 Solution

Use $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. Since $d\tau =dt/\gamma$, the four-velocity and four-momentum areThese use the proper time and Lorentz factor; the mass shell identity is $E^2-c^2|\mathbf{p}{|}^2=M^2{c}^4$.

Initially the parent four-momentum is $(Mc,0)$. By four-momentum conservation, the remaining single particle has energy $M{c}^2-E$ and three-momentum $-\mathbf{p}$. Applying its mass shell identity gives $m^2{c}^4=(M{c}^2-E{)}^2-c^2|\mathbf{p}{|}^2$, henceThe positive root is selected for the rest mass; physically admissible decay data also give nonnegative remaining energy.