On Sunday, September 17, 2023 at 1:24:27 AM UTC-5, Luigi Fortunati wrote:
> If clocks A and B are stopped, they remain synchronized.
Not for a moving observer per SR! Unless the two clocks are at the same
location, different observers will see different times on the two clocks.
What each observer "sees" via light from each clock, and what each
observer calculates as "now" at each clock are two very different things.
> If clock B stands still and clock A moves (land reference), every time
> they meet clock A lags behind.
Correct. And all observers would agree, no matter their state of motion
because the two clocks are at the same position. Clock A on the carousel
lags because it is always moving relative to clock B.
> If clock A stands still and clock B moves (reference of the carousel),
> every time they meet clock B lags behind.
Clock A, on the carousel, is not in an inertial frame. SR is not really the
correct theory for describing the appearance of clock B in this rotating
frame. Never the less, when clock A and clock B "meet", the time on
clock A will lag that on clock B.
Again, I think you are getting tripped up on the difference between the
times shown on the clocks "now" (which you cannot observe directly)
and the times that can be seen via light from the clocks. As the
observer at clock A rotates away from clock B, they will observe (via
light) clock B going very much slower. As observer A rotates around
the carousel and starts to move towards B again they will see clock
B speed up rapidly and overtake the time on clock A.
The times "now" are very much more complicated. When A is moving
away from B, the "now" time on B, as calculated by A, will slow down
and possibly almost stop when accelerating away from B. As A comes
around the back side of the carousel, the time A calculates as "now"
at B will rapidly advance to what A calculates as the present, then as
A accelerates towards B the calculated "now" will advance to a
future time, which the A clock will gradually catch up to when they
meet. The time shown on clock B will be advanced relative to that
on clock A because clock A has followed an accelerated path while
clock B has been inertially at rest the whole time.
I'm afraid the new animation does not illuminate this problem.
Rich L.