Post by: Lluis Olle on 13/09/2022 21:13:52
In a nutshell and in layman's words, at any time the speed yet accomplished by the leading edge acceleration will be greater than the speed reached by the trailing edge, and so the thread will have to continually stretch due to the relative non-zero velocity of the leading edge vs the trailing one.
I was doing some preliminary numbers, considering a 9.8 m/s2 acceleration, and the relative speeds and forces involved are really tiny, but of course if the proper acceleration is constant, the string breaks for sure.
So, what about considering instead a constant thrust force?. For a spaceship of mass M, the engine is set to produce a thrust force equal to M*9,8. That would make the proper acceleration constant, if the thrust force is constant (for sake of simplicity, I ignore any mass change due to burning propellant, or the thrust can adjust continually to compensate the change in mass).
Then, the conditions of the paradox are slightly different, because when the thread stretches then the trust force will remain constant, and so the proper acceleration of the leading spaceship has to decrease, although the Center of Mass of the 2 spaceships system will remain with the same proper acceleration.
I'm just an retired Electrical Engineer, so take the following reasoning with a grain of salt.
From the theory of constant acceleration in Special Relativity, for an observer located in the CM frame, both spaceships will have a relative outwards velocity, that will make the distance to steadily increase between them, and between the CM. Remember that such measures are always done from the CM observer’s perspective.
I can argue that this relative velocity and due acceleration, is a relativistic effect that can be somehow considered as a kind of “gravitational” field, with the following differences:
- It’s repulsive, as opposed to gravity that is attractive.
- It increases with the distance, as opposed with gravity that decreases with the square of the distance. At the CM position, is just zero.
- Any way, that relativistic “field” produces a relativistic force over any mass, and that force follows Newton’s second Law.
From Born rigid motion concept, we have the formula that gives the required proper acceleration at a distance D’ from the observer, so that distance keeps always constant from that observer is:
Then, from the point of view of the CM Observer, if somehow the acceleration of a point located at D’, that initially is moving with the same proper acceleration as the observer, decreases to the required acceleration value of the Born formula, the relativistic force would cease.
So, the “relativistic field” could be reasonably described as:
The dynamic equation of motion from the CM point of view, considering that the other spaceship behaves symmetrically, counterbalancing the elastic force of the thread (same tension at both ends), is:
being E the Young modulus and A the section area of the string.
I've done a Python program, to solve and graph the differential equation, and the result is that even considering a really fragile string and some heavy spaceships, the string doesn't break, but there's a bouncing of the spaceships in which the string is stretched but stops the outward movement very soon, pulls back returning the initial kinetic energy absorbed, until the "relativistic force" after some time moves the spaceships outwards again ... bouncing forever. The amazing result is that the string is loose - without tension - most of the time!
Or I'm totally wrong?
Post by: Halc on 13/09/2022 22:24:34
Welcome to TNS! Somebody with Tex skills no less.
I understand that if both spaceships maintain the same proper acceleration, then the string will break because of the non-simultaneity of the acceleration effects at both ends of the thread.I'd not have described it that way. Identical proper acceleration means they both measure the same g force, and what other acceleration effects are there besides what the accelerometer reads?
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In a nutshell and in layman's words, at any time the speed yet accomplished by the leading edge acceleration will be greater than the speed reached by the trailing edgeThe words 'at any time' are fairly meaningless without a frame of reference. You can talk about local things, but nothing like 'the other end of the string' using language that doesn't make the frame explicit.
In the frame of either ship, the ship is stationary, so there is no nonzero 'speed yet accomplished' relative to that frame. In the original inertial frame relative to which the ships were initially at rest and simultaneously commenced identical proper acceleration, both ships will always be moving at the same speed and will maintain constant separation. In this frame, explaining the string breaking is because the string contracts with speed and can no longer reach across that constant separation.
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So, what about considering instead a constant thrust force?Constant thrust/force results in constant proper acceleration iff the two ships are identical in mass. The usual wording of the problem ignores this irrelevant detail since the point is illustrated easily without worrying about the mass of the accelerating objects. It's their motion we care about, not what they must do to achieve that motion.
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For a spaceship of mass M, the engine is set to produce a thrust force equal to M*9,8 (assuming M is constant, which seems unreasonable for any spaceship, which again is why we ignore this part and just specify the motion.for sake of simplicity, I ignore any mass change due to burning propellant[/quote]Right. You know this part then.
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or the thrust can adjust continually to compensate the change in massThe proper mass of the ship doesn't change (per your statement above), and proper acceleration is absolute, so is not dependent on relativistic (frame dependent) mass. So no matter how fast your 1 kg object goes, it takes only one newton of force to accelerate it at a proper 1 m/sec² in any direction.
Bottom line, the thrust on either ship in either the ship frame or the original inertial frame is always the same if the proper mass doesn't change.
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and so the proper acceleration of the leading spaceship has to decreaseThis violates the terms of the scenario which says it is constant for both objects.
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although the Center of Mass of the 2 spaceships system will remain with the same proper acceleration.Center of mass of two separated objects and its center of mass is frame dependent and not meaningful without its specification. In the original inertial frame, your statement is true. In either ship frame, it is not true.
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From the theory of constant acceleration in Special RelativityThere's a named theory about that?
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for an observer located in the CM frameI am willing to accept such a frame, an accelerating frame in which both objects (with identical proper acceleration) are always equidistant. My apologies for breaking up your sentence like this.
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both spaceships will have a relative outwards velocity, that will make the distance to steadily increase between them, and between the CM. Remember that such measures are always done from the CM observer’s perspective.Yes, agree.
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It’s repulsiveIf it was, it would be a function of the mass of the objects, but this scenario assumes negligible mass for any object, hence gravitational attraction or repulsion seems the incorrect way to describe what's going on. Still, I see where you're going with it.
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It increases with the distance, as opposed with gravity that decreases with the square of the distance.Try a big separation. Put the two ships 10 light years apart and have them both accelerate at a proper 1g. Now in the frame of the CoM, what are the ships doing? The distance is so large that (in the accelerating frame of the CoM) the 'repulsion' causes the distance from the CoM to either ship to increase at a rate greater than c. This isn't violating any rules that I can see. Just trying to grok what you're describing here.
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Any way, that relativistic “field” produces a relativistic force over any mass, and that force follows Newton’s second Law.A force not felt by either ship? I mean the trailing one is being pushed back faster than light, and yet they experience only a positive 1g proper acceleration. If it followed Newton's laws, they'd feel it.
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From Born rigid motion concept, we have the formula that gives the required proper acceleration at a distance D’ from the observer, so that distance keeps always constant from that observer is:So the distance stays constant in the accelerating frame of the object exhibiting Born rigid motion. It would not remain constant in any other frame. Just reminding you that frame dependent statements like that require frame specification.
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Then, from the point of view of the CM Observer, if somehow the acceleration of a point located at D’, that initially is moving with the same proper acceleration as the observer, decreases to the required acceleration value of the Born formula, the relativistic force would cease.OK. Again, what happens to our ships 10 LY apart if the CM accelerates at a proper 1g? Answer: The three objects do not exhibit Born rigid motion.
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I've done a Python program, to solve and graph the differential equation, and the result is that even considering a really fragile string and some heavy spaceships, the string doesn't break, but there's a bouncing of the spaceships in which the string is stretched but stops the outward movement very soon, pulls back returning the initial kinetic energy absorbed, until the "relativistic force" after some time moves the spaceships outwards again ... bouncing forever.I think you're describing sound waves travelling along the string.
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The amazing result is that the string is loose - without tension - most of the time!If the string (and the ships and CM tied to it) exhibit Born rigid motion, then yes, the string need not break.
Or I'm totally wrong?
Post by: Lluis Olle on 13/09/2022 23:24:09
First of all, it's not the Bell's Paradox, but a new one, in which I introduce an slight and subtle change.
The change is consider a constant thrust force, instead of a constant proper acceleration. It seems the same, but it's not the same!
Lets say that if you have to keep a constant proper acceleration, no matter what, then I can argue that I will be using strings each time stronger, a Titanium cable with a diameter of 1 meter for example (seems more like a rod, but all is Relative, isn't it?). And then what? No problem, because as you "magically" keep the proper acceleration constant, then the string breaks. You don't play fair with the string!
But if you trust force is constant, then the scenario is different! You play with different rules.
When the string begin to stretch, it will pull both spaceships (by a tiny amount, if you wish). The proper acceleration of the Center of Mass will keep constant, because an internal force could not change that, but the proper acceleration of the trailing spaceship will increase, and the proper acceleration of the leading spaceship will decrease. You don't have the "extra whatever" needed to keep the constant proper acceleration, but you only have you initial constant thrust force.
Comment that always that I refer to time, I mean the proper time of the spaceship, and when I compare two proper times, I'm referencing the simultaneous proper times t11/t12 observed by spaceship (1), and the pair t21/t22 observed by spaceship (2).
For trailing spaceship (1) t12>t11 - both times simultaneous for (1) -, and for leading spaceship (2) t22>t21 - both times simultaneous for (2). You can take a look at the Python program that I appended (as a pdf file) in the original post, and see how I do it, right or not, I can be mistaken.
Post by: Halc on 14/09/2022 01:19:13
The change is consider a constant thrust force, instead of a constant proper acceleration. It seems the same, but it's not the same!If a constant force is applied to two identical masses attached by a strong string (maintaining constant proper distance), the tension on the string will cancel some of that force for the front object and add to the force applied to the rear object, resulting in the rear one experiencing a greater proper acceleration than the rear one. You can't have your cake (constant proper acceleration) and eat it too (constant proper separation between the two).
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But if you trust force is constant, then the scenario is different! You play with different rules.Will it? Make the string really long and you'll see it isn't 1g for born-rigid motion and the force being enough to accelerate each endpoint at 1g.
When the string begin to stretch, it will pull both spaceships (by a tiny amount, if you wish). The proper acceleration of the Center of Mass will keep constant
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but the proper acceleration of the trailing spaceship will increase, and the proper acceleration of the leading spaceship will decrease.Right. So we have essentially rigid motion now, or at least close to it if the acceleration is small and the separation isn't too big.
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Comment that always that I refer to time, I mean the proper time of the spaceshipThere's two spaceships, one at each end of the string. Time for one doesn't stay in sync with the other.
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and when I compare two proper times, I'm referencing the simultaneous proper times t11/t12 observed by spaceship (1), and the pair t21/t22 observed by spaceship (2).Sorry, but lacking say a spacetime diagram with labels, these notations are unclear. I'm guessing that t12 is time 2 (later) for ship 1, and thus t12-t11 is some duration of acceleration for the trailing ship.
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For trailing spaceship (1) t12>t11 - both times simultaneous for (1)Again unclear. If t12 is greater than t11 (makes sense), how are those two times simultaneous?
Looking at the python code. Never seen phython, but how hard can it be? Nicely commented, which helps.
Post by: Lluis Olle on 14/09/2022 13:38:47
Then, I want to consider what happens if instead of a proper acceleration, what you keep is a constant thrust, lets say a constant "proper" thrust. Its not the scenario of the Bell's Paradox, but another slightly changed. We don't want to maintain a constant proper acceleration... we don't know yet how the proper acceleration will evolve, if what we consider is a constant thrust force.
To keep a constant proper acceleration is not the same as to keep a constant proper thrust. With a constant proper thrust, you apply a constant force always. With a constant proper acceleration, you must apply the exact force needed to maintain your constant proper acceleration, and that force is then potentially variable in nature.
Now, my statement of the problem, likewise Bell did, is as follows:
Two spaceships tied by a string, which has known mechanical characteristics, have its engines adjusted to give a fixed thrust. Knowing the mass of the spaceships, the engine thrust is calculated to produce initially a comfortable acceleration during the journey. Lets say, 9,8 m/s2.
The string is winded into a reel in the leading spaceship (S2), and is fixed in the trailing spaceship (S1). At the start of the journey, the string goes from S2 to S1 without any tension and covers a length of 10.000 meters, which is the initial separation. The string can freely, and with no resistance - unwind if needed to adjust to the required distance between the spaceships, if that distance changes.
The only assumption we make is that the mass of the string is negligible, compared with the mass of the spaceships.
The journey starts, both engines gently reach its nominal and constant thrust value, the initial distance between the spaceships is measured and adjusted to 10,000 meters from the point of view of the CM observer.
I'll consider the Center of Mass (CM) of the two spaceship system. The CM will be located, at least for the Earth observer, halfway between both spaceships. After 24 hours of proper time as measured by CM observer, it signals to the S2 spaceship to gently brake the reel, and can't no longer unwind. What happens next?
For the CM observer, each spaceships have a relative outwards speed, that increases its distance to the CM. This is the reason of the tension force that finally breaks any string in the original Bell constant proper acceleration scenario. If this was not the case, the Bell's string would not break - there's no magic.
I'll refer to the following YT video, for the proper acceleration concepts: youtu.be/O92pQXZaEnw
(https://llober.neocities.org/Bell/Constant%20acceleration%20hyperbolas.PNG)
So, first thing first, which are the relative speeds of the spaceships and CM when we brake the reel at proper time 24 hours (CM time)? In the Earth frame reference, we can describe the movement of the CM as:
For the trailing spaceship S1, the equations are the same, but the x-coordinate is shifted by D0/2 to the left.
For the leading spaceship S2, the equations are the same, but the x-coordinate is shifted by D0/2 to the right.
And so, the relative speed of the CM in the Earth frame at a given proper time tc of the CM, is given by:
The CM observer has a proper time of tc. To get the corresponding simultaneous proper time in the leading S2 spaceship, we know that the line of simultaneous events for the CM is given by a line that passes by the (0,0) origin of the space-time diagram and the corresponding event-point in the CM world-line at time tc. Refer to the YT video.
Now, we have to do the intersection of the simultaneity line from the CM observer, with the S2 leading spaceship world-line. In the Python program, this is done using a multi precision package called "mpmath", because the world-line hyperbolas are so really close, and is not enough to use normal 16 digit float precision. This is the first clue that we're playing with really tiny, tiny values in front of c or c2!
We have tc CM proper time and the corresponding simultaneous observed proper time t2 in spaceship S2. Both times are simultaneous for the CM observer - lets say that S2 has a rear display that shows the S2 proper time in big numbers, and CM observer, at proper time tc, sees that the S2 display reads t2.
Having vc and v2, we can use the relativistic velocity addition formula to get the v2c speed of S2 relative to the CM:
In our case, this gives v2c = 1.66 mm/hour at the 24 hours CM proper time. For our 1000,000 Tons spaceship, it results in a whooping 0.000106 J kinetic energy relative to the CM :( This is another clue!
From the CM point of view, we can draw the following diagram of forces. Note that the CM is an accelerating reference frame, and that for the CM system, the internal forces due to tension in the string will not affect the proper acceleration of CM.
(https://llober.neocities.org/Bell/Fuerzas.png)
The relativistic force Fr is derived from the Born rigid motion concept. In no case I mean that the spaceship reaches the Born condition, I only say that we can consider a Relativistic field that behaves like "gravity", but:
- It’s repulsive, as opposed to gravity that is attractive.
- It increases with the distance, as opposed with gravity that decreases with the square of the distance. At the CM position, is just zero.
- Any way, that relativistic “field” produces a relativistic force over any mass, and that force over the mass follows Newton’s second Law
and that from the Born rigid body concept, could be described as a field that interacts with any mass, producing a force over that mass, that per unit-mass is:
Note that the S2 proper acceleration is a variable, not a constant!. The equation for the leading S2 spaceship, considering that the other spaceship S1 behaves in a symmetrically way, counterbalancing the tension of the string, is a second-order differential non-linear equation:
where E is the Young modulus and A the area section of the string.
Solving the differential equation, applying as initial condition the relative speed of S2 to CM, we get this graphic result:
(https://llober.neocities.org/Bell/Eq_06.png)
IMHO, I can explain the result as follows:
When the reel is braked and stops unwinding, the S2 spaceship has a relative speed with respect to the CM observer. This speed is very tiny, something like 1.66 mm/hour. So, the relative kinetic energy is only 0,000106 J.
The "Relativistic force" is very, very small, ... the elastic force easily outruns it, so the tension of the string quickly stops the forward motion and pulls back, returning the kinetic energy that was stored under elastic energy in the string.
As the "Relativistic force" is so small, it takes long time for its effect to reverse the motion... and we have a bouncing equilibrium, in which the string is loose - with no tension - most of the time!
And in no moment not the string nor the spaceships reaches the Born rigid motion condition!!!
Post by: Halc on 14/09/2022 23:49:48
Then, I want to consider what happens if instead of a proper acceleration, what you keep is a constant thrust, lets say a constant "proper" thrust.You should bring up the reeling-out of the string right away because it took me forever to figure out what you’re doing. I presumed either a fixed string (rigid motion, constant string tension) or a broken one (identical acceleration of end objects), instead of the reel-out and then brake scenario.
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The string is winded into a reel in the leading spaceship (S2), and is fixed in the trailing spaceship (S1). At the start of the journey, the string goes from S2 to S1 without any tension and covers a length of 10.000 meters, which is the initial separation.OK, you finally introduced something new, which is a string on a spool. We seem to be back at Bell’s scenario now, except the string is being reeled out as needed as the proper separation between the objects changes. There’s just enough tension on the string to accelerate the string and exert zero force on the trailing ship? In reality, this cannot work. The string must stretch (deform) if it is not independently accelerated at every point along its length, keeping it in rigid motion with the rear ship. At your numbers, this problem can be ignored.
The effects are going to be hard to see if the acceleration and separation are both so low. It’s almost a Newtonian exercise at those numbers. I tend to go for high g and big separations to make the effects obvious.
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the initial distance between the spaceships is measured and adjusted to 10,000 meters from the point of view of the CM observer.OK, the CM point is halfway up the string in the proper frame of the CM, and has to out-pace the accelerating string over half the rate that the lead ship is reeling it out in order to stay midway between the ships. He’s not hanging on to the string. Keep in mind that the CM in the Earth frame is not the CM of the objects in its own frame. This is not obvious, especially with such small numbers.
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I'll consider the Center of Mass (CM) of the two spaceship system. The CM will be located, at least for the Earth observer, halfway between both spaceships.Why is Earth (an arbitrary inertial frame) special? Why can’t the CM be halfway between the ships in its own frame? That would seem more useful/meaningful. Otherwise you could have 20 different CM observers, each halfway between the ships in a different inertial frame, all in different places.
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After 24 hours of proper time as measured by CM observer, it signals to the S2 spaceship to gently brake the reel, and can't no longer unwind. What happens next?Tension on the string of course. It breaks if it cannot take the strain. If the string takes it, the tension alters the acceleration of both objects.
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So, first thing first, which are the relative speeds of the spaceships and CM when we brake the reel at proper time 24 hours (CM time)?At 10 km, 1g and 1 day, CM time is almost the same as either ship time and Earth time. It’s why I like to really put the hammer down. You’ve already discovered that small floating point numbers cannot simulate the differnces.
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The CM observer has a proper time of tc. To get the corresponding simultaneous proper time in the leading S2 spaceship, we know that the line of simultaneous events for the CM is given by a line that passes by the (0,0) origin of the space-time diagram and the corresponding event-point in the CM world-line at time tc. Refer to the YT video.This seems wrong. The origin of your spacetime diagram seems arbitrarily assigned along the x axis. The time assignment is not arbitrary since it is the time at which all objects are stationary in that frame. But the x assignment can’t be arbitrary for this to work. Given your description, it needs to be at the Rindler horizon of the CM worldline, which in your scenario is about a lightyear to the left.
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for the CM system, the internal forces due to tension in the string will not affect the proper acceleration of CM.I would hope that the CM wasn’t in any way touching the string, but instead accelerating on its own. So of course tension shouldn’t affect the CM. The string merely serves as a tape measure of sorts and can even be marked like one.
You kind of lost me after this because the scenario changes so much I don’t know which one you’re using. The video depicts constant proper acceleration at both ends and perhaps a constantly reeled-out string that exhibits born-rigid motion with the rear object. But you also talk about tension on the string, which is a different scenario, and I don’t know which one is being referenced.
In the rigid scenario, the end objects stay at a constant proper distance from the CM, and thus don’t get ‘repulsed’ at all. In the proper-acceleration scenario, there’s no tension on the string, but you are using that tension in your force diagrams. I don’t know what you’re doing. I presume something like constant proper acceleration everywhere at first, but then clamping down on the string which simply breaks it, stretches it, or puts infinite force on the string. ‘Gently’ braking allows the string to continue to reel out, but also exerts a force on the objects at either end, thus altering their accelerations. But your description says it stops unwinding, not just begins to resist the unwinding.
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The "Relativistic force" is very, very small, ... the elastic force easily outruns it, so the tension of the string quickly stops the forward motion and pulls back, returning the kinetic energy that was stored under elastic energy in the string.OK, so we allow the string to deform.
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As the "Relativistic force" is so small, it takes long time for its effect to reverse the motion... and we have a bouncing equilibrium, in which the string is loose - with no tension - most of the time!We’re not just going to let it find an equilibrium? Are you taking into consideration the speed of sound in the string where force from the braking at the reel is not immediately felt at the trailing object?
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And in no moment not the string nor the spaceships reaches the Born rigid motion condition!!!Not if all these interactions are completely elastic, no. It will bounce forever without somewhere for the heat to dissipate.
Post by: Lluis Olle on 15/09/2022 16:38:20
I'll try to do my best, and answer your comments...
You should bring up the reeling-out of the string right away
Sorry for that! Its a poetic license, to make the story more appealing :). Not really, I introduce the unwinding at an arbitrary proper time because I want the spaceships to have some kinetic energy as a "worst case scenario" for the string, and because to be as realistic as possible, I wanted to give time for the spaceships system to settle down after departure.
For example, at 24 hours there's a relative speed of the S2 spaceship with respect the CM of 1,66 mm/hour... is very little, but in one week (proper) time, if such speed keeps constant - actually is increasing - it means than the string has elongated by at least 28x2=56 cm.
OK, you finally introduced something new, which is a string on a spool. We seem to be back at Bell’s scenario now, except the string is being reeled out as needed as the proper separation between the objects changes. There’s just enough tension on the string to accelerate the string and exert zero force on the trailing ship? In reality, this cannot work. The string must stretch (deform) if it is not independently accelerated at every point along its length, keeping it in rigid motion with the rear ship. At your numbers, this problem can be ignored.
I don't get why this means that I get back to the Bell's original scenario... I keep a constant engine thrust always, I never change the scenario.
As I stated, I assume the string to have a negligible mass. In the Python program, I used real steel cable technical data, with 1 mm diameter and a break tension of 840 N. It's a very lightweight string.
And yes, the string initially extends from leading ship S2 to trailing spaceship S1, and is being pulled by S2 with a tension force enough to accelerate the mass of the string, and it unwinds because there's a relativistic effect that makes the relative distance between the spaceships to increase with time, and that translates into an "unwinding force". The distance from spaceships will increase, but the string adapts to the required length.
Remember that is a known and accepted fact that Bell's string breaks, but because it's fixed at both ends. While the string can unwind freely from one end, the string would not break.
To consider the mass of such string, and the stretching due to that string mass will add nothing to the solution.
The effects are going to be hard to see if the acceleration and separation are both so low. It’s almost a Newtonian exercise at those numbers. I tend to go for high g and big separations to make the effects obvious.
As I said previously, the relative speed at 24 hour is 1,66 mm/hour ... You can easily measure that just with a wrist watch, a normal ruler, an ink marker and some patience. And the exercise is what it is, there's nothing wrong with Newtonian mechanics if applied in the right scenario.
Why is Earth (an arbitrary inertial frame) special? Why can’t the CM be halfway between the ships in its own frame?
The CM is a conceptual point. In the Bell's original scenario, the Earth observer always sees the distance between the spaceships to be a constant, so being both spaceships identical it would deduce that the CM is always located halfway between.
In the constant proper acceleration scenario, or the constant thrust scenario, I would say that the CM is located such it observes the same outwards simultaneous relative speed for both spaceships, but I don't dare to say that this will be halfway (I'm pretty sure that if its not centered, the offset is not more than a tenth of a milliliter or so - compared with the length of 10,000 meters of the string, the offset is negligible).
Tension on the string of course. It breaks if it cannot take the strain. If the string takes it, the tension alters the acceleration of both objects.
That's right, the whole exercise is to see if the string breaks or not. As I said, my example string breaks at 840 N, but anyway, consider whatever breaking value you want, but being a realistic one of course.
At 10 km, 1g and 1 day, CM time is almost the same as either ship time and Earth time. It’s why I like to really put the hammer down. You’ve already discovered that small floating point numbers cannot simulate the differnces.
As I said, we can consider any realistic value. The precision problems are related to the disproportion between the speed of light, normally used squared, with tiny relativistic effects. But anyway, I use a multi precision math package that operates with arbitrary precision (Python mpmath), and so, no problem.
And no, I have not discovered about float precision yesterday, I know about floating point IEEE precision from about 45 years now.
This seems wrong. The origin of your spacetime diagram seems arbitrarily assigned along the x axis. The time assignment is not arbitrary since it is the time at which all objects are stationary in that frame. But the x assignment can’t be arbitrary for this to work. Given your description, it needs to be at the Rindler horizon of the CM worldline, which in your scenario is about a lightyear to the left.
My coordinate system starts where I want - this is a democratic forum :) -, and I can arbitrarily set that the S1 spaceship at Earth (rest frame) time = 0 is located at position
, S2 spaceship is located at 
, and the CM is located at 
, being D0 the initial separation. If those numbers seem to big, is just a relative perception problem.Considering the spaceships at (0,0) position at time = 0 will only make the maths much, much harder to follow and understand.
I would hope that the CM wasn’t in any way touching the string, but instead accelerating on its own. So of course tension shouldn’t affect the CM. The string merely serves as a tape measure of sorts and can even be marked like one.
The Center of Mass of a system is a conceptual point. Also, to say that "the CM observer whatever..." is just a conceptual way of describing things - there's no such "observer" hanging of the string at more or less halfway with a walkie-talkie!
And internal forces in a system can't alter the dynamics of the CM of the system, that's why is a good reference frame.
You kind of lost me after this because the scenario changes so much I don’t know which one you’re using. The video depicts constant proper acceleration at both ends and perhaps a constantly reeled-out string that exhibits born-rigid motion with the rear object. But you also talk about tension on the string, which is a different scenario, and I don’t know which one is being referenced.
I don't understand. The YT video is not about this scenario, but about the general concepts of constant proper acceleration from a Relativistic point of view. It's referenced just for the casual reader to see from where some formula or concept comes from, because I can't explain everything from the very beginning every time. My scenario is NOT Bell's one, is NOT a constant proper acceleration one, is a different constant thrust force scenario, but of course I have to reference somehow the theory behind constant proper acceleration that is absolutely related.
In the rigid scenario, the end objects stay at a constant proper distance from the CM, and thus don’t get ‘repulsed’ at all. In the proper-acceleration scenario, there’s no tension on the string, but you are using that tension in your force diagrams. I don’t know what you’re doing. I presume something like constant proper acceleration everywhere at first, but then clamping down on the string which simply breaks it, stretches it, or puts infinite force on the string. ‘Gently’ braking allows the string to continue to reel out, but also exerts a force on the objects at either end, thus altering their accelerations. But your description says it stops unwinding, not just begins to resist the unwinding.
Again, I don't understand... which is the "rigid scenario"? When the string can freely unwind from the spool, the tension in the string is due to: 1) the inertia of the mass of the string 2) the relative speed between ends, which appears as a relativistic and counter-intuitive effect.
The tension due to the inertial mass is negligible as stated. The relativistic effect, makes the string to unwind, but will not stress the string too much, because we don't consider the spool to oppose any significant resistance... until is braked.
So yes, there's a little initial stretching (tension) of the string due to exposed reasons, but is negligible compared with the stretching that will start developing when the spool is braked.
To say that we brake the spool, its to say that from that time on, both ends of the string are fixed at each spaceship. And from the very first moment that this happens, the string will begin to stretch due to the elastic tension force. It's that easy.
The force diagram describes what happens when both ends of the string are fixed. The starting time=0 of the "simulation" is when this event happens, arbitrarily 24 hours after departure (or whatever value you want to start with). This lets us play with an initial condition for the relative velocity between the CM and the S2 spaceship.
And the accelerations at both ends are altered... but this is the whole point of this scenario!!!
And this is what makes the difference with the Bell's Paradox: in the statement of the Bell's Paradox, proper accelerations for both spaceships is constant, so the accelerations can't change, because is what the statement of the Paradox reads... it's "set in stone" and that's it.
I'll write it again in bold uppercase letters (lets hope not to offend somebody):
IN BELL'S SPACESHIP STRING PARADOX, THE PROPER ACCELERATIONS OF THE SPACESHIPS CAN'T CHANGE, NOT EVEN A TINY BIT, NOT EVEN FOR AN INSTANT, IT'S FORBIDDEN BY THE STATEMENT OF THE PARADOX!!!
This is not fair play for the string... but who cares.
And in Bell's Paradox, for the proper accelerations at both ends to don't change, then what? Then you have to apply somehow an enough force... and enough is whatever is needed, because for you the force is a variable, to maintain a proper acceleration constant no matter what happens. Frankly, I don't know how to express it more clearly.
We’re not just going to let it find an equilibrium? Are you taking into consideration the speed of sound in the string where force from the braking at the reel is not immediately felt at the trailing object?
To consider the speed of sound is an exercise left for the advanced readers. And the same way Bell didn't take that fact into account, I'm only interested in the special relativity effects.
Not if all these interactions are completely elastic, no. It will bounce forever without somewhere for the heat to dissipate.
Yes, right. Actually, will radiate some heat and finally stop the bouncing, but it will take some time.
By the way, this is the result I get if the string is fixed at both ends at start (so, no spool):
(https://llober.neocities.org/Bell/Eq_06c.png)
And this is the result I get if the spool brakes at proper time 30 days after departure:
(https://llober.neocities.org/Bell/Eq_06d.png)
And the same graphic as above, but zoomed at the starting time to see what's going on:
(https://llober.neocities.org/Bell/Eq_06e.png)
Post by: Halc on 16/09/2022 01:59:50
As I stated, I assume the string to have a negligible mass. In the Python program, I used real steel cable technical data, with 1 mm diameter and a break tension of 840 N. It's a very lightweight string.Not that lightweight. At 10 km, the tension needed to accelerate it at 1 g is already over half the rated load on it. Drag 20 km of it with nothing tied to the far end and it will break. But if it's long enough, no load rating is enough to prevent it from breaking. Hence my suggestion for the string to have independent thrust applied to it along its length, but that's a different scenario than the realistic one you're going for.
Is the 50kg force the string puts on S2 subtracted from the force being applied to S2, or is the force on S2 higher than the force on S1 to compensate for the fact that it's dragging 50 kg of string?
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And yes, the string initially extends from leading ship S2 to trailing spaceship S1, and is being pulled by S2 with a tension force enough to accelerate the mass of the string, and it unwinds because there's a relativistic effect that makes the relative distance between the spaceships to increase with time, and that translates into an "unwinding force". The distance from spaceships will increase, but the string adapts to the required length.Right. I got that from the last post.
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To consider the mass of such string, and the stretching due to that string mass will add nothing to the solution.If you use pre-stressed string, the stretching of it doesn't come into play during the reeling-out phase. Simplifies the simulation of it. Either way, the string exerts force on S2, and that very much does alter the solution.
The effects are going to be hard to see if the acceleration and separation are both so low. It’s almost a Newtonian exercise at those numbers. I tend to go for high g and big separations to make the effects obvious.
As I said previously, the relative speed at 24 hour is 1,66 mm/hour ... You can easily measure that just with a wrist watch, a normal ruler, an ink marker and some patience.[/quote]From a spaceship putting out a g for a whole day? No, I don't think you'd get the thrust precision accurate enough to being able to measure a couple mm per hour.
Put S2 out at Mu Pegasi (a star just over 100 LY away) and as measured by S1, after 24 hours, S2 will be reeling out that string at a rate higher than c. Of course your steel wire won't like that, being limited to about 160 km. But it's a thought experiment. Like I said, I like to make my scenario larger to see the effects more clearly.
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In the Bell's original scenario, the Earth observer always sees the distance between the spaceships to be a constant, so being both spaceships identical it would deduce that the CM is always located halfway between.Sure, but only in that frame. The CM is elsewhere in a different frame, including the frame of a third ship in the middle that also has that identical constant proper acceleration. In the frame of that middle ship, the rear ship will recede slower than the lead ship. The center of mass of the system would be elsewhere. It seems wrong to measure the location of the CoM of a system in a frame other than the frame of the CoM of the system. The CoM of a rotating symmetrical wheel for instance is off-center in all other frames, but is centered on the axis in the frame of the axis.
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In the constant proper acceleration scenario, or the constant thrust scenario, I would say that the CM is located such it observes the same outwards simultaneous relative speed for both spaceshipsAgree, but that's not where the Earth guy puts it, which is why I pushed back on using that somewhat arbitrary frame.
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I'm pretty sure that if its not centered, the offset is not more than a tenth of a milliliter or so - compared with the length of 10,000 meters of the string, the offset is negligibleOf course it's negligible. Make the string longer or the g forces higher and it won't be negligible. Like I said, the effects become obvious if you scale it up, otherwise you might as well just use Newtonian physics, which was fine for putting guys on the moon. Nothing ever move fast or far enough to matter.
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That's right, the whole exercise is to see if the string breaks or not.OK, I guess my 100 LY string is unrealistic then. Pure hypoThe exercise isn't about relativity, it's about the ability to drag a long string. It breaks at about 170 km at 1 g without anything attached to the other end. 500 N to drag it, leaving around 300 N for the 'gentle braking'. If the ships are massive enough, that 300N will not be enough to halt the increasing separation, or cancel this apparent repulsive force between them.
My coordinate system starts where I want - this is a democratic forum :) -, and I can arbitrarily set ...[/quote]No, the choice of origin can't be arbitrary, else the line passing the arbitrary origin and the corresponding event-point in the CM world-line at time tc will not correspond to the line of simultaneous events for the CM frame. It would be an arbitrary line, not one specific to the CM frame.To get the corresponding simultaneous proper time in the leading S2 spaceship, we know that the line of simultaneous events for the CM is given by a line that passes by the (0,0) origin of the space-time diagram and the corresponding event-point in the CM world-line at time tc.This seems wrong. The origin of your spacetime diagram seems arbitrarily assigned along the x axis.
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Again, I don't understand... which is the "rigid scenario"? When the string can freely unwind from the spool, the tension in the string is due to: 1) the inertia of the mass of the string 2) the relative speed between ends, which appears as a relativistic and counter-intuitive effect.If the (ideal) string is being reeled out with only enough tension to accelerate the string to a speed matching S1 at the rear, then the entire string (except the part still on the reel) exhibits rigid motion, and one can apply rigid rules to anything stationary relative to the string, which neither S2 nor the CoM is.
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The tension due to the inertial mass is negligible as stated.Over half the breaking tension is negligible?
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The force diagram describes what happens when both ends of the string are fixed. The starting time=0 of the "simulation" is when this event happens, arbitrarily 24 hours after departure (or whatever value you want to start with). This lets us play with an initial condition for the relative velocity between the CM and the S2 spaceship.OK, so not gentle braking. You clamp on and let the string take up the strain until the two speeds are matched. Then it 'bounces' and for a bit the rear of the string goes slack.
Not if all these interactions are completely elastic, no. It will bounce forever without somewhere for the heat to dissipate.
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And this is the result I get if the spool brakes at proper time 30 days after departure:Oh... No slack time. There's always negative force (pulling S2 back) exerted by the string except that wee hump you show in the closeup. What is that? Why would the string push the ship away for a brief moment when it clamps onto the string?
Does the positive hump also exist at each peak like the one at time 5¼?
What are the time units in the pictures? Surely not all the same.
Post by: Lluis Olle on 17/09/2022 10:49:57
Not that lightweight. At 10 km, the tension needed to accelerate it at 1 g is already over half the rated load on it.
OK, that's right, my fault. I'll consider the string-mass for the 10,000 meters steel cable, but I could have chosen a 100 meters Kevlar thread instead, that would make it enough mass-less I guess.
Nevertheless, I think that the string-mass would only play a little role, because we describe our system using an accelerated reference frame. So, we would only measure the relative acceleration of the elongated mass, i.e. the mass of the "increasing" length of the string due to elastic tension.
Done... I think that the differential equation should be like that:
where ρ is the lineal density of the cable (0,005 Kg/m, for example).
But it's a thought experiment. Like I said, I like to make my scenario larger to see the effects more clearly.
I'll see your point.If you want to teach SR, then large scenarios are better.
The CM is elsewhere in a different frame, including the frame of a third ship in the middle that also has that identical constant proper acceleration.
Great idea! Then, the reference frame for the observers will be a third spaceship, located intially halfway between.
As you suggest, I consider a third spaceship labeled as CM. Is not as big as S1 or S2, but the thrust of its engines is adjusted the same way as the other spaceships, to give initially the same acceleration of 9.8 m/s2.
The string just goes through Scm, and the CM observer can see the string passing along without any friction, and measure some ruler markings carved on the string.
We'll use the Rindler coordinates to describe the world lines, as follows:
, S1 starts at (0, 
, Scm starts at (0, 
, S2 starts at (0, 
No, the choice of origin can't be arbitrary, else the line passing the arbitrary origin and the corresponding event-point in the CM world-line at time tc will not correspond to the line of simultaneous events for the CM frame. It would be an arbitrary line, not one specific to the CM frame.
The choice of the starting x-position of the spaceships in the coordinate system is arbitrary, and I always can put the (0,0) origin mark where I consider appropriate for the problem studied. In this case, just contrary to your commentary, if the world line of the Scm spaceship starts at (0,
, then for any event-point of that world line, the corresponding line of simultaneous events is the line that connects the (0,0) origin with the event-point ,and that makes the math much easier.If you don't believe, then we do the derivative of (ct,x) parametric world line with respect to Ԏ:
[
]This vector is tangent to the world line event-points, and so it represents the ct' axis for the accelerating spaceship. The x' Minkowski-orthogonal aixs is the reciprocal, and then:
[
]And so, a line that passes by the (0,0) origin and connects with the event-point in the world line, is just the x' axis/line of simultaneous events for that event-point. This is a known property of the Rindler horizon, and at any proper time, the (0,0)-event is always simultaneous with the proper time in the spaceship.
OK, so not gentle braking. You clamp on and let the string take up the strain until the two speeds are matched. Then it 'bounces' and for a bit the rear of the string goes slack.
The "gently braking" can be simulated, considering a time variable Young modulus for example. I program the E Young modulus to vary linearly from 0 to E in 60 seconds.
To take into account that now the string has mass, the spooling system is designed in the following way:
In S2 there's a telemetry laser system, that continually monitors the relative distance from S2 to S1, which is about 10 Km. Also, through this laser system, gets information from a tension sensor located at the hook point of the string in the S1 spaceship. Remember that the string is fixed in S1.
With this data and taking into account the effects of the speed of sound, the relativistic proper time differences between spaceships, and the dynamics of the spool assembly, the computer system controls the spool to continually release enough string as to keep a zero tension in the S1 hook point.
Oh... No slack time. There's always negative force (pulling S2 back) exerted by the string except that wee hump you show in the closeup. What is that? Why would the string push the ship away for a brief moment when it clamps onto the string?
Does the positive hump also exist at each peak like the one at time 5¼?
What are the time units in the pictures? Surely not all the same.
The units are always SI units, but the graphics scale automatically, and in the upper or right part of the graphic - depending on the axis - prints the scale if needed. To make things a little clearer, here you have the simulation for braking the spool after 1 year proper time in the CM reference frame. At that proper time, and as observed by CM:
- Length from S1 to CM is 7,900.618106 meters
- Length from S2 to CM is 7,900.618106 meters, no difference with 6 decimal places
- Total length of the deployed string (distance) at braking time is 15,801.24 meters, so the spool has deployed 5,801.24 meters of string during the 1 Year time.
- Relative S2 speed to the CM: 719.88 mm/hour.
- Relativistic acceleration is about 8.6 10-12 m/s2.
(https://llober.neocities.org/Bell/S1_b365_1.png)
The dotted green line and the corresponding right axis, is the acceleration (in the reference frame). In the zoomed graphic above, the acceleration scale is 1.0e-8. There's always a nearly constant outward acceleration - you hardly see it when the elastic force is loose - and is about 8.6 10-12 m/s2. This is the action of the "relativistic force" that breaks the Bell's string - if I'm right of course.
A more zoomed graphic...
(https://llober.neocities.org/Bell/S1_b365_3.png)
When the string can't deploy any more (brake action) and has to stretch, there's an initial relative velocity of 719.88 mm/hour (this are not the units of the graphic, it's only in the title of the graphic), and the elastic force of the string beings to pull the spaceship.
So you have into play the "relativistic force" pushing a little, the initial relative velocity condition - which means a kinetic energy - and the elastic force of the string that easily overcomes both the "relativistic force" and the kinetic energy, quickly reversing the movement.
As soon as the x-elongation reaches zero or negative, the elastic tension stops (there's an "if x<0 then..." in the program), and so only the "relativistic force" is left... is very small, takes a long time to bounce. As you can see in the following graphic, it takes about two years to recover.
(https://llober.neocities.org/Bell/S1_b365_2.png)
As you see, the forces are really tiny. That's why I think that the Bell's string paradox is a little bit misleading, because it uses the "always constant proper acceleration" trick.
Post by: Halc on 17/09/2022 14:44:54
OK, that's right, my fault. I'll consider the string-mass for the 10,000 meters steel cable, but I could have chosen a 100 meters Kevlar thread instead, that would make it enough mass-less I guess.That makes a negligible relativistic effect even more negligible. This is not an engineering exercise. You're already giving your rockets far more delta-v (about 40x) than can be done by any current technology. So it's a thought experiment, and so your cable can be as light and strong as you want, barring relativistic limits. Bell's spaceship scenario isn't a problem at all with human rockets because the scenario just won't come up.
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I'll see your point.If you want to teach SR, then large scenarios are better.If you're not worried about SR specifically, and it is your goal to see what accelerating massive objects do when tied to each other and are moving at trivial different speeds, then just start on engine a millisecond after the other and save that 24 hours of burn that doesn't change how the string bounces or breaks. The prior 24 hours of acceleration does nothing that the millisecond delay doesn't.
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As you suggest, I consider a third spaceship labeled as CM. Is not as big as S1 or S2, but the thrust of its engines is adjusted the same way as the other spaceships, to give initially the same acceleration of 9.8 m/s2.Initially 9.8, yes, and trivially more at braking time since your string is so short.
The string just goes through Scm, and the CM observer can see the string passing along without any friction, and measure some ruler markings carved on the string.
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We'll use the Rindler coordinates to describe the world lines, as follows:What are these? Coordinates in the Earth frame? Certainly not the frame of the string and the markings along it (which represents the Rindler frame), and since the Earth frame is inertial, I don't see how these are Rindler coordinates. The string ceases to represent the original Rindler frame once the friction on the reel commences.
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The choice of the starting x-position of the spaceships in the coordinate system is arbitrary, and I always can put the (0,0) origin mark where I consider appropriate for the problem studied. In this case, just contrary to your commentary, if the world line of the Scm spaceship starts at (0,(0,
is not arbitrary then. What if the origin was assigned so Scm starts at (0, 1000 km)? The line passing through that (much closer) origin would not be a line of simultaneity for events on the Scm worldline. This is what I mean by it not being arbitrary.Quote
In S2 there's a telemetry laser system, that continually monitors the relative distance from S2 to S1, which is about 10 Km.Can't just compute it? I mean it works quite well at just 10 km, but I'd never have thought to do it that way for my longer scenarios. But this being a thought-experiment, sensory equipment is only hypothetical and doesn't actually measure anything.
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With this data and taking into account the effects of the speed of sound, the relativistic proper time differences between spaceships, and the dynamics of the spool assembly, the computer system controls the spool to continually release enough string as to keep a zero tension in the S1 hook point.If the string isn't pre-stressed, this isn't going to work. At time zero, S2 starts moving but the string at S1 does not, so tension drops below zero there until the speed of sound can go from S2 to S1 and commence the acceleration of the string. Then there's the issue of how one goes about pre-stressing a stationary string with stress that is variable along its length.
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The middle graph shows time going from 1 to 8, with a 1e6 printed below the 8. I take it from later discussion that this means each number is 1e6 seconds.Quote from: HalcWhat are the time units in the pictures? Surely not all the same.The units are always SI units
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after 1 year proper time in the CM reference frame. At that proper time, and as observed by CM:I got 7908.8 myself, but I might have done it incorrectly. I take it the 10km string still applies.
- Length from S1 to CM is 7,900.618106 meters
- Length from S2 to CM is 7,900.618106 meters, no difference with 6 decimal places
If the CM is where it is supposed to be, there cannot be a difference in distances of the end ships by definition.
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There's always a nearly constant outward acceleration - you hardly see it when the elastic force is loose - and is about 8.6 10-12 m/s2. This is the action of the "relativistic force" that breaks the Bell's string - if I'm right of course.Bell doesn't consider force (or mass) at all. He has a string of a certain length attached to objects that move apart in the frame of the string, or, in another frame, a string that contracts while the attachment points don't move together. The motion is defined. The force required to achieve that motion is irrelevant. The string breaks because it isn't long enough to reach between the two defined points.
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As soon as the x-elongation reaches zero or negative, the elastic tension stops (there's an "if x<0 then..." in the program), and so only the "relativistic force" is left... is very small, takes a long time to bounce. As you can see in the following graphic, it takes about two years to recover.Let's see: Shortly after the braking, S2 slows and S1 speeds up. They're now moving together at around 720 m/hour. Does S2 reel in the line during this? If it took a year for their difference in velocity to get to 720, then it takes a year for them to match speeds again, and another year to achieve their original separation. So yea, the period of the bounce will resemble twice the time before the initial braking takes place.
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As you see, the forces are really tiny. That's why I think that the Bell's string paradox is a little bit misleading, because it uses the "always constant proper acceleration" trick.Bell didn't use a speed change of only 845 km/sec. You can actually see curves in his spacetime diagrams. Your string would definitely break in the 1-year scenario if it wasn't being reeled out. Constant acceleration isn't a trick. The exercise illustrates a point about the contradiction between rigid motion of an extended object and equal acceleration of its parts.
Post by: Lluis Olle on 17/09/2022 15:19:57
That makes a negligible relativistic effect even more negligible. This is not an engineering exercise. You're already giving your rockets far more delta-v (about 40x) than can be done by any current technology. So it's a thought experiment, and so your cable can be as light and strong as you want, barring relativistic limits. Bell's spaceship scenario isn't a problem at all with human rockets because the scenario just won't come up.
If I say that the string has a negligible mass, wrong. If then I consider the string to have a mass, wrong. Now, I'm using "rockets" with exceeding delta-v's, which I did not mention anywhere in my exposition or comments, and I can't talk of a generic spaceship, because as a thought experiment I can say whatever and then, as those scenarios would never come up, is useless. Well, for me with this first comment is the end of the thread, or if you wish, breaks the thread.
You can delete the thread it if you wish.
Anyway, for me has been a nice time answering. Thanks
Post by: MikeFontenot on 02/10/2022 21:41:20
About this classical paradox, I understand that if both spaceships maintain the same proper acceleration, then the string will break because of the non-simultaneity of the acceleration effects at both ends of the thread.
[...]
So, what about considering instead a constant thrust force?. For a spaceship of mass M, the engine is set to produce a thrust force equal to M*9,8. That would make the proper acceleration constant, if the thrust force is constant [...]
Why is that different from your first scenario?
Post by: paul cotter on 03/10/2022 14:10:31
Post by: Halc on 03/10/2022 16:01:37
in the frame of the trailing craft surely the distance to the leading craft is foreshortened to the same extent as the foreshortening of the string?In the frame of the trailing craft, both craft and string are completely stationary and there is no foreshortening of either. The lead craft is slowly pulling away in that frame, and, per the OP, reeling out additional string as needed.
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assuming both are traveling at relativistic speeds.All speeds are relativistic speeds, but I admit that the OP has his spacing so close (10 km) and speed so slow (about ¼% of c) that the string only needs about 4cm reeled out by the end of the day.
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Or am I just stoopid?I've seen your posts. Plenty of stoopid ones here, but you're not one of them.
Post by: paul cotter on 03/10/2022 20:56:15
Post by: paul cotter on 21/10/2022 13:11:46
Post by: Halc on 21/10/2022 13:29:38
If the two rockets were tethered with a taut string with their propulsion off and an observer travelling at 0•5c and accelerating made a close pass would the string break?Well, a passing observer (accelerating or not) should have zero effect on the system with which it doesn't interact, so the string breaks only if the two rocks (a rocket with no propulsion is a rock) are moving relative to each other. If they're not, the string doesn't break. Looking at it from a different frame cannot change that physical fact. The distance between the rocks is frame dependent, but the string breaking or not breaking is a frame independent thing in any scenario.
Post by: paul cotter on 21/10/2022 14:55:35
Post by: Halc on 21/10/2022 15:11:52
What I meant to say is that in the frame of reference of the high speed accelerating observer the two rockets ( or rocks if you prefer! ) will appear to be in same scenario as the original paradox was formulated.No, the rockets were the things accelerating in the first scenario. Proper acceleration is absolute, and not frame dependent. So in the frame of the accelerating observer, the rocks merely appear to be falling in unison and it is the observer that experiences the acceleration. The difference is proper acceleration vs coordinate acceleration. Proper acceleration of the rockets breaks the string in the first scenario. The rockets have an accelerometer which reads something nonzero. That's what proper acceleration means. Coordinate acceleration is what make you accelerate to the ground if you jump off a cliff. If you had an accelerometer on you, it would say 1g up while you're standing on the cliff, and 0g while falling. You gain coordinate speed only relative to the accelerated reference frame (that of the cliff), but it's the cliff being properly accelerated by force applied from below.
Free falling rocks attached with a string have no reason to break that string. They would in something like Earth's gravitational field due to tidal forces, a secondary effect, but there are no tides for the falling rocks in your observer's frame. Measuring tidal effects is considered a non-local experiment, and thus isn't a violation of the equivalence principle.
I hope I've not rambled too the point of adding to the confusion.
Post by: paul cotter on 21/10/2022 15:32:14