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I read John Rennie's answer to "Reality" of length contraction in SR. He concludes about the length of a stick and length contraction that

... it's just that due to the rotation in spacetime we are viewing the two ends at different times.

This answer seems to describe it similarly:

In a relatively moving frame, the coordinate length of the rod is smaller than the proper length.

Both answers sound to me as if length contraction is a matter of the observer's "relativistic perspective" alone.

Does this not contradict the (afaik accepted) explanation of Bell's spaceship paradox, according to which the thread between the spaceships breaks? I construe the breaking thread as evidence that length contraction is not only a matter of the observer's "relativistic perspective", but that contraction is "real" in the moving system.

Can someone piece together the "relative perspective only" and "breaking thread" views? If the answer includes some "there is acceleration in the spaceship paradox", it would be great to learn how this is relevant.

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  • $\begingroup$ Not a full answer, but: The Wikipedia page on the Bell paradox contains an explanation of the physical process that makes the thread break. But anyway, this is why we use math. Words can be ambiguous; math produces clear predictions no matter what words you want to use to describe them. $\endgroup$ – Javier Dec 27 '17 at 15:05
  • $\begingroup$ Sounds great, no hurry. Physics will not change while you are out :-) $\endgroup$ – Harald Dec 27 '17 at 16:04
  • $\begingroup$ Here's a simple analogy that might help. In Newtonian mechanics, velocity is dependent on the coordinate system we use. The paperweight on my desk can have any velocity you like, if you pick the appropriate frame of reference. But just because something is coordinate-dependent, that doesn't mean it's not meaningful or can't be measured. Once you pick a coordinate system, you can measure velocities of different objects and compare them, and the comparison is meaningful. Same thing for the accelerating spaceship. You can compare velocities (velocities of the spaceship at different times). $\endgroup$ – user4552 Dec 27 '17 at 17:46
  • $\begingroup$ Well let's say that we have a spaceship that extends across space only, and thus its motion is across the time dimension only, and that synchronized clocks are located at the opposite ends. Say this one of two twin spaceships. If it gains spatial velocity, takes off, turns around and heads back to the other spaceship, its clocks are now ticking slower and are no longer synchronized. Its spatial length has contracted due to 4D rotation in space-time. Because of these changes, it will see the other ship as though it is shorter, its clocks are slower, and its clocks are no longer synchronized. $\endgroup$ – Sean Jan 2 '18 at 5:32
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Some four-positions

We consider two points in the frame $S'$ which is moving relative to the frame $S$ at speed $v$. In the coordinates of $S'$, these points have four-positions \begin{equation} \mathbf{x}_A' = (t_A',x_A')^\intercal, \quad \mathbf{x}_B' = (t_B',x_B')^\intercal. \end{equation} The proper distance between these points is given by \begin{equation} \Delta s^2 = -\Delta t^{\prime 2} + \Delta x^{\prime 2}. \end{equation} Expanding the primed coordinates in terms of the unprimed coordinates, using the Lorentz transformations, we have \begin{equation} -(t_A'-t_B')^2 + (x'_A-x'_B)^2 = -(\gamma(t_A-vx_A)-\gamma(t_B-vx_B))^2 + (\gamma(x_A - v t_A) - \gamma(x_B - vt_B))^2, \end{equation} where we have deliberately not expanded out the brackets, so that we recall that the left squares and right squares correspond on both sides of the equation.

Bell's spaceships

Now we can state the Bell spaceship problem as follows. Consider two spaceships that are uniformly accelerated in $S$ so that the distance between them in $S$ remains constant. Between them lies a string that just spans the distance. The question is whether the string will break due to this acceleration.

Mathematically we return to our four-position analysis. We wish to define a length $L' := x_A'-x_B'$ in $S'$ and so we are required to set $t_A'=t_B'$. This cancels the left-hand square on both sides of the equation. To define a length in $S$ we must do the same, defining $L := x_A - x_B$ at $t_A = t_B$. Crucially, we now set $L$ to be constant. This implies that $L'$, i.e. the distance that the observer in $S'$ experiences, as $L' = \gamma L$. We see that the distance between these two points from the point of view of $S'$ increases as the velocity increases. Note that there is no rod here, just two-comoving points and two perspectives on their absolute distance. To see length contraction, we simply fix $L'$ as the proper distance. This means that $L$ is going to see the rod obeying $L = L'/\gamma$.

Discussion

The solution to the paradox comes when we realise that fixing the distance $L$ is going to cause real stress on the object when it wants to contract, at least from the point of view of $S$. This fixing in $S$ would be viewed in $S'$ as the gradual distancing of the two spaceships as the they accelerate, causing equal stress and equally causing the string to break. The entire paradox does not crucially rely on accelerating frames. The hidden statement of the paradox is in the rewording of the setup:

What happens to a rod if we boost it to a velocity without allowing it to contract?

Bell found that many physicists did not have good intuition for this situation and put it down to the fact that they had never been taught what I call the 'ether intuition'. That is to say that relativity can be consistently understood by invoking an ether which is a preferred state of rest but entirely undetectable. This ether imposes physical length contraction that can have physical effects like breaking the string. Both interpretations of relativity provide the same result but Bell argued that the former could provide pedagogical power.

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  • $\begingroup$ "Ether" interpretation provides not only "pedagogical power" but extreme simplicity of resolution. "Actual" distance between spaceships doesn't change; but measuring tools of the astronauts do. Due to contraction of their measuring rulers these astronauts would measure that their spaceships scatter from each other. Also their clocks would go out of sync. This way if they wish to get back into their previous position and would start reverse thrust of engines according to Einsteinian simultaneity they will not come back but would scatter and scatter again during their each "return trip". $\endgroup$ – Albert Nov 12 '19 at 21:21
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The answer is easy enough. The spaceships are accelerating, so their rest frame in not the frame in which the distance between them (ie the fixed length of the string) is set. The problem goes away if you say that the two rockets accelerate in such a way that the proper distance between them stays constant- in that case the string won't break. But the proper distance seems to become smaller and smaller in any inertial reference frame from which their accelerating motion is viewed. If you want to fix the distance between them in such an other frame, even though they are accelerating, you must increase the proper distance between them, and hence the string snaps.

To avoid conceptual traps in connection with SR it is a good idea to try to bear in mind that its effects are entirely reciprocal. If an object appears length contracted because it is passing us at speed, we seem length contracted by exactly the same amount from its perspective. In our own rest frame our dimensions are unchanged, and so are its in its rest frame. If you remember that in connection with Bell's spaceship paradox you will realise that if the rope is to break it must be because the spaceships are getting further apart in their own frame.

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  • $\begingroup$ The problem goes away if you say that the two rockets accelerate in such a way that the proper distance between them stays constant I don't think that works. If A says that the proper distance between A and B stays constant, then I don't think B says the proper distance between them stays constant. $\endgroup$ – user4552 Nov 12 '19 at 20:08
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    $\begingroup$ Hi Ben, by definition the proper distance is the same for all observers. Perhaps it would be clearer if we eliminate the spaceships and the string and simply say that if the distance between two spatially separated points is to appear fixed to an observer in an accelerating reference frame, then the proper distance between the two points must be increasing. $\endgroup$ – Marco Ocram Nov 12 '19 at 20:20
  • $\begingroup$ @MarcoOcram: just a small question about "reciprocity" of length contraction - feynmanlectures.caltech.edu/I_34.html, chapter 34–8 Aberration. Could you please to clarify - why, according to Feynman, the man on earth would have found that the horizontal distance $vt$ (and apparent distance to the star) would have increased $1/ \sqrt{1-v^2/c^2}$ times? It turns out, that apparent distance to the star is not getting shorter from the point of view of moving observer, but longer. Is there such an impression? $\endgroup$ – Albert Nov 12 '19 at 21:55
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Math might not lie, but math can't tell the difference between duration contraction and length contraction. The final answer is the same.

If an object travels from A to B and accelerates, the trip gets shorter. Not distance shorter, duration shorter. This is where the confusion arises. Length contraction is real, but it's not the contraction of physical length. It's length contraction of duration.

Time dilation means less time passes on the clock of a traveller moving at relativistic speeds. At 87% of light speed, it's 50% slower than normal. For the traveler, a ten year journey, only takes 5 years. The distance doesn't change, only the duration from the traveller's perspective. It's real. Both are real. The clock on the traveler ship will confirm that the trip only took 5 years, but, it's a clock, not an odometer.

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