The twin Paradox revisited

[Lorentz Jr]

5 minutes ago, martillo said:

As md65536 posted, Lorentz transform are linear transforms preserving midpoints.

Except it's not that simple, martillo. They also transform t, so "midpoint" preservation only applies to equal values of t (simultaneous in A's frame), not equal values of t' (simultaneous in B's frame).

Edited February 19, 2023 by Lorentz Jr

[martillo]

17 minutes ago, Lorentz Jr said:

 

I think I'm getting it...

Only the values doesn't match well: 1/γv << 2γv. If I look at the projections of the final point and the point C into the time axis they don't match well with those numerical values.

Edited February 19, 2023 by martillo

[Lorentz Jr]

9 minutes ago, martillo said:

Only the values doesn't match well: 1/γv << 2γv.

For v = 0.866, γ = 2. So 1/γv = 0.577 and 2γv = 3.46.

You're right martillo, the diagram is quantitatively inaccurate.

Edited February 19, 2023 by Lorentz Jr

[martillo]

14 minutes ago, Lorentz Jr said:

You're right martillo, the diagram is quantitatively inaccurate.

It's fine Lorentz Jr. I got a much better understanding of what is going on. I will try some calculations based on the diagram but I would late too much. May be in some future I could post a thread with the results to reevaluate things if necessary.

I appreciate very much all this discussion and I hope it could have resulted positive for all of you too.

Edited February 19, 2023 by martillo

[Lorentz Jr]

42 minutes ago, Lorentz Jr said:

For v = 0.866, γ = 2. So 1/γv = 0.577 and 2γv = 3.46.

The ratio of 1/γv to 2γv in the diagram is about 6, so we have

1/γv = 6 * 2γv

1 + v² = 12 v²

v² = 1/11, so v = 0.30 and γ = 1.05.

Point A should be a little bit higher, so the signal paths are closer to 45 degrees (or at least so the slope of the signal lines is 0.3 times the slope of the line from C to the meeting point).

Edited February 19, 2023 by Lorentz Jr

[martillo]

I see a visual problem in the diagram. The projections of the points onto the vertical axis shows the times of their events, right? Well, projecting point C to the vertical axis I see that the elapsed time to the final point will be bigger than the B point to the final point. This would mean the reading of the clock C bigger than that of the clock B (clock C > clock B) while it was supposed to be the inverse. Time would pass slower for C so his clock should show a smaller value (clock C < clock B). Seems this would happen whatever the values of v and distances would be considered.

Am I wrong in something?

1 hour ago, Lorentz Jr said:

 

 

Edited February 19, 2023 by martillo

[Lorentz Jr]

5 minutes ago, martillo said:

projecting point C to the vertical axis I see that the elapsed time to the final point will be bigger than the B point to the final point. This would mean the reading of the clock C bigger than that of the clock B

The graph shows t'. That's the reading on B's clock, not on C's clock.

5 minutes ago, martillo said:

Time would pass slower for C so his clock should show a smaller value (clock C < clock B).

The problem is symmetric in A's reference frame, so the elapsed times for B and C are equal.

Time passes more slowly on C's clock, and that makes up for the longer elapsed time from C's signal on B's clock.

Edited February 19, 2023 by Lorentz Jr

[martillo]

12 minutes ago, Lorentz Jr said:

t' is the reading on B's clock, not on C's clock.

The problem is symmetric in A's reference frame, so the elapsed times for B and C are equal.

Time passes more slowly on C's clock, and that makes up for the longer elapsed time from C's signal on B's clock.

Sorry, I'm a bit tired and got confused again with an old bad concept. I need to have a rest now... 

Edited February 19, 2023 by martillo

[Genady]

6 hours ago, martillo said:

Sorry, I'm a bit tired and got confused again with an old bad concept. I need to have a rest now... 

While you're resting, may I recommend these free and fun online presentations, which give good explanation for and practice in spacetime diagrams and Lorentz transformations (with Spanish subtitles available):

Understanding Einstein: The Special Theory of Relativity | Coursera

 

[martillo]

40 minutes ago, Genady said:

While you're resting, may I recommend these free and fun online presentations, which give good explanation for and practice in spacetime diagrams and Lorentz transformations (with Spanish subtitles available):

Understanding Einstein: The Special Theory of Relativity | Coursera

 

Thanks for the link, I will take a look.

I think I'm understanding well enough the space-time diagrams now. Just one thing remains for me to understand in the twins paradoxes and is what you pointed out: dist(B,A) ≠ dist(A,C). I think is something I must study on my own now considering both, Lorentz transform and space-time diagrams to get the key points properly. I accepted the thing before but some postings later make me doubt and so I must study the subject deeper. I will do it but I will late some time of course.

As for now here my diagram of the problem we were analyzing (Point A looks at the midpoint but it is not, I know now, it just should be near I think):

The dashed black line at the bottom represent the initial time of the emission of the signals from A to B and C in frame B. Points D and E are the locations of traveler C (on line red) when he receives the signal and when traveler B (on line blue) receives the signal respectively. Any comment about is welcome.

Designing the diagram made me understand some things better... 

Edited February 19, 2023 by martillo

[Genady]

6 minutes ago, martillo said:

Any comment about is welcome.

The red line should not be parallel to the yellow line of the signal. The yellow lines have to be at 45⁰.

[Lorentz Jr]

10 minutes ago, Genady said:

The yellow lines have to be at 45⁰.

By convention. For quantitative accuracy, the important thing is that the ratio of the slopes is equal to the reciprocal of the ratio of the speeds. r_slopes * r_speeds = 1.

[martillo]

19 minutes ago, Genady said:

The red line should not be parallel to the yellow line of the signal. The yellow lines have to be at 45⁰.

 

6 minutes ago, Lorentz Jr said:

By convention. For quantitative accuracy, the important thing is that the ratio of the slopes is equal to the reciprocal of the ratio of the speeds. r_slopes * r_speeds = 1.

Ok, here the diagram now:

Thanks for the correction. I think I got it! It's just about the slopes of the lines of the signals... Thanks very much...

Edited February 19, 2023 by martillo

[Genady]

13 minutes ago, Lorentz Jr said:

By convention. For quantitative accuracy, the important thing is that the ratio of the slopes is equal to the reciprocal of the ratio of the speeds. r_slopes * r_speeds = 1.

x=ct

ct/x = 1

equivalently, c=1, v in units of c

Edited February 19, 2023 by Genady
fixed the expressions

[martillo]

26 minutes ago, Genady said:

t=cv

t/cv = 1

equivalently, c=1, v in units of c

Got it. Just one more question: would the slope in my diagram be to big? I mean the velocity of the traveler C must never be higher than light velocity: v < c. Seems to be right, the time axis is on the vertical and so Δx/Δt < 1 implies Δt/Δx > 1 for v in units of c.

Edited February 19, 2023 by martillo

[Genady]

11 minutes ago, martillo said:

the time axis is on the vertical and so Δx/Δt < 1 implies Δt/Δx > 1.

Yes, it's correct.

Notice the corrected expressions in my post. Sorry for that.

[martillo]

1 hour ago, Genady said:

The red line should not be parallel to the yellow line of the signal. The yellow lines have to be at 45⁰.

For v in units of c. Got it! 😄 

The final question: can the elapsed times in each frame be obtained geometrically with the length on the lines of the travelers? I mean do the lengths on the lines represent elapsed times en each own frame?

If so, the triangle B-D-F must be isosceles with base B-D and height to F which would mean equal times for B and C in their own frames.

Or not?

I looked for that on google and found that if the frame is inertial then the length between two points on the lines represent the elapsed time between the events in the frame.

Link: https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_9HB__Special_Relativity_and_Thermal_Statistical_Physics/2%3A_Kinematics_and_Dynamics/2.1%3A_Spacetime_Diagrams

The above is right then because the travelers are moving at constant velocity v, not experimenting accelerations and so their frames would be inertial.

Edited February 19, 2023 by martillo

[Genady]

33 minutes ago, martillo said:

can the elapsed times in each frame be obtained geometrically with the length on the lines of the travelers?

Generally, not. The metric of the diagram is not Euclidean. It has Minkowski metric. IOW, the "length" squared between two events on the diagram is not dt²+dx². It is dt²-dx². For example, the "length", called "interval", between any two events on a light line (45⁰ line) is 0. IOW, the units on the lines belonging to different frames are not equal.

[martillo]

4 minutes ago, Genady said:

Generally, not. The metric of the diagram is not Euclidean. It has Minkowski metric. IOW, the "length" squared between two events on the diagram is not dt²+dx². It is dt²-dx². For example, the "length", called "interval", between any two events on a light line (45⁰ line) is 0. IOW, the units on the lines belonging to different frames are not equal.

I edited the post above:

I looked for that on google and found that if the frame is inertial then the length between two points on the lines represent the elapsed time between the events in the frame.

Link: https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_9HB__Special_Relativity_and_Thermal_Statistical_Physics/2%3A_Kinematics_and_Dynamics/2.1%3A_Spacetime_Diagrams

The above is right then because the travelers are moving at constant velocity v, not experimenting accelerations and so their frames would be inertial.

Edited February 19, 2023 by martillo

[Genady]

3 minutes ago, martillo said:

I edited the post above:

I looked for that on google and found that if the frame is inertial then the length between two points on the lines represent the elapsed time between the events in the frame.

Link: https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_9HB__Special_Relativity_and_Thermal_Statistical_Physics/2%3A_Kinematics_and_Dynamics/2.1%3A_Spacetime_Diagrams

The above is right then because the travelers are moving at constant velocity v, not experimenting accelerations and so their frames would be inertial.

Yes. The interval between two events on B's timeline is the B's elapsed time between these two events. Likewise, the interval between two events on C's timeline is the C's elapsed time between these two events.

[martillo]

1 minute ago, Genady said:

Yes. The interval between two events on B's timeline is the B's elapsed time between these two events. Likewise, the interval between two events on C's timeline is the C's elapsed time between these two events.

The good thing of this is that calculations can be made with the space-time diagrams just with geometrical considerations. I think that was intuitively the approach in my calculations although making some things wrongly of course. I will try to remake them some day and if I find something to discuss may be I post a new thread about.

[Genady]

2 minutes ago, martillo said:

The good thing of this is that calculations can be made with the space-time diagrams just with geometrical considerations. I think that was intuitively the approach in my calculations although making some things wrongly of course. I will try to remake them some day and if I find something to discuss may be I post a new thread about.

Please, do. I'd prefer a fresh thread for this.

[martillo]

3 minutes ago, Genady said:

Please, do. I'd prefer a fresh thread for this.

Yes, it would be much better.

[Boltzmannbrain]

3 hours ago, martillo said:

 

Ok, here the diagram now:

Thanks for the correction. I think I got it! It's just about the slopes of the lines of the signals... Thanks very much...

I read that you did not want the distance from B to A to equal the distance from A to C, but this diagrams shows exactly that.  If you want to understand Spacetime diagrams, you should get familiar with the Minkowski metric, if you aren't already.  This is the metric you have to use with spacetime diagrams. 

It is just like a 4 dimension Euclidean metric ds² = db² + dx² + dy² + dz², but you will see that the time vector is negative in this Minkowski metric which is ds² = -dt² + dx² + dy² + dz².   The negative sign only matters at angles more than 0 degrees from the point that you want to know the distance to because -dt² = 0.  Then in the case of spacetime diagrams, you just have ds² = -dt² + dx².  And in the case from B to A or from B to C it is just ds² = 0 + dx²

So, as you see in your graph, from B to A and from B to C, you would use just the normal Euclidean/Pythagorean metric.  

Knowing this will save you a lot of pain and frustration, especially as angles are more than 45 degrees.  For angles more than 45 degrees, you have to totally through out all intuitive notions of distance and angles.  For example, the angle that B and C make with F is not Tan(opposite/adjacent) like we were taught in school.  And the distance from C to F is not found using Pythagoras a² = b² + c², also like we were taught in school.  

Edited February 19, 2023 by Boltzmannbrain

[martillo]

7 minutes ago, Boltzmannbrain said:

I read that you did not want the distance from B to A to equal the distance from A to C, but this diagrams shows exactly that. 

Yes, the diagrams, Genady and Lorentz Jr have shown me that. Is not that I don't want it to be, I just want the real thing. That is why I considered for instance beards in the twins as a way to show that the times showed by the clocks is not so relative as some ones use to think. They must match and this goes against the intuitive notion of the other twin aging less. I know Physics does not need to be intuitive and can give counterintuitive results, I just wanted to be sure about this subject. Is something I need to process on my own way now, no other way. Someone telling me that it is just that way is not enough. I need to refute my own intuition with some proper arguments...

18 minutes ago, Boltzmannbrain said:

If you want to understand Spacetime diagrams, you should get familiar with the Minkowski metric, if you aren't already.  This is the metric you have to use with spacetime diagrams. 

It is just like a 4 dimension Euclidean metric ds² = dt + dx² + dy² + dz², but you will see that the time vector is negative in this Minkowski metric which is ds² = -dt² + dx² + dy² + dz².   The negative sign only matters at angles more than 0 degrees from the point that you want to know the distance to because -dt² = 0.  Then in the case of spacetime diagrams, you just have ds² = -dt² + dx².  And in the case from B to A or from B to C it is just ds² = 0 + dx²

So, as you see in your graph, from B to A and from B to C, you would use just the normal Euclidean/Pythagorean metric.  

Knowing this will save you a lot of pain and frustration, especially as angles are more than 45 degrees.  For angles more than 45 degrees, you have to totally through out all intuitive notions of distance and angles.  For example, the angle that B and C make with F is not Tan(opposite/adjacent) like we were taught in school.  And the distance from C to F is not found using Pythagoras a² = b² + c², also like we were taught in school.  

 I appreciate your meaningful advices. They prevent me to make several mistakes. The problem is that I'm working in some other very important subject for me and it needs attention too. Some things take time, you know. I think could be another important subject to discuss here in this forum where I have already found the very right answers other times. Important subject and not only for me, I think. May be I post something soon, I don't know...