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

In summary, two MM-devices A and B are at rest in the same frame with equal arm lengths of 1 light-second. B is accelerated away along the positive x-axis at a relative velocity of 0.5 c. According to Special Relativity, B's milestone and ship length remain unchanged in its own frame, but they appear shorter in A's frame due to the Lorentz transformation. This is not a physical change, but rather a frame-dependent measurement. The same principle applies to the synchronization of clocks, where the Einstein convention is used for A and B to have synchronized clocks when at rest, but after acceleration, B's clock appears desynchronized in its own frame.
  • #1
Saw
Gold Member
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Two MM-devices A and B (=two perpendicular arms x and y being the paths for light signals) are at rest in the same frame. Both arms are of equal length, say, 1 light-second long. B is accelerated away along the positive x-axis and reaches relative velocity, say, 0.5 c. I asked: will B have to move the milestone marking 1 light-second to another location, according to SR? The answer was negative. B's milestone is still where it was before, in B frame B ship is still measured as 1 light-second long, although in A frame it starts being measured as 1/gamma = 0.86654 light-second long. Reciprocally, in B frame A ship starts being measured as 0.86654 light-second long.

I have difficulties in understanding why. Why should A change its view on the length of B ship? I only see two possible explanations:

(i) the "Lorentzian" view: B's length has physically changed or
(ii) what we could call a "conventional" approach: B has "recalibrated" its measurement instrument (the MM device) in the x arm so that the two light signals meet after their respective round trips.

In the absence of any of these "physical" changes (one automatic and given by nature, the other conventional and carried out by a human hand), I insist: why should A start measuring a change in the length of B?

The same applies to synchronisation of clocks. When they were at rest with each other, A and B synchronised the clocks located at both ends of their ships using the Einstein convention. After some time, all their A and B clocks mark 10 s. At this instant, B is accelerated away along the positive X axis. I would also expect that B's clock located at the right edge becomes desynchronised in B frame, unless "something" happens: nature's or human intervention...
 
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  • #2
Saw said:
Two MM-devices A and B (=two perpendicular arms x and y being the paths for light signals) are at rest in the same frame. … B is accelerated away along the positive x-axis and reaches relative velocity, say, 0.5 c.

I have difficulties in understanding why. Why should A change its view on the length of B ship? I only see two possible explanations:

(i) the "Lorentzian" view: B's length has physically changed or
(ii) what we could call a "conventional" approach: B has "recalibrated" its measurement instrument (the MM device) in the x arm so that the two light signals meet after their respective round trips.

In the absence of any of these "physical" changes (one automatic and given by nature, the other conventional and carried out by a human hand), I insist: why should A start measuring a change in the length of B?

Hi Saw! :smile:

But the MM apparatus relies on the two arms staying together …

you have one of the arms disappearing at c/2.

And if the other arm goes with B also, then the stationary observer sees the light along that arm follow a zigzag path, while the light along the 'B' arm doesn't bother to return to where it started.
 
  • #3
Saw said:
Two MM-devices A and B (=two perpendicular arms x and y being the paths for light signals) are at rest in the same frame. Both arms are of equal length, say, 1 light-second long. B is accelerated away along the positive x-axis and reaches relative velocity, say, 0.5 c. I asked: will B have to move the milestone marking 1 light-second to another location, according to SR? The answer was negative. B's milestone is still where it was before, in B frame B ship is still measured as 1 light-second long, although in A frame it starts being measured as 1/gamma = 0.86654 light-second long. Reciprocally, in B frame A ship starts being measured as 0.86654 light-second long.

I have difficulties in understanding why. Why should A change its view on the length of B ship? I only see two possible explanations:

(i) the "Lorentzian" view: B's length has physically changed or
(ii) what we could call a "conventional" approach: B has "recalibrated" its measurement instrument (the MM device) in the x arm so that the two light signals meet after their respective round trips.

In the absence of any of these "physical" changes (one automatic and given by nature, the other conventional and carried out by a human hand), I insist: why should A start measuring a change in the length of B?

The same applies to synchronisation of clocks. When they were at rest with each other, A and B synchronised the clocks located at both ends of their ships using the Einstein convention. After some time, all their A and B clocks mark 10 s. At this instant, B is accelerated away along the positive X axis. I would also expect that B's clock located at the right edge becomes desynchronised in B frame, unless "something" happens: nature's or human intervention...

The fact that something is longer as measured from one reference frame than another doesn't mean that "something" happened to it, it just means that length is frame dependent. Physics has had frame dependent quantities since Newton. For example, kinetic energy is frame dependent. The fact that an object's kinetic energy will increase if I increase my relative velocity doesn't mean that "something" happened to the object. Ditto for clocks.

Al
 
  • #4
tiny-tim said:
But the MM apparatus relies on the two arms staying together … you have one of the arms disappearing at c/2

And if the other arm goes with B also, then the stationary observer sees the light along that arm follow a zigzag path, while the light along the 'B' arm doesn't bother to return to where it started.

Thanks, tiny-tim. I wasn’t precise enough. Of course, I assume that the two arms of each apparatus stay together: A’s y and x arms stay in A frame and B’s x and y arms are accelerated, thus becoming B frame.

Al68 said:
The fact that something is longer as measured from one reference frame than another doesn't mean that "something" happened to it, it just means that length is frame dependent. Physics has had frame dependent quantities since Newton. For example, kinetic energy is frame dependent. The fact that an object's kinetic energy will increase if I increase my relative velocity doesn't mean that "something" happened to the object.

AI68, thanks, too. I see that is the point. It is just that I find it hard to understand. That is why my mind looked for the “conventional” explanation. I do not mean I am right in not seeing it. Just that I do not see it, so far…

Anyhow, forget about my above speculations.

So I assume that, if two MM apparatuses A and B are at rest with each other and working properly (light signals sent in both directions along arms of equal length return to the origins at the same time) and B is accelerated away, B will not have to recalibrate the length of any arm and the apparatus will keep working as originally. Right?

Al68 said:
Ditto for clocks.

Does it mean that, if B had an array of clocks along, say, the x arm, which clocks had been synchronized when B was at rest with A, B does not have to resynchronize them after the acceleration? Is it so? I had thought that at least this “recalibration” would have to be done.
 
  • #5
I would like to retake this thread, in case someone can help, because the doubt still bothers me.

Let us forget about the rest and concentrate on this issue: you are on a ship and have synchronized distant clocks of your frame with yours; then you are accelerated; don't you have to resynchronize your distant clocks by means of light signals, following the Einstein convention?

Take for example a twin paradox scenario. In the standard solution, it is said that, for the twin who left the Earth and returned, there are two frames, one for the outward trip and another for the return trip. These two frames have different simultaneity lines. This change of simultaneity lines..., doesn't it require a new synchronization?

For example:

Relative v between ships A and B is 0.5 c. For A, B moves left-to-right; for B, A moves right-to-left. A1 and B1 (observers at the left edges of the ships) synchronize their clocks to zero when they meet and synchronize clocks in their respective frames through light signals.
In A frame, A’s ship is 1.732 ls long, B’s ship is half of that, 0.866 ls long.
In B frame, B’s ship is 1 ls long, A’s ship is 1.5 ls long.
When B1 reaches Am (mid-point of A), B ship turns around.
At that event, Am’s clock reads 1.732 s, while B1’s clock reads 1.5 s. No discrepancy with regard to this.
What about clocks distant to that event, like B2 or A1?
There is discrepancy:
In A frame, at the time of the turnaround, B2, whose clock is unsynchronised and reads 1s, is by A2.
In B frame, B2 has already been passed by A2, whose clock reads 2.165 s and is already by B3/4.
That is confusing enough.
Anyhow, let us take A’s version. As I said, coinciding with the turnaround, B2 was reading B1’s time at the turnaround event (1.5 s) minus (since B was receding away) vx/c^2 = 1.5s – 0.5s = 1s. It seems that, from the turnaround onwards, since now B is approaching, the time of B2 should instantaneously “jump” from minus to plus vx/c^2 = 1.5s + 0.5 s = 2s…

How do you handle the twin paradox for the distant clocks? Not through a new synchronization?
 
  • #6
Saw said:
you are on a ship and have synchronized distant clocks of your frame with yours; then you are accelerated; don't you have to resynchronize your distant clocks by means of light signals, following the Einstein convention?
Why would you even bother? The distant clock isn't going to stay synchronized with yours anyway. If you want to know what the distant clock displays in your "now" (in the new co-moving inertial frame), just use a Lorentz transformation to calculate it.
 
  • #7
Fredrik said:
Why would you even bother? The distant clock isn't going to stay synchronized with yours anyway. If you want to know what the distant clock displays in your "now" (in the new co-moving inertial frame), just use a Lorentz transformation to calculate it.

Fredrik, I don´t follow you. Why should I bother? Because they are my clocks, they are in my frame, comoving with me and I may need them for the usual purposes: to measure the speed of passing by objects, to ring my wife and remind her it's time to go and fetch the children..., whatever. I should not need a LT for that: I would use a LT if I transformed from the time reading of a passing frame, but that would be quite cumbersome. I prefer my clocks and for this purpose I need to know if, due to my recent acceleration, they have gone out-of-sync in this new context or, on the contrary, they have automatically adjusted to the new situation...
 
  • #8
Learn about Rindler coordinates. I am not 100% sure, but I think that your clocks will wind up automatically synchronized in the new frame. But you can derive it yourself from Rindler coordinates.
 
  • #9
Hello all.

Sorry to butt in but this may be relevant. I asked a question in the forum some time ago but unfortunately I cannot now find it and so cannot give the exact wording but the idea was: Given two clocks, synchronized by the standard procedure, mounted on a ship, do they need to be resynchronized if, after a period of acceleration they are returned to an inertial frame. The answer was they do not need resynching.

Later I read about the equivalence principle which suggested that clocks in a gravitational field run at different rates and so presumably become out of synch, depending on the gravitational potential, and that as gravitation is indistinguishable from acceleration the same is true for accelerated clocks. So accelerated clocks go out of synch.

Assuming both parts are correct and putting them together, would it be correct or not to say that clocks go out of synch during acceleration but return to synch when returned to an inertial frame?

I am also assuming that the same applies to the global acceleration of a frame to which saw seems to be referring. Would that be correct?

Matheinste
 
  • #10
Saw said:
Fredrik, I don´t follow you. Why should I bother? Because they are my clocks, they are in my frame, comoving with me
They are? You didn't say that. If you meant that they are both in the rocket, e.g. one in the back and one in the front, then I think DaleSpam is right about the Rindler coordinates.
 
  • #11
matheinste said:
I asked a question in the forum some time ago but unfortunately I cannot now find it
Fortunately you suck at typing :wink:, so it was easy to find. It's the only thread with "umsynchronization" in the title :biggrin:.

matheinste said:
the idea was: Given two clocks, synchronized by the standard procedure, mounted on a ship, do they need to be resynchronized if, after a period of acceleration they are returned to an inertial frame. The answer was they do not need resynching.
I think I interpreted your question that time as "if they're at rest in some inertial frame, and then accelerated for a while, and finally brought back to rest in that same frame..." And even in that case, I wasn't sure if it's always true that they will be back in synch. (It would take some work to check it for an arbitrary acceleration, and I haven't done it).

matheinste said:
Later I read about the equivalence principle which suggested that clocks in a gravitational field run at different rates and so presumably become out of synch, depending on the gravitational potential, and that as gravitation is indistinguishable from acceleration the same is true for accelerated clocks. So accelerated clocks go out of synch.
It's much easier to do the math the other way round. Do it in SR and use the EP to carry the result over to GR.

matheinste said:
Assuming both parts are correct and putting them together, would it be correct or not to say that clocks go out of synch during acceleration but return to synch when returned to an inertial frame?
I haven't done the math on this one.

matheinste said:
I am also assuming that the same applies to the global acceleration of a frame to which saw seems to be referring. Would that be correct?
I don't think I understand this question.
 
  • #12
If you accelerate clocks (e.g. at the tip and at the tail of a rocket), they get out of synch and stay so.
 
  • #13
Hello Ich,

Quote:-
---If you accelerate clocks (e.g. at the tip and at the tail of a rocket), they get out of synch and stay so. -----

Does this mean that they appear out of synch to a comoving observer to who they appeared to be in synch before the acceleration.

Matheinste
 
  • #14
Does this mean that they appear out of synch to a comoving observer to who they appeared to be in synch before the acceleration.
Not sure I understand correctly, so let me re-state:
If two clocks are in synch, as measured in their common rest frame, and then accelerate to an different common v, they are no longer in synch as measured in their new common rest frame.
If they had both exactly the same acceleration profile, they still would be in synch in the original frame, but not in the new one.
 
  • #15
Thanks Ich.

So i can say that the clocks as you described them will be out of synch as measured by an observer in their new common rest frame, but in synch if measured by an observer in their original common rest frame. Assuming a common acceleration profile.

Matheinste
 
  • #16
Exactly.
 
  • #17
Hello Ich.

Thats clear then.

Now let me be an observer in an inertial frame. When i see clocks in another inertial frame moving relative to me, i would normally measure them out of synch with respect to each other and running slow with respect to my clocks. So does this not apply if i view clocks moving relative to me that were originally synched in the frame from which i am measuring.

Matheinste.
 
  • #18
They would still run slow, but in synch.
I you made them artificially run faster, you can introduce a common coordinate time that clocks in both states of motion show. A similar procedure is used to define http://en.wikipedia.org/wiki/Terrestrial_Time" .
 
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  • #19
Hello Ich.

So in the case of clocks originally synched in a common frame with my clocks, they would, when moving inertially relative to my frame, of course run slow but would, as measured by me, be in synch with each other in their common frame, but not in synch with my clocks because they are running at a different rate.

Matheinste.
 
  • #20
...but not in synch with my clocks because they are running at a different rate.
That's correct in a sense, but might cause some cofusion.
"In synch" would usually mean that if I measure two events to be simultaneous, clocks in another frame would measure them to be simultaneous, too. And that is the case in your setup.
But this is only semantics, where different people have different views, and it might well be possible that my understanding differs from some mainstream definition that I'm unaware of.
 
  • #21
Ich said:
If they had both exactly the same acceleration profile, they still would be in synch in the original frame, but not in the new one.
This is definitely correct, but it should be mentioned (again) that two clocks at the front and rear of an accelerating rocket can't have the same acceleration profile, since the rear accelerates faster than the front.

matheinste said:
Now let me be an observer in an inertial frame. When i see clocks in another inertial frame moving relative to me, i would normally measure them out of synch with respect to each other and running slow with respect to my clocks. So does this not apply if i view clocks moving relative to me that were originally synched in the frame from which i am measuring.
A clock isn't "in" an inertial frame. Its motion is a curve in Minkowski space and every point on that curve has coordinates in all inertial frames. If your question means what I think it does, then the answer is what Ich said in the quote above (in this post).
 
  • #22
So can we conclude, now that the facts of the question are clear, that the clocks do have to be resynched? No objection?

To avoid any doubt, let me refine even more the question. I think we should omit any consideration about the effects of the acceleration, in terms of GR or otherwise. Let us assume that those effects are negligible or simply, for discussion purposes, inexistent. The only effect of the acceleration is that ship B has suffered a change of relative velocity wrt to A: initially it was at rest with A, now it moves relative to it. Thus it is a "new frame". I think this way the discussion is simpler and cleaner. Of course, we have thus established an ideal model that may never exist in real life, but it is just a question of introducing afterwards the applicable corrections in view of real life conditions.
 
  • #23
DaleSpam said:
Learn about Rindler coordinates. I am not 100% sure, but I think that your clocks will wind up automatically synchronized in the new frame. But you can derive it yourself from Rindler coordinates.
I have changed my mind about this being correct. The rear accelerates faster than the front, and a faster acceleration means a bigger dx contribution in the calculation of the proper time integral. The dx contribution is negative, so at two events that are simultaneous in the original rest frame, the clock in the rear is displaying an earlier time than the clock in the front. Now consider the same event at the rear, and the event at the front that's simultaneous with it in the inertial frame that's co-moving with the rear. This is an event that occurs later in the original rest frame (than the event at the front that we considered before), so the clock in the front will be even more ahead of the clock in the rear when we compare these two events.
 
  • #24
Hello Fredrik.

Quote:-
---A clock isn't "in" an inertial frame. Its motion is a curve in Minkowski space and every point on that curve has coordinates in all inertial frames.

I don't quite understand the first sentence. If have always assumed, possibly incorrectly, that if an object is moving inertially there is an inertial frame "in" which it is at rest. IF this is so why can this not apply to a clock?

Matheinste.
 
  • #26
For clocks in rectilinearly moving frame of reference value unsynchronization does not depend on the law of motion frame of reference, but only to the magnitude of its velocity at a given time (formules (4.6), (4.9)).
 
  • #27
Thanks Omega i'll take a look at that.

Now the original poster, saw, is back I'll leave the thread to get back to its original purpose but will watch with interest.

Thanks Matheinste.
 
  • #28
Saw said:
I think we should omit any consideration about the effects of the acceleration, in terms of GR or otherwise. Let us assume that those effects are negligible or simply, for discussion purposes, inexistent.
Those effects are nothing but a different representation of the effect that a change of relative velocity has. You don't have to use this representation, but they certainly must be existent. But you do nothing wrong if you make all calculations in an inertial frame.
There is one fine point you should consider: you can tweak the acceleration profile, or, in the case of instantaneous acceleration, the time when it occurs, such that the clocks stay in synch in their eventual rest frame.
It's just that without fine tuning, in the general case, they will end up out of synch.

To clarify: This includes Rindler-like acceleration as well. There, the same interval of coordinate time means different proper time intervals depending on the clock's position. AKA "gravitational time dilatation".
 
  • #29
matheinste said:
---A clock isn't "in" an inertial frame. Its motion is a curve in Minkowski space and every point on that curve has coordinates in all inertial frames.

I don't quite understand the first sentence. If have always assumed, possibly incorrectly, that if an object is moving inertially there is an inertial frame "in" which it is at rest. IF this is so why can this not apply to a clock?
Your assumption is correct. I was just nitpicking the language you used. :smile: For example, when you said "clocks in another inertial frame", you should have said "clocks that are at rest in another inertial frame". Maybe I shouldn't be nitpicking your statements, but I seem to have developed some sort of allergy against statements like that after lots of discussions with people who don't understand SR nearly as well as you do. They often talk about one event that "happens in one frame", and some other event that "happens in another frame". I think it's important for those of us who understand the basics to use correct language, because if we don't, how are we going to convince those who don't understand this stuff to express themselves carefully?
 
  • #30
matheinste said:
Now the original poster, saw, is back I'll leave the thread to get back to its original purpose but will watch with interest.

Well, my question, as finally simplified, is not different from what you, matheinste, have asked, except for a detail: as noted above, I would like to leave out, if this makes any sense, the implications of the acceleration, other than the fact that (as an ultimate consequence of it) the frame in question (B, I call it) is moving relative to its original rest frame (A).

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. 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?

There are two approaches for this purpose:

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. This is an ideal “model”, but that is what SR is after all. I’ve simply added a single and tiny complication: frame B has shifted from rest to relative motion without any further consequence.

That should suffice, although if we are not convinced with such bold theoretical approach, we can look for an experimental display that matches our simplified approach. This may even be a way to “solve” the problem.

For example, frame B was already inertially moving wrt to A. Please consider the numbers that I stated above, when I retook the thread.

When A1, left edge of A, meets with B1, left edge of B, A1 transfers its time to B1 through a light signal (they are so close to each other that the trip time of the signal is negligible). Just by sheer chance, it happens to be the same reading that B1’s clock marked = 12 am! Hence if A1 had physically jumped to B frame in order to “marry” B1, it would not need to change its time.

When Am, mid-point of A and 0.866 ls from A1 in A frame, meets with B2, right edge of B and 1 ls away from B1 in B frame, they do the same operation: Am’s time is communicated to B2.

Here there is forcefully a discrepancy. If the clocks of A1 and B1 showed the same time, then it’s impossible that the clocks of Am and B2 show the same time. The two events (meeting of A1-B1, on the one hand, and meeting of Am-B2, on the other hand) happen at different times for each observer.

In particular, B2, when meeting Am, will read 12 am – 0.5 s.

Thus, from the perspective of A frame, if Am had jumped to B frame in order to “marry” B2, it would have to change its reading in order to adapt to B2’s time, which is 12 am -0.5s.

Certainly, B would claim that the two visiting clocks, A1 and Am, must jump simultaneously to B frame, if things are to be done properly. In particular, if we fix as acceptable the “marriage” of A1 and B1, then Am must jump to B “at the same time” and this same time for B means something else, a later time: it means that Am would already be by B3/4 and its clock would be showing 12 am + 0.433 s. Therefore, from B’s perspective, Am would have to correct its time into 12 am (by cancelling the additional +0.433 s), although Am would thus jump to a different place of A frame (namely B3/4, instead of B2)...

And so what? I confess I have no idea…

OMEGA, thanks, but when I click on your link I get a page in Russian and see no way to switch to English...
 
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  • #31
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.
 
  • #32
Saw said:
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. :smile: (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).

Saw said:
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.

Saw said:
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.
 
  • #33
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.
 
  • #34
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.)
 

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  • #35
DrGreg said:
(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?
 

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