Immediate acceleration

Minkowski diagram: Length ${\displaystyle L'}$ between the ships in S′ after acceleration is longer than the previous length ${\displaystyle L'_{old}}$ in S′, and longer than the unchanged length ${\displaystyle L}$ in S. The thin lines are "lines of simultaneity".

Loedel diagram of the same scenario

Similarly, in the case of Bell's spaceship paradox the relation between the initial rest length ${\displaystyle L}$ between the ships (identical to the moving length in S after acceleration) and the new rest length ${\displaystyle L'}$ in S′ after acceleration, is:^{[3][6][8][16]}

${\displaystyle L'=\gamma L}$.

This length increase can be calculated in different ways. For instance, if the acceleration is finished the ships will constantly remain at the same location in the final rest frame S′, so it's only necessary to compute the distance between the x-coordinates transformed from S to S′. If ${\displaystyle x_{A}}$ and ${\displaystyle x_{B}=x_{A}+L}$ are the ships' positions in S, the positions in their new rest frame S′ are:^[3]

${\displaystyle {\begin{aligned}x'_{A}&=\gamma \left(x_{A}-vt\right)\\x'_{B}&=\gamma \left(x_{A}+L-vt\right)\\L'&=x'_{B}-x'_{A}\\&=\gamma L\end{aligned}}}$

Another method was shown by Dewan (1963) who demonstrated the importance of relativity of simultaneity.^[8] The perspective of frame S′ is described, in which both ships will be at rest after the acceleration is finished. The ships are accelerating simultaneously at ${\displaystyle t_{A}=t_{B}}$ in S (assuming acceleration in infinitesimal small time), though B is accelerating and stopping in S′ before A due to relativity of simultaneity, with the time difference:

${\displaystyle {\begin{aligned}\Delta t'&=t'_{B}-t'_{A}=\gamma \left(t_{B}-{\frac {vx_{B}}{c^{2}}}\right)-\gamma \left(t_{A}-{\frac {vx_{A}}{c^{2}}}\right)\\&={\frac {\gamma vL}{c^{2}}}\end{aligned}}}$

Since the ships are moving with the same velocity in S′ before acceleration, the initial rest length ${\displaystyle L}$ in S is shortened in S′ by ${\displaystyle L'_{old}=L/\gamma }$ due to length contraction. From the frame of S′, B starts accelerating before A and also stops accelerating before A. Due to this B will always have higher velocity than A up until the moment A is finished accelerating too, and both of them are at rest with respect to S′. The distance between B and A keeps on increasing till A stops accelerating. Although A's acceleration timeline is delayed by an offset of ${\displaystyle \Delta t'}$, both A and B cover the same distance in their respective accelerations. But B's timeline contains acceleration and also being at rest in S` for ${\displaystyle \Delta t'}$ till A stops accelerating. Hence the extra distance covered by B during the entire course can be calculated by measuring the distance traveled by B during this phase. Dewan arrived at the relation (in different notation):^[8]

${\displaystyle {\begin{aligned}L'&=L'_{old}+v\Delta t'={\frac {L}{\gamma }}+{\frac {\gamma v^{2}L}{c^{2}}}\\&=\gamma L\end{aligned}}}$

It was also noted by several authors that the constant length in S and the increased length in S′ is consistent with the length contraction formula ${\displaystyle L=L'/\gamma }$, because the initial rest length ${\displaystyle L}$ is increased by ${\displaystyle \gamma }$ in S′, which is contracted in S by the same factor, so it stays the same in S:^[6][14][18]

${\displaystyle L_{contr.}=L'/\gamma =\gamma L/\gamma =L}$

Summarizing: While the rest distance between the ships increases to ${\displaystyle \gamma L}$ in S′, the relativity principle requires that the string (whose physical constitution is unaltered) maintains its rest length ${\displaystyle L}$ in its new rest system S′. Therefore, it breaks in S′ due to the increasing distance between the ships. As explained above, the same is also obtained by only considering the start frame S using length contraction of the string (or the contraction of its moving molecular fields) while the distance between the ships stays the same due to equal acceleration.

Constant proper acceleration

The world lines (navy blue curves) of two observers A and B who accelerate in the same direction with the same constant magnitude proper acceleration (hyperbolic motion). At A′ and B′, the observers stop accelerating.

Two observers in Born rigid acceleration, having the same Rindler horizon. They can choose the proper time of one of them as the coordinate time of the Rindler frame.

Two observers having the same proper acceleration (Bell's spaceships). They are not at rest in the same Rindler frame, and therefore have different Rindler horizons

Instead of instantaneous changes of direction, special relativity also allows to describe the more realistic scenario of constant proper acceleration, i.e. the acceleration indicated by a comoving accelerometer. This leads to hyperbolic motion, in which the observer continuously changes momentary inertial frames^[34]

${\displaystyle {\begin{aligned}x&={\frac {c^{2}}{\alpha }}\left({\sqrt {1+\left({\frac {\alpha t}{c}}\right)^{2}}}-1\right)={\frac {c^{2}}{\alpha }}\left(\cosh {\frac {\alpha \tau }{c}}-1\right)\\c\tau &={\frac {c^{2}}{\alpha }}\operatorname {asinh} {\frac {\alpha t}{c}},\quad ct={\frac {c^{2}}{\alpha }}\sinh {\frac {\alpha \tau }{c}}\end{aligned}}}$

where ${\displaystyle t}$ is the coordinate time in the external inertial frame, and ${\displaystyle \tau }$ the proper time in the momentary frame, and the momentary velocity is given by

${\displaystyle v={\frac {\alpha t}{\sqrt {1+\left({\frac {\alpha t}{c}}\right)^{2}}}}=c\tanh {\frac {\alpha \tau }{c}}}$

The mathematical treatment of this paradox is similar to the treatment of Born rigid motion. However, rather than ask about the separation of spaceships with the same acceleration in an inertial frame, the problem of Born rigid motion asks, "What acceleration profile is required by the second spaceship so that the distance between the spaceships remains constant in their proper frame?"^[35][34][36] In order for the two spaceships, initially at rest in an inertial frame, to maintain a constant proper distance, the lead spaceship must have a lower proper acceleration.^[3][36][37]

This Born rigid frame can be described by using Rindler coordinates (Kottler-Møller coordinates)^[34][38]

${\displaystyle {\begin{aligned}ct&=\left(x'+{\frac {c^{2}}{\alpha }}\right)\sinh {\frac {\alpha t'}{c}},&y&=y',\\x&=\left(x'+{\frac {c^{2}}{\alpha }}\right)\cosh {\frac {\alpha t'}{c}}-{\frac {c^{2}}{\alpha }},&z&=z'.\end{aligned}}\ (t'=\tau )}$

The condition of Born rigidity requires that the proper acceleration of the spaceships differs by^[38]

${\displaystyle \alpha _{2}={\frac {\alpha _{1}}{1+{\frac {\alpha _{1}L'}{c^{2}}}}}}$

and the length ${\displaystyle L'=x_{2}^{\prime }-x_{1}^{\prime }}$ measured in the Rindler frame (or momentary inertial frame) by one of the observers is Lorentz contracted to ${\displaystyle L=x_{2}-x_{1}}$ in the external inertial frame by^[38]

${\displaystyle L={\frac {L'}{\cosh {\frac {\alpha t'}{c}}}}=L'{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$

which is the same result as above. Consequently, in the case of Born rigidity, the constancy of length L' in the momentary frame implies that L in the external frame decreases constantly, the thread doesn't break. However, in the case of Bell's spaceship paradox the condition of Born rigidity is broken, because the constancy of length L in the external frame implies that L' in the momentary frame increases, the thread breaks (in addition, the expression for the distance increase between two observers having the same proper acceleration becomes also more complicated in the momentary frame^[17]).