Past exam of the short options course of the University of Oxford / 2023 / 8 / b / i

Table of contents
- Solution i

Solution

For the collinear boosts specified in the question, multiplication of the two

2

-by-

2

blocks and the hyperbolic addition formulas give

(cosh θ sinh θ sinh θ cosh θ) (cosh θ^{'} sinh θ^{'} sinh θ^{'} cosh θ^{'}) = (cosh (θ + θ^{'}) sinh (θ + θ^{'}) sinh (θ + θ^{'}) cosh (θ + θ^{'})) .

Thus the resulting rapidity is

θ^{''} = θ + θ^{'}

. Collinearity matters: boosts in different spatial directions do not in general combine into a pure boost of this form.

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