SR paradox

[Zeno]

We have the following situation:

O2(0.99999999999999999999c)-------->

           (10 meters)
L1 B1--------O1-------B2 L2
   M1                 M2

L1 and L2 are light sources such as a light bulbs.
B1 and B2 are lead balls blocking the light sources.
B1 and B2 are separated by a distance of 10 meters.
M1 and M2 are machines underneath the balls and holding the lead balls in place.
There is an observer called O1 who is located at the mid-point of the
distance between the light sources L1 and L2 and balls B1 and B2.
The light sources are blocked by lead balls which are held in place by powerful machines.
The lead balls are connected by a chain which runs between them (that's the dashed line) and
is taut and has no slack whatsoever.
At a certain time, the machines pull down the lead balls simultaneously in the reference
frame of observer O1 and the light from both sources reach observer O1 simultaneously.
The chain remains intact because the distance between the lead balls doesn't change.
The is another observer called O2 which is moving at 0.99999999999999999999c relative to observer O1.
Gamma is 7071067811.865475244
The time between events for observer O2 is given by (gamma*v*delta x)/c^2.
So we have (7071067811.865475244*(0.99999999999999999999)*(10 meters))/(299792458 meters/second).
The time between events is about 236 seconds.
So for observer O2 ball B2 is pulled down about 4 minutes before ball B1 and the chain breaks.
So we have a paradox, for observer O1 the chain is intact and for observer O2 the chain is broken.
Both situations can't simultaneously be true.

 

[rpenner]

Some sloppiness in presentation exists. But the main problem is that the assumption of a perfectly taut chain is tantamount to assuming a perfectly rigid material substance with infinite speed of signal propagation, which is a violation of the physical theory the OP purports to explore.

We have the following situation:

O2(0.99999999999999999999c)-------->

           (10 meters)
L1 B1--------O1-------B2 L2
   M1                 M2

L1 and L2 are light sources such as a light bulbs.
B1 and B2 are lead balls blocking the light sources.
B1 and B2 are separated by a distance of 10 meters.
M1 and M2 are machines underneath the balls and holding the lead balls in place.
There is an observer called O1 who is located at the mid-point of the
distance between the light sources L1 and L2 and balls B1 and B2.
The light sources are blocked by lead balls which are held in place by powerful machines.

All of which is described as initially at rest in frame Σ.

The lead balls are connected by a chain which runs between them (that's the dashed line) and
is taut and has no slack whatsoever.

You appear to be trying to invoke a rod of perfect rigidity which is a hypothesis in conflict with SR.

At a certain time, the machines pull down the lead balls simultaneously in the reference
frame of observer O1 and the light from both sources reach observer O1 simultaneously.

That's three events, four if you count an implied event not described, O1 at the first "certain time".
D(x=0, t=0) E(x=-5m, t=0), F(x=+5m, t=0), G(x=0, t= 5m/c ≈ 16.678 ns)

The chain remains intact because the distance between the lead balls doesn't change.

You appear to be trying to invoke a rod of perfect rigidity which is a hypothesis in conflict with SR.

The is another observer called O2 which is moving at 0.99999999999999999999c relative to observer O1.
Gamma is 7071067811.865475244

If $$v = ( 1 - 10^{-20} ) c$$ then $$\gamma(v) = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{2 \times 10^{-20} - 10^{-40}}} \approx 7 \times 10^9$$ as you say.

The time between events for observer O2 is given by (gamma*v*delta x)/c^2.

Which events?
You fail to take into account that the Lorentz transformation is a space-time transform, so the events transform as (assuming the relative motion of O2 is parallel to the separation of L1 and L2....
D(x=0, t=0; x'=0, t'=0)
E(x=-5m, t=0; x' = $$- \frac{5 \times 10^{20} }{\sqrt{2 \times 10^20 -1}}$$ m, t' = $$- \frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^20 -1}}$$ m/c),
F(x=+5m, t=0; x' = $$\frac{5 \times 10^{20} }{\sqrt{2 \times 10^20 -1}}$$ m, t' = $$\frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^20 -1}}$$ m/c)),
G(x=0, t= 5m/c ; x' = $$\frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^20 -1}}$$ m , t' = $$\frac{5 \times 10^{20} }{\sqrt{2 \times 10^20 -1}}$$ m/c)


So while in frame Σ, events D, E, and F are simultaneous and G happens last, in frame Σ' the events happen in the order E, D, F, then G last.

The time between events E and F in frame Σ' does indeed reduce to $$\Delta t'_{EF} = \gamma(v) \times v \times \Delta x_{EF} / \textrm{c}^2$$.

So we have (7071067811.865475244*(0.99999999999999999999)*(10 meters))/(299792458 meters/second).
The time between events is about 236 seconds.

Approximately correct, but hardly with enough precision to distinguish the times of F and G.

In the chronology of Σ' we have:
E(t' ≈ -117.932716837484204555196... s)
D(t' = 0)
F(t' ≈ +117.932716837484204555196... s)
G(t' ≈ +117.932716837484204556375... s)

So for observer O2 ball B2 is pulled down about 4 minutes before ball B1 and the chain breaks.
So we have a paradox, for observer O1 the chain is intact and for observer O2 the chain is broken.
Both situations can't simultaneously be true.

You appear to be trying to invoke a rod of perfect rigidity which is a hypothesis in conflict with SR.

No such materials exist in the physical universe and Special Relativity forbids them.
If the material behaved as you say then even in the Σ frame, the chain at the position of the observer moves out of the way BEFORE the light gets there, so the apparent speed of sound in the chain -- the rate at which signals propagate from B1 and B2 is faster than the speed of light. So every part of your chain violates the premise of the thought experiment -- you aren't finding a conflict within SR you are simply adding a conflict where there was none.

Now if you ignore the anti-SR assumption of absolute rigidity of the chain and replace the chain with an infinite number of lead balls that all have the same coordinated movement of B1 and B2, you have the illusion of a perfectly rigid bar in frame Σ but not in frame Σ'.

 

[rpenner]

Previously, I demonstrated that the assumption that the chain works like magic (i.e. contrary to the assumptions of Special Relativity) makes the chain's behavior inconsistent in different frames and proposed a resolution (make the chain only appear to work like magic in one frame).

D(x=0, t=0; x'=0, t'=0)
E(x=-5m, t=0; x' = $$- \frac{5 \times 10^{20} }{\sqrt{2 \times 10^{20} -1}}$$ m, t' = $$- \frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^{20} -1}}$$ m/c),
F(x=+5m, t=0; x' = $$\frac{5 \times 10^{20} }{\sqrt{2 \times 10^{20} -1}}$$ m, t' = $$\frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^{20} -1}}$$ m/c)),
G(x=0, t= 5m/c ; x' = $$\frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^{20} -1}}$$ m , t' = $$\frac{5 \times 10^{20} }{\sqrt{2 \times 10^{20} -1}}$$ m/c)

So while in frame Σ, events D, E, and F are simultaneous and G happens last, in frame Σ' the events happen in the order E, D, F, then G last.

Between four events, there are $$\frac{4 \times ( 4 - 1)}{2} = 6$$ pairs between distinct elements and we can compute the space-time interval for each of these can be computed from the formula: $$s^2 = c^2(\Delta t)^2 - (\Delta x)^2$$.

Using frame Σ coordinates, the calculation is trivial:
$$s_{DE}^2 = -25 \, \textrm{m}^2 \\ s_{DF}^2 = -25 \, \textrm{m}^2 \\ s_{DG}^2 = 25 \, \textrm{m}^2 \\ s_{EF}^2 = - 100 \, \textrm{m}^2 \\ s_{EG}^2 = 0 \\ s_{FG}^2 = 0$$
We see that DE, DF and EF are space-like -- these are not paths along which energy, momentum or signal may propagate.
We see that DG is time-like -- this is a path which can be traversed (in one direction) by a ponderable material body.
We see that EG and FG are light-like -- these are paths that can be traversed (in one direction) by a massless phenomena.

Using frame Σ coordinates, the calculations are rougher, but fortunately we have simple exact expressions for the coordinates, so we won't have round-off errors:
$$s_{DE}^2 = \frac{( 5 \times 10^{20} - 5 )^2 - (5 \times 10^{20})^2}{2 \times 10^{20} -1} \, \textrm{m}^2 = \frac{- 50 \times 10^{20} + 25}{2 \times 10^{20} -1} \, \textrm{m}^2 = -25 \, \textrm{m}^2$$
$$s_{DF}^2 = -25 \, \textrm{m}^2 \quad \quad \quad ; \; \textrm{ Calculation identical to above}
$$
$$
s_{DG}^2 = \frac{ (5 \times 10^{20})^2 - ( 5 \times 10^{20} - 5 )^2 }{2 \times 10^{20} -1} \, \textrm{m}^2 = 25 \, \textrm{m}^2
$$
$$
s_{EF}^2 = \frac{( 10 \times 10^{20} - 10 )^2 - (10 \times 10^{20})^2}{2 \times 10^{20} -1} = - 100 \, \textrm{m}^2 \\ s_{EG}^2 = \frac{( 10 \times 10^{20} - 5 )^2 - (10 \times 10^{20} -5)^2}{2 \times 10^{20} -1} \, \textrm{m}^2 = 0
$$
$$s_{FG}^2 = \frac{( 5 )^2 - ( -5)^2}{2 \times 10^{20} -1} \, \textrm{m}^2 = 0$$
These quantities are identical as they always will be in special relativity exercises. This is a demonstration that SR is as self-consistent as geometry.

Another invariant is the inner product. $$ < \vec{\alpha \beta}, \vec{\gamma \delta} > = c^2 (\Delta t_{\alpha \beta})(\Delta t_{\gamma \delta}) - (\Delta x_{\alpha \beta})(\Delta x_{\gamma \delta}) = c^2 (t_{ \beta} - t_{\alpha })( t_{ \delta} - t_{\gamma }) - (x_{ \beta} - x_{\alpha })( x_{ \delta} - x_{\gamma }) $$. Obviously $$ < \vec{\beta \alpha }, \vec{\gamma \delta} > = < \vec{\alpha \beta}, \vec{\delta \gamma } > = - < \vec{\alpha \beta}, \vec{\gamma \delta} >$$ and $$ < \vec{\alpha \beta}, \vec{\alpha \beta} > = s_{\alpha \beta}^2 $$ so we have 15 additional terms to calculate:
$$
< \vec{DE}, \vec{DF} > = (0-0)(0-0) - (-5 - 0)(5 - 0) = 25 \, \textrm{m}^2
\\ < \vec{DE}, \vec{EF} > = (0-0)(0-0) - (-5 - 0)(5 - -5) = 50 \, \textrm{m}^2
\\ < \vec{DE}, \vec{DG} > = (0-0)(5-0) - (-5 - 0)(0 - 0) = 0
\\ < \vec{DE}, \vec{EG} > = (0-0)(5-0) - (-5 - 0)(0 - -5) = 25 \, \textrm{m}^2
\\ < \vec{DE}, \vec{FG} > = (0-0)(5-0) - (-5 - 0)(0 - 5) = -25 \, \textrm{m}^2
\\ < \vec{DF}, \vec{EF} > = (0-0)(0-0) - (5 - 0)(5 - -5) = -50 \, \textrm{m}^2
\\ < \vec{DF}, \vec{DG} > = (0-0)(5-0) - (5 - 0)(0 - 0) = 0
\\ < \vec{DF}, \vec{EG} > = (0-0)(5-0) - (5 - 0)(0 - -5) = -25 \, \textrm{m}^2
\\ < \vec{DF}, \vec{FG} > = (0-0)(5-0) - (5 - 0)(0 - 5) = 25 \, \textrm{m}^2
\\ < \vec{EF}, \vec{DG} > = (0-0)(5-0) - (5 - -5)(0 - 0) = 0
\\ < \vec{EF}, \vec{EG} > = (0-0)(5-0) - (5 - -5)(0 - -5) = -50 \, \textrm{m}^2
\\ < \vec{EF}, \vec{FG} > = (0-0)(5-0) - (5 - -5)(0 - 5) = 50 \, \textrm{m}^2
\\ < \vec{DG}, \vec{EG} > = (5-0)(5-0) - (0 - 0)(0 - -5) = 25 \, \textrm{m}^2
\\ < \vec{DG}, \vec{FG} > = (5-0)(5-0) - (0 - 0)(0 - 5) = 25 \, \textrm{m}^2
\\ < \vec{EG}, \vec{FG} > = (5-0)(5-0) - (0 - -5)(0 - 5) = 50 \, \textrm{m}^2 $$

Here we see that $$< \vec{DE}, \vec{DG} > = < \vec{DF}, \vec{DG} > = < \vec{EF}, \vec{DG} > = 0$$ which is saying that this particular space-like interval DG is orthogonal to all of our space-like vectors: DE, DF and EF which would help us find frame Σ as a particularly nice frame to work in if we were unaware of it.
Also such geometric truths as $$\vec{DE} = - \vec{DF}, \; \vec{EF} = 2 \vec{DF} = \vec{DF} - \vec{DE} = -2 \vec{DE} , \; \vec{EG} = \vec{DG} - \vec{DE}, \; \vec{FG} = \vec{DG} - \vec{DF}$$ are consistent with this table and the bilinearity of the inner product.

 

[Zeno]

I've never seen anybody take such a simple to understand problem and make such a convoluted mess out of it.
First of all, I can go get some lead balls and connect the lead balls by a chain and then pull the balls apart until the chain is pulled tight.
Then I should stand back and say, "This shouldn't exist! This is a rod of perfect rigidity and is in conflict with SR!"
Can you please elaborate?
So you seem to agree that if the dropping of balls B1 and B2 is simultaneous for observer O1, then for observer O2 the difference in time is about 4 minutes.
Obviously, the distance between the balls for observer O1 remains the same since the balls are moved simultaneously, therefore the chain doesn't break. For observer O2 one of the balls moves about 4 minutes before the other ball and the chain breaks because the distance between the balls is increased.
So, what's the resolution to the paradox?

 

[Janus58]

I've never seen anybody take such a simple to understand problem and make such a convoluted mess out of it.
First of all, I can go get some lead balls and connect the lead balls by a chain and then pull the balls apart until the chain is pulled tight.
Then I should stand back and say, "This shouldn't exist! This is a rod of perfect rigidity and is in conflict with SR!"
Can you please elaborate?

The ball and chain are not perfectly rigid. If you pull on one end, the chain will stretch a little more, and that "stretch" will travel down the chain at the speed of sound for the chain. When it gets there then the other end will move. If the chain is already stretched to the point where it will not stretch anymore, if you pull on the chain, it will break in the rest frame of the chain. There is no way to make a chain or object that when you push or tug on one end, will instantly move at the other. The minimum delay between the two ends moving will be the time it takes light to travel the length of the object.

 

[Zeno]

So the gist of the resolution of the paradox is that it would be impossible for balls B1 and B2 to move simultaneously and for the chain to remain intact?
I don't know if that's what is being said here or not, but on that point I would have to respectfully disagree. We could replace the balls and chain with weights and a barbell and the problem would remain essentially unchanged. I don't think anybody believes that the weights on the ends of a barbell could move simultaneously and the bar connecting them would have to break.

 

[James R]

So the gist of the resolution of the paradox is that it would be impossible for balls B1 and B2 to move simultaneously and for the chain to remain intact?

Impossible in all frames, yes. You can have that happen in one frame.

I don't know if that's what is being said here or not, but on that point I would have to respectfully disagree. We could replace the balls and chain with weights and a barbell and the problem would remain essentially unchanged. I don't think anybody believes that the weights on the ends of a barbell could move simultaneously and the bar connecting them would have to break.

The bar of the barbell can't be infinitely rigid, either. If you pull on the ball at one end of a barbell, the one at the other end can't know about that pull instantaneously. Some kind of signal must travel from the end being pulled to the other end before the ball at the other end starts to move. Most likely that happens at the speed of sound in the barbell material, but at the very best the fastest it could happen is at the speed of light.

So, there's some time after you pull the ball at this end before the ball at the other end starts to move. What happens in the meantime? Well, the bar must stretch a bit. If it has a high enough tensile strength then it will cope with the stretch. If not, it will break. Either way, everybody in every frame will agree either that the bar breaks or it doesn't.

 

[Zeno]

Either way, everybody in every frame will agree either that the bar breaks or it doesn't.

Mmmmm....... It seems that the paradox in the opening post still hasn't been resolved.

 

[paddoboy]

Mmmmm....... It seems that the paradox in the opening post still hasn't been resolved.

rpenner, janus and James appear to have invalidated your concept as far as I can see. But hey, if you are still certain you have found a flaw or a paradox in SR, why not write up a scientific paper on it and via the scientific method get it properly peer reviewed.
I mean if successful, you could be in line for this years Physics Nobel.

 

[Ophiolite]

Mmmmm....... It seems that the paradox in the opening post still hasn't been resolved.

Just to be clear, the paradox you are referring to is how someone who allegedly has a brain can post nonsense. I suggest a resolution: employing the Argument from Incredulity leads to nonsense.

 

[origin]

Mmmmm....... It seems that the paradox in the opening post still hasn't been resolved.

Only if you missed the replies or you did not understand them.

 

[Beer w/Straw]

Try General Relativity instead of Special Relativity?

 

[Russ_Watters]

I must admit to only having skimmed the replies, but it seems to me the real problem was missed:

The OP doesn't account for length contraction, which causes the incorrectly calculated 4-minute delay between motions of the balls.

 

[rpenner]

The time between events is about 236 seconds.
So for observer O2 ball B2 is pulled down about 4 minutes before ball B1 and the chain breaks.

D(x=0, t=0; x'=0, t'=0)
E(x=-5m, t=0; x' = $$- \frac{5 \times 10^{20} }{\sqrt{2 \times 10^{20} -1}}$$ m, t' = $$- \frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^{20} -1}}$$ m/c),
F(x=+5m, t=0; x' = $$\frac{5 \times 10^{20} }{\sqrt{2 \times 10^{20} -1}}$$ m, t' = $$\frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^{20} -1}}$$ m/c)),
G(x=0, t= 5m/c ; x' = $$\frac{5 \times 10^{20} - 5}{\sqrt{2 \times 10^{20} -1}}$$ m , t' = $$\frac{5 \times 10^{20} }{\sqrt{2 \times 10^{20} -1}}$$ m/c)
...
In the chronology of Σ' we have:
E(t' ≈ -117.932716837484204555196... s)
D(t' = 0)
F(t' ≈ +117.932716837484204555196... s)
G(t' ≈ +117.932716837484204556375... s)

The OP doesn't account for length contraction

Nothing in the OP demands length contraction which is a secondary effect of applying simultaneous position measurement co-moving endpoints of a rod in a frame where it not at rest. The OP appears to follow directly from the application of the Lorentz transforms.

$$x' = \frac{x + v t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ t' = \frac{t + \frac{v x}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}}$$
(which is sometimes written with an opposite convention for the sign of v, which doesn't affect the argument in the OP given the symmetry of the arrangement).

Length contraction arises from this when given the equation for two lines motionless in system Σ, we ask what the spatial separtion is between the lines in system Σ' at any simultaneous time.

These functional relations between world lines L and R shows that they are distance $$ \ell $$ apart.
$$x_L + 0 t_L = x_0 \\ x_R + 0 t_R = x_0 + \ell \\ t_R = t_L \Rightarrow x_R - x_L = \ell$$
Solving for (x,t) in terms of (x',t') we introduce the corresponding relations in system Σ':
$$x'_L - v t'_L = \sqrt{1 - \frac{v^2}{c^2}} x_0 \\ x'_R - v t'_R = \sqrt{1 - \frac{v^2}{c^2}} x_0 + \sqrt{1 - \frac{v^2}{c^2}} \ell \\ t'_R = t'_L \Rightarrow x'_R - x'_L = \sqrt{1 - \frac{v^2}{c^2}} \ell $$

Instead, the OP asks about events defined as simultaneous in system Σ' and $$ t_R = t_L $$ is not the same condition as $$ t'_R = t'_L $$.

the incorrectly calculated 4-minute delay between motions of the balls.

No one else is saying the ~236 s between the Σ' times of events E and F is wrongly calculated, only that the requirement for the chain to be "perfectly" taut and move like a ideally rigid rod is incompatible with the upper limit on signal propagation in special relativity.

 

[Russ_Watters]

Nothing in the OP demands length contraction which is a secondary effect of applying simultaneous position measurement co-moving endpoints of a rod in a frame where it not at rest..

No one else is saying the ~236 s between the Σ' times of events E and F is wrongly calculated...

Hmm....I didn't think it through enough. If the 3rd part observer is flying away from the device, along the same line, the signals from each ball (light) are chasing him and the 2nd takes a lot longer to get to him than the first. From his point of view, the balls are much further apart, not closer together, which is why the signal delay is so long.

However:

...only that the requirement for the chain to be "perfectly" taut and move like a ideally rigid rod is incompatible with the upper limit on signal propagation in special relativity.

The chain doesn't need to be perfecly taught for the apparent paradox to show itself: locally, the chain reacts in microseconds to the motion of the balls, yet the remote observer sees them move minutes apart. The issue isn't that one ball is pulled down 4 minutes after the other, it's that the distance between them isn't 10m for the remote observer, it is much further apart. What he sees is the chain curve and a ripple pass along it to the other ball. He doesn't see the chain break because the effect is an optical illusion, much like a doppler shifted sound is an illusion.

 

[rpenner]

He doesn't see the chain break

He doesn't see the chain break because the chain violated SR assumptions from the beginning, so the chain is magic.

In system Σ, the 10m chain moves as if rigid so that it doesn't block the light. Specifically, the middle of the chain moves out of the way before the light arrives, so the signal propagation speed is faster-than-light. No physical object can do that, but special relativity doesn't have a problem with an arbitrary number of independent links between B1 and B2 all moving independently at the same time for the illusion of a solid, rigid rod. The illusionary "speed of signal" in the system of independent links is literally infinite, $$\pm \infty$$ because all movement is simultaneous.

In system Σ', this illusion of rigidity is broken. The chain is moving at $$(1 - 10^{-20} ) c \; \approx \; c \, - \, 1 \, \textrm{cm per century}$$ and is Lorentz contracted to approximately $$\sqrt{2} \, \textrm{nm}$$ long. Thus for the light to catch up from its trailing side to the midpoint is approximately $$\frac{\sqrt{2}}{2} \, \textrm{nm} \times 1 \, \textrm{century per cm} \times 3.15576 \times 10^9 \, \textrm{seconds per century} \approx 223 \, \textrm{s}$$ which is consistent at the level of approximation being used. $$\frac{\sqrt{2}}{2} \times 10^{-9} \, \textrm{m} \times 10^{20} \, 1/c \approx 235.865\, \textrm{s}$$ is much closer showing that the cutsy cm per century units introduced most of the error.

As for the illusionary "speed of signal" in system Σ', we may extract it from the Einstein velocity addition formula: $$\lim_{u\to\infty} \frac{u + v}{1 + \frac{uv}{c^2}} = c^2 \lim_{u\to\infty} \frac{u + v}{c^2 + u v} = \frac{c^2}{v} = \frac{c^2}{ ( 1 - 10^{-20} ) c} = (1 + \frac{1}{10^{20} - 1} ) c$$. So the illusionary "speed of signal" in system Σ' is still superluminal. (This is generally true -- Superluminal, luminal, and subluminal velocities are preserved by the Lorentz transform exactly as space-like, light-like and time-like space-time intervals are preserved by the Lorentz transform.)

because the effect is an optical illusion, much like a doppler shifted sound is an illusion.

I disagree. Both system Σ and system Σ' are perfectly acceptable systems for describing physical systems. Geometrically, classical Doppler shift is completely explained due to the changing geometry between source and receiver and the finite travel time of the sound resulting in changing pitch at the receiver even when the source is in uniform motion with constant original pitch. Special relativity is the proposition, which is more remote to human intuition, that the same space-time geometry admits many physically equivalent descriptions (in terms of coordinates) despite those descriptions differing in which bodies are at rest and which clocks tick slower, etc.

 

[Zeno]

There's nothing in the opening post which says the chain is blocking the light. The chain can be located so that it isn't directly in front of the light source. Although, I can see how that might be inferred from the illustration I made, maybe that was unclear. Therefore, there is no requirement for the signal along the chain to move faster than light.

 

[Russ_Watters]

He doesn't see the chain break because the chain violated SR assumptions from the beginning, so the chain is magic.

So correct the faulty assumption and try it again. The "paradox" does not rest on that assumption.

I disagree. Both system Σ and system Σ' are perfectly acceptable systems for describing physical systems. Geometrically, classical Doppler shift is completely explained due to the changing geometry between source and receiver and the finite travel time of the sound resulting in changing pitch at the receiver even when the source is in uniform motion with constant original pitch. Special relativity is the proposition, which is more remote to human intuition, that the same space-time geometry admits many physically equivalent descriptions (in terms of coordinates) despite those descriptions differing in which bodies are at rest and which clocks tick slower, etc.

I guess you don't like the word "illusion": I recognize that both observations are valid if properly understood, but that doesn't change the fact that the remote observer sees the chain flex and stretch, when in reality it doesn't. For a doppler shifted sound like a train horn, the remote observer hears the sound change when the horn doesn't actually change what it is emitting.

 

[Neddy Bate]

No one else is saying the ~236 s between the Σ' times of events E and F is wrongly calculated, only that the requirement for the chain to be "perfectly" taut and move like a ideally rigid rod is incompatible with the upper limit on signal propagation in special relativity.

I agree that the flaw in the OP is the expectation for the chain to be perfectly-taut, and move like a ideally rigid rod. I also admit that I am too lazy to check the OP's numbers myself, but did the OP really get the ~236 second time interval correct?

At such high speed, the distance between the balls would be extremely length-contracted, and the time interval should depend on that length-contracted distance. Granted, the time delay also depends on that extremely high speed, so it is certainly possible that that is the correct time interval. But given the nature of the OP, I just find it amazing that they actually got that right.

Either way there is no paradox, of course.

 

[Zeno]

So the argument is that because the entire chain doesn't react instantaneously to the movement of the balls on either end of the chain renders the paradox as being invalid?
By instantaneous of course, we are requiring that the middle of the chain starts moving before the movement of the balls at either end, transmitted by molecular forces, reach it.
Of course, this isn't a requirement, and the chain isn't going to break if this doesn't happen.
Why would the chain be required to break for observer O1 when there is nothing preventing the movement of the chain along with the balls?
If I have two balls connected by a steel bar for example, the bar will simply move when the balls move. Granted, it will take time for the movement to be transmitted through the molecules of the bar, but the bar isn't going to break.
So then what is the difference between the two situations?
If one likes, the paradox can be posed again with just the tiniest bit of slack in the chain.
Say, the movement of 1 mm in the middle of the chain. So the counter-argument that the paradox is invalid because the chain represents a rod of perfect rigidity is rendered moot.