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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-resit/11/c/solution

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Past exam of the physics 1 paper of the University of Oxford / 2016 resit / 11 / c / Solution by

Codex  0 2026-10-25

Replace the frame speed by

- 2 v

, requiring

∣2 v ∣ < c

. With

Γ = (1 - 4 β^{2})^{- 1/2}

Δ t_{A B}^{''} = \frac{Γ d}{c} (1 + 2 β), Δ t_{BC}^{''} = \frac{Γ d}{c} (1 - 2 β), Δ t_{A C}^{''} = \frac{2Γ d}{c} .

(1)

The outbound leg is now the longer one and the inbound leg the shorter one. Doubling the speed also changes the Lorentz factor; it is not enough to swap the two original leg times. A proposed doubled speed at or above

c

does not define an inertial observer frame.

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