 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-2/7b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

With coincident origins and aligned axes, the Lorentz transformation is

x^{'} = γ_{v} (x - v t), t^{'} = γ_{v} (t - \frac{vx}{c ^{2}}), γ_{v} = (1 - v^{2} / c^{2})^{- 1/2} .

Its inverse replaces

v

- v

. Along the worldline of the origin of

S^{''}

d x^{'} / d t^{'} = u

. Differentiating the inverse Lorentz transformation gives

d x = γ_{v} (d x^{'} + v d t^{'}), d t = γ_{v} (d t^{'} + v, d x^{'} / c^{2}) .

V = \frac{u + v}{1 + uv / c ^{2}} .

For

∣ u ∣, ∣ v ∣ ≪ c

, the denominator tends to one, recovering $V ≃ u + v$ . Simultaneously

γ_{v} \to 1

, so

x^{'} ≃ x - v t

and

t^{'} ≃ t

: these are the Galilean transformation and its Newtonian velocity rule.

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