Bell’s Spaceship Paradox. Does the string break?

[Neddy Bate]

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.

Awhile ago, I tried to teach you that there was a completely different scenario than Bell's in which the separation distance remains constant according to the people on the trailing rocket (so the string does not break). That scenario requires the rockets to get closer together in the initial inertial frame. But I don't see anything like that in your paper.

As far as I can tell, your paper is claiming the string does not break in a scenario in which the rockets remain a constant distance apart in the initial inertial frame. But today you have been saying that as long as that is the case, then the separation distance must increase according to the people on the trailing rocket, as per the LCE. That seems to contradict the claim in the paper.

 

[Mike_Fontenot]

Awhile ago, I tried to teach you that there was a completely different scenario than Bell's in which the separation distance remains constant according to the people on the trailing rocket (so the string does not break). That scenario requires the rockets to get closer together in the initial inertial frame. But I don't see anything like that in your paper.

The best thing to do at this point is for me to quote the important part of that proof. ("He" is the traveling twin, and "She" is the home twin):

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 and motionless 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. I'm making that assumption, because that is what a lot of people think. I know that isn't true, but I'm making that assumption so that I can then show it leads to an absurdity. 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.
(end quote)

What I actually believe is that the two rockets, with equal accelerometer readings, will maintain a constant separation. I base that on what Einstein said in his 1907 paper on the subject, and also on Einstein's gravitational equivalence principle. Einstein DID get the gravitational time dilation equation wrong (he said it was an exponential), as I showed in my first paper:

https://vixra.org/pdf/2109.0076v2.pdf .

I gave the corrected equation in my second paper:

https://vixra.org/pdf/2201.0015v3.pdf .

(Time for supper ... lets talk tomorrow.)

 

[Neddy Bate]

I'm making that assumption, because that is what a lot of people think. I know that isn't true, but I'm making that assumption so that I can then show it leads to an absurdity.

As far as I can tell, your "assumption" is that the people on the trailing rocket say that the separation distance increased. We have agreed all day today that is in fact what SR says happens. You have been telling me that you are "okay with that" and that it is REQUIRED by the length contraction equation LCE. And never have you shown that any Absurdity results from it. So what is the paper for?

 

[Mike_Fontenot]

(Keep reading ... you're going to like the ending!)

It's definitely true that, according to the people in the trailing rocket (whose accelerometer shows the same constant reading as the accelerometer on the leading rocket), the separation of the two rockets is constant (as they can conform by looking at the tape measure reading that I have previously described). And it is also true that the initial inertial observers must conclude that the rockets get closer together as the acceleration progresses ... the length contraction equation (LCE) requires that.

But think about what happens if we do a series of SEPARATE experiments, with the acceleration "A" getting greater and greater with each iteration of the experiment, and with the time duration "T" of the acceleration getting smaller and smaller with each iteration, so that the product of "A" and "T", "A * T" is the same from iteration to iteration. The product "A * T" is the "rapidity" theta..

So suppose we start with A = 1 ly/y, and T = 1 year. So A*T = 1.0. The velocity "v" during that acceleration, for 0<=t<=1, is v(t) = tanh[ theta(t) ], and the distance moved is d(t) = ln { cosh[ theta(t) ] }.

Next, we repeat the above, with "A" ten times larger, and "T" ten times smaller, so that theta = A*T is that same as in the first iteration. The velocity and the distance moved at "T" are the same as for the first iteration. And likewise for each iteration.

So, in the limit as T goes to zero and A goes to infinity, we get the same value for v(T) and the same value for d(T). But that means that in that limit, the distance instantaneously changes from zero to the finite, non-zero result. So I was wrong when I said in my paper that for an inertial observer to see an instantaneous non-zero displacement of some material object is "absurd" and "impossible" ... that happens every time we replace a finite acceleration "A" and a finite duration "T" with an infinite acceleration and an infinitesimal duration! My proof was incorrect.

I'm NOT saying that I was wrong to say that two rockets with equal accelerometer readings will have a constant separation. I still believe THAT, based merely on Einstein's 1907 paper, and based merely on Einstein's gravitational equivalence principle. But I just didn't prove it as I thought I had done.

 

[Mike_Fontenot]

I (Mike) had said:

"I'm making that assumption, because that is what a lot of people think. I know that isn't true, but I'm making that assumption so that I can then show it leads to an absurdity."

The assumption I was talking about in the above quote is that a lot of people say that the separation of two rockets with equal readings on their accelerometers increases with time, according to the people on the trailing rocket. So I definitely haven't "agreed all day" that that is what SR says.

When I have talked about the length contraction result, what I've said is that if the people on the trailing rocket say the separation of the rockets is constant, then the initial inertial observers will way that the separation decreases with time, because that is what the length contraction equation requires. Or, alternatively, if the inertial observers say the separation is constant, the people on the trailing rocket will say that the separation of the rockets is increasing. There is no mention of accelerometers in those two cases.

As far as I can tell, your "assumption" is that the people on the trailing rocket say that the separation distance increased. We have agreed all day today that is in fact what SR says happens. You have been telling me that you are "okay with that" and that it is REQUIRED by the length contraction equation LCE. And never have you shown that any Absurdity results from it. So what is the paper for?

 

[Mike_Fontenot]

(Keep reading ... you're going to like the ending!)

It's definitely true that, according to the people in the trailing rocket (whose accelerometer shows the same constant reading as the accelerometer on the leading rocket), the separation of the two rockets is constant (as they can conform by looking at the tape measure reading that I have previously described). And it is also true that the initial inertial observers must conclude that the rockets get closer together as the acceleration progresses ... the length contraction equation (LCE) requires that.

But think about what happens if we do a series of SEPARATE experiments, with the acceleration "A" getting greater and greater with each iteration of the experiment, and with the time duration "T" of the acceleration getting smaller and smaller with each iteration, so that the product of "A" and "T", "A * T" is the same from iteration to iteration. The product "A * T" is the "rapidity" theta..

So suppose we start with A = 1 ly/y, and T = 1 year. So A*T = 1.0. The velocity "v" during that acceleration, for 0<=t<=1, is v(t) = tanh[ theta(t) ], and the distance moved is d(t) = ln { cosh[ theta(t) ] }.

Next, we repeat the above, with "A" ten times larger, and "T" ten times smaller, so that theta = A*T is that same as in the first iteration. The velocity and the distance moved at "T" are the same as for the first iteration. And likewise for each iteration.

So, in the limit as T goes to zero and A goes to infinity, we get the same value for v(T) and the same value for d(T). But that means that in that limit, the distance instantaneously changes from zero to the finite, non-zero result. So I was wrong when I said in my paper that for an inertial observer to see an instantaneous non-zero displacement of some material object is "absurd" and "impossible" ... that happens every time we replace a finite acceleration "A" and a finite duration "T" with an infinite acceleration and an infinitesimal duration! My proof was incorrect.

I'm NOT saying that I was wrong to say that two rockets with equal accelerometer readings will have a constant separation. I still believe THAT, based merely on Einstein's 1907 paper, and based merely on Einstein's gravitational equivalence principle. But I just didn't prove it as I thought I had done.

I think my conclusions in my above post (post #264) may be wrong. I've just realized that in that post, I used the time quantity "T" without saying WHO's time that was. I THINK it was the initial inertial people's time, and I THINK it SHOULD have been the trailing person's time.

To elaborate, I need to show a chart that was last shown on page 4 in post #75. I'll see if I can upload that chart again:




Along the lower curve, there are "tic marks" showing the age of the person on the trailing rocket. Note that they are different from the tic marks on the horizontal axis (which correspond to the age of the initial inertial observers).

So to get the view of the person on the trailing rocket (about the distance he's covered and the velocity he's achieved, we would need to do a plot of his distance, versus HIS age, not versus the age of the initial inertial observers. I haven't done that yet, but I suspect that changes the conclusions I came to in my last post ... in particular, I suspect I will now NOT conclude that he sees an instantaneous change in his position, and if so, I will no longer conclude that he directly sees something with his own eyes that is absurd ... i.e., I think my last paper may still be correct.

 

[Halc]

Along the lower curve, there are "tic marks" showing the age of the person on the trailing rocket.

Both curves have that. You also drew lines of simultaneity for two events on the lead ship worldline, complete with tic marks for the x' axis, showing how far away the start event is 'according to' the lead rocket. You're on record for being in denial of the correctness of that picture.


Anyway, Hi, I'm back and caught up. Would like to ask some questions of my own, since I'm trying to understand your stance.

1) You may not be able to do this one, but in frame S at time zero, a lone ship stationary in S at location X (pick a value if it helps) takes off, accelerating continuously at a proper 1g. What is the location of that ship in S after 2 months in S? It would be a function of X of course. There are calculators online to do it, and my program that I showed you does it, but you seem to be in denial of all the mathematics of both of those, so I'm asking where you put the ship.

2) Two ships this time, both with identical proper acceleration after a simultaneous (in S) launch. That's the scenario. After two months in S, is the velocity of the two ships identical or different? This would really help me understand your assertions.

 

[Mike_Fontenot]

1) You may not be able to do this one, but in frame S at time zero, a lone ship stationary in S at location X (pick a value if it helps) takes off, accelerating continuously at a proper 1g.
[...]

I now think ANY use of the upper curve (which has the same shape as the lower curve) gives erroneous results, because when the upper and lower curves have a constant separation (according to the initial inertial observers), you get the Bell scenario, which is a DIFFERENT scenario from the scenario I am interested in. In the Bell scenario, the accelerometers show different readings.

I am interested in the scenario where the people on the trailing rocket conclude (using their own measurements with a tape measure) that the separation of the rockets is constant, and that the readings on the two accelerometers are identical. In that case, the initial inertial observers will say that the separation between the two rockets decreases, by the gamma factor. I've previously given a chart showing the initial inertial observers' view ... I'll try to repeat that here:

 

[Halc]

You didn't answer either of my questions.

in frame S at time zero, a lone ship stationary in S at location X (pick a value if it helps) takes off, accelerating continuously at a proper 1g. What is the location of that ship in S after 2 months in S?

I now think ANY use of the upper curve

There is no 'upper curve' in the question I asked. There is but the one ship in that question.

I am interested in the scenario where the people on the trailing rocket conclude (using their own measurements with a tape measure) that the separation of the rockets is constant, and that the readings on the two accelerometers are identical.

OK, my second question concerned the case where the readings on the two accelerometers are constant and identical, and yet you didn't answer the question. It is a simple question. Please answer it. I asked if the velocity in S is the same after some time has passed in S. Your answer gave no mention of velocity.

2) Two ships this time, both with identical and constant proper acceleration after a simultaneous (in S) launch. That's the scenario. After two months in S, is the velocity of the two ships in S identical or different?

 

[Mike_Fontenot]

Text is red are mine (Mike's):

You didn't answer either of my questions.

There is no 'upper curve' in the question I asked. There is but the one ship in that question.

But that one ship STARTS from a position above the origin. It's still "an upper curve". You've just eliminated the lower curve.

OK, my second question concerned the case where the readings on the two accelerometers are constant and identical, and yet you didn't answer the question. It is a simple question. Please answer it. I asked if the velocity in S is the same after some time has passed in S. Your answer gave no mention of velocity.

2) Two ships this time, both with identical and constant proper acceleration after a simultaneous (in S) launch. That's the scenario. After two months in S, is the velocity of the two ships in S identical or different?

If the accelerometers show the same constant readings, the people on the trailing rocket will say the separation is constant. The initial inertial observers will say the upper curve decreases because it must get closer to the lower curve by the factor gamma. That is required by the length contraction equation (LCE). Since the upper curve decreases according to the initial inertial observers, they will say the velocities of the two ships are different, because the velocities are just the slopes of the curves.

 

[Mike_Fontenot]

Glad you're back, though, Halc.

 

[Halc]

But that one ship STARTS from a position above the origin. It's still "an upper curve". You've just eliminated the lower curve.

Your evasiveness tells me that I am asking the correct question. Your answer seems to suggest that one's choice of where to assign the origin matters, like there's a preferred location in space or something, so let me reword the question in a way that makes no reference to it.

There is a lone ship and lone launchpad both relatively stationary and colocated. At time zero on the lauchpad clock, the ship takes off, accelerating continuously at a constant proper 1g. How far away from the launchpad is the ship after 2 months, time measured by the launchpad clock?
Extra credit, how far away from the launchpad is the ship after 2 months, time measured by the ship clock?

Again, there are online calculators that do this for you, but they probably use physics of which you are in denial. Still, it can be done with a couple button pushes on a calculator.

If it makes it easier, we can alter the proper acceleration down 3% to 9.51 m/sec², which is a nice natural acceleration of 1 ly/lr²

If the accelerometers show the same constant readings, the people on the trailing rocket will say the separation is constant. The initial inertial observers will say the upper curve decreases because it must get closer to the lower curve by the factor gamma. That is required by the length contraction equation (LCE). Since the upper curve decreases according to the initial inertial observers, they will say the velocities of the two ships are different, because the velocities are just the slopes of the curves.

Ah, you actually answered this one in only 2 tries! That's great.
If at some time (2 months say) in S the velocities of the two ships are unequal, what is the velocity of the string? The ends, attached to the respective ships, are moving at different velocities in that frame. So how is the LCE applicable to a string that isn't all moving at the same speed? Keep in mind that I'm fine with the string not moving at the same speed everywhere. But if it is moving at different speeds relative to S at various points along its length, the LCE cannot be applied to just one speed, but rather must be integrated along its length. How do you know the contracted length of the string then?

Second point: The first derivative of velocity is (coordinate) acceleration. True or false?

 

[Mike_Fontenot]

My (Mike's comments) are in red.

You didn't answer either of my questions.

You ask so many different questions that I can't keep up with them. I'll just address the most important issue. It's really the only issue.

There is no 'upper curve' in the question I asked. There is but the one ship in that question.

So, with only one ship, why do you not start it from the origin? I can see no reason to start a single ship from a non-zero distance above the origin. So, with a single rocket, I'll start it from the origin. In that case, it's position (according to the initial inertial observers) is just given by the equation

X(t) = ln(cosh(A* t).

Without another ship (the "leading ship"), there isn't anything else to say.

To have anything to talk about, we need two separated ships, one starting at the origin, and the other starting at some arbitrary distance above the origin. Both are accelerating at "A" ly/y, as reported by the two accelerometers. According to the initial inertial observers, the lower curve is given by the above equation [ln(cosh(A*t)]. According to the initial inertial observers, the upper curve is a distance (D / gamma) above the lower curve. "D" is the original separation of the ships. Gamma is

Gamma = 1 / {sqrt[ 1 - (v * v) ] },

where "v" is the velocity of the trailing ship, according to the initial inertial observers. "v" is a function of the initial inertial observers' time "t", and so "v(t)" is just the slope of the lower curve at "t".

That's all there is to it.

 

[Mike_Fontenot]

My (Mike's) changes to this previous post of mine will be in red.

[...]

Before, I supposedly showed that a person (call him Tom) who instantaneously changes his speed (from zero to some large fraction of the speed of light) will, as a consequence of that, also instantaneously move a finite distance. I then used that result to argue that absurdities happen all over the place, thus diminishing the importance of the absurdity described in my paper.

But that conclusion was incorrect, as can be seen by comparing it to the traveling twin (Jim) in the twin paradox, when Jim instantaneous turns around (modified by starting the turnaround when he is STATIONARY wrt the home twin [Jane]). Jim DOES instantaneously changes his speed, but he DOESN'T instantaneously move a finite distance toward Jane. The same thing is what happens to Tom in the current scenario: Tom DOESN'T move during his instantaneous speed change. So there is no absurdity in this case (as I had incorrectly contended), but there IS absurdity in the scenario of my paper, where the (incorrect) assumption that the two rockets move apart results in the absurd instantaneous finite displacement of the leading rocket, as directly observed by the home twin.

 

[Mike_Fontenot]

My (Mike's) responses are in red:

As far as I can tell, your "assumption" is that the people on the trailing rocket say that the separation distance increased.

No, it's the home twin (she), who has been standing next to the leading rocket for a long time before it fired and did its instantaneous finite position change. She sees that happen with her own eyes. And she would regard that as absurd. Hence, the assumption that the separation of the trailing and leading rockets, whose accelerometer readings are the same, increases, is incorrect, and must be rejected.

 

[Halc]

it's position (according to the initial inertial observers) is just given by the equation

X(t) = ln(cosh(A* t).

Where does this come from? I mean, you're in denial of relativity and even Newtonian physics, so is this something you worked out? It gives reasonable values for low speeds, but really diverges from SR as time goes on. I ran your program (using the above formula) for 5 years, and the gamma value at each year is 1.54, 3.76, 10.07, 27.31, 74.21 where SR says it should be 1.41, 2.24, 3.16, 4.12, 5.10. In other words, your rapidity is increasing at a phenomenal rate after a couple years, over 10 times that which relativity says.

Without another ship (the "leading ship"), there isn't anything else to say.

So you've been saying it depends on what other things are doing, events taking place outside the causal light cone of the ship taking off. And you 'stand behind this'.

OK, so suppose there is another ship doing the same thing, taking off at the same time, same direction in S. But nobody at our launchpad is aware of it since it is taking place light years away on a planet we don't even know about. How far away from the launchpad does our ship get after 2 months? You don't know I presume since I didn't say where the other ship is.

Second scenario: Suppose there is a large number of ships taking off all at once, one every 10 light years. You know the series of ships is there, but you don't know where you are in the list. There might be an unbounded number of ships in the line. Now how far away from the launchpad does our ship get after 2 months?


You replied twice to post 269 twice, which is fine since the first reply didn't answer the first question.
I reworded the first question in 272, immaterial since you sort of answered it. But I asked some new questions there, all ignored. See 272 for full text.
If it's coming too fast, take your time. This isn't a school test with a time limit.

This concerns the case with two ships with identical constant finite proper acceleration.
1) If at some time (2 months say) in S the velocities of the two ships are unequal as you claim, what is the velocity of the string? The ends, attached to the respective ships, are moving at different velocities in that frame. So how is the LCE meaningful to a string that isn't all moving at the same speed?

2) The first derivative of velocity is (coordinate) acceleration. True or false?

Before, I supposedly showed that a person (call him Tom) who instantaneously changes his speed (from zero to some large fraction of the speed of light) will, as a consequence of that, also instantaneously move a finite distance. I then used that result to argue that absurdities happen all over the place, thus diminishing the importance of the absurdity described in my paper.

You're saying that teleportation is no longer absurd? It's kind of hard to parse out the gist of your last post. Maybe a picture is in order if you think the ones we drew are wrong. There's no discontinuities in any worldlines, so nobody experiences anything teleporting.

 

[Mike_Fontenot]

I (Mike) previously said:


Its position (according to the initial inertial observers) is just given by the equation

X(t) = ln(cosh(A* t). (wrong)

And my response now is to say that I accidentally left off a factor that needs to be in that equation. I should have written

X(t) = (1 / A) * ln [ cosh ( A * t ) ]. (correct)

Sorry for that careless mistake.

The initial inertial observers say the velocity "v" of the trailing rocket is given by

v(t) = tanh ( A * t ),

where "A" is the acceleration in ly/y/y .

So the distance traveled by the trailing rocket, according to the initial inertial observers, is just the integral of v(t). For years I did that integration numerically, but fairly recently I discovered that the integral has the closed-form equation that I give above.

[...]

 

[phyti]

Mike;

You could have studied SR from one of many sources, and within a year learned the basics. You chose to invent your own version and after all those years, how much progress have you made?

If you can't solve either problem, 'twins' or "Bell', mixing them together won't help.
Your drawings are NOT Minkowski diagrams. There were reasons why the convention is vertical ct time axis and horizontal x space axis. One is any spatial axis can be ±, but ct is always positive (time accumulates), which allows the use of symmetry when needed.

Your acceleration notation 1 ly/y=ct/t=c, is an unattainable value. If acceleration is g then use g. In summation, you have a weak foundation for your ideas.

Your plot of a spaceship as a point doesn't allow an illustration of length contraction. If an extended object (cylinder) was used, you could track the center of mass, a conventional method. Then you could show the length contraction of each ship, while their centers remained a constant separation, and the gap (back of one to front of the other) increases, and the string breaks. The ships operate independently of each other, thus an equivalent acceleration produces an equivalent velocity (postulate 1 in SR).

When the pair of rockets acquire their instantaneous velocity of -.866 at the reversal +10, they will require a clock synchronization since Biff trailing behind has changed course 180º. When Biff makes a measurement between centers, he gets 69.5.
That would have been their separation at launch. If both move at any constant velocity they measure the same separation which is the 'proper' distance in the 4D jargon.

SR was not developed with any intent or purpose of looking into the future. Cause and effect mean you can't see/perceive an event until it happens. The essential difference of SR from the Newtonian era is the finite speed of light, resulting in time dilation and length contraction. SR predicts a delay between an event occurring and an observer detecting the light from it. That's looking at the past, not the future.

Example:
You observe sunspots for a minute. You want know when on your clock they occurred. You already have sufficient data (distance to sun & speed of light) to conclude you saw them as they appeared approx. 8 min ago.

 

[Halc]

Its position (according to the initial inertial observers) is just given by the equation
X(t) = ln(cosh(A* t). (wrong)
And my response now is to say that I accidentally left off a factor that needs to be in that equation. I should have written

X(t) = (1 / A) * ln [ cosh ( A * t ) ]. (correct)

Again, where does this come from? I was using A=1, so your change makes no difference.

fairly recently I discovered that the integral has the closed-form equation that I give above.

OK, you 'discovered' it, but it contradicts SR, so where did you find this?

The initial inertial observers say the velocity "v" of the trailing rocket is given by
v(t) = tanh ( A * t ) where "A" is the acceleration in ly/y/y .

That also contradicts SR. So where are you getting all this stuff since you don't seem to be up to the task of deriving anything yourself.

How does an engineer know to use the equation for a trailing rocket or not? How does NASA get anything into space if the mathematics requires knowledge of unknowable things? It would be a lot easier to get something to the moon if it was a leading rocket and they pointed it straight down.

Another unanswered question:
Suppose there is a large number of ships taking off all at once, one every 10 light years. You know the series of ships is there, but you don't know where you are in the list. There might be an unbounded number of ships in the line. Now how far away from the launchpad does our ship get after 2 months?

3rd unanswered question: The first derivative of velocity is (coordinate) acceleration. True or false?

 

[Mike_Fontenot]

The trailing rocket's position (the bottom curve), in every chart we've been discussing in this VERY LONG thread, has been computed from the equation

X(t) = (1 / A) * ln [ cosh ( A * t ) ],

or in some cases from the numerical integration of

v = tanh(A * t) .

You've never complained about that bottom curve before.

So I think you're doing something wrong in your calculations. Are you using "g's" for your unit of acceleration? If so, that is your problem ... you need to use units of ly/y/y for acceleration, ly/y for velocity, ly for distance, and "y" (years) for time.

[...]