The twin Paradox revisited

[martillo]

21 minutes ago, Genady said:

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.

Let we try. I'm not very good with space-time diagrams but good to know your reasoning.

Thinking now that the travel is spatially symmetrical in a straight line with A at the midpoint. This is spatially symmetrical. May be there is something asymmetric in time due to that in the frame considered of B, C is moving while B not... If we switch to the frame in A the travel is perfectly symmetrical in everything.

 

Edited February 14, 2023 by martillo

[Genady]

8 minutes ago, martillo said:

Let we try. I'm not very good with space-time digrams but good to know your reasoning.

Here it is. Time axis goes up, position axis horizontal. I know we calculate for the B frame, but do you agree that this is how symmetrically travelling B and C would look in the frame of their meeting point?

Point A is where the signals go off. Point Z is the meeting point. It is in the same position as A, but some time later, thus higher on the diagram. B comes toward the Z from the left, C comes toward Z from the right.

They are always equally distanced from the meeting point. OK?

[martillo]

6 minutes ago, Genady said:

Here it is. Time axis goes up, position axis horizontal. I know we calculate for the B frame, but do you agree that this is how symmetrically travelling B and C would look in the frame of their meeting point?

Point A is where the signals go off. Point Z is the meeting point. It is in the same position as A, but some time later, thus higher on the diagram. B comes toward the Z from the left, C comes toward Z from the right.

They are always equally distanced from the meeting point. OK?

Right. You are considering frame A...

Edited February 14, 2023 by martillo

[Genady]

Just now, martillo said:

Right. You are considering frame A isn't it?

Yes, the drawing is how A would see the situation. But we do not calculate for A, just establish what symmetrical means. On the next step, we consider B. OK?

[martillo]

1 minute ago, Genady said:

Yes, the drawing is how A would see the situation. But we do not calculate for A, just establish what symmetrical means. On the next step, we consider B. OK?

Go on...

[Lorentz Jr]

3 hours ago, martillo said:

I calculate the lectures in the clocks

3 hours ago, Genady said:

Maybe I am confused because of the word "lectures"

"readings"

Edited February 14, 2023 by Lorentz Jr

[martillo]

3 minutes ago, Lorentz Jr said:

"readings"

Yes, right. The word didn't come to my mind... By the way, English is not my natural language...

Edited February 14, 2023 by martillo

[Genady]

4 minutes ago, martillo said:

Go on...

In the frame B, the moment when the signals go off in A is simultaneous with spacetime events P and Q. P is where in the spacetime B is at this moment. Q is where in the spacetime C is at this moment. Again, this moment in the B's time. Questions?

7 minutes ago, Lorentz Jr said:

"readings"

Thanks.

 

5 minutes ago, martillo said:

By the way, English is not my natural language

Neither it is mine.

[martillo]

4 minutes ago, Genady said:

In the frame B, the moment when the signals go off in A is simultaneous with spacetime events P and Q. P is where in the spacetime B is at this moment. Q is where in the spacetime C is at this moment. Again, this moment in the B's time. Questions?

I'm getting it for now, I think...

[Genady]

4 minutes ago, martillo said:

I'm getting it for now, I think...

At this moment (in B frame), PA is a distance from B to A (in B frame), and QA is a distance from C to A (in B frame). In your notation, PA = dist__B(B,A) and QA = dist__B(A,C).    

You see on the diagram that PA and QA are not, and cannot be, equal.  

[martillo]

2 minutes ago, Genady said:

At this moment (in B frame), PA is a distance from B to A (in B frame), and QA is a distance from C to A (in B frame). In your notation, PA = dist__B(B,A) and QA = dist__B(A,C).    

You see on the diagram that PA and QA are not, and cannot be, equal.  

I have some of questions on the second diagram.

1) For me the diagram is related to frame in A because the vertical line passes through A. Now you are talking about B frame but still over the same diagram. I think in B frame the vertical line passes through B. I mean the space-time line of B would be vertical.

2) I'm not sure if the points P and Q you mention are related to the instant of emission of the signals by A or to the reception of the signals by B and C. Seems they are the points at the reception of the signals which I know that in the B frame do not happen at the same time.

3) I forgot the third...

The third issue: What I know is that A, B and C are all travelling in a straight line with A as a midpoint. This makes dist(B,A) = dist(A,C) always at least considering spatial distances. This mean a symmetrical travel in space...

[Genady]

6 minutes ago, martillo said:

I have some of questions on the second diagram.

1) For me the diagram is related to frame in A because the vertical line passes through A. Now you are talking about B frame but still over the same diagram. I think in B frame the vertical line passes through B. I mean the space-time line of B would be vertical.

2) I'm not sure if the points P and Q you mention are related to the instant of emission of the signals by A or to the reception of the signals by B and C. Seems they are the points at the reception of the signals which I know that in the B frame do not happen at the same time.

3) I forgot the third...

1) This is why the spacetime diagrams are so useful. You can draw it in one frame, which will appear orthogonal, and draw and analyze on the same diagram any other frame, which will appear tilted relative to the first. This allows for comparison between frames.

2) They are points simultaneous with the emission of the signals in point A, in frame B. There is a straightforward rule, how to find simultaneous points for any frame on the diagram. The points P, Q, and A are simultaneous in the frame B, and A is the emission. There is nothing about the reception, yet. Do you want me to show the reception points?

[martillo]

2 minutes ago, Genady said:

Do you want me to show the reception points?

Yes, I'm curious on the subject... 😄 

[Genady]

16 minutes ago, martillo said:

Yes, I'm curious on the subject... 😄 

Here it is. R is where C receives the signal, and S is where B receives the signal:

[martillo]

Well, let me say that my little expertise in space-time diagrams doesn't let me analyze them appropriately. I know they are two dimensional diagrams with the spatial axis and the timing axis and sometimes they can be worked separately as the projections of the  events involved. My calculations are over the spatial coordinates only and seemed to be enough to find a solution to the problem, right or not, it shows it can be worked that way. 

As I already mentioned in a post I'm sure, by the specifications, that A, B and C  are all travelling in a straight line with A as a midpoint always. This makes dist(B,A) = dist(A,C) always considering spatial distances. This mean a symmetrical travel in space. I don't understand how can there be a problem with that.

I will continue reviewing the calculations thinking in what you said but I'm not sure to be able to work in the space-time diagrams appropriately as I said above.

[Genady]

Well, thank you for letting me know. I'll take a note and will not waste my time anymore.

[Lorentz Jr]

1 hour ago, martillo said:

My calculations are over the spatial coordinates only and seemed to be enough to find a solution to the problem, right or not, it shows it can be worked that way.

Your calculations are incorrect. No relativity problem can be solved without taking the relativity of time into account.

1 hour ago, martillo said:

This mean a symmetrical travel in space. I don't understand how can there be a problem with that.

It's a problem because it's only true in A's reference frame. You were analyzing the problem in B's reference frame, where it's not true.

Edited February 14, 2023 by Lorentz Jr

[martillo]

20 minutes ago, Lorentz Jr said:

Your calculations are incorrect. No relativity problem can be solved without taking the relativity of time into account.

May be I didn't exprese the things properly. The factor γ appears in the calculations related to the time dilation effect. The relativity of time is considered in the calculations.

[Lorentz Jr]

26 minutes ago, martillo said:

The factor γ appears in the calculations

That's not enough. You must use the full Lorentz transformations. They're what Genady's diagrams represent.

Edited February 15, 2023 by Lorentz Jr

[martillo]

36 minutes ago, Genady said:

Well, thank you for letting me know. I'll take a note and will not waste my time anymore.

I wasn't a waste of time for me. I had a strong work based in your posts about the key points to consider in the problem which inspired me in a possible approach to the problem. I apologize to not be able to work with space-time diagrams appropriately. Unfortunately I arrived to something that doesn't match with some relativistic prediction and that is a big problem I know. You pointed out a possible error on which I disagree or just don't understand. I will continue reviewing and looking for my probable errors.

[md65536]

On 2/14/2023 at 4:59 PM, martillo said:

Unfortunately I arrived to something that doesn't match with some relativistic prediction and that is a big problem I know. You pointed out a possible error on which I disagree or just don't understand. I will continue reviewing and looking for my probable errors.

Looking over the thread, my opinion is that it is far too complicated, with too many aspects you haven't figured out yet, but which you would need to describe all the things you're trying to (eg. relativity of time, relativity of distance).

Instead of looking at the things that don't make sense to you, why not start with at least one thing you do understand, and build off of that? I suggest the following:

1) Start with a simpler scenario. Use just 2 observers at first. The basic twin paradox is good because it is concrete; the twins start and end at the same place and can see each other's immediate age without worrying about synchronizing with a distant clock.

Or if you prefer it, both observers moving inertially and setting a start event using a light signal from one to the other, is also simple enough. You could choose, but stick to one. You can add more to it later, to figure out other things.

2) Understand it first from at least ONE inertial frame of reference only, to start. Don't start trying to figure what a second observer measures, if you can't made sense of what the first observer does.

3) Write out the equations as you go. I don't see how you could understand how the numbers add up without calculating them.

 

I bet that SR could be figured out from scratch by amateurs using just the 2 postulates, some geometry, and the timing of light signals, if you're interested... But you really have to understand more of the basics to figure out what you're trying to in this thread.

[Markus Hanke]

3 hours ago, md65536 said:

Start with a simpler scenario

Yes, precisely +1. The basic twin scenario is simple, I don’t understand why people feel the need to add so many extra complications to it that do nothing to illuminate the underlying physics. This seems to be a problem with SR in general.

[martillo]

4 hours ago, md65536 said:

Looking over the thread, my opinion is that it is far too complicated, with too many aspects you haven't figured out yet, but which you would need to describe all the things you're trying to (eg. relativity of time, relativity of distance).

Instead of looking at the things that don't make sense to you, why not start with at least one thing you do understand, and build off of that? I suggest the following:

1) Start with a simpler scenario. Use just 2 observers at first. The basic twin paradox is good because it is concrete; the twins start and end at the same place and can see each other's immediate age without worrying about synchronizing with a distant clock.

Or if you prefer it, both observers moving inertially and setting a start event using a light signal from one to the other, is also simple enough. You could choose, but stick to one. You can add more to it later, to figure out other things.

2) Understand it first from at least ONE inertial frame of reference only, to start. Don't start trying to figure what a second observer measures, if you can't made sense of what the first observer does.

3) Write out the equations as you go. I don't see how you could understand how the numbers add up without calculating them.

 

I bet that SR could be figured out from scratch by amateurs using just the 2 postulates, some geometry, and the timing of light signals, if you're interested... But you really have to understand more of the basics to figure out what you're trying to in this thread.

I find very appropriated your final comment: "I bet that SR could be figured out from scratch by amateurs using just the 2 postulates, some geometry, and the timing of light signals, if you're interested.." This is the way my calculations were made.

It is considered the simple scenario of just two travelers B and C travelling in opposite directions to meet at the midpoint A. A is just the midpoint where the synchronizing signals are emitted at an initial time and is the final crossing point of them where A, B and C reunite. This is the specification of the problem to analyze. In my calculations I assumed the stationary frame of observation in B, the travelling one in C and I gave particular values to velocities and initial distances. I posted the calculations to discuss them if they were right or wrong and discuss posible errors.

Genady pointed out that a possible problem in my calculations would be the initial assumption of A be a midpoint in frame  B:

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

He said that would not be right and I disagreed.

Let us consider now the Lorentz transform to see how the midpoint is preserved in the different frames:

x' = γ(x - vt)

t' = γ(t - vx/c²)

Consider frame A with coordinates x, t and frame B with the coordinates x', t' in such a way that initially their origins coincide for the equations apply correctly.

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...

Edited February 16, 2023 by martillo

[Lorentz Jr]

49 minutes ago, martillo said:

x' = γ(x - vt)

t' = γ(t - vx/c²)

Good job! This is the right way to start. 🙂

49 minutes ago, martillo said:

Consider frame A with coordinates x, t and frame B with the coordinates x', t' in such a way that initially their origins coincide for the equations apply correctly.

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.

What do you mean by "initially", martillo? A's and B's clocks can't be synchronized if A is at x=0 and B is at x=-L.

Edited February 16, 2023 by Lorentz Jr

[martillo]

8 minutes ago, Lorentz Jr said:

What do you mean by "initially", martillo? A's and B's clocks can't be synchronized if A is at x=0 and B is at x=-L.

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

Edited February 16, 2023 by martillo