B The synchronization of clocks and the relativity of motion

[Ibix]

Diagram 1.0 shows that does not happen.

The two-way convention calculates the one-way as a mean of each leg which sets the one-way time of flight the same for both observers. But as is shown in diagram 1.0 the one way time is not the same for both observers.

Once again, you are failing to understand the difference between asymmetries in experimenal results that are due to asymmetries you chose to put in your experimental setup, and the symmetry of the physical laws. (Which is the same point @Dale makes about boundary conditions in the post above this one).

 

[AlMetis]

What you call reciprocal kinematics does not follow from the principle of relativity. I have already explained why, but I will do so again in more depth. ...

Thank you, that was an excellent explanation.
I agree with everything except your claim that the boundary conditions are different in my diagrams.
Time=0 shows identical boundary conditions except the arrow indicating the relative motion of A and B.

Please point out what boundary condition/s differ.

 

[Dale]

except the arrow indicating the relative motion of A and B.

That is a different boundary condition

Please point out what boundary condition/s differ.

The initial velocities of A, B, C, and D differ between the frames. Also, although it is not relevant for this problem, the initial times on the clocks would be different as would the length of the train. All of those are boundary conditions as described above

 

[Nugatory]

The entire discussion is just besides the point. The problem with Einstein's original paper of 1905 is that at this time nobody was aware about the mathematical structure which is adequate to formulate spacetime models, and that's why Einstein has to use pretty subtle gedanken experiments to establish the synchronization convention.

And it is for this reason that reading Einstein's papers is an exercise in the history of science, not the science itself. History of science is a fascinating discipline in its own right (and arguably more valuable to a well-rounded layperson than the science itself) but it is a different discipline with different goals and methods.

 

[AlMetis]

That is a different boundary condition

Relative motion is a single condition, how can it differ between itself?

 

[AlMetis]

The initial velocities of A, B, C, and D differ between the frames.

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.

 

[Ibix]

There is only one velocity, the relative velocity of A and B.

No, there are the velocities of the various things relative to your frames.

A, C and D are the same frame, how can they differ from themselves.

If I am in a car I might consider myself at rest. A pedestrian watching me drive by says I'm doing 30mph. Do you understand that? If so, what's the problem with your observers having different velocities with respect to different frames?

 

[Sagittarius A-Star]

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.

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.

 

[PeterDonis]

There is only one velocity, the relative velocity of A and B.

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. They have the same magnitude, but opposite directions.

I think you have not really thought through what actually changes when you change frames.

 

[AlMetis]

No, there are the velocities of the various things relative to your frames.

There are 2 frames A and B, one relative velocity.

 

[AlMetis]

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.

 

[AlMetis]

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.

 

[Sagittarius A-Star]

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.

 

[PeterDonis]

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.

 

[AlMetis]

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.

 

[Ibix]

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.

 

[AlMetis]

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?

 

[AlMetis]

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.

 

[PeterDonis]

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:

There is only one velocity, the relative velocity of A and B.

 

[AlMetis]

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.

 

[Ibix]

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?

 

[PeterDonis]

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.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed.

 

[Dale]

Sorry about the late post, but I wanted to follow up on the statements about boundary conditions.

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.

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