[MikeFontenot (OP)]

To the above 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".

The original (incorrect) curve (which shows a constant distance between the two rockets, according to the initial inertial observers who are stationary wrt the rockets before the rockets are fired), violates the length contraction equation (LCE) of Special Relativity.  The curve below shows what the initial inertial observers conclusions must be, according to the LCE.


Scan 2023-7-19 13.34.42.jpg (781.69 kB . 1700x2338 - viewed 1048 times)

[MikeFontenot (OP)]

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.

[MikeFontenot (OP)]

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. THE CHART FOR THAT SOLUTION IS IN POST #68. A MORE DETAILED CHART IS GIVEN IN:
   https://vixra.org/abs/2307.0151

 * But the answer it gives when there are only two rockets is still incorrect, even though it doesn't produce an obvious inconsistency.  The only correct method is the previous one, where the trajectory of a sufficiently distant rocket has a negative slope on the chart initially, before curving upward eventually .

[MikeFontenot (OP)]

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: [removed by mod]

____________________________________________________________________________________


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.

[MikeFontenot (OP)]

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.

[pzkpfw]

If you have accelerometers at the front and back of a single spaceship, will they show the same acceleration?

[MikeFontenot (OP)]

If you have accelerometers at the front and back of a single spaceship, will they show the same acceleration?

That's not a question that interests me.

[MikeFontenot (OP)]

A related question is of interest to me though:

What if a rigid rod connects the two separated spaceships.  Perhaps with a rigid connection to the leading spaceship, but with a sliding connection to the trailing spaceship.  And with distance markings printed on the rod, like a tape measure.