Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 1 / c

The proper acceleration is $\alpha =\sqrt{-A^2}$, the magnitude of the instantaneous rest-frame three-acceleration. A pure force is a four-force preserving rest mass, or equivalently satisfying $F\cdot U=0$. Indeed $F=d(mU)/d\tau$ gives $F\cdot U=c^{2}dm/d\tau$.

For the right-hand branch of the hyperbola, use $c=1$, $x>0$ and $x=\sqrt{L^2+t^2}$, taking $L>0$. Direct differentiation givesThe sign of $v$ follows the sign of $t$, whereas the coordinate acceleration on this branch is positive. ConsequentlyThe proper-time parameterization $t=L\sinh (\tau /L)$, $x=L\cosh (\tau /L)$ independently gives constant proper acceleration. Restoring dimensions, a hyperbola $x^2-c^2{t}^2=L^2$ has $\alpha =c^2/L$. Constant proper acceleration does not mean that all inertial-frame components of the four-force are constant.