Bell’s Spaceship Paradox. Does the string break?

Discussion in 'Physics & Math' started by Mike_Fontenot, Mar 18, 2023.

  1. Mike_Fontenot Registered Senior Member

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    622
    Title: A Resolution of Bell’s Spaceship Paradox
    Author: Michael Leon Fontenot
    email: PhysicsFiddler@gmail.com ___________________________________________________________________________

    Abstract:

    There is not unanimous agreement about the outcome of Bell’s Spaceship Paradox. Some say the thread breaks, and some say it doesn’t. Who is correct? Does the string break or not? The answer depends on how the acceleration is determined. If the acceleration is determined by calculations and measurements made by various inertial observers, the string WILL break. But if the acceleration is what accelerometers attached to the spaceships display, the string will NOT break.

    ___________________________________________________________________________

    Two perpetually inertial observers (IO1 and IO2), perpetually mutually stationary with one another, are initially co-located with two separated observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to begin a constant (according to them) acceleration "A" (with the separation in the direction of their acceleration). AO1 and AO2 KNOW that their acceleration is "A", because they each are carrying an accelerometer that confirms it. IO1 and IO2 will conclude that AO1 and AO2 maintain the separation "L" during the accelerations. And AO1 and AO2 will agree with that: AO1 and AO2 conclude that their separation remains constant at "L" during the acceleration. But two other inertial observers, IO3 and IO4, who are momentarily co- located with AO1 and AO2 at any time later in the trip, will NOT agree that the separation "L" is constant: they will say that it has increased since the start of the trip.

    The separation of the two people undergoing the acceleration (AO1 and AO2), ACCORDING TO THOSE TWO PEOPLE THEMSELVES, can be referred to as the “PROPER” separation.

    The REFERENCE FRAME of an accelerating observer, say AO1's reference frame, is constructed in the same manner as an inertial observer constructs his reference frame (as Einstein explained to us). AO1's helpers just lay out yardsticks, end-to-end, in the direction of the acceleration. To keep the yardsticks in place, they each are attached to an accelerometer- controlled rocket, so they each are accelerating at "A" lightyears/year/year. And between every pair of yardsticks, there is a clock. The clocks were initially synchronized before the constant acceleration started. Once the acceleration starts, the clocks don't remain synchronized. Clocks located farther in the direction of the acceleration tic faster, by the rate ratio “R”. Einstein gave an exponential equation for “R” ( https://einsteinpapers.press.princeton.edu/vol2-trans/319 )
    but I have previously shown that his exponential equation is incorrect
    ( https://vixra.org/abs/2109.0076 ). I later gave the corrected equation for the rate ratio “R”
    ( https://vixra.org/abs/2201.0015 ).

    Although Einstein’s exponential equation is incorrect, he was correct in his belief that the separation “L” is constant during the acceleration (according to the people actually undergoing the acceleration). Einstein clearly believed that "L" is constant, because he didn't give an equation for how "L" varies with time. If he had thought "L" varied with time, he would have needed to include an L(t) equation as part of his solution. He did NOT do that.

    Each accelerometer directs its attached rocket to accelerate at exactly "A" lightyears per year per year. NEWTONIAN physics would say that the velocity of AO1 and AO2 would increase linearly with time, forever:

    v = A * t. (incorrect)

    That equation means that "v" would go to infinity as "t" goes to infinity, which we know can't be true in special relativity. So the above equation is clearly wrong. Special relativity says the quantity "(A * t)", which it calls the "rapidity" (denoted by the variable "theta"), is related to the velocity "v" by

    velocity = v = tanh(rapidity) = tanh(theta) = tanh(A * t). (correct)

    That says that with a constant acceleration "A", "v" approaches (but never equals) the speed of light, "c", as "t" goes to infinity.

    The distance "D" each rocket moves, according to AO1 and AO2, is

    D = integral {from 0 to tau} [ v dt ] .

    = integral {from 0 to tau} [ tanh(A*t) dt ] .

    The integral of tanh(x) is equal to log[cosh(x)], so

    D = log[cosh(tau)] / A - log[cosh(0)] / A .

    D = log[cosh(tau)] / A .

    So "D" grows forever, but it's RATE of growth decreases as tau increases.

    The distance "D" each rocket moves during the acceleration is EXACTLY the same, so the separation "L" between AO1 and AO2, according to THEM, can't change during the acceleration.

    The INERTIAL observers (IO1 and IO2) also conclude that the separation "L" between AO1 and AO2 stays constant during the acceleration. But two inertial observers (IO3 and IO4) who are momentarily co-located with AO1 and AO2 at some later instant in the trip will conclude that the separation between AO1 and AO2 is LARGER than it was when the acceleration started. And that larger separation continues to increase as the trip progresses, according to the INERTIAL observers momentarily co-located with AO1 and AO2 later in the trip.

    So the accelerating observers (AO1 and AO2) say that their separation is CONSTANT during their trip. The inertial observers (IO3 and IO4) say that the separation of AO1 and AO2 INCREASES during the trip. Those two groups of observers DISAGREE. That's just the way special relativity IS.

    But it's normal in special relativity for an accelerating observer to agree with the inertial observer who is momentarily co-located with him at some instant ... that's what the Co-Moving-Inertial-Frame (CMIF) simultaneity method IS. The inertial observer IO3 is momentarily co-located with AO1, and IO3 tells AO1 that the separation between AO1 and AO2 is larger than it was when the separation began. Does that contradict my above argument? No, it doesn't, because the scenarios are themselves different: the actual accelerations are slightly different. How do the people producing the scenario with all the inertial observers achieve the acceleration "A"? It is based on the CALCULATIONS by the inertial people: they measure positions of AO1 and AO2 versus the time on their own watches, and COMPUTE the acceleration. It is NOT based on any accelerometer, and it differs from what AO1 and AO2 read on their accelerometers.

    Note that the above description is relevant to the well-known (and much misunderstood) Bell's Spaceship Paradox:

    https://en.wikipedia.org/wiki/Bell's_spaceship_paradox .

    The above Wiki article should be read in its entirety, including its list of references. There have been MANY papers written on Bell’s paradox over the years, and different conclusions have been drawn. There has definitely not been a unanimous conclusion.

    Does the string break or not? The answer depends on how the acceleration is measured. If the acceleration is determined by calculations and measurements made by various inertial observers, the string WILL break. But if the acceleration is what accelerometers attached to the spaceships display, the string will NOT break.

    (The above paper is on the online viXra repository at https://vixra.org/abs/2303.0049 .)
     
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  3. Neddy Bate Valued Senior Member

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    Anyone who understands SR should agree that the string breaks, given the description of the scenario. It may take them a while to figure it out, but once they do, they should all agree. That is what Bell found when he questioned people at CERN. At first they got it wrong, but then they changed their minds once they analysed it properly.

    Sorry, but this cannot be correct. The spaceships have rocket engines on them, and while the engines are firing, the spaceships experience proper acceleration. This means accelerometers attached to the spaceships will definitely display an acceleration. Yet the string must break, contrary to your statement.

    On the other hand, if I accelerate past the two spaceships before their rocket engines fire, I could say they have coordinate acceleration relative to me. Yet their accelerometers display zero, and the string between them does not break while I do so. This is completely opposite to your claim.

    Perhaps what you mean to say is that if the distance between the spaceships is constant according to an inertial frame, then the string breaks. But if the distance between the spaceships is constant according to the spaceships, then the string will not break.
     
    Last edited: Mar 19, 2023
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  5. Mike_Fontenot Registered Senior Member

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    That's exactly what I said.

    But regarding your first sentence, the only inertial frames that say the separation is constant are the frames I labeled IO1 and IO2 (the observers who are stationary wrt AO1 and AO2 before the acceleration starts). The inertial frames that are momentarily stationary later in the acceleration (IO3 and IO4) say that the separation is greater than it was before the acceleration started.

    And your second sentence is exactly what I said. The separation will be constant in the case where the accelerometers attached to the spaceships show the same readings. And that is what the accelerating observers AO1 and AO2 say is the case. In the case of a constant acceleration according to IO1 and IO2, the accelerometers will NOT read the same value ... the leading accelerometer will show a slightly greater value than the trailing accelerometer. The two viewpoints (inertial versus accelerating) produce two slightly different SCENARIOS. The string breaks in one scenario, and it does NOT break in the other scenario.
     
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  7. Neddy Bate Valued Senior Member

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    Mike,

    Sorry but I don't think you are being very clear. In post #1 you said, "If the acceleration is what accelerometers attached to the spaceships display, the string will NOT break." But in the basic Bell scenario, the string does break, and yet the accelerometers do show acceleration. So, what did you mean to write there? Something like, "If ________________ the string will NOT break." But the _______________ cannot be simply "if the acceleration is what accelerometers attached to the spaceships display" because that is by definition what the accelerometers do, and yet the string does break in the standard Bell's Spaceships. Are you trying to say something about the acceleration being the same on both accelerometers simultaneously? Even then, they do show the same acceleration simultaneously in the inertial frame, and yet the string does break. So you still have to clarify something for that to make any sense.

    It is much clearer to just state what I stated in post #2,"If the distance between the spaceships remains constant according to the spaceships, then the string will not break." That will only be the case if the inertial frame says the spaceships get closer together at the same rate that length contraction happens. But that is a modified version, and not the original Bell's Spacehips.
     
    Last edited: Mar 21, 2023
  8. Mike_Fontenot Registered Senior Member

    Messages:
    622
    Yes, that's true (assuming that the two accelerometers display exactly the same reading).

    (The red addition in square brackets was made my me, mlf ... I think that's what you intended to say, but didn't).

    No. In the case where the string breaks, if there IS an accelerometer on each spaceship, they will NOT show exactly the same acceleration (the leading one will show slightly greater acceleration). In the "string breaks" scenario, the acceleration of each ship is NOT determined by accelerometers. The determination of the acceleration of each spaceship is based on POSITION OBSERVATION, DISTANCE MEASUREMENTS, and CALCULATIONS by IO1 and IO2 that tell THEM that the accelerations are exactly the same. But accelerometers on each spaceship would disagree with that determination.

    In the case where the string DOESN'T break, the acceleration shown on the accelerometers on each spaceship WILL be the same. And in that case, IO1 and IO2 will each say that the accelerations of the two spaceships are NOT the same (according the their DISTANCE MEASUREMENTS and their CALCULATIONS). They will say that the leading rocket has slightly less acceleration.

    The two scenarios are NOT the same, and the fate of the string is different for the two scenarios.
     
  9. Neddy Bate Valued Senior Member

    Messages:
    2,546
    Okay, at least you are being clear now. But still, what you wrote there is wrong.

    1. In the standard Bell's Spaceships, the two identical spaceships accelerate at the same rate as measured by the inertial frame of their pre-launch state.
    2. The two ships achieve this by firing their identical rocket engines simultaneously as measured by the inertial frame of their pre-launch state.
    3. The two ships thus undergo identical proper acceleration which would make their accelerometers show identical readings.
    4. The two ships maintain a constant distance apart as measured by the inertial frame.
    5. The string breaks.
     
  10. Mike_Fontenot Registered Senior Member

    Messages:
    622
    True. Call that inertial frame IO1.

    The people in the inertial frame IO1 only know that, according to their measurements, the acceleration is exactly what they wanted. But if the rockets DO have associated accelerometers, the IO1 people will see that the leading accelerometer reads a slightly higher acceleration than the trailing accelerometer, and will assume that one or both of the accelerometers are inaccurate.

    The people in the IO1 frame can't say anything about "proper" acceleration. They only know about their measurements and calculations. "Proper acceleration" ONLY refers to the conclusions of the two people (AO1 and AO2) who are accelerating.

    That's what the inertial frame IO1 concludes. It's NOT what the inertial frames like IO3 conclude, that are moving relative to IO1.

    If the accelerometers show exactly the same readings, the string WON'T break. If the leading accelerometer shows a slightly greater acceleration, the string WILL break. Those are two different scenarios, and they have two different outcomes.
     
  11. Neddy Bate Valued Senior Member

    Messages:
    2,546
    Okay.

    Where are you getting this crazy idea from? It is not in any standard SR description of Bell's Spaceships, so I am thinking this is your own theory gone wrong.

    Whatever number is displayed on the accelerometers is frame-invariant. That means if it constantly says for example "1 gee" in the rocket's frame, then it also says "1 gee" in all frames. The people in IO1 can be located right next to the accelerometers, and see them with their own eyes.

    So you are claiming two identical rocket engines fired simultaneously in frame IO1 will produce two identical coordinate accelerations (maintaining a constant distance between the spaceships in IO1) but will magically produce two different proper acceleration readings on the accelerometers! What causes that, the infinitely weak string? Would you still make that bizarre claim if there were no string?

    Wrong. The basic Bell's Spaceships scenario is exactly that, and the string does break.

    There is a modified version of Bell's Spaceships scenario in which the string does not break, but in that one, the spaceships get closer together in frame IO1 at just the right rate so that the string is allowed to be length-contracted as it naturally must in order to not break. In that case, the coordinate accelerations are not equal, and the string does not break, but that is still different than what you are claiming.
     
    Last edited: Mar 22, 2023
  12. Mike_Fontenot Registered Senior Member

    Messages:
    622
    First of all, there IS no "standard description of Bell's Spaceship paradox" ... read the Wiki page I referenced, including the Wiki page's list of references. There have been MANY different explanations put foreword over the years, and LOTS of arguments.

    I contend that there are two DIFFERENT scenarios, depending on WHO is controlling the experiment :

    When the inertial people are in control, they get a scenario in which the leading rocket accelerates more than the trailing rocket accelerates, so the separation between the rockets increases, and the string breaks. The inertial observers at the instant the acceleration begins (IO1 and IO2) DON'T believe that, but the the inertial observers later in the acceleration (IO3 and IO4) DO believe that. The accelerating observers (AO1 and AO2) also agree that their separation increases, and the string breaks. But AO1 and AO2 also observed that the accelerometers DIDN'T read the same acceleration during the acceleration. AO1 and AO2 weren't controlling the scenario, the inertial observers were.

    In the second scenario, it is AO1 and AO2 who are controlling the scenario, not the inertial observers. In that case, the two accelerometers read exactly the same value, and the separation of the two rockets is constant, and the string doesn't break.

    Two different scenarios. Two different outcomes.
     
  13. Neddy Bate Valued Senior Member

    Messages:
    2,546
    This is right from the wiki:

    " A delicate thread hangs between two spaceships. 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. "

    " So, calculations made in both frames show that the thread will break; in S′ due to the non-simultaneous acceleration and the increasing distance between the spaceships, and in S due to length contraction of the thread. "

    https://en.wikipedia.org/wiki/Bell's_spaceship_paradox

    But the scenario is specified as being equal acceleration, and yet the string breaks. It is not necessary for the front to accelerate more than the rear for it to break.

    You are way off on this, I assure you. Read the wiki again. Or better yet, watch the first 6 minutes of this video:



    Both variations of the Bell's Spaceship scenarios are explained fully. It covers the standard Bell's Spaceships which have equal acceleration, and the string breaks. It also covers the modified Bell's Spaceships which have unequal acceleration in just the right way so that the string does not break. In that case, the front must accelerate LESS than the rear.

    The rest of the video goes on to develop Rindler coordinates, so it is really only the first 6 minutes that are pertinent here.
     
    Last edited: Mar 22, 2023
  14. Mike_Fontenot Registered Senior Member

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    622
    There is one piece of the puzzle that you may not have seen yet. Einstein (in the reference I've given in several of my vixra papers) gave an equation that gives the ratio of the tic rates of two separated clocks, as a function of the product L*A, where "L" is the separation and "A" is the acceleration. He clearly considered "L" to be constant during the acceleration ... otherwise, he would have needed to specify the way "L" varies with time (in order to have a complete solution for the tic rate ratio), and he did NOT do that.
     
  15. Mike_Fontenot Registered Senior Member

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    622
    Go to 1:15 on that video. The narrator says that Einstein said the separation between the spaceships never changes. Q.E.D.
     
  16. Neddy Bate Valued Senior Member

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    2,546
    If Einstein was talking about gravity, then L can remain constant over time, but that does not make it so for the two spaceships in SR.

    I had a feeling you might be trying to rewrite SR to suit your other ideas. So you came up with the idea that even though the two spaceships both start accelerating simultaneously and equally according to inertial frame IO1, and even though they maintain a constant separation distance in inertial frame IO1, and even though both have the same instantaneous velocity at all times in inertial frame IO1, that somehow their accelerometers could display different readings? Because otherwise you can't explain how the string breaks?

    Have you tried calculating what the proper length of the string would have to be in order for it not to break? Choose any time during the acceleration, and find the instantaneous velocity of the spaceships at that time. Calculate gamma for that velocity. Now multiply that gamma by the distance between the spaceships, and that is what the proper length of the string would have to be in order to not break. But the proper length of the string is not changing, it is fixed from the beginning. The proper length of the string is shorter than the proper length calculation I just asked you to do, because the spaceships are growing farther apart in their own reference frames. It is simple SR length contraction. Or at least it was before you tried to rewrite it.
     
  17. Neddy Bate Valued Senior Member

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    2,546
    In that video, "Einstein" is the inertial observer, or in other words "Einstein" is your IO1. Saying the separation between the spaceships never changes according to Einstein is the same as saying it never changes in IO1. That is not to say it never changes in the spaceships own frame!

    Watch the rest of the 6 minutes of the video, please.
     
  18. Mike_Fontenot Registered Senior Member

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    No, he wasn't talking about the gravitational case, he was talking about the special relativity acceleration case. He knew how to handle the SR case, and he hoped to use that, together with the equivalence principle, to tell him something about gravitation (which he was just beginning to try to solve).
     
  19. Mike_Fontenot Registered Senior Member

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  20. phyti Registered Senior Member

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    mike;

    There are two identical ships A and B in the launch frame U.

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    Each is tracked in time by the front of the ship.
    They are connected by a flexible cord which is taut between the front of A and the back of B, a distance L.
    Each ship accelerates simultaneously using the same computer program to approx. .4c. The acceleration phase results in length contraction of each ship to 1.85 units.
    Since A and B experience length contraction, L increases to 2.15, the tension breaks the cord.
    A sends blue signals to B to synchronize clocks.
    This establishes the green Ax axis, aka axis of simultaneity.
    The perception of A and B is the front clocks ahead of the back clocks and the back clocks behind the front clocks. U declares the A and B clocks synchronized.

    The accelerometers verify the acceleration. Td and lc depend on speed, thus require measurements and calculations. In the Wiki article, Dewan and Beran state the space between the ships remains constant. Not possible if they are independent/not rigidly connected.
    This is in the SR environment, so the EP doesn’t apply. They are not stacked in a gravitational gradient. That means equivalent accelerations.
     
  21. Mike_Fontenot Registered Senior Member

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    Yes, you're right about that. But later in the video, Einstein's perspective is according to people on the two spaceships. I.e., Einstein is accelerating with the spaceships.

    I just did. The only important stuff is from 1:07 to 1:30 . Here is that segment:

    "The spaceships will always accelerate in the same manner, so that they are always traveling at the same velocity, according to Einstein. And the distance D between the ships is always the same, according to Einstein. The question is, 'Does the string between the ships break?'. Some people might argue that the string will not break, because the distance between the ships does not change, according to Einstein."

    THAT is exactly what Einstein DOES argue. And that completely agrees with my results.
     
  22. phyti Registered Senior Member

    Messages:
    732
    mike;

    Treating the two ships connected by a cord as an integrated object,
    install a more powerful engine in A.
    A accelerates, the thrust is not transferred via the cord to B.
    The cord folds until A contacts B.
    There are two independent ships.

    From Wiki article:
    "This means, by definition, that with respect to S the distance between the two rockets does not change even when they speed up to relativistic velocities."
    This phrase is true if applied to center of mass for the ships, i.e. the same location relative to a ship
    This still allows lc of the ships and expansion of the gap between.
     
  23. Mike_Fontenot Registered Senior Member

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    622
    More than 20 years ago, I plotted a chart showing two separated objects undergoing the same constant acceleration "A". (That chart still hangs on the wall above my desk, and I've never questioned it before). The plot supposedly shows the view of things according to an inertial reference frame (the IRF) that is stationary wrt the two objects immediately before the acceleration begins. One curve starts from the origin with slope zero at the origin, but then curves upward with a curvature that monotonically decreases as time increases, and asymptotically approaching a slope of "c", the speed of light. I use units where "c" equals 1.0, so the curve approaches a slope of 1.0 on the chart.

    The other curve has exactly the same shape, but starts at some distance "D" above the origin. The two curves are always separated by a vertical distance of "D".

    The idea, I think, was that the two curves must have exactly the same shape because of "the Principle of Relativity" ... i.e., it shouldn't matter where in space you start the curve, the curves should always have the same shape.

    But here's the quandary: An observer in the inertial frame IRF is told by the chart that the two objects always have the same distance apart. But the length contraction equation (LCE) of special relativity says that an inertial observer should conclude that a moving yardstick should get shorter and shorter as its speed wrt the inertial observer increases. That seems to contradict what the chart says, and it seems to contradict the Principle of Relativity. The LCE seems to require that the two curves get closer together as time increases. Does the upper curve slowly get closer to the lower curve? Or does the lower curve approach the upper curve? Or is there some combination of those two movements? Any of those movements contradicts what the chart says, and it thus seems to contradict the Principle of Relativity.

    Any ideas? I'm really stuck ... I don't know the answer.
     

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