 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/12c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Along the particle's worldline,

x = u_{x} t, y = u_{y} t .

The boost from

S

S^{'}

gives

x^{'} = γ (x - V t), y^{'} = y, t^{'} = γ (t - \frac{V x}{c ^{2}}) .

Therefore

\frac{d t ^{'}}{d t} = γ (1 - \frac{V u _{x}}{c ^{2}}),

and division of the transformed coordinate velocities by this factor gives the relativistic velocity-addition formula

u_{x}^{'} = \frac{u _{x} - V}{1 - V u _{x} / c ^{2}}, u_{y}^{'} = \frac{u _{y}}{γ ( 1 - V u _{x} / c ^{2} )}, u_{z}^{'} = 0 .

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