 Solution

ID: past-exam-of-the-relativity-course-of-the-university-of-cambridge/2022/section-b/6/b/ii/solution

t^{'} = γ (t - vx)

x^{'} = γ (x - v t)

in units

c = 1

. Adding and subtracting gives

p^{'} = γ (1 + v) (t - x), q^{'} = γ (1 - v) (t + x) .

Since

γ (1 + v) = (1 + v) / (1 - v)

and the other factor is its reciprocal,

p^{'} = α p, q^{'} = q / α, α = \frac{1 + v}{1 - v} = e^{η} .

Here

η

is the boost rapidity. Their product is unchanged, as required by invariance of the Minkowski interval.

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