I'd like to address MacM's latest scenario according to SR.
MacM said:
Three clocks "A", "B" & "C" are at common rest at location "C" and "A" & "B" are
launched out to the ends of a prearranged course where they have equal acceleration
range from points 'X' to the test start lines "S" which are equal distance from "C" in
opposite directions.
A------------->!................................................ ...><............................................. ......!<-------------B
X---------------S------------------------------------------C------------------------------------------S--------------X
Fig 1
Once there "A" & "B" send "Ready" light signals to "C". Upon receipt of both
signals "C" then simultaneously sends "Start" light signals to both "A" & "B". They
immediately upon receipt launch simultaneously according to "C" and accelerate equally
until they simultaneously reach the "S" line. Both set their clocks to t=0, cut their rockets and go inertial and send "Confirmation" light signals to "C".
Upon the simultaneous receipt of the "A" & "B" signals "C" then sets his clock according
to the time required for the light signals to travel from the "S" positions based on known
distance and the speed of light 'c'.
Assuming "S" is 0.6 light hours from "C" and "A" & "B" achieve a 0.6c velocity then
"C" knows to set his clock to 0.6 times 60 minutes = 36 minutes so that all three clocks will
have been set to t=0 at the same time according to "C"
This procedure synchronises sets all the clocks to zero at the same time in C's frame, but not in the frame of A or B, as I will show below.
MacM said:
At the 0.6c velocity "A" & "B" will require 1 hour to make the trip according to "C".
But as "A" & "B" pass by "C" simultaneously they transmit their clock data to "C" by
digital signal and due to time dilation they will have only accumulated:
t' = t(1 - v^2/c^2)^0.5 = 0.8 x 60 minutes = 48 minutes. This is because "A" & "B" had
symmetrical acceleration and equal velocity to the common rest frame.
This is a correct conclusion, but it has nothing to do with acceleration. This "test" starts when the clocks are set to zero, after the acceleration period. Therefore, the accelerations are irrelevant to the problem. There is no need to specify a "common rest frame", which is a concept that is not required in relativity.
What is important here is that in spite of having had relative velocity to each other "A" & "B" clocks accumulate the same amount of time. So relative velocity did not
generate time dilation between them.
It is true that A and B accumulated the same amount of time. MacM's statement about time dilation is hard to decode, so let's ignore it and do the problem properly.
Let's denote three sets of coordinates as follows:
Frame C: (x,t)
Frame A: (x', t')
Frame B: (x'', t'')
The three clocks start "accumulating time" (i.e. like a stopwatch starting) when A and B are at the start of the course (positions S). A's speed is 0.6c in the positive x direction, B's speed is 0.6c in the negative x direction (i.e. B's velocity is -0.6c).
Let's look at the coordinates of the relevant events in the C frame coordinates:
A starts his clock: (x,t) = (-0.6 light hours, 0)
B starts his clock: (x,t) = (0.6 light hours, 0)
C starts his clock: (x,t) = (0,0)
A passes C: (x,t) = (0, 1 hour)
B passes C: (x,t) = (0, 1 hour)
Now, let us use the Lorentz transformations of special relativity to calculate the cooordinates of these same events in A's frame:
A starts his clock: (x',t') = (-0.75 light hours, +0.45 hours)
B starts his clock: (x',t') = (0.75 light hours, -0.45 hours)
C starts his clock: (x',t') = (0,0)
A passes C: (x',t) = (-0.75 light hours, +1.25 hours)
B passes C: (x',t) = (-0.75 light hours, +1.25 hours)
Note that according to A's clock, A's clock was started 0.45 hours (or 27 minutes)
after C's clock was started. i.e. the starting of A's and C's clocks was not simultaneous in A's frame.
The elapsed time for the trip according to A's clock is: +1.25 hours - (+0.45) hours = 0.8 hours, or 48 minutes.
The "actual" time taken for the trip by B, as observed by A using A's clocks is (1.25 - (-0.45) = 1.7 hours. Therefore, A concluded that B's clocks were running slower than A's clocks during the trip.
Now B's frame:
A starts his clock: (x'',t'') = (-0.75 light hours, -0.45 hours)
B starts his clock: (x'',t'') = (+0.75 light hours, +0.45 hours)
C starts his clock: (x'',t'') = (0,0)
A passes C: (x'',t'') = (+0.75 light hours, +1.25 hours)
B passes C: (x'',t'') = (+0.75 light hours, +1.25 hours)
Note that according to B's clock, B's clock was started 0.45 hours (or 27 minutes)
after C's clock was started. i.e. the starting of B's and C's clocks was not simultaneous in B's frame.
The elapsed time for the trip according to B's clock is: +1.25 hours - (+0.45) hours = 0.8 hours, or 48 minutes.
The "actual" time taken for the trip by A, as observed by B using B's clock is (1.25 - (-0.45) = 1.7 hours. Therefore, B concluded that A's clocks were running slower than B's clocks during the trip.
One more point: let's look at the ORDER of events here.
In C's frame, all three clocks were started at t=0 - i.e. simultaneously.
In A's frame, the clocks were started at intervals of 27 minutes: first B's clock, then C's, then A's.
In B's frame, the clocks were started at intervals of 27 minutes: first A's clock, then C's, then B's.