Past exam of the relativity course of the University of Cambridge / 2019 / Section A / 2 / 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 isA 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 yieldsThe 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.