 Solution

ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2022/3/b/solution

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

Divide each spatial displacement by

d t^{'} = γ_{v} d t (1 - v u_{x} / c^{2})

w_{x} = \frac{u _{x} - v}{1 - v u _{x} / c ^{2}}, w_{y} = \frac{u _{y}}{γ _{v} ( 1 - v u _{x} / c ^{2} )}, w_{z} = \frac{u _{z}}{γ _{v} ( 1 - v u _{x} / c ^{2} )} .

The same result follows by transforming the four-velocity and using

w_{i} = c U^{' i} / U^{'0}

. The transverse velocities change because the transformed time increment depends on

d x

, even though the transverse spatial coordinates are unchanged.

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