The twin Paradox revisited

[martillo]

40 minutes ago, Genady said:

Yes. 5-10 minutes...

@martillo:

Thanks very much. I expected that but wasn't sure. In the graphic the travelers B and C are initially at the x axis isn't it? I will think about the timings now... Taking into account the Lorentz transforms...

[Genady]

5 minutes ago, martillo said:

the travelers B and C are initially at the x axis isn't it?

I don't know what you mean, initially. It just shows that at some moment in B frame, they are on the x axis.

[Lorentz Jr]

3 hours ago, martillo said:

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.

As @Genady said, there are two signals, so those are two separate events. They're not simultaneous in B's reference frame.

B receives its signal at t'=0, but C already received its signal at t' = -2γv. So, in B's reference frame, C's clock is slower but there's more time between C's signal and the meeting. Those two effects exactly cancel each other out.

3 hours ago, martillo said:

I'm still confused with some things...

That's okay, martillo. Everyone is confused about relativity. It doesn't make sense in the simple way that old-fashioned physics does.

20 minutes ago, Genady said:

at some moment in B frame, they are on the x axis.

Can you show the events? C's signal received at -2γv, B's signal received at 0, and the crossing at 1/γv?

Edited February 18, 2023 by Lorentz Jr

[martillo]

13 minutes ago, Lorentz Jr said:

As @Genady said, there are two signals, so those are two separate events. They're not simultaneous in B's reference frame.

B receives its signal at t'=0, but C already received its signal at t' = -2γv. So, in B's reference frame, C's clock is slower but there's more time between C's signal and the meeting. Those two effects exactly cancel each other out.

The problem in the calculations I made is that they do not cancel. Thinking about that now...

53 minutes ago, Genady said:

I don't know what you mean, initially. It just shows that at some moment in B frame, they are on the x axis.

I'm assuming t = 0 the time of the emission of the synchronizing signals from the midpoint to B and C. I think this matches with your diagram with B and C at the x axis at time t = 0.

Edited February 18, 2023 by martillo

[Lorentz Jr]

36 minutes ago, martillo said:

I'm assuming t = 0 the time of the emission of the synchronizing signals from the midpoint to B and C. I think this matches with your diagram with B and C at the x axis at time t = 0.

But you said B and C change their clocks to t = 0 when they receive their signals. The signal is sent at t = 0 in A's reference frame, but that has nothing to do with Genady's diagram, because it's in B's reference frame.

36 minutes ago, martillo said:

I think this matches with your diagram with B and C at the x axis at time t = 0.

No, that line doesn't mean anything important for this problem. The diagram is confusing. It should show when the signals are received, not when t' = 0 on C's world line.

Edited February 18, 2023 by Lorentz Jr

[martillo]

2 minutes ago, Lorentz Jr said:

But you said B and C change their clocks to t = 0 when they receive their signals. The signal is sent at t = 0 in A's reference frame, but that has nothing to do with Genady's diagram, because it's in B's reference frame.

You are right (me wrong). Genady's diagram must be reviewed to locate appropriately the travelers locations at the times of the receiving signals as you asked.

[Genady]

41 minutes ago, martillo said:

I'm assuming t = 0 the time of the emission of the synchronizing signals from the midpoint to B and C. I think this matches with your diagram with B and C at the x axis at time t = 0.

No, it does not. Why would you think so? 

 

7 minutes ago, Lorentz Jr said:

The diagram is confusing.

There is nothing confusing about this diagram, unless one reads into it more than it shows. 

 

B and C on the diagram are not events. They mark world lines of observer B and observer C, i.e., the lines, one vertical and the other one tilted. There is no even a mark for t=0. The horizontal line is just a line of events simultaneous in B. I shouldn't even call it x axis, just x line. At some moment in B, the observers B and C are on this line, in the events where it intersects with the world lines of B and C.

7 minutes ago, martillo said:

Genady's diagram must be reviewed to locate appropriately the travelers locations at the times of the receiving signals as you asked.

There is nothing to review, but there is a lot to add to this diagram.

Better be done step-by-step. Poco poco.

[Lorentz Jr]

24 minutes ago, martillo said:

Genady's diagram must be reviewed to locate appropriately the travelers locations at the times of the receiving signals

C received its signal at t' = -2γv (t'' = 0).
B receives its signal at t' = 0.
C and B meet at t' = 1/γv.

Edited February 19, 2023 by Lorentz Jr

[martillo]

9 minutes ago, Genady said:

There is nothing to review, but there is a lot to add to this diagram.

Better be done step-by-step. Poco poco.

Right! 😄 

Edited February 19, 2023 by martillo

[Genady]

6 minutes ago, Lorentz Jr said:

C received its signal at t' = -2γv.

What is v in the current formulation?

[martillo]

1 minute ago, Genady said:

What is v in the current formulation?

The velocity of the travelers.

[Genady]

Just now, martillo said:

The velocity of the travelers.

The velocity relative to what?

[Lorentz Jr]

1 minute ago, Genady said:

The velocity relative to what?

Relative to A.

The relative speed between B and C is v' = 2v/(1 + v²).

[Genady]

Just now, Lorentz Jr said:

Relative to A.

The relative speed between B and C is v' = 2v/(1 + v²).

What is A?

[martillo]

3 minutes ago, Genady said:

The velocity relative to what?

Good question. I will wait Lorentz Jr answer because I would prefer to maintain the convention of his calculations.

[Lorentz Jr]

1 minute ago, Genady said:

What is A?

The observer at the midpoint.

[Genady]

1 minute ago, martillo said:

I will wait Lorentz Jr answer

I think it will be better if you answer, make mistakes, discuss, and fix them. We already know that Lorentz Jr knows how to solve this.

1 minute ago, Lorentz Jr said:

The observer at the midpoint.

What midpoint? In B frame?

[Lorentz Jr]

3 minutes ago, Genady said:

I think it will be better if you answer, make mistakes, discuss, and fix them. We already know that Lorentz Jr knows how to solve this.

That's true, martillo. We need to know what assumptions you're making.

3 minutes ago, Genady said:

What midpoint? In B frame?

In all frames. The positions of B and C are basically synchronized ahead of time in A's frame. (They're equidistant from A.)

A sends the signal, and B and C are moving toward A at speed v from opposite directions during the experiment.

[martillo]

10 minutes ago, Genady said:

I think it will be better if you answer, make mistakes, discuss, and fix them. We already know that Lorentz Jr knows how to solve this.

I think it must be chosen the velocity that makes the calculations easier. At a first approach seems now, as we are working the problem now, would be to consider v as the relative velocity between the travelers. The velocity of the midpoint A doesn't matter to me. The intention is to compare just the observations between the travelers.

Edited February 19, 2023 by martillo

[Genady]

2 minutes ago, Lorentz Jr said:

(They're equidistant from A.

If they are equidistant from A in A then they are not equidistant from A in B, at any moment in B.

5 minutes ago, martillo said:

I think it must be chosen the velocity that makes the calculations easier. At a first approach seems now, as we are working the problem now, would be to consider v as the relative velocity between the travelers. The velocity of the midpoint A doesn't matter to me. The intention is to compare just the observations between the travelers.

So, what is v? Velocity of which frame relative to which frame? C relative to B? Something else?

[martillo]

10 minutes ago, Genady said:

So, what is v? Velocity of which frame relative to which frame? C relative to B? Something else?

Right, C relative to B. This way it is not needed to use the relativistic addition of velocities I think. The movement of A doesn't matter to me, just its initial position when the signals are emitted from this point A matters.

Edited February 19, 2023 by martillo

[Lorentz Jr]

10 minutes ago, Genady said:

If they are equidistant from A in A then they are not equidistant from A in B, at any moment in B.

Maybe you're right. They're equidistant in A's frame.

3 minutes ago, martillo said:

This way it is not needed to use the relativistic addition of velocities I think.

So how do you know when C received its signal?

[Genady]

1 minute ago, martillo said:

Right, C relative to B. This way it is not needed to use the relativistic addition of velocities I think.

OK, but note that it is not the same v that appeared in the Lorentz Jr's calculations.

So, we got the diagram with observers B and C which meet at some event up there and with C moving toward B with velocity v. What else do we know?

2 minutes ago, Lorentz Jr said:

They're equidistant in A's frame

Exactly. But in B frame, C moves toward A slower than A moves toward B and yet they meet all together. Thus, C has to be closer to A than B.

[Lorentz Jr]

12 minutes ago, Genady said:

But in B frame, C moves toward A slower than A moves toward B and yet they meet all together. Thus, C has to be closer to A than B.

Good point. Anyway, the problem is that @martillo keeps trying to apply the simultaneity of the signals in A's frame to the analysis in B's frame. Or maybe he's gotten beyond that now, but he hasn't decided when C received its signal.

Edited February 19, 2023 by Lorentz Jr

[martillo]

2 minutes ago, Lorentz Jr said:

So how do you know when C received its signal?

Calculating: first I determine the distance at which it receives the signal. At some distance X of A the signal travels at c velocity and encounters traveler C who comes traveling from initial point C at velocity V. It just needed to solve a simple linear equation in X. Just give me some time to show it. Having the distances is just dividing by the velocity v to find the time traveled..