1. According to special relativity, when a stationary observer looks at a line of moving clocks that are synchronized in their own frame, they will be progressively out of sync along the line. One might say there is a sync gap that increases with distance. Suppose there are four such lines that are joined to form a square that the clocks continually move round. How do the sync gaps match up?

Please only post comments on this thread if they deal directly with observations arising from the experiment I've described.

2.

3. Originally Posted by Andrew?
According to special relativity, when a stationary observer looks at aline of moving clocks that are synchronized in their own frame, they will beprogressively out of sync along the line. One might say there is a sync gap that increases with distance. Suppose there are four such lines that arejoined to form a square that the clocks continually move round. How do the sync gaps match up?
For clocks moving around a square track, it's not clear to me that one can actually synchronise all the clocks in that frame of reference. Bear in mind that you are now asking about an accelerated frame of reference and this is much more difficult than an inertial frame of reference. It should be remarked that the relationship between an accelerated frame of reference and an inertial frame of reference is not symmetric.

But, I must ask: What does the notion of simultaneity mean to you?

4. Originally Posted by KJW
For clocks moving around a square track, it's not clear to me that one can actually synchronise all the clocks in that frame of reference.
Maybe it would help to state the problem as follows:

Suppose four lines of clocks form a rectangle having identically rounded corners. The clocks are synchronized at rest and then they continually move round the rectangle at a constant speed. How do the sync gaps match up?

It is true that the clocks have been accelerated, but this should be irrelevant in SR as the effects of the predicted non-simultaneity only depend on the current relative speed and distance. The observer remains at rest, and the clocks’ acceleration can be analysed using co-moving frames if necessary.

5. Originally Posted by Andrew?
Originally Posted by KJW
For clocks moving around a square track, it's not clear to me that one can actually synchronise all the clocks in that frame of reference.
Maybe it would help to state the problem as follows:

Suppose four lines of clocks form a rectangle having identically rounded corners.
Actually the problem wasn't the sharp corners but the accelerated frame of reference which complicates simultaneity because things that one may take for granted in an inertial frame of reference or even a constantly accelerated frame of reference might not apply in more general accelerated frames of reference. For example, X being instantaneous to Y doesn't imply that Y is instantaneous to X. In this case, it is not possible to synchronise X and Y because they would not agree on being synchronised.

Originally Posted by Andrew?
The clocks are synchronized at rest and then they continually move round the rectangle at a constant speed. How do the sync gaps match up?
If the clocks are synchronised at rest, then they will remain synchronised in that frame of reference even though they are in motion. The clocks won't remain synchronised in their own frame of reference, so the question of how the sync gaps match up does not apply.

Originally Posted by Andrew?
It is true that the clocks have been accelerated, but this should be irrelevant in SR as the effects of the predicted non-simultaneity only depend on the current relative speed and distance. The observer remains at rest, and the clocks’ acceleration can be analysed using co-moving frames if necessary.
Just because an accelerated frame of reference can be dealt with in special relativity by considering inertial frames of reference that are at instantaneous rest, it doesn't mean that one can ignore the acceleration and treat the frame of reference as if it is inertial.

By the way, you didn't answer my question of what the notion of simultaneity means to you.

6. Originally Posted by KJW
If the clocks are synchronised at rest, then they will remain synchronised in that frame of reference even though they are in motion. The clocks won't remain synchronised in their own frame of reference,
OK, thanks for that. If the initially synchronized clocks become out of sync in their accelerated frames, what is their observed pattern of sync gaps within each line and around the corners?

A related situation occurs if evenly spaced clocks around the edge of a circular disc are synchronized at rest and then the disc rotates at a constant rate. Symmetry suggests the clocks remain synchronized, both for each clock and for an observer at rest. The disc can be made as large as we wish to locally approximate a straight line of moving clocks.

Originally Posted by KJW
By the way, you didn't answer my question of what the notion of simultaneity means to you.
Using the last situation as an example, I would say the clocks are simultaneous because they show the same time when seen from the disc's centre.

7. Originally Posted by Andrew?
Originally Posted by KJW
If the clocks are synchronised at rest, then they will remain synchronised in that frame of reference even though they are in motion. The clocks won't remain synchronised in their own frame of reference,
OK, thanks for that. If the initially synchronized clocks become out of sync in their accelerated frames, what is their observed pattern of sync gaps within each line and around the corners?
Without having done a detailed analysis, I would suggest that in the frame of any given clock, all the clocks ahead of the given clock are earlier than the given clock, and all the clocks behind the given clock are later than the given clock.

Originally Posted by Andrew?
A related situation occurs if evenly spaced clocks around the edge of a circular disc are synchronized at rest and then the disc rotates at a constant rate. Symmetry suggests the clocks remain synchronized, both for each clock and for an observer at rest.
Symmetry does indicate that all the clocks are equivalent, but equivalence doesn't imply that the clocks are synchronised in their own frames of reference. I suspect that you think there is a transitive relation being violated, but that is not the case because each clock has its own frame of reference.

Originally Posted by Andrew?
Originally Posted by KJW
By the way, you didn't answer my question of what the notion of simultaneity means to you.
Using the last situation as an example, I would say the clocks are simultaneous because they show the same time when seen from the disc's centre.
My question was about the notion of simultaneity in general. Your example requires an observer at the disc's centre and only applies to that observer. In principle, every possible observer has to have a rule for determining what events are simultaneous to themselves. This rule must be uniform for all possible observers, but its application by any particular observer only applies to that observer.

8. Originally Posted by Andrew?
Originally Posted by KJW
If the clocks are synchronised at rest, then they will remain synchronised in that frame of reference even though they are in motion. The clocks won't remain synchronised in their own frame of reference,
OK, thanks for that. If the initially synchronized clocks become out of sync in their accelerated frames, what is their observed pattern of sync gaps within each line and around the corners?
Without having done a detailed analysis, I would suggest that in the frame of any given clock, all the clocks ahead of the given clock are earlier than the given clock, and all the clocks behind the given clock are later than the given clock.

Originally Posted by Andrew?
A related situation occurs if evenly spaced clocks around the edge of a circular disc are synchronized at rest and then the disc rotates at a constant rate. Symmetry suggests the clocks remain synchronized, both for each clock and for an observer at rest.
Symmetry does indicate that all the clocks are equivalent, but equivalence doesn't imply that the clocks are synchronised in their own frames of reference. I suspect that you think there is a transitive relation being violated, but that is not the case because each clock has its own frame of reference.

Originally Posted by Andrew?
Originally Posted by KJW
By the way, you didn't answer my question of what the notion of simultaneity means to you.
Using the last situation as an example, I would say the clocks are simultaneous because they show the same time when seen from the disc's centre.
My question was about the notion of simultaneity in general. Your example requires an observer at the disc's centre and only applies to that observer. In principle, every possible observer has to have a rule for determining what events are simultaneous to themselves. This rule must be uniform for all possible observers, but its application by any particular observer only applies to that observer.

9. Originally Posted by KJW
Without having done a detailed analysis, I would suggest that in the frame of any given clock, all the clocks ahead of the given clock are earlier than the given clock, and all the clocks behind the given clock are later than the given clock.
It seems easier to consider a large spinning disc with clocks that are spaced evenly around the edge. The clocks are then synchronized using signals sent out from the centre of the disc. A stationary observer is at the edge of the disc looking at the moving clocks. A sync gap should be seen between the two nearest clocks on the disc. The same will be found between the next pair of clocks as they move round and soon.

What happens when the observer sees a given clock for the second time? Do the cumulative sync gaps mean that the clock is out of sync with its former self? Meanwhile there is another stationary observer above the centre of the disc for whom symmetry dictates there are no sync gaps. How can the clocks be both synchronized and not synchronized in the same inertial frame of reference?

The disc can be made as large as necessary, but there seems no reason why a moderate deviation from linear motion would eliminate the sync gaps. The gaps are meant to arise from the distance moved by a clock whilst light travels through the observer’s frame. Distance is still travelled whether this is along an arc or a straight line.

10. Originally Posted by Andrew?
Originally Posted by KJW
Without having done a detailed analysis, I would suggest that in the frame of any given clock, all the clocks ahead of the given clock are earlier than the given clock, and all the clocks behind the given clock are later than the given clock.
It seems easier to consider a large spinning disc with clocks that are spaced evenly around the edge. The clocks are then synchronized using signals sent out from the centre of the disc. A stationary observer is at the edge of the disc looking at the moving clocks. A sync gap should be seen between the two nearest clocks on the disc. The same will be found between the next pair of clocks as they move round and soon.

What happens when the observer sees a given clock for the second time? Do the cumulative sync gaps mean that the clock is out of sync with its former self? Meanwhile there is another stationary observer above the centre of the disc for whom symmetry dictates there are no sync gaps. How can the clocks be both synchronized and not synchronized in the same inertial frame of reference?

The disc can be made as large as necessary, but there seems no reason why a moderate deviation from linear motion would eliminate the sync gaps. The gaps are meant to arise from the distance moved by a clock whilst light travels through the observer’s frame. Distance is still travelled whether this is along an arc or a straight line.
Let's be clear about how simultaneity is defined. Consider an observer in an arbitrary frame of reference that emits a light-pulse at time t1. This light-pulse travels through a vacuum to a mirror at some arbitrary location which reflects the light-pulse back to the observer. The returning light-pulse travels through a vacuum to the observer who receives it at time t2. Then, in the frame of the observer, the event at which the light-pulse is reflected by the mirror is simultaneous to the time that is precisely halfway between t1 and t2.

In your example of the spinning disc with equally-spaced clocks that are synchronised by a signal sent from the centre of the disc, the clocks remain synchronised in the frame of the observer at the centre. This observer also sees the clocks as reading the same time because he is equidistant from all the clocks. For the stationary observer at the edge of the disc, because this observer is at rest relative to the inertial observer at the centre, the clocks are also synchronised in his frame of reference. However, he no longer sees the same time on each clock because the clocks are no longer the same distance from this observer. Thus, simultaneity is not about seeing the clocks at the same time, due to the time it takes light to travel from the clock to the observer.

The moving clocks are not synchronised in their own frame of reference. The clock immediately in front reads later and the clock immediately behind reads earlier (I think I mistakenly had this back-to-front in previous posts). Although this is true for all clocks around the disc, the "sync gaps" aren't cumulative because each clock has its frame of reference, and it is not true that in the frame of a single clock, each subsequent clock around the disc is ahead to the previous clock (leading to a clock being ahead of itself).

11. Originally Posted by KJW
The moving clocks are not synchronised in their own frame of reference. The clock immediately in front reads later and the clock immediately behind reads earlier (I think I mistakenly had this back-to-front in previous posts). Although this is true for all clocks around the disc, the "sync gaps" aren't cumulative because each clock has its frame of reference, and it is not true that in the frame of a single clock, each subsequent clock around the disc is ahead to the previous clock (leading to a clock being ahead of itself).
Yes but I am not talking about times within the frame of the moving clocks. I'm comparing the measurements of a stationary observer in the plane of the disc with a stationary one on the disc's axis.

12. Originally Posted by Andrew?
Originally Posted by KJW
The moving clocks are not synchronised in their own frame of reference. The clock immediately in front reads later and the clock immediately behind reads earlier (I think I mistakenly had this back-to-front in previous posts). Although this is true for all clocks around the disc, the "sync gaps" aren't cumulative because each clock has its frame of reference, and it is not true that in the frame of a single clock, each subsequent clock around the disc is ahead to the previous clock (leading to a clock being ahead of itself).
Yes but I am not talking about times within the frame of the moving clocks. I'm comparing the measurements of a stationary observer in the plane of the disc with a stationary one on the disc's axis.
Which I dealt with in the previous paragraph.

13. Originally Posted by KJW
However, he no longer sees the same time on each clock because the clocks are no longer the same distance from this observer. Thus, simultaneity is not about seeing the clocks at the same time, due to the time it takes light to travel from the clock to the observer.
I didn't see the significance of this as the observer is comparing the times on successive pairs of clocks as they move round, and the distance to each clock could be equalised. Anyway, if we agree that the clocks are synchronized for both stationary observers then, as you rightly say, the issue is the times seen by each moving clock.

Originally Posted by KJW
The moving clocks are not synchronised in their own frame of reference. The clock immediately in front reads later and the clock immediately behind reads earlier
It still seems to me that these clocks are synchronized in their own frame. Symmetry shows that each clock would receive the central timing signal at the same time, and as the speed of light is meant to be constant in their frame, each clock would see identical times on adjoining clocks.

If the hindmost clock in your last sentence is clock A then B-A is positive and so is C-B. From C's perspective clock B is similarly earlier, so C-B is positive as is D-C. This continues until Z-Y is positive and A-Z is positive. I don't see how this avoids a cumulative time gap between the frames of the second and first instances of A.

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