= Solution
For motion along one spatial axis, introduce the <rapidity> $\eta$ so that the <four-velocity> is $u^\mu=c(\cosh\eta,\sinh\eta,0,0)$. Its <four-acceleration> is
$$
a^\mu=c\frac{d\eta}{d\tau}(\sinh\eta,\cosh\eta,0,0),\qquad a_\mu a^\mu=-c^2\left(\frac{d\eta}{d\tau}\right)^2.
$$
A constant signed <proper acceleration> $g$ therefore gives $d\eta/d\tau=g/c$, hence $\eta=\eta_0+g\tau/c$. Differentiating the <four-velocity> twice yields
$$
\boxed{\frac{d^2u^\mu}{d\tau^2}=\frac{g^2}{c^2}u^\mu.}
$$
The zero transverse components also satisfy this equation. The constant acceleration measured on board is the <proper acceleration>, rather than a constant coordinate acceleration in the laboratory.
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