231 Appendix F: Heavy Quark Production Radiation Derivation (because k · p = 0 ⇒ k 0 = 0 for p µ = ( m, ⃗ 0) $^{µ}$), and Eq. (F.35) becomes For t > 0 we will ignore the k · p ′ = 0 pole because that merely gives the Coulomb field; radiation is produced by the k 0 = | ⃗ k | and k 0 = −| ⃗ k | poles. Since k 2 = ( k 0 − | ⃗ k | )( k 0 + | ⃗ k | ) and k · p = k $^{0}$p0 − ⃗ k · ⃗ p we have that A $^{µ}$( x ) = ∫ d $^{3}$k (2 π ) $_{3}$e i ⃗ k · ⃗ x e | ⃗ k | $_{2}$(1 , ⃗ 0) µ = 2 π (2 π ) $_{3}$e$^{∫}$ ∞ 0 dk ∫ π 0 sin θdθe ikr cos θ ︸ ︷︷ ︸ π/r (1 , ⃗ 0) µ = 1 4 πr (1 , ⃗ 0) $^{µ}$. (F.37) A $^{µ}$( x ) = ∫ d $^{3}$k (2 π ) 3 ( − 2 πi ) 2 π ( − ie )$^{[}$ e − i | ⃗ k | $^{t}$ei ⃗ k · ⃗ x 1 2 | ⃗ k | ( p ′ µ | ⃗ k | p ′ 0 − ⃗ k · ⃗ p ′ − p µ | ⃗ k | p 0 − ⃗ k · ⃗ p$^{)}$ + e i | ⃗ k | $^{t}$ei ⃗ k · ⃗ x 1 − 2 | ⃗ k | ( p ′ µ −| ⃗ k | p ′ 0 − ⃗ k · ⃗ p ′ − p µ −| ⃗ k | p 0 − ⃗ k · ⃗ p$^{)]}$ = − e$^{∫}$ d $^{3}$k (2 π ) 3 1 2 | ⃗ k | [ e − i | ⃗ k | $^{t}$ei ⃗ k · ⃗ x ( p ′ µ | ⃗ k | p ′ 0 − ⃗ k · ⃗ p ′ − p µ | ⃗ k | p 0 − ⃗ k · ⃗ p$^{)}$ + e i | ⃗ k | $^{t}$e − i ⃗ k · ⃗ x ( p ′ µ | ⃗ k | p ′ 0 − ⃗ k · ⃗ p ′ − p µ | ⃗ k | p 0 − ⃗ k · ⃗ p$^{)]}$ , (F.38) where, to get to the last line, we were able to freely change ⃗ k → − ⃗ k because