239 Appendix F: Heavy Quark Production Radiation Derivation potential in the rest frame. For t < 0 we pick up only the k · p = 0 pole. Then Let’s now examine the t > 0 solution. Again we’ll ignore the uninteresting k · p ′ = 0 Yukawa pole contribution. We find that A $^{µ}$( x ) = ie$^{∫}$ d $^{3}$k (2 π ) $_{3}$e − i ( ⃗ k · ⃗ p p 0 $^{$^{)}$t}$e i ⃗ k · ⃗ x $^{(}$2 πi 2 π ) 1 − ⃗ k 2 − µ 2 ( p $^{0}$, ⃗ 0) µ p 0 = e (1 , ⃗ 0) µ 2 π (2 π ) 3 ∫ ∞ 0 k $^{2}$dk ∫ π 0 sin θdθ 1 k 2 + µ $_{2}$e ikr cos θ ︸ ︷︷ ︸ πe $^{µr}$/r = e 4 πr e − $^{µr}$(1 , ⃗ 0) $^{µ}$. (F.66) A $^{µ}$( x ) = − ie − 2 πi 2 π ∫ d $^{3}$k (2 π ) 3 { 1 2 k 0 + ( p ′ µ k 0 $_{+}$p ′ 0 − ⃗ k · ⃗ p ′ − p µ k 0 $_{+}$p 0 − ⃗ k · ⃗ p$^{)}$ e − ik 0 $^{$_{+}$t}$e i ⃗ k · ⃗ x 1 2 k 0 − ( p ′ µ k 0 $_{−}$p ′ 0 − ⃗ k · ⃗ p ′ − p µ k 0 $_{−}$p 0 − ⃗ k · ⃗ p$^{)}$ e − ik 0 $^{$_{−}$t}$e i ⃗ k · ⃗ x } . (F.67) A$_{Proca}$ = − e k 0 ( p ′ µ k · p ′ − p µ k · p$^{)}$ , (F.68) We can therefore write A µ Proca in the form of Eq. (F.39) with where, from now on, we will implicitly take k 0 = k 0 $_{+}$. However because k $^{µ}$k$_{µ}$ = µ 2 instead of 0 as in the E&M case we cannot immediately write | ⃗ E | $^{2}$+ | ⃗ B | 2 as ⃗ E · ⃗ E ∗ as we did before. Additionally, since we do not have gauge invariance like in the E&M case, we do not have transversality of the fields; there are three