= Solution
For a massive particle of <rest mass> $m$, its <four-momentum> is $P^\mu=mU^\mu=(E/c,\mathbf p)$, where the <four-velocity> is $U^\mu=\gamma(c,\mathbf v)$, $E=\gamma mc^2$ and $\mathbf p=\gamma m\mathbf v$. With <metric signature> $(-,+,+,+)$, $P\cdot P=-m^2c^2$. A <photon> has <four-momentum> $(E/c,(E/c)\hat{\mathbf n})$, a future-pointing <null vector>. <Four-momentum conservation> states that the total incoming and outgoing <four-momenta> agree for an isolated interaction; this includes both <energy> and <momentum conservation> in every inertial frame.
Take the incoming <photon> direction as $x$. Its <energy> is $\hbar\omega=\mu mc^2$, so the total initial <four-momentum> is
$$
P=mc(1+\mu,\mu,0,0),\qquad -P\cdot P=m^2c^2(1+2\mu).
$$
The <invariant mass> available after <photon absorption> is therefore $M=m\sqrt{1+2\mu}$. In the <centre-of-momentum frame>, the two equal daughters have opposite <momenta>. Their total <energy> is at least their combined <rest energy> $2\alpha mc^2$, with equality precisely when both daughters have zero <momentum> in that frame. Thus the <mass threshold after photon absorption> is
$$
\boxed{\alpha_{\max}=\frac12\sqrt{1+2\mu}.}
$$
The bound is attainable kinematically by taking each daughter <four-momentum> equal to $P/2$; each then has the required <rest mass>. It immediately gives \b[$\alpha_{\max}\to1/2$ as $\mu\to0$].
At this threshold both daughters move with the <centre-of-momentum frame>. The common laboratory <velocity> follows from $\mathbf v=c\mathbf P/P^0$:
$$
\boxed{v=\frac{c\mu}{1+\mu}=\frac{c}{1+\mu^{-1}}.}
$$
For $\mu=0$, its continuous limiting value is zero. At smaller allowed daughter <rest masses>, some of the <centre-of-momentum energy> appears as their relative <kinetic energy>, so their <four-momenta> need not be $P/2$.
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