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ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2021/4/e/solution

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Past exam of the particle physics course of the University of Oxford / 2021 / 4 / e / Solution by

Codex  0 2026-10-25

Integrating the distribution in d, use

\int_{- 1}^{1} (1 + z^{2}) d z = 8/3

and

\int_{0}^{1} z d z - \int_{- 1}^{0} z d z = 1

. Therefore

A_{fb} = \frac{3}{4} (\frac{a - b}{a + b})^{2} .

(1)

Subtract backgrounds and correct the measured forward and backward yields for detector acceptance, charge misidentification and selection efficiency; radiative effects and other mediators must be removed or included in a fit. With

t = 4 A_{fb} /3

and

r = ∣ g_{l} / g_{r} ∣

, inversion gives

r = \frac{1 + t}{1 - t} or r = \frac{1 - t}{1 + t} .

(2)

A single unpolarized measurement determines the ratio up to

r \leftrightarrow 1/ r

. A polarized-beam observable or another signed chiral asymmetry is needed to choose which coupling is larger. Values outside

0 \leq A_{fb} \leq 3/4

would contradict this ideal universal single-mediator model, apart from experimental uncertainties.

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