Bell’s Spaceship Paradox. Does the string break?

[Mike_Fontenot]

From the Wiki article:

"They start accelerating simultaneously and equally as measured in the inertial frame S, thus having the same velocity at all times as viewed from S. Therefore, they are all subject to the same Lorentz contraction, so the entire assembly seems to be equally contracted in the S frame with respect to the length at the start. At first sight, it might appear that the thread will not break during acceleration. This argument, however, is incorrect as shown by Dewan and Beran, and later Bell.[1][2] The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S, it is effectively defined to remain the same, due to the equal and simultaneous acceleration of both spaceships in S." (The bold emphasis is mine).

I think that last sentence is very badly worded, and not even consistent. I think they should have said "If the spaceships are forced to maintain a constant separation according to the S inertial frame, the leading spaceship would HAVE to accelerate faster than the trailing spaceship, so that the combination of an increased separation due to the higher acceleration would exactly cancel the Lorentz contraction that must occur". IF my wording better describes their scenario, then the two scenarios (Bell's and mine) WOULD be different, which would be a good thing.

 

[phyti]

Mike;

Wikipedia

"so the entire assembly seems to be equally contracted in the S frame with respect to the length at the start."

That is the error. The ships operate independently of each other. Each would use the same program for equivalent accelerations. Each would experience length contraction, which results in an increasing gap between the two. They are not physically joined as a rigid object.

Because anybody can be an editor on Wiki, the articles may not be correct.

Now that I see your plot in #1, you show the ships moving at v>c! (v=Δx/Δt)
Swap axes and it would be more correct. The hyperbolic curves would vary as a function of speed.

 

[Mike_Fontenot]

It is possible to show that the string doesn't break in Bell's Paradox (assuming that the accelerations of the two spaceships are confirmed with accelerometers to be equal), without referring to ANY inertial observers.

In 1907, Einstein derived the gravitational time dilation equation (GTD), which says that if two clocks are separated by a fixed distance "L", in a constant gravitational field (with the separation along the direction of the field), such that the field strength doesn't vary with distance from the source of the field, then the clock farther from the source of the field will run faster than the clock closer to source of the field, by the factor f(L*g).

From the equivalence principle, that means that two spaceships (with no gravitational fields anywhere), each accelerating with a constant acceleration "A" (as confirmed by accelerometers), and separated by the distance "L" in the direction of the acceleration at some instant, will always maintain that separation (for as long as that acceleration equals "A"). And clocks in the leading spaceship will run faster than clocks in the trailing spaceship, by the factor f(L*A).

Since the distance between the two spaceships is constant, a thread stretched between them never breaks.

 

[phyti]

Mike;

I automatically see vertical as time and horizontal as space, which is the standard convention. The most likely history, in a standing position the ground is a horizontal xy plane where most activity occurrs and z is vertical.

My last response was wrong, (most likely due to no resolution of this example), so I pasted in your 20 yr old drawing (which I hadn't seen prior, most likely due to not signing in), reoriented it, and reread your text. Now it's clear. You scenario is two ships A and B stacked in a g-field.
The A clock rate is slower than the B clock rate prior to launch, being lower in the g-field. The A ship will require slightly more power to ascend than the B ship. They can still maintain a constant spacing of center of mass, but will experience length contraction resulting in an increasing gap between them. This also supports Einstein's exponential formula with the gradient between two levels being insignificant as they move away from the center of earth.


The Bell scenario involves A and B in tandem, requiring a launch from space initially in that configuration.
The caption 1.0 ly/y/y is confusing since it equals c/y, and followed by .97g?
I guessed your intention was acceleration at g.

Am also tired of defending SR against all the anti relativity posts. I don’t judge people, only the statements they make.
Probably time for me to get off the bus.

 

[Mike_Fontenot]

I automatically see vertical as time and horizontal as space, which is the standard convention.

You're right ... that is the way relativity diagrams are usually drawn. But I've always chosen to use the same orientation as is used in newtonian physics, and also in most diagrams in other fields for plotting distance and time. Most often, time is thought of as the independent variable, and distance is the dependent variable, and the independent variable is usually plotted on the horizontal axis. I think relativity is already different enough from newtonian physics as it is, without making it seem even more different by reversing the two axes. So I choose to plot time horizontally and distance vertically.

My last response was wrong, (most likely due to no resolution of this example), so I pasted in your 20 yr old drawing (which I hadn't seen prior, most likely due to not signing in), reoriented it, and reread your text. Now it's clear. Your scenario is two ships A and B stacked in a g-field.

Why do you say "g-field" in the above sentence? The scenario I'm analyzing is purely a special relativity scenario. Special relativity assumes no gravitational fields anywhere in space.

It's true that the equivalence principle is sometimes used to connect a gravitational scenario consisting of a gravitational field and no motion, to a special relativity scenario with motion and no gravitational fields. But my interest is purely in special relativity, with motion but without any gravitational fields. (When Einstein was first trying to come up with his general relativity theory (explaining gravitation), he made use of his existing special relativity PLUS the equivalence principle to give him some guidance about how general relativity has to work.)

The caption 1.0 ly/y/y is confusing since it equals c/y, and followed by .97g?
I guessed your intention was acceleration at g.

I'm choosing to use "years" for the time coordinate, and "lightyears" ("ly") for the spatial coordinate. So velocity than has the unit "ly/y", and acceleration has the unit "ly/y/y". One ly/y/y just coincidentally equals about 0.970g. (The acceleration 1.0 ls/s/s, which I'm not using here, is about 32 MILLION "g's"!!) I included the fact that 1.0 ly/y/y is about 0.970 "g's" just to give people who aren't familiar with the unit ly/y/y some feel for the magnitude I was choosing for the acceleration. I chose to set the acceleration at 1.0 ly/y/y, just like I did for the original (incorrect) diagram.

 

[phyti]

Mike;
Then your inertial ref. frame is a low mass object in space, and the two ships can follow the same acceleration program. The curves are the same.
Your scenario is still different from Bell’s.

distance/time=vt/t=speed v
ly/y=c*1y/y=c!

 

[Mike_Fontenot]

Somewhere (I don't remember where), you (Neddy Bate) described a situation where a rod between two people would show a shrinkage in the one-foot markings on the rod, causing the rod itself to move with respect to the trailing person, so that additional rod that was previously behind him would now be co-located with him (while the leading end of the rod remaining rigidly connected to the leading person) . That didn't make any sense to me at the time, but I've encountered a situation where that seems to be happening, and I'm stuck on any further progress in my analysis. If you can recall that discussion, it would help me a lot if you could elaborate a little.

 

[Neddy Bate]

I don't recall discussing that particular scenario.

However, it sounds like you have an SR scenario with a "leading person," and a "trailing person," and a "rod". You say the front end of the rod is rigidly connected to the 'leading person' but the rear end of the rod is not connected to 'trailing person'.

So, for simplicity, let's start from an inertial frame. Plot the trajectory of the 'leading person' whilst including the length of the trailing rod. If the leading person is accelerating, the length of the trailing rod will get shorter as the speed increases, due to length contraction (SR). But we know the front end will still be co-located wit the 'leading person' by definition.

Now, let's think about the 'trailing person'. Where are they? You could have stipulated that the trailing person will always be co-located with the back end of the rod, but you didn't. In the inertial frame, it is trivial to see that would require a greater acceleration for the 'trailing person' than the 'leading person'.

 

[Mike_Fontenot]

I don't recall discussing that particular scenario.

The problem I'm working (where the long tape measure needs to slide past the trailing person) is part of a VERY long discussion on one of the "Naked Scientists" forums ... too long to repeat here. But if you're interested, you can get to it on this link:

https://www.thenakedscientists.com/forum/index.php?topic=86264.0

 

[Mike_Fontenot]

I have finished my correction of the viewpoint of the initial inertial observers who are stationary wrt the two rockets before the rockets are ignited (thus producing equal accelerations according to attached accelerometers). Those initial inertial observers are said to have concluded that the rocket separation is constant during the accelerations. I've previously shown a chart that shows that conclusion. But that conclusion violates the length contraction equation (LCE) of Special Relativity (SR). The LCE says that inertial observers will always measure a moving yardstick to have a shorter length that their own yardsticks. Therefore, the distance between the two rockets must decrease during the acceleration, according to those initial inertial observers.

I have produced a diagram that shows the correct conclusions of the initial inertial observers, that I will try to upload below. First, I need to give a few explanations of some labels I use in the diagram.

To the diagram, I've added the two lines (for the leading and trailing rockets) for the case where D = 2 ly. (I had only done D = 1 ly before). The four lines are labeled U2, L2, U1, and L1, standing for "upper curve, D = 2", "lower curve, D = 2", "upper curve, D = 1", and "lower curve, D = 1".

I'll try to upload the diagram now:

 

[Mike_Fontenot]

I forgot to say, in the above information, that in the motion involved in reducing the separation of the rockets (according to the initial inertial observers), my derivation stipulates that the center of mass of the two rockets doesn't take part in that motion. I.e., the two rockets both move by equal amounts toward their center of mass, at each instant of time.

 

[Mike_Fontenot]

I've realized that the above solution is incorrect. It works nicely when there are only two rockets (a trailing rocket and a leading rocket). But when there are three or more rockets, it doesn't work, because each of the rockets that are distant from the trailing rocket require a DIFFERENT behavior for the trailing rocket, which is impossible ... there is only one trailing rocket, and it obviously can't be simultaneously behaving in the different ways required by the different distant rockets. The previous solution (which has the initial DECREASING trajectory for the distant rockets) is the correct one.

 

[Mike_Fontenot]

Here is the correct diagram, showing the case where some of the rockets are distant enough to have a negative slope initially. The negative slope does NOT mean that those rockets have reversed the direction of their thrust ... their rockets are always accelerating away from their starting point. It is just is an effect produced by the length contraction equation.



Each of the curves starts from a given point of the vertical axis, called "D". D is zero for the bottom curve, and D = 0.5 for the next curve up, then D = 1 for the next one, and then D = 2, and then D = 3. Note that, for any value of "t", the vertical distance between each of the upper curves and the lowest curve is just equal to indicated "D" for that upper curve, divided by the gamma factor of the bottom curve at that value of "t". That gamma factor is determined from the velocity of the lower curve at that point (which is the slope of the lower curve at that point), and gamma is given by

gamma = 1 / [sqrt{ 1 - (v*v)}].

The lower curve is computed using the equation

x = (1/A) * ln{cosh(A * t)},

where "A" is the constant acceleration. (A = 1 ly/y/y in this example, about 0.97 "g's".)

 

[Mike_Fontenot]

Here's my latest: https://vixra.org/abs/2308.0045

Title: A Proof that the Separation Between Accelerating Rockets is Constant

Author: Michael Leon Fontenot

email: PhysicsFiddler@gmail.com

____________________________________________________________________________________


Abstract:

For rockets whose accelerometers show identical, constant readings, their separation is constant. The proof of that fact makes use of the limit of a sequence of accelerations "A", lasting for a time "delta_t", such that the total change in rapidity "A*delta_t", and therefore the total change in the velocity, don't vary for each iteration of the sequence of accelerations. In the limit, as "A" goes to infinity, and "delta_t" goes to zero, the velocity of the rockets changes instantaneously, and their separation doesn't change. The result is analogous to the CoMoving Inertial Frame (CMIF) simultaneity method of Special Relativity, which says that, according to the traveling twin (him), the home twin (she) instantaneously gets older during his instantaneous turnaround. Likewise, the ages of the people on the leading rocket instantaneously get older during their instantaneous velocity change.
____________________________________________________________________________________


The above abstract really says all that needs to be said about the proof that the separation between accelerating rockets is constant. The only thing that would be useful to add, is to elaborate a bit about the CoMoving Inertial Frame (CMIF) simultaneity method used to resolve the twin paradox, and to give the "delta_CADO" equation that makes the CMIF method especially easy and quick to use.

The CMIF simultaneity method says that the accelerating person (the "AP") must agree with the inertial person ("IP") who is momentarily stationary with respect to the accelerating person at any given instant. In the case of the instantaneous turnaround, there is an IP1 immediately before the turnaround, and an IP2 immediately after the turnaround. For each of those IP's, their line of simultaneity ("LOS") can be plotted on a Minkowski diagram. Where those two LOS's intersect the home twin's world line then give her age, according to the AP, immediately before and after the turnaround. I.e., that gives the amount by which she instantaneously ages during his turnaround, according to him.

It's even easier to get that instantaneous age change by using the "delta_CADO" equation:

delta_CADO = - L * delta(v),

where

delta(v) = v_after_turnaround - v_before_turnaround,

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

 

[Mike_Fontenot]

The answer is "Yes and No", depending upon the scenario.

There are two DIFFERENT scenarios:

Scenario 1: The two accelerometers always display the same value during the trip, and the thread DOESN'T break.

Scenario 2: The initial inertial observers REQUIRE that the separation THEY measure is constant, and therefore those inertial observers conclude that, in the rocket frame, the separation must be increasing, and so the thread WILL break.

In the WIKI article on the Bell's Paradox, it says (for the initial inertial frame "S")

"The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S, it is effectively DEFINED to remain the same, due to the equal and simultaneous acceleration of both spaceships in S."

So they didn't specify that the rockets had accelerometers that showed equal constant readings. And, most important, they specified that the separation was constant, ACCORDING TO THE INITIAL INERTIAL FRAME. That rules out the separation being constant in the trailing rocket's frame ... it REQUIRES that the separation increases in the rocket's frame and it REQUIRES that the accelerations are different, as measured by the accelerometers.

The two scenarios are DIFFERENT, and it's not surprising that their conclusions about the survival of the thread are different.

Acknowledgement: Neddy Bate was the first one to recognize that the two scenarios are different, and I am grateful for his insight.

 

[James R]

Moderator note: I have merged four threads, all talking about essentially the same problem.

I can't see any good reason for having a number threads endlessly re-asking the same question. Let's keep it all in one place.

 

[Halc]

Moderator note: I have merged four threads, all talking about essentially the same problem.

You missed the original

sciforums.com/threads/a-quandary-about-accelerated-motion-in-special-relativity.165975

which is a copy of the fruitless discussion in TNS to which Mike provided a link:

But if you're interested, you can get to it on this link:
https://www.thenakedscientists.com/forum/index.php?topic=86264.0

in which he claims Neddy is at least partially on his side on this.

Scenario 1: The two accelerometers always display the same value during the trip, and the thread DOESN'T break.

Scenario 2: The initial inertial observers REQUIRE that the separation THEY measure is constant, and therefore those inertial observers conclude that, in the rocket frame, the separation must be increasing, and so the thread WILL break.
...
The two scenarios are DIFFERENT, and it's not surprising that their conclusions about the survival of the thread are different.

Acknowledgement: Neddy Bate was the first one to recognize that the two scenarios are different, and I am grateful for his insight.

I do not see that recognition anywhere in this topic. Perhaps elsewhere. No link was given. But it was this comment by Mike that prompts me to reply here.

Both are Bell's scenario, and thus not different. The thread breaks.
Scenario 2 seems to say that the acceleration (apparently not identical) results from the requirement that both ships take identical trajectories relative to the IRF.
More mistakes: There is no 'the rocket frame'. There is the frame of one rocket or the other, but they're not the same frame.
Mike's assertions lead to the lead rocket moving backwards despite its accelerometer indicating forward acceleration. The assertions require faster than light motion and causality. None of these contradictions deter the assertions.


So I wanted to comment on what Neddy's last post, admittedly from a month ago.

However, it sounds like you have an SR scenario with a "leading person," and a "trailing person," and a "rod". You say the front end of the rod is rigidly connected to the 'leading person' but the rear end of the rod is not connected to 'trailing person'.

Yes, a pre-stressed tape measure if you will, with the marking applied in the pre-stressed state. Without the stress, the acceleration of the rocket pulling it will no affect the far end of the tape since speed of sound cannot be anywhere near c. With the pre-stressed tape, this is not a problem and the entire tape can exhibit B0rn-rigid motion so long as the proper acceleration of the rocket is constant. I say all this because without these details, the tape cannot exhibit rigid motion without violating causality, invalidating the exercise.

So, for simplicity, let's start from an inertial frame. Plot the trajectory of the 'leading person' whilst including the length of the trailing rod. If the leading person is accelerating, the length of the trailing rod will get shorter as the speed increases, due to length contraction (SR).

Yes, but not by λ (the dilation factor from speed of ship) since relative to that IRF, the tape measure is not all moving at the same speed along its length (such would result in an immediate contradiction). So the tape is actually shorter than L/λ

Now, let's think about the 'trailing person'. Where are they? You could have stipulated that the trailing person will always be co-located with the back end of the rod, but you didn't.

Mike specified quite explicitly that both rockets have identical proper acceleration.

In the inertial frame, it is trivial to see that would require a greater acceleration for the 'trailing person' than the 'leading person'.

Trivial to somebody that has a nodding acquaintance with physics. Not trivial to Mike who is in direct denial of exactly that. Given identical accelerations, the rod/tape-measure either breaks, or the read guy sees it out-pacing his ship. If it breaks, it's because it cannot be pre-stressed to more than infinite force. The acceleration of the lead ship forms a Rindler horizon behind it (about a light year at 1 g acceleration), and if the rod is longer than that, it cannot exhibit rigid acceleration without moving faster than c.

 

[Mike_Fontenot]

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 the two separated rockets in the Bell's Paradox (whose accelerometers show the same constant readings) DON'T maintain a constant separation, that contradicts the Twin Paradox resolution.

Here's how to see that contradiction:

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

Imagine that, at the instant before he instantaneously increases his speed toward her, he is colocated and stationary wrt the trailing rocket. And suppose that the leading rocket is co-located and stationary with her then. (The rockets are unaccelerated before and at that instant).

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

Immediately after his instantaneous speed change, suppose the leading rocket is assumed to instantaneously increase its separation from the trailing rocket. THAT would result in her seeing the leading rocket INSTANTANEOUSLY move a finite distance away from her, WHICH IS ABSURD. So the assumption that the separation of the rockets isn't constant can't be correct.

Q.E.D.

 

[DaveC426913]

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.

What citation are you referring to that concludes this?

 

[Mike_Fontenot]

I do not believe this is true.

What citation are you referring to that concludes this?

How do YOU describe the twin paradox?