= Solution
With metric signature $(+,-,-,-)$, <four-momentum> is $p^\mu=(E/c,\mathbf p)$; for a particle of <rest mass> $m$ it satisfies $p^\mu p_\mu=m^2c^2$. For a massive particle $p^\mu=mu^\mu$, where $u^\mu$ is its four-velocity. The total four-momentum is the sum of the individual four-momenta. If that sum is timelike, its center-of-momentum frame exists and the total energy in that frame is
$$
\boxed{W=\sqrt{(E_1+E_2)^2-c^2|\mathbf p_1+\mathbf p_2|^2}.}
$$
This follows because the spatial momentum is zero in that frame and the Minkowski norm is invariant under a <Lorentz boost>. A null total momentum, such as parallel photons, would not have such a rest frame.
For the photon striking an electron initially at rest, the initial invariant gives $W^2=m_e^2c^4+2m_ec^2E_\gamma$. The three final particles have total rest energy $3m_ec^2$. Their center-of-momentum energy is at least this value, with equality when all have zero relative momentum. Energy-momentum conservation therefore requires
$$
m_e^2c^4+2m_ec^2E_\gamma\geq9m_e^2c^4,
\qquad
\boxed{E_\gamma\geq4m_ec^2.}
$$
The <photon-induced triplet pair-production threshold> is consequently \b[four electron rest-energy units]. This is attainable as a kinematic threshold: at equality the center-of-momentum speed in the laboratory is $4c/5$, and boosting the three resting final particles gives each energy $5m_ec^2/3$ and momentum $4m_ec/3$. Their totals are the initial $5m_ec^2$ and $4m_ec$. Recoil of the original electron is why twice the rest energy is insufficient.
Back to article page
