The synchronization of clocks and the relativity of motion

In summary, the relativity of simultaneity states that two clocks that are moving relative to each other will not always be synchronized.
  • #71
Sagittarius A-Star said:
C and D are at rest in the train's rest-frame and move in the ground's rest-frame. This creates different boundary conditions for the (frame-dependent) travel distances of the light and does not violate the 1st postulate.
Sorry, my mistake.
B,C and D are the same frame.
 
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  • #72
PeterDonis said:
No. Suppose that, relative to A, B is moving to the right. Then, relative to B, A is moving to the left. These are two different relative velocities.
If that is the case, then the two diagrams show two velocities, the same velocity, one from each of two frames.
 
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  • #73
AlMetis said:
Sorry, my mistake.
B,C and D are the same frame.
In your diagram, A, C and D are at rest in the the same frame.
 
  • #74
AlMetis said:
If that is the case, then the two diagrams show two velocities, the same velocity, one from each of two frames.
If you mean your two diagrams in post #29, no, they don't. They show what I described, except that you have A moving to the right relative to B (in your Diagram 1.0) and B moving to the left relative to A (in your Diagram 2.0).

You really need to take a step back and think carefully. You are repeatedly making wrong claims and this thread cannot go on indefinitely with you simply continuing to repeat your wrong claims even after they have been repeatedly corrected.
 
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  • #75
Sagittarius A-Star said:
In your diagram, A, C and D are at rest in the the same frame.
Yes, they are in the same frame. It was my last post that was the mistake. I misunderstood your post.
To your last post - A, C and D do not change between frames of reference at time=0.
The motion between the two frames is what causes the different times of flight, but C and D remain in the rest frame of A.
 
  • #76
AlMetis said:
There are 2 frames A and B, one relative velocity.
There are also velocities relative to frames, which is what you're transforming.

Do you understand that my velocity in a car can be zero in one frame and 30mph in another? Because I'm really not convinced you accept anything that basic.
 
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  • #77
PeterDonis said:
except that you have A moving to the right relative to B (in your Diagram 1.0) and B moving to the left relative to A (in your Diagram 2.0).
If A moves to the right relative to, they observe B moving to the left and vise versa. How is that not correct?
 
  • #78
Ibix said:
Do you understand that my velocity in a car can be zero in one frame and 30mph in another? Because I'm really not convinced you accept anything that basic.
I do understand it. I don't understand why you think there is any conflict with that basic premise and what is described in the diagrams.
 
  • #79
AlMetis said:
If A moves to the right relative to, they observe B moving to the left and vise versa. How is that not correct?
I didn't say that wasn't correct. It is. But it contradicts your earlier claim here:

AlMetis said:
There is only one velocity, the relative velocity of A and B.
 
  • #80
PeterDonis said:
I didn't say that wasn't correct. It is. But it contradicts your earlier claim here:
The velocity of B relative to A is the same as the velocity of A relative to B. There is not a velocity “attached” to each, there is an observed velocity between them. They agree on this velocity and it is identical. If I was wrong in saying it is the one and the same, I stand corrected.
 
  • #81
AlMetis said:
I do understand it. I don't understand why you think there is any conflict with that basic premise and what is described in the diagrams.
So you understand that the receivers have different velocities in the two frames, and hence that the distances from emission to reception events can be different in the two frames? And hence that the reception events can be simultaneous in one frame and not in the other?
 
  • #82
AlMetis said:
The velocity of B relative to A is the same as the velocity of A relative to B.
No, it is not. You have already stated otherwise and your own diagrams show otherwise. You are contradicting yourself.
 
  • #84
After moderator review, the thread will remain closed.
 
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  • #85
Sorry about the late post, but I wanted to follow up on the statements about boundary conditions.

AlMetis said:
Relative motion is a single condition, how can it differ between itself?
No, relative motion is not the boundary condition. The boundary conditions include the initial position and initial velocity of each object of interest in the frame in which you are solving the equations of motion, as I mentioned in post 60.

The solutions to the differential equations don’t care about the relative velocity, they care about the variables (position), their initial values (initial positions), and their initial first derivatives (initial velocities), all in the specified frame. All of your recent comments about relative velocity are irrelevant regarding the boundary conditions and the equations of motion (your kinematics).

Hence, the boundary conditions are different in the two frames, but the laws of physics are the same. This is consistent with the principle of relativity.

AlMetis said:
There is only one velocity, the relative velocity of A and B.
A, C and D are the same frame, how can they differ from themselves.
Sure, that is a bit of a personal preference. A, B, C, and D all have different positions, so their positions are separate variables. So I would use their initial positions and initial velocities as separate boundary conditions and use constraint equations to set the appropriate velocities equal.

Directly using the reduced boundary conditions without explicit constraints will get you equivalent equations of motion either way, but it is a less systematic approach in my opinion.

Also in my opinion, you could use a bit more of a systematic approach. Your current approach seems very unproductive. You learn very little and very slowly. And the cost of what little you do learn is the good-will of this community of helpful experts. I hope you will be willing to modify your approach when you return
 
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