PS3-solutions - Physics 253a 1 Problem Set 3 Solutions October 6 2010 1(a In the center of mass frame the e and e approach each other with opposite cm

PS3-solutions - Physics 253a 1 Problem Set 3 Solutions...

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Physics 253a 1 Problem Set 3 Solutions October 6, 2010 1. (a) In the center of mass frame, the e + and e - approach each other with opposite momenta and total energy E cm : p e + = E cm 2 (1 , 1 , 0 , 0) and p e - = E cm 2 (1 , - 1 , 0 , 0). The muon and anti-muon look the same, but with their momenta rotated by θ : p μ + = E cm 2 (1 , cos θ, sin θ, 0) and p μ - = E cm 2 (1 , - cos θ, - sin θ, 0). Now, s = ( p e + + p e - ) 2 = ( E cm , 0 ) 2 = E 2 cm t = ( p μ - - p e - ) 2 = - 2 p μ - · p e - = - E 2 cm 2 (1 - cos θ ) u = ( p μ + - p e - ) = - 2 p μ + · p e - = - E cm 2 (1 + cos θ ) (b) Note that s + t + u = 0. (c) In class we found d Ω = e 4 64 π 2 E 2 cm (1 + cos 2 θ ) Note that 4 tu s 2 = 1 - cos 2 θ . So we can rewrite the above as d Ω = e 4 32 π 2 s 1 - 2 tu s 2 = e 4 32 π 2 s 3 ( t 2 + u 2 ) where we’ve used s 2 = ( t + u ) 2 = t 2 + 2 tu + u 2 . (d) We have s + t + u = (2 m 2 e + 2 p e - · p e + ) + ( m 2 e + m 2 μ - 2 p e - · p μ - ) +( m 2 e + m 2 μ - 2 p e - · p μ + ) = 4 m 2 e + 2 m 2 μ - 2 p e - · ( p μ + + p μ - - p e + ) = 4 m 2 e + 2 m 2 μ - 2 p e - · p e - = 2( m 2 e + m 2 μ ) Where we’ve used conservation of momentum in the last line. In general, we’d have s + t + u = i m 2 i .
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Physics 253a 2 2. (a) (Almost) straight off Wikipedia: d Ω = (2 Z ) 2 α 2 mv 2 2 1 sin 4 ( θ/ 2) where m is the mass of the α particle, v is its incoming velocity, Z is the nucleon charge, and α = e 2 4 π is the fine structure constant (the Wikipedia article forgets the product of the α particle and nucleon charge 2 · Z ). This formula treats the nucleus N as having infinite mass, so that the α particle scatters with the same kinetic energy E kin = 1 2 mv 2 as it started with. It also assumes the α particle is a nonrelativistic point particle moving in a classical Coulomb potential. (b) i. Let the α particle have initial momentum p i = ( E, p i ) and final momentum p f = ( E, p f ). Note that the initial and final energies are the same in the limit that the nucleon mass is infinite. Also, | p | = 2 mE kin in the nonrelativistic limit. The relevant Feynman diagram is γ N α, p i N α, p f By momentum conservation, the virtual photon has k = p f - p i = (0 , p f - p i ) = p 2 mE kin (0 , 1 - cos θ, - sin θ, 0) ii. & iii. Rewriting m e m , and taking e 2 2 Ze 2 , we have d Ω = (2 Z ) 2 e 4 4 π 2 m 2 k 4 = (2 Z ) 2 e 4 4 π 2 m 2 4 m 2 E 2 kin ((1 - cos θ ) 2 + sin θ 2 )
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