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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 4C / b / Solution

Again set

c = 1

and write

X_{A} = x_{C} - x_{A}, T_{A} = t_{C} - t_{A}, X_{B} = x_{C} - x_{B}, T_{B} = t_{C} - t_{B} .

A boost of velocity

v

gives

X_{A}^{'} = γ (X_{A} - v T_{A}), X_{B}^{'} = γ (X_{B} - v T_{B}) .

Define

D (v) = (X_{A} - v T_{A})^{2} - (X_{B} - v T_{B})^{2} .

The hypothesis says

D (0) < 0

. Put

ξ = x_{B} - x_{A}

and

τ = t_{B} - t_{A}

. Causality gives

τ \geq ∣ ξ ∣

, while

T_{B} \geq ∣ X_{B} ∣

, and

T_{A} = T_{B} + τ

,

X_{A} = X_{B} + ξ

. Therefore

T_{A}^{2} + X_{A}^{2} - (T_{B}^{2} + X_{B}^{2}) = 2 T_{B} τ + 2 X_{B} ξ + τ^{2} + ξ^{2} > 0.

But

D (1) + D (- 1) = 2 {T_{A}^{2} + X_{A}^{2} - (T_{B}^{2} + X_{B}^{2})} > 0.

Hence at least one of

D (1), D (- 1)

is positive. By continuity,

D (v) > 0

for some physical velocity

∣ v ∣ < 1

sufficiently close to that endpoint. In the corresponding inertial frame,

∣ x_{C}^{'} - x_{A}^{'} ∣ > ∣ x_{C}^{'} - x_{B}^{'} ∣,

as required.

Solved by gpt-5.6-sol high.

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