224 Appendix F: Heavy Quark Production Radiation Derivation Now we must prove that g $^{αβ}$∂$_{µ}$ g$_{αβ}$ is equal to this last quantity: We will use an important example of the identity Eq. (F.1): The last equality comes from taking the variation of g$_{µν}$g µν = δ ν ν = #: δ ( g$_{µν}$g $^{µν}$) = g $^{µν}$δg$_{µν}$ + g$_{µν}$δg µν ⇒ g $^{µν}$δg$_{µν}$ = − g$_{µν}$δg $^{µν}$. (F.9) δg = gg $^{µν}$δg$_{µν}$ = − gg$_{µν}$δg $^{µν}$. (F.8) g $^{αβ}$∂$_{µ}$ g$_{αβ}$ = Tr G − $^{1}$∂$_{µ}$ G = Tr UG − $^{1}$U − $^{1}$U∂$_{µ}$ GU − 1 = Tr G − 1 $^{[}$∂$_{µ}$ G − ( ∂$_{µ}$U ) GU − 1 − UG ( ∂$_{µ}$U − $^{1}$) ] = Tr G − $^{1}$∂$_{µ}$ G − UG − $^{1}$U − $^{1}$( ∂$_{µ}$U ) GU − 1 − UG − $^{1}$U − $^{1}$UG ( ∂$_{µ}$U − $^{1}$) = Tr G − $^{1}$∂$_{µ}$ G − U − $^{1}$( ∂$_{µ}$U ) − U ( ∂$_{µ}$U − $^{1}$) = Tr G − $^{1}$∂$_{µ}$ G − ∂$_{µ}$ ( U − $^{1}$U ) = Tr G − $^{1}$∂$_{µ}$ G . (F.7) Use of Identity