Unfortunately, that is not the right answer--both Alice's conclusion and reasons are actually correct. I got this problem from Bell's book,
Speakable and unspeakable in quantum mechanics, although he does not give any derivation for the answer. The whole thing seems out of place from the rest of the book, although it's interesting enough by itself. Bell argues that those with a classical physics background have a stronger intuition, both in general and in regards to this question in particular.
Darth Holbytlan's reasoning is pretty clear as to why (although Glass Pearl Player also stated the right answer as a particular case before that, and others stated pretty much the same thing afterward as well):
Darth Holbytlan wrote:From your point of view, Alice and Bob maintain the same velocity at all times. Therefore they maintain the same distance at all times in your frame of reference. But this means the distance is increasing between Alice and Bob's vessels in their frame of reference.
This is pretty direct: if the contracted length L/γ is constant, where L is the proper length of the string, then since γ is increasing with velocity, L must be increasing. Thus, the string is being stretched, and therefore eventually breaks. (Note: DH's footnote caveat is equivalent to saying that the acceleration is small, although the conclusion is true regardless). What's different from the ships themselves?
Terralthra wrote:If we were to consider a double-hulled spaceship with an engine on each hull, and Alice and Bob are on separate hulls, we would not even be asking the question of whether or not the ship tears itself apart through Lorentz contraction.
In in your frame, the ships are Lorentz-contracting. Thus, to not have additional stress, the string should be Lorentz-contracting--but it's not! bz249, after posting his answer, also makes the following comment:
bz249 wrote:BTW note that Alice and Bob are two different frames, because of the different origins, thus the two Lorentz transformations are different.
This can be viewed as another way to understand the problem. The Lorentz transformation is t' = γ(t - vx/c²), x' = γ(x-vt). Note the vx/c² term. It means that even though the two ships' clocks tick as the same rate (both dilated by γ in your frame), they are out of phase even if initially synchronized: according to one ship, it will have spent more time accelerating than the other. If I understood GPP and Steel correctly, they seems to have arrived at the correct answer through essentially this kind of reasoning.
In terms of spacetime diagrams, a uniformly accelerated object has a hyperbolic worldline. So in spacetime, the trajectories of the two ships are identical hyperbolas, side by side (a spatial translation). Picking an arbitrary point on the one belonging to the trailing ship, the direction tangent to the hyperbola represent the ship's time axis (light blue on the diagram). Orthogonal to this (inverse slope in the diagram, in light green) it the ship's line of simultaneity at that instant. It overshoots
your line of simultaneity (light red, horizontal), intersecting the other's worldline at an event with higher velocity. Thus, the trailing ship sees the front ship pulling further and further ahead, as some have noted. The string stretches, and eventually breaks.
