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ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2019/2/a/solution

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Past exam of the symmetry and relativity course of the University of Oxford / 2019 / 2 / a / Solution by

Codex  0 2026-10-25

Use the inverse standard Lorentz boost,

t = γ_{v} (t^{'} + v x^{'} / c^{2})

and

x = γ_{v} (x^{'} + v t^{'})

. Label two rod endpoints by a separation

Δ x^{'} = L

; their world lines satisfy

y^{'} = u t^{'}

. A rod's orientation in

S

must compare endpoint events simultaneous in

S

, not in

S^{'}

. Consequently

0 = Δ t = γ_{v} (Δ t^{'} + \frac{vL}{c ^{2}}), Δ t^{'} = - \frac{vL}{c ^{2}} .

Then

Δ x = γ_{v} (L + v Δ t^{'}) = \frac{L}{γ _{v}}, Δ y = u Δ t^{'} = - \frac{uvL}{c ^{2}} .

The signed counterclockwise angle therefore satisfies

tan θ = - γ_{v} \frac{uv}{c ^{2}} .

For positive

u, v

, the rod slopes downward to the right, so the acute clockwise angle is

arctan (γ_{v} uv / c^{2})

. This is a relativity of simultaneity effect in addition to the longitudinal length contraction; the rod is not being mechanically bent by the coordinate change.

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