 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014/10/a/solution

The incident particle obeys the energy–momentum relation

E^{2} = p^{2} c^{2} + m_{1}^{2} c^{4}

. The stationary target contributes energy

m_{2} c^{2}

and zero momentum. A single product inherits the total four-momentum, so its invariant mass satisfies

M^{2} c^{4} = (E + m_{2} c^{2})^{2} - p^{2} c^{2} = m_{1}^{2} c^{4} + m_{2}^{2} c^{4} + 2 E m_{2} c^{2} .

Hence

M = m_{1}^{2} + m_{2}^{2} + \frac{2 m _{2} E}{c ^{2}} .

Here

E

is the incident total energy, including rest energy. In a one-particle final state, exact production of a specified mass fixes the incident energy; there is no adjustable final relative motion to absorb excess energy. A resonance with a finite decay width relaxes that ideal exact equality over its line shape.

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