I Some thoughts about Bell's string paradox alternatives

[member 728827]

Obviously that depends on the masses of the spaceships, the proper accelerations involved, the tensile strength of the string, and how much the string is stretched before it breaks. What's the issue?

I've no issue.

What I did is to chose a very weak real steel cable (a really ridiculous one, with 1 mm diameter), with a break tension of 840 N, consider a mass 10⁹ Kg (yes, that much) for each spaceship, and simulate an scenario of 1g acceleration and 10 km of cable... and the cable doesn't break, not even close to break, not even considering that the 50 kg cable, pulled by the leading spaceship, is at a load of 500 N just to start with, for being pulled by the leading spaceship.

One reason for that, is that the cable gets to the Born rigidity condition.

All this is what is explained in the post. The cable data is in the "huge" post, just in the "irrelevant" section at the end.

 

[member 728827]

Yes, it is: @Ibix did it in just a few lines of algebra. The worldlines that the ships have to follow for the two conditions (Bell condition, with string stretching more and more until it breaks, vs. Born rigid condition, where the string's proper length is constant) are well known and easily written down.

@Ibix just expressed the formula of the Born rigidity condition... that doesn't tell if the spaceships system at a proper time meet that condition due the action of a force, like the string pulling. It just tells you which should be the proper acceleration for keeping a given constant proper distance to another point with other proper acceleration.

 

[DAH]

Does time dilation have to be taken into account here? I think for an accelerating ship clocks at the front of the ship will tick faster than the back, so presumably, clocks will tick faster on the leading ship.

 

[member 728827]

The first mention of tension in the string is in the 11th paragraph (including the summary and your list of numbers at 1 hour). And even this mention does not explicitly point this out as the key difference from the standard scenario. The reader is expected to infer that. No wonder you have repeatedly experienced people thinking the scenario is no different from the standard one. The difference should be explicitly stated in the first paragraph, not hidden in the 11th.

The statement “fragile thread” is not ambiguous. It means a thread with a negligible ultimate or breaking strength. Just as “smooth” means negligible friction, and “flexible” means negligible bending stiffness, and “light” means negligible mass. All of these are standard terms with unambiguous meaning. The thread being fragile indicates that we can neglect the tension in the thread compared to the other effects in the problem. I.e. the tension force is taken to be 0 in Bell’s formulation. There is nothing ambiguous about his description of the string as “fragile”, but it differs from yours as your string can sustain a non-zero tension and therefore is not “fragile”.

Yes, for a “fragile” thread any non-zero stress is “intolerable” and it breaks.

Yes, you are. You are putting in a non-fragile string. That is the key difference.

By Newton’s 2nd law Bell’s assertion that the accelerations are equal means that the net forces are equal. With a fragile string, the tension is 0 so the net force on each rocket is equal to the thrust. Thus, with a fragile string, you asserting equal thrust is the same as Bell asserting equal acceleration.

It is only with your additional change to a non-fragile string that the tension force is non-negligible, the net forces differ, and therefore the accelerations differ. This additional change is therefore essential to your analysis of the scenario and should be explicitly stated in the initial statement.

I apologize for any perceived impoliteness. It was intended as a clear critique of the description, rather than a personal attack. You have received a lot of feedback from a lot of experts on a lot of forums. By your own description, most believe your scenario to be identical to Bell’s. If one person had made that mistake you could attribute it to their sloppy reading or lack of expertise, but since this has happened repeatedly with many experts the communication problem lies with your description: the description does not clearly state the important change of using a non-fragile string, and it includes a lot of unnecessary details.

Technical writing is not easy because the concepts themselves are not easy. I am not being derogatory or insulting of you, I am showing you the issue so that you can understand why you have received the feedback you have received and (hopefully) become a better technical writer in the future.

When I got my first rough draft back from my advisor for my first technical paper, I remember thinking that it looked like he had bled all over my manuscript there were so many red marks. The feedback was clear, and it hurt, but it was without malice and it made me a better writer.

Ok, thanks Dale

 

[Ibix]

@Ibix just expressed the formula of the Born rigidity condition...

...which is manifestly the equilibrium condition, from boost symmetry. Thus anything else is not an equilibrium and the system will either oscillate around the equilibrium or simply tend towards it.

 

[Ibix]

Does time dilation have to be taken into account here? I think for an accelerating ship clocks at the front of the ship will tick faster than the back, so presumably, clocks will tick faster on the leading ship.

Time dilation isn't relevant here, no. Which clock ticks faster depends quite a lot on who is doing the comparing, and details depend on which acceleration scenario you're considering - I'd suggest starting a separate thread if you want to discuss it.

 

[PeterDonis]

@Ibix just expressed the formula of the Born rigidity condition... that doesn't tell if the spaceships system at a proper time meet that condition due the action of a force, like the string pulling.

That's because the kinematics of the situation are the same regardless of how they are achieved, and given that the kinematics are valid, there is no need to "prove" that they can be achieved, which is what you were saying was "not easy". It is easy, because showing that the kinematics are valid is showing that they can be achieved.

 

[member 728827]

Does time dilation have to be taken into account here? I think for an accelerating ship clocks at the front of the ship will tick faster than the back, so presumably, clocks will tick faster on the leading ship.

You can always take into account time dilation. The lead spaceship @ 365 days proper time, is about 41μs ahead.

As a curiosity fact, being the proper acceleration 1g, if the distance from ship to ship was 8.849 m (it's not the actual case, but ...), I think that 41μs should be similar to the time difference between an atomic clock at sea level, and an atomic clock at the top of the Everest, after 1 year.

 

[member 728827]

That's because the kinematics of the situation are the same regardless of how they are achieved, and given that the kinematics are valid, there is no need to "prove" that they can be achieved, which is what you were saying was "not easy". It is easy, because showing that the kinematics are valid is showing that they can be achieved.

The term "is easy" or "not is easy" is a relativistic term, that's influenced by the mass of the knowledge you gained, and the proper time you dedicated to the topic.

In the post, I just wait a time (lets say, 1 hour), and then I fix the string - it's just for fun. So, then the stretching begins at that moment.

Being that the silly scenario I chose, I have a "little" problem with the initial relative kinetic energy. If you do that not @1 hour, but @12 hours, it's not possible reach a static equilibrium at the Born point, because the energy balance will not fit, and the spaceship will bounce.

 

[Dale]

showing that the kinematics are valid is showing that they can be achieved

Where “can be achieved” is a statement of logical consistency with the known laws of physics rather than a statement of engineering or economic feasibility.

 

[Nugatory]

Does time dilation have to be taken into account here?

Not specifically time dilation, but the more general notion of relativity of simultaneity is involved.

The “paradox” in Bell’s original statement of the problem is that in a naive analysis, the string breaks in the ground frame but not the ship frame. The paradox goes away when we recognize the simultaneity assumption hidden in the statement that both ships fly the same acceleration profile.

 

[member 728827]

But why the complex maths to learn that? A pair of Bell rockets have acceleration $g$ and are connected by a string of length $l$. If the front rocket instead follows the Born rigid trajectory let its proper acceleration be $g-\delta g$. If we make the observations that the $x$ coordinate of such a family of observers is related to its proper acceleration by $x=c^2/a$ and that the difference in $x$ coordinates of the rockets must be $l$, then $\delta g$ can be found by $$\begin{eqnarray*}
\frac{c^2}g+l&=&\frac{c^2}{g-\delta g}\\
&\approx&\frac{c^2}g\left(1+\frac{\delta g}g\right)\\
\delta g&\approx&\frac{lg^2}{c^2}
\end{eqnarray*}$$which is manifestly tiny for any realisable $l$ and $g$. Thus, only a tiny force is needed.

Thanks Ibix.

As I'm calculating the proper times, relative speeds and accelerations from the coordinate parametric hyperbolic functions, using the intersection of the line of simultaneous events with the hyperbolic curves, and using a multi-precision math package that allows arbitrary precision, the question is:

the $x=\frac{c^2}{a}$ is an exact equality, or is a simplified version of $x=\frac{c^2}{a} \cdot \cosh(\frac{a \cdot \tau}{c})$ or similar, because although the results I get to compare are quite similar, are not enough "equal" with the precision I'm using.

I'm trying to see if my code is working correctly, and the calculus are being done with the required precision.

 

[PeterDonis]

the $x=\frac{c^2}{a}$ is an exact equality

For the case of constant proper acceleration, this formula is exact for the value of $x$ that the spaceship starts at, at time $t = 0$ in the inertial frame in which the spaceships start out at rest. If you want an equation that is valid along the entire worldline, it would be $x^2 - c^2 t^2 = c^4 / a^2$.

For the standard Bell spaceship paradox and Rindler cases, $a$ is a constant along each spaceship's worldline.

For your modified case in which the string tension exerts a non-negligible force and the proper acceleration along a spaceship's worldline is not constant, none of the equations we have discussed are valid as they stand.

 

[member 728827]

For the case of constant proper acceleration, this formula is exact for the value of $x$ that the spaceship starts at, at time $t = 0$ in the inertial frame in which the spaceships start out at rest. If you want an equation that is valid along the entire worldline, it would be $x^2 - c^2 t^2 = c^4 / a^2$.

Thanks a lot! Cosh(0) = 1 so, that explains it.

For your modified case in which the string tension exerts a non-negligible force

Well, not exactly, I know that I get out of the coordinate equations validity as son as I act on the acceleration with an additional tension (the string do have an initial tension, because is being pulled). For that reason, what I describe in the post is:

1. Nominal acceleration g, given by the constant F/M ratio, is reached @1 minute.
2. @1 hour proper time of S₂, then we fix the cable in S₂. Until then, S₂ only pulls string - it's mass is included in spaceship mass -, but S₁ tension in the trailing spaceship is kept to zero.

So, 1 minute that I don't really know what, but let's think it does not affect too much.

But @1 hour, I only get the initial conditions from the coordinate equations. After that, I use Classical Mechanics. Of course is a bit of "a guess", to consider the Born rigidity formula at that time, but given that the change in acceleration is so small, I think the conclusion is correct.

Why I do this? because in classical Mechanics, there's nothing to start with, there's no relative velocity between spaceships in the first place, so nothing to do. I need some realistic initial condition, and, after all, @1 hour nobody can't say me not to use Classical Mechanics, isn't it?

And that is one of the reasons because is "huge" post (not my words), because I explain those little details.

Thanks.

 

[PeterDonis]

@1 hour nobody can't say me not to use Classical Mechanics, isn't it?

I can. In classical (Newtonian) mechanics, the distinction between the two kinds of motion--the "Bell" motion, in which the string is stretched, and the "Born rigid" motion, in which the string is not stretched, does not exist. In Newtonian mechanics, if both spaceships start out with the same proper acceleration in the same direction, the proper distance between them never changes and a string between them does not stretch.

 

[member 728827]

I can. In classical (Newtonian) mechanics, the distinction between the two kinds of motion--the "Bell" motion, in which the string is stretched, and the "Born rigid" motion, in which the string is not stretched, does not exist. In Newtonian mechanics, if both spaceships start out with the same proper acceleration in the same direction, the proper distance between them never changes and a string between them does not stretch.

And that is what happens, until @1 hour then the string pulls and changes a little bit the acceleration. As I said, from that moment I use the classical mechanics, but predicting (is what I say literally) that the Born condition can be meet, and if so all number fit - with a non-Newtonian condition over the acceleration (Born), the static equilibrium is plausible.

If the change in acceleration needed was, so to speak, 1%, I would be very suspicious. The change in acceleration is from 9.8 to 9.799999999995.

PD. And after the Born condition is meet, I'm again playing by the rules.

Thanks for you comments.

 

[PeterDonis]

And that is what happens, until @1 hour then the string pulls and changes a little bit the acceleration

No, in your scenario the spaceships do separate until you put tension on the string. But Newtonian mechanics predicts that they would not.

from that moment I use the classical mechanics, but predicting (is what I say literally) that the Born condition can be meet, and if so all number fit - with a non-Newtonian condition over the acceleration (Born), the static equilibrium is plausible.

No. You can't use classical (Newtonian) mechanics with a "non-Newtonian condition over the acceleration". That makes your model inconsistent.

In Newtonian mechanics, the motion is Born rigid before you put tension in the string--i.e., when the proper acceleration of both spaceships is the same. And it is not Born rigid after you put tension in the string.

In relativity, the motion is Born rigid after you put tension in the string and allow time for a new equilibrium to be reached--i.e., when the proper acceleration of the front spaceship is less than the proper acceleration of the rear spaceship. But it is not Born rigid before you put tension in the string.

There is no way to reconcile these two models or mix them in this scenario. Using Newtonian mechanics is simply wrong.

 

[PeterDonis]

If the change in acceleration needed was, so to speak, 1%, I would be very suspicious.

Then you would be wrong. It is perfectly possible, using the correct laws--i.e., relativity--to set up a scenario where the change is 1%, or 10%, or even larger. You just use larger proper accelerations and/or wait longer before you put tension in the string.

I think you should carefully consider the comments that have been made in this thread about your approach being too complicated. Adding all of the complications has not helped you; it has confused you.

 

[member 728827]

Hi all,

Just to do some recap and not get lost, because part of the discussion has been in another thread about the Born rigidity, and IMHO:

The summary, to avoid any confusion, should read:

"Doing two changes in the statement of the Bell's paradox: to consider spaceship's engines with constant and same F/M ratio (being F the propulsion force), instead of a constant and same proper acceleration (as is deduced from the Bell scenario), and a real string with a weak ultimate tensile strength, I think that string doesn't break (is my own conclusion, not being an expert in SR)."

Then, an excerpt of the original Bell's statement should have been quoted as a reference:

"Suppose that a fragile thread is tied initially between projections from B and C (Fig. 3). If it is just long enough to span the required distance initially, then as the rockets speed up, it will became too short, because of his need to the Fitzgerald contract, and must finally break. It must break when, at sufficiently high velocity, the artificial prevention of the natural contraction imposes intolerable stress".

Then, I should had to comment that "fragile" in the Bell's sense means that the string has really no ultimate tensile strength, and so, will break at any tension. And then, for this alternate scenario, remark that I do consider a real string, although with a very weak ultimate tensile strength in relation to the scale of the mass of the spaceships.

I should have made clearer why is the constant F/M ratio considered in this alternate, not Bell's, scenario, and perhaps include some link to an example to show what it means.

And, my conclusions are:

- @Dale has deduced from the metric of the Rindler coordinates in the Born rigidity condition and the Christoffel symbols of the metric, that the ultimate tensile strength limit of the string is given by the following formula:

$\Delta f=m\frac{g^2}{c^2}h$

- In the Bell's scenario, this does not to seem relevant, because any "fragile" string breaks, as such "fragile" adjective in the Scientific and Physics community is understood as that the string has no ultimate tensile strength, and breaks at any tension. So, it breaks.

- In the alternate scenario, the values I considered for the string where: steel, 1mm diameter, 130,000 N/mm2 Young modulus, 0.05Kg/m lineal density, 840 N ultimate tensile strength, with a length of 10Km. For the spaceships, each with a weight of 10⁹ Kg (so, one million tons, give or take).

Putting the considered values in the @Dales formula, if the string can oppose an ultimate tensile string of at least 0,01 N, will not break, and can enter Born rigidity condition. The considered weak string ultimate tensile strength, considering that being pulled and accelerated has already a 500 N tension force to start with, is 34,000 times greater than the calculated limit.

Thanks to everybody

 

[PeterDonis]

spaceship's engines with constant and same F/M ratio (being F the propulsion force), instead of a constant and same proper acceleration

This is not a good description of what you are changing. The key point you appear to want to make is that the string exerts a non-negligible force on the spaceships, whereas in Bell's formulation of the scenario, it doesn't. That is the change you need to specify. Talking about the spaceship engines is irrelevant since the specification for what they do does not have to change at all from Bell's scenario in order to investigate what you say you want to investigate.

a real string with a weak ultimate tensile strength, I think that string doesn't break

In order for the string to exert a non-negligible force on the spaceships, so that the motion ends up becoming Born rigid and the string doesn't break, the string would need to have an ultimate tensile strength that was not weak. So this doesn't look like a good description either of what you say you are interested in investigating.

 

[PeterDonis]

Putting the considered values in the @Dales formula, if the string can oppose an ultimate tensile string of at least 0,01 N, will not break, and can enter Born rigidity condition. The considered weak string ultimate tensile strength, considering that being pulled and accelerated has already a 500 N tension force to start with, is 34,000 times greater than the calculated limit.

This doesn't make sense. If the string has an ultimate tensile strength of 0.01 N, and it is pulled with a force of 500 N, it will break. That's what "ultimate tensile strength" means: that any force greater than the ultimate tensile strength breaks the string.

Perhaps what you mean to say is that, with the given numbers, if the string can exert a force on the spaceships of only 0.01 N, it can make the motion Born rigid; and since the string has to withstand a pulling force from the ships of 500 N, it clearly must have an ultimate tensile strength of much greater than 0.01 N, so it can indeed exert a force of 0.01 N back on the spaceships and pull them into a Born rigid condition.

If the latter is what you meant, I agree it's correct; but it isn't what you said.

 

[member 728827]

This doesn't make sense.

If a have a leading spaceship, that moves at 1g acceleration, and pulls a 0,05 kg/m 10 Km steel cable, at the hook point of the cable with the spaceship the tension is 500 N. Of course, this is Newton 2L, and you seem to dislike this, but its what it is.

So, the cable, call it weak o whatever, has an initial 500 N tension, for a 840 ultimate tensile tension, and it's not yet pulling the trailing spaceship.

This let's 340 N of tension margin, until the ultimate tension that breaks the string. This is 34,000 times greater than the "at least 0,01 N that must withstand" to go Born. It withstands the 0,01, and 34,000 - 0,01 more.

This is where I should insert a "home work" problem of some string pulling weights, but I'm tired.

Thanks

 

[Dale]

The key point you appear to want to make is that the string exerts a non-negligible force on the spaceships, whereas in Bell's formulation of the scenario, it doesn't.

He needs both the thrust and the non-fragile string. I think the updated description is fine

 

[PeterDonis]

He needs both the thrust and the non-fragile string.

My point is that the "thrust" part is unchanged from Bell's original scenario. The only change to Bell's original scenario is the "non-fragile string (that exerts a non-negligible force on the spaceships)" part. But the OP keeps saying that the "thrust" part is a change to the original scenario. It's not, and since we have already pointed out that the specification of the scenario should be clear on what changes from Bell's original scenario, I think it's appropriate to point that out.

 

[member 728827]

But the OP keeps saying that the "thrust" part is a change to the original scenario. It's not,

The OP said me if I can share my sad string experience in this forum. So this is my relative point of view story, as a string, a simple and weak string. No more, no less.

So, I'm the string, I have some ultimate tensile tension - not a lot because I'm weak, I stretch until a point if there's tension, and I do the stuff is supposed that a string will do. I'm tied between two spaceships, I don't know why, but there I'm.

The two spaceships began to move, reach their nominal proper acceleration, and I, the string, begin to feel some additional tension - I had some to start with, because the leading spaceship is pulling me at 1g -, but this feeling is new for me ... what is this extra tension I feel, what is this? For now, is bearable, but wait, seems that little by little is increasing.

I oppose a tension force, I don't want to stretch more, but I do, I have to. How could I stop that, what can I do? My string father told me that I have to stretch to adapt, but this is what I'm doing right now, and the tension grows, and grows, the spaceships seems unaffected by any tension I'm able to exert.

I remember now, old stories written with the ancient Christoffel symbols, somehow predicted my fate and the fate of all string tied to spaceships. I can't do anything, my resistance is futile. Not even my Grandpa, which was very strong and by any means weak, could cope with this tension, because the spaceships can put any needed force to stretch me... this is not fair play, I break. Someone has to put an end to this.

So, which's the morale of this short fiction piece? if you consider written into stone that the proper acceleration of the spaceships is always constant - which is the statement of the Bell's paradox -, then it's all said. To keep always a constant proper acceleration, means that the spaceships engines will exert whatever the force is needed to do that. And then, the "weakness" of the string is irrelevant, because any non-infinite ultimate tensile strength string breaks.

And if the ultimate tensile strength is infinite, then you have another paradox.

 

[Dale]

My point is that the "thrust" part is unchanged from Bell's original scenario.

No. Bell’s original scenario specifies “identical acceleration programmes” and “fragile thread”:

With the fragile thread “identical acceleration programmes” is the same as “same F/M ratio (being F the propulsion force)”. But with a non-fragile thread “identical acceleration programmes” differs from “same F/M ratio” because now the differing tension forces produce different “acceleration programmes” given the “same F/M ratio”. The OP indeed needs to include both changes for the scenario they are considering.

 

[member 728827]

No. Bell’s original scenario specifies “identical acceleration programmes” and “fragile thread”:

View attachment 314874
With the fragile thread “identical acceleration programmes” is the same as “same F/M ratio (being F the propulsion force)”. But with a non-fragile thread “identical acceleration programmes” differs from “same F/M ratio” because now the differing tension forces produce different “acceleration programmes” given the “same F/M ratio”. The OP indeed needs to include both changes for the scenario they are considering.

I would like to remark that no string was hurt during this post, as it was only a "what if...".

Thanks Dale.

 

[PeterDonis]

with a non-fragile thread “identical acceleration programmes” differs from “same F/M ratio”

I think that depends on what Bell meant by "identical acceleration programmes". I think he meant "same F/M ratio". In other words, he was specifying what the spaceships' engines were programmed to do: put out the same constant F/M. Note that Bell specifies "identical acceleration programmes" before he even mentions the fragile thread. So he seems to me to be specifying "identical acceleration programmes" to be what it would be in the absence of any thread or any other connection between the spaceships, i.e., to be same F/M ratio.

To put it another way: if we compare Bell's original scenario with the OP's scenario, do the spaceships' engines do anything different? I think the answer is no; more to the point, I think the OP thinks the answer is no.

 

[PAllen]

I think that depends on what Bell meant by "identical acceleration programmes". I think he meant "same F/M ratio". In other words, he was specifying what the spaceships' engines were programmed to do: put out the same constant F/M. Note that Bell specifies "identical acceleration programmes" before he even mentions the fragile thread. So he seems to me to be specifying "identical acceleration programmes" to be what it would be in the absence of any thread or any other connection between the spaceships, i.e., to be same F/M ratio.

To put it another way: if we compare Bell's original scenario with the OP's scenario, do the spaceships' engines do anything different? I think the answer is no; more to the point, I think the OP thinks the answer is no.

I think the OP agrees with @Dale:

"So, which's the morale of this short fiction piece? if you consider written into stone that the proper acceleration of the spaceships is always constant - which is the statement of the Bell's paradox -, then it's all said. To keep always a constant proper acceleration, means that the spaceships engines will exert whatever the force is needed to do that. And then, the "weakness" of the string is irrelevant, because any non-infinite ultimate tensile strength string breaks."

That has also been my interpretation of Bell in the case where one claims any string will eventually break - this means that thrust force will need to increase to compensate for string tension to maintain constant proper acceleration.

 

[PeterDonis]

I think the OP agrees with @Dale:

For the scenario stated as given in what you quote, I agree. I just don't think that is the same scenario as Bell's original scenario. See below.

That has also been my interpretation of Bell in the case where one claims any string will eventually break

But, as has already been shown, Bell's original scenario did not claim that "any" string will eventually break. His original scenario specified a "fragile" thread. As @Dale said in an earlier post, "fragile" implies zero (or at least negligible) tensile strength, which in turn implies zero (or at least negligible) effect on the rockets' motion without the rocket engines having to make any compensation for string tension.

I agree that if we extend his original scenario to apply to any string whatever and specify that proper acceleration remains constant under those conditions, then obviously the rocket thrust needs to increase to compensate for increasing string tension until the string breaks. One way to explain why that is possible in principle would be to observe that, while relativity puts a finite limit on the tensile strength of materials, it does not put a finite limit on rocket thrust or proper acceleration.

Perhaps the upshot of this is that Bell's original scenario is ambiguous as to how it can be extended to conditions beyond what he originally specified.