 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019/10/c/solution

At threshold, the pion is at rest in the final centre-of-momentum frame, which moves in the laboratory with

β = \frac{E _{γ}^{th}}{m _{p} c ^{2} + E _{γ}^{th}} = 0.1336592 \dots, γ = \frac{m _{p} c ^{2} + E _{γ}^{th}}{( m _{p} + m _{π} ) c ^{2}} = 1.0090539 \dots .

m_{π} c^{2} /2

. The specified forward and backward directions remain collinear under this boost, so

E_{\pm} = \frac{m _{π} c ^{2}}{2} γ (1 \pm β) .

At threshold,

γ (1 + β) = (m_{p} + m_{π}) / m_{p}

and

γ (1 - β) = m_{p} / (m_{p} + m_{π})

. Hence

E_{+} = 77.2148 MeV, E_{-} = 59.0075 MeV .

The forward photon is the more energetic one. Their product remains

(m_{π} c^{2} /2)^{2}

, as required by the invariant mass of an oppositely directed photon pair, while their sum equals the laboratory pion energy.

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