234 Appendix F: Heavy Quark Production Radiation Derivation Let’s use all our knowledge to find the energy radiated by this motion. Recall that Consider first the electric piece: E$_{EM}$ = ∫ d $^{3}$xT 00 EM = 1 2 ∫ d $^{3}$x $^{(}$| ⃗ E ( ⃗ x ) | 2 + | ⃗ B ( ⃗ x ) | 2 ) . (F.50) E$_{E}$ = 1 2 ∫ d $^{3}$x | ⃗ E ( ⃗ x ) | 2 = 1 8 ∫ d $^{3}$x$^{∫}$ d $^{3}$k (2 π ) 3 ∫ d $^{3}$k ′ (2 π ) 3 $^{[}$⃗ E ( ⃗ k ) e − ik · x + ⃗ E $^{∗}$( ⃗ k ) e ik · x ] · $^{[}$⃗ E ( ⃗ k $^{′}$) e − ik $^{′}$· x + ⃗ E $^{∗}$( ⃗ k $^{′}$) e ik $^{′}$· x ] = 1 8 ∫ d $^{3}$k (2 π ) 3 $^{[}$⃗ E ( ⃗ k ) · ⃗ E ( − ⃗ k ) e − 2 ik $^{0}$t + ⃗ E $^{∗}$( ⃗ k ) · ⃗ E $^{∗}$( − ⃗ k ) e 2 ik $^{0}$t +2 ⃗ E ( ⃗ k ) · ⃗ E $^{∗}$( ⃗ k ) ] . (F.51)