content_image/305538.jpg content_image/305539.jpg
(because k · p = 0 ⇒ k 0 = 0 for p µ = ( m, ⃗ 0) $^{µ}$), and Eq. (F.35) becomes 

<img src='content_image/305538.jpg'>

For t > 0 we will ignore the k · p ′ = 0 pole because that merely gives the Coulomb field; radiation is produced by the k 0 = | ⃗ k | and k 0 = −| ⃗ k | poles. Since k 2 = ( k 0 − | ⃗ k | )( k 0 + | ⃗ k | ) and k · p = k $^{0}$p0 − ⃗ k · ⃗ p we have that 

<img src='content_image/305539.jpg'>

where, to get to the last line, we were able to freely change ⃗ k → − ⃗ k because 

