Hint 6

The minimum total energy is required for this process when the proton and anti-proton are produced at rest relative to each other. In the c.m. frame this means that both particles are produced at rest and so the total 4-momentum is $(0,0,0,i2m_pc)$. This is conserved and so is equal to the total 4-momentum of the electron and positron. In this frame the electron and positron have equal and opposite 3-momenta and (since their masses are the same) equal energies of $m_pc^2$. The kinetic energy of one of them is thus $T=(m_p-m_e)c^2=937.8$ MeV.

The square of the total 4-momentum can be easily calculated in the c.m. frame, where the protons are produced at rest. It is equal to $-(2m_pc)^2$. This quantity is an invariant and so its value is the same in all inertial frames. In the frame where the electron is at rest, the 4-momenta of the positron and electron are, respectively,

$${\sf p}_{e+}=(p_{e+},\,0,\,0,\,iE_{e+}/c)\quad\hbox{and}\quad {\sf p}_{e-}=(0,\,0,\,0,\,im_e c).$$

Squaring their sum, and using $p^2c^2-E^2=-m^2c^4$ (or ${\sf p}^2 =-m^2c^2$ and ${\sf p}_{e+}\cdot{\sf p}_{e-}=-E_{e+}m_e$), gives

$$-2E_{e+}m_e-2m_e^2c^2=-4m_p^2c^2.$$

This can now be solved to get

$$E_{e+}={2m_p^2-m_e^2\over m_e}c^2=3.45\ \hbox{TeV},$$

which is essentially all kinetic energy.