 Solution

ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2022/2/b/solution

p_{π} = p_{1} + p_{2}

. With the

(-, +, +, +)

p_{π}^{2} = - M_{π}^{2}

, while each photon has

p_{i}^{2} = 0

and

∣ p_{i} ∣ = E_{i}

. Taking the Minkowski norm gives

- M_{π}^{2} = 2 p_{1} \cdot p_{2} = 2 (- E_{1} E_{2} + p_{1} \cdot p_{2}) = - 2 E_{1} E_{2} (1 - cos A) .

The half-angle identity therefore yields

sin^{2} (A /2) = \frac{M _{π}^{2}}{4 E _{1} E _{2}} .

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