Clock synchronising by clock transport?

• B
• Peter Strohmayer
Ibix said:
You need an initial velocity, as I keep saying.
If you want to calculate the proper time τ of an accelerated object from a coordinate time T, yes.

#30 deals with the opposite case ("any inertial system"): the absolute values of the proper time τ between two events and the acceleration profile are given. - The initial velocity is only a relative quantity that depends on the initial system whose time T is to be calculated.

PeterDonis said:
Really? Says who?

There are lots of practical cases that do not meet your specifications.
Ibix said:
better to junk all but one of the array of clocks
Yes, one transported clock is sufficient for the synchronization process.

Based on this, I wanted to understand the acceleration of extended bodies (lIbix #27 "they drift out of sync and back into sync") as the most important case of acceleration occurring in nature.

But I don't want to test your patience anymore. Thank you @Ibix for your patient response to my arguments.

Peter Strohmayer said:
#30 deals with the opposite case ("any inertial system"): the absolute values of the proper time τ between two events and the acceleration profile are given. - The initial velocity is only a relative quantity that depends on the initial system whose time T is to be calculated.
That doesn't change that you need the arc length, not the path curvature history. You'll need to integrate twice. One constant is the clock setting at the start of the experiment, which you are free to set to zero. The other is an initial velocity with respect to whatever frame you are using to record the acceleration history. (You are, of course, free to record acceleration history as a function of proper time - but then you don't need to calculate proper time, you recorded it. And you can't synchronise clocks without a synchronised convention.)
Peter Strohmayer said:
Based on this, I wanted to understand the acceleration of extended bodies (lIbix #27 "they drift out of sync and back into sync") as the most important case of acceleration occurring in nature.
Note that, as stated upthread, extended bodies do not typically behave like a set of Bell's spaceships. They typically behave more like Rindler observers, plus transient effects when the acceleration changes. So clocks on a ruler will typically desynchronise in all frames.

PeterDonis
Ibix said:
you don't need to calculate proper time, you recorded it
Of course. From the point of view of the transported clock, I measured the proper acceleration as a function of the proper time τ between two events (acceleration profile). The choice of an inertial system gives a world line and the coordinate time T between these events.
Ibix said:
Note that, as stated upthread, extended bodies do not typically behave like a set of Bell's spaceships. They typically behave more like Rindler observers, plus transient effects when the acceleration changes. So clocks on a ruler will typically desynchronise in all frames.
I completely agree. But regardless of whether the parts of an accelerated body behave like Bell's spaceships or like Rindler observers, the clocks distributed over a solid body, which were synchronized in the unaccelerated state, will essentially maintain their synchronization when they return to their original system.
With an undamaged body, it does not matter, if the aging of the individual atoms (= clocks) shows slight differences because they have traveled slightly different world lines (in the relation between the amount of proper time τ and the coordinate time T). The age of an undamaged rigid body can be regarded as homogeneous.

For the acceleration of an extended body, the choice of the initial condition of the above acceleration profile is limited. The proper acceleration must be zero for both events. So a uniform time can be assigned to the transported clocks (the solid body). The complicated desynchronization phases during acceleration can be ignored.

Peter Strohmayer said:
The age of an undamaged rigid body can be regarded as homogeneous.
You have already been told multiple times that this is not correct. This thread will remain closed

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