235 Appendix F: Heavy Quark Production Radiation Derivation Now consider the magnetic piece: We will now prove that ( ˆ k × ⃗ E ) · ( ˆ k × ⃗ E ) = ⃗ E · ⃗ E . E$_{B}$ = 1 2 ∫ d $^{3}$x | ⃗ B ( ⃗ x ) | 2 = 1 8 ∫ d $^{3}$x$^{∫}$ d $^{3}$k (2 π ) 3 ∫ d $^{3}$k ′ (2 π ) 3 $^{[}$ˆ k × ⃗ E ( ⃗ k ) e − ik · x + ˆ k × ⃗ E $^{∗}$( ⃗ k ) e ik · x ] · $^{[}$ˆ k ′ × ⃗ E ( ⃗ k $^{′}$) e − ik $^{′}$· x + ˆ k ′ × ⃗ E $^{∗}$( ⃗ k $^{′}$) e ik $^{′}$· x ] = 1 8 ∫ d $^{3}$k (2 π ) 3 { $^{[}$ˆ k × ⃗ E ( ⃗ k ) ] · $^{[}$( − ˆ k ) × ⃗ E ( − ⃗ k ) $^{]}$e − 2 ik $^{0}$t + $^{[}$ˆ k × ⃗ E $^{∗}$( ⃗ k ) ] · $^{[}$( − ˆ k ) × ⃗ E $^{∗}$( − ⃗ k ) $^{]}$e 2 ik $^{0}$t +2 $^{[}$ˆ k × ⃗ E ( ⃗ k ) ] · $^{[}$ˆ k × ⃗ E $^{∗}$( ⃗ k ) ] } . (F.52) ϵ $^{ijk}$k$^{j}$E$^{k}$ϵ$^{ilm}$k$^{l}$Em = ( δ $^{jl}$δkm − δ $^{jm}$δ$^{kl}$) k $^{j}$E$^{k}$k$^{l}$Em = ( ˆ k · ˆ k ) ( ⃗ E · ⃗ E ) − ( ˆ k · ⃗ E )( ˆ k · ⃗ E ) = ⃗ E · ⃗ E . (F.53) E$_{EM}$ = 1 2 ∫ d $^{3}$k (2 π ) 3 ⃗ E ( ⃗ k ) · ⃗ E $^{∗}$( ⃗ k ) . (F.54) Therefore we find that the energy radiated is