PS8-solutions

PS8-solutions - Phys 253a 1 Problem Set 8 Solutions...

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Phys 253a 1 Problem Set 8 Solutions November 22, 2010 1. (a) The t and u channels both contribute to Moller scattering. In Feynman gauge ( ξ = 1), we have i M = ( - ie ) 2 u 3 γ μ u 1 u 4 γ ν u 2 - ig μν t - ( - ie ) 2 u 4 γ μ u 1 u 3 γ ν u 2 - ig μν u M = e 2 t u 3 γ μ u 1 u 4 γ μ u 2 - e 2 u u 4 γ μ u 1 u 3 γ μ u 2 . The minus sign between the two channels comes from switching the external fermion lines. Squaring and summing over spins, we have 1 e 4 X spins |M| 2 = X spins ± 1 t 2 ( u 3 γ μ u 1 u 1 γ ν u 3 )( u 4 γ μ u 2 u 2 γ ν u 4 ) - 1 tu ( u 3 γ μ u 1 u 1 γ ν u 4 u 4 γ μ u 2 u 2 γ ν u 3 ) - 1 tu ( u 4 γ ν u 1 u 1 γ μ u 3 u 3 γ ν u 2 u 2 γ μ u 4 ) + 1 u 2 ( u 4 γ μ u 1 u 1 γ ν u 4 )( u 3 γ μ u 2 u 2 γ ν u 3 ) ² = 1 t 2 Tr(( 6 p 3 + m ) γ μ ( 6 p 1 + m ) γ ν )Tr(( 6 p 4 + m ) γ μ ( 6 p 2 + m ) γ ν ) - 1 tu Tr(( 6 p 3 + m ) γ μ ( 6 p 1 + m ) γ ν ( 6 p 4 + m ) γ μ ( 6 p 2 + m ) γ ν ) - 1 tu Tr(( 6 p 4 + m ) γ ν ( 6 p 1 + m ) γ μ ( 6 p 3 + m ) γ ν ( 6 p 2 + m ) γ μ ) + 1 u 2 Tr(( 6 p 4 + m ) γ μ ( 6 p 1 + m ) γ ν )Tr(( 6 p 3 + m ) γ μ ( 6 p 2 + m ) γ ν ) Now we have some traces to evaluate. First, Tr(( 6 p 3 + m ) γ μ ( 6 p 1 + m ) γ ν ) = Tr( 6 p 3 γ μ 6 p 1 γ ν ) + m 2 Tr( γ μ γ ν ) = 4( p μ 3 p ν 1 + p μ 1 p ν 3 + ( m 2 - p 1 · p 3 ) g μν ) Thus Tr(( 6 p 3 + m ) γ μ ( 6 p 1 + m ) γ ν )Tr(( 6 p 4 + m ) γ μ ( 6 p 2 + m ) γ ν ) = 16[2 p 3 · p 1 ( m 2 - p 2 · p 4 ) + 2 p 2 · p 4 ( m 2 - p 1 · p 3 ) +2( p 1 · p 2 )( p 3 · p 4 ) + 2( p 1 · p 4 )( p 3 · p 2 ) + 4( m 2 - p 1 · p 3 )( m 2 - p 2 · p 4 )] = 64 m 4 - 64 um 2 + 8 t 2 + 16 u 2 + 16 tu
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Phys 253a 2 The numerator of the u 2 term differs just by t u . The other kind of trace looks like Tr(( 6 p 3 + m ) γ μ ( 6 p 1 + m ) γ ν ( 6 p 4 + m ) γ μ ( 6 p 2 + m ) γ ν ) = m 4 Tr( γ μ γ ν γ μ γ ν ) + Tr( 6 p 3 γ μ 6 p 1 γ ν 6 p 4 γ μ 6 p 2 γ ν ) m 2 X ( i,j )=(3 , 1) , (1 , 4) , (4 , 2) , (2 , 3) Tr( 6 p i γ μ 6 p j γ ν γ μ γ ν ) + m 2 X ( i,j )=(3 , 4) , (1 , 2) Tr( 6 p i γ μ γ ν 6 p j γ μ γ ν ) = - 32 m 4 + 16 m 2 X i<j p i · p j - 32( p 1 · p 2 )( p 3 · p 4 ) = 32 m 4 - 8( t + u ) 2 In getting to the third line, we’ve used our identity γ ν γ μ γ ν = - 2 γ μ several times. This is symmetric under p 3 p 4 , so the other cross term gives the same thing. Putting everything together, we have 1 e 4 X spins |M| 2 = 64 m 4 - 64 um 2 + 8 t 2 + 16 u 2 + 16 tu t 2 + 64 m 4 - 64 tm 2 + 8 u 2 + 16 t 2 + 16 tu u 2 - 64 m 4 - 16( t + u ) 2 tu = 16(4( t 2 - tu + u 2 ) m 4 - 4( t 3 + u 3 ) m 2 + ( t 2 + ut + u 2 ) 2 ) t 2 u 2 For the total cross section, we should divide by 4 to account for the average over initial spins, and multiply by our usual phase-space factor
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PS8-solutions - Phys 253a 1 Problem Set 8 Solutions...

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