= Solution
Use the $+---$ <metric signature>, with the time component of <four-momentum> equal to $E/c$. The incident particle has <energy> $\gamma mc^2$ and <momentum> of magnitude $mc\sqrt{\gamma^2-1}$. Adding the stationary target's <four-momentum> gives total laboratory quantities
$$
E_{\rm tot}=(\gamma+2)mc^2,\qquad |\mathbf p_{\rm tot}|=mc\sqrt{\gamma^2-1}.
$$
The <centre-of-momentum energy> $W$ follows from the invariant <Minkowski norm> of the total <four-momentum>:
$$
W^2=E_{\rm tot}^2-c^2|\mathbf p_{\rm tot}|^2
=m^2c^4\big[(\gamma+2)^2-(\gamma^2-1)\big]
=(4\gamma+5)m^2c^4.
$$
At the <relativistic collision mass threshold>, the two final particles have zero relative <velocity> in their <centre-of-momentum frame>. Their total <energy> there is therefore $2\alpha mc^2$. <Four-momentum conservation> requires $W\geq2\alpha mc^2$, so
$$
\boxed{\gamma_{\min}=\alpha^2-\frac54.}
$$
The condition $\alpha>3/2$ ensures $\gamma_{\min}>1$. This is the \b[minimum required] <Lorentz factor>; larger values allow final <kinetic energy> in the <centre-of-momentum frame>. Equivalently, the incident laboratory <kinetic energy> at threshold is $(\alpha^2-9/4)mc^2$.
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