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* “Confused” where you speak of “cumulative Time Dilation.” I have no idea what you mean as Time Dilation is of the form “3 ticks of the moving clock per 5 ticks of stationary clock I am holding” or more simply : The time dilation is “3 to 5” or “3/5”. I have no idea how you “accumulate” 3/5 time dilation. AFAIK, to “accumulate” you need well defined “Start accumulation” and Stop accumulation” events, but that is hard to do (or impossible) “simultaneously” two different frames and drags in the “simultaneity problems,” which I avoid by COMPUTING the time dilation. You told how to compute that in post 93 for “MacM SR” and I know how to do that computation for standard SR.
You are wasting our time Billy T I am NOT going through this again with you. You post is misleading, false and simply put BS and is not a proper description of what has been said. End of converation.
I'll only note as to this last comment by you. Pathetic. I'll give my scenario just this one last time.
A, B & C are at common rest and have laid out a test course.
The course consists of miles markers equal distance and in opposite direction from the rest location.
A and B are launched to go out to the last mile marker and signal they are prepared for the test. When C gets both their ready signals he transmits a launch you craft light signal. When A & B receive that signal Which is simultaneous to C, then they accelerate according to plan with equal rate because their craft are identical and are equipped with precisoin accelerometer controls.
According to plan when trhe both cross mile marker number 1 lyr they throttle down and go inertial at 0.6c, setting their clocks to "0". They instantly transmit to C a Start your clock light signal which happens to also be simultaneous to C..
When C receives those signals since C knows the distance involved and the speed of light he knows to pre-set his clock for the delay in receipt of the signals such that he sets his clock to 1 year so his start "0" conforms with the amount of time by his clock that has passed since they set their clocks to "0".
Since they launched simultneously to C and have an equal velocity to C and were equal distance from C they will pass C simultaneously after (1.6666 years) 1 year and 8 months or 56,764,800 seconds by C's clock.
At the moment they pass they each transmit to the other a digital signal saying what their clock accumulated time is. They also transmit what they computed each others clocks should read according to relativity.
At the juncture the test is over and all clocks are stopped for comparison. The entire test has been conducted only during inertial relative veloicty, has been simultaneously synchronized at C.
They can delerate and return to C for comparison but that is not necessary. C has all the test data. and A & B have the same data.
What is found is that the physical accumulated times are:
C = 56,764,800 seconds
A = 45,411,840 seconds
B = 45,411,840 seconds
This is in complete agreement with the predictions of the common rest frame.
A & B according to SR believe their relative velocity to be 0.88235c not the logical 1.2 based on .6c in opposite directions to C.
Based on that A predicted:
C = 45,411,840 seconds (this assumes SR's ludricrus assumption that A will see 0.6c to C and not have a dilated clock, even though it is traversing 1 lyr in 1.44 years by his clock which would compute to be 0.6944c not 0.6c.?
B = 12,885,884 seconds.
B predicts the same results for A and C.
But the ONLY valid prediction is the one that is made reference to the initial common rest frame and NOT any made while in motion.
Further it remains to be proven that A & B would compute 0.6c when they are seeing mile markers pass in 0.8 times as fast as they should or they would think they are going 0.75c regardless if the lesser accumulated time is due to TD or LC.
I do hope you got itv now because I WILL NOT repeat it again.
DO NOT come back with but A & B don't thik the other started simultaneously because what they think is irrelevant to the issue. The defacto simultaneous starting time of clocks is.
In contrast to your assertions that my view is silly consider this folks.
If A or B use any legitimate thoughtvprocess (assuming they actually see 0.6c not o,75c that I argue) then they mightvwell conclude that since I see C approaching at 0.6c and B approaching at 0.88235c then B must be moving 0.28235 relative to C hence B will have accumulated 54,455,123 seconds.
Hmmmm. Silly concept indeed.
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