= Solution
Let the fixed direction of motion be the unit vector $\mathbf n$ and decompose $\mathbf s=s_\parallel\mathbf n+\mathbf s_\perp$. In <2/c/ii> the two parallel contributions cancel, whereas the transverse equation is
$$
\dot s_\parallel=0,\qquad
\dot{\mathbf s}_\perp=-\frac{\gamma^2v\dot v}{c^2}\mathbf s_\perp
=-\frac{\dot\gamma}{\gamma}\mathbf s_\perp.
$$
For the stated speed law, $1-v^2/c^2=e^{-2\Gamma\tau}$, hence $\gamma=e^{\Gamma\tau}$. Since the particle is initially at rest, $\gamma(0)=1$, and
$$
\boxed{\mathbf s(\tau)=s_\parallel(0)\mathbf n
+e^{-\Gamma\tau}\mathbf s_\perp(0).}
$$
The corresponding <spin four-vector> has components $s^0=\gamma\beta s_\parallel(0)$, $s^{\rm spatial}_\parallel=\gamma s_\parallel(0)$, and $\mathbf s^{\rm spatial}_\perp=\mathbf s_\perp(0)$. Its squared norm is therefore $s_\parallel(0)^2+|\mathbf s_\perp(0)|^2$, independent of time. The transverse decay of the source's laboratory three-vector is a boost-normalization effect, not a loss of proper <angular momentum> or a torque.
For a real increasing speed take $\Gamma>0$ and $\tau\geq0$. The prescribed profile has a singular initial acceleration: $v\sim c\sqrt{2\Gamma\tau}$ near zero. The differential transport calculation applies for $\tau>0$ and the solution has the continuous initial limit displayed above. This trajectory should not be silently described as motion at constant <proper acceleration>.
Back to article page
