# Maintaining the appearance of fixed distance in spite of length contraction?

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So I got completely confused in this other thread...

Suppose an observer moves along the x axis in some known way at speeds where length contraction is significant.

Is it possible to synchronize the movement of another object on the x axis, at some distance from the observer, such that the object always appears the same distance away according to the observer?

For example, suppose the observer and object are relatively at rest separated by 1 light second, and then at a predetermined time the observer moves in the direction of the object at v=0.866c for one second (with negligible acceleration time) and then stops. Is it possible to move the object in such a way that it always appears 1 light second away according to the observer, despite any length contraction that the observer might experience?

Edited by md65536
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It will always be 1 light-second away if it stays in the same reference frame, i.e. it is at rest with respect to the observer.

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It will always be 1 light-second away if it stays in the same reference frame, i.e. it is at rest with respect to the observer.

Is this true even if the frame isn't an inertial frame?

If the observer and object change velocity (relative to some other frame) at "the same time", even if separated by say a light year, does their relative simultaneity remain fixed as long as they don't move relative to each other? And thus we can talk about "simultaneously" for the entire frame, no matter how big it is?

If the object and observer moved independently, but in sync... Suppose the object was allowed to choose whether to move along with the observer's reference frame, or stay stationary relative to the other frame.

If the observer moved at 0.866 c for 1 second and then stopped, but the object was 1 LY away, the observer would not be able to tell whether or not the object moved, until about a year later. Does this mean that any possible visible length-contraction effects would appear one year later?

I assume the answer is "no", and that length contraction would be apparent immediately. In which case, for the object to be synchronized with the observer, it would actually have to start its movement early... (by half a year in this case, with gamma=2, I think).

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It will always be 1 light-second away if it stays in the same reference frame, i.e. it is at rest with respect to the observer.

I don't think they can, even in the theory, coordinate to maintain the distance and stay at rest wrt each other at all times during accelerations.

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I don't think they can, even in the theory, coordinate to maintain the distance and stay at rest wrt each other at all times during accelerations.

If they synchronize clocks and perform the same actions, I don't see why not.

Is this true even if the frame isn't an inertial frame?

I'm not sure if it's generally true in an accelerating frame.

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If they synchronize clocks and perform the same actions, I don't see why not.

If they synchronized clocks and performed the same actions the distance between them would have changed. (unless they are back to the initial rest frame)

Edited by J.C.MacSwell
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If they synchronized clocks and performed the same actions the distance between them would have changed.

If we are neglecting the acceleration time as per the OP, why?

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If we are neglecting the acceleration time as per the OP, why?

Because of length contraction. Until they are back to the initial rest frame the distance in their frame of rest would have changed. An extra displacement of one is required to maintain the separation distance as measured in their current rest frame. So they cannot be at rest with respect to each other and maintain that same distance at all times.

A distance can be maintained, from the viewpoint of the observer or the object, but not both.

Is this true even if the frame isn't an inertial frame?

If the observer and object change velocity (relative to some other frame) at "the same time", even if separated by say a light year, does their relative simultaneity remain fixed as long as they don't move relative to each other? And thus we can talk about "simultaneously" for the entire frame, no matter how big it is?

If the object and observer moved independently, but in sync... Suppose the object was allowed to choose whether to move along with the observer's reference frame, or stay stationary relative to the other frame.

If the observer moved at 0.866 c for 1 second and then stopped, but the object was 1 LY away, the observer would not be able to tell whether or not the object moved, until about a year later. Does this mean that any possible visible length-contraction effects would appear one year later?

I assume the answer is "no", and that length contraction would be apparent immediately. In which case, for the object to be synchronized with the observer, it would actually have to start its movement early... (by half a year in this case, with gamma=2, I think).

The farther apart the observer object are, the more gradual the observers acceleration must be to allow enough time for the extra displacement of the object.

For a light year distance to change to 1/2 a light year in the changing frame of the observer, the object would have to displace that much distance again, and of course this cannot exceed c in any inertial frame.

So even if he left early to allow for the lag, the distance is to great in this set up. It has to be much closer, slower, and at a more gradual acceleration.

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If we are neglecting the acceleration time as per the OP, why?

Because to maintain clock synchronization, the object would have to get a head start on the observer.

A signal to begin a synchronized start would have to come at the same time as the observation of the object beginning that start, and these would be delayed by the speed of light.

So you're changing the rest distance between object and observer, and are thus forced to change the lead time that the object acts with.

But I think this reasoning is wrong!???

I'll try to work through an example with "signals" instead of "observations", which usually gives some insights.

Suppose observer O and remote point P are 1 LY apart, and P is sending 1 signal every day.

At some arbitrary time, P will move in the direction away from O, at a speed of 0.866c, for a given number of days, and then stop.

O knows this much information and will attempt to remain stationary relative to O.

At rest, O is receiving signals sent "a year ago" from P. There are currently 365 signals "en route" from P to O.

When it happens, P begins moving away from O at 0.866c.

This length contracts the initial distance between them to 0.5 LY.

Due to an update in simultaneity, only a half year has passed at P relative to O. There are now only about 365/2 ~= 182 signals currently "en route" from P to O.

According to causality, O will not be able to observe any change in the signals, including their rate, or their source distance, for at least half a year. It must receive those 182 signals exactly as if no length contraction has taken place.

O then receives the signal to begin moving, 0.5 years after P begins moving.

O instantly accelerates and is now at rest relative to P.

Let's suppose this can happen in much less than a day, and timed so that O never observes any of the daily signals being length-contracted.

I would assume that the rest distance between them will now be greater than the original rest distance, because P has been moving away during its head start.

But is this a mistake??? The rest distance now applies to a different frame, so I don't think I can easily claim this.

Okay, I can't figure my way past this part.

In a "stationary observer's frame", P has had a 0.5 year head start at 0.866c, so the rest distance between O and P is now 1.433 LY, however this stationary observer would not currently observe O or P at rest, so would probably observe that distance under length contraction.

What is the rest distance according to O and P, who are currently at rest relative to each other?

It would have to be the same distance for both of them, so how is this possible, when P observes O starting much later???

I'm guessing it would have to still be 1 LY, but I can't figure out how this is so according to P's perspective!

Okay so according to P's perspective...

P starts moving. Distance to O contracts to 0.5 LY, and O moves away at 0.866c.

O continues moving away until it is 1 LY away, and then it remains at rest relative to P for as long as P remains in this inertial frame.

I don't have a firm grasp on why this is.

But yes, I now agree:

1. If an observer can sync its movements perfectly with a distant object, the observer will always appear to be at rest relative to the object.

2. In such cases, the rest distance between object and observer seems to remain fixed.

A distance can be maintained, from the viewpoint of the observer or the object, but not both.

Agreed, but I was wondering if it was even possible from just one of those viewpoints, because it didn't seem intuitive in my example.

I suppose the solution is typical: If one looks at it in the simplest way, the right answer may come easily. If one looks at it in the most complicated way, the wrong answer seems to come out, until all the effects of SR are properly accounted for, at which point the right answer again works out.

Edited by md65536
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Agreed, but I was wondering if it was even possible from just one of those viewpoints, because it didn't seem intuitive in my example.

I suppose the solution is typical: If one looks at it in the simplest way, the right answer may come easily. If one looks at it in the most complicated way, the wrong answer seems to come out, until all the effects of SR are properly accounted for, at which point the right answer again works out.

It won't work in the second case when they are a light year apart.

At a light second apart it's closer, but I think they are still too far apart for it to work at 0.866c, unless the observers accelerations were more gradual.

The "appearance", makes it a little more complicated. It is somewhat tricky just maintaining the displacement wrt the observer alone.

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Because to maintain clock synchronization, the object would have to get a head start on the observer.

A signal to begin a synchronized start would have to come at the same time as the observation of the object beginning that start, and these would be delayed by the speed of light.

So you're changing the rest distance between object and observer, and are thus forced to change the lead time that the object acts with.

But I think this reasoning is wrong!???

In clock synchronization you account for this delay. If we are 1 light-second apart, and I send you a signal at noon, you know you receive it at 12:00:01

Because of length contraction. Until they are back to the initial rest frame the distance in their frame of rest would have changed. An extra displacement of one is required to maintain the separation distance as measured in their current rest frame. So they cannot be at rest with respect to each other and maintain that same distance at all times.

A distance can be maintained, from the viewpoint of the observer or the object, but not both.

I still don't see where the problem is. There is no length contraction in the observer frame, and if the object matches the observer's speed, then they are in the same frame. You do not see length contraction inside your rocket, for example, because that's in the rest frame of the observer. What length is being contracted? You don't see length contraction inside a rocket, because that's all in the rest frame. You see physical compression, because the structural forces still propagate at c, but it's as if we're firing a rocket at the far end too, to circumvent that.

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I still don't see where the problem is. There is no length contraction in the observer frame, and if the object matches the observer's speed, then they are in the same frame. You do not see length contraction inside your rocket, for example, because that's in the rest frame of the observer. What length is being contracted? You don't see length contraction inside a rocket, because that's all in the rest frame. You see physical compression, because the structural forces still propagate at c, but it's as if we're firing a rocket at the far end too, to circumvent that.

It's usually not significant because no rockets are a light year in length, or even anything close to a light second, nor do they accelerate instantaneously to .866c.

A relative rocket avoids contracting because the front end and back end accelerate differently, and until they stop accelerating they are actually in slightly different rest frames This effect would be superimposed, or additional to the other physical/structural effects you are suggesting, so setting them aside for now there would still be additional stresses and strains on the rocket.

All points along the rocket cannot remain at rest with respect to each other while accelerating. If the front and back synchronized to an agreed on 1 second acceleration to .866c, and started at the agreed on time, would they later, after the acceleration, agree that they had both started at the same time? Finished the acceleration at the same time? If they cannot, then how could they have always shared the same rest frame?

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I agree with swansont. I can't find a way to force a single observer to see the ship's length change, without there being a simple way to compensate.

Most other observers would see the length between the observer and the object change. It's easy for me to mix up frames and get stuck thinking that their rest length must change.

Is a rest length between two relatively moving points even defined?

Sharing a frame with something that changes velocity, will depend on timing. Due to lack of simultaneity, there is no absolute sharing of a frame by multiple objects. Whether 2 objects are in the same frame or not depends on how they are observed.

In clock synchronization you account for this delay. If we are 1 light-second apart, and I send you a signal at noon, you know you receive it at 12:00:01

If at noon, you begin moving away from me at v=.866c, and then immediately (negligible delay) you send a signal to tell me to start, I will receive that signal after 0.5s by your clock, which is after 0.5 *gamma = 1 second according to my clock, as observed by you. So again it works out.

But from my perspective, if my clock is synchronized to yours, and I know that at noon you'll begin moving...

Suppose for now I'm just observing signals from you, and not trying to stay in your rest frame.

If gamma = 1 (no movement, just a signal), I would receive it at 12:00:01.

But with gamma = 2, your frame switch would change the synchronization of our clocks. When it's 12:00:00.05 for me, I think it's noon for your clocks, according to me??? (This is only for an instant where you've instantly accelerated to 0.866c but before you've covered a considerable distance.) I would receive your signal after 0.5s, again at 12:00:01.

In previous posts I must have been incorrectly dealing with the synchronization change.

So it seems the details of a complicated viewpoint balance to match the outcome of a simple viewpoint, which should be expected.

But there are always further complications to consider!

According to the above,

When it's noon my time, it's noon your time and you are 1 LS away.

When it's 12:00:00.05 my time, it's noon your time (in your new frame), and you are 0.5 LS away.

However, due to the travel time of light, I won't actually observe any disruption in your passage of time, and I will see you instantly jump from 1 LS away to 0.5 LS away, at 12:00:01.

I think...

Edited by md65536
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All points along the rocket cannot remain at rest with respect to each other while accelerating. If the front and back synchronized to an agreed on 1 second acceleration to .866c, and started at the agreed on time, would they later, after the acceleration, agree that they had both started at the same time? Finished the acceleration at the same time? If they cannot, then how could they have always shared the same rest frame?

Let's say we have a Docking Station 1 light year long, with an additional extention on each end, 2 light seconds long.

Brand New Spaceship tied to it, also 1 light year in length. Sister ship to Old Model Spaceship which is identical but has "old technology" propulsion, that limits the acceleration capability.

The new rockets on Brand New Spaceship are so powerful that they can accelerate mass to 0.866c in 1 second.

To minimize the stresses during acceleration Brand New Spaceship has many of the new rockets spread out along it's length and will be synchronized to work together by a number of well trained Operators.

For the maiden voyage they have decided to have an impressive start with a full acceleration take off.

Chief Operator Captain "Bert" is at the back of the ship waving to Dockyard Worker "Betty". Some months earlier they had noticed the front of Old Model Spaceship pass by in the direction of the front of their new craft. It was still passing by even though they estimated her speed to be 0.866c.

First Mate Operator "Frank" is at the front of the ship waving to Dockyard Worker "Fiona". They had still seen no sign of Old Model Spaceship even though they new she was scheduled to fly by.

All the Operators had their watches synchronized and were ready to engage there rockets at the agreed upon time. They understood the importance of the timing to the structure of Brand New Spaceship.

As the start time approached, Bert and Betty noticed the stern end of Old Model Spaceship finally approaching.

Still no sign of Old Model Spaceship for Frank and Fiona.

The start time came, and all operators engaged their rockets for a full and smooth 1 second "burn". Each rocket accelerated perfectly and after 1 second "rocket time" each portion of Brand New Spaceship was at 0.866c, as measured from the Dock. Betty and Fiona, after allowing for the transmission time, considered this to have taken more than 1 second, but still less than 2, so they considered this at least a partial, if not overwhelming, success.

Bert was ecstatic. Referencing the Dock, he knew he had reached 0.866c. Not only that, but he glanced over to see a number of cheering passengers on the stern end of Old Model Spaceship. In only 1 second, from a dead start, he had matched their speed. He was now in the same rest frame as Old Model Spaceship.

Frank was also ecstatic. He also knew he was at 0.866c relative to the extension at his end of the Dock. But he still saw no sign of Old Model Spaceship. Nor did Fiona. For them it was no where in sight. Not that it surprised them, as Fiona still wasn't expecting to see it for months.

What happened? Is Frank in the same rest frame as Old Model Spaceship? Is he in the same rest frame as Bert? Is Brand New Spaceship still intact?

Is there an absolute limit on acceleration that depends on length? Can anything accelerate while all it's parts stay in exactly the same shared rest frame at all times?

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Can anything accelerate while all it's parts stay in exactly the same shared rest frame at all times?

Yes, iff it can be treated as a point particle.

In general, no. I think we all agree that the sharing of frames is relative, and that if all parts of the velocity-changing ship share the same frame according to one observer (point) location on the ship, then other observers on the same ship will not see all the parts of the ship remain in the same frame.

The instant acceleration is just to simplify the thought experiments.

If you could build a ship where all of its parts can accelerate the same way (ie every point location on the ship is part of a propulsion system that accelerates the same way as every other point), and you synchronize the time at all locations on the ship to one point location's clock, and you have all parts of the ship coordinated to accelerate and decelerate at the same specific times according to their own local clocks, then the whole ship will remain in the rest frame of that one point location, according to that one point location. But the same observation won't be seen by any other part of the ship.

Is there an absolute limit on acceleration that depends on length?

I don't think so.

If it's possible to build a ship that stays in the rest frame of a single observer's location on the ship, then according to that location, the ship need not experience any stresses while accelerating, at any rate.

Therefore, the ship need not be torn apart due to the impossibility of synchronizing its parts.

If it doesn't get torn apart according to one observer, it won't be torn apart according to another observer. Other observers will necessarily see the ship affected by length contraction, and its various parts moving at different times, however it must be that other observers also observe the ship experiencing no stress. Any deformations (of space or material) observed will balance each other to allow no stress.

There are likely engineering reasons that make such a ship impossible or infeasible to build, but I don't think SR itself limits acceleration.

What happened? Is Frank in the same rest frame as Old Model Spaceship? Is he in the same rest frame as Bert? Is Brand New Spaceship still intact?

Yes, Frank is in the same rest frame as Old Model Spaceship, and the same rest frame as Bert. The ship can remain intact.

When you say the ship was synchronized, that must be done according to a specific clock.

In this example it seems like it's synchronized to Bert's clock.

Your example is certainly puzzling, but I think you're failing to consider changes to relative simultaneity that occur during frame switches.

I'm sure that if this was considered, then you'd find that if Bert was next to the stern of Old Model Spaceship immediately after finishing accelerating, then Frank would be next to the bow immediately after accelerating, even though the timing of various events would be different for the different viewpoints.

I think your outcome implies that Frank and Bert remained synchronized both to each other and to the Old ship???, which is impossible.

Edited by md65536
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I think your outcome implies that Frank and Bert remained synchronized both to each other and to the Old ship???, which is impossible.

I would say it implies that Frank and Bert can not remained synchronized.

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I would say it implies that Frank and Bert can not remained synchronized.

But Bert can remain synchronized to Frank, according to Bert.

Then they are not synchronized, according to Frank.

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But Bert can remain synchronized to Frank, according to Bert.

Then they are not synchronized, according to Frank.

No. Bert would require Frank to finish adjacent to the bow of the Old Model Spaceship, which would have required him to move in the opposite direction at much higher than light speed, yet remain at rest with respect to him at the same time.

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If at noon, you begin moving away from me at v=.866c, and then immediately (negligible delay) you send a signal to tell me to start, I will receive that signal after 0.5s by your clock, which is after 0.5 *gamma = 1 second according to my clock, as observed by you. So again it works out.

This description and calculation is so wrong its embarrassing.

I got the relativity of motion and of light wrong.

I applied time dilation backwards.

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No. Bert would require Frank to finish adjacent to the bow of the Old Model Spaceship, which would have required him to move in the opposite direction at much higher than light speed, yet remain at rest with respect to him at the same time.

I defy you to find any rate of acceleration that wouldn't produce a similar contradiction (ie. Frank moving backward or seen moving backward by Bert, at any speed) by your reasoning.

Meanwhile I'll try to explain your hypothetical scenario differently...

In the scenario, the New ship is synchronized to Bert at the back's clocks.

Bert sees Frank at the front begin to move at the same time that Bert does.

Bert also sees the ship reach full speed just as he's pulling alongside the back of Old ship.

Bert writes: "Cabooseman's log, stardate... whatever. We have begun our journey at the same time that Old ship came by, and we are side-by-side."

Old ship was contracted to 0.5 LY length, but now it is back to 1 LY length. The front of Old ship should be beside the front of New ship, just as full speed has been reached.

However, Bert will not immediately see this, because light from the front of the ship is still delayed by 1 LY.

What Bert will see might be described as a wave traveling up the length of the length of the Old ship, changing the ship from length-contracted to rest length, but not all at once.

I believe this would take a year to complete. Only after a year, does Bert see the front of Old ship coincide with the front of New ship. He sees Frank wave and give a thumbs up, 1 year into the journey.

In other words, Bert sees the front of the ship continuing to move forward, catching up to Frank after a year of Bert's time.

This is consistent with what Frank sees.

Frank's clocks are synchronized to Bert's. According to Frank, he starts moving long before Bert does.

Frank might write in his log: "Captain's log... I have begun our journey long before the back of Old ship is scheduled to catch up to Bert."

Frank matches the Old ship's velocity, but this happens a very long way away, so Frank doesn't observe this happening immediately.

Frank will observe the Old ship continuing to appear to be moving toward Frank, even though Frank knows he's moving forward too.

I don't know how to explain this properly. One might say: If Frank were now at rest relative to Old ship, the light from the observations of Old ship would still be catching up to Frank at a velocity of c. No matter what velocity Frank takes on, the observations will still catch up to Frank, so Frank can never see himself outrunning (or at rest relative to) those observations... In other words, Frank will still see the Old ship continue to catch up.

I think it's valid to say that Frank does not actually reach full speed AND enter the rest frame of the distant Old ship simultaneously. This is not a contradiction, because the 2 are separated by a great distance when Frank reaches full speed.

I'll stop with the details here because I know I'm going to screw them up.

Frank sees himself start the journey early, and after a long time (1 year I guess it must be) he sees Old ship's front catch up and come to rest next to him.

He waves to Bert and gives a thumbs up.

Different observers see different events happening simultaneously.

There is no contradiction or impossible situation here.

Addendum: Bert does not join the rest frame of the back of Old ship at the same time he joins the rest frame of the front of Old ship, even if it's the same frame.

Frank does not join the rest frame of the distant front of Old ship at the same time that Frank reaches full speed.

Things from one frame can switch to a different shared frame at different times, if those things are separated by distance.

Edited by md65536
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I defy you to find any rate of acceleration that wouldn't produce a similar contradiction (ie. Frank moving backward or seen moving backward by Bert, at any speed) by your reasoning.

That was my point, though I don't see it as a contradiction.

Any nonzero distance (as measured in every current rest frame) between two objects cannot be maintained while they are accelerating (in the direction of the distance between them) while they stay in shared rest frames.

A distance (a maximum of which is limited by their separation and rate of acceleration...ie. Frank can't break the laws of physics) can be maintained with respect to the rest frame of one or the other, but not both.

Edited by J.C.MacSwell
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That was my point, though I don't see it as a contradiction.

Any nonzero distance (as measured in every current rest frame) between two objects cannot be maintained while they are accelerating (in the direction of the distance between them) while they stay in shared rest frames.

A distance (a maximum of which is limited by their separation and rate of acceleration...ie. Frank can't break the laws of physics) can be maintained with respect to the rest frame of one or the other, but not both.

If Bert is in the front seat of their spaceship and Frank is in the back seat, Frank will measure the same distance between himself and Bert no matter how much they accelerate or how fast they go. They share the same frame of reference.

Chris

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If Bert is in the front seat of their spaceship and Frank is in the back seat, Frank will measure the same distance between himself and Bert no matter how much they accelerate or how fast they go. They share the same frame of reference.

Chris

(just to stay consistent; Bert is in the back and Frank is in the front)

They are at rest in the same inertial frame at the start and start to accelerate simultaneously in that frame. At that point they would agree on the time and distance.

Immediately, as they accelerate, this agreement is lost.

After both have finished accelerating (same time as measured in the original or Dock frame, different times in their new frame) they will again agree on simultaneity but will disagree on the passage of time (Frank will have aged more, having waited for Bert to finish accelerating)

Their Ship will have stretched to approximately twice it's length in their new rest frame.(or most likely come apart, depending on it's physical properties)

All "theoretically" possible. Staying the same length (one light year), in the frames of Bert or Frank while accelerating at that rate is not. It would require breaking the rules of SR.

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(just to stay consistent; Bert is in the back and Frank is in the front)

They are at rest in the same inertial frame at the start and start to accelerate simultaneously in that frame. At that point they would agree on the time and distance.

Immediately, as they accelerate, this agreement is lost.

After both have finished accelerating (same time as measured in the original or Dock frame, different times in their new frame) they will again agree on simultaneity but will disagree on the passage of time (Frank will have aged more, having waited for Bert to finish accelerating)

Their Ship will have stretched to approximately twice it's length in their new rest frame.(or most likely come apart, depending on it's physical properties)

All "theoretically" possible. Staying the same length (one light year), in the frames of Bert or Frank while accelerating at that rate is not. It would require breaking the rules of SR.

The thought experiment you propose sounds like "Bell's Spaceship Paradox":

Bell's spaceship paradox is a thought experiment in special relativity. Two spaceships, which are initially at rest in some common inertial reference frame are connected by a taut string. At time zero in the common inertial frame, both spaceships start to accelerate, with a constant proper acceleration g as measured by an on-board accelerometer. Question: does the string break - i.e. does the distance between the two spaceships increase?

Is this the situation you're thinking about?

Chris

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The thought experiment you propose sounds like "Bell's Spaceship Paradox":

Is this the situation you're thinking about?

Chris

I made it up to make a point as others were claiming that a set distance between 2 points (or an object of non zero dimension) could be maintained while accelerating and at the same time the 2 points could at all times share a rest frame.

I'm sure there are a few similar paradox themes around that are similar though I don't recall one directly for that purpose. It's just the way simultaneity is broken with acceleration in SR-when one moves into the others past.

Edit: I read the Wiki and interestingly it looks like it's the same Bell from Quantum physics that came up with "Bell's Inequality", so this could be "Bells Inequality II", although it originated earlier with someone else who was pointing out the actual physical contraction required to accelerate (or stresses produced as I mentioned earlier)

Edited by J.C.MacSwell

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