238 Appendix F: Heavy Quark Production Radiation Derivation For a conserved current, ∂$_{β}$j β = 0, we have a consistency equation Our EOM, Eq. (F.61), become Fourier analysis immediately leads one to ∂$_{β}$ $^{(}$∂$_{α}$ ∂ $^{α}$Aβ − ∂$_{α}$∂ $^{β}$Aα + µ $^{2}$A β ) = ∂$_{β}$j β ⇒ ∂$_{α}$∂ $^{α}$∂$_{β}$ A β − ∂$_{β}$∂ $^{β}$∂$_{α}$ A α + µ $^{2}$∂$_{β}$ A β = 0 ⇒ ∂$_{β}$A β = 0 . (F.62) ∂ $^{2}$A β + µ $^{2}$A β = j $^{β}$. (F.63) A $^{µ}$( x ) = ∫ d $^{4}$k (2 π ) $_{4}$e − ik · x − ie k 2 − µ 2 ( p ′ µ k · p ′ + iϵ − p µ k · p − iϵ$^{)}$ . (F.64) The poles in this occur at k · p ′ = 0, k · p = 0, and k 2 − µ 2 = 0; the pole structure is exactly the same as for the massless case, Fig. F.1, but with the radiation poles giving a dispersion relation of Let’s examine the t < 0 case and make sure we recover the correct Yukawa k 0 ± = ± $^{√}$⃗ k 2 + µ $^{2}$. (F.65)