# Bell’s Spaceship Paradox. Does the string break?

Thanks for posting that, Mike (# 239 and # 240). It demonstrates exactly what I was trying to say in post # 235. When I type something that says "Bell's description has identical acceleration in the inertial frame which means identical accelerometer readings" your brain filters that through your so-called paper and you state the opposite in your reply. I'm sorry but you cannot make any progress if you can't even recognize any of the corrections to your mistakes.

Neddy, why didn't you address the SUBSTANCE of my last two postings? What is your response to my comments about the use of the length contraction equation in this discussion?

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.
Yet again, the rockets are not inertial, so a reference to them does not identify a frame.

You seem to have indirectly conceded that your 'proof' has errors in it, and its conclusion is faulty, if nothing else by the fact that you incorrectly say that Bell's scenario is irrelevant, and yet it drags in Bell's scenario. You're contradicting yourself.

So I will attempt to fix some of the errors (in red, the way you like it. The first one references only frame S in which Alice is always stationary,
First of all, we're going to reference the case where he's already out there, since it just isn't Bell's unless they're relatively at rest at first, and Alice's rocket undergoes the same acceleration as does Bob's. This isn't true at all if there's an abrupt turnaround without a pause, until he's 30 in this case.

A Constant Separation of the Rockets in S does not Contradict anything

During the traveler's (his) instantaneous acceleration homeward when he turns 30, his home twin's (Alice) age of 50 in S is unchanged. But IF it's true that the two separated rockets in the Bell's Paradox (whose accelerometers show equal constant readings) DON'T maintain a constant separation in S during Bob's acceleration, that CONTRADICTS Alice's empirical observations.

Here's how to see that contradiction:

Bob's acceleration duration is zero, so in any inertial frame, including that of S, the duration of Alice's worldline is also zero. Therefore she should experience no change in location of the ship in her vicinity during that period of zero time. 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 in S isn't constant during that one moment in time CAN’T be correct.

I will do it again, this time using the S" frame in which the rockets are stationary after their respective acceleration events:

A Non-Constant Separation of the Rockets in S" does not Contradict anything

During the traveler's (his) instantaneous coming to a halt in S" when he turns 30, his home twin's (Alice) age of 80 in S is unchanged, so similar to above, Alice notices no teleporting of anything at any time. The two separated rockets maintaining a constant separation in S" would contradict relativity of simultaneity.

Here's how to see that contradiction:

In frame S, the two rocket accelerations are simultaneous. Thus in frame S" (or any other frame in which the x component of pre-acceleration velocity is nonzero), those two events cannot be simultaneous. If they're not simultaneous in S", then one acceleration occurs before (in S") the other, and that must change the separation distance in S.

In this particular instance, Alice's ship stops (in S") first, and Bob's ship stops (in S") 60 years (in S") later.
My apologies, but most of the text got removed since it was all either wrong or irrelevant. There's just no inertial frame (or any other kind of frame for that matter) where the worldline of any object (person, rocket, clock, whatever) is discontinuous. You agree with this because you drew the same picture as us, at least for this abrupt acceleration case. Your conclusion seems to rely on such a discontinuity, which has just not been demonstrated, mostly due to you fooling yourself by your own lack of frame references.

Either wording shows that a string stationary in S connecting the rockets while the rockets are stationionary in S would be of length 34.6. It also shows that after the acceleration events both take place, a string stationary in S" connecting the rockets after the rockets both become stationary in S", would be longer.

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"

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

Mike responds:
I thought that IS what you believe.
It, lacking a frame reference, is meaningless, and nobody who knows their relativity would make such a frame-dependent statement sans frame reference. They maintain constant separation in S, so your comment as worded is incorrect, and only utter novices would say such a thing. There are a lot of utter novices, so I didn't deny the truth of the comment, I just said that those who know their relativity wouldn't have said that.

The string won't break if the separation of the rockets (according to the people on the rockets) is constant.
For that to happen, the front rocket would have to have less acceleration in order to not be moving faster than the rear rocket in the inertial frame of either rocket. A denial of that is a denial of relativity of simultaneity. If they're constantly accelerating and moving at the same velocity in S, they can't be moving at identical velocity in another frame. That would imply that the velocity is the same at two different times, contradicting the fact that each is supposedly accelerating.

The people on the rockets certainly have their own frame.
Since all people and other objects are in every frame, any particular frame is not necessarily 'theirs'. Almost nobody uses such a convention. When I get in my car, I consider myself to be going fast, not the road. That's me choosing to use a frame other than the one in which I am stationary.

Specify the frame, especially since the people are not stationary, so saying 'Bob's frame' does not identify an particular frame at all. It's easier and far clearer to say 'in S' than it is to say 'according to accelerating Bob'.

When they say "the separation between our two rockets is constant", they are NOT talking about ANY inertial frame.
Obviously not, since no frame is mentioned. So be better than that and mention the frame.

There are many forms of accelerating coordinate systems. Saying 'Bob's accelerating frame' does not identify which of those systems you're using, or which Bob is using. For instance, you have zero clue which coordinate system Einstein was using when he wrote about that equation that you called the GTD. He was using Lass coordinates, which makes sense since it most closely resembles the way simultaneity conventions are used in his earlier works.

Yes, if Bob switches abruptly from using S to using S" (something he's free to do anytime, not requiring acceleration at all), then he changes his LoS to a different event on the lead rocket's worldline. It's a change in abstraction only, not a physical change to the worldline of the rocket. But your 'proof' (omitting all frame references) totally misses that.

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Yet again, the rockets are not inertial, so a reference to them does not identify a frame.

Your statement makes no sense to me at all. It is ridiculous to claim that a person who is accelerating doesn't have his own frame.

Someone (Tom) who is accelerating DOES have his own frame. He still can wonder "How old is Jane RIGHT NOW (a friend whom he met several years ago, and who has been moving in an arbitrary manner, at near the speed of light wrt him for a long time, and who is vary distant from him)"?

And there is a specific answer to that question that Tom has asked. The answer is, "Jane is the age that an inertial person (Sam), who happens to momentarily colocated with her right now, says she is. Sam just asks Jane how old she is, right at that instant, and relays her answer to Tom.

You seem to have indirectly conceded that your 'proof' has errors in it

No, there are no errors in it.

I'm done arguing with you and Neddy ... it's like trying to reason with a rock.

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Neddy;
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.

Can't agree here. The question asked by the observer is, 'what is my clock event that corresponds to the distant event of interest e?'. That implies they are aware of the event e. The los does not allow looking into the future. Her los is horizontal.
When she is 10 the last clock image from him is Bt=2.7. She is only aware of events within her light cone. If she makes periodic measurements, she will calculate his clock was Bt=5 when she was 10. His clock Bt=20 is 30 yr in her future!
The transformation only works IF light from At=10 reaches B at Bt=37.3. That does NOT happen since B instantaneously reverses at Bt=20. An los is never formed for At10 to Bt20!
It's an experiment that can't be done.

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

B sends a signal to A at BT=10.7, requiring 9.3 yr. He is much closer because of the
instantaneous reversal.
There is no constant velocity for the experiment. Constant speed is not the same.

My (Mike Fontenot's) statements are all in red below:

I (Mike Fontenot) had said:

"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."

I do not believe this is true.

I THINK Neddy believed it, at least some years ago, and maybe he still does. Halc may have never believed it, but I'm not sure either way. There ARE a lot of people who DON'T believe it, but Brian Greene (the famous physicist at Columbia University) DOES believe it , as evidenced by his NOVA program about the alien at the other side of our universe. To anyone (including you, Dave) who doesn't believe it, I would ask "How do YOU explain the twin paradox?".

What citation are you referring to that concludes this?

I think Taylor and Wheeler's book "Spacetime Physics" might give that resolution.

Neddy;Can't agree here.
I actually saw nothing in your post that differed from what Neddy said.

The question asked by the observer is, 'what is my clock event that corresponds to the distant event of interest e?'
No, in the twins scenario, the only question asked by any observer is to compare ages once reunited, and that can be done without light since they are in each other's presence. That's the whole point of the scenario, to demonstrate differential aging. There is no need for anybody anywhere to communicate during any part of the separation.

That implies they are aware of the event e. The los does not allow looking into the future.
I am aware of my next birthday, despite the fact that I cannot see it. For that matter, I cannot see my prior one either (not directly anyway), but I am quite aware of both events. Likewise, everybody knows the itinerary, and since it is a thought experiment, the only person who might actually care about events other than the final comparison is the narrator of the story.

Her los is horizontal.
Her los is our gain. Sorry, couldn't resist such low hanging fruit.

Her LoS is entirely abstract and is wherever she decides it goes. There is nothing physical about a LoS. She likely chooses the LoS of the solar system, which seems most pragmatic for the situation. But then, the traveler probably also chooses that one for the same reason, at least when considering what's going on back home, but using a different one (still not his own) for local tasks like navigating to the head.

Not sure what you're trying to accomplish with all the posts which seem unrelated to what the OP is pushing. Your drawings are full of light and LoS lines, all of which are pretty unnecessary to explain the measured time differential, especially in this simplified SR example.

My apologies, Mike. I missed this post of yours:

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.

Yes, I agree with that. So does that mean you agree that the string breaks in Bell's scenario? I never really heard you agree it breaks, and that would really help me understand your position if you do agree that it does break.

And I claim that the accelerometers do NOT display the same readings. Do you agree with that?

I know you claim the accelerometers show different readings in Bell's scenario, but please allow me to keep this part of your claim separate for now. We can go back to the accelerometers once we have some agreement on the string breaking, at least.

My (Mike's) responses are in red.

"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 Neddy just now responded:

Yes, I agree with that. So does that mean you agree that the string breaks in Bell's scenario? I never really heard you agree it breaks, and that would really help me understand your position if you do agree that it does break.

Yes, given that the separation has increased, according to the people on the trailing rocket, the string must break in the the Bell scenario. But that is a different scenario than the one I've been addressing.

I know you claim the accelerometers show different readings in Bell's scenario,

Yes. If the separation is increasing according to the people on the trailing rocket, the leading accelerometer must show a greater acceleration than the trailing accelerometer.

but please allow me to keep this part of your claim separate for now. We can go back to the accelerometers once we have some agreement on the string breaking, at least.

Yes, given that the separation has increased, according to the people on the trailing rocket, the string must break in the the Bell scenario. But that is a different scenario than the one I've been addressing.

Great! So let me just clarify one more more thing on the Bell scenario. As the separation increases according to the people on the trailing rocket, surely you do not think that this causes the leading rocket to instantaneously relocate in the initial inertial frame, correct? The scenario requires that the separation must be constant in the initial inertial frame, and that is what lead us to conclude the separation increases according to the people on the trailing rocket. So hopefully you agree that we do not go back to the initial inertial frame and say the separation must also increase there, correct?

The text in red is mine (Mike's):

I (Mike) had previously said:

"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 then you (Neddy) said:

"So does that mean you agree that the string breaks in Bell's scenario?"

And then I (Mike) said:

"Yes, given that the separation has increased, according to the people on the trailing rocket, the string must break in the the Bell scenario. But that is a different scenario than the one I've been addressing."

And then you (Neddy) said:

"As the separation increases according to the people on the trailing rocket, surely you do not think that this causes the leading rocket to instantaneously relocate in the initial inertial frame, correct?"

In the above, we've been discussing the case where the accelerations are constant and FINITE, and last for a finite DURATION "T". But your (Neddy's) above question then asks whether "this causes the leading rocket to instantaneously relocate in the initial inertial frame?" In the FINITE case that we have been discussing, there are NO instantaneous relocations ... the leading rocket, during the FINITE duration "T" moves forward by a given distance during a FINITE time. It is only in the case where we let the duration "T" approach zero, and the acceleration "A" approach infinity (in such a way that the DISTANCE MOVED remains the same for each iteration) that we get the INSTANTANEOUS relocation. And an instantaneous relocation IS impossible, thus proving that the assumed increase in the separation of the rockets (whose accelerometers show the same readings) is incorrect.

In the above, we've been discussing the case where the accelerations are constant and FINITE, and last for a finite DURATION "T". But your (Neddy's) above question then asks [/COLOR]whether "this causes the leading rocket to instantaneously relocate in the initial inertial frame?" In the FINITE case that we have been discussing, there are NO instantaneous relocations ... the leading rocket, during the FINITE duration "T" moves forward by a given distance during a FINITE time. It is only in the case where we let the duration "T" approach zero, and the acceleration "A" approach infinity (in such a way that the DISTANCE MOVED remains the same for each iteration) that we get the INSTANTANEOUS relocation. And an instantaneous relocation IS impossible, thus proving that the assumed increase in the separation of the rockets (whose accelerometers show the same readings) is incorrect.

Right now we are discussing the Bell scenario. Please stick to one topic at a time. Does the fact that the separation distance increases in the rocket frame also mean the separation distance increases in the initial inertial frame?

The red is by me (Mike).

Right now we are discussing the Bell scenario.

I didn't realize that we hadn't switched to the scenario in my viXra paper ... I thought we were in agreement that the Bell scenario is a DIFFERENT scenario from the scenario in my viXra paper. And I thought we were in complete agreement about the Bell scenario itself.

Does the fact that the separation distance increases in the rocket frame also mean the separation distance increases in the initial inertial frame?

No. The length contraction equation (LCE) requires that, if the separation is constant in the initial inertial frame, the separation must be INCREASING in the frame of the trailing rocket.

Mike;
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).

There is no acceleration. It's magic, i.e. not according to the laws of physics.
The state of the rockets changes without any change of energy!
When they instantly acquire their motion relative to her, their motion alters their perception and measurements. Their perception is a greater measured separation.
The reality is there is no change in separation.
Erom the rocket pair ref. frame, she moving at .866 will measure their separation as
.69/2=34.5.

I didn't realize that we hadn't switched to the scenario in my viXra paper ... I thought we were in agreement that the Bell scenario is a DIFFERENT scenario from the scenario in my viXra paper. And I thought we were in complete agreement about the Bell scenario itself.

I just wanted to make sure I understood your position on the Bell scenario, and I think I am very close now.

No. The length contraction equation (LCE) requires that, if the separation is constant in the initial inertial frame, the separation must be INCREASING in the frame of the trailing rocket.

Okay, perfect. So, if we have two rockets which accelerate in such a way that they maintain a constant separation distance in the initial inertial frame, then you would have no problem with the people on the trailing rocket saying the separation distance has increased in the rockets frame, even though the separation distance is constant in the initial inertial frame.

So, before we go to the scenario in your paper, let's do one as similar as possible to it, but without the infinite acceleration. Let the acceleration be finite but great enough that we have two rockets reaching 0.866c very quickly as measured by the initial inertial frame. They maintain a constant separation distance in the initial inertial frame. What is your position on the separation distance according to the people on the trailing rocket?

I just wanted to make sure I understood your position on the Bell scenario, and I think I am very close now.

I (Mike) had said:

"The length contraction equation (LCE) requires that, if the separation is constant in the initial inertial frame, the separation must be INCREASING in the frame of the trailing rocket."

Okay, perfect. So, if we have two rockets which accelerate in such a way that they maintain a constant separation distance in the initial inertial frame, then you would have no problem with the people on the trailing rocket saying the separation distance has increased in the rockets frame, even though the separation distance is constant in the initial inertial frame.

Yes, not only would I "have no problem with that", I would say that the length contraction equation doesn't allow the initial inertial people and the trailing rocket people to BOTH say the separation is constant. The separation according to the rocket people MUST be greater than the separation according to the initial inertial people. The LCE REQUIRES that.

So, before we go to the scenario in your paper, let's do one as similar as possible to it, but without the infinite acceleration. Let the acceleration be finite (and the duration of the acceleration be finite) but with the acceleration great enough that we have two rockets reaching 0.866c very quickly as measured by the initial inertial frame. They maintain a constant separation distance in the initial inertial frame. What is your position on the separation distance according to the people on the trailing rocket in that case? If the initial inertial people say the separation is constant, then the trailing rocket people MUST say that the separation is increasing. The LCE requires that.

Yes, not only would I "have no problem with that", I would say that the length contraction equation doesn't allow the initial inertial people and the trailing rocket people to BOTH say the separation is constant. The separation according to the rocket people MUST be greater than the separation according to the initial inertial people. The LCE REQUIRES that.

And I would agree. Progress!

in that case? If the initial inertial people say the separation is constant, then the trailing rocket people MUST say that the separation is increasing. The LCE requires that.

Indeed. We are in agreement!

We can make the acceleration even greater, and the rockets still maintain a constant separation distance in the initial inertial frame. In this case, in the initial inertial frame, the only difference I can see is that the rockets reach the velocity 0.866c even sooner. And for the people on the trailing rocket, the only difference I can see is that the rockets reach their greater separation distance even sooner. Agreed?

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And I would agree. Progress!
[...]
We can make the acceleration even greater, and the rockets still maintain a constant separation distance in the initial inertial frame. In this case, in the initial inertial frame, the only difference I can see is that the rockets reach the velocity 0.866c even sooner. And for the people on the trailing rocket, the only difference I can see is that the rockets reach their greater separation distance even sooner.

For the people on the trailing rocket, they IMMEDIATELY see a greater separation than the starting separation, by using my description of HOW they measure their separation (having a tape measure whose tip is attached to the rear of the leading rocket, and with the rear of the case attached to the front of the trailing rocket, but with the roll of tape free to move in or out of the case).

I think I see where you're going to go with this, though: You're trying to show that seeing an instantaneous FINITE displacement ISN'T absurd, and therefore my conclusion in my paper is invalid. But there is a difference: In your example, the people seeing the instantaneous finite displacement are the people on the trailing rocket. In my proof, the person supposedly seeing the infinite displacement is the home twin, inertial and stationary wrt the leading rocket. To HER, seeing a finite instantaneous displacement IS absurd, and impossible. IF YOU saw an instantaneous displacement of your coffee cup while sitting at your dining room table, you'd say that was absurd.

For the people on the trailing rocket, they IMMEDIATELY see a greater separation than the starting separation, by using my description of HOW they measure their separation (having a tape measure whose tip is attached to the rear of the leading rocket, and with the rear of the case attached to the front of the trailing rocket, but with the roll of tape free to move in or out of the case).

If the acceleration is finite, then I would NOT agree that the people on the trailing rocket IMMEDIATELY see a greater separation distance. Take the slowest possible acceleration you can imagine, and compare the separation distance before and after it starts. What is the gamma factor for v=0.000000 compared to v=0.0000001? Now realize that the velocity does not jump from v=0.000000 to v=0.0000001, it is continuously changing. No immediate change in separation distance.

I think I see where you're going to go with this, though: You're trying to show that seeing an instantaneous FINITE displacement ISN'T absurd, and therefore my conclusion in my paper is invalid. But there is a difference: In your example, the people seeing the instantaneous finite displacement are the people on the trailing rocket. In my proof, the person supposedly seeing the infinite displacement is the home twin, inertial and stationary wrt the leading rocket. To HER, seeing a finite instantaneous displacement IS absurd, and impossible. IF YOU saw an instantaneous displacement of your coffee cup while sitting at your dining room table, you'd say that was absurd.

First, for the case of finite acceleration, I do not agree that the people on the trailing rocket see an instantaneous finite displacement, as I just stated above. To clarify, I maintain the separation distance would have an infinitesimal increase during the first moment of acceleration.

Second, there is nothing in the case of finite acceleration which we are discussing that suggests that there will ever be an instantaneous finite displacement in the initial inertial frame, which is where SHE is stationary. The greater the acceleration, the sooner she sees the rocket reach 0.866c, that is all. In the limit where the acceleration tends toward infinite, the time needed for the rocket to reach 0.866c tends toward zero. But the rocket does not move at infinite speed in the initial inertial frame regardless of the magnitude of the acceleration.

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Sorry, Neddy. I completely misinterpreted your comments. I thought you were "pulling out all the stops" and allowing a speed arbitrarily close to the speed of light, in order to try to produce an instantaneous finite displacement in the Bell scenario, like the one I give for the leading rocket, as seen by the home twin, in my paper. I'm getting paranoid in my old age (80 years old now). I guess I still don't understand why you're spending so much time on the Bell scenario ... I think we agree about everything there, and we agree that it is a different scenario than the scenario that I address in my paper ... YOU were the one who taught me that.