228 Appendix F: Heavy Quark Production Radiation Derivation Using the E-L equations ( ∂$_{α}$∂ L /∂∂$_{α}$A$_{β}$ − ∂ L /∂A$_{β}$ = 0), we find that where we get from the second to the third line be choosing to be in the Lorentz gauge ( ∂$_{α}$A α = 0). For the Beta decay problem, following Peskin 6.1 (pgs. 177-180) [251], our external current takes the form where, for the trajectory of our particle, we take We will work with the Fourier transform of the pertinent quantities: ∂$_{α}$F αβ = j β (F.23) ∂$_{α}$∂ $^{α}$Aβ − ∂ $^{β}$∂$_{α}$ A α = j β ∂$_{α}$∂ $^{α}$Aβ = j $^{β}$, (F.24) j $^{µ}$( x ) = e$^{∫}$ dτ dy $^{µ}$( τ ) dτ δ (4) $^{(}$x − y ( τ ) $^{)}$, (F.25) y $^{µ}$( τ ) =      ( p $^{µ}$/m ) τ for τ < 0; ( p ′ $^{µ}$/m ) τ for τ > 0 . (F.26) A $^{µ}$( x ) = ∫ d $^{4}$k (2 π ) $_{4}$e − ik · x ˜ A $^{µ}$( k ) (F.27) j $^{µ}$( x ) = ∫ d $^{4}$k (2 π ) $_{4}$e − ik · x ˜ j $^{µ}$( k ) . (F.28)