PS10-solutions

PS10-solutions - Phys 253a 1 Problem Set 10 Solutions...

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Unformatted text preview: Phys 253a 1 Problem Set 10 Solutions December 2, 2010 1. (a) The phase space integral for 1 → 2 is 1 2 m φ Z d 3 p (2 π ) 3 2 E p d 4 q (2 π ) 3 E q (2 π ) 4 δ 4 ( P- p- q ) This is easiest to evaluate in the center of mass frame, P = ( m φ , ). If we first do the integral over q , the spatial part of the delta function enforces- p- q = 0, so q =- p and E q = E p . We get 1 2(2 π ) 2 m φ Z d 3 p 4 E 2 p δ ( m φ- 2 p m 2 e + p 2 ) This is spherically symmetric in p , so we can switch to spherical coordinates 4 π 8(2 π ) 2 m φ Z p 2 dp p 2 + m 2 e δ ( m φ- 2 p m 2 e + p 2 ) = 1 8 πm φ p 2 p 2 + m 2 e p m 2 e + p 2 2 p √ m 2 e + p 2 = 1 2 m φ = 1 16 πm φ s 1- 4 m 2 e m φ (b) i. The amplitude is i M = ig S u ( p ) v ( q ), which satisfies X spins |M| 2 = g 2 S Tr[( 6 p + m e )( 6 q- m e )] = 4 g 2 S ( p · q- m 2 e ) = 2 g 2 S ( m 2 φ- 4 m 2 e ) Γ = g 2 S m φ 8 π (1- 4 x 2 ) 3 / 2 ii. The amplitude is i M =- g P u ( p ) γ 5 v ( q ), so we get X spins |M| 2 =- g 2 S Tr[ γ 5 ( 6 p + m e ) γ 5 ( 6 q- m e )] =- 4 g 2 S (- p · q- m 2 e ) = 2 g 2 S m 2 φ Γ = g 2 S m φ 8 π p 1- 4 x 2 Phys 253a 2 iii. The amplitude is i M = ig V μ u ( p ) γ μ v ( q ). We get 1 3 X spins |M| 2 = 1 3 g 2 V Tr[( 6 p + m e ) γ μ ( 6 q- m e ) γ ν ] X i i μ i ν = 4 g 2 V 3 ( p μ q ν + p ν q μ- ( p · q + m 2 e ) η μν )- η μν + P μ P ν m 2 φ ! = 4 g 2 V 3- 2 p · q + 2 ( p · P )( q · P ) m 2 φ + 3( p · q + m 2 e ) ! = 4 g 2 V 3 (2 m 2 e + m 2 φ ) Γ = g 2 V m φ 12 π (1 + 2 x 2 ) p 1- 4 x 2 iv. The amplitude is- g A μ u ( p ) γ μ γ 5 v ( q ), so we have 1 3 X spins |M| 2 = 1 3 g 2 A Tr[( 6 p + m e ) γ μ γ 5 ( 6 q- m e ) γ ν γ 5 ] X i i μ i ν = 4 g 2 A 3 (( m 2 e- p · q ) η μν + p μ q ν + q μ p ν )- η μν + P μ P ν m 2 φ ! = 4 g 2 A 3 ( m 2 φ- 4 m 2 e ) Γ = g 2 A m φ 12 π (1- 4 x 2 ) 3 / 2 v. The tricky part of this one is to write down the polarization sum for an antisymmetric tensor. We’ll assume it’s transverse to pick out the spin 1 part. Then we know the polarization sum Σ μνρσ = X i i μν i * ρσ must satisfy P μ Σ μνρσ = 0 (and similarly if we dot p into any index), along with Σ μνρσ =- Σ νμρσ = Σ ρσμν . The only objects we have to build Σ with are η μν and P μ , and the only tensors we can build out of these with the correct index symmetries are η μρ η νσ- η μσ η νρ and P μ P ρ η νσ- P μ P σ η νρ + P ν P σ η μρ- P ν P ρ η μσ Phys 253a 3 The transverseness condition implies that we should take the combination X i i μν i ρσ = η μρ η νσ- η μσ η νρ- 1 m 2 φ ( P μ P ρ η νσ- P μ P σ η νρ + P ν P σ η μρ- P ν P ρ η μσ ) (Actually, this only fixes the spin sum up to an overall scale. We must additionally check a special case, say ...) Alternatively, we can write this as X i i μν i * ρσ = C μρ C νσ- C μσ C νρ Where C μρ := η μρ- P μ P ρ m 2 φ is manifestly transverse. The amplitude isis manifestly transverse....
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PS10-solutions - Phys 253a 1 Problem Set 10 Solutions...

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