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

[Saw]

I've seen other posts were made while I was drafting mine. Thanks. I apologize if my latest post does not consider your comments, which I will study.

 

[Fredrik]

As Fredrik points out, the acceleration takes time to do its job. If the acting force acts at the rear, it will take time to be transmitted to the front and vice versa. That is, certainly, a complication we would face in real life.

I don't think I said anything about that in this thread, but I have talked about it in other threads. (What I was talking about earlier and in Matheinste's thread is how the rear will accelerate faster than the front even if the engine pushes so gently that the shape of the rocket won't be significantly distorted because of the finite propagation speed of the force from the engine).

Furthermore, the clocks, material as they are, will be shaken and their mechanism distorted in an unpredictable way. But can’t we avoid those complications, for the sake a clean and clear analysis?

Yes, let's do that.

One is the purely theoretical one, which is quite straightforward: I simply stipulate that the acceleration is instantaneous and the two ends of the rod accelerate at the same time, as measured in the original rest frame, without any mechanical distortion.

You need to be very careful when you make an assumption like this. Note that your specifications make sure that both endpoints are at the same distance from each other at all times in the original rest frame. So the rod is the same length when its velocity (in the original rest frame) is v as when it's 0, even though it's Lorentz contracted by a factor of gamma in the first case. This means that you have forcefully stretched the rod to a longer rest length.

Things won't be any better if that the boost is simultaneous in the other frame. In that case, a similar argument shows that the rod is getting forcefully compressed.

Another way to deal with this is to assume that the rod is doing Born rigid motion. A solid object will accelerate in an approximately Born rigid way if the push/pull is gentle enough. That's the assumption I've been making in this thread and in Matheinste's thread.

 

[matheinste]

Hello Fredrik.

I think the way you desribe clocks at rest in an inertial frame is much better and less likely to lead to misunderstandings and I will try to get out of my ingrained bad habits. I always hope that anyone at any time will put me right if I make mistakes, I do not consider constructive criticism to be nitpicking.

I still have a lot of thinking to do about accelerated clocks.

Thanks Matheinste.

 

[DrGreg]

The attached spacetime diagram might cast some light on the situation.

It shows 3 yellow rods, each the same length in their own co-moving inertial frame. The one on the left is stationary relative to observer. The one in the middle is moving at constant velocity relative to observer. The one on the right is accelerating in such a way to keep the rod's length constant in the co-moving inertial frame. (This is called "Born rigid acceleration".)

In all three cases:
- observer time is vertically up
- observer distance is horizontal
- the red line is the front clock and the red dots represent ticks of the clock, at one-second intervals
- the blue line is the back clock and the blue dots represent ticks of the clock, at one-second intervals
- the green lines represent simultaneity relative to the comoving inertial frame; the clocks are synchronised relative to the comoving inertial frame whenever they show the same value at both ends of the same green line

This is a scale diagram plotted mathematically, so horizontal distances and vertical times are accurate. (Make sure you view the picture at 100% scale, or else you might not be able to see the dots clearly.)

Spend some time studying these diagrams and making sense of them and this may answer your questions. (I'm not going to answer any specific question because the question keeps changing from one post to the next and I've lost the plot.)

 

[Saw]

(I'm not going to answer any specific question because the question keeps changing from one post to the next and I've lost the plot.)

My fault. 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?

 

[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[/color] 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.