229 Appendix F: Heavy Quark Production Radiation Derivation With this normalization for the transforms we have that Solving the EOM (Eq. (F.24)) in momentum space we find that ˜ A $^{µ}$( k ) = ∫ d $^{4}$ke ik · $^{x}$A$^{µ}$( x ) (F.29) ˜ j $^{µ}$( k ) = ∫ d $^{4}$ke ik · $^{x}$j$^{µ}$( x ) . (F.30) ˜ A $^{µ}$( k ) = − 1 k $_{2}$˜ j $^{µ}$( k ) . (F.31) j $^{µ}$( x ) = e$^{∫}$ ∞ 0 dτ p ′ µ m δ (4) ( x − p ′ m τ$^{)}$ + e$^{∫}$ 0 −∞ dτ p µ m δ (4) $^{(}$x − p m τ ) (F.32) ⇒ ˜ j $^{µ}$( k ) = ∫ d $^{4}$xe ik · x $^{)}$j $^{µ}$( x ) = e$^{∫}$ ∞ 0 dτ p ′ µ m e i ( k · p $^{′}$/m + iϵ ) τ + e$^{∫}$ 0 −∞ dτ p µ m e i ( k · p/m − iϵ ) τ = ie$^{(}$ p ′ µ k · p ′ + iϵ − p µ k · p − iϵ$^{)}$ . (F.33) We can find ˜ j from the given trajectory (Eq. (F.26)) and the definition of the Fourier transform: Therefore we have the field in position space as an integral over momentum: The integral on the right hand side (RHS) has the pole structure as shown A $^{µ}$( x ) = ∫ d $^{4}$k (2 π ) $_{4}$e − ik · x − ie k 2 [ p ′ µ k · p ′ + iϵ − p µ k · p − iϵ$^{]}$ . (F.34)