Bell’s Spaceship Paradox. Does the string break?

[Halc]

1. The typical anaut cartoon has him in a ship moving from Earth to the nearest star 4 ly distant. He accelerates in a large orbit until reaching his target speed of .6c prior to departure.

The typical telling of the twins scenario involves an orbit?? The scenario is an SR exercise and under SR there is no gravity and thus no orbit. Any speed over over about 0.000025 c is not an orbit, at least not around Earth.

Moving at a constant velocity, SR allows him to assume a pseudo rest frame.

Nothing pseudo about it. Any inertial frame is as good as another, per the first premise of SR. This was presumed centuries before Einstein, with Galilean relativity.

The star arrives at ship time 5.34 yr, which he reports to Earth.

Earth can be watching (or even not if they know the itinerary). Him reporting anything is not wrong, but Earth learns nothing it doesn't already know.

He verified his velocity while leaving Earth. To reconcile the time difference, he assumes the universe outside his ship was moving in the opposite direction at .6c (the reciprocity of the principle of relativity), and was length contracted, relying on visual cues for his conclusion.

He can for instance see Mars go by at 0.6c

If a 2nd anaut used the same method, and simultaneously departed in the opposite direction at .6c, his report sent at 5.34 yr (predicted by SR) should arrive at Earth simultaneously with the other report.

Simultaneous with the arrival of the first 'anaut'. Apparently you don't want the first guy to just turn around using a trampoline or something. All fine. Just commenting.

The Earth monitor would conclude, if the universe moved at .6c in opposite directions simultaneously, it didn't move!

This makes no sense. The Earth is the universe (or it is at least approximately stationary with most local objects within it), and yes, they conclude from lack of acceleration that it has never moved significantly in its frame. No 0.6c figures into that. Sure, there are these spaceships moving that fast relative to that frame, but that's the spaceships moving, not the rest of the universe.

Why/how can the motion of a tiny ship modify the motion of the universe?

In no inertial frame does this occur, so answer: It doesn't.

Why is the anaut the only person who experiences this

He experiences his own acceleration, or if not, Earth was always moving fast in his frame, so either way, the velocity of Earth and the rest of the universe wasn't modified.

Why are there no astronomical observations of the specified star changing its motion?

If I float a km from an asteroid holding a Go-Pro camera fixed to a crossbow quarrel, The camera image of the asteroid appears stationary relative to it. If I suddenly fire the projectile, it records (observes) the asteroid it suddenly getting closer. That's the camera changing motion, never the asteroid changing. A proper change in motion can measured by an accelerometer, and the accelerometer on the rock says zero the whole time, but not so for the Go-Pro. Likewise, it ain't the star what's motion is changing.

His perception is altered by his motion.

A direct contradiction with the first premise of SR. His perception is unaltered. Physics is the same in all frames.

The length contraction of material objects is a physical process.

For the most part, it is an abstract process, a byproduct of geometry. I say 'most part' because there are objective ways to demonstrate length contraction which are frame independent: Everybody agrees that the same objects got shorter.

When NASA wants to land a probe on an asteroid or Mars, etc., they need to know its location.

Mars isn't moving at relativistic speed relative to Earth. Nobody made relativistic corrections when putting somebody on the moon, but they do need to do it for GPS, whose motion is very much sensitive to relativistic effects.

Center of mass is an abstract calculation and has no physical location.

Of course it has a physical location, even if there's no material there. I can for instance still point to it or give the coordinates of it, things I cannot do to something with no physical location.

It was omitted to emphasize the lc of the ships affects the length of the gap.

That it does, but the ships were considered to have zero length for simplicity. It's a needless complication that detracts from the point of Bell's exercise. The string exhibits relativistic length contraction because by definition it is an extended object. The contracted length of the gradually accelerating string is not as easy to compute as Mike would have it. His speculation directly leads to several contradictions, some of which were identified in my prior post.

The motion outbound ends at At=10 with instantaneous deceleration to 0.

This does not parse. It is a reference to your 2nd diagram posted above , with 20 years total and the traveler going at 0.5c. 'At=10' says that there is an undefined variable called 'At' that is set to 10. Syntax error...
I'm guessing you mean 'at t=10 (Earth frame)' the turnaround occurs. So the diagram shows.

The motion inbound begins at At=10 with instantaneous acceleration to .5

Well, acceleration to -0.5
Velocity is changed from 0.5 to -0.5, so that's a coordinate change in velocity of -1c, The proper change in velocity is a bit more than that.

On second reading, the return ship isn't the same as the outbound one, so it's a tag team. Picture doesn't convey that.

Yes, there are two ships to complete the B round trip. The essence of the problem is the accumulated time on the A-clock and B-clock, for a closed course.

There isn't just one B clock if there's two ships, unless the first guy tosses the clock onto the return ship, in which case it's the clock that counts and it accelerates as per above. If it's not the same clock, then there's no real time differential, but the the mathematics still works.

You have all these blue light-lines and green connections between some of the points. Don't know what those depict.

You said:

the spacetime path is discontinuous, meaning there is no connection of the outbound to inbound paths

The picture shows connections of each spacetime path, with the outbound connecting to the inbound path at the X=5,ct=10 event where you put the '8.66' time label.
So I still don't have any idea what you're trying to convey with those words.

There is no requirement for people or even synchronization. Einstein's 1905 paper, par.4 described 1 of 2 clocks departing on a closed course and returning to the 2nd, indicating less elapsed time.
I think it was Paul Langevin who proposed twin life forms, for the purpose of emphasizing biological clocks.[/QUOTE]No argument with any of that. They actually did the experiment, two clocks on a closed course, and verified both SR and GR (and Sagnac effect for good measure) in doing so.

The effects from SR only have meaning to humans

Ah, so if squirrels put up GPS satellites, they would work with normal clocks without relativistic corrections. We should contract the squirrels to implement the simpler physics for the navigation industry.

Sorry phyti, but the physics of the universe is not different for humans than for everything else. We describe it with our theories, but don't prescribe it as you seem to suggest.

 

[phyti]

Mike;

The typical spacetime graphics show a history of events, like the one on the left. These can be confusing relative to simultaneous events. We have discussed this the past involving 'now' and instant knowledge. Here is another example.

With A as the ref. frame, consider the two events At=10 and Bt=8.66.

Each observer is at the center of their perception space (blue light cones), and can only be aware of events that have occurred within their light cone. An observer is coincident with the emission and detection of a round trip light signal. When the reflection occurs is uncertain. The SR solution is to divide the round trip time by 2. The reason being, if the observer can assume a pseudo rest frame when moving at constant velocity, his expectation would be half the time for light with constant independent velocity. Additionally, there are no clock images being communicated, but electronically encoded time signals.

On that basis:
in the center graphic, A is only aware of Bt=5.77 occurring at At=6.67 and the A clock was pinged for a signal at At=5,
and on the right, B is only aware of At=5 occurring at Bt=5.77.

 

[Mike_Fontenot]

This is the proof I gave that is very important, because it proves that two rockets with equal constant accelerometer readings will have a constant separation, according to people on the trailing rocket:
_____________________________________________________________________

A Non-Constant Separation of the Rockets Contradicts the Resolution of the Twin Paradox

The resolution of the Twin Paradox is well-known: during the traveler's (his) instantaneous turnaround, he must conclude that his home twin's (her) age instantaneously increases. (And the amount of her instantaneous ageing, according to him can be easily calculated). But IF it is assumed that the two separated rockets in the Bell's Paradox (whose accelerometers show equal constant readings) DON'T maintain a constant separation, that CONTRADICTS the resolution of the Twin Paradox.

Here's how to see that contradiction:

Suppose we start out with him being separated and stationary with respect to her.

Imagine that, at the instant before he instantaneously increases his speed toward her, he is colocated and stationary with respect to the TRAILING rocket. And suppose that the LEADING rocket is colocated and stationary with HER then. (The rockets are unaccelerated before and at that instant).

When he instantaneously changes his speed with respect to her from zero to some large non-zero value, the two rockets instantaneously do the same thing.

During his instantaneous speed change, suppose that the leading rocket is ASSUMED to instantaneously INCREASE its separation from the trailing rocket. THAT would result in HER seeing the leading rocket INSTANTANEOUSLY move a finite distance away from her, WHICH IS ABSURD! So the ASSUMPTION that the separation of the rockets isn't constant CAN’T be correct. The separation of the rockets is constant.

Q.E.D.

 

[Halc]

With A as the ref. frame, consider the two events At=10 and Bt=8.66

Those are times, not events. You can say A at time At=10 or B at Bt=8.66, or you can say At = 10, x = 0 and At = 10, x = 5 respectively. Those are events (or at least they are in a 2D spacetime diagram).

Each observer is at the center of their perception space (blue light cones), and can only be aware of events that have occurred within their light cone.

The outcome of the twins scenario has nothing to do with perceptions, light cones, or awareness. All that is needed is a pair of clocks, sealed in a box if you like and compared only when reunited. The reflection of light signals seems totally unnecessary.

This is the proof I gave

Your 'proofs' never state their assumptions. Proofs are quite formal, an are only valid if the premises are valid. You have a lot of unstated premises, most of with which I cannot agree.

it proves that two rockets with equal constant accelerometer readings will have a constant separation

That's pretty different from having identical proper acceleration. I can put broken accelerometers in any ships and send them anywhere I want. You've very clearly shown that your accelerometers indicate a different reading than the actual acceleration of the ships.
Integrating the accelerometer reading over time yields velocity (a very different one than in your pictures). Integrating velocity over time gives position, a very different position than what your pictures show.
These are basic calculus skills. Perhaps you unaware of the technique.
You've given the excuse of it being the difference between proper acceleration and coordinate acceleration, but the former is just the latter multiplied by λ which isn't going to change the sign.

I notice you still didn't correct any of my numbers several posts back. The string breaks, and your (nonexistent) numbers don't show otherwise. No error in my calculations was identified. (I need a gif of Kirk goading Khan: "I'm laughing at the superior intellect")


I will allow your 'proof' to go on so I can point out the inevitable errors. Magnanimous of me, no?
We will assume that the Bell rockets have identical proper acceleration despite the fact that you didn't actually say that above. It just wouldn't be Bell without that.

I will state another premise of yours:
When you say "he must conclude", those words mean that he chooses to reference the inertial frame in which he is presently stationary.
I say this because he is very much not under any obligation to use that coordinate system and hence [he must conclude something about events not in his presence] just doesn't follow.

Kindly let me know if I'm in error with any of these premises, because none of your assertions follow without them..

A Non-Constant Separation of the Rockets Contradicts the Resolution of the Twin Paradox

The resolution of the Twin Paradox is well-known: during the traveler's (his) instantaneous turnaround, he must conclude that his home twin's (her) age instantaneously increases.

Simultaneous with the turnaround event and relative to his associated non-inertial frame, I agree.
The part with which I don't agree is that this is the 'well known resolution' to the twins paradox. Someone well versed in relativity has a far better solution that generalizes to any situation, and there are resolutions that are far simpler and more intuitive to a layman. Your 'proof' seems not to hinge on this obfuscating resolution being 'the well known' one, so we let it go.
Yes, I've seen it explained with changing lines of simultaneity, but never with this 'must conclude' wording that leaves so much unsaid.

But IF it is assumed that the two separated rockets in the Bell's Paradox (whose accelerometers show equal constant readings) DON'T maintain a constant separation

We don't assume that. The separation is constant in the orginal IRF (call it U), and you didn't specify a different frame (or any frame for that matter) in the above statement, rendering it meaningless. It is not an assumption. It is readily computed since the exact worldline of each ship can be plotted in U simply by knowing its acceleration profile and when it starts. The code I supplied to you does exactly that.

Suppose we start out with him being separated and stationary with respect to her.

Imagine that, at the instant before he instantaneously increases his speed toward her

This is not Bell's. Bell scenario has continuous acceleration, not just some abrupt change. It doesn't matter since the string breaks either way (as shown by my number and the lack of yours), but if it isn't Bell, and your 'proof' should be about Bell, not some scenario where the ship ceases acceleration immediately.

he is colocated and stationary with respect to the TRAILING rocket. And suppose that the LEADING rocket is colocated and stationary with HER then.

When he instantaneously changes his speed with respect to her from zero to some large non-zero value, the two rockets instantaneously do the same thing.

OK, per the example I chose (since you didn't supply any numbers), which was 2 LY separation in U, and .866c velocity change.

During his instantaneous speed change, suppose that the leading rocket is ASSUMED to instantaneously INCREASE its separation from the trailing rocket.

It does in its new inertial frame, which my numbers show. It was computed. There was no assumption about it. They're 4 LY apart in that frame, at all times when both ships are stationary in that frame. So far so good except you calling every bit of valid arithmetic an 'ASSUMPTION'. Getting closer to the end and no fatal error in the proof so far, but no contradiction either.

THAT would result in HER seeing the leading rocket INSTANTANEOUSLY move a finite distance away from her

Non-sequitur. He might choose to conclude that her age changes like that, but his choices (well outside her past light cone) have no effect on her experiences.

Proof spirals down the toilet. So sad...

In fact, you've perfectly disproved your own scenario since both observers would experience exactly that absurdity: the rocket instantly moving (teleporting) elsewhere, whether they're riding it or not. Disagree? Then where are the rockets (in U coordinates) immediately after they do that? You can't do that trivial task? Or you just won't because it show the exact absurdity that you fallaciously claim for the correct numbers?

Funny that your punch line is exactly one way to disproves your assertions.

 

[Mike_Fontenot]

[...]

Instead of using the shotgun or grenade approach, please just tell me the first sentence I wrote in my most recent post (post #83) that you believe to be incorrect, and state clearly why you believe it is incorrect.

 

[DaveC426913]

Your 'proofs' never state their assumptions. Proofs are quite formal, an are only valid if the premises are valid. You have a lot of unstated premises, most of with which I cannot agree.

...most of which with I cannot agree.

...with most of which I cannot agree.

 

[Neddy Bate]

Instead of using the shotgun or grenade approach, please just tell me the first sentence I wrote in my most recent post (post #83) that you believe to be incorrect, and state clearly why you believe it is incorrect.

For me, that would be this:

During his instantaneous speed change, suppose that the leading rocket is ASSUMED to instantaneously INCREASE its separation from the trailing rocket. THAT would result in HER seeing the leading rocket INSTANTANEOUSLY move a finite distance away from her, WHICH IS ABSURD!

I would say "in HIS frame" for the first sentence, and then "in HER frame" for the second sentence, and then you have to show how you think the second is a consequence of the first, which it isn't.

Here a better way to do it:

In your post # 83, the simplest way to approach the problem is from HER frame, just like it is with the Twin Scenario. In her frame the rockets instantaneously accelerate simultaneously at some time (say t=0) when both her clock and the clocks on the rockets all display t=0. She says they just move at constant speed (v) for all time after t=0, maintaining constant distance between them, as per Bell’s scenario. She has no contradictions or problems accounting for any of this, and she easily explains why the traveling twin returns younger than herself without any “time jump” since she never accelerated.

Your “resolution” to the twin’s paradox comes from HIS frame where he has to account for her getting older though a “time jump”. So, since you want to incorporate that, you have to use HIS frame instead of HERS. That is a much more complicated approach, but you cannot get contradicting results by simply looking at a defined scenario from a different frame, unless of course you make an error.

We know from HIS frame after the instantaneous acceleration that HER clock has jumped well ahead of t=0 to some much greater time, (say t=70). So, the leading rocket must now be located a known distance behind her, having traveled all that time from t=0 to t=70 in her frame at one speed (v) relative to her. So during the instant HE accelerated, the leading clock has moved from being co-located with her to well behind her in HIS frame. The other rocket remains co-located with him. So, both rocket’s locations in his frame should be able to be calculated from those simple equations, just don’t forget length contraction of HER whole x axis in HIS frame.

There is still a problem with the time behind her being even more advanced than where she is, in HIS frame. So just use HER frame.

 

[Halc]

...most of which with I cannot agree.

...with most of which I cannot agree.

That did sort of get stuck in the pipe, but I'll be darned if I write 'most of which I cannot agree with'.
I'm old enough to know a preposition is not something you end a sentence with.

Thx for the suggestions.

Instead of using the shotgun or grenade approach, please just tell me the first sentence I wrote in my most recent post (post #83) that you believe to be incorrect, and state clearly why you believe it is incorrect.

Well, it opens with "This is a proof ...", and I disagree with that, for the reasons stated in my post.
Neddy points out the core problem, which is closer to the last sentence than the first one.

There's no mathematics, so it is essentially an asserted solipsistic moving spotlight philosophical position, and being solipsistic, nobody else (her in particular) experiences anything, so no contradiction in the end. Wigner ran into the same problem with a similar stance in his quantum interpretation, and eventually abandoned it precisely because it implied solipsism and he wasn't comfortable with that.


The lack of response to the numbers I posted actually displays your failure. Your assertions don't hold up to the simplest arithmetic and you refuse to demonstrate where my numbers are wrong, and refuse different numbers.

I would say "in HIS frame" for the first sentence, and then "in HER frame" for the second sentence, and then you have to show how you think the second is a consequence of the first, which it isn't.

Mike has a long history of neglecting his frame references, and then drawing conclusions from that omission, as he does here. See some of his TNS topics and try to count how many times I say his statements are meaningless without such a reference.
He also thinks in terms of perspectives instead of frames, which is why he seems to require a human at every clock, perhaps to read it to you. It reminds me of Sigourney Weaver's role in Galaxy Quest (a Star Trek fandom spoof) where her only task on the ship was to repeat what the ship computer just said, making her the computer's helper friend (HF).

 

[DaveC426913]

a preposition is not something you end a sentence with.

That is the kind of arrant pedantry up with which i will not put!
- W.Churchill

 

[Mike_Fontenot]

[...]

I (Mike Fontenot) said:
"During his instantaneous speed change, suppose that the leading rocket is ASSUMED to instantaneously INCREASE its separation from the trailing rocket." (That is according to the people on the trailing rocket).

That is what everybody has been telling me is the true situation: Specifically, that two separated rockets with accelerometers that show the same readings will get farther and farther apart as the acceleration progresses. So I start out by ASSUMING what everyone else thinks is true, and then I show that THAT assumption is inconsistent with the resolution of the twin paradox. That assumption implies that the home twin will see a rocket, that is stationary wrt her at one instant, instantaneously move a finite distance from her, and that is absurd. So the original assumption can't be correct.

 

[Neddy Bate]

I (Mike Fontenot) said:
"During his instantaneous speed change, suppose that the leading rocket is ASSUMED to instantaneously INCREASE its separation from the trailing rocket." (That is according to the people on the trailing rocket).

That is what everybody has been telling me is the true situation: Specifically, that two separated rockets with accelerometers that show the same readings will get farther and farther apart as the acceleration progresses. So I start out by ASSUMING what everyone else thinks is true, and then I show that THAT assumption is inconsistent with the resolution of the twin paradox. That assumption implies that the home twin will see a rocket, that is stationary wrt her at one instant, instantaneously move a finite distance from her, and that is absurd. So the original assumption can't be correct.

Except you can't jump from the trailing rocket's frame to HER frame as if they are the same thing. That would be like saying since the trailing rocket claims her age instantaneously jumps ahead, that her age also must instantaneously jump ahead in her own frame. That does not follow at all. Jeez, it's like you've forgotten everything you used to (sort of) know about relativity.

 

[Mike_Fontenot]

Except you can't jump from the trailing rocket's frame to HER frame as if they are the same thing.

By (my) design, she is colocated and stationary with the leading rocket at the instant before the leading rocket ignites. That is an event, needing no further definition. And I've shown that she sees that rocket instantaneously move a finite distance away from her. That is absurd, so the assumption that produced that absurdity can't be correct. And that assumption was that separated rockets with equal accelerometer readings will move apart, according to the people on the trailing rocket.

 

[Neddy Bate]

By (my) design, she is colocated and stationary with the leading rocket at the instant before the leading rocket ignites. That is an event, needing no further definition. And I've shown that she sees that rocket instantaneously move a finite distance away from her. That is absurd, so the assumption that produced that absurdity can't be correct. And that assumption was that separated rockets with equal accelerometer readings will move apart, according to the people on the trailing rocket.

No, the assumption that caused the absurdity is you thinking that what happens in the rocket frame at a distance is also what happens in her frame. The rocket says she ages instantaneously, but she does not say that. She says she ages normally. The rocket says the other rocket moved instantaneously, but she does not say that. She says it moves at constant speed over time.

 

[Neddy Bate]

Mike,

Let me walk you thru this with some questions.

1. What is her age according to the trailing rocket just before it accelerates? Giver me a number.
2. What is her age according to the trailing rocket just after it accelerates? Giver me a number.

From those two answers I can walk you through this easily.

 

[Mike_Fontenot]

The (trailing) rocket says the other (leading) rocket moved instantaneously, but she does not say that. She says it moves at constant speed over time. (additions in red added by me (Mike Fontenot))

You seem to have missed the fact that, in this scenario, the rockets are NOT moving at a constant finite acceleration "A" for a finite amount of time "t". In this scenario, I have let the acceleration "A" go to infinity, and the time "t" go to zero, in such a way as to keep the total velocity change constant. I.e., the accelerations are Dirac delta functions. I do this because that is the simplified way that the twin paradox is usually stated, with instantaneous speed changes, not finite accelerations.

 

[Mike_Fontenot]

Mike,

Let me walk you thru this with some questions.

1. What is her age according to the trailing rocket just before it accelerates? Giver me a number.
2. What is her age according to the trailing rocket just after it accelerates? Giver me a number.

From those two answers I can walk you through this easily.

I haven't specified a specific twin paradox scenario, because my proof didn't require it. But I can give you a specific twin-paradox-type scenario.

You should make a sketch as I go along, and maybe check my results by plotting a Minkowski diagram. I know you know how to do that.

Let his outbound speed be 0.866 ly/y, lasting for 40 years, according to her. So gamma is 2.0. She says they are 34.64 ly apart at that instant. There, he changes his speed to zero wrt her. He is 20 years old then (according to him AND her, because that is an event).
He coasts for 10 years, at the end of which he is 30 and she is 50.

Then, he instantaneously changes his speed to 0.866 ly/y, toward her. When they are reunited, he is 50 and she is 90.

So just before his returning speed change, he is 30, and she is 50. Just after he changes speed, he is 30, and she is 80. She instantaneously aged by 30 years, according to him, during his second speed change.

You can check that using a Minkowski disagram. I got it from my delta_CADO equation:

delta_CADO = -L * delta_v,

where

delta_CADO is her change in age (positive when increasing),

and

delta_v = v_after_speed_change - v_before_speed_change,

and L is their (positive) separation, according to her. Velocities are positive when directed away from her, and negative when directed toward her.

So we get

delta_CADO = ( - 34.64 ) ( -0.866 ) = +30.

So she instantaneously gets 30 years older during his speed change from zero to -0.866 ly/y.

If I haven't made a careless mistake anywhere, that's your answer.

 

[Neddy Bate]

You seem to have missed the fact that, in this scenario, the rockets are NOT moving at a constant finite acceleration "A" for a finite amount of time "t". In this scenario, I have let the acceleration "A" go to infinity, and the time "t" go to zero, in such a way as to keep the total velocity change constant. I.e., the accelerations are Dirac delta functions. I do this because that is the simplified way that the twin paradox is usually stated, with instantaneous speed changes, not finite accelerations.

You are saying the speed is constant, right? And my words were "She says it (meaning the leading rocket) moves at constant speed over time." So we agree. What did I say that made you think I was using constant acceleration?

 

[Neddy Bate]

So just before his returning speed change, he is 30, and she is 50. Just after he changes speed, he is 30, and she is 80. She instantaneously aged by 30 years, according to him, during his second speed change.

Okay, so your answers to my questions are here:

1. What is her age according to the trailing rocket just before it accelerates? 50
2. What is her age according to the trailing rocket just after it accelerates? 80

Alright, now the event where the leading rocket is co-located with her (and instantaneously accelerates) occurs when her age is 50, not when she is 80. This is important for you to remember. Think about it, please.

When she is 80, the leading rocket has long since past her, and is located far away, as it was moving smoothly away from her all that time. This is why when the trailing rocket says she changes from 50 to 80, we are all telling you the rocket has to become located far away from her. You are telling us the rocket should still be next to her when she is 80 years old. Your error is in thinking that whatever the trailing rocket says, such as, "the leading rocket instantaneously changed position," is also what she has to say. But she does not say her age jumped from 50 to 80, and she does not say anything teleported to a distant position. She says she lived 30 years of life between 50 and 80, and the leading rocket had a full 30 years to change position.

 

[Mike_Fontenot]

I've (Mike Fontenot) added (in red color) some things to your statement below:

No, the assumption that caused the absurdity is you thinking that what happens in the (trailing) rocket frame at a distance is also what happens in her (leading rocket) frame. The (trailing) rocket says she ages instantaneously, but she (and the people on the leading rocket) does not say that. She says she ages normally (of course, how could she not?). The (people on the trailing) rocket says the other rocket (the leading rocket) moved instantaneously, but she does not say that (yes, she does say that ... she is standing right there next to that rocket, and she can't help but see it instantaneously move a finite distance away from her). She says it moves at constant speed over time. (I have no idea what you're saying about anything moving at constant speed over time ... it certainly wasn't the leading rocket. The leading rocket was initially stationary wrt her, but then instantaneously moved a finite distance away from her. THAT is an absurdity, produced by the original incorrect assumption that two separated rockets with equal readings on their accelerometers will move farther apart, according to the people on the trailing rocket.)

Before your response above, I (Mike Fontenot) had said:
"By (my) design, she is colocated and stationary with the leading rocket at the instant before the leading rocket ignites. That is an event, needing no further definition. And I've shown that she sees that rocket instantaneously move a finite distance away from her. That is absurd, so the assumption that produced that absurdity can't be correct. And that assumption was that separated rockets with equal accelerometer readings will move apart, according to the people on the trailing rocket."

 

[Neddy Bate]

I've (Mike Fontenot) added (in red color) some things to your statement below:

Please don't add to my statements, reply to them using the quote feature as normal. I do not agree with the following statement from you, and yet you've made it look like I said it:

(I have no idea what you're saying about anything moving at constant speed over time ... it certainly wasn't the leading rocket. The leading rocket was initially stationary wrt her, but then instantaneously moved a finite distance away from her. THAT is an absurdity, produced by the original incorrect assumption that two separated rockets with equal readings on their accelerometers will move farther apart, according to the people on the trailing rocket.)

Perhaps you need to remind yourself of the scenario you created. You have the leading rocket next to her, and then in instantly accelerates to a constant velocity v, which causes it to move away from her at constant speed while she ages from 50 to 80. By the time she is 80, it is no longer next to her anymore, it is a great distance away, right?