Bell’s Spaceship Paradox. Does the string break?

[Mike_Fontenot]

I (Mike Fontenot) am responding in red:

Are you seriously thinking that when the traveling twin changes his LoS it will instantaneously affect the home twin and cause her to see illusions?

Of course not. My recent comments haven't mentioned the traveling twin at all. My focus is only on the home twin, and on the leading rocket that is stationary by her side when she is 50- and for some time before. Then, when she is 50+, by prior agreement, the leading rocket does its delta-function acceleration (i.e., it instantaneously changes its speed from zero to 0.866 ly/y, in the direction away from the traveling twin). But much more important than its speed change is that, by assumption, the distance between the trailing rocket and the leading rocket instantaneously increases by a finite amount. That means that she directly SEES the leading rocket instantaneously leave her side and instantly reappear a finite distance away, in the direction opposite to the traveling twin. (All of that happens at the instant 50+ in her life). THAT instantaneous displacement is impossible and ABSURD, so we must conclude that the assumption that the two rockets get farther apart during their acceleration is incorrect ... we have proven that the separation of the two rockets during their acceleration is constant.

[...]

 

[Mike_Fontenot]

As before, I will address your post by responding in red:

Are you seriously thinking that when the traveling twin changes his LoS it will instantaneously affect the home twin and cause her to see illusions?

No, of course not. In this scenario, the leading rocket is stationary wrt the home twin when she is 50-, and continuously has been for ten years before that. And the trailing rocket has also been stationary wrt the home twin for the last 10 years of her life. Both of those rockets have been programmed to instantaneously change their speeds from zero to 0.866 ly/y at the same instant, ACCORDING TO THE HOME TWIN, when she is 50+. I don't need to refer to the traveling twin at all, and I don't.

 

[Neddy Bate]

Of course not. My recent comments haven't mentioned the traveling twin at all. My focus is only on the home twin, and on the leading rocket that is stationary by her side when she is 50- and for some time before. Then, when she is 50+, by prior agreement, the leading rocket does its delta-function acceleration (i.e., it instantaneously changes its speed from zero to 0.866 ly/y, in the direction away from the traveling twin). But much more important than its speed change is that, by assumption, the distance between the trailing rocket and the leading rocket instantaneously increases by a finite amount. That means that she directly SEES the leading rocket instantaneously leave her side and instantly reappear a finite distance away, in the direction opposite to the traveling twin. (All of that happens at the instant 50+ in her life). THAT instantaneous displacement is impossible and ABSURD, so we must conclude that the assumption that the two rockets get farther apart during their acceleration is incorrect ... we have proven that the separation of the two rockets during their acceleration is constant.

Interesting that you don't even make any distinction between the traveling twin and the home twin. It is the traveling twin that says the leading rocket instantly relocates, not the home twin.

No, of course not. In this scenario, the leading rocket is stationary wrt the home twin when she is 50-, and continuously has been for ten years before that. And the trailing rocket has also been stationary wrt the home twin for the last 10 years of her life. Both of those rockets have been programmed to instantaneously change their speeds from zero to 0.866 ly/y at the same instant, ACCORDING TO THE HOME TWIN, when she is 50+. I don't need to refer to the traveling twin at all, and I don't.

Okay, enjoy your day.

 

[Neddy Bate]

Here is the latest update on my Minkowski diagram:

Please note that I added coordinates for the event where the leading rocket accelerates away from the home twin. The coordinates for that event in the S'' double primed frame are (x'',t'')=(86.60,100.00) and the coordinates for the trailing rocket (or traveling twin) in the S'' double primed frame are (x'',t'')=(155.88,160.00). So even though those events are not simultaneous in the S'' double primed frame, they are still 155.88-86.60=69.28 light years apart in that frame. There is no relative motion between the rockets in that frame, so it does not matter whether the events are simultaneous or not. The string breaks simply because it is not long enough.

I found that interesting because this shows that the reason the string breaks is simply due to it being denied the natural need to become length contracted. It does not even have to do with the LoS pointing to the event at the upper left corner (which is the same x'' coordinate anyway, x''=86.60.) Neat!

So Mike, if you insist on using the wrong LoS, that is okay. Even using the one that you prefer (the one that points to her being 50 years old instead of 80 years old), the string is still shown to break by these x'' coordinates. You're welcome! Have a great life.

 

[phyti]

Mike;

You don't have a (green) los from Bt20 to At10.
B would have to continue (magenta) past Bt20 to make the required measurement.
You have enough info to calculate an los for At5.4.
You claim to see the future of A, but you only see the past. At Bt30 you see 5.4+10=15.4, the last image from A!

You don't get something for nothing, and there is no instant knowledge.

 

[phyti]

For those who don't think the triangle is discontinuous at the reversal,
refer to ‘classification of discontinuities’ at Wiki.
The scenario is fantasy physics.

 

[Mike_Fontenot]

My (Mike Fontenot's) comments are in red:

Interesting that you don't even make any distinction between the traveling twin and the home twin.

I have no idea what you mean by the above.

It is the traveling twin that says the leading rocket instantly relocates, not the home twin.

Why in the world do you say that? The home twin and the leading rocket are colocated for YEARS before the leading rocket instantaneously changes its location. Who is better able to say what happens to the leading rocket when it instantaneously changes its speed than the person who has been colocated and stationary with it for years? She directly SEES the leading rocket fire. No one else does.

 

[Neddy Bate]

Mike,

I will try to explain this one more time using a different analogy. The worldlines are like spaghetti strings that are baked into a loaf of bread (spacetime). Nothing can ever move faster than c in any inertial frame, so that defines how the spaghetti strings are laid out in the loaf. You will note that in the Minkowski diagrams we all drew, nothing ever exceeds c in the worldlines shown.

The lines of simultaneity are like laser knife cuts through the loaf of bread. The laser cuts do not and can not change the string's paths. No matter how they cut the loaf, there will still never be anything that moves faster than c in any inertial frame. You will note that the only place in our Minkowski diagrams where the leading rocket instantaneously changes its location is when the LoS that points to her being 50 changes to the LoS that points to her being 80. That is the non-inertial frame of the traveling twin when he accelerates. He can calculate things moving faster than c because he is not inertial. That is not going to affect what she sees, because she is inertial. It is simply a calculation that the traveling twin makes on his calculator or with pen and paper. It is a concept. The traveling twin might do the calculation right, or wrong, either way it does not affect her life at all.

To say that the the calculation he makes would result in her seeing the rocket relocate is like saying that if you drew a map of the United States without California on it, that would result in people in L.A. drowning in the Pacific ocean. The map is not the territory.

Look, you and I have discussed scenarios where her age goes forwards, backwards, and then forwards again. How would you feel if a novice jumped into our converation and said, "If she sees her own age go backwards that is an Absurdity!" Do we just congratulate the novice on being smarter than all of us? No, we correct them and tell them it does not work that way. If you want to believe otherwise, go ahead. But remember, the map is not the territory. The people of L.A. are not going to drown if you draw the US map without California on it. And she is not going to see the rocket move faster than light no matter what LoS anyone uses. Peace out brother.

 

[Mike_Fontenot]

Neddy, your latest post is completely incoherent. I think you are beyond hope.

 

[Halc]

Are you seriously thinking that when the traveling twin changes his LoS it will instantaneously affect the home twin and cause her to see illusions?

Of course not.

says the guy who concludes:

That means that she directly SEES the leading rocket instantaneously leave her side and instantly reappear a finite distance away

Indeed of course not, but your "proof" relies on exactly that presumption, it all being a non-sequitur without it.

Are you seriously thinking that when the traveling twin changes his LoS it will instantaneously affect the home twin and cause her to see illusions?

Of course not.

Indeed of course not, but your "proof" relies on exactly that presumption, it all being a non-sequitur without it. To illustrate, consider I can use your own logic to disprove any statement.

I still stand by everything I've stated ... and I'm certain that I'm correct.

OK, I am going to disprove that you are certain about being correct, using the very 'proof' you have provided. I will take it word for word except some critical parts which I will highlight in bold.

MF believing he is correct 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.But IF it's true that MF believes he is correct, 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 speedtoward 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 it is ASSUMED that Mike believes he is correct. THAT would result in HER seeing the leading rocket INSTANTANEOUSLY move a finite distance away from her, WHICH IS ABSURD! So the ASSUMPTION that Mike believes he is correct CAN’T be correct

Now I think this 'proof' has many of the same errors as I've pointed out earlier, but you stand by the logic of it. Therefore you cannot believe that you're correct. What's wrong with the proof? I mean, he concludes from his instantaneous speed change that she suddenly experiences being 80 and there being 30 years of travel between her and the leading rocket, which is ABSURD, therefore the assumption (about your believing you are correct) must be wrong.

Back to your post. It is full of errors, same as the 'proof'.

the leading rocket does its delta-function acceleration (i.e., it instantaneously changes its speed from zero to 0.866 ly/y, in the direction away from the traveling twin).

The speed of everything is frame dependent. Your statement is true in S and only S, but no frame mentioned. That's an error.

But much more important than its speed change is that, by assumption, the distance between the trailing rocket and the leading rocket instantaneously increases by a finite amount.

Separation distance between worldlines is a frame dependent quantity. In no inertial frame does the separation distance abruptly change. It cannot. In S it never changes at all. In S" it finishes changing when the trailing ship does it's return acceleration, but it had been growing (from 17.3) for 30 years in that frame. Again, in no inertial frame is there an abrupt change in separation distance. That would indeed be absurd since nothing moves faster than c in any inertial frame. You of course explicitly deny this, defending FTL motion, but then 'you stand by your personal beliefs'.

That means that she directly SEES the leading rocket instantaneously leave her side and instantly reappear a finite distance away

So you insist. That assertion proves (via the proof submitted above) that you don't believe your own crap.

 

[Mike_Fontenot]

Halc, I think you've caught what Neddy's got. My response to you is the same as my response to Neddy:

Your latest post is completely incoherent. I think you are beyond hope.

 

[Neddy Bate]

Neddy, your latest post is completely incoherent. I think you are beyond hope.

Okay, well sorry you can't understand it. I thought post # 208 was one of my clearest posts in this thread. It's a simple analogy where spacetime is a loaf of bread! Slicing bread in different ways doesn't change what is inside the bread. The slices are, by analogy, LoS.

But I like my post # 204 even more. Isn't it neat how the x'' coordinates show the string would break even when the leading rocket is still co-located with the home twin? That is because once that string is moving 0.866c relative to her, the string is prevented from becoming length contracted due to the other end of the string being held a fixed distance away by the trailing rocket. So, even if your proof were right, it still would not be enough to disprove the thread breaking. The thread breaking does not even require the leading rocket to move any distance away from the home twin, it just has to achieve relativistic speed, which it does instantaneously in your example. That is so cool!

 

[exchemist]

A bit off-topic, but I see Mike actually has a book (or pamphlet) out on Amazon, which one can even buy in the UK for £5.59 : https://www.amazon.co.uk/Simultaneity-Accelerated-Observers-Special-Relativity/dp/B08C98YX3X

 

[Halc]

Halc, I think you've caught what Neddy's got. My response to you is the same as my response to Neddy:

Your latest post is completely incoherent. I think you are beyond hope.

Expected that. When cornered, bury head in sand. Your assertions don't stand up to analysis, and you abort (claim incoherency) rather than defend your contradictions.

Let me put things a little simpler since we all drew the same picture, but without the outbound. Reminder: Frame S is depicted, the one in which Alice is permanently stationary. S' is the frame in which Bob's ship is stationary while outbound. S" is the frame in which the two ships are stationary, running in parallel as Bob returns.

The contention seems to be if the string between those two ships breaks. It does, as everybody's diagram shows, faulty 'proof' otherwise notwithstanding.

In frame S, the two ships are equidistant at all times after t=40. They were colocated at t=0, and distanced themselves from each other as Bob departed for hist destination 34.6 LY away. Thereafter their separation in S is constant, as all our diagrams show. After the acceleration events at t=50, any string moving with those ships would need to be 69 LY long to span the 34.6 LY gap since it would be length contracted. So it breaks.

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.

All this is shown on the diagrams, especially the ones that put S" coordinates for each of the events.
Let me know which parts you disagree with, or if, as usual, you find this all too complicated and you go for the ostrich option. You will notice that at not time in any frame does any object teleport or otherwise move faster than c. All our lastest pictures show that. Your older pictures are another story.

A bit off-topic, but I see Mike actually has a book (or pamphlet) out on Amazon, which one can even buy in the UK for £5.59

Yes, he's spammed those links on TNS now and then, removed since selling products is forbidden there. The papers are all available for free elsewhere, and exhibited more understanding of SR than recently, even if they're still a crock. It's now a full embrace of the inner troll.

The paper contradicts the more recent assertions. In this topic, he says that the accelerating guy 'must conclude' this and that, but the pamphlet encourages him concluding other things in order to avoid distasteful things like distant people getting younger when you accelerate away from them, so the same disconnect between abstract coordinate choice and physical reality.

 

[Mike_Fontenot]

A bit off-topic, but I see Mike actually has a book (or pamphlet) out on Amazon, which one can even buy in the UK for £5.59 : https://www.amazon.co.uk/Simultaneity-Accelerated-Observers-Special-Relativity/dp/B08C98YX3X

My later results showed that that method is incorrect. The correct result is the CMIF result. It is of value ONLY for people who can't stand the correct result, which says the traveling twin (he) will say that the home twin (she) instantaneously gets much older when he reverses course. Even more intolerable for some people is the result that, if he instantaneously goes faster AWAY from her, he says she instantaneously gets YOUNGER.

 

[James R]

Mike,

Why aren't you addressing any of the criticisms made against your position?

They are not "incoherent", as you claim. They are quite clear. And yet, we hear nothing in response from you, other than additional assertions, or restatements of the claims you have made that are debunked by the responses.

Are you hoping to make money by selling your faulty pamphlet? That might explain why you don't seem to care whether you're right or wrong. Or is there some other reason you can't or won't address objections to your work?

You came to us, remember. We didn't come to you. You asked for feedback on your theory about the breaking string, but when people gave you feedback, you suddenly became uninterested in listening to any of it, or in attempting to respond. Instead, you just complain weakly that you don't understand the responses you got because you think, for some reason, that they are "incoherent". As far as I can tell, you have given no reasons for calling them incoherent, except that you don't want to think about them.

Is this the best you can do?

 

[exchemist]

Mike,

Why aren't you addressing any of the criticisms made against your position?

They are not "incoherent", as you claim. They are quite clear. And yet, we hear nothing in response from you, other than additional assertions, or restatements of the claims you have made that are debunked by the responses.

Are you hoping to make money by selling your faulty pamphlet? That might explain why you don't seem to care whether you're right or wrong. Or is there some other reason you can't or won't address objections to your work?

You came to us, remember. We didn't come to you. You asked for feedback on your theory about the breaking string, but when people gave you feedback, you suddenly became uninterested in listening to any of it, or in attempting to respond. Instead, you just complain weakly that you don't understand the responses you got because you think, for some reason, that they are "incoherent". As far as I can tell, you have given no reasons for calling them incoherent, except that you don't want to think about them.

Is this the best you can do?

Moneymaking won't be the motive. From the resumés I have found on line, Mike seems to be another retired autodidact with a hobby, that's all.

It does seem a bit grandiose to self-publish (unreviewed) stuff on Amazon, but he's not the first amateur to do that either. That notorious (and much-banned) crank Farsight, a.k.a. John Duffield, has a book on Amazon too: https://www.amazon.co.uk/RELATIVITY-Theory-Everything-John-Duffield/dp/0956097804 .

P.S. Even I understood Neddy's last post, by the way.

 

[Mike_Fontenot]

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.

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.

 

[phyti]

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.
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 fantasy triangle case changes direction during the B trip, thus it doesn't qualify as an inertial frame.
Notice in the B frame, gamma = 2 for the 1st and 4th quarter, but changes for the middle half.
Notice in the A frame, B is at the same distance (18.6) when sending and receiving his signal.


Fig.1 shows a real world transition of the velocity from positive to negative.
Fig.2 shows the fantasy instantaneous transition, where for the common instant the velocity is 0, i.e. B is at rest relative to A, and the los is horizontal.

 

[Halc]

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. It is exactly the lack of them that seems to be driving your confusion. It is a mistake made multiple times in almost all your posts, including the one I'm quoting.
A frame reference is of the form "relative to S, S', or S"". It would at least help if you said 'according to Bob', which is better than nothing, and despite a frame not being identified without knowing the beliefs/conclusions of Bob, it allows us to presume you're referencing the inertial frame in which Bob is stationary at some event on his worldline. But you don't even do that, rendering meaningless any statement about a frame dependent value.

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

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, which would be a meaningless statement as worded. That's a physical and thus frame independent fact. The separation distance is neither.

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.

Not even wrong, for the same reason. Neddy's drawing, (the one all three of us drew) has correct geometry, and all three drawings demonstrate the doubled length of string needed to maintain a connection between the rockets. Your drawing shows this as much as ours.

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).

Yes. That's one thing you've neglected to put on your drawing is that lead-rocket worldline. We presume you would put it in the same place, but if not, please say where you think it goes, and then the three drawings will no longer match.

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.

That would be absurd, as your 'proof' states. There is no teleporting going on in any valid SR scenario.
If your assertion is that you posit such an abrupt teleport, then your assertions are falsified because you rightly declare such teleporting to be absurd.