# “Reality” of length contraction in SR vs. Bell's spaceship paradox

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.

• 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. – Javier Dec 27 '17 at 15:05
• Sounds great, no hurry. Physics will not change while you are out :-) – Harald Dec 27 '17 at 16:04
• 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). – Ben Crowell Dec 27 '17 at 17:46
• 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. – Sean Jan 2 '18 at 5:32

## 1 Answer

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.