ph135cS7 - Ph135c. Solution set #7, 5/20/10 1. 8.7 We can...

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Unformatted text preview: Ph135c. Solution set #7, 5/20/10 1. 8.7 We can write: d Γ = 2 π | A fi | 2 δ ( E e + E ν- Δ) d 3 p e (2 π ) 3 d 3 p ν (2 π ) 3 = 1 (2 π ) 5 | A fi | 2 δ ( E e + E ν- Δ) p e 2 dp e d Ω e p ν 2 dp ν d Ω ν where Δ = m n- m p . To get the electron momentum distribution, we integrate over all variables except p e . This gives: d Γ dp e = 1 2 π 3 | A fi | 2 Z ∞ dp ν p ν 2 p e 2 δ ( E e + E ν- Δ) = 1 2 π 3 | A fi | 2 Z ∞ dE ν q E ν 2- m ν 2 p e 2 δ ( E e + E ν- Δ) = 1 2 π 3 | A fi | 2 Z ∞ dE ν E ν q E ν 2- m ν 2 p e 2 δ ( E e + E ν- Δ) = 1 2 π 3 | A fi | 2 (Δ- p p e 2 + m e 2 ) p e 2 q (Δ- p p e 2 + m e 2 ) 2- m ν 2 8.8 This time we have: d Γ = 2 π | A fi | 2 δ ( E e + E 1 + E 2- Δ) d 3 p e (2 π ) 3 d 3 p 1 (2 π ) 3 d 3 p 2 (2 π ) 3 = | A fi | 2 (2 π ) 8 δ ( E e + E 1 + E 2- Δ) d Ω e d Ω 1 d Ω 2 E e q E e 2- m 2 dE e E 1 2 dE 1 E 2 2 dE 2 Now we integrate over all variables except E e : d Γ dE e = | A fi | 2 4 π 5 Z δ ( E e + E 1 + E 2- Δ) E e q E e 2- m 2 E 1 2 dE...
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This note was uploaded on 12/04/2011 for the course PHY 7070 taught by Professor Smith during the Spring '11 term at University of Wisconsin.

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ph135cS7 - Ph135c. Solution set #7, 5/20/10 1. 8.7 We can...

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