# When and how can I apply Born rigidity condition?

• I
• member 728827
In summary: Born rigidity condition, until the very instant after I-T ends, when the probe is in Born rigidity condition.In summary, the I-T causes a transient increase in the distance between the space probes (δ), which then disappears when the Born rigidity condition is reached.
member 728827
TL;DR Summary
Can Born rigidity condition be used at not relativistic speeds, and if so, can we use thereafter the Newton's laws with the reached Born acceleration, and do Energy before/after balances in the Classical way?
Hello,

I try to better understand how and when I can apply the Born rigidity condition.

So, for the following example:

We've two space probes (Pa and Pb), that travel at an exact equal and same proper acceleration. At a given time tb0 in Pb, and as measured by Pb, the distance is Lba0 (it's quite large).

For our experiment, we need the proper distance to keep exactly constant as observed by Pa/Pb, so a very precise ion thruster starts in Pb at tb0, and adjusts the αb proper acceleration to the required value to the Born rigidity condition. It has enough precision to do that (even at the pico-Newton scale).

The speed, relative to Earth-frame stationary starting position, is about 10 Km/s at tb0 (in Sb). It's the order of magnitude what's important. The Lba0 length exact value doesn't matter either, but what's critical is that the final proper distance after going to Born rigidity condition keeps constant thereafter, or our experiment would not work reliably.

The questions are:

- Is the Born rigidity condition applicable in situations where the speeds are clearly not relativistic, or that equation only holds when talking about relativistic speeds. If so, what is the minimum v/c ratio to use it?

If the Born condition holds at this non-relativistic speeds,

##\alpha_{bbo} = \frac{c^2 \cdot \alpha_a}{c^2 + \alpha_a \cdot L_{ba0}'}##

then

- After the Born rigidity condition has been reach, can we apply locally Newton's laws to Pb, and write :

##M \cdot (\alpha_b - \alpha_{bbo}) = F_{ion}##

- After the Born rigidity condition has been reach, do the following Energy balance between just before and just after the Born condition be considered locally at Pb, and write:

##\frac{1}{2} \cdot M \cdot v_{ba}^2 = \int_0^{\delta} F'_{rel} \cdot d\delta - \int_0^{\delta} F'_{ion} \cdot d\delta##

where vba is the speed Pb has relative to Pa just when the ion-thruster starts and due to the non-simultaneity of accelerations at both ends. F'rel is like a "fictitious" relativistic force, much like the Coriolis force can be for an observer in a non-inertial rotating frame, and δ is the proper length measured in Pb, traveled by Pb during the transient phase.

I think that's mostly Ok, and there's no problem doing that, but...?. Any help or suggestions will be appreciated.

Thanks

Lluis Olle said:
- Is the Born rigidity condition applicable in situations where the speeds are clearly not relativistic, or that equation only holds when talking about relativistic speeds.
The answer to this question as you ask it is that the condition is applicable in any scenario.

However, the question itself suggests a fundamental misunderstanding on your part. Non-relativistic Newtonian mechanics is an approximation to relativity, not the other way around. So any equation that is valid "at relativistic speeds" and is therefore a relativistic equation must also be valid at non-relativistic speeds. There is no way to have an equation that is only valid "at relativistic speeds".

Lluis Olle said:
After the Born rigidity condition has been reach, can we apply locally Newton's laws
The law you write down here is not just "Newton's laws", it is relativistically correct for the particular case under discussion (motion and acceleration all along a single line, in a single direction).

Lluis Olle said:
After the Born rigidity condition has been reach, do the following Energy balance between just before and just after the Born condition
After this point I don't understand what you are trying to do or why.

vanhees71, PeroK and member 728827
Lluis Olle said:
Summary: Can Born rigidity condition be used at not relativistic speeds, and if so, can we use thereafter the Newton's laws with the reached Born acceleration, and do Energy before/after balances in the Classical way?

F'rel is like a "fictitious" relativistic force, much like the Coriolis force can be for an observer in a non-inertial rotating frame, and δ is the proper length measured in Pb, traveled by Pb during the transient phase.
Is your goal just to find this fictitious force? If so, there is a much easier way.

vanhees71 and member 728827
PeterDonis said:
the question itself suggests a fundamental misunderstanding on your part.
That's why people normally do questions, to try to understand things. Could be , could be that I have a fundamental misunderstanding, and so I question to the people that has the fundamental understanding I don't.
PeterDonis said:
After this point I don't understand what you are trying to do or why.
To see if the following reasoning is correct:

There's an initial relative (to the other space probe) Kinetic energy, due to a relativistic "drift" speed, vba just before the ion-thruster (I-T) starts. If the action of the I-T makes the acceleration go to the Born condition, the relative speed vba vanishes, so where that Kinetic energy went?

During that transient phase, the distance between the space probes grow a tiny bit (initially, there was a relative speed of vba, and after the transient phase this speed vanished). This is my δ for the length used.

So, it's just an Energy balance, from the very instant before I-T starts, when the probe is not in the Born condition, to the very instant afterwards in which the probe is in the Born condition.

Whatever the reason for doing the balance, the underlaying question is if it's valid to use a Born condition - that comes from the SR theory -, to calculate the desired acceleration, but for the rest, just use normal Classical mechanics, if we're not any close to a relativistic speed.

Last edited by a moderator:
Lluis Olle said:
There's an initial relative (to the other space probe) Kinetic energy, due to a relativistic "drift" speed, vba just before the ion-thruster (I-T) starts. If the action of the I-T makes the acceleration go to the Born condition, the relative speed vba vanishes, so where that Kinetic energy went?
Since both space probes have nonzero proper acceleration, kinetic energy is not conserved to begin with: you have the energy stored in the fuel carried by the probes being converted to kinetic energy by the probes' engines. You have to take that into account.

That said, the effect of the string tension will not just be to slow down the front probe; it will also be to speed up the rear probe. So obviously kinetic energy is being transferred from the front probe to the rear probe.

The usual way to make things like this as clear as possible is to do the analysis in the center of mass frame of the system.

Lluis Olle said:
a Born condition - that comes from the SR theory
No, it doesn't. It is perfectly possible to define Born rigid motion in Newtonian mechanics. It's just that Newtonian mechanics makes different predictions from relativity about what specific motions will be Born rigid. That's why you can't mix relativistic and Newtonian equations in this context.

member 728827
PeterDonis said:
That said, the effect of the string tension will not just be to slow down the front probe; it will also be to speed up the rear probe. So obviously kinetic energy is being transferred from the front probe to the rear probe.
Don't mix scenarios. In this Born question - in anyway related to the "other" :) -, there're two space probes, not tied by anything, no strings here. The slowdown is due to an ion thruster (I-T). So no transfer of energy is possible to the other space probe.

So, if the equations were written from an accelerated reference frame that is midpoint between both probes then the Kinetic energy would be conserved for that "cm" observer?
PeterDonis said:
The usual way to make things like this as clear as possible is to do the analysis in the center of mass frame of the system.
Just is what I did in the "other" scenario. But I don't call it "Center of Mass", because I first Googled the term, and it seems that the definition of "Center of Mass" is SR is not so trivial.

As a way to avoid to get into troubles with something I don't know, what I did is to consider a conceptual midpoint called "cm", which follows the hyperbolic motion represented by the coordinate equation where the X-coordinate has no "+L", and so, the line that connects the (0,0)-event with any event in that worldline is the line of simultaneous events (all this explained in the huge post).

And other advantage of this "cm": it's never disturbed by any action of the string, because that string doesn't touch that accelerated reference frame. In that scenario, the equations I write here, are there considered from the "cm" accelerated frame point of view.

For example, what I do there:

Simultaneous proper times in S1, S2 from cm midpoint observer:
    # Used to find intersection of simultaneity line from tc in CM to corresponding t1 in S1
tc11 = lambda t1,k: (C2/A0)*(k*sinh(A0*tc/C)-sinh(A0*t1/C))
tc12 = lambda t1,k: (C2/A0)*(k*cosh(A0*tc/C)-cosh(A0*t1/C))+(0.5*L0)

# Used to find intersection of simultaneity line from tc in CM to corresponding t2 in S2
tc21 = lambda t2,k: (C2/A0)*(k*sinh(A0*tc/C)-sinh(A0*t2/C))
tc22 = lambda t2,k: (C2/A0)*(k*cosh(A0*tc/C)-cosh(A0*t2/C))-(0.5*L0)

t1,k=findroot([tc11,tc12],(tc,1.0))
t2,k=findroot([tc21,tc22],(tc,1.0))

# Coordinates (ct,x) for Earth frame
ct1,x1=ct_x(t1,A0,-0.5*L0)
ctc,xc=ct_x(tc,A0,0.0)
ct2,x2=ct_x(t2,A0,+0.5*L0)

# Proper (=real) distance from CM to...
lc1=sqrt(power(x1-xc,2)-power(ct1-ctc,2))
lc2=sqrt(power(x2-xc,2)-power(ct2-ctc,2))

Thanks.

Sorry @Lluis Olle I may have missed it somewhere, but can you please clarify your goal here. I am getting lost in the minutia.

Is your goal simply to find the fictitious force? If so there is a HUGE shortcut.

member 728827
Lluis Olle said:
Summary: Can Born rigidity condition be used at not relativistic speeds, and if so, can we use thereafter the Newton's laws with the reached Born acceleration, and do Energy before/after balances in the Classical way?

Hello,

I try to better understand how and when I can apply the Born rigidity condition.

So, for the following example:

We've two space probes (Pa and Pb), that travel at an exact equal and same proper acceleration. At a given time tb0 in Pb, and as measured by Pb, the distance is Lba0 (it's quite large).

For our experiment, we need the proper distance to keep exactly constant as observed by Pa/Pb, so a very precise ion thruster starts in Pb at tb0, and adjusts the αb proper acceleration to the required value to the Born rigidity condition. It has enough precision to do that (even at the pico-Newton scale).

The speed, relative to Earth-frame stationary starting position, is about 10 Km/s at tb0 (in Sb). It's the order of magnitude what's important. The Lba0 length exact value doesn't matter either, but what's critical is that the final proper distance after going to Born rigidity condition keeps constant thereafter, or our experiment would not work reliably.

The questions are:

- Is the Born rigidity condition applicable in situations where the speeds are clearly not relativistic, or that equation only holds when talking about relativistic speeds. If so, what is the minimum v/c ratio to use it?

If the Born condition holds at this non-relativistic speeds,

##\alpha_{bbo} = \frac{c^2 \cdot \alpha_a}{c^2 + \alpha_a \cdot L_{ba0}'}##

then

- After the Born rigidity condition has been reach, can we apply locally Newton's laws to Pb, and write :

##M \cdot (\alpha_b - \alpha_{bbo}) = F_{ion}##

- After the Born rigidity condition has been reach, do the following Energy balance between just before and just after the Born condition be considered locally at Pb, and write:

##\frac{1}{2} \cdot M \cdot v_{ba}^2 = \int_0^{\delta} F'_{rel} \cdot d\delta - \int_0^{\delta} F'_{ion} \cdot d\delta##

where vba is the speed Pb has relative to Pa just when the ion-thruster starts and due to the non-simultaneity of accelerations at both ends. F'rel is like a "fictitious" relativistic force, much like the Coriolis force can be for an observer in a non-inertial rotating frame, and δ is the proper length measured in Pb, traveled by Pb during the transient phase.

I think that's mostly Ok, and there's no problem doing that, but...?. Any help or suggestions will be appreciated.

Thanks
xx
PeterDonis said:
Since both space probes have nonzero proper acceleration, kinetic energy is not conserved to begin with: you have the energy stored in the fuel carried by the probes being converted to kinetic energy by the probes' engines. You have to take that into account.

That said, the effect of the string tension will not just be to slow down the front probe; it will also be to speed up the rear probe. So obviously kinetic energy is being transferred from the front probe to the rear probe.

The usual way to make things like this as clear as possible is to do the analysis in the center of mass frame of the system.No, it doesn't. It is perfectly possible to define Born rigid motion in Newtonian mechanics. It's just that Newtonian mechanics makes different predictions from relativity about what specific motions will be Born rigid. That's why you can't mix relativistic and Newtonian equations in this context.

Dale said:
Sorry @Lluis Olle I may have missed it somewhere, but can you please clarify your goal here. I am getting lost in the minutia.

Is your goal simply to find the fictitious force? If so there is a HUGE shortcut.
No, it's not my goal.

I just was interested in the sign of the work done by this force during the Born transition, because if negative I argued that then an static equilibrium was not possible, because that would mean that the relativistic force in the transient phase would have to pull the spaceship, and so the equilibrium would be a bouncing spaceship pair.

But, anyway, if there's a way to have that fictitious force, good to know.

Thanks,

Lluis Olle said:
The slowdown is due to an ion thruster
Ah, sorry, I missed that in your OP to this thread. Then obviously the energy change comes from the ion thruster's fuel. And, as I said, since we have nonzero proper acceleration involved, you should not expect kinetic energy to be conserved even if we restrict to using a single inertial frame. Kinetic energy is only conserved in purely elastic collisions between freely moving particles.

Lluis Olle said:
So, if the equations were written from an accelerated reference frame that is midpoint between both probes then the Kinetic energy would be conserved for that "cm" observer?
No, because in the accelerated frame freely falling objects do not move in straight lines and their speed is not constant. You have a "fictitious force" in that frame that changes the kinetic energies of objects even if they have zero proper acceleration.

Lluis Olle said:
As a way to avoid to get into troubles with something I don't know, what I did is to consider a conceptual midpoint called "cm"
In other words, you got rid of one thing you don't know by substituting another thing you don't know. How do you know this "conceptual midpoint called cm" is valid? How do you know the equations you wrote down in this frame are correct? Answer: you don't.

Lluis Olle said:
I just was interested in the sign of the work done by this force during the Born transition, because if negative I argued that then an static equilibrium was not possible, because that would mean that the relativistic force in the transient phase would have to pull the spaceship, and so the equilibrium would be a bouncing spaceship pair.
In the scenario you give in this thread, you only have an ion thruster on the front spaceship, and you have specified that you can adjust its thrust to whatever precision is necessary to exactly realize the Born rigidity condition. So in your scenario in this thread, it is impossible for any "bounce" to occur, because your specification of the scenario rules it out.

In the scenario in your previous thread, with a string between the two ships, yes, it is obvious that there will be oscillations about the Born rigid equilibrium condition; the only question is whether you include damping, which would mean the oscillations would gradually die out and the spaceships would end up in the Born rigid equilibrium condition, or whether you ignore damping, which would mean the spaceships would oscillate forever about a Born rigid equilibrium. Obviously including damping would be more realistic.

But in any case, it should be obvious that what the equilbrium will be will depend on how you specify the scenario.

Lluis Olle said:
Just after the Born rigidity condition has been reach, can we apply Newton's law from the <c> observer point of view ?
No, because the frame is non-inertial.

Lluis Olle said:
Just after the Born rigidity condition has been reach, is the following Energy balance between just before and just after the Born condition as observed by the <c> observer correct ?
Impossible to answer because you have not given a formula for what you call ##F'_{rel}##. Also, you have not explained why there are primes on the forces in this equation, but not in the previous one.

As a general comment, since the frame is non-inertial, work done does not necessarily equal kinetic energy change.

Lluis Olle said:
I just was interested in the sign of the work done by this force during the Born transition, because if negative I argued that then an static equilibrium was not possible
I do not see how any of the questions you ask either in the OP to this thread or in your revised scenario in post #11 are related to this. My remarks in post #12 apply to your revised scenario in post #11 just as well as to your scenario in the OP of this thread, as compared with the "string" scenario in your previous thread.

PeterDonis said:
No, because the frame is non-inertial.
PeterDonis said:
As a general comment, since the frame is non-inertial, work done does not necessarily equal kinetic energy change.
These comments of mine of course raise the question: why are you not simply doing your analysis in an inertial frame?

As I'm turning mad with the "quoting" system of the forum, and losing continually all the formatting in the math eq, I answer in a "non-quoted fashion". Also, this is my last post in this thread, to don't make it too long, and take too much of your attention.

Considering then a conceptual accelerated reference frame <c> observer, that follows the hyperbolic motion described by the initial proper acceleration of the probes, and that starts midpoint from both. We denotate as αc this initial proper acceleration common to the probes and the <c> observer. The motion of this <c> point is perfectly described by the same equations that describe the probes, except that starts midpoint both.

Also, now there's an I-T engine in each space probe, in Pa just to accelerate a tiny bit the probe, and in Pb just to decelerate a tiny bit the probe.

Now, considering the leading probe Pb, if the Born condition with respect to <c> can be used at this non-relativistic speeds...

##\alpha_{bbo} = \frac{c^2 \cdot \alpha_c}{c^2 + \alpha_c \cdot L_{cb}}##

- Just after the Born rigidity condition has been reach, can we apply Newton's law from the <c> observer point of view ?

##M \cdot (\alpha_c - \alpha_{bbo}) = F_{ion}##

Remark: The fictitious force due to the non-inertial frame is taken out, because that fictitious force is the thrust of the Pb main engine, that gives exactly the αc acceleration of the observer's <c> frame, which remains undisturbed by any action of the I-T engines. The thrust of the main engines is constant (or... the F/M ratio).

- Just after the Born rigidity condition has been reach, is the following Energy balance between just before and just after the Born condition as observed by the <c> observer correct ?

##\frac{1}{2} \cdot M \cdot v_{cb}^2 = \int_0^{\delta_{cb}} F_{rel} \cdot d\delta - \int_0^{\delta_{cb}} F_{ion} \cdot d\delta##

where vcb is the speed Pb has relative to <c> just when the ion-thruster starts, and due to the non-simultaneity of accelerations at both ends. Frel is like a "fictitious" relativistic force, much like the Coriolis force can be for an observer in a non-inertial rotating frame, and δ is the proper length elongation from <c> to Pb, as observed by <c> during the transient phase.

The "primed" in the integrals is an typo, and does not mean anything. It's gone.

Remark that the motion of <c> observer remains always undisturbed by the action of the I-T engines in the probes, and is at αc constant proper acceleration motion always.

Thanks

Lluis Olle said:
Just after the Born rigidity condition has been reach, can we apply Newton's law from the <c> observer point of view ?
Same answer as I gave before in post #12.

Lluis Olle said:
ust after the Born rigidity condition has been reach, is the following Energy balance between just before and just after the Born condition as observed by the <c> observer correct ?
Same response as I gave before in post #12, because, while you did take the primes out, you still have not given a formula for ##F_{rel}##.

Lluis Olle said:
this is my last post in this thread
Since your post still contains unresolved issues, do you really want to make it your last post in this thread?

PeterDonis said:
Since your post still contains unresolved issues, do you really want to make it your last post in this thread?
Ok, thanks, I think I got it. Let's see.

Then after the probe is in the Born condition,

##\text{<}B_b\text{>}: F_{thr} - F_{ion} = M \cdot \alpha_{bbo}##
##\text{<}C_b\text{>}: F_{thr} - F_{ion} + F_{rel} = M \cdot \alpha_c##
##\text{<}c_b\text{>}: F_{thr} - F_{ion} + F_{rel} - F_{ine} = M \cdot (0)##
##F_{ine} = M \cdot \alpha_c##
##\Rightarrow F_{ion} = F_{rel} = M \cdot (\alpha_c - \alpha_{bbo})##

where the uppercase letter denotes a inertial observer, and the lowercase letter what is observed. B/b is the leading probe, C/c is the midpoint <c>, that always travels with a proper acceleration of αc.

For the energy, the just before / after balance from the <c> observer point of view

##\frac{1}{2} \cdot M \cdot v_{bc}^2 + \int_0^{\delta_{bc}} F_{rel}(\delta) \cdot d\delta - \int_0^{\delta_{bc}} F_{ion}(\delta) \cdot d\delta = 0##

Now, trying to see how to describe the Frel force. Seems something much related to the geometry (distance). Investigating.

Thanks.

Lluis Olle said:
Then after the probe is in the Born condition,

##\text{<}B_b\text{>}: F_{thr} - F_{ion} = M \cdot \alpha_{bbo}##
##\text{<}C_b\text{>}: F_{thr} - F_{ion} + F_{rel} = M \cdot \alpha_c##
##\text{<}c_b\text{>}: F_{thr} - F_{ion} + F_{rel} - F_{ine} = M \cdot (0)##
##F_{ine} = M \cdot \alpha_c##
##\Rightarrow F_{ion} = F_{rel} = M \cdot (\alpha_c - \alpha_{bbo})##

where the uppercase letter denotes a inertial observer, and the lowercase letter what is observed.
What is observed by whom?

Also, how are you defining the forces? In relativity the ##F = ma## equation is not as simple as it is in Newtonian mechanics even in inertial frames.

Lluis Olle said:
For the energy, the just before / after balance from the <c> observer point of view

##\frac{1}{2} \cdot M \cdot v_{bc}^2 + \int_0^{\delta_{bc}} F_{rel}(\delta) \cdot d\delta - \int_0^{\delta_{bc}} F_{ion}(\delta) \cdot d\delta = 0##
What frame is this in? And what is ##v_{bc}##?

Lluis Olle said:
Now, trying to see how to describe the Frel force.
In an inertial frame this force doesn't exist.

You seem to be trying to wave your hands and use Newtonian formulas and hope they sort of work, instead of actually doing the correct math using relativity. That's not a good approach.

Vanadium 50, member 728827 and vanhees71
I would say that Vba or Vbc is so low that one can use Newton's formulas. But this is not an exact calculation.

member 728827
The summary of the question:

Can Born rigidity condition be used at not relativistic speeds, and if so, can we use thereafter the Newton's laws with the reached Born acceleration, and do Energy before/after balances in the Classical way?
PeterDonis said:
You seem to be trying to wave your hands and use Newtonian formulas and hope they sort of work, instead of actually doing the correct math using relativity
So if you get the impression that I'm trying to use Classical mechanics in this case, then yes, you hit the nail on the head.

I'm pretty sure that there's a correct and elegant way of doing it, using always relativity math, but as you said relativity always applies, but classical mechanics has to do the job also if not at relativistic scenarios, and the scenario I'm talking about is not pure relativistic. And at no-relativistic speeds, both ways should give pretty similar results, if done properly which seems I don't do.
PeterDonis said:
What frame is this in? And what is vbc?
For the energy balance, is done by the <c>observer in the non-inertial reference frame, located midpoint - at start - between the space probes -, and that follows a perfect hyperbolic accelerated motion, with a proper acceleration αc, that is always the same acceleration that can exert the space probe main engine over the mass of the probe, and is always constant by definition.

The vbc is the relative speed of Pb with respect to the <c> non-inertial frame, just when the Ion-Thruster starts. As @externo has mentioned, it's tiny at the relativistic speeds I want to consider. But is there.

As for the nomenclature used for myself, I explain:

<Bb> is the B inertial frame (B uppercase = inertial), that observes Pb probe, when that probe passes by B, and the (not accelerated) speed of B is just equal to the speed of the Pb probe at that moment.

<Cb> is the C inertial frame, that observes Pb probe, when <c>midpoint passes by C, and the (not accelerated) speed of C is just equal to the speed of <c>midpoint at that moment.

<cb> is the <c>midpoint non-inertial frame (c lowercase = non-inertial), that observes the Pb probe.

And to derive that formulas, I used the following reasoning:

- For the B inertial observer, Pb has an acceleration αbbo, that is the acceleration that corresponds to the Born condition respect <c>midpoint frame. It observes forces Fthr and Fion.

- For the C inertial observer, Pb has an acceleration αc. At the Born condition, C doesn't observe any relative motion between <c> and Pb , so both accelerations are the same for him. Because C is not local to Pb, but at a certain distance L, C introduces an "unknown" force, and calls it Frel, and deduces it's somehow geometrically related to the fact it does the observation at a distance of L from Pb.

- For the <c> non-inertial midpoint frame, the relative acceleration that observes to Pb is zero, as <c> and Pb follow a Born rigid motion. Being a non-inertial frame, introduces a fictitious force that he calls "inertia". For the same reason as the C inertial observer, because <c>observer is not local to Pb, but at a certain distance L, <c>observer introduces an "unknown" force, and calls it Frel, and deduces it's somehow geometrically related to the fact it does the observation at a distance of L from Pb.

As always, thank you very much.

If the two space probes had no speed at the start, they accelerated simultaneously in the Earth frame.
If the two space probes already had a certain inertial speed relative to the Earth before accelerating, the rear probe accelerated before the front probe in the Earth frame.

member 728827
Lluis Olle said:
the scenario I'm talking about is not pure relativistic
What does "not pure relativistic" mean?

Lluis Olle said:
at no-relativistic speeds, both ways should give pretty similar results
This is true in the sense that, if all of the speeds involved, in the inertial frame in which the objects are all initiaily at rest, are much less than the speed of light, relativistic corrections will be very small numerically.

However, if your purpose is to show a transition from "Bell spaceship" motion (in which the ships are moving apart) to Born rigid motion (in which the ships maintain a constant proper distance), as I have already noted, Newtonian mechanics simply cannot model this correctly, because in Newtonian mechanics, the initial motion, in which the spaceships both have the same proper acceleration, is Born rigid, and the final motion, in which the front ship has less proper acceleration than the rear ship, is not. And this is, of course, exactly the opposite of the correct relativistic prediction (in which the initial motion is not Born rigid--the spaceships are moving apart--and the final motion is). But even that assumes that your measurements are accurate enough that you can see the difference between the Newtonian and relativistic predictions, which means you can't possibly be using an approximation in which Newtonian mechanics is valid.

In other words, if you restrict yourself to speeds that are small enough for Newtonian mechanics to be an acceptable approximation numerically, the thing you are trying to investigate cannot even be modeled at all. There is not even a detectable difference between the "initial" and "final" states of motion: the ion thruster, to this approximation, does not even fire because there is nothing for it to do. The initial motion is already Born rigid so no adjustment needs to be made to make it Born rigid.

vanhees71 and member 728827
externo said:
If the two space probes had no speed at the start, they accelerated simultaneously in the Earth frame.
If the two space probes already had a certain inertial speed relative to the Earth before accelerating, the rear probe accelerated before the front probe in the Earth frame.
It's not exactly that scenario, but here you'll see what I'm talking about more precisely.

Thanks @externo

PeterDonis said:
The initial motion is already Born rigid so no adjustment needs to be made to make it Born rigid.
Then, returning for a moment to the Bell's thread, which summary reads:

"Doing a small change in the statement of the Bell's paradox - to consider engines with constant and same F/M ratio (F is the propulsion force), instead of a constant and same proper acceleration, I think that string does not break (is my own conclusion, not being an expert in SR)."

And with respect to the original Bell Paradox, published in the book "Speakable and unspeakable in Quantum mechanics", Chapter 9 "How to teach SR", page 67, from which I literally copy this paragraph:

"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".

Now, let me abuse of your patience and write a little fantasy, just to clarify in a funny manner why my alternative scenario, different to Bell's one, needs to introduce the statement about constant F/M ratio, and that nobody seems to get to the point... Is a little satiric, but then I practice a little of English, which is not my native language (and with some help from Google translator also)...

"So, I'm the string, I have some tensile break resistance - not a lot because I'm fragile, 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 accelerate, reach the 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 -, 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. 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.

Now that I'm near my tensile limit, the relative acceleration between the spaceships is not decreasing, but it's indeed increasing, is unaffected by any tension that I could provide, not even my Grandpa, which was very strong and by any means fragile, could cope with this tension, because seems that the spaceships can put infinite force - if needed - to stretch me... this is not fair play, I break. Someone has to put an end to this."

So, where's the point? The point is that 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 "fragility" of the string is irrelevant, because any non-infinite tensile strength string breaks.

And then, at last, it seems that someone agrees. Thank you! As you said, being, at the start the values required to be in Born condition so tiny, the string obviously does not break, even if it's a ridiculous weak string, if you don't let the spaceship engines to exert an arbitrary amount of force to break the string.

And, last but not least, I attach a simulation of the situation of the space probe, @1 week after departure, traveling at 0.1g, if both probes are separated by a 1" light distance.

Thanks a lot for you patience.

Lluis Olle said:
As you said, being, at the start the values required to be in Born condition so tiny
That doesn't have to be the case. Suppose the two spaceships started out with a proper acceleration of ##10^{10}## g instead of ##1## g.

Lluis Olle said:
the string obviously does not break, even if it's a ridiculous weak string, if you don't let the spaceship engines to exert an arbitrary amount of force to break the string
No, it doesn't have to be an arbitrary amount of force. Only a finite force is needed to break any string, because any string only has a finite breaking strength. With an appropriate selection of string vs. spaceship proper acceleration, it's perfectly possible to have the breaking strength of the string be small enough compared to the spaceship proper acceleration that the string does not have any significant effect on the motion of the spaceships, even though the force exerted by the spaceships is finite. That is basically what Bell assumed.

It is also, of course, possible to choose a string and a spaceship acceleration for which the string does have a significant effect on the motion of the spaceships. We could, for example, have a string made out of carbon nanotubes and spaceships with small rockets and limited thrust.

Lluis Olle said:
Then, returning for a moment to the Bell's thread, which summary reads:

"Doing a small change in the statement of the Bell's paradox - to consider engines with constant and same F/M ratio (F is the propulsion force), instead of a constant and same proper acceleration, I think that string does not break (is my own conclusion, not being an expert in SR)."
As long as force is measured in the rocket momentary rest frame, and the mass is momentary rest mass of the rocket (important, since a rocket expelling thrust has changing rest mass), then the ratio is proper acceleration (even in SR), and the scenario is no different than the Bell case.

Lluis Olle said:
Now, let me abuse of your patience and write a little fantasy, just to clarify in a funny manner why my alternative scenario, different to Bell's one, needs to introduce the statement about constant F/M ratio, and that nobody seems to get to the point
I don't think that anyone is unaware of your point at this point. We get your point. You have been heard and understood. We understand your scenario and Bell’s original scenario and the differences.

The issue is not that we believe you don't need to introduce your constant F/M thrust for a different outcome. The issue is that you seem to not understand that in addition to fixing F/M you ALSO must indicate that the string is not fragile. If the string is fragile then your modified scenario is the same as Bell's scenario. It is only by specifying BOTH that you get a different scenario.

Btw, I mentioned earlier that there is a shortcut for finding the fictitious force. Ships that are accelerating and keeping a constant proper-distance are at rest in a Rindler coordinate system. Such a system has the metric $$ds^2=-\frac{g^2}{c^2} X^2 \ dT^2+dX^2+dY^2+dZ^2$$

The shortcut is that once you have the metric, then fictitious forces are given by the Christoffel symbols. In this case the relevant Christoffel symbol is $$\Gamma^X {}_{TT}=\frac{g^2}{c^2}X$$ which gives a fictitious force in the ##X## direction of ##-m\frac{g^2}{c^2}X##.

Now, if one rocket is at ##X=\frac{c^2}{g}## and the other is at ##X=\frac{c^2}{g} + h## then the fictitious force on one is ##f_1=-mg## and the fictitious force on the other is ##f_2=-mg-m\frac{g^2}{c^2}h##. So the difference in the fictitious forces at the front and back rockets is $$\Delta f=m\frac{g^2}{c^2}h$$

This is the force that is the threshold. If the ultimate tension in the string is larger than ##\Delta f## then the string holds in your scenario and if the ultimate tension is less than ##\Delta f## then the string breaks in your scenario. Since by definition a fragile string has an ultimate tension less than ##\Delta f## a fragile string breaks.

Lluis Olle said:
I have some tensile break resistance - not a lot because I'm fragile
Fragile means that any tension breaks the string. A fragile string does not have any tensile break resistance.

Last edited:
member 728827
Dale said:
if one rocket is at ##X=\frac{c^2}{g}## and the other is at ##X=\frac{c^2}{g} + h## then the fictitious for
I don't think this is correct.

If both rockets are at constant Rindler ##X## coordinates, then we have the Born rigid condition, not the Bell spaceship paradox condition; there is no force on the string because the rockets are maintaining constant proper distance. To properly model the Bell spaceship paradox in Rindler coordinates, the front rocket should be moving in the positive Rindler ##X## direction, not stationary. (Or we could have the rear spaceship moving in the negative ##X## direction, but I won't analyze that case here.)

If the front rocket is moving in the positive ##X## direction, then the proper distance between the rockets, which will simply be the difference in their Rindler ##X## coordinates, is an increasing function ##h(T)##. If we assume that the string's normal unstretched length is ##h(0)## (i.e., the value of ##h(T)## at ##T = 0##), then the increase in length of the string will be ##\Delta h = h(T) - h(0)##, and if we assume the string obeys Hooke's Law, the tensile force on the string will be ##F = k \Delta h##, where ##k## is a constant that characterizes the "stiffness" of the string (similar to the spring constant of a spring). When ##\Delta h## gets large enough that ##F## exceeds the tensile strength of the string, the string breaks.

The proper accelerations involved, or, equivalently, the "fictitious forces" in the non-inertial Rindler coordinates, are irrelevant if we want to know the value of ##\Delta h## at which the string breaks. The proper accelerations involved are relevant if we want to know the time it will take to reach that point; the larger the proper acceleration on the rockets (which in the Bell case is the same for both of them), the more quickly a given value of ##\Delta h## will be reached. To see that, we can calculate the expansion scalar of the Bell congruence, which is easier to do in Minkowski coordinates. The 4-velocity field of the Bell congruence in Minkowski coordinates is:

$$U = \sqrt{1 + a^2 t^2} \ \partial_t + at \ \partial_x$$

where ##a## is the proper acceleration. Since this is only a function of ##t## and does not vary with ##x##, the expansion scalar is ##\theta = \partial_t U^t##, which gives

$$\theta = \frac{a^2 t}{\sqrt{1 + a^2 t^2}}$$

Heuristically, the expansion scalar is proportional to ##d \Delta h / dt##, so the larger the proper acceleration ##a## is, the larger ##d \Delta h / dt## is and the more quickly a given value of ##\Delta h## will be reached. (I have glossed over a number of fine points here, but none of them affect the basic conclusion.)

PeterDonis said:
If both rockets are at constant Rindler X coordinates, then we have the Born rigid condition,
Yes. This is the difference in force required to produce the Born rigid condition. If the difference in force is less than this amount then the proper distance between the ships expands.

PeterDonis said:
not the Bell spaceship paradox condition
That is correct. This is not the Bell spaceship condition. In the Bell spaceship condition the distance expands at a given rate. But it is not necessary for it to expand at that rate for the string to break. If it expands for long enough at any rate it will break. So the threshold force of interest is indeed the difference in force for maintaining the Born rigid condition.

PeterDonis said:
The proper accelerations involved, or, equivalently, the "fictitious forces" in the non-inertial Rindler coordinates, are irrelevant if we want to know the value of Δh at which the string breaks.
I was not attempting to find that. I was only attempting to find if it does break. That does not require knowing the stiffness, only if the ultimate tension is sufficient to halt the expansion. If ##\Delta f## is less than the ultimate tension then at some point (assuming a quasi static expansion) the tension in the expanding string reaches ##\Delta f## and the expansion stops. Otherwise at some point the tension in the expanding string reaches the ultimate tension and the string breaks.

I should state explicitly that I am assuming that the expansion is quasi static and also that the string doesn’t stretch much before reaching its ultimate tension.

PeterDonis said:
To see that, we can calculate the expansion scalar of the Bell congruence
In the OP’s scenario the material of the string is not necessarily described by the Bell congruence. Depending on the tension in the string it can be any congruence between the Bell congruence and a Born rigid congruence.

Your analysis would be correct for the usual Bell’s paradox, but not for the OP’s modification.

Last edited:
Dale said:
This is the difference in force required to produce the Born rigid condition.
Ah, I see. I was misinterpreting what you were trying to do. Yes, any difference in force less than that will produce a positive expansion scalar (though not necessarily of the same magnitude as the Bell congruence expansion scalar) and will therefore break a sufficiently fragile string.

Dale
PAllen said:
As long as force is measured in the rocket momentary rest frame, and the mass is momentary rest mass of the rocket (important, since a rocket expelling thrust has changing rest mass), then the ratio is proper acceleration (even in SR), and the scenario is no different than the Bell case.
EXAM PROBLEM.-

A vehicle's engine exerts at every moment an exact force, that always keeps the same proportion to the current mass of the vehicle, and so we can say that the Force/mass ratio is a constant. Call it F/M ratio.

We do this in order to compensate the mass loss due to the engine burning fuel, and keep the same acceleration initially.

The vehicle burns fuel, and losses 10 gr of mass every second when running. No propulsion force is considered from the exhaust gases.

The constant F/M ratio is adjusted to 0.25, so the engine will exert 0.25 N of force per 1 Kg of mass of the vehicle, which means that the vehicle will accelerate initially at 0.25 m/s².

The vehicle starts moving, and after a short transient time, it reaches the initial acceleration of 0.25 m/s². We start the timer watch at that moment, and also we measure the distances from that position. Also, the mass of the vehicle is 1000 Kg and the initial speed is 3 m/s. We don't consider any aerodynamic air resistance.

QUESTIONS.-

1. Which distance has the vehicle traveled @1 minute? (1 point).

2. Which is the total mass of the vehicle @1 minute? (0.5 points).

3. Which is the acceleration of the vehicle @1 minute? (0.5 points).

4. After 1 minute, the vehicle enters a steep road, with a constant slope of 5%. Which is the acceleration of the vehicle 10 seconds after? (8 points).

Dale,

Thanks, I'll read carefully you very interesting and instructive answer, that has led to some very interesting and instructive comments by others, but I'll disagree fundamentally in one detail, not really related to you exposition, which has nothing to do with the "fragility of the string", and holds its validity undisturbed by that minor detail.
Dale said:
Fragile means that any tension breaks the string. A fragile string does not have any tensile break resistance.
What you describe then is not a string, is a mathematical line than extends from point to point with a defined length, and some properties defined by an Euclidian geometry or whatever, but has no physical material properties - such as tensile strength. That line can't neither break because lines don't break and then you have two lines, or a lot of bits of lines.

"fragile thread" that "finally breaks" due to "intolerable stress" ... is Bell's wording, not mine, in the statement of the original paradox. He's not talking about a zero tensile resistance thread, he's talking about a string than has some relatively weak resistance, but obviously not zero, because then it's not a string... such thing doesn't exists, a line do exists as a mathematical object, a zero tensile string doesn't exists, not even conceptually, because breaks spontaneously, just in the action to try to tie it to the hook in the spaceship.

I agree that if it's a community consensus that when in a paper you write "we consider a fragile string", every Physics literate will read "ah, it's a zero tensile break resistance string, nothing to worry about, will not molest us with any force", then I should have to include that fact in the summary. At the very first paragraphs of my huge post, no need to deep further, you already get immediately that I consider a string with some mechanical characteristics, it's not hidden anywhere.

But, what a problem! Consider modified the summary of the post, and read also "considering a weak string with a small tensile break resistance", and that's it.

PD.
Putting the numbers into you equation, and for a 10 Km string, and for a 109 Kg spaceship pulling, I get that the force threshold is 0.01 N.

I was considering as "fragile" a cable with a tensile strength limit of 840 N, and at the proper scale of the huge spaceship - with 109 Kg mass - that's fragile enough for me, but as Groucho Marx said: if that's not enough fragile for you, I have others, because from 840 N to 0.01 I have some margin to play.

And, by the way, not everybody gets what the F/M constant ration means, or what effect it has in the Bell's paradox. Really. I'll try using sock puppets to explain me better :)

Thanks

Last edited by a moderator:
Motore
PeterDonis said:
That doesn't have to be the case. Suppose the two spaceships started out with a proper acceleration of ##10^{10}## g instead of ##1## g.
Then I would use a ##10^{11}## Km diameter cable, made of Vibranium, bought in the Borg Amatzon galactic page. And, by the way, I guess such an acceleration would break, not the string, but something in the very current standard model of physics... just a guess.

Last edited by a moderator:
Lluis Olle said:
What you describe then is not a string
You seem to want to rail against idealizations. You will presumably rail against "test particles" as well. Such particles have a mass that is strictly positive but negligible. Strictly positive so that the particle may follow a space-like trajectory. Negligible so that it will not significantly affect the geometry of the space within which its motion is analyzed.

Bell's string has neglible tensile strength. It is a string so that it has a defined length and gives us a mental picture. It has sufficient tensile strength to hold itself together (in a Born rigid fashion, one assumes) but insufficient tensile strength to significantly alter the trajectories of the ships to which it is attached.

I see no need to attack the idealization instead of attacking the problem.

• Special and General Relativity
Replies
36
Views
3K
• Special and General Relativity
Replies
8
Views
2K
• Special and General Relativity
Replies
75
Views
4K
• Special and General Relativity
Replies
29
Views
1K
• Mechanics
Replies
1
Views
1K
• Thermodynamics
Replies
1
Views
3K