Deriving length of the rope in bells spaceship paradox I want to derive the distance between spaceships in Bell's paradox as a function of proper time of the trailing spaceship, in the case of constant proper acceleration of both space ships.
My attempt didn't work, as it gave a distance that gets shorter with time. I would like to know why it didn't work, and what the actual solution is.
My attempt:
First, define the trajectory of the trailing spaceship as a hyperbolic trajectory with a rindler coordinate frame. In the inertial frame: $x_T = \frac1\alpha \cosh(\alpha \bar t)$
In the inertial frame, the leading spaceship has the same hyperbolic trajectory except it is spatially shifted to the right by the length of the rope $l$. Hence, the difference in the spatial coordinates is always $x_L - x_T = l$ as measured in the inertial frame.
Next, use the Rindler coordinate transformation to express the difference in coordinates in terms of $\bar x$, given that the time of measurement is $\bar t$ .
$$x = \bar x \cosh(\alpha \bar t) \implies x_L - x_T = (\bar x_L - \bar x_T)\cosh(\alpha \bar t) \implies \bar x_L - \bar x_T = \frac l{\cosh(\alpha \bar t)}$$
This expression says that the distance between the spaceships in the trailing spaceship's MCRF is decreasing with time, meaning the rope will not break which is wrong. What is the mistake?
Thank you
 A: The issue when measuring lengths is that you need to define a line (hyperplane of spacetime in general, but in your case this is a line) of simultaneity that is space-like. In this plane, you can then use the metric and define the distance unambiguously.
For each time-like observer, there is a natural line of simultaneity to consider at each event of its worldline, you simply take the line passing by the event orthogonal to the $4-$velocity of the observer at the event.
For the intertial observer, this corresponds to the lines $t=t_0$, and by construction, the rope always has length $L$.
For the accelerated observers, a Rindler coordinate transform does the job by construction. You just got mixed in the change coordinate. You ned to be careful how time transforms as well. I'll fix the proper acceleration $\alpha=1$ (up to change of timescale). Say the first observer has worldline:
$$
\begin{align}
x &= \cosh \tau_1 \\
t &= \sinh \tau_1
\end{align}
$$
and the second one:
$$
\begin{align}
x &= L+\cosh \tau_2 \\
t &= \sinh \tau_2
\end{align}
$$
both parametrised by their respective proper times, synchronised with the inertial observer's at $t=0$.
If you want to go the the first one's frame, you need to apply the transformation:
$$
\begin{align}
x &= X_1\cosh T_1 \\
t &= X_1\sinh T_1
\end{align}
$$
so the respective worldlines become:
$$
\begin{align}
X_1 &= 1 \\
T_1 &= \tau_1
\end{align}
$$
and:
$$
\begin{align}
X_1 &= \sqrt{1+2L\cosh \tau_2+L^2} \\
T_1 &= \text{arctanh }\frac{\tanh \tau_2}{1+L/\cosh \tau_2}
\end{align}
$$
In particular, you get the diverging length in this frame:
$$
L_1 = \sqrt{1+2L\cosh \tau_2+L^2}-1
$$
Note that while it diverges exponentially, you have agreement with the inertial frame when you are close to the synchronisation:
$$
\begin{align}
L_1 &\sim L & \tau_2 &\to 0 \\
L_1 &\sim \sqrt{L}e^{\tau_2/2} & \tau_2 &\to +\infty
\end{align}
$$
For the second observer thongs are more interesting. While formally, you just need to switch the sign of $L$, there are small caveats. The coordinate change is given by:
$$
\begin{align}
x &= L+X_2\cosh T_2 \\
t &= X_2\sinh T_2
\end{align}
$$
so the worldline of the second is:
$$
\begin{align}
X_2 &= 1 \\
T_2 &= \tau_2
\end{align}
$$
and the one of the first one is:
$$
\begin{align}
X_2 &= \pm\sqrt{1-2L\cosh \tau_1+L^2} \\
T_2 &= \text{arctanh }\frac{\tanh \tau_1}{1-L/\cosh \tau_1}
\end{align}
$$
with the sign of square root being $+$ when $L<1$ and $-$ when $L>1$.
The issue is when:
$$
\tau_1\notin (-\tau_1^*,\tau_1^*) \\
\tau_1^*=\text{arccosh }\frac{1+L^2}{2L}
$$
Note that in particular that when $L=1$, $\tau_1^*=0$ so you have an issue for the whole wordline.
This is because the Rindler coordinates cover only two opposing quadrants of spacetime, giving two horizons at the asymptotes of the wordlines. The first observer crosses these worldines exactly at his proper times $\pm \tau_1^*$.
The added subtlety is that the first observer crosses the horizons at infinite proper time of the second one $T_2(\pm\tau_1^*)=\pm\infty$. The second observer is therefore incapable of witnessing the diverging stretching of the rope.
For the second observer, the length is given by:
$$
L_2= \begin{cases}
1-\sqrt{1-2L\cosh \tau_1+L^2} & L<1\\
\sqrt{1-2L\cosh \tau_1+L^2}+1 & L>1
\end{cases}
$$
So in the first case the rope stretches to length $1$ and in the second case it even contracts to length $1$.
Note that there is no contradiction since the lines of simultaneity do not coincide except at the synchronisation time. It is therefore expected that they measure different lengths outside this moment. Additionally, everyone agrees at the time of synchronisation that the length is indeed $L$ as expected.
Hope this helps.
A: Your mistake is assuming the leading spaceship is displaced by $L$ in Rindler coordinates at equal proper-time. Because of gravitational time dilation from the uniform acceleration, according to the rear ship, the front ship has had more proper time and has thus accelerated more, so it's moving faster, hence it is moving away from rear ship.
See:
https://en.wikipedia.org/wiki/Rindler_coordinates#A_%22paradoxical%22_property
Now you can just add $L$ if you stay in the initial rest-frame ($S$), find its position at time $t$, and then figure out where its hyper-plane of simultaneity intersects the front ship's word line (which is in the relative future according to $S$, and hence, it's moving faster). I think you get transcendental equations, though.
Of course, the accelerations don't have to be uniform (they can't be in $S$), they just need to be the same functions of $t$ in $S$.
You can pick a simple profile like:
$$ w_{rear}(t, x) = (t, \beta\delta(t))$$
$$ w_{front}(t, x) = (t, L+\beta\delta(t))$$
and solve for any $t>0$, which is just an exercise in trig, showing Lorentz dilation of $x'$. That is, the instant after the acceleration, according to the rear ship, the front ship had already started a while ago...string is already broken.
