jk B.79 Identities involving the transverse/longitudinal decomposition Note: u is a constant unit vector,Aa≡0·A,a/u≡aV,A≡A-aAa,V V-ua/8 A=A,+uAu V=V;+ (B.81) B.82 (·V)φ=0 (B.83) (B.84) Vt·(φ)=0 (B87) V×(的)=-0×Vφ (B.88) V:×(×A)=avA2 89 ax(V×A)=VAn (B.90) 0×(V×A1)=0 (B.91) 0.(0×A)=0 (B.92) a×(×A) (B.93) (B.94) V·A=VA2+ (B.95) V×A=VxA+ax/4 vA a-o (B.97) V×V×A=|V,×V×A *EdA an|+a(V1·A)-V2A (B.98) vA=V(v·A,) v×V1×A|+av2A ②2001 by CRC Press LLC∇ × Ee− jk·r  = − jk × Ee− jk·r (B.78) ∇2 Ee− jk·r  = −k2 Ee− jk·r (B.79) Identities involving the transverse/longitudinal decomposition Note: uˆ is a constant unit vector, Au ≡ uˆ · A, ∂/∂u ≡ uˆ · ∇, At ≡ A − uˆ Au, ∇t ≡ ∇ − uˆ∂/∂u. A = At + uˆ Au (B.80) ∇=∇t + uˆ ∂ ∂u (B.81) uˆ · At = 0 (B.82) (uˆ · ∇t) φ = 0 (B.83) ∇tφ = ∇φ − uˆ ∂φ ∂u (B.84) uˆ · (∇φ) = (uˆ · ∇)φ = ∂φ ∂u (B.85) uˆ · (∇tφ) = 0 (B.86) ∇t · (uˆφ) = 0 (B.87) ∇t × (uˆφ) = −uˆ × ∇tφ (B.88) ∇t × (uˆ × A) = uˆ∇t · At (B.89) uˆ × (∇t × A) = ∇t Au (B.90) uˆ × (∇t × At) = 0 (B.91) uˆ · (uˆ × A) = 0 (B.92) uˆ × (uˆ × A) = −At (B.93) ∇φ = ∇tφ + uˆ ∂φ ∂u (B.94) ∇ · A = ∇t · At + ∂ Au ∂u (B.95) ∇ × A = ∇t × At + uˆ ×  ∂At ∂u − ∇t Au  (B.96) ∇2 φ = ∇2 t φ + ∂2φ ∂u2 (B.97) ∇×∇× A =  ∇t × ∇t × At − ∂2At ∂u2 + ∇t ∂ Au ∂u  + uˆ  ∂ ∂u (∇t · At) − ∇2 t Au  (B.98) ∇2 A =  ∇t(∇t · At) + ∂2At ∂u2 − ∇t × ∇t × At  + uˆ∇2Au (B.99)