 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-4/10a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

P = (P^{0}, P) = (γm c, γm u), γ = (1 - ∣ u ∣^{2} / c^{2})^{- 1/2} .

Thus

P^{0} = m c + \frac{m ∣ u ∣ ^{2}}{2 c} + \frac{3 m ∣ u ∣ ^{4}}{8 c ^{3}} + O (∣ u ∣^{6} / c^{5}) .

c

identifies the first two terms as the rest energy

m c^{2}

and the Newtonian kinetic energy

m ∣ u ∣^{2} /2

For the fixed-target reaction, the initial invariant is

(P_{1} + P_{2})^{2} = 2 m^{2} c^{2} (1 + γ) .

At threshold the four final particles are at rest in their centre-of-momentum frame, giving invariant

(4 m c)^{2}

. Hence

γ = 7

and the least laboratory speed is

u = c 1 - \frac{1}{49} = \frac{4 3}{7} c .

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