Problem Set 5 Solutions

# 24 may be obtained from eqn23 by working in the

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Unformatted text preview: q ) ￿p (1 − γ 5 ) v (k )) (21) (22) (b.) In order to obtain the decay rate we need to ﬁnd the square of the absolute value of the matrix element (22) and sum it over all polarizations of the outgoing leptons. This involves working out trace relationships. It is easy to see that: ￿ ￿ ￿ |M|2 = 8 (GF fπ )2 2 (k · p) (q · p) − (k · q ) p2 (23) pol = 4 m2 m2 (GF fπ )2 π l ￿ 1− m2 l m2 π ￿ (24) Eqn.(24) may be obtained from eqn.(23) by working in the center of mass frame so that p = (mπ , 0) and − − = −→. Since the neutrino is practically massless one also uses the relationship q 2 = 0. → q k Finally we obtain the decay rate by performing the 2-particle (ﬁnal state) phase space integral: Γ(π → l ν ) = + + = − → dΩcm 1 2| k | ￿ |M|2 4 π 8 π Ecm pol ￿ ￿ 2 22 mπ ml ml (GF fπ )2 1 − 2 4π mπ 1 2 mπ ￿ (25) (26) It is evident from (26) that this rate vanishes in the limit ml → 0. The relative rate of pion decay to electrons and muons is (using me = 0.5 M eV , mµ = 106 M eV , and mπ = 140 M eV ): Γ(π + → e+ ν ) Γ(π + → µ+ ν ) = ￿ me mµ ￿2 (1 − m2 /m2 )2 e π = 10−4 (1 − m2 /m2 )2 µ π (27) Eqn.(27) tells us that the decay of pion is dominated by the channel π + → µ+ ν . Using this approxiamtion in eqn.(26) and Γπ = 1/τπ ≈ 2.53 × 10−14 M eV , we obtain: fπ = ￿ 4 π m3 Γπ π m2 G2 (m2 − m2 )2 µF π µ where we have used: GF = 1.166 × 10−11 M eV...
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