ū(e )(1 γ 5 )γ α v( ν e ) v( ν e )γ β (1 + γ 5 )u(e ) tr (1 γ 5 )γ α ( p ν m ν )γ β (1 + γ 5 )( p e + m e ).
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1 PHY 396 K. Soluion for problem e #. Problem (a: A poin of noaion: In he oluion o problem, he indice µ, e, ν ν µ, and ν ν e denoe he paricle. For he Lorenz indice, I hall ue α, β, γ, δ, σ, and ρ, bu never µ or ν. Thu, p µα denoe he α componen of he muon 4 momenum, ec., ec. Le ar wih he muon decay ampliude M(µ e ν µ ν e = G F ū(ν µ ( γ 5 γ α u(µ ū(e ( γ 5 γ α v( ν e. ( I complex conjugae can be wrien a M = G F ū(µ γ β ( + γ 5 u(ν µ v( ν e γ β ( + γ 5 u(e, (S. where he ( γ 5 facor become ( + γ 5 becaue γ 5 γ 0 (γ 5 γ 0 = γ 5. Conequenly, M = G F ū(ν µ ( γ 5 γ α u(µ ū(µ γ β ( + γ 5 u(ν µ (S. ū(e ( γ 5 γ α v( ν e v( ν e γ β ( + γ 5 u(e and hence all M = 4 G F (( r γ 5 γ α ( p µ + M µ γ β ( + γ 5 ( p ν + m ν ( r ( γ 5 γ α ( p ν m ν γ β ( + γ 5 ( p e + m e. (S.3 Having derived eq. (S.3, we now need o evaluae he race. For he fir race, we eliminae erm conaining odd number of γ ρ marice and wrie ( r ( γ 5 γ α ( p µ + M µ γ β ( + γ 5 ( p ν + m ν = = r (( γ 5 γ α p µ γ β ( + γ 5 p ν + r (( γ 5 γ α M µ γ β ( + γ 5 m ν
2 ( ( = r ( γ 5 γ α p µ γ β p ν ( γ 5 + M µ m ν r ( γ 5 γ α γ β ( + γ 5 = r (( γ 5 γ α p µ γ β p ν + M µ m e r (( + γ 5 ( γ 5 γ α γ β = r (( γ 5 γ α p µ γ β p ν + 0 = r (γ α p µ γ β p ν r (γ 5 γ α p µ γ β p ν = 8 p α µp β ν + p β µp α ν g αβ (p µ p ν + 8iǫ αγβδ p µγ p νδ. (S.4 Similarly, he econd race evaluae o r ( ( γ 5 γ α ( p e + m e γ β ( + γ 5 ( p ν m ν = (S.5 = 8 (p eα p νβ + p eβ p να g αβ (p e p ν + 8iǫ αρβσ p ρ νp σ e. I remain o ubiue he race formulæ (S.4 and (S.5 back ino eq. (S.3 and conrac he Lorenz indice. Thu, ( M = 6G F p α µ pβ ν + pβ µ pα ν gαβ (p µ p ν + iǫ αγβδ p µγ p νδ all ( p eα p νβ + p eβ p να g αβ (p e p ν + iǫ αρβσ p ρ νp σ e uing ymmery/aniymmery of facor under α β ( = 6G F p α µ pβ ν + pβ µ pα ν gαβ (p µ p ν p eα p νβ + p eβ p να g αβ (p e p ν ǫ αγβδ p µγ p νδ ǫ αρβσ p ρ νp σ e ( = 6G F (p µ p e (p ν p ν + (p µ p ν(p ν p e (p µ p ν (p e p ν (p µ p ν (p e p ν + 4(p µ p ν (p e p ν + (p µ p ν (p ν p e (p µ p e (p ν p ν Q.E.D. = 64G F (p µ p ν (p ν p e. (S.6
3 Problem (b: A explained in he Pekin & Schroeder exbook, he parial rae of a decay proce (in he re frame of he iniial paricle i given by dγ = M 0 M dp (S.7 where M i he decay ampliude, M i M averaged over he unknown iniial and ummed over he unmeaured final, and dp i he infinieimal phae pace facor for he final paricle. For hree final paricle, dp = d 3 p (π 3 (E d 3 p (π 3 (E d 3 p 3 (π 3 (E 3 (π3 δ (3 (p +p +p 3 (πδ(e +E +E 3 M 0 (S.8 where he energy-momenum conervaion law apply in he re frame, hu p + p + p 3 = p o = 0 and E + E + E 3 = E o = M 0. We ar by uing he momenum-conervaion δ funcion o eliminae eliminae he p 3 a independen variable, hu dp = d3 p d 3 p δ(e + E + E 3 E o 56π 5 E E E 3 Nex, we ue pherical coordinae for he wo remaining momena, p3= (p +p. (S.9 d 3 p = p dp d Ω, d 3 p = p dp d Ω, (S.0 and hen replace he d Ω decribing he direcion of he econd paricle momenum relaive o he fixed exernal frame wih d Ω ( = dθ in θ dφ ( decribing he ame direcion of p relaive o he frame cenered on he p. Conequenly, d Ω d Ω = d Ω d Ω ( = d Ω dφ ( dθ in θ d 3 Ω d(coθ (S. 3
4 and hence dp = d3 Ω 56π 5 p p dp dp d(coθ δ(e + E + E 3 E o E E E 3 p3= (p +p. (S. Nex, we ue he coine heorem which give p 3 = (p + p = p + p + p p co θ (for fixed p, p and herefore d(coθ = p 3 dp 3 p p dp = d3 Ω 56π 5 p p p 3 E E E 3 dp dp dp 3 δ(e + E + E 3 E o. (S.3 Finally, we noice ha for a relaiviic paricle of any ma, pdp = EdE and hence dp = d3 Ω 56π 5 de de de 3 δ(e + E + E 3 E o. (S.4 Subiuing hi formula ino eq. (S.7 give eq. (3 for he parial decay rae. I remain o deermine he limi of kinemaically allowed way o diribue he ne energy E o = M 0 of he proce among he hree final paricle. Such limi follow from he riangle inequaliie for he hree momena, p p + p 3, p p + p 3, p 3 p + p, (S.5 which look imple bu produce raher complicaed inequaliie for he energie E = p + m, E = p + m, and E 3 = p 3 + m3. However, when all hree final paricle are male, he kinemaic rericion become imply E E + E 3 = M 0 E (S.6 and dio for he oher wo inequaliie, or equivalenly 0 E, E, E 3 M 0, while E + E + E 3 = M 0. (5 4
5 Problem (c: In ligh of eq. ( and (3, he parial decay rae of he muon a re i given by dγ(µ e ν µ ν e = G F 8π 5 M µ (p µ p ν (p e p ν de e de ν de ν d 3 Ωδ(E e +E ν +E ν M µ. (S.7 Specializing o he muon frame, we have (p µ p ν = M µ E ν (S.8 while (p e p e = E e E ν p e p ν coθ eν neglecing m e, m ν, m ν = E e E ν + p e + p ν p ν = E e E ν + E e + E ν E ν (S.9 = (E e + E ν E ν uing E e + E ν = M µ E ν = M µ(m µ E ν. Hence, dγ(µ e ν µ ν e = G F 6π 5 M µe ν (M µ E ν de e de ν de ν d 3 Ωδ(E e +E ν +E ν M µ. (S.0 A hi poin we are ready o inegrae over he final-ae variable. In ligh of d 3 Ω = 8π and he kinemaic limi (5, we immediaely obain Γ(µ e ν µ ν e = G F M µ π 3 = G F M µ π 3 = G F M µ π 3 Mµ de e de ν de ν E ν (M µ E ν δ(e e + E ν + E ν M µ 0 Mµ 0 Mµ 0 de e Mµ de ν E ν(m µ E ν Mµ Ee de e E e ( M µ 3 E e. (S. In oher word, he parial muon decay rae wih repec o he final elecron energy i given 5
6 by dγ de e = G F M µ π 3 E e(3m µ 4E e (S. or raher { dγ G F π M 3 µ Ee (3M µ 4E e for E e < M µ, de e 0 for E e > M µ. (S.3 Graphically, dγ/de e E e M µ Noe how hi curve moohly reache i maximum a E e = M µ and hen abruply fall down o zero. I remain o calculae he oal decay rae of he muon by inegraing he parial rae (S.3 over he elecron energy. The reul i Γ o (µ eν ν = G F M5 µ 9π 3. (S.4 6
7 Problem : A he ree level, here are wo Feynman diagram for he Bhabha caering e + e e + e : ougoing e e + e e + + (S.5 incoming e e + e e + Conequenly, he ree-level ampliude i M ree = M + M, M = e ū(p γ µ u(p v(kγ µ v(k, M = + e v(kγµ u(p ū(p γ µ v(k, (S.6 where p and p are momena of he iniial / final elecron e while k and k are momena of he iniial / final poiron e +. Noe he ign of he wo ampliude: The channel diagram ha a direc poiron-in-poiron-ou line, hence an overall minu ign; he u channel diagram doe no have hi ign facor becaue all i fermionic line involve an elecron e on one end or he oher. Le u um over he ; pleae bear eq. ( in mind! Saring wih he fir diagram, we have M = = ū(p γ µ u(pū(pγ ν u(p v(kγ µ v(k v(k γ ν v(k r ( p + mγ µ ( p + mγ ν r ( k mγ µ ( k mγ ν r p γ µ pγ ν r kγ µ k γ ν (S.7 7
8 = 4 p µ p ν + p ν p µ g µν (p p 4 k µk ν + k νk µ g µν (k k = 6 (k p (k p + (k p(k p (k k(p p (k k(p p + 4(k k(p p = 3 (k p (k p + (k p(k p = 8 + u (S.7 where he la equaliy follow from he kinemaic relaion (k p = (k p = m, (p p = (k k = m, (S.8 (k p = (p k = m u u. In a imilar manner, he econd diagram yield M = v(kγ µ u(pū(pγ ν v(k ū(p γ µ v(k v(k γ ν u(p = r( k mγ µ ( p + mγ ν r ( p mγ µ ( k mγ ν r kγ µ pγ ν r p γ µ k γ ν (S.9 = 4 k µ p ν + k ν p µ g µν (k p 4 k µ p ν + k ν p µ g µν(k p = 6 (k k(p p + (k p(k p. (k p (k p (k p (k p + 4(k p (k p = 3 (k k(p p + (k p(k p = 8 + u. (S.9 8
9 The inerference erm i more inereing: M M = e4 ū(pγ µ u(p ū(p γ ν v(k v(k γ µ v(k v(kγ ν u(p = e4 r ( p + mγ µ ( p + mγ ν ( k mγ µ ( k mγ ν e4 r pγ µ p γ ν k γ µ kγ ν uing γ ν k γ µ kγ ν = kγ µ k = + e4 r pγ µ p kγ µ k uing γ µ p kγ µ = 4(p k = +8 e4 (p k r p k = +3 e4 (p k (p k = +8 e4 u. (S.30 Subiuing he pin um (S.7, (S.9 and (S.30 ino eq. ( and dividing by 4 (for he 4 pin ae of he wo iniial paricle, we arrive a 4 all M = e 4 + u + + u = e 4 ( + ( = e u u ( u + u (S.3 where he la equaliy follow from + = 4m u u. In he cener of ma frame = 4E 4p, = p ( co θ, (S.3 u = p ( + coθ, 9
10 and hence u 4 = 6 + ( coθ4 + ( + co θ 4 4( coθ = (3 + co θ ( coθ. (S.33 Therefore, dσ dω cm = 64π 4 M = α 4 (3 + co θ ( coθ. (S.34 all 0
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