023 by martillo

[Genady]

I've noticed the edit:

28 minutes ago, martillo said:

C travelling at some velocity towards A as A travels to B as seen by a stationary B

This confuses me even more regarding what you are calculating and what you expect in result. 

Velocity of C towards A relative to A?

What about velocity of A towards B?

I guess I will understand all this when I see your calculations.

[martillo]

7 minutes ago, Genady said:

I've noticed the edit:

This confuses me even more regarding what you are calculating and what you expect in result. 

Velocity of C towards A relative to A?

What about velocity of A towards B?

I guess I will understand all this when I see your calculations.

Both doesn't matter for the calculations. They just don't appear. As I said (may be I was editing) the three A, B and C reunite in the same position at the end so, tu calculate the time of C travelling to A is the same as to calculate his travel to B. The travel of A doesn't matter, what matters is the relative travel between B and C.

[Genady]

4 minutes ago, martillo said:

Both doesn't matter for the calculations. They just don't appear. As I said (may be I was editing) the three A, B and C reunite in the same position at the end so, tu calculate the time of C travelling to A is the same as to calculate his travel to B. The travel of A doesn't matter, what matters is the relative travel between B and C.

OK. The end point of both clocks is when they meet. What is a starting point of each?

[martillo]

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.

 

 

 

32 minutes ago, Genady said:

OK. The end point of both clocks is when they meet. What is a starting point of each?

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.

Edited February 14, 2023 by martillo

[Genady]

Here is a problem:

1 hour ago, martillo said:

dist__B(B,A) = dist__B(A,C)

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.

[martillo]

15 minutes ago, 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.

That counts in the initial state only. Just in the initial instant (when A emits the signals) dist__B(B,A) = dist__B(A,C). Not after of course. My fault is to have not commented that or to have not made it visible in the notation. The title of the paragraph says: INITIAL CONSIDERATIONS. I thought that was enough. Or is that this is not the point?

Edited February 14, 2023 by martillo

[Genady]

Just now, martillo said:

That counts in the initial state only. Just in the initial instant (when A emits the signals) dist__B(B,A) = dist__B(A,C). Not after of course. My fault is to have not commented that or to have not made it visible in the notation.

No, no, I understand that. I am talking about the initial state, here:

 

1 hour ago, martillo said:

dist__B(B,A) = dist__B(A,C) = 3x10⁸ mts

If these distances are equal in frame B in the initial state, then B and C are not travelling symmetrically between the source of the signal and their meeting point.

If you want, I can show this on a symmetrical spacetime diagram.

I think it is not possible to work with relativistic situations using a space diagram only. You need to show it on a spacetime diagram.

[martillo]

12 minutes ago, Genady said:

No, no, I understand that. I am talking about the initial state, here:

 

If these distances are equal in frame B in the initial state, then B and C are not travelling symmetrically between the source of the signal and their meeting point.

If you want, I can show this on a symmetrical spacetime diagram.

I think it is not possible to work with relativistic situations using a space diagram only. You need to show it on a spacetime diagram.

Thinking about... Seems to me now that the equality is valid always in the entire travel. I mean A moves such a way is always the midpoint between B and C in the entire travel. Why do you say this is not a symmetrical travel? I don't get it.

Edited February 14, 2023 by martillo

[Genady]

2 minutes ago, martillo said:

A moves such a way is always the midpoint between B and C in the entire travel

It is not so in the B frame. I can show it to you on the spacetime diagram. Step-by-step, if you want. Should I show step 1? If yes, it will take me a few minutes to draw. If we agree on step 1, I'll draw step 2.