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ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2021/section-b/5/c/solution

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Past exam of the particle and nuclear physics course of the University of Cambridge / 2021 / Section B / 5 / c / Solution by

Codex  0 2026-10-18

In the simple box-state counting used here, the final three-momentum lies in a spherical shell:

d n = C p^{2} d p

, where the common constant

C

contains the normalization volume and angular factor. Momentum conservation fixes the other final particle's momentum. Hence the density of states is

ρ_{ℓ} (E_{0}) = C p_{ℓ}^{2} \frac{d p _{ℓ}}{d E _{0}} .

(1)

Combining parts (a) and (b) gives

\frac{ρ _{μ}}{ρ _{e}} = [\frac{m _{π}^{2} - m _{μ}^{2}}{m _{π}^{2} - m _{e}^{2}}]^{2} \frac{m _{π}^{2} + m _{μ}^{2}}{m _{π}^{2} + m _{e}^{2}} .

(2)

With the paper's rounded masses,

R_{naive} = (\frac{19600 - 11236}{19600 - 0.25})^{2} \frac{19600 + 11236}{19600 + 0.25} = 0.28650.

(3)

By Fermi's golden rule, equal squared matrix elements make the rate ratio equal to this density ratio. Thus the naive constant-matrix-element estimate is $R ≃ 0.287$ , which favours the electronic decay by about

3.49

to one.

This is the particular noncovariant state-counting convention in the question. A calculation with the Lorentz-invariant phase-space measure places factors

1/ (2 E)

in the measure and correspondingly normalizes its decay amplitude differently. Only the complete product of phase space and squared matrix element is convention-independent; mixing the two conventions would give a wrong rate.

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