The twin Paradox revisited

[Genady]

Just now, Lorentz Jr said:

the problem is that @martillokeeps trying to apply the simultaneity in A's frame to the analysis in B's frame.

Yes, this is the problem. But we keep trying not to let him do this  .

PS. I have to go now. We'll be back in about an hour, I think. Otherwise, good night to all.

[martillo]

Point D : when C encounters the signal.

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 )/v = (L – X)/v = time__B(signal(A,D)) = X/c

Substituting values:

(L – X)/v) = X/c

X = Lc/(v + c)

Edited February 19, 2023 by martillo

[Lorentz Jr]

6 minutes ago, martillo said:

time__B(C,D) = (dist__B(A,C) - X )/v = (L – X)/v = time__B(signal(A,D)) = X/c

You can't do this, martillo. You're assuming dist_B(A,C) = L, but it's not.

[martillo]

3 minutes ago, Lorentz Jr said:

You can't do this, martillo. You're assuming dist_B(A,C) = L, but it's not.

Well, that could be my problem. How would you find the distance (not the time) where C encounters its signal?

[Lorentz Jr]

1 minute ago, martillo said:

How would you find the distance (not the time) where C encounters its signal?

This is the distance from A in A's reference frame:

On 2/16/2023 at 12:21 PM, Lorentz Jr said:

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.

 

[martillo]

2 minutes ago, Lorentz Jr said:

This is the distance from A in A's reference frame:

No, is the distance I'm assuming now in B frame. I don't use frame A...

Edited February 19, 2023 by martillo

[Lorentz Jr]

4 minutes ago, martillo said:

No, is the distance I'm assuming now in B frame. I don't use frame A...

In B's frame:

On 2/16/2023 at 10:40 AM, 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. 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.

x_i' = 2γ.

Edited February 19, 2023 by Lorentz Jr

[martillo]

20 minutes ago, Lorentz Jr said:

On 2/16/2023 at 1:40 PM, 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. 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.

No, please consider B always at X = 0 to match with the diagrams you and Genady have made. B is stationary, doesn't move.

I'm assuming initially (time when the signals are emitted) and in B frame dist(B,A) = dist (A,C) = L. After sometime B and C receive their signals, B still at x = 0 and C at some point L + X where X as I calculated above. May be I calculated it wrongly as you said. I don't know why... 

Edited February 19, 2023 by martillo

[Lorentz Jr]

55 minutes ago, martillo said:

I'm assuming initially (time when the signals are emitted) and in B frame dist(B,A) = dist (A,C) = L.

You can't do that. The only way A can be placed midway between B and C is if their positions are all measured at the same value of t (i.e. at the same time in A's frame), but then they would correspond to different values of t' (i.e. different times in B's frame).

Edited February 19, 2023 by Lorentz Jr

[martillo]

29 minutes ago, Lorentz Jr said:

You can't do that.

Yes, you all have already told me that. I just showed my calculation to illustrate how to calculate X with a simple equation. I just want to know which values to use for the two distances. I thought I could deduce that with the help of the diagrams. Point D needs to be located and calculated...

Edited February 19, 2023 by martillo

[Lorentz Jr]

13 minutes ago, martillo said:

I just want to know which values to use for the two distances.

Forget about them, martillo! They're worthless! You can't analyze relativity problems in terms of objects. You have to focus on events, and there are no events involving A after it sends the signals.

The only thing you can do is calculate the time and positions when the signals arrive in A's reference frame, and then transform them to get x' and t' for the signal's arrival at C in B's frame.

Edited February 19, 2023 by Lorentz Jr

[Genady]

54 minutes ago, martillo said:

I'm assuming initially (time when the signals are emitted) and in B frame dist(B,A) = dist (A,C) = L.

If you are assuming this, then B and C are not symmetrical, and I can prove it.

Since they are not symmetrical, when they meet their clocks and beards will not be equal.

[martillo]

Well, I tried to improve Lorentz Jr diagram. Are the points A and D well situated or not? A will also belong to the midline between the lines of B and C...

Edited February 19, 2023 by martillo

[Genady]

7 minutes ago, martillo said:

Well, I tried to improve the diagram. Are the points A and D well situated or not?

I understand that event D is "Observer C receives the signal". What are the events A, B, and C on this diagram?

[Lorentz Jr]

5 minutes ago, Genady said:

understand that event D is "Observer C receives the signal". What are the events A, B, and C on this diagram?

B and C are signal receptions at t = 0 and x = 1 or -1.

Edited February 19, 2023 by Lorentz Jr

[Genady]

1 minute ago, Lorentz Jr said:

B and C are signal receptions.

And A?

Wait, D is signal reception...

 

[Lorentz Jr]

6 minutes ago, Genady said:

And A?

More confusion. @martillo keeps trying to analyze the signals from A to B and from A to C in B's frame, and I'm starting to lose patience with it.

Notice how he puts A midway between B and C, ignoring the time axis.

6 minutes ago, Genady said:

Wait, D is signal reception...

D is more confusion.

Edited February 19, 2023 by Lorentz Jr

[martillo]

3 minutes ago, Lorentz Jr said:

More confusion. @martillo keeps trying to analyze the signals from A to B and from A to C in B's frame, and I'm starting to lose patience with it.

I'm trying to locate points A (signal emission) and point D (signal reception by C) in the diagram. Is it so complex?

[Genady]

3 minutes ago, Lorentz Jr said:

More confusion.

@martillo,

The spacetime diagrams are good for qualitative comparisons between events and frames, but to get quantitative comparisons you will have to calculate Lorentz transformations.

[martillo]

13 minutes ago, Lorentz Jr said:

Notice how he puts A midway between B and C, ignoring the time axis.

It is just an intuitive location of events. I expect help to improve the diagram... May be I'm wrong I know...

5 minutes ago, Genady said:

@martillo,

The spacetime diagrams are good for qualitative comparisons between events and frames, but to get quantitative comparisons you will have to calculate Lorentz transformations.

Right. I think the same. I thought that with just lines I could locate points A and D. May be I'm wrong.

Edited February 19, 2023 by martillo

[Genady]

4 minutes ago, martillo said:

I'm trying to locate points A (signal emission) and point D (signal reception by C) in the diagram. Is it so complex?

It looks incorrect. Why the three points A, B, and C are on a straight line? And why A is in the middle of that line?

[Lorentz Jr]

21 minutes ago, Genady said:

What are the events A, B, and C on this diagram?

 

[Genady]

Just now, Lorentz Jr said:

 

Yes, this is clear.

[martillo]

4 minutes ago, Genady said:

It looks incorrect. Why the three points A, B, and C are on a straight line? And why A is in the middle of that line?

As md65536 posted, Lorentz transform are linear transforms preserving midpoints. A will always be the midpoint. He commented that distances and times would be different. I'm trying to view that.

[Genady]

1 minute ago, martillo said:

As md65536 posted, Lorentz transform are linear transforms preserving midpoints. A will always be the midpoint. He commented that distances and times would be different. I'm trying to view that.

It does not matter. A is not a midpoint in the B frame.