Why should A change its view on the length of B ship during relative motion?

[Fredrik]

Let us make it simple. All clocks along my ship are synched. I accelerate my ship. A good time later (when the physical effect of acceleration has vanished and my ship is inertial again) I check if my clocks are still synched (through Einstein convention). Are they or aren't they?

My answer to this is what I said in #23. What you need to know to understand #23 is the definition of proper time, and that a clock measures the proper time of the curve in spacetime that represents its motion. The latter is not a derived result. It's one of the axioms that tell us how to interpret the mathematics of SR as predictions about the results of experiments.

Also see my posts in Matheinste's thread for comments (but not a definitive answer) about the synchronization in the case where the ship is accelerated for a while and eventually brought back to its original velocity.

 

[DrGreg]

All clocks along my ship are synched. I accelerate my ship. A good time later (when the physical effect of acceleration has vanished and my ship is inertial again) I check if my clocks are still synched (through Einstein convention). Are they or aren't they?

No. Here's a diagrammatic explanation.

At the bottom of the picture the ship is stationary relative to the observer and the red & blue clocks are synchronised. Half way up the diagram the ship is coasting at constant velocity relative to the observer and the clocks are out of sync in their own rest frame (follow the green lines). They're also out of sync in the starting frame (follow horizontal lines I haven't drawn).

If the ship subsequently decelerates in an appropriate way symmetrical to the original acceleration then, by the top of the diagram, the clocks are back in sync again. However this isn't true for arbitrary acceleration/deceleration where the symmetry is broken. (For example, if you braked twice as hard as you accelerated, that would break the symmetry and sync would not be restored.) All of the above assumes "Born rigid" acceleration -- the answer could be different otherwise.

Remember that whether two clocks are synchronised depends on the cumulative history of what has happened to those two clocks since they were last synchronised. Whereas the relative ticking rates of two clocks depends only on what is happening right now ("now" as defined by the observer).

 

[Ich]

That's a very good illustration, DrGreg!

@Saw: the green lines mark simultaneity in the comoving frame, while the numbers show proper time, i.e. the reading of the clocks. During acceleration, you see that the right clock "runs twice as fast" as the left clock. This is derived by SR alone, but is nothing else than "gravitational time dilatation".

 

[Saw]

That's a very good illustration, DrGreg!

And very good accompanying explanations! Thanks to all.

I will check my understanding: Let us call A the original rest frame of my ship and B the new frame I’ve changed to due to the acceleration. After the acceleration has finished, the proper time of my blue clock is 4s. The green slanting line is my simultaneity line in B. It is slanting because the picture is drawn in A frame. If the picture had been drawn in the CS of B frame, the green line would be horizontal and I would expect that the red clock read the same time as the blue clock, if synched according to Einstein convention. As it does not (it reads 4s more, i.e. 8s), I find that my clocks need what I called at the beginning of the thread “re-calibration” = I have to carry out a new synch operation. Right?

On top of that you give me plenty of interesting information and precise calculations about a particular type of acceleration model that has particular effects (Born rigid acceleration), but we could say that the simple question has a simple answer: whenever I accelerate, regardless how, the clocks at rest in my frame will go out-of-synch and I’ll have to re-synch them. Is this general conclusion correct?

 

[Ich]

whenever I accelerate, regardless how, the clocks at rest in my frame will go out-of-synch and I’ll have to re-synch them. Is this general conclusion correct?

No, see https://www.physicsforums.com/showpost.php?p=2091879&postcount=28".
You can always make a clock run slower like in the twin paradox. Doing this, you can restore synchronisatzion.

 

[DrGreg]

whenever I accelerate, regardless how, the clocks at rest in my frame will go out-of-synch and I’ll have to re-synch them. Is this general conclusion correct?

No, see https://www.physicsforums.com/showpost.php?p=2091879&postcount=28".
You can always make a clock run slower like in the twin paradox. Doing this, you can restore synchronisatzion.

Just to clarify, Ich is disagreeing with the phrase "regardless how".

As a general rule the answer is yes, but there are special cases where the answer is no.

My example in post #37 is an example of "no", in the cumulative effect for the whole journey from bottom to top. But if you break down my journey into smaller parts the answer is "yes" for each part, and I think I'm right in saying that will always be the case for any journey where the clocks are "rigidly attached" to each other. This loss of synchronisation is really just another name for "gravitational time dilation" -- an accelerating ship in empty space is (locally) equivalent to a ship hovering stationary relative to a gravitating planet.

(If the clocks are not "rigidly attached" to each other, then all bets are off and anything could happen -- it depends on the specific motion in question. I put the phrase "rigidly attached" in quotation marks because there are no rigid substances, and "Born rigid" motion could be achieved in practice only by applying pre-arranged forces to each clock. Simply pushing one end of the ship wouldn't maintain rigidity. Nevertheless, when the push does not change suddenly (i.e. the jerk is small) this would be a reasonable approximation of rigidity.)

 

[Saw]

My example in post #37 is an example of "no", in the cumulative effect for the whole journey from bottom to top. But if you break down my journey into smaller parts the answer is "yes" for each part

"Each smaller part" in your example, I understand, is (a) inertial motion, (b) progressive POSITIVE acceleration, (c) inertial motion with a different velocity, (d) progressive NEGATIVE acceleration and (e) inertial motion with the same initial velocity as in (a).

Thus I understand that a SINGLE acceleration would always make clocks out-of-sync. Loss of synch is always present, although the original simultaneity line can be “restored” (after being lost) if a SECOND acceleration of opposite sign is applied. But even that is not sufficient: this restoration would require that the two clocks were attached to a thoroughly rigid rod, which does not exist in practice, although a good approximation can be obtained through either (i) applying different forces evenly along all parts of the rod or (ii) very gentle acceleration following Born rigid model. More or less ok now?

 

[DrGreg]

"Each smaller part" in your example, I understand, is (a) inertial motion, (b) progressive POSITIVE acceleration, (c) inertial motion with a different velocity, (d) progressive NEGATIVE acceleration and (e) inertial motion with the same initial velocity as in (a).

Thus I understand that a SINGLE acceleration would always make clocks out-of-sync. Loss of synch is always present, although the original simultaneity line can be “restored” (after being lost) if a SECOND acceleration of opposite sign is applied. But even that is not sufficient: this restoration would require that the two clocks were attached to a thoroughly rigid rod, which does not exist in practice, although a good approximation can be obtained through either (i) applying different forces evenly along all parts of the rod or (ii) very gentle acceleration following Born rigid model. More or less ok now?

Pretty much, yes.

Except that my comment about rigidity wasn't specifically about the restoration of sync, it was about the whole analysis. If the two clocks could move independently of each other, then the problem would be ill-defined and the two clocks wouldn't even agree with each other on a definition of Einstein-simultaneity. Nevertheless, in general sync would be lost. For non-inertial motion, retention of sync is the exception, not the rule. You can assume that sync is always going to be lost unless there's a good reason why not.

The condition for restoration of sync is really symmetry. If you rotate my spacetime diagram through 180 degrees (i.e. reverse both space and time) it looks identical, and that's what restores the sync.

I haven't worked it out but it might perhaps be possible to come up with some motion of a set of disconnected clocks such that, from the point of view of one clock, the other clocks remained synchronised during acceleration. But the other clocks would disagree, and the clocks wouldn't be a fixed distance apart.