Help understanding Bell's spaceship paradox
"To an observer in the original rest frame, the spaceships stay the same distance, d, apart.". But why do they stay the same distance apart to an observer in the original rest frame?
The spaceships move with constant mutual distance in the original rest frame, since their corresponding parts have the same velocity function of time. The description of the situation in the original question directly implies this.
Shouldn't the distance between the spaceships undergo length contraction, as they are connected by a rope?
No, this would be contrary to the specified situation.
Bell's thought experiment is set up in such a way that the distance between the ships, call it $d$, remains the same in the stationary frame; after all, both ships have the same velocity $v$ at the same time $t$, so their distance never changes. Let's use $(x,t)$ as coordinates in the stationary frame and $(x',t')$ in the space ships' frame, we have $\Delta x = d$ if the positions are measured simultaneously, i.e. $\Delta t = 0$. Applying the Lorentz transform, we find $$ \begin{align} \Delta x' &= \gamma\left(\Delta x - v\Delta t\right) = \gamma d,\\ \Delta t' &= \gamma\left(\Delta t - \frac{v}{c^2}\Delta x\right) = -\gamma\frac{vd}{c^2}. \end{align} $$ So the distance between the ships in the moving frame does increase: $d'=\gamma d$. Notice also that there is a simultaneity issue: in the moving frame, the space ships are at rest at different times. You can argue that this complicates the notion of a distance in the moving frame. However, we can solve this if we switch off the accelerations simultaneously in the stationary frame; then both ships will have the same constant $v$, and both ships will stay at rest in the moving frame, so it doesn't matter at which time their position is measured: their distance will be $d'=\gamma d$, so the rope between them will break.
Note that there's no real paradox here: the distance $d'$ between the ships, measured in their rest frame, is Lorentz-contracted to $d$ when the ships are observed to be moving with respect to a stationary frame. It only seems odd because this thought experiment is set up in such a way that $d$ doesn't change; whereas in the usual discussion of length contraction one defines $d'$ as the constant length. It's simply a matter of different conditions.
Now, one can ask if it's possible to accelerate the space ships in such a way that the rope doesn't break. The answer is yes: it's called Born rigid motion. A discussion is given in this paper, which I will summarize here. I first need to introduce the concept of proper acceleration. Unlike normal coordinate acceleration, proper acceleration is Lorentz invariant: see this post and this post for more information. In the absence of perpendicular velocities, the proper acceleration $\alpha$ is given by $$ \alpha = \gamma^3\frac{\text{d}v}{\text{d}t} = \frac{1}{(1 - v^2/c^2)^{3/2}}\frac{\text{d}v}{\text{d}t}.\tag{1} $$ Now, let's call the first space ship $A$ and the second $B$, and we each give them a different but constant proper acceleration, $\alpha_A$ and $\alpha_B$. Since the ships have different accelerations, they will have the same velocity $v$ at different times $t_A$ and $t_B$. It's straightforward to derive the velocity of the ships from (1) as $$ \begin{align} v &= \frac{\alpha_A t_A}{\sqrt{1 + \alpha_A^2 t_A^2/c^2}} = \frac{\alpha_A t_A}{\gamma},\\ v &= \frac{\alpha_B t_B}{\sqrt{1 + \alpha_B^2 t_B^2/c^2}} = \frac{\alpha_B t_B}{\gamma}, \end{align} $$ where we used the property $$ \gamma = (1-v^2/c^2)^{-1/2} = \sqrt{1 + \alpha_A^2 t_A^2/c^2} = \sqrt{1 + \alpha_B^2 t_B^2/c^2}. $$ Their positions, initially a distance $d$ apart, are then $$ \begin{align} x_A &= \frac{c^2}{\alpha_A}\left(\sqrt{1 + \alpha_A^2 t_A^2/c^2} - 1\right) +d = \frac{c^2}{\alpha_A}(\gamma -1) +d,\\ x_B &= \frac{c^2}{\alpha_B}\left(\sqrt{1 + \alpha_B^2 t_B^2/c^2} - 1\right) = \frac{c^2}{\alpha_B}(\gamma -1). \end{align} $$ If we use the shorthand notation $$ \delta = \frac{c^2}{\alpha_A} - \frac{c^2}{\alpha_B}, $$ we find, $$ \begin{align} \Delta x &= x_A - x_B = (\gamma - 1)\delta + d,\\ \Delta t &= t_A - t_B = \frac{\gamma v}{c^2}\delta. \end{align} $$ The corresponding distance in the moving frame is $$ \Delta x' = \gamma\left(\Delta x - v\Delta t\right) = \delta - \gamma\delta + \gamma d. $$ Now, we impose the condition $\Delta x' = d$, such that the rope doesn't break. This implies a relation between the accelerations $$ (1-\gamma)\delta = (1-\gamma)d, $$ in other words, $$ \delta = \frac{c^2}{\alpha_A} - \frac{c^2}{\alpha_B} = d. $$ Therefore, the second space ship has to accelerate more than the first, and their distance will appear shortened in the stationary frame (if measured at the same time $t$). We also see that $$ \Delta t' = \gamma\left(\Delta t - \frac{v}{c^2}\Delta x\right) = \gamma\left( \frac{\gamma v}{c^2}d - \frac{v}{c^2}(\gamma - 1)d - \frac{v}{c^2}d \right) = 0. $$
I've read your question a number of times in order to try and understand precisely what is puzzling you about this well known 'paradox'. It seems to me that this is it:
Well the two ends of the rope are moving at the same velocity,
Let's stipulate that, in the inertial frame of reference in which the two spacecraft are initially at rest, the two ends of the rope have the same instantaneous velocity. Then, it follows that the length of the rope, as measured in this frame, is constant and equal to the length of the rope when the rope was initially at rest: $L = L_0$.
This must be the case since the world lines of the ends of the rope are, by stipulation, congruent.
Now, imagine another inertial reference frame with relative speed $v$ in which the rope is momentarily at rest* and that, in this moment, the rope is measured to have length $L'$.
According to the Lorentz transformations, it must be the case that
$$L' = \gamma_vL > L_0$$
In other words, in a momentarily co-moving reference frame, the measured rope length must be longer than the rope's initial rest length.
Let's summarize:
- We stipulate that the ends of the rope have the same velocity in the initial rest reference frame
- Thus, the length of the rope in the initial rest reference frame is constant and equal to the initial rest length of the rope
- Thus, the length of the rope in a momentarily co-moving reference frame is longer than the initial rest length of the rope
So, to answer this question:
Shouldn't the distance between the spaceships undergo length contraction, as they are connected by a rope?
As I've shown above, if we stipulate that the ends of the rope have the same velocity in the initial rest frame, it follows that the length of the rope (and therefore the distance between the spaceships) must be greater in a momentarily co-moving reference frame than the initial rest length of the rope.
*In fact, this is not quite possible. Only in the initial rest frame do the two ends of the rope have the same velocity. In a relatively moving inertial frame, the ends of the rope have different velocities so the statement "the rope is momentarily at rest" isn't well defined. This is a subtle point that is discussed more completely here.