MadSci Network: Physics |
Actually, the formula you give is an approximation. The short answer is that in order to conserve energy AND momentum, the nucleus must recoil. Consider the case where the initial photon is traveling in the z-direction, the nucleus is initially at rest, and the created electron and positron have zero momentum. The photon is destroyed in the reaction, the electron/positron are not moving, so the final nucleus momentum must be equal to that of the photon. However, since the nucleus is moving, it also gained some kinetic energy. Thus the initial photon needs to have energy equal to twice the electron rest mass PLUS a little extra for the nucleus. Here is the mathematical derivation for the situation described above. I use units such that c=1. p_g == initial photon momentum == E_g == initial photon energy p_N == final nucleus momentum E_N == final nucleus momentum = sqrt(M^2+p_N^2) M == nucleus rest energy, m==electron rest energy Conservation of momentum in z-direction: p_g = p_N, or E_g = p_N (remember c=1) Conservation of energy: M + E_g = sqrt(M^2+p_N^2) + 2m Substituting from the momentum eqn: M + E_g = sqrt(M^2+E_g^2) + 2m Solving for E_g: E_g = 2m(1-m/M) ------- (1-2m/M) This is the exact answer. Note that M>>m, so we can use the binomial expansion: 1/(1-2m/M) = 1 + 2m/M + ... E_g = 2m(1-m/M)(1+2m/M+...) = 2m(1 + 2m/M - m/m - 2m^2/M^2 + ...) Dropping all terms of order m^2/M^2 and higher, we arrive at E_g ~= 2m(1+m/M) Putting c back in E_g ~= 2mc^2(1+m/M)
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