233 Appendix F: Heavy Quark Production Radiation Derivation we then have Let’s now prove some nice relations regarding E and B that will be useful later on: transversality and orthogonality. To prove that ⃗ E is orthogonal to ⃗ k recall that A µ ∝ ( p ′ $^{µ}$/k · p $^{′}$) − ( p $^{µ}$/k · p ). Then ⃗ E ( ⃗ k ) = ik 0 ⃗ A ( ⃗ k ) − i ⃗ k A $^{0}$( ⃗ k ) (F.45) ⃗ B ( ⃗ k ) = i ⃗ k × ⃗ A ( ⃗ k ) . (F.46) ⃗ k · ⃗ E ( ⃗ k ) ∝ − i | ⃗ k | $^{2}$( ˆ k · ⃗ A − A $^{0}$) ∝ − i | ⃗ k |$^{[(}$ ⃗ k · ⃗ p ′ k · p ′ − ⃗ k · ⃗ p k · p$^{)}$ −$^{(}$ k $^{0}$p ′ 0 k · p ′ − k $^{0}$p0 k · p$^{)]}$ ∝ − i | ⃗ k |$^{[}$ k · p ′ k · p ′ − k · p k · p$^{]}$ = 0 . (F.47) ⃗ k × ⃗ E = ⃗ k × $^{[}$ik 0 ⃗ A − i ⃗ k A 0 ] = ik 0 ⃗ k × ⃗ A ; (F.48) Also since k 0 = ±| ⃗ k | , and we won’t be interested in the sign, we have that ⃗ B = ˆ k × ⃗ E . (F.49)