Symmetrical motion and resolving differences in FoR

[PAllen]

Since the velocity change is instantaneous, so is the rotation of the sim axis. Events 4-28 are simultaneous with blue(14, 16).
https://www.physicsforums.com/attachments/65577
The left pic in the attachment is your case with blue light paths and green sim axes.
The right pic is modified with 6 time units of velocity change. This allows a continuous rotation of the sim axis. Each blue event is simultaneous with a different red event.

post#27:

In a rest frame nothing can move slower, therefore you will not see time contraction!

You could save some effort in symmetrical cases like drawing 2 and 3. One rotated 180 ° equals the other. Analyzing the 2nd is redundant.

Which case is related to drawing 4 (the nipple)?

Your drawing horizontal position does not correspond to proper distance along the given simultaneity line. Normally, you want such correspondence. Gwellsjr's 'nipple' diagram corresponds to your smooth turnaround case. His shows proper distance, yours shows a meaningless spatial coordinate. The metric for the picture as gwellsjr shows would be the Fermi-Normal metric; for yours, who knows? something more complex to scale the distance.

 

[ghwellsjr]

Since the velocity change is instantaneous, so is the rotation of the sim axis. Events 4-28 are simultaneous with blue(14, 16).
https://www.physicsforums.com/attachments/65577

I don't understand why you are saying that. I can see from this diagram:

...that blue's event 15 is almost simultaneous with red's event 7. Why are you lumping 7 in with all the other ones that I put during the transition?

The left pic in the attachment is your case with blue light paths and green sim axes.
The right pic is modified with 6 time units of velocity change. This allows a continuous rotation of the sim axis. Each blue event is simultaneous with a different red event.

Sorry, I just can't see your reasoning without a lot more explanation.

post#27:

In a rest frame nothing can move slower, therefore you will not see time contraction!

That's true for an Inertial Reference Frame as I emphasized several times in my previous posts but it's not true for a non-inertial reference frame.

You could save some effort in symmetrical cases like drawing 2 and 3. One rotated 180 ° equals the other. Analyzing the 2nd is redundant.

You can take drawing 2 and load it into Paint and rotate 180 degrees and it will come out the same (with the colors interchanged). I think you meant that you can flip it vertically.

Nevertheless, it doesn't really matter how I produce drawing 3, I did it for a purpose. I guess I didn't make the purpose very clear.

Which case is related to drawing 4 (the nipple)?

As I said, drawing 4 is the non-inertial frame for the "instantly co-moving inertial observer". In order to draw it, I did a lot of copying and pasting. I told you it was a complex process. I started with drawing 2 which was transformed from drawing 1 to a speed of -0.6c. I copied a rectangular block that included the first eleven blue events that were at rest on the x=0 axis along with the first seven red events onto a new drawing. Then I did another transform at -0.5 and copied just one more pair of blue and red events. I continued doing this with transforms approaching a speed of zero and then going in the positive direction and ending with the top eleven blue events and top seven red events. Finally, I connected the red dots with straight line segments. I don't plan to ever do this again.

The reason for all this is to see as the blue observer is changing his speed, what would the simultaneous red events for a bunch of different inertial observer's frames look like that corresponded to each blue event. I'm probably not saying that clearly enough but you can do a search on MCIF (momentarily comoving inertial frame) as PAllen pointed out in post #26 if you want more explanation.

And to repeat, I want to emphasize that even though the gradual acceleration got rid of the instantaneous "jump", there still is a rapid transition in red's events, so much so, that it results in a Time Compression. You can see in the 4th drawing that between the Coordinate Times of 13 and 14 years, 3 years has gone by for the red observer. That sure sounds like Time Contraction to me.

 

[phyti]

I don't understand why you are saying that. I can see from this diagram:

...that blue's event 15 is almost simultaneous with red's event 7. Why are you lumping 7 in with all the other ones that I put during the transition?

My reference to 7 is for the drawing with the embedded events in one point.

As to this drawing drawing:
Simultanous in the the U (rest) frame, but it isn't in the blue frame since he receives it after acceleration. Here is a similar drawing. Notice R8 is simultaneous with B10 while B is sitting in the U frame, but not after he accelerates. I.e. the emission and detection don't always occur at the same speed.

Fig.4a shows the space-time paths in the U frame, using instant speed change for simplification.
Roy moves for 16 days at v=.6c, then 10 days at v=0. Bob moves for 10 days at v=0, then 16 days at v= .6c. Using the radar method, Bob pings Roy's ship periodically and assigns a position using the SR definition of synchronous (divide the round trip time for light in half). One instance is a signal from B10 to R18 to B22. B assigns a position of B(6, 16). The path of Roy according to Bob is developed geometrically over the actual path.
Fig 4b is the result showing the R clock initially slow, then fast, then slow, compared to the B clock.
https://www.physicsforums.com/attachments/65703

 

[ghwellsjr]

My reference to 7 is for the drawing with the embedded events in one point.

That would be this drawing:

But I still don't understand since blue 15 is almost simultaneous with red 7. Your comment was:

Since the velocity change is instantaneous, so is the rotation of the sim axis. Events 4-28 are simultaneous with blue(14, 16).

I think you got your units digits interchanged and meant to say "Events 8-24 are simultaneous with blue(14, 16)".

But note that I have rejected that diagram in favor of the second one in post #25.

As to this drawing drawing:
Simultanous in the the U (rest) frame, but it isn't in the blue frame since he receives it after acceleration.

I'm not sure what you are saying here with regard to my diagram in your previous post. I stated in post #8 that it was "for the IRF in which the blue observer is at rest at the beginning". My next drawing was for the IRF in which the blue observer is at rest at the end. There is no IRF in which the blue (or the red) observer is at rest during the entire scenario which is why we have to revert to non-inertial reference frames if we insist on having a frame that the observer can call his own.

Here is a similar drawing. Notice R8 is simultaneous with B10 while B is sitting in the U frame, but not after he accelerates. I.e. the emission and detection don't always occur at the same speed.

Fig.4a shows the space-time paths in the U frame, using instant speed change for simplification.
Roy moves for 16 days at v=.6c, then 10 days at v=0. Bob moves for 10 days at v=0, then 16 days at v= .6c. Using the radar method, Bob pings Roy's ship periodically and assigns a position using the SR definition of synchronous (divide the round trip time for light in half). One instance is a signal from B10 to R18 to B22. B assigns a position of B(6, 16). The path of Roy according to Bob is developed geometrically over the actual path.
Fig 4b is the result showing the R clock initially slow, then fast, then slow, compared to the B clock.
https://www.physicsforums.com/attachments/65703

I concur with your drawing of the non-inertial rest frame for Bob. And you will note that you have shown Ray's clock as ticking faster than the Coordinate Time, in other words, "contracted". In post #30, with regard to my my drawing of a non-inertial rest frame, you said:

In a rest frame nothing can move slower, therefore you will not see time contraction!

I'm glad you have taken the initiative to make your own non-inertial rest frame drawing and come to the same conclusion.

 

[phyti]

That would be this drawing:

In this drawing why didn't you assign 16 to the reversal for red?