 Solution

Along a massive particle's world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}$. A pure force preserves its rest mass. Differentiating $P=mU$ and $U^2=c^2$ givesso $\boxed{F\cdot U=0}$ for a pure force. This is a restriction on rest-mass transfer, not a statement that its work in every frame vanishes.

Write $\mathbf{a}=d\mathbf{v}/dt$. Since $\dot{\gamma }={\gamma }^3\mathbf{v}\cdot \mathbf{a}/c^2$, the four-acceleration isIf $\mathbf{v}\cdot \mathbf{a}=0$, this reduces to $A=(0,{\gamma }^2\mathbf{a})$, whenceHere $\alpha$ is the proper acceleration. The positive expression printed in the source is $A^2$ in the opposite metric signature, or the squared acceleration magnitude in ours. For example, at rest with nonzero acceleration, $A=(0,\mathbf{a})$ has $A^2=-a^2$ in our convention. Thus one must not combine the printed positive sign with a mostly-minus norm.