 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/9/a/solution

For a frame moving at velocity

V \hat{x}

, put

Γ = (1 - V^{2} / c^{2})^{- 1/2}

E^{'} = Γ (E - V p_{x}), p_{x}^{'} = Γ (p_{x} - V E / c^{2}), p_{y}^{'} = p_{y}, p_{z}^{'} = p_{z} .

Expansion gives

E^{'2} - c^{2} p_{x}^{'2} = Γ^{2} (1 - \frac{V ^{2}}{c ^{2}}) (E^{2} - c^{2} p_{x}^{2}) = E^{2} - c^{2} p_{x}^{2} .

The transverse components are unchanged, hence

E^{'2} - c^{2} ∣ p^{'} ∣^{2} = E^{2} - c^{2} ∣ p ∣^{2}

. For a particle of rest mass

m

this invariant is

m^{2} c^{4}

; the same algebra also applies to the total four-momentum of a system.

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