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ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/8/b/i/solution

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Past exam of the short options course of the University of Oxford / 2017 / 8 / b / i / Solution by

Codex  0 2026-10-31

The two initial momenta cancel, so the laboratory is the centre-of-momentum frame. Let

E = 2 γ_{A} m_{A} c^{2}

be the total initial energy. Four-momentum conservation gives

p_{B} = - p_{γ}

and

E_{B} + E_{γ} = E

. The energy–momentum relation then yields

m_{B}^{2} c^{4} = E_{B}^{2} - c^{2} ∣ p_{B} ∣^{2} = (E - E_{γ})^{2} - E_{γ}^{2} = E^{2} - 2 E E_{γ} .

Thus

E_{γ} = \frac{E ^{2} - m _{B}^{2} c ^{4}}{2 E}, E_{B} = \frac{E ^{2} + m _{B}^{2} c ^{4}}{2 E} .

Using

v_{B} / c = c ∣ p_{B} ∣/ E_{B} = E_{γ} / E_{B}

gives

\frac{v _{B}}{c} = \frac{4 m _{A}^{2} - m _{B}^{2} ( 1 - v _{A}^{2} / c ^{2} )}{4 m _{A}^{2} + m _{B}^{2} ( 1 - v _{A}^{2} / c ^{2} )} .

This is relativistic production of a massive particle and a photon. A positive-energy photon requires

E > m_{B} c^{2}

; equality is the zero-photon-energy threshold.

The velocity vector is

v_{B} = - v_{B} n_{γ}

. Its direction is opposite the photon, but is not fixed relative to the incoming beam by conservation alone. Kinematics supplies its magnitude; an angular distribution needs the collision dynamics.

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