= 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 $U^2=c^2$ gives
$$
F\cdot U=\frac{dm}{d\tau}U^2+m\frac{dU}{d\tau}\cdot U
=c^2\frac{dm}{d\tau},
$$
so $\boxed{F\cdot U=0}$ 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> is
$$
A^\mu=\left(\frac{\gamma^4\mathbf v\cdot\mathbf a}{c},\;
\gamma^2\mathbf a+\frac{\gamma^4\mathbf v\cdot\mathbf a}{c^2}\mathbf v\right).
$$
If $\mathbf v\cdot\mathbf a=0$, this reduces to $A=(0,\gamma^2\mathbf a)$, whence
$$
\boxed{A^\mu A_\mu=-\gamma^4a^2,\qquad
\alpha^2\equiv-A^2=\gamma^4a^2.}
$$
Here $\alpha$ 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.
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