This article explains Bell's spaceship paradox using a simple example involving ants and a thread.
E. Dewan and M. Beran first introduced the two-spaceship paradox in 1959. J. S. Bell later revisited the thought experiment in 1987, and it is now widely known as Bell's spaceship paradox.
The paradox is usually described using two spaceships. However, because the speed of light is so enormous, the situation can be difficult to visualize. To make the idea more intuitive, let us imagine instead that the speed of light is only 30 millimeters per second.
First, consider a caterpillar on a desk. Its body length is 30 millimeters.
Suppose the caterpillar accelerates for 10 seconds and reaches a speed of 24 millimeters per second. How long does it appear after Lorentz contraction?
The Lorentz contraction formula is:
$$ L = L_0 \sqrt{1-\frac{v^2}{c^2}} $$Substituting L0 = 30 mm, v = 24 mm/s, and c = 30 mm/s gives:
$$ L = 30 \sqrt{1-\frac{24^2}{30^2}} = 18 $$The contracted length is therefore L = 18 mm. In the caterpillar's own rest frame S′, however, its length is still 30 mm.
Now place two ants on the desk, 30 millimeters apart.
Both ants accelerate for 10 seconds, beginning and ending their acceleration at the same time in the desk frame S. They then reach a speed of 24 mm/s. What is the distance between them in frame S?
One might expect the distance to shrink because of Lorentz contraction. But since both ants start and stop accelerating simultaneously in frame S, their separation remains 30 mm in that frame.
Suppose each ant accelerates at a constant rate a = 2.4 mm/s2 for 10 seconds in frame S. Their motion can then be described as follows.
Let ant A start at x(0) = 0 mm at t = 0. Its position after time t is:
$$ x(t) = \tfrac{1}{2} a t^2 $$After 10 seconds, x(10) = ½·a·(10 s)2 = 120 mm.
If ant B starts at x(0) = 30 mm, its position y(t) is:
$$ y(t) = 30 + \tfrac{1}{2} a t^2 $$After 10 seconds, y(10) = 30 + ½·a·(10 s)2 = 150 mm.
Thus the separation is still 150 mm − 120 mm = 30 mm. Once the acceleration ends, both ants move at the same constant speed, v = 24 mm/s, so their separation remains 30 mm in frame S.
Now imagine that the two ants are connected by a thread. What happens to the thread when the ants begin to accelerate?
The thread breaks. This seems puzzling in frame S, where the distance between the ants stays constant. So why does the thread snap?
In ant A's rest frame S′, the relativity of simultaneity changes the order of events: ant B finishes accelerating before ant A does. As a result, B temporarily moves faster than A, so the distance between the ants increases. In frame S′, the separation grows to 50 mm, and the thread breaks.
The paradox arises because, although the separation between the ants remains unchanged in frame S, the thread cannot adjust its internal stresses instantaneously. From the thread's point of view, it is being stretched, and that tension causes it to snap.
Lorentz contraction can also be understood geometrically using Minkowski spacetime.
The Lorentz transformation equations are:
$$ x' = \gamma (x - \beta ct) $$ $$ ct' = \gamma (ct - \beta x) $$where γ and β are defined by:
$$ \gamma = \frac{1}{\sqrt{1-\beta^2}} $$ $$ \beta = \frac{v}{c} $$With v = 24 mm/s and c = 30 mm/s, these become:
$$ x' = \frac{5}{3} \bigl(x - \tfrac{4}{5} ct\bigr) $$ $$ ct' = \frac{5}{3} \bigl(ct - \tfrac{4}{5} x\bigr) $$In the spacetime diagram, ant A follows world line A, and ant B follows world line B. In frame S, the distance between them is represented by segment OR. In frame S′, the corresponding distance is represented by segment OR′.
Similarly, the caterpillar's proper length is represented by segment OQ in its rest frame, while its contracted length is represented by segment OQ′ in the desk frame.
The important geometric feature is the cross-shaped structure formed by segments PP′ and QQ′. This structure captures the essence of how the Lorentz transformation reshapes space and time.
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