γ+ M → e^- + e^+ + M In pair-production, a photon of energy is absorbed in the neighborhood of a particle of mass M to make an e^-e^+ pair, each with mass m_e. The threshold photon energy for this process is known to be h*nu_min = 2m_e*c^2 for M→∞ h*nu_min = 4m_e*c^2 for M→m_e Derive a formula for h*nu_min that is valid for any M, and show that the result gives agreement in these two special cases. Hint: Consider the simplest case of all three particles moving the Lab frame in the same direction with the same speed u. From the hint it sounds like I should be using conservation of momentum or energy for the derivation. What I've come to for momentum is: h*nu/c + M*u*c = 2m_e*u*c + M*u*c which goes to h*nu + M*u*c^2 = 2m_e*u*c^2 + M*u*c^2 Based upon my class work, the velocity of the particle with mass M is not changed in the pair production (it is an "essential spectator"). This gets me to something close to the first equation as the M*u*c drops from each side of the equation, but I have an extra u hanging around on the RHS. So I tried energy: h*nu + K_m + M*c^2 = 2(m_e*c^2 + K_m_e) + K_m Where K_m is the kinetic energy of the particle with mass M and K_m_e is the kinetic energy of the electrons after the pair-production. I can sub in the usual 0.5mu^2 for the kinetic energies (with respectively correct masses, M and m_e, for m). But after canceling K_m from both sides, I don't have something that can satisfy conditions for M as above. I feel like I'm missing something rather straightforward, but since I am a mathematician and not a physicist I feel like I don't have the intuition to make the leap. Can someone provide a way ahead for this problem? Thanks.