 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017/9/a/i/solution

m

v

p = γm v, E = γm c^{2}, γ = (1 - v^{2} / c^{2})^{- 1/2} .

E^{2} - c^{2} ∣ p ∣^{2} = γ^{2} m^{2} c^{4} (1 - v^{2} / c^{2}) = m^{2} c^{4} .

This is the squared Minkowski norm of the four-momentum multiplied by

c^{2}

. The Lorentz transformation preserves that norm, so it has the same value in every inertial frame. In the particle's rest frame, the momentum is zero and the total energy is its rest energy

m c^{2}

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