 Solution

ID: past-exam-of-the-flows-fluctuations-and-complexity-and-symmetry-and-relativity-paper-of-part-b-of-the-physics-course-of-the-university-of-oxford/2011/7/1/solution

The official notes give only “Part 1: as in textbooks” (PDF p. 13, handwritten p. 8). The requested derivation is supplied here.

S

and emit two pulses simultaneously toward its ends, each a distance

ℓ

away. In

S

, the reflection events are

(t, x) = (ℓ / c, \pm ℓ)

x^{'} = A (x - v t)

t^{'} = A (t - vx / c^{2})

. Applying the inverse transformation with speed

- v

gives

A^{2} (1 - v^{2} / c^{2}) = 1

, hence

A = γ

The transformed reflection events have

x_{+}^{'} = γ ℓ (1 - β), t_{+}^{'} = \frac{γ ℓ}{c} (1 - β), x_{-}^{'} = - γ ℓ (1 + β), t_{-}^{'} = \frac{γ ℓ}{c} (1 + β) .

These events are not simultaneous in

S^{'}

. Both rod ends move at

- v

there, so extrapolating each endpoint to

t^{'} = 0

gives

x_{\pm}^{'} (0) = x_{\pm}^{'} + v t_{\pm}^{'} = \pm ℓ / γ

. Measuring simultaneous endpoints therefore yields length contraction:

L^{'} = x_{+}^{'} (0) - x_{-}^{'} (0) = \frac{2 ℓ}{γ} = 2 ℓ 1 - β^{2} .

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