236 Appendix F: Heavy Quark Production Radiation Derivation With two transverse polarizations ⃗ ϵ$_{λ}$ ( ⃗ k ) we have that This leads us finally to We would now like to show that one may replace the three vector quantities with their corresponding four vectors, then take ∑ ϵ$_{µ}$ϵ ∗ ν → − g$_{µν}$ . If we orient k µ = ( k, 0 , 0 , k ) µ along the z -direction, then we may take our two transverse polarizations as ϵ µ 1 = (0 , 1 , 0 , 0) µ and ϵ µ 2 = (0 , 0 , 1 , 0) $^{µ}$. It is now clear that we may freely take the three vectors to four, as the zeroth components of the polarization vectors is zero. To turn the sum over polarizations into a metric we need that the Ward identity be satisfied, k$_{µ}$j µ = k$_{µ}$ A µ = 0. Then, with our k oriented along z , ⃗ E ( ⃗ k ) · ⃗ E $^{∗}$( ⃗ k ) = E$_{1}$E ∗ 1 + E$_{2}$E ∗ 2 = ∑ λ =1 , 2 E$_{λ}$E ∗ λ = ∑ λ =1 , 2 | ⃗ ϵ$_{λ}$ ( ⃗ k ) · ⃗ E ( ⃗ k ) | $^{2}$, = k 0 2 ∑ λ =1 , 2 | ⃗ ϵ$_{λ}$ ( ⃗ k ) · ⃗ A ( ⃗ k ) | 2 = | ⃗ k | 2 ∑ λ =1 , 2 | ⃗ ϵ$_{λ}$ ( ⃗ k ) · ⃗ A ( ⃗ k ) | $^{2}$. (F.55) E$_{EM}$ = ∫ d $^{3}$k (2 π ) 3 e 2 2 ∑ λ =1 , 2 ∣ ∣ ⃗ ϵ$_{λ}$ ( ⃗ k ) · ( ⃗ p ′ k · p ′ − ⃗ p k · p$^{)}$$^{∣}$ ∣ ∣ 2 . (F.56) k $^{µ}$A$_{µ}$ = kA$_{0}$ + kA$_{3}$ = 0 ⇒ A$_{0}$ = − A$_{3}$. (F.57)