Bell’s Spaceship Paradox. Does the string break?

[Neddy Bate]

In frame S", Alice and Bob 17.3 LY apart (with Bob in the lead) and are moving at 0.866c, colocated with ships connected by a 34.6 LY string contracted to 17.3 due to the motion. The ship by Alice suddenly stops, breaking the string. Alice continues on at 0.866c. 60 years later, Bob stops, along with his ship. After 60 years of the trailing ship moving away and the lead ship stopped, the distance has grown from 17.3 LY to 69.3 LY. The string broke 60 years prior, as soon as the one ship stopped and the one receding did not.

I like your description of the events as they happen in the S'' frame. I had not thought of these particular details before.

 

[Neddy Bate]

Neddy, I have a question about your most recent diagram ... it is still not correct. Recall that the reason for this whole discussion is the widely-held (but incorrect, according to me) view that two separated rockets with equal accelerometer readings will get farther apart as the acceleration proceeds. If that were true, then delta-function-type accelerations which we are using here (where the acceleration "A" is infinite, but lasts only for an infinitesimal time) would cause an instantaneous FINITE increase in the separation of the two rockets.

Luckily I think I understand what you mean, even though as Halc points out, you did not specify in which frame the rockets get farther apart. The distance between the rockets is 34.64 light years when the rockets are stationary relative to the home twin (frame S). Once they start moving at v=-0.866c relative to the home twin (frame S), the distance between them remains fixed at 34.64 light years as measured by frame S. This is analogous to Bell's scenario where they remain a fixed distance apart in the initial inertial frame, even as they accelerate.

But after the acceleration, the distance between the rockets is 69.28 light years in the rockets' own frame (frame S'').

Here's my question: when I look at your latest diagram (post number 204), I can see the world-line of the leading rocket, heading off to the upper left in a straight line (you've labeled it). It shows that world-line STARTING from the point "t = 50" and "x = 0". It SHOULD show an initial horizontal line extending to the left, corresponding to the required sudden finite displacement of the leading rocket, before changing to a straight line heading to the upper left.

A horizontal worldline extending to the left would represent an infinite speed on the leading rocket in frame S. My diagram does not show the leading rocket moving at infinite speed in frame S because that does not happen. You should still be able to figure out from the unprimed coordinates (x, t) that the distance is 34.64 in frame S, and you should also be able to figure out from the double-primed coordinates (x'', t'') that the distance is 69.28 in frame S''.

 

[Neddy Bate]

Neddy;

The coordinate transforms apply to inertial (constant velocity) systems.
Even if the route is planned, all plans don't succeed. The outcome is conditional on experience agreeing with predictions for every detail.

Yes, the hypothetical scenario is one where everything goes as planned. It is a thought experiment, not a real life space mission.

The transform for event A(0, 10) yields B(-17.3, 20).
Yet measurements by B yield B(-9.3, 20). Why the difference?

The coordinates (x,t)=(0.00,10.00) represents the home twin (she is always located at x=0) when she is celebrating her 10th birthday. As she blows out the candles on her cake, they take a picture of her. In the background of the photo there is a train (an inertial space-train) moving v=0.866c, and on the train there is a clock showing t'=20. That particular train car is labeled x'=-17.32. That is what the transformation (x,t)=(0.00,10.00) to (x',t')=(-17.32,20.00) represents.

That train is the one the traveling twin jumped on when the twins were born. He landed on the train car marked x'=0.00, and the clock in that car showed t'=0.00 at that time. He accepts that all clocks on the train are Einstein synchronised in that train's frame. When he is 20 years old he calculates that his sister is 10 years old as per SR. The photo supports that, because it shows the train time t'=20.00 and her time t=10.00 in the same photo.

I don't know what your measurement (-9.3, 20) is. Perhaps you could explain, and then I can offer more help.

 

[Mike_Fontenot]

Neddy, I DID carefully read your critique above. It is too contorted to respond to, through no fault of your own ... getting down to that detail is just very contorted, but fortunately unnecessary. I think the wording I have in my paper is completely adequate, as is. But I will elaborate a little here:

The important thing to understand is that, from the home twin's (her) point of view, she has no concern about what the traveling twin (he) does, or what the trailing rocket does, after the trailing rocket does its instantaneous speed change. (She knows he'll eventually get back home). All that is important to her from moment to moment is what the leading rocket does, partly because she spends a lot of time standing right next to it. And we know (and she knows) exactly what the leading rocket then does: for the (incorrect) assumption that two rockets with equal accelerometer readings will get farther apart during their accelerations, the leading rocket will instantaneously move a finite distance from the home twin (in the direction opposite to the direction to the traveling twin). And that is absurd ... it isn't possible. Therefore, the assumption that two separated rockets, with equal accelerometer readings, will have an increasing (or a decreasing) separation, is false. Their separation will be constant during the acceleration. And their separation will be constant during an instantaneous speed change.

 

[Halc]

Neddy, I have a question about your most recent diagram ... it is still not correct.

I actually did find something in it with which I disagree.
The event of Bob's 20th birthday is captioned "before/after deceleration to v=0.00". There is no 'deceleration' in physics. There is just a negative change in velocity, and thus an acceleration to v=0 from v=0.866, or from v'=0 to v'=-0.866. Deceleration is an everyday term for a decrease in scalar speed, whereas acceleration in physics is a vector change in velocity.

Just saying... Numbers are all great, but that choice of wording makes it sound like there's a preferred frame, adding to the confusion.

 

[Mike_Fontenot]

I (Mike Fontenot) said:

[...] the widely-held (but incorrect, according to me) view that two separated rockets with equal accelerometer readings will get farther apart as the acceleration proceeds.


Halc said:

Nobody who knows relativity would say that, so it is a stretch to call it a 'widely held view'. The consensus view is that the string between the ships breaks. That does not imply that they get further apart as the acceleration proceeds,[...]

I (Mike Fontenot) respond:

Your explanation (that the two rockets maintain the same separation, but the thread between them shrinks and breaks) is bizarre! Why would a tape measure (with it's tip fixed rigidly to the leading rocket, and with the casing rigidly attached to the trailing rocket, but with the roll of tape inside free to unwind or wind at will) behave differently than a thread? THAT description of the tape measure IS HOW the people on the trailing rocket would conclude that the separation between the two rockets was constant. They WOULDN'T see the tape moving either INTO, or OUT OF, the casing. The tape would be showing them a CONSTANT reading ... that's what it means to say that the people on the trailing rocket say the separation is constant.

 

[Halc]

Your explanation (that the two rockets maintain the same separation, but the thread between them shrinks and breaks) ...

I never said that. Please quote what I actually say.

I said the two rockets maintain the same separation in S. That's a very different statement than the one you attribute to me. The two rockets are separated by twice that distance in S" once Bob stops his rocket at age 30 and the distance between him and Alice's rocket (stationary in S" a long time ago in S") ceases to increase over time.

Why would a tape measure (with it's tip fixed rigidly to the leading rocket, and with the casing rigidly attached to the trailing rocket, but with the roll of tape inside free to unwind or wind at will) behave differently than a thread?

It doesn't. The tape measure, if moving with the rockets, is also length contracted in S. Since the separation is 34.6 in S, it would take 69.3 LY of tape measure (moving at 0.866c in S) length contracted down to 34.6 in S, to span that distance.

THAT description of the tape measure IS HOW the people on the trailing rocket would conclude that the separation between the two rockets was constant.

Again, a meaningless statement. If the tape measure is stationary in S", it measures the separation distance only in S", and it measures 69.3, but only after both ships have come to rest in S".

They WOULDN'T see the tape moving either INTO, or OUT OF, the casing.

Well, in S", Alice's ship suddenly stops. That alone will break the tape measure since it cannot withstand the instant acceleration, even if it is only a one meter plumb bob. But the reeling out tape measure thingy mathematically works if the acceleration is low enough, which in this case would require at most about 0.25 m/sec² of acceleration of the two ships, and this is assuming that the speed of sound is infinite, a violation of causality. So one must consequently arrange for all the parts of the string/tape-measure to be independently accelerated.

So rather than doing all that, we can just assume that the tape measure is already stationary in S" and the ships just latch onto it when they become stationary in S" as well.

The tape would be showing them a CONSTANT reading ... that's what it means to say that the people on the trailing rocket say the separation is constant.

The distance between the two ships in S" is not constant over time. Drawings by both Neddy and I show this. There is a 60 year (relative to S") period where the separation distance in S" is changing at a rate of 0.866c. It's all described in post 214, and even drawn in a picture I posted quite some time back, before I knew the exact scenario (20 out, 10 pause, 20 back), so the numbers are different.

 

[Mike_Fontenot]

I (Mike Fontenot) had said:

"Your explanation (that the two rockets maintain the same separation, but the thread between them shrinks and breaks) ...

I never said that. Please quote what I actually say.
I said the two rockets maintain the same separation in S.

OK. S is, I think, the home twin's frame (the frame in which she is at rest). So what do you say about the separation between the two rockets, according to the people on the rockets?

 

[Mike_Fontenot]

Luckily I think I understand what you mean, even though as Halc points out, you did not specify in which frame the rockets get farther apart.

In the rocket frame.

This is analogous to Bell's scenario where they remain a fixed distance apart in the initial inertial frame, even as they accelerate.

The Bell scenario is a different scenario from the the one we are discussing. In the Bell scenario, the initial inertial observers FORCE the distance between the rockets to be constant (according to their frame). That means that people on the trailing rocket will say that the separation between the rockets is INCREASING as the acceleration progresses. If there are accelerometers on the rockets, they don't show the same readings. (You're the one who taught me that, Neddy).

In contrast, in the scenario we are discussing (where the accelerometers DO show the same readings), the people on the trailing rocket will say that the separation between the two rockets is constant. (And I thought you were the one who taught me that, also).

 

[phyti]

Neddy;

The Lorentz transformations describe the simultaneity or the "now" of the events happening simultaneously with the current time in the reference frame. The traveler can know, without waiting for any light signals to be reflected back, the coordinates of hypothetical events that are happening "now" in his reference frame. It is not images of events from the past, it is the current events that are considered to be happening "now", at a certain time and place, even though the eye cannot see them yet. Different reference frames can have different simultaneities when they are in relative motion. It is all defined in the LT equations.

We don't get images from GPS satellites either, just electronic encoded signals.

Minkowski made graphical plots more usable by scaling the time variable from t to ct, enabling analysis of small time intervals. Many don't realize his spacetime graphics are the geometric equivalent of the coordinate transformations. Here is an example.
U frame:
A is moving at .5 relative to U, and sends a (blue) light signal that reflects from an object (event e). The inverse of γ is √(1-v2), which is a circular arc. The arc of radius 16 centered on the origin intersects the distance 8. The height of the arc subtracted from the radius is 13.9, the dilated A time. It is simpler than pasting in a hyperbola and there are only two clocks. The small circle on line At is where A thinks he is, i.e. his time and space have contracted by .866. He would assign the time of event e as (8-3.5)/2=2.3.
Using the red line, any clock event on At can be transferred to Ut.
A frame:
Transferring the A-times to the A frame, A's description can be formed.
Applying the coordinate transforms, provides a means of verifying the graphics.
U(8, 6) yields A(5.8, 2.3).

 

[Halc]

OK. S is, I think, the home twin's frame (the frame in which she is at rest). So what do you say about the separation between the two rockets, according to the people on the rockets?

The pictures depict S, and the separation between rockets in S is constant at any time past t=40. The picture shows this.
As to 'according to people on rockets', that all depends on which coordinate system they arbitrarily choose to label remote events. It's best to reference the frame, and not the people.

In the rocket frame.

'The rocket' doesn't define an inertial frame since it isn't inertial.

The Bell scenario is a different scenario from the the one we are discussing.

Actually, if you read the scenario, it isn't different. The scenario (per D&B, quoted by me much earlier) says simply 'identically constructed rockets', which means they have identical acceleration profiles. There's no specification of it being for instance constant over time. In our simplified scenario, the identical construction is a howitzer, shooting a ballistic object at 0.866c. That qualifies it as the same scenario. Read the actual scenario and not all the web sites that discuss it, with all the errors they often add. Wiki has several errors for instance, but their quote of D&B has no errors.

In the Bell scenario, the initial inertial observers FORCE the distance between the rockets to be constant (according to their frame).

Most importantly, this is wrong. First of all, 'their frame' doesn't define a frame. Secondly, no such wording is in the scenario. The constant separation in S is not forced, but mathematically (trivially) derived (by logic I quoted in post 145, and which you declared 'too convoluted') from the actual specification which is identical acceleration profiles.

That means that people on the trailing rocket will say that the separation between the rockets is INCREASING as the acceleration progresses.

They can say what they want, but without a frame reference, they'd be not even wrong. Yes, given identical acceleration profiles, the distance between rockets in any inertial frame in which one of the accelerating rockets is momentarily stationary must increase. In that frame, the other rocket might not even have started accelerating at all, such as if the lead rocket (34.6 LY apart) accelerates at at least 0.03g. In any inertial frame in which that rocket is momentarily stationary, the trailing rocket hasn't even launched yet, so of course it is pulling away.

If there are accelerometers on the rockets, they don't show the same readings. (You're the one who taught me that, Neddy).

By specification, both ships exhibit identical accelerations at the same proper times on their respective clocks. That means identical proper accelerations, identical coordinate accelerations in S, but not necessarily identical acceleration at any time relative to a frame other than S.

Where did Neddy 'teach you'that the rockets, specified to have identical proper accelerations, show different accelerometer measurements at the same time on their respective clocks? I mean, they're not co-located, so relative to other frames, sure they don't necessarily read the same value, but it was never specified that they do.

All this is pretty irrelevant. You've drawn the pictures of your above assertions. An observer (not on any rocket) can see it disappear from one location and then reappear a considerable distance away, and you say that is absurd, so you've falsified your assertions. One can observe the same rocket in three separate locations at once. That is also absurd.

 

[Mike_Fontenot]

My (Mike Fontenot's) writing is in red:

Apologies to both for answering a question not directed at me.

Mike: Yet again, so much of what you say would make so much more sense if you included frame references.

Sorry, I always INTEND to specify which frame I'm talking about. I suspect that usually I meant that the frame is that of the people on the rockets.

[...]

I (Mike) had said:

"the widely-held (but incorrect, according to me) view that two separated rockets with equal accelerometer readings will get farther apart as the acceleration proceeds"

And Halc responds:

Nobody who knows relativity would say that, so it is a stretch to call it a 'widely held view'.

I thought that IS what you believe. I thought I am the only person who says two rockets showing equal accelerometer readings will have a constant separation, according to the people on the rockets.

Do you really believe that?

The consensus view is that the string between the ships breaks.

The string won't break if the separation of the rockets (according to the people on the rockets) is constant. Why would you think that?

[...]

 

[Mike_Fontenot]

Again, my (Mike's) writing is in red.

The pictures depict S, and the separation between rockets in S is constant at any time past t=40. The picture shows this.
As to 'according to people on rockets', that all depends on which coordinate system they arbitrarily choose to label remote events. It's best to reference the frame, and not the people.

The people on the rockets certainly have their own frame. When they say "the separation between our two rockets is constant", they are NOT talking about ANY inertial frame. They have little or no interest in any inertial frame. They are talking about THEIR frame.

[...]

 

[Neddy Bate]

The Bell scenario is a different scenario from the the one we are discussing. In the Bell scenario, the initial inertial observers FORCE the distance between the rockets to be constant (according to their frame). That means that people on the trailing rocket will say that the separation between the rockets is INCREASING as the acceleration progresses. If there are accelerometers on the rockets, they don't show the same readings. (You're the one who taught me that, Neddy).

I have tried in vain to tell you many times that the two accelerometers in Bell's scenario both read the same. Bell's setup is "two rockets that accelerate identically in the initial inertial frame, keeping constant distance between them in that frame". I told you only "nonsense physics" would interpret that to mean two different accelerometer readings. Accelerometers measure acceleration, after all.

In contrast, in the scenario we are discussing (where the accelerometers DO show the same readings), the people on the trailing rocket will say that the separation between the two rockets is constant. (And I thought you were the one who taught me that, also).

The two accelerometers display the same reading in your setup also.

 

[Neddy Bate]

I think I have figured out the problem here. At first I thought the reason Mike was saying our replies to him are things like "incoherent" was simply because he was choosing to believe his own paper rather than even try to consider if what we are saying is true. There is still an element of that in play here, but it goes even further. His own paper has literally made it impossible for him to understand Bell's scenario. There is a filter in his brain that changes the words from what is written on the screen into what he thinks they should be, preventing him from understanding.

 

[Neddy Bate]

I actually did find something in it with which I disagree.
The event of Bob's 20th birthday is captioned "before/after deceleration to v=0.00". There is no 'deceleration' in physics. There is just a negative change in velocity, and thus an acceleration to v=0 from v=0.866, or from v'=0 to v'=-0.866. Deceleration is an everyday term for a decrease in scalar speed, whereas acceleration in physics is a vector change in velocity.

Just saying... Numbers are all great, but that choice of wording makes it sound like there's a preferred frame, adding to the confusion.

You are correct, but the velocities I am using are all measured in the S frame for sake of consistency. Is this better?

 

[Mike_Fontenot]

Neddy, in the Bell paradox, the trajectories of the two rockets, according to the initial inertial observers, would be identical, except that the curve for the leading rocket would be higher than the trailing rocket on the chart, by the initial separation. Do you agree with that? That means the separation, according to the people on the trailing rocket, CAN'T be constant. The length contraction equation (LCE) won't allow that. The people on the trailing rocket will conclude that the separation is increasing. And I claim that the accelerometers do NOT display the same readings. Do you agree with that?

 

[Neddy Bate]

Neddy;


We don't get images from GPS satellites either, just electronic encoded signals.

Minkowski made graphical plots more usable by scaling the time variable from t to ct, enabling analysis of small time intervals. Many don't realize his spacetime graphics are the geometric equivalent of the coordinate transformations. Here is an example.
U frame:
A is moving at .5 relative to U, and sends a (blue) light signal that reflects from an object (event e). The inverse of γ is √(1-v2), which is a circular arc. The arc of radius 16 centered on the origin intersects the distance 8. The height of the arc subtracted from the radius is 13.9, the dilated A time. It is simpler than pasting in a hyperbola and there are only two clocks. The small circle on line At is where A thinks he is, i.e. his time and space have contracted by .866. He would assign the time of event e as (8-3.5)/2=2.3.
Using the red line, any clock event on At can be transferred to Ut.
A frame:
Transferring the A-times to the A frame, A's description can be formed.
Applying the coordinate transforms, provides a means of verifying the graphics.
U(8, 6) yields A(5.8, 2.3).

View attachment 5609

I know that Minkowski diagrams are the graphical equivalent of the Lorentz transforms. I have both the graphics and the coordinate transformations in my diagram. There are no light signals on my diagram. The only thing light is used for is to Einstein synchronise all the clocks in all the inertial systems.

I'm sorry but I don't understand anything else in your post.

 

[Mike_Fontenot]

[...]
His (Mike's) own paper has literally made it impossible for him to understand Bell's scenario.
[...]

I understand the Bell scenario just fine. The Bell scenario is different from my scenario. By making the rocket separation constant according to the initial inertial observers, in their scenario the trailing rocket people MUST say that that the separation is increasing. The length contraction equation demands that. That means the accelerometers will show different readings.

In my scenario, the trailing rocket people say that the separation is constant. That means that the accelerometers will show the same readings. And the initial inertial observers will say that the separation decreases with time. The length contraction equation demands that.

 

[Mike_Fontenot]

I have tried in vain to tell you many times that the two accelerometers in Bell's scenario both read the same.

That's not possible, with their scenario description. If the initial inertial observers say the separation of the rockets is constant, then the length contraction equation requires that the separation of the rockets in the rocket frame must be increasing. And if the separation of the rockets in the rocket frame is increasing, then the accelerometers can't show the same readings.