Some thoughts about Bell's string paradox alternatives

[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.

 

[member 728827]

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.

Bell writes: "Then, as reckoned by an observer in A - the rest observer at the starting position -, they will have at every moment the same velocity, and so remain displaced one from the other by a fixed distance". This is a constant acceleration motion description.

Wikipedia comments: "Bell's spaceship paradox is not about preserving the rest length between objects (as in Born rigidity), but about preserving the distance in an inertial frame relative to which the objects are in motion...".

Bell scenario is in the Chapter 9 of the book "Speakable and unspeakable in Quantum Mechanics". The title of the chapter is "How to teach Special Relativity". So, is an example chosen to teach SR. As such, is not a very precise description, and needed to put some drama in the plot.

I think that some pages after, Bell's says something like "actually, the thread will pull the spaceships and break if it can't change the acceleration of the heavy masses of the spaceships" (I don't have now the exact quote, but I remember more or less). All very imprecise.

But, of course, what seems to happen is that there's really no need for the string to be "weak", because the relativistic acceleration due to the time dilation in accelerated motion is so tiny, that at 1g, a spaceship of 10⁹ Kg, would not be able to break a 0,02 N ultimate tensile strength thread, and so the students would probably say: bah, and what's the point then? It introduced some dramatic effects for the string.

 

[PeterDonis]

This is a constant acceleration motion description.

Yes, I understand that. But, as has already been noted, he also specified a "fragile" string, which means that the string exerts no force on the spaceships, so "constant acceleration" is the same as "the same F/M ratio".

As I remarked earlier, Bell's scenario is ambiguous about how to handle extensions to cases where the string is not "fragile".

what seems to happen is that there's really no need for the string to be "weak", because the relativistic acceleration due to the time dilation in accelerated motion is so tiny

As I commented before, that's only because you picked a very small proper acceleration. With a much larger proper acceleration, even a real, not "fragile" string would break quickly. You would need a string made of a very special material (I suggested carbon nanotubes in an earlier post) to withstand much larger acceleration forces long enough to have an effect on the motion of the ships.

Bell, of course, did not specify the magnitude of the proper acceleration in his scenario either. So this is yet another ambiguity in how to extend the scenario to cover more cases.

 

[PeterDonis]

Bell's scenario is ambiguous about how to handle extensions to cases where the string is not "fragile".

Bell, of course, did not specify the magnitude of the proper acceleration in his scenario either. So this is yet another ambiguity in how to extend the scenario to cover more cases.

@Lluis Olle I should emphasize that I am not saying your proposed scenarios are not valid scenarios. They are. I am only pointing out that, given what Bell left out in his specification of his scenario, the range of possible ways to extend it is extremely wide.

 

[PeterDonis]

So, is an example chosen to teach SR. As such, is not a very precise description, and needed to put some drama in the plot.

It might be worth looking at why Bell proposed this scenario as an example of how to teach SR. He did it because he wanted to emphasize that using frame-dependent concepts like "length contraction" to reason about what would happen in a particular scenario was perfectly valid. In other words, he was protesting, in a way, against claims by other physicists that said you should avoid using such frame-dependent concepts at all and only look at invariants. He was simply giving an example of a scenario where a simple use of the concept of "length contraction" gives the right answer, whereas Bell was able to confuse many other physicists who refused to use such concepts into giving the wrong answer when he posed his scenario to them.

Given the above, it's easy to see why Bell included the "fragile" string in his scenario: he wanted to rule out any kind of objection that the string might exert a force that would affect the motion of the spaceships. And even with that very strong constraint, he was able to confuse other physicists into giving the wrong answer, that the "fragile" string would not break.

You appear to be using Bell's scenario as a basis for illustrating something different, so it is to be expected that there will be information that is simply not specified in Bell's scenario that is important to your scenario.

 

[member 728827]

Is just for fun. I'm not going to discover the onion soup.

 

[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.

Yes absolutely, I change my original answer a little. Time dilation between the trailing and leading spaceships is the key concept to understand why the string in the paradox "breaks" (if it does so... conditions apply^(*)).

Why the string "breaks"^(*) from the viewpoint of a non-inertial observer that travels in the trailing spaceship (for example)? Because for him, the leading spaceship has a relative speed, is physically moving away from this observer. Then, if the string can't cope stretching, and can't change the accelerations of the spaceships with its elastic tension, breaks.

And why is the leading spaceship moving away? Because if the observer with its own t₁ proper time, watches a rear display of the leading spaceship, that with big characters displays continually its t₂ proper time, then he can do a simple reasoning, and say that leading spaceship has a relative speed such as:

$v_{21}=g*(t_2-t_1)$

This is not quite so, because in the accelerated frame there's also an acceleration that depends on the position of the point considered with respect to the observer $\frac{g^2}{c^2} \cdot x$. An approximate correction is:

$\int_{t_1}^{t_2}\frac{g^2}{c^2} \cdot x \cdot dt \simeq \int_{t_1}^{t_2}\frac{g^2}{c^2} \cdot \left[\frac{1}{2} \cdot g \cdot (t-t_1)^2 \right] \cdot dt$

So

$v_{21} \simeq g*(t_2-t_1)+\frac{\left[g*(t_2-t_1)\right]^3}{6 \cdot c^2}$

 

[PeterDonis]

Time dilation between the trailing and leading spaceships is the key concept to understand why the string in the paradox "breaks"

No, it isn't. The key concept to understand why the string breaks (under conditions in which it does) is that, in SR, unlike in Newtonian mechanics, a congruence of worldlines all having the same proper acceleration in the same direction is expanding (in more technical language, its expansion scalar is positive). This is the invariant that explains why the string breaks.

I really think you need to stop speculating about such things and learn the correct concepts from SR.

 

[member 728827]

No, it isn't. The key concept to understand why the string breaks (under conditions in which it does) is that, in SR, unlike in Newtonian mechanics, a congruence of worldlines all having the same proper acceleration in the same direction is expanding (in more technical language, its expansion scalar is positive). This is the invariant that explains why the string breaks.

At least, it sounds very impressive.

But then, it's not true that $g*(t_2-t_1)$ gives a very close value to the relative speed of the spaceships in the Bell's paradox ?, and $g*(t_2-t_1)+\frac{\left[ g \cdot (t_2-t_1) \right]^3}{6 \cdot c^2}$ is a better approximation ?, because numerically gives the correct value up to 15 decimal places.

I really think you need to stop speculating about such things and learn the correct concepts from SR.

Then, think also about not being so impolite with people. I'm absolutely insensitive to such comments, but is just because I think you would be happier in life.

 

[Ibix]

But then, it's not true that $g*(t_2-t_1)$ gives a very close value to the relative speed of the spaceships in the Bell's paradox ?

It may well do. But it's a coordinate dependent claim and probably specific to this scenario, and these are precisely the kind of things that make Bell's paradox a paradox in the first place. The lesson we learn from Bell's paradox, where a lot of physicists made wrong predictions from quick coordinate-based reasoning, is that it's a mistake because it very rarely generalises well.

It's like the people who say "acceleration causes the traveling twin to be younger", which is a bad generalisation of a rule that only applies to some versions of the scenario. It may give correct predictions in certain cases, but it's hard to determine what cases those are and when the rule fails completely. Does your rule work if the accelerations vary? If I choose to work in Milne coordinates?

The way that works in general is to study invariants - the expansion scalar is the relevant one here. Then you can derive approximations and rules of thumb as much as you like, with the added bonus of understanding when they will go wrong.

Then, think also about not being so impolite with people. I'm absolutely insensitive to such comments, but is just because I think you would be happier in life.

Peter (and I am sure he will correct me if Ixm wrong) will tell you that he's not unhappy. He's just giving you his opinion (and on the second page of your third (fourth?) thread on the topic, so not without evidence available. That is, an expert in a topic is telling you that you are exploring blind alleys. Being insensitive to such comments is probably a mistake. There's nothing wrong with exploring blind alleys if that's what makes you happy, but you should be aware you're doing it.

 

[member 728827]

It may well do. But it's a coordinate dependent claim and probably specific to this scenario, and these are

It's a claim from the point of view of the tip of the string. And if you feel more confortable, is the viewpoint of an observer in the trailing spaceship, or do you mean that's not a valid because is "coordinate dependent", or it's not applicable talking about the Bell's paradox?

Peter (and I am sure he will correct me if Ixm wrong) will tell you that he's not unhappy

I said "happier". If you're not confortable with this thread, close it and that's it.