Twin watches

[Luigi Fortunati]

If clocks A and B are stopped, they remain synchronized.

If clock B stands still and clock A moves (land reference), every time
they meet clock A lags behind.

If clock A stands still and clock B moves (reference of the carousel),
every time they meet clock B lags behind.

The animation
https://www.geogebra.org/m/mxg5xnzy
is correct?

Luigi

[[Mod. note -- All observers can agree on
(a) The events when the two clocks are next to each other, i.e., the events
when the carousel has made 0 revolutions (the starting event), and when
the carousel has made 1 complete revolution.
(b) The two clock readings at an event when the two clocks are next to each
other. At the starting event both readings are 0. After one complete
revolution the readings are carousel=9, ground=10.

So, the question to figure out is, how does the carousel observer obtain
consistent results? I must confess that the answer isn't immediately
obvious to me, but it's late a night in my time zone and I'm tired. :)
Clearly the carousel reference frame is not an inertial reference frame.
Accelerations don't affect the rates of (ideal) clocks in relativity,
but maybe we need to consider the Sagnac effect?
-- jt]]

[Mike Fontenot]

[Luigi Fortunati]

....

I didn't quite understand your answer. lol

In any case, I clarify my question further with this other animation
https://www.geogebra.org/m/ar3dqcwk

There is a standing electromagnetic wave in a carousel that rotates
360=B0 in 2 time.

It works exactly like a light clock.

In the terrain reference, is the time of one complete rotation the same
or is it more or less?

Luigi

[Mike Fontenot]

Post by Luigi Fortunati

....

I didn't quite understand your answer. lol
[[Mod. note -- All observers can agree on
(a) The events when the two clocks are next to each other, i.e., the events
when the carousel has made 0 revolutions (the starting event), and when
the carousel has made 1 complete revolution.
(b) The two clock readings at an event when the two clocks are next to each
other. At the starting event both readings are 0. After one complete
revolution the readings are carousel=9, ground=10.
So, the question to figure out is, how does the carousel observer obtain
consistent results? I must confess that the answer isn't immediately
obvious to me, [...]

I (Mike Fontenot, ***@gmail.com ) said:

I don't see why you say there were any inconsistent results. The
results all looked fine to me. Could you elaborate?

[Richard Livingston]

Post by Luigi Fortunati
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.

Post by Luigi Fortunati
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.

Post by Luigi Fortunati
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.

[Tom Roberts]

Post by Luigi Fortunati
If clocks A and B are stopped, they remain synchronized.

No. A stopped "clock" is not a clock, so "synchronized" is meaningless.
Fortunately you did not attempt to use this silliness.

You omitted important information about the physical situation. I
presume you mean clock B is located on the ground next to a rotating
carousel, and clock A is fixed to the carousel on its rim, so the two
clocks repeatedly meet, once per rotation of the carousel. For
simplicity, I also presume the ground is an inertial frame. (Your
animation confirms most of this.)

Post by Luigi Fortunati
If clock B stands still and clock A moves (land reference), every
time they meet clock A lags behind.

Yes, between meetings clock A experiences less elapsed proper time
than does clock B. This is just a demonstration of the twin paradox.

Post by Luigi Fortunati
If clock A stands still and clock B moves (reference of the
carousel), every time they meet clock B lags behind.

No! The elapsed proper time between meetings for each clock is an
invariant. So it does not matter which coordinates one uses as a
reference, clock A always has less elapsed proper time between meetings
than does clock B.

You seem to have fallen into the trap of believing that "moving clocks
run slow". That oft-repeated sound bite is FALSE in several different ways:
1. The moving clock ITSELF does not "run slow", because all
clocks ALWAYS tick at their usual rate [#]. It's just that
it is OBSERVED to run slow by an inertial frame relative to
which it is moving.
2. This only applies to a moving clock being observed by an
inertial frame. For a non-inertial observer, that sound
bite may or may not apply (it takes a real calculation).

[#] If this were not true, Einstein's first postulate
of SR could not be valid.

Your animation is woefully incorrect.

[[to the moderator]]
This is not the Sagnac effect, this is merely the fact that different
paths between a given pair of endpoints can have different path lengths
(aka elapsed proper time for timelike paths).

Tom Roberts

[Luigi Fortunati]

Post by Tom Roberts

Post by Luigi Fortunati
If clocks A and B are stopped, they remain synchronized.

No. A stopped "clock" is not a clock, so "synchronized" is meaningless.

Be careful, when the carousel is stopped with respect to the ground, clock
A is also stopped with respect to the ground and with respect to clock B:
they are all stopped with respect to each other.

But clocks, even when they are stopped somewhere, their hands continue to
move, otherwise what kind of clocks would they be?

If you had clicked on the "No rotation" button of my animation
https://www.geogebra.org/m/mxg5xnzy
you would have seen the synchronization between the two clocks A and B
stopped (relative to each other) with your own eyes.

Post by Tom Roberts
You omitted important information about the physical situation. I
presume you mean clock B is located on the ground next to a rotating
carousel, and clock A is fixed to the carousel on its rim, so the two
clocks repeatedly meet, once per rotation of the carousel. For
simplicity, I also presume the ground is an inertial frame. (Your
animation confirms most of this.)

Yes, my animation confirms this.

Post by Tom Roberts

Post by Luigi Fortunati
If clock B stands still and clock A moves (land reference), every
time they meet clock A lags behind.

Yes, between meetings clock A experiences less elapsed proper time
than does clock B. This is just a demonstration of the twin paradox.

And that's exactly what you see in my animation by choosing the terrain
reference, where clock A lags behind (time=9 instead of 10).

Post by Tom Roberts

Post by Luigi Fortunati
If clock A stands still and clock B moves (reference of the
carousel), every time they meet clock B lags behind.

No! The elapsed proper time between meetings for each clock is an
invariant. So it does not matter which coordinates one uses as a
reference, clock A always has less elapsed proper time between meetings
than does clock B.

Ok: this means that the rotation of the carousel in the ground reference
is a real rotation and the rotation of the ground in the carousel
reference is an apparent rotation and not a real one.

On this objection of yours, I have already modified my animation, here it
is:
https://www.geogebra.org/m/b9dxtzmp

Now, in both references, the clock that lags behind is always clock A.

Post by Tom Roberts
Your animation is woefully incorrect.

As I showed you, the previous animation was correct in two out of three
cases: the one with the carousel stopped (no clock lags behind) and the
one in the ground reference (the carousel clock lags behind).

And in my latest animation it is also correct in the reference of the
carousel: clock A (time=9) is always behind clock B (time=10).

Luigi.