 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-4/4c/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

With

γ (v) = (1 - v^{2} / c^{2})^{- 1/2}

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

Adding and subtracting

c t^{'}

gives

x_{+}^{'} = γ (1 - v / c) x_{+} = \frac{c - v}{c + v}, x_{+},

x_{-}^{'} = γ (1 + v / c) x_{-} = \frac{c + v}{c - v}, x_{-} .

The product of the two multipliers is one, so

x_{+}^{'} x_{-}^{'} = x_{+} x_{-}

, which is

x^{'2} - c^{2} t^{'2} = x^{2} - c^{2} t^{2}

Successive transformations multiply the

x_{+}

factors. Equating

λ (v_{3}) = λ (v_{2}) λ (v_{1})

and solving gives the relativistic velocity-addition law

v_{3} = \frac{v _{1} + v _{2}}{1 + v _{1} v _{2} / c ^{2}} .

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